Problem

Solutions For An Introduction to Genetic Analysis Chapter 4 Problem 70P

Step-by-Step Solution

Solution 1

An RFLP analysis is done by two pure lines with the genotypes:

A/A B/B x a/a b/b

The analysis showed that the first parent was homozygous, for a long RFLP allele symbolized by l. The second parent is homozygous, for a short RFLP allele symbolized by s. These two parents were crossed to give the F1 progeny. The F1 progeny was then backcrossed to the second parent.

The progeny was then tabulated as shown:

Picture 1

The details of the cross are:

A/A B/B l/l x a/a b/b s/s

The F1 progeny was then backcrossed to the second parent. The progeny indicates that all the three genes are linked.

b)

By comparing the most frequent class of parentals with the least frequent class of the double crossovers, we can deduce that the RFLP allele is between the genes A/a and B/b.

The distance between these markers can be calculated by the recombination frequency:

Between gene A/a and RFLP marker:

Between RFLP marker and gene B/b :

Hence, the distance between the two genes would beapart. The map showing the distance between the linked genes is as follows:

Picture 7

c)

While, incorporating the RFLP fragments into the gene map. The map obtained to be:

Picture 8

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