Problem

Solutions For An Introduction to Genetic Analysis Chapter 4 Problem 49P

Step-by-Step Solution

Solution 1

A cross was done in mice to experiment on few alleles. One of the allele is “W”, which refers for waltzing gait, “w,” represents non-waltzing gait, “G,” represents normal gray color, “g,” represents albino, “B,” represents bent tail, and “b,” represents straight tail.

A waltzing gray bent-tailed mouse is crossed with a non-waltzing albino straight-tailed mouse. Over several years, the progeny details are as follows:

Picture 4

When a testcross of a triple heterozygous parent is done, three outcomes are expected. If all the three genes are linked, pairs of approximately equal number of progeny in four different frequencies will be seen. If all the genes are unlinked, eight classes of progeny will be seen in approximately equal numbers. If two of the genes are linked and the third is unlinked, then two groups of four of two different frequencies are seen. This pattern is observed in the progeny details given above.

When observing the data, we find that the four most common classes are waltzing bent or non-waltzing straight with gray or albino segregating equally. This indicates that the W/w and B/b genes are linked and the G/g gene is unlinked to either of these genes.

Hence, the genotypes of the two parental mice are as follows:

The chromosomes of the parents can be diagrammatically represented in the following way:

Picture 1

Since the genes of W and B are linked, any progeny of the testcross that are Wb or wB are recombinant.

The map distance can be calculated in the following way:

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