Biology: AP Biology: Hardy-Weinberg Equilibrium Worksheet

Question 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?

Accepted Answer

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.

Question 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?

Accepted Answer

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.

Question 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?

Accepted Answer

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.

Question 4

List the five major conditions that must be met for a population to remain in Hardy-Weinberg equilibrium.

Accepted Answer

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.

Question 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?

Accepted Answer

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.

Question 6

A population has 250 AA individuals, 500 Aa individuals, and 250 aa individuals. What are the allele frequencies p and q?

Accepted Answer

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.

Question 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?

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Question 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?

Accepted Answer

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.

Question 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.

Accepted Answer

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.

Biology: AP Biology: Hardy-Weinberg Equilibrium

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