Ascospores from a cross a $\times$ b c (all auxotrophic markers)are analyzed.If genes b and c are linked,10 map units apart,but gene a is located on a different chromosome,what proportion of ascospores is expected to be a⁺b⁺c⁺? A)about 12.5% B)about 22.5% C)about 25% D)about 45% E)about 50%

Question

Quizplus · Accepted Answer

Since genes b and c are linked with a distance of 10 map units, they will be separated by crossing over events in about 10% of the meiotic divisions. This means that we can expect about 90% of ascospores to have parental combinations for genes b and c. Gene a is located on a different chromosome, which means that it will independently assort in meiosis. Therefore, we can multiply the probability of getting the parental combination for genes b and c (0.9) with the probability of getting the desired alleles for gene a (0.5, since the F1F1 parent is heterozygous for the a locus). So the expected proportion of ascospores with the genotype a⁺b⁺c⁺ is: 0.9 × 0.5 × 0.5 × 0.5 = 0.1125 or about 11.25%, which is closest to choice B (about 22.5%).

Ascospores from a Cross A ×\times× B C (All Auxotrophic Markers)are Analyzed

Ascospores from a Cross A $\times$ B C (All Auxotrophic Markers)are Analyzed