Problem

Solutions For An Introduction to Genetic Analysis Chapter 5 Problem 39P

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Solution 1

Jin generalized-transduction experiment, phages are collected from an E. coli donor strain of genotype cys+ leu+ thr+. These genotypes are used to transduce a recipient of genotype cys-leu-thr-. The treated recipient population is plated on a minimal medium supplemented with leucine and threonine and respective colonies are obtained. From the given colonies the results obtained are:

a) The minimal medium is supplemented with leucine and threonine, owing to the medium used cys+ will present in all colonies. Whereas, the other two genes leu and thr are either + or -. The possible genotypes of these colonies are,

cys + leu + thr +

cys + leu + thr -

cys + leu - thr +

cys + leu - thr -

b) When these colonies are plated onto three different media,

The possible genotypes grown on the minimal media supplemented with threonine are cys+ leu+ thr+ and cys+ leu+ thr-

The possible genotypes grown on the minimal media supplemented with leucine are cys+ leu+ thr+ and cys+ leu- thr+

The possible genotypes grown on the minimal media without any supplement is cys+ leu+ thr+

c) The actual genotypes of the colonies on medium (1) had cys+ leu+ thr- and medium (2) had cys+ leu- thr+. The remaining cultures were cys+ leu- thr-. Since, none of the genotypes grew on minimal medium and no colony was leu+ thr+. The remaining percentage of genotype occurred in 100% - 56% = 39% of the colonies.

d) Since, cys and leu are co-transduced, 56 percent of the time. In addition, cys and thr are co-transduced only 5 percent of the time.

Hence, this indicates that the gene cys is closer to leu than it is to thr. cys gene is in the middle because no cys+ leu+ thr+ co-transduced are found.

Therefore, the genetic map showing the order of the three genes is as follows,

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