In C. elegans, lon-2 and unc-2 are X-linked genes. A lon-2 male is mated to a...

In C. elegans, lon-2 and unc-2 are X-linked genes. A lon-2 male is mated to a homozygous unc-2 hermaphrodite. Individual wild-type hermaphrodite offspring were then crossed to wild-type males. Among 359 male offspring from these matings, 9 were wild-type for both traits. What is the approximate map distance between lon-2 and unc-2?

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Expert Solution

In the first cross; a lon-2 male is crossed with a homozygous unc-2 hermaphrodite. Thus, the genotypes of the two worms are:

lon-2, +/Y x +, unc-2/+, unc-2

The hermaphrodite offspring from this cross are: lon-2, +/+, unc-2

Now, this dihybrid hermaphrodite was crossed to wild-type males. This cross can be depicted as:

lon-2, +/+, unc-2 x +,+/Y

From this cross, 9 offspring of the 359 male offspring were Wild-type. These would have received only one chromosome from the hermaphrodite, which was the result of a crossing over event. This crossing over would produce +, +, and lon-2, unc-2 gametes.

Therefore, map distance between lon-2 and unc-2
= Recombination frequency between lon-2 and unc-2 = 9*2 * 100/339 = 5.309 %

The map distance between the two loci is 5.31 map units.

Note: The number of recombinants is taken as 9*2 because the 9 Wild type gametes produced represent only half of the gametes produced by recombination in the F1 hermaphrodite. Thus, to adjust the number to the number of recombinant gametes, the number of wild-type gametes was multiplied by 2.

gladiator answered 1 year ago

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