Answer to Question #339264 in Genetics for lisa

Question

A phenotypically wild-type female fruit fly that was heterozygous for genes controlling body color and wing length was crossed to a homozygous mutant male with black body (allele b) and vestigial wings (allele vg). The cross produced the following progeny: gray body, normal wings 126; gray body, vestigial wings 24; black body, normal wings 26; black body, vestigial wings 124. Do these data indicate linkage between the genes for body color and wing length? What is the frequency of recombination?

Answer 1

According to the rules of two-hybrid crossing, the ratio of phenotypes in this case should be as follows:

1:1:1:1

In order to find out if the above genes are inherited independently, let's do a chi-square test.

If there are 300 offspring in total, then in each phenotypic class, respectively, there should be:

• 300/4 = 75 offspring.

We calculate the chi-square using the following formula:

$χ^2 =Σ(E−T)^2/T$

χ² = (126-75)^2/75 + (24-75)^2/75 + (26-75)^2/75 + (124-75)^2/75 = 34.68 + 34.68 + 32.013 + 32.013 = 69.36 + 64.026 = 133.386.

According to the chi-square table, with a probability greater than 0.05 and with a degree of freedom of 3, the chi-square value will be less than 7.81, and with a probability less than 0.01 and with a degree of freedom of 3, the chi-square value will be greater than 11.34.

The calculated value is greater than 11.34, which means that the null hypothesis is refuted. This means that genes for body color and wing length are linked.

We calculate the recombination frequency:

(24+26)*100%/300 = 16.7%.