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It is found that in cha-cha chickens, the allele for a double comb (A) is dominant to the allele for a single comb (a). A mothe...Question
It is found that in cha-cha chickens, the allele for a double comb (A) is dominant to the allele for a single comb (a). A mother with the genotype Aa and a father with the genotype Aa produce an offspring. What is the percent chance that the offspring will have a single comb? Word Bank: 7505010025
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GPT-4o mini
To determine the percent chance that the offspring will have a single comb (genotype aa), we can use a Punnett square to analyze the genotypes of the parents.
Given the genotypes of the parents:
- Mother: Aa
- Father: Aa
We can set up the Punnett square as follows:
| | A | a |
|---|---|---|
| A | AA | Aa |
| a | Aa | aa |
From the Punnett square, we can see the possible genotypes for the offspring:
- AA (double comb): 1
- Aa (double comb): 2
- aa (single comb): 1
Counting the genotypes:
- Total combinations = 4 (1 AA + 2 Aa + 1 aa)
- Combinations for aa = 1
Now, we can calculate the probability of having a single comb (aa):
- Probability of aa = (Number of aa) / (Total combinations) = 1/4
To convert this probability into a percentage:
- (1/4) × 100 = 25%
Thus, the percent chance that the offspring will have a single comb is **25**.
Given the genotypes of the parents:
- Mother: Aa
- Father: Aa
We can set up the Punnett square as follows:
| | A | a |
|---|---|---|
| A | AA | Aa |
| a | Aa | aa |
From the Punnett square, we can see the possible genotypes for the offspring:
- AA (double comb): 1
- Aa (double comb): 2
- aa (single comb): 1
Counting the genotypes:
- Total combinations = 4 (1 AA + 2 Aa + 1 aa)
- Combinations for aa = 1
Now, we can calculate the probability of having a single comb (aa):
- Probability of aa = (Number of aa) / (Total combinations) = 1/4
To convert this probability into a percentage:
- (1/4) × 100 = 25%
Thus, the percent chance that the offspring will have a single comb is **25**.
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