Category: Class 11

Some Basic Concepts of Chemistry Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 1 Some Basic Concepts of Chemistry Textbook Exercise Questions and Answers.

Class 11 Chemistry Chapter 1 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 1 Exercise Solutions

1. Choose the most correct option.

Question A.
A sample of pure water, whatever the source always contains …………. by mass of oxygen and 11.1 % by mass of hydrogen.
a. 88.9
b. 18
c. 80
d. 16
Answer:
a. 88.9

Question B.
Which of the following compounds can NOT demonstrate the law of multiple proportions?
a. NO, NO₂
b. CO, CO₂
c. H₂O, H₂O₂
d. Na₂S, NaF
Answer:
d. Na₂S, NaF

Question C.
Which of the following temperature will read the same value on celsius and Fahrenheit scales.
a. – 40°
b. + 40°
c. – 80°
d. – 20°
Answer:
a. – 40°

Question D.
SI unit of the quantity electric current is
a. Volt
b. Ampere
c. Candela
d. Newton
Answer:
b. Ampere

Question E.
In the reaction N₂ + 3H₂ → 2NH₃, the ratio by volume of N₂, H₂ and NH₃ is 1 : 3 : 2 This illustrates the law of
a. definite proportion
b. reciprocal proportion
c. multiple proportion
d. gaseous volumes
Answer:
d. gaseous volumes

Question F.
Which of the following has maximum number of molecules ?
a. 7 g N₂
b. 2 g H₂
c. 8 g O₂
d. 20 g NO₂
Answer:
b. 2 g H₂

Question G.
How many g of H₂O are present in 0.25 mol of it ?
a. 4.5
b. 18
c. 0.25
d. 5.4
Answer:
a. 4.5

Question H.
The number of molecules in 22.4 cm³ of nitrogen gas at STP is
a. 6.022 × 10²⁰
b. 6.022 × 10²³
c. 22.4 × 10²⁰
d. 22.4 × 10²³
Answer:
a. 6.022 × 10²⁰

Question I.
Which of the following has the largest number of atoms ?
a. 1g Au_(s)
b. 1g Na_(s)
c. 1g Li_(s)
d. 1g Cl_2(g)
Answer:
c. 1g Li_(s)

2. Answer the following questions.

Question A.
State and explain Avogadro’s law.
Answer:
i. In the year 1811, Avogadro made a distinction between atoms and molecules and thereby proposed Avogadro’s law.

ii. Avogadro proposed that, “Equal volumes of all gases at the same temperature and pressure contain equal number of molecules”.
e.g. Hydrogen gas combines with oxygen gas to produce water vapour as follows:

According to Avogadro’s law, if 1 volume contains n molecules, then 2n molecules of hydrogen combine with n molecules of oxygen to give 2n molecules of water, i.e., 2 molecules of hydrogen gas combine with 1 molecule of oxygen to give 2 molecules of water vapour as represented below:

Question B.
Point out the difference between 12 g of carbon and 12 u of carbon.
Answer:
12 g of carbon is the molar mass of carbon while 12 u of carbon is the mass of one carbon atom.

Question C.
How many grams does an atom of hydrogen weigh ?
Answer:
The mass of a hydrogen atom is 1.6736 × 10^-24 g.

Question D.
Calculate the molecular mass of the following in u.
a. NH₃
b. CH₃COOH
c. C₂H₅OH
Answer:
i. Molecular mass of NH₃ = (1 × Average atomic mass of N) + (3 × Average atomic mass of H)
= (1 × 14.0 u) +(3 × 1.0 u)
= 17 u

ii. Molecular mass of CH₃COOH = (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12.0 u) + (4 × 1.0 u) + (2 × 16.0 u)
= 60 u

iii. Molecular mass of C₂H₅OH = (2 × Average atomic mass of C) + (6 × Average atomic mass of H) + (1 × Average atomic mass of O)
= (2 × 12.0 u) + (6 × 1.0 u) + (1 × 16.0 u)
= 46 u
Ans: i. The molecular mass of NH₃ = 17 u
ii. The molecular mass of CH₃COOH = 60 u
iii. The molecular mass of C₂H₅OH = 46 u

Question E.
How many particles are present in 1 mole of a substance ?
Answer:
The number of particles in one mole is 6.0221367 × 10²³.

Question F.
What is the SI unit of amount of a substance ?
Answer:
The SI unit for the amount of a substance is mole (mol).

Question G.
What is meant by molar volume of a gas ?
Answer:
The volume occupied by one mole of a gas at standard temperature (0 °C) and pressure (1 atm) (STP) is called as molar volume of a gas. The molar volume of a gas at STP is 22.4 dm³.

Question H.
State and explain the law of conservation of mass.
Answer:
Law of conservation of mass:

• The law of conservation of mass states that, “Mass can neither be created nor destroyed” during chemical combination of matter.
• Antoine Lavoisier who is often referred to as the father of modem chemistry performed careful experimental studies for various combustion reactions, namely burning of phosphorus and mercury in the presence of air.
• Both his experiments resulted in increased weight of products.
• After several experiments, in burning of phosphorus, he found that the weight gained by the phosphoms was exactly the same as the weight lost by the air. Hence, total mass of reactants = total mass of products.
• When hydrogen gas bums and combines with oxygen to form water, the mass of the water formed is equal to the mass of the hydrogen and oxygen consumed. Thus, this is in accordance with the law of conservation of mass.

Question I.
State the law of multiple proportions.
Answer:
The law states that, “When two elements A and B form more than one compounds, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers”.

3. Give one example of each

Question A.
Homogeneous mixture
Answer:
Homogeneous mixture: Solution (An aqueous solution of sugar)

Question B.
Heterogeneous mixture
Answer:
Heterogeneous mixture: Suspension (of sand in water)

Question C.
Element
Answer:
Element: Gold

Question D.
Compound
Answer:
Compound: Distilled water.

4. Solve problems :

Question A.
What is the ratio of molecules in 1 mole of NH₃ and 1 mole of HNO₃.
Answer:
One mole of any substance contains particles equal to 6.022 × 10²³.
1 mole of NH₃ = 6.022 × 10²³ molecules of NH₃
I mole of HNO₃ = 6.022 × 10²³ molecules of HNO₃
∴ Ratio = \(\frac{6.022 \times 10^{23}}{6.022 \times 10^{23}}\) = 1 : 1
Ans: The ratio of molecules is = 1 : 1.

Question B.
Calculate number of moles of hydrogen in 0.448 litre of hydrogen gas at STP.
Answer:
Given: Volume of hydrogen at STP = 0.448 L
To find: Number of moles of hydrogen

Molar volume of a gas = 22.4 dm³ mol^-1 = 22.4 L at STP

Ans: Number of moles of hydrogen = 0.02 mol

Question C.
The mass of an atom of hydrogen is 1.008 u. What is the mass of 18 atoms of hydrogen.
Answer:
Mass of 1 atom of hydrogen = 1.008 u
∴ Mass of 18 atoms of hydrogen = 18 × 1.008 u = 18.144 u
Ans: The mass of 18 atoms of hydrogen = 18.144 u

Question D.
Calculate the number of atom in each of the following (Given : Atomic mass of I = 127 u).
a. 254 u of iodine (I)
b. 254 g of iodine (I)
Answer:
a. 254 u of iodine (I) = x atoms
Atomic mass of iodine (I) = 127 u
∴ Mass of one iodine atom = 127 u
∴ x = \(\frac{254 \mathrm{u}}{127 \mathrm{u}}\) = 2 atoms

b. 254 g of iodine (I)
Atomic mass of iodine = 127 u
∴ Molar mass of iodine = 127 g mol^-1
Now,

Now,
Number of atoms = Number of moles × Avogadro’s constant
= 2 mol × 6.022 × 10²³ atoms/mol
= 12.044 × 10²³ atoms
= 1.2044 × 10²⁴ atoms
Ans. i.Number of iodine atoms in 254 u = 2 atoms
ii. Number of iodine atoms in 254 g = 1.2044 × 10²⁴ atoms

Question E.
A student used a carbon pencil to write his homework. The mass of this was found to be 5 mg. With the help of this calculate.
a. The number of moles of carbon in his homework writing.
b. The number of carbon atoms in 12 mg of his homework writting.
Answer:
a. 5 mg carbon = 5 × 10^-3 g carbon
Atomic mass of carbon = 12 u
∴ Molar mass of carbon 12 g mol^-1

b. 12 mg carbon = 12 × 10^-3 g carbon

Number of atoms = Number of moles × Avogadro’s constant
Number of atoms of carbon = 1 × 10^-3 mol × 6.022 × 10²³ atoms/mol
= 6.022 × 10²⁰ atoms
Ans: Number of moles of carbon in his homework writing = 4.167 × 10^-4 mol
Number of atoms of carbon in 12 mg homework writing = 6.022 × 10²⁰ atoms

Question F.
Arjun purchased 250 g of glucose (C₆H₁₂O₆) for Rs 40. Find the cost of glucose per mole.
Answer:
Given: Mass of urea = 250 g, cost for 250 g glucose = Rs 40, molecular formula of glucose = C₆H₁₂O₆
To find: Cost per mole of glucose
Calculation: Molecular formula of glucose is (C₆H₁₂O₆).
Molecular mass of glucose
= (6 × Average atomic mass of C) + (12 × Average atomic mass of H) + (6 × Average atomic mass of O)
= (6 × 12 u) + (12 × 1 u) + (6 × 16 u)
=180 u
∴ Molar mass of glucose = 180 g mol^-1

Now,
\(\frac {250}{180}\) mol of glucose cost = Rs 40
1 mol glucose cost = x
∴ x = \(\frac{40 \times 180}{250}\) = Rs 28.8/mol of glucose
Ans. The cost of glucose per mole is Rs 28.8.

[ Calculation using log table:
\(\frac{40 \times 180}{250}\)
= Antilog₁₀ [log₁₀(40) + log₁₀(180) + log₁₀(250)]
= Antilog₁₀ [1.6021 + 2.2553 – 2.3979]
= Antilog₁₀ [1.4595] = 28.80 ]

Question G.
The natural isotopic abundance of ¹⁰B is 19.60% and ¹¹B is 80.40 %. The exact isotopic masses are 10.13 and 11.009 respectively. Calculate the average atomic mass of boron.
Answer:
Average atomic mass of Boron(B)

Ans. Average atomic mass of boron = 10.84 u

Question H.
Convert the following degree Celsius temperature to degree Fahrenheit.
a. 40 °C
b. 30 °C
Answer:
a. Given: Temperature in degree Celsius =40°C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 40 °C in the formula,
°F = \(\frac {9}{5}\) (°C)+32
= \(\frac {9}{5}\) (40) + 32
= 72 + 32
= 104 °F

b. Given: Temperature in degree Celsius = 30 °C
To find: Temperature in degree Fahrenheit
Formula: °F = \(\frac {9}{5}\) (°C) + 32
Calculation: Substituting 30 °C in the formula,
°F = \(\frac {9}{5}\)(°C) + 32
= \(\frac {9}{5}\)(30) + 32
= 54 + 32
= 86 °F
Ans: i. The temperature 40 °C corresponds to 104 °F.
ii. The temperature 30 °C corresponds to 86 °F.

Question I.
Calculate the number of moles and molecules of acetic acid present in 22 g of it.
Answer:
Given: Mass of acetic acid = 22 g
To find: The number of moles and molecules of acetic acid
Formulae: Number of moles = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
ii. Number of molecules = Number of moles × Avogadro’s constant
Calculator: Mass of acetic acid = 22 g
Molecular mass of acetic acid, CH₃COOH
= (2 × Average atomic mass of C) + (4 × Average atomic mass of H) + (2 × Average atomic mass of O)
= (2 × 12 u) + (4 × 1 u) + (2 × 16 u) = 60 u
∴ Molar mass of acetic acid = 60 g mol^-1

Now,
Number of molecules of acetic acid = Number of moles × Avogadro’s constant
= 0.367 mol × 6.022 × 10²³ molecules/mol
= 2.210 × 10²³ molecules
Ans: Number of moles = 0.367 mol
Number of molecules of acetic acid = 2.210 × 10²³ molecules

Question J.
24 g of carbon reacts with some oxygen to make 88 grams of carbon dioxide. Find out how much oxygen must have been used.
Answer:
Given: Mass of carbon (reactant) = 24 g, mass of carbon dioxide (product) = 88 g
To find: Mass of oxygen (reactant)
Calculation: 12 g of carbon combine with 32 g oxygen to form 44 g of carbon dioxide as follows:

Hence, (2 × 12 = 24 g) of carbon will combine with (2 × 32 = 64 g) of oxygen to give (2 × 44 = 88 g) carbon dioxide.
Ans: Mass of oxygen used = 64 g

Question K.
Calculate number of atoms is each of the following. (Average atomic mass : N = 14 u, S = 32 u)
a. 0.4 mole of nitrogen
b. 1.6 g of sulfur
Answer:
a. 0.4 mole of nitrogen (N)
Number of atoms of N = Number of moles × Avogadro’s constant
= 0.4 mol × 6.022 × 10²³ atoms/mol
= 2.4088 × 10²³ atoms of N

b. 1.6 g of Sulphur (S)
Molar mass of sulphur = 32 g mol^-1

Number of atoms of S = Number of moles × Avogadro’s constant
= 0.05 mol × 6.022 × 10²³ atoms/mol
= 0.3011 × 10²³ atoms
= 3.011 × 10²² atoms of S
Ans: a. Number of nitrogen atoms in 0.4 mole = 2.4088 × 10²³ atoms of N
b. Number of sulphur atoms in 1.6 g = 3.011 × 10²² atoms of S

Question L.
2.0 g of a metal burnt in oxygen gave 3.2 g of its oxide. 1.42 g of the same metal heated in steam gave 2.27 of its oxide. Which law is verified by these data ?
Answer:
Here, metal oxide is obtained by two different methods; reactions of metal with oxygen and reaction of metal with water vapour (steam).
In first reaction (reaction with oxygen),
The mass of oxygen in metal oxide = 3.2 – 2.0 = 1.2 g
% of oxygen = \(\frac{1.2}{3.2}\) × 100 = 37.5%
% of metal = \(\frac{2.0}{3.2}\) × 100 = 62.5%
In second reaction (reaction with steam),
The mass of oxygen in metal oxide = 2.27 – 1.42 = 0.85 g
% of oxygen = \(\frac{0.85}{2.27}\) × 100 = 37.44 ≈ 37.5%
% of metal = \(\frac{1.42}{2.27}\) × 100 = 62.56 ≈ 62.5%
Therefore, irrespective of the source, the given compound contains same elements in the same proportion. The law of definite proportions states that “A given compound always contains exactly the same proportion of elements by weight”. Hence, the law of definite proportions is verified by these data.
Ans: The law of definite proportions is verified by given data.

Question M.
In two moles of acetaldehyde (CH₃CHO) calculate the following
a. Number of moles of carbon
b. Number of moles of hydrogen
c. Number of moles of oxygen
d. Number of molecules of acetaldehyde
Answer:
Molecular formula of acetaldehyde: C₂H₄O
Moles of acetaldehyde = 2 mol
a. Number of moles of carbon atoms = Moles of acetaldehyde × Number of carbon atoms
= 2 × 2
= 4 moles of carbon atoms

b. Number of moles of hydrogen atoms = Moles of acetaldehyde × Number of hydrogen atoms
= 2 × 4
= 8 moles of hydrogen atoms

c. Number of moles of oxygen atoms = Moles of acetaldehyde × Number of oxygen atoms
= 2 × 1
= 2 moles of oxygen atoms

d. Number of molecules of acetaldehyde = Moles of acetaldehyde × Avogadro number
= 2 mol × 6.022 × 10²³ molecules/mol
= 12.044 × 10²³ molecules of acetaldehyde
Ans: i. Number of moles of carbon, hydrogen and oxygen are 4, 8, 2 respectively,
ii. Number of molecules of acetaldehyde = 12.044 × 10²³

Question N.
Calculate the number of moles of magnesium oxide, MgO in
i. 80 g and
ii. 10 g of the compound.
(Average atomic masses of Mg = 24 and O = 16)
Answer:
Given: i. Mass of MgO = 80 g
ii. Mass of MgO = 10 g
To find: Number of moles of MgO
Formulae: Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
Calculation: i. Molecular mass of MgO = (1 × Average atomic mass of Mg) + (1 × Average atomic mass of O)
= (1 × 24u) + (1 × 16 u)
= 40 u
∴ Molar mass of MgO = 40 g mol^-1
Mass of MgO = 80 g
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{80 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 2 mol

ii. Mass of MgO = 10 g, Molar mass of MgO = 40 g mol^-1
Number of moles (n) = \(\frac{\text { Mass of a substance }}{\text { Molar mass of a substance }}\)
= \(\frac{10 \mathrm{~g}}{40 \mathrm{~g} \mathrm{~mol}^{-1}}\)
= 0.25 mol
Ans: i. The number of moles in 80 g of magnesium oxide, MgO = 2 mol
ii. The number of moles in 10 g of magnesium oxide, MgO = 0.25 mol

Question O.
What is volume of carbon dioxide, CO₂ occupying by i. 5 moles and ii. 0.5 mole of CO₂ gas measured at STP.
Answer:
Given: i. Number of moles of CO₂ = 5 mol
ii. Number of moles of CO₂ = 0.5 mol
To find: Volume at STP
Formula: Number of moies of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
Calculation: Molar volume of a gas 22.4 dm³ mol^-1 at STP.
Number of moles of a gas (n) = \(\frac{\text { Volume of a gas at STP }}{\text { Molar volume of a gas }}\)
∴ i. Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 5mol × 22.4 dm³ mol^-1 = 112 dm³
ii. Volume of the gas at STP Number of moles of a gas (n) × Molar volume of a gas
= 0.5 mol × 22.4 dm³ mol^-1 = 11.2 dm³
Ans: i. Volume of 5 mol of CO₂ = 112 dm³
ii. Volume of 0.5 mol of CO₂ = 11.2 dm³

Question P.
Calculate the mass of potassium chlorate required to liberate 6.72 dm³ of oxygen at STP. Molar mass of KClO₃ is 122.5 g mol^-1.
Answer:
The molecular formula of potassium chlorate is KClO₃.
Required chemical equation:

2 moles of KClO₃ = 2 × 122.5 = 245 g
3 moles of O₂ at STP occupy = (3 × 22.4 dm³) = 67.2 dm³
Thus, 245 g of potassium chlorate will liberate 67.2 dm³ of oxygen gas.
Let ‘x’ gram of KClO₃ liberate 6.72 dm³ of oxygen gas at S.T.P.
∴ x = \(\frac{245 \times 6.72}{67.2}\) = 24.5 g
Ans: Mass of potassium chlorate required = 24.5 g

Question Q.
Calculate the number of atoms of hydrogen present in 5.6 g of urea, (NH₂)₂CO. Also calculate the number of atoms of N, C and O.
Answer:
Given: Mass of urea = 5.6 g
To find: The number of atoms of hydrogen, nitrogen, carbon and oxygen
Calculation: Molecular formula of urea: (NH₂)₂CO
Molar mass of urea = 60 g mol^-1

∴ Moles of urea = 0.0933 mol
Number of atoms = Number of moles × Avogadro’s constant
Now, 1 molecule of urea has total 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C and 1 of O.
∴ Number of H atoms in 5.6 g of urea = (4 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 2.247 × 10²³ atoms of hydrogen
∴ Number of N atoms in 5.6 g of urea = (2 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 1.124 × 10²³ atoms of nitrogen
∴ Number of C atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of carbon
∴ Number of O atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of oxygen
Ans: 5.6 g of urea contain 2.247 × 10²³ atoms of H, 1.124 × 10²³ atoms of N, 0.562 × 10²³ atoms of C and 0.562 × 10²³ atoms of O.

Question R.
Calculate the mass of sulfur dioxide produced by burning 16 g of sulfur in excess of oxygen in contact process. (Average atomic mass : S = 32 u, O = 16 u)
Answer:
Given: Mass of sulphur (reactant) = 16 g
To find: Mass of sulphur dioxide (product)
Calculation: 32 g of sulphur combine with 32 g oxygen to form 64 g of sulphur dioxide as follows:

Hence, (0.5 × 32 = 16 g) of sulphur will combine with (0.5 × 32 = 16 g) of oxygen to give (0.5 × 64 = 32 g) sulphur dioxide.
Ans: Mass of sulphur dioxide produced = 32 g

5. Explain

Question A.
The need of the term average atomic mass.
Answer:

• Several naturally occurring elements exist as a mixture of two or more isotopes.
• Isotopes have different atomic masses.
• The atomic mass of such an element is the average of atomic masses of its isotopes.
• For this purpose, the atomic masses of isotopes and their relative percentage abundances are considered.

Hence, the term average atomic mass is needed to express atomic mass of elements containing mixture of two or more isotopes.

Question B.
Molar mass.
Answer:
i. The mass of one mole of a substance (element/compound) in grams is called its molar mass.
ii. The molar mass of any element in grams is numerically equal to atomic mass of that element in u.
e.g.

iii. Similarly, molar mass of polyatomic molecule, in grams is numerically equal to its molecular mass or formula mass in u.
e.g.

Question C.
Mole concept.
Answer:

• Even a small amount of any substance contains very large number of atoms or molecules. Therefore, a quantitative adjective ‘mole’ is used to express the large number of sub-microscopic entities like atoms, molecules, ions, electrons, etc. present in a substance.
• Thus, one mole is the amount of a substance that contains as many entities or particles as there are atoms in exactly 12 g (or 0.012 kg) of the carbon -12 isotope.
• One mole is the amount of substance which contains 6.0221367 × 10²³ particles/entities.

Question D.
Formula mass with an example.
Answer:

• The formula mass of a substance is the sum of atomic masses of the atoms present in the formula.
• In substances such as sodium chloride, positive (sodium), and negative (chloride) entities are arranged in a three-dimensional structure in a way that one sodium (Na⁺) ion is surrounded by six chlorides (Cl^–) ions, all at the same distance from it and vice versa. Thus, sodium chloride does not contain discrete molecules as the constituent units.
• Therefore, NaCl is just the formula that is used to represent sodium chloride though it is not a molecule.
• In such compounds, the formula (i.e., NaCl) is used to calculate the formula mass instead of molecular mass.

e.g. Formula mass of sodium chloride = atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u

Question E.
Molar volume of gas.
Answer:
i. It is more convenient to measure the volume rather than mass of the gas.
ii. It is found from Avogadro law that one mole of any gas occupies a volume of 22.4 dm³ at standard temperature (0 °C) and pressure (1 atm) (STP).
iii. The volume of 22.4 dm³ at STP is known as molar volume of a gas.
iv. The relationship between number of moles and molar volume can be expressed as follows:

[Note: IUPAC has recently changed the standard pressure to 1 bar. Under these new STP conditions the molar volume of a gas is 22.71 L mol^-1]

Question F.
Types of matter (on the basis of chemical composition).
Answer:
Matter on the basis of chemical composition can be classified as follows:
i. Pure substances: They always have a definite chemical composition. They always have the same properties regardless of their origin.
e.g. Pure metal, distilled water, etc.

They are of two types:
a. Elements: They are pure substances, which cannot be broken down into simpler substances by ordinary chemical changes.
Elements are further classified into three types:
1. Metals:

• They have a lustre (a shiny appearance).
• They conduct heat and electricity.
• They can be drawn into wire (ductile).
• They can be hammered into thin sheets (malleable).
• e.g. Gold, silver, copper, iron. Mercury is a liquid metal at room temperature.

2. Nonmetals:

• They have no lustre, (except diamond, iodine)
• They are poor conductors of heat and electricity, (except graphite)
• They cannot be hammered into sheets or drawn into wire, because they are brittle. e.g. Iodine

3. Metalloids: Some elements have properties that are intermediate between metals and nonmetals and are called metalloids or semimetals.
e.g. Arsenic, silicon and germanium.
b. Compounds: They are the pure substances which are made up of two or more elements in fixed proportion.
e.g. Water, ammonia, methane, etc.

ii. Mixtures: They have no definite chemical composition and hence no definite properties. They can be separated by physical methods.
e.g. Paint (mixture of oils, pigment, additive), concrete (a mixture of sand, cement, water), etc.

Mixtures are of two types:

• Homogeneous mixture: In homogeneous mixture, constituents remain uniformly mixed throughout its bulk.
  e.g. Solution, in which solute and solvent molecules are uniformly mixed throughout its bulk.
• Heterogeneous mixture: In heterogeneous mixture, constituents are not uniformly mixed throughout its bulk.
  e.g. Suspension, which contains insoluble solid in a liquid.

11th Chemistry Digest Chapter 1 Some Basic Concepts of Chemistry Intext Questions and Answers

Can you tell? (Textbook Page No. 1)

Question 1.
Which are mixtures and pure substances from the following?
i. Sea water
ii. Gasoline
iii. Skin
iv. A rusty nail
v. A page of textbook
vi. Diamond
Answer:

Can you tell? (Textbook Page No. 2)

Question 1.
Classify the following as element and compound.
i. Mercuric oxide
ii. Helium gas
iii. Water
iv. Table salt
v. Iodine
vi. Mercury
vii. Oxygen
viii. Nitrogen
Answer:

Can you tell? (Textbook Page No. 6)

Question 1.
If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:

If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, then 10 volumes of water vapour would be produced.

Can you recall? (Textbook Page No. 6)

Question 1.
What is an atom and molecule? What is the order of magnitude of mass of one atom? What are isotopes?
Answer:

• The smallest indivisible particle of an element is called an atom.
• A molecule is an aggregate of two or more atoms of definite composition which are held together by chemical bonds.
• Every atom of an element has definite mass. The order of magnitude of mass of one atom is 10^-27 kg.
• Isotopes are the atoms of the same element having same atomic number but different mass number.

Try this (Textbook Page No. 8)

Question 1.
Find the formula mass of CaSO₄, if atomic mass of Ca = 40.1 u, S =32.1 u and O = 16.0 u.
Solution:
Formula mass of CaSO₄
= Average atomic mass of Ca + Average atomic mass of S + Average atomic mass of four O
= (40.1) + 32.1 + (4 × 16.0) = 136.2 u
Ans: Formula mass of CaSO₄ = 136.2 u

Can you recall? (Textbook Page No. 8)

Question 1.
i. One dozen means how many items?
ii. One gross means how many items?
Answer:
i. One dozen means 12 items.
ii. One gross means 144 items.

Try this (Textbook Page No. 10)

Question 1.
Calculate the volume in dm³ occupied by 60.0 g of ethane at STP.
Solution:
Given: Mass of ethane at STP = 60.0 g
To find: Volume of ethane
Formulae:

Calculation: Molar volume of a gas = 22.4 dm³ mol^-1 at STP
Molecular mass of ethane = 30 g mol^-1

∴ Volume of the gas at STP = Number of moles of a gas (n) × Molar volume of a gas
= 2 mol × 22.4 dm³ mol^-1 = 44.8 dm³
Ans: Volume of ethane = 44.8 dm³

Activity :

Activity 1.
Collect information from various scientists and prepare charts of their contributions to chemistry.
Answer:

Maharashtra State Board Class 11 Chemistry Textbook Solutions

• Some Basic Concepts of Chemistry Class 11 Chemistry Textbook Solutions
• Introduction to Analytical Chemistry Class 11 Chemistry Textbook Solutions
• Basic Analytical Techniques Class 11 Chemistry Textbook Solutions
• Structure of Atom Class 11 Chemistry Textbook Solutions
• Chemical Bonding Class 11 Chemistry Textbook Solutions
• Redox Reactions Class 11 Chemistry Textbook Solutions
• Modern Periodic Table Class 11 Chemistry Textbook Solutions
• Elements of Group 1 and 2 Class 11 Chemistry Textbook Solutions
• Elements of Group 13, 14 and 15 Class 11 Chemistry Textbook Solutions

Skeleton and Movement Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 16 Skeleton and Movement Textbook Exercise Questions and Answers.

Class 11 Biology Chapter 16 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 16 Exercise Solutions

1. Choose the correct option

Question (A).
The functional unit of striated muscle is …………..
a. cross-bridges
b. myofibril
c. sarcomere
d. z-band
Answer:
c. sarcomere

Question (B).
A person slips from the staircase and breaks his ankle bone. Which bones are involved?
a. Carpals
b. Tarsal
c. Metacarpals
d. Metatarsals
Answer:
b. Tarsal

Question (C).
Muscle fatigue is due to the accumulation of ……..
a. pyruvic acid
b. lactic acid
c. malic acid
d. succinic acid
Answer:
b. lactic acid

Question (D).
Which one of the following is NOT antagonistic muscle pair?
a. Flexo-extensor
b. Adductor-abductor
c. Levator-depressor
d. Sphinetro-suprinater
Answer:
d. Sphinetro-suprinater

Question (E).
Swelling of sprained foot is reduced by soaking in hot water containing a large amount of common salt,
a. due to osmosis
b. due to plasmolysis
c. due to electrolysis
d. due to photolysis
Answer:
a. due to osmosis

Question (F).
Role of calcium in muscle contraction is ……….
a. to break the cross bridges as a cofactor in the hydrolysis of ATP
b. to bind with troponin, changing its shape so that the actin filament is exposed
c. to transmit the action potential across the neuromuscular junction.
d. to re-establish the polarisation of the plasma membrane following an action potential
Answer:
b. to bind with troponin, changing its shape so that the actin filament is exposed

Question (G).
Hyper-secretion of parathormone can cause which of the following disorders?
a. Gout
b. Rheumatoid arthritis
c. Osteoporosis
d. Gull’s disease
Answer:
c. Osteoporosis

Question (H).
Select correct option between two nasal bones

Answer:
(c)

Question 2.
Answer the following questions

Question (A).
What kind of contraction occurs in your neck muscles while you are reading your class assignment?
Answer:

1. Isometric contractions occur in the neck muscles while reading class assignment.
2. These contractions are important for supporting objects in a fixed position.

Question (B).
Observe the diagram and enlist importance of ‘A’, ‘B’ and ‘C’.

Answer:

1. A – Posterior portion of vertebral foramen of atlas vertebrae; Importance – The spinal cord runs through this portion of vertebral foramen
2. B – Anterior portion of vertebral foramen of axis vertebrae; Importance – In this portion, the odontoid process of axis vertebrae forms ‘NO’ joint.
3. C – Inferior articular facet; Importance – It articulates with superior articular facet of axis and permits rotatory movement of head.

Question (C).
Raju intends to train biceps; while exercising using dumbbells, which joints should remain stationary and which should move?
Answer:
While performing exercise of biceps using dumbbells, the joint which should remain stationary are wrist joint or radiocarpal joint, ball and socket joint of shoulder. The only joint which should move is hinge joint of elbow.

Question (D).
In a road accident, Moses fractured his leg. One of the passers by, tied a wodden plank to the fractured leg while Moses
was rushed to the hospital Was this essential? Why?
Answer:

1. Fracture is a significant and traumatic injury which requires medical attention however, getting timely first aid is important.
2. If any bone is fractured, it is essential that the fractured part be immobilized to prevent further injury. It can be done with the help of any available wooden plank or batons or rulers. Thus, a wooden plank was tied to Moses’s fractured leg as a first aid for fracture.
3. A fractured bone is immobilized to prevent the sharp edges of the fractured bone from moving and cutting tissue, muscle, blood vessels and nerves. Immobilization can also help reduce pain or control shock.

Question (E).
Sprain is more painful than fracture. Why?
Answer:

1. A sprain is an injury that involves the ligaments (tissues that connect bones at joints), whereas a fracture is an injury that involves bones.
2. Sprains can be of three degree: 1^st degree: Mild with micro-tears, 2^nd degree: Partial with visible tear in ligament, 3^rd degree: Completely torn ligament.
3. If a sprain is 3^rd degree, it will be more painful than a fracture. It usually requires a surgery to fix this injury, while breaking a bone, most of the time does not require surgery.
4. Breaks or Fractures also vary greatly. Minor fractures (like stress/ hairline fractures) are much less painful than compound/ complex fractures in which the bone may be cracked into half.
5. Blood supply is essential for growth and regeneration. Bones are highly vascularized whereas, ligaments are not. This causes the bones to heal comparatively faster than severe sprains. Thus, the duration of enduring pain until the injury heals also differs.
6. Also, ligaments have a rich supply of sensory nerves, which may also be responsible for an elevated sense of pain during severe sprains.

[Note: 1^st and 2^nd degree sprains are not very serious and may be lesser painful than a fracture. Depending on the severity of the injury, intensity of pain will vary.]

Question (F).
Why a red muscle can work for a prolonged period whereas white muscle fibre suffers from fatigue after a shorter work? (Refer to chapter animal tissues.)
Answer:

1. Red muscle fibres contain large amount of myoglobin and mitochondria (site of aerobic respiration), whereas white muscles fibres contain lesser amount of myoglobin and mitochondria.
2. Myoglobin is an iron-containing pigment that carries oxygen molecules to muscle tissues. Abundance of these pigments in red muscle fibres supports higher rate of aerobic respiration, whereas white muscle fibres have less mitochondria and depend upon anaerobic respiration.
3. Anaerobic respiration in muscle white fibres leads to the production of lactic acid and accumulation of higher of levels lactic acid can result in fatigue in white muscle fibres.

Thus, red muscle fibres can perform prolonged work and show less fatigue due to accumulation of negligible amount loss or of lactic acid, whereas white muscle fibres suffer from fatigue after a shorter work due to accumulation of higher amount of lactic acid.

3. Answer the following questions in detail

Question (A).
How is the structure of sarcomere suitable for the contractility of the muscle? Explain its function according to sliding
filament theory. (Refer to chapter animal tissues.)
Answer:
i. Sarcomere is the functional unit of myofibril. It has specific arrangement of actin and myosin filaments. The components of sarcomere are organized into variety of bands and zones. Actin and myosin are referred as contractile proteins. Actin is called as thin filament whereas myosin in called as thick filament. The structure of sarcomere:

ii. ‘A’ band – dark bands present at the centre of sarcomere and contain myosin as well as actin.
‘H’ zone or Hensen’s zone – light area present at the centre of ‘A’ band
‘M’ line – present at the centre of ‘H’ zone
‘I’ band – light bands present on the either side of ‘A’ band containing only actin
Z’ line – adjacent ‘I’ bands are separated by ‘Z’ line.

iii. Sliding filament theory: It was put forth by H.E Huxley and A.F Huxley. It is also known as ‘Walk along theory’ or Ratchet theory.

• According to the sliding filament theory, the interaction between actin and myosin filaments is the basic cause of muscle contraction. The actin filaments are interdigitated with myosin filaments.
• The head of the myosin is joined to the actin backbone by a cross bridge forming a hinge joint. From this joint, myosin head cannot tilt forward or backward. This movement is an active process as it utilizes ATP.
• Myosin head contains ATPase activity. It can derive energy by the breakdown of ATP molecule. This energy can be used for the movement of myosin head.
• During contraction, the myosin head gets attached to the active site of actin filaments and pull them inwardly so that the actin filaments slide over the myosin filaments. This results in the contraction of muscle fibre.
  

Question (B).
Ragini, a 50 year old office goer, suffered hair-line cracks in her right and left foot in short intervals of time. She was worried about minor jerks leading to hair line cracks in bones. Doctor explained to her why it must be happening and prescribed medicines.

What must be the cause of Ragini’s problem? Why has it occurred? What precautions she should have taken earlier? What care she should take in future?
Answer:

1. Considering Ragini’s age, she may be undergoing menopause. After menopause, oestrogen level declines resulting in lower bone density.
2. Osteoporosis:
   • In this disorder, bones become porous and hence brittle. It is primarily age related disease and is more common in women than men.
   • Osteoporosis may be caused due to decreasing estrogen secretion after menopause, deficiency of vitamin D, low calcium diet, decreased secretion of sex hormones and thyrocalcitonin.
3. As age advances, bone resorption outpaces bone formation. Hence, the bones lose mass and become brittle. More calcium is lost in urine, sweat, etc., than it is gained through diet. Thus, prevention of disease is better than treatment by consuming adequate amount of calcium and exercise at young age.
4. A person with previous hairline fractures is more susceptible to reoccurrence of fractures. Hence, Ragini needs to take her medications and supplements properly, avoid jerky movements and maintain body weight.

Question (C).
How does structure of actin and myosin help muscle contraction?
Answer:
i. Myosin filament:

1. Each myosin filament is a polymerized protein.
   Many meromyosins (monomeric proteins) constitute one thick filament.
2. Myosin molecule consists of two heavy chains (heavy meromyosin / HMM) coiled around each other forming a double helix. One end of each of these chains is projected outwardly is known as cross bridge. This end folds to form a globular protein mass called myosin head.
3. Two light chains are associated with each head forming 4 light chains/light meromyosin / LMM.
4. Myosin head has a special ATPase activity. It can split ATP to produce energy.
5. Myosin contributes 55% of muscle proteins.
6. In sarcomere, myosin tails are arranged to point towards the centre of the sarcomere and the heads point to the sides of the myofilament band.
   

ii. Actin filament: It is a complex type of contractile protein. It is made up of three components:

1. F actin: It forms the backbone of actin filament. F actin is made up of two helical strands. Each strand is composed of polymerized G actin molecules. One ADP molecule is attached to G actin molecule.
2. Tropomyosin: The actin filament contains two additional protein strands that are polymers of tropomyosin molecules. Each strand is loosely attached to an F actin. In the resting stage, tropomyosin physically covers the active myosin-binding site of the actin strand.
3. Troponin: It is a complex of three globular proteins, is attached approx. 2/3^rd distance along each tropomyosin molecule. It has affinity for actin, tropomyosin and calcium ions. The troponin complex is believed to attach the tropomyosin to the actin. The strong affinity of troponin for calcium ions is believed to initiate the contraction process.
   

Question (D).
Justify the structure of atlas and axis vertebrae with respect to their position and function.
Answer:
i. Atlas vertebrae:

1. Atlas is the ring-like, 1^st cervical vertebrae. It has anterior, posterior arches and large lateral masses.
2. It lacks centrum and spinous process. The superior surfaces of the lateral masses are concave and are known as superior articular facets.
3. These facets articulate with the occipital condyles of the occipital bone thereby forming atlanto-occipital joints. This articulation permits ‘YES movement’ or nodding movement.
4. The inferior surfaces of the lateral masses known as inferior articular facets articulate with axis vertebrae,

ii. Axis vertebrae:

1. It is the 2^nd cervical vertebrae.
2. A peg-like process called odontoid process projects superiorly through the anterior portion of the vertebral foramen of the atlas.
3. The odontoid process forms a pivot on which the atlas and head rotate. This arrangement allows ‘NO movement’ or side to side movement of the head.
4. The articulation formed between the anterior arch of atlas, the odontoid process of the axis and between their articular facets is called as atlanto-axial joint.

Question (E).
Observe the blood report given below and diagnose the possible disorder.

Answer:
On observing Report D, it is clear that the level of uric acid is more than normal, thus the patient must be suffering from gouty arthritis.

Also, the elevated blood urea nitrogen (BUN) indicates dysfunctional liver and/ or kidneys. It generally occurs due to decrease in GFR, caused by renal disease or obstruction of urinary tract.

Question 4.
Write short notes on following points

Question (A).
Actin filament
Answer:
Actin filament: It is a complex type of contractile protein. It is made up of three components:

1. F actin: It forms the backbone of actin filament. F actin is made up of two helical strands. Each strand is composed of polymerized G actin molecules. One ADP molecule is attached to G actin molecule.
2. Tropomyosin: The actin filament contains two additional protein strands that are polymers of tropomyosin molecules. Each strand is loosely attached to an F actin. In the resting stage, tropomyosin physically covers the active myosin-binding site of the actin strand.
3. Troponin: It is a complex of three globular proteins, is attached approx. 2/3rd distance along each tropomyosin molecule. It has affinity for actin, tropomyosin and calcium ions. The troponin complex is believed to attach the tropomyosin to the actin. The strong affinity of troponin for calcium ions is believed to initiate the contraction process.
   

Question (B).
Myosin filament
Answer:
i. Myosin filament:

1. Each myosin filament is a polymerized protein.
   Many meromyosins (monomeric proteins) constitute one thick filament.
2. Myosin molecule consists of two heavy chains (heavy meromyosin / HMM) coiled around each other forming a double helix. One end of each of these chains is projected outwardly is known as cross bridge. This end folds to form a globular protein mass called myosin head.
3. Two light chains are associated with each head forming 4 light chains/light meromyosin / LMM.
4. Myosin head has a special ATPase activity. It can split ATP to produce energy.
5. Myosin contributes 55% of muscle proteins.
6. In sarcomere, myosin tails are arranged to point towards the centre of the sarcomere and the heads point to the sides of the myofilament band.
   

Question (C).
Role of calcium ions in contraction and relaxation of muscles.
Answer:
Calcium ions play a major role in contraction and relaxation of muscles.

1. Calcium ions are released from the sarcoplasm during muscle contraction and stored in sarcoplasmic reticulum during muscle relaxation.
2. When a skeletal muscle is excited and an action potential travels along the T tubule, the concentration of calcium ions increases.
3. These calcium ions bind to troponin which in turn undergoes a conformational change that causes tropomyosin to move away from the myosin-binding sites on actin. Once these binding sites are free, myosin heads bind to them to form cross-bridges and the muscle fiber contracts.
4. The decrease in calcium ion concentration in the sarcoplasmic reticulum causes tropomyosin to slide back and block the myosin binding sites on actin. This causes the muscle to relax.

Question 5.
Draw labelled diagrams

Question (A).
Synovial joint.
Answer:
i. Synovial joints / freely movable joints / diarthroses:

1. It is characterized by presence of a space called synovial cavity between articulating bones that renders free movement at the joint.
2. The articulating surfaces of bones at a synovial joint are covered by a layer of hyaline cartilage. It reduces friction during movement and helps to absorb shock.
3. Synovial cavity is lined by synovial membrane that forms synovial capsule. Synovial membrane secretes synovial fluid.
4. Synovial fluid is a clear, viscous, straw coloured fluid similar to lymph. It is viscous due to hyaluronic acid. The synovial fluid also contains nutrients, mucous and phagocytic cells to remove microbes.
   Synovial fluid lubricates the joint, absorbs shocks, nourishes the hyaline cartilage and removes waste materials from hyaline cartilage cells (as cartilage is avascular). Phagocytic cells destroy microbes and cellular debris formed by wear and tear of the joint.
5. If the joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
6. The joint is provided with capsular ligament and numerous accessory ligaments. The fibrous capsule is attached to periosteum of articulating bones. The ligament helps in avoiding dislocation of joint.
7. The types of synovial joints are on follows:

1. Pivot joint: In this type of joint, the rounded or pointed surface of one bone articulates with a ring formed partly by another bone and partly by the ligament. Rotation only around its own longitudinal axis is possible. e.g. in joint between atlas and axis vertebrae, head turns side ways to form ‘NO’ joint.

2. Ball and socket joint: The ball like surface of one bone fits into cup like depression of another bone forming a movable joint. Multi-axial movements are possible. This type of joint allows movements along all three axes and in all directions. e.g. Shoulder and hip joint.

3. Hinge joint: In a hinge joint, convex surface of one bone fits into concave surface of another bone. In most hinge joints one bone remains stationary and other moves. The angular opening and closing motion (like hinge) is possible. In this joint only mono-axial movement takes place like flexion and extension. e.g. Elbow and knee joint.

4. Condyloid joint: It is an ellipsoid joint. The convex oval shaped projection of one bone fits into oval shaped depression in another bone. it is a biaxial joint because it permits movement along two axes viz, flexion, extension, abduction, adduction and circumduction is possible. e.g. Metacarpophalangeal joint.

5. Gliding joint: It is a planar joint, where the articulating surfaces of bones are flat or slightly curved. These joints are non-axial because the motion they allow does not occur along an axis or a plane. e.g. Intercarpal and intertarsal joints.

6. Saddle joint: This joint is a characteristic of Homo sapiens. Here the articular surface of one bone is saddle-shaped and that of other bone fits into saddle (each bone forming this joint have both concave and convex areas). It is a modified condyloid joint in which movement is somewhat more free. It is a biaxial joint that allows flexion, extension, abduction, adduction and circumduction.
e.g. Carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb.

Question (B).
Different cartilagenous joints.
Answer:
Cartilaginous / slightly movable joints / amphiarthroses:
These joints are neither fixed nor freely movable. Articulating bones are held together by hyaline or fibrocartilages. They are further classified as

a. Synchondroses: The two bones are held together by hyaline cartilage. They are meant for growth. On completion of growth, the joint gets ossified, e.g. Epiphyseal plate found between epiphysis and diaphysis of a long bone, Rib – Sternum junction.

b. Symphysis: In this type of joint, broad flat disc of fibrocartilage connects two bones. It occurs in mid-line of the body. e.g. Intervertebral discs, manubrium and sternum, pubic symphysis.

Practical / Project :

Identify the following diagrams and demonstrate the concepts in classroom.

Answer:
The diagrams A, B and C represent Class I, Class II and Class III lever respectively.
For description:

1. Class I lever: The joint between the first vertebra and occipital condyle of skull is an example of Class I lever. The force is directed towards the joints (fulcrum); contraction of back muscle provides force while the part of head that is raised acts as
   
2. Class II lever: Human body raised on toes is an example of Class II lever. Toe acts as fulcrum, contracting calf muscles provide the force while raised body acts as resistance.
   
3. Class III lever: Flexion of forearm at elbow exhibit lever of class III. Elbow joint acts as fulcrum and radius, ulna provides resistance. Contracting bicep muscles provides force for the movement.
   

[Students are expected to perform the given activity on their own]

12th Biology Digest Chapter 16 Skeleton and Movement Intext Questions and Answers

Movements And Locomotion (Textbook Page No. 193)

Question 1.
Streaming of protoplasm, peristalsis, walking, running, etc. Which of the above-mentioned movements are internal? Which are external? Can you add few more examples?
Answer:

1. Streaming of protoplasm, peristalsis are internal movements. Walking and running are external movements.
2. Examples of internal movement: Contraction and relaxation heart, inspiration and expiration, contraction of blood vessels, etc.
3. Examples of external movement: Swimming; movement tongue, jaws, snout, tentacles, movement of ear pinna, etc.

Can you recall? (Textbook Page No. 193)

Question 1.
Which are different types of muscular tissues?
Answer:

1. Smooth / non-striated / visceral / involuntary muscles
2. Cardiac muscles
3. Skeletal / straited / voluntary muscles.

Question 2.
Name the type of muscles which bring about running and speaking.
Answer:
Skeletal muscles (Voluntary muscles)

Question 3.
Name the muscles which do not contract as per our will.
Answer:
Involuntary muscles (smooth muscles and cardiac muscles)

Question 4.
Which type of muscles show rhythmic contractions?
Answer:
Cardiac muscles

Question 5.
Which type of muscle is present in the diaphragm of the respiratory system?
Answer:
Skeletal muscle

Question 6.
State the functions of:

1. Smooth muscles
2. Cardiac muscles
3. Striated muscles

Answer:

1. Smooth muscles: They bring about involuntary movements like peristalsis in the alimentary canal, constriction and dilation of blood vessels.
2. Cardiac muscles: They bring about contraction and relaxation of the heart.
3. Striated muscles: They control voluntary movements of limbs, head, trunk, eyes, etc.

Can you recall? (Textbook Page No. 193)

Question 1.
Name the part of human skeleton situated along the vertical axis.
Answer:
Axial skeleton

Question 2.
Give an account of bones of human skull.
Answer:
Skull is made up of 22 bones. It is located at the superior end of vertebral column. The bones of skull are
joined by fixed or immovable joints except for jaw.

Skull consists of cranium or brain box and facial bones.

i. Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1^st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

ii. Facial Bones: Fourteen facial bones give a characteristic shape to the face. The growth of face stops of the
age of 16.
Following bones comprise the facial bones:

1. Nasals: These are paired bones that form the bridge of nose.
2. Maxillae: These form the upper jaw bones. They are paired bones that join with all facial bones except mandible. Upper row of teeth are lodged maxillae.
3. Palatines: These are paired bones forming the roof of buccal cavity or floor of the nasal cavity.
   
4. Zygomatic bones: They are commonly called as cheek bones.
5. Lacrimal bones: These are the smallest amongst the facial bones.
   These bones form the medial wall of each orbit. They have lacrimal fossa that houses lacrimal sacs. These sacs gather tears and send them to nasal cavity.
6. Inferior nasal conchae: They form the part of lateral wall of nasal cavity. They help to swirl and filter air before it passes to lungs.
7. Vomer: The median, roughly triangular bone that forms the inferior portion of nasal septum.
8. Mandible: This median bone forms the lower jaw. It is the largest and strongest facial bone. It is the only movable bone of skull. It has curved horizontal body and two perpendicular branches i.e. rami. These help in attachment of muscles. It has lower row of teeth lodged in it.

Think about it. (Textbook Page No. 193)

Question 1.
Did you ever feel tickling in muscles?
Answer:
Yes, the tickling sensation in muscles can be felt and sometimes it is also accompanied by itching sensation.

Question 2.
What is locomotion?
Answer:
The change in locus of whole body of living organism from one place to another place is called locomotion.

Question 3.
State the four basic types of locomotory movements seen in animals.
Answer:
The four basic types locomotory movements seen in animals are:

1. Amoeboid movement: It is performed by pseudopodia, e.g. leucocytes.
2. Ciliary movement: It is performed by cilia, e.g. ciliated epithelium. In Paramoecium, cilia help in passage of food through cytopharynx.
3. Whirling movement: It is performed by flagella, e.g. sperms.
4. Muscular movement: It is performed by muscles, with the help of bones and joints.

Question 4.

Question 1.
Why do muscles show spasm after rigorous contraction?
Answer:

1. Rigorous contraction of muscles occurs during strenuous activities (swimming. running, cycling, aerobics. etc.)
2. Muscle contraction requires energy. Glucose in muscle cells breakdown during anerobic respiration resulting in accumulation of lactic acid.
3. This lactic acid buildup triggers muscle spasm around muscle cells.

Question 2.
Why do we shiver during winter?
Answer:

1. Humans are homeotherms as the can regulate their body temperature with respect to the surrounding temperature. During winter, when temperature falls, the thermoreceptors detect the change in temperature and send signals to the brain.
2. Shivering reflex i.e. rapid contraction of muscles is triggered by the brain to generate heat and raise the body temperature.

Can you tell? (Textbook Page No. 194)

Question 1.
Why are movement and locomotion necessary among animals?
Answer:

• Movement is one of the important characteristics of all the living organisms. Animals exhibit wide range of
  movements like rhythmic beating of heart, movement of diaphragm during respiration, ingestion of food,
  movement of eyeballs, etc.
• Locomotion results into change in place or location of an organism. Animals locomote in search of food, mate, shelter, breeding ground. while escaping from the enemy, etc.
• Thus, locomotion and movement are necessary to support the living of animals.

Question 2.
All locomotions are movements but all movements are not locomotion. Justify
Answer:
Locomotion occurs when body changes its position, however all movements may not result in locomotion. Thus, all locomotions are movements but all movements are not locomotion.

Question 3.
Kriti was diagnosed with knee tendon injury. She asked the doctor whether she will be able to walk due to the injury? If not then state the reason.
Answer:
Knee tendon injury affects the ability to walk. Kriti may not be able to walk freely as the tendons attached to the bones help in the movement of the parts of skeleton.

Question 4.
What are antagonistic muscles? Explain with example.
Answer:

1. The muscles that work in pairs and produce opposite action are known as antagonistic muscles, e.g. biceps and triceps of upper arm.
2. The biceps (flexors) bring flexion (folding) and triceps (extensors) bring extension of the elbow joint.
3. One member from a pair is capable of bending the joint by pulling of bones the other member is capable of straightening the same joint by pulling.
4. In antagonistic pair of muscles, one member is stronger than the other, e.g. The biceps are stronger than the triceps.

Question 5.
Describe the antagonistic muscles in detail.
Answer:
Following are the important antagonistic muscles:

1. Flexor and extensor: Flexor muscle on contraction results into bending or flexion of joint, e.g. Biceps. Extensor muscle on contraction results in straightening or extension of a joint, e.g. Triceps.
2. Abductor and adductor: Abductor muscle moves a body part away from the body axis. e.g. Deltoid muscle of shoulder moves the arm away from the body. Adductor muscle moves a body part towards the body axis, e.g. Latissimus dorsi of shoulder moves the arm near the body.
3. Pronator and supinator: Pronator turns the palm downwards and supinator turns the palm upward.
4. Levator and depressor: Levator raises a body part and the depressors lower the body part.
5. Protractor and retractor: Protractor moves forward, whereas the retractor moves backward.
6. Sphincters: Circular muscles present in the inner walls of anus, stomach, etc., for closure and opening.

Question 6.
Differentiate between:
i. Flexor and extensor muscles
Answer:

ii. Pronator and supinator: Pronator turns the palm downwards and supinator turns the palm upward.

Can you recall? (Textbook Page No. 194)

Question 1.
Comment on contraction of skeletal muscles.
Answer:
Skeletal muscles show quick and strong voluntary contractions. They bring about voluntary movements of the body. For mechanism of muscle contraction:

When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of acto-myosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Do You Know How (Textbook Page No. 195)

Question 1.
Do skeletal muscles contract and bring about movement and locomotion?
Answer:
When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of acto-myosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Internet my friend. (Textbook Page No. 196)

Question 1.
Collect information about‘T’ tubules of sarcoplasmic reticulum.
Answer:

1. T tubules or the transverse tubules are invaginations of the sarcolemma penetrating into the myocyte interior, forming a highly branched and interconnected network that makes junctions with the sarcoplasmic reticulum.
2. These tubules are selectively enriched with specific ion channels and proteins crucial in the development of calcium transients necessary in excitation-contraction coupling, thereby facilitating a fast-synchronous contraction of the entire cell volume.
3. They are unique to straited muscle cells.
   [Source: https://www. uniprot. org/locations]

Can you tell? (Textbook Page No. 197)

Question 1.
Explain the chemical changes taking place in muscle contraction.
Answer:
The muscle undergoes various chemical changes during contraction, they are as follows:

1. A nerve impulse arrives at the motor nerve. The neurotransmitter – acetylcholine is released at the neuromuscular junction (N-M junction or motor endplate) enters into the sarcomere.
2. This leads to inflow of Na+ inside the sarcomere and generates an action potential in the muscle fibre.
3. The action potential passes down the T tubules and activates calcium channels in the T tubular membrane. Activation of calcium channel allows calcium ions to pass into the sarcoplasm. These Ca⁺⁺ ions binds to the specific sites on troponin of actin filaments and a conformational change occurs in the troponin – tropomyosin complex, thereby removing the masking of active sites for myosin on the actin filament.
4. In the myosin head, the enzyme ATPase gets activated in the presence of Ca⁺⁺ and converts ATP into ADP and inorganic phosphate.
5. This energy from ATP hydrolysis is utilized by myosin bridges or myosin heads to bind with active sites of actin and form actomyosin complex pulling the actin filaments towards the centre of sarcomere. The myosin heads are now tilted backwards and pull the attached actin filament inwardly towards them. The actin filament slides over mysosin and contraction occurs.
6. Also, the ADP needs to be converted back to ATP immediately as they required for muscular contraction, This is achieved in the muscles by the presence of another high energy compound, creatine phosphate.
7. ADP combines with creatine phosphate to produce ATP and creatinine due to which the supply of ATP for muscle contraction is restored but the level of creatine phosphate keeps decreasing and the level of creatinine keeps on increasing.
8. The creatinine formed needs to be reconverted to creatine phosphate. This is done by ATP produced during oxidation of glycogen through glycolysis.

Question 2.
Why are muscle rich in creatine phosphate?
Answer:

1. Creatine phosphate or phosphocreatine is formed from ATP, when the muscle is in relaxed state. It is a phosphorylated form of creatine.
2. Muscle cells contain creatine phosphate which acts as energy reserve as this high energy compound acts as a phosphate donor for ATP formation.
3. ATP acts as an immediate source of energy for contraction

Question 3.
Explain the mechanism of muscle contraction and relaxation.
Answer:
Mechanism of muscle contraction:

When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of actomyosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Muscle relaxation: During relaxation, all the events occur in reverse direction as that of muscle contraction.

1. When the stimulation is terminated, the actomyosin complex is broken down and myosin head gets detached from actin filaments. This process utilizes ATP.
2. Also, the Ca⁺⁺ ions are pumped back into the sarcoplasmic reticulum. This process too is an energy dependent process and utilizes ATP.
   
3. As a result, the troponin-tropomyosin complex is restored again which covers the active sites of act in filament, due to disappearance of the Ca⁺⁺ ions. The interaction between actin and myosin ceases and the actin filaments return back to their original position.
4. This results in muscle relaxation.

Question 4.
What do you understand by muscle twitch?
Answer:
Single muscle twitch:

Single muscle twitch: It is a muscle contraction initiated by a single brief-stimulation. It occurs in 3 stages: a latent period of no contraction, a contraction period and a relaxation period.

1. The involuntary contraction of muscle fibers is known as muscle twitch.
2. Muscle twitch is also known as fasciculation.
3. It is caused due accumulation of lactic acid in muscles.

Do You Know How (Textbook Page No. 198)

Question 1.
Exoskeletal components change from lower to higher group of animals. These include chitinous structures, nails, horns, hooves, scales, hair, shell, plates, fur, muscular foot, tube feet, etc.

Question 1.
Do you know any of these exoskeletal structures help in movement and locomotion?
Answer:
Nails, hooves, scales, plates, muscular foot and tube feet help in movement and locomotion.

Question 2.
How do scales and plates help in movement and locomotion?
Answer:
Scales and plates in reptiles like snakes provide grip to move on rough edgy surfaces.

Question 3.
Are scales of a fish and that of snake similar?
Answer:
Fishes have dermal scales (bony scales), whereas reptiles like snakes have epidermal scales or scutes (horny, tough extensions of outer layer of skin i.e., stratum corneum).

Question 4.
Find out more information about exoskeletal structures and their role in movement and locomotion.
Answer:
Exoskeletal structures: Exoskeleton provide support, help in movement and also provides protection from predators. The exoskeletal structures vary from organism to organism. Echinoderms have tube feet for locomotion whereas molluscs (e.g. Chiton) have muscular foot for movement and locomotion

[Students are expected to find out more information about exoskeletal structures on their own.]

Question 2.
Name the tissues that form the structural framework of the body.
Answer:
Cartilage and bone

Do you remember? (Textbook Page No. 198)

Question 1.
What are the components of skeletal system?
Answer:
The components of skeletal system are bones, tendons, ligaments and joint.

Question 2.
What type of bones are present in our body?
Answer:
Long bones, short bones, flat bones, irregular bones and sesamoid bones.

Question 3.
How do bones help in various ways?
Answer:

1. Bones form the framework of our body and thus provide shape to the body.
2. They protect vital organs thus help in the smooth functioning of body.
3. The joints between the bones help in movement and locomotion.
4. They provide firm surface for attachment of muscles.
5. They are reservoirs of calcium and form important site for hemopoiesis.

Use your brain power. (Textbook Page No. 198)

Question 1.
Can you compare bone, muscle and joint which help in locomotion with any simple machines you have studied earlier?
Answer:
Bone, muscle and joint can be compared to the simple machines called levers. Joints act as fulcrum, respective muscle generates the force required to move the bone associated with joint.

Question 2.
Explain the three types of lever found in human body.
Answer:
The three types of lever are as follows:

1. Class I lever: The joint between the first vertebra and occipital condyle of skull is an example of Class I lever. The force is directed towards the joints (fulcrum); contraction of back muscle provides force while the part of head that is raised acts as resistance.
   
2. Class II lever: Human body raised on toes is an example of Class II lever. Toe acts as fulcrum, contracting calf muscles provide the force while raised body acts as resistance.
   
3. Class III lever: Flexion of forearm at elbow exhibit lever of class III. Elbow joint acts as fulcrum and radius, ulna provides resistance. Contracting bicep muscles provides force for the movement.
   

Use your brain power. (Textbook Page No. 199)

Question 1.
Why are long bones slightly bent and not straight?
Answer:

1. Long bones include tibia, fibula, femur, humerus, radius, ulna, etc.
2. They have greater length than width. They consist of a shaft and variable number of epiphysis.
3. They are slightly bent or curved to absorb the stress of the body’s weight and evenly distribute the body weight at several different points.
4. If long bones were straight, the weight of the body would be unevenly distributed and the bone would fracture more easily.

Identify and label. (Textbook Page No. 199)

Question 1.
Identify the different bones.
Answer:

Identify and label. (Textbook Page No. 200)

Question 1.
Name A, B, C and D from the given figure and discuss in group.

Answer:
A – Coronal suture,
B – Sagittal suture,
C – Lambdoidal suture,
D – Lateral / Squamous suture

Skull has many sutures (type of immovable joints) present, out of which four prominent ones are:

1. Coronal suture: Joins frontal bone with parietals.
2. Sagittal suture: Joins two parietal bones.
3. Lambdiodal suture: Joins two parietal bones with occipital bone.
4. Lateral/squamous sutures: Joins parietal and temporal bones on lateral side.

Can you tell? (Textbook Page No. 201)

Question 1.
Give schematic plan of human skeleton.
Answer:

[Note: Numbers in the bracket indicate the number of bones.]

Can you tell? (Textbook Page No. 201)

Question 1.
Enlist the bones of cranium.
Answer:
Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

Can you tell? (Textbook Page No. 201)

Question 1.
Write a note on structure and function of skull.
Answer:
i. Structure of skull:
Skull is made up of 22 bones. It is located at the superior end of vertebral column. The bones of skull are joined by fixed or immovable joints except for jaw.

Skull consists of cranium or brain box and facial bones.

i. Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1^st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

ii. Facial Bones: Fourteen facial bones give a characteristic shape to the face. The growth of face stops of the age of 16.
Following bones comprise the facial bones:

1. Nasals: These are paired bones that form the bridge of nose.
2. Maxillae: These form the upper jaw bones. They are paired bones that join with all facial bones except mandible. Upper row of teeth are lodged maxillae.
   
3. Palatines: These are paired bones forming the roof of buccal cavity or floor of the nasal cavity.
4. Zygomatic bones: They are commonly called as cheek bones.
5. Lacrimal bones: These are the smallest amongst the facial bones. These bones form the medial wall of each orbit. They have lacrimal fossa that houses lacrimal sacs. These sacs gather tears and send them to nasal cavity.
6. Inferior nasal conchae: They form the part of lateral wall of nasal cavity. They help to swirl and filter air before it passes to lungs.
7. Vomer: The median, roughly triangular bone that forms the inferior portion of nasal septum.
8. Mandible: This median bone forms the lower jaw. It is the largest and strongest facial bone. It is the only movable bone of skull. It has curved horizontal body and two perpendicular branches i.e. rami. These help in attachment of muscles. It has lower row of teeth lodged in it.

ii. Functions of skull:

1. It protects the brain.
2. It provides sockets for ear, nasal chamber and eyes.
3. Mandible bone of the skull helps in opening and closing of the mouth.

Internet my friend. (Textbook Page No. 201)

Question 1.
Cleft palate and cleft lip
Answer:

1. Cleft palate and cleft lip are the birth defects that occur when a baby’s lip or mouth does develop properly.
2. Cleft palate happens when the tissue that forms the roof of the mouth does not join together completely during pregnancy.
3. Cleft lip happens when the tissue that makes up the lip does not join completely before birth. This leads to formation of an opening in the upper lip.

[Students are expected to find more information about Cleft Palate and lip on internet.]

Can you tell? (Textbook Page No. 202)

Question 1.
Why skull is important for us? Enlist few reasons.
Answer:
Functions of skull:

• It protects the brain.
• It provides sockets for ear, nasal chamber and eyes.
• Mandible bone of the skull helps in opening and closing of the mouth.

Internet my friend. (Textbook Page No. 202)

Question 1.
Find out information about sinuses present in skull, functions of skull and disorder ‘sinusitis’.
Answer:
Sinuses are the hollow cavities present in the skull. They humidify the air we breathe.

i. The four types of sinuses present in the skull:

• Frontal sinuses: They are located above each eye. There are right and left frontal sinuses.
• Maxillary sinuses: They are the largest among all sinuses, located just behind the cheekbones near to upper jaws.
• Sphenoid sinuses: These are present just behind the nose.
• Ethmoid sinuses: These are present between the eyes.

ii. Functions of skull:

Functions of skull:

• It protects the brain.
• It provides sockets for ear, nasal chamber and eyes.
• Mandible bone of the skull helps in opening and closing of the mouth.

iii. Sinusitis: It is the inflammation of tissue lining the sinuses. Healthy sinuses when get blocked with mucus and germs causing infection which may lead to sinusitis.

[Students are expected to find more information about sinusitis, using the internet.]

Something interesting. (Textbook Page No. 202)

Question 1.
If police suspect strangulation, they carefully inspect hyoid bone and cartilage of larynx. These get fractured during strangulation. V arious such investigations are done in case of suspicious death of an individual where ossification of sutures in skull, width of pelvic girdle, etc. are examined to find out approximate age of victim or gender of victim, etc. You may find out information about forensic science.
Answer:
Forensic science is an application of science which is used in the matter of criminal determination and civil law. It is generally used in investigation of crimes. Forensic scientists collect, preserve and analyze the evidence during the course of investigation.
[Students are expected to find more information about forensic science on internet.]

Try this. (Textbook Page No. 202)

Question 1.
Feel your spine (vertebral column). Is it straight or curved?
Answer:
Our spinc shows four slight curves which are visible when viewed from the sides.

Question 2.
Find information about slipped disc. (Textbook page no.202)
Answer:

1. The bones of vertebral column are supported by the intervertebral discs.
2. These intervertebral discs act as shock absorbers due to which they are constantly compressed.
3. The disc consists of two parts – soft gelatinous inner part (nucleus pulposus) and tough outer ring.

If the ligaments of the intervertebral discs become injured, the pressure developed in the nucleus pulposus protrudes posteriorly or into one of the adjacent vertebrae. This is known as slipped disc.

[Students are expected to find more information using the internet.]

Can you tell? (Textbook Page No. 204)

Question 1.
Write a note on curvatures of vertebral column and mention their importance.
Answer:

1. The four curvatures in human spine are cervical, lumbar, thoracic and sacral curvatures.
2. The cervical and lumbar curvatures are secondary and convex whereas the thoracic and sacral curvatures are
   primary and concave.
3. Importance: Curvatures help in maintaining balance in upright position. They absorb shocks while walking
   and also protect the vertebrae from fracture.

Question 2.
Explain the structure of typical vertebra.
Answer:

1. Each vertebra has prominent central body called centrum.
2. The centra of human vertebrae are flat in anterio-posterior aspect. Thus, human vertebrae are amphiplatyan.
3. From the either side of the centrum are two thick short processes which unite to form an arch like structure called neural arch, posterior to centrum.
4. Neural arch forms vertebral foramen which surrounds the spinal cord.
5. Vertebral foramina of all vertebrae form a continuous ‘neural canal’. Spinal cord along with blood vessels and protective fatty covering passes through neural canal.
6. The point where two processes of centrum meet, the neural arch is drawn into a spinous process called neural spine.
   
7. From the base of neural arch, two articulating processes called zygapophyses are given out on either side. The anterior is called superior zygapophyses and posterior called inferior zygapophyses.
8. In a stack of vertebrae, inferior zygapophyses of one vertebra articulates with superior zygapophyses of next vertebra. This allows slight movement of vertebrae without allowing them to fall.
9. At the junction of zygapophyses, a small opening is formed on either side of vertebra called intervertebral foramen that allows passage of spinal nerve.
10. From the base of neural arch, lateral processes are given out called transverse processes. Neural arch, neural spine and transverse processes are meant for attachment of muscles.

Question 2.
How will you identify a thoracic vertebra?
Answer:
Thoracic vertebrae can be identified on the basis of centrum, as the centrum of the thoracic vertebrae is heart shaped.

Can you recall? (Textbook Page No. 206)

Question 1.
How does humerus form ball and socket joint? Where is it located?
Answer:
The head of humerus fits into the glenoid cavity of scapula and forms ball and socket joint. it is located in shoulder and hips.

Can you tell? (Textbook Page No. 208)

Question 1.
Differentiate between the skeleton of palm and foot.
Answer:

Question 2.
Explain the longest bone in human body.
Answer:
Femur: The thigh bone is the longest bone in the body. The head is joined to shaft at an angle by a short neck. It forms ball and socket joint with acetabulum cavity of coxal bone. The lower one third region of shaft is triangular flattened area called popliteal surface. Distal end has two condyles that articulate with tibia and fibula.

Internet my friend. (Textbook Page No. 212)

Question 1.
Find out information about types of fractures and how they heal.
Answer:

1. Fractures are classified based on their severity, shape or position of the fracture line or the physician who first described them.
2. Types of fractures:
   1. Open fractures: The broken ends of the bone protrude through skin.
   2. Comminuted fractures: The bone is splintered, crushed or broken into pieces at the site of impact and smaller bone fragments lie between the two main fragments.
   3. Greenstick fractures: A partial fracture in which one side of the bone is broken and the other side bends.
   4. Impacted fractures: One end of the fractured bone is forcefully driven into the inferior of the other.
   5. Pott fractures: Fracture of the distal end lateral leg bone with serious injury of the distal tibial articulation.
   6. Codes fractures: Fracture of the distal end of the lateral forearm in which the distal fragment is displaced posteriorly.
3. A fractured bone heals in four phases viz, reactive phase, fibrocartilaginous formation phase, bony callus formation phase and bone remodeling phase.

[Source: Tortora, G., Derrickson, B. Principles of Anatomy and Physiology. 15^th Edition]

[Students are expected to find out more information about healing of fractures using the internet.]

Do you remember? (Textbook Page No. 208)

Question 1.
What are joints? What are the types?
Answer:
i. A point where two or more bones get articulated is called joint or articulation or athrosis.
They are classified based on degree of flexibility or movement they permit into lastly synovial or freely movable or diarthroses type of joints.

ii. Synarthroses / fibrous joints / movable joints:
In this joint, the articulating bones are held together by means of fibrous connective tissue. Bones do not exhibit movement. Hence, it is immovable or fixed type of joint. Synarthroses are further classified into sutures, syndesmoses and gomphoses.

• Sutures: It is composed of thin layer of a dense fibrous connective tissue. Sutures are places of growth. They remain open till growth is complete. On completion of growth, they tend to ossify. Sutures may permit some moulding during childhood. Sutures are further classified into butt joint, scarf joint, lap joint and serrate joint.
• Syndesmoses: It is present where there is greater distance between articulating bones. At such locations, fibrous connective tissue is arranged as a sheet or bundle, e.g. Distal tibiofibular joint, interosseous membrane between tibia and fibula and that between radius and ulna.
• Gomphoses: In this type of joint, a cone shaped bone fits into a socket provided by other bone,
  e. g. Tooth and jaw bones.
  

iii. Cartilaginous / slightly movable joints / amphiarthroses:
These joints are neither fixed nor freely movable. Articulating bones are held together by hyaline or
fibrocartilages. They are further classified as

• Synchondroses: The two bones are held together by hyaline cartilage. They are meant for growth.
  On completion of growth, the joint gets ossified, e.g. Epiphyseal plate found between epiphysis and diaphysis of a long bone, Rib – Sternum junction.
• Symphysis: In this type of joint, broad flat disc of fibrocartilage connects two bones. It occurs in mid-line of the body. e.g. Intervertebral discs, manubrium and sternum, pubic symphysis.
  

iv. Synovial joints / freely movable joints / diarthroses:

1. It is characterized by presence of a space called synovial cavity between articulating bones that renders free movement at the joint.
2. The articulating surfaces of bones at a synovial joint are covered by a layer of hyaline cartilage. It reduces friction during movement and helps to absorb shock.
   
3. Synovial cavity is lined by synovial membrane that forms synovial capsule. Synovial membrane secretes synovial fluid.
4. Synovial fluid is a clear, viscous, straw coloured fluid similar to lymph. It is viscous due to hyaluronic acid. The synovial fluid also contains nutrients, mucous and phagocytic cells to remove microbes. Synovial fluid lubricates the joint, absorbs shocks, nourishes the hyaline cartilage and removes waste materials from hyaline cartilage cells (as cartilage is avascular). Phagocytic cells destroy microbes and cellular debris formed by wear and tear of the joint.
5. If the joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
6. The joint is provided with capsular ligament and numerous accessory ligaments. The fibrous capsule is attached to periosteum of articulating bones. The ligament helps in avoiding dislocation of joint.

g. The types of synovial joints are on follows:

1. Pivot joint: In this type of joint, the rounded or pointed surface of one bone articulates with a ring formed partly by another bone and partly by the ligament. Rotation only around its own longitudinal axis is possible. e.g. in joint between atlas and axis vertebrae, head turns side ways to form ‘NO’ joint.

2. BaIl and socket joint: The ball like surface of one bone fits into cup like depression of another bone forming a movable joint. Multi-axial movements are possible. This type of joint allows movements along all three axes and in all directions. e.g. Shoulder and hip joint.

3. Hinge joint: In a hinge joint, convex surface of one bone fits into concave surface of another bone. In most hinge joints one bone remains stationary and other moves. The angular opening and closing motion (like hinge) is possible. In this joint only mono-axial movement takes place like flexion and extension. e.g. Elbow and knee joint.

4. Condyloid joint: It is an ellipsoid joint. The convex oval shaped projection of one bone fits into oval shaped depression in another bone. It is a biaxial joint because it permits movement along two axes viz, flexion, extension, abduction, adduction and circumduction is possible. e.g. Metacarpophalangeal joint.

5. Gliding joint: It is a planar joint, where the articulating surfaces of bones are flat or slightly curved. These joints are non-axial because the motion they allow does not occur along an axis or a plane. e.g. Intercarpal and intertarsal joints.

Saddle joint: This joint is a characteristic of Homo sapiens. Here the articular surface of one bone is saddle-shaped and that of other bone fits into saddle (each bone forming this joint have both concave and convex areas). It is a modified condyloid joint in which movement is somewhat more free. It is a biaxial joint that allows flexion, extension, abduction, adduction and circumduction. e.g. Carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb.

Imagine. (Textbook Page No. 208)

Question 1.
If your elbow joint would be a fixed type of joint and joint between teeth and gum would be freely movable.
Answer:

1. If the elbow joint would be fixed the flexion and extension of the forearm won’t be possible. Also, rotation of the forearm and wrist would not be not possible.
2. Gomphoses is the type of joint that holds the teeth in the jaw bone. If this joint would be freely movable, we would not be able to chew and all our teeth would fall out.

Use your brain power. (Textbook Page No. 210)

Question 1.
Why are warming up rounds essential before regular exercise?
Answer:

1. Warming up before exercise stimulates the production and secretion of synovial fluid which reduces the stress on joints during exercise.
2. Also, if a joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
3. Warming up increases the blood circulation, loosening the joints and increasing the blood flow. It also prepares the muscles for physical activity and prevents injuries.

Can you tell? (Textbook Page No. 211)

Question 1.
Classify various types of joints found in human body. Present the information in the form of chart. Give example of each type.
Answer:

i. A point where two or more bones get articulated is called joint or articulation or athrosis.
They are classified based on degree of flexibility or movement they permit into lastly synovial or freely movable or diarthroses type of joints.

ii. Synarthroses / fibrous joints / movable joints:
In this joint, the articulating bones are held together by means of fibrous connective tissue. Bones do not exhibit movement. Hence, it is immovable or fixed type of joint. Synarthroses are further classified into sutures, syndesmoses and gomphoses.

1. Sutures: It is composed of thin layer of a dense fibrous connective tissue. Sutures are places of growth. They remain open till growth is complete. On completion of growth, they tend to ossify. Sutures may permit some moulding during childhood. Sutures are further classified into butt joint, scarf joint, lap joint and serrate joint.
2. Syndesmoses: It is present where there is greater distance between articulating bones. At such locations, fibrous connective tissue is arranged as a sheet or bundle, e.g. Distal tibiofibular joint, interosseous membrane between tibia and fibula and that between radius and ulna.
3. Gomphoses: In this type of joint, a cone shaped bone fits into a socket provided by other bone,
   e. g. Tooth and jaw bones.
   

iii. Cartilaginous / slightly movable joints / amphiarthroses:
These joints are neither fixed nor freely movable. Articulating bones are held together by hyaline or
fibrocartilages. They are further classified as

• Synchondroses: The two bones are held together by hyaline cartilage. They are meant for growth.
  On completion of growth, the joint gets ossified, e.g. Epiphyseal plate found between epiphysis and diaphysis of a long bone, Rib – Sternum junction.
• Symphysis: In this type of joint, broad flat disc of fibrocartilage connects two bones. It occurs in mid-line of the body. e.g. Intervertebral discs, manubrium and sternum, pubic symphysis.
  

iv. Synovial joints / freely movable joints / diarthroses:

1. It is characterized by presence of a space called synovial cavity between articulating bones that renders free movement at the joint.
2. The articulating surfaces of bones at a synovial joint are covered by a layer of hyaline cartilage. It reduces friction during movement and helps to absorb shock.
   
3. Synovial cavity is lined by synovial membrane that forms synovial capsule. Synovial membrane secretes synovial fluid.
4. Synovial fluid is a clear, viscous, straw coloured fluid similar to lymph. It is viscous due to hyaluronic acid. The synovial fluid also contains nutrients, mucous and phagocytic cells to remove microbes.
   Synovial fluid lubricates the joint, absorbs shocks, nourishes the hyaline cartilage and removes waste materials from hyaline cartilage cells (as cartilage is avascular). Phagocytic cells destroy microbes and cellular debris formed by wear and tear of the joint.
5. If the joint is immobile for a while, the synovial fluid becomes viscous and as joint movement starts, it becomes less viscous.
6. The joint is provided with capsular ligament and numerous accessory ligaments. The fibrous capsule is attached to periosteum of articulating bones. The ligament helps in avoiding dislocation of joint.

g. The types of synovial joints are on follows:

1. Pivot joint: In this type of joint, the rounded or pointed surface of one bone articulates with a ring formed partly by another bone and partly by the ligament. Rotation only around its own longitudinal axis is possible. e.g. in joint between atlas and axis vertebrae, head turns side ways to form ‘NO’ joint.

2. BaIl and socket joint: The ball like surface of one bone fits into cup like depression of another bone forming a movable joint. Multi-axial movements are possible. This type of joint allows movements along all three axes and in all directions. e.g. Shoulder and hip joint.

3. Hinge joint: In a hinge joint, convex surface of one bone fits into concave surface of another bone. In most hinge joints one bone remains stationary and other moves. The angular opening and closing motion (like hinge) is possible. In this joint only mono-axial movement takes place like flexion and extension. e.g. Elbow and knee joint.

4. Condyloid joint: It is an ellipsoid joint. The convex oval shaped projection of one bone fits into oval shaped depression in another bone. It is a biaxial joint because it permits movement along two axes viz, flexion, extension, abduction, adduction and circumduction is possible. e.g. Metacarpophalangeal joint.

5. Gliding joint: It is a planar joint, where the articulating surfaces of bones are flat or slightly curved. These joints are non-axial because the motion they allow does not occur along an axis or a plane. e.g. Intercarpal and intertarsal joints.

Saddle joint: This joint is a characteristic of Homo sapiens. Here the articular surface of one bone is saddle-shaped and that of other bone fits into saddle (each bone forming this joint have both concave and convex areas). It is a modified condyloid joint in which movement is somewhat more free. It is a biaxial joint that allows flexion, extension, abduction, adduction and circumduction. e.g. Carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb.

[Students are expected to prepare a chart on their own.]

Can you tell? (Textbook Page No. 211)

Question 1.
Human beings can hold an object in a better manner than monkeys. Why?
Answer:

1. Humans and monkeys both have five fingers including thumb, however humans can hold an object in better manner than monkeys because humans have highly developed opposable thumbs. The opposable thumb allows better grip.
2. The saddle joint in thumb allows free and independent movement to the thumb the carpometacarpellar joint between carpal (trapezium) and metacarpal of thumb makes the thumb opposable. It allows biaxial movements, i.e. flexion – extension and adduction – abduction but not rotation.

[Note: Gorillas, chimpanzees, orangutans and some other variants of apes have opposable thumb.]

Internet my friend. (Textbook Page No. 211)

Question 92.
Now a days we hear from many elderly people that they are undergoing knee replacement surgery. Find out why one has to undergo knee replacement; how it is carried out and how it can be prevented.
Answer:
Knee replacement is done in following cases:

1. Osteoarthritis: The cartilage in the knee undergoes degradation. It is caused by many factors such as muscle weakness, aging, obesity, etc.
2. Rheumatoid arthritis: It is characterised by inflammation of the synovial membrane, where it starts secreting excess of synovial fluid in the joint. This fluid exerts extensive pressure on the joint and causes severe pain.
3. Post-traumatic arthritis: This is caused due to breakage of ligament or cartilage. The breakage can be due to severe injury or accident. It causes severe pain and requires knee replacement.
4. Procedure:
   The procedure involves removal of the damaged cartilage or ligament and replaces it with artificial implant made up of either metal, plastic or both. Metal or plastic knee caps are used to cover the knees. The implant is connected to the bone and an artificial knee joint is made between them.
5. Prevention: Maintaining body weight, exercising regularly, consuming appropriate medications and supplements, etc.

[Students are expected to find out more information about knee replacement on internet]

Find out. (Textbook Page No. 212)

Question 1.
You must have heard of Sachin Tendulkar suffering from ‘tennis elbow’, a cricketer suffering from a disorder named after another game. Can common people too suffer from this disorder? Find out more information about this disorder.
Answer:

1. Tennis elbow is caused due to inflammation of tendon which joins muscles of forearm to the bone of upper arm (humerus). It is known as lateral epicondylitis.
2. It causes severe pain in the elbow. It occurs due to extensive repetitive movement of hand. This damages the tendon and increases the tenderness of the elbow joint.
3. This disorder develops not only in athletes but also in other common people whose job involves extensive movement of hand such as carpenter, painter, plumber, etc.

[Students are expected to find more information about tennis elbow on their own.]

Maharashtra State Board Class 11 Biology Textbook Solutions

• Plant Tissues and Anatomy Class 11 Biology Textbook Solutions
• Morphology of Flowering Plants Class 11 Biology Textbook Solutions
• Animal Tissue Class 11 Biology Textbook Solutions
• Study of Animal Type : Cockroach Class 11 Biology Textbook Solutions
• Photosynthesis Class 11 Biology Textbook Solutions
• Respiration and Energy Transfer Class 11 Biology Textbook Solutions
• Human Nutrition Class 11 Biology Textbook Solutions
• Excretion and Osmoregulation Class 11 Biology Textbook Solutions
• Skeleton and Movement Class 11 Biology Textbook Solutions

Excretion and Osmoregulation Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 15 Excretion and Osmoregulation Textbook Exercise Questions and Answers.

Class 11 Biology Chapter 15 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 15 Exercise Solution

1. Chose the correct option

Question (A).
Which one of the following organisms would spend maximum energy in the production of nitrogenous waste?
a. Polar bear
b. Flamingo
c. Frog
d. Shark
Answer:
b. Flamingo

Question (B).
In human beings, uric acid is formed due to metabolism of __________.
a. amino acids
b. fatty acids
c. creatinine
d. nucleic acids
Answer:
d. nucleic acids

Question (C).
Visceral layer : Podocytes :: PCT : _______
a. Cilliated cells
b. Squamous cells
c. Columnar cells
d. Cells with brush border
Answer:
d. Cells with brush border

Question (D).
Deproteinised plasma is found in __________.
a. Bowman’s capsule
b. Descending limb
c. Glomerular capillaries
d. Ascending limb
Answer:
a. Bowman’s capsule, b. Descending limb, d. Ascending limb

Question (E).
Specific gravity of urine would _______ if level of ADH increases.
a. remain unaffected
b. increases
c. decreases
d. stabilise
Answer:
b. increases

Question (F).
What is micturition?
a. Urination
b. Urine formation
c. Uremia
d. Urolithiasis
Answer:
a. Urination

Question (G).
Which one of the following organisms excrete waste through nephridia?
a. Cockroach
b. Earthworm
c. Crab
d. Liver Fluke
Answer:
c. Crab

Question (H).
Person suffering from kidney stone is advised not to have tomatoes as it has _______.
a. seeds
b. lycopene
c. oxalic acid
d. sour taste
Answer:
c. oxalic acid

Question (I).
Tubular secretion does not take place in ________.
a. DCT
b. PCT
c. collecting duct
d. Henle’s loop
Answer:
b. PCT

Question (J).
The minor calyx ____________.
a. collects urine
b. connects pelvis to ureter
c. is present in the cortex
d. receives column of Bertini
Answer:
a. collects urine

Question (K).
Which one of the followings is not a part of human kidney?
a. Malpighian body
b. Malpighian tubule
c. Glomerulus
d. Loop of Henle
Answer:
b. Malpighian tubule

Question (L).
The yellow colour of the urine is due to presence of ___________
a. uric acid
b. cholesterol
c. urochrome
d. urea
Answer:
c. urochrome

Question (M).
Hypotonic filtrate is formed in _______
a. PCT
b. DCT
c. LoH
d. CT
Answer:
a. PCT

Question (N).
In reptiles, uric acid is stored in _____
a. cloaca
b. fat bodies
c. liver
d. anus
Answer:
a. cloaca

Question (O).
The part of nephron which absorbs glucose and amino acid is______
a. collecting tubule
b. proximal tubule
c. Henle’s loop
d. DCT
Answer:
b. proximal tubule

Question (P).
Bowman’s capsule is located in kidney in the ________
a. cortex
b. medulla
c. pelvis
d. pyramids
Answer:
a. cortex

Question (Q).
The snakes living in desert are mainly__________
a. aminotelic
b. ureotelic
c. ammonotelic
d. uricotelic
Answer:
d. uricotelic

Question (R).
Urea is a product of breakdown of ___________
a. fatty acids
b. amino acids
c. glucose
d. fats
Answer:
b. amino acids

Question (S).
Volume of the urine is regulated by__________
a. aldosterone
b. ADH
c. both a and b
d. none
Answer:

Question 2.
Answer the following questions

Question (A).
Doctors say Mr. Shaikh is suffering from urolithiasis. How it could be explained in simple words?
Answer:
Urolithiasis is the condition of having calculi in the urinary tract (which also includes the kidneys), which may pass into urinary bladder.

Question (B).
Anitaji needs to micturate several times and feels very thirsty. This is an indication of change in permeability of certain part of nephron. Which is this part?
Answer:

1. Need to micturate several times (polyuria) and feeling very thirsty (polydipsia) is a symptom of diabetes insipidus (imbalance of fluids in the body).
2. ADH prevents diuresis and due to absence of ADH, large amount of dilute urine is excreted.
3. ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
4. If the permeability of these cells changes, it will result in increase in urine volume (frequent micturition) and increase in the osmolarity of blood. An imbalance in volume and osmolarity of body fluids increases thirst.

[Note: Water is reabsorbed by osmosis in PCT, DCT and descending limb of loop of Henle)

Question (C).
Effective filtration pressure was calculated to be 20 mm Hg; where glomerular hydrostatic pressure was 70 mm of Hg. Which other pressure is affecting the filtration process? How much is it?
Answer:
The other pressure affecting the filtration process is osmotic pressure of blood and filtrate hydrostatic pressure. Commonly effective filtration pressure (EFP) is represented as;
EFP = Glomerular Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
If EFP = 20 mmHg and Glomerular Hydrostatic pressure = 70 mmHg
20 = 70 – (Osmotic pressure of blood + Filtrate hydrostatic pressure)
∴ (Osmotic pressure of blood + Filtrate hydrostatic pressure) = 70-20
Then (Osmotic pressure of blood + Filtrate Hydrostatic pressure) = 50 mmHg .

[Note: Given values are insufficient to calculate the exact osmotic pressure of blood and filtrate hydrostatic pressure. The sum of the two values can be calculated to be 50 mmHg ]

Question (D).
Name any one guanotelic organism.
Answer:
Spiders, scorpions and penguins are guanotelic organisms as they excrete guanine.

Question (E).
Why are kidneys called ‘retroperitoneal’?
Answer:
Kidneys are located in abdomen. Kidneys are not surrounded by peritoneum instead they are located posterior to it. Thus, kidneys are called retroperitoneal.

Question (F).
State role of liver in urea production.
Answer:

1. Ammonia formed during the breakdown of amino acids is converted into urea in the liver of ureotelic animals.
2. This conversion takes place by the help of the ornithine / urea cycle.
3. 3 ATP molecules are used to produce one molecule of urea using the ornithine/ urea cycle. Since, the liver contains carrier molecules and enzymes necessary for urea cycle, it plays a major role in urea production.

Question (G).
Why do we get bad breath after eating garlic or raw onion?
Answer:

1. Raw onion and garlic contain volatile sulphur-containing compounds.
2. Sulphur-containing compounds have a distinctive odour which remain in the mouth after consumption of onion and garlic.
3. Also, volatile compounds (like certain sulphur containing compounds) in foodstuffs are generally excreted through the lungs and may result in bad breath.

3. Answer the following questions

Question (A).
John has two options as treatment for his renal problem : Dialysis or kidney transplants. Which option should he choose? Why?
Answer:

1. If John has two options of dialysis and kidney transplant, readily available he must opt for kidney transplant.
2. A kidney transplant, if successful, can improve the quality of life of a patient and reduce the risk of death.
3. The patient would not have to endure frequent dialysis procedures. Repeated visits for dialysis takes time and may not allow the patient to perform normal activities or go to office regularly.
4. Dialysis is regarded as a holding measure until kidney transplant can be performed or a supportive measure in those for whom a transplant would be inappropriate. However, dialysis cannot replace all the functions of a normal kidney such as production of hormones like erythropoietin, calcitriol and renin. Hence, if John has an option of kidney transplant, he must opt for it.

Question (B).
Amphibian tadpole can afford to be ammonotelic. Justify.
Answer:

1. Tadpole (larval stage of life cycle of amphibian) is aquatic. They are ammonotelic as they excrete nitrogenous waste in the form of ammonia.
2. Ammonia is very toxic and requires large amount of water for its elimination.
3. It is readily soluble in water and diffuses across the body surface and into the surrounding water.
4. Also, the water lost during excretion can be made up through the surrounding water in ammonotelic organisms.

Hence, amphibian tadpole can afford to be ammonotelic.

Question (C).
Birds are uricotelic in nature. Give reason.
Answer:

1. Birds are capable of converting ammonia into uric acid by ‘inosinic acid pathway’ in their liver.
2. Uric acid is least toxic and hence, it can be retained in the body for some time.
3. It is least soluble water hence, negligible amount of water is required for its elimination.
4. This mode of excretion can also help reduce body weight (an adaptation for flight) and those animals which
   need to conserve more water follow uricotelism.

Hence, in order to conserve water as an adaptation for flight, birds are uricotelic in nature.

Question 4.
Write the explanation in your word

Question (A).
Nitya has been admitted to hospital after heavy blood loss. Till proper treatment could be given; how did Nitya’s body must have tackled the situation?
Answer:

1. Heavy blood loss is called haemorrhage. In case of haemorrhage or severe dehydration, the osmoreceptors stimulate Antidiuretic hormone (ADH) secretion.
2. ADH is important in regulating water balance through the kidneys. For detailed mechanism of reabsorption by ADH:

Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

Another regulatory mechanism that must have been activated is RAAS. For detailed mechanism of electrolyte reabsorption:

Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin-converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

Question 5.
Complete the diagram / chart with correct labels / information. Write the conceptual details regarding it

Question (A).

Answer:

The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Question (B).

Answer:

1. Nephrons are structural and functional units of kidney.
2. Each nephron consists of a 4 – 6 cm long, thin-walled tube called the renal tubule and a bunch of capillaries known as the glomerulus.
3. The wall of the renal tubule is made up of a single layer of epithelial cells.
4. Its proximal end is wide, blind, cup-like and is called as Bowman’s capsule, whereas the distal end is open.
5. The nephron is divisible into Ilowman’s capsule, neck, proximal convoluted tubule (PCT), Loop of Henle (LoH), distal convoluted tubule (DCT) and collecting tubule (CT).
6. The glomerulus is present in the cup-like cavity of Bowman’s capsule and both are collectively known as renal corpuscle or Malpighian body.

Question (C)

Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:

i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.

a. Glomerulus:
Glomerulus is a bunch of fine blood capillaries located in the cavity of Bowman’s capsule.
A small terminal branch of the renal artery, called as afferent arteriole enters the cup cavity (Bowman capsule) and undergoes extensive fine branching to form network of several capillaries. This bunch is called as glomerulus.
The capillary wall is fenestrated (perforated).

All capillaries reunite and form an efferent arteriole that leaves the cup cavity.
The diameter of the afferent arteriole is greater than the efferent arteriole. This creates a high hydrostatic pressure essential for ultrafiltration, in the glomerulus.

b. Bowman’s capsule:
It is a cup-like structure having double walls composed of squamous epithelium.
The outer wall is called as parietal wall and the inner wall is called as visceral wall.
The parietal wall is thin consisting of simple squamous epithelium.
There is a space called as capsular space / urinary space in between two walls.
Visceral wall consists of special type of squamous cells called podocytes having a foot-like pedicel. These podocytes are in close contact with the walls of capillaries of glomerulus.
There are small slits called as filtration slits in between adjacent podocytes.

ii. Renal tubule:

a. Neck:
The Bowman’s capsule continues into the neck. The wall of neck is made up of ciliated epithelium. The lumen of the neck is called the urinary pole. The neck leads to proximal convoluted tubule.

b. Proximal Convoluted Tubule :
This is highly coiled part of nephron which is lined by cuboidal cells with brush border (microvilli) and surrounded by peritubular capillaries. Selective reabsorption occurs in PCT. Due to convolutions (coiling), filtrate flows slowly and remains in the PCT for longer duration, ensuring that maximum amount of useful molecules are reabsorbed.

c. Loop of Henle :
This is ‘U’ shaped tube consisting of descending and ascending limb.
The descending limb is thin walled and permeable to water and lined with simple squamous epithelium.
The ascending limb is thick walled and impermeable to water and is lined with simple cuboidal epithelium.
The LoH is surrounded by capillaries called vasa recta.
Its function is to operate counter current system – a mechanism for osmoregulation.
The ascending limb of Henle’s loop leads to DCT.

d. Distal convoluted tubule:
This is another coiled part of the nephron.
Its wall consists of simple cuboidal epithelium.
DCT performs tubular secretion / augmentation / active secretion in which, wastes are taken up from surrounding capillaries and secreted into passing urine.
DCT helps in water reabsorption and regulation of pH of body fluids.

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Question (D).

Answer:

The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.
This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.

[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).
Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin-converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na⁺ and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Question (E).

Answer:

1. When renal function of a person falls below 5 – 7 %, accumulation of harmful substances in blood begins. In such a condition the person has to go for artificial means of filtration of blood i.e. haemodialysis.
2. In haemodialysis, a dialysis machine is used to filter blood. The blood is filtered outside the body using a dialysis unit.
3. In this procedure, the patients’ blood is removed; generally from the radial artery and passed through a cellophane tube that acts as a semipermeable membrane.
4. The tube is immersed in a fluid called dialysate which is isosmotic to normal blood plasma. Hence, only excess salts if present in plasma pass through the cellophane tube into the dialysate.
5. Waste substances being absent in the dialysate, move from blood into the dialyzing fluid.
6. Filtered blood is returned to vein.
7. In this process it is essential that anticoagulant like heparin is added to the blood while it passing through the tube and before resending it into the circulation, adequate amount of anti-heparin is mixed.
8. Also, the blood has to move slowly through the tube and hence the process is slow.

Question 6.
Prove that mammalian urine contains urea.
Answer:

1. Urea is a nitrogenous waste formed by breakdown of protein (deamination of amino acids).
2. During this process, amino groups are removed from the amino acids present in the proteins and converted to highly toxic ammonia. The ammonia is finally converted to area through ornithine cycle. Thus, the urea formed is passed to kidneys and excreted out of the body through urine.
3. Reabsorption of urea (proximal tubule, collecting ducts) and active secretion of urea (Henle loop) leads to a urea circulation (urea recycling) between the lumen of the nephron and renal medulla, which is an important element of the renal urine concentration.
4. About 54 g of urea is filtered per day in the glomerular capsule, of which approximately 30 g is excreted in the urine and 24 g is reabsorbed into blood (assuming GFR is 180 litres/day).
5. Urinalysis can help detect the amount of urea in urine (Urine urea nitrogen test, urease test, etc.).

Practical / Project :

Visit to a nearby hospital or pathological laboratory and collect detailed information about different blood and urine tests.
Answer:
Testing the urine is known as urinalysis. It generally has three parts:

1. Visual examination: Check sample colour and clearness.
2. Dipstick examination: Checks for abnormal amounts of glucose, protein, etc.
3. Microscopic examination: Check for presence of RBCs. WBCs, bacteria, crystals, etc.
4. Apart from routine urine examination, specific tests may also be done. They are as follows:
   • BUN (Blood Urea Nitrogen) Test: It measures the amount of nitrogen in blood and evaluates kidney function.
   • Urease Test/ Urea Nitrogen Test: It is done to check the amount of urea in urine sample.
   • Urine albumin to creatine ratio (UACR) test: Estimates the amount of albumin in urine.

[Students are expected to collect more information and perform the given activity on their own]

12th Biology Digest Chapter 15 Excretion and Osmoregulation Intext Questions and Answers

Can you recall? (Textbook Page No. 174)

Question 1.
Why are various waste products produced in the body of an organism?
Answer:
Metabolism produces a variety of by-products, some of which need to be eliminated. Such by-produçts are called metabolic waste products.

Question 2.
How are these waste eliminated?
Answer:
Depending on the type of waste product, they are eliminated through various organs of the body:

The various excretory products produced by the human body are as follows:

1. Fluids such as water; gaseous wastes like CO₂ nitrogenous wastes like ammonia, urea and uric acid, creatinine; minerals; salts of sodium, potassium. calcium, etc. if present in body in excess are excreted through urine, faeces and sweat.
2. Pigments formed due to breakdown of haemoglobin like bilirubin (excreted through faeces) and urochrome (eliminated through urine).
3. The pigments present in consumed foodstuffs like beet root or excess of vitamins, hormones and drugs.
4. Volatile substances present in spices (eliminated through lungs).

Have you ever observed? (Textbook Page No. 174)

Question 1.
When does urine appear deeply coloured?
Answer:
Urine can appear deeply coloured due to various reasons:

• Severe dehydration resulting in production of concentrated urine.
• Consumption of foodstuff like beet root, which contain coloured pigments.
• Some medications can also cause the urine to appear deeply coloured.

Think about it. (Textbook Page No. 174)

Question 1.
Do organisms differ in type of metabolic wastes they produce?
Answer:
Yes, organisms differ in the type of metabolic wastes they produce. Some organisms excrete ammonia while some excrete urea or uric acid as metabolic wastes.

Question 2.
Do environment or evolution have any effect on type of waste produced by an organism?
Answer:

• The theory of evolution proposes that life started in an aquatic environment.
• Aquatic organisms are generally ammonotelic. It is believed that the urea cycle evolved to adapt to a changing environment when terrestrial life forms evolved.
• Arid conditions probably led to the evolution of the uric acid pathway as a means of conserving water.
• However, the correlation between evolution and type of waste production is uncertain.

Question 3.
How do thermoregulation and food habits affect saste production?
Answer:

1. To generate heat. endotherms convert the food that they eat into energy through a process called metabolism. Hence, they consume more tì.od in order to meet their energy requirements.
2. Also, carnivorous diet contains more proteins than herbivores.
3. Consumption of high protein or more food containing proteins can result in production of large amount of nitrogenous waste
4. These animals would also require more energy to eliminate the high levels oF nitrogenous wastes which build up when animal protein is digested.

Use your brain power. (Textbookpage No. 175)

Question 1.
Why ammonia is highly toxic?
Answer:

1. Ammonia is basic in nature and its retention in the body would disturb the pH of the body.
2. An increase in pH would disturb all enzyme catalysed reactions in the body and also make the plasma membrane unstable.

Hence, ammonia is highly toxic to the body.

Find out. (Textbook page No. 175)

You will study about a type of arthritis called gouty arthritis caused due to accumulation of uric acid in joints. Where does uric acid come from in case of ureotelic human beings?
Answer:

1. Uric acid produced as a waste product during the normal breakdown of nucleic acids (purines) and certain naturally occurring substances found in foods such as mushrooms. Mackerel, dried beans. etc.
2. This uric acid is generally excreted out along with urine.
3. If uric acid is produced in excess or not excreted, it accumulates in joints causing gouty arthritis.

Think about it. (Textbook Page No.175)

Endotherms consume more food in order to meet energy requirements. Also, carnivorous diet contains more proteins than herbivorous. Does it affect excretion of nitrogenous waste?
Answer:

1. To generate heat, endotherms convert the food that they eat into energy through a process called metabolism. Hence, they consume more food in order to meet their energy requirements.
2. Also, carnivorous diet contains more proteins than herbivores.
3. Consumption of high protein or more food containing proteins can result in production of large amount of nitrogenous waste.
4. These animals would also require more energy to eliminate the high levels of nitrogenous wastes which build up when animal protein is digested.

Observe and Discuss (Textbook Page No. 176)

Question 1.
These are blood reports of patients undergoing investigations for kidney function. What is creatinine? What is your observation and opinion about the findings? Why is it used as an index of kidney function?


Answer:
i. Creatinine:

• Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
• It provides a ready source of high energy phosphate.

ii. Observations and Opinion:
Report A indicates a value of creatinine which is higher than the normal range. This would indicate impaired kidney function.
Report B indicates high fasting blood sugar and detection of sugar in the blood is known as glucosuria. High fasting blood sugar (>126 mg/dL) is usually indicative of diabetes.

iii. Creatinine used as an index of kidney function:

• Normally blood creatinine levels remain steady because the rate of production matches his excretion in urine.
• Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.

Think about it. (Textbook Page No. 176)

Question 1.
During summer, we tend to produce less urine, is it so?
Answer:

1. Generally, excess water containing wastes is lost from the body in the time of urine. sweat and faeces.
2. During summer when the surrounding temperature is high. we also lose a considerable amount of water in the form of sweat.

Thus, the kidneys retain water for maintaining the concentration of body fluids and reduce the amount of water lost through urine.

Question 2.
Marine birds like Albatross spend their life on the sea. That means the water the, drink is salty. how do they manage osnioregulation then?
Answer:

1. Marine birds like Albatross have special glands called salt glands near their nostrils.
2. These glands are capable of secreting salts by active transport and help to manage osmotic balance,

[Note: The salt glands in Albatross are located in or on the skull in the area of eyes.]

Question 3.
like ectothermic and endothermic animals, do organisms differ in the way they maintain salt balance?
Answer:
Yes, organisms differ in the way they maintain salt balance.

1. Animals can be either isosmotic to the surrounding (osmoconformers or control the internal environment independent of external environment (osmoregulators).
2. Marine organisms are mostly osmoconfòrmers because their body fluids and external environment are isosmotic in nature while fresh water forms and terrestrial organisms are osmoregulators,
3. Generally, most organisms can tolerate only a narrow range of salt concentrations. Such organism are known as stenohaline organisms.
4. Organisms which are capable of handling a wide change in salinity are called euryhaline organisms.e.g. Barnacles, clams etc.

Find out. (Textbook Page No 176)

Question 1.
How do freshwater fishes and marine fishes carry out osmoregulation?
Answer:
Osmoregulation is the process of maintaining an internal concentration of salt and water in the body of fishes.

i. Freshwater fishes:
The salt concentration inside the body of freshwater fishes is higher than their surrounding water. Due to this, water enters the body due to osmosis.
If the flow of water into the body is not regulated. fishes would swell and get bigger.
To compensate this, the kidneys produce a large amount of urine,
Excretion of large amounts of urine regulates the level of water in the body hut results in the loss of salts.
Thus, in order to maintain a sufficient salt level, special cells in the gills (chloride cells) take tip ions from
the water, which are then directly transported into the blood.

ii. Marine fishes:
Since the salt content in blood of marine fishes is much lower than that of seawater, they constantly tend to lose water and build up salt.
To replace the water loss, they continually need to drink seawater.
Since their small kidney can only excrete a relatively small amount of urine, salt is additionally excreted through gills, where chloride cells work in reverse as in freshwater fishes.

Make a table. (Textbook Page No. 178)

Question 1.
The details of modes of excretion of nitrogenous wastes.
Answer:
The three main modes of excretion in animals are as follows:

i. Ammonotelism
ii. Ureotelism
iii. Uricotelism

i. Ammonotelism:

1. Elimination of nitrogenous wastes in the form of ammonia is called as ammonotelism.
2. Ammonia is basic in nature and hence it can disturb the pH of the body, if not eliminated immediately.
3. Any change in pH would disturb all enzyme catalyzed reactions in the body and would also make the plasma membrane unstable.
4. Ammonia is readily soluble in water and needs large quantity of water to dilute and reduce its toxicity.
5. This is however an energy saving mechanism of excretion and hence all animals that have plenty of water available for dilution of ammonia, excrete nitrogenous wastes in the form of ammonia.
6. Animals that follow this mode of excretion are known as ammonotelic animals.
7. 1 gm ammonia needs about 300 – 500 ml of water for elimination.
8. Ammonotelic animals excrete ammonia through general body surface (skin), gills and kidneys.
   e.g. Ammonotelism is found in aquatic invertebrates, bony fishes, and aquatic / larval amphibians. Animals without excretory system (Protozoa) are also ammonotelic.

ii. Ureotelism:

1. Elimination of nitrogenous wastes in the form of urea is called as ureotelism.
2. Urea is comparatively less toxic and less water-soluble than ammonia. Hence, it can be concentrated to some extent in body.
3. The body requires less water for elimination.
4. Since it is less toxic and less water soluble, ureotelism is suitable for animals that need to conserve water to some extent. Hence, ureotelism is common in terrestrial animals, as they have to conserve water.
5. It takes about 50 ml H₂O for removal of 1 gm NH₂ in form of urea.
6. Ureotelic animals generally convert ammonia to urea in the liver by operating ornithine / urea cycle in which 3 ATP molecules are used to produce one molecule of urea.
   e.g. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most of the adult amphibians, etc. are ureotelic.

iii. Uricotelism:

1. Elimination of nitrogenous wastes in the form of uric acid is called as uricotelism.
2. Uric acid is least toxic and hence, it can be retained in the body for some time in concentrated form.
3. It is least soluble in water. Hence there is minimum (about 5 – 10 ml for 1 gm) or no need of water for its elimination.
4. Those animals which need to conserve more water follow uricotelism. However, these animals need to spend more energy.
5. Ammonia is converted into uric acid by ‘inosinic acid pathway’ in the liver of birds, e.g. Birds, some insects, many reptiles, land snails, are uricotelic.

[Students can Refer these and make a chart on their own.]

Use your brain power. (Textbook Page No. 178)

Question 1.
Creatinine is considered as index of kidney function. Give reason.
Answer:

1. Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
2. It provides a ready source of high energy phosphate.
3. Normally blood creatinine levels remain steady because the rate of production matches its excretion in urine.
4. Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.

[Note: Plasma creatinine is a waste product produced by muscles from the breakdown of a compound called ‘creatine phosphate ’.]

Make a table. (Textbook Page No. 178)

Question 1.
The excretory organs found in various animal phyla.
Answer:

Observe and complete. (Textbook Page No. 178)

Question 1.
Label the diagram and complete following paragraphs.

i. Kidney: A pair of ____ shaped kidneys are present on either side of the ____ from 12^th thoracic to 3^rd lumbar vertebra. Kidneys are present behind ___. Hence are called retroperitoneal. Dimensions of
each kidney are 10 × ____ × ____ cms. Average weight is ___ g in males and 135 g in ____. Outer surface is ___ and inner is concave. Notch on the inner concave surface is called ___. Renal artery enters and renal vein as well as ureter leave the kidney through hilus. Each kidney has almost 1 million functional units called ___.

ii. Ureters: A pair of ureters arise from ___ of each kidney. Each ureter is a long muscular tube 25 – 30 cm in length. Ureters open into ___ by separate openings, which are not guarded by valves. They pass obliquely through the wall of urinary bladder. This helps in prevention of ___ of urine due to compression of ureters while bladder is filled.

iii. Urinary bladder: It is a median ___ sac. A hollow muscular organ, the bladder is situated in pelvic cavity posterior to pubic symphysis. At the base of the ___ there is a small inverted triangular area called trigone. At the apex of this triangle is opening of urethra. At the two points of the base of the triangle are openings of ureters. Urinary bladder is covered externally by peritoneum. Inner to peritoneum is muscular layer. It is formed by detrusor muscles which consist of three layers of smooth muscles. Longitudinal – circular – longitudinal respectively. Innermost layer is made up of transitional ___. It helps bladder to stretch.

iv. Urethra: It is a ___ structure arising from urinary bladder and opening to the exterior of the body.
There are ___ urethral sphincters between urinary bladder and urethra.
a. Internal sphincter: Made up of ___ muscles, involuntary in nature.
b. External sphincter: Made up of ___ muscles, voluntary in nature.
Answer:

i. Kidney: A pair of bean shaped kidneys are present on either side of the backbone from 12^th thoracic to 3^rd lumbar vertebra. Kidneys are present behind peritoneum. Hence are called retroperitoneal. Dimensions of each kidney are 10 × 5 × 4 cms. Average weight is 150 g in males and 135 g in females. Outer surface is convex and inner is concave. Notch on the inner concave surface is called hilum. Renal artery enters and renal vein as well as ureter leave the kidney through hilus. Each kidney has almost 1 million functional units called nephron.

ii. Ureters: A pair of ureters arise from hilum of each kidney. Each ureter is a long muscular tube 25 – 30 cm in length. Ureters open into urinary bladder by separate openings, which are not guarded by valves. They pass obliquely through the wall of urinary bladder. This helps in prevention of backward flow of urine due to compression of ureters while bladder is filled.

iii. Urinary bladder: It is a median pear-shaped sac. A hollow muscular organ, the bladder is situated in pelvic cavity posterior to pubic symphysis. At the base of the urinary bladder there is a small inverted triangular area called trigone. At the apex of this triangle is opening of urethra. At the two points of the base of the triangle are openings of ureters. Urinary bladder is covered externally by peritoneum. Inner to peritoneum is muscular layer. It is formed by detrusor muscles which consist of three layers of smooth muscles. Longitudinal – circular – longitudinal respectively. Innermost layer is made up of transitional epithelial tissue. It helps bladder to stretch.

iv. Urethra: It is a fibromuscular tube-like structure arising from urinary bladder and opening to the exterior of the body. There are two urethral sphincters between urinary bladder and urethra.
a. Internal sphincter: Made up of detrusor muscles, involuntary in nature.
b. External sphincter: Made up of striated muscles, voluntary in nature.
If this valve is not functioning properly during inflammation of bladder, it can lead to kidney infection.

Internet is my friend. (Textbook Page No. 179)

Question 1.
Find out what is floating kidney.
Answer:

1. Floating kidney or nephroptosis, is an inferior displacement or dropping of the kidney.
2. This condition occurs when the kidney slips from its normal position because it is not held securely in place by the adjacent organs or its fat covering.
3. It generally develops in extremely thin people whose adipose capsule or renal fascia is deficient.
4. It may result in twisting of the ureter and cause blockage of urine flow. The resulting backup of urine would put pressure on the kidney and damage the tissues.
5. Twisting of the ureter may also cause pain and discomfort.
6. This condition is more common in females than males and happens commonly among one in four people.
7. Weakening of the fibrous bands that hold the kidney in place can predispose to floating kidney.

Can you recall? (Textbook Page No. 179)

Question 1.
Observe the figure carefully and label various regions of L.S. of kidney.

Answer:

Can you tell? (Textbook Page No.182)

Question 1.
Why are kidneys called ‘retroperitoneal’?
Answer:
Kidneys are located in abdomen. Kidneys are not surrounded by peritoneum instead they are located posterior to it. Thus, kidneys are called retroperitoneal.

Question 2.
Why urinary tract infections are more common in females than males?
Answer:

• The urethra in women (4 cm) is much shorter than that of males (20 cm).
• This allows easy passage of bacteria into the urinary bladder.

Hence, urinary tract infections are more common in females than males.

Question 3.
What is nephron? Which are its main parts? Why are they important?
Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:

i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.

a. Glomerulus:
Glomerulus is a bunch of fine blood capillaries located in the cavity of Bowman’s capsule.
A small terminal branch of the renal artery, called as afferent arteriole enters the cup cavity (Bowman capsule) and undergoes extensive fine branching to form network of several capillaries. This bunch is called as glomerulus.
The capillary wall is fenestrated (perforated).

All capillaries reunite and form an efferent arteriole that leaves the cup cavity.
The diameter of the afferent arteriole is greater than the efferent arteriole. This creates a high hydrostatic pressure essential for ultrafiltration, in the glomerulus.

b. Bowman’s capsule:
It is a cup-like structure having double walls composed of squamous epithelium.
The outer wall is called as parietal wall and the inner wall is called as visceral wall.
The parietal wall is thin consisting of simple squamous epithelium.
There is a space called as capsular space / urinary space in between two walls.
Visceral wall consists of special type of squamous cells called podocytes having a foot-like pedicel. These podocytes are in close contact with the walls of capillaries of glomerulus.
There are small slits called as filtration slits in between adjacent podocytes.

ii. Renal tubule:

a. Neck:
The Bowman’s capsule continues into the neck. The wall of neck is made up of ciliated epithelium. The lumen of the neck is called the urinary pole. The neck leads to proximal convoluted tubule.

b. Proximal Convoluted Tubule :
This is highly coiled part of nephron which is lined by cuboidal cells with brush border (microvilli) and surrounded by peritubular capillaries. Selective reabsorption occurs in PCT. Due to convolutions (coiling), filtrate flows slowly and remains in the PCT for longer duration, ensuring that maximum amount of useful molecules are reabsorbed.

c. Loop of Henle :
This is ‘U’ shaped tube consisting of descending and ascending limb.
The descending limb is thin walled and permeable to water and lined with simple squamous epithelium.
The ascending limb is thick walled and impermeable to water and is lined with simple cuboidal epithelium.
The LoH is surrounded by capillaries called vasa recta.
Its function is to operate counter current system – a mechanism for osmoregulation.
The ascending limb of Henle’s loop leads to DCT.

d. Distal convoluted tubule:
This is another coiled part of the nephron.
Its wall consists of simple cuboidal epithelium.
DCT performs tubular secretion / augmentation / active secretion in which, wastes are taken up from surrounding capillaries and secreted into passing urine.
DCT helps in water reabsorption and regulation of pH of body fluids.

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Think about ¡t. (Textbook Page No. 182)

Question 1.
How much blood ¡s supplied to kidney?
Answer:
Around 600 ml of blood passes through each kidney per minute.

Do this. (Textbook Page No. 183)

Question 1.
Check blood reports of patients and comment about possibility of glucosuria.
Answer:
Glucosuria is the presence of glucose sugar in urine. High glucose in urine is usually indicative of diabetes mellitus.

[Students can get access to sample reports on the internet and refer the above table to comment on blood reports of patients on their own.]

Use your brain power. (Textbook Page No. 185)

Question 1.
In which regions of nephron the filtrate will he isotonic to blood?
Answer:
Filtrate leasing the proximal convoluted tubule (PCT) is isotonic to the blood plasma.

Can you tell? (Textbook Page No. 185)

Question 1.
Explain the process of urine formation in details.
Answer:
Process of urine formation is completed in three steps, namely;

i. Ultrafiltration/ Glomerular filtration,
ii. Selective reabsorption,
iii. Tubular secretion / Augmentation

i. Ultrafiltration / Glomerular filtration :
Diameter of afferent arteriole is greater than the efferent arteriole. The diameter of capillaries is still smaller than both arterioles. Due to the difference in diameter, blood flows with greater pressure through the glomerulus. This is called as glomerular hydrostatic pressure (GHP) and normally, it is about 55 mmHg. GIIP is opposed by osmotic pressure of blood (normally, about 30 mm Hg) and capsular pressure (normally, about 15 mm Hg).

Hence net / effective filtration pressure (EFP) is 10 mm Hg.
EFP = Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
= 55 – (30 + 15)
= 10 mm Hg

Under the effect of high pressure, the thin walls of the capillary become permeable to major components of blood (except blood cells and macromolecules like protein).
Thus, plasma except proteins oozes out through wall of capillaries.
About 600 ml blood passes through each kidney per minute.

The blood (plasma) flowing through kidney (glomeruli) is filtered as glomerular filtrate, at a rate of 125 ml / min. (180 L/d).
Glomerular filtrate / deproteinized plasma / primary urine is alkaline, contains urea, amino acids, glucose, pigments, and inorganic ions.
Glomerular filtrate passes through filtration slits into capsular space and then reaches the proximal convoluted tubule.

ii. Selective reabsorption :
Selective reabsorption occurs in proximal convoluted tubule (PCT). It is highly coiled so that glomerular filtrate passes through it very slowly. Columnar cells of PCT are provided with microvilli due to which absorptive area increases enormously.
This makes the process of reabsorption very effective.
These cells perform active (ATP mediated) and passive (simple diffusion) reabsorption.

Substances with considerable importance (high threshold) like – glucose, amino acids, vitamin C, Ca⁺⁺, K⁺, Na⁺, Cl^– are absorbed actively, against the concentration gradient. Low threshold substances like water, sulphates, nitrates, etc., are absorbed passively.
In this way, about 99% of glomerular filtrate is reabsorbed in PCT and DCT.

iii. Tubular secretion / Augmentation :
Finally filtrate reaches the distal convoluted tubule via loop of Henle. Peritubular capillaries surround DCT. Cells of distal convoluted tubule and collecting tubule actively absorb the wastes like creatinine and ions like K⁺, H⁺ from peritubular capillaries and secrete them into the lumen of DCT and CT, thereby augmenting the concentration of urine and changing its pH from alkaline to acidic.
Secretion of H⁺ ions in DCT and CT is an important homeostatic mechanism for pH regulation of blood. Tubular secretion is the only process of excretion in marine bony fishes and desert amphibians.

Question 2.
How does counter current mechanism help concentration of urine?
Answer:
Under the conditions like low water intake or high water loss due to sweating, humans can produce concentrated urine. This urine can be concentrated around four times i.e. 1200 mOsm/L than the blood (300 mOsm/L). Hence, a mechanism called countercurrent mechanism is operated in the human kidneys. The countercurrent mechanism operating in the Limbs of Henle’s loop of juxtamedullary nephrons and vasa recta is as follows:

1. It involves the passage of fluid from descending to ascending limb of Henle’s loop.
2. This mechanism is called countercurrent mechanism, since the flow of tubular fluid is in opposite direction through both limbs.
3. In case of the vasa recta, blood flows from ascending to descending parts of itself.
4. Wall of descending limb is thin and permeable to water, hence, water diffuses from tubular fluid into tissue fluid due to which, tubular fluid becomes concentrated.
5. The ascending limb is thick and impermeable to water. Its cells can reabsorb Na⁺ and Cl^– from tubular fluid and release into tissue fluid.
6. Due to this, tissue fluid around descending limb becomes concentrated. This makes more water to move out from descending limb into tissue fluid by osmosis.
7. Thus, as tubular fluid passes down through descending limb, its osmolarity (concentration) increases gradually due to water loss and on the other hand, progressively decreases due to Na⁺ and Cl^– secretion as it flows up through ascending limb.
8. Whenever retention of water is necessary, the pituitary secretes ADH. ADH makes the cells in the wall of collecting ducts permeable to water.
9. Due to this, water moves from tubular fluid into tissue fluid, making the urine concentrated.
10. Cells in the wall of deep medullar part of collecting ducts are permeable to urea. As concentrated urine flows through it, urea diffuses from urine into tissue fluid and from tissue fluid into the tubular fluid flowing through thin ascending limb of Henle’s loop.
11. This urea cannot pass out from tubular fluid while flowing through thick segment of ascending limb, DCT and cortical portion of collecting duct due to impermeability for it in these regions.
12. However, while flowing through collecting duct, water reabsorption is operated under the influence of ADII. Due to this, urea concentration increases in the tubular fluid and same urea again diffuses into tissue fluid in deep medullar region.
13.  Thus, same urea is transferred between segments of renal tubule and tissue fluid of inner medulla. This is called urea recycling; operated for more and more water reabsorption from tubular fluid and thereby excreting small volumes of concentrated urine.
14. Osmotic gradient is essential in the renal medulla for water reabsorption by counter current multiplier system.
15. This osmotic gradient is maintained by vasa recta by operating counter current exchange system.
16. Vasa recta also have descending and ascending limbs. Blood that enters the descending limb of the vasa recta has normal osmolarity of about 300 mOsm/L.
17.  As it flows down in the region of renal medulla where tissue fluid becomes increasingly concentrated, Na⁺, Cl^– and urea molecules diffuse from tissue fluid into blood and water diffuse from blood into tissue fluid.
18. Due to this, blood becomes more concentrated which now flows through ascending part of vasa recta. This part runs through such region of medulla where tissue fluid is less concentrated.
19. Due to this, Na⁺, Cl^– and urea molecules diffuse from blood to tissue fluid and water from tissue fluid to blood. This mechanism helps to maintain the osmotic gradient.
    

Try this. (Textbook Page No. 185)

Question 1.
Read the given urine report and prepare a note on composition of normal urine.

Answer:
The composition of normal urine is as follows:

1. A volume of 1 – 2 litres of urine in 24 hours is normal. This volume can however vary considerably as it depends on fluid intake, physical activity, temperature, etc.
2. The colour of normal urine is generally pale yellow due to urochrome (pigment produced by breakdown of bile). The colour of urine may vary slightly due to urochrome concentration and diet.
3. The appearance of urine is generally clear and transparent.
4. Any form of deposits (sediments/ crystals) is generally absent in normal urine.
5. The pH of normal urine is acidic and is generally around 6.0 (Range: 4.6 to 8.0). The pH varies considerably with the diet of a person.
6. The specific gravity of urine is an average of 1.02 ( Range : 1.001 to 1.035).
7. Albumin, sugar, bile salts bile pigments, ketone bodies and casts are absent in normal urine.
8. Occult blood is generally not seen in normal urine.

Think (Textbook Page No. 185)

Question 1.
What would happen if ADH secretion decreases due to any reason?
Answer:
In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus. Frequent excretion of large amount of dilute urine may cause a person to feel thirsty.

Think and appreciate. (Textbook Page No. 185)

Question 1.
How do kidneys bring about homeostasis? Is there any role of neuro endocrine system in it?
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.
If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.
Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.
[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).
Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

No. Both ADH and RAAS are essential for homeostasis.

1. Only ADH can lower blood Na⁺ concentration by way of water reabsorption in DCI and collecting duct. whereas RAAS stimulates Na⁺ reabsorption and maintains osmolarity of body fluid.
2. Action of ADH and RAAS leads to increase in blood volume and osmolarity.
3. For mechanism of Atrial natriuretic peptide:

Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

ADH is produced by the hypothalamus and is stored and released by the posterior pituitary or the neurohypophysis in response to appropriate trigger. Hence, there is a role of the neuroendocrine system in homeostasis.

Use your brain power. (Textbook Page No. 186)

Question 1.
Can we use this knowledge in treatment of high blood pressure? Why high BP medicines are many a times diuretics?
Answer:

1. Yes, the knowledge of homoeostasis is used in the treatment of high blood pressure.
2. Some commonly used theories for treatment of high blood pressure are as follows:
   • Angiotensin II receptor blockers (ARBs) are used as medications to treat high blood pressure. These medications block the action of angiotensin II by preventing angiotensin II from binding to angiotensin II receptors on the muscles surrounding blood vessels. As a result, blood vessels enlarge (dilate), and blood pressure is reduced.
   • Another method is the use of ‘Angiotensin converting enzyme’ ACE blockers. These inhibitors inhibit activity of ACE and therefore decrease the production of angiotensin II. As a result, these medications cause the blood vessels to enlarge or dilate, and this reduces blood pressure.
3. Vasodilation reduces arterial pressure. Reduced angiotensin II leads to natriuresis (increased excretion of Na⁺ in urine) and diuresis, thereby reducing blood pressure.
4. Too much salt can cause extra fluid to build up in the blood vessels, raising blood pressure. Diuretics are substances that slow renal absorption of water and thereby cause diuresis (elevated urine flow rate) which in turn reduces blood volume and blood pressure by flushing out salt and extra fluid. Hence, high BP medicines are many a times diuretics.

Can you tell? (Textbook page no. 186)
How do skin and lungs help in excretion?
OR
Can you tell? (Textbook page no. 187)
Explain role of lungs and skin in excretion.
Answer:
Yes, various organs other than the kidney participate in excretion. They are as follows:

i. Skin:

Skin acts as an accessory excretory organ. The skin of many organisms is thin and permeable. It helps in diffusion of waste products like ammonia.
Human skin however is thick and impermeable. It shows presence of two types of glands namely, sweat glands and sebaceous glands.

• Sweat glands are distributed all over the skin. They are abundant in the palm and facial regions.
  These simple, unbranched, coiled, tubular glands open on the surface of the skin through an opening called sweat pore. Sweat is primarily produced for thermoregulation but it also excretes substances like water, NaCl, lactic acid and urea.
• Sebaceous glands are present at the neck of hair follicles. They secrete oily substance called sebum.
  It forms a lubricating layer on skin making it softer. It protects skin from infection and injury.

ii. Lungs:

Lungs are the accessory excretory organs. They help in excretion of volatile substances like CO₂ and water vapour produced during cellular respiration. Along with CO₂, lungs also remove excess of H₂O in the form of vapours during expiration. They also excrete volatile substances present in spices and other food stuff.

Can you tell? (Textbook Page No. 187)

Question 1.
When does kidney produce renin? Where is it produced in kidney?
Answer:
Kidney produces renin whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration).
The juxtaglomerular Apparatus (JGA) cells secrete renin.

Question 2.
Explain how electrolyte balance of blood plasma maintained.
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

i. Regulating water reabsorption through ADH
ii. Electrolyte reabsorption though RAAS
iii. Atrial Natriuretic Peptide

i. Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.
Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.
In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.
[Note: Hypothalamus is a part of forebrain]

ii. Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

iii. Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

No. Both ADH and RAAS are essential for homeostasis.

1. Only ADH can lower blood Na⁺ concentration by way of water reabsorption in DCI and collecting duct. whereas RAAS stimulates Na⁺ reabsorption and maintains osmolarity of body fluid.
2. Action of ADH and RAAS leads to increase in blood volume and osmolarity.
3. For mechanism of Atrial natriuretic peptide:

Atrial natriuretic peptide (ANP): A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na⁺ and Cl^– reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na⁺ in urine) and diuresis.

Can you tell? (Textbook Page No. 187)

Question 1.
What is the composition of sweat?
Answer:
Sweat is composed of water, NaCl, lactic acid and urea.

Internet my friend. (Textbook Page No. 189)

Question 1.
Treatments other than surgical removal of kidney stone like Lithotripsy. (Breaking down of kidney stones using shock waves).
Answer:
a. Cystoscopy and ureteroscopy:
During cystoscopy, the doctor uses a cystoscope to look inside the urethra and bladder to find a stone in the urethra or bladder.
During ureteroscopy, the doctor uses a ureteroscope, which is longer and thinner than a cystoscope, to see detailed images of the lining of the ureters and kidneys.

The doctor inserts the cystoscope or ureteroscope through the urethra to see the rest of the urinary tract. Once the stone is found, the doctor can remove it or break it into smaller pieces.
The doctor performs these procedures in the hospital with anesthesia.

b. Percutaneous nephrolithotomy:
The doctor uses a thin viewing tool, called a nephroscope, to locate and remove the kidney stone.
The doctor inserts the tool directly into your kidney through a small cut made in your back.
For larger kidney stones, the doctor also may use a laser to break the kidney stones into small pieces. The doctor performs percutaneous nephrolithotomy in a hospital with anesthesia.

c. Generally for smaller stones doctors recommend drinking lots of water, consuming pain relievers and consuming medicines like alpha blocker to relax the ureter muscles, and help pass the kidney stones more quickly and with less pain

[Students are expected to find more information using the internet.]

Question 2.
Dietary restrictions suggested for kidney patients.
Dietary restrictions for kidney patients include the following:

1. Drinking large amounts of water.
2. Reduce consumption of oxalate rich food like rhubarb, beets, okra, spinach, Swiss chard, sweet potatoes, nuts, tea, chocolate and soy products.
3. Follow a diet low in salt and animal protein.
4. Reduce consumption of calcium supplements (if any) but consume appropriate amount of calcium in food.

[Students are expected to find more information using the internet.]

Maharashtra State Board Class 11 Biology Textbook Solutions

• Plant Tissues and Anatomy Class 11 Biology Textbook Solutions
• Morphology of Flowering Plants Class 11 Biology Textbook Solutions
• Animal Tissue Class 11 Biology Textbook Solutions
• Study of Animal Type : Cockroach Class 11 Biology Textbook Solutions
• Photosynthesis Class 11 Biology Textbook Solutions
• Respiration and Energy Transfer Class 11 Biology Textbook Solutions
• Human Nutrition Class 11 Biology Textbook Solutions
• Excretion and Osmoregulation Class 11 Biology Textbook Solutions
• Skeleton and Movement Class 11 Biology Textbook Solutions

Human Nutrition Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 14 Human Nutrition Textbook Exercise Questions and Answers.

Class 11 Biology Chapter 14 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 14 Exercise Solutions

1. Choose the correct option

Question A.
Acinar cells are present in ……………..
a. liver
b. pancreas
c. gastric glands
d. intestinal glands
Answer:
b. pancreas

Question B.
Which type of teeth is maximum in number in the human buccal cavity?
a. Incisors
b. Canines
c. Premolars
d. Molars
Answer:
d. Molars

Question C.
Select odd one out on the basis of digestive functions of tongue.
a. Taste
b. Swallowing
c. Talking
d. Mixing of saliva in food
Answer:
c. Talking

Question D.
Complete the analogy:
Ptyalin: Amylase : : Pepsin : …………….. .
a. Lipase
b. Galactose
c. Proenzyme
d. Protease
Answer:
d. Protease

2. Answer the following questions

Question A.
For the school athletic meet, Shriya was advised to consume either Glucon-D or fruit juice but no sugarcane juice. Why it must be so?
Answer:
Sugarcane juice contain disaccharides. Disaccharides take time to digest i.e. breaking into monosaccharides, Glucon — D and fruit juices contain monosaccharide. Therefore, for instant supply of energy during athletic meet Glucon – D or fruit juices are preferred and not sugarcane.

Question B.
Alcoholic people may suffer from liver disorder. Do you agree? Explain your answer.
Answer:

1. Liver disorder in alcoholic people may occur after years of heavy drinking.
2. Most of the alcohol in the body is broken down in the liver by an enzyme called alcohol dehydrogenase, which transforms ethanol into a toxic compound called acetaldehyde (CH₃CHO).
3. ver consumption of alcohol leads to cirrhosis (distorted or scarred liver) and eventually to liver failure.
   Therefore, alcoholic people may suffer from liver disorder.

Question C.
Digestive action of pepsin comes to a stop when food reaches small intestine. Justify.
Answer:
Pepsin acts in acidic medium thus it is active in stomach. There is alkaline condition in the small intestine. pH of small intestine is very high for pepsin to work. Therefore, pepsin gets denatured in the small intestine.

Question D.
Small intestine is very long and coiled. Even if we jump and run, why it does not get twisted? What can happen if it gets twisted?
Answer:

1. Mesentery is a tissue that is located in the abdomen. It attaches the small intestine to the wall of the abdomen and keeps it in place and therefore it does not get twisted while running and jumping.
2. If small intestine gets twisted, the affected spot may block the food, liquid passing through it. It may sometimes cut off the blood flow if the twist is very severe. If this happens the surrounding tissue may die and can cause serious problems.

3. Write down the explanation

Question A.
Digestive enzymes are secreted at appropriate time in our body. How does it happen?
Answer:

1. The digestive enzymes and juices are produced in sequential manner and at a proper time.
2. These secretions are under neurohormonal control.
3. Sight, smell and even thought of food trigger saliva secretion.
4. Tenth cranial nerve stimulates secretion of gastric juice in stomach.
5. Even the hormone gastrin brings about the same effect.

B. Explain the structure of tooth. Explain why human dentition is considered as thecodont, diphydont and heterodont.
Answer:

1. Structure of tooth:
   • A tooth consists of the portion that projects above the gum called crown and the root that is made up of two or three projections which are embedded in gum.
   • A short neck connects the crown with the root.
   • The crown is covered by the hardest substance of the body called enamel which is made up of calcium phosphate and calcium carbonate.
   • Basic shape of tooth is derived from dentin which is a calcified connective tissue.
   • The dentin encloses the pulp cavity. It is filled with connective tissue pulp. It contains blood vessels and nerves.
   • Pulp cavity has extension in the root of the tooth called root canal.
   • The dentin of the root of tooth is covered by cementurn which is a bone like substance that attaches the root to the surrounding socket in the gum.
2. Human dentition is described as thecodont, diphyodont and heterodont.
3. It is called the codont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
4. It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
5. It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.
   

Question C.
Explain heterocrine nature of pancreas with the help of histological structure.
Answer:
Pancreas:

1. Pancreas is a leaf shaped heterocrine gland present in the gap formed by bend of duodenum under the stomach.
2. Exocrine part of pancreas is made up of acini, the acinar cells secrete alkaline pancreatic juice that contains various digestive enzymes.
3. Pancreatic juice is collected and carried to duodenum by pancreatic duct.
4. The common bile duct joins pancreatic duct to form hepato-pancreatic duct. It opens into duodenum.
5. Opening of hepato-pancreatic duct is guarded by sphincter of Oddi.
6. Endocrine part of pancreas is made up of islets of Langerhans situated between the acini.
7. It contains three types of cells a-cells which secrete glucagon, P-cells which secretes insulin and 5 cells secrete somatostatin hormone.
8. Glucagon and insulin together control the blood-sugar level.
9. Somatostatin hormone inhibits glucagon and insulin secretion.

4. Write short note on

Question A.
Position and function of salivary glands.
Answer:
Salivary Glands:

• There are three pairs of salivary glands which open in buccal cavity.
• Parotid glands are present in front of the ear.
• The submandibular glands are present below the lower jaw.
• The glands present below the tongue are called sublingual.
• Salivary glands are made up of two types of cells.
• Serous cells secrete a fluid containing digestive enzyme called salivary amylase.
• Mucous cells produce mucus that lubricates food and helps swallowing.

Question B.
Jaundice
Answer:

1. Jaundice is a disorder characterized by yellowness of conjunctiva of eyes and skin and whitish stool.
2. It is a sign of abnormal bilirubin metabolism and excretion.
3. Jaundice develops if excessive break down of red blood cells takes place along with increased bilirubin level than the liver can handle or there is obstruction in the flow of bile from liver to duodenum.
4. Bilirubin produced from breakdown of haemoglobin is either water soluble or fat soluble.
5. Fat soluble bilirubin is toxic to brain cells.
6. There is no specific treatment to jaundice.
7. Supportive care, proper rest are the treatments given to the patient.
   [Note: Treatment ofjaundice will depend on the underlying cause of it. For example, hepatitis-induced jaundice would require treatment which includes antiviral or steroid medications ]

Question 5.
Observe the diagram. This is histological structure of stomach. Identify and comment on significance of the layer marked by arrow.

Answer:
The layer marked in the diagram represents glandular epithelium of mucosa.
Significance of the glandular epiihelium of mucosa:
Goblet cells of the epithelial layer of a mucous membrane secrete mucus which lubricates the lumen of the alimentary canal. This helps in movement of food through the gastrointestinal tract.

Question 6.
Find out pH maxima for salivary amylase, trypsin, nucleotidase and pepsin and place on the given pH scale

Answer:
Salivary amylase = 6.8
Trypsin = 8
Nucleotidase = 7.5
Pepsin = 2

Question 7.
Write the name of a protein deficiency disorder and write symptoms of it.
Answer:

1. Kwashiorkor is a protein deficiency disorder.
2. This protein deficiency disorder is found generally in children between one to three years of age.
3. Children suffering from Kwashiorkor are underweight and show stunted growth, poor brain development, loss of appetite, anaemia, protruding belly, slender legs, bulging eye, oedema of lower legs and face, change in skin and hair colour.

Question 8.
Observe the diagram given below label the A, B, C, D, E and write the function of A, C in detail.

Answer:
A- Bile duct, B- Stomach, C- Common hepatic duct, D- Pancreas, E- Gall Riadder

Functions: Bile duct: It carries hile from the gall bladder and empties it into the tipper part of the small intestine. Common hepatic duct: It drains bile from the liver. It helps in transportation of waste from liver and helps in digestion by releasing bile.
[Note: Labels (A) and (O) have been modified for the better understanding of the students]

Practical / Project : Here are the events in the process of digestion. Fill in the blanks and complete the flow chart.

Answer:

11th Biology Digest Chapter 14 Human Nutrition Intext Questions and Answers

Can you recall? (Textbook Page No. 161)

Question 1.
What is nutrition?
Answer:

1. Nutrition is the sum of the processes by which an organism consumes and utilizes food substances,
2. WHO defines nutrition as the intake of food, considered in relation to the body’s dietary needs.
3. The term nutrition includes the process like ingestion, digestion, absorption, assimilation and egestion.

Question 2.
Enlist life processes that provide us energy to perform different activities.
Answer:
The life processes which are essential and provide us energy are nutrition and respiration.

Think about it (Textbook Page No. 161)

Question 1.
Our diet includes all necessary nutrients. Still we need to digest it. Why is it so?
Answer:

1. Digestion is a very important process of converting complex, noil-diffusible and non-absorbable food substances into simple, diffusible and assimilable substances.
2. Our diet includes all necessary nutrients, which are in the form of complex substances like carbohydrates, proteins, fats and vitamins.
3. These complex substances are converted into simple, diffusible and assimilable substances through the process of digestion.
   Hence, there is a need for digestion of food.

Human Digestive System (Textbook Page No. 161)

Question 1.
Label the diagram
Answer:

Do you know? (Textbook Page No. 162)

Question 1.
Who controls the deglutition?
Answer:
The process of swallowing is called deglutition. Medulla oblongata controls the deglutition.

Question 2.
Is deglutition voluntary or involuntary?
Answer:

• Deglutition consists of three phases: oral phase, pharyngeal phase and oesophagal phase.
• The oral phase is voluntary whereas the pharyngeal and oesophagal phases are involuntary.
  [Source: Goya!, R. K., & Mashimo, H. (2006,.). Physio!o’ of oral, pharyngeal, and esophageal motility. GI Motility online.]

Use your brain power (Textbook Page No. 165)

Question 1.
Draw a neat labelled diagram of human alimentary canal and associated glands in situ.
Answer:

Question 2.
Write a note on human dentition.
Answer:

1. Human dentition is described as thecodont, diphyodont and heterodont.
2. It is called thecodont type because each tooth is fixed in a separate socket present in the jaw bones by gomphosis type of joint.
3. It is called diphyodont type because we get only two sets of teeth, milk teeth and permanent teeth.
4. It is called heterodont type because humans have four different type of teeth like incisors, canines, premolars and molars.

Question 3.
Muscularis layer in stomach is thicker than that in intestine. Why is it so?
Answer:
Muscularis layer in stomach is thicker than that of intestine because food is churned and gastric juices are mixed in the stomach whereas in intestine only absorption takes place.

Question 4.
Liver is a vital organ. Justify.
Answer:

1. Kupffer cells of liver destroy toxic substances, dead and worn-out blood cells and microorganisms.
2. Bile juice secreted by liver emulsifies fats and makes food alkaline.’
3. Liver stores excess of glucose in the form of glycogen.
4. Deamination of excess amino acids to ammonia and its further conversion to urea takes place in liver.
5. Synthesis of vitamins A, D, K and BI2 takes place in liver.
6. It also produces blood proteins like prothrombin and fibrinogen.
7. During early development, it acts as haemopoietic organ.
   Therefore, liver is a vital organ.

Internet my friend: (Textbook Page No. 171)

Question 1.
Collect the different videos of functioning of digestive system,
Answer:
[Note: Students can scan the adjacent Q.R code to get conceptual clarity with the aid of a relevant video.]

Find out (Textbook Page No. 162)

Question 1.
What will be the dental formula of a three years old child?
Answer:
The dental formula of a three-year-old child will be: I \(\frac{2}{2}\), C \(\frac{1}{1}\), M \(\frac{2}{2}\) = \(\frac{2,1,2}{2,1,2}\)
i. e. 5 × 2 = 10 teeth in each jaw = 20 teeth.
As a child has 20 teeth by the age of three.

Question 2.
What is dental caries and dental plaque? How can one avoid it?
Answer:

• Dental caries are tooth decay or cavities caused by acids secreted by bacteria. Dental caries may be yellow or black in color.
• Dental plaques also known as tooth plaque is a soft, sticky film which forms on the teeth regularly. It is colourless to pale yellow in colour.
• Tooth decay and dental plaque can be prevented by brushing teeth twice a day with a fluoride containing tooth paste.
• Rinsing mouth thoroughly with a mouth wash and use of dental floss or interdental cleaners to clean teeth daily can help to avoid dental caries and dental plaque.

Internet my friend (Textbook Page No. 162)

Question 1.
Find out the role of orthodontist and dental technician.
Answer:
a. Orthodontics is a specialization in dental profession. Orthodontist straightens the crooked teeth, locates problem in patients’ teeth and their overall oral development. They might use X-rays, plaster molds or dental appliances like retainers and space maintainers to correct the problems,

b. Dental technicians are the ones which improves patients’ appearance, ability to chew and speech. They make dentures, crowns, bridges and dental braces.

Question 2.
What is a root canal treatment?
Answer:

• Root canal treatment is also known as endodontic treatment.
• It is a dental treatment of removing infection from inside of a tooth.
• Root canal is hollow section of tooth which contains the nerve tissue, blood vessels and other cells, this is also known as pulps.
• Crown and root are a part of tooth. Crown is present above the gum while root is embedded in the gum.
  e. Pulp which is present inside the root canal nourishes the tooth and provides moisture to the surrounding material.
• The nerves present inside the pulp sense hot cold temperatures as pain.
• First step of a root canal treatment is removal of dead pulp tissues by making a hole on the surface of tooth.
• In second step, the dentist cleans and decontaminates the area and fills the hollow area with adhesive cement in order to seal the canal completely.
• The tooth is dead after the therapy and the patient no longer feel any pain but the tooth becomes more fragile than ever.
• The last step of root canal is adding a crown or filling. Until the crown or filling is complete, patient is not supposed to chew or bite using that tooth. After the crown or filling patient can use that tooth as before.

Find out (Textbook Page No. 163)

Question 1.
You must have heard about appendicitis. It is inflammation of appendix. Find more information about this disorder.
Answer:

1.  Appendicitis is a condition where there is inflammation of appendix.
2. Appendix is a vestigial organ. It is a linger shaped pouch that projects from colon on the lower right side of the abdomen.
3. Appendicitis pain is very severe. It initially starts from the navel and then moves.
4. It occurs in the people of age group between 10 to 30.
5. Surgical removal is the standard treatment for appendicitis.
6. Symptoms: Nausea and vomiting, loss of appetite, low grade fever, constipation, abdominal bloating, severe pain in the right side of the abdomen.
7. Appendicitis is caused when there is blockage in the lining of the appendix that results in infection. The bacteria multiply rapidly and causes inflammation and it is then filled with pus.
8. If not treated properly appendix can rupture which can lead to further complications.
   [Students can use above answer for reference and find more information about appendicitis.]

Question 2.
What is heartburn? Why do we take antacids to control it?
Answer:
Heart burn is a problem created when stomach contents (acid) are forced back up to oesophagus. It causes a burning pain in lower chest.

Antacids are bases and help to treat heartburn by neutralizing the stomach acid. The key ingredients of antacids are calcium carbonate, magnesium hydroxide, aluminium hydroxide or sodium bicarbonate.

Activity (Textbook Page No. 163)

Make a model of human digestive system in a group.
Answer:
[Students are expected to perform this activity on their own.]

Always Remember (Textbook Page No. 166)

Question 1.
Food remains for a very short time in mouth but action of salivary amylase continues for further IS to 30 minutes till gastric juice mixes with food in the stomach. Why do you think it stops after the food gets mixed with gastric juice?
Answer:

1. The gastric juices are mixed with food in the stomach.
2. The pH of the stomach is 1.0-2.0 which is very acidic. Such high level of acidity leads to denaturation of salivary amylase’s protein structure.
3. On the other hand, pH 6.8 is required for salivary amylase to carry out the activity which is not found in stomach. Thus, activity of salivary amylase is stopped when food is mixed with gastric juice.

Internet my friend (Textbook Page No. 167)

Question 1.
How are bile pigments formed?
Answer:

1. When old and worn out red blood cells are destroyed by macrophages in liver, the globin portion of hemoglobin is split off and heme is converted to biliverdin.
2. Most of this biliverdin is converted to bilirubin, which gives bile its major pigmentation.
   [Source http://www.biologydiscussion.com/human-physiology/digestive-system/bile-pigments/bile-pigments-origin-and-formation-digestive-juice-human-biology/81803]

Think about it (Textbook Page No. 167)

Question 1.
How can I keep my pancreas healthy? Can a person live without pancreas?
Answer:

1. Pancreas can be kept healthy by:
   • Eating proper balanced and low-fat diet, with plenty of whole grains, fruits and vegetables.
   • Regular exercise and maintaining a healthy weight.
   • Limiting alcohol consumption and avoid smoking.
   • Adequate intake of water.
   • Regular checkups.
2. The pancreas is a gland that secretes digestive enzymes and insulin which is needed for a person to survive.
3. Without pancreas the person will develop diabetes and will have to take insulin for the rest of the life.
4. Without pancreas the body’s ability to absorb nutrients also decreases.
   Hence, though a person can survive without pancreas he may have to remain dependent on the medicines for survival.

Do it yourself? (Textbook Page No. 167)

Question 1.
You have studied the representation of enzymatic actions in the form of reactions.
Write the reactions of pancreatic enzymes.
Answer:

Do it yourself (Textbook Page No. 168)

Question 1.
Observe the following reactions and explain in words.

Answer:

1. Maltase acts on maltose to form glucose.
2. Sucrase acts on sucrose to form glucose and fructose.
3. Lactase acts on lactose to form glucose and galactose.
4. Dipeptidase acts on dipeptides to form amino acids.
5. Emulsified fats are converted into fatty acids and glycerol by lipase.

Use your brain power (Textbook Page No. 168)

Question 1.
Make a flow chart for digestion of carbohydrate.
Answer:

Question 2.
What is a proenzyme? Enlist various proenzymes involved in process of digestion and state their function.
Answer:
Proenzymes are synthesized in cells as an inactive precursor that undergo some modification before becoming catalytically active.
The various proenzymes involved in process of digestion are as follows:

• Pepsinogen: Pepsinogen when converted into its active form pepsin acts on proteins to form peptones and proteoses.
• Trypsinogen: Trypsinogen when converted to it active form trypsin converts proteins, proteoses and peptones to polypeptides.
• Chymotrypsinogen: Chymotrypsinogen when converted to active form chymotrypsin it converts polypeptides to dipeptides.

Question 3.
Differentiate between Chyme and Chyle.
Answer:

Question 4.
Digestion of fats take place only after the food reaches small intestine. Give reason.
Answer:
Digestion of fats takes place in small intestine because the presence of fats in small intestine stimulates the release of pancreatic lipase from pancreas and bile from liver. Pancreatic lipases hydrolyze fat molecules into fatty acids and monoglycerides and bile brings about emulsification of fats. Therefore, digestion of fats occur when food reaches small intestine.

Observe and Discuss (Textbook Page No. 169)

Question 1.
Action of digestive juice in your group.
Answer:

Can you recall? (Textbook Page no. 170)

Question 1.
What is balanced diet?
Answer:
Balanced diet is a diet which contains proper amount of carbohydrates, fats, vitamins, proteins and minerals to maintain a good health.

Question 2.
Explain the terms undernourished, over-nourished and malnourished in details.
Answer:

• Undernourished: When supply of nutrients is less than the minimum amount of nutrients or food required for good health is called undernourished.
• Over-nourished: The intake of nutrients is excessive. In over-nourished the amount of nutrients exceeds the amount required for normal growth.
• Malnourished: Malnourished is a condition where a person’s diet does not contain right amount of nutrients.

Do you know? (Textbook Page No. 170)

Question 1.
What is gross calorific value?
Answer:
The amount of heat liberated by complete combustion of lg food in a bomb calorimeter is termed as gross calorific (gross energy) value.

Question 2.
What is physiological value?
Answer:
The actual energy produced by 1 g food is its physiological value.

Question 3.
Name the following
Energy content of food in animals is expressed in terms of?
Answer:
Heat Energy

Question 4.
Complete the following table representing Gross calorific value and physiological value of food component.

Answer:

Find out (Textbook Page No. 171)

Question 1.
Find out the status of nialnutrition among children in Maharashtra and efforts taken by the government to overcome the situation. Search for various NGOs working in this field.
Answer:
93,783 children have been diagnosed with severe acute malnutrition and 5.7 lakh with moderate acute malnutrition in Maharashtra.
Steps taken by government to overcome malnutrition:

1. Promotion of infant and young child feeding practices.
2. Management of malnutrition at community and facility level by trained service providers.
3. Treatment of children with severe acute malnutrition at special units called the Nutrition Rehabilitation Centres (NRCs), set up at public health facilities.
4. A special program to combat micronutrient deficiencies of Vitamin A, Iron and Folic acid.
5. The initiatives like Mother and Child protection card, village health and nutrition days, are taken by the government for addressing the nutrition concerns in children, pregnant women and lactating mothers.

Various NCOs working in this field:

1. Akshay Patra
2. Fight Hunger Foundation,
3. Feeding India,
4. No Hungry child
   [Source: http://pib.nic.in/newsite/PrintRelease.aspx?relid=l 13725; https://yourstory.com/2016/10/world- food-day-ngosj
   [Note: Students can use above answer as reference and find more information from the internet.]

Question 2.
Are jaundice and hepatitis same disorders?
Answer:
Jaundice and Hepatitis are two different disorders.

Jaundice: Jaundice occurs when the rate of bilirubin production exceeds the rate of its elimination. It causes yellowing of skin and eyes.

Hepatitis: It is a disease where there is inflammation of liver. It may be caused because of infection, over alcohol consumption, immune system disorder etc.

Do you know (Textbook Page No. 171)

Question 1.
Alcoholism causes different disorders of liver like steatosis (fatty liver), alcoholic hepatitis, fibrosis and cirrhosis. Collect more information on these disorders and try to increase awareness against alcoholism in society. Collect information about NGOs working against alcoholism.
Answer:
Steatosis (fatty liver): Steatosis is accumulation of fat in the liver. Treatment can help but it cannot be cured. Major risk factors are obesity and Diabetes type II, it is also associated with excessive alcohol consumption. Fatigue, weight loss and abdominal pain are some symptoms. It is a benign condition but in very smaller number of patients it can lead to liver failure. Treatment involves diet and exercise to reduce obesity.

Alcoholic hepatitis: Alcoholic Hepatitis is liver inflammation caused by excessive consumption of alcohol. It occurs in people who drink heavily for many years. Symptoms like yellowing of skin and eye, accumulation of fluid in stomach which leads to increase in stomach size. Treatments like completely stopping of alcohol consumption, hydration and nutrition care are carried out. Administration of steroid drugs reduces liver inflammation.

Fibrosis: There is significant scarring of liver tissue in this condition. Fibrosis itself does not cause any symptoms. Diagnosis includes doctor’s evaluation, blood tests and imaging tests, liver biopsy. Treatments include stopping the consumption of alcohol. There are no such effective drugs for curing of fibrosis.

Cirrhosis: It is a chronic liver damage caused due to various reasons which leads to irreversible scarring of liver and liver failure. Causes of cirrhosis are chronic alcohol abuse and hepatitis. Patients may experience fatigue, weakness and weight loss. In later stages, patients may develop jaundice, abdominal swelling and gastrointestinal bleeding. In advanced stage, a liver transplant is required.

NGOs working against alcoholism:

1. Muktangan Rehabilitation Centre
2. Anmol Jeevan Foundation
3. Sankalp Rehabilitation Trust
4. Kripa Foundation
5. Harmony Foundation
6. Hands for you Rehab Centre

Maharashtra State Board Class 11 Biology Textbook Solutions

• Plant Tissues and Anatomy Class 11 Biology Textbook Solutions
• Morphology of Flowering Plants Class 11 Biology Textbook Solutions
• Animal Tissue Class 11 Biology Textbook Solutions
• Study of Animal Type : Cockroach Class 11 Biology Textbook Solutions
• Photosynthesis Class 11 Biology Textbook Solutions
• Respiration and Energy Transfer Class 11 Biology Textbook Solutions
• Human Nutrition Class 11 Biology Textbook Solutions
• Excretion and Osmoregulation Class 11 Biology Textbook Solutions
• Skeleton and Movement Class 11 Biology Textbook Solutions

Respiration and Energy Transfer Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 13 Respiration and Energy Transfer Textbook Exercise Questions and Answers.

Class 11 Biology Chapter 13 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 13 Exercise Solutions

1. Choose the Correct option.

Question (A)
The reactions of the TCA cycle occur in
(A) ribosomes
(B) grana
(C) mitochondria
(D) endoplasmic reticulum
Answer:
(C) mitochondria

Question (B)
In eukaryotes the complete oxidation of a molecule of glucose results in the net gain of
(A) 2 molecules of ATP
(B) 36 molecules of ATP
(C) 4 molecules of ATP
(D) 38 molecules of ATP
Answer:
(D) 38 molecules of ATP

Question (C)
Which step of Krebs cycle operates substrate-level phosphorylation?
(A) ∝-ketoglutarate → succinyl CoA.
(B) Succinyl CoA → succinate
(C) Succinate → fumarate
(D) Fumarate → malate
Answer:
(B) Succinyl CoA → succinate

2. Fill in the blanks with suitable words.

Question 1.
A. Acetyl CoA is formed from __________ and co-enzyme A.
B. In the prokaryotes ________ molecules of ATP are formed per molecule of glucose oxidised.
C. Glycolysis takes place in ________ .
D. F₁ – F₀ particles participate in the synthesis of _________ .
E. During glycolysis _________ molecules of NADH⁺H⁺ are formed.
Answer:
A. pyruvic acid
B. 2/38
C. cytoplasm
D. ATP
E. 2
[Note: ii. In prokaryotes, during anaerobic respiration 2 ATPs are formed per glucose and 38 ATPs are formed during aerobic respiration.]

3. Answer the following questions

Question (A)
When and where does anaerobic respiration occur in man and yeast?
Answer:
1. In absence of oxygen, anaerobic respiration takes place in skeletal muscles of man during vigorous exercise.
2. Anaerobic respiration occurs in the cytoplasm of the yeast cell.

Question (B)
Why is less energy produced during anaerobic respiration than in aerobic respiration?
Answer:
Anaerobic respiration produces less energy because:

1. Incomplete breakdown of respiratory substrate takes place.
2. Some of the products of anaerobic respiration can be oxidised further to release energy which shows that anaerobic respiration does not liberate the whole energy contained in the respiratory substrate.
3. NADH₂ does not produce ATP, as electron transport is absent.
4. Only 2 ATP molecules are generated from one molecule of glucose during anaerobic respiration.

Question (C)
Which is the site for ETS in mitochondrial respiration?
Answer:
The inner mitochondrial membrane is the site for ETS in mitochondrial respiration.

Question (D)
Which compound is the terminal electron acceptor in aerobic respiration?
Answer:
Molecular oxygen is the terminal electron acceptor in aerobic respiration.

Question (E)
What is RQ.? What is its value for fats?
Answer:
1. Respiratory quotient (R.Q.) or respiratory ratio is the ratio of volume of CO₂ released to the volume of O₂ consumed in respiration.
2. R.Q. = Volume of CO₂ released / Volume of O₂ consumed

Question (F)
What are respiratory substrates? Name the most common respiratory substrate.
Answer:
Respiratory substrates are the molecules that are oxidized during respiration to release energy which can be used for ATP synthesis. Carbohydrates, fats and proteins are the common respiratory substrate. Glucose is the most common respiratory substrate.

Question (G)
Write explanatory notes on:

Question (i)
Glycolysis
Answer:
Glycolysis is a process where glucose is broken down into two molecules of pyruvic acid, hence called glycolysis (glucose-breaking). It is common to both aerobic and anaerobic respiration. It occurs in the cytoplasm of the cell. It involves ten steps.
Glycolysis consists of two major phases:
1. Preparatory phase (1-5 steps).
2. Payoff phase (6-10 steps).
1. Preparatory phase:
a. In this phase, glucose is phosphorylated twice by using two ATP molecules and a molecule of fructose 1,6-bisphosphate is formed.
b. It is then cleaved into two molecules of glyceraldehyde-3-phosphate and dihydroxy acetone phosphate. These two molecules are 3-carbon carbohydrates (trioses) and are isomers of each other.
c. Dihydroxy acetone phosphate is isomerised to second molecule of glyceraldehyde-3-phosphate.
d. Therefore, two molecules of glyceraldehyde-3- phosphate are formed.
e. Preparatory phase of glycolysis ends.

2. Payoff phase:
a. In this phase, both molecules of glyceraldehyde-3-phosphate are converted to two molecules of 1,3- bisphoglycerate by oxidation and phosphorylation. Here, the phosphorylation is brought about by inorganic phosphate instead of ATP.
b. Both molecules of 1, 3-bisphosphoglycerate are converted into two molecules of pyruvic acid through series of reactions accompanied with release of energy. This released energy is used to produce ATP (4 molecules) by substrate-level phosphorylation.

Question (ii)
Write explanatory notes on: Fermentation by yeast
Answer:
Alcoholic fermentation is a type of anaerobic respiration where the pyruvate is decarboxylated to acetaldehyde. The acetaldehyde is then reduced by NADH+H⁺ to ethanol and Carbon dioxide. Since ethanol is produced during the process, it is termed alcoholic fermentation.

Question (iii)
Write explanatory notes on: Electron transport chain
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. During oxidation of NADH+H⁺ and FADH₂ , electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
7. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a₃. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
8. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
9. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F₀ part of oxysome (complex V) present in inner mitochondrial membrane.
10. This proton channelling by F₀ is coupled to catalytic site of F₁ which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
11. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Significance of ETS:

1. Major amount of energy is generated through ETS or terminal oxidation in the form of ATP molecules.
2. Per glucose molecule 38 ATP molecules are formed, out of which 34 ATP molecules are produced through ETS.
3. Oxidized coenzymes such as NAD and FAD are regenerated from their reduced forms (NADH+H⁺ and FADH₂) for recycling.
4. In this process, energy is released in a controlled and stepwise manner to prevent any damage to the cell.
5. ETS produces water molecules.

Question (H)
How are glycolysis, TCA cycle and electron transport chain-linked? Explain.
Answer:
Glycolysis, TCA cycle and electron transport chain are linked in the following manner:

1. The coenzymes are initially present in the form of NAD⁺ and FAD⁺ which latter get reduced to NADH+H⁺ and FADH+H⁺ by accepting the hydrogen from organic substrate during glycolysis, link reaction and Krebs cycle.
2. During glycolysis, glucose is oxidised to two molecules of pyruvic acid with net gain 2 molecules of NADH+H⁺.
3. This pyruvic acid undergoes link reaction to form two molecules of acetyl CoA and two molecules of NADH+H⁺.
4. Acetyl CoA, thus formed enters into the Krebs cycle and it gets completely oxidised to C0₂ and H₂0; with a net gain of 6 NADH+H+ and 2 FADH+H⁺ are formed.
5. During ETS, reduced coenzymes are reoxidized to NAD⁺ and FAD⁺ with a net gain of 34 ATPs. The ATPs thus formed are used during glycolysis.
6. The oxidized NAD⁺ and FAD⁺ will again accept the hydrogen from organic substrate. Thus, reduced coenzymes are converted back to their oxidized forms by dehydrogenation to keep the process going.

Question (I)
How would you demonstrate that yeast can respire both aerobically and anaerobically?
Answer:
Respiration in yeast can be demonstrated with the help of an experiment.
Anaerobic respiration in yeast:

1. A pinch of dry baker’s yeast suspended in water containing 10ml of 10% glucose in a test tube (test tube A).
2. The surface of the liquid is covered with oil to prevent entry of air and the test tube is closed tightly with rubber stopper to prevent leakage.
3. One end of a short-bent glass tube is inserted through it to reach the air inside the tube.
4. Other end of the glass tube is connected by a polyethylene or rubber tubing to another bent glass tube fitted into a stopper.
5. The open end of the glass tube (delivery tube) is dipped into lime water containing in a test tube
   (Tube B).
6. Stoppers of both the tubes are fitted tightly to prevent leakage of gases. First test tube is placed in warm water (37° C-38° C) in a beaker.
7. Lime water gradually turns milky, indicating the evolution of carbon dioxide from the yeast preparation.
8. Level of the lime water in the delivery tube does not rise, showing that there is no decline in volume of gas in test tube A and consequently no utilization of oxygen by yeast. Preparation is stored for a day or two.
9. When we open the stopper of tube A we will notice a smell of alcohol indicating the formation of ethanol.
10. From this activity it may be inferred that yeast respires anaerobically to ferment glucose to ethanol and carbon dioxide.

Aerobic respiration in yeast: Experiment explained can be carried out for demonstrating aerobic respiration in yeast.

1. If the level of the lime water in the test tube B rises, indicating intake of oxygen, hence the level of volume of gas rises.
2. The preparation tube is stored for a day or two, if no smell of alcohol is noticed it indicates that the yeast respires aerobically.

Question (J)
What is the advantage of step wise energy release in respiration?
Answer:
In ETS energy is released in step wise manner to prevent damage of cells.

1. A stepwise release of the chemical bond energy facilitates the utilization of a relatively higher proportion of that energy in ATP synthesis.
2. Activities of enzymes for the different steps may be enhanced or inhibited by specific compounds. This provides a means of controlling the rate of the pathway and the energy output according to need of the cell.
3. The same pathway may be utilized for forming intermediates used in the synthesis of other biomolecules like amino acids.

Question (K)
Explain ETS.
Answer:

1. NADH₂ and FADH₂ produced during glycolysis, connecting link reaction and Krebs cycle are oxidized with the help of various electron carriers and enzymes.
2. These carriers and enzymes are arranged on inner mitochondrial membrane in the form of various complexes as complex I, II, III, VI and V.
3. NADH+H⁺ is oxidised by NADH dehydrogenase (complex I) and it’s electrons are transferred to ubiquinone (coenzyme Q-CoQ) present on inner membrane of mitochondria. Reduced ubiquinone is called as ubiqunol.
4. FADH₂ is oxidised by complex II (Succinate dehydrogenase) and these electrons are also transferred to CoQ.
5. During oxidation of NADH+H⁺ and FADH₂ , electrons and protons are released but only electrons are canned forward whereas protons are released into outer chamber of mitochondria (intermembrane space).
6. Ubiquinol is oxidised by complex-III (Cytochrome bcl complex) and it’s electrons are transferred to cytochrome C. Cytochrome C is a small, iron-containing protein, loosely associated with inner membrane. It acts as a mobile electron carrier, transferring the electrons between complex III and IV.
7. Cytochrome C is oxidised by complex IV or cytochrome C oxidase consisting of cytochrome a and a₃. Electrons are transferred by this complex to the molecular oxygen. This is terminal oxidation.
8. Reduced molecular oxygen reacts with protons to form water molecule called as metabolic water.
9. Protons necessary for this are channelled from outer chamber of mitochondria into inner chamber by F₀ part of oxysome (complex V) present in inner mitochondrial membrane.
10. This proton channelling by F₀ is coupled to catalytic site of F₁ which catalyses the synthesis of ATP from ADP and inorganic phosphate. This is oxidative phosphorylation.
11. As transfer of protons is accompanied with synthesis of ATP, this process is named as ‘Chemiosmosis’ by Peter Mitchell.

Question (L)
Discuss “The respiratory pathway is an amphibolic pathway”.
OR
Question (M)
Why is Krebs cycle referred as amphibolic pathway?
Answer:

1. Respiration is considered as a catabolic process; however, it is not entirely correct in case of Krebs cycle.
2. Many reactions of Krebs cycle involve oxidation of acetyl CoA to release energy and C02.
3. However, the breakdown of respiratory substrates provides intermediates like a-ketoglutarate, oxaloacetate are used as precursors for synthesis of fatty acids, glutamic acid and aspartic acid respectively.
4. Thus, as the same respiratory process acts as catabolic as well as anabolic pathway for synthesis of various intermediate metabolic products, it is called amphibolic pathway.

Question (N)
The common pathway for both aerobic and anaerobic respiration is
(A) Krebs cycle
(B) Glycolysis
(C) ETS
(D) Terminal oxidation
Answer:
(B) Glycolysis

4. Compare

Question (A)
Photosynthesis and respiration.
Answer:

Question (B)
Aerobic respiration and Anaerobic respiration
Answer:

5. Differentiate between

Question (A)
Respiration and combustion.
Answer:

Question (B)
Distinguish between Glycolysis and Krebs cycle.
Answer:

Question (C)
Aerobic respiration and fermentation.
Answer:

Question 6.
Identify the cycle given below. Correct it and fill in the blanks and write description of it in your own
Answer:

1. Krebs cycle or citric acid cycle is the second phase of aerobic respiration which takes place in the matrix of the mitochondria.
2. The acetyl CoA formed during the link reaction undergoes aerobic oxidation.
3. This cycle serves a common oxidative pathway for carbohydrates, fats and proteins.
4. In mitochondria pyruvic acid is decarboxylated and the remaining 2-carbon fragment is combined with a molecule of coenzyme A to form acetyl-CoA.
5. This reaction is an oxidative decarboxylation process and produces H⁺ ions and electrons along with carbon dioxide. During the process NAD⁺ is reduced to NADH+H⁺.
6. P-oxidation of fatty acids also produces acetyl-CoA as the end product.
7. Acetyl-CoA from both sources is condensed with oxaloacetic acid to form citric acid. Citric acid is oxidized step-wise by mitochondrial enzymes, releasing carbon dioxide.
8. Regeneration of oxaloacetic acid occurs to complete the cycle.
9. There are four steps of oxidation in this cycle, catalyzed by dehydrogenases (oxidoreductases) using NAD⁺ or FAD⁺ as the coenzyme.
10. The coenzymes are consequently reduced to NADH+H⁺ and FADH₂ respectively. These transfer their electrons to the mitochondrial respiratory chain to get reoxidised.
11. One molecule of GTP (ATP) is also generated for every molecule of citric acid oxidized.

Practical / Project:

Question 1.
Make Powerpoint Presentation on Glycolysis, Krebs Cycle and Conduct the group discussion on it in classroom.
[Note: Students are expected to perform above activity on their own.]

11th Biology Digest Chapter 13 Respiration and Energy Transfer Intext Questions and Answers

Can you recall? (Textbook Page No. 151)

(i) Which nutrients are used for energy production?
Answer:
Nutrients like carbohydrates, fats and proteins are used for energy production.

(ii) Why do organisms take up oxygen and release carbon dioxide?
Answer:
a. At cellular level, organisms require energy to carry out different metabolic activities.
b. The energy is made available by oxidizing/breaking the food.
Therefore, oxygen is required by aerobic organisms for breaking the food and carbon dioxide is released as a byproduct of oxidation.

Use your brainpower (Textbook Page No. 152)

Why is glycolysis considered as biochemical proof of evolution?
Answer:

1. Glycolysis does not require oxygen. Hence it might have been used by earlier organisms for energy production, as there was no free oxygen in atmosphere of primitive earth.
2. Glycolysis is the first metabolic pathway, an ancient pathway which is common to both aerobic and anaerobic organisms.
3. All cells have glycolysis in their metabolic pathway.
4. Upto pyruvate the pathway is similar to all aerobic and anaerobic organisms. Later, the fate of pyruvic acid can be either C0₂ or ethanol or lactic acid depending upon the type of organism.
5. Hence it is considered as a biochemical proof of evolution.

Use your brainpower (Textbook Page No. 152)

(i) What is role of Mg⁺⁺, Zn⁺⁺ in various steps of glycolysis?
Answer:
a. Mg⁺⁺ and Zn⁺⁺ are the cofactors that are tightly bound to enzymes and helps the enzymes to perform their functions.
b. They regulate the activity of the most important enzymes like Hexokinase, Phosphoffuctokinase, Triose phosphate dehydrogenase, Phosphoglycerate kinase, Enolase, Pyruvate kinase.

(ii) Why some reactions of glycolysis are reversible and some irreversible?
Answer:
Irreversible chemical reactions:
Some chemical reactions can occur in only one direction i.e. these reactions are irreversible reactions. The reactants can change to the products, but the products cannot change back to the reactants.

Reversible chemical reactions:

1. Some chemical reactions can occur in both directions i.e. these reactions are reversible reactions. In this case the reactants change to the products and the products can change back to the reactants, atleast under specific conditions.
2. Out of ten, four are irreversible reactions which involves the enzyme kinase that is required for phosphorylation reactions, these reactions involve large negative energy AG, hence the reactions are irreversible.
3. Other reversible reactions do not involve high negative energy hence are reversible.

Use your brainpower (Textbook Page No. 152)

Why do athletes like sprinters have higher proportion of white muscle fibers?
Answer:
1. The white muscle fibres produce energy in a very short period of time that is required for fast and severe work. Thus, the energy becomes immediately available to the athletes.
2. On the other hand, the red muscle produce energy over a prolonged period of time, hence athletes have higher proportion of white muscle fibers.

Can you recall? (Textbook Page No. 151)

Which steps are involved in aerobic respiration?
Answer:
It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Can you recall? (Textbook Page No. 151)

What is aerobic and anaerobic respiration?
Answer:
For anaerobic respiration: Anaerobic respiration is the cellular respiration that does not involve the atmospheric oxygen. It is also called as fermentation. It involves glycolysis where the product of glycolysis i.e. pyruvate is converted to either lactic acid or ethanol and for aerobic respiration.
1. Aerobic respiration occurs in the presence of free molecular oxygen during oxidation of glucose.
2. In this type of respiration, the glucose is completely oxidized to C0₂ and H₂0 with release of large amount of energy. It involves glycolysis, acetyl CoA formation (connecting link reaction), Krebs cycle, electron transfer chain reaction and terminal oxidation.

Use your brainpower (Textbook Page No. 157)

Do the plants breath like animals? If yes, how and why?
Answer:

1. Yes, plants breath like animals because they also require energy to carry out different metabolic activities. Hence, plants take up oxygen required for respiration and they also give out C0₂.
2. Plants take care of their gas exchange needs. The stomata and lenticels are important for this purpose.
3. Leaves are well adapted for gaseous exchange during photosynthesis.
4. Large amount of gases is exchanged. In plants, each living cell is located quite close to the surface of the plants.

Internet my friend (Textbook Page No. 155)

What is effect of carbon monoxide poisoning on cytochromes?
Answer:

1. At sub-cellular level, carbon monoxide is toxic for mitochondria.
2. It alters the mitochondrial respiratory chain at the cytochrome c oxidase level (complex IV of the mitochondrial respiratory chain) and causes inhibition of ETS.
3. This inhibition leads to the development of symptoms observed in acute CO poisoning, and in some diseases related to smoking.
4. These symptoms include headache, nausea, vomiting, dizziness, weakness, difficulty in concentration or confusion, visual changes, syncope, seizures, abdominal pain and muscle cramping.

Can you recall? (Textbook Page No. 151)

Which is most preferred nutrient among carbohydrate, protein and fat for energy production? Why?
Answer:

1. The preferred nutrient is carbohydrate because it quickly supplies energy as compared to other nutrients.
2. Carbohydrates are easy to digest as compared to fats.
3. The RQ of carbohydrate is 1. Hence allows complete oxidation of food. Thus, the preferred nutrient is carbohydrate.

Internet my friend (Textbook Page No. 158)

Calculate the RQ for different respiratory substrates using appropriate formula.
Answer:
The RQ for different respiratory substrates are:
1. Carbohydrates (R.Q. is 1)
When carbohydrates are used as substrate, equal volumes of C0₂ and 0₂ are released and consumed respectively, thus its R.Q. is 1.
C₆ H₁₂ O₆ + 6O₂ → 6 C0₂ + 6H₂0
R.Q. = 6C0₂ / 60₂ = 1.0

2. Fats (R.Q. is less than 1)
Substrates like fats are poorer in oxygen than carbohydrates. Thus, more oxygen is utilized for its complete oxidation.
2(C₅₁ H₉₈ O₆) + 145O₂ → 102CO₂ + 98H₂O + Energy
R.Q. = C0₂ / 0₂ = 102 / 145 = 0.7

3. Protein respiration (R.Q. is less than 1)

1. When proteins serve as respiratory substrate, they are first degraded to amino acids.
2. Then, amino acids are converted into various intermediates of carbohydrates.
3. However, amino acids have low proportion of O₂ as compared to carbohydrates.
4. Thus, they require more O₂ during their complete oxidation and value of R.Q. becomes less than 1.
5. In case of proteins, the R.Q. is approximately 0.9.

Maharashtra State Board Class 11 Biology Textbook Solutions

• Plant Tissues and Anatomy Class 11 Biology Textbook Solutions
• Morphology of Flowering Plants Class 11 Biology Textbook Solutions
• Animal Tissue Class 11 Biology Textbook Solutions
• Study of Animal Type : Cockroach Class 11 Biology Textbook Solutions
• Photosynthesis Class 11 Biology Textbook Solutions
• Respiration and Energy Transfer Class 11 Biology Textbook Solutions
• Human Nutrition Class 11 Biology Textbook Solutions
• Excretion and Osmoregulation Class 11 Biology Textbook Solutions
• Skeleton and Movement Class 11 Biology Textbook Solutions

Photosynthesis Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 12 Photosynthesis Textbook Exercise Questions and Answers.

Class 11 Biology Chapter 12 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 12 Exercise Solutions

1. Choose the correct option.

Question (A)
A cell that lacks chloroplast does not
(a) evolve carbon dioxide
(b) liberate oxygen
(c) require water
(d) utilize carbohydrates
Answer:
(b) liberate oxygen

Question (B)
Energy is transferred from the light reaction step to the dark reaction step by
(a) chlorophyll
(b) ADP
(c) ATP
(d) RuBP
Answer:
(c) ATP

Question (C)
Which one is wrong in photorespiration?
(a) It occurs in chloroplasts
(b) It occurs in day time only
(c) It is characteristic of C₄-plants
(d) It is characteristic of C₃-plants
Answer:
(c) It is characteristic of C₄-plants

Question (D)
Non-cyclic phosphorylation differs from cyclic photophosphorylation in that former
(a) involves only PS
(b) Include evolution of 02
(c) involves formation of assimilatory power
(d) both (b) and (c)
Answer:
(d) both (b) and (c)

Question (E)
For fixation of 6 molecules of C0₂ and formation of one molecule of glucose in Calvin cycle, requires
(a) 3 ATP and 2 MADPPE
(b) 18 ATP and 12 NADPH₂
(c) 30 ATP and 18 NADPH₂
(d) 6 ATP and 6 NADPIT₂
Answer:
(b) 18 ATP and 12 NADPH₂

Question (F)
In maize and wheat, the first stable products formed in bundle sheath cells respectively are
(a) OAA and PEPA
(b) OAA and OAA
(c) OAA and 3PGA
(d) 3PGA and OAA
Answer:
(c) OAA and 3PGA

Question (G)
The head and tail of chlorophyll are made up of
(a) porphyrin and phytin respectively
(b) pyrrole and tetrapyrrole respectively
(c) porphyrin and phytol respectively
(d) tetrapyrole and pyrrole respectively
Answer:
(c) porphyrin and phytol respectively

Question (H)
The net result of photo-oxidation of water is release of ……………. .
(a) electron and proton
(b) proton and oxygen
(c) proton, electron and oxygen
(d) electron and oxygen
Answer:
(c) proton, electron and oxygen

Question (I)
For fixing one molecule of C0₂ in Calvin cycle are required.
(a) 3ATP + 1NADPFE
(b) 3ATP + 2NADPH₂
(c) 2ATP + 3NADPH₂
(d) 3ATP + 3NADPFE
Answer:
(b) 3ATP + 2NADPH₂

Question (J)
In presence of high concentration of oxygen, RuBP carboxylase converts RuBP to …………… .
(a) Malic acid and PEP
(b) PGA and PEP
(c) PGA and malic acid
(d) PGA and phosphoglycolate
Answer:
(d) PGA and phosphoglycolate

Question (K)
The sequential order in electron transport from PSII to PSI of photosynthesis is
(a) FeS, PQ, PC and Cytochrome
(b) FeS, PQ, Cytochrome and PC
(c) PQ, Cytochrome, PC and FeS
(d) PC, Cytochrome, FeS, PQ
Answer:
(c) PQ, Cytochrome, PC and FeS

2. Answer the following questions

Question (A)
Describe the light-dependent steps of photosynthesis. How are they linked to dark reactions?
Answer:
The light dependent steps of photosynthesis include cyclic and non-cyclic photophosphorylation,
1. Cyclic photophosphorylation:
a. Illumination of photosystem-I causes electrons to move continuously out of the reaction center of photosystem-I and back to it.
b. The cyclic electron-flow is accompanied by the photophosphorylation of ADP to yield ATP. This is termed as Cyclic photophosphorylation.
c. Since this process involves only pigment system I, photolysis of water and consequent evolution of oxygen does not take place.

2. Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

3. Link between light-dependent and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C0₂ is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

(B)

Question (a)
Distinguish between Respiration and Photorespiration
Answer:

Question (b)
Distinguish between Cyclic photophosphorylation and Non-cyclic photophosphorylation
Answer:

Question (C)
Answer the following questions.
1. What are the steps that are common to C₃ and C₄ photosynthesis?
2. Differentiate between C₃ and C₄ plants.
Answer:
Steps that are common to C₃ and C₄ photosynthesis are Carboxylation, Reduction, Glucose synthesis, Regeneration.
[Note: Students are expected to refer the given Q.R code for detail understanding the common steps between C₁ and C₄ plants.]

Question (D)
Are the enzymes that catalyze the dark reactions of carbon fixation located inside the thylakoids or outside the thylakoids?
Answer:
Carbon fixation occurs in the stroma by series of enzyme catalyzed steps. The enzymes that catalyze the dark reactions of carbon fixation are located outside the thylakoids.

Question (E)
Calvin cycle consists of three phases, what are they? Explain the significance of each of them.
Answer:
The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.

The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.
Significance:
1. Carboxylation: RuBisCO is the most abundant enzyme in the world. It is responsible for fixing carbon in the form of C0₂ into sugar. As a result of Carboxylation, the first stable product of carbon fixation i.e. 3- PGA is synthesized.
2. Reduction/Glycolytic reversal: NADPH2 donates electrons to 1, 3-Bisphoshoglycerate to form 3- phosphoglyceraldehyde molecules. During this process ADP and NADP are generated which are used in light reaction.
3. Regeneration of RuBP: Some 3-phosphoglyceraldehyde molecules are involved in production of glucose while others are recycled to regenerate the 5-carbon compound RuBP which used to accept new carbon molecules. Thus, regeneration of RuBP is required for Calvin cycle to run continuously.

Question (F)
Why are plants that consume more than the usual 18 ATP to produce 1 molecule of glucose favoured in tropical regions?
Answer:

1. C₄ plants are favoured in tropical regions as they require 30 ATP to produce 1 molecule of glucose.
2. High temperature in tropical regions leads to closure of stomata to reduce rate of transpiration. Due to this availability of C0₂ decreases.
3. PEP carboxylase present in mesophyll cells can fix C0₂ even at low concentration. This helps the plant in efficient assimilation of atmospheric carbon dioxide.
4. C₄ plants contain a special leaf anatomy called Kranz anatomy which minimizes the losses due to photorespiration.
5. It helps C₄ plants to survive in conditions of high daytime temperatures, intense sunlight and low moisture.

Question (G)
What is the advantage of having more than one pigment molecule in a photocenter?
Answer:
Advantages of having more pigment molecules in a photocenter are as follows:
1. Having more than one pigment molecule in photocenter means more sunlight being captured and thus facilitating more effective light reaction.
2. It will provide protection to chlorophyll molecule against photo-oxidation.
3. More pigments will capture more energy to start the initial reactions, which is not possible by single pigment.

Question (H)
Why does chlorophyll appear green in reflected light and red transmitted light? Explain the significance of these phenomena in terms of photosynthesis.
Answer:
1. Chlorophyll is a light absorbing pigment. It absorbs light in red and blue regions of the visible light spectrum. Absorption spectrum of chlorophyll for red light is maximum so chlorophyll appears red in transmitted light. Green light is not absorbed but reflected so chlorophyll appear green in reflected light.
2. Chlorophyll predominantly absorbs red and violet-blue light and it allows plants to use this light as a form of energy for photosynthesis process.
3. It is most effective wavelength of light in photosynthesis as it has exactly right amount of energy to excite electrons of chlorophyll and boost them out of their orbits to higher energy level.

Question (I)
Explain why photosynthesis is considered the most important process in the biosphere.
Answer:
Photosynthesis is considered to be the most important process in the biosphere due to following reasons:
1. Photosynthesis is the biochemical process through which all plants (primary producers) produce food.
2. It is responsible for release of oxygen in the atmosphere.
3. Heterotrophs are directly or indirectly dependent on autotrophs for energy and other related resources. Therefore, photosynthesis is considered the most important process in the biosphere.

Question (J)
Why is photolysis of water accompanied with non-cyclic photophosphorylation?
Answer:
1. Photolysis of water provides new electrons to Photosystem – II.
2. The water molecule is lysed into three components:
a. Protons (H⁺) which are used as part of reactions that makes NADPH.
b. Second component formed is electrons which replaces the electrons lost by PS-II.
c. The third component is oxygen (0₂) which is released into the atmosphere.
3. Photosystem I sends electrons to reduce NADP+.
4. Then, Photosystem II sends replacement electrons to Photosystem I.
5. Finally, photolysis of water replaces the electrons lost by Photosystem II.
6. Water is the ultimate source of electrons for photosynthesis.
Therefore, photolysis of water is accompanied with non – cyclic photophosphorylation.

Question (K)
In C-4 plants, why is C-3 pathway operated in bundle sheath cells only?
Answer:

1. Decarboxylation of malic acid occurs in bundle sheath cells of C₄ plants. Due to which concentration of C0₂ increases in bundle sheath cells.
2. The enzymes required for Calvin cycle i.e. RuBisCO is present in bundle sheath cells.
3. In presence of high concentration of C0₂, RuBisCO acts as carboxylase and bring about carboxylation of RuBP.
4. Hence, in C-4 plants, C-3 pathway is operated in bundle sheath cells only.

Question (L)
What would have happened if C-4 plants did not have Kranz anatomy?
Answer:
Photorespiration would occur if C₄ plants did not have Kranz anatomy.

Question (M)
Why does RuBisCO carry out preferentially carboxylation than oxygenation in C₄ plants?
Answer:

1. In C₄ plants, C0₂ taken from the atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells.
2. This leads to the formation of 4-carbon compound oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
3. It is converted to another 4-carbon compound called malate.
4. Malate is transported to chloroplasts of bundle sheath cells where malate is converted to pyruvate and releases C0₄ in the cytoplasm thus increasing the concentration of C0₂ in the bundle sheath cells.
5. Chloroplasts of bundle sheath cells contains enzymes of Calvin cycle.
6. Thus, due to high concentration of C0₂, RuBisCO participates in carboxylation and not in oxygenation.

Question (N)
What would have happened if plants did not have accessory pigments?
Answer:

1. Accessory pigments are light absorbing molecules which are found in photosynthetic organisms.
2. They transfer the absorbed light to chlorophyll-a and thus increasing the photosynthetic rate.
3. In absence of accessory pigments less amount of light will be absorbed and also there would be no protection provided to chlorophyll molecule from photo-oxidation.

Question (O)
How can you identify whether the plant is C₃ or C₄? Explain / Justify.
Answer:

1. By observing the cross section of a leaf we can identify whether the plant is a C₃ plant or a C₃ plant.
2. C₄ plants possess a special anatomy of leaves called Kranz anatomy. In Kranz anatomy two types of chloroplasts are present, agranal in bundle sheath cells and granal in mesophyll cells.
3. In C₃ plants Kranz anatomy is absent.

Question (P)
In C₄ plants, bundle sheath cells carrying out Calvin cycle are very few in number. Then also, C₄ plants are highly productive. Explain.
Answer:

1. C₄ plants have special type of leaf anatomy called Kranz anatomy.
2. In C₄ plants, C0₂ fixation occurs twice.
3. In these plants, chloroplasts of mesophyll cells contain enzyme PEP carboxylase which fixes atmospheric C0₂.
4. Thus, first C0₂ fixation occurs in mesophyll cells.
5. Decarboxylation of malic acid in bundle sheath cells results in increase in C0₂ concentration.
6. Thus, RuBisCO acts as carboxylase and brings about carboxylation of RuBP.
7. Due to this oxygenation of RuBP and photorespiration is prevented.
8. Thus, despite of having less number of bundle sheath cells carrying out Calvin cycle, C₄ plants are highly productive.

Question (Q)
What is functional significance of Kranz anatomy?
Answer:

1. Leaves of C₄ plants show some structural peculiarities called Kranz anatomy.
2. The chloroplast of mesophyll cells contain enzyme PEP Carboxylase, which can fix C0₂ at low concentration.
3. Thus, light reaction and evolution of 0₂ occurs in mesophyll cells.
4. Decarboxylation of malate occurs in bundle sheath cells, which results in release of C0₂, due to which concentration of C0₂ in bundle sheath cells increases.
5. Enzyme RuBisCO present in bundle sheath cells acts as carboxylase in presence of high C0₂ concentraion and catalyses carboxylation of RuBP.
6. Thus, possibility of oxygenation of RuBP is avoided and photorespiration does not take place.

3. Correct the pathway and name it.

Question 1.
Correct the pathway and name it.
Answer:

1. The pathway shown is C₄ pathway.
2. M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
3. It occurs in tropical and sub-tropical grasses and some dicotyledons.
4. The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.

Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid.
3. It is transported to the chloroplasts of bundle sheath cells.
4. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
5. Thus, concentration of C0₂ increases in the bundle sheath cells.
6. Chloroplasts of these cells contain enzymes of Calvin cycle.
7. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
8. Sugar formed in Calvin cycle is transported into the phloem.
9. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
10. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP.
11. Thus, C₄ pathway needs 12 additional ATP.
12. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP.
13. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.

4. Is there something wrong in following schematic presentation? If yes, correct it so that photosynthesis will be operated.

Question 1.
Is there something wrong in following schematic presentation? If yes, correct it so that photosynthesis will be operated.
Answer:

Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Practical/ Project:

Question 1.
Draw schematic presentation of different processes/ cycles/ reactions related to photosynthesis.
Answer:
Cyclic photophosphorylation:
a. Illumination of photosystem-I causes electrons to move continuously out of the reaction center of photosystem-I and back to it.
b. The cyclic electron-flow is accompanied by the photophosphorylation of ADP to yield ATP. This is termed as Cyclic photophosphorylation.
c. Since this process involves only pigment system I, photolysis of water and consequent evolution of oxygen does not take place.

Non-cyclic photophosphorylation::
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Interdependence of light and dark reactions:

1. The light reaction gives rise to two important products, a reducing agent NADPH₂ and an energy rich compound ATP. Both these are utilized in the dark phase of photosynthesis.
2. ATP and NADPH₂ molecules function as vehicles for transfer of energy of sunlight into dark reaction leaving to carbon fixation. In this reaction C0₂ is reduced to carbohydrate.
3. During dark reaction, ATP and NADPH₂ are transformed into ADP, iP and NADP which are transferred to the grana in which light reaction takes place.

Calvin cycle: The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.
The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.

Photorespiration: Mechanism:
1. Photorespiration involves three organelles chloroplast, peroxisomes and mitochondria and occurs in a series of cyclic reactions which is also called PCO cycle. (Photosynthetic Carbon Cycle)
2. Enzyme Rubisco acts as oxygenase at higher concentration of O₂ and photorespiration begins.
3. When RuBP reacts with 0₂ rather than C0₂ to form a 3-carbon compound (PGA) and 2-carbon compound phosphoglycolate.
4. Phosphoglycolate is then converted to glycolate which is shuttled out of the chloroplast into the peroxisomes.
5. In Peroxisomes, glycolate is converted into glyoxylate by enzyme glycolate oxidase.
6. Glyoxylate is further converted into amino acid glycine by transamination.
7. In mitochondria, two molecules of glycine are converted into serine (amino acid) and C0₂ is given out.
8. Thus, it loses 25% of photosynthetically fixed carbon.
9. Serine is transported back to peroxisomes and converted into glycerate.
10. It is shuttled back to chloroplast to undergo phosphorylation and utilized in formation of 3-PGA, which get utilized in C₃ pathway.
Hatch-Slack pathway: M. D. Hatch and C. R. Slack while working on sugarcane found four carbon compounds (dicarboxylic acid) as the first stable product of photosynthesis.
It occurs in tropical and sub-tropical grasses and some dicotyledons.
The first product of this cycle is a 4-carbon compound oxaloacetic acid. Hence it is also called as C₄ pathway and plants are called C₄ plants.
Mechanism:

1. C0₂ taken from atmosphere is accepted by a 3-carbon compound, phosphoenolpyruvic acid in the chloroplasts of mesophyll cells, leading to the formation of 4-C compound, oxaloacetic acid with the help of enzyme phosphoenolpyruvate carboxylase.
2. It is converted to another 4-C compound, malic acid.
3. It is transported to the chloroplasts of bundle sheath cells.
4. Malic acid (4-C) is converted to pyruvic acid (3-C) with the release of C0₂ in the cytoplasm.
5. Thus, concentration of C0₂ increases in the bundle sheath cells.
6. Chloroplasts of these cells contain enzymes of Calvin cycle.
7. Because of high concentration of C0₂, RuBP carboxylase participates in Calvin cycle and not photorespiration.
8. Sugar formed in Calvin cycle is transported into the phloem.
9. Pyruvic acid generated in the bundle sheath cells re-enter mesophyll cells and regenerates
   phosphoenolpyruvic acid by consuming one ATP.
10. Since this conversion results in the formation of AMP (not ADP), two ATP are required to regenerate ATP from AMP.
11. xi. Thus, C₄ pathway needs 12 additional ATP.
12. The C₃ pathway requires 18 ATP for the synthesis of one glucose molecule, whereas C4 pathway requires 30 ATP. Thus, C₄ plants are better photosynthesizers as compared to C₃ plants as there is no photorespiration in these plants.
13. CAM Pathway: In CAM plants, malic acid accumulates during night, which is formed from Oxaloacetic acid in presence of the enzyme malate dehydrogenase.

Question 2.
Check the effects of different factors on photosynthesis under the guidance of teacher.
Answer:
External factors which affect photosynthesis are as follows:
1. Light:
a. It is an essential factor as it supplies the energy necessary for photosynthesis.
b. Quality and intensity of light affects the photosynthesis.
c. Highest rate of photosynthesis takes place in red light followed by blue light.
d. The rate of photosynthesis considerably decreases in plants which are growing under a forest canopy.
e. In most of the plants, photosynthesis is maximum in bright diffused sunlight.
f. Uninterrupted and continuous photosynthesis for a very long period of time may be sustained without any visible damage to the plant.

2. Carbon dioxide:
The main source of C0₂ in land plants is the atmosphere, which contains only 0.3% of the gas.
b. Under normal conditions of temperature and light, carbon dioxide acts as a limiting factor in photosynthesis.
c. Increase in concentration of CO₂ increases the photosynthesis.
d. Increase in C0₂ to about 1% is advantageous to most of the plants.
e. Higher concentration of the gas has an inhibitory effect on photosynthesis.

3. Temperature:
a. Like all other physiological processes, photosynthesis also needs a suitable temperature.
b. The optimum temperature at which the photosynthesis is maximum is 25-30 °C. Except in plants like Opuntia, photosynthesis takes place at as high as 55 °C.
c. This is the maximum temperature. Minimum temperature is temperature at which photosynthesis process just starts.
d. In the presence of sufficient light and CO₂, photosynthesis increases with the rise of temperature till it becomes maximum. After that there is a decrease or fall in the rate of the process.

4. Water:
a. Water is necessary for photosynthetic process.
b. An increase in water content of the leaf results in the corresponding increase in the rate of photosynthesis.
c. Thus, the limiting effect of water is not direct but indirect.
d. It is mainly due to the fact that it helps in maintaining the turgidity of the assimilatory cells and the proper hydration of their protoplasm.
[Students can refer the given information and perform this activity on their own]

11th Biology Digest Chapter 12 Photosynthesis Intext Questions and Answers

Can you recall? (Textbook Page No. 138)

(i) Why energy is essential in different life processes?
Answer:
a. Energy is the basic requirement of life.
b. Without energy no work can be done.
c. All living organisms need energy for reproduction and survival.
d. Sun is the main source of energy, and that energy should be transformed into the usable forms for living organisms to carry out life processes.
Therefore, energy is essential in different life processes.

(ii) How do we get energy?
Answer:
a. Sun is the main source of energy.
b. Plants utilize sunlight, carbon dioxide and water for the process called photosynthesis to produce sugars.
c. Animals make use of these sugars provided by the plants in their own cellular energy factories called mitochondria. Thus, energy is obtained.

Use your brainpower (Textbook Page No. 138)

Justify: All life on earth is ‘bottled solar energy’.
Answer:

1. Life on earth is dependent on solar energy directly or indirectly.
2. Plants by carrying out photosynthesis converts solar energy into chemical energy by producing carbohydrates.
3. Humans and animals depend on plants for food. Basically, life on earth depends totally on photosynthesis for food and energy.
4. Therefore, all life on earth is bottled solar energy.

Can you tell? (Textbook Page No. 140)

Draw well labeled diagram of ultrastructure of chloroplast.
Answer:

1. The chloroplasts are discoid and lens shaped in higher plants. Chloroplast is bounded by a double membrane.
   System of chlorophyll bearing a double-membrane sac is present inside the stroma.
2. These are stacked one above the other to form grana.
3. Individual sacs in each granum is are known as thylakoid.
4. All the pigments chlorophylls, carotenes and xanthophylls are located in thylakoid membranes.
5. These pigments are fat soluble and are present in lipid part of membrane also they absorb light of specific spectrum in the visible regions.

Use your brainpower (Textbook Page No. 140)

The photosynthetic lamellae taken out from a chloroplast and suspended in a nutrient medium in the presence of C0₂ and light. Will they synthesize sugar or not?
Answer:
Photosynthetic ladmellae will not synthesize sugar because sugar synthesis occurs only in stroma, as all the enzymes required for sugar synthesis are present there. In photosynthetic lamellae only light reactions occur. Thus, lamellae cannot synthesize sugar even when C0₂, light and other nutrients are provided.

Internet my friend (Textbook Page No. 139)

Collect information: Why does chlorophyll appear red in reflected light and green in transmitted light?
Answer:
Chlorophyll is a light absorbing pigment. It absorbs light in red and blue regions of the visible light spectrum. Absorption spectrum of chlorophyll for red light is maximum so chlorophyll appears red in transmitted light. Green light is not absorbed but reflected so chlorophyll appear green in reflected light. [Note: Chlorophyll appear red in transmitted light and green in reflected light.[

Activity 1 (Textbook Page No. 139)

Grind the spinach leaves in small quantity of acetone / nail paint remover. Mix the contents properly and filter with filter paper in test tube. Test tube contains green filtrate. Take the test tube in dark-room and put a flash of torch on it. Now, solution appears red. Why does this occur? Which phenomenon is this? Discuss this with your physics, chemistry and biology teachers.
Answer:
Chlorophyll is the green pigment present in chloroplast. It absorbs light in red and blue region of visible spectrum. It does not absorb green light and thus the green light is reflected which is why it appears green. In this experiment, the chlorophyll in test tube appears red when a flash torch is put on it in the dark room.

This is because when the electrons of the chlorophyll molecule are excited in dark in the absence of electron transport chain the electrons release their energy in the form of red light as they return to ground state. This phenomenon observed here is transmission of light.

Activity 2 (Textbook Page No. 139)

To separate the chloroplast pigments by paper chromatography. Concentrate the extracted chlorophyll solution by evaporation. Apply a drop of it at one end, 2cm away from edge of a strip of chromatography paper and allow it to dry thoroughly. Take a mixture of petroleum ether and acetone in the ratio of 9:1 at temperature of 40 to 60°C. Hang the strip in the jar with its loaded end dipping in the solvent. Close the jar tightly and keep it for an hour. The pigments separate into distinct green and yellow bands of chlorophyll and carotenoid respectively.
Answer:
Pigments are the molecules which reflects only certain wavelengths of visible light. Chromatography is the technique used to separate the chloroplast pigments. Carotenes form yellow-orange band, chlorophyll forms blue-green band, chlorophyll b forms yellow-green bands.

Can you tell? (Textbook Page No. 139)

Tomatoes, carrots and chillies are red in colour due to the presence of pigments. Name the pigment. Answer:
Red colour pigment present in tomatoes, carrots and chillies is lycopene.

Think about it (Textbook Page No. 140)

Large number of gas bubbles are evolved during day time in a pond of water.
Answer:
Photosynthesis occurs in the presence of sunlight. During photosynthesis, plants give out oxygen and take in carbon dioxide. The plants present underwater carryout photosynthesis and release oxygen. Hence, large number of gas bubbles are evolved during day time in a pond.

Think about it (Textbook Page No. 141)

Does moonlight support photosynthesis?
Answer:
The reactions of photosynthesis take place in the presence of sunlight. The intensity of moonlight is several thousand times less than that of direct sunlight which is insufficient for the light dependent phase of photosynthesis. As the sun sets, rate of photosynthesis also decreases. Therefore, moonlight does not support photosynthesis.

Can you tell? (Textbook Page No. 145)

How chlorophyll – a is excited? Show it with a diagram.
Answer:

1. Chlorophyll-a is an essential photosynthetic pigment as it converts light energy into chemical energy and acts as a reaction centre.
2. Initially, it lies at ground state or singlet state but when it absorbs or receives photons (solar energy), it gets activated and goes in excited state or excited second singlet state.
3. In the excited state, chlorophyll-a emits an electron. The emitted electron is energy rich, i.e. has extra amount of energy.
4. Due to the loss of electron (e_–), chlorophyll-a becomes positively charged. This is the ionized state.
5. Chlorophyll-a molecule cannot remain in the ionized state for more than 10‘9 seconds. Hence the photo-chemical reaction or electron transfer occurs very fast.
6. The energy rich electron is then transferred through various electron acceptors and donors (carriers).
7. During the transfer, the electron emits energy which is utilized for the synthesis of ATP. This shows that light energy is converted into chemical energy in the form of ATP.

Can you tell? (Textbook Page No. 140)

What made Hill to perform his experiment?
Answer:
Robert Hill proved that the source of oxygen evolved during photosynthesis is water and not carbon dioxide. Hence, it is called Hill’s Reaction.

1. In this experiment, Hill cultured isolated chloroplasts in a medium containing C0₂ free water, haemoglobin and ferric compound.
2. Ferric salts and haemoglobin were added in the medium as hydrogen and oxygen acceptors respectively.
3. When the suspension was illuminated, he observed that haemoglobin turned into oxyhaemoglobin (red colour).
4. This confirmed that water must have oxidized releasing 0₂, that reacted with haemoglobin. Reduction of ferric compound was also indicated by change in colour.
5. The H₂O molecule oxidized to evolve 0₂ as a by-product. Thus, Hill proved that the source of evolving 0₂ is H₂0 and not C0₂.
6. This process of splitting up of water molecules under the influence of light in the presence of chlorophyll is called Photolysis of water or Hill Reaction.
7. Hill’s reaction can be represented as follows:

Can you tell? (Textbook Page No. 145)

Draw a flowchart of non-cyclic photophosphorylation.
Answer:
Non-cyclic photophosphorylation:
a. It involves both photosystems- PS-I and PS-II.
b. In this case, electron transport chain starts with the release of electrons from PS-II.
c. In this chain high energy electrons released from PS-II do not return to PS-II but, after passing through an electron transport chain, reach PS-I, which in turn donates it to reduce NADP to NADPH.
d. The reduced NADP⁺ (NADPH) is utilized for the reduction of CO₂ in the dark reaction.
e. Electron-deficient PS-II brings about oxidation of water-molecule. Due to this, protons, electrons and oxygen atom are released.
f. Electrons are taken up by PS-II itself to return to reduced state, protons are accepted by NADP⁺ whereas oxygen is released.
g. As in this process, high energy electrons released from PS-II do not return to PS-II and it is accompanied with ATP formation, this is called Non-cyclic photophosphorylation.

Can you tell? (Textbook Page No. 145)

Describe Calvin’s cycle.
Answer:
The entire process of dark reaction was traced by Dr. Melvin Calvin along with his co-worker, Dr. Benson. Hence, the process is called as Calvin cycle or Calvin- Benson cycle. Since the first stable product formed is a 3-carbon compound, it is also called as C₃ pathway and the plants are called C₁₄ plants.
Calvin carried out experiments on unicellular green algae (Chlorella), using radioactive isotope of carbon, C₁₄ as a tracer.
It is also called synthesis phase or second phase of photosynthesis.
The cycle is divided into the following phases:
1. Carboxylation phase:
a. Carbon dioxide reduction starts with a five-carbon sugar ribulose-l,5-bisphosphate (RuBP). It is a 5- carbon sugar with two phosphate groups attached to it.
b. RuBP reacts with CO₂ to produce an unstable 6 carbon intermediate in the presence of Rubisco.
c. It immediately splits into 3 carbon compounds called 3-phosphoglyceric acid.
d. RuBisCO is a large protein molecule and comprises 16% of the chloroplast proteins.

2. Glycolytic reversal:
a. 3-phosphoglyceric acid form 1,3-diphosphoglyceric acid by utilizing ATP molecule.
b. These are then reduced to glyceraldehyde-3-phosphate (3-PGA) by NADPH supplied by the light reactions of photosynthesis.
c. In order to keep Calvin cycle continuously running there must be sufficient number of RuBP and regular supply of ATP and NADPH.
d. Out of 12 molecules of 3-phosphoglyceraldehyde, two molecules are used for synthesis of one glucose molecule.

3. Regeneration of RuBP:
a. 10 molecules of 3-phosphoglyceraldehyde are used for the regeneration of 6 molecules of RuBP at the cost of 6 ATP.
b. Therefore, six turns of Calvin cycle are needed to get one molecule of glucose.

Can you tell? (Textbook Page No. 147)

Summarise the photosynthetic reaction.
Answer:
6C0₂ + 12H₂0 → C₂H₁₂0₆ + 60₂ + 6H₂0
1. Photosynthesis is a two step process.
The light dependent reactions convert the light energy from the sun into chemical energy.
The light independent reactions convert the chemical energy to synthesize carbohydrates.
2. Light dependent reactions: Light is absorbed by chlorophyll which results in the production of ATP. Photolysis of water produce oxygen and hydrogen. The hydrogen and ATP are used in the light independent reactions and the oxygen is released from stomata.
3. Light independent reactions: ATP and hydrogen are transferred to the site of light independent reactions. The hydrogen is combined with carbon dioxide to form complex organic compounds.
The ATP provides the required energy to power these anabolic reactions and fix the carbon molecules.

Can you tell? (Textbook Page No. 147)

Summarise the photosynthetic reaction.
Answer:
6C0₂ + 12H₂0 → C₂H₁₂0₆ + 60₂ + 6H₂0
1. Photosynthesis is a two step process.
The light dependent reactions convert the light energy from the sun into chemical energy.
The light independent reactions convert the chemical energy to synthesize carbohydrates.
2. Light dependent reactions: Light is absorbed by chlorophyll which results in the production of ATP. Photolysis of water produce oxygen and hydrogen. The hydrogen and ATP are used in the light independent reactions and the oxygen is released from stomata.
3. Light independent reactions: ATP and hydrogen are transferred to the site of light independent reactions. The hydrogen is combined with carbon dioxide to form complex organic compounds.
The ATP provides the required energy to power these anabolic reactions and fix the carbon molecules.

Can you tell? (Textbook Page No. 147)

C₄ plants are more productive. Why?
Answer:

1. Photorespiration is considered to be a wasteful process in plants. It is an energy consuming process in plants which ultimately leads to reduction in final yield of plants.
2. During C₃ photosynthesis, 25% of the carbon dioxide fixed has to pass through photorespiratory process.
3. This decreases the photosynthetic productivity.
4. In C₄ plants, photorespiration is absent and hence they have better productivity.

Can you tell? (Textbook Page No. 147)

Xerophytic plants survive in high temperature. How?
Answer:

1. Xerophytic plants are those that have adapted to dry environments.
2. They have adapted to arid conditions by storing water in stems.
3. Stomata of these plants remain closed during day time to reduce the rate of transpiration to bare minimum.
4. Leaves are modified into spines or are reduced in size to check the loss of water due to transpiration.
5. The waxy surfaces of xerophytic plants prevent the loss of moisture.
6. Thus, they are able to survive in high temperature.

Can you tell? (Textbook Page No. 147)

Compare C₄ plants and CAM plants.
Answer:

Maharashtra State Board Class 11 Biology Textbook Solutions

• Plant Tissues and Anatomy Class 11 Biology Textbook Solutions
• Morphology of Flowering Plants Class 11 Biology Textbook Solutions
• Animal Tissue Class 11 Biology Textbook Solutions
• Study of Animal Type : Cockroach Class 11 Biology Textbook Solutions
• Photosynthesis Class 11 Biology Textbook Solutions
• Respiration and Energy Transfer Class 11 Biology Textbook Solutions
• Human Nutrition Class 11 Biology Textbook Solutions
• Excretion and Osmoregulation Class 11 Biology Textbook Solutions
• Skeleton and Movement Class 11 Biology Textbook Solutions

Study of Animal Type Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 11 Study of Animal Type: Cockroach Textbook Exercise Questions and Answers.

Class 11 Biology Chapter 11 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 11 Exercise Solutions

Question (A)
Chemical nature of chitin is ____________ .
(A) protein
(B) carbohydrate
(C) lipid
(D) glycoprotein
Answer:
(B) carbohydrate & (D) glycoprotein

Question (B)
Cockroach has ___________ type of mouthparts.
(A) sponging
(B) chewing and biting
(C) piercing and sucking
(D) lapping
Answer:
(B) chewing and biting

Question (C)
Spiracle is a part of ________ system of cockroach.
(A) circulatory
(B) respiration
(C) reproductive
(D) nervous
Answer:
(B) respiration

Question (D)
________ is a part of digestive system.
(A) Trachea
(B) Hypopharynx
(C) Haemocyte
(D) Seminal vesicle
Answer:
(B) Hypopharynx

Question (E)
_________ is also called as brain of cockroach.
(A) Supra-oesophageal ganglion
(B) Sub-oesophageal ganglion
(C) Hypo-cerebral ganglion
(D) Thoracic ganglion
Answer:
(A) Supra-oesophageal ganglion

2. Answer the following questions

Question (A)
Describe the digestive system of cockroach.
OR
With the help of neat and labelled diagram, describe the digestive system of cockroach.
Answer:
1. Digestive system of cockroach consists of mouthparts, alimentary canal and salivary glands.
2. Mouthparts: Pre-oral cavity present in front of the mouth receives food. It is bounded by chewing and biting type of mouth parts.
These are movable, segmented appendages that help in ingestion of food. The mouthparts of cockroach comprises of:
(a) Labrum: It forms the upper lip. It is a single flap-like movable part which covers the mouth from upper side. It forms an anterior wall of pre¬oral cavity.
Function: It is useful in holding the food during feeding.

(b) Mandibles: These are two dark, hard, chitinous structures with serrated median margins.They are true jaws present on either side, behind the labrum.
Function: They perform co-ordinated side-wise movements with the help of adductor and abductor muscles to cut and crush the food.

(c) Maxillae: These are the accesssory jaws. They are also called as first pair of maxillae. These are situated on the either side of mouth behind the mandibles. Each maxilla consists of sclerites like cardo, stipes, galea, lacinia and maxillary palps.
Functions: Maxillae hold food, help mandibles for mastication. They are also used for cleaning the antennae and front legs. Maxillary palps act as tactile organs.

(d) Labium: It forms the lower lip. Labium is also known as second maxilla which covers the pre-oral cavity from the ventral side. It is firmly attached to the posterior part of head. It has three jointed labial palps which are sensory in function.
Function: It is useful in pushing the chewed food in the pre-oral cavity. It prevents the loss of food falling from the mandibles, while chewing.

(e) Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food.

3. Alimentary canal: It is long a (6 – 7cm) tube of different diameters with two openings.

4. The alimentary canal is divisible into three parts: foregut, midgut and hindgut
(a) Foregut or stomodaeum: It consists of pharynx, oesophagus, crop and gizzard.
1. Pharynx: It is very short, narrow but muscular tube that opens into oesophagus.
Function: Conduction of food into the oesophagus.
2. Oesophagus: It is slightly long and narrow tube which opens into crop.
3. Crop: Crop is a large, pear shaped and sac- like organ.
Function: It temporarily stores the food and then sends it to gizzard.
4. Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

(b) Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

(c) Hindgut or proctodaeum : It consists of ileum, colon and rectum.
1 Ileum: It is short and narrow part of hindgut. Malpighian tubules open in the anterior lumen of ileum, near the junction of midgut and hindgut. Posterior region of ileum contains sphincter.
Ileum directs the nitrogenous wastes and undigested food towards colon.
2. Colon: It is a longer and wider part of the hindgut. It directs waste material towards the rectum. It reabsorbs water from wastes as per the need.
3. Rectum: It is oval or spindle-shaped, terminal part of the hindgut. It contains six rectal pads along the internal surface for absorption of water. Rectum opens into anus. Anus is present on the ventral side of the 10th segment. It is the last or posterior opening of the digestive system. The undigested food is released out of the body through anus.

5. Salivary glands:
a. Cockroach has a pair of salivary glands which secrete saliva.
b. Each salivary gland has two glandular lobes and a receptacle or reservoir.
c. The glandular lobes consists of several irregular-shaped white coloured lobules which secrete saliva.
d. Each gland has a salivary duct.
Both the ducts unite to form a common salivary duct.
e. Receptacle of each salivary gland is thin-walled, elongated, sac-like structure. Each receptacle has a duct. These ducts unite to form common reservoir duct.
f. Common salivary duct and common reservoir duct unite together to form a common efferent salivary duct. The efferent salivary duct opens at the base of tongue or hypopharynx.

Question (B)
Give an account on tracheal system of cockroach.
Answer:
1. Cockroach has an internal respiratory system of air tubes called tracheal system by which the air is brought into the body and is in contact with every part of the body. It allows the exchange of gases directly between the air and tissues without the need of blood.
These air tubes of internal respiratory system begin at the opening on body surface called spiracles.

2. Spiracles: They are paired respiratory openings. Spiracles are present on the ventro-lateral side of the body, in pleural membrane. Cockroaches have two pairs of thoracic and eight pairs of abdominal spiracles.The spiracles open into a series of air sacs from which arise the tubes called trachea. The spiracles let the air into and out of trachea.

3. Trachea: The trachea form a definite pattern of branching tubes arranged transversely as well as longitudinally. They are about 1mm thick and have spiral or annular thickening of chitin. The inner lining of chitin prevents the trachea from collapsing. Each trachea further branches into smaller tubes called tracheoles.

4. Tracheoles: These are fine intracellular tubes that penetrate deep into tissues. They are thin and not lined by chitin. They end blindly in the cells. Each tracheole at the blind end is filled with a watery fluid through which exchange of gases takes place. The content of this fluid keeps changing. At high muscular activity, part of fluid part is drawn into the tissues to enable more and rapid oxygen intake.

Question (C)
Describe the nervous system of cockroach.
Answer:
Nervous system in cockroach:
Nervous system of cockroach is ventral, solid and ganglionated. It consists of central nervous system (CNS), peripheral nervous system (PNS) and autonomous nervous system (ANS).
Central nervous system (CNS): Central nerv ous system consists of nerve ring and ventral nerve cord.
1. Nerve ring consists of:
a. a pair of supra-oesophageal ganglia
b. a pair of circum-oesophageal connectives
c. a pair of sub-oesophageal ganglia
(a) Supra-oesophageal ganglia or cerebral ganglia: A pair of supra-oesophageal ganglia is collectively known as the brain. Brain is present in head, above the oesophagus and between antennal bases. Each supra-oesophageal ganglion is formed by the fusion of three small ganglia – protocerebram, deutocerebrum and tritocerebrum.
(b) Circum-oesophageal connectives: Supra-oesophageal ganglia are connected to sub-oesophageal ganglion by a pair of lateral nerves called as circum-oesophageal connectives. Connectives arise from supra-oesophagial ganglia.
(c) Sub-oesophageal ganglia: It is a bilobed and present below the oesophagus, in head. It is also formed by the fusion of three pairs of ganglia.

2. Ventral nerve cord:
a. It arises from the sub-oesophageal ganglion. It is present along mid-ventral position, in perineural sinus.
b. It is double ventral nerve cord and consists of nine segmental, paired ganglia.
c. First three pairs of segmental ganglia are large and known as thoracic ganglia. The other six pairs of segmental ganglia are in abdomen (abdominal ganglia).
d. 6th abdominal ganglion is the largest and it is present in 7th abdominal segment.
e. There is no ganglion in 6th segment.

Peripheral nervous system (PNS):

1. The peripheral nervous system comprises of nerves that arise from various ganglia of CNS.
2. Six pairs of nerves arise from the supra-oesophageal ganglia.They supply to the eyes, antenna and labrum.
3. Nerves arising from the sub-oesophageal ganglion supply to the mandibles, maxillae and labium.
4. Nerves arising from the thoracic ganglia supply to the wings, legs and internal thoracic organs.
5. Nerves from abdominal ganglia go to the abdominal organs of respective abdominal segments.
6. Autonomic nervous system (ANS): It consists of four ganglia and a retrocerebral complex.

The ganglia are as follows:

1. Frontal ganglion: It is present above the pharynx and in front of brain.
2. Hypocerebral ganglion: It is present on the anterior region of oesophagus.
3. Ingluvial ganglion: It is present on crop. It is also called as visceral ganglion.
4. Ventricular ganglion: It is present on gizzard.

Question (D)
With the help of neat labelled diagram, describe female reproductive system of cockroach.
Answer:

1. Female reproductive system of cockroach consists a pair of ovaries, a pair of oviducts, vagina, spermatheca and accessory glands.
2. Ovaries are primary reproductive organs. They are paired and lie lateral in position in 2nd – 6lh abdominal segments. Each ovary is formed of a group of 8 ovarian tubules or ovarioles, containing a chain of developing ova. All ovarioles of an ovary open in lateral oviduct of respective side.
3. The lateral oviducts unite to form a common oviduct or vagina.
   Common oviduct or vagina opens into the Bursa copulatrix (genital chamber), the female organ of copulation.
4. Spermatheca, is a sperm storing structure present in the 6th segment opens into genital chamber. It receives the sperms during copulation and store them for fertilization.
5. Collaterial glands are accessory paired glands that open in genital chamber.
6. Female gonapophyses consists of six chitinous plates surrounding the genital pore.
7. In males, genital pouch or genital chamber lies at the hind end of abdomen which is bounded dorsally by 9th and 10th terga and ventrally b; male genital pore and gonapophysis.

Question (E)
Draw a labelled diagram of digestive system of a cockroach.
Answer:
1. Digestive system of cockroach consists of mouthparts, alimentary canal and salivary glands.
2. Mouthparts: Pre-oral cavity present in front of the mouth receives food. It is bounded by chewing and biting type of mouth parts.
These are movable, segmented appendages that help in ingestion of food. The mouthparts of cockroach comprises of:
(a) Labrum: It forms the upper lip. It is a single flap-like movable part which covers the mouth from upper side. It forms an anterior wall of pre¬oral cavity.
Function: It is useful in holding the food during feeding.
(b) Mandibles: These are two dark, hard, chitinous structures with serrated median margins.They are true jaws present on either side, behind the labrum.
Function: They perform co-ordinated side-wise movements with the help of adductor and abductor muscles to cut and crush the food.
(c) Maxillae: These are the accesssory jaws. They are also called as first pair of maxillae. These are situated on the either side of mouth behind the mandibles. Each maxilla consists of sclerites like cardo, stipes, galea, lacinia and maxillary palps.
Functions: Maxillae hold food, help mandibles for mastication. They are also used for cleaning the antennae and front legs. Maxillary palps act as tactile organs.
(d) Labium: It forms the lower lip. Labium is also known as second maxilla which covers the pre-oral cavity from the ventral side. It is firmly attached to the posterior part of head. It has three jointed labial palps which are sensory in function.
Function: It is useful in pushing the chewed food in the pre-oral cavity. It prevents the loss of food falling from the mandibles, while chewing.
(e) Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food,

3. Alimentary canal: It is long a (6 – 7cm) tube of different diameters with two openings.

4. The alimentary canal is into three parts: foregut, midgut and hindgut
(a) Foregut or stomodaeum: It consists of pharynx, oesophagus, crop and gizzard.
1. Pharynx: It is very short, narrow but muscular tube that opens into oesophagus.
Function: Conduction of food into the oesophagus.
2. Oesophagus: It is slightly long and narrow tube which opens into crop.
3. Crop: Crop is a large, pear shaped and sac- like organ.
Function: It temporarily stores the food and then sends it to gizzard.
4. Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

(b) Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

(c) Hindgut or proctodaeum : It consists of ileum, colon and rectum.
1 Ileum: It is short and narrow part of hindgut. Malpighian tubules open in the anterior lumen of ileum, near the junction of midgut and hindgut. Posterior region of ileum contains sphincter.
Ileum directs the nitrogenous wastes and undigested food towards colon.
2. Colon: It is a longer and wider part of the hindgut. It directs waste material towards the rectum. It reabsorbs water from wastes as per the need.
3. Rectum: It is oval or spindle-shaped, terminal part of the hindgut. It contains six rectal pads along the internal surface for absorption of water. Rectum opens into anus. Anus is present on the ventral side of the 10th segment. It is the last or posterior opening of the digestive system. The undigested food is released out of the body through anus.

5. Salivary glands:
a. Cockroach has a pair of salivary glands which secrete saliva.
b. Each salivary gland has two glandular lobes and a receptacle or reservoir.
c. The glandular lobes consists of several irregular-shaped white coloured lobules which secrete saliva.
d. Each gland has a salivary duct.
Both the ducts unite to form a common salivary duct.
e. Receptacle of each salivary gland is thin-walled, elongated, sac-like structure. Each receptacle has a duct. These ducts unite to form common reservoir duct.
f. Common salivary duct and common reservoir duct unite together to form a common efferent salivary duct. The efferent salivary duct opens at the base of tongue or hypopharynx.

Question (F)
A student observed that the cockroaches are killed for dissection by simply putting them in soap water. He inquired whether soap is so poisonous. Teacher said it is due to its peculiar respiratory system. How?
Answer:
Cockroaches when put in soap solution, the solution enters into their body through the small respiratory openings called spiracles. The spiracles lead to trachea which further branches into smaller tubes called tracheoles. Each of these tracheoles has body fluid which acts as a stationary medium for diffusion. The soap solution rapidly diffuses through the entire respiratory system which may result in suffocation and eventually lead to the death of cockroach.

Question (G)
Describe the circulatory system of cockroach.
Answer:

1. Haemolymph: Haemolymph is colourless as it is without any pigment. It consists of plasma and seven types of blood cells/haemocytes. Plasma consists of water with some dissolved organic and inorganic solutes. It is rich in nutrients and nitrogenous wastes like uric acid.Cockroach has open circulatory system. It consists of colourless blood (haemolymph), a dorsal blood vessel (heart and dorsal aorta) and haemocoel.

2. Haemocoel: The body cavity of cockroach (haemocoel) can be divided into three sinuses due to two diaphragms i.e. dorsal and ventral diaphragm. These diaphragms are thin, fibromuscular septa (sing.septum)
which remain attached to terga along lateral sides at intermittent points.
(a) Dorsal diaphragm: It has 12 pairs (10 abdominal and 2 thoracic) of fan-like alary muscles. Alary muscles are triangular with pointed end attached to terga at lateral side and broad end lies between the heart and dorsal diaphragm.
(b) Ventral diaphragm: It is flat and present just above the ventral nerve cord. Laterally, it is attached to sterna at intermittent points.
(e) Sinuses: The coelom of cockroach is divided into three sinuses – pericardial sinus, perivisceral sinus and perineural sinus.

1. Pericardial sinus: It is dorsal, very small and contains dorsal vessel.
2. Perivisceral sinus: It is middle and largest sinus. It contains fat bodies and almost all major visceral organs of alimentary canal and reproductive system.
3. Perineural sinus: It is ventral, small and contains ventral nerve cord. It is continuous into legs. All the three sinuses communicate with each other through the pores present between two successive points of attachments of diaphragms.
4. Dorsal blood vessel: This is present in pericardial sinus, just below the tergum. It is divisible into posterior heart and anterior aorta (dorsal aorta/cephalic vessel).
(a) Heart: It is about 2.5 cm long, narrow, muscular tube that is open anteriorly and closed posteriorly. It starts from 9th abdominal segment and extends anteriorly upto 1st thoracic segment. Heart of cockroach is 13 chambered, out of which 10 chambers are in abdominal region and 3 chambers are in thoracic region. Each chamber has a pair of vertical slit-like incurrent aperture or opening called ostium (plural: ostia). Ostia are present along lateral side in the posterior region of first 12 chambers. Each ostium has lip-like valves that allow the flow of blood from sinus to heart only.
(b) Anterior aorta: Heart is continued by a short, thin-walled vessel called dorsal aorta. It lies in head region and opens in haemocoel.

3. Answer the following questions

Question (A)
How will you identify male or female cockroach?
Answer:
Male and female cockroach can be identified with the help of following differences:

Question (B)
Write a note on: Gizzard of cockroach.
Answer:
Gizzard: Gizzard or proventricuius is a small spherical organ. It is provided internally with a circlet of six chitinous teeth and backwardly directed bristles.The foregut ends with gizzard.
Function: The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.

Question (C)
Give the systematic position of cockroach.
Answer:
Systematic position of cockroach:

Question (D)
What would have happened if cockroach did not have gizzard?
Answer:
1. The gizzard in cockroach is a spherical organ which has chitinous teeth and bristles.
2. The chitinous teeth present in gizzard are responsible for crushing the food and the bristles help to filter the food.
3. If the cockroach did not have gizzard, the food will not be crushed into small particles and unfiltered food will enter the hindgut. Thus, digestion will be affected in the absence of gizzard.

Question (E)
What is the functional difference between eyes of cockroach and human being?
Answer:
1. Cockroaches have compound eyes whereas humans have simple eyes.
2. Eyes of cockroach possess several ommatidia that collectively form an image and help them to detect even the slightest movement of its predator. They provide mosaic or hazy vision.
3. Human eyes contain single lens and a clear image is formed on the retina. Humans have binocular vision which provides an improved perception of depth and gives a three-dimensional image of their surroundings.

Question (F)
What is the functional difference between respiratory systems of cockroach and human being?
Answer:
The functional difference between the respiratory systems of cockroach and human being is that in respiratory system of cockroach transport of gases does not occur via. blood whereas in human respiratory system transport of gases takes place via blood. In cockroach, the circulatory system has no role in respiratory process whereas in humans, circulatory system plays an important in respiratory process.

4. Explain the following in short.

Question (A)
What are anal cerci?
Answer:
1. Anal cerci are a pair of appendages at the end of the abdomen that arise from the 10th segment of the body of both male and female cockroach.
2. They are sensitive to wind movements and detect vibrations.

Question (B)
What is ganglion?
Answer:
1. Ganglion is a group of nerve cell bodies.
2. It represents the brain in advanced invertebrates.

Question (C)
Write a short note on hypopharynx.
Answer:
Hypopharynx: Hypopharynx is also known as lingua. It is a somewhat cylindrical single structure, located in front of the labium and between first maxillae. The salivary duct opens at the base of hypopharynx. Hypopharynx bears comb-like plates called super-lingua on either side. Hypopharynx is present at the centre of the mouth.
Function: It is useful in the process of feeding and mixing saliva with food.

Question (D)
What is mesentron?
Answer:
Midgut or mesenteron: It consists of stomach and hepatic caeca.
1. Ventriculus or stomach: It is straight, short and narrow. Stomach is lined by glandular epithelium which secretes digestive enzymes.
Function: It is mainly responsible for digestion and absorption.
2. Hepatic caeca: These are thin, transparent, short, blind (closed) and hollow tubules.
Function: They secrete digestive enzymes.

Question (E)
Location of tergum.
Answer:
1. Tergum is a chitinous plate located in the abdomen of cockroach.
2. The abdomen is elongated and made up of ten segments. Each segment has a dorsal tergum and ventral sternum. Tergum is jointed to the sternum laterally by a soft cuticle called pleura.

Question (F)
What is ootheca?
Answer:
1. The secretion of collaterial glands forms a capsule around them is called as ootheca or egg case.
2. It is about 8 mm long and ranges from dark reddish to blackish brown.
3. Ootheca contains 14 to 16 fertilized eggs in two rows.
4. They are dropped or glued to a suitable surface, like a crack or crevice with good humidity near a food source.
5. A female cockroach on an average, produces 9 to 10 oothecae during its lifespan.

Question (G)
How many chambers are present in heart of a cockroach?
Answer:
13 chambers are present in heart of a cockroach, out of which 10 chambers are in abdominal region and 3 are in thoracic region.

Practical/Project:

Question 1.
Visit to nearest sericulture farm and study the life cycle of silk worm.
Answer:

1. The life cycle of the silk moth consists of four stages namely, egg, larva, pupa and adult.
2. Thousands of eggs deposited by female moths are incubated artificially to reduce the incubation period.
3. Larvae hatching out of eggs are released on mulberry plants to obtain nourishment from mulberry leaves.
4. After feeding for 3 – 4 weeks, larvae move to branches of mulberry plant.
5. The silk thread is formed from the secretion of salivary glands of larvae.
6. Larvae spin this thread around themselves forming a cocoon, which may be spherical in shape.
7. Ten days before the pupa turns into an adult, all the cocoons are transferred into boiling water.
8. Due to the boiling water, the pupa dies in the cocoon and silk fibres become loose.
9. These fibres are then unwound, processed and reeled.
10. Different kinds of fabric are woven from silk threads.

[The life cycle of silkworm is given for reference. Students are expected to visit the nearest sericulture farm and attempt this activity on their own.]

11th Biology Digest Chapter 11 Study of Animal Type: Cockroach Intext Questions and Answers

Can you recall? (Textbook Page No. 127)

How many different types of animals are present around us?
Answer:
Animals on earth show great diversity. The different types of animals present around us are;
a. Unicellular and multicellular
b. Prokaryotic and eukaryotic
c. Vertebrates and invertebrates
d. Unisexual and hermaphrodite
e. Aquatic, terrestrial, amphibian, reptilian, aerial, etc.

Can a person do a complete detailed study of each of those animals?
Answer:
Yes, a person can do a complete detailed study of each of those animals. Classification of animals based on characteristics into various groups has made it easier to study them.

Which phylum is most diverse and populous?
Answer:
Phylum Arthropoda is most diverse and populous.

Curiosity box: (Textbook Page No. 127)

Why do insects need moulting?
Answer:
a. Insects undergo metamorphosis (change of form or structure in an individual after hatching or birth). Each time an insect enters the next growth stage it has to molt.
b. Moulting is the process in which formation of new chitinous exoskeleton and subsequent shedding of the old one occurs.
c. The insects need moulting as their exoskeleton is rigid unlike the skin and does not allow the body to grow.

What is the difference between simple and compound eyes?
Answer:

Use your brainpower. (Textbook Page No. 131)

Why do body cavity of cockroach is called as haemocoel?
Answer:
The body cavity of cockroach is known as haemocoel as it is filled with haemolymph (blood). Cockroaches have open type of circulation thus; the body cavity is filled with haemolymph.

Internet my friend. (Textbook Page No. 136)

Collect the information about techniques and objectives of rearing the cockroaches in countries like China and make a Powerpoint presentation including video clips.
Answer:
1. Cockroach rearing industry is a booming industry in China. Cockroaches are reared in more than hundred farms in China. A giant farm in China produces around 6 billion cockroaches.
2. It is believed that cockroaches can be used to prepare a medicine that can prevent stomach cancer. They are also used to treat compost waste.
[Students can search on internet for more information about the techniques and objectives of rearing the cockroaches]

Maharashtra State Board Class 11 Biology Textbook Solutions

• Plant Tissues and Anatomy Class 11 Biology Textbook Solutions
• Morphology of Flowering Plants Class 11 Biology Textbook Solutions
• Animal Tissue Class 11 Biology Textbook Solutions
• Study of Animal Type : Cockroach Class 11 Biology Textbook Solutions
• Photosynthesis Class 11 Biology Textbook Solutions
• Respiration and Energy Transfer Class 11 Biology Textbook Solutions
• Human Nutrition Class 11 Biology Textbook Solutions
• Excretion and Osmoregulation Class 11 Biology Textbook Solutions
• Skeleton and Movement Class 11 Biology Textbook Solutions

Animal Tissue Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 10 Animal Tissue Textbook Exercise Questions and Answers.

Class 11 Biology Chapter 10 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 10 Exercise Solutions

1. Choose the correct option

Question (A)
The study of structure and arrangement of tissue is called _______.
(a) anatomy
(b) histology
(c) microbiology
(d) morphology
Answer:
(b) histology

Question (B)
_______ is a gland which is both exocrine and endocrine.
(a) Sebaceous
(b) Mammary
(c) Pancreas
(d) Pituitary
Answer:
(c) Pancreas

Question (C)
_______ cell junction is mediated by integrin.
(a) Gap
(b) Hemidesmosomes
(c) Desmosomes
(d) Adherens
Answer:
(b) Hemidesmosomes

Question (D)
The protein found in cartilage is _______ .
(a) ossein
(b) haemoglobin
(c) chondrin
(d) renin
Answer:
(c) chondrin

Question (E)
Find the odd one out.
(a) Thyroid gland
(b) Pituitary gland
(c) Adrenal gland
(d) Salivary gland
Answer:
(d) Salivary gland

2. Answer the following questions

Question (A)
Identify and name the type of tissues in the following:

1. Inner lining of the intestine
2. Heart wall
3. Skin
4. Nerve cord
5. Inner lining of the buccal cavity

Answer:

1. Epithelial tissue (Columnar epithelium)
2. Cardiac muscles (Muscular tissue)
3. Epithelial tissue (Stratified epithelium)
4. Nervous tissue
5. Epithelial tissue (Ciliated epithelium)

Question (B)
Why do animals in cold regions have a layer of fat below their skin?
Answer:
1. In adipose tissues, fats are stored in the form of droplets.
2. The adipose tissue acts as good insulator and helps retain heat in the body. This helps in survival of animals in the colder regions. Hence, animals in cold regions have a layer of fat below their skin.

Question (C)
What enables the ear pinna to be folded and twisted while the nose tip can’t be twisted?
Answer:
1. The ear pinna (outer ear) is made up of a thin plate of elastic cartilage and is connected to the surrounding.
2. The nose tip is made up of elastic cartilage. However, several bones and cartilage make up the bony- cartilaginous framework of the nose.
Hence, even though the tip of the nose is made up of elastic cartilage, it cannot be twisted like the ear pinna due to presence of bony-cartilaginous framework.

Question (D)
Sharad touched a hot plate by mistake and took away his hand quickly. Can you recognize the tissue and its type responsible for it?
Answer:
1. Nervous and muscular tissues are responsible for this action
2. Nervous tissue recognizes the stimuli whereas muscular tissue allows responding to the stimuli.

Question (E)
Priya got injured in an accident and hurt her long bone and later on she was also diagnosed with anaemia. What could be the probable reason?
Answer:
1. The centre of long bones (diaphysis) contains bone marrow, which is a site of production of blood cells (red blood cells).
2. Any severe injury to the bone marrow can affect rate of haematopoiesis (formation of blood cells).
3. A low count of erythrocytes (red blood cells) is characterised as anaemia. Hence, an injury to Priya’s long bone might have resulted in anaemia.

Question (F)
Supriya stepped out into the bright street from a cinema theatre. In response, her eye pupil shrunk. Identify the muscle responsible for the same.
Answer:
Smooth muscles (Involuntary muscles) are responsible for shrinking of eye pupil.

3. Answer the following questions

Question (A)
What is cell junction? Describe different types of cell junctions.
Answer:
1. Cell junctions: The epithelial cells are connected to each other laterally as well as to the basement
membrane by junctional complexes called cell junctions.
2. The different types of cell junctions are as follows:
a. Gap Junctions (GJs): These are intercellular connections that allow the passage of ions and small molecules between cells as well as exchange of chemical messages between cells.
b. Adherens Junctions (AJs): They are involved in various signalling pathways and transcriptional regulations.
c. Desmosomes (Ds): They provide mechanical strength to epithelial tissue, cardiac muscles and meninges.
d. Hemidesmosomes (HDs): They allow the cells to strongly adhere to the underlying basement membrane. These junctions help maintain tissue homeostasis by signalling.
e. Tight junctions (TJs): These junctions maintain cell polarity, prevent lateral diffusion of proteins and ions.

Question (B)
Describe in brief about areolar connective tissue with the help of suitable diagram.
Answer:
Areolar tissue is a loose connective tissue found under the skin, between muscles, bones, around organs, blood vessels and peritoneum. It is composed of fibres and cells.
The matrix of areolar tissues contains two types of fibres i.e. white fibres and yellow fibres.
a. White fibres: They are made up of collagen and give tensile strength to the tissue.
b. Yellow fibres: They are made up of elastin and are elastic in nature.
The four different types of cells present in this tissue are as follows:
a. Fibroblast: Large flat cells having branching processes. They produce fibres as well as polysaccharides that form the ground substance or matrix of the tissue.
b. Mast cells: Oval cells that secrete heparin and histamine.
c. Macrophages: Amoeboid, phagocytic cells.
d. Adipocytes (Fat cells): These cells store fat and have eccentric nucleus.

Question (C)
Describe the structure of multipolar neuron.
Answer:
A neuron is the structural and functional unit of the nervous tissue. A neuron is made up of cyton or cell body and cytoplasmic extensions or processes.
1. Cyton:
The cyton or cell body contains granular cytoplasm called neuroplasm and a centrally placed nucleus. The neuroplasm contains mitochondria, Golgi apparatus, RER and Nissl’s granules.
2. Cytoplasmic extensions or processes:
(a) Dendron: They are short, unbranched processes.
The fine branches of a dendron are called dendrites.
Dendrites carry an impulse towards the cyton.

(b) Axon: It is a single, elongated and cylindrical process.

1. The axon is bound by the axolemma.
2. The protoplasm or axoplasm contains large number of mitochondria and neurofibrils.
3. The axon is enclosed in a fatty sheath called the myelin sheath and the outer covering of the myelin sheath is the neurilemma. Both the myelin sheath and the neurilemma are parts of the Schwann cell.
4. The myelin sheath is absent at intervals along the axon at the Node of Ranvier.
5. The fine branching structure at the end of the axon (terminal arborization) is called telodendron.

Question (D)
How to differentiate the skeletal and the smooth muscles based on their nucleus?
Answer:
Skeletal muscles contain nucleus arranged at periphery. Striated or smooth muscles are with centrally placed single large oval nucleus therefore, skeletal and smooth muscle fibres can be identified.

Question 4.
Complete the following table.
Answer:

Question 5.
Match the following.

Answer:

Practical / Project:

Question 1.
To study the different tissues with the help of permanent slides in your college laboratory.
Answer:
Students may observe permanent slides of different tissues like epithelial tissue, connective tissue, muscular tissue and nervous tissue slides in laboratory.
[Students are expected to perform this activity on their own.]

Question 2.
Collect the information about the exercise to keep muscles healthy and strong.
Answer:

1. Muscles become stronger when we are physically active.
2. Physical activities like walking, jogging, lifting weights, playing tennis, climbing stairs, jumping, and dancing are good ways to exercise our muscles.
3. Apart from this, swimming and biking can also be considered as good workouts for muscles.
4. Different kinds of activities, work different muscles. Hence, it is essential to perform various types of physical activities.
5. Also, activities that increase our breath rate, help in exercising our heart muscle as well.
   [Students are expected to collect more information on their own.]

11th Biology Digest Chapter 10 Animal Tissue Intext Questions and Answers

Can you recall? (Textbook Page No. 116)

What is tissue?
Answer:
A group of cells having the same origin, same structure and same function is called ‘tissue’.

Do you know? (Textbook Page No. 116)

Number of cells in human body.
Answer:
There are about 100 trillion of 200 different types of cells in the human body.

Can you tell? (Textbook Page No. 119)

Explain basic structure of epithelial tissue and mention its types.
Answer:
The characteristics of epithelial tissues are as follows:

1. Epithelial tissue forms a covering on inner and outer surface of body and organs.
2. The cells of this tissue are compactly arranged with little intercellular matrix.
3. The cells rest on a non-cellular basement membrane.
4. The epithelial cells are polygonal, cuboidal or columnar in shape.
5. A single nucleus is present at the centre or at the base of the cell.
6. The tissue is avascular and has a good regeneration capacity.
7. The major function of the epithelial tissue is protection. It also helps in absorption, transport, filtration and secretion.

The different types of epithelial tissues are as follows:
1. Simple epithelium: Epithelial tissue made up of single layer of cells is known as simple epithelium. Simple epithelium is further classified into:
a. Squamous Epithelium
b. Cuboidal Epithelium
c. Columnar Epithelium
d. Ciliated Epithelium
e. Glandular Epithelium
f. Sensory epithelium
g- Germinal epithelium

2. Compound epithelium: Epithelium composed of several layers is called compound epithelium. Compound epithelium is further classified into:
a. Stratified epithelium
b. Transitional epithelium

Epithelial tissue has good capacity of regeneration. Give reason.
Answer:
Epithelial tissue rests on a basement membrane which acts as a scaffolding on which epithelium can grow and regenerate after injuries.

Can you recall? (Textbook Page No. 116)

Where is squamous epithelium located?
Answer:
Location: It is present in blood vessels, alveoli, coelom, etc.

Can you tell? (Textbook Page No. 119)

Write a note on glandular epithelium.
Answer:
Structure:
1. The cells of the glandular epithelium can be columnar, cuboidal or pyramidal in shape.
2. The nucleus of these cells is large and situated towards the base.
3. Secretory granules are present in the cell cytoplasm.
4. Glands consist of glandular epithelium. The glands may be either unicellular (goblet cells of intestine) or multicellular (salivary gland), depending on the number of cells.
5. Types: Depending on the mode of secretion, multicellular glands can be further classified as duct bearing glands (exocrine glands) ad ductless glands (endocrine glands).
a. Exocrine glands: These glands pour their secretions at a specific site. e.g. salivary gland, sweat gland, etc.
b. Endocrine glands: These glands release their secretions directly into the blood stream, e.g. thyroid gland, pituitary gland, etc.
6. Function: Glandular epithelium secretes mucus to trap the dust particles, lubricate the inner surface of respiratory and digestive tracts, secrete enzymes and hormones, etc.
Heterocrine glands
1. Heterocrine glands or composite glands have both exocrine and endocrine function.
2. Pancreas is called a heterocrine gland because it secretes the hormone insulin into blood which is an endocrine function and enzymes into digestive tract which is an exocrine function.

Use your brain power? (Textbook Page No. 118)

When do the transitional cells change their shape?
Answer:
Transitional cells change their shape depending on the degree of distention (stretch) needed. As the tissue stretches, the transitional cells start changing shape from round and globular to thin and flat.

Can you tell? (Textbook Page No. 119)

How do cell junctions help in functioning of epithelial tissue?
Answer:
1. Cell junctions: The epithelial cells are connected to each other laterally as well as to the basement
membrane by junctional complexes called cell junctions.
2. The different types of cell junctions are as follows:
a. Gap Junctions (GJs): These are intercellular connections that allow the passage of ions and small molecules between cells as well as exchange of chemical messages between cells.
b. Adherens Junctions (AJs): They are involved in various signalling pathways and transcriptional regulations.
c. Desmosomes (Ds): They provide mechanical strength to epithelial tissue, cardiac muscles and meninges.
d. Hemidesmosomes (HDs): They allow the cells to strongly adhere to the underlying basement membrane. These junctions help maintain tissue homeostasis by signalling.
e. Tight junctions (TJs): These junctions maintain cell polarity, prevent lateral diffusion of proteins and junctions.

Can you tell? (Textbook Page No. 122)

Give reason.
As we grow old, cartilage becomes rigid.
Answer:
Calcified cartilage is a type of cartilage that becomes rigid due to deposition of salts in the matrix. This reduces the flexibility of joints in old age and cartilage becomes rigid.

Can you recall? (Textbook Page No. 116)

Enlist functions of bone.
Answer:
Bones support and protect different organs and help in movement.

Can you tell? (Textbook Page No. 122)

(i) Give reason. Bone is stronger than cartilage.
Answer:
a. Bone is rigid, non-pliable, dense connective tissue characterised by the hard matrix called ossein (made up of calcium salt hydroxyapatite). An outer tough membrane called periosteum encloses the matrix. The matrix is arranged in the form of concentric layers called lamellae. Bones are well vascularized and possess blood vessels and nerves that pierce through the periosteum,
b. Cartilage is a pliable supportive connective tissue. On comparison with bones, cartilage is thin, avascular and flexible. In cartilage, a sheath of collagenous fibres called perichondrium encloses the matrix.
Hence, a bone is stronger than a cartilage.

(ii) Explain histological structure of mammalian bone.
Answer:
a. The bone is characterised by hard matrix called ossein which is made up of mineral salt hydroxy apatite (Ca₁₀ (P0₄)₆ (OH)₂).
b. An outer tough membrane called periosteum encloses the matrix.
c. Blood vessels and nerves pierce through the periosteum.
d. The matrix is arranged in the form of concentric layers called lamellae.
e. Each lamella contains fluid filled cavities called lacunae from which fine canals called canaliculi radiate.
f. The canaliculi of adjacent lamellae connect with each other as they traverse through the matrix.
g. Active bone cells called osteoblasts and inactive bone cells called osteocytes are present in the
lacunae.
h. The mammalian bone shows the peculiar haversian system.
i. The haversian canal encloses an artery, vein and nerves.

Can you recall? (Textbook Page No. 122)

How can exercise improve your muscular system?
Answer:
1. Exercise can improve both muscular strength and stamina endurance.
2. Exercises are commonly grouped into two types depending on the effect they have on the body:
a. Aerobic exercises: such as cycling, walking, and running. They increase muscular endurance and cardiovascular health, etc.
b. Anaerobic exercises: such as weight training or sprinting, increase muscle strength, etc.
3. Anaerobic exercies: It comprises brief periods of physical exertion and high-intensity, strength-training activities.
Anaerobic exercise is a physical exercise intense enough to cause lactate to form.
It is used by athletes to promote strength, speed and power; and by body builders to build muscle mass.

Can you recall? (Textbook Page No. 122)

How many skeletal muscles are present in human body?
Answer:
There are over 650 named skeletal muscles in the human body.

Can you tell? (Textbook Page No. 125)

Differentiate between medullated and non-medullated fibre.
Answer:

Internet is my friend. (Textbook Page No. 125)

Learn about transmission of impulse from one neuron to another.
Answer:

1. A nerve impulse is transmitted from one neuron to another through junctions called synapses.
2. A synapse is formed by the membranes of a pre-synaptic neuron and a post-synaptic neuron, which may or may not be separated by a gap called synaptic cleft.
3. There are two types of synapses, namely, electrical synapses and chemical synapses.
4. Electrical synapses: The membranes of pre- and post-synaptic neurons are in very close proximity.
   Thus, electrical current can flow directly from one neuron into the other across these synapses.
   Impulse transmission across an electrical synapse is faster.
5. Chemical synapse: The membranes of the pre- and post-synaptic neurons are separated by a fluid- filled space called synaptic cleft.
6. Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses.
7. The axon terminals contain vesicles filled with these neurotransmitters.
8. When an impulse arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and release their neurotransmitters into the synaptic cleft.
9. The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane.
10. This binding opens ion channels and allows the entry of ions which can generate a new potential in the post-synaptic neuron.

[Students are expected to refer the given information and collect more information from the internet.]
[Note: Students can scan the adjacent QR code to get conceptual clarity with the aid of a relevant video ]

Observe and Discuss (Textbook Page No. 125)

Explain the structure of nerve.

Answer:

1. Each spinal nerve consists of many axons and contains layers of protective connective tissue coverings.
2. Axons are enclosed in a fatty sheath called myelin sheath.
3. Individual axons within a nerve are wrapped in an endoneurium (innermost layer).
4. Groups of axons with their endoneurium are arranged in bundles called fascicles.
5. Each fascicle is wrapped in perineurium (middle layer).
6. The outermost covering over the entire nerve is the epineurium. The epineurium extends between fascicles.
7. Many blood vessels nourish the nerve and are present within the perineurium and epineurium.
   [Source: Tortora. G, Derrickson. B. Principles of Anatomy and Physiology. 11th Edition.]

Maharashtra State Board Class 11 Biology Textbook Solutions

• Plant Tissues and Anatomy Class 11 Biology Textbook Solutions
• Morphology of Flowering Plants Class 11 Biology Textbook Solutions
• Animal Tissue Class 11 Biology Textbook Solutions
• Study of Animal Type : Cockroach Class 11 Biology Textbook Solutions
• Photosynthesis Class 11 Biology Textbook Solutions
• Respiration and Energy Transfer Class 11 Biology Textbook Solutions
• Human Nutrition Class 11 Biology Textbook Solutions
• Excretion and Osmoregulation Class 11 Biology Textbook Solutions
• Skeleton and Movement Class 11 Biology Textbook Solutions

Morphology of Flowering Plants Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 9 Morphology of Flowering Plants Textbook Exercise Questions and Answers.

Class 11 Biology Chapter 9 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 9 Exercise Solutions

1. Choose the correct option

Question (A)
Which one of the following will grow better in the moist and shady regions?
(a) Opuntia
(b) Orchid
(c) Mangrove
(d) Lotus
Answer:
(b) Orchid

Question (B)
A particular plant had a pair of leaves at each node arranged in one plane. What is the arrangement called?
(a) Alternate phyllotaxy
(b) Decussate phyllotaxy
(c) Superposed phyllotaxy
(d) Whorled phyllotaxy
Answer:
(c) Superposed phyllotaxy

Question (C)
In a particular flower the insertion of floral whorls was in such a manner, so the ovary was below other three whorls, but its stigma was taller than other three whorls. What will you call such flower?
(a) Hypogynous
(b) Perigynous
(c) Inferior ovary
(d) Half superior – half inferior
Answer:
(c) Inferior ovary

Question (D)
Beet and Arum both store food for perennation.
Are the examples for two different types?
(a) Beet is a stem but Arum is a root
(b) Beet is a root but Arum is a stem
(c) Beet is a stem but Arum is a leaf
(d) Beet is a stem but Arum is an inflorescence
Answer:
(b) Beet is a root but Arum is a stem

2. Answer the following questions

Question (A)
Two of the vegetables we consume are nothing but leaf bases. Which are they?
Answer:
Onion, Garlic

Question (B)
Opuntia has spines but Carissa has thorns. What is the difference?
Answer:

1. In Opuntia, stem is modified into leaf like photosynthetic organ known as phylloclade.
2. Spines growing on phylloclade of Opuntia are leaves, modified to reduce the loss of water through transpiration.
3. Thoms in Carissa are modified apical buds. They provide protection against browsing animals.
4. Thus, spines in Opuntia and thorns in Carissa have different origin and function.

Question (C)
Teacher described Hibiscus as solitary Cyme. What it means?
Answer:
1. In Cymose inflorescence, growth of peduncle is finite and it terminates into flower.
2. In Hibiscus, flower is borne singly at the tip of peduncle. Hence, teacher described Hibiscus as solitary cyme.

3. Write notes on

Question (A)
Fusiform root.
Answer:
Fusiform root:
1. Fusiform root is the modification of tap root for food storage.
2. Fusiform root:
The fusiform root is swollen in the middle and tapering towards both ends forming spindle shaped structure, e.g. Radish (Raphanus sativus)

Question (B)
Racemose inflorescence.
Answer:
Racemose inflorescence

Question (C)
Fasciculated tuberous root.
Answer:
Fasciculated tuberous root:
1. Fasciculated tuberous roots are modification of adventitious roots for storage of food.
2. Fasciculated tuberous roots do not develop any definite shape like modified tap roots.
3. a. A cluster of roots arising from one point which becomes thick and fleshy due to storage of food is known as fasciculated tuberous root.
b. These clusters are seen at the base of the stem, e.g. Dahlia, Asparagus, etc.

Question (D)
Region of cell maturation.
Answer:
Region of maturation/region of differentiation:
a. It is the uppermost major part of the root.
b. The cells of this region are quite impermeable to water due to thick wall.
c. The cells show differentiation and form different types of tissues.
d. This region helps in fixation of plant and conduction of absorbed substances.
e. Development of lateral roots also takes place from this region.

Question (E)
Rhizome.
Answer:
Rhizome:

1. Rhizome is a modification of underground stem for storage of food.
2. It is prostrate, dorsiventrally thickened and brownish in colour.
3. It grows either horizontally or obliquely beneath the soil.
4. Rhizome shows nodes and intemodes. It bears terminal and axillary buds at nodes.
5. Terminal bud under favourable conditions produces aerial shoot which degenerates at the end of favourable condition.
6. Growth of rhizome takes place with lateral buds, such growth is known as sympodial growth, e.g. Ginger (Zingiber officinale), Turmeric {Curcuma domestica), Canna etc.
7. In plants where rhizomes grow obliquely, terminal bud brings about growth of rhizomes. This is known as monopodial growth, e.g. Nymphea, Nelumbo (Lotus), Pteris (Fern) etc.
8. Rhizomes perform functions like storage of food, vegetative propagation and perennation.

Question (F)
Stolon.
Answer:
Stolons:
1. The slender lateral branch arising from the base of main axis is known as stolon.
2. In some plants it is above ground (wild strawberry).
3. Primarily stolon shows upward growth in the form of ordinary branch, but when it bends and touches the ground terminal bud grows into new shoot and develops adventitious roots.
e.g. Wild Strawberry, Jasmine, Mentha, etc. [Any one example]

Question (G)
Leaf venation.
Answer:
Leaf venation:

1. Arrangement of veins and veinlets in leaf lamina is known as venation.
2. Veins are responsible for conduction of water and minerals as well as food.
3. The structural framework of the lamina is developed by veins.
4. There are two types of leaf venation: parallel venation which is found in monocot leaves and reticulate venation which is found in dicot leaves.

Question (H)
Cymose inflorescence.
Answer:
Cymose inflorescence.

Question (I)
Perianth.
Answer:
Perianth (P):
a. Many times, calyx and corolla remain undifferentiated. Such member is known as tepal.
b. The whorl of tepals is known as Perianth.
c. It protects other floral whorls.
d. If all the tepals are free the condition is called as polyphyllous and if they are fused the condition is called as gamophyllous.
e. Sepaloid perianth shows green tepals, while petaloid perianth shows brightly coloured tepals. e.g. Lily, Amaranthus, Celosia, etc.
f. Petaloid tepal helps in pollination and sepaloid tepals can perform photosynthesis.

Question (J)
Write a short note on vexillary aestivation.
Answer:
Vexillary: Corolla is butterfly shaped and consists of five petals. Outermost and largest is known as standard or vexillum, two lateral petals are wings and two smaller fused forming boat shaped structures keel. e.g. Pisum sativum

Question (K)
Write a short note on axile placentation.
Answer:
Axile placentation: Placentation: The mode of arrangement of ovules on the placenta within the ovary is called placentation.
Axile: Ovules are placed on the central axis of a multilocular ovary, e.g. China rose, Cotton, etc.

Question 4.
Identify the following figures and write down the types of leaves arrangement.

Answer:
1. The given figures represent phyllotaxy. It is the arrangement of leaves on the stem and branches in a specific manner.
2. Figure ‘a’ and ‘b’ represents, alternate phyllotaxy. In this type of phyllotaxy, single leaf arises from each node of a stem. e.g. Mango
3. Figure ‘c’ represents opposite decussate phyllotaxy. In this type of phyllotaxy, a pair of leaf arise from each node and the consecutive pair at right angle to the previous one. e.g. Calotropis.

5. Students were on the excursion to a botanical garden. They noted following observation. Will you be able to help them in understanding those conditions?

Question (A)
A wiry outgrowth was seen on a plant arising from in between the leaf and stem.
Answer:
A wiry outgrowth on a plant arising from in between the leaf and stem can be an axillary stem tendril. Stem tendrils:
a. Tendrils are thin, wiry, photosynthetic, leafless coiled structures.
b. They give additional support to developing plant.
c. Tendrils have adhesive glands for fixation.

Question (B)
There was a green plant with flat stem, but no leaves. The entire plant was covered by soft spines.
Answer:
Student must have observed phylloclade, which is a modification of stem.
Phylloclade:
a. Modification of stem into leaf like photosynthetic organ is known as phylloclade.
b. Being stem it possesses nodes and internodes.
c. It is thick, fleshy and succulent, contains mucilage for retaining water e.g. Opuntia, Casuarina (Cylindrical shaped phylloclade) and Muehlenbeckia (ribbon like phylloclade).

Question (C)
Many oblique roots were given out from the lower nodes, apparently for extra support.
Answer:
a. Students must have observed adventitious roots in monocotyledonous plants like maize, sugarcane, wheat, etc.
b. Adventitious roots develop from any part of a plant other than radicle.
c. In such plants, adventitious roots arise from the lower node of a stem and provide extra support to the plant. These roots are also called as stilt roots.

Question (D)
Many plants in the marshy region had upwardly growing roots. They could be better seen during low tide.
Answer:
a. Plants growing in marshy region (halophytes) produce upwardly growing roots called as
pneumatophores or respiratory roots.
b. The main root system of these plants does not get sufficient air for respiration as soil is water logged.
c. Due to this, mineral absorption of plant also gets affected.
d. To overcome this problem underground roots, develop special roots which are negatively geotropic; growing vertically upward.
e. These roots are conical projections present around main trunk of plant.
f. Respiratory roots show presence of lenticels which helps in gaseous exchange.

Question (E)
A plant had leaves with long leaf apex, which was curling around a support.
Answer:
a. Students must have observed leaf tip tendril.
b. In some weak stems, leaf apex modifies into thin, green, wiry, coiled structure called as leaf tendril.
c. Such leaf tendrils, help in climbing by curling around a support.

Question (F)
A plant was found growing on other plant. Teacher said it is not a parasite. It exhibited two types of roots.
Answer:
a. Student must have observed an epiphytic plants like Dendrobium, Vanda growing on other plant.
b. The two types of roots exhibited by this plant must be clinging roots and epiphytic roots.
c. Clinging roots:
1. Clinging roots are tiny roots develop along intemodes, show disc at tips.
2. It exudes sticky substance which enables plant to get attached to the substratum without damaging it.

d. Epiphytic roots:
1. Epiphytic plants like Vanda, Dendrobium grow on branches of trees in dense rain forests and are unable to obtain moisture from soil.
2. Such plants produce epiphytic roots which hang in the air.
3. The roots are provided with a spongy membranous absorbent covering of the velamen tissue.
4. The cells of velamen tissue are hygroscopic and have porous walls, thus they can absorb moisture from air.
5. Epiphytic roots can be silvery white or green and are without root cap.

Question (G)
While having lunch onion slices were served to them. Teacher asked which part of the plant are you eating?
Answer:
a. The edible part of an onion is fleshy leaves.
b. Onion is a bulb, in which stem is highly reduced, discoid and possesses adventitious roots at the base.
c. This stem bears a whorl of fleshy leaves which store food material.
d. The scale leaves or fleshy leaves are arranged in concentric manner over the stem. Some outer scale leaves become thin and dry. Thus, it is also called as tunicated or layered bulb.

Question (H)
Students observed large leaves of coconut and small leaves of Mimosa. Teacher asked it what way they are similar?
Answer:
a. Both large leaves of coconut and small leaves of Mimosa show pinnately compound leaves.
b. In both plants, leaf lamina is divided into number of leaflets.
c. Leaflets are present laterally on a common axis called rachis, which represents the midrib of the leaf.

Question (I)
Teacher showed them Marigold flower and said it is not one flower. What the teacher meant?
Answer:
a. Marigold flower is an inflorescence in which flowers are produced in a definite manner on a peduncle.
b. In Marigold, racemose type of inflorescence can be observed.
c. In this, peduncle condenses to form a flat rounded structure called receptacle.
d. Opening of flower centripetal i.e. younger flowers are towards the centre and open later, while older flowers towards the periphery and open first.

Question (J)
Students cut open a Papaya fruit and found all the seeds attached to the sides. Teacher inquired about the possible placentation of Papaya ovary.
Answer:
a. In Papaya, seeds are attached to the sides of a fruit. Thus, parietal placentation is possible in papaya ovary,
b. In parietal placentation, ovules are placed on the inner wall of unil unilocular ovary of multicarpellary, syncarpus gynoecium.

Question 6.
Match the following.

Answer:
(i-c-1), (ii-e-3), (iii-a-4), (iv-b-5), (v-d-2)
[Note: Another example of palmately compound leaf (Bifoliate) is Balanites roxburghii.]

Question 7.
Observe the following figures and label the different parts.

Answer:

8. Differentiate with diagrammatic representation.

Question (A)
Differentiate with diagrammatic representation: Racemose and cymose inflorescence.
Answer:

Question (B)
Differentiate with diagrammatic representation: Reticulate and parallel venation
Answer:

Question (C)
Differentiate with diagrammatic representation: Taproot and Adventitious roots
Answer:

Practical / Project:

Question 1.
Collect different leaves from nearby region and observe variation in margin, leaf base, apex etc.
[Note: Students can scan the given Q.R code to study the different le
af margin, leaf base and apex.]

Question 2.
Find out and make a note of economically important plant from family Fabaceae, Solanaceae and Liliaceae.
Answer:
1. Economically important plant from family Fabaceae:
Family Fabaceae includes many pulses like gram, arhar, moong, soybean; edible oil seeds like soybean, groundnut; dye (lndigofera); fibres which can be obtained from Sun hemp, Sesbania trifolium which can be used as fodder; some plants are ornamental like lupin, sweet pea; some medicinal plants like muliathi.

2. Economically important plant from family Solanaceae:
Family Solanaceae includes many plants which are good source of food e.g. tomato, brinjal, potato; Spice e.g. chilli; Medicine e.g. belladonna, ashwagandha; Ornamental plants like Petunia.

3. Economically important plant from family Liliaceae:
Family Liliaceae includes many ornamental plants like tulip, Gloriosa, Medicinal plants like Aloe vera. Asparagus and source of colchicine, e.g. Colchicum autumnale.

Question 3.
Collect different leaves from garden and observe their veins and classify it.

11th Biology Digest Chapter 9 Morphology of Flowering Plants Intext Questions and Answers

Use your brainpower. (Textbook Page No. 102)

Why underground stem is different from roots?
Answer:
Underground stems are modified to perform different functions like storage of food, perennation and vegetative propagation. However, they differ from root in having nodes and intemodes.

Use your brainpower. (Textbook Page No. 104)

Why the stem has to perform photosynthesis in xerophytes?
Answer:
1. Xerophytes are the plants which grow in regions with scanty or no rainfall like desert.
2. In Xerophytes, leaves get modified into spines or get reduced in size to check the loss of water due to transpiration.
3. As the leaves are modified into spines, the stem becomes green in colour to do the function of photosynthesis.

Internet My Friend. (Textbook Page No. 106)

Collect information of types of leaf venation.
Answer:
1. Figure ‘R’ shows types of reticulate venation. WTien the veins and veinlets form a network, it is called
reticulate venation.
On the basis of number of mid-veins, reticulate venation is of two types:
a. Pinnate or unicostate: It is with single midrib e.g. Peepal, Mango.
b. Palmate or multicostate: It is with two or more prominent veins. It is further divided into convergent or divergent.
1. Multicostate convergent reticulate: Many prominent veins appear from the base of leaf lamina and converged in a curved manner towards the leaf apex. e.g. Zizyphus
2. Multicostate divergent reticulate: Prominent veins arise from the single point at the base of leaf lamina
and then diverge from one another towards the leaf margin, e.g. Cucurbita

2. Figure ‘P’ shows types of parallel venation. When veins run almost parallel to one another it is called parallel venation. It is of two types:
a. Unicostate: In this, lamina has single prominent mid vein from which many lateral parallel veins arise at regular intervals, e.g. Banana
b. Multicostate: In this, two or more mid veins run parallel to each other. It is further divided into convergent or divergent.

1. Multicostate convergent parallel:
Many prominent veins arise from the leaf base and then converge at leaf apex. e.g. Grasses
2. Multicostate divergent parallel:
Many prominent veins arise from the leaf base and then diverge towards margin, e.g. Borassus flabellifer (Toddy palm)

Observe and Discuss. (Textbook Page No. 112)

Observe and Discuss.
Answer:
1. Figure ‘a’ shows fruit of tomato.

• It is a simple fruit as it develops from a single flower with bicarpellary syncarpous gynoecium.
• It is a berry, because it has fleshy endocarp and many seeds.

2. Figure ‘b’ shows fruit of Custard apple.

• It is an aggregate fruit, because it develops from a single flower with polycarpellary, apocarpous gynoecium.
• Here, the ovary of each carpel gives rise to a part of the fruit called fruitlet. Hence, it is called an aggregation of fruitlets.
• Custard apple can be further described as Etaerio of berries.

3. Figure ‘c’ shows fruit of pineapple.

• It is a composite fruit, because it develops from a complete inflorescence.
• Pineapple can be further described as Sorosis, as it develops from catkin type of inflorescence.

4. Figure ‘d’ shows fruit of milkweed.

• It is a simple dehiscent dry fruit.
• It has many seeds. When pericarp becomes dry and thin, it breaks open by one ventral suture.

Activity. (Textbook Page No. 113)

Study the family Liliaceae, prepare a table of following characteristics.
Answer:

Maharashtra State Board Class 11 Biology Textbook Solutions

• Plant Tissues and Anatomy Class 11 Biology Textbook Solutions
• Morphology of Flowering Plants Class 11 Biology Textbook Solutions
• Animal Tissue Class 11 Biology Textbook Solutions
• Study of Animal Type : Cockroach Class 11 Biology Textbook Solutions
• Photosynthesis Class 11 Biology Textbook Solutions
• Respiration and Energy Transfer Class 11 Biology Textbook Solutions
• Human Nutrition Class 11 Biology Textbook Solutions
• Excretion and Osmoregulation Class 11 Biology Textbook Solutions
• Skeleton and Movement Class 11 Biology Textbook Solutions

Plant Tissues and Anatomy Class 11 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 11th Biology Textbook Solutions Chapter 8 Plant Tissues and Anatomy Textbook Exercise Questions and Answers.

Class 11 Biology Chapter 8 Exercise Solutions Maharashtra Board

Biology Class 11 Chapter 8 Exercise Solutions

1. Choose the correct option.

Question (A)
Location or position of meristematic regions is divided into _______ types.
(A) one
(B) two
(C) three
(D) none of the above
Answer:
(C) three

Question (B)
Cambium is also called
(A) apical meristem
(B) intercalary meristem
(C) lateral meristem
(D) none of the above
Answer:
(C) lateral meristem

Question (C)
Collenchyma is a type of ________ tissue.
(A) living
(B) dead
(C) living and dead
(D) none of the above
Answer:
(A) living

Question (D)
_______ is a complex permanent tissue.
(A) Parenchyma
(B) Sclerenchyma
(C) Chlorenchyma
(D) Xylem
Answer:
(D) Xylem

Question (E)
Mesophyll tissue is present in ________ .
(A) root
(B) stem
(C) leaf
(D) flower
Answer:
(C) leaf

2. Answer the following questions

Question (A)
A fresh section was taken by a student but he was very disappointed because there were only few green and most colourless cells. Teacher provided a pink colour solution. The section was immersed in this solution and when observed it was much clearer. What is the magic?
Answer:
1. The pink coloured solution given by teacher must be a saffanin stain.
2. Saffanin is used to stain plant tissues, especially lignified tissues such as cell wall and xylem.

Question (B)
While observing a section, many scattered vascular bundles could be seen. Teacher said, in spite of this large number the stem cannot grow in girth. Why?
Answer:

1. Students must have observed monocot stems.
2. It is because, monocot stem shows scattered vascular bundles.
3. In monocot stem, vascular bundles are closed i.e. without cambium.
4. Thus, secondary growth does not occur which is required for increase in girth. Hence, in spite of having large number of scattered vascular bundles, monocot stems do not grow in girth.

Question (C)
A section of the stem had vascular bundles, where one tissue was wrapped around the other. How will you technically describe it?
Answer:
Concentric vascular bundle:
a. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
b. When phloem is encircled by xylem, it is called as leptocentric vascular bundle, whereas when xylem is encircled by phloem, it is called as hadrocentric vascular bundle.
c. When xylem is encircled by phloem on both faces, it is called as amphicribral vascular bundle. When phloem is encircled by xylem on both faces it is called as amphivasal vascular bundle.

Question (D)
There were two cut logs of wood lying in the campus. One had growth rings and other didn’t. Teacher said it is due to differences in their pattern of grow th which is dependent on season. How?
Answer:
1. It is possible that one of the cut logs was of a tropical tree, whereas the other was of a temperate tree. Since tropical trees grow in a similar manner all year, growth rings are not apparent. Another explanation for this could be that the log which had growth rings must be of an old tree which has experience many seasons, whereas the log without growth rings must be of younger tree, that has not been subjected to seasonal changes and hence not developed prominent growth rings.

2. Growth rings are formed due cambial activity during favourable and non-favourable climatic conditions.

3. During favourable conditions, spring wood (early wood) is formed which has broader xylem bands, lighter colour, tracheids with thin wall and wide lumen, fibres are less in number, low density. Whereas, during unfavourable conditions, autumn wood (late wood) is formed which has narrow xylem band, darker in colour, lumen is narrow and walls are thick with abundant fibres, high density.

4. Spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.

5. These growth rings can be used to estimate the age of the tree. These are found more in older trees as compare to younger tree.

Question (E)
While on the trip to Kashmir, Pintoo observed that cut portions of large trees show distinct rings, which he never found in Maharashtra. Why is so?
Answer:
1. Cut portions of large tress show distinct rings which are annual rings formed due to activity of cambium during favourable and non-favourable climatic conditions.
2. Kashmir falls under temperate region where the climatic conditions are not uniform through the year. In the spring season, conditions are favourable due to which cambium is active, whereas in autumn season, conditions are unfavourable due to which cambium is less active. This leads to formation of spring wood and autumn wood that appear as alternate light and dark concentric rings, constitute an annual ring or growth ring.
3. Maharashtra falls under tropical region where climatic conditions are favourable throughout the year. In tropical areas, continuous growth of secondary xylem occurs. Thus, trees growing in tropical regions show less or no annual rings as compared to trees in temperate region.

Question (F)
A student was observing a slide with no label under microscope. The section had some vascular bundles scattered in the ground tissue. It is section of a monocot stem! He exclaimed. No! it is section of fern rachis, said the teacher. Teacher told to observe vascular bundle again. Student agreed, Why?
Answer:

1. In fern rachis, the number of vascular bundles is less as compared to number of vascular bundles in monocot stem. In monocot stem, vascular bundles are numerous.
2. In fern rachis, xylem consists of only tracheids whereas in monocot stem, xylem consists of vessels (protoxylem and metaxylem) as well as tracheids. Monocot stem shows presence of lysigenous cavity just below protoxylem.
3. In fern rachis, phloem consists of only sieve cells whereas in monocot stem, phloem consists of sieve tubes and companion cells. Thus, a student must hav e observed these differences in the given section and agreed to teacher’s statement that the given section is of fern rachis and not of monocot stem.

Question (G)
Student found a wooden stopper in lab. He was told by an old lab attendant that it is there for many years. He kept thinking how it did not rot?
Answer:
1. Wooden stopper or cork is obtained from the phellem (cork) part of a bark.
2. Phellem (cork) is impervious in nature and does not allow entry of water due to suberized walls.
3. Due to this it does not rot and remains as it is for many years.

Question (H)
Student while observing a slide of leaf section observed many stomata on the upper surface. He thought he has placed slide upside down. Teacher confirmed it is rightly placed. Explain.
Answer:
1. In a dicot leaf, stomata are generally absent on upper epidermis but are present on lower epidermis. Thus, the student must have thought that he has placed slide upside down.
2. According to teacher, the section was placed rightly, thus the given section must be of monocot leaf.
3. It is because, in monocot leaf stomata are present on both upper and lower epidermis.

3. Write short notes on the following points.

Question (A)
Structure of stomata.
Answer:

1. Small gateways in the epidermal cells are called as stomata.
2. Stoma is controlled or guarded by specially modified cells called guard cells.
3. These guard cells may be kidney-shaped (dicot) or dumbbell-shaped (monocot), collectively called as stomata.
4. Guard cells have chloroplasts to carry out photosynthesis.
5. Change in turgor pressure of guard cells causes opening and closing of stomata, which enables exchange of gases and water vapour.
6. Stomata are further covered by subsidiary cells.
7. Stoma, guard cells and subsidiary cells form a unit called stomatal apparatus.

Question (B)
Write a short note on secondary growth.
OR
With the help of neat and labelled diagram explain the secondary growth in dicot stem.
Answer:
Secondary growth:

1. Dicotyledonous plants and gymnosperms exhibit increase in girth of root and stem.
2. In dicot stem, secondary growth begins with the formation of a continuous cambium ring.
3. The cambium present between the primary xylem and primary phloem of a vascular bundle is called intrafascicular cambium.
4. The cells of medullary rays adjoining these intrafascicular cambium strips become meristematic (regain the capacity to divide) and form the interfascicular cambium.
5. Thus, a complete and continuous ring of vascular cambium is formed.
6. The cambium ring cuts off new cells, towards both inner and outer sides.
7. The cells that are cut-off towards pith (inner side) mature into secondary xylem and cells that are cut-off towards periphery mature into secondary phloem.
8. Generally, amount of secondary xylem is more than the secondary phloem.

Question (C)
Write a short note on peculiarity of a sclerenchyma cell wall.
Answer:
Peculiarity of a sclerenchyma cell wall:
1. Cell wall of sclerenchyma is evenly thickened due to uniform deposition of lignin.
2. Cell wall of sclereids is extremely thick and strongly lignified.

4. Differentiate

Question (A)
Differentiate between vascular bundles of monocot and dicot.
Answer:

1. Vascular bundle of monocot and dicot root.
2. Vascular bundle of monocot and dicot stem.
3. Vascular bundle of monocot and dicot leaf.

Question (B)
Differentiate between xylem and phloem.
Answer:

5. Draw neat labelled diagrams

Question (A)
T.S. of dicot leaf.
Answer:
1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.

4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Question (B)
T.S. of Monocot root.
Answer:

Question (C)
Draw neat labelled diagrams of T.S. of dicot stem.
Answer:

Question 6.
Write the information related to diagram given below.

Answer:


[Note: The labelled part can be considered as the ‘region of maturation ’ of root apical however, the region of maturation does not contain meristematic tissue ]
Classification of meristematic tissue based on its position:
1. Apical meristem:
a. It is produced from promeristem and forms growing point of apices of root, shoot and their lateral branches.
b. It brings about increase in length of plant body and is called as apical initials.
c. Shoot apical meristem is terminal in position whereas in root it is subterminal i.e. located behind the root cap.

2. Intercalary meristem:
a. Intercalary meristematic tissue is present in the top or base area of node.
b. Their activity is mainly seen in monocots.
c. These are short lived.

3. Lateral meristem:
a. It is present along the sides of central axis of organs.
b. It takes part in increasing girth of stem or root, e.g. Intrafascicular cambium.
c. It is found in vascular bundles of gymnosperms and dicot angiosperms.

Question 7.
Identify the following diagrams, label it and prepare a chart of characteristics.
Answer:
1. Figure ‘c’

Question 8.
Distinguish between dicot and monocot leaf on the basis of following characters.
Answer:

Practical/ Project:

Question 1.
Prepare detail anatomical charts with diagrammatic representation of dicot and monocot plants.
Answer:
Anatomy of dicot root: The transverse section of a typical dicotyledonous root shows following anatomical features:
1. Epiblema: It is the outermost single layer of cells without cuticle. Some epidermal cells prolong to form unicellular root hairs.
2. Cortex: It is made up of many layers of thin walled parenchyma cells. Cortical cells store food and water.
3. Exodermis: After the death of epiblema, outer layer of cortex become cutinized and is called Exodermis.

4. Endodermis:
The innermost layer of cortex is called Endodermis.
The cells are barrel-shaped and their radial walls bear Casparian strip or Casparian bands composed of suberin. Near the protoxylem, there are unthickened passage cells.

5. Stele: It consists of pericycle, vascular bundles and pith.
a. Pericycle: Next to the endodermis, there is a single layer of thin walled parenchyma cells called pericycle. It forms outermost layer of stele or vascular cylinder.
b. Vascular bundle: Vascular bundles are radial. Xylem and Phloem occur in separate patches arranged on alternate radii. Xylem is exarch in root that means protoxylem vessels are towards periphery and metaxylem elements are towards centre. Xylem bundles vary from two to six number, i.e. they may be diarch, triarch, tetrarch, etc.
Connective tissue: A parenchymatous tissue is present in between xylem and phloem.
c. Pith: The central part of stele is called pith. It is narrow and made up of parenchymatous cells, with or without intercellular spaces.
6. At a later stage cambium ring develops between the xylem and phloem causing secondary growth.

Anatomy of monocot stem: A transverse section of maize (monocot) stem shows the following structures:

1. Epidermis: It is single-layered and without trichomes.
2. Hypodermis: It is sclerenchymatous.
3. Ground tissue: It consists of thin walled parenchyma cells. It extends from hypodermis to the centre. It is not differentiated into cortex, endodermis, pericycle and pith.
4. Vascular bundles: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and closed (without cambium). Xylem is endarch and shows lysigenous cavity.
5. Pith: Pith is absent.

Anatomy of dicot leaf:

1. Structure of dorsiventral leaf: The mesophyll tissue is differentiated into palisade and spongy parenchyma in a dorsiventral leaf. This type is very common in dicot leaf. The different parts of this leaf are as follows:
2. Upper epidermis: It consists of a single layer of tightly packed rectangular, barrel shaped, parenchymatous cells which are devoid of chloroplast. A distinct layer of cuticle lies on the outside of the epidermis. Stomata are generally absent.
3. Mesophyll: Between upper and lower epidermis, there is chloroplast-containing photosynthetic tissue called mesophyll It is differentiated into Palisade parenchyma and Spongy parenchyma.
a. Palisade parenchyma:
Palisade parenchyma is present below upper epidermis and consists of closely packed elongated cells. The cells contain abundant chloroplasts and help in photosynthesis.
b. Spongy parenchyma:
Spongy parenchyma is present below palisade tissue and consists of loosely arranged irregularly shaped cells with intercellular spaces. The spongy parenchyma cells contain chloroplast and are in contact with the atmosphere through stomata.
4. Vascular system: It is made up of a number of vascular bundles of varying size depending upon the venation. Each one is surrounded by a thin layer of parenchymatous cells called bundle sheath. Vascular bundles are closed. Xylem lies towards upper epidermis and phloem towards lower epidermis. Cambium is absent, hence there is no secondary growth in the leaf.
5. Lower epidermis: It consists of a single layer of compactly arranged rectangular, parenchymatous cells. A thin layer of cuticle is also present. The lower epidermis contains a large number of microscopic pores called stomata. There is an air-space called substomatal chamber at each stoma.

Anatomy of monocot leaf:

1.It is single layered, present on both sides of the leaf.
It consists of compactly arranged rectangular transparent parenchymatous cells.
Both the surfaces contain stomata.
Both the surfaces have a distinct layer of cuticle.
2. Mesophyll:
Mesophyll is not differentiated into palisade and spongy tissue.
3. Vascular bundle:
These are conjoint, collateral and closed.

Question 2.
Observe different slides related to anatomy of flowering plants under the guidance of teacher.
[Students are expected to perform this practical own their own.]

11th Biology Digest Chapter 8 Plant Tissues and Anatomy Intext Questions and Answers

Can you recall? (Textbook Page No. 85)

(i) Which component brings about important processes in the living organisms?
Answer:
Cell is the component that brings about important processes in the living organisms.

(ii) What is tissue?
Answer:
A group of cells having essentially a common function and origin is called as tissue.

(iii) Explain simple and complex tissue.
Answer:
a. Simple tissue:
1. They are made up of only one type of cells.
2. They are found in all the plant parts.
3. They perform many functions.
4. Simple tissues in plants are Parenchyma, Collenchyma, Sclerenchyma.

b. Complex tissue:
1. They are made up of many types of cells.
2. They are found only in the vascular regions of the plant.
3. They mainly perform the function of conduction of food and water.
4. Complex tissues in plants are Xylem and Phloem.

(iv) Complete the flow chart.
Organisms → Organs → Cells
Answer:
Organism → Organ system → Organs → Tissue system → Tissue → Cells

Can you tell? (Textbook Page No. 86)

Enlist the characteristics of meristematic tissue.
Answer:
Characteristics of meristematic tissue:

1. It is a group of young, immature cells.
2. These are living cells with ability to divide in the regions where they are present.
3. These are polyhedral or isodiametric in shape without intercellular spaces.
4. Cell wall is thin, elastic and mainly composed of cellulose.
5. Protoplasm is dense with distinct nucleus at the centre and vacuoles if present, are very small.
6. Cells show high rate of metabolism.

Can you tell? (Textbook Page No. 86)

Classify meristematic tissue on the basis of origin.
Answer:
Classification of meristematic tissue on the basis of origin:
1. Promeristem / Primordial meristem:
a. It is also called as embryonic meristem.
b. It usually occupies very minute area at the tip of root and shoot.

2. Primary meristem:
a. It originates from the primordial meristem and occurs in the plant body from the beginning, at the root and shoot apices.
b. Cells are always in active state of division and give rise to permanent tissues.

3. Secondary meristem:
a. These tissues develop from living permanent tissues during later stages of plant growth hence are called as secondary meristems.
b. This tissue occurs in the mature regions of root and shoot of many plants.
c. Secondary meristem is always lateral (to the central axis) in position e.g. Fascicular cambium, inter fascicular cambium, cork cambium.

Can you tell? (Textbook Page No. 89)

Write a note on parenchyma.
Answer:
Parenchyma:

1. It is a type of simple permanent tissue.
2. Cells in this tissue are thin walled, isodiametric, round, oval to polygonal or elongated in shape.
3. Cell wall is composed of cellulose.
4. Cells are living with prominent nucleus and cytoplasm with large vacuole.
5. Parenchyma has distinct intercellular spaces. Sometimes, cells may show compact arrangement.
6. The cytoplasm of adjacent cells is interconnected through plasmodesmata and thus forms a continuous tissue.
7. This is less specialized permanent tissue.
8. Occurrence:
   These cells are distributed in all the parts of a plant body viz. epidermis, cortex, pericycle, pith, mesophyll cells, endosperm, xylem and phloem.
9. Functions:
   These cells store food, water, help in gaseous exchange, increase buoyancy, perform photosynthesis and different functions in plant body.
10. Dedifferentiation in parenchyma cells develops vascular cambium and cork cambium at the time of
    secondary growth.

Can you tell? (Textbook Page No. 89)

Describe sclerenchyma fibres.
Answer:
Sclerenchyma fibres:
1. Fibres are thread-like, elongated and narrow structures with tapering and interlocking end walls.
2. Fibres are mostly in bundles. Pits are narrow, unbranched and oblique.
They provide mechanical strength.

Can you tell? (Textbook Page No. 89)

Sketch and label T.S. of phloem tissue.
Answer:
T.S. of phloem tissue: Structure of phloem:
1. Phloem is a living tissue. It is also called as bast.
2. It is responsible for conduction of organic food material from source (generally leaf) to a sink (other plant parts).
3. On the basis of origin, it can be protophloem (first formed) and metaphloem (latterly formed).
4. It is composed of sieve elements (sieve cells and sieve tubes), companion cells, phloem parenchyma and phloem fibres.

2. Sieve elements:
a. Sieve tubes are long tubular conducting channel of phloem.
b. These are placed end to end with bulging at end walls.
c. The sieve tube has sieve plate formed by septa with small pores.
d. The sieve plates connect protoplast of adjacent sieve tube cells.
e. The sieve tube cell is a living cell with a thin layer of cytoplasm, but loses its nucleus at maturity.
f. The sieve tube cell is connected to companion cell through phloem parenchyma by plasmodesmata.
g. Sieve cells are found in lower plants like pteridophytes and gymnosperms and sieve tubes are found in angiosperms.
h. The cells are narrow, elongated with tapering ends and sieve area located laterally.

3. Companion cells:
a. These are narrow elongated and living.
b. Companion cells are laterally associated with sieve tube elements.
c. Companion cells have dense cytoplasm and prominent nucleus.
d. Nucleus of companion cell regulates functions of sieve tube cells through simple pits.
e. From origin point of view, sieve tube cells and companion cell are derived from same cell. Death of the one result in death of the other type.

4. Phloem parenchyma:
a. Cells of phloem parenchyma are living, elongated found associated with sieve tube and companion cells.
b. Their chief function is to store food, latex, resins, mucilage, etc.
c. The cells carry out lateral conduction of food material.
d. These cells are absent in most of the monocots.

5. Phloem fibres (Bast fibres):
a. Phloem fibres are the only dead tissue among this unit.
b. They are sclerenchymatous.
c. They are generally absent in primary phloem, but present in secondary phloem.
d. These cells have with lignified walls and provide mechanical support.
e. They are used in making ropes and rough clothes.

Can you tell? (Textbook Page No. 92)

Concentric vascular bundles are always closed. Describe.
Answer:

1. When one vascular tissue is completely encircling the other, it is called as concentric vascular bundle.
2. When cambium is not present between xylem and phloem, it is known as closed vascular bundle.
3. Due to absence of cambium between xylem and phloem, concentric vascular bundles are always closed.

Can you tell? (Textbook Page No. 92)

How is the structure of vascular bundles of the root?
Answer:

1. Vascular bundles of the root are radial.
2. In radial vascular bundles, complex tissues are situated separately on separate radius as separate bundle.
3. The xylem and phloem bundles are arranged alternating with each other.

Can you tell? (Textbook Page No. 92)

Why vascular bundles of dicot stem are described as conjoint collateral and open?
Answer:
Vascular bundles of dicot stem are described as conjoint collateral and open because;
1. In dicot stem, the complex tissue is collectively present as neighbours of each other on the same radius in the form of xylem inside and phloem outside. Such type of vascular bundles are called as conjoint and collateral.
2. In dicot stem, a strip of cambium is present between xylem and phloem. Hence, it is called as open vascular bundle.

Can you tell? (Textbook Page No. 92)

How is the arrangement of vascular bundles in dicot and monocot stem?
Answer:
1. Vascular bundle in dicot stem: Vascular bundles are conjoint, collateral, open, and are arranged in a ring. Each one is composed of xylem, phloem and cambium. Xylem is endarch. A strip of cambium is present between xylem and phloem.
2. Vascular bundle in monocot stem: Vascular bundles are numerous and are scattered in ground tissue. Each vascular bundle is surrounded by a sclerenchymatous bundle sheath. Vascular bundles are conjoint, collateral and cloused (without cambium). Xylem is endarch and shows lysigenous cavity.

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