Final answer:
Using the product rule of probability, one can determine that the proportion of offspring from the cross Aa Bb Cc Dd x Aa bb Cc Dd that are homozygous recessive for all three genes A, B, C, and D is 1/128.
Step-by-step explanation:
The question asks what proportion of offspring will be homozygous recessive for all three genes in a cross of the genotypes Aa Bb Cc Dd x Aa bb Cc Dd, assuming complete dominance and independent assortment. To find this, we can use the product rule of probability, which states that the probability of two independent events both occurring is the product of their individual probabilities.
For a gene to be homozygous recessive (aa, bb, cc, dd), the offspring must inherit a recessive allele from both parents. Since Aa Bb Cc Dd can produce a recessive allele half of the time (1/2 for a, b, c, and d), and Aa bb Cc Dd can also produce a recessive allele half of the time for genes A, C, and D, and always for gene B (bb), the probability for each gene is calculated separately and then multiplied together:
- A: 1/2 (from Aa) x 1/2 (from Aa) = 1/4
- B: 1 (from bb) x 1/2 (from Bb) = 1/2
- C: 1/2 (from Cc) x 1/2 (from Cc) = 1/4
- D: 1/2 (from Dd) x 1/2 (from Dd) = 1/4
Therefore, the probability that an offspring will be homozygous recessive for all three genes is:
(1/4) × (1/2) × (1/4) × (1/4) = 1/128