Question

scalloped (sd) is an X-linked recessive and ebony (e) is an autosomal recessive mutation. What proportion of scalloped, ebony females (relative to whole population) is expected in the F2 starting with a true breeding scalloped female which is wild type for ebony mating with a true breeding male mutant only for ebony.

Answer 1

1/16

Step-by-step explanation:

• scalloped (Xsd) is an X-linked recessive trait to Xsd+ (wild type)
• ebony (e) is an autosomal mutation recessive to e+ (wild type).

The two genes are independent because they are located on different chromosomes.

Parental generation:

True breeding scalloped female wild type for ebony (Xsd Xsd e+e+) mates with a true breeding male mutant only for ebony (Xsd+ Y ee).

The female only produces Xsd e+ gametes. The male produces Xsd+ e or Y e gametes. Therefore, the F1 females will have the genotype Xsd Xsd+ e e+ and the F1 males will have the genotype Xsd Y e e+.

If you complete a Punnett Square with the gametes the two F1 individuals can produce, you will get all the F2 proportions. The scalloped, ebony females have a Xsd Xsd e e genotype and appear in a 1/16 proportion.