Question

When a mouse having a long tail (Mm) is crossed with a mouse having short tail (mm) what would be phenotypic and genotypic characters expressed in the F1 progeny. Show the schematic diagram. (ii) Now if the one of the long-tailed progeny of F1 is crossed with another long tailed mouse (MM). Show the next generation with the Punnet square. ​

Answer 1

Answer:

You are studying three autosomal recessive mutations in the fruit fly Drosophila

melanogaster. Flies that are homozygous for the hb– mutation are “humpbacked” (wildtype flies are straight-backed). Flies that are homozygous for the bl– mutation are

“blistery-winged” (wild-type flies are smooth-winged). Flies that are homozygous for the

st– mutation are “stubby-legged” (wild-type flies are long-legged).

You mate flies from two true-breeding strains, and the resulting F1 flies are all are

straight-backed, smooth-winged, and long-legged. F1 females are then mated to males

that are humpbacked, blistery-winged, and stubby-legged. In the F2 generation, among

1000 progeny resulting from this cross, you observe the following phenotypes:

Explanation:

Phenotype Number

humpbacked, blistery-winged, and stubby-legged (26 flies)

humpbacked, blistery-winged, and long-legged (455 flies)

humpbacked, smooth-winged, and long-legged (24 flies)

straight-backed, blistery-winged, and stubby-legged (27 flies)

straight-backed, blistery-winged, and long-legged (4 flies)

straight-backed, smooth-winged, and stubby-legged (442 flies)

straight-backed, smooth-winged, and long-legged (22 flies)

(a) The male flies that were bred to the F1 generation in order to produce the F2

generation were humpbacked, blistery-winged, and stubby-legged. On each of their

chromosomes, they have the alleles hb– bl– st–. Using this notation, state the

genotype of each of the two true-breeding parental strains (i.e. the two strains in the P