In humans, the rh blood factor is dominant to the rh- blood factor. in a certain human population in hardy-weinberg equilibrium, 4% are homozygous recessive for the rh blood factor. what percentage of the population would be heterozygous for the rh blood factor

0.32

Explanation:

Hardy Weinberg equilibrium states that the frequency of allele will remain the same in the absence of disturbances such mutation,no gene flow, no selection, random mating and infinite population size

the equilibrium is defined the equation

p² + 2pq + q²

where q² = f, frequency of the homozygous recessive allele, p² = f, of the homozygous dominant allele and 2pq is the frequency of the carrier of the dominant  + rh blood factor

q² = 0.04

q = √0.04 = 0.2

and p + q = 1

p = 1 - 0.2 = 0.8

therefore 2pq = 2 × 0.8 ×0.2 = 0.32

complete the square on the right side of the equation;

-5(x + 0)^2 + 15

answer: d. y = -x^2 + 15

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