If 72 out of 200 individuals express the recessive phenotype, what percentage will be homozygous dominant? !

16%

Explanation:

If the population is in Hardy Weinberg equilibrium:

p + q = 1

p² + 2pq + q² = 1

where, p= frequency of dominant allele

q= frequency of recessive allele

p² = frequency of homozygous dominant population

q² = frequency of homozygous recessive population

2pq= frequency of heterozygous dominant population

Given, q²= 72/200 = 0.36

q= √0.36 = 0.6

Hence, p = 1-q

= 0.4

p² = 0.16 = 16%

Hence, 16% of population will be homozygous dominant.

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