NCERT Exemplar Class 12 Biology Solutions Chapter 13 Organisms and Population

NCERT Exemplar Class 12 Biology Solutions Chapter 13 Organisms and Population

Edited By Priyanka kumari | Updated on Aug 26, 2022 03:21 PM IST | #CBSE Class 12th

NCERT exemplar Class 12 Biology solutions chapter 13 is one of the easiest yet most practical chapters from the book that investigated different aspects of ecology and various unique phenomena taking place around us. The NCERT exemplar Class 12 Biology solutions chapter 13 have been carefully prepared by experts in this field keeping in mind the guidelines and marking scheme of CBSE that will help and aid students for their Board exams. The professionals have much experience over years and have given their best in forming these solutions of NCERT. These solutions can also be accessed offline through NCERT exemplar Class 12 Biology solutions chapter 13 pdf download to provide you with the best study and reference material for your exams.
Also read - NCERT Class 12 Biology Solutions.

NCERT Exemplar Class 12 Biology Solutions Chapter 13 Is Given Below:

Multiple Choice Questions:

Question:1

Autecology is the:
a. Relation of heterogeneous populations to its environment
b. Relation of an individual to its environment
c. Relation of a community to its environment
d. Relation of a biome to its environment

Answer:

The answer is the option (b) Relation of an individual to its environment
Explanation: Relation of an individual to its environment is called auto ecology. An individual is closest to the species hence ‚’ b’ is the correct answer.

Question:2

Ecotone is:
a. A polluted area
b. The bottom of a lake
c. A zone of transition between two communities
d. A zone of developing community

Answer:

The answer is the option (c) A zone of transition between two communities
Explanation: (c) A zone of transition between two communities

Question:3

Biosphere is:
a. a component in the ecosystem
b. composed of the plants present in the soil
c. life in the outer space
d. composed of all living organisms present on earth which interact with the physical environment

Answer:

The answer is the option (d) composed of all living organisms present on earth which interact with the physical environment
Explanation: Biosphere is the thin zone of interaction of hydrosphere, atmosphere and troposphere where living beings are found.

Question:4

The ecological niche is:
a. the surface area of the ocean
b. an ecologically adapted zone
c. the physical position and functional role of a species within the community
d. formed of all plants and animals living at the bottom of a lake

Answer:

The answer is the option (c) the physical position and functional role of a species within the community
Explanation: (c) the physical position and functional role of a species within the community

Question:5

According to Allen’s Rule, the mammals from colder climates have:
a. shorter ears and longer limbs
b. longer ears and shorter limbs
c. longer ears and longer limbs
d. shorter ears and shorter limbs

Answer:

The answer is the option (d) shorter ears and shorter limbs
Explanation: Joel Asaph and Allen proposed this rule in 1877. According to this rule, the body shape and proportions of endotherms vary according to climatic temperature. In colder climate, the surface area is minimized to prevent heat loss and it is maximized in warm climate to promote heat loss. Endotherms in cold climate usually have small ears and small legs.

Question:6

Salt concentration (Salinity) of the sea measured in parts per thousand is:
a. 10 – 15
b. 30 – 70
c. 0 – 5
d. 30 – 35

Answer:

The answer is the option (d) 30-35

Question:7

Formation of tropical forests needs mean annual temperature and mean annual precipitation as:
a. 18 – 25°C and 150 – 400 cm
b. 5 – 15°C and 50 – 100 cm
c. 30 – 50°C and 100 – 150 cm
d. 5 – 15°C and 100 – 200 cm

Answer:

The answer is the option (a) 18- 25°C and 150- 400 cm

Question:8

Which of the following forest plants controls the light conditions at the ground?
a. Lianas and climbers
b. Shrubs
c. Tall trees
d. Herbs

Answer:

The answer is the option (c) Tall trees
Explanation: Tall trees form dense canopy and thus control the amount of light, which reaches the ground.

Question:9

What will happen to a well-growing herbaceous plant in the forest if it is transplanted outside the forest in a park?
a. It will grow normally
b. It will grow well because it is planted in the same locality
c. It may not survive because of a change in its micro-climate
d. It grows very well because the plant gets more sunlight

Answer:

The answer is the option (c) It may not survive because of change in its microclimate

Question:10

If a population of 50 Paramoecium present in a pool increases to 150 after an hour, what would be the growth rate of the population?
a. 50 per hour
b. 200 per hour
c. 5 per hour
d. 100 per hour

Answer:

The answer is the option (d) 100 per hour

Question:11

What would be the per cent growth or birth rate per individual per hour for the same population mentioned in the previous question (Question 10)?
a. 100
b. 200
c. 50
d. 150

Answer:

The answer is the option (b) 200
Explanation: New population - Old population =150-50=100
Percentage growth =\left ( \frac{100}{50} \right )\times =200\; ^{o}/_{o}

Question:12

A population has more young individuals compared to older individuals. What would be the status of the population after some years?
a. It will decline
b. It will stabilise
c. It will increase
d. It will first decline and then stabilize

Answer:

The answer is the option (c) It will increase
Explanation: A higher portion of young population means birth rate is higher than mortality and hence population will grow.

Question:13

What parameters are used for tiger census in our country’s national parks and sanctuaries?
a. Pugmarks only
b. Pugmarks and faecal pellets
c. Faecal pellets only
d. Actual headcounts

Answer:

The answer is the option (b) Pug marks and faecal pellets
Explanation: (b) Pug marks and faecal pellets

Question:14

Which of the following would necessarily decrease the density of a population in a given habitat.
a. Natality > mortality
b. Immigration > emigration
c. Mortality and emigration
d. Natality and immigration

Answer:

The answer is the option (c) Mortality and emigration

Question:15

A protozoan reproduces by binary fission. What will be the number of protozoans in its population after six generations?

a. 128
b. 24
c. 64
d. 32

Answer:

The answer is the option (c) 64
Explanation: After six generations, population =2^{6}=64

Question:16

In 2005, for each of the 14 million people present in a country, 0.028 were born and 0.008 died during the year. Using the exponential equation, the number of people present in 2015 is predicted as:
a. 25 million
b. 17 million
c. 20 million
d. 18 million

Answer:

The answer is the option (b) 17 millions
Explanation: Use the formula for compound interest.
Here, P = 14 million, r = 2% and time = 10 years
Population after 10 years = =10\left ( \frac{102}{100} \right )^{10}=17.06 million

Question:17

Amensalism is an association between two species where:
a. one species is harmed and other is benefitted
b. one species is harmed and other is unaffected
c. one species is benefitted and other is unaffected
d. both the species are harmed.

Answer:

The answer is the option (b) one species is harmed and other is unaffected

Question:18

Lichens are an association of:
a. bacteria and fungus
b. alga and bacterium
c. fungus and alga
d. fungus and virus

Answer:

The answer is the option (c) fungus and algae
Explanation: This is a kind of mutualism. Fungus provides water and minerals, while algae prepare the food.

Question:19

Which of the following is a partial root parasite?
a. Sandalwood
b. Mistletoe
c. Orobanche
d. Ganoderma

Answer:

The answer is the option (a) Sandal wood
Explanation: Orobanche is an obligate parasitic herbaceous plant and is found in temperate climate in the Northern hemisphere. However, Sandal wood is a partial root parasite.

Question:20

Which one of the following organisms reproduces sexually only once in its lifetime?
a. Banana
b. Mango
c. Tomato
d. Eucalyptus

Answer:

The answer is the option (d) Eucalyptus

Very Short Answer Type Questions:

Question:1

Species that can tolerate a narrow range of temperature are called ______.

Answer:

Stenothermic

Organisms and Population Excercise: 1.4

Question:2

What are Eurythmic species?

Answer:

Species, which can tolerate and thrive in a wide range of temperatures, are called eurythermic species.

Question:4

Define stenohaline species.

Answer:

Species, which thrive in a narrow range of salinity, are called stenohaline species.


Question:5

What is the interaction between two species called?

Answer:

The interaction between two species is called interspecific interaction.

Question:6

What is commensalism?

Answer:

Relationship between two organisms in which one organism is benefitted, and there is a neutral effect on another organism; is called commensalism.
Many commensals live in the alimentary canal of humans to get shelter and food, but they have neither a positive nor negative impact on humans.

Question:8

What is Mycorrhiza?

Answer:

Mycorrhiza is a mutualistic association between the fungus and the roots of higher plants. The fungi get shelter and food, and lieu of that helps the plant in absorption of phosphorous from the soil. It also provides resistance to the pant from root-borne pathogens.

Question:10

Why do high altitude areas have brighter sunlight and lower temperatures as compared to the plains?

Answer:

When sunlight reaches us, it undergoes refraction through several layers of the atmosphere. The relative degree of refraction at higher altitude is less than at lower altitude. Due to this, sunlight appears brighter at high altitudes than in plains. Temperature reduces with increase in altitude. Due to this, higher altitudes have a lower temperature compared to the plains.

Question:11

What is homeostasis?

Answer:

The process by which an organism maintains constancy in its internal environment is called homeostasis. Osmoregulation and thermoregulation are parts of homeostasis.

Question:12

Define aestivation.

Answer:

When an organism goes into summer sleep to avoid unfavourable environment; this process is called aestivation.

Question:13

What are diapause and its significance?

Answer:

The period of suspended development in order to tide over the unfavourable conditions is called diapause.

Question:14

What would be the growth rate pattern, when the resources are unlimited?

Answer:

When resources are unlimited, the growth rate of population is exponential. This gives a J- shaped curve on the graph.

Question:15

What are the organisms that feed on plant sap and other plant parts called?

Answer:

Organisms that feed on plant sap and other plant parts are called phytophagous.

Question:16

What is high altitude sickness? Write its symptoms.

Answer:

A range of symptoms due to the low atmospheric pressure of high altitudes, the body does not get enough oxygen is called high altitude sickness. The symptoms of altitude sickness are nausea, fatigue and heart palpitations.

Question:17

Give a suitable example for commensalism.

Answer:

Egrets ride on the back of cattle. When cattle forage for grass, insects fly away abruptly. This gives the egrets a chance to feed on those insects. In this relationship, cattle get no benefit from egrets. This can be considered a suitable example of commensalism.

Question:18

Define ectoparasite and endoparasite and give suitable examples.

Answer:

Ectoparasite: Parasites which live on the body of hosts are called ectoparasites, e.g. leech and ticks.
Endoparasites: Parasites which live inside the body of hosts are called endoparasites, e.g. roundworms and hookworms.

Question:19

What is brood parasitism? Explain with the help of an example.

Answer:

In this type of parasitism, a bird lays its eggs in the nest of another bird. The eggs resemble those of the host bird. The host bird incubates the eggs, unaware of its actual identity. This is seen in Koyal (cuckoo) which lays its eggs in a crow’s nest.

Short Answer Type Questions:

Question:1

Why are coral reefs not found in the regions from West Bengal to Andhra Pradesh but are found in Tamil Nadu and on the east coast of India?

Answer:

In West Bengal and Andhra Pradesh, freshwater is released from the Ganges and some other rivers. Due to this, coral reefs are not found in regions from West Bengal to Andhra Pradesh. From Tamil Nadu to the southern tip of the east coast, coral reefs are present because of no interference from freshwater.

Question:2

If a freshwater fish is placed in an aquarium containing seawater, will the fish be able to survive? Explain giving reasons.

Answer:

Freshwater is adapted to maintain its homeostasis in freshwater. It is unable to do so in saline water. Hence, the fish will not survive in an aquarium, which contains seawater.

Question:3

Why do all the freshwater organisms have contractile vacuoles whereas the majority of marine organisms lack them?

Answer:

In freshwater organisms, the concentration of solute inside the cell is higher than that in the external environment. This means that the external environment is hypotonic. This leads to a movement of water inside the cells because of osmosis. Contractile vacuoles work like safety devices and prevent lysis of the cell. In marine organisms, the concentration of solute inside the cell is lower than that in the external environment.
This means that the external environment is hypertonic. Hence, the cell does not have to face the risk of lysis. This explains the absence of contractile vacuoles in marine organisms.

Question:4

Define heliophytes and sciophytes. Name a plant from your locality that is either heliophyte or sciophyte.

Answer:

Heliophyte: Plants, which are adapted to live under plenty of sunlight, are called heliophytes, e.g. banyan. Heliophytes form the roof of a forest.
Sciophyte: Plants, which are adapted to live under low light conditions, are called sciophtyes, e.g. herbs. Sciophytes form the understory of a forest.

Question:5

Why do submerged plants receive weaker illumination than exposed floating plants in a lake?

Answer:

Most of the colours from the visible spectrum fail to reach deep into a water body. This happens because of the refraction of light through different layers of water. Due to this, floating plants get plenty of sunlight, but submerged plants get weaker illumination.

Question:6

In a seashore, the benthic animals live in sandy, muddy, and rocky substrata and accordingly developed the following adaptations.
a. Burrowing
b. Building cubes
c. Holdfasts / peduncle
Find suitable substratum against each adaptation.

Answer:

(a) These animals live in the sandy substratum and make burrows, e.g. lobsters, crabs, etc. (b) These animals live in muddy substratum as cube-like aggregates. The small size of cube helps in passive feeding mechanisms, e.g. zooplanktons.
(c) These animals live in the rocky substratum and attach to the substratum with the help of holdfast, e.g. sponges.

Question:7

Categorise the following plants into hydrophytes, halophytes, mesophytes and xerophytes. Give reasons for your answers.
a. Salvinia
b. Opuntia
c. Rhizophora
d. Mangifera

Answer:

Plants

Adaption

Reason

Salvinia

Hydrophyte

Lives in water and is a floating plant

Opuntia

Xerophytes

Lives in the desert

Rhizophora

Halophyte

Lives in mangrove forests and is adapted to saline marshes

Mangifera

Mesophyte

It is the scientific name of mango. Mango plants are found in a moderate climate.


Question:8

In a pond, we see plants which are free-floating; rooted-submerged; rooted emergent; rooted with floating leaves. Write the type of plants against each of them.

Answer:

Plant Name - Type
(a) Hydrilla - Rooted submerged
(b) Typha - Rooted emergent
(c) Nymphaea - Rooted with floating leaves
(d) Lemna -Free-floating
(e) Vallisnaria- Rooted emergent

Question:9

The density of a population in a habitat per unit area is measured in different units. Write the unit of measurement against the following:
a. Bacteria _______________
b. Banyan _______________
c. Deer _______________
d. Fish _______________

Answer:

(a) Number unit volume
(b) Biomass/ square kilometer (Area)
(c) Number/of square kilometer (Area)
(d) Weight/square kilometer (Area)

Question:10


Label the three tiers 1, 2, 3 given in the above age pyramid.
b. What type of population growth is represented by the above age pyramid?

Answer:

(a) 1: Pre-reproductive, 2: Reproductive, 3: Post-reproductive
(b) This age pyramid shows an expanding population.

Question:11

In an association of two animal species, one is a termite which feeds on wood and the other is a protozoan Trichonympha present in the gut of the termite. What type of association do they establish?

Answer:

Termite and Trichonympha show mutualism; a relationship in which both the organisms are benefited from each other. Trichonympha breaks down cellulose in the wood and thus help the termite in the digestion of wood. Trichonympha also feeds on fragments of wood.

Question:12

Lianas are vascular plants rooted in the ground and maintain erectness of their stem by making use of other trees for support. They do not maintain a direct relationship with those trees. Discuss the type of association the lianas have with the trees.

Answer:

Lianas are in competition with other trees. While taking support from other trees, lianas usually strangulate the other tree. They also compete for sunlight and water and in the process, harm the host tree.

Question:14

What is a tree line?

Answer:

The edge of the habitat at which trees are capable of growing is called the tree line. Trees show stunted growth at the tree line. They cannot grow beyond the tree line because of unfavourable environmental conditions.
Treeline shows a gradual transition from dense cover to sparse cover.

Question:15

Define ‘zero population growth rate’. Draw an age pyramid for the same.

Answer:

Zero population growth rate is seen when the population of a location neither grows nor declines. The following pyramid shows a population with zero growth rate.

Organisms and Population Exercise: 1.4

Question:16

List any four characters that are employed in the human population census.

Answer:

Following are the four characters that are employed in human population census:

  1. Sex-ratio

  2. Age group

  3. Occupation

  4. Income

Question:17

Give one example for each of the following types.
a) Migratory animal
b) Camouflaged animal
c) Predator animal
d) Biological control agent
e) Phytophagous animal
f) Chemical defence agent

Answer:

(a) Siberian crane (b) Chameleon (c) Lion (d) Dragonfly (e) Honey bee (f) Capsaicin produced by bell pepper

Question:18

Fill in the blanks


Species A

Species B

Type of interaction

Example

+

-

......B......

.....D....

+

+

.......C.......

......E.....

+

......A......

Commensalism

....F.....


Answer:

Species A

Species B

Type of interaction

Example

+

.

Predation

Rabbit and lion

+

+

Mutualism

Lichen

+

0

Commensalism

Egrets nd cattle


Question:19

Observe the set of 4 figures A, B, C and D and, answer the following questions
i. Which one of the figures shows mutualism?
ii. What kind of association is shown in D?
iii. Name the organisms and the association in C.
iv. What role is the insect performing in B?

Answer:

(a) Figure ‚’ A’ shows mutualism because the butterfly gets the food from the plant and helps the plant in pollination.
(b) This shows predation in which a carnivore is eating herbivore.
(c) This shows the association between egrets and buffaloes. This shows commensalism.
(d) Insect is performing the role ectoparasite.

Long Answer Type Questions:

Question:1

Comment on the following figures: 1, 2 and 3:


Answer:

Figure 1 shows a population of species A. All the individuals are interacting among themselves.
Figure 2 shows a community with three populations of A, B and C. They are interacting with each other and their environment.
Figure 3 shows a biome with three communities. One of the communities is in the climax, and the other two are at different stages of development. All the three communities are in the same environment, and they interact with each other and with the environment.

Question:2

An individual and population have certain characteristics. Name these attributes with definitions.

Answer:

A population has certain characteristics that and individual does not have. Following are the main attributes of the population:

  1. Birth Rate: An individual is born, but a population shows the birth rate. The birth rate is usually expressed in terms of the number of births per thousand population.

  2. Death Rate: An individual dies, but a population shows the death rate. The death rate is usually expressed in terms of a number of deaths per thousand population.

  3. Sex Ratio: An individual is either a male or a female, but a population shows sex ratio. The ratio of a number of males to the number of females is called sex ratio. For example, the sex ratio in India is 934 females to 1000 males.

  4. Age Distribution: An individual can be of a certain age, but a population has people of different ages. For convenience, age groups are taken into consideration. Age group is normally categorized as pre-reproduction, reproduction, and post-reproduction age groups.

  5. Population Density: Population density is usually taken in terms of the number of individuals per square km. Population density depends on various abiotic factors. The plains usually have extremely high population density while mountains have very low population density.

Question:3

The following diagrams are the age pyramids of different populations. Comment on the status of these populations.

Answer:

Figure A shows a population, in which the ratio of individuals in pre-reproductive age is highest, following by reproductive age and post-reproductive age. This pyramid shows a growing population.
Figure B shows a population in which the number of individuals in pre-reproductive and reproductive age is equal. A number of individuals in post-reproductive age is less. This pyramid shows a stable population, i.e. population growth, is negligible.
Figure C shows a population in which the number of individuals in pre-reproductive age is the least. The number of individuals in reproductive age is the highest. Less number of individuals in pre-reproductive age shows a negative growth rate in the population. This pyramid shows a population in de-growth.

Question:4

Comment on the growth curve given below.

Answer:

The growth curve in this figure shows logistic growth of population. We know that resources are limited in nature, and they cannot support unlimited growth in population. Limited resources mean tough competition for resources among individuals. Only those survive which are fit to survive in such a condition. This is what happens in a real-life situation.
In the initial phase, the population shows a lag phase, i.e. it grows at a slow pace.
The lag phase is followed by a phase of acceleration, during which population growth is rapid. The acceleration phase is followed by a phase of deceleration, during which population growth slows down.
Finally, a phase comes when population density reaches the carrying capacity. Carrying capacity defines the load limit of nature; in terms of sustaining a particular population density. Once the population density reaches the carrying capacity, it reaches the phase of the asymptote. This is the phase when population beings to show negative growth.

Question:5

A population of Paramoecium caudatum was grown in a culture medium. After 5 days, the culture medium became overcrowded with Paramoeium and had depleted nutrients. What will happen to the population and what type of growth curve will the population attain? Draw the growth curve.

Answer:

The population of Paramoecium will show logistic growth which can be shown by a sigmoid curve. The population growth will show the following phases:
(a) Lag phase
(b) Acceleration phase
(c) Deceleration phase
(d) Asymptote phase
The asymptote phase shall be reached on the fifth day. The following figure shows the growth curve:

Question:6

Discuss the various types of positive interactions between species.

Answer:

Following are the positive interactions between species:
(a) Mutualism: This is a relationship between two organisms in which both the organisms are benefitted. Lichen is a particularly good example of mutualism. Lichen is an association of fungi and algae. The fungal part attaches to the substratum and gets water and minerals for the algal part. The algal part carries out photosynthesis and provides food to the fungal part.
(b) Commensalism: This is a relationship between two organisms in which one organism is benefitted, while another organism is neither a gain nor a loss. Egrets keep on sitting on the back of herbivore animals, like buffalo, cow, etc. when the cattle forage for grass, the insects under the grass fly away haphazardly. This gives a chance to the egrets to feed on insects. Without the help of cattle, it will be difficult for egrets to find insects in the grass. In this relationship, there is no benefit to the cattle.

Question:7

In an aquarium, two herbivorous species of fish are living together and feeding on phytoplankton. As per the Gause’s Principle, one of the species is to be eliminated in due course of time, but both are surviving well in the aquarium. Give possible reasons.

Answer:

Gause proposed that when resources are limited then out of the two competing species, one of the species will eventually become extinct. The species with superior predatory skills will survive in the end. But the evidence for such conclusive extinction is rare in nature. Many recent studies do not support Gause’s Principle of competitive exclusion. One of the means to survive is called resource partitioning.
For example, if two species compete for the same resource; they partition their time of foraging or different foraging patterns. MacArthur showed that five species of warblers living on the same tree effectively change their feeding patterns to survive in limited resources. The same thing can happen in the given aquarium. Either the resources are unlimited, or two herbivorous species of fish may have changed their feeding times or feeding pattern to survive in the aquarium.

Question:8

While living in and on the host species, the animal parasite has evolved certain adaptations. Describe these adaptations with examples.

Answer:

Parasites show the following types of adaptations:
(a) Morphological Adaptations: Many parasites are bigger in size than their free-living counterparts are. This helps in producing more eggs, necessary for survival. Endoparasites lose unnecessary sense organs to conserve resources. Endoparasites lose their gut or gut is reduced in size. They absorb nutrients through the skin. Special organs for attachment are present, e.g. suckers.
(b) Life Cycle Adaptations: An endoparasite needs some way to enter another host. Many parasites use a secondary host for this purpose. For example, plasmodium uses mosquitoes as secondary hosts so that they can easily infect another primary host. Some parasites lay their eggs in the lower portion of the alimentary canal of the host so that eggs or cysts can go out along with stool. Later on, the cysts get transmitted to another host through contaminated food and water, e.g. Entamoeba histolytica.
(c) Immunological Adaptations: Some parasites engulf antigens from the host. Some parasites disturb the immunological function of the host. These mechanisms help the parasite survive immunological response of the host.
(d) Biochemical Adaptations: Many endoparasites do not produce digestive enzymes. They rely on the host’s digestive enzymes and absorb the simple nutrients. Endoparasites do not carry out the synthetic reaction. This is possible because their energy need is very low, which is fulfilled from the host’s resources.

Question:9

Do you agree that regional and local variations exist within each biome? Substantiate your answer with suitable example.

Answer:

A biome is a climatically and geographically contiguous area with similar climatic conditions. Both organic and inorganic components are similar in a particular type of biome. However, regional and local variations exist within each biome. To understand this, let us consider the following examples.
Coniferous forest is a type of biome which mainly contains trees which bear cones with sees inside. In all the coniferous forests around the world, coniferous trees are the main plant species. But regional variations exist in coniferous forests. The coniferous forests of North America are dominated by black spruce. On the other hand, the coniferous forests of northeast Europe are dominated by Norway spruce. Forests of Siberia are dominated by pine. This shows regional variations in a particular type of biome.
In the case of coniferous forests, the canopy is so thick that negligible sunlight reaches the ground. This results in sparse vegetation on the ground. The dense canopy on top and sparse vegetation at ground show local variations in a particular type of biome.

Question:10

Which element is responsible for causing soil salinity? At what concentration does the soil become saline?

Answer:

Minerals are responsible for causing soil salinity. Following are some of the cause for the increase in soil salinity:
High level of salt in the soil.
Water table movement because of landscape features.
Human activities; like clearing of land and excess use of synthetic fertilizers.
Salt runoff from streets can happen in cold climates because salt is applied on the road to prevent snow accumulation.
Measurement of Soil salinity: Soil salinity is measured in terms of a gram per litre of soil solution or electrical conductivity in dS/m. Soil salinity is usually determined from an extract of the saturated paste of soil, and then EC is termed as ECe. When 4 < ECe < 8, the soil is termed as slightly saline. When 8 < ECe < 16, the soil is termed as moderately saline. When ECe > 16, the soil is termed as highly saline.

Question:11

Does light factor affect the distribution of organisms? Write a brief note giving suitable examples of either plants or animals.

Answer:

Light is an important factor in determining the distribution of organisms. We know that green plants need sunlight to carry out photosynthesis, and hence the distribution of green plants is highly dependent on the amount of illumination. This is the reason; tropics have dense plant cover, but poles have negligible plants. Moreover, plants also need sunlight for flowering. This phenomenon is called photoperiodism.
Hence, based on photoperiodism, plants are categorized as short-day plants, long-day plants and day-neutral plants. The area near equator gets a long duration of light as compared to the area near poles. Thus, this affects the distribution of plants on earth. Animals too depend on light and show certain behavioural pattern based on diurnal and seasonal variations in light. Many animals get up early in the morning and start foraging. The nocturnal animals remain active during the night only, e.g. owl. Reproductive and migratory activities are also governed by the amount period of illumination in many animals.

Question:12

Give one example for each of the following:
i. Eurythermal plant species _______________
ii. A hot water spring organism _______________
iii. An organism seen in deep ocean trenches _______________
iv. An organism seen in compost pit _______________
v. A parasitic angiosperm _______________
vi. A stenothermal plant species _______________
vii. Soil organism _______________
viii. A benthic animal _______________
ix. Antifreeze compound seen in antarctic fish _______________
x. An organism which can conform _______________

Answer:

(a) Gracilaria (b) Alycyclobacillus (c) Giant tube worms (d) Earthworm (e) Misteltoe (f) Pine (g) Earthworm (h) Sponge (i) Antifreeze glycoproteins (j) Mangifera

Introduction to NCERT Exemplar Class 12th Biology Chapter 13 Organisms and Population

NCERT exemplar Class 12 Biology solutions chapter 13 deals with a variety of events happening around us. It ends the curiosity of students regarding various things happening around related to living organisms and their physical environment and their interaction. The world around us is full of diversity and complexity, which is not easy to understand. The NCERT exemplar Class 12 Biology chapter 13 solutions mainly focus on ecology that studies the interaction of organisms with other organisms and their abiotic environment and how all of these affect one another.

Class 12 Biology NCERT exemplar solutions chapter 13 discovers organism and population levels of ecological study and how these organisms in nature are affected or react to the outside environment including land, water, light and soil. The chapter also focuses on different adaptations and responses to various abiotic factors.

Major Subtopics in NCERT Exemplar Class 12 Biology Solutions Chapter 13 Organisms and Population

  • Organism and its Environment
  • Major Abiotic Factors
  • Temperature
  • Water
  • Light
  • Soil
  • Responses to Abiotic Factors
  • Regulate
  • Conform
  • Migrate
  • Suspend
  • Adaptations
  • Populations
  • Population Attributes
  • Population Growth
  • Natality
  • Mortality
  • Immigration
  • Emigration
  • Life History Variation
  • Population Interactions
  • Predation
  • Competition
  • Parasitism
  • Commensalism
  • Mutualism

NCERT Exemplar Class 12 Biology Solutions Chapter 13 Organisms and Population – Learning Outcomes

The Class 12 Biology NCERT exemplar solutions chapter 13 will give students a knowledgeable insight on ecology at orgasmic and population levels. The students will learn about new methods and complexities of various things happening around them between different organisms and their interaction with the physical environment. The chapter is important for students who want to establish their career in zoology and the concepts defined here also aid in different methods of biotechnology.

NCERT exemplar solutions for Class 12 Biology chapter 13 also help students to learn and understand different concepts of ecology in an effective and easy manner. The chapter also deals with brief study of different population attributes, and population growth that will help students to relate to their outer environment even more nicely.

NCERT Exemplar Class 12 Biology Chapter Wise Links:

Important Topics in NCERT Exemplar Class 12 Biology Solutions

The students through NCERT exemplar Class 12 Biology solutions will learn a lot more about interactions and phenomenon happening around them in between different organisms, and organisms and their physical environment. Some of the most important topics to study and read are organisms and their environment, their interaction with abiotic factors, their response to abiotic factors, the concept of adaptation, population, their growth and their attributes.

NCERT exemplar Class 12 Biology solutions chapter 13 is an important lesson to learn for students that have decided to establish their careers in zoology, biotechnology and also as a vet, or doctor. The chapter talks about various important happenings around us that are not easily understandable, but can be effectively be explained through science.

NCERT Exemplar Class 12 Solutions Subject Wise:

Check NCERT solutions for questions given in the book:

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Frequently Asked Questions (FAQs)

1. 1. Is the study material reliable from the point of view of exams?

The solutions provided under NCERT exemplar Class 12 Biology solutions chapter 13 are fully reliable for exams and are created by trained professionals to help you score great in exams.

2. 2. Is the chapter important for Board exams?

Yes, the chapter 13 of Biology, organisms and population is an easy as well a very interesting chapter that can help you score marks through thorough study and understanding.

3. 3. Are the solutions available offline?

Yes, the NCERT exemplar Class 12 Biology solutions chapter 13 can be access offline by you according to your convenience by clicking and downloading the solutions in pdf format through NCERT exemplar Class 12 Biology solutions chapter 13 pdf download.

4. 4. What are some of the important topics to focus on from the chapter?

Chapter 13 of biology is very scoring from the point of view of exams and covers topics such as organisms and their environment, their interaction with abiotic factors, their response to abiotic factors, the concept of adaptation, population, their growth and their attributes that should be understood and learned thoroughly.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.
  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.
  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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