Leader: Kirsten Karkow
Course: Biol 211 - Wilsey
Date: 3/26/2012
Worksheet 22 Key
Supplemental Instruction
Iowa State University
Practice Hardy-Weinberg problems:
You sample a population and determine that the percentage of the homozygous recessive
genotype “aa” is 36%. Calculate the following:
1. The frequency of the “a” allele.
Answer: q2 = 0.36 so q = 0.6. The frequency of “a” is 60%.
2. The frequency of the “A” allele.
Answer: q = 0.6 and p + q =1, so p = 0.4. The frequency is 40%.
3. The frequencies of the genotypes “AA” and “Aa”.
Answers: p = 0.4 so the frequency of AA = p2 = 0.16 (16%).
Aa is calculated as 2pq = 2(0.6)(0.4) = 0.48 (48%).
4. The frequencies of each phenotype if “A” is completely dominant over “a”.
Answers: The frequency of the dominant phenotype will equal the frequency of AA (p 2) plus
the frequency of Aa (2pq) , since both result in the dominant phenotype.
p2 + 2pq = 0.16 + 0.48 = 0.64 (64% dominant).
p2 + 2pq + q2 = 1, so q2 = 1 – 0.64 = 0.36 (36% recessive)
Sickle-cell anemia is a genetic disease in which individuals homozygous for the sickle-cell trait
(ss) have distorted red blood cells that lead to a variety of complications. Individuals
heterozygous (Ss) for the condition, however, have only limited sickling and are resistant to
malaria. If 9% of an African population is born with a severe sickle-cell anemia (ss), what
percentage of the population will be more resistant to malaria because they are heterozygous?
Answer: 9% = .09 = q2, so q = 0.3
q + p = 1, so p must equal 0.7
The frequency of heterozygotes is 2pq = 2(0.7)(0.3) = 0.42 (42%)
Supplemental Instruction
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A bird population is composed of 396 red-sided individuals and 557 tan-sided individuals.
Assume that red is completely recessive. Calculate the following:
A. The allele frequencies of each allele.
Answer: The total number of individuals is 396 + 557 = 953.
The red-sided individuals are recessive, so q2 = 396/953 = 0.416.
Therefore, q = 0.645. Since p + q = 1, p must equal 1 – 0.645 = 0.355.
B. The expected genotype frequencies.
Answer: AA = p2 = 0.3552 = 0.126.
Aa = 2pq = 2(0.355)(0.645) = 0.458.
aa = q2 = 0.416 (above).
C. The number of heterozygous individuals you would expect in this population.
Answer: Just times frequency of Aa by total population size (N). (0.458)(953) = about 436.
D. Assume that next year there are 1245 individuals. If all of the Hardy-Weinberg
conditions are met, how many of these would you expect to be red-sided and how
many would you expect to be tan-sided?
Answer: The number of red-sided birds (recessive) is just q2 x N = 0.416(1245) = 518. All of the
rest are tan-sided: 1245 – 518 = 727.
Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian
population of the United States. Calculate the following:
A. The frequency of the recessive allele in the population.
Answer: q2 is given as 1/2,500 or 0.0004. Therefore, q = 0.02 (2%)
B. The frequency of the dominant allele in the population.
Answer: The frequency of the dominant (normal) allele is simply 1 - 0.02 = 0.98 (or 98%).
C. The percentage of heterozygous individuals (carriers) in the population.
Answer: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are carriers.