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 =  q² = 1/10,000 = 0.0001.  Thus, q = square root of q² = 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 = p²   = (0.6)(0.6) = 0.36; the frequency of Ee = 2pq = 2(0.6)(0.4) = 0.48; and the frequency of ee =  q² = (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 (q²) = 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  p²  = (0.8)²   = 0.64.