Answer:
1/64
Explanation:
The following genotypes were crossed: AaBbCc X+Xr and AaBBcc X+Y
If we asume that the autosomal genes are in different chromosomes, then they will assort independently during meiosis and gametogenesis. In that case, we can use the Multiplication Rule of Probability to obtain the probability of having a specific genotype in the progeny. This rule states that when two or more events are independent, the probability of all them happening at the same time will be the result of the multiplication of the individual probabilities of each event.
We can separate each gene in the cross to determine the genotypic ratios in the offspring, and then multiply the probabilities of the aaBbCc X+X+ genotypes to obtain the overall probability of having progeny with that genotype.
1/4 AA
2/4 Aa
1/4 aa
1/2 BB
1/2 Bb
1/2 Cc
1/2 cc
1/4 X+X+
1/4 X+Y
1/4 X+Xr
1/4 XrY
The probability of having offspring with the aaBbCc X+X+ genotype will be: 1/4 × 1/2 × 1/2 × 1/4 = 1/64