Question

Please explain why the answer is h.

1. Consider the cross: Hfr: str erys arg+ met+ gal-lac-lys- x F:stre eryk arg-met-gal+lac+ lys+ where the genetic map is shown below (the < indicates the Origin of transfer in the Hfr strain). arg gal ery str met lac Circle the media below which would select for the most exconjugants. a) Minimal media + arginine b) Minimal media + streptomycin c) Minimal media + galactose + methionine + lysine d) Minimal media + streptomycin + erythromycin e) Minimal media + galactose + streptomycin f) Minimal media + streptomycin+ methionine g) Minimal media + methionine + lactose + lysine h) Minimal media + streptomycin + lysine + arginine i) Minimal media + streptomycin+ erythromycin + methionine j) Minimal media + erythromycin + lactose + methionine + arginine k) Minimal media + streptomycin + lactose + methionine + lysine + arginine 1) Minimal media + erythromycin + streptomycin + arginine + lysine + methionine m) Minimal media + streptomycin + arginine + lysine + methionine +lactose +galactose n) None of these above

Answer 1

Exconjugates are members of a conjugating pair of protozoan ciliates after separation and prior to a mitotic division. So for most of Exconjugates, we will use Minimal media + streptomycin + Lysine + arginine.

From the given genetic map of Hfr strain, the argine is the first one to enter the cell because it is present just after the origin of transfer then streptomycin and then lysine. If all these three genes are present in the same loci with minimum distance between them the conjugates will surely have these three genes. other genes methionine, erythromycin, lactose, galactose, genes were very far from the origin of transfer so the chances of transferring these genes are very much less.

Hfr strain is sensitive to streptomycin and F- is resistant to streptomycin and this is used as a selectable marker to confirm the cross between these two strains.

So to select the Exconjugates the combination of media + streptomycin + lysine + arginine was used.