Question

two mice are heterozygous for albinism (aa). the dominant allele (a) codes for normal pigmentation, and the recessive allele (a) codes for no pigmentation. what percentage of their offspring would have an albino phenotype? view available hint(s) 100 50 75 25

Explanation:

Step1: Set up Punnett Square

For two heterozygous (Aa) mice, the Punnett square is:

	A	a
a	Aa	aa

Step2: Determine Genotypes and Phenotypes

• AA: normal (dominant)
• Aa: normal (dominant)
• aa: albino (recessive)

From the square, out of 4 possible genotypes, 1 is aa (albino).

Step3: Calculate Percentage

Percentage = $\frac{\text{Number of albino genotypes}}{\text{Total genotypes}} \times 100 = \frac{1}{4} \times 100 = 25$

Answer:

25