An individual is heterozygous for two linked genes, but whether its genotype is A B/a b or A b/a B is not known. The individual is crossed with a b/a b, and among the progeny are the following:
62 A B/a b
13 A b/a b
18 a B/a b
51 a b/a b
These results imply that the genotype of the doubly heterozygous parent was A B/a b.

Question

Quizplus · Accepted Answer

The key to solving this problem is to understand how linked genes behave during meiosis. Linked genes are genes that are located close to each other on the same chromosome and tend to be inherited together. In the case of the individual in question, we know that it is heterozygous for two linked genes, but we don't know which alleles are on the same chromosome. Let's consider the two possibilities:

1. A B/a b: In this case, the two dominant alleles (A and B) are on one chromosome, and the two recessive alleles (a and b) are on the other chromosome. When this individual is crossed with a b/a b, the following gametes can be produced: AB, a b, Ab, and a B. Because A and B are on the same chromosome, they tend to be inherited together, as do a and b. Therefore, the gametes AB and a b will be produced more frequently than the gametes Ab and a B.

2. A b/a B: In this case, the dominant allele A and the recessive allele b are on one chromosome, and the recessive allele a and the dominant allele B are on the other chromosome. When this individual is crossed with a b/a b, the following gametes can be produced: AB, a b, Ab, and a B. However, because A and b are on different chromosomes, they can be inherited independently of each other. Therefore, the gametes AB, Ab, a B, and a b will be produced in roughly equal frequencies.

Now let's look at the results of the cross. We see that there are 62 progeny with the genotype A B/a b and 51 progeny with the genotype a b/a b. This suggests that the two dominant alleles (A and B) tend to be inherited together, which is consistent with the first possibility. If the individual were actually A b/a B, we would expect to see roughly equal numbers of progeny with the genotype AB, Ab, a B, and a b. Therefore, we can conclude that the genotype of the doubly heterozygous parent was A B/a b.

An Individual Is Heterozygous for Two Linked Genes, but Whether