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# Organisms and Populations Class 12 Biology Important Questions

Please refer to Organisms and Populations Class 12 Biology Important Questions with answers below. These solved questions for Chapter 13 Organisms and Populations in NCERT Book for Class 12 Biology have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these solved problems properly as these will help them to get better marks in your class tests and examinations. You will also be able to understand how to write answers properly. Revise these questions and answers regularly. We have provided Notes for Class 12 Biology for all chapters in your textbooks.

## Important Questions Class 12 Biology Chapter 13 Organisms and Populations

All Organisms and Populations Class 12 Biology Important Questions provided below have been prepared by expert teachers of Standard 12 Biology. Please learn them and let us know if you have any questions.

Case Based MCQs

Read the following passage and answer the questions from 46 to 50 given below.
Growth of a population with time shows specific and predictable patterns. Two types of growth patterns of population are exponential and logistic growth. When resources in the habitat are unlimited each species has the ability to realise fully its innate potential to grow in number. Then the population grows in exponential fashion.
When the resources are limited growth curve shows an initial slow rate and then it accelerates and finally slows giving the growth curve which is sigmoid.

46. Which of the following statement is incorrect?
(a) Exponential growth occurs in organism such as lemmings.
(b) Logistic growth is more realistic.
(c) Exponential growth has two phases lag and log.
(d) In logistic growth, population passes well beyond the carrying capacity of ecosystem.

Question. Which of the following equations correctly represents Verhulst-Pearl logistic growth?
(a) dN/dt = rN(K − N) K
(b) dN/dt = rN /K
(c) dN/dt = N(K N)/K
(d) dN/dt = r(K N)/K

Question. Study the population growth curves (A and B) in the given graph and select the incorrect option.

(a) Curve ‘A’ shows exponential growth,represented by equation dN /dt = rN. .
(b) Curve ‘B’ shows logistic growth, represented by equation dN /dt = rN (K – N)/K.
(c) Exponential growth curve is considered as more realistic than the logistic growth curve.
(d) Curve ‘A’ can also be represented by equation Nt = N0ert.

Question. The population growth is generally described by the following equation:
dN /dt = rN ( K – N) /K
What does ‘r’ represent in the given equation?
(a) Population density at time ‘t’
(b) Intrinsic rate of natural increase
(c) Carrying capacity
(d) The base of natural logarithm

Question. Which of the following equations correctly represents the exponential population growth curve?
(a) dN/dt = rN
(b) dN/dt=rN(K −N) K
(c) Nt = N0ert
(d) Both (a) and (c)

Question. Give one example where population estimation of an organism is done indirectly without actually counting the organism.
Answer : Sometimes population size is indirectly estimated without actually counting them, for example, tiger census in our National parks and tiger reserves is often based on pug (animal’s foot print) marks and faecal pellets.

Question. Plants like Calotropis have evolved adaptations for defence against grazers. Explain.
Answer :  In order to keep predators or herbivores away from grazing, certain plants like Calotropis produce highly poisonous cardiac glycosides that can make herbivores sick when eaten. Hence, grazers like cattle, goats and sheep do not graze it.

Question. In a population, per capita birth rate is 0.025 and per capita death rate is 0.008 during a unit time period. What is the value of intrinsic rate of natural increase, ‘r’ for the population?
Answer :  Intrinsic rate of natural increase, r = Per capita birth rate – Per capita death rate = 0.025 – 0.008 = 0.017.

Question. Name the type of interaction that exists between barnacles and whale.

Question. Give an example of drought escapers xerophytic plant.
Answer :  Emphemerals are drought escapers xerophytic plants, for example: Tribulus terrestris.

Question. When and why do some animals like frogs hibernate?
Answer : When animals are exposed to low temperatures, hibernation is necessary for cold-blooded animals like frogs to prevent their metabolic rate from getting slow down.

Question. When and why do some animals like snails go into aestivation?
Answer : When animals like snails are exposed to lethal high temperatures, they go into aestivation to avoid the heat of summer.

Question. What is an interaction called when an orchid grows on a mango plant?

Question. Soil horizons A and B represent solum. Why?
Answer :  Only A and B horizons of soil represents solum or true soil because they have weathered products of the parent rock.

Question. Give example of an organism that enters ‘diapause’ and why.
Answer : Bombyx mori (silk moth) is an insect that enters diapause due to some adverse environmental conditions such as drought, extreme temperature, reduced food availability which, in turn, delays the overall development. The physiological and metabolic activities diminish at this particular time.

Question. Differentiate between the mutualism and competition.

Question. Why the plants that inhabit a desert are not found in a mangrove? Give reasons.
Answer : Plants inhabiting desert (xerophytes) are not found in mangroves, because xerophytic plants are adapted to dry and hot environment. They possess various physical modifications to tolerate extreme water scarcity and heat, like extensive root system, succulent organs, leaf reduced to spine, etc. Mangrove swamp is a region of vegetation where soil is highly saline and water logged. Only halophytes can survive in such regions as they possess aerial roots called pneumatophores through which gaseous exchange occurs. Roots of xerophytes are positively geotropic and will suffocate and die in such badly aerated soil ultimately leading the whole plant to death.

Question. What does the given age pyramid signify about the status of a population? (The bar at the base represents pre-reproductive individuals.)

Answer : Given is a bell-shaped age pyramid which signifies that the population is stable. Such age pyramid is formed when the number of pre-reproductive and reproductive individuals is almost equal and the post-reproductive individuals are comparatively fewer. It implies that the population is neither decreasing nor increasing, instead is maintained at a stable level.

Question. Define the following:
(a) Ecological niche
(b) Gause’s competitive exclusion principle
(c) Mimicry
Answer : (a) Ecological niche (Grinnel, 1917) is a specific part of habitat occupied by individuals of a species which is circumscribed by its range of tolerance, range of movement, microclimate, type of food and its availability, shelter, type of predator and timing of activity. (b) Gause’s competitive exclusion principle states that two or more species with similar niche requirements cannot coexist indefinitely in the same area and one of the two gets eliminated. (c) Animals develop strategies to live better in their environment. Mimicry is the resemblance of one species with another in order to obtain advantage, especially against predation.

Question. Shark is eurythermal while polar bear is stenothermal. What is the advantage the former has and what is the constraint the later has?
Answer : Sharks being eurythermal can tolerate wide range of temperature variations and thus have wider distribution on earth, on the other hand, polar bear being stenothermal can tolerate only narrow range of temperature and is restricted to specific regions only.

Question. (a) List any three ways of measuring population density of a habitat.
(b) Mention the essential information that can be obtained by studying the population of an organism.
Answer :  (a) The different methods to study population size are as follows: – Quadrat method : It is a method which involves the use of square of particular dimension to measure number of organisms. For example the number of Parthenium plants in a given area can be measured using the quadrat method. – Direct observation: It involves counting of organisms. For example, in order to determine the number of bacteria growing in a petri dish, their colonies are counted. – Indirect method : The number of fishes caught per trap gives the measure of their total density in a given water body.
(b) Whether competition of survival exists or not, whether the population is increasing or decreasing, natality, mortality, emigration and immigration.

Question. Explain why very small animals are rarely found in polar region.
Answer : Very small animals have large surface area to body volume ratio. It results in excessive heat loss from exposed body surface. Such a great extent of heat loss makes it impossible for very small animals to survive in cold polar regions. Large animals have small surface area to body volume ratio, reducing heat loss and making temperature maintenance easier for them. This effect of temperature on the absolute size of an animal and the relative proportions of various body parts is also known as Bergmann’s rule.

Question. Explain parasitism and coevolution with the help of one example of each.
Answer : Parasitism is the interspecific interaction where one of species (called parasite) depends on the other species (host) for food and shelter and damages the host. E.g., malarial parasite in blood cells of humans. Coevolution in parasitism refers to the process in which parasite evolves mechanism to interact and neutralise the mechanism evolved by the host to reject or resist parasite.

Question. (a) How are herbs able to grow on forest floors?
(b) What are osmoconformers?
Answer : (a) Sciophytes or shade plants like herbs and shrubs survive under the shadow of big canopied trees or sun plants or heliophytes in forests as they are perfect shade tolerant plants which show better growth in lower level of light intensity. They grow in a manner, that they are arranged in different strata according to their shade tolerance.
(b) Osmoconformers are those organisms which cannot maintain constant osmolarity of their body fluids and it varies according to their surrounding medium.

Question. Why do clown fish and sea anemone pair up? What is this relationship called?
Answer : Mutualism is the interaction between clown fish and sea anemone. The clownfish lives among the stinging tentacles of sea anemone and gets protection from its predators. Clownfish has a slimy mucus covering that protects it from the poisonous tentacles of sea anemone. Also clown fish makes its meals from the anemone’s leftover. In return clown fish helps anemone in catching its prey by luring other fish towards it. It also eats the dead tentacles keeping the anemone and the area around it clean.

Question. Draw and explain expanding age pyramids of human population. Why is it so called?
Answer : An age pyramid for expanding human population is as follows:

Pyramid with broad base or triangular shape indicates a rapidly expanding population with a high percentage of prereproductive individuals followed by reproductive then postreproductive individuals. Thus, in rapidly growing population, birth rate is high and population keeps growing.

Question. During a school trip to ‘Rohtang Pass’, one of your classmates suddenly developed ‘altitude sickness’. But, she recovered after sometime.
(a) Mention one symptom to diagnose the sickness.
(b) What caused the sickness?
(c) How could she recover by herself after sometime?
Answer : (a) Heart palpitation (b) Sickness is due to low atmospheric pressure of high altitudes, as body does not get enough oxygen.
(c) After sometimes, body compensates for low oxygen availability by increasing red blood cell production, decreasing binding capacity of haemoglobin and by increasing breathing rate.

Question. Refer to the given graph and answer the following questions.

(a) What does the graph represent? Identify A, B and C.
(b) Differentiate between A and B.
(c) How organism C regulates their body functions?
Answer :  (a) The given graph represents various ways of organismic response i.e., possibilities of living organisms to cope with stressful conditions. A represents the conformers, B represents regulators and C represents partial regulators. (b) Differences between A (conformers) and B (regulators) are as follows:

(c) C i.e., partials regulators have the ability to regulate body functions to a limited extent. Beyond that limit they become conformers.

Question. Name and explain the kind of interaction in the following:
(i) Algae and fungi in lichens
(ii) Hermit crab and sea anemone
Answer : (i) Mutualism is found between algae and fungi in lichen. Lichen is a composite entity which is formed jointly by an alga and a fungus. The fungus provides water, minerals and shelter to the alga. In return alga provides food to the fungus.
(ii) Interaction between sea anemone and hermit crab is considered as an example of mutualism (or as protocooperation by some). Sea anemone uses hermit crab as a portable home and is able to find more food. Hermit crab in turn gets protection from its enemies. Thus it is a mutually beneficial association. Recent studies reveal that it is obligate mutualism not proto-cooperation.
(iii) Parasitism is found between head louse and humans. Head louse is an ectoparasite that sucks the blood of man or feed on living tissues of head’s skin.

Question. Predation is usually referred to as detrimental association. State any three positive roles that a predator plays in an ecosystem.
Answer : Predators play important role in ecosystem. These are discussed as follows: (i) Maintaining prey population : In nature, the population of predator is quite small as compared to that of the prey. The prey has high reproductive potential. If, for some time, the prey population is allowed to grow without predation, then it would grow beyond the carrying capacity of the environment. The predator keeps the population of the prey under check so that an equilibrium is maintained. Example, the prickly pear cactus introduced in Australia in the early 1920’s caused havoc by spreading rapidly into millions of hectares of rangeland. Finally, the invasive cactus was brought under control only after a cactus-feeding predator (a moth) from its natural habitat was introduced into the country. (ii) Maintaining species diversity : Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among competing prey species. Example, in the rocky intertidal communities of the American Pacific Coast, the starfish Pisaster is an important predator. When all the starfish were removed from an enclosed intertidal area, more than 10 species of invertebrates became extinct within a year because of interspecific competition. (iii) Vegetation : Predation helps in growth of vegetation all over the globe by restricting population of herbivores.

Question. Study the graph given below and answer the questions that follow :

(i) The curve ‘b’ is described by the following equation:

What does ‘K ’ stand for in this equation? Mention its significance.
(ii) Which one of the two curves is considered a more realistic one for most of the animal populations?
(iii) Which curve would depict the population of a species of deer if there are no predators in the habitat? Why is it so?
Answer : (i) In the given graph, ‘a’ represents exponential or J-shaped growth and ‘b’ represents logistic or sigmoid growth. The equation dN /dt = rN (K – N) K , represents logistic growth form and ‘K’ represents carrying capacity for a particular species in the given habitat. In nature, a given habitat has resources to support a certain number of individuals of a population, beyond which no further growth is possible. This limit is called nature’s carrying capacity (K) for that species in that habitat.
(ii) The curve ‘b’ is considered to be more realistic growth model for most of the animal population because resources are limited in this type of growth curve. Whereas, in case of curve ‘a’, the resources (such as food, space, etc.) are unlimited.
(iii) The curve ‘a’ would depict the population of a species of deer in absence of predators in the habitat as the population increases exponentially. In absence of predators, the resources will be unlimited for the deer population and it can reach high population densities in a short time. This type of growth pattern of a population results in J-shaped curve.

Question. A species is introduced into a new habitat with favourable environment. What kind of population growth does it undergo on a long term basis? Describe briefly.
Answer : When a species is introduced into a new habitat with favourable environment (unlimited resources) it undergoes exponential growth. It has two phases-lag and log phase. In lag phase, species shows poor growth as trying to establish in the new environment. After its establishment, species shows maximum growth in the log or exponential phase, resulting in J-shape curve.

Question. Name and explain the type of interaction that exists in mycorrhizae and between cattle egret and cattle.
Answer :  Mycorrhiza is a mutualistic interaction between fungus and roots of higher plants. The root provides food and shelter to the fungus. The fungus helps the plant in solublisation and absorption of minerals, water uptake and protection against pathogenic fungi. The egret and grazing cattle in close association is an example of commensalism. Commensalism is the interaction in which one organism is benefitted and other organism is neither harmed nor benefitted. The egrets always forage close to where the cattle are grazing because the cattle, as they move, stir up and flush out from the vegetation insects that otherwise might be difficult for the egrets to find and catch.

Question. If 8 individuals in a population of 80 butterflies die in a week, calculate the death rate of population of butterflies during that period.
Answer : Death rate is defined as the number of deaths per 1000 individuals of a population. Since, total number of butterflies = 80, Number of butterflies that died = 8 Death rate = 8 = 80 0.1 butterflies per week.

Question. Explain Verhulst-Pearl Logistic Growth of a population.
Answer : According to Verhulst-Pearl logistic growth, population increases in size in sigmoid fashion. S-shaped growth form is found in stable population. It shows population growth in a habitat with limited resources. Population shows initially a lag phase, followed by phases of increase and decrease and finally the population density reaches the carrying capacity. 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 as explained by the following equation : dN / dt = rN (K – N )K  Where N = population density at a time t; r = intrinsic rate of natural increase and; K = carrying capacity.

Question. What type of interaction is seen when koel lays eggs in crow’s nest?
Answer :  Koel or cuckoo laying its eggs in crow’s nest is an example of brood parasitism. In brood parasitism, the parasitic bird lays its eggs in the nest of its host, and the host incubates them. Here, the eggs of the parasitic bird have evolved to resemble the host’s egg (in size and colour) so as to avoid being detected.

Question. When you go for a trek/trip to any high altitude places, you are advised to take it easy and rest for the first two days. Comment, giving reasons.
Answer : Atmospheric pressure is low at higher altitudes as compared to plains. When we go for a trek/trip on high altitude, then due to low atmospheric pressure our body does not get enough oxygen, as a result of which we experience nausea, fatigue and heart palpitation (altitude sickness). But by taking rest for first two days, body gets acclimatised to high altitude conditions. The body compensates low oxygen availability by increasing red blood cell production, decreasing binding capacity of haemoglobin and increasing breathing rate. Hence, we automatically stop experiencing altitude sickness.

Question. Water is very essential for life. Write any three features both for plants and animals which enable them to survive in water scarce environment.
Answer : Water is very essential for life. Plants and animals show modifications according to availability of water in the area. Some of the adaptations seen in plants which enable them to survive water scarce environment are as follows: – Plants of hot deserts are adapted to survive in dry conditions of soil and high temperatures. The plant which evade dry conditions are known as ephemerals. Some plants have deep tap root which are capable of absorbing water from deep soil e.g., Prosopis, Acacia, etc. – In case of cacti and succulents, the presence of fleshy leaves and stems to store water (succulence) is an adaptation to dry environment. In cacti, leaves are reduced to spines, where stems are modified to fleshy structures. – Many tropical plants, particularly grasses which grow in hot and arid climates possess C4 pathway of photosynthesis. So, these plants perform better in low soil water environments. Such plants, use less water to achieve higher rates of photosynthesis. Some of the adaptations seen in animals which enable them to survive water scarce environment are as follows: – Desert lizards lack the physiological ability that mammals have to deal with the high temperature. They keep their body temperature fairly constant by behavioural means. They enjoy in the sun and absorb heat when their body temperature drops below the comfort zone but move into shade when the surrounding temperature starts increasing. – The Kangaroo rats conserves water by excreting solid urine and can live from birth to death without even drinking water. Loss of water is minimised by producing nearly solid urine and faeces. – The camels show tolerance to wide fluctuations in body temperature and are able to maintain blood stream moisture even during extreme heat stress.

Question. Differentiate between hibernation and aestivation. Give one example of each.
Answer : The differences between hibernation and aestivation are as follows:

Question. Some organisms suspend their metabolic activities to survive in unfavourable conditions.Explain with the help of any four examples.
Answer : To tide over unfavourable conditions, some organisms suspend their metabolic activities. These are discussed as follows :
(i) Bacteria, fungi and lower plants develop thick walled spores, which germinate during suitable conditions.
(ii) Polar bears go into hibernation during winter season to escape cold.
(iii) Some snails and fish undergo aestivation to avoid summer related problems like heat and dessication.
(iv) During unfavourable conditions, zooplanktons in lakes and ponds are known to enter diapause, i.e., stage of suspended development.

Question. List the different ways by which organisms cope or manage with abiotic stresses in nature. Explain any three ways listed.
Answer : Living organisms cope with stressful conditions by various methods :
(i) Hibernation and aestivation : Hibernation is winter sleep in which animal passes the winter period in dormant condition in a warm place. Polar bears hibernate during winters. Aestivation is summer sleep in which animal rests in a cool/ shady and moist place during extreme heat period. Ground squirrels of South-Western deserts undergo aestivation and lie in torpid state inside burrows during hot dry periods.
(ii) Camouflage : It is the ability to blend with the surroundings or background. It is protective to animals which are preyed upon by others and it is also advantageous to predators as it eases predation, e.g., it is difficult to distinguish leaf like grasshopper from the surrounding foliage.
(iii) Mimicry : It is resemblance of one species with another in order to obtain advantage specially against predation. The species which is imitated is called model while the species which imitates is known as mimic. E.g., Viceroy butterfly mimics unpalatable, toxic Monarch butterfly.
(iv) Migration : The organisms can migrate temporarily from the unfavourable habitat to more favourable area and return when unfavourable period is over. Many animals, particularly birds, during winter undergo long-distance migrations to more favourable areas.
(v) Perennating structures : Various kinds of thick walled spores are formed in bacteria, fungi and lower plants which help them survive under unfavourable conditions. These germinate on return of suitable conditions.
(vi) Diapause : Under unfavourable conditions many zooplanktons in lakes and ponds are known to enter diapause i.e., a stage of suspended development.

Question. (a) What is “population” according to you as a biology student?
(b) “The size of a population for any species is not a static parameter”. Justify the statement with specific reference to fluctuations in the population density of a region in a given period of time.
Answer : (a) According to me as a biology student, population is defined as the total number of interbreeding individuals of a species found in a geographical area who share and compete for similar resources. (b) The population density is the number of individuals of a species per unit area/space at a given time. The size of a population (population density) is not a static parameter. It keeps changing with time, depending upon a number of factors : abiotic and biotic, food availability, predation pressure, etc. The density of a population changes due to four basic processes: (i) Natality : Number of births during a given period per unit population. (ii) Mortality : Number of deaths in the population during a given period.
(iii) Immigration : Number of individuals of the same species moving inside a population during the time period. (iv) Emigration : Number of individuals moving outside from a habitat during the time period. Therefore, if N is the population density at time t, then its density at time t + 1 can be explained by the given equation : Nt+1 = Nt + [(B + I) – (D + E)] Where, B represents natality or number of births; I represents number of immigrants; D represents mortality or number of deaths; E represents number of emigrants. From the above equation it is clear that population density increases if the number of births plus the number of immigrants (B + I) is more than the number of deaths plus the number of emigrants (D + E). Otherwise it will decrease.

Question. Study the table given below in regard to population interactions and answer the questions that follow :

[Note: (+) plus = beneficial interaction; (–) minus = detrimental interaction; (0) zero = neutral interaction]
(i) Identify the interactions.
(ii) Explain each one of them.
Answer : (i) (a) = Amensalism (b) = Parasitism/Predation (c) = Competition (d) = Mutualism (e) = Commensalism
(ii) (a) Amensalism is an association between two species that is detrimental to one of the species but has no effect on the other. A common example of amensalism is the release of chemical toxins by plants that can inhibit the growth of the other plant species (allelopathy).
(b) Parasitism is an association in which one organism (the parasite) lives on or in the body of another organism (host), from which it obtains its nutrients. It is an one sided relationship in which parasite is benefitted and host is harmed.
(c) Competition is a rivalry between two or more organisms for obtaining the same resources. It may be between individuals of same species (intraspecific) or different species (interspecific).
(d) Mutualism is an interaction between two organisms of different species where both the partners are benefitted and are obligatory to each other.
(e) Commensalism is the interaction between two individuals of different species in which one is benefitted while other remains unaffected.

Question. (a) Compare, giving reasons, the J-shaped and S-shaped models of population growth of a species.
(b) Explain “fitness of a species” as mentioned by Darwin.
Answer : (a) The comparison between J-shaped or exponential growth and S-shaped or logistic growth is as follows :

(b) The fitness, according to Darwin, refers ultimately and only to reproductive fitness. Hence, those who are better fit in an environment, leave more progeny than others. These, therefore, will survive more and hence are selected by nature. He called it natural selection and implied it as a mechanism of evolution.

Question. (a) Study the flow chart given below and complete the equation that follows by identifying 1, 2, 3 and 4.

Nt + 1 = Nt + {(1 + 2) – (3 + 4)}
(b) Mention the different ways by which the population density of different species can be measured.
Answer : (a) In the given equation, 1, 2, 3 and 4 respectively are B, I, D and E. Therefore, the equation will be Nt + 1 = Nt + [(B + I) – (D + E)] (b) Population density is defined as number of individuals of a species per unit area or per unit volume of environment. Population density may be measured by : (i) Numerical density calculated by number of individuals per unit area or volume. For example, if in a pond there were 20 lotus plants last year and through reproduction 8 new plants are added, taking the current population to 28, the birth rate will be calculated as 8/20 = 0.4 offspring per lotus per year.
(ii) Biomass density calculated as biomass per unit area or volume. For example if in an area, there are 200 Parthenium plants but only a single huge banyan tree, then the percent cover or biomass is more meaningful measure of the population size.
(iii) Abundance or absolute number of population. For ecological investigations, population density is measured as absolute population densities or relative densities. For example the tiger census in our National parks and tiger reserves is often based on pug marks and fecal pellets.

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