Dec 9

A geneticist is mapping three mutants (call them a, b, and c) found on the 3^rd chromosome of mice. He crosses an F1 mouse that is heterozygous for all three genes to a recessive tester (homozoyous recessive for all three) and gets the following data in the F2: (+ means wild type, a means homozogous recessive for a, etc).

Phenotype	Number
abc	300
+++	310
a++	90
+bc	95
ab+	20
++c	25
a+c	3
+b+	2
total	535

Which are the parentals, single, and double crossovers?

What is the correct gene order?

Draw a linkage map showing the genes and the map distance between them.

The parental classes have the highest frequency and the double crossover are the rarest. So, abc and +++ are the parental types and a+c and +b+ are the double crossovers.

The other four phenotypes are the result of single crossovers.

To get the gene order, compare the double crossovers to the parentals and see which one gene "changes place". In this example it is b (compare abc with a+c or +++ with +b+).

So, b must be in the middle.

A single crossover between genes a and b will produce offspring a++ and +bc.

Get the total recombination distance between a and b by summing all of the recombination events in that interval (including double crossovers) and dividing by the total:

(90+95+2+3) / 535 = 0.355 or 35.5 cM

A single crossover between b and c will produce ab+ and ++c.

The recombination distance will be (20+25+2+3)/535 = 0.093 = 9.3 cM.