Question

. Explain fully the following observations on three genes in Drosophila. In all cases invent allelesymbols.

a.

Twenty units of enzyme E are normally found in homozygous wild-type cells. A homozygous mutant shows zero units. A heterozygote is mutant in appearance and has 10 units of E per cell.

b.

Thirty units of enzyme F are normally found in homozygous wild-type cells. A homozygous mutant shows zero units. A heterozygote is wild-type in appearance and has 15 units of E per cell.

Answer 1

A- When two alleles of wild-type enzymes are present they lead to production of 20 units of enzymes which means that each allele is responsible for 10 unit of enzymes which is further confirmed when the wild-type allele is present with the mutant allele in heterozygous condition

Suppose E is for wild allele and e for mutant allele

So cross of EE (20 unit )and ee ( 0 unit) will give

Ee progeny which produce 10 unit of enzymes.

B-

A- When two alleles of wild-type enzymes are present they lead to production of 30 units of enzymes which means that each allele is responsible for 15 unit of enzymes which is further confirmed when the wild-type allele is present with the mutant allele in heterozygous condition

Suppose E is for wild allele and e for mutant allele

So cross of EE(30 unit) and ee (No unit) will give

Ee progeny which produce 15 unit of enzymes.