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Question
clyde has the recessive, no tongue (tt) trait. amelia has a heterozygous genotype for her tongue (tt).
clyde
amelia
phenotype probability (%)
tongue:
no tongue:
Step1: Determine Gametes
Clyde's genotype is \( tt \), so he can only produce \( t \) gametes. Amelia's genotype is \( Tt \), so she can produce \( T \) and \( t \) gametes.
Step2: Complete Punnett Square
| \( t \) | \( t \) | |
|---|---|---|
| \( t \) | \( tt \) | \( tt \) |
Step3: Calculate Phenotype Probabilities
- For "Tongue" (genotype \( Tt \)): There are 2 out of 4 squares, so probability is \( \frac{2}{4} \times 100\% = 50\% \).
- For "No tongue" (genotype \( tt \)): There are 2 out of 4 squares, so probability is \( \frac{2}{4} \times 100\% = 50\% \).
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Tongue: \( 50\% \)
No tongue: \( 50\% \)