Solutions For An Introduction to Genetic Analysis Chapter 4 Problem 64P

The second cross of pure line genotypes are:

A/A B/B C/C D/D E/E a/a B/B c/c D/D e/e

The F₁ progeny is crossed with a recessive tester and the progeny is tabulated.

Given that the genes D and E assort independently.

a) In the first cross, since the genes C and E are same for all the progeny, the parental genotypes of the first cross can be written as:

Parents: A/A B/B D/D a/a b/b d/d

The F₁ genotype: A/a B/b D/d a/a b/b d/d [tester]

The testcross progeny shows that these three genes are linked.

The distance between the genes A and B:

The distance between the genes B and D:

In the second cross, since the genes B and D are the same for all progeny, the parental genotypes can be written as:

Parents: A/A C/C E/E a/a c/c e/e

The F₁ genotype: A/a C/c E/e a/a c/c e/e [tester]

The testcross progeny shows that these three genes are linked.

The distance between the genes A and C:

The distance between C and E:

The map showing the gene distances:

b) Interference can be calculated using the following formula:

For cross 1:

Therefore, cross 1 shows no interference.

For cross 2:

Therefore, cross 2 shows no interference.