Question

1. clyde has a homozygous dominant genotype for his hand (hh), and amelia has the recessive trait of no clyde has a homozygous dominant genotype for his hand (hh), and amelia has the recessive trait of no 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 \( H \) is dominant for hand, \( h \) recessive for no hand).

Step2: Set up the Punnett square

Clyde's genotype: \( HH \) (homozygous dominant), so he can only produce gametes with \( H \).
Amelia's genotype: \( hh \) (homozygous recessive), so she can only produce gametes with \( h \).

The Punnett square will have all offspring with genotype \( Hh \) (since \( H \) from Clyde and \( h \) from Amelia).

Step3: Determine phenotypes

The dominant allele \( H \) is expressed in \( Hh \), so all offspring will show the dominant trait (hand). Thus, the probability of "hand" phenotype is \( \frac{4}{4} \times 100\% = 100\% \), and "no hand" is \( 0\% \), which matches the given table. (If the question was to confirm or complete this, the Punnett square analysis shows the results are correct.)

Answer:

The Punnett square analysis confirms that all offspring will have the genotype \( Hh \), resulting in a 100% probability for the "hand" phenotype and 0% for "no hand", consistent with the provided table. (If the task was to verify or explain the phenotype probabilities, this is the conclusion.)