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):
Loading ...
Correct Answer: C. frequency of homozygote BB is q2 .
This question requires just a simple understanding of the According to the Hardy-Weinberg equation, (p+q)2=1, , or rather: and p2 + 2pq + q2 = 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 = p2
predicted frequency of homozygote BB = q2
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.
Get it right? Tweet at us:
Subscribe below to receive the MCAT Question of the Day delivered straight to your inbox every morning.
