Two autosomal genes, J and K, are 60 map units apart. You perform the following testcross: J K / j k x j k / j k. At what frequencies would you find the following progeny types: J K / j k ; j k / j k ; J k / j k ; j K / j k ?

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.

The map unit is the distance between the pair of genes for which every 100 meiotic products, one of them results in a recombinant one.

So, en the exposed example:

J and K are autosomal genes

J and K are separated by 60 M.U.

60 M.U. means that there is 60% of recombination.

Cross) J K / j k x j k / j k

Gametes) JK Parental jk, jk, jk, jk

jk Parental

Jk Recombinant

jK Recombinant

One map unit equals 1% of recombination frequency. This means that every 100 meiotic products, one of them is a recombinant one.

1 M.U. -------------- 1% recombination

60 M.U. ------------ 60% recombination

30% Jk + 30% jK

100 M.U. - 60 M.U. = 40 M.U.

40M.U.--------------40 % Parental (Not recombinant)

20% JK + 20% jk

Punnet Square) JK jk Jk jK

jk JK/jk jk/jk Jk/jk jK/jk

J K / j k = 20%

j k / j k = 20%

J k / j k = 30%

j K / j k = 30%