Hardy Weinberg Equations

A rather large population of foxes has 555 individuals with tan fur and 390 individuals with red fur. Assume that red is recessive. Calculate the following: 

1. The frequencies of each allele. 

2. The frequencies of each genotype.

3. How many foxes are of the heterozygous genotype?

Answer explanation

Tags

NGSS.HS-LS3-3

NGSS.HS-LS2-2

If the frequency of recessive allele is 0.2, what is the frequency of the homozygous recessive individuals?

• 0.04

• 0.16

• 0.4

• 0.8

Answer explanation

The frequency of homozygous recessive individuals is 0.04, which is calculated by squaring the frequency of the recessive allele (0.2 x 0.2 = 0.04).

What is the Hardy-Weinberg genotypic frequency equation?

• p³/q³+pq=1

• 1/2ab sin c

• p²+2pq+q²=1

• a²+b²=c²

Tags

NGSS.HS-LS4-3

What are two of the five assumptions about the Hardy-Weinberg equilibrium?

• no mutation

• natural selection

• random mating

• small population

Is the Hardy-Weinberg equilibrium realistic?

• Yes

• No

• I don't know

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-LS3-3

If the frequency of allele $A$ is 0.8 in a population at Hardy-Weinberg equilibrium, what is the frequency of the genotype $aa$?

• 0.04

• 0.16

• 0.64

• 0.32

Answer explanation

In Hardy-Weinberg equilibrium, the frequency of genotype $aa$ is given by $q^2$. If the frequency of allele $A$ (p) is 0.8, then $q = 1 - p = 0.2$. Thus, $q^2 = (0.2)^2 = 0.04$. Therefore, the frequency of genotype $aa$ is 0.04.

Tags

NGSS.HS-LS4-3

If the frequency of the recessive phenotype is 0.09 in a population at Hardy-Weinberg equilibrium, what is the frequency of the dominant allele?

• 0.7

• 0.3

• 0.81

• 0.9

Answer explanation

In Hardy-Weinberg equilibrium, the frequency of the recessive phenotype (q^2) is 0.09. Thus, q = √0.09 = 0.3. The frequency of the dominant allele (p) is 1 - q = 1 - 0.3 = 0.7, which is the correct answer.

Tags

NGSS.HS-LS4-3

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-LS3-3

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-LS4-3