Question

Suppose the A and B genes are on the same chromosome but separated by 75 map units. What fraction of the progeny from the cross AB/ab x ab/ab would be phenotypically Ab (genotype-A bb)?

A) 25%
B) 37.5%
C) 50%
D) 75%
E) None of the above

Answer 1

The fraction of progeny phenotypically Ab from the cross AB/ab x ab/ab is 37.5%, as genes A and B are on the same chromosome separated by 75 map units, which typically leads to recombinant phenotypes in 37.5% of cases.

Step-by-step explanation:

The subject of this question is Biology, and it deals with genetics and recombination frequency. Specifically, it explores the outcome from crossing organisms with linked genes. Given that genes A and B are on the same chromosome and separated by 75 map units, the fraction of the progeny from the cross AB/ab x ab/ab that would be phenotypically Ab (genotype A bb) would be 37.5% (option B). This is because recombinants are expected to occur in half of the cases when genes are separated by 50 or more map units (indicating independent assortment), but since only one recombinant type (Ab) contributes to the phenotype in question, only half of these (37.5%) would show the phenotype. Therefore, we expect 37.5% of the progeny to have the recombinant Ab genotype.