A Given genes A and B are linked and are 5 cM apart Parental

A.) Given: genes A and B are linked and are 5 cM apart

Parental Generation: AABB X aabb

F1: AaBb

If you were to perform a testcross with the F1 generation and observed 10,000 progeny, how many progeny of each of the 4 genotypes would you expect to observe?

The answer is:

AaBb - 4750

aabb – 4750

Aabb - 250

aaBb - 250

I\'m not sure how to get to the answer, please show steps.

B.) A and B are two linked genes 25 cM apart.

Parental Generation: AAbb X aaBB

F1 Generation: AaBa X aabb (test cross)

F2 Generation:

Out of 10,000 offspring, how many would you expect to have the following phenotypes:

The answer is: AaBb 1,250

Aabb 3,750

Please show steps

Solution

A: The recombination frequency is directly proportional to distance between genes. Here, the two genes are 5cM apart i.e. the recombination frequency is 5%. Since the parent generation has A and B genes together, the AaBb genotype would produce 95% parental gametes (AB and ab) and 5% recombinant gametes (Ab and aB). And the other parent in the test cross would produce only gamtes with ab genotype. So, the F2 generation would have 95% parental genotype (AaBb and aabb) and 5% recombinant genotype (Aabb and aaBb).
Out of 10,000 number of parental genotype= (95/100) x 10,000= 9500
So, Number of AaBb and aabb genotypes= 9500/2= 4750 each.
Out of 10,000 number of recombinant genotype= (5/100) x 10,000= 500
So, Number of Aabb and aaBb genotypes= 500/2= 250 each.

B: Here, the parental generation has AAbb and aaBB genotype.
And, the two genes are 25cM apart i.e. the recombination frequency is 25%. Since the parent generation has (A/ b and a/B) genes together, the AaBb genotype would produce 75% parental gametes (Ab and aB) and 25% recombinant gametes (AB and ab). And the other parent in the test cross would produce only gamtes with ab genotype. So, the F2 generation would have 75% parental genotype (Aabb and aaBb) and 25% recombinant genotype (AaBb and aabb).
Out of 10,000 number of parental genotype= (75/100) x 10,000= 7500
So, Number of (Aabb and aaBb) genotypes= 7500/2= 3750 each.
Out of 10,000 number of recombinant genotype= (25/100) x 10,000= 2500
So, Number of (AaBb and aabb) genotypes= 1250/2= 250 each.