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Organisms and Populations Class 13 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 13 Organisms and Populations Textbook Exercise Questions and Answers.

Class 12 Biology Chapter 13 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 13 Exercise Solutions

1. Multiple choice questions

Question 1.
Which factor of an ecosystem includes plants, animals, and microorganisms?
(a) Biotic factor
(b) Abiotic factor
(c) Direct factor
(d) Indirect factor
Answer:
(a) Biotic factor

Question 2.
An assemblage of individuals of different species living in the same habitat and having functional interactions is ……………….
(a) Biotic community
(b) Ecological niche
(c) Population
(d) Ecosystem
Answer:
(a) Biotic community

Question 3.
Association between sea anemone and Hermit crab in gastropod shell is that of ………………..
(a) Mutualism
(b) Commensalism
(c) Parasitism
(d) Amensalism
Answer:
(b) Commensalism

Question 4.
Select the statement which explains best parasitism.
(a) One species is benefited.
(b) Both the species are benefited.
(c) One species is benefited, other is not affected.
(d) One species is benefited, other is harmed.
Answer:
(d) One species is benefited, other is harmed.

Question 5.
Growth of bacteria in a newly inoculated agar plate shows ………………….
(a) exponential growth
(b) logistic growth
(c) Verhulst-Pearl logistic growth
(d) zero growth
Answer:
(c) Verhulst-Pearl logistic growth

2. Very short answer questions.

Question 1.
Define the following terms
a. Commensalism
Answer:
The interaction between two species in which one species gets benefits and the other is neither harmed nor benefited is called commensalism.

b. Parasitism
Answer:
The interaction between two species in which one parasitic species derives benefit from the other host species by harming it is called parasitism.

c. Camouflage
Answer:
Camouflage is the disguising colouration or behaviour to merge with the surrounding so that prey or predator can remain hidden.

Question 2.
Give one example for each
a. Mutualism
b. Interspecific competition
Answer:
a. Lichen is composed of alga (cyanobacteria) and fungus. They cannot survive independently. Their association is mutualistic alga synthesises food by photosynthesis and fungus does the absorption of moisture.

b. Leopard and lion competing for a same prey. Sheep and cow competing for grazing in the same land.

Question 3.
Name the type of association:
a. Clown fish and sea anemone
b. Crow feeding the hatchling of Koel
c. Humming birds and host flowering plants
Answer:
a. Commensalism
b. Brood parasitism
c. Mutualism

Question 4.
What is the ecological process behind the biological control method of managing with pest insects?
Answer:

1. Pest insects act as prey to predator birds or frogs.
2. The biological control method consists of releasing the predators in the farms so that they can control the pest population in the natural way.
3. This also eliminates the use of chemical pesticides.
4. Frogs are natural predators of locust, therefore the population of this hazardous insect is controlled by frogs and the produce from agricultural farm can be saved.

Protocooperation:

1. Protocooperation is a type of population interaction where two species interact with each other.
2. Both are benefited but they have no need to interact with each other.
3. They can survive and grow even in the absence of other species.
4. Therefore this interaction is purely for the gain that they receive in such type of interaction.
5. The interaction that occurs can be between different kingdoms.

3. Short answer questions.

Question 1.
How is the dormancy of seeds different from hibernation in animals?
Answer:
In dormancy seed is not showing any metabolic activities. It can come back to life if and only if it gets suitable moisture and sunlight. Hibernation is suspended state, in which metabolic reactions do take place but at a very reduced pace. Animal arouses on its own after the winter sleep is over. This arousal is spontaneous and depends upon the ambient temperature. Dormant seed does not show such change unless it is planted or thrown in to moist place.

Question 2.
If a marine fish is placed in a fresh water aquarium, will it be able to survive? Give reason.
Answer:
Marine fish has its own osmoregulation which is different from the osmoregulation seen in fresh water fish. In marine water, the ambient salinity is more than the concentration of ions in the body. But in fresh water reverse is the case. Therefore, marine fish has different machinery to cope up with high saline environment. Therefore, it cannot survive in fresh water as its osmoregulation is not possible in less saline waters.

Question 3.
How is the dormancy of seeds different from hibernation in animals?
Answer:
In dormancy seed is not showing any metabolic activities. It can come back to life if and only if it gets suitable moisture and sunlight. Hibernation is suspended state, in which metabolic reactions do take place but at a very reduced pace. Animal arouses on its own after the winter sleep is over. This arousal is spontaneous and depends upon the ambient temperature. Dormant seed does not show such change unless it is planted or thrown into moist place.

Question 4.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?
Answer:

1. Orchid is an epiphyte. It gets the support from the mango tree. But it does not cause any harm to the mango tree.
2. Mango tree does not derive any benefit from this association. Therefore, this interaction is of type of commensalism.

Question 5.
Distinguish between the following:
a. Hibernation and Aestivation
Answer:

b. Ectotherms and Endotherms
Answer:

c. Parasitism and Mutualism
Answer:

Question 6.
Write a short note on
a. Adaptations of desert animals
Answer:

1. Animals which are well-adapted to live in deserts are called xerocoles. These animals show adaptations for water conservation or heat tolerance.
2. These animals show low basal metabolic rate. They obtain moisture from succulent plants and rarely drink water. E.g Gazella and Oryx.
3. Desert animals like camel produce concentrated urine and dry dung.
4. Many other hot desert animals are nocturnal, seeking out shade during the day or dwelling underground in burrows.
5. Smaller animals from desert, emerge from their burrows at night.
6. Mammals living in cold deserts have developed greater insulation through warmer body fur and insulating layers of fat beneath the skin.
7. Few adaptations to desert life are unable to cool themselves by sweating so they shelter during the heat of the day. Many desert reptiles are ambush predators and often bury themselves in the sand, waiting for prey to come within range.
8. Other animals have bodies designed to save water. Scorpions and wolf spiders have a thick outer covering which reduces moisture loss. The kidneys of desert animals concentrate urine, so that they excrete less water.

b. Adaptations of plants to water scarcity
Or
Adaptations in desert plants.
Answer:

1. Thick cuticle on their leaf surfaces
2. Stomata of desert plants is sunken that is it is in deep pits to minimize loss of water through transpiration.
3. Desert plants also have a special photosynthetic pathway (CAM -Crassulacean acid metabolism) that enables their stomata to remain closed during daytime.
4. Some desert plants like Opuntia, have their leaves reduced or they are modified to spines. Loss of leaf surface helps in prevention of transpiration.
5. Photosynthetic function is taken over by the flattened stems called as phylloclade.

c. Behavioural adaptations in animals
Answer:

1. Behavioural responses to cope with variations in their environment are shown by few animals.
2. Desert lizards manage to keep their body temperature fairly constant by behavioural adaptations. They bask in the sun and absorb heat, when their body temperature drops below the comfort zone, but move into shade, when the ambient temperature starts increasing. Even snakes also show basking during winter months.
3. Since they are ectothermic, this kind of behaviour saves them from extreme temperatures.
4. Many smaller animals show burrowing behaviour to adapt to the temperature extremes.
5. Some species burrow into the sand to hide and escape from the heat.
6. Migrations shown by the birds and mammals are also behavioural responses for adapting to severe winter temperatures.

Question 7.
Define Population and Community.
Answer:
Population:
Group of organisms belonging to same species that can potentially interbreed with each other and live together in a well-defined geographical area by sharing or competing for similar resources, is called population.

Community:
Several populations of different species in a particular area makes a community.

4. Long answer questions.

Question 1.
With the help of suitable diagram, describe the logistic population growth curve.
Answer:

1. Naturally all populations of any species always have limited resources to permit exponential growth. Due to this there is always competition between individuals for limited resources. The most fit organisms succeed by survival and reproduction.
2. A given habitat has enough resources to support a maximum possible number, but beyond a particular limit the further growth is impossible.
3. This limit is called nature’s carrying capacity (K) for that species in that habitat.
4. A population growing in a habitat with limited resources show following phases in a sequential manner, (a) A lag phase (b) Phase of acceleration (c) Phase of deceleration (d) An asymptote, when the population density reaches the carrying capacity.
5. A plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth.
6. Since resources for growth for most animal populations are finite and become limiting sooner or later, the logistic growth model is considered as a more realistic one.
7. Logistic growth thus always shows sigmoid curve.

Question 2.
Enlist and explain the important characteristics of a population.
Answer:
Important characteristics of a population are as follows:
1. Natality:

1. Natality is the birth rate of a population. Due to increased natality the population density rises.
2. Natality is a crude birth rate or specific birth rate.
3. Crude birth rate : Number of births per 1000 population/year gives crude birth rate. Crude birth rate is helpful in calculating population size.
4. Specific birth rate : Crude birth rate is relative to a specific criterion such as age. E.g. If in a pond, there were 200 carp fish and their population rises to 800. Then, taking the current population to 1000, the birth rate becomes 800/200 = 4 offspring per carp per year. This is specific birth rate.
5. Absolute Natality : The number of births under ideal conditions when there is no competition and the resources such as food and water are abundant, then it give absolute natality.
6. Realized Natality : The number of births under different environmental pressures give realized natality. Absolute natality will be always more than realized natality.

2. Mortality:

1. Mortality is the death rate of a population. It gives a measure of the number of deaths in a particular population, in proportion to the size of that population, per unit of time.
2. Mortality rate is typically expressed in deaths per 1,000 individuals per year.
   A mortality rate of 9.5 (out of 1,000) in a population of 1,000 would mean 9.5 deaths per year in that entire population or 0.95% out of the total.
3. Absolute Mortality : The number of deaths under ideal conditions when there is no competition, and all the resources such as food and water are abundant, then it gives absolute mortality.
4. Realized Mortality : The number of deaths under environmental pressures come into play gives realized mortality.
5. It must be remembered that absolute mortality will always be less than realized mortality.

3. Density:
The density of a population in a given habitat during a given period fluctuates due to changes in four basic processes, viz.

1. Natality i.e. birth rate (The number of births during a given period in the population that are added to the initial density).
2. Mortality i.e. death rate (The number of deaths in the population during a given period).
3. Immigration i.e. number of individuals of the same species that have come into the habitat from elsewhere during the time period under consideration.
4. Emigration i.e. the number of individuals of the population who left the habitat and gone elsewhere during the time period under consideration.
5. Natality and immigration increase in population density whereas mortality and emigration decrease it.

4. Sex ratio : Ratio of the number of individuals of one sex (male) to that of the other sex (female) is called sex ratio. In nature male, female ratio is always 1 : 1. This 1 : 1 ratio is called evolutionary stable strategy of ESS for each population.

5. Age distribution and age pyramid : This parameter is important for human population. Each population is composed of individuals of different ages. The age distribution is plotted for the population, the resulting structure is called an age pyramid. For making the age pyramid, the entire population is divided into three age groups as Pre-Reproductive (age 0-14 years), Reproductive (age 15-44 years) and Post-reproductive (age 45 -85+ years).

6. Growth : Growth of a population causes rise in its density. The size and density are dynamic parameters as they keep on changing with time, and various factors including food, predation pressure and adverse weather. From the density, one comes to know if the population is flourishing or declining.

Maharashtra State Board 12th Std Biology Textbook Solutions 

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• Enhancement of Food Production Class 12 Biology Textbook Solutions
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Std 7 Science Chapter 17 Effects of Light Question Answer Maharashtra Board

Balbharti Maharashtra State Board Class 7 Science Solutions Chapter 17 Effects of Light Notes, Textbook Exercise Important Questions and Answers.

Class 7 Science Chapter 17 Effects of Light Question Answer Maharashtra Board

1. Fill in the blanks.

Question a.
When the beams from the headlights of a car fall on an object in the night, the shadows – called ………. and ………… . can be seen.
Answer:
umbra, penumbra

Question b.
During a lunar eclipse the shadow of the …………. falls on the ………. .
Answer:
earth, moon

Question c.
During a solar eclipse the shadow of the ………. falls on the ………… .
Answer:
moon, earth

Question d.
Various shades of colour are seen in the sky at sunrise and sunset due to ……….. .
Answer:
scattering of light

2. Give reasons.

Question a.
Space beyond the earth’s atmosphere appears dark.
Answer:

1. Space beyond the earth’s atmosphere does have some gas and cosmic dust but there is not any atmosphere.
2. As there are no particles to scatter the sunlight, the space appears black.

Question b.
We are able to read while sitting in the shade.
Answer:

1. We are able to read because the sun light which falls on the book is scattered and reaches our eyes.
2. While sitting in the shade our eyes adjust to the environment and to amount of light available. That is how we are able to read.

Question c.
We should not observe the solar eclipse with naked eyes.
Answer:

1. During a solar eclipse ultra-violet rays which are harmful to us reach the earth and may lead to eclipse blindness or retinal bums and cataracts.
2. In order to protect our eyes a solar eclipse should never be watched with the naked eye.
3. A special type of goggles should be used for this.

3. Give some examples of scattering of light that we come across in day to day life.

Question a.
Give some examples of scattering of light that we come across in day to day life.
Answer:

1. The formation of rainbow, shift in position of stars, increased day time, mirage, inverted image, glittering of diamond, the working of lens and prism, bending of pen in water are examples of scattering of light.
2. The blue colour of the sky is due to the scattering of sunlight by the molecules of air.
3. During sunrise and sunset, sunlight has to travel greater distance, so shorter wavelength gets scattered off and removed and only orange and red with longer wavelengths reach us.
4. So during sunrise and sunset, sky appears fed and orange.

4. Why is the shadow of a bird flying high not seen on the earth?

Question a.
Why is the shadow of a bird flying high not seen on the earth?
Answer:

1. Birds flying high in the sky do cast their shadow but because they are shading an area that is very tiny the shadow is not visible.
2. The higher the bird flies, the smaller the shadow it casts.
3. Also when the bird flies high, the dark part of the shadow that is called Umbra does not reach the ground so we do not see its shadow.

5. Why is a penumbra not obtained from a point source?

Question a.
Why is a penumbra not obtained from a point source?
Answer:

1. The umbra, penumbra are the distinct parts of a shadow, created by any light source after striking on an opaque object.
2. For a point source, only the umbra is cast sharp dark shadow and not penumbra, because all the light of the point source will be blocked by any shadowing object.
3. Whereas penumbra forms only when some of the light from the source gets blocked by the shadowing object and not all of it does.

6. Answer the following questions in your own words.

Question a.
What is meant by scattering of light?
Answer:

1. Scattering of light is the deviation of light rays from its straight path.
2. As light propagates through the atmosphere, it travels in a straight path until it is obstructed by bits of dust or gas molecules in the atmosphere
3. The process in which light gets deflected by the particles in the medium through which the light passes is called scattering, e.g. The blue colour of the sky is due to the scattering of sunlight by the molecules of air.

Question b.
Does the shadow really vanish in the zero shadow condition?
Answer:
1. Yes, the day on which the sun reaches exactly overhead, at noon, shadow completely disappears.
‘This event can be seen in the region between the Tropic of Cancer (23.5°N) and Tropic of Capricorn (23.5°S).
2. The shadow diminishes and eventually disappears for a while only to reappear later.

Question c.
Will the laser beam be seen if it passes through a glass box which contains a lighted incense stick?
Answer:
Yes, it will be seen.

7. Discuss and write:

Question a.
Write a science based paragraph on ‘What if the sun did not rise’?
Answer:

1. The sun is a star and the centre of our solar system. Everything in our solar system revolves around the sun.
2. If the sun were to suddenly disappear, Earth and the other planets would retain their forward motion, effectively flying off into outer space in a straight line.
3. If the Sun didn’t rise means the Earth’s rotation had come to a screeching halt.
4. Sunrise and sunset are a result of the earth’s rotation so we will not get to see sunrise or sunset. Earth’s spinning generates the magnetic field at the core and it is saving us from harmful rays from the sun.
5. Without sun it would be very dark. No sun means no plants and no animals. Of course, without sun none of us would even exist.
6. Moon will disappear, because the moon dose not produce light. We only see the moon because sunlight is reflected by the moon.
7. Without the sun’s warmth, Earth would quickly become a much colder place. Life would be difficult, oceans will freeze.
8. Without sun rays, all photosynthesis on earth would stop. All plants would die.
9. All animals that rely on plants for food, including humans would die.

Question b.
What efforts will you make to remove the misconceptions about eclipses?
Answer:
Solar, lunar eclipse have been traditionally observed as an ominous sign and therefore superstitions are prevalent in association with these natural phenomena.
1. All these misconceptions should be removed by explaining scientifically the process of eclipse how it happens: (a) Explain with the help of diagram and models of sun, moon and earth, (b) Explain how special glasses which will protect us from UV rays, and excess heat can be used to observe eclipses safely.
2. Also inform that eclipses are natural phenomena and can be predicted in advance.

Question c.
Various eclipses and the conditions during that period.
Answer:
There are two eclipses:

1. Solar eclipse
2. Lunar eclipse

1. Solar eclipse:
There are two types of Solar eclipse, (a) Total solar eclipse (b) Partial solar eclipse

a. Total solar eclipse: In total solar eclipse, the moon is between the sun and the earth and the moon shadow covers the sun disc fully by perfect alignment. The part of the earth that lies in the umbra of the moon experiences total solar eclipse.

b. Partial solar eclipse: In partial solar eclipse, the moon is between the sun and the earth arid the moon shadow does not cover the sun disc fully, because of imperfect alignment. The part of the earth that lies in the penumbra of the moon experiences partial solar eclipse.

c. The solar eclipse occur on a new moon day.
d. Solar eclipse last for few minutes.

2. Lunar eclipse:
There are two types of Lunar eclipse, (a) Total lunar eclipse (b) Partial lunar eclipse

a. Total lunar eclipse: In total lunar eclipse, the earth comes in between the sun and the moon and the earth’s shadow covers the moon’s surface fully because of perfect alignment.

b. Partial lunar eclipse: In partial lunar eclipse, the earth’s shadow does not cover the moon’s surface fully because of inperfect alignment. A small part of the moon’s surface is covered by umbra part of earth’s shadow.

c. The lunar eclipse occurs on a full moon night.
d. Lunar eclipse last for few hours.

8. Explain the difference:

Question a.
Point sources and Extended sources.
Answer:

Question b.
Umbra and Penumbra.
Answer:

Project:

Question a.
Obtain information about the special goggles used to watch a solar eclipse.

Class 7 Science Chapter 17 Effects of Light Important Questions and Answers

Fill in the blanks.

Question 1.
During solar eclipse ………….. comes between the sun and earth.
Answer:
moon

Question 2.
A solar eclipse is seen only on a ………….. day.
Answer:
new moon day

Question 3.
During lunar eclipse ……………. comes between the sun and the moon.
Answer:
earth

Question 4.
A lunar eclipse is seen only on a ………….. night.
Answer:
full moon

Question 5.
The day on which the sun reaches exactly overhead is called the …………… .
Answer:
zero shadow day

Question 6.
As seen from the earth, when a planet or star passes behind the moon, that state is called a ………….. .
Answer:
occultation

Give scientific reasons:

Question 1.
Sky appears blue to us.
Answer:

1. Sunlight is scattered by the molecules of gases like nitrogen, oxygen in the atmosphere.
2. The blue colour in the sunlight which is at shorter wavelength is scattered the most than other colours and therefore the sky appears blue.

Question 2.
Solar eclipse is either partial or total.
Answer:

1. When the moon comes in between the sun and the earth and the solar disc is corripletely covered by the moon, it is called total solar eclipse.
2. When the solar disc is not covered fully by the moon, it is partial solar eclipse.

Explain the difference:

Question 1.
Solar eclipse and Lunar eclipse.
Answer:

Solar eclipse	Lunar eclipse
1. When the moon comes between the sun and the earth, a shadow of the moon is cast on the earth and sun cannot be seen from the part in shadow. This is called a solar eclipse.	1. When the earth comes between the sun and the moon a shadow of the earth is cast on the moon and a part of the moon in covered this is called the lunar eclipse.
2. A solar eclipse is seen only on a new moon day.	2. A lunar eclipse is seen only on a full moon night.
3. A solar eclipse should never be watched with the naked eye. because ultra violate rays which are harmful to us reach the earth	3. A lunar eclipse can be seen with the naked eye.
4. It can be seen for a few minutes only.	4. It can be seen over a period of a few hours.

Explain diagrams:

Question 1.
Explain Solar eclipse with diagram.
Answer:
Solar eclipse:

1. During its revolution, when the moon comes between the sun and the earth, a shadow of the moon is cast on the earth and the sun cannot be seen from the part in the shadow. This is called a solar eclipse
2. A solar eclipse is seen only on a new moon day.
3. The solar eclipse may be either partial or total,
4. Sometimes the solar disc is completely covered by the moon. This is the total solar eclipse.
5. When the solar disc is not covered fully by the moon, we have a partial solar eclipse.
6. During a solar eclipse, ultra-violet rays which are harmful to us reach the earth.
7. A solar eclipse should never be watched with the naked eye.
8. A special type of goggles should be used for this purpose.
9. Solar eclipse can be seen for a few minutes only.

Question 2.
Explain Lunar eclipse with diagram
Answer:
Lunar eclipse:

1. When the earth comes between the sun and the moon a shadow of the earth is cast on the moon and a part of the moon is covered. This is called the lunar eclipse.
2. A lunar eclipse is seen only on a full moon night. If the whole moon comes in the shadow of the earth, it is a total lunar eclipse.
3. When the shadow of the earth is cast only on a part of the moon, it is a partial lunar eclipse. You can watch a lunar eclipse with the naked eye.
4. A lunar eclipse can be seen over a period of a few hours.

Question 3.
Write a short note on Zero Shadow Day.
Answer:

1. The day on which the sun reaches exactly overhead is called zero shadow day.
2. On this day, at noon, shadow completely disappears.
3. This event can be seen in the region between the Tropic of Cancer (23.5°N) and at tropic of Capricon (23.5°S).
4. The shadow diminishes and eventually disappears for a while only to reappear later
5. This phenomenon occurs twice every year Mumbai got to witness it on May 14, 2018 last year.

Question 4.
Explain the phenomenon of scattering of light with the help of an experiment.
Answer:

1. When the sun rises our surroundings appear illuminated. The entire sky appears bright.
2. This happens because of the dust and other tiny particles in the air. This is the scattering of sunlight by the tiny particles of the various constituents of air.
3. Had there been no atmosphere, the sky would have appeared dark during the day and of course, the sun would be directly seen.
4. This has been verified by observations from the rockets and satellites which go out of the earth’s atmosphere.

Apparatus: A table lamp with a 60 or 100 W milky bulb (LED will not do), thick black paper, sticking tape, a packing needle, 100/200 ml. glass beaker, milk or milk powder, dropper, spoon, etc.

Procedure: Cover the mouth of the lampshade properly with black paper, using sticking tape. Prick a hole of 1 to 2mm diameter in the center of the paper with the help of the packing needle.

1. Take clear water in the beaker. Light the bulb and place the beaker in contact with the hole.
2. Observe from the front and at an angle of 90°.
3. Now add 2-3 drops of milk to the water and stir. Observe again.
4. A few more drops of milk may have to be added to make the water turbid.
5. A blue tinge is seen when observed along the 90° angle. This is the scattered blue light.
6. Because the blue light is scattered, a red-yellow light is seen from the front, and the hole appears reddish.

Question 5.
Short note on Shadow.
Answer:

1. Shadow is a dark patch formed behind an opaque object when it is placed in the path of light.
2. A shadow is formed only when a light source, an opaque object and a screen are present, e.g. during a lunar eclipse we see a part of the earth’s shadow on the surface of the moon.
3. This happens when the earth, the sun, and the moon are in a straight line with the earth between sun and the moon.
4. Here the sun acts as the light source, the earth as the opaque object and moon as the screen.
5. Shadows are formed due the rectilinear propagation of light.
6. The size and the shape of the shadow depend on the position and orientation of the opaque object between the source of light and the screen.
7. If the distance of the object from the source is decreased, then the size of the shadow increases.
8. If the object is moved away from the source, then the size of the shadow decreases.
9. In older days shadows caused by objects placed in the sun were used to measure time. Such a device is called sun dial.

Use your brain power!

Answer the following questions:

Question 1.
Eclipses and transits which will occur recently.
Answer:

• January 6, 2019 – Partial solar eclipse
• January 21, 2019 – Total Lunar eclipse
• July 2, 2019 – Total solar eclipse
• July 16, 2019 – Partial Lunar eclipse
• December 26, 2019 – Annular solar eclipse.

Find Out:

Answer the following questions:

Question 1.
If a few drops of milk are added in the experiment given the reddish colour seen from the front becomes an intense red. However, if many more drops are added the reddish colour is not seen. Why is this so?
Answer:

1. As more and more milk is added, more particles of protein and fat scatter the light and the blue colour is scattered more and more than orange and red light and the beam appears blue from the sides.
2. If few drops of milk are added, along with blue colour, orange and yellow also are scattered and only the intense red is seen from the front.
3. But when many more drops are added, even the red colour is scattered and we do not see any colour from the front.

Question 2.
What is Occultation?
Answer:
As seen from the earth, when a planet or a star passes behind the moon, that state is called an Occultation.

Complete the Chart.

Question 1.

Answer:

• A – Solar B – Lunar
• C – Total Solar
• D – Partial Solar
• E – Total Lunar
• F – Partial Lunar
• G – Moon is not in a straight line between earth and sun
• H – It is in the penumbra region of the moon
• I – The earth comes in between the sun and moon and they are not in a straight line.

Maharashtra State Board Class 7 Science Textbook Solutions

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Biotechnology Class 12 Exercise Question Answers Solutions Maharashtra Board

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

Class 12 Biology Chapter 12 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 12 Exercise Solutions

1. Multiple choice questions

Question 1.
MU The bacterium which causes a plant disease called crown gall is ………………..
(a) Helicobacter pylori
(b) Agrobacterium tumifaciens
(c) Thermophilus aquaticus
(d) Bacillus thuringienesis
Answer:
(b) Agrobacterium tumtfaciens

Question 2.
The enzyme nuclease hydrolyses ……………….. of polynucleotide chain of DNA.
(a) hydrogen bonds
(b) phosphodiester bonds
(c) glycosidic bonds
(d) peptide bonds
Answer:
(b) phosphodiester bonds

Question 3.
In vitro amplification of DNA or RNA segment is known as ………………..
(a) chromatography
(b) southern blotting
(c) polymerase chain reaction
(d) gel electrophoresis
Answer:
(c) polymerase chain reaction

Question 4.
Which of the following is the correct recognition sequence of restriction enzyme hind III.
(a) 5′ —A-A-G-C-T-T— 3′
3′ —T-T-C-G-A-A—5′
(b) 5′ — G-A-A-T-T-C—3′
3′ — C-T-T-A-A-G—5′
(c) 5′ — C-G-A-T-T-C—3′
3′ — G-C-T-A-A-G—5′
(d) 5′ — G-G-C-C—3′
3′ — C-C-G-G—5′
Answer:
(a) 5’ —A-A-G-C-T-T—3’
3’ —T-T-C-G-A-A—5’

Question 5.
Recombinant protein ……………….. is used to dissolve blood clots present in the body.
(a) insulin
(b) tissue plasminogen activator
(c) relaxin
(d) erythropoietin
Answer:
(b) tissue plasminogen activator

Question 6.
Recognition sequence of restriction enzymes are generally ……………….. nucleotide long.
(a) 2 to 4
(b) 4 to 8
(c) 8 to 10
(d) 14 to 18
Answer:
(b) 4 to 8

2. Very short answer questions

Question 1.
Name the vector which is used in production of human insulin through recombinant DNA technology.
Answer:
PBR 322

Question 2.
Which cells from Langerhans of pancreas do produce a peptide hormone insulin?
Answer:
cells of islets of Langerhans of a peptide hormone insulin.

Question 3.
Give the role of Ca⁺⁺ ions in the transfer of recombinant vector into bacterial host cell.
Answer:
Ca⁺⁺ ions promotes binding of plasmid DNA to lipo polysaccharides on bacterial cell surface. Then plasmid can enter the cell on heat shock.

Question 4.
Expand the following acronyms which are used in the held of biotechnology:

1. YAC
2. RE
3. dNTP
4. PCR
5. GMO
6. MAC
7. CCMB.

Answer:

1. YAC : Yeast Artificial chromosome
2. RE : Restriction Endonuclease
3. dNTP : Deoxyribonucleoside triphosphates
4. PCR : Polymerase Chain Reaction
5. GMO : Genetically Modified Organisms
6. MAC : Mammalian Artificial Chromosome
7. CCMB : Centre for Cellular and Molecular Biology

Question 5.
Fill in the blanks and complete the chart.

Answer:

3. Short answer type questions.

Question 1.
Explain the properties of a good or ideal cloning vector for r-DNA technology.
Answer:
Desired characteristics of ideal cloning vector are as follows:

1. Vector should be able to replicate independenly (through ori gene), so that as vector replicates, multiple copies of the DNA insert are also produced.
2. It should be able to easily transferred into host cells.
3. It should have suitable control elements like promoter, operator, ribosomal binding sites, etc.
4. It should have marker genes for antibiotic resistance and restriction enzyme recognition sites within them.

Question 2.
A PCR machine can rise temperature up to 100 °C but after that it is not able to lower the temperature below 70 °C automatically. Which step of PCR will be hampered first in this faulty machine? Explain why?
Answer:

1. If the faulty machine is not able to lower the temperature below 70 °C, then the primer annealing step will be hampered first.
2. Each primer has a specific annealing temperature, depending upon its A, T, G, C content.
3. For most of the primers annealing temperature is about 40-60 °C.
4. Hence, if temperature is more than primers annealing temperature, it will be able to pair with its complementary sequence in ssDNA.

Question 3.
In the process of r-DNA technology, if two separate restriction enzymes are used to cut vector and donor DNA then which problem will arise in the formation of r-DNA or chimeric DNA? Explain.
Answer:
In the process of r-DNA technology, if two separate restriction enzymes are used to cut vector and donor DNA, then it will result in fragments with different sticky ends which will not be complementary to each other.

Question 4.

Answer:

4. Long answer type questions.

Question 1.
(i) Define and explain the terms Bioethics.
Answer:

1. Bioethics is the study of moral vision, decision and policies of human behaviour in relation to biological phenomena or events.
2. Bioethics deals with wide range of reactions on new developments like cloning, transgenic, gene therapy, eugenics, r-DNA technology, in vitro fertilization, sperm bank, gene therapy, euthanasia, death, maintaining those who are in comatose state, prenatal genetic selection, etc.
3. Bioethics also includes the discussion on subjects like what should and should not be done in using recombinant DNA techniques.

Ethical aspects pertaining to the use of biotechnology are:

1. Use of animals cause great sufferings to them.
2. Violation of integration of species caused due to transgenosis.
3. Transfer of human genes into animals and vice versa.
4. Indiscriminate use of biotechnology pose risk to the environment, health and biodiversity.
5. The effects of GMO on non-target organisms, insect resistance crops, gene flow, the loss of diversity.
6. Modification process disrupting the natural process of biological entities.

(ii) Define and explain the term Biopiracy.
Answer:

1. Biopiracy is defined as ‘theft of various natural products and then selling them by getting patent without giving any benefits or compensation back to the host country’.
2. It is unauthorized misappropriation of any biological resource and traditional knowledge.
3. It is bio-patenting of bio-resource or traditional knowledge of another nation without proper permission of the concerned nation or unlawful exploitation and use of bioresources without giving compensation.

Following are the examples of biopiracy:
(a) Patenting of Neem (Azadirachta indica):

1. Pirating India’s traditional knowledge about the properties and uses of neem, the USDA and an American MNC W.R. Grace sought a patent from the European Patent Office (EPO) on the “method for controlling on plants by the aid of hydrophobic extracted neem oil,” in the early 90s.
2. The patenting of the fungicidal properties of Neem, was an example of biopiracy.

(b) Patenting of Basmati:

1. Texmati is a trade name of “Basmati rice line and grains” for which Texas based American company Rice Tec Inc was awarded a patent by the US Patent and Trademark Office (USPTO) in 1997.
2. This is a case of biopiracy as Basmati is a long-grained, aromatic variety of rice indigenous to the Indian subcontinent.
3. Very broad claims about “Inventing” the said rice was the basis of patent application.
4. The UPSTO has rejected all the claims due to people movement against Rice Tec in March 2001.

(c) Haldi (Turmeric) Biopiracy:

1. A patent claim about the healing properties of Haldi was made by two American researchers of Indian origin of the University of Mississippi Medical Center, to the US Patent and Trademark Office.
2. They were granted a patent in March 1995.
3. This is an example of biopiracy because healing properties of Haldi is not a new discovery, but it is a traditional knowledge in ayurvedas for centuries.
4. The Council of Scientific and Industrial Research (CSIR) applied to the US Patent Office for a reexamination and they realized the mistake and cancelled the patent.

(iii) Define and explain the term Biopatent.
Answer:

1. Biopatent is a biological patent awarded for strains of microorganisms, cell lines, genetically modified strains, DNA sequences, biotechnological processes, product processes, product and product applications.
2. It allows the patent holder to exclude others from making, using, selling or importing protected invention for a limited period of time.
3. Duration of biopatentis five years from the date of the grant or seven years from the date of filing the patent application, whichever is less.
4. Awarding biopatents provides encouragement to innovations and promote development of scientific culture in society. It also emphasizes the role of biology in shaping human society.
5. First biopatent was awarded for genetically engineered bacterium ‘Pseudomonas’ used for clearing oils spills.
6. Patent jointly issued by Delta and Pineland company and US department of agriculture having title ‘control of plant gene expression’, is based on a gene that produces a protein toxic to plant and thus prevents seed germination.

This patent was not granted by Indian government. Such a patent is considered morally unacceptable and fundamentally unequitable. Such patents would pose a threat to global food security as financially powerful corporations would acquire monopoly over biotechnological process.

Question 2.
Explain the steps in process of r-DNA technology with suitable diagrams.
Answer:

The steps involved in gene cloning are as follows:
(1) Isolation of DNA (gene) from the donor organism:

• To obtain the desired gene to be cloned, the cells of the donor organism are sheared with the blender and treated with suitable detergent. Genetic material is then isolated and purified.
• Isolated purified DNA is then cleaved using restriction Endonucleases.
• Restriction fragment containing desired gene is isolated and selected for cloning. This is now called foreign DNA or passanger DNA.
• A desired gene can also be obtained directly from genomic library or c-DNA library.

(2) Insertion of desired foreign gene into a cloning vector (vehicle DNA):

• The foreign DNA or passanger DNA is inserted into a cloning vector (vehicle DNA) like bacterial plasmids and the bacteriophages like lamda phage and M13. The most commonly used plasmid is pBR 322.
• Plasmids are isolated from the bacteria and are cleaved by using same RE which is used in the isolation of the desired gene from the donor.
• Enzyme DNA ligase is used to join foreign DNA and the plasmid DNA.
• Plasmid DNA containing foreign DNA is called recombinant DNA (r-DNA) or chimeric DNA.

(3) Transfer of r-DNA into suitable competent host or cloning organism:

• The r-DNA is introduced into a competent host cell, which is mostly a bacterium.
• Host cell takes up naked r-DNA by process of ‘transformation’ and incorporates it into its own chromosomal DNA which finally expresses the trait controlled by passenger DNA.
• The transfer of r-DNA into a bacterial cell is assisted by divalent Ca⁺⁺.
• The cloning organisms are E.coli and Agrobacterium tumifaciens.
• The competent host cells which have taken up r-DNA are called transformed cells.
• By using techniques like electroporation, microinjection, lipofection, shot gun, ultrasonification, biolistic method, etc. Foreign DNA can also be transferred directly into the naked cell or protoplast of the competent host cell, without using vector.
• In plant biotechnology the transformation is through Ti plasmids of A. tumifaciens.

(4) Selection of the transformed host cell:

• For isolation of recombinant cell from non-recombinant cell, marker gene of plasmid vector is employed.
• For example, pBR322 plasmid vector contains different marker genes like ampicillin resistant gene and tetracycline resistant gene. When pstl RE is used, it knocks out ampicillin resistant gene from the plasmid, so that the recombinant cells become sensitive to ampicillin.

(5) Multiplication of transformed host cell:

• The transformed host cells are introduced into fresh culture media where they divide.
• The recombinant DNA carried by them also multiplies.

(6) Expression of gene to obtain desired product. Then desired products like enzymes, antibiotiocs etc. separated and purified through down stream processing using bioreactors.

Question 3.
Explain the gene therapy. Give two types of it.
Answer:
Gene therapy is the treatment of genetic disorders by replacing, altering or supplementing a gene that is absent or abnormal and whose absence or abnormality is responsible for the disease.
Types of gene therapy:
(a) Germ line gene therapy:

1. In this germ cells are modified genetically to correct a genetic defect.
2. Normal gene is introduced into germ cells like sperms, eggs, early embryos.
3. It allows transmission of the modified genetic information to the next generation.
4. Although it is highly effective in treatment of the genetic disorders, its use is not preferred in human beings because of various technical and ethical reasons.

(b) Somatic cell gene therapy:

1. In this somatic cells are modified genetically to correct a genetic defect.
2. Healthy genes are introduced in somatic cells like bone marrow cells, hepatic cells, fibroblasts endothelium and pulmonary epithelial cells, central nervous system, endocrine cells and smooth muscle cells of blood vessel walls.
3. Modification of somatic cells only affects the person being treated and the modified chromosomes cannot be passed on the future generations.
4. Somatic cell gene therapy is the only feasible option and the clinical trials have already employed for the treatment of disorders like cancer, rheumatoid arthritis, SCID, Gaucher’s disease, familial hypercholesterolemia, haemophilia, phenylketonuria, cystic fibrosis, sickle-cell anaemia, Duchenne muscular dystrophy, emphysema, thalassemia, etc.

Question 4.
How are the transgenic mice used in cancer research?
Answer:

1. Transgenic mice are used in various research areas of cancer research.
2. Transgenic mice containing a particular oncogene (cancer causing gene) develop specific cancer.
3. They are used to study the relationship between oncogenes and cancer development, cancer treatment and prevention of malignancy.
4. The transgenic mouse model for the investigation of the breast cancer was developed in the laboratory of Philip Leder in Harvard (USA).
5. Transgenic mice containing oncogenes myc and ras were analyzed to find out role of these genes in the development of breast cancer.

Question 5.
Give the steps in PCR or polymerase chain reaction with suitable diagrams.
Answer:

(1) The DNA segment and excess of two primer molecules, four types of dNTPs, the thermostable DNA polymerase are mixed together in ‘eppendorf tube’.

(2) One PCR cycle is of 3-4 minutes duration and it involves following steps:

• Denaturation : The reaction mixture is heated at 90-98°C. Due to this hydrogen bonds in the DNA break and two strands of DNA separate. This is called denaturation.
• Annealing of primer : When the reaction mixture is cooled to 40-60°C, the primer pairs with its complementary sequences in ssDNA. This is called annealing.
• Extension of primer : In this step, the temperature is increased to 70-75°C. At this temperature thermostable Taq DNA polymerase adds nucleotides to 3’end of primer using single-stranded DNA as template. This is called primer extension. Duration of this step is about two minutes.

(3) In an automatic thermal cycler, the above three steps are automatically repeated 20-30 times.
(4) Thus, at the end of ‘n’ cycles 2n copies of DNA segments, get synthesized.

Question 6.
What is a vaccine? Give advantages of oral vaccines or edible vaccines.
Answer:

1. A vaccine is a biological preparation that provides active acquired immunity against a certain disease.
2. Vaccine is often made from a weakened or killed form of the microorganism, its toxins or one of its surface protein antigens.
3. Edible vaccine is an edible plant part engineered to produce an immunogenic protein, which when consumed gets recognized by immune system.
4. Immunogenic protein of certain pathogens are active when’administered orally.
5. When animals or mainly humans consume these plant parts, they get vaccinated against certain pathogen.
6. Oral or edible vaccines have low cost, they are easy to administer and store.

Question 7.
Enlist different types of restriction enzymes commonly used in r-DNA technology? Write on their role.
Answer:

1. Different restriction enzymes commonly used in r-DNA technology are Alu I, Bam HI, Eco RI, Hind II, Hind III, Pst I, Sal I, Taq I, Mbo II, Hpa I, Bgl I, Not I, Kpn I, etc.
2. They are the molecular scissors which recognize and cut the phosphodiester back bone of DNA on both strands, at highly specific sequences.
3. The sites recognized by them are called recognition sequences or recognition sites.
4. Different restriction enzymes found in different organisms recognize different nucleotide sequences and therefore cut DNA at different sites.
5. Restriction cutting may result in DNA fragments with blunt ends or cohesive or sticky ends or staggered ends (having short, single stranded projections).
6. Restriction endonucleases like Bam HI and EcoRI produce fragments with sticky ends.
7. Restriction endonucleases like Alu I, Hind III produce fragments with blunt ends.
8. Type I restriction endonucleases fuction simultaneously as endonuclease and methylase e.g. EcoK.
9. Type II restriction endonucleases have separate cleaving and methylation activities. They are more stable and are used in r-DNA technology e.g. EcoRI, Bgll. They cut DNA at specific sites within the palindrome.
10. Type III restriction endonucleases cut DNA at specific non-palindromic sequences e.g. Hpal, MboII.
11. In bacterial cells, REs destroy various viral DNAs that might enter the cell, thus restricting the potential growth of the virus.

Question 8.
Enlist and write in brief about the different biological tools required in r-DNA technology.
Answer:
The biological tools used in r-DNA technology are various enzymes, cloning vectors and competent hosts.
(1) Enzymes:

• Enzymes like lysozymes, nucleases (exonucleases and endonucleases), DNA ligase, reverse transcriptase, DNA polymerase, alkaline phosphatases, etc. are used in r-DNA technology.
• The restriction endonucleases are used as biological or molecular scissors. They are able to cut a DNA molecule at a specific recognition site.

(2) Vectors:

• Vectors are DNA molecules which carry foreign DNA segment and replicate inside the host cell.
• Vectors may be plasmids, bacteriophages (M13, lambda virus), cosmid, phagemids, BAC (bacterial artificial chromosome), YAC (yeast artificial chromosome), transposons, baculoviruses and mammalian artificial chromosomes (MACs).
• Most commonly used vectors are plasmid vectors (pBR 322, pUC, Ti plasmid) and bacteriophages (lamda phage, M13 phage).

(3) Competent host cells:

1. They are bacteria like Bacillus haemophilus, Helicobacter pyroliand E. coli.
2. Mostly E. coli is used for the transformation with recombinant DNA.

Maharashtra State Board 12th Std Biology Textbook Solutions 

• Control and Co-ordination Class 12 Biology Textbook Solutions
• Human Health and Diseases Class 12 Biology Textbook Solutions
• Enhancement of Food Production Class 12 Biology Textbook Solutions
• Biotechnology Class 12 Biology Textbook Solutions
• Organisms and Populations Class 12 Biology Textbook Solutions
• Ecosystems and Energy Flow Class 12 Biology Textbook Solutions
• Biodiversity, Conservation and Environmental Issues Class 12 Biology Textbook Solutions

Enhancement of Food Production Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 11 Enhancement of Food Production Textbook Exercise Questions and Answers.

Class 12 Biology Chapter 11 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 11 Exercise Solutions

1. Multiple choice questions

Question 1.
Antibiotic Chloromycetin is obtained from ………………….
(a) Streptomyces erythreus
(b) Penicillium chrysogenum
(c) Streptomyces venezuelae
(d) Streptomyces griseus
Answer:
(c) Streptomyces venezuelae

Question 2.
Removal of large pieces of floating debris, oily substances, etc. during sewage treatment is called ………………….
(a) primary treatment
(b) secondary treatment
(c) final treatment
(d) amplification
Answer:
(a) primary treatment

Question 3.
Which one of the following is free living bacterial biofertilizer?
(a) Azotobacter
(b) Rhizobium
(c) Nostoc
(d) Bacillus thuringiensis
Answer:
(a) Azotobacter

Question 4.
Most commonly used substrate for industrial production of beer is ………………….
(a) barley
(b) wheat
(c) corn
(d) sugar cane molasses
Answer:
(a) barley

Question 5.
Ethanol is commercially produced through a particular species of ………………….
(a) Aspergillus
(b) Saccharomyces
(c) Clostridium
(d) Trichoderma
Answer:
(b) Saccharomyces

Question 6.
One of the free-living anaerobic nitrogen- fixers is ………………….
(a) Azotobacter
(b) Beijerinckia
(c) Rhodospirillum
(d) Rhizobium
Answer:
(c) Rhodospirillum

Question 7.
Microorganisms also help in production of food like ………………….
(a) bread
(b) alcoholic beverages
(c) vegetables
(d) pulses
Answer:
(a) bread

Question 8.
MOET technique is used for ………………….
(a) production of hybrids
(b) inbreeding
(c) outbreeding
(d) outcrossing
Answer:
(a) production of hybrids

Question 9.
Mule is the outcome of ………………….
(a) inbreeding
(b) artificial insemination
(c) interspecific hybridization
(d) outbreeding
Answer:
(c) interspecific hybridization

2. Very Short Answer Questions

Question 1.
What makes idlis puffy?
Answer:
During preparation of idlis, rice and black gram flour is fermented by air borne Leuconostoc and Streptococcus bacteria. CO₂ produced during fermentation makes them puffy.

Question 2.
Bacterial biofertilizers.
Answer:
Rhizobium, Frankia, Pseudomonas striata, Bacillus polymyxa, Agrobacterium, Microccocus, Azotobacter, Costridium, Beijerinkia, Klebsiella.

Question 3.
What is the microbial source of vitamin B₁₂?
Answer:
The microbial source of vitamin B₁₂ is Pseudomonas denitrificans.

Question 4.
What is the microbial source of enzyme invertase?
Answer:
The microbial source of enzyme invertase is Saccharomyces cerevisiae.

Question 5.
Milk starts to coagulate when Lactic Acid Bacteria (LAB) are added to warm milk as a starter. Mention any two other benefits of LAB.
Answer:
Lactic Acid Bacteria (LAB) check the growth of disease causing microbes and produce vitamin B.

Question 6.
Name the enzyme produced by Streptococcus bacterium. Explain its importance in medical sciences.
Answer:
The enzyme produced by Streptococcus spp. is streptokinase. It is used as a ‘clot buster’ for clearing blood clots in the blood vessels in heart patients.

Question 7.
What is breed?
Answer:
Breed is a group of animals related by descent and similar in most characters like general appearance, features, size, configuration, etc.

Question 8.
Estuary
Answer:
Estuary is a place where river meets the sea.

Question 9.
What is shellac?
Answer:
Shellac is the pure form of lac obtained by washing and filtering.

3. Short Answer Questions.

Question 1.
Many microbes are used at home during preparation of food items. Comment on such useful ones with examples.
Answer:

1. Many food preparations made at home involves the use of microorganisms.
2. The microbes Lactobacilli are used in the preparation of dhokla from gram flour and buttermilk by the process of fermentation.
3. Dosa and idlis are prepared by using batter of rice and black gram which is fermented by air borne Leuconostoc and Streptococcus bacteria.
4. Large, fleshy fruiting bodies of some mushrooms and truffles are directly used as food. It is sugar free, fat free food rich in proteins, vitamins, minerals and amino acids. It is food with low calories.
5. Curd is prepared by inoculating milk with Lactobacillus acidophilus. Lactic acid produced during fermentation causes coagulation and partial digestion of milk protein casein and milk turns into curd.
6. Buttermilk is the acidulated liquid left after churning of butter from curd, is called buttermilk.

Question 2.
What is biogas? Write in brief about the production process.
Answer:
Biogas is a mixture of methane CH₄ (50-60%), CO₂ (30-40%), H₂S (0-3%) and other gases (CO, N₂, H₂) in traces.

Biogas production process:
a. A typical biogas plant consists of digester (made up of concrete bricks and cement or steel and is partly buried in the soil) and gas holder (a cylindrical gas tank to collect gases).
b. Raw materials like cow dung is mixed with water in equal proportion to make slurry which is fed into the digester’ through a side opening (charge pit).

Anaerobic digestion involves following processes:
i. Hydrolysis or solubilization:
Anaerobic hydrolyzing bacteria like Clostridium and Pseudomonas hydrolyse carbohydrates into simple sugars, proteins into amino acids and lipids into fatty acids.

ii. Acidogenesis:
Facultative and obligate anaerobic, acidogenic bacteria convert simple organic substances into acids like formic acid, acetic acid, H₂ and CO₂

iii. Methanogenesis:
Anaerobic methanogenic bacteria like Methanobacterium, Methanococcus convert acetate, H₂ and CO₂ into Methane, CO₂ and H₂O and other products.
12 mol CH₃COOH → 12CH₄ + 12CO₂ 4mol H.COOH → CH₄ + 3CO₂ + 2H₂O CO₂ + 4H₂ → CH₄ + 2H₂O

Question 3.
Biocontrol agents.
Answer:
(1) Biocontrol agents are the organisms like (bacteria, fungi, viruses and protozoans) act which are employed for controlling pathogens, pests and weeds.

(2) They cause the disease to the pest or compete or kill them.

(3) The use of biocontrol measures greatly reduces use of toxic chemicals and pesticides that are harmful to human beings and also pollute our environment.

(4) Biocontrol agents and their hosts.

• Bacteria (Bacillus thuringiensis, B. papilliae and B. lentimorbus Hosts : Caterpillars, cabbage worms, adult beetles
• Fungi (Beauveria bassiana, Entomophothora, pallidaroseum, Zoophthora radicans) Host : Aphid crocci, A. unguicilata, mealy bugs, mites, white flies, etc.
• Protozoans (Nosema locustae) Host: Grasshoppers, caterpillars, crickets
• Viruses (Nucleopolyhedro virus-NPV, Granulovirus-GV) Host : Caterpillars, Gypsy moth, ants and beetles.

(5) Some examples:

• Bacillus thuringiensis (Bt) is a microbiai pesticide used to get rid of butterfly, caterpillars.
• Trichoderma fungus is an effective biocontrol agent against soil borne fungal plant pathogens which infect roots and rhizomes.
• Phytophthora palmiuora is a mycoherbicide that controls milk weed in orchards.
• Pseudomonas spp. is a bacterial herbicide that attacks several weeds.
• Tyrea moth controls the weed Senecio jacobeac.

Question 4.
Name any two enzymes and antibiotics with their microbial source.
Answer:

1. Microbial source of Chloromycetin. – Streptomyces venezuelae
2. Microbial source of Erythromycin. – Streptomyces erythreus
3. Microbial source of Penicillin. – Penicillium chrysogenum
4. Microbial source of Streptomycin. – Streptomyces griseus
5. Microbial source of Griseofulvin. – Penicillium griseojulvum
6. Microbial source of Bacitracin. – Bacillus licheniformis
7. Microbial source of Oxytetracyclin / Terramycin. – Streptomyces aurifaciens
8. The enzyme produced by Streptococcus bacterium. – Streptokinase
9. Microbial source of Invertase. – Saccharomyces cerevisiae
10. Microbial source of Pectinase. – Sclerotinia libertine, Aspergillus niger
11. Microbial source of Lipase. – Candida lipolytica
12. Microbial source of Cellulase. – Trichoderma konigii

Question 5.
Write the principles of farm management.
Answer:
The principles of farm management are as follows:

1. Selection of high-yielding breeds.
2. Understanding the feed requirements of farm animals.
3. Supply of adequate nutritional sources for the animals.
4. Maintaining the cleanliness of environment.
5. Maintenance of health with the help of veterinary supervision.
6. Undertaking vaccination programmes.
7. Development of high-yielding cross-bred varieties.
8. Making various products and their preservation.
9. Distribution and marketing of the farm produce.

Question 6.
Give the economic importance of fisheries.
Answer:
Economic importance of fisheries is as follows:

1. Fish is a nutritious food and thus is a source of many vitamins, minerals and nutrients.
2. Commercial products such as fish oil, fish meal and fertilizers, fish guano, fish glue, isinglass are prepared from fish.
3. These by-products are used in paints, soaps, oils and medicines.
4. Some organisms like prawns and lobsters have high export value and market price.
5. Fish farming and other fishery trades provide job opportunity and self-employment
6. Productivity and national economy is improved through fishery practices.

Question 7.
Enlist the species of honey bee mentioning their specific uses.
Answer:
(1) The four species of honey bees commonly found in India : Apis dorsata (rock bee, or wild bee), Apis jlorea (little bee), Apis mellifera (European bee) and Apis indica (Indian bee).

(2) Uses:

• Rock bee : They produce 36 kg of honey per comb per year. They produce bee wax.
• Little bee : They produce half kg of honey per hive per year.
• European bee : The average production per colony per year is 25 to 40 kg.
• Indian bee : The average production per colony per year is 6 to 8 kg.

Question 8.
What are A, B, C, D in the table given below.

Answer:
A : Penicillium chrysogenum
B : Vinegar (Acetic acid)
C : Fungus
D : Sachharomyces cerevisiae var. ellipsoidis

4. Long Answer Questions.

Question 1.
Explain the process of sewage water treatment before it can be discharged into natural bodies. Why is this treatment essential?
Answer:
Sewage treatment includes following steps:
(1) Preliminary Treatment:

• Screening: The larger suspended or floating objects are filtered and removed in screening chambers by passing the sewage through screens or net in the chambers.
• Grit Chamber : Filtered sewage is passed into series of grit chambers which contain large stones (pebbles) and brick-ballast. Coarse particles which settle down by gravity are removed.

(2) Primary treatment (physical treatment):

• The sewage water is pumped into the primary sedimentation tank where 50-70% of the suspended solid or organic matter get sedimented and about 30-40% (in number) of coliform organisms are removed.
• The organic matter which is settled down is called primary sludge.
• Primary sludge is removed by mechanically operated devices.
• Dissolved organic matter and micro-organisms in the supernatant (effluent) are then removed by the secondary treatment.

(3) Secondary treatment (biological treatment):

• The primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it.
• The mesh like masses of aerobic bacteria, slime and fungal hyphae, known as floes are formed.
• Aerobic microbes consume most of the organic matter and this reduces BOD (Biochemical Oxygen Demand) of the effluent.

(4) Tertiary treatment:

• Once the BOD is sufficiently reduced, waste water is passed into a settling tank where the floes are allowed to sediment.
• The sediment is called activated sludge.
• Small part of activated sludge is transferred to aeration tank and the major part is pumped in to large anaerobic sludge digesters.
• In these tanks, anaerobic bacteria grow and digest the bacteria and fungi in the sludge and gases like methane, hydrogen sulphide, CO₂, etc. are released.
• Effluents from these digesters are released in natural water bodies like rivers and streams after chlorination which kills pathogenic bacteria.
• Digested sludge is then disposed.

Question 2.
Lac culture.
Answer:

1. Lac is a pink coloured resin secreted by dermal glands of female lac insect (Trachardia lacca) that hardens on coming in contact with air forming lac.
2. Lac is a complex substance having resin, sugar, water, minerals and alkaline substances.
3. Lac insect is colonial in habit and it feeds on succulent twigs like ber, peepal, palas, kusum, babool,
4. These plants are artificially inoculated in order to get better and regular supply of good quality and quantity of lac.
5. Natural lac is always contaminated and pure form of lac obtained by washing and filtering is called as shellac.
6. Lac is used to make bangles, toys, woodwork, inks, mirrors, etc.
7. India’s share is 85% of total lac produced in the world.

Question 3.
Describe various methods of fish preservation.
Answer:

1. Fish is a highly perishable commodity.
2. After catching the fish it immediately starts spoilage process.
3. In order to prevent this process, the fish preservation is done.

The different methods of fish preservation are as follows:

1. Chilling : This involves covering the fish with layers of ice. Ice is effective for short term preservation. It inhibits the activity of autolytic enzymes.
2. Freezing : It is a long duration preservation method. Fish are freezed at 0°C to -20°C. This also inhibit autolytic enzyme activities and slows down bacterial growth.
3. Freeze drying : The deep frozen -fish at -20°C are dried by direct sublimation of ice to water vapour with any melting into liquid water. This is achieved by exposing the frozen fish to 140°C in a vacuum chamber. The fish is then packed or canned in dried condition.
4. Sun drying : This inhibits the growth of microorganisms that spoil the fish.
5. Smoke drying : Smoke is prepared by burning woods with less resinous matter. Bacteria are destroyed by the acid content of the smoke. Smoking also give the characteristic colour, taste and odour to fish.
6. Salting : Salt removes the moisture from the fish tissues by osmosis. High salt concentration destroys autolytic enzymes and halts bacterial activity.
7. Canning : Canning involves sealing the food in a container, heat ‘sterilising’ the sealed unit and cooling it to ambient temperature for subsequent storage.

Question 4.
Give an account of poultry diseases.
Answer:
Various poultry diseases are as follows:

1. Viral diseases : Ranikhet, Bronchitis, Avian influenza (bird flu), etc. Bird flu had serious impact on poultry farming and also caused human infection.
2. Bacterial diseases : Pullorum, Cholera, Typhoid, TB, CRD (chronic respiratory disease), Enteritis, etc.
3. Fungal diseases : Aspergillosis, Favus and Thrush.
4. Parasitic diseases : Lice infection, round worm, caecal worm infections, etc.
5. Protozoan diseases : Coccidiosis.

Question 5.
Give an account of mutation breeding with examples.
Answer:

1. Mutations are sudden heritable changes in the genotype.
2. Natural mutations occur at a very slow rate.
3. Natural physical mutagens include exposure to high temperature, high concentration of C02, X-rays, UV rays.
4. Mutations can be induced by using various mutagens.
5. Mutagens cause gene mutations and chromosomal aberrations.
6. Chemical mutagens include nitrous acid, EMS (Ethyl – Methyl – Sulphonate), mustard gas, colchicine, etc.
7. Seedlings or seeds are irradiated by using CO60 or UV bulbs or X-ray machines.
8. The mutated seedlings are then screened for resistance to diseases/pests, high yield, etc.
9. Examples of mutant varieties in different crops are Jagannath (rice), NP 836 (rust resistant wheat variety), Indore-2 (cotton variety resistant to bollworm), Regina-II (cabbage variety resistant to bacterial rot).

Question 6.
Describe briefly various steps of plant breeding methods.
Answer:
The main steps of the plant breeding program (Hybridization) are as follows:

(1) Collection of variability:

• Germplasm collection is the entire collection of all the diverse alleles for all genes in a given crop.
• Wild species and relatives of the cultivated species having desired traits are collected and preserved.
• Forests and natural reserves are the means of in situ conservation of germplasm.
• Botanical gardens, seed banks, etc. are means of ex situ conservation of germplasm.

(2) Evaluation and selection of parents:

• The collected germplasm is evaluated to identify healthy and vigorous plants with desirable and complementary characters.
• Selected parents are selfed for three to four generations to increase homozygosity.
• Only pure lines are selected, multiplied and used in the hybridization.

(3) Hybridization:

• The variety showing maximum desirable features is selected as female (recurrent) parent and the other variety which lacks good characters found in recurrent parent is selected as male parent (donor).
• The pollen grains from anthers of male parent are artificially dusted over stigmas of emasculated flowers of female parent.
• Hybrid seeds are collected and sown to grow F₁ geneartion.

(4) Selection and Testing of Superior Recombinants:

• The F₁ hybrid plants which are superior to both the parents and having high hybrid vigour, are selected and selfed for few generations to make them homozygous for the said desirable characters.
• This ensures that there is no further segregation of the characters.

(5) Testing, release and commercialization of new cultivars:

• The newly selected lines are evaluated for the productivity and desirable features like disease resistance, pest resistance, quality, etc.
• They are initially grown under controlled conditions of water, fertilizers, etc. and their performance is recorded.
• The selected lines are then grown for at least three generations in natural field, in different agroclimatic zones.
• Finally variety is released as new variety for use by the farmers.

Maharashtra State Board 12th Std Biology Textbook Solutions 

• Control and Co-ordination Class 12 Biology Textbook Solutions
• Human Health and Diseases Class 12 Biology Textbook Solutions
• Enhancement of Food Production Class 12 Biology Textbook Solutions
• Biotechnology Class 12 Biology Textbook Solutions
• Organisms and Populations Class 12 Biology Textbook Solutions
• Ecosystems and Energy Flow Class 12 Biology Textbook Solutions
• Biodiversity, Conservation and Environmental Issues Class 12 Biology Textbook Solutions

Human Health and Diseases Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 10 Human Health and Diseases Textbook Exercise Questions and Answers.

Class 12 Biology Chapter 10 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 10 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Which of the following is NOT caused by unsterilized needles?
(a) Elephantiasis
(b) AIDS
(e) Malaria
(d) Hepatitis
Answer:
(a) Elephantiasis

Question 2.
Opium derivative is …………………
(a) Codeine
(b) Caffeine
(c) Heroin
(d) Psilocybin
Answer:
(c) Heroin

Question 3.
The stimulant present in tea is …………………
(a) tannin
(b) cocaine
(C) caffeine
(d) crack
Answer:
(c) caffeine

Question 4.
WhIch of the following Is caused by smoking?
(a) Liver cirrhosis
(b) Pulmonary tuberculosis
(c) Emphysema
(d) Malaria
Answer:
(c) Emphysema

Question 5.
An antibody is …………………
(a) molecuic that binds specifically an antigen
(b) WBC which invades bacteria
(c) secretion of mammalian RBC
(d) cellular component of blood
Answer:
(a) molecule that binds specifically an antigen

Question 6.
The antiviral proteins released by a virus-infected cell are called …………………
(a) histamines
(b) interferons
(c) pyrogens
(d) allergens
Answer:
(b) interferons

Question 7.
Both B-cells and T-cells are derived from …………………
(a) lymph nodes
(b) thymus glands
(c) liver
(d) stem cells in bone marrow
Answer:
(b) thymus glands

Question 8.
Which of the following diseases can be contracted by droplet infection?
(a) Malaria
(b) Chicken pox
(c) Pneumonia
(d) Rabies
Answer:
(c) Pneumonia

Question 9.
Confirmatory test used for detecting HIV infection is …………………
(a) ELISA
(b) Western blot
(c) Widal test
(d) Eastern blot
Answer:
(b) Western blot

Question 10.
Elephantiasis is caused by …………………
(a) W. barterofti
(b) P. vivax
(c) Bedbug
(d) Elephant
Answer:
(a) W. bancrofti

Question 11.
Innate immunity is provided by …………………
(a) phagocytes
(b) antibody
(c) T-lymphocytes
(d) B-lymphocytes
Answer:
(c) T-lymphocytes

2. Short Answer Questions

Question 1.
What is the source of cocaine?
Answer:
Source of cocaine is coca plant – Erythroxylum coca.

Question 2.
Name one disease caused by smoking.
Answer:
Emphysema. (Damaged and enlarged lungs causing breathlessness)

Question 3.
Which cells stimulate B-cells to form antibodies ?
Answer:
Helper T-cells stimulate B-cells to form antibodies.

Question 4.
What does the abbreviation AIDS stand for?
Answer:
AIDS stands for Acquired Immuno Deficiency Syndrome.

Question 5.
Name the causative agent of typhoid fever.
Answer:
Salmonella typhi

Question 6.
What is Rh factor?
Answer:
Antigen ‘D’ present on the surface of RBCs is known as Rh factor.

Question 7
What is schizont?
Answer:
Schizont is a ring-like form produced from merozoites inside the erythrocytes of human beings, infected by Plasmodium, which again forms new merozoites.

Question 8.
Name the addicting component found in tobacco.
Answer:
Nicotine

Question 9.
Name the pathogen causing Malaria.
Answer:
Plasmodium vivax

Question 10.
Name the vector of Filariasis.
Answer:
Female Culex mosquito

Question 11.
Name of the causative agent of ringworm.
Answer:
Trichophyton

Question 12.
Health
Answer:
Health is defined as the state of complete physical, mental and social well¬being and not merely the absence of disease or infirmity.

3. Short Answer Questions

Question 1.
What are acquired diseases?
Answer:
Diseases which are developed after the birth of an individual are called acquired diseases. These are of two types, viz. (a) Communicable or infectious diseases and (b) Non- communicable or Non-infectious diseases. Communicable or infectious diseases are transmitted from infected person to another healthy person either directly or indirectly. They are caused due to pathogens like viruses, bacteria, fungi, helminth worms, etc. Non-communicable or Non-infectious diseases cannot be transmitted from infected person to another healthy one either directly or indirectly.

Question 2.
Antigen and antibody.
Answer:

Question 3.
Name the infective stage of Plasmodium. Give Symptoms of malaria
Answer:
Sporozoite
I. Symptoms of malaria:

1. Fever accompanied by shivering.
2. Joint pain or arthralgia.
3. Vomiting.
4. Anaemia caused due to rupture of RBCs or haemolysis.
5. Haemoglobinuria.
6. Retinal damage.
7. Convulsions.
8. Cyclical occurrence of sudden coldness followed by rigor and then fever and sweating lasting for four to six hours. This is called a classic symptom of malaria.
9. Splenomegaly or enlarged spleen, severe headache, cerebral ischemia, hepatomegaly, i. e. enlarged liver, hypoglycaemia and haemoglobinuria with renal failure may occur in severe infections.

II. Spread / Transmission of malaria:

1. Malaria parasite is transmitted through the female Anopheles mosquito and hence it is known as mosquito-borne disease. Mosquito acts as a vector.
2. There are four species of Plasmodium, viz., P. vivax, P. falciparum, P. ovale and P. malariae which transmit malaria.

Question 4.
Explain the mode of infection and cause of elephantiasis.
Answer:
Mode of infection, i.e. transmission:

1. The parasite Wuchereria bancrofti is transmitted from a patient to other normal human being by female Culex mosquito.
2. The filarial larvae leave mosquito body and arrive on the human skin where they penetrate the skin and enter inside.
3. They undergo two moultings to become adults. Later they settle in the lymphatic system. They incubate for about 8-16 months.
4. When they settle in lymphatic system, this infection is called lymphatic filariasis.
5. The worms start infecting lymphatic circulation resulting into enlargement of lymph vessels and lymph nodes. The extremities like legs or limbs become swollen which resembles elephant legs. Therefore it is called elephantiasis.
6. This condition is lymphoedema, i.e. accumulation of lymph fluid in tissue causing swelling.

Question 5.
Why is smoking a bad habit?
Answer:

1. Smoking involves inhaling the cigarette smoke which contains nicotine and other toxic substances like N-nitrosodimethlene. There is some amount of carbon monoxide.
   All these substances affect the normal respiratory health.
2. Smoking invites problems like asthma, hypertension, heart disease, stroke, lung damage.
3. The worst impact is that these substances are carcinogenic and hence can cause cancer of larynx, trachea, lung, etc.
4. Smoking not only affects the smokers but also has bad effect on others due to passive smokers.
5. In women, smoking is still hazardous as their ovaries can undergo mutations due to mutagenic chemicals found in smoke.
6. Therefore, smoking is a very bad habit.

Question 6.
What do the abbreviations AMIS and CMIS denote?
Answer:
AMIS is Antibody-mediated immune system or humoral immunity and CMIS is cell- mediated immune system.

Question 7.
What is a carcinogen? Name one chemical carcinogen with its target tissue.
Answer:

1. Carcinogen is the substance or agent that causes cancer.
2. Urinary bladder cancer caused by 2-naphthylamine and 4-aminobiphenyl.

Question 8.
Active immunity and passive immunity.
Answer:

4. Short Answer Questions

Question 1.
B-cells and T-cells.
Answer:

Question 2.
What are the symptoms of malaria? How does malaria spread?
Answer:
Symptoms of malaria:

1. Fever accompanied by shivering.
2. Joint pain or arthralgia.
3. Vomiting.
4. Anaemia caused due to rupture of RBCs or haemolysis.
5. Haemoglobinuria.

Question 3.
AIDS.
Answer:
(1) AIDS or the acquired immuno deficiency syndrome, is fatal viral disease caused by a retrovirus (ss RNA) known as the human immuno deficiency virus (HIV) which weakens the body’s immune system. It is called a modern pandemic.

(2) The HIV attacks the immune system which in turn causes many opportunistic infections, neurological disorders and unusual malignancies ultimately leading to death.

(3) AIDS was first noticed in USA in 1981 whereas in India, first confirmed case of AIDS was in April 1986 from Tamil Nadu.

(4) HIV is transmitted through body fluids such as saliva, tears, nervous system tissue, spinal fluid, blood, semen, vaginal fluid and breast milk. However, only blood, semen, vaginal secretions and breast milk generally transmit infection to others.

(5) The transmission of HIV occurs by sexual contact, through blood and blood products and by contaminated syringes, needles, etc. There is also transplacental transmission or through breast milk at the time of nursing.

(6) Accidental needle injury, artificial insemination with infected donated semen and transplantation with infected organs are some of the rare occasions of transmission of HIV.

(7) HIV infection is not spread by casual contact such as hugging, bite of mosquitoes or using other objects touched by a patient.

(8) Acute HIV infection progresses over time to asymptomatic HIV infection and then to early symptomatic HIV infection. Later, it progresses to full blown AIDS when patient shows advanced HIV infection with CD4 T-cell count below 200 cells/mm.

Question 4.
Give the symptoms of cancer.
Answer:
Symptoms of cancer:

1. Presence of lump or tumour.
2. White patches in the mouth.
3. Change in a wart or mole on the skin.
4. Swollen or enlarged lymph nodes.
5. Vertigo, headaches or seizures if cancer affect the brain.
6. Coughing and shortness of breath if lungs are affected due to cancer.

Question 5.
Antigens on blood cells.
Answer:

1. There are about 30 known antigens on the surface of human red blood cells. They decide the type of blood group such as ABO, Rh, Duffy, Kidd, Lewis, P MNS, Bombay.
2. The different blood groups are determined genetically due to presence of a particular antigen.
3. Landsteiner found two antigens or agglutinogens on the surface of human red blood cells which are named as antigen A and antigen B.
4. There is another antigen called Antigen D which decides the Rh status of the blood. If Antigen D is present, the person is said to be RH positive and when it is lacking, the person is Rh negative.
5. These antigens are responsible for types of blood group and the specific transfusions.
6. Antigens present on the RBCs and antibodies present in the serum can cause agglutination reactions if they are non-compatible. Therefore, at the time of transfusion blood groups are checked properly.

Question 6.
Antigen-antibody complex:
Img 1
Answer:

1. Between antigen and antibody there is specificity.
2. Each antibody is specific for a particular antigen.
3. On the antigens there are combining sites which are called antigenic determinants or epitopes.
4. Epitopes react with the corresponding antigen binding sites of antibodies which are called paratopes.
5. The antigen binding sites are located on the variable regions of the antibody. Variable regions have small variations which make each antibody highly specific for a particular antigen.
6. Owing to variable region the antibody can recognize the specific antigen.
7. Antibody thus binds to specific antigen in a lock and key manner, forming an antigen- antibody complex.

Question 7.
What are the various public health measures, which you would suggest as safeguard against infectious diseases?
Answer:
Infectious diseases spread through pathogens, therefore, it is an important duty of each person to decrease the risk of infecting our own self or others. This can be achieved by

1. Washing hands often, especially whenever, we are in contact with food and water. Before and after preparing food, before eating and after using the toilet, hand wash is a must.
2. Vaccinations : Immunization helps us to protect against contracting many diseases. Therefore, timely vaccination should be taken. Especially at the time of epidemic, one must keep distance from infected area or get vaccinated.
3. One must be at home if there are signs and symptoms of an infection. By going out, we may infect other healthy persons.
4. Proper diet and exercise should be followed to improve one’s own immunity.
5. Hygiene should be utmost in the kitchen and dining area. One must take care while eating uncovered and leftover food.
6. Bathroom and toilet should be cleaned daily as there can be a high concentration of bacteria or other infectious agents in these areas.
7. One should have responsible sexual behaviour to avoid sexually transmitted diseases.
8. Personal items such as toothbrush, comb, towel, undergarments or razor blade should never be shared.
9. Travelling should be avoided because we may infect other passengers during travel. Moreover, our illness can be aggravated. Some special immunizations are needed during certain travels, such as anti-cholera vaccine while going to Pandharpur during Ashadhi.

Question 8.
How does the transmission of each of the following diseases take place?
(a) Amoebiasis:
Answer:
Amoebiasis is usually transmitted by the following ways:

1. The faecal-oral route.
2. Through contact with dirty hands or objects.
3. By anal-oral contact.
4. Through contaminated food and water.

(b) Malaria:
Answer:
Symptoms of malaria:

1. Fever accompanied by shivering.
2. Joint pain or arthralgia.
3. Vomiting.
4. Anaemia caused due to rupture of RBCs or haemolysis.
5. Haemoglobinuria.

(c) Ascariasis:
Answer:

1. Unsafe and unhygienic food and drinks contaminated with the eggs of Ascaris are the main mode of transmission.
2. Eggs hatch inside the intestine of the new host.
3. The larvae pass through various organs and settle as adults in the digestive system.

(d) Pneumonia:
Answer:

1. Pneumonia usually spreads by direct person to person contact.
2. It is also spread via droplet infection, i.e. droplets released by infected person.
3. Using clothes and utensils of the patient.

Question 9.
What measures would you take to prevent water-borne diseases?
Answer:

1. To prevent water-borne diseases, use of safe, clean and potable water is a must. Water should be filtered, then boiled and stored in covered container. If possible water purifier systems should be installed at home.
2. One should preferably use bottled water or carry our own water container while travelling.
3. Cleaning of water containers and maintaining personal hygiene near water storage is a must.
4. Megacities offer chlorinated and purified water for citizens. But villages and smaller rural set ups use river water which may be highly contaminated with pathogens. Such water should be purified before consumption to prevent water-borne diseases.

Question 10.
Typhoid.
Answer:
Typhoid is an infective disease caused by Gram-ve bacterium, Salmonella typhi.
(1) It is food and water-borne infection. In the intestinal lumen of infected person this bacteria is found.

(2) The bacterium has “O” – antigen, which is a lipopolysaccharide (LPS), present on surface coat and its flagella has “H” – antigen. Thus it becomes pathogenic.

(3) Signs and Symptoms of typhoid are as follows:
Prolonged and high fever with nausea, fatigue, headache.
Abdominal pain, constipation or diarrhoea. In severe cases rose-coloured rash is seen on skin. Tongue shows white coating and there is cough. Anorexia or loss of appetite is seen. In chronic cases there is breathlessness, irregular heartbeats and haemorrhage.

(4) Poor hygiene habits and poor sanitation and insects like houseflies and cockroaches spread typhoid.

(5) Typhoid is diagnosed by Widal test.

(6) Antibiotics like Chloromycetin can cure typhoid. Preventive vaccines such as oral Ty21a vaccine and injectable typhim vi and typherix against typhoid are also available. Chronic cases need surgical removal of gall bladder.

5. Match the following.

Answer:

6. Long Answer Questions

Question 1.
Describe the structure of antibody.
Answer:
Img 2

1. Antibodies are highly specific to specific antigens. They are glycoprotein called immunoglobulins (Igs.).
2. They are produced by plasma cells. Plasma cells are in turn formed by B-lymphocytes.
3. About 2000 molecules of antibodies are formed per second by the plasma cells.
4. Antibody is a ‘Y’-shaped molecule. It has four polypeptide chains, two heavy or H-chains and two light or L-chains.
5. Disulfide bonds (-s-s-) hold the polypeptide chains together to form a ‘Y’-shaped structure.
6. The region holding arms and stem of antibody is termed as hinge. Each chain of the antibody has two distinct regions, the variable region and the constant region.
7. Variable regions have a paratope which is an antigen-binding site. This part of antibody recognizes and binds to the specific antigen forming an antigen-antibody complex.
8. Antibodies are called bivalent as they carry two antigen binding sites.

Question 2.
Vaccination.
Answer:

1. Vaccines are prepared from inactivated pathogen, in the form of protein or sugar from pathogen or dead form of pathogen or toxoid from pathogens or attenuated pathogen.
2. These when they are administered to a person to protect against a particular pathogen, it is called vaccination.
3. Vaccination ’teaches’ the immune system to recognize and eliminate pathogenic organism. Because, already in the body the vaccine is injected and body has made antibodies in response to it. Thus, body is prepared before the attack, if at all it is exposed to pathogen.
4. Thus, it is an important form of primary prevention, which reduces the chances of illness by protecting people. It works by exposing the pathogen in a safe form.
5. Vaccinations control spread of diseases like measles, polio, tetanus and whooping cough that once threatened many lives.
6. Vaccination controls the epidemic outbreak of diseases, if all the people Eire pre-vaccinated.
7. Some hazardous diseases like small, pox and polio have been completely eradicated by the vaccination.

Question 3.
What is cancer? Differentiate between benign tumour and malignant tumour. The main five types of cancer
Answer:
I. Cancer : Cancer is a disease caused by uncontrolled cell division due to disturbed cell cycle.

II. Difference between benign tumour and malignant tumour:

III. The main five types of cancer:

Types of Cancer : According to the tissue affected, the cancers are classified into five main types. These are as follows:

1. Carcinoma : Cancer of epithelial tissue covering or lining the body organs is known as carcinoma. E.g. breast cancer, lung cancer, cancer of stomach, skin cancer, etc.
2. Sarcoma : Cancer of connective tissue is called sarcoma. Following are the types of sarcoma osteosarcoma (bone cancer), myosarcoma (muscle cancer),
   chondrosarcoma (cancer of cartilage) and liposarcoma (cancer of adipose tissue).
3. Lymphoma : Cancer of lymphatic tissue is called lymphoma. Lymphatic nodes, spleen and tissues of immune system are affected due to lymphoma.
4. Leukaemia : Leukaemia is blood cancer. In this condition, excessive formation of leucocytes take place in the bone marrow. There are millions of abnormal immature leucocytes which cannot fight infections. Monocytic leukaemia, lymphoblastic leukaemia, etc. are the types of leukaemia.
5. Adenocarcinoma : Cancer of glandular tissues such as thyroid, pituitary, adrenal, etc. is called adenocarcinoma.

Question 4.
Describe the different type of immunity.
Answer: There are two basic types of immunity, viz. innate immunity and acquired immunity.
(A) Innate immunity:

1. Innate immunity is natural, inborn immunity, which helps the body to fight against the invasion of microorganisms.
2. Innate immunity is non-specific because it does not depend on previous exposure to foreign substances.
3. Innate immunity mechanisms consist of various types of barriers such as anatomical barriers, physiological barriers, phagocytic barriers and inflammatory barriers. They prevent entry of foreign agents into the body.

(B) Acquired immunity:

1. The immunity that an individual acquires during his life is called acquired immunity or adaptive immunity or specific immunity. It helps the body to adapt by fighting against specific antigens hence it is called adaptive immunity. Since it is produced specifically against an antigen, it is called specific immunity.
2. Acquired immunity takes long time for its activation.
3. This type of immunity is seen only in vertebrates.
4. Due to acquired immunity, the body is able to defend against any invading foreign agent.

Question 5.
Describe the ill-effects of alcoholism on health.
Answer:

1. Alcohol in any form is toxic for the body. Hence as soon as alcohol is consumed, the liver tries to detoxify it.
2. In low doses it acts as a stimulant but in high dose, it acts on central nervous system, especially the cerebrum and cerebellum. Still higher dose can induce a comatose condition.
3. Alcohol affect the gastrointestinal tract by causing inflammation and damage to gastric 4 mucosa. Ulceration and painful condition arises in alcoholics.
4. Excessive doses of alcohol induce vomiting.
5. The worst effect of alcohol is on liver causing diseases like cirrhosis.
6. Alcohol induces hypertension and cardiac problems.
7. Apart from physical effect, it causes deterioration of mental health and emotional well-being.
8. Alcoholic person cannot think due to numbness in his/her cerebrum.
9. The social health is greatly affected as the alcoholic can cause problems to his family, friends and society in general.

Question 6.
In your view, what motivates the youngsters to take to alcohol or drugs and how can this be avoided?
Answer:
I. Taking drugs or alcohol:

1. Youngsters are at the vulnerable age, where they lack the planning about their future.
2. If they fall into bad company or are facing parental neglect, they get hooked on to alcohol or drugs.
3. Some common causes for addiction among youngsters are insufficient parental supervision and monitoring or excessive pressure and expectations from them. Lack of communication between an adolescent and parents.
4. Poorly defined rules for the family. Continuous family conflicts.
5. Favourable parental attitudes towards alcohol and drug uses. Many a times, at home children are exposed to such habits.
6. Inability to cope up with present and hence switching to the addictions. Risk taking behaviour which is common among youngsters.

II. Methods/measures to avoid drug abuse:

1. There should be complete acceptance for the child, because the adolescent phase is the most crucial phase when the children should be treated with love, care and respect.
2. Many physical, hormonal and psychological transformations are taking place in this phase. Therefore child suffers from stressful situation.
3. Wrong company and bad influence of peer group can trap the child in bad addictive habits. Thus, family should be supportive and communicative to help such children.
4. The sexual thoughts should be sublimed by channelizing energy into healthy pursuits like sports, reading, music, yoga and other extracurricular activities.
5. Ill-effects of drugs or alcohol should be told to youngsters.
6. Education and counselling can control the children from getting hooked on to the addictions.

Question 7.
Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?
Answer:
Friends can influence one to take alcohol and drugs, if a boy or girl is timid and non-communicative with his or her parents and teachers. It also depends on the personality of the indtvidual. In the adolescent age, many fall in trap due to such peer pressure. The confusion in the mind and role of hormones playing on the psyche and thought process makes one unable to understand the hazards of such habits. Also there is curiosity to do these experimentations due to bad influence of media.

If there is complete trust and friendship with sensible parents, then such influence does not work. One should protect himself or herself by a strong denial. Communicating such incidents to an elder in whom a boy or girl can confide, is very important. One should tell his or her friends about the ill-effects of alcohol and drugs. He should be made aware of these aspects that he or she has learnt in this lesson.

Maharashtra State Board 12th Std Biology Textbook Solutions 

• Control and Co-ordination Class 12 Biology Textbook Solutions
• Human Health and Diseases Class 12 Biology Textbook Solutions
• Enhancement of Food Production Class 12 Biology Textbook Solutions
• Biotechnology Class 12 Biology Textbook Solutions
• Organisms and Populations Class 12 Biology Textbook Solutions
• Ecosystems and Energy Flow Class 12 Biology Textbook Solutions
• Biodiversity, Conservation and Environmental Issues Class 12 Biology Textbook Solutions

Control and Co-ordination Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 9 Control and Co-ordination Textbook Exercise Questions and Answers.

Class 12 Biology Chapter 9 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 9 Exercise Solutions

1. Multiple choice questions

Question 1.
The nervous system of mammals uses both electrical and chemical means to send signals via neurons. Which part of the neuron receives impulse?
(a) Axon
(b) Dendron
(c) Nodes of Ranvier
(d) Neurilemma
Answer:
(b) Dendron

Question 2.
……………. is a neurotransmitter.
(a) ADH
(b) Acetyl CoA
(c) Acetyl choline
(d) Inositol
Answer:
(c) Acetyl choline

Question 3.
The supporting cells that produce myelin sheath in the PNS are …………….
(a) Oligodendrocytes
(b) Satellite cells
(c) Astrocytes
(d) Schwann cells
Answer:
(d) Schwann cells

Question 4.
A collection of neuron cell bodies located outside the CNS is called …………….
(a) tract
(b) nucleus
(c) nerve
(d) ganglion
Answer:
(d) ganglion

Question 5.
Receptors for protein hormones are located …………….
(a) in cytoplasm
(b) on cell surface
(c) in nucleus
(d) on Golgi complex
Answer:
(b) on cell surface

Question 6.
If parathyroid gland of man Eire removed, the specific result will be …………….
(a) onset of aging
(b) disturbance of Ca⁺⁺
(c) onset of myxoedema
(d) elevation of blood pressure
Answer:
(b) disturbance of Ca⁺⁺

Question 7.
Hormone thyroxine, adrenaline and non¬adrenaline are formed from ……………
(a) Glycine
(b) Arginine
(c) Ornithine
(d) Tyrosine
Answer:
(d) Tyrosine

Question 8.
Pheromones are chemical messengers produced by animals and released outside the body. The odour of these substance affects …………….
(a) skin colour
(b) excretion
(c) digestion
(d) behaviour
Answer:
(d) behaviour

Question 9.
Which one of the following is a set of discrete endocrine gland?
(a) Salivary glands, thyroid, adrenal, ovary
(b) Adrenal, testis, ovary, liver
(c) Pituitary, thyroid, adrenal, thymus
(d) Pituitary, pancreas, adrenal, thymus
Answer:
(c) Pituitary, thyroid, adrenal, thymus

Question 10.
After ovulation, Graafian follicle changes into …………….
(a) corpus luteum
(b) corpus albicans
(c) corpus spongiosum
(d) corpus callosum
Answer:
(a) corpus luteum

Question 11.
Which one of the following pairs correctly matches a hormone with a disease resulting from its deficiency?
(a) Parathyroid hormone – Diabetes insipidus
(b) Luteinising hormone – Diabetes mellitus
(c) Insulin – Hyperglycaemia
(d) Thyroxine – Tetany
Answer:
(c) Insulin – Hyperglycaemia

Question 12.
……………. is in direct contact of brain in humans.
(a) Cranium
(b) Dura mater
(c) Arachnoid
(d) Pia mater
Answer:
(d) Pia mater

2. Very very short answer questions.

Question 1.
What is the function of red nucleus?
Answer:
Red nucleus plays an important role in controlling posture and muscle tone, modifying some motor activities and motor coordination.

Question 2.
What is the importance of corpora quadrigemina?
Answer:
Corpora quadrigemina consists of 4 solid rounded structures, viz. superior and inferior colliculi. Superior colliculi control visual reflexes while inferior colliculi control auditory reflexes.

Question 3.
What does the cerebellum of brain control?
Answer:
Cerebellum of brain is an important centre which maintains equilibrium of body, posture, balancing orientation, moderation of voluntary movements and maintenance of muscle tone.

Question 4.
Name the three ear ossicles.
Answer:
Malleus [hammer], incus [anvil] and stapes [stirrup].

Question 5.
Name the anti abortion hormone.
Answer:
Progesterone.

Question 6.
Name an organ which acts as temporary endocrine gland.
Answer:
Placenta. Corpus luteum in ovary.

Question 7.
Name the type of hormones which bind to the DNA and alter the gene expression.
Answer:
Steroid hormones.

Question 8.
What is the cause of abnormal elongation of long bones of arms and legs and of lower jaw.
Answer:
Hypersecretion of growth hormones in adults causes abnormal elongation of long bones of arms and legs and of lower jaw i.e. acromegaly.

Question 9.
Name the hormone secreted by the pineal gland.
Answer:
Melatonin.

Question 10.
Which endocrine gland plays important, role in improving immunity?
Answer:
The endocrine gland, thymus plays an important role in improving immunity.

3. Match the organism with the type of nervous system found in them.

Answer:

4. Very short answer questions.

Question 1.
Describe the endocrine role of islets of Langerhans.
OR
Islets of Langerhans.
Answer:
Endocrine cells of pancreas form groups of cells called Islets of Langerhans. There are four kinds of cells in islets of Langerhans which secrete hormones.

1. Alpha (α) cells : They are 20% and secrete glucagon. Glucagon is a hyperglycemic hormone. It stimulates liver for glucogenolysis and increases the blood glucose level.
2. Beta (β) cells : They are 70% and secrete insulin. Insulin is a hypoglycemic hormone. It stimulates liver and muscles for glycogenesis. This lowers blood glucose level.
3. Delta (δ) cells : They are 5% and secrete somatostatin. Somatostatin inhibits the secretion of glucagon and insulin. It also decreases the gastric secretions, motility and absorption in digestive tract. In general it is a growth inhibiting factor.
4. PP cells or F cells : They form 5%. They secrete pancreatic polypeptide (PP) which inhibits the release of pancreatic juice.

Question 2.
Mention the function of testosterone?
Answer:
Testosterone is a steroid sex hormone secreted by testes and cortex of adrenal glands. It controls the secondary sexual characters in males.

Question 3.
Give symptoms of the disease caused by hyposecretion of ADH.
Answer:
Polydipsia, i.e. frequent thirst and polyuria, i.e. frequent urination are the symptoms of the disease caused by hyposecretion of ADH.

5. Short answer questions

Question 1.
Rakesh got hurt on his head when he fell down from his motorbike. Which inner membranes must have protected his brain? What other roles do they have to play
Answer:

1. When Rakesh fell down from his motorbike, the inner membranes that protected his brain were meninges, viz. dura mater, arachnoid membrane and pia mater. Morevover, CSF must have also acted as a shock absorber.
2. Dura mater : It is the outer tough membrane protective in function.
3. Arachnoid membrane : It is the middle web-like membrane which communicates with fluids of upper sub dural space and lower sub arachnoid space.
4. Pia mater : It is the innermost highly vascularised nutritive membrane in close contact with brain and spinal cord.

Question 2.
Injury to medulla oblongata may prove fatal.
OR
Injury to medulla oblongata causes sudden death. Explain.
Answer:

1. Medulla oblongata is the region of the brain that controls all the involuntary activities.
2. Vital activities such as heartbeats, respiration, vasomotor activities, peristalsis, etc. are under the control of medulla oblongata.
3. When medulla oblongata is injured, all these vital functions are instantly stopped.
4. Therefore, injury to medulla oblongata causes sudden death.

Question 3.
Distinguish between the sympathetic and parasympathetic nervous system on the basis of the effect they have on:
Heartbeat andUrinary Bladder.
Answer:

Question 4.
While holding a tea cup Mr. Kothari’s hands rattle. Which disorder he may be suffering from and what is the reason for this?
Answer:

1. This condition is due to Parkinson’s disease.
2. It is due to degeneration of dopamine- producing neurons in the CNS.
3. 80% of the patients develop this condition along with stiffness, difficulty in walking, balance and coordination.

Question 5.
List the properties of the nerve fibres.
Answer:

1. Excitability / irritability
2. Conductivity
3. Stimulus
4. Summation
5. All or none
6. Refractory period
7. Synaptic delay
8. Synaptic fatigue
9. Velocity.

Question 6.
How does tongue detect the sensation of taste?
Answer:

1. The surface of tongue is with gustatoreceptors.
2. These receptors are sensitive to the chemicals [sweet, salt, sour, bitter and umami (savory)] present in the food.
3. The receptor cells get stimulated, generate the impulse which is given to the sensory neuron.

Question 7.
State the site of production and function of Secretin, Gastrin and Cholecystokinin.
Answer:

Question 8.
An adult patient suffers from low heart rate, low metabolic rate and low body temperature. He also lacks alertness, intelligence and initiative. What can be this disease? What can be its cause and cure ?
Answer:

1. The above symptoms indicate that the person is suffering from Myxoedema.
2. Myxoedema is condition caused due to hypothyroidism.
3. Hypothyroidism causes deficiency of thyroid hormones like T₃ and T₄ (thyroxine). This results in low BMR.
4. This condition can be cured by giving injections of thyroxine or tablets containing hormone preparation.

Question 9.
Where is the pituitary gland located? Enlist the hormones secreted by anterior pituitary.
Answer:
The pituitary gland is attached to hypothalamus on the ventral surface of brain. It is lodged in a bony depression called sella turcica of sphenoid bone.
For names of hormones:

1. GH : [Growth Hormone/STH : Somatotropic Hormone]
2. TSH/TTH – [Thyroid Stimulating Hormone/ Thyrotropic Hormone]
3. ACTH – [adrenocorticotropic hormone]
4. PRL – [prolactin]

Gonadotropins-

1. FSH [follicle stimulating hormone]
2. LH/ICSH – [leutinizing hormone/ interstitial cells stimulating hormone]

Question 10.
Explain how the adrenal medulla and sympathetic nervous system function as a closely integrated system.
Answer:

1. Adrenal medulla originates from embryonic neuro – ectoderm.
2. It consists of rounded group of large granular cells called chromaffin cells. They are modified post-ganglionic cells of sympathetic nervous system which have lost normal processes and acquired glandular function.
3. These cells are connected with pre-ganglionic fibres of sympathetic nervous system.
4. Hence adrenal medulla is an extension of sympathetic nervous system.
5. Thus adrenal medulla and sympathetic nervous system functions as a closely integrated system.

Question 11.
Name the secretion of alpha, beta and delta cells of islets of Langerhans. Explain their role.
Answer:

Question 12.
Which are the two types of goitre? What are their causes?
Answer:
(1) Goitre is the enlargement of thyroid gland. It is easily visible at the base of neck when a person is suffering from it.

(2) Goitre is of two types.

1. Simple goitre : It is also called endemic goitre. This is due to iodine deficiency in the food. This causes iodine deficit in blood. In an attempt to take more iodine from blood, the blood supply to the gland increases. This results in swelling on the thyroid.
2. Exophthalmic goitre : It is also called toxic goitre. This is due to hyperactive thyroid gland. This can happen if there is overstimulation of thyroid due to excess of ACTH. This disorder is also called Grave’s disease or hyperthyroidism.

Question 13.
Name the ovarian hormone and give their functions.
Answer:

6. Answer the following.

Complete the table.

Answer:

7. Long answer questions.

Question 1.
Explain the process of conduction of nerve impulses up to development of action potential.
Answer:

1. The origin and maintenance of resting potential depends on the original state of no stimulation.
2. Any stimulus or disturbance to the membrane will make the membrane permeable to Na⁺ ions. This causes rapid influx of Na⁺ ions.
3. The voltage gated Na⁺/K⁺ channels are unique. They can change the potential difference of the membrane as per the stimulus received and also the gates operate separately and are self closing.
4. During resting potential, both gates are closed and resting potential is maintained.
5. However during depolarization, the Na⁺ channels open but not the K⁺ channels. This causes Na⁺ to rush into the axon and bring about a depolarisation. This condition is called action potential.
6. Extra cellular fluid (ECF) becomes electronegative with respect to the inner membrane which becomes electropositive.

Question 2.
Draw the neat labelled diagrams.
a. Human ear.
Answer:

b. Sectional view of human eye.
Answer:

c. Draw the neat labelled diagram of sagittal section or L.S. of human brain
Answer:

d. Draw the neat labelled diagram of Multipolar Neuron.
Answer:

Question 3.
Answer the questions after observing the diagram given below.

a. What do the synaptic vesicles contain?
Answer:
Synaptic vesicles contain a neurotransmitter – acetyl choline.

b. What process is used to release the neurotransmitter ?
Answer:
Exocytosis.

c. What should be the reason for the next impulse to be conducted?
Answer:
Removal of neurotransmitter by the action of acetyl cholinesterase.

d. Will the impulse be carried by post synaptic membrane even if one pre-synaptic neuron is there?
Answer:
As far as impulse is transmitted by pre-synaptic neuron, it will be received by post-synaptic neuron.

e. Can you name the channel responsible for their transmission?
Answer:
Ca⁺⁺ channel

Question 4.
Explain the Reflex Pathway with the help of a neat labelled diagram.
OR
With the help of a neat and labelled diagram, describe reflex arc.
Answer:

I. Reflex action : Reflex action Is defined as a quick, automatic involuntary and often unconscious action brought about when the receptors are stimulated by external or internal stimuli.

II. Reflex arc : Reflex actions are controlled by CNS. Reflex arc is the structural or functional unit of reflex action. Simple reflex arc is formed of the following five components.
(1) Receptor organ : The sensory part that receives the stimulus is called receptor organ. It can be any sense organ that receives the stimulus and converts it into the impulse, e.g. skin, eye, ear, tongue, nasal epithelium, etc.

(2) Sensory neuron or afferent neuron:
Sensory part carrying impulse from receptor organ to CNS is called sensory neuron. Its cyton is located in dorsal root ganglion. Its dendron is long and connected to receptor while the axon enters in the grey matter of spinal cord to form a synapse.

(3) Association, adjustor or intermediate neuron : It is present in the grey matter of spinal cord. Receiving impulse from sensory neuron, interpreting it and generating motor impulse are done by association neuron.
(4) Motor neuron (effector) : The cyton of motor neuron is present in the ventral horn of grey matter and axon travels through ventral root. It conducts motor impulse from spinal cord to effector organ.

(5) Effector organ : Effector organ is a specialized part of the body which is excited by receiving the motor impulse. It gives proper response to the stimulus, e.g. muscles or glands. The path of reflex action is followed by the unidirectional impulse. It originates in the receptor organ and ends in effector organ through CNS.

Question 5.
Krishna was going to school and on the way he saw a major bus accident. His heartbeat increased and hands and feet become cold. Name the part of the nervous system that had a role to play in this reaction.
Answer:

1. The symptoms observed in Krishna were due to sympathetic nervous system. Emergency conditions trigger sympathetic nervous system to stimulate adrenal medulla.
2. The cells of adrenal medulla secrete catecholamines like adrenaline and nor¬adrenaline.
3. These hormones have direct effect on the pacemaker of the heart which causes increase in the heart rate and other associated symptoms.
4. This is a typical fright reaction caused by intervention of sympathetic nervous system.

Question 6.
What will be the effect of thyroid gland atrophy on the human body?
Answer:

1. Atrophy means degeneration. Atrophy of thyroid gland will result in deficient secretion of thyroid hormones leading to hypothyroidism. Deficiency of thyroid hormones [T₃ and T₄] and thyrocalcitonin will cause following effects on the body.
2. Decrease in BMR i.e. basal metabolic rate, decrease in the blood pressure, heart beat, body temperature, etc.
3. Occurrence of myxoedema in which there is abnormal deposition of fats under the skin giving puffy appearance in adults.
4. Irregularities in menstrual cycle in case of female patients.
5. Hair become brittle and fall.
6. Calcium metabolism also disturbs due to lack of thyrocalcitonin.

Question 7.
Write the names of hormones and the glands secreting them for the regulation of following functions
(a) Growth of thyroid and secretion of thyroxine.
Answer:
TSH by adenohypophysis.

(b) Helps in relaxing pubic ligaments to facilitate easy birth of young ones.
Answer:
Relaxin by degenerating corpus luteum of the ovary.

(c) Stimulate intestinal glands to secrete intestinal juice.
Answer:
Secretin by duodenal mucosa.

(d) Controls calcium level in the blood.
Answer:
Calcitonin [hypocalcemic hormone] by thyroid and parathormone [ hypercalcemic hormone] by parathyroid glands.

(e) Controls tubular absorption of water in kidneys.
Answer:
ADH by hypothalamus.

(f) Urinary elimination of water.
Answer:
Atrial natriuretic factor by atria of heart.

(g) Sodium and potassium ion metabolism.
Answer:
Aldosterone by adrenal cortex.

(h) Basal Metabolic rate.
Answer:
T₃ and T₄ by thyroid gland.

(I) Uterine contraction.
Answer:
Oxytocin by hypothalamus.

(j) Heartbeat and blood pressure.
Answer:
Adrenaline, non-adrenaline [stimulation] and acetylcholine [inhibition] by adrenal medulla.

(k) Secretion of growth hormone.
Answer:
GHRF by hypothalamus.

(l) Maturation of Graafian follicle.
Answer:
FSH by anterior pituitary.

Question 8.
Explain the role of hypothalamus and pituitary as a coordinated unit in maintaining homeostasis.
Answer:

1. Homeostasis is maintenance of constant internal environment of the body.
2. When certain hormones from any endocrine glands are secreted in excess quantity, the : inhibiting factors from hypothalamus, automatically exert negative feedback and stop the production of stimulating hormones from pituitary.
3. Similarly, if any hormone is in deficit, then j the concerned gland is given message through releasing factor. This way the hormone production remains in a balanced state or homeostasis.
4. E.g. If thyroxine from thyroid gland is secreted in excess, the secretion of TSH from pituitary is stopped by stopping the production of TRF from hypothalamus.
5. Though most of the endocrine glands are under the influence of pituitary gland, it is in turn controlled by hypothalamus.
6. Hypothalamus secretes releasing factors and inhibiting factors and hence regulate the secretions of pituitary (hypophysis).
7. There is negative feedback mechanism in controlling the secretions of the endocrine glands.
8. Hypothalamus forms the hypothalamo- hypophyseal axis through which transportation of neurohormones take place.

Following are the releasing and inhibiting factors produced by hypothalamus:

1. Somatotropin/GHRF : It stimulates release of growth hormone.
2. Somatostatin/GHRIF : It inhibits the release of growth hormone.
3. Adrenocorticotropin Releasing Hormone / CRF : It stimulates the release of ACTH by the anterior pituitary gland.
4. Thyrotropin Releasing Hormone /TRF : It stimulates the release of TSH by anterior pituitary gland.
5. Gonadotropin Releasing Hormone (GnRH) : It stimulates pituitary to secrete gonadotropins.
6. Prolactin Inhibiting Hormone (Prolactostatin) : It inhibits prolactin released by anterior pituitary gland.
7. Gastrin Releasing Peptide (GRP).
8. Gastric Inhibitory Polypeptide (GIP).

Question 9.
What is adenohypophysis ? Name the hormones secreted by it.
Answer:

1. Adenohypophysis is the large anterior lobe of pituitary gland.
2. It is derived from embryonic ectoderm in the form of Rathke’s pouch which is a small outgrowth from the roof of embryonic stomodaeum.
3. It is made up of epitheloid secretory cells.

It secretes following hormones:

1. GH : [Growth Hormone/STH : Somatotropic Hormone]
2. TSH/TTH – [Thyroid Stimulating Hormone/ Thyrotropic Hormone]
3. ACTH – [adrenocorticotropic hormone]
4. PRL – [prolactin]

Gonadotropins-

1. FSH [follicle stimulating hormone]
2. LH/ICSH – [leutinizing hormone/ interstitial cells stimulating hormone]

Question 10.
Describe, in brief, an account of disorders of adrenal gland.
Answer:
(1) Disorders of adrenal cortical secretions are caused due to hyposecretion and hypersecretion of adrenal corcoid hormones.

(2) Hyposecretion of corticosteroids causes Addison’s disease.

(3) The symptoms of Addison’s disease are low blood sugar, low body temperature, feeble heart action, low BR acidosis, low Na⁺ and K⁺ concentration in plasma, excessive loss of Na⁺ and water in urine, impaired kidney functioning and kidney failure, etc. it leads to weight loss, general weakness, nausea, vomiting and diarrhoea.

(4) Hypersecretion of corticoids causes Cushing’s disease.

(5) The symptoms of Cushing’s disease are high blood sugar level, glucosuria, alkalosis, enhancement of total quantity of electrolytes in extracellular fluid, polydipsia, increased BR muscle paralysis, obesity, wasting of limb muscles, etc.

Question 11.
Explain action of steroid hormones and proteinous hormones.
OR
Explain the mode of action of steroid hormones.
Answer:
The hormones always act on their target organs or tissues to induce their effects. The target tissues have specific binding sites or receptor sites which contain hormone receptors.
I. Steroid hormones:

1. The steroid hormones are lipid soluble and can easily cross the lipoproteinous plasma membrane.
2. The hormone receptors for steroid hormones are present in cytoplasm or in nucleus.
3. Hormone-receptor complex formed in cytoplasm enters the nucleus and regulate the gene expression or chromosome function.
4. In some cases the receptors are present inside the nucleus where hormone receptor complex is formed.
5. These complexes interact with the genome to evoke biochemical changes that result in physiological and developmental functions.

II. Protein hormones:

1. The hormone receptors for protein hormones are present on the cell membrane (i.e. membrane bound receptors).
2. When the hormone binds to its receptor, it forms hormone-receptor complex. Each receptor is specific to a specific hormone.
3. The hormones which interact with membrane bound receptors normally do not enter the target cell but generate second messengers. Such as cyclic AMP Ca⁺⁺ or IP (Inositol triphosphate), etc.
4. This leads to certain biochemical changes : in the target tissue.
5. Thus, the tissue metabolism and consequently the physiological functions are regulated by hormones.

Question 12.
Describe in brief an account of disorders of the thyroid.
OR
What are the functional disorders of thyroid gland? Describe in brief.
Answer:
Disorders of thyroid gland are of three types, viz. hypothyroidism, hyperthyroidism and simple goitre.
(1) Hypothyroidism : Hypothyroidism is deficient secretion of thyroxine. This hyposecretion causes two types of disorders, viz. cretinism in children and myoxedema in adults.
(i) Cretinism : Hyposecretion of thyroxine in childhood causes cretinism. The symptoms of cretinism are retardation of physical and mental growth.

(ii) Myxoedema : Deficiency of thyroxine in adults causes this disorder. It is also referred to as Gull’s disease. Symptoms are thickening and puffiness of the skin and subcutaneous tissue particularly of face and extremities. Patients with low BMR. It also causes mental dullness, loss of memory, slow action.

(2) Hyperthyroidism : Excessive secretion of thyroxine causes exophthalmic goitre or Grave’s disease. There is slight enlargement of thyroid gland. It increases BMR, heart rate, pulse rate and BE Reduction in body weight due to rapid oxidation, nervousness, irritability. Peculiar symptom is exophthalmos, i.e. bulging of eyeballs with staring look and less blinking. This is caused by deposition of fats behind the eye balls in eye sockets. There is muscular weakness and loss of weight.

(3) Simple goitre (Iodine deficiency goitre) : Simple goitre occurs due to deficiency of iodine in diet or drinking water. Simple goitre causes enlargement of thyroid gland. Thyroid gland in an attempt to get more iodine from the blood, swells due to increased blood supply. Prevention of goitre can be done by administering iodized table salt. It is also called endemic goitre as it is common in hilly areas.

Maharashtra State Board 12th Std Biology Textbook Solutions 

• Control and Co-ordination Class 12 Biology Textbook Solutions
• Human Health and Diseases Class 12 Biology Textbook Solutions
• Enhancement of Food Production Class 12 Biology Textbook Solutions
• Biotechnology Class 12 Biology Textbook Solutions
• Organisms and Populations Class 12 Biology Textbook Solutions
• Ecosystems and Energy Flow Class 12 Biology Textbook Solutions
• Biodiversity, Conservation and Environmental Issues Class 12 Biology Textbook Solutions

Plant Growth and Mineral Nutrition Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 7 Plant Growth and Mineral Nutrition Textbook Exercise Questions and Answers.

Class 12 Biology Chapter 7 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 7 Exercise Solutions

1. Multiple choice questions

Question 1.
Which of the hormone can replace vernalization?
(a) Auxin
(b) Cytokinin
(c) Gibberellins
(d) Ethylene
Answer:
(c) Gibberellins

Question 2.
The principle pathway of water translocation in angiosperms is ………………..
(a) Sieve cells
(b) Sieve tube elements
(c) Xylem
(d) Xylem and phloem
Answer:
(c) Xylem

Question 3.
Abscissic acid controls ………………..
(a) cell division
(b) leaf fall and dormancy
(c) shoot elongation
(d) cell elongation and wall formation
Answer:
(b) leaf fall and dormancy

Question 4.
Which is employed for artificial ripening of banana fruits?
(a) Auxin
(b) Ethylene
(c) Cytokinin
(d) Gibberellin
Answer:
(b) Ethylene

Question 5.
Which of the following is required for stimulation of flowering in plants?
(a) Adequate oxygen
(b) Definite photoperiod
(c) Adequate water
(d) Water and minerals
Answer:
(b) Definite photoperiod

Question 6.
For short day plants, the critical period is ………………..
(a) light
(b) dark/night
(c) UV rays
(d) Both (a) and (c)
Answer:
(b) dark/night

Question 7.
Which of the following is NOT day neutral plant?
(a) Tomato
(b) Cotton
(c) Sunflower
(d) Soybean
Answer:
(d) Soybean

Question 8.
Essential macro elements are ………………..
(a) manufactured during photosynthesis
(b) produced by enzymes
(c) obtained from soil
(d) produced by growth hormones
Answer:
(c) obtained from soil

Question 9.
Function of Zinc is ………………..
(a) closing of stomata
(b) biosynthesis of 3-IAA
(c) synthesis of chlorophyll
(d) oxidation of carbohydrates
Answer:
(b) biosynthesis of 3-LAA

Question 10.
Necrosis means ………………..
(a) yellow spot on the leaves
(b) death of tissue
(c) darkening of green colour in leaves
(d) wilting of leaves
Answer:
(b) death of tissue

Question 11.
Conversion of nitrates to nitrogen is called ………………..
(a) ammonification
(b) nitrification
(c) nitrogen fixation
(d) denitrification
Answer:
(d) denitrification

Question 12.
How many molecules of ATP are required to fix one molecule of nitrogen?
(a) 12
(b) 20
(c) 6
(d) 16
Answer:
(d) 16

2. Very short answer questions

Question 1.
Enlist the phases of growth in plants.
Answer:
The three phases of growth are phase of cell division, phase of cell enlargement and phase of cell maturation.

Question 2.
Give full form of IAA.
Answer:
Full form is Indole Acetic Acid.

Question 3.
What does it mean by ‘open growth’?
Answer:
In plants the growth is indeterminate and takes place throughout the life at specific regions having meristems.

Question 4.
Plant stress hormone.
Answer:
Abscissic acid.

Question 5.
What is denitrification?
Answer:
Anaerobic bacteria can convert nitrates of soil back into nitrogen gas. That process performed by denitrifying bacteria is denitrification.

Question 6.
Bacteria responsible for conversion of nitrite to nitrate.
Answer:
Nitrobacter.

Question 7.
What is the role of gibberellins in rosette plants?
Answer:
In rosette plants like beet and cabbage, bolting, i.e. elongation of internodes before flowering is observed due to effect of gibberellins.

Question 8.
Vernalization
Answer:
The response of plant to the influence of low temperature on flowering in plants is called vernalization.

Question 9.
Photoperiodism
Answer:
The response of plant to the influence of light for initiation of flowering is known as photoperiodism.

Question 10.
What is grand period of growth?
Answer:
There are three phases of growth and the total time required for all phases to occur is called grand period of growth.

3. Short Answer Questions

Question 1.
(i) Differentiation
Answer:

1. It is a process of maturation of cells derived from apical meristems.
2. Differentiation is a permanent change in structure and function of cells that leads to its maturation.
3. Cell undergoes major anatomical and physiological change during differentiation process.
4. In hydrophytic plants parenchyma cells develop large schizogenous cavities which help them in aeration, buoyancy and mechanical support.

(ii) Redifferentiation
Answer:

1. It is a process in which cells produced by de-differentiation lose their capacity of division and become mature.
2. The cells mature to perform specific function.
3. Interfascicular cambium is formed by process of dedifferentiation loses its capacity to divide.
4. Secondary xylem and secondary phloem is formed form this cambium in vascular cylinder.

Question 2.
Arithmetic growth and Geometric growth
Answer:

Question 3.
Enlist the role and deficiency symptoms of: (a) nitrogen (b) phosphorus (c) potassium.
Answer:
(a) Nitrogen:
Role : Constituent of proteins as amino acids, nucleic acids, vitamins, hormones, coenzymes, ATP and chlorophyll molecule.
Deficiency symptoms : stunted growth and chlorosis.

(b) Phosphorus:
Role : Constituent of cell membrane, certain proteins, nucleic acids and nucleotides, required for all phosphorylation reactions.
Deficiency symptoms : Poor growth, leaves dull green

(c) Potassium :
Role : Determination of anion – cation balance in cell, necessary for protein synthesis, involved in formation of cell membrane, opening and closing of stomata, activates enzymes, helps in maintenance of turgidity of cells.
Deficiency symptom : Yellow edges in leaves, premature death.

Question 4.
What is short day plant? Give any two examples.
Answer:
The plants which flower when the day length or light period is shorter than the critical photoperiod are called short day plants or SDP
SDPs usually flower during winter and late summer.
Examples – Dahlia, Aster, Tobacco, Chrysanthemum, Soybean (Glycine max) and Cocklebur (Xanthium).

Question 5.
What is vernalization? Give its significance.
Answer:
A low temperature or chilling treatment that induces early flowering in plants is known as vernalization.

Significance:

1. Due to chilling treatment crops can be produced earlier.
2. Crops can be grown in areas where they do not grow naturally.

4. Long answer questions

Question 1.
Explain sigmoid growth curve with the help of diagram.
Answer:

1. When growth occurs in plants three distinct phases of growth are noticed.
2. Phase of cell formation is first phase where meristematic cells divide and new cells added.
3. In phase of cell enlargement newly formed cells elongate and with turgidity there is cell enlargement.
4. In phase of cell maturation cells get differentiated.
5. When we compare the growth rate it differs in these three phases.
6. In first phase or lag phase it is slow, while in log phase or exponential phase, growth rate accelerates and it reaches maximum.
7. In stationary phase of maturation growth rate slows down and comes to steady state.
8. When this changing rate of growth is plotted against time duration in a graph a sigmoid or S-shaped growth curve is obtained.

Question 2.
Describe the types of plants on the basis of photoperiod required, with the help of suitable examples.
Answer:

1. Effect of light duration on flowering of plants is known as photoperiodism.
2. Depending on photoperiodic response, plants are categorised into three types – Short day plants, long day plants and day neutral plants.

1. Short day plants : Plants that flower under short day length conditions are called short day plants. Plants such as Dahlia, Xanthium, Soybean, Aster, Tobacco and Chrysanthemum are short day plants or SDR. Short day plants require a long uninterrupted dark period for flowering. Therefore, they are also called long night plants.

2. Long day plants : Plants that flower only when they are exposed to light period longer than their critical photoperiod are called long day plants or LDP Long day plants require a short dark or night period for flowering. Hence, they are also called short night plants. Plants such as radish, spinach, wheat, poppy, cabbage, pea, sugar beet, etc. are long day plants.

3. Day neutral plants : Plants in which the flowering is not affected by the day length period are called day neutral plants or DNP or photoneutral plants. Plants such as cucumber, sunflower, cotton, balsam, maize, tomato, etc. are day neutral plants.

Question 3.
Explain biological nitrogen fixation with example.
Answer:

1. Conversion of atmospheric nitrogen into nitrogenous salts to make it available to plants for its update is described as nitrogen fixation.
2. When living organisms are involved in nitrogen fixation process it is known as biological nitrogen fixation.
3. The process is mainly carried out by prokaryotic organisms, i.e. different kinds of bacteria present in soil.
4. The nitrogen fixing organisms are known as diazotrophs or nitrogen fixers and about 70% nitrogen is fixed by them.
5. The nitrogen fixers are either free living bacteria or symbiotic associated with other higher plants e.g. Rhizobium.
6. The cyanobacteria have specialized cells heterocysts which help in process of nitrogen fixation.
7. Nitrogen fixation is high energy requiring process and 16 ATP molecules are needed for fixation of one molecule of nitrogen to ammonia.
8. Soil bacteria like Nitrosomonas, Nitrosocyccus convert ammonia to nitrate and the Nitrobacter convert nitrite to nitrate. This is known as nitrification, biological oxidation.
9. These bacteria are chemoautotrophic and utilize these processes for their metabolism.
10. Fabaceae plants like pea, bean have root nodules which harbour symbiotic bacterium Rhizobium which fixes nitrogen. It is host specific, soil bacterium, Nitrogen is made available to host plant.

Question 4.
Write on macro and micro nutrients required for plant growth.
Answer:

1. Plants absorb mineral nutrients from their surroundings.
2. For a proper growth of plants about 35 to 40 different elements are required.
3. Plants absorb these nutrients in ionic or dissolved form from soil with their root system e.g. Phosphorus as PO₄^–, Sulphur as SO₄^2- etc.
4. Based on their requirement in quantity, they are classified as major nutrients or macronutrients and those needed in small amounts Eire minor or micronutrients.
5. Macroelements are required in large amounts, as they play nutritive and structural roles e.g. C, H, O, R Mg, N, K, S and Ca. – Ca pectate cell wall component, Mg component of chlorophyll.
6. C, H, O are non-mineral major elements obtained from air and water e.g. CO₂ is source of carbon, Hydrogen from water.
7. Microelements are required in traces as they mainly have catalytic role as co-factors or activators of enzymes.
8. Microelements may be needed for certain activity in life cycle of plant e.g. B for pollen germination, Si has protective role during stress conditions and fungal attacks, Al enhances availability of phosphorus.
9. The important micronutrients for plant growth are Mn, B, Cu, Zn, Cl.

Maharashtra State Board 12th Std Biology Textbook Solutions

• Reproduction in Lower and Higher Plants Class 12 Biology Textbook Solutions
• Reproduction in Lower and Higher Animals Class 12 Biology Textbook Solutions
• Inheritance and Variation Class 12 Biology Textbook Solutions
• Molecular Basis of Inheritance Class 12 Biology Textbook Solutions
• Origin and Evolution of Life Class 12 Biology Textbook Solutions
• Plant Water Relation Class 12 Biology Textbook Solutions
• Plant Growth and Mineral Nutrition Class 12 Biology Textbook Solutions
• Respiration and Circulation Class 12 Biology Textbook Solutions

Plant Water Relation Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 6 Plant Water Relation Textbook Exercise Questions and Answers.

Class 12 Biology Chapter 6 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 6 Exercise Solutions

1. Multiple Choice Questions

Question 1.
In soil, water available for absorption by root is ……………..
(a) gravitational water
(b) capillary water
(c) hygroscopic water
(d) combined water
Answer:
(b) capillary water

Question 2.
The most widely accepted theory for ascent of sap is ……………..
(a) capillarity theory
(b) root pressure theory
(c) diffusion
(d) transpiration pull theory
Answer:
(d) transpiration pull theory

Question 3.
Water movement between the cells is due to ……………..
(a) T.E
(b) W.P
(c) D.P.D.
(d) incipient plasmolysis
Answer:
(c) D.P.D.

Question 4.
In guard cells, when sugar is converted into starch, the stomata pore ……………..
(a) closes almost completely
(b) opens partially
(c) opens fully
(d) remains unchanged
Answer:
(a) closes almost completely

Question 5.
Surface tension is due to ……………..
(a) diffusion
(b) osmosis
(c) gravitational force
(d) cohesion
Answer:
(d) cohesion

Question 6.
Which of the following type of solution has lower level of solutes than the solution?
(a) Isotonic
(b) Hypotonic
(c) Hypertonic
(d) Anisotonic
Answer:
(b) Hypotonie

Question 7.
During rainy season wooden doors warp and become difficult to open or to close because of ……………..
(a) plasmolysis
(b) imbibition
(c) osmosis
(d) diffusion
Answer:
(b) imbibition

Question 8.
Water absorption takes place through ……………..
(a) lateral root
(b) root cap
(c) root hair
(d) primary root
Answer:
(c) root hair

Question 9.
Due to low atmospheric pressure the rate of transpiration will ……………..
(a) increase
(b) decrease rapidly
(c) decrease slowly
(d) remain unaffected
Answer:
(a) increase

Question 10.
Osmosis is a property of ……………..
(a) solute
(b) solvent
(c) solution
(d) membrane
Answer:
(c) solution

2. Very short answer question

Question 1.
What is osmotic pressure?
Answer:
The pressure exerted due to osmosis is osmotic pressure.

Question 2.
Name the condition in which protoplasm of the plant cell shrinks.
Answer:
Plasmolysis

Question 3.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Answer:
When a pressure greater than the atmospheric pressure is applied to pure water or a solution then water potential of pure water or solution increases.

Question 4.
Which type of solution will bring about deplasmolysis ?
Answer:
Placing a plasmolysed cell in hypotonic solution will bring about deplasmolysis.

Question 5.
Which type of plants have negative root pressure?
Answer:
Plants showing excessive transpiration have negative root pressure.

Question 6.
In which conditions transpiration pull will be affected?
Answer:
Due to temperature fluctuations during day and night gas bubbles may be formed which affects transpiration pull.

Question 7.
Mention the shape of guard cells in Cyperus.
Answer:
Kidney shaped and dumbbell shaped guard cells are seen.

Question 8.
Why do diurnal changes occur in osmotic potential of guard cells?
Answer:
Enzyme activity of phosphorylase converts starch into sugar during daytime and sugar is converted to starch during night. This causes changes in osmotic potential of guard cells.

Question 9.
What is symplast pathway?
Answer:
When water is absorbed by root hair it passes across from one living cell to other living cell through the plasmodesmatal connections between them, then it is called symplast pathway across the root.

3. Answer the Following Questions

Question 1.
Describe mechanism of absorption of water.
Answer:

1. The absorption of water takes place by two modes, i.e. active absorption and passive absorption.
2. Passive absorption is the chief method of absorption (98%).
3. There is no expenditure of energy in passive absorption.
4. Transpiration pull is a driving force and water moves depending upon concentration gradient. Water is pulled upwards.
5. It occurs during daytime when there is active transpiration.
6. Active absorption occurs usually during night time as due to closure of stomata transpiration stops.
7. Water absorption is against D.ED. gradient, A.T.R energy is required which is available from respiration.
8. Active absorption may be osmotic or non- osmotic type.
9. For osmotic absorption root pressure has a role.

Question 2.
Discuss theories of water translocation.
Answer:

1. Translocation of water is transport of water along with dissolved minerals from roots to aerial parts.
2. The movement is against the gravity and described as ascent of sap.
3. The translocation occurs through lumen of water conducting tissue xylem mainly vessels and tracheids.
4. Different theories have been discussed for translocation mechanism like vital force theory (Root pressure), relay pump, physical force (capillary), etc.
5. Cohesion tension theory or transpiration pull theory is most widely accepted theory.

Question 3.
What is transpiration? Describe mechanism of opening and closing of stomata.
Answer:

1. The loss of water in the form of vapour is called transpiration.
2. Stomatal transpiration is a main type of transpiration where minute pores are concerned with it.
3. Stomata are bounded by two guard cells which in turn are surrounded by accessory cells.
4. Opening and closing of stomata is controlled by turgidity of guard cells.
5. When guard cells become turgid due to endosmosis their lateral thin and elastic wall bulges or stretch out.
6. The inner thick and inelastic wall is pulled apart, thus the stoma opens during daytime.
7. At night when guard cells become flaccid due to exosmosis the wall relaxes and stoma closes.
8. Endosmosis and exosmosis takes place due to changes in osmotic potential of guard cells.

Question 4.
What is transpiration? Explain role of transpiration.
Answer:
Transpiration : The loss of water from plant body in the form of vapour is called transpiration.

Role of transpiration:

1. Removal of excess of water
2. Helps in passive absorption of water and minerals
3. Helps in ascent of sap – transpiration pull
4. Maintains turgor of cells
5. Imparts cooling effect by reducing temperature 90% – 93% is stomatal transpiration and hence when stomata are open gaseous exchange takes place.

Question 5.
Explain root pressure theory and its limitations.
Answer:

1. Root pressure theory is proposed by J. Pristley.
2. For translocation of water, activity of living cells of root is responsible.
3. Absorption of water by root hair is a constant and continuous process and due to this a hydrostatic pressure is developed in cortical cells.
4. Owing to this hydrostatic pressure i.e. root pressure, water is forced into xylem and further conducted upwards.
5. Root pressure is an osmotic phenomenon.

Limitation of this theory:

1. Not applicable to tall plants above 20 metres.
2. Even in absence of root pressure ascent of sap is noticed.
3. In actively transpiring plants, root pressure is not developed.
4. In taller gymnosperms, root pressure is zero.
5. Xylem sap is under tension and shows negative hydrostatic pressure.

Question 6.
Explain capillarity theory of water translocation.
Answer:

1. Capillarity theory of water translocation is proposed by Bohem.
2. Capillarity is because of surface tension and cohesive forces and adhesive forces of water molecules.
3. Xylem vessels and tracheids are tubular elements having their lumen.
4. In these elements water column exists due to combined action of cohesive and adhesive forces of water and lignified wall.
5. As a result of this capillarity water is raised upwards.

Question 7.
Why is transpiration called ‘a necessary evil’?
Answer:

1. The loss of water in the form of water vapour is called transpiration.
2. About 90 – 93% of transpiration occurs through stomata, small apertures located in the epidermis of leaves.
3. For this process stomata must remain open and then only gaseous exchange by diffusion takes places.
4. Gaseous exchange is necessary for respiration and photosynthesis. If stomata remain closed then it will affect productivity of plant.
5. The process is necessary evil because water which is important for plant is lost in the process.
6. At the same time it helps in absorption of water and its translocation. Hence it cannot be avoided.
   So Curtis has rightly called it as necessary evil.

Question 8.
Explain movement of water in the root.
Answer:

1. Root hairs absorb water by imbibition then diffusion which is followed by osmosis.
2. As water is taken inside the root hair cell it becomes turgid i.e. increase in turgor pressure (T.E)
3. Root hair cell has less D.ED. but adjacent cortical cell has more D.PD.
4. The inner cortical cell has more osmotic potential so it will suck water from root hair cell.
5. Root hair cell becomes flaccid and ready to absorb soil water.
6. Water is passed on similarly in inner cortical cells.
7. Water moves rapidly through loose cortical cells up to endodermis and through passage cells in pericycle.
8. From pericycle due to hydrostatic pressure developed it is forced into protoxylem.

Question 9.
(i) Osmosis
Answer:
It is a special type of diffusion of solvent through a semipermeable membrane.

(ii) Diffusion
Answer:
It is the movement of ions/ atoms/molecules of a substance from the region of higher concentration to the region of their lower concentration.

(iii) Plasmolysis
Answer:
Exo-osmosis in a living cell when placed in hypertonic solution is called plasmolysis.

(iv) Imbibition
Answer:
It is swelling up of hydrophilic colloids due to adsorption of water.

(v) Guttation
Answer:
The loss of water in the form of liquid is called guttation.

(vi) Transpiration
Answer:
The loss of water from plant body in the form of vapour is called transpiration.

(vii) Ascent of sap
Answer:
The transport of water with dissolved minerals in it from root to other aerial parts of plant against the gravity is called ascent of sap.

(viii) Active absorption
Answer:
Water absorption by activity of root which is against the D.PD. gradient along with expenditure of A.T.E energy generated by respiration is the process of active absorption.

(ix) Diffusion Pressure Deficit (D.P.D.)
Answer:
The difference in the diffusion pressures of pure solvent and the solvent in a solution is called diffusion pressure deficit.

(x) Turgor pressure
Answer:
It is the pressure exerted by turgid cell sap on to the cell membrane and cell wall.

(xi) Water potential
Answer:
Chemical potential of water is called water potential.

(xii) Wall pressure
Answer:
Thick and rigid cell wall exerts a counter pressure to turgor pressure developed on the cell sap is called wall pressure that operates in opposite direction.

(xiii) Root pressure
Answer:
As absorption of water by root hair being a continuous process, a sort of hydrostatic pressure is developed in living cells of root, this is called root pressure.

Question 10.
Osmotic Pressure (O.P) and Turgor Pressure (T.P)
Answer:

Question 11.
How are the minerals absorbed by the plants ?
Answer:

1. Soil is the chief source of minerals for the plants.
2. Minerals get dissolved in the soil water.
3. Minerals are absorbed by the plants in the ionic form mainly through roots.
4. Absorption of minerals is independent of water.
5. Absorbed minerals are pulled upwards along with xylem sap.
6. Mineral ions can be remobilized in the plant body form older parts to young plants E.g. Ions of S, P and N.

4. Long answer questions

Question 1.
Describe structure of root hair.
Answer:

1. Water from soil is absorbed by plants with the help of root hairs.
2. Root hairs are present in zone of absorption.
3. Epidermal cells form unicellular extensions which are short lived (ephemeral) structures i.e. root hairs.
4. Root hairs are nothing but cytoplasmic extensions of epiblema cell.
5. Root hairs are long tube like structures of about 1 to 10 mm.
6. They are colourless, unbranched and very delicate structures.
7. A large central vacuole is surrounded by thin layer of cytoplasm, plasma membrane and outer cell wall.
8. The cell wall of root hair is thin and double layered with outer layer of pectin and inner layer of cellulose which is freely permeable.

Question 2.
Write on journey of water from soil to xylem in roots.
Answer:

1. Unicellular root hairs which are tubular extensions of epiblema cells absorb readily available capillary water from soil.
2. The three physical processes imbibition, diffusion and osmosis are concerned with absorption of water.
3. Water molecules get adsorbed on cell wall of root hair (imbibition).
4. They enter the root hair cell by diffusion through cell wall which is freely permeable.
5. By process of osmosis they enter through plasma membrane which is semipermeable.
6. The root hair cell becomes turgid and hence its turgor pressure increases and D.ED. value decreases.
7. The adjacent cell of cortex has more D.ED. value as its osmotic potential is more.
8. The cortical cell thus takes water from epidermal cell which is turgid. This process goes on due to gradient of suction pressure developed from cell to cell till thin walled passage cells of endodermis.
9. From endodermis it will enter pericycle and then due to hydrostatic pressure it is forced in protoxylem cell.
10. The pathway of water is by apoplast and symplast.
11. When water passes through cell wall and intercellular spaces of cortex it is apoplast pathway.
12. When water passes across living cells through their plasmodesmatal connections it is symplast pathway.

Question 3.
Explain cohesion theory of translocation of water.
Answer:

1. This is very widely accepted theory of ascent of sap proposed by Dixon and Joly.
2. It is based on principles of adhesion and cohesion of water molecules and transpiration by plants.
3. A strong force of attraction existing between water molecules is cohesion and the force of attraction between water molecules and lignified walls of xylem elements is adhesion.
4. Ascent of sap occurs through lumen of xylem elements.
5. Owing to cohesive and adhesive forces a continuous water column is maintained in xylem from root to aerial parts i.e. leaves.
6. Transpiration occurs through stomata and transpiration pull is developed in leaf vessels.
7. This tension or pull is transmitted downwards through vein to roots which triggers ascent of sap.
8. In transpiration, water is lost in vapour form and this increases D.PD. of mesophyll cells that are near guard cells.
9. Mesophyll cells absorb water from xylem in leaf and a gradient of D.PD. or suction pressure (S. E) is set.
10. Owing to this gradient from guard cell to xylem in leaf, a transpiration pull or tension is created in xylem.
11. Hence water column is pulled upward passively against gravity.

Question 4.
Write on mechanism of opening and closing of stomata.
Answer:

1. Transpiration takes place through stomata. Turgidity of guard cells controls opening and closing of stomata
2. Turgor pressure exerted on unevenly thickened wall of guard cell is responsible for the movement.
3. The outer thin wall which is elastic is stretched out which pulls inner thick inelastic wall and thus stomata open.
4. When guard cells are flaccid that results in closure of stomata.
5. According to starch-sugar in ter conversion theory enzyme phosphorylase converts starch to sugar during daytime.
6. Sugar being osmotically active, the O.E of guard cells is increased. The water is absorbed from subsidiary cells. Due to turgidity walls are stretched and stoma opens.
7. During night-time sugar is converted to starch and hence guard cells loose water and become flaccid. Hence there is closure of stomata.
8. According to proton transport theory, the movement is due to transport of H⁺ and K⁺ ions.
9. Subsidiary cells are reservoirs of K⁺ ions. Starch is converted to malic acid which dissociate into malate and proton (H⁺) during day.
10. Proton transported to subsidiary cells and K⁺ ions are taken from it. This forms potassium malate in guard cells.
11. Potassium malate increases osmotic potential and endo osmosis occurs hence turgidity of guard cells. → stomata opens,
12. The uptake of K⁺ and Cl^– ions is stopped by abscissic acid formed during night. This changes permeability. Guard cells become hypotonic and loose water as they become flaccid stomata close.

Question 5.
What is hydroponics? How is it useful in identifying the role of nutrients?
Answer:
(1) Growing plants in aqueous (soilless) medium is known as hydroponics. [Greek word hudor = water and ponos = work]

(2) It is technique of growing plants by supplying all necessary nutrients in the water supply given to plant.

(3) A nutrient medium is prepared by dissolving necessary salts of micronutrients and macronutricnts In desired quantity and roots of plants are suspended in this liquid with appropriate support.

(4) Hydroponics is of great use in studying the deficiency symptoms of different mineral nutrients.

(5) The plants uptake mineral nutrients in the form of dissolved ions with the help of root hairs from the surrounding medium or nutrient solution supplied.

(6) While preparing the required nutricnt medium particular nutrient can be totally avoided and then the effect of lack of that nutrient can be studied in variation of plant growth.

(7) Any visible change noticed from normal structure and function of the plant is the symptom or hunger sign considered.

(8) For e.g. Yellowing of leaf is observed due to loss of chlorophyll pigments or Chiorosis is noticed if Magnesium is lacking as it is a structural componen of chlorophyll pigment.

Question 6.
Explain the active absorption of minerals.
Answer:

1. Plants absorb minerals from the soil with their root system.
2. MInerals are absorbed from the soil In the form of charged particles, positively charged cations and negatively charged anions.
3. The absorption of minerals against the concentration gradient which requires expenditure of metabolic energy is called active absorption.
4. The ATP energy derived from resp’ration in root cells Is utilized for active absrption.
5. Ions get accumulated in the root hair against the concentration gradient.
6. These ions pass into cortical cells and finally reach xylem of roots.
7. Along with the water these minerals are carried to other parts of plant.

Maharashtra State Board 12th Std Biology Textbook Solutions

• Reproduction in Lower and Higher Plants Class 12 Biology Textbook Solutions
• Reproduction in Lower and Higher Animals Class 12 Biology Textbook Solutions
• Inheritance and Variation Class 12 Biology Textbook Solutions
• Molecular Basis of Inheritance Class 12 Biology Textbook Solutions
• Origin and Evolution of Life Class 12 Biology Textbook Solutions
• Plant Water Relation Class 12 Biology Textbook Solutions
• Plant Growth and Mineral Nutrition Class 12 Biology Textbook Solutions
• Respiration and Circulation Class 12 Biology Textbook Solutions

Origin and Evolution of Life Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 5 Origin and Evolution of Life Textbook Exercise Questions and Answers.

Class 12 Biology Chapter 5 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 5 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Who proposed that the first form of life could have come from pre-existing non-living organic molecules?
(a) Alfred Wallace
(b) Oparin and Haldane
(c) Charles Darwin
(d) Louis Pasteur
Answer:
(b) Oparin and Haldane

Question 2.
The sequence of origin of life may be
(a) Organic materials – inorganic materials – Eobiont – colloidal aggregates – cell.
(b) Inorganic materials – organic materials – colloidal aggregates – Eobiont – cell.
(c) Organic materials – inorganic materials – colloidal aggregates – cell.
(d) Inorganic materials – organic materials – Eobiont – colloidal aggregates – cell.
Answer:
(b) Inorganic materials – organic materials- colloidal aggregates – Eobiont – cell.

Question 3.
In Hardy-Weinberg equation, the frequency of homozygous recessive individual is represented by-
(a) p²
(b) pq
(c) q²
(d) 2pq
Answer:
(c) q²

Question 4.
Select the analogous organs.
(a) Forelimbs of whale and bat
(b) Flippers of dolphins and penguin
(c) Thorn and tendrils of bougainvillea and Cucurbita
(d) Vertebrate hearts or brains
Answer:
(b) Flippers of dolphins and penguin

Question 5.
Archaeopteryx is known as missing link because it is a fossil and share characters of both
(a) Fishes and amphibians
(b) Annelida and Arthropoda
(c) Reptiles and birds
(d) Chordates and non-chordates
Answer:
(c) Reptiles and birds

Question 6.
Identify the wrong statement regarding evolution.
(a) Darwin’s variations are small and directional.
(b) Mutations are random and non- directional.
(c) Adaptive radiations leads to divergent evolution.
(d) Mutations are non-random and directional.
Answer:
(d) Mutations are non-random and directional

Question 7.
Gene frequency in a population remain constant due to ……………….
(a) Mutation
(b) Migration
(c) Random mating
(d) Non-random mating
Answer:
(c) Random mating

Question 8.
Which of the following characteristic is not : shown by the ape?
(a) Prognathous face
(b) Tail is present
(c) Chin is absent
(d) Forelimbs are longer than hind limbs
Answer:
(b) Tail is present

Question 9.
………………. can be considered as connecting link between ape and man.
(a) Australopithecus
(b) Homo habilis
(c) Homo erectus
(d) Neanderthal man
Answer:
(a) Australopithecus

Question 10.
The Cranial capacity of Neanderthal man was ……………….
(a) 600 cc
(b) 940 cc
(c) 1400 cc
(d) 1600 cc
Answer:
(c) 1400 cc

2. Very short answer questions

Question 1.
Define
(i) Gene pool
Answer:
The sum total of genes of all individuals of interbreeding population or Mendelian population is called gene pool.

(ii) Gene frequency
Answer:
The proportion of an allele in the gene pool as compared with other alleles at the same locus is termed as gene frequency.

(iii) Organic evolution
Answer:
Organic evolution can be defined as slow, gradual, continuous and irreversible changes through which the present-day complex forms of the life developed (or evolved) from their simple pre-existing forms.

(iv) Population
Answer:
All individuals of the same species form a group which is called a population.

(v) Speciation
Answer:
Formation of new species from the pre-existing single group of organisms is called speciation.

Question 2.
What is adaptive radiation?
Answer:
The process of evolution which results in transformation of original species to many different varieties is called adaptive radiation.

Question 3.
If the variation occurs in population by chance alone and not by natural selection and bring change in frequencies of an allele, what is it called?
Answer:
If the variation occurs in population by chance alone and not by natural selection to bring change in frequencies of an allele, it is called genetic drift.

Question 4.
State the Hardy-Weinberg equilibrium law.
Answer:
The Hardy-Weinberg equilibrium law states that at equilibrium point both the allelic frequency and genotypic frequency remain constant from generation to generation, in the diploid, sexually reproducing, large, free interbreeding population in which mating is random and there is absence of any other factors that change the allele frequency.

Question 5.
What is homologous organs?
Answer:
Homologous organs are those organs, which are structurally similar but perform different functions.

Question 6.
What is vestigial organ?
Answer:
Vestigial organs are imperfectly developed and non-functional organs which are in degenerate form, they may be functional in some related and other animals or in ancestor.

Question 7.
What is the scientific name of modern man?
Answer:
Homo sapiens sapiens is the scientific name of modern man.

Question 8.
What is coacervate?
Answer:
Coacervates are colloidal aggregations of hydrophobic proteins and lipids which grew in size by taking up material from surrounding aqueous medium.

Question 9.
Which period is known as ‘age of Reptilia’?
Answer:
Jurassic period from Mesozoic era is known as age of Reptilia.

Question 10.
Name the ancestor of human which is described as man with ape brain.
Answer:
Australopithecus, the ancestor of human which is described as man with ape brain.

Short Answer Questions

Question 1.
Genetic drift.
Answer:

1. Genetic drift is random, directionless fluctuation that takes place in allele frequency.
2. It occurs by pure chance, in small sized population.
3. Genetic drift becomes an evolutional factor as it can change the gene frequency.
4. Sewall wright has given this concept and hence it is also known as Sewall wright effect.
5. Due to genetic drift, some alleles of a population are lost or reduced by chance and some others may be increased.
6. Some time, a few individuals become isolated from the large population and they produce new population in new geographical area.
7. Genetic drift is also called founders’ effect because original drifted population becomes ‘founders’ in the new area.
   E.g. Non-adaptive character of huge horns in Antelope is fixed due to genetic drift.

Question 2.
Enlist the different factors that are responsible for changing gene frequency.
Answer:
Gene flow, genetic drift, gene mutations, chromosomal aberrations such as deletion, duplication, inversion and translocation, genetic recombinations, natural selection, isolation are some of the factors which are responsible for changing the gene frequency.

Question 3.
Draw a graph to show that natural selection leads to disruptive change.
Answer:

Question 4.
Significance of fossils
Answer:

1. Fossils are studied under palaeontology. They are used in reconstruction of phylogeny.
2. Fossil study helps in studying various forms and structures of extinct animals.
3. By understanding the structure of fossil, record of missing link between two groups of organisms can be deduced.
4. By studying fossils various body forms and their evolution can be understood. They also help to understand the habit and habitat.
5. Some fossils provide the evolutionary evidences such a connecting links.

Question 5.
Write the objections to Mutation theory of Hugo de Vries.
Answer:
Objections to Mutation Theory:

1. Hugo de Vries observed the large and discontinuous variation. But these were chromosomal aberrations. Only gene mutations usually bring about minor changes.
2. Rate by which mutations take place is very slow as compared to the requirement of evolution.
3. Chromosomal aberrations are very unstable.
4. The organisms with chromosomal aberration are usually sterile and thus chromosomal aberrations have little significance in evolution.

Question 6.
What is disruptive selection? Give example.
Answer:
Disruptive selection:

1. The natural selection that disrupts the mean characters of the population, is called disruptive selection.
2. Greater number of individuals acquire peripheral character value at both ends of the distribution curve. E.g. Finches with large size or small size, both will be selected.
3. Extreme phenotypes are selected in evolutionary process and intermediate forms are eliminated.
4. When distribution curve is plotted it shows two peaks for two extremes.
5. Disruptive selection is rare because, nature always tries to balance the characters.
6. It ensures the effect on the entire gene pool of a population, considering all mating types or systems.

Example of disruptive selection:
African seed cracker finches are types of seed-feeder birds which have different sizes of beak. The seeds available to them were of small and large sized. Large beak sized birds feeds on large seeds while small beak sized birds feed on small seeds.

Such large and small birds thus thrive well. However, intermediate beak sized birds are unable to feed on either type of seeds so they starve and their population was decreased gradually. Natural selection eliminated them and thus the population of finches appear disrupted.

4. Match the columns

Question 1.

Answer:

5. Long Answer Questions

Question 1.
Would you consider wings of butterfly and bat as homologous or analogous and why?
Answer:
Wings of butterfly are made up of chitin. They neither have bones, nor muscles in the wings. The bat’s wings are actually patagium. They have muscles and bones just as those seen in all vertebrate limb series. Therefore, these two examples cannot be homologous. However, both the animals use the wings for flight. This is an indication that their function is similar but structure is different, hence they are analogous organs.

Question 2.
What is adaptive radiation? Explain with suitable example.
Answer:

1. Adaptive radiation is the process of evolution which results in transformation of original species to many different varieties.
2. The well-known example of adaptive radiation is Darwin’s Finches. When Charles Darwin went on his voyage to Galapagos islands, he noticed finches which is a variety of small birds.
3. According to Darwin’s observations, the American main land species of finches was the original one which must have migrated to the different islands of Galapagos.
4. Since environmental conditions here were different, they adapted in various ways to the differing environmental conditions of these islands.
5. Original bird had a beak suited for eating seeds, but the changed feeding pattern has changed the shape of beaks too. Some birds also show altered beaks for insectivorous mode. Thus, this demonstrated adaptive radiation.
6. Adaptive radiation in Australian Marsupials is also well studied. In Australia, there are many marsupial mammals who evolved from common ancestor.
7. Adaptive radiation leads to divergent evolution.

Question 3.
By talking industrial melanism as one example, explain the concept of natural selection.
OR
Explain natural selection in action by quoting the example of industrial melanism.
Answer:
1. Industrial melanism is the best example of natural selection which was studied by Kettlewell. In U.K. there are two varieties of peppered moths, Biston betularia and Biston carbonaria.

2. Before industrialisation, in Great Britain, Biston betularia were more in number than Biston carbonaria. B. Betularia is greyish white while B.carbonaria is melanic form.

3. These nocturnal moths rest on tree trunk during day. White-winged moth can camouflage well with the lichen covered whitish barks of trees. They thus escaped the attention of the predatory birds. But at the same time melanic forms were visible due to white barks of the trees. Their number was thus reduced as they were preyed upon by birds.

4. Later there was an industrial revolution, which ultimately resulted in air pollution causing dark soot to settle on the barks of the trees. Lichens too were destroyed and the melanic forms were now at advantage. Melanic forms could camouflage with black tree trunks and their number increased. White-winged moth become clearly seen in changed colours of the trees and thus they were easily caught by predatory birds. This caused decrease in their number.

5. Natural selection thus acted in changed environmental conditions and helped in the establishment of a phenotypic traits. The changed traits were more adaptive and hence were selected. Natural selection encourages those genes or traits that assure highest degree of adaptive efficiency between population and its environment.

Question 4.
Describe the Urey and Miller’s experiment.
Answer:

1. Urey and Miller performed an experiment to prove Oparin’s theory of chemical evolution.

2. They selected a spark discharge apparatus that consisted of closed system of glass having tungsten electrodes, flask for water boiling, a side tube connected to a vacuum pump, a cooling jacket and U-shaped trap.

3. The entire apparatus was first evacuated and made sterile and pre-biotic atmosphere was created in it.

4. The flask was filled with some water and mixture of methane, ammonia and hydrogen in the ratio of 1 : 2 : 2 were slowly passed through the stopcock, without allowing air.

5. Heat was supplied to the flask at very low temperature causing water to boil. The flask simulated the ocean present on primitive earth. Process of evaporation and precipitation was simulated by using heating mantle and condenser respectively.

6. Water vapours along with other gases were circulated continuously through continuous electric sparks. These sparks were given to the mixture for several days causing the gases to interact. This too simulated lightning.

7. Mixture of CH₄, NH₃ and H₂ gases passed through a condenser and was condensed to liquid.

8. The liquefied mixture was collected in the U-shaped trap, present at the bottom of the apparatus. It was found that variety of simple organic compounds (urea, amino acids, lactic acid and sugars) were formed in the apparatus.

This experiment provides the evidence in support to the fact that simple molecules present in the earth’s early atmosphere combined to form the organic building blocks of life.

Question 5.
What is Isolation? Describe the different types of reproductive isolations.
Answer:
1. Isolation means separation of the population of a particular species into smaller units. The organisms belonging to these subunits are prevented from interbreeding due to some barrier. These barriers are called isolating mechanisms.

2. They prevent the genetic exchange and gene flow.

3. Due to isolating mechanisms in nature the divergence among organisms takes place gradually leading to speciation. The isolating mechanisms are of two types namely, geographical isolation and reproductive isolation.

I. Geographical Isolation : The barrier in the form of physical distance or geographical barrier is called geographical isolation. The original population gets divided into two or more groups by geographical barriers such as river, ocean, mountain, glacier, etc. Organisms cannot cross the barriers on their own and hence interbreeding is prevented between isolated groups.

The separated groups experience different environmental factors and they acquire new traits by mutations. The separated populations develop distinct gene pool and they do not interbreed. Each subgroup then evolves differently which results into formation of new species. E.g. Darwin’s Finches, African elephant, Loxodonta and Indian elephant, Elephas.

II. Reproductive Isolation : Two populations may be occupying the same area, they may not be separated by geographical barrier, but then also they are reproductively isolated. Such reproductive isolation occurs due to change in genetic material, gene pool and structure of genital organs. Such differences prevent interbreeding between population. Such isolation later leads to speciation.

III. Different types of reproductive isolations : Reproductive isolation is of two types, viz. pre-zygotic and post-zygotic isolating mechanisms.

1. Pre-zygotic or pre-mating isolating mechanisms do not allow individuals to mate with each other at all.
2. By various mechanisms the two groups remain isolated.
3. In post-zygotic or post-mating isolating mechanisms, the two individuals can mate but the result of mating is not favourable.
4. Thus the populations remain isolated without the actual genetic exchange.

Question 6.
What is Genetic variations? Explain the different factors responsible for genetic variations.
Answer:
Genetic variations : The change in gene and gene frequencies is known as genetic variation. Genetic variations are caused by following factors:
(i) Mutations : Sudden permanent heritable change is called mutation. Mutation can occur in the gene, in the chromosome structure and in chromosome number. Mutation that occurs within the single gene is called point mutation or gene mutation. This leads to the change in the phenotype of the organism, causing variations.

(ii) Genetic recombination : In sexually reproducing organisms, during gamete formation, exchange of genetic material occurs between non-sister chromatids of homologous chromosomes. This is called crossing over. It produces new genetic combinations which result in variation. Fertilization between opposite mating gametes leads to various recombinations resulting into the phenotypic variations. These result in change in the frequencies of alleles.

(iii) Gene flow : Gene flow is movement of genes into or out of a population. Gene movement may be in the form of migration of organism, or gametes (dispersal of pollens) or segments of DNA (transformation). Gene flow also alters gene frequency causing evolutionary changes.

(iv) Genetic drift : Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift. For example, when the size of a population is severely reduced due to natural disasters like earthquakes, floods, fires, etc. elimination of particular alleles from a population becomes possible. Smaller populations have greater chances for genetic drift. It results in the change in the gene frequency. Genetic drift is also an important factor for evolutionary change.

(v) Chromosomal aberrations : The structural, morphological change in chromosome due to rearrangement of genes is called chromosomal aberrations. Due to changes in the gene arrangement or gene sequence variations are caused.

6. Complete the chart

Answer:

Maharashtra State Board 12th Std Biology Textbook Solutions

• Reproduction in Lower and Higher Plants Class 12 Biology Textbook Solutions
• Reproduction in Lower and Higher Animals Class 12 Biology Textbook Solutions
• Inheritance and Variation Class 12 Biology Textbook Solutions
• Molecular Basis of Inheritance Class 12 Biology Textbook Solutions
• Origin and Evolution of Life Class 12 Biology Textbook Solutions
• Plant Water Relation Class 12 Biology Textbook Solutions
• Plant Growth and Mineral Nutrition Class 12 Biology Textbook Solutions
• Respiration and Circulation Class 12 Biology Textbook Solutions

Molecular Basis of Inheritance Class 12 Exercise Question Answers Solutions Maharashtra Board

Balbharti Maharashtra State Board 12th Biology Textbook Solutions Chapter 4 Molecular Basis of Inheritance Textbook Exercise Questions and Answers.

Class 12 Biology Chapter 4 Exercise Solutions Maharashtra Board

Biology Class 12 Chapter 4 Exercise Solutions

1. Multiple Choice Questions

Question 1.
Griffith worked on ………………..
(a) Bacteriophage
(b) Drosophila
(c) Frog eggs
(d) Streptococci
Answer:
(d) Streptococci

Question 2.
The molecular knives of DNA are ………………..
(a) Ligases
(b) Polymerases
(c) Endonucleases
(d) Transcriptase
Answer:
(c) Endonucleases

Question 3.
Translation occurs in the ………………..
(a) Nucleus
(b) Cytoplasm
(c) Nucleolus
(d) Lysosomes
Answer:
(b) Cytoplasm

Question 4.
The enzyme required for transcription is ………………..
(a) DNA polymerase
(b) RNApolymerase
(c) Restriction enzyme
(d) RNase
Answer:
(b) RNA polymerase

Question 5.
Transcription is the transfer of genetic information from ………………..
(a) DNA to RNA
(b) t-RNA to m-RNA
(c) DNA to m-RNA
(d) m-RNA to t-RNA
Answer:
(a) DNA to RNA

Question 6.
Which of the following is NOT part of protein synthesis?
(a) Replication
(b) Translation
(c) Transcription
(d) All of these
Answer:
(a) Replication

Question 7.
In the RNA molecule, which nitrogen base is found in place of thymine?
(a) Guanine
(b) Cytosine
(c) Thymine
(d) Uracil
Answer:
(d) Uracil

Question 8.
How many codons are needed to specify three amino acids?
(a) 3
(b) 6
(c) 9
(d) 12
Answer:
(a) 3

Question 9.
Which out of the following is NOT an example of inducible operon?
(a) Lactose operon
(b) Histidine operon
(c) Arabinose operon
(d) Tryptophan operon
Answer:
(d) Tryptophan operon

Question 10.
Place the following event of translation in the correct sequence ………………..
i. Binding of met-t-RNA to the start codon.
ii. Covalent bonding between two amino acids.
iii. Binding of second t-RNA.
iv. Joining of small and large ribosome subunits.
(a) iii, iv, i, ii
(b) i, iv, iii, ii
(c) iv, iii, ii, i
(d) ii, iii, iv, i
Answer:
(b) i, iv, iii, ii

2. Very Short Answer Questions

Question 1.
What is the function of an RNA primer during protein synthesis?
Answer:
During DNA replication, RNA primer provides 3’ OH to which DNA polymerase enzyme can add nucleotides to synthesize new strand using parental strand of DNA as template.
[Note : RNA primer has no direct role in protein synthesis.]

Question 2.
Why is the genetic code considered as commaless?
Answer:
The triplet codon are arranged one after the other on m-RNA molecule without any gap or space and therefore genetic code is considered as commaless.

Question 3
Genome
Answer:
Genome is the total genetic constitution of an organism or a complete copy of genetic information (DNA) or one complete set of chromosomes (monoploid or haploid) of an organism.

Question 4.
Which enzyme does remove supercoils from replicating DNA?
Answer:
Super-helix relaxing enzyme (Topoisomerase) removes supercoils from replicating DNA.

Question 5.
Why are Okazaki fragments formed on lagging strand only?
Answer:
Okazaki fragments are formed only on lagging template as only short stretch of lagging template becomes available for replication at one time.

Question 6.
When does DNA replication take place?
Answer:
In eukaryotes DNA-replication takes place during S-phase of interphase of cell cycle and in prokaryotes. DNA replicates prior to cell division.

Question 7.
Define term Codogen and Codon
Answer:
Codogen is a triplet of nucleotides present on the DNA which specifies one particular amino acid.
Codon is a triplet of nucleotides present on the m-RNA which specifies one particular amino acid.

Question 8.
What is degeneracy of genetic code?
Answer:
Genetic code is degenerate as 61 codons code for 20 amino acids, that is two or more codons can specify the same amino acid. E.g. Cysteine has two codons, while isoleucine has three codons.

Question 9.
Which are the nucleosomal ‘core’ histones?
Answer:
Two molecules each of histone proteins, viz. H₂A. H₂B, H₃ and H₄ are the nucleosomal ‘core’ histones.

3. Short Answer Questions

Question 1.
DNA packaging in eukaryotic cell.
Answer:

1. In eukaryotic cells, DNA (2.2 metres) is condensed, coiled and supercoiled to be packaged efficiently in the nucleus (10^-16 m).
2. DNA is associated with histone and non-histone proteins.
3. Histones are a set of positively charged, basic proteins, rich in basic amino acid residues lysine and arginine.
4. Nucleosome consists of nucleosome core (two molecules of each of histone proteins viz. H₂A, H₂B, H₃ and H₄ forming histone octamer) and negatively charged DNA (146 bps) that wraps around the histone octamer by 1 3/4 turns.
5. H₁ protein binds the DNA thread where it enters and leaves the nucleosome.
6. Adjacent nucleosomes are linked with linker DNA (varies in length from 8 to 114 bp, average length of linker DNA is about 54 bp).
7. Each nucleosome contains 200 bp of DNA.
8. Packaging involves formation of – Beads on string (10 nm diameter), Solenoid fibre (looks like coiled telephone wire, 30 nm diameter/300Å), Chromatin fibre and Chromosome.
9. Non-Histone Chromosomal Proteins (NHC) contribute to the packaging of chromatin at a higher level.

Question 2.
Enlist the characteristics of genetic code.
Answer:
The characteristics of genetic code are

1. Genetic code is triplet, commaless and non-overlapping.
2. It is degenerate and non-ambiguous.
3. It is universal
4. It has polarity.

Question 3.
Applications of DNA fingerprinting.
Answer:
Applications of DNA fingerprinting are as follows:

1. In forensic science to solve rape and murder cases.
2. Finds out the biological father or mother or both, of the child, in case of disputed parentage.
3. Used in pedigree analysis in cats, dogs, horses and humans.

Question 4.
Explain the role of lactose in ‘Lac Operon’.
Answer:

1. A small amount of beta-galactoside permease enzyme is present in cell even when Lac operon is switched off and it allows a few molecules of lactose to enter into the cell.
2. Lactose binds to repressor and inactivates it.
3. Repressor – lactose complex cannot bind with the operator gene, which is then turned on.
4. RNA polymerase transcribes all the structural genes to produce lac m-RNA which is then translated to produce all enzymes.
5. Thus, lactose acts as an inducer.
6. When the inducer level falls, the operator is blocked again by repressor and structural genes are repressed again. This is negative feedback.

4. Short Answer Questions

Question 1.
Human genome project.
Answer:
1. Human Genome Project (HGP) was initiated in 1990 under the International administration of the Human Genome Organization (HUGO) and it was completed r in 2003.

2. The main aims:

• To sequence 3 billion base pairs of DNA in human genome and to map an estimated 33000 genes.
• To store the information collected from the project in databases.
• To develop tools and techniques for analysis of the data.
• Transfer of the related technologies to the private sectors, such as industries.
• Taking care of the legal, ethical and social issues which may arise from project.
• To sequence the genomes of several other organisms such as bacteria e.g. E.coli, Caenorhabditis elegans, Saccharomyces cerevisiae, Drosophil, rice, Arabidopsis), Mus musculus, etc.

3. Significance:

1. HGP has a major impact in the fields like Medicine, Biotechnology, Bioinformatics and the Life sciences.
2. More understanding of functions of genes, proteins and human evolution.

Question 2.
Describe the structure of operon.
Answer:

1. An operon is a unit of gene expression and regulation.
2. It includes the structural genes and their control elements. Control elements are promoters and operators.
3. The structural genes code for proteins, r-RNA and t-RNA that are necessary for all the cells.
4. Promoters are signal sequences in DNA. They start the RNA synthesis. They also act as sites where the RNA polymerases are bound during transcription.
5. Operators are present between the promoters and structural genes.
6. There is repressor protein that binds to the operator region of the operon.
7. There are regulatory genes which are responsible for the formation of repressors which interact with operators.

Question 3.
In the figure below A, B and C are three types of

Answer:
Answer: A, B and C are A : m-RNA, B : r-RNA, C : t-RNA

Question 4.
Identify the labelled structures on the following diagram of translation.

Part A is the ………………
Part B is the ………………
Part C is the ………………
Answer:
Part A is the anti-codon.
Part B is the amino acid.
Part C is the larger subunit of ribosome.

Question 5.
Match the entries in Column I with those of Column II and choose the correct answer.

Answer:

5. Long Answer Questions

Question 1.
Explain the process of DNA replication.
Answer:

DNA replication is semi-conservative replication. It involves following steps:
Activation of Nucleotides:

1. Nucleotides (dAMP dGMR dCMP and dTMP) present in the nucleoplasm, are activated by ATP in presence of an enzyme phosphorylase.
2. This phosphorylation results in the formation of deoxyribonucleotide triphosphates i.e. dATE dGTR dCTP and dTTE

Point of Origin or Initiation point:

1. Replication begins at specific point ‘O- Origin and terminates at point ‘T’.
2. At the point ‘O’, enzyme endonuclease nicks (breaks the sugar-phosphate backbone or the phosphodiester bond) one of the strands of DNA, temporarily.

Unwinding of DNA molecule:

1. Enzyme DNA helices breaks weak hydrogen bonds in the vicinity of ‘O’.
2. The strands of DNA separate and unwind. This unwinding is bidirectional.
3. SSBP (Single strand binding proteins) remains attached to both the separated strands and prevent them from recoiling (rejoining).

Replicating fork:

1. Y-shape replication fork is formed due to unwinding and separation of two strands.
2. The unwinding of strands results in strain which is released by super-helix relaxing enzyme.

Synthesis of new strands:

1. Each separated strand acts as a template for the synthesis of new complementary strand.
2. A small RNA primer (synthesized by activity of enzyme RNA primase) get attached to the 3′ end of template strand and attracts complementary nucleotides from surrounding nucleoplasm.
3. These nucleotides bind to the complementary nucleotides on the template strand by hydrogen bonds (i.e. A = T or T = A; G = C or C = G, CEG).
4. The phosphodiester bonds are formed between nucleotides of new strand to form a polynucleotide strand.
5. The enzyme DNA polymerase catalyses synthesis of new complementary strand always in 5′ – 3′ direction.

Leading and Lagging strand:

1. The template strand with free 3′ is called the leading template.
2. The template strand with free 5′ end is called the lagging template.
3. The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
4. New strands are always formed in 5′ → 3′ direction.
5. The new strand which develops continuously towards replicating fork is called the leading strand.
6. The new strand which develops discontinuously away from the replicating fork is called the lagging strand.
7. Maturation of Okazaki fragments : The lagging strand is synthesized in the form of small Okazaki fragments which are joined by enzyme DNA ligase.
8. Later RNA primers are removed by the combined action of RNase H, an enzyme that degrades the RNA strand of RNA-DNA hybrids, and polymerase I.
9. Gaps formed are filled by complementary DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-a in eukaryotes.
10. Finally, DNA gyrase (topoisomerase) enzyme forms double helix to form daughter DNA molecules.

Formation of two daughter DNA molecules:

1. In each daughter DNA molecule, one strand is parental and the other one is newly synthesized.
2. Thus, 50% part (i.e. one strand of the helix) is contributed by mother DNA. Hence, it is described as semiconservative replication.

Question 2.
Describe the process of transcription in protein synthesis.
Answer:

Transcription involves three stages, viz. Initiation, Elongation and Termination.
(1) Initiation:

1. RNA polymerase binds to promoter site.
2. It then moves along the DNA and causes local unwinding of DNA duplex into two strands in the region of the gene.
3. Only antisense strand functions as template.

(2) Elongation:

• The complementary ribonucleoside tri-phosphates get attached to exposed bases of DNA template chain.
• As transcription proceeds, the hybrid DNA-RNA molecule dissociates and makes m-RNA molecule free.
• As the m-RNA grows, the transcribed region of DNA molecule becomes spirally coiled and regains double helical form.

(3) Termination:
When RNA polymerase reaches the terminator site on the DNA, both enzyme and newly formed m-RNA (primary transcript) gets released.

Question 3.
Describe the process of translation in protein synthesis.
Answer:

Translation involves the following steps:
1. Activation of amino acids and formation of charged t-RNA (t-RNA – amino acid complex):
i. In the presence of an enzyme amino acyl t-RNA synthetase, the amino acid is activated and then attached to the specific t-RNA molecule at 3’ end to form charged t-RNA (t-RNA – amino acid complex).

ii. ATP is essential for the reaction.

2. Initiation of Polypeptide chain:

• Small subunit of ribosome binds to the m-RNA at 5’ end.
• Start codon is positioned properly at P-site.
• Initiator t-RNA, (carrying amino acid methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA, by its anticodon (UAC) through hydrogen bonds.
• The large subunit of ribosome joins with the smaller subunit in the presence of Mg⁺⁺.
• Thus, initiator charged t-RNA occupies the P-site and A – site is vacant.

3. Elongations of polypeptide chain:
Addition of amino acid occurs in 3 Step cycle-
i. Codon recognition.
Anticodon of second (and subsequent) amino acyl t-RNA molecule recognizes and binds with codon at A-site by hydrogen bonds.

ii. Peptide bond formation.

1. Ribozyme catalyzes the peptide bond formation between amino acids on the initiator t-RNA at P-site and t-RNA at A-site.
2. It takes less than 0.1 second for formation of peptide bond.
3. Initiator t-RNA at ‘P’ site is then released from E-site.

iii. Translocation.

1. Translocation is the process in which sequence of codons on m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
2. Due to this A’-site becomes vacant to receive next charged t-RNA molecule.
3. The events like arrival of t-RNA – amino acid complex, formation of peptide bond, ribosomal translocation and release of previous t-RNA, are repeated.
4. As ribosome move over the m-RNA, all the codons on m-RNA are exposed oiie by one for translation.

4. Termination and release of polypeptide:
When stop codon (UAA, UAG, UGA) gets exposed at the A-site, the release factor binds to the stop codon, thereby terminating the translation process
The polypeptide gets released in the cytoplasm.
Two subunits of ribosome dissociate and last t-RNA and m-RNA are released in the cytoplasm.
m-RNA gets denatured by nucleases immediately.

Question 4.
Describe Lac ‘Operon’.
Answer:

Lac operon consists of the following components:
(1) Regulator gene:

• Regulator gene precedes the promoter gene.
• It may not be present immediately adjacent to operator gene.
• Regulator gene codes for a repressor protein which binds with operator gene and represses (stops) its action.

(2) Promoter gene:

• It precedes the operator gene.
• It is present adjacent to operator gene.
• RNA Polymerase enzyme binds at promoter site.
• Promoter gene base sequence determines which strand of DNA acts a template.

(3) Operator gene:

• It precedes the structural genes.
• When operator gene is turned on by an inducer, the structural genes get transcribed to form m-RNA.

(4) Structural gene:

• There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.
• Enzymes produced are β-galactosidase, β-galactoside permease and transacetylase respectively.
  Inducer Allolactose acts as an inducer. It inactivates the repressor by binding with it.

Question 5.
Justify the statements. If the answer is false, change the underlined word(s) to make the statement true.
(i) The DNA molecule is double stranded and the RNA molecule is single stranded.
Answer:

1. DNA as the genetic material has to be chemically and structurally stable.
2. It should be able to generate its replica.
3. Sugar-phosphate backbone and complementary base pairing between the two strands, give stability to DNA.
4. Both the strands of DNA act as template for synthesis of their complementary strands. This allows accurate replication of DNA.
5. Single stranded RNA can be folded to form complex structures and perform specific functions such as synthesis of proteins.

(ii) The process of translation occurs at the ribosome.
Answer:

1. Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.
2. Ribosome has one binding site for m-RNA. It orients m-RNA molecule in such a way that all the codons are properly read.
3. Ribosome has three binding sites for t-RNA : P-site (peptidyl t-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
4. t-RNAs place the required amino acids in correct sequence and translate the coded message of RNA.
5. In eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.
6. Thus ribosome plays an important role in translation.

(iii) The job of m-RNA is to pick up amino acids and transport them to the ribosomes.
Answer:
The job of t-RNA is to pick up amino acids and transport them to ribosomes. t-RNA is an adapter molecule. It reads the codons of m-RNA and also simultaneously transfer specific amino acid to m-RNA Ribosome complex. It binds with amino acid at its 3′ end.

(iv) Transcription must occur before translation may occur.
Answer:
In prokaryotes, translation can start before transcription is complete, as both these processes occur in the same compartment, i.e. cytoplasm. But in eukaryotes, transcription and processing of hnRNA occurs in nucleus. hnRNA then comes out of the nucleus through nuclear pores and then it is translated at ribosomes in the cytoplasm.

Question 6.
Guess
(i) the possible locations of DNA on the collected evidence from a crime scene and
(ii) the possible sources of DNA.

Answer:

Maharashtra State Board 12th Std Biology Textbook Solutions

• Reproduction in Lower and Higher Plants Class 12 Biology Textbook Solutions
• Reproduction in Lower and Higher Animals Class 12 Biology Textbook Solutions
• Inheritance and Variation Class 12 Biology Textbook Solutions
• Molecular Basis of Inheritance Class 12 Biology Textbook Solutions
• Origin and Evolution of Life Class 12 Biology Textbook Solutions
• Plant Water Relation Class 12 Biology Textbook Solutions
• Plant Growth and Mineral Nutrition Class 12 Biology Textbook Solutions
• Respiration and Circulation Class 12 Biology Textbook Solutions