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Organisms and Populations Test - 75

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Organisms and Populations Test - 75
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  • Question 1
    1 / -0
    According to the law of population growth, if an initial population of yeast cells is 10 in number, was allowed to grow for 6 hours, the expected number of individuals in the final population will be about 
    Solution
    The projection of population growth in yeast is given by
    $$N = N_O e^{rt}$$ 
    Where e = national logarithmic base = 2.72, $$N_o$$ = initial population, r = intrinsic rate of natural increase (0.5 for yeast), t = time. 
    N = $$10 \times 2.72 ^{(0.5 \times 6)}$$ 
    Hence, N = 200.86, i.e., after 6 hours, the population is expected to have about 200 individuals.
    Therefore, the correct answer is option D.
  • Question 2
    1 / -0
    Elephants, man and mountain sheep;
    Solution
    R-strategist produces offsprings which have a low probability of surviving into adulthood. (e.g., Insects and Spiders)
    In K-strategy the species are strong competitors in crowded niches, and invest more heavily in fewer offsprings, each of which has a relatively high probability of surviving into adulthood. (e.g., Elephants and Lions).
    Thus, the correct answer is option C.
  • Question 3
    1 / -0
    Which of the following statements regarding species interdependence are true?
    A.     A relationship between two species of organism where both the partners are benefitted from each other is called mutualism.
    B. An interspecific association where both partners derive benefit from each other is called commensalism.
    C. A direct food relation between two species of animals in which one animal kills and feeds on another is referred as predation.
    D.     An association of two species where one is benefitted and other remains unaffected is called symbiosis.

  • Question 4
    1 / -0
    The population density can be calculated by
    Solution
    The formula for finding the population density is PD = N/S or population density = a number of organisms/area. 
    In the formula, PD = N/S, PD stands for population density, N stands for the number of organisms, S stands for the area to get the PD, divide the number of organisms by the area. 
    So, the correct answer is option B.
  • Question 5
    1 / -0
    Match list I with list II and choose the correct options.
    List IList II
    (a) Pacific salmon fish1. Verhulst-Pearl logistic growth
    (b) $$N_t = N_0 e^{et}$$2. Breeds only once in lifetime
    (c) Oyster3. Exponential growth
    (d) $$ \displaystyle dN/dt = rN \left( \frac{K - N}{K} \right)$$4. A large number of small sized offspring
    Solution
    (a)- 2: Pacific salmon fish is called as a semelparous type of fish. These fishes live in the ocean water. They swim and then lay eggs in the fresh water. They spawn and then die. These type of animals spawn only once in their lifetime and die after spawning.
    (b)- 3: The exponential function appears when original quantity grows exponentially over time. If a quantity N increases at a rate proportional to the amount present at time t, then the quantity can be written as N(t) = N$$_0$$ekt, where N$$_0$$ is the value of N at time t = 0 and N increases as t increases and k is called an exponential growth function.
    (c)- 4: There are variations found in the production of offspring in different animals. Some of the animals produce small sized offsprings which are many in number. The examples of such animals are oysters and pelagic fishes.
    (d)- 1: In a finite world, no population can grow exponentially for very long. Sooner or later every population must encounter either difficult environmental conditions or shortages of its requisites for reproduction. This is explained by the Verhulst-Pearl Logistic growth equation which is given in the form of dN/dt=rN(K−N/K), where N denotes population, K is the carrying capacity and r shows the rate of increase in the population. 
    So, the correct answer is option D.
  • Question 6
    1 / -0
    Without our present day atmosphere the average surface temperature of the earth would be
    Solution
    The atmosphere is composed of several gasses like water vapour and carbon dioxide that trap heat from the sun. This maintains average temperature on the planet's surface. 
    Without it, the surface temperature of the Earth would be –18 degrees Celsius and life, even if possible, would be quite different. Hence option A is correct.
  • Question 7
    1 / -0
    Two different species can not live for long duration in the same niche or habitat. This law is 
    Solution
    • Allen's rule states that in a warm-blooded animal species having distinct geographic populations, the limbs, ears, and other appendages of the animals living in cold climates tend to be shorter than in animals of the same species living in warm climates.
    • Gause's law of competitive exclusion states that two species competing for the same resource cannot coexist at constant population values if other ecological factors remain constant. 
    • According to the theory of germplasm or Weismann's theory, which is independent from all other cells of the body, is the essential element of germ cells and is the hereditary material that is passed from generation to generation.
    Thus, the correct answer is option B.
  • Question 8
    1 / -0
    Match the following and choose the correct combination from the options given.
    Population InteractionExamples
    (a) Mutualism                    (1) Ticks on dogs
    (b) Commensalism(2) Balanus and Chanthamalus
    (c) Parasitism(3) Sparrow and any seed
    (d) Competiton(4) Epiphyte on a mango branch
    (e) Predation(5) Orchid, Ophrys, and bee
    Solution
    • In mutualism, both the host and the predator will get benefitted.
    • In commensalism the one get benefited by other but do not harm the other.
    • In parasitism it totally depends on the host and may harm the host.
    • Competition is not a relationship but fights for something.
    • Predation is harming others for the food. 
    • Thus, the correct answer is option E.
  • Question 9
    1 / -0
    After exponential increase, population growth declines and stagnates. The growth curve is
    Solution
    There are three different sections to an S-shaped growth curve. Initially, growth is exponential, because there are few individuals and ample resources available. Then, as resources begin to become limited, the growth rate decreases. Finally, growth levels off at the carrying capacity of the environment, with little change in population size over time. 
    When resources are unlimited, populations exhibit exponential growth, resulting in a J-shaped curve. 
    When resources are limited, populations exhibit logistic growth. In logistic growth, population expansion decreases as resources become scarce, leveling off when the carrying capacity of the environment is reached, resulting in an S-shaped curve.
    Thus, the correct answer is option A.
  • Question 10
    1 / -0
    Hyperparasite is one
    Solution
    A hyperparasite is a parasite whose host is a parasite. This form of parasitism is especially common among entomophagous parasites. The most common examples are insects that lay their eggs inside or near parasitoid larvae, which are themselves parasitizing the tissues of a host, again usually an insect larva. 
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