In a popluation that meets the Hardy-Weingber assumptions (i.e. is in equilibriu

Question

In a popluation that meets the Hardy-Weingber assumptions (i.e. is in equilibrium), 81% of the individuals are homozygous for a recessive mutation. Assuming there are two alleles for the affected gene, what percentage of the individuals in this population would be expected to be heterozygous?

A. 18

B. 9

C. 81

D. 1

Polycystic kidney disease is a rare autosomal recessive trait in humans. For this trait, assume that the allele 'A' is dominant and the allele 'a' is recessive. If there are 100 affected individuals (aa) in a population of 1 million, what is the expected frequency of homozygous (AA) individuals if the population is in Hardy-Weinberg equilibrium?

A. 0.98

B. 0.0196

C. 0.01

D. 0.0001

Explanation / Answer

(1)

Answer: A. 18

Let’s q be the recessive allele and p be the dominant normal allele

q2=0.81

q=0.9

Hardy-Weinberg equilibrium

p+q=1

p=1-0.9

p=0.1

p2(homozygous dominant) = 0.1 x 0.1= 0.01=1%

2pq (heterozygous)=2 x 0.9 x 0.1=0.18=18%

q2(homozygous recessive)= 81%

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