Origin of Life on Earth and Various Related Evidences
Very Short Answer Type Questions
Question. Define Diapause.
Answer : Diapause is a stage of suspended development in zooplanktons species under unfavourable conditions.
Question. How do seed – bearing plants tide over dry and hot weather conditions ?
Answer : In higher plants, seeds and some other vegetative reproductive structures serve to tide over periods of stress. They reduce their metabolic activity and go into a state of ‘dormancy’. They germinate under favourable moisture and temperature.
Question. Very small animals are rarely found in polar regions. Give two reasons.
Answer : Small animals have larger surface area relative to their volume, loose heat very fast, due to small size, expend much energy to generate body heat through metabolism.
Question. Mention how do bears escape from stressful time in winter ?
Answer : Bears undergo hibernation to get rid of the stressful time in winter.
Question. Mention any two activities of animals which get cues from diurnal and seasonal variations in light intensity.
Answer : Many plants depend on sunlight to meet their photoperiod requirements for flowering. For many animals too, light is important. They use the diurnal and seasonal variations in light intensity and duration (photoperiod) as cues for timing their foraging, reproductive and migratory activities.
Question. How do snails escape from stressful time in summers ?
Answer : They undergo aestivation to avoid summers.
Question. Give an example of an organism that enters ‘diapause‘ and why ?
Answer : Many species of Zooplankton, unfavourable condition.
Short Answer Type Questions – l
Question. Why are the plants that inhabit a desert are not found in a mangrove? Give reasons.
Answer : Desert plants are not adapted to survive in saline / aquatic conditions. Plants are conformers / stenothermal / cannot maintain constant internal environment / temperature / osmotic concentration of the body fluids affects kinetics of enzymes through basal metabolism / activity and other physiological functions of the organisms.
Question. Some organisms suspend their metabolic activities to survive in unfavourable conditions.
Explain with the help of any four examples.
Answer : (i) The frogs hibernate during very cold season and reduce their metabolic activities.
(ii) The kangaroo rat has the ability to concentrate its urine and can live without drinking water, thereby conserving water.
(iii) Barnacles and molluscs, living in very cold inter-tidal zones of northern shores, show their reduce activities.
(iv) Some animals go to summer sleep.
Question. Shark is eurythermal while polar bear is stenothermal. What advantage does the former have and what is the constraint the later has ?
Answer : Shark : Tolerates wide range of temperature so wide spread / survives in all waters.
Polar bear : Restricted occurrence in narrow range of temperature so constraint to live in very cold icy environment.
Question. Many fresh water animals cannot survive in marine environment. Explain.
Answer : Fresh water animals are not able to maintain their osmotic concentration in marine conditions. If they are transferred in marine conditions, their body will shrink due to exosmosis. Hence, they cannot survive in marine environment.
Question. Why do people suffer from altitude sickness after reaching the high altitude regions ? How does their body acclimatized after a couple of days ?
Answer : “Altitude sickness” is because of low atmospheric pressure at high altitude, the body does not get sufficient oxygen.
The body compensates low oxygen availability by increasing RBCs production, decreasing the binding capacity of haemoglobin, by increasing breathing rate.
Question. Between amphibians and birds which are more likely to be able to cope with global warming ? Give reasons.
Answer : The birds are more likely to be able to cope with the global warming because they are eurythermic organisms and as such show a wide range of temperature tolerance.
Question. A moss plant is unable to complete its life cycle in a dry environment. State two reasons.
Answer : Mosses cannot complete their life cycle in a dry environment because of the following reasons :
(i) They need water for sexual reproduction as water acts as a medium for flagellated sperm to reach the egg and undergo fertilization.
(ii) Since their roots are rudimentary, they cannot absorb water. Therefore, they need to grow in moist environment for their survival.
Question. How do mammals living in colder regions and seals living in polar regions able to reduce the loss of their body heat ?
Answer : Mammals from colder climates generally have shorter ears and limbs which minimise heat loss (Allen’s rule).
In polar region, seals have thick layer of fat (blubber) below their skin that acts as insulator and reduce loss of body heat.
Question. Why are some organisms called as eurythermals and some others stenohalines ?
Answer : Eurythermals are the organisms which can tolerate and live in a wide range of temperature while stenohalines can tolerate only a narrow range of salinities.
Question. Why do algae and fungi shift to sexual mode of reproduction just before the onset of adverse conditions ?
Answer : For survival during unfavourable conditions / Fusion of gametes helps to pool their resources for survival (hunger theory of sex) / Zygote develops a thick wall that is resistant to dessication and damage, undergoes a period of rest before germination.
Question. Heat loss or heat gain depends upon the surface area of the organism’s body. Explain with the help of a suitable example.
Answer : Small animal like humming bird / shrew, have a larger surface area relative to their volume, they tend to loose heat when it is cold outside, hence spend much energy to generate body heat through metabolism.
Short Answer Type Questions – ll
Question. Why are certain organisms called regulators or conformers ? Explain with the help of one example of each.
Answer : Regulators maintain / regulate constant body temperature irrespective of external condition. Conformers changing / varying body temperature as per external condition.
Regulators : Humans maintain a constant body temperature of 37°C. In summer when temperature is more outside we sweat, when evaporates causes cooling / in winter when outside temperature is much lower than 37°C, we start shivering / a kind of exercise which produce heat and raises body temperature (any other suitable appropriate example).
Conformers : In aquatic animals, the osmotic concentration of body fluid changes with that of the ambient water osmotic concentration / Internal body temperature of reptiles, amphibians, fishes change with that of external temperature. (Any other suitable appropriate example.)
Question. In certain seasons we sweat profusely while in some other season we shiver. Explain.
Why do we experience shivering during winters when temperature is very low ?
Answer : To regulate body temperature.
In summer, outside temperature is higher than body temperature, sweating causes cooling by evaporation of sweat and thus lowering the body temperature.
In winter, outside temperature is much lower than body temperature, shivering is an (involuntary) exercise which produces heat and raises the body temperature in winter.
Question. Are humming birds and fish regulators or conformers ? Give reasons in support of your answer.
Answer : Conformers.
Heat loss or gain is a function of surface area. Since small animals have a larger surface area (relative to their volume), they tend to lose body heat very fast when it is cold outside, they have to expend much energy, to generate body heat through metabolism (cannot maintain a constant body temperature).
Question. (i) ”Organisms may be conformers or regulators.”
Explain this statement and give one example of each.
(ii) Why are there more conformers than regulators in the animal world ?
Answer : (i) Conformers : Organisms which cannot maintain a constant internal environment under varying external environmental conditions or change body temperature and osmotic concentration with change in external environment E.g. all plants, fishes, amphibians, reptiles.
Regulators : Organisms which can maintain homeostasis (by physiological means or behavioural means), maintain constant body temperature and osmotic concentration E.g. birds, mammals.
(ii) Thermoregulation is energetically expensive for animals.
Question. Explain by taking three different examples how do certain organisms pull through the adverse conditions when unable to migrate under stressful period.
Answer : Hibernation—winter sleep to escape cold weather e.g. bears.
Aestivation—summer sleep to avoid heat and desiccation e.g. snails / fish.
Diapause—suspended development / activity e.g. zooplanktons.
Spore formation—to tide over unfavourable conditions e.g. fungi / bacteria / lower plants.
Dormancy—By reducing metabolic activity e.g. seeds.
Question. How do kangaroo rats and desert plants adapt themselves to survive in their extreme habitat?
Answer : Kangaroo rats : Internal fat oxidation where water
is a byproduct, excretes concentrated urine.
esert Plants : Thick cuticle / sunken stomata
/ leaves reduced to spines / deep roots / Special
photosynthetic pathway / CAM (Any four)
Question. The graph given below shows the distribution of biomes:
a) What do the ‘X’ and ‘Y’ axes represent?
b) Identify the ‘grassland’ and ‘coniferous forest’ biomes, from the above figure.
c) Why is ‘F’ located at the given position in the graph?
Answer : (i) ‘X’ axis–Mean annual precipitation (cm)
‘Y’ axis–Mean annual temperature (0°C)
(ii) Grassland – B
Coniferous forest – E
(iii) The mean annual temperature ranges from – 12 to 20°C (error accepted ± 2) and mean annual precipitation ranges from 10 – 125 cm, these are the optimum conditions in tundra biome.
Question. Explain with the help of suitable examples the three different ways by which organisms overcome their stressful conditions lasting for short duration.
How do organisms cope with stressful external environmental conditions that are localized or of short duration ?
Answer : (i) Migration : The organisms (animals) can move away temporarily from stressful habitat to a more hospitable area and return when stressful period is over.
e.g – Humans moving from Delhi to Shimla during summer / many animals or birds undertake long distance migration to hospitable area. (Any one example)
(ii) Spore formation : Various kind of thick walled spores are formed which germinate on availability of suitable environment. e.g – bacteria / fungi / lower plants. (Any one)
(iii) Dormancy : Seeds or vegetative reproductive structures help to tide over stress by reducing their metabolic activity.
e.g seeds or vegetative reproductive structures of higher plants.
(iv) Hibernation : It takes place during winter. e.g bears or any other correct relevant example.
Aestivation : It takes place during summer to avoid heat and dessication in animals.
e.g snails / fish or any other correct relevant example.
Diapause : Under unfavourable conditions zooplanktons enter a stage of suspended metabolic activity.
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 : Plants : Ephemeral mode (complete life cycle in short period) / Deep tap roots / Deciduous leaves / Waxy cuticle / Sunken stomata / Succulence to store
water / C4 Pathway of photosynthesis. (Any three) Animals : No sweating / uricotelic / deposition of at in sub-epidermal layer / burrowing nature / thick skin / body covered with scales.
Question. When you go for a trip to any high altitude places you are advised to take it easy and rest for first two days. Comment giving reasons.
Why do tribes who live in high altitude of Himalayas experience discomfort in respiration?
How do they get adapted to survive in such a situation?
Answer : At high altitude, the availability of oxygen is low. Therefore, it is advised to take it easy and take rest during initial period of high altitude trip because during this period, the early rest of the body compensates the low O2 availability by increasing RBC production and increasing the breathing rate.
Question. (i) State how the constant internal environment is beneficial to organisms.
(ii) Explain any two alternatives by which organisms can overcome stressful external conditions.
Answer : (i) (a) A relatively constant internal (within the body) environment permits all biochemical reaction and physiological function to proceed with maximum efficiency and thus, enhance the overall ‘fitness’ of the species. Hence, the living system always tends to remain in a steady state with the help of a self–regulatory mechanism. Such a phenomenon, which involves the maintenance of a constant internal environments, is known as homeostasis.
(ii) (a) Regulate : Maintain omeostasis by ensuring constant body temperature (thermoregulation) and constant osmotic concentration (osmoregulation).
(b) Conform : Change the internal environment with the external environment.
Question. During the school trip to ‘Rohtang Pass’, one of your classmates suddenly developed ‘altitude sickness’. But, she recovered after sometime.
(i) Mention one symptom to diagnose the sickness.
(ii) What caused the sickness ?
(iii) How could she recover by her self after sometime?
Answer : (i) The symptom of altitude sickness include difficulty in breathing, nausea, fatigue and heart palpitations.
(ii) The sickness was caused because of lack of sufficient oxygen and comparatively low RBC count.
(iii) She got acclimatised and stopped experiencing altitude sickness because her body compensated low oxygen availability by increasing the red blood cell production.
Long Answer Type Questions
Question. (a) Following are the responses of different animals to various abiotic factors. Describe each one with the help of an example
(b) If 18 individuals in a population of 80 butterflies die in a week, calculate the death rate of population of butterflies during the period
Answer : (a) (i) Regulate: Maintain constant internal temperature / osmotic concentration / homeostasis.
e.g. birds / mammals.
(ii) Conform : Do not maintain constant internal temperature / osmotic concentration / No homeostasis.
e.g. any one example of animal other than birds and mammals.
(iii) Migrate: Temporary movement of organisms from the stressful habitats to hospitable areas and return when stressful period is over. example: Bar headed geese
(iv) Suspend: Reducing / minimising the metabolic activities during unfavourable conditions.
e.g. Polar bear / amphibian / snails / fish / any other example of animals .
(b) Death rate = 18/80 = 0.225 therefore, death rate percentage will be = 0.225 × 100 = 2.25% butterfly death per week [4+
Question. (a) The graph given below represents the organisms response to temperature as an environmental condition.
i) Which one of the two lines represents conformers and why ?
(ii) What does the other line in the graph represent and why ?
(b) Mention the different adaptations the parasites have evolved with, to be able to successfully complete their life cycles in their hosts.
Answer : (a) (i) A is a conformer Cannot maintain homeostasis / constancy of
internal environment by physiological means / their body temperature and osmotic concentration of body fluids changes with the ambient temperature.
Maintain homeostasis by physiological means / capable of thermo regulation / maintain a constant body internal environment.
(b) Loss of unnecessary sense organs, presence of adhesive organs / suckers to cling on to the host, loss of digestive system, high reproductive capacity.
Population and Population Interactions
Very Short Answer Type Questions
Question. Provide an instance where the population size of a species can be estimated indirectly, without actually counting them or seeing them.
Answer : Tiger census in National parks and Tiger reserves was done on the basis of counting pug marks / faecal pellets.
Question. Name the type of interaction that exists between barnacles and whale.
Name the interaction between a whale and the barnacles growing on its back.
Answer : Commensalism.
Question. State Gause’s competitive exclusion principle.
Answer : According to this principle, the two closely related species competing for the same but limited resources cannot co-exist continuously for a long time . Eventually the competitively inferior one will be eliminated.
Question. Name the interaction that exists between cuscuta and shoe-flower plant.
Answer : Parasitism.
Question. How are closely related species of warblers able to co-exist in a competitive environment ?
Answer : They can co–exist due to behavioural differences in their foraging activities.
Question. Name the type of association that the genus Glomus exhibits with higher plants.
Answer : Glomus exhibits arbuscular mycorrhizal (AM) association.
Question. Name the interaction that exists between sucker fish and shark.
Answer : Commensalism.
Question. In a pond there were 200 frogs. 40 more frogs were born in a year. Calculate the birth rate of the population
Answer : Birth rate = 0.2 frogs/yr or 20%.
Question. Name two intermediate hosts which the human liver fluke depends on to complete its life cycle so as to facilitate parasitisation of its primary host.
Answer : Snail and Fish.
Question. State the type of interaction that exists between ticks and dogs.,
Answer : Ecto (Parasitism).
Question. What does nature’s carrying capacity for a species indicate ?
Answer : (In nature) a given habitat has enough or limited resources to support a maximum possible number and nature’s carrying capacity indicates that no further growth in population is possible.
Question. Name two inter specific interactions where one partner is neutral i.e. not affected.
Answer : Amensalism and commensalism.
Question. Name the type of interaction seen between fig and wasps.
Answer : Mutualism.
Question. Why are cattle and goats not seen browsing on Calotropis growing in the field ?
Answer : Cattle and goats are not seen browsing on Calotropis because Calotropis produces a highly poisonous chemical called cardiac glycosides.
Question. Pollinating species of wasps shows mutualism with specific fig plants. Mention the benefits the female wasps derive from the fig trees from such an interaction.
Answer : The female wasp uses the fruit not only as an oviposition (egg-laying) site but uses the developing seeds within the fruit for nourishing its larvae. The wasp pollinates the fig inflorescence while searching for suitable egg-laying sites. In return for the favour of pollination, the fig offers the wasp some of its developing seeds as food for the developing wasp larvae.
Short Answer Type Questions – l
Question. Differentiate between commensalism and mutualism by taking one example each from plants only.
Answer : Commensalism : In this interaction, one species is benefitted and the other species is neither benefitted nor harmed.
e.g. an orchid growing as an epiphyte on the branch of a mango.
Mutualism : In this interaction, both the interacting species are benefitted.
e.g. Lichens exhibit mutualistic relationship between a fungus that absorbs water and nutrients from soil and photosynthesizing algae / cyanobacteria.
Question. Apart from being part of the food chain, predators play other important roles. Mention any two such roles supported by examples.
Explain the role played by predators in a community.
Answer : (i) Predators act as conduits for energy transfer across trophic levels.
(ii) They keep prey population under control.
(iii) They help in maintaining species diversity in a community by reducing intensity of competition among competing prey species.
(iv) An efficient predator may cause extinction of prey species.
Question. What is mutualism ? Mention any two examples where the organisms involved are commercially exploited in agriculture.
Answer : Interaction between two species in which both are benefitted
(i) Rhizobium in the roots (nodules) of legumes
(ii) Mycorrhiza / Glomus with the roots of higher plants.
Question. Explain brood parasitism with the help of an example.
Answer : Koel is a parasitic bird (which has lost the instinct to make its own nest to lay eggs). It has evolved the technique of laying eggs in the nest of a crow. Its eggs bear resemblances to those of crow.
Question. Explain Verhulst – Pearl Logistic Growth of a population.
Answer : A population growing in a habitat with limited resources show initially a lag phase, followed by phases of acceleration and deceleration, finally an asymptote when the population density reaches the carrying capacity.
dN / dt = rN (K – N / K)
Question. Koel is clever enough to lay eggs in a crow’s nest.
Write the reason for this peculiar behaviour. Name the type of interaction.
Answer : So that the crow can incubate the Koel’s eggs.
Interaction – Brood parasitism.
Question. How does Monarch butterfly defend itself from predators ? Explain.
Answer : Monarch butterflies are highly distasteful to their predators. This butterfly species accumulates a chemical by feeding on a poisonous weed during its caterpillar stage.
Question. Besides acting as ‘conduits’ for energy transfer across trophic levels, predators play other important roles. Justify.
Answer : Besides acting as ‘conduits’ of energy transfer across trophic levels, predator play other important roles like :
(i) They keep prey population under control.
(ii) Predators also help in maintaining species diversity in a community by reducing the intensity of competition among competing prey species.
Question. Describe the mutual relationship between fig tree and wasp and comment on the phenomenon that operates in their relationship.
Answer : Wasp helps in pollination while fig tree helps in oviposition and nourishment of larva.
The phenomenon seen here is mutualism (two organisms existing symbiotically or in mutual cooperation and benefits to each other) and Coevolution (the evolution of fig tree and wasp in accordance with the changes in each other) and also coextinction.
Question. Why do clown fish and sea anemone pair up ?
What is this relationship called ?
Answer : Clown fish gets protection from its predators
by moving around the stinging tentacles of the sea anemone. The sea anemone is neither helped nor harmed by the interaction with the fish. This relationship is called Commensalism.
Short Answer Type Questions – ll
Question. Name and explain the type of interaction that exists in mycorrhizae and between cattle egret and cattle.
Answer : Mycorrhiza is an association between fungi and the roots of higher plants. It is an example of mutualism in which both fungi and plants are dependent on each other for nutritional needs.
The fungi help the plant in the absorption of essential nutrients from the soil while the plant provides the fungi with carbohydrates.
The interaction that exists between cattle egret and cattle is known as commensalism. In this type of interaction, one species is benefitted whereas the other is neither benefitted nor harmed.
The cattle egret (bird) usually forages in close proximity to the grazing cattle. As cattle moves in the grass they stir up the grass and flush out the insects which then become an easy target for the egret. In this way, the cattle is neither benefitted nor harmed but the egret is benefitted.
Question. When do you describe the relationship between two organisms as mutualistic, competitive and parasitic ? Give one example of each type.
Answer : Mutualistic : Both the interacting organisms are benefitted from each other e.g., Lichens – Algae and fungi mutually help each other. (any other appropriate example) Competition : When two organisms belonging to closely related species / unrelated groups compete for the same resources that are limited, both are losers e.g. superior barnacle dominates and excludes the small barnacles / in some South American lakes visiting flamingoes and resident fishes compete for their common food
(zooplankton) in the lake / any other appropriate example.
Parasitic : One of the two organisms is dependent on the other (host) for nutrition and support / the host is harmed and the parasite is benefitted e. g.
Malarial parasite and human / Cuscuta on host plant / or any other appropriate example.
Question. Highlight the differences and a similarity between the following population interactions, competition, predation and commensalism.
Question. Highlight the differences between the population interactions given below. Give an example of each.
Answer : Parasitism : Only one species benefits e.g.,
Cuscuta / Tape worm
Amensalism : One species is harmed whereas the other is unaffected e.g., Penicillium growing on bacterial culture / Trichoderma – biological control agent and plant pathogen
Mutualism : Both species are benefitted E. g.,
lichens exhibit mutualistic relationship with fungus that absorbs water and nutrients from soil and photosynthesizing algae / cyanobacteria
Question. (i) Explain any two defence mechanisms plants have evolved against their predators.
(ii) How does predation differ from parasitism ?
Answer : Plants develop following defence mechanisms :
(i) (a) Thorns are (morphological) means of defence.
(b) Plants may produce / store chemicals such as nicotine, strychnine etc. for defence which inhibit digestion / disrupts reproduction / kill the predator / calotropis produces highly poisonous cardiac glycosides / plants may produce chemicals such as nicotine/ caffeine/ quinine/ strychnine/ opium are produced as defence.
Question. Explain parasitism and co-evolution with the help of one example of each.
Answer : Mode of interaction between two species in which one species (parasite) depends on the other species (host) for food and shelter / one organism is benefitted, the other is harmed.
e.g. Human liver fluke / Malarial parasite / Cuscuta.
Co-evolution is the relationship between two interacting organisms where both organisms failed to survive in the absence of the other.
e.g. Fig and Fig wasp / Ophrys and bumble bee. (or any other suitable example)
Question. (i) Write the parasitic adaptations the parasites have evolved in accordance with their lifestyles.
(ii) Hosts and parasites tend to co–evolve. Explain.
Answer : (i) Parasites have evolved the following adaptation :
(a) Loss of unnecessary sense organs.
(b) Presence of hook / adhesive organs and suckers.
(c) Loss of digestive system.
(d) High reproductive capacity.
(ii) In some cases, a parasitic species may evolve with its host taxa. Long–term co-evolution sometimes leads to a relatively stable relationship, tending to commensalism or mutualism, as it is in the evolutionary interest of the parasite that its host thrives. A parasite may evolve to become less harmful for its host or a host may evolve to cope with the unavoidable presence of a parasite, to the point that the parasite’s absence causes the host harm.
Question. (i) A parasite has to adapt to be able to live in a host. Write the various parasite adaptations?
(ii) Mention an adaptive feature exhibited in brood parasitism in Koel and Crow.
Answer : (i) The various parasitic adaptation enabling it to live in host are as follows :
(a) Loss of unnecessary sense organs
(b) Adhesive organs or suckers to cling on to the host.
(c) Loss of digestive system.
(d) High reproductive capacity.
(e) Loss of chlorophyll and leaves.
(ii) The eggs of the parasitic bird (Koel) resemble the host’s egg (Crow) in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them out from the nest.
Question. Name the type of interaction seen in each of the following examples :
(i) Ascaris worms living in the intestine of human
(ii) Wasp pollinating fig inflorescence
(iii) Clown fish living among the tentacles of seaanemone
(iv) Mycorrhizae living on the roots of higher plants
(v) Orchid growing on a branch of a mango tree
(vi) Disappearance of smaller barnacles when Balanus dominated in the coast of Scotland.
Answer : (i) Parasitism
Question. Study the graph given below and answer the question that follow :
(i) Write the status of food and space in the curves (a) and (b).
(ii) In the absence of predators, which one of the two curves would appropriately depict the prey population ?
(iii) Time has been shown on X-axis and there is a parallel dotted line above it. Give the significance of this dotted line.
Answer : (i) a – Unlimited food and space.
b – Limited food and space.
(ii) Curve a.
(iii) Carrying capacity / a given habitat has enough resources to support maximum possible
number – beyond which no further growth is possible.
Question. Co-evolution is a spectacular example of mutualism between an animal and a plant. Describe coevolution with the help of an example.
Answer : Co-evolution is the mutual relationship between two interacting organisms where both the organisms are unable to survive in the absence of the other. The co-evolution of fig and wasp as a pollinator is highly linked with one another. Fig and wasp is a good example of mutualism and coevolution between a plant species and an animal species. The female wasp uses the fruit and fig for oviposition / egg laying uses seeds within the fruit (developing seeds) for nourishing its larvae. In return the wasp pollinates the fig inflorescence. The given fig species can be pollinated only by its ‘partner’ wasp species and no other species.
Question. Draw and explain a logistic curve for a population of density (N) at time (t) whose intrinsic rate of natural increase is (r) and carrying capacity is (K).
Answer : A population growing in a habitat with limited resources show initially a lag phase, this is followed by phases of acceleration and deceleration and finally an asymptote when the population density reaches carrying capacity (K). A plot of N in relation to time (t) result in a sigmoid curve (Verhulst – Pearl Logistic Growth).
Question. (i) Explain the birth rate and death rate in population with the help of an example.
(ii) What is age pyramid ?
Answer : (i) In a population, the birth rate and death rate refer to per capita births and deaths, respectively.
Examples of birth rate and death rate are :
Birth rate : 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, we calculate the birth rate as 8/20 = 0.4 offspring per lotus per year.
Death rate : If 4 individuals in a laboratory population of 40 fruit flies died during a specified time interval, say a week, the death rate in the population during that period is 4/40 = 0.1 individuals per fruit fly per week.
(ii) A population at any given time is composed of individuals ages. If the age distribution (per cent individuals of a given age or age group) is plotted for these population, the resulting structure is called an age pyramid.
Question. Explain co-evolution with reference to parasites and their hosts. Mention any four special adaptive features evolved in parasites for their parasitic mode of life.
Answer : Co-evolution can be defined as reciprocal adaptations in two interacting organisms that brings about the evolutionary change in both of them. In terms of the relation of host and parasite, it can be explained as follows : Parasite is an organism that is totally dependent on the host organism for its survival but in doing so, it also harms the host. The host evolved over a long period of time to protect itself from parasite, while parasite evolved so that it can find other way to derive nutrition from the host and hence, the cycle continues.
The various parasitic adaptation enabling it to live in host are as follows :
(i) Loss of unnecessary sense organs
(ii) Adhesive organs or suckers to cling on to the host.
(iii) Loss of digestive system.
(iv) High reproductive capacity.
(v) Loss of chlorophyll and leaves.
Question. a) In a pond there were 200 frogs. 40 more were born in the year. Calculate the birth rate of the population.
b) Population in terms of number is not always a necessary parameter to measure population density. Justify with two examples.
Answer : a) Birth rate: No of individuals/ Total no. of individuals = 40 /200 = 0.2 frogs per year.
b) Number is not always a necessary parameter to measure population density.
1. If there are 200 Parthenium plants but only a single huge banyan tree with a large canopy, the population density of banyan is low relative to that of Parthenium which amounts to underestimating the enormous role of the Banyan in that community. In such cases, the per cent cover or biomass is a more meaningful measure of the population size.
2. In a dense laboratory culture of a microbial population in a petri dish, the total number of microbes is again not an easily adoptable measure because population is huge, counting is impossible and time-consuming.
Question. (i) List the different attributes that a population has and not an individual organism.
(ii) What is population density ? Explain any three different ways the population density can be measured, with the help of an example each.
Answer : (i) Following are the attributes that a population has but an individual organism does not have :
(a) Birth rate : Per capita births.
(b) Death rate : Per capita deaths.
(c) Sex ratio : Ratio of number of males to females in a population.
(ii) Population density : It means number of individuals present per unit area. Population density can be measured by determining the population size. The different methods to study population size are as follows :
(a) Quadrat method : It is a method that involves the use of square of particular dimensions to measure the number of organisms e.g. The number of parthenium plants in a given area can be measured using the quadrat method.
(b) Direct observation : It involves the counting of organisms in a given area e.g. In order to determine the number of bacteria growing in a petridish, their colonies are counted.
(c) Indirect method : In this method, there is no need to count the organisms individually e.g. number of fishes caught per trap gives the measure of their total density in a given water body.
Question. What is Predation ? Explain with the help of suitable examples why is it required in a community with rich biodiversity?
Answer : Organism of higher trophic level (predator) feeds on organism of lower trophic level (prey) is called the predation.
Importance of predation :
(i) It helps in transfer of energy from one trophic level to the next.
(ii) It keeps the prey population under control.
(iii) It helps in biological control, helps maintain species diversity.
Long Answer Type Questions
Question. (i) What is population density ? Why are ecologists interested in measuring it ?
(ii) Write the different ways of measuring population density. Explain any two with the help of specific examples.
Answer : (i) The population density is the number of individuals of a population found per unit area at a given time.
Whatever ecological process we wish to investigate in a population (competition / pesticide applicable) we always evaluate in terms of any change in population size (numbers / biomass).
(ii) Number of organism
Biomass of organism
Example : Three ways of measuring population density of a habitat
(a) Per cent cover for trees with larger canopy.
(b) Number of fishes caught per trap.
(c) Pug marks or faecal pellets for tiger census.
(i) Which of the above represents the increase or decrease of population ?
(ii) If N is the population density at time t, then what would be its density at time (t + 1) ? Give the formula.
(iii) In a barn there were 30 rats. 5 more rats enter the barn and 6 out of the total rats were eaten by the cats. If 8 rats were born during the time period under consideration and 7 rats left the barn, find out the resultant population at time (t + 1).
(iv) If a new habitat is just being colonized, out of the four factors affecting the population growth which factor contributes the most ?
Answer : (i) Natality (B) and Immigration (I) represents increase of population and Emigration (E) and Mortality (D) represent decrease of population.
(ii) Nt + 1 = Nt + [(B + I) – (D + E)]
(iii) Here Nt = 30; I = 5; E = 7; D = 6; B = 8
Putting the value in Nt + 1
= Nt + [(B + I) – (D + E)]
Nt + 1 = 30 + [(8 + 5) – (6 + 7)]
= 30 + [13 – 13]
= 30 + 0 = 30 rats
(iv) Immigration contributes the most.
Question. (i) Compare, giving reason, the J-shaped and S-shaped models of population growth, of a pecies.
(ii) Explain ”fitness of a species” as mentioned by Darwin.
Answer : (i)
Note : Marks to be awarded only if the corresponding difference is written.
(ii) When resources are limited, competition occurs between individuals, fittest will survive, which will reproduce to leave more progeny.
Question. (i) Explain the equation.
Nt + 1 = Nt + [(B + I) – (D + E)]
on the basis of the flow chart given below :
(ii) Mention the different ways by which the population density of different species can be measured.
Answer : (i) In the given equation, N is the population density at time t and its density at time t + 1 is
Nt + 1 = Nt + [(B + I) – (D + E)]
The above equation shows 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. Under normal conditions, births and deaths are the most important factors. The other two factors are important only under special conditions. For instance, if a new habitat is just being colonized, immigration may contribute more significantly to population growth than birth rates.
(ii) Population density of a species is the number of individuals of a species per unit area or volume e.g. number of animals per square kilometer, number of trees per hectare, number of phytoplanktons per cubic liter of water. Population density (PD) can be calculated as P.D. = N/S
where, N = Number of individuals in a region.
S = Number of unit areas in a region or total unit land area of the region. Population of an area is described on the basis of three parameters.
(a) Number and kind of individuals of a species.
(b) A given space or an area.
Population density reflects the success of a species in a given area.