Solutions For An Introduction to Genetic Analysis Chapter 4 Problem 61P

b) When the plant Abc/aBC is crossed with the abc/abc plant, the progeny genotypes can be explained in the following way:

Since the parentals class includes those plants which have not experienced a crossover, their frequency can be calculated as:

Since each parental class should be represented equally:

Abc – 0.28 [half of 0.56]

aBC – 0.28 [half of 0.56]

The frequency of the ABC gamete would be:

The frequency of the ABc gamete would be:

This will also be the frequency of the other gamete abC.

Hence, the expected genotype results for 1000 progeny would be:

• Abc – 280

• aBC – 280

• ABC – 70

• Abc – 70

• AbC – 120

• aBc – 120

• ABc – 30

• abC – 30

c) Assuming 20% interference, we need to calculate the observed double crossover value:

The A-B distance:

The B-C distance:

The Parental:

Hence, for 1000 progeny, the expected genotypes are:

o Abc – 274

o aBC – 274

o ABC – 76

o Abc – 76

o AbC – 126

o aBc - 126

o ABc - 24

o abC - 24