A fruit fly of genotype AaBb parent 1 is crossed to another

A fruit fly of genotype A/a.B/b (parent 1) is crossed to another fruit fly of genotype a/a.b/b (parent 2). The progeny of this cross were: What gametes were produced by parent 1 and in what proportions? What can be deduced from these proportions? What is the recombination frequency (RF)? For the example in Q4, what is the RF between A gene ang B gene? For the example in Q4, if one sets up a selfing for parent 1 (A/b.B/b X A/a.B/b), wh fraction of the offpring will have the genotype (a/a.b/b)?

Solution

Answer:

4. (a) The genotype of parent 1 is AaBb.

So, gametes produced by parent1 are, AB, ab, Ab, aB.The proportions were 32% AB, 33% ab, 17% Ab, and 18% aB.  

(b) From the above proportions, it can be deduced that the genes does not assort independently, since we expect 25% of each for independent assortment to take place.

5. Number of recombinants = 17 + 18 = 35

Total progeny = 100

So, RF = 35/100 = 0.35 = 35%

6. If we cross AaBb * AaBb, 1/16 offspring will have the genotype aabb.