Three Drosophila mutations on the third chromosome were being mapped in a geneti

Question

Three Drosophila mutations on the third chromosome were being mapped in a genetic cross by a disorganized graduate student.. The mutant phenotypes were rosy eyes, groucho (bristles clumped above the eyes) and ebony (black body color). Unfortunately he forgot to write down the genotypes of the parental flies. So he took the F1 females, which were wild type in phenotype, and did a testcross with males that were homozygous for rosy, groucho, and ebony. The testcross results are shown below:

Testcross progeny phenotype                 Number

Wild type                                                            51

Groucho                                                              144

Ebony                                                                   54

Rosy                                                                      10

Rosy & Ebony                                                    130

Rosy & Groucho                                               43

Ebony & Groucho                                            7

Rosy, Ebony, & Groucho                                               61

a. What is the genotype of the wild type F1 females?

b. Give the order and map distances between these genes.

Explanation / Answer

The genotype of the F1 flies carrying wild-type and rosy, groucho and ebony alleles is RrGrEe and the testcross flies are rrggee.
The F1 cross is:
r --- g --- e x RrGgEe
r --- g --- e
It is unknown until we analyze the offspring data as to which chromosome the r, g or e alleles are on, the same or different homologue of the dominant alleles. For instance, the genotype could be Rge//rGE, or other combinations
Testcross progeny phenotype                              Number
Wild type                                                             51
Groucho                                                               144
Ebony                                                                   54
Rosy                                                                     10
Rosy & Ebony                                                       130
Rosy & Groucho                                                      43
Ebony & Groucho                                                     7
Rosy, Ebony, & Groucho                                           61
So, the parental types are groucho (144) and ebony, rosy (130). So, the chromsomes in the unknown F1 female are:
R --- g --- E
r --- G --- e
The single crossovers are the ebony, rosy, groucho (61), ebony (54), wildtype (51), groucho, rosy (43).
The double crossovers are the rosy (10) and ebony, groucho (7). So, to position, the genes, the double crossovers help us there. So, ebony must be in the middle and groucho and rosy on the ends. The F1 female's chromosome will be:
R --- E --- g
r --- e --- G
The single and double crossovers help us determine the map distances:
SCO between E and G ebony, rosy, and groucho (61) Wild Type (51)

Also, take half of the double crossovers to determine map distance. The distance between ebony and groucho is:
[(61 + 51) + (10 + 7)/2]/(144 + 130 + 61 + 54 + 51 + 43 + 10 + 7) = (112 + 17/2)/500 = (112 + 8.5)/500 = 120.5/500 = 0.241
So, 0.241 x 100 = 24.1% or 24.1 map units is the distance between ebony and groucho
SCO between R and E
ebony (54) groucho and rosy (43)
[(54 + 43) + (10 + 7)/2]/500 = (97 + 8.5)/500 = 105.5/500 = 0.211
So, 0.211 x 100 = 21.1% or 21.1 map units is the distance between groucho and rosy
r --- [21.1] --- e --- [24.1] --- g

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.