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Evolution Test - 41

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Evolution Test - 41
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  • Question 1
    1 / -0
    The frequency of a gene in an ideal population is not affected by ______________.
    Solution

  • Question 2
    1 / -0
    Select the correct pair from the following.
    Solution
    In evolutionary biology, adaptive radiation is a process in which organisms diversify rapidly into a multitude of new forms, particularly when a change in the environment makes new resources available, creates new challenges and opens environmental niches. One famous example where adaptive radiation is seen is with Darwin's finches. It has been observed by many evolutionary biologists that fragmented landscapes often times are a prime location for adaptive radiation to occur. The differences in geography throughout disjointed landscapes such as islands are believed to promote such diversification. Darwin's finches occupy the fragmented landscape of the Galapagos Islands and are diversified into many different species which differ in ecology, song, and morphology, specifically the size and shapes of their beaks.
    Therefore, the correct answer is option A.
  • Question 3
    1 / -0
    Which of the following process gives rise to an array of species occupying different niches from an ancestral species?
  • Question 4
    1 / -0
    The Hardy-Wein-berg equation used to reflect genotypic frequencies in a population, is
                     $${p}^{2}+2pq+{q}^{2}=1$$
    Which of the following is indicated by the term $$2pq$$?
    Solution
    • Hardy-Weinberg equation is $$p^2$$ + 2pq + $$q^2$$ = 1. 
    • Here p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. 
    • Thus, $$p^2$$ represents the frequency of the homozygous dominant genotype AA, $$q^2$$ is the frequency of the homozygous recessive genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. 
    • Further, it also states that the sum of the allele frequencies for all the alleles at the locus should be 1, so p + q = 1. 

    Thus, the correct answer is option E.
  • Question 5
    1 / -0
    Which of the following helps early protobionts to evolve into living cells?
    Solution
    Competition for resources and the development of hereditary mechanisms helps early protobionts to evolves into living cells. A thin limiting membrane was developed around the cytoplasm by the folding of a monolayer of phospholipids to form cell-membrane. These are precursors for the early life which evolved into a living cell. The organic and inorganic molecules present in the protobionts, later on, develops into the genetic material which acts as a carrier of hereditary characters.
    So, the correct answer is option B.
  • Question 6
    1 / -0

    Directions For Questions

    Brachiopods are the animals belonging to phylum mollusca but are of rare existence. They show resemblance to clams and live in very cold waters and in deep ocean. Plaeontologists find great interest in Brachiopods as they are useful in providing information related to time sequence and past environmental conditions. The given chart represent the geological time scale for different genera of Brachiopods.

    ...view full instructions

    Which time period of the geological scale is responsible for mass extinction of Brachiopods?

    Solution
    Permian - Triassic
    The extinction event, also known as the P-T r extinction. approximately 252 million years ago.
  • Question 7
    1 / -0
    What is true about evolution?
  • Question 8
    1 / -0
    All of the following are the requirement for a population to be maintained in Hardy-Weinberg equilibrium through several generations of intermating except
    Solution
    According to the Hardy-Weinberg law, the allele and genotype frequencies in a large randomly mating population remain constant under the absence of factors responsible for evolution. These factors are namely mutations, migration and natural selection. Thus, the correct answer is option A.
  • Question 9
    1 / -0
    The frequency of the dominant allele in a turtle population living in a local pond showing $$4$$% of recessive phenotype would be
    Solution
    Option (C) is correct 
    Given,
    4% of population has recessive group 
    $$ \therefore $$ frequency of recessive allele
    must be square root of $$ 0.04 $$
    Then, $$ \sqrt{0.04} = 0.20 $$
    Since, 
    $$  p+q = 1 $$
    where,
    $$ p = $$ frequency of dominate  allele
    $$ q= $$ frequency of recessive allele
    we know,
    $$ p+q = 1 $$
    $$ p = 1-q $$
    $$ p = 1-0.20 $$
    $$ p = 0.80 $$ 
  • Question 10
    1 / -0
    The Hardy-Wein-berg equation used to reflect genotypic frequencies in a population, is
                     $${p}^{2}+2pq+{q}^{2}=1$$
    The value of $${p}^{2}$$ reflects
    Solution
    Hardy-Weinberg equation is $$p^2$$ + 2pq + $$q^2$$ = 1. Here p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. Thus, $$p^2$$ represents the frequency of the homozygous dominant genotype AA, $$q^2$$ is the frequency of the homozygous recessive genotype aa, and 2pq represents the frequency of the heterozygous genotype Aa. Further, it also states that the sum of the allele frequencies for all the alleles at the locus should be 1, so p + q = 1. Thus, the correct answer is option B.
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