The survival to sexual maturity rates for genotypes A1A1, A1A2, and A2A2 are 90%, 85%, &...

The survival to sexual maturity rates for genotypes A₁A₁, A₁A₂, and A₂A₂ are 90%, 85%, & 75%, and their fecundities are 50, 55, and 70 offspring, respectively. a. what are the absolute fitnesses (W) of these genotypes? b. Using A₁A₁ as the fitness reference, what are the genotype’s relative fitnesses (w)? c. If the frequency of the A₂ allele is p = 0.5, what will be its frequency one generation later?

Expert Solution

Answer:-

(a) The survival to sexual maturity rates of genotype A1A1 is 90%, A1A2 is 85% and A2A2 is 75%. The fecundity of genotype A1A1 is 50, A1A2 is 55 and A2A2 is 70.

So, the absolute fitness i.e W of these genotypes can be determined by multiplying their survival rates by their fecundities.

Absolute fitness of genotype A1A1 = 90/100 × 50 = 45
Absolute fitness of genotype A1A2 = 85/100 × 55 = 46.75
Absolute fitness of genotype A2A2 = 75/100 × 70 = 52.5

Therefore, the absolute fitness for genotypes A1A1, A1A2, A2A2 are 45, 46.75 and 52.5 respectively.

(b) By taking fitness reference of genotype A1A1 i.e 45, the relative fitness of the genotypes can be determined by dividing the absolute fitness of the genotypes by fitness reference.

Relative fitness of genotype A1A1 = 45/45 = 1
Relative fitness of genotype A1A2 = 46.75/45 = 1.03
Relative fitness of genotype A2A2 = 52.5/45 = 1.16

Therefore, the relative fitness for genotypes A1A1, A1A2, A2A2 are 1, 1.03, 1.16 respectively.

(c) If the frequency of the A2 allele is p = 0.5. This frequency will remain same after one generation without the presence of evolutionary factors according to Hardy Weinberg principle.

Hope the above answer comes up to your expectation. Please do come up with your feedback.

gladiator answered 2 years ago

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