Consider two linked traits in a population in which half of the individuals are
ID: 33930 • Letter: C
Question
Consider two linked traits in a population in which half of the individuals are
double heterozygotes with genotype AB/ab and the other half are double homozygotes (AB/AB). Assuming no selective advantage of any combination of alleles for these traits and no preferential mating, after recombination brings the population to equilibrium, what will be the ratio of AB/AB, Ab/aB = aB/Ab and ab/ab individuals?
Hint: . Calculate the overall allelic composition of the equilibrium population and distribute the alleles randomly among individuals.
Please help
Explanation / Answer
Ans.
Given,
Double Homozygotes = AABB
Double Heterozygotes = AaBb
( only these 2 types of Genotypes
and thier in Equal Proportion )
Now,
AABB will give, = AB and AB
AaBb will give, = AB , Ab , aB and ab
Allele frequency =
AB = 3 AB / 3 AB + Ab + aB + ab = 3/6 = 0.50
Ab = 1 Ab / 3 AB + Ab + aB + ab = 1/6 = 0.16
aB = 1 aB / 3 AB + Ab + aB + ab = 1/6 = 0.16
ab = 1 ab / 3 AB + Ab + aB + ab = 1/6 = 0.16
Now On Recombination,
AB x AB = AABB = 0.5 * 0.50 = 0.25
AB x Ab = AABb = 0.5 * 0.16 = 0.08
AB x aB = AaBB = 0.5 * 0.16 = 0.08
AB x ab = AaBb = 0.5 * 0.16 = 0.08
Ab x Ab = AABb = 0.16 * 0.16 = 0.02
Ab x aB = AaBb = 0.16 * 0.16 = 0.02
Ab x ab = Aabb = 0.16 * 0.16 = 0.02
aB x aB = aaBb = 0.16 * 0.16 = 0.02
aB x ab = aaBb = 0.16 * 0.16 = 0.02
ab x ab = aabb = 0.16 * 0.16 = 0.02
Overall genotypes ,
AABB = 0.25
AaBb = 0.08 + 0.02 = 0.1
aabb = 0.02
The Above will be their Ratio.