Question

sex - linked traits
day 17
colorblindness is an x - linked recessive trait. the normal vision allele is x^n, and the colorblind allele is x^n. a carrier female (x^nx^n) is crossed with a normal male (x^ny).

1. complete the punnett square for this cross.
2. write the genotypes and phenotypes of all possible offspring.
3. what percent of their sons will be colorblind?
4. what percent of their daughters will be carriers?

Explanation:

Step1: Determine the gametes

The female (XⁿXⁿ) can produce gametes Xⁿ and Xⁿ. The male (XⁿY) can produce gametes Xⁿ and Y.

Step2: Complete the Punnett square

	Xⁿ	Xⁿ
Y	XⁿY	XⁿY

Step3: Genotypes and phenotypes

• Genotypes: XⁿXⁿ (female - normal), XⁿXⁿ (female - carrier), XⁿY (male - normal), XⁿY (male - colorblind)
• Phenotypes: Females - normal or carrier; Males - normal or colorblind

Step4: Calculate for sons

Sons have genotypes XⁿY and XⁿY. So, 50% of sons are colorblind.

Step5: Calculate for daughters

Daughters have genotypes XⁿXⁿ and XⁿXⁿ. So, 50% of daughters are carriers.

Answer:

1. Punnett square as above.
2. Genotypes: XⁿXⁿ (female - normal), XⁿXⁿ (female - carrier), XⁿY (male - normal), XⁿY (male - colorblind); Phenotypes: Females - normal or carrier; Males - normal or colorblind.
3. 50%
4. 50%