Question

A and B genes are linked at 20 cM apart. A genotype with known linkage phase AB/ab is testcrossed to aa bb. What proportion of the offspring is expected to be dominant for both genes?

Select one:

A. 80%

B. 40%

C. 10%

D. 20%

E. 0%

Answer 1

Since the A and B genes are 20 cM apart, the recombination frequency between the two genes is 20%. This means 20% of gametes produced are recombinant for A and B and 80% of gametes are non-recombinant. The genotype of the dihybrid individual is given to be AB/ab, and therefore progeny that has a dominant characteristic for both alleles is produced from a non-recombinant gamete.

Now, two gametes are non-recombinant: AB and ab, both totalling 80% of all gametes. Therefore AB gametes, which result in progeny that is dominant for both alleles, are expected to be 80/2 = 40% of the total progeny.

B) 40% is the correct response.