Solutions For An Introduction to Genetic Analysis Chapter 4 Problem 17P

A female with the genotype A/a B/b is crossed with a homozygous recessive male with the genotype a/a b/b.

The progeny obtained are:

• 442 - A/a B/b.

• 458 - a/a b/b.

• 46 - A/a b/b.

• 54 - a/a B/b.

When an organism is crossed with a homozygous recessive organism in a testcross, the progeny ratio is always 1:1:1:1. This ratio is not seen in the progeny of the above testcross. Hence, the genes are linked.

Moreover, on observing the progeny, we see that the parental genotypes are more in number. The other two genotypes are the recombinants. By totaling the numbers of the parental genotypes in the progeny, we get 900 (442 - A/a B/b: 458 - a/a b/b). The recombinants constitute 100 in number (46 - A/a b/b: 54 - a/a B/b).

Hence, the distance between the two genes is 10 m.u. and they are on the same chromosome.

The distance can be calculated in this way: