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   $X^bX^b$

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

                 

Answer 2

Following are the solution to the given points:

For question A:

• Assume that the bar mutant allele is represented by b; when a bar female is married to a wild-type male for the first time, all of the F1 flies are the bar.
• The wildtype male genotype will be X Y, while the bar female genotype will be $X^b X^b$

                  $\frac{\text{wildtype male}}{\text{bar female}} \ \ \ \ \ \ \ \ \ \ \ X^b \ \ \ \ \ \ \ \ \ \ \ X^b\\\\ X \ \ \ \ \ \ \ \ \ \ \ X X^b\ \ \ \ \ \ \ \ \ \ \ X X^b\\\\ Y \ \ \ \ \ \ \ \ \ \ \ Y X^b \ \ \ \ \ \ \ \ \ \ \ Y X^b$

• Because of the presence of $X^b$, all offspring genotypes in this hybrid will exhibit the bar phenotype.
• When a crossbar male is mated to a wild-type female in the second cross, 857 bar females and 905 wild-type men are seen.
• The ratio of bar and wild flies is roughly 16:17, which might be interpreted as 1:1 when evaluating the cross result.

                         $\frac{\text{wildtype male}}{\text{bar female}} \ \ \ \ \ \ \ \ \ \ \ X \ \ \ \ \ \ \ \ \ \ \ X\\\\ X^b \ \ \ \ \ \ \ \ \ \ \ X X^b\ \ \ \ \ \ \ \ \ \ \ X X^b\\\\ Y \ \ \ \ \ \ \ \ \ \ \ Y X \ \ \ \ \ \ \ \ \ \ \ Y X$

• This cross has 50% of the flies with the bar phenotype and 50% with the typical wild type.
• Therefore, the method of inheritance will be X-linked inheritance.

For question B:

The genotype of flies for the first cross will be$\to \text{male (XY) \ and\ female} (X^b X^b)$

The genotype of flies for the second cross will be:

$\to \text{male} \ (X^b \ Y)\ \text{and female \ (XY)}$

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