Assuming independent assortment for all gene pairs, what is the probability that a cross between the following parents, AABbCc × AaBbCc, will produce an AaBbCc offspring?

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BlackTom AI · Accepted Answer

We start by listing the parental genotypes and the target offspring genotype:
- Parent 1: AABbCc
- Parent 2: AaBbCc
- Desired offspring genotype: AaBbCc

Now we evaluate each genetic locus under independent assortment, then combine the probabilities:

Option 1: 1/2
- A locus: Parent 1 can only contribute A (no a alleles), while Parent 2 can contribute A or a with equal probability. To obtain Aa, the offspring must inherit a from Parent 2. Probability = 1/2.
- B locus: Both parents are Bb, so the chance of the offspring being Bb is 1/2.
- C locus: Both parents are Cc, so the chance of the offspring being Cc is 1/2.
- If we considered just one locus, 1/2 is the result; but to get the full AaBbCc genotype across all three loci under independent assortment, we must multiply the locus probabilities. The overall probability would be (1/2) × (1/2) × (1/2) = 1/8, not 1/2. Therefore, 1/2 is too large as a final result.

Option 2: 1/16
- This would imply multiplying (1/2) × (1/2) × (1/4) or some other mismatch among loci. However, for each of B and C loci, the chance to get the desired heterozygote Bb and Cc is exactly 1/2 for each, not 1/4. The A locus contributes 1/2, so the correct combined product remains (1/2)³ = 1/8, not 1/16. Hence this option undercounts the probability.

Option 3: 3/4
- A large probability like 3/4 would require very high likelihoods at loci, which is inconsistent with the actual heterozygote outcomes we need (Aa, Bb, Cc). Since each locus contributes at most a 1/2 chance to the desired heterozygote state, multiplying across three loci cannot exceed (1/2) × (1/2) × (1/2) = 1/8. Therefore 3/4 is incorrect.

Option 4: 1/8
- A locus: 1/2 probability to contribute the a allele from Parent 2 (to form Aa).
- B locus: 1/2 probability to obtain Bb from Bb × Bb.
- C locus: 1/2 probability to obtain Cc from Cc × Cc.
- Since the loci assort independently, multiply: (1/2) × (1/2) × (1/2) = 1/8. This matches the required target AaBbCc offspring genotype across all three gene pairs.

In summary, each locus contributes a 1/2 chance toward the desired heterozygous state, and the independent assortment across three loci yields a combined probability of (1/2)³ = 1/8. The other options fail because they either miscount locus-specific probabilities or ignore the independence across loci.

Bio 93 Section CDE 2025 Final Exam Practice, Part 1

Assuming independent assortment for all gene pairs, what is the probability that a cross between the following parents, AABbCc × AaBbCc, will produce an AaBbCc offspring?

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