In a cross of AaBbCc x AaBbCc, what is the probability of producing an offspring with the genotype aabbcc?单项选择题
A
a. 1/2
B
b. 1/16
C
c. 1/64
D
d. 1/8

Question

In a cross of AaBbCc x AaBbCc, what is the probability of producing an offspring with the genotype aabbcc?

BlackTom AI · Accepted Answer

Each locus segregates independently: an Aa x Aa cross produces the genotype aa with probability 1/4.

1) Relevant knowledge points and concepts
- Mendel’s law of independent assortment: alleles at different loci segregate independently in gamete formation (assuming loci are unlinked).
- For a single heterozygote × heterozygote cross (Aa × Aa), the genotype probabilities are 1/4 AA, 1/2 Aa, and 1/4 aa.
- Independent events: the probability of multiple independent outcomes all occurring equals the product of their individual probabilities.

2) Complete reasoning step by step
- We want the probability that offspring is aabbcc. This requires aa at locus A, bb at locus B, and cc at locus C simultaneously.
- Compute each locus probability from an Aa × Aa cross.

3) Intermediate calculation steps
- Probability(aa from Aa × Aa) = 1/4.
- Probability(bb from Bb × Bb) = 1/4.
- Probability(cc from Cc × Cc) = 1/4.
- Because loci assort independently, multiply: (1/4) × (1/4) × (1/4) = 1/64.

4) Why the correct answer is correct
- Each required homozygous recessive genotype has probability 1/4 independently; the offspring must have all three recessive homozygotes simultaneously, so the probabilities multiply to give 1/64.

5) Why the other choices are incorrect
- a. 1/2: far too large; 1/2 is typical for a single heterozygote producing a particular allele in a gamete but not for producing homozygous recessive genotype from Aa × Aa.
- b. 1/16: equals (1/4) × (1/4) for two loci only; it ignores the third locus.
- d. 1/8: would correspond to (1/2)^3 (e.g., getting a specific allele from each parent) but does not reflect genotype probabilities for Aa × Aa producing a homozygous genotype.

6) Conclusion
Multiplying the three independent 1/4 probabilities gives 1/64. This matches the provided correct answer: c. 1/64.

In a cross of AaBbCc x AaBbCc, what is the probability of producing an offspring with the genotype aabbcc?单项选择题

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