Question

Question 2: In an outbreeding diploid population, an autosomal locus has two alleles, A1 and A2. You observe the allele frequencies of A1 and A2 are both equal to 0.5, and the relative fitnesses of the three genotypes (A1A1, A1A2, and A2A2) are equal to 2, 1.5, and 1, respectively.

A) Which allele will become fixed in this population? (1 point)

**The recessive beneficial allele A2A2 will eventually become fixed in the population (A2A2=1)

B) What is the allele frequency of A1 in the next generation? (2 points)

C) What is the allele frequency of A1 in the next generation, if the population is inbreeding and the inbreeding coefficient is 0.5? (2 points)

Answer 1

GIVEN -

Frequency of allele A1 and A2 is equal, 0.5.

Thus,

p = 0.5

q= 0.5

Genotype	A1A1	A1A2	A2A2
Relative fitness	2	1.5	1

Relative fitness is used to measure how fit a genotype is , as compared to that of least fit genotype. Here, the least fit is A2A2 (since it has lowest relative fitness value). Fitness reflects reproductive capacity of a genotype.

Also, A1A1 individuals are most fit, twice as that of A2A2 individuals and A1A2 individuals have a fitness of 1.5 times that of A2A2 individuals.

A) Eventually allele A1 should get fixed, as it mas maximum relative fitness.

B) Allele frequency of A1 in next generation:

Average fitness, W'

= p^2 (W_A1A1 ) + 2pq (W_WA1A2​​​​​​) + q^2 (W _A2A2)

= (0.5×0.5) (2) + )(2×0.5×0.5) (1.5) + (0.5×0.5) (1)

= 1.5

Thus, frequency of A1

= p^2 (WA1A1) /W' + pq(WA1A2) /W'

= (0.5×0.5) (2) /1.5 + (0.5×0.5) (1.5) /1.5

=0.58

(C) Inbreeding coefficient, F = 0.5

Allele frequency of A1 in next generation,

= p^2 + Fpq + pq(1-F)

= (0.5) ² + {(0.5)× 0.5 × 0.5} + { (0.5) ×(0.5) }(1-0.5)

= 0.25 + 0.125 + 0.125

= 0. 50