Question

If the a and b loci are 20 m.u. apart in humans and an AB/ab woman mates with an ab/ab man, what is the probability that their first child will be Ab/ab

Answer 1

Answer:

10%

Explanation:

The distance between the given two loci is 20 mu. This means that the recombination frequency between these two loci is 20% and the heterozygote will form 20% recombinant gametes and rest 80% parental type gametes. Therefore, the heterozygous woman AB/ab will form a total of 20% recombinant gametes. The proportion of each recombinant gamete will be 20/2= 10%.

Gametes formed by the woman= 10% Ab, 10% aB, 40% AB, and 40% ab

Gametes formed by the man= All ab (since the man is homozygous for both loci).

Therefore, proportion of Ab/ab genotype= 10% Ab (female gamete) x ab (male gamete)= 10% Ab/ab