The answer is-None of these above (Minimal media + streptomycin + lactose + meth

Question

The answer is-None of these above (Minimal media + streptomycin + lactose + methionine + lysine + arginine + galactose)

Can someone please walk me through how you would solve this question. Thank you!

1. (7 Points) Consider the cross: Hfr strs arg- met gal+lac+lys- x F: str arg met- gal-lac lys. where the genetic map is shown below (the indicates the Origin of transfer in the Hfr strain). gal arg lys lac str met Circle the media below which would select for the most exconjugants. a) Minimal media streptomycin b) Minimal media galactose c) Minimal media arginine galactose d) Minimal media streptomycin lactose e) Minimal media galactose streptomycin Minimal media lactose methionine arginine g) Minimal media streptomycin+ methionine h) Minimal media methionine galactose lysine i) Minimal media galactose arginine lysine methionine j) Minimal media streptomycin galactose lactose methionine lysine k) Minimal media streptomycin lactose methionine lysine arginine l) None of these above

Explanation / Answer

During conjugation process,the part of DNA is being inserted in the F- cells from Hfr cells thus giving part of the characters of Hfr to F- cells. But according to genetic map of Dna of bot Hfr and F- cells, the Streptomycin resistence or sensitive part is located on the other part from other genes that is arg, met, gal, lac & lys, which are positive for Hfr while negative for F-. After conjugtion, the genetical part for streptomycin resistence will be present while the other genes will be replaced by Hfr genes. Thus we should select minimal media containing Streptomycin, lactose, methionine, lysine, arginine & galactose.

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