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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 shiny fur and skeletal abnormalities, and 1 was wild type.
(a) Diagram the cross described, and calculate the distance between the two genes.
(b) What would have been the results if you had counted 100 female progeny of the cross?

Respuesta :

Answer:

A) 3 mu

B) Female progeny characteristics and type would be similar to that of male progeny.

Explanation:

A)

Given

Dominant allele is "Gs" which is X linked and produces shiny fur

Recessive allele is "Gs+" which is wild type and produces normal fur.

Bhd is dominant allele and produces skeletal abnormalities

Bhd+ is recessive allele and produces normal skeletal structure

Genotype of normal homozygous male is GsGs/ Bhd Bhd

Genotype of heterozygous female is Gs Gs+/Bhd Bhd+

The cross would be as shown in the attached image

The distance between these two genes

[tex]\frac{2+1}{100} \\=0.03 \\= 3 mu[/tex]

B) It would have been a  true cross where the daughter would have Gs+ Bhd+ X  received from their father. Therefore, the number of genotype and phenotype and frequencies of all genotype for female offspring  would be same to  the male offspring

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