Question

In a hardy-weinberg population, 16% of individuals display the homozygous recessive phenotype. What proportion of the population will be heterozygous carriers of the recessive allele?.

Answer 1

The proportion of the population will be heterozygous carriers of the recessive allele is 0.48.

What is Hardy-Weinberg principle?

• Hardy-Weinberg principle says that allele frequencies in a population are stable and is constant from generation to generation. The gene pool (total genes and their alleles in a population) remains a constant. This is called genetic equilibrium. Sum total of all the allelic frequencies is 1.
• Individual frequencies, for example, can be named p, q, etc. In a diploid, p and q represent the frequency of allele A and allele a.
• The frequency of AA individuals in a population is simply p². This is simply stated in other ways, i.e., the probability that an allele A with a frequency of p appear on both the chromosomes of a diploid individual is simply the product of the probabilities, i.e., p².
• Similarly. that of aa is q², of Aa 2pq.
• Hence, p²+2pq+q²=1, is a binomial expansion of (p+q)².

By given data

$q2=16%$%

$q2=\frac{16}{100}=0.16$

$q=0.4$

$p=1-0.4=0.6$

$2pq=2(0.6)(0.4)=0.48$

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