A female mouse from a true-breeding wild-type strain was crossed to a male mouse with apricot eyes (ap)and grey body (gy).The F₁ mice were wild-type for both traits.When the F₁ were interbred, the F₂ were distributed as follows: Females $$\begin{array} { | l | l | } \hline \text { all wild type } & 200 \ \hline \end{array}$$ Males $$\begin{array} { | l | r | } \hline \text { wild type } & 91 \ \hline \text { apricot } & 11 \ \hline \text { grey } & 9 \ \hline \text { apricot, grey } & 89 \ \hline \end{array}$$ Which of the following statements is correct? A)ap and gy are unlinked B)ap and gy are linked on an autosome and 10 map units apart C)ap and gy are linked on an autosome and 20 map units apart D)ap and gy are X-linked and 10 map units apart E)ap and gy are X-linked and 20 map units apart

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The distribution of phenotypes, especially the difference in phenotype distribution between males and females, suggests that the traits are X-linked. The fact that all females are wild type indicates that they must receive one wild-type allele from their father, suggesting that the traits are linked to the X chromosome, which males only have one of. The recombination frequency (11 apricot + 9 grey out of 200 total males) indicates a 10% recombination rate, which corresponds to 10 map units apart on the X chromosome.

A Female Mouse from a True-Breeding Wild-Type Strain Was Crossed all wild type 200\begin{array} { | l | l | } \hline \text { all wild type } & 200 \\\hline\end{array} all wild type ​200​​

A Female Mouse from a True-Breeding Wild-Type Strain Was Crossed $\begin{array} { | l | l | } \hline \text { all wild type } & 200 \\\hline\end{array}$