To determine the fraction of offspring that are homozygous recessive for all three traits from a cross between two animals with the genotype \(AaBbCc\), we need to consider the probability of each trait being homozygous recessive.
Each trait follows Mendelian inheritance, where:
- \(A\) and \(a\) are alleles for the first trait,
- \(B\) and \(b\) are alleles for the second trait,
- \(C\) and \(c\) are alleles for the third trait.
For each trait, the probability of an offspring being homozygous recessive (aa, bb, or cc) from a heterozygous cross (Aa x Aa, Bb x Bb, Cc x Cc) is \( \frac{1}{4} \).
Since the traits are independent, we multiply the probabilities for each trait:
\[ P(aa) \times P(bb) \times P(cc) = \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64} \]
Therefore, the fraction of the offspring that are expected to be homozygous recessive for all three traits is:
\[ \frac{1}{64} \]
The answer is the fourth one: \( \frac{1}{64} \).
Explanation for each option:
- \( \frac{1}{16} \): Incorrect. This would be the probability for two traits being homozygous recessive.
- \( \frac{1}{4} \): Incorrect. This would be the probability for one trait being homozygous recessive.
- \( \frac{1}{8} \): Incorrect. This does not correspond to any correct combination of probabilities for homozygous recessive traits.
- \( \frac{1}{64} \): Correct. This is the probability for all three traits being homozygous recessive.
In summary, the fraction of the offspring expected to be homozygous recessive for all three traits is \( \frac{1}{64} \).
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