Question

In mice, the dominant allele Gs of the X-linked gene Greasy produces shiny fur, while the recessive wild type Gs+ allele determines normal fur. The dominant allele Bhd of the X-linked Broadhead gene causes skeletal abnormalities including broad heads and snouts, while the recessive wild-type Bhd+ allele yields normal skeletons. Female mice heterozygous for the two alleles of both genes were mated with wild-type males. Among 100 male progeny of this cross, 49 had shiny fur, 48 had skeletal abnormalities, 2 had shiney fur and skeletal abnormalities, and 1 was wild type. Diagram the cross described and calculate the distance between the two genes.

Answer 1

In question it is given that, the fur related allele and skeletal abnormalities related allele are X-linked.

Gs = shiny fur -----> Dominant

Gs+ = normal fur ------> recessive, wild (+)

Bhd = cause skeletal abnormalities -----> dominant

Bhd+ = normal skeleton ------> recessive, wild (+)

Female mice heterozygous = XGs+Bhd/ XGsBhd+ = Gs+GsBhd+Bhd

Male mice (wild) = XGs+Bhd+, Y = Gs+Bhd+

When male and female cross

Gs+Gs/Bhd+Bhd X Gs+/Bhd+

Female Gametes = Gs+Bhd+, Gs+Bhd, GsBhd+, GsBhd

Male Gamete = Gs+Bhd+ , Y

	Gs+Bhd+	Gs+Bhd	GsBhd+	GsBhd
Y	Gs+Bhd+, Y	Gs+Bhd, Y	GsBhd+,Y	GsBhd, Y

Among 100 males, the progenies obtained are as follows:

49 = shiny fur ; 48 = skeletal abnormalities

2 = shiny fur and skeletal abnormalities ; 1 = wild type

The 49 and 48 progenies are parental gametes, so progenies 2 and 1 are the recombinant Gametes.

Recombination frequency %= recombinants / total progeny

RF % = (2+1) / 100 * 100

3/100 *100

= 3% or 3 m.u.

So the distance between 2 genes is 3 m.u. or 3 cM.

Thank you....

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