If a gene has two alleles, A and B, and its frequencies are p and q, respectively, which statement holds true (suppose that population has reached the Hardy-Weinberg equilibrium):

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Correct Answer: **C.** *frequency of homozygote BB is q*^{2} .

This question requires just a simple understanding of the According to the Hardy-Weinberg equation, (p+q)^{2}=1, , or rather: and p^{2} + 2pq + q^{2} = 1.

If we have a gene with two alleles, A and B, and its frequencies are p and q, then we can know that the frequencies of the genotypes are as follows, by plugging into our Hardy Weinberg equation:

predicted frequency of homozygote AA = p^{2}

predicted frequency of homozygote BB = q^{2}

predicted frequency of heterozygote AB = 2pq.

The only answer choice that fits this solution is answer choice C.

Looking at the other answers, A is incorrect because it is an inaccurate representation of the Hardy Weinberg equation. Answer B is also incorrect because if one assumes that a gene has only two alleles, then then the frequency of p+q combined has to be 100%, or rather p + q = 1.

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