Question

question 3/14
a cross between a white rooster and a black hen results in 100% erminette (checkerboard) offspring. when two of these erminette offspring are mated, the predicted phenotypic percentages that could be seen in their offspring would be ____. (work out a punnett square as needed to determine the answer.) (aks 7e1)
a 25% black, 50% erminette, 25% white
b 75% erminette, 25% white
c 100% erminette
d 75% black, 25% white

Explanation:

Step1: Analyze Inheritance Pattern

Erminette (checkerboard) is a result of codominance or incomplete dominance? Wait, when white (let's say WW) and black (BB) cross to get 100% erminette (WB, assuming codominance where both alleles are expressed). So erminette is heterozygous (WB).

Step2: Set Up Punnett Square

Cross two erminette (WB x WB). The Punnett square will have:

• Top row: W, B
• Left column: W, B
• Cells: WW (white), WB (erminette), WB (erminette), BB (black)

Step3: Calculate Phenotypic Ratios

Count the phenotypes: WW (white) = 1, WB (erminette) = 2, BB (black) = 1. Total = 4.

• Black (BB): \( \frac{1}{4} = 25\% \)
• Erminette (WB): \( \frac{2}{4} = 50\% \)
• White (WW): \( \frac{1}{4} = 25\% \)

So the percentages are 25% black, 50% erminette, 25% white.

Answer:

A. 25% black, 50% erminette, 25% white