AP Biology Hardy Weinberg

A population is in Hardy-Weinberg equilibrium with allele frequencies $p = 0.4$ for $A$ and $q = 0.6$ for $a$. What is the expected frequency of the genotype $Aa$?

• 0.24

• 0.48

• 0.36

• 0.16

Answer explanation

In Hardy-Weinberg equilibrium, the frequency of the heterozygous genotype Aa is given by 2pq. Here, p = 0.4 and q = 0.6, so 2pq = 2(0.4)(0.6) = 0.48. Thus, the expected frequency of genotype Aa is 0.48.

Tags

NGSS.HS-LS4-2

In a population, the frequency of the genotype $AA$ is 0.36, and the population is in Hardy-Weinberg equilibrium. What is the frequency of allele $A$?

• 0.6

• 0.36

• 0.4

• 0.64

Answer explanation

In Hardy-Weinberg equilibrium, the frequency of genotype $AA$ (p^2) is 0.36. Thus, p = sqrt(0.36) = 0.6. The frequency of allele A (p) is 0.6, and the frequency of allele a (q) is 1 - p = 0.4.

Tags

NGSS.HS-LS4-3

In a population at Hardy-Weinberg equilibrium, the frequency of allele $a$ is 0.2. What is the expected frequency of the homozygous recessive genotype $aa$?

• 0.04

• 0.16

• 0.64

• 0.36

Answer explanation

In Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype $aa$ is given by $q^2$. Here, $q = 0.2$, so $q^2 = 0.2^2 = 0.04$. Thus, the expected frequency of genotype $aa$ is 0.04.

Tags

NGSS.HS-LS3-3

If the frequency of the homozygous recessive genotype $aa$ is 0.16 in a population, what is the frequency of the recessive allele $a$?

• 0.4

• 0.16

• 0.2

• 0.8

Answer explanation

The frequency of the homozygous recessive genotype $aa$ is given by $q^2 = 0.16$. To find the frequency of the recessive allele $a$, take the square root: $q = \sqrt{0.16} = 0.4$. Thus, the frequency of allele $a$ is 0.4.

Tags

NGSS.HS-LS4-3

A population is in Hardy-Weinberg equilibrium with allele frequencies $p = 0.7$ for $A$ and $q = 0.3$ for $a$. What is the expected frequency of the homozygous dominant genotype $AA$?

• 0.21

• 0.49

• 0.09

• 0.42

Answer explanation

In Hardy-Weinberg equilibrium, the frequency of the homozygous dominant genotype (AA) is calculated as p^2. Here, p = 0.7, so p^2 = 0.7 * 0.7 = 0.49. Thus, the expected frequency of AA is 0.49.

Tags

NGSS.HS-LS4-2

NGSS.HS-LS4-3

In a population at Hardy-Weinberg equilibrium, the frequency of allele $A$ is 0.6. What is the expected frequency of the heterozygous genotype $Aa$?

• 0.24

• 0.36

• 0.48

• 0.72

Answer explanation

In Hardy-Weinberg equilibrium, the frequency of heterozygous genotype $Aa$ is given by 2pq. Here, p (frequency of A) = 0.6 and q (frequency of a) = 0.4. Thus, 2pq = 2(0.6)(0.4) = 0.48, which is the correct answer.

Tags

NGSS.HS-LS4-3

In a population of 1000 individuals, the frequency of the recessive allele $a$ is 0.3. What is the frequency of the dominant allele $A$?

• 0.3

• 0.7

• 0.5

• 0.9

Answer explanation

The frequency of the recessive allele $a$ is 0.3. Since the total frequency of alleles must equal 1, the frequency of the dominant allele $A$ is 1 - 0.3 = 0.7. Thus, the correct answer is 0.7.

Tags

NGSS.HS-LS3-3

• 0.25

• 0.5

• 0.75

• 0.1

Tags

NGSS.HS-LS2-2

NGSS.HS-LS4-3

• 0.3

• 0.09

• 0.7

• 0.81

Tags

NGSS.HS-LS4-2

• 0.64

• 0.16

• 0.32

• 0.04