Genetics Fitness and Selection Practice Questions and 80 out a hundred individua

Question

Genetics Fitness and Selection Practice Questions and 80 out a hundred individuals with the AA and Aa genotypes survive to reproductive age and 80 out of 100 individuals with the aa genotype survive to reproductive age what are the fitnesses of the three genotypes? otype is 0.1. The selective coefficient for the aa 2) The selective coefficient for the AA gen genotype is 0. There is complete dominance of the A allele over the a allele. What are the fitnesses of the three genotypes? 3) Ectrodactyly fused. Otherwise, individuals with this condition are healthy. Suppose the frequency of ectrodactyly among newborns is approximately 1/10,000 (when the parents are unrelated). What is an autosomal recessive condition in humans in which the fingers and toes are

Explanation / Answer

1)

Since "aa" individuals have highest survival rate, its relative fitness will be 1 and of AA and Aa individuals will be 0.5.

2)

Selection coefficient (s), measures the degree to which a genotype is selected against.
s = 1 ? W

where W = relative fitness.

Selection coefficient for AA genotype = 0.1

Relative fitness = 1 - 0.1 = 0.9

Selection coeff. for aa genotype = 1

Relative fitness = 1 - 1 = 0

Since "A" is dominant over "a", selection coeff. for Aa genotype will also be 0.1, and

Relative fitness of Aa = 1 - 0.1 = 0.9

3)

Frequency of allele "q" responsible for disease = 1/10,000 = 0.01

Frequency of allele "p" = 1 - q = 0.99

Inbreeding coefficient (F) when parents are siblings = 25% or 0.4

Frequency of homozygous recessives for full siblings mating = q2 + pq(1/4)= (0.01)2 +(0.01)(0.99)(1/4 )

= 0.0026

Risk of a child with ectrodactyly if the parents are siblings is 26 in every 10,000 individuals OR approximately 3 in every 1000 individuals.

Genotype AA Aa aa # before selection 100 100 100 # after selection 40 40 80 Survival rate 0.4 0.4 0.8 Relative fitness (w) 0.5 0.5 1

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