Concepts of Biology
(BIOL115) - Dr. S.G. Saupe (ssaupe@csbsju.edu); Biology Department, College of St. Benedict/St. John's
University, Collegeville, MN 56321 |

**Answers to Hardy-Weinberg Problems **(click
here to return to problems)

__Problem 1:__

We
know that the frequency of the homo recessive allele = q^{2} =
1/10,000 = 0.0001. Thus, q = square
root of q^{2} = 0.01. The frequency of
the dominant allele, p, must be equal to 0.99 since p = 1-q = 1 -
0.01 = 0.99. To calculate the
frequency of the heterozygote, 2pq, simply plug in the allele frequencies:
2pq = 2(.99)(.01) = .0198 = approx 2%.

__Problem 2:__

The frequency of EE = p^{2}^{
}= (0.6)(0.6) = 0.36; the frequency of Ee = 2pq = 2(0.6)(0.4) =
0.48; and the frequency of ee = q^{2}
= (0.4)(0.4) = 0.16. If the population has 10,000 individuals, then there
will be 3,600 EE (0.36 x 10,000), 4,800 Ee (0.48 x 10,000) and 1,600 ee (0.16 x
10,000).

__Problem 3:__

The frequency of homozygous recessive individuals (q^{2})
= 0.04. Thus, the frequency of the green allele, q equals 0.2, and the
frequency of p = 1 - q = 0.8. The frequency of heterozygous frogs = 2pq =
2(0.2)(0.8) = 0.32 and the frequency of homozygous dominant frogs equals p^{2}
= (0.8)^{2} = 0.64.

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