Question

Three autosomal recessive mutations in yeast, all producing the same phenotype \((m 1, m 2, \text { and } m 3),\) are subjected to complementation analysis, Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown-that is, \(m 2 \times m 3\) Cross \(1: \quad m I \times m 2 \longrightarrow P_{1}=\) all wild-type progeny Cross \(2: \quad m I \times m 3 \longrightarrow P_{1}:\) all mutant progeny

Short answer

Answer: Mutations m1 and m3 are alleles of the same gene. The cross m2 x m3 is predicted to result in all wild-type progeny.

Step-by-step solution

Interpret results of Cross 1

The first cross results in all wild-type progeny. This indicates that \(m1\) and \(m2\) complement each other and are most likely not alleles of the same gene.

Interpret results of Cross 2

The second cross results in all mutant progeny. This suggests that \(m1\) and \(m3\) do not complement one another, which means they are alleles of the same gene.

Predict results of Cross 3 (\(m 2 \times m 3\))

Since \(m1\) and \(m3\) are alleles of the same gene, and \(m1\) and \(m2\) complement each other, it is likely that \(m2\) and \(m3\) will interact similarly. Therefore, the cross \(m 2 \times m 3\) should result in all wild-type progeny. In conclusion, \(m1\) and \(m3\) are alleles of the same gene, and the cross \(m 2 \times m 3\) should result in all wild-type progeny.