Question

14. Given below are the genotypic frequencies for a single gene with two alleles for three
different populations:
Population 1
AA
0.25

Aa
0.50
aa
0.25
Population 2 0.35 0.56 0.09
Population 3 0.49 0.42 0.09

Which of the following is NOT a correct statement about these three populations?
A) Only two of the populations are in Hardy-Weinberg equilibrium
B) Population 1 is in Hardy-Weinberg equilibrium; the frequency of allele A is 0.5
C) Population 2 is not in Hardy-Weinberg equilibrium; the frequency of allele a is 0.37
D) Population 3 is in Hardy-Weinberg equilibrium; the frequency of allele A is 0.7
E) Population 3 is in Hardy-Weinberg equilibrium; the frequency of allele a is 0.3

14. Given below are the genotypic frequencies for a single gene with two alleles for three different populations: АА ga Population 1 Population 2 Population 3 0.25 0.35 0.49 Aa 0.50 0.56 0.42 0.25 0.09 0.09 Which of the following is NOT a correct statement about these three populations? A) Only two of the populations are in Hardy-Weinberg equilibrium B) Population 1 is in Hardy-Weinberg equilibrium; the frequency of allele A is 0.5 C) Population 2 is not in Hardy-Weinberg equilibrium; the frequency of allele a is 0.37 D) Population 3 is in Hardy-Weinberg equilibrium; the frequency of allele A is 0.7 E) Population 3 is in Hardy-Weinberg equilibrium; the frequency of allele a is 0.3

Answer 1

c. Population 2 is not in Hardy Weinberg equilibrium the frequency of allele a is 0.37

This statement is partially true as population 2 is not in Hardy Weinberg equilibrium but the frequency of allele a is not 0.37 but is 0.3.

p² + 2pq + q² = 1 and p + q = 1

p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population

p² = percentage of homozygous dominant individuals
q² = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals

Population 1

AA or genotypic frequency or p2 = 0.25

Therefore the frequency of allele A or p is 0.5

aa or genotypic frequency or q2 = 0.25

Therefore the frequency of allele a or q = 0.5

If this population follows Hardy Weinberg`s equilibrium then p + q = 1

here 0.5 +0.5 = 1.0

This population follows Hardy Weinberg`s equilibrium

Population 2

AA or genotypic frequency or p2 = 0.35

Therefore the frequency of allele A or p is 0.59

aa or genotypic frequency or q2 = 0.09

Therefore the frequency of allele a or q = 0.3

If this population follows Hardy Weinberg`s equilibrium then p + q = 1

here 0.59 + 0.3 = 0.89

which is not equal to 1 hence this population does not follow Hardy Weinberg`s equilibrium.

Population 3-

AA or genotypic frequency or p2 = 0.49

Therefore frequency of allele A or p is 0.7

aa or genotypic frequency or q2 = 0.09

Therefore the frequency of allele a or q = 0.3

If this population follows Hardy Weinberg`s equilibrium then p + q = 1

here 0.7+ 0.3 = 1.0

Hence it follows Hardy Weinberg`s equilibrium