In a dihybrid cross (Aa;Bb x Aa;Bb) involving two independently assorting genes, what proportion of offspring would be heterozygote at both loci?

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Question: In a dihybrid cross (Aa;Bb x Aa;Bb) involving two independently assorting genes, what proportion of offspring would be heterozygote at both loci?

Option 75%: This overestimates the joint heterozygosity. While each gene independently has a 1/2 chance of being heterozygous (Aa) in the cross Aa x Aa, the probability of being Aa for both genes at the same time is not 1/2 × 1/2 that yields 25%, so 75% is too high and misinterprets the joint likelihood.

Option 50%: This would imply that half of the offspring are heterozygous at both loci, but the independent assortment of Aa vs Aa and Bb vs Bb does not produce a simple 1/2 chance for both loci simultaneously. The joint probability requires multiplying the individual locus probabilities, which gives a smaller value than 50%.

Option 25%: The correct joint probability arises from treating the loci independently. For each gene, the cross Aa × Aa yields 1/4 AA, 1/2 Aa, 1/4 aa, so a 1/2 chance of Aa at each locus. Since the loci assort independently, the chance of Aa at both loci is (1/2) × (1/2) = 1/4 = 25%. This reflects the multiplicative rule for independent events in genetics.

Option 6.25%: This would correspond to multiplying two 1/4 probabilities (1/4 × 1/4), which would describe being homozygous recessive at one locus and homozygous recessive at the other, or another misinterpretation. It does not represent heterozygosity at both loci, so it is not the correct joint outcome.

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

In a dihybrid cross (Aa;Bb x Aa;Bb) involving two independently assorting genes, what proportion of offspring would be heterozygote at both loci?

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