In the Hardy-Weinberg equation,the frequency | vedclass

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(B) The correct answer is $2pq$.In a stable population,for a gene with two alleles,'$A$' (dominant) and '$a$' (recessive),if the frequency of '$A$' is

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$2pq$

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(B) The correct answer is $2pq$.In a stable population,for a gene with two alleles,'$A$' (dominant) and '$a$' (recessive),if the frequency of '$A$' is $p$ and the frequency of '$a$' is $q$,then the frequencies of the three possible genotypes ($AA$,$Aa$,and $aa$) are expressed by the Hardy-Weinberg equation:$p^2 + 2pq + q^2 = 1$Where:$p^2$ = Frequency of $AA$ (homozygous dominant) individuals.$q^2$ = Frequency of $aa$ (homozygous recessive) individuals.$2pq$ = Frequency of $Aa$ (heterozygous) individuals.

In the Hardy-Weinberg equation,the frequency of heterozygous individuals is represented by:

Similar Questions

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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If mating is random,the population is large,and mutation does not occur,then the gene frequencies of the population from generation to generation remain constant. This principle was proposed by:

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