Hardy-Weinberg Practice Problems (Assignment 11)
1) An investigator has determined by inspection that 16% of a human population has a recessive
trait (t). Complete all the genotype and allele frequencies for this population, assuming that it is
in Hardy-Weinberg equilibrium.
a.) p = 0.6
q= 0.4
b.) TT= 0.36
Tt= 0.48
tt= 0.16
a.) q: t2 = 0.16, √t2 = √16 = 0.4;
p + q = 1, p + 0.4 = 1, p = 0.6
2
2
b.) TT = p = 0.6 = 0.36;
Tt = 2pq = 2(0.6)(0.4) = 0.48;
tt = q2 = 0.42 =0.16
2) A large population of laboratory animals has been allowed to breed randomly for a number of
generations. After several generations, 36% of the animals display a recessive trait (aa), the
same percentage as at the beginning of the breeding program. The rest of the animals show the
dominant phenotype, with heterozygotes indistinguishable from the homozygous dominants.
(complete dominance)
a. What is the estimated frequency of allele a in the gene pool?
q2 = 0.36, √q2 = √0.36, q = 0.6
(p + q = 1, p + 0.6 = 1, p = 0.4)
b. What proportion of the population in probably heterozygous (Aa) for this trait?
Aa = 2pq = 2(0.4)(0.6) = 0.48
3) In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency
of the allele a is 0.7.
a.) What is the percentage of the population that is homozygous for this allele?
aa = q2 = 0.72 = 0.49 = 49%
(p + q = 1, p + 0.7 = 1, p = 0.3)
b.) What is the percentage of the population that is heterozygous for this allele?
Aa = 2pq = 2(0.3)(0.7) = 0.42 = 42%
4) In a Hardy-Weinberg population with two alleles, A and a, that are in equilibrium, the frequency
of allele a is 0.2. What is the frequency of individuals with Aa genotype?
a = q = 0.2;
p + q = 1, p + 0.2 = 1, p = 0.8
Aa = 2pq = 2(0.8)(0.2) = 0.32
5) In a population with two alleles, A and a, the frequency of a is 0.50. What would be the
frequency of heterozygotes if the population is in Hardy-Weinberg equilibrium?
a = q = 0.5;
p + q = 1, p + 0.5 = 1, p = 0.5
Aa = 2pq = 2(0.5)(0.5) = 0.50
6) In a hypothetical population of 1,000 people, tests of blood-type genes show that 160 have the
genotype AA, 480 have the genotype AB, and 360 have the genotype BB. (incomplete
dominance)
160 AA
480 AB
360 BB
1000
a. What is the frequency of the A allele?
p = [2(AA)] + [(AB)] = [2(160)] + [(480)] = 320 + 480 = 800 = 0.4
1000(2)
A = p = 0.4
2000
2000
2000
b. What is the frequency of the B allele?
q = [2(BB)] + [(AB)] = [2(360)] + [(480)] = 720 + 480 = 1200 = 0.6
1000(2)
2000
2000
2000
B = q = 0.6
c. If there are 4,000 children born to this generation, how many would be expected to
have AB blood under the conditions of Hardy-Weinberg equilibrium?
AB = 2pq = 2(0.4)(0.6) = 0.48;
4,000(0.48) = 1920
7) In peas, a gene controls flower color such that R = purple and r = white. In an isolated pea patch,
there are 36 purple flowers and 64 white flowers. Assuming Hardy-Weinberg equilibrium, what
is the value of q for this population?
64 = 0.64, q2 = 0.64, √q2 = √0.64, q = 0.8
100
8) A population of 200,000 aphids in a poppy field is initially in HW equilibrium with respect to the
alleles for a particular digestive enzyme. The frequency of the dominant allele (E) is 0.6.
a. What is the frequency of the recessive allele (e)?
p + q = 1, 0.6 + q = 1, q = 0.4
b. What are the initial genotypic and phenotypic frequencies (complete dominance)?
p2 + 2pq + q2 = 1, 0.62 + 2(0.6)(0.4) + 0.42 = 1
Genotypic: EE = 0.36
Phenotypic: E _= 0.84
Ee = 0.48
ee = 0.16
e = 0.16
9) In a population of 10,000 people, 4897 have blood type MM, 4210 have blood type MN, and 893
have blood type NN. What are the allelic frequencies, the genotypic frequencies, and the
phenotypic frequencies for this population? (incomplete dominance)
4897 MM
4210 MN
893 NN
10,000
p2 = 0.72 = 0.49
p = [2(4897) + (4210) = (9794) + (4210) = 14,004 = 0.7002 = 0.7
20,000
20,000
20,000
q = [2(893) + (4210) = (1786) + (4210) = 5,996 = 0.2998 = 0.3
20,000
20,000
20,000
2pq = 2(0.7)(0.3) = 0.42
q2 = 0.32 = 0.09
Allelic: M = 0.7, N = 0.3
Genotypic: MM = 0.49, MN = 0.42, NN = 0.09
Phenotypic: same as genotypic