Question

1. show the cross of a red eyed female (heterozygous) and a red eyed male. what are the genotypes of the parents?

____ & ___________
how many are
white eyed, male__white eyed, female __
red eyed, male ___ red eyed, female ___

Explanation:

Step1: Determine Parent Genotypes

In fruit flies, eye color (red/white) is X - linked. Let \( X^{R} \) be the allele for red eyes (dominant) and \( X^{r} \) be the allele for white eyes (recessive). A heterozygous red - eyed female has genotype \( X^{R}X^{r} \), and a red - eyed male has genotype \( X^{R}Y \) (since males have one X and one Y chromosome, and red eyes are dominant, so they have \( X^{R} \) on the X and Y is non - sex - linked for this trait).

Step2: Set Up Punnett Square

The female (\( X^{R}X^{r} \)) can produce gametes \( X^{R} \) and \( X^{r} \). The male (\( X^{R}Y \)) can produce gametes \( X^{R} \) and \( Y \).
The Punnett square is:

	\( X^{R} \)	\( Y \)
\( X^{r} \)	\( X^{R}X^{r} \) (red - eyed female)	\( X^{r}Y \) (white - eyed male)

Step3: Count Phenotypes

• White - eyed male: From the Punnett square, there is 1 out of 4 (or 25% chance, but in terms of the Punnett square cells, 1 cell) with genotype \( X^{r}Y \), so the number is 1.
• White - eyed female: For a female to be white - eyed, she needs \( X^{r}X^{r} \). Looking at the Punnett square, there are no such cells, so the number is 0.
• Red - eyed male: The genotype \( X^{R}Y \) is red - eyed male, and there is 1 cell with this genotype, so the number is 1.
• Red - eyed female: The genotypes \( X^{R}X^{R} \) and \( X^{R}X^{r} \) are red - eyed females. There are 2 cells with these genotypes, so the number is 2.

Answer:

Parent genotypes: \( X^{R}X^{r} \) & \( X^{R}Y \)
White - eyed, male: 1; White - eyed, female: 0; Red - eyed, male: 1; Red - eyed, female: 2