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Chapter 13 Organisms and Populations Biology Worksheet for Class 12
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Class 12 Biology Chapter 13 Organisms and Populations Worksheet Pdf
Very Short Answers Type Questions
Question. What are eurythermic species?
Answer : Eurythermic species are those species which possess or show a wide range of temperature tolerance.
Question. Species that can tolerate wide range of salinity are called........
Answer : Euryhaline species Water, another major abiotic component affects the life of organism. The quality of water like pH, salinity (salt concentration) are water related problems faced by aquatic organisms, organisms or species that can tolerate wide range of salinity are regarded as euryhaline Species.
Question. Define stenohaline species.
Answer : Stenohaline species are those species which show a narrow range of salinity tolerance.
Question. Why do high altitude areas have brighter sunlight and lower temperatures as compared to the plains?
Answer : At high altitude the sun light is brighter as compared to plains, it is because of reduced distance from the sun and particles free air. Similarly lower temperatures are because of lower atmospheric pressure which is more in plains as compared to high altitude.
Question. What is homeostasis?
Answer : Homeostasis is the tendency of the organism to maintain a constant internal environment despite varying external environmental conditions like temperature.
Question. What is high altitude sickness? Write its symptoms.
Answer : High altitude sickness is experienced by the people going to high altitudes, where oxygen concentrations are low and the body system reacts by developing the symptoms like nausea, headache and heart palpitations.
Question. Give a suitable example for commensalism.
Answer : An interaction between blue whale and the barancle growing on its back is an example of commensalism (interspecific relation) between them.
Question. Define ectoparasite and endoparasite and give suitable examples.
Answer : Ectoparasite feeds on the external surface of the host organism, e.g., Lice on humans and ticks on dogs. Many marine fish are infested with ectoparasitic copepods. Endoparasites live inside the host body at different sites (liver, kidney, lungs, red blood cells, etc.). Such as malarial parasites P. vivax, gut parasites, i.e., tapeworm. Their morphological and anatomical features of endoparasites are greatly simplified while emphasising their reproductive potential.
Question. What is brood parasitism? Explain with the help of an example.
Answer : Brood parasitism is the phenomenon in which an organism (parasite) lays eggs on the nest of other organism (host). e.g., Cuckoo (koel) bird lays its eggs in the nest of its host and lets the host incubate them. The eggs of the parasitic bird resemble the host’s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest.
Short Answer Type Questions
Question. Categorise the following plants into hydrophytes, halophytes, mesophytes and xerophytes.
Give reasons for your answers.
(a) Salvinia
(b) Opuntia
(c) Rhizophora
(d) Mangifera
Answer : (a) Salvinia is a hydrophytes – Partially or completly submerged in water.
(b) Opuntia is a xerophyte – Dry habitat succulent leaves.
(c) Rhizophora is a halophyte – Saline habitat
(d) Mangifera is a mesophyte – Terrestrial habitat.
Question. 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 : (a) Hydrilla is submerged hydrophyte
(b) Typha is rooted emergent
(c) Nymphaea is rooted with floating leaves
(d) Lemna is free floating hydrophyte
(e) Vallisnaria is rooted submerged hydrophytes.
Question. 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 : The density of population per unit area is measured in following units
(a) The density of a bacterial population in a habitat per unit area is measured in volume/number unit.
(b) Biomass/area/region is measuring unit for density of population of banyan.
(c) Number/area is measuring unit for the density of population of deer.
(d) Weight/area is measuring unit for density of population of fish.
Question. 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 defense agent
Answer : (a) Migratory animal-American buffalo and dolphin.
(b) Camouflaged animal-Grasshopper and chameleon.
(c) Predatoranimal-Lion.
(d) Biological control agent-Myxoma virus to kill European rabbit and Gambusia fish to check growth of mosquito larvae.
(e) Phytophagous animal-Insects (beetle, butterfly, etc.).
(f) Chemical defense agent-Cardiac glycosides.
Question. Fill in the blanks
Answer :
Question. 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? (Image 237)
Answer : I. Figure A show a pollinator (bee) on a flower. The association between pollinator insect and plant is termed as mutulalism.
II. Predation.
III. Egrets and grazing cattle is good example of commensalism.
IV. Scavenging- The insect is playing the role of an scavenger.
Question. (a) 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) The three tiers are to be labelled as
(i) Pre-reproductive phase
(ii) Reproductive phase
(iii) Post-reproductive phase
(b) The given age pyramid represents the expanding type of population growth.
Question. 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 they establish?
Answer : The termite provides shelter and space for the protozoan Trichonympha to live. The Protozoa present in gut digests the wood, which termite feeds upon. In the absence of Trichonympha the termite is unable to digest wood and hence dies. Thus, the association of two given animal species represent mutualism.
Question. Define ‘zero population growth rate’. Draw a age pyramid for the same.
Answer : When the pre-reproductive age group individuals are comparatively fewer and both reproductive and post-reproductive stages are almost in equal stage, i.e., at same level. It is zero population growth rate. An inverted bell shaped age pyramid is obtained for zero population growth rate.
Question. List any four characters that are employed in human population census.
Answer : A population has the following characterstics that are employed in human population census.
(i) Natality and mortality
(ii) Sex ratio
(iii) Population density
(iv) Age distribution
(v) Population growth
Long Answer Type Questions
Question. Does light factor affect the distribution of organisms? Write a brief note giving suitable examples of either plants or animals.
Answer : Plants require sunlight for photosynthesis. Therefore, light is an important factor that affects the distribution of plants. e.g.,
(i) Many species of small plants (herbs and shrubs) growing in forests are adapted to photosynthesise optimally under very low light conditions so they will be seen distributed in shady areas under tall, canopied trees.
(ii) Many plants in the shade will grow vertically to gain access to light. These plants will appear to have smaller leafs and smaller than others of the same species of the same age found in conditions with better sunlight.
(iii) Large sized trees will be present in areas that get abundant sunlight.
(iv) Plants dependent on sunlight to meet their photoperiodic requirements for flowering, will try to be distributed in area, where this requirement is being met for their reproductive success.
Question. An individual and a population has certain characteristics. Name these attributes with definitions.
Answer : An individual and a population has following certain attributes like pattern of distribution. dispersal biotic potential and gene pool. Phenomenon of distribution of individual within geographical boundaries of the population is termed as interapopulation dispersion or internal distribution patterns or dispersion. Dispersal an individual is dispersed at one or another time during their life in a population which is revealed by immigration or emigration. (i) Immigration is the number of individuals of the same species that have come into the habitat from elsewhere during a specified time period. (ii) Emigration is the number of individuals of the population who exit or leave the habitat and go elsewhere during a specified time period. Biotic Potential Biotic potential is the natural capacity of a population to increase its size under ideal environmental conditions. Gene pool All the genotypes of all individuals in a breeding population is referred to as gene pool.
Question. In an aquarium two herbivorous species of fish are living together and feeding on phytoplanktons. 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 : Competition is a rivalry relationship between two or more organisms. A competition between individual of same species (intraspecific) is more acute than the competition between individual of different species as all the members in a intraspecific competition have same basic requirements like food, water, light, space, mating and shelter. But this is true only when resources are limited. According to Gause's principle, one of the species is to be eliminated. But studies recently have revealed that species facing intraspecific competition may evolve mechanism to pencourage co-existence rather than exclusion. This can also be done by a method known as ‘ resource partitioning’.
Question. While living in and on the host species, the animal parasite has evolved certain adaptations. Describe these adaptations with examples.
Answer : Parasites have evolved special adaptations such as
(i) The loss of unnecessary sense organs as in lice, mites and fleas don't have wings.
(ii) Presence of adhesive organs or suckers to cling on to the host-in tapeworms and leeches.
(iii) Loss of digestive system i e . ., tapeworm.
(iv) High reproductive capacity i e . ., roundworm produces large progeny.
Question. Do you agree that regional and local variations exist within each biome? Substantiate your answer with suitable example.
Answer : A biome can be define as the large communities of the world and shows that area with similar climate have communities of same type. Climate is the main factor that determine the type of soil which in turn determines the type of vegetation. The type of vegetation and climate together determine the kind of microorganisms and animals. The other determining factors are latitude and altitude, intensity and duration of winter and summer days, water mass and topography. The main biome of the world does not show boundary of any country and regional and local variations exist in each biome. e.g., temperate deciduous forest receive an annual precipitation between 75-150 cm and tropical rain forest show a rainfall above 140 cm/yr which may reach upto 400 cm/yr.
Question. 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 : I. Mango, Acacia
II. Archaebacteria
III. Jelly fishes
IV. Earthworm
V. Cuscuta
VI. Cocos nucifera
VII. Bacteria
VIII. Octopus
IX. Salt content (osmotic regulation)
X. All plant and fish like large mouth bass (temperature conformer)
Question. Plants that inhabit a rainforest are not found in a wetland. Explain.
Answer : Plants that inhabit rainforest are not able to germinate in wetland due to presence of excess water and anaerobic conditions (due to water logging). Wetlands are marshy areas and plants growing there have negatively geotropic roots, called pneumatophores which help in gaseous exchange. Pneumatophores are not present in plants inhabiting rainforests
Question. Many freshwater animals cannot survive in marine environment. Explain.
Answer : If a freshwater animal is placed in marine environment, then it will not be able to survive because of osmoregulation problem. The freshwater animal is adapted to live in fresh environment, so, if it is kept in saline water, it will not be able to cope with outside hypertonic environment and it would face death.
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. How does a desert plant adapt to the dry, warmer environmental conditions?
Answer : Desert plants or xerophytes have various adaptations to cope with dry, hot environmental conditions such as leaves with thick waxy, hairy coating to reduce transpiration, leaves reduced to spines and photosynthetic stems, fleshy organs to store water, sunken stomata that open only during night and deep penetrating roots that reach water table.
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. How are mammals living in colder regions and seals living in polar regions able to reduce the loss of their body heat?
Answer : Animals inhabiting cold areas possess thick coat of hairs, feathers and subcutaneous fat to reduce loss of body heat.
Question. A student on a school trip started sneezing and wheezing soon after reaching the hill station for no explained reasons. But, on return to the plains, the symptoms disappeard. What is such a response called? How does the body produce it?
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. Explain why very small animals are rarely found in polar region.
Answer : Small animals have large surface area relative to volume, so they tend to lose body heat very fast in cold environment as compared to large animals. They have to spend more energy to generate body heat through metabolism. Thus, considering the difficulty of maintaining constant internal temperature, small animals are rarely found in polar regions.
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. Explain with the help of suitable examples the three different ways by which organisms overcome their stressful conditions lasting for short duration.
Answer : Physiological and behavioural adaptations help organisms to manage stressful conditions. These include migration, hibernation, aestivation, camouflage, mimicry etc. Caribou migrate during winter to warmer places for search of food. Some animals undergo hibernation during winter to avoid low temperature whereas others undergo aestivation to avoid extreme heat. Viceroy butterfly mimics unpalatable toxic monarch butterfly in order to get protection against predator.
Question. How do snails, seeds, bears, zooplanktons, fungi and bacteria adapt to conditions unfavourable for their survival?
Answer : Ecological adaptations are special characteristics evolved or developed by organisms in order to live comfortably and successfully under a prevailing set of environmental conditions. Adaptations may be morphological, physiological or behavioural or a combination of them. The ultimate aim of all adaptations is to make the individual fit to obtain food and space for its survival, opportunities for its reproduction and rearing of young ones.
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. Some organisms retard their metabolic activities under stress conditions and undergo hibernation or aestivation. For example, polar bears go into hibernation during winter season to escape extreme cold. Some snails and fish undergo aestivation to avoid summer-related problems like heat and dessication.
Under unfavourable conditions many zooplanktons in lakes and ponds are known to enter diapause i.e., a stage of suspended growth and development. Seeds remain dormant in unfavourable conditions. They break dormancy and germinate in favourable environmental conditions.
Question. (a) State how the constant internal environment is beneficial to organisms.
(b) Explain any two alternatives by which organisms can overcome stressful external conditions.
Answer : (a) Regulators are organisms which maintain constant internal environment despite changes in external conditions through thermoregulation and osmoregulation. Such organisms generally have wide range of distribution.
(b) Organisms can overcome stressful external conditions by following adaptation :
(i) Migration - Birds of colder areas of northern hemisphere begin their southward migration as the day length begins to shorten.
(ii) Aestivation - Ground squirrels undergo aesti- vation to avoid heat by spending dry hot period in burrows.
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.
OR
How do organisms cope with stressful external environmental conditions which are localised or of short duration?
Answer : Water is very essential for life. Plants and animals show various adaptations to cope up with water scarcity in the area where they are found. Some of the adaptations seen in plants which enable them to survive in water scarce environment are as follows:
– Some plants have deep tap root system which is capable of absorbing water from deep soil e.g., Prosopis, Acacia etc.
– Cacti and succulents, have fleshy leaves and stems to store water.
– Many tropical plants, 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 in water scarce environment are as follows:
– Desert lizards 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 rat conserves water by excreting nearly solid urine and can live from birth to death without even drinking water.
– The camels shows tolerance to wide fluctuations in body temperature and are able to maintain blood stream moisture even during extreme heat stress.
OR
Living organisms cope with stressful conditions by any of the following methods:
(i) Migration : The organism can migrate temporarily from the unfavourable habitat to more favourable area and return when unfavourable period is over e.g., Siberian birds migrate from Siberia to other parts every winter.
(ii) Hibernation : The phenomenon of spending extreme cold period of the year in an inactive stage by an animal, e.g., polar bears undergo hibernation during winter season to escape extreme cold.
(iii) Aestivation : The phenomenon of spending dry hot period of the year in an inactive stage by an animal e.g., snails and fish.
(iv) Diapause : It is a dormant stage of suspended development of an organism. During this period there is reduction in the amount of free water.
Question. (a) Explain giving reasons why the tourists visiting Rohtang pass or Mansarovar are advised to resume normal ‘active life’ only after a few days of reaching there.
(b) It is impossible to and small animals in the polar regions. Give reasons.
Answer : (a) Tourists visiting high altitude areas such as Rohtang Pass or Mansarovar, experience altitude sickness. Its symptoms include nausea, fatigue and heart palpitations. This is because in the low atmospheric pressure of high altitudes, the body does not get enough oxygen. But, gradually it gets acclimatised and stops experiencing altitude sickness. The body compensates low oxygen availability by increasing red blood cell production, decreasing the binding affinity of haemoglobin and increasing breathing rate. Thus, the visitors are advised to resume their normal active life involving heavy works only after few days because for doing heavy tasks our body needs energy and this energy is obtained by the oxidation of glucose. This oxygen is carried to the cells by haemoglobin present in RBCs. As oxygen is the limiting factor in high altitudes, more carrier molecules i.e., haemoglobins are needed to provided sufficient amount of oxygen to cells. Thus, increased RBC production that starts in a few days after reaching altitudes prepares the tourist to lead a normal active life.
(b) According to Bergman’s rule, temperature affects the absolute size of an animal and the relative proportion of various body parts. Birds and mammals attain greater body size in cold regions than in warm areas . Thus, smaller animals are rarely found in polar regions. It can be explained as that small animals have a larger surface area relative to their volume and they tend to loose body heat very fast when it is cold outside. Due to higher heat loss they have to spend much energy to generate body heat through metabolism. Greater body size reduces surface area to volume ratio and decreases heat loss which is very essential for surviving in cold polar regions.
Question. State Gause’s competitive exclusion principle.
Answer : 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.
Question. Explain with the help of an example each of the three population interactions where the organisms live closely together.
Answer : Population interaction is the interaction among individuals of the same species or of different species, in which individual may get benefitted, harmed or remain unaffected.
Three population interactions are -
(i) Mutualism - It is an interaction between individuals of two different species where both are benefitted and none is capable of living separately. E.g., association of Rhizobium bacterium with root nodules of leguminous plants for nitrogen fixation.
(ii) Commensalism - It is an interaction where two animals live together without any physiological dependence between them, where one gets benefit and other remains unaffected e.g., close association between cattle egret and cattle.
(iii) Proto-cooperation - Two organisms are mutually benefitted by each other. But the association is non-obligatory. E.g. association between crocodile and bird. Crocodile bird enters the open mouth of crocodile and feeds on leeches present there and helps crocodile in getting rid of leeches.
Question. (a) Explain the birth rate and death rate in the population with the help of an example each.
(b) What is age-pyramid? Draw an agepyramid of an expanding population.
Answer : (a) The birth rate (natality) of a population refers to the average number of young ones produced by birth, hatching or germination per unit time (usually per year). In the case of humans, it is commonly expressed as the number of births per 1,000 individuals in the population per year
The death rate (mortality) of a population is the average number of individuals that die per unit time (usually per year). In humans, it is commonly expressed as the number of deaths per 1,000 persons in a population per year.
(b) Age pyramid is a model representing geometrically the proportion of different age groups in the population of any organism. It is a vertical bar graph in which the number or proportion of individuals in various age ranging at any given time is shown from youngest at the bottom of the graph to oldest at the top
Question. Why is predation required in a community of different organisms?
Answer : Predators plays an important role in a community :
(i) They act as conduits for energy transfer across trophic levels.
(ii) Predators keep prey population under control. They are used for biological control of weeds and pests.
(iii) Predators help in maintaining species diversity.
(iv) They help in growth of vegetation by controlling population of herbivores.
Question. What is population density? Explain any three different ways the population density can be measured, with the help of an example each.
Answer : 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.
Question. (a) List any three parameters used by ecologists under different situations to measure the population size in a habitat.
(b) Mention what do the following stand for in the equation given below:
(i) Nt+1
(ii) B and
(iii) E.
Nt + 1 = Nt + [(B + I) – (D + E)]
Give an explanation for the above equation.
Answer : (a) The equation dN/dt= rN , represents expo- nential growth form. ‘r’ represents intrinsic rate of
natural increase. It is a very important parameter chosen for assessing impacts of any biotic or abiotic factor on population growth. Its value depends upon the birth rates and death rates.
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 backgroud. 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.
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Chapter 13 Organisms and Populations CBSE Class 12 Biology Worksheet
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