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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 ?

This question was previously asked in
MH SET 2023 Paper 2 Computer Science and Applications Question Paper (26-Mar-2023)
The correct answer is

ser–, gly+, leu+; ser+, gly+, leu+; ser+, gly+, leu–

In transduction, a phage carries a fragment of donor DNA into the recipient, where recombination can replace any of the recipient markers (ser–, gly–, leu–) with the donor alleles. Which transductants survive depends only on what the plating medium lacks.

The minimal medium is supplemented with leucine and serine but not with glycine. So a cell grows only if it can make glycine itself, that is, only if it has received the donor gly+ allele.

Because leucine and serine are supplied, the ser and leu loci are not selected: a gly+ transductant may carry ser+ or ser– and leu+ or leu– (depending on how much of the donor fragment was co-transduced) and still form a colony.

Therefore every genotype listed for the colonies must be gly+. The sets that include any gly– genotype are impossible, since a glycine auxotroph cannot grow on this plate.

Check: ser– gly+ leu+, ser+ gly+ leu+ and ser+ gly+ leu– are all gly+ and their ser/leu needs are met by the supplements, so all three grow; every other set contains at least one gly– genotype.

Hence, ser–, gly+, leu+; ser+, gly+, leu+; ser+, gly+, leu–.

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