Question

In Drosophila, the bar mutant is characterized by eyes that are restricted to a narrow, vertical bar. When a bar female is mated to a wildtype male, all the F1 flies are bar. However, when a bar male is mated to a wildtype female, 857 bar females and 905 wildtype males are observed.
a. What is the mode of inheritance of the bar mutant?
b. What is the genotype of each of the flies in the two crosses above? Include the parents and the F1 generations.

Answer 1

Answer:

See the answer below

Explanation:

a) The mode of inheritance of the bar mutant is X-linked and dominant.

Assuming that barness is represented by the allele B.

Bar female = $X^BX^B$ or $X^BX^b$

Wild type male = $X^bY$

Bar female x wild type male:

                 $X^BX^B$    x    $X^bY$

          $X^BX^b$   $X^BY$  $X^BX^b$    $X^BY$

All of the F1 flies are bar because a single B allele is required for the trait to be expressed.

Bar male x wildtype female

                            $X^BY$   x   $X^bX^b$

                 $X^BX^b$    $X^BX^b$   $X^bY$    $X^bY$

All the female offspring are barred while all the males are wildtype barred resulting in a 50/50 (857/905 is approximately 50/50) ratio of barred/wildtype offspring.

b) Genotypes of each of the cross above, including the parents and the F1 generations.

1.             $X^BX^B$    x    $X^bY$

          $X^BX^b$   $X^BY$  $X^BX^b$    $X^BY$

2.               $X^BY$   x   $X^bX^b$

           $X^BX^b$    $X^BX^b$   $X^bY$    $X^bY$