VBQs Organisms and Populations Class 12 Biology with solutions has been provided below for standard students. We have provided chapter wise VBQ for Class 12 Biology with solutions. The following Organisms and Populations Class 12 Biology value based questions with answers will come in your exams. Students should understand the concepts and learn the solved cased based VBQs provided below. This will help you to get better marks in class 12 examinations.
Organisms and Populations VBQs Class 12 Biology
Very Short Answer Type Questions
Question. Give an example of an organism that enters ‘diapause‘ and why ?
Answer : Many species of Zooplankton, unfavourable condition.
Detailed Answer :
During unfavourable conditions, many zooplankton species in lakes and ponds are known to enter diapause, a stage of suspended development.
Example – silk moth.
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. 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.
Short Answer Type Questions
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.
Detailed Answer :
The desert plants are adapted to survive in water deficient conditions. They possess adaptations to conserve water such as thick cuticle, reduced leaf and low rate of transpiration.
The mangrove plant, on the other hand are adapted to survive in water logged and anoxic conditions.
They possess prominent leaves with high rate of transpiration. Thus, the desert plants have different adaptations to survive, which cannot be found in a mangroves.
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. 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. 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. 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.
Detailed Answer :
(i) Polar bear : They hibernate during winter to escape the cold weather.
(ii) Snails / fishes : They go into aestivation during summer to avoid heat related problems and desiccation.
(iii) Seeds of higher plants / spores of bacteria / fungi : They become dormant in unfavourable conditions and in case of Amoeba, cyst formation takes place.
(iv) Some species of zooplankton : They undergo diapause.
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. 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.
(any three with corresponding examples)
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. Different animals respond to changes in their surroundings in different ways. Taking one example each, explain ”some animals undergo aestivation while some others hibernation”. How do fungi respond to adverse climatic conditions ?
Answer : Some animals go into aestivation to avoid summer related problems (heat and dessication) e.g. snails / fish.
Some animals go into hibernation to avoid winter related problem (extreme cold) e.g. bear.
Fungi form thick walled spores and suspend their activities to respond to adverse climatic condition.
Detailed Answer :
Aestivation means summer sleep or inactivity during the summer season, undertaken by organisms of warm climates during hot seasons. This is a type of response shown by organisms when the stressful conditions persists for a short time. Organisms like snails, fishes etc. aestivate to avoid summer related problems like heat, dessication etc.
Hibernation is inactivity during cold conditions, when the organism cannot carry out its normal functions due to extreme conditions. It is known as winter sleep. Organisms of polar regions such as polar bears hibernate.
Fungi respond to adverse climatic conditions by forming thick walled spores, which are resistant to such conditions. Encystation / spore formation enable them to overcome unfavourable conditions, on availability of favourable conditions they germinate.
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).
Detailed Answer :
Humming birds and fishes are conformers.
Conformers are not able to bear the energetic expenses to maintain the constant body temperature.
Heat loss or gain is the surface phenomenon. The conformers have more surface area in comparison to their volume.
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
(i) Regulate
(ii) Conform
(iii) Migrate
(iv)) Suspend
(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
Detailed Answer:
(a) (i) Regulate: Certain animals have the ability to maintain a constant temperature and constant osmolarity to keep up their homeostasis. For example, mammals have a constant body temperature (37˚C) irrespective of the outside temperature. In summers, to maintain the temperature, they sweat and in winters they shiver to produce heat.
(ii) Conform : About 99% of animals and nearly all plants cannot maintain a constant internal environment. Their body temperature or osmotic concentration change with the surrounding conditions. They are called conformers. Hence, the internal environment of conformers changes with external environment. For example in aquatic animals, osmotic concentration of body fluids changes with that of the ambient osmotic concentration.
(iii) Migrate: It is the temporary movement from a stressful habitat to a more hospitable area and return when stressful condition is over.
For example during winter, Keoladeo National Park (Bharatpur, Rajasthan) hosts migratory birds coming from Siberia and other extremely cold northern regions.
(iv) Suspend: In bacteria, fungi and lower plants, thick walled spores are formed which help them to overcome stressful, unfavourable conditions. Spores germinate when conditions are favourable.
(b) Death rate in the population during that period is calculated as:
Death rate = Number of organism died/ number of organism present initially.
= 18/80
= 0.225 deaths/week therefore, death rate percentage will be
= 0.225 × 100 = 2.25%