Question

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

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: \( HH \) (produces only \( H \) gametes).
Amelia's genotype: \( hh \) (produces only \( h \) gametes).

The Punnett square cross is \( HH \times hh \). Each offspring will inherit one \( H \) from Clyde and one \( h \) from Amelia, so all offspring have genotype \( Hh \).

Step3: Determine phenotypes

The dominant allele \( H \) (hand) is present in all \( Hh \) genotypes, so all offspring show the "hand" phenotype. The "no hand" phenotype (recessive) requires \( hh \), which no offspring have.

Step4: Calculate probabilities

Total offspring in Punnett square: 4 (from 2x2 square).

• Hand phenotype: 4 out of 4. Probability \( = \frac{4}{4} \times 100\% = 100\% \).
• No hand phenotype: 0 out of 4. Probability \( = \frac{0}{4} \times 100\% = 0\% \).

Answer:

Hand: \( 100\% \)
No hand: \( 0\% \)