Question

The allele se gives Drosophila a sepia body and se+ gives brown, the wild-type phenotype. The allele ro of a separate gene gives rough eyes and ro+ gives smooth eyes, the wild-type phenotype. The allele ss of a third gene gives flies without thorax spines and ss+ gives flies with thorax spines, the wild-type phenotype.

When females heterozygous for each of these genes were testcrossed with sepia, rough, spineless males, the following classes and numbers of progeny were obtained:

wild type 338 sepia 96 spineless 46 rough 156 sepia, spineless 173 sepia, rough 43 spineless, rough 114 sepia, spineless, rough 370

Total 1336

a. Construct a linkage map with the genes in their correct order and indicate the map distance between the genes. Show your calculations.

b. Draw the alleles in their proper positions (i.e. in the parental arrangement) on the chromosomes of the triple heterozygote

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The allele se gives Drosophila a sepia body and set gives brown, the wild-type phenotype. The allele ro of a separate gene gives rough eyes and rot gives smooth eyes, the wild-type phenotype. The allele ss of a third gene gives flies without thorax spines and sst gives flies with thorax spines, the wild-type phenotype. When females heterozygous for each of these genes were testcrossed with sepia, rough, spineless males, the following classes and numbers of progeny were obtained: wild type 338 sepia 96 spineless 46 rough 156 173 sepia, spineless sepia, rough 43 spineless, rough 114 sepia, spineless, rough 370 Total 1336 a. Construct a linkage map with the genes in their correct order and indicate the map distance between the genes. Show your calculations. b. Draw the alleles in their proper positions (i.e. in the parental arrangement) on the chromosomes of the triple heterozygote.

Answer 1

a) We first need to write down the parental genotypes:

Female: se ro ss / se+ ro+ ss+

Male: se ro ss / se ro ss

Now, we have to classify each phenotype in the F1 generation as either Parental or Recombinant. The Parentals are either se ro ss or se+ ro+ ss+, and any other combination is recombinant. Let us beggin and classify the phenotypes but we are going to address the loci by pairs:

- Se and ro:

wild type 338 P 96 R sepia spineless rough 46 P 156 R 173 R 43 P sepia, spineless sepia, rough spineless, rough sepia, spineless, rough 114 R 370P Total 1336

Now, the distance is calculated by divding the total number of recombinants by the total number of offspring and then multiply the whole thing by 100:

D = ((96 + 156 + 173 + 114)/1336)(100) = 40.34 mu

- Se and ss:

wild type 338 P 96 R sepia spineless 46 R rough 156 P 173 P sepia, spineless sepia, rough spineless, rough sepia, spineless, rough 43 R 114 R 370 P Total 1336

D = ((96 + 46 + 43 + 114)/1336)(100) = 22.38 mu

- Ro and ss:

338 P 96 P 46 R 156 R wild type sepia spineless rough sepia, spineless sepia, rough spineless, rough sepia, spineless, rough 173 R 43 R 114 P 370 P Total 1336

D = ((46 + 156 + 173 + 43)/1336)(100) = 31.28 mu

We already have the distances, now let us map them:

se SS ro 22.38 31.28 40.34

b) In this case we have to draw the diploid genotype of the female parent. Remember that she has all the wildtype allele in one chromosome and all the mutated alleles in the other, it looks like this: