NEET Sample Paper NEET Sample Test Paper-14

  • question_answer 135) At a particular locus, frequency of 'A' allele is 0.6 and that of 'a' is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium?

    A)  0.36                           

    B)  0.16

    C)  0.24                           

    D)  0.48

    Correct Answer: D

    Solution :

    The mathematical expression to calculate allele frequency can be given as p + q = 1. Where p represents the frequency of dominant allele and q represents the frequency of recessive allele. The binomial expression of Hardy-Weinberg's law to calculate genotype frequency is \[{{p}^{2}}+2pq+{{q}^{2}}=1.\] Where\[{{p}^{2}}=\]homozygous dominant 2pq = heterozygous dominant \[{{q}^{2}}=\]homozygous recessive Frequency of heterozygous individuals \[=2\times 0.6\times 0.4=0.48\]

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