Question

A large study of DNA from people in the USA determined that the frequency for the recessive allele for ear shape .2 and the dominant is .8. If 10 % of the population shows the recessive phenotype is the population in HW equilibrium for this trait? Note that the population of the USA is over 200 million!

		no
		yes
		cannot be determined from the data given

Answer 1

Ans:

Yes, the population is in HW equilibrium

Explanation: Given recessive allele frequency (p) =0.2

dominant allele frequency (q) =0.8

Now population of USA =200 million

10% of population shows recessive phenotype

i.e recessive phenotype (p) =0.1

dominant phenotype (q) =0.9

According to HW equilibrium

p² +2pq +q² =1

substitute value of p and q in HW equilibrium equation given below

p² +2pq +q² =1

[ p²= frequency of rare homozygotes; 2pq = Frequency of heterozygotes; q² = Frequency of common homozygotes]

(0.1)² + 2x0.1x0.9 + (0.9)² =1

p² = 0.01 or 1%

2pq = 0.18 or 18%

q² = 0.81 or 81%

Now Common Homozygotes (Genotype)

Expected (CH) = 82.36

Observed (CH) = 81

Heterozygotes (Genotype)

Expected (CH) = 33.27

Observed (CH) = 36

Rare Homozygotes (Genotype)

Expected (CH) = 3.36

Observed (CH) = 2

Chi-squared value = 0.79 (Non-significant)

p Allele Frequency = 0.83

q Allele Frequency = 0.16

Since Chi-square value is non-significant and recessive and dominant allele frequencies are almost same as the observed allele frequencies determined (0.2 amd 0.8). Therefore data is in HW equilibrium as their is no significant deviation of allele frequencies in the population for the ear shape.