Discussion

Phages are collected from an infected E.coli donor strain of genotype ser⁺, gly⁺, leu⁺ and used to transduce a recipient of genotype ser^-, gly^-, leu^-. The treated recipient population is plated on a minimal medium supplemented with leucine and serine. Many colonies grew. Which one of the following combination of genotypes is appropriate for the colonies that grew ?

A.ser\(^{-}\), gly\(^{-}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{-}\)
B.ser\(^{+}\), gly\(^{+}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{-}\)
C.ser\(^{-}\), gly\(^{+}\), leu\(^{+}\); ser\(^{+}\), gly\(^{+}\), leu\(^{+}\); ser\(^{+}\), gly\(^{+}\), leu\(^{-}\)
D.ser\(^{-}\), gly\(^{+}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{+}\); ser\(^{+}\), gly\(^{-}\), leu\(^{-}\)

Answer: C

Every colony that grew must be gly⁺, so the answer is the set in which all three genotypes carry gly⁺: ser^-, gly⁺, leu⁺; ser⁺, gly⁺, leu⁺; ser⁺, gly⁺, leu^-.

The recipient is ser^- gly^- leu^-. The plate is minimal medium with leucine and serine added, so the only nutrient a cell still has to make for itself is glycine. Only transductants that received the gly⁺ allele from the donor can grow; whether they also picked up ser⁺ or leu⁺ does not matter, because both amino acids are supplied.

So the colonies are a mixture of gly⁺ cells that may or may not be co-transduced for the other two markers. Each of the other three options includes at least one gly^- genotype, which could not form a colony on this medium.

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