Question

In Drosophila melanogaster, the map positions of the three following genes are known in centiMorgans. The three genes are striped thorax (3-62), str, white ocelli (3-76), wo, and claret colored eyes (3-100), cl.

A. If str wo cl+/ str+ wo+ cl females are crossed to males that are homozygous recessive for all three genes, what will the types of offspring be and what will their numbers be if 1,000 progeny are scored? Assume that there is no interference in this region of the chromosome. Show your calculations neatly.

B. How would the classes of individuals change if the interference were 0.6 in the region between str and cl? Which classes would increase, decrease or remain the same? Only a qualitative answer is necessary.

Answer 1

In Drosophila melanogaster, the map positions of the three following genes are known in centiMorgans. The three genes are striped thorax (3-62), str, white ocelli (3-76), wo, and claret colored eyes (3-100), cl.
A. If str wo cl+/ str+ wo+ cl females are crossed to males that are homozygous recessive for all three genes, what will the types of offspring be and what will their numbers be if 1,000 progeny are scored? Assume that there is no interference in this region of the chromosome. Show your calculations neatly.
B. How would the classes of individuals change if the interference were 0.6 in the region between str and cl? Which classes would increase, decrease or remain the same? Only a qualitative answer is necessary.

The parents are (str wo cl+/ str+ wo+ cl) x (str wo cl/ str wo cl)

The cross over can occur between these genes and recombinants will arise. The recombination frequency corresponds to the map distance. The map distance between str and wo = 76-62 = 14 mU = 14% recombinants.
The map distance between wo and cl= 100-76 = 24mU = 24% recombinants.
The double recombinants can also be produced by recombination between both str-wo and wo-cl.
expected double recombinants = 100 * (0.14 * 0.28) = 3.92 = 4%
There will be two type of recombination in each cross over, so 14% = 7% and 7%; 28 = 14 and 14%, 4% = 2 and 2%.
Remaining are parental types = 100 - (14+28+4) = 54
So, out of 1000 progeny,
Recombination between str-wo = str wo+ cl 7
Recombination between str-wo = str + wo cl+ 7
Recombination between wo - cl = str wo cl 14
Recombination between wo- cl = str+ wo+ cl+ 14
Double cross overs           = str wo+ cl+ 2
Double cross overs           = str+ wo cl 2
parental                    = str wo cl+ 27
parental                    = str+ wo+ cl 27

B. Interfernce = 0.6
interference reduces the expected double cross overs. they will decrease by 1 - 0.6 = 0.4 factor.