Question

An Hfr strain with the genotype met+ arg+ eryR strS was mated with an F- strain with the genotype met- arg- eryS strR. These genes are transferred in the order given, with met+ entering F- cells first. Exconjugants were plated on a medium containing the antibiotic, streptomycin (Str), to kill theHfr cells as well as the antibiotic erythromycin (Ery), the last marker transferred. The exconjugants that grew on this medium were then tested for growth on media supplemented with Arg (arginine) or Met (methionine) as indicated below. The following results were obtained for growth on minimal media with the following supplements:

Ery only: 263 colonies

Ery and Arg: 264 colonies

Ery and Met: 290 colonies

Ery, Arg, and Met: 300 colonies

Determine the map distances from arg to met, from arg to ery, and from met to ery. Show your work.

Answer 1

Answer :)

Ery gene reaches at the end of the transfer because we get the least number of colonies. Ery and Arg are closely linked because Ery+Arg cells are almost equal to Ery transferred cells. Ery and Met are the second most numbered cells therefore; Methionine would transfer first among all three genes. Therefore, the order of the three genes is as follows:

Met- Arg- Ery

Map distance between Met and Arg = (Number of colonies of Met⁺ Arg^– only / Total colonies)x 100

Number of colonies of Met⁺ Arg^– = 290

Total number of colonies = 263+ 264+ 290+ 300

Total number of colonies = 1117

Map distance between Met and Arg = (290/ 1117) x 100

Map distance between Met and Arg = 0.26 x 100

Map distance between Met and Arg = 26 cM

Map distance between Ery and Arg = (Number of colonies of Arg⁺Ery^–/ Total colonies)x 100

Number of colonies of Arg⁺Ery^– = 264 – 263

Number of colonies of Arg⁺Ery^- = 1

Map distance between Ery and Arg = (1/1117)x 100

Map distance between Ery and Arg = 0.09 cM