19. In Drosophila the genes for normal bristles and normal eye color are known t

Question

19. In Drosophila the genes for normal bristles and normal eye color are known to be about 20 units apart on the same chromosome. Individuals homozygous for these genes are mated with homozygous recessive individuals. The Fl progeny were then test crossed (mated with homozygous recessive flies). If there were 1000 offspring from the test cross, how many of them would you predict to show the crossover phenotypes? 20a. In Drosophila, the allele for miniature wing (m) is recessive to the allele for normal wing(M), and the gene for vermilion eye (v) is recessive to the allele for normal eye (V). A female heterozygous for vermilion cye and miniature wing was mated to a vermilion-eyed, miniature-winged male. The following offspring were collected. 140 normal wing, normal cyes 3 normal wing, vermilion eyes 151 miniature wing, vermilion eyes 6 miniature wing, normal eyes a. Are these genes assorting independently? Explain. b. What is the crossover rate for these genes? 20b. A female Drosophila heterozygous for the recessive alleles sable body (s) and miniature wing (m), was mated with a sable-bodied, miniature-winged male and the following offspring were collected: 250 normal body, normal wings 215 sable body, miniature wings 15 normal body, miniature wings 20 sable body, normal wings

Explanation / Answer

Answer 19) Though we can directly give the answer via considering the genetic distance between two genes. but, let's first understand the given crosses;

Given alleles are; b+ = normal bristles, b = mutated bristles, e+ = normal eye and e = mutant eye.

First cross: b+ e+ / b+ e+ x b e / b e

F1 progenies = b+ e+ / b e

Test cross = F1 progeny x homozygous recessive fly ( b+ e+ / b e x b e / b e)

In such case, we can observe recombination (crossing over) only in the F1 progeny (as parent).

also, we know that the recombination frequency is directly related with the genetic distance between the two genes such that 1mU distance gives 1 % recombinant progenies.

similarly, 20mU genetic distance between the given two genes will give 20 % recombinant phenotype.

20% of 1000 = 200

therefore, 200 progenies will show crossover phenotype.

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