Biology: AP Biology: Hardy-Weinberg Equilibrium
Using allele frequencies to model population genetics
Biology: AP Biology: Hardy-Weinberg Equilibrium
Using allele frequencies to model population genetics
Biology - Grade 9-12
- 1
In a population of 1,000 beetles, the recessive phenotype appears in 90 individuals. Assume the population is in Hardy-Weinberg equilibrium. What are q, p, and the expected number of heterozygous beetles?
Start by recognizing that individuals with the recessive phenotype represent q^2, not q.
The recessive phenotype frequency is q^2 = 90/1000 = 0.09, so q = 0.30. Since p + q = 1, p = 0.70. The heterozygote frequency is 2pq = 2(0.70)(0.30) = 0.42, so the expected number of heterozygous beetles is 420. - 2
A flower color trait has two alleles, R and r. In a population at Hardy-Weinberg equilibrium, the frequency of the r allele is 0.25. What are the expected genotype frequencies for RR, Rr, and rr?
If q = 0.25, then p = 0.75. The expected frequency of RR is p^2 = 0.75^2 = 0.5625, the expected frequency of Rr is 2pq = 2(0.75)(0.25) = 0.375, and the expected frequency of rr is q^2 = 0.25^2 = 0.0625. - 3
A diagram shows a gene pool with allele A frequency 0.6 and allele a frequency 0.4. If the population is in Hardy-Weinberg equilibrium, what percentage of the next generation is expected to be heterozygous?
Heterozygotes have one copy of each allele, so use the 2pq term.
The heterozygote frequency is 2pq. With p = 0.6 and q = 0.4, 2pq = 2(0.6)(0.4) = 0.48. Therefore, 48% of the next generation is expected to be heterozygous. - 4
List the five major conditions that must be met for a population to remain in Hardy-Weinberg equilibrium.
A population remains in Hardy-Weinberg equilibrium when there is no mutation, random mating, no natural selection, a very large population size, and no gene flow through immigration or emigration. - 5
In a population of moths, 16% have a recessive wing pattern. If the population is in Hardy-Weinberg equilibrium, what percentage of the population is expected to be homozygous dominant?
Take the square root of the recessive phenotype frequency to find q.
The recessive phenotype frequency is q^2 = 0.16, so q = 0.40. Then p = 1 - 0.40 = 0.60. The homozygous dominant frequency is p^2 = 0.60^2 = 0.36, so 36% of the population is expected to be homozygous dominant. - 6
A population has 250 AA individuals, 500 Aa individuals, and 250 aa individuals. What are the allele frequencies p and q?
There are 1,000 total alleles because 500 individuals each have 2 alleles. The number of A alleles is 2(250) + 500 = 1,000, and the number of a alleles is 500 + 2(250) = 1,000. The total number of alleles is 2,000, so p = 1000/2000 = 0.50 and q = 1000/2000 = 0.50. - 7
A population of 400 fish is in Hardy-Weinberg equilibrium. The frequency of the dominant allele B is 0.7. How many fish are expected to have the recessive phenotype?
The recessive phenotype appears only in individuals with the bb genotype.
If p = 0.7, then q = 0.3. The recessive phenotype frequency is q^2 = 0.3^2 = 0.09. In a population of 400 fish, 0.09 times 400 = 36 fish are expected to have the recessive phenotype. - 8
In a sample of 200 plants, the observed genotypes are 98 TT, 84 Tt, and 18 tt. Calculate the allele frequencies p and q for the T and t alleles.
There are 400 total alleles. The number of T alleles is 2(98) + 84 = 280, so p = 280/400 = 0.70. The number of t alleles is 84 + 2(18) = 120, so q = 120/400 = 0.30. - 9
Using the allele frequencies from the plant population in problem 8, calculate the expected numbers of TT, Tt, and tt individuals under Hardy-Weinberg equilibrium for a population of 200 plants.
Multiply each expected genotype frequency by the total population size.
With p = 0.70 and q = 0.30, the expected TT frequency is p^2 = 0.49, the expected Tt frequency is 2pq = 0.42, and the expected tt frequency is q^2 = 0.09. For 200 plants, the expected counts are 98 TT, 84 Tt, and 18 tt. - 10
A population of birds has genotype counts of 45 BB, 30 Bb, and 25 bb. A student claims the population is definitely in Hardy-Weinberg equilibrium because p + q = 1. Explain why this claim is not enough.
The claim is not enough because p + q = 1 only confirms that the allele frequencies add to 100% for a two-allele gene. Hardy-Weinberg equilibrium also requires that genotype frequencies match p^2, 2pq, and q^2 and that the population meets the assumptions of the model, such as random mating and no selection. - 11
A graph compares observed and expected genotype frequencies in a lizard population. Observed frequencies are AA = 0.30, Aa = 0.60, and aa = 0.10. Expected Hardy-Weinberg frequencies are AA = 0.36, Aa = 0.48, and aa = 0.16. Which genotype shows the largest difference, and what might this suggest?
Compare the absolute difference between observed and expected frequencies for each genotype.
The heterozygous genotype Aa shows the largest difference because 0.60 - 0.48 = 0.12. This may suggest that the population is not in Hardy-Weinberg equilibrium, possibly because of nonrandom mating, selection, population structure, or another evolutionary force. - 12
Cystic fibrosis is caused by a recessive allele. In a large randomly mating population, 1 in 2,500 newborns has cystic fibrosis. Assuming Hardy-Weinberg equilibrium, estimate the carrier frequency for this allele.
Carriers are heterozygous, so calculate 2pq after finding q.
The affected frequency is q^2 = 1/2500 = 0.0004, so q = 0.02. Then p = 0.98. The carrier frequency is 2pq = 2(0.98)(0.02) = 0.0392, or about 3.92%, which is approximately 1 in 25 people.