Question

1. you are presented with the following incomplete dominance scenario, where there ar3 possible phenotypes: curly hair, wavy hair, and straight hair. if a curly haired squrel (dd) is crossed with a wavy haired squirrel (dd), what are the offspring’s genotypand phenotypes of the f₁ generation?a. what are the f₁ generation genotypes?b. what are the f₁ generation phenotypes?

Explanation:

Step1: Analyze the Punnett Square

The Punnett square shows the combinations from the cross of \( DD \) (curly - haired) and \( Dd \) (wavy - haired). The cells in the Punnett square are \( DD \), \( DD \), \( Dd \), and \( Dd \).

Step2: Determine Genotypes for Part (a)

Count the number of each genotype. There are 2 cells with \( DD \) and 2 cells with \( Dd \). So the genotypes of the \( F_1 \) generation are \( DD \) and \( Dd \). The ratio of \( DD:Dd = 2:2=1:1 \) (or we can say 50% \( DD \) and 50% \( Dd \)).

Step3: Determine Phenotypes for Part (b)

Given incomplete dominance, \( DD \) gives curly hair and \( Dd \) gives wavy hair (since in incomplete dominance, the heterozygote shows an intermediate phenotype). From the genotypes, \( DD \) individuals will have curly hair and \( Dd \) individuals will have wavy hair. Since there are 2 \( DD \) and 2 \( Dd \), the phenotypes are curly hair and wavy hair, with a ratio of 1:1 (50% curly - haired and 50% wavy - haired).

Answer:

a. The \( F_1 \) generation genotypes are \( DD \) (50%) and \( Dd \) (50%) (or in ratio \( DD:Dd = 1:1 \)).
b. The \( F_1 \) generation phenotypes are curly hair (50%) and wavy hair (50%) (or in ratio curly hair : wavy hair \( = 1:1 \)).