Let’s complete the Punnett square for a cross between two heterozygous parents (Tt x Tt) for eye color traits where brown eyes (T) are dominant over blue eyes (t).
Here is the completed Punnett square:
T | t
----------------
T | TT | Tt
----------------
t | Tt | tt
Now we can interpret the results:
1. What is the percent chance (probability) of genotype TT?
- There is 1 TT out of 4 total squares.
- Probability of TT = \( \frac{1}{4} \times 100 = 25% \)
2. What is the percent chance (probability) of genotype Tt?
- There are 2 Tt out of 4 total squares.
- Probability of Tt = \( \frac{2}{4} \times 100 = 50% \)
3. What is the percent chance (probability) of genotype tt?
- There is 1 tt out of 4 total squares.
- Probability of tt = \( \frac{1}{4} \times 100 = 25% \)
4. What is the probability of the phenotype of brown eyes?
- The phenotype for brown eyes can be expressed by either TT or Tt. So, adding those probabilities:
- Probability of TT = 25%
- Probability of Tt = 50%
- Total = 25% + 50% = 75%
5. What is the probability of the phenotype of blue eyes?
- The phenotype for blue eyes is only represented by the tt genotype.
- Probability of tt = 25%
To summarize:
- Probability of TT = 25%
- Probability of Tt = 50%
- Probability of tt = 25%
- Probability of brown eyes = 75%
- Probability of blue eyes = 25%
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