Question

clyde has a homozygous dominant genotype for his hand (hh), and amelia has the recessive trait of no
clyde
amelia
phenotype probability (%)
hand: 100%
no hand: 0%

Explanation:

Step1: Determine Amelia's genotype

Since Amelia has the recessive trait (no hand), her genotype must be homozygous recessive, so \( hh \) (assuming the recessive allele is \( h \) and dominant is \( H \)). Clyde is homozygous dominant \( HH \).

Step2: Perform the Punnett square cross

When crossing \( HH \) (Clyde) with \( hh \) (Amelia), each of Clyde's gametes is \( H \), and each of Amelia's gametes is \( h \). The Punnett square will have all offspring with genotype \( Hh \).

Step3: Determine phenotype probabilities

The dominant trait (hand) is expressed in \( Hh \) (since \( H \) is dominant). So all offspring (100%) will show the hand phenotype, and 0% will show the no - hand (recessive) phenotype.

Answer:

The Punnett square analysis shows that the probability of the hand phenotype is 100% and no - hand is 0%, which matches the given values. The offspring genotypes are all \( Hh \), and the phenotype for hand is dominant, so the probabilities are as stated.