In a population of 10,000 cattle,6400 have ho | vedclass

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(B) According to the Hardy-Weinberg principle,the sum of genotypic frequencies is given by the equation: $p^2 + 2pq + q^2 = 1$.Here,$p^2$ represents t

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$3200$

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(B) According to the Hardy-Weinberg principle,the sum of genotypic frequencies is given by the equation: $p^2 + 2pq + q^2 = 1$.Here,$p^2$ represents the frequency of homozygous dominant $(AA)$,$q^2$ represents the frequency of homozygous recessive $(aa)$,and $2pq$ represents the frequency of heterozygous $(Aa)$.Total population $(N)$ = $10,000$.Number of $AA$ $(p^2 	imes N)$ = $6400$. Therefore,$p^2 = 6400 / 10000 = 0.64$. Thus,$p = \sqrt{0.64} = 0.8$.Number of $aa$ $(q^2 	imes N)$ = $400$. Therefore,$q^2 = 400 / 10000 = 0.04$. Thus,$q = \sqrt{0.04} = 0.2$.We know that $p + q = 1$ $(0.8 + 0.2 = 1)$,which confirms the values.The frequency of heterozygous genotype $(Aa)$ is $2pq = 2 	imes 0.8 	imes 0.2 = 0.32$.Number of cattle with heterozygous genotype $(Aa)$ = $0.32 	imes 10,000 = 3200$.

In a population of $10,000$ cattle,$6400$ have homozygous dominant genotype $(AA)$ and $400$ have homozygous recessive genotype $(aa)$. Find out the number of cattle having heterozygous genotype $(Aa)$ as per Hardy-Weinberg's principle.

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