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Evolution question

2005 · Q79
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Evolution question

2005 · Q79

NEETBiologyEvolutionMCQ+4 / −1
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 ?
  1. A
    0.48
  2. B
    0.24
  3. C
    0.36
  4. D
    0.16
View written solutionFree

Correct answer: A

In a stable population, for a gene with two alleles, A (dominant) and a (recessive), if the frequency of A is p and the frequency of a is q, then the frequencies of the three possible genotypes (AA, Aa, and aa) can be expressed by the given HardyWeinberg equation :
p2 + 2pq + q2 = 1.
where p2 = frequency of AA (homozygous dominant) individuals, 2pq = frequency of Aa (heterozygous) individuals, and q2 = frequency of aa (homozygous recessive) individuals. The equation can be used to calculate allele frequencies if the numbers of homozygous recessive individuals in the population is known. The equation and the equilibrium are named after British mathematician G.H. Hardy and German physician W. Weinberg.
So p = 0.6 and q = 0.4 (given)
∴\therefore∴ 2pq (frequency of heterozygote)
= 2 × 0.6 × 0.4 = 0.48

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