If a cross produced a 12:3:1 phenotypic ratio in 320 offspring, how many offspring would you expect to be homozygous recessive at both loci?

Question

BlackTom AI · Accepted Answer

We are told that the cross yields a phenotypic ratio of 12:3:1 in 320 offspring. To translate this into a count, think of the ratio as 12 parts : 3 parts : 1 part, totaling 16 parts.

Option analysis:
- Option 3 (32): If 32 were correct, that would correspond to 32/320 = 1/10, which does not match any 12:3:1 fraction of 16 parts. This is inconsistent with the given ratio.
- Option 16 (16): This would be 16/320 = 1/20, again not aligning with a 1/16 fraction from the 12:3:1 breakdown.
- Option 20 (20): This equals 20/320 = 1/16, which is exactly one part out of the 16 parts in the 12:3:1 ratio. In many dihybrid cross scenarios that produce a 12:3:1 phenotypic ratio, the smallest class (1 of 16 parts) corresponds to the double recessive genotype (homozygous recessive at both loci, aa bb). Therefore, about 1/16 of the offspring are expected to have that double recessive phenotype, which is 320 × 1/16 = 20.
- Option 3 (rechecking): As noted, 3/16 would correspond to 60 offspring, which would not match aa bb if that class is the 1/16 portion in this ratio. Since the question asks for the homozygous recessive at both loci, the 3/16 portion does not fit the interpretation of the double recessive class.

In summary, using the 12:3:1 ratio to assign the smallest 1/16 portion to the double recessive genotype (aa bb) yields an expected count of 320 × 1/16 = 20. The other options (3/16 of 320, or 60; or 1/20 of 320, or 16; or 1/10 of 320, or 32) do not align with the 1/16 fraction associated with the aa bb class in this ratio.

AGRI10051_2025_SM2 Supplementary or Special Exam: Genetics for Agriculture (AGRI10051_2025_SM2)- Requires Respondus LockDown Browser

If a cross produced a 12:3:1 phenotypic ratio in 320 offspring, how many offspring would you expect to be homozygous recessive at both loci?

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