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(B) According to the Hardy-Weinberg principle,the genetic equilibrium is represented by the equation $(p + q)^2 = p^2 + 2pq + q^2 = 1$,where $p^2$ is

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$42$

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(B) According to the Hardy-Weinberg principle,the genetic equilibrium is represented by the equation $(p + q)^2 = p^2 + 2pq + q^2 = 1$,where $p^2$ is the frequency of dominant homozygous individuals,$2pq$ is the frequency of heterozygous individuals,and $q^2$ is the frequency of recessive homozygous individuals.Given that the percentage of dominant homozygous individuals $(p^2)$ is $49\%$,we have $p^2 = 0.49$.Taking the square root,we get $p = \sqrt{0.49} = 0.7$.Since $p + q = 1$,we can find $q$ as $q = 1 - p = 1 - 0.7 = 0.3$.The frequency of heterozygous individuals is represented by $2pq$.Substituting the values,$2pq = 2 	imes 0.7 	imes 0.3 = 0.42$.Converting this to a percentage,$0.42 	imes 100 = 42\%$.Therefore,the percentage of heterozygous individuals in the population is $42\%$.

In a randomly mating population of $28,800$ individuals,the percentage of dominant homozygous individuals is $49\%$. Find the percentage of heterozygous individuals. (in $\%$)

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