Question

In mice,the dominant allele Gs of the X-linked gene Greasy produces shiny fur,while the recessive wild type Gs+ produces 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 produces normal skeletons. Female mice heterozygous for both genes were mated to wild type males.Of the 100 male offspring scored in this,49 had shiny fur,48 had skeletal abnormalities,2 had shiny fur and skeletal abnormalities,and 1 was wild type.What is the genotype of the heterozygous female?

a) Gs/Gs+;Bhd/Bhd+
b) GsBhd/Gs+Bhd
c) Gs Bhd+/Gs+Bhd
d) there's not enough data to tell.

Answer 1

The correct option in the final answer is c. Based on the inheritance patterns of X-linked genes and the phenotypic outcomes of the offspring, the correct genotype of the heterozygous female mouse is Gs Bhd+/Gs+Bhd, which allows for the observed distribution of traits in the male offspring.

Step-by-step explanation:

The question asks us to determine the genotype of a heterozygous female mouse that is mated to a wild type male, based on the phenotypic outcomes of their offspring. Given that the female mouse is heterozygous for two X-linked genes, we must consider the inheritance pattern of X-linked traits. These traits are more frequently expressed in males as they inherit only one X chromosome from their mothers. This means that the traits associated with the genes on the X chromosome will be expressed if they are dominant, or even if they are recessive, since there is no corresponding allele on the Y chromosome from the father to mask their expression.

The fact that the offspring include males with shiny fur (49 out of 100), with skeletal abnormalities (48 out of 100), with both traits (2 out of 100), and one wild type confirms that the female must have been heterozygous for each gene, with one wild type and one mutant allele for each of the two genes in question. Therefore, the correct genotype of the heterozygous female is Gs Bhd+/Gs+Bhd (option c). This configuration would result in approximately half the male offspring receiving the dominant allele for one trait and the wild type allele for the other, and a rare occurrence of males receiving both dominant alleles or both wild type alleles due to recombination. The fact that we see these phenotypic ratios closely aligns with these expectations, allowing for the natural variation in random segregation and recombination events.