Respuesta :
Answer:
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.
The proportion of the female progeny having the mutant phenotype is 0%.
X-linked recessive mutations are mutations which only occur in the X-chromosome and are only expressed when present in the h0m0zygous state. Thus, X-linked recessive mutations are only expressed in females.
A female Drosophila that is heterozygous for a recessive X-linked mutation will only carry the mutant allele but it will not be expressed.
If this female Drosophila is crossed with a wild-type male, i.e. a male lacking the mutation and therefore, carrying the normal gene, the offspring of the cross can be shown below:
Let the genotype of the female Drosophila be XcX; where Xc is the mutant allele and X is the normal allele.
Let the genotype of the wild-type male Drosophila be XY; where X and Y are the normal X and Y chromosomes.
XcX x XY will produce; XcX, XX, XcY and XY
The progeny include two males and two females.
The female progeny include one h0m0zygous normal and one heterozygous for the recessive X-linked mutation with both showing the normal phenotype.
Thus, the proportion of the female progeny having the mutant phenotype is 0%.
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