If a female Drosophila that is heterozygous for a recessive X-linked mutation is crossed to a wild-type male, what proportion of female progeny will have the mutant phenotype?

0%.

Explanation:

The disease might be dominant or recessive depending on its expression and inheritance pattern. The recessive X linked mutation means the disease is expressed only in homozygous condition and located on the X chromosome.

The female Drosophila is heterozygous (XXh) is crossed with the wild type male (XY). The cross is as follows:

XXh * XY

progeny - XX, XhX, XY and XhY.

From the cross it is clear that no female drospohila shows the mutant phenotype.

Thus, the answer is 0%.

0% (none)

Explanation:

The female drosophila is heterozygous for X-linked recessive mutation. An X-linked recessive mutation is expressed in females only when present in the homozygous state. Therefore, the female progeny should have two copies of the mutant allele to express this mutation. Let's suppose that the genotype of the heterozygous female is X^cX and that of the male is XY. The cross would obtain progeny in following ratio=

X^cX x XY = 1/4 X^cX: 1/4 XX: 1/4 X^cY: 1/4 XY

Since none of the female progeny is homozygous recessive (X^cX^c), the cross would not obtain any female progeny expressing the X-linked recessive mutation.