Question

In Drosophila, the mutant alleles vermillion (r), on the X chromosome, and cinnabar (n), on an autosome, cause the same phenotype: bright-red eye color instead of the wild type purple. A male from a true-breeding vermillion stock and a female from a true breeding cinnabar stock are crossed. Predict phenotypic ratios among their F1 and F2 progeny, individually for sons and daughters

Answer 1

Answer:

1. F1: Sons: 100% wild-type, daughters: 100% wild-type.

2. F2: Sons: 5:3 bright-red eye color to wild-type, daughters: 1:3 bright-red eye color to wild-type.

Explanation:

In this example two genes in different chromosomes produce the same phenotype: bright-red eye color. First, vermillion gene is in X chromosome, and the vermillion color is a recessive trait, so vermillion mutants should be:

Mutant Females: $X^{r}X^{r}$

Mutant Males: $X^{r}Y$

Wild-type Females: $X^{R} X^{R} /X^{R} X^{r}$

Wild-type Males: $X^{R}$

Second, cinnabar gene is in an autosome and it's recessive too, so mutants should be:

Mutant males or females: nn  

Wild-type: NN

If a vermillion male ($X^{r}YNN$) is crossed with a cinnabar female $X^{R}X^{R} nn$, it will produce an F1 as in the first Punnett Square. In this case, all sons and daughters are going to be wild-type.

When F1 is crossed, an F2 is produced as in the second and third Punnett Squares. In this case, in daughters, it will produce a phenotypic ratio of 1:3 red eyes to wild-type eyes.

In sons, it will produce a phenotypic ratio of 5:8 red eyes to wild-type eyes.