Chapter 5

19.  An Hfr strain has the fertility factor F integrated into the chromosome. An F+ strain has the fertility factor free in the cytoplasm. An F− strain lacks the fertility factor.

23.  Although the interrupted-mating experiments will yield the gene order, it will be relative only to fairly distant markers. Thus, the mutation cannot be precisely located with this technique. Generalized transduction will yield information with regard to very close markers, which makes it a poor choice for the initial experiments because of the massive amount of screening that would have to be done. Together, the two techniques allow, first, for localization of the mutant (interrupted mating) and, second, for the precise determination of the location of the mutant (generalized transduction) within the general region.

28.  The best explanation is that the integrated F factor of the Hfr looped out of the bacterial chromosome abnormally and is now an F′ that contains the pro+ gene. This F′ is rapidly transferred to F− cells, converting them into pro+ (and F+).

33.  The expected number of double recombinants is (0.01)(0.002)(100,000) = 2. Interference = 1 − (observed double cross-over/expected double crossover) = 1 − 5/2 = −1.5. By definition, the interference is negative.

836

37. 

  1. This process appears to be specialized transduction. It is characterized by the transduction of specific markers based on the position of the integration of the prophage. Only those genes near the integration site are possible candidates for misincorporation into phage particles that then deliver this DNA to recipient bacteria.

  2. The only media that supported colony growth were those lacking either cysteine or leucine. These media selected for cys+ or leu+ transductants and indicate that the prophage is located in the cys–leu region.

42.  No. Closely linked loci would be expected to be cotransduced; the greater the contransduction frequency, the closer the loci are. Because only 1 of 858 metE+ was also pyrD+, the genes are not closely linked. The lone metE+ pyrD+ could be the result of cotransduction, it could be a spontaneous mutation of pyrD to pyrD+, or it could be the result of coinfection by two separate transducing phages.

47. 

  1. To determine which genes are close, compare the frequencies of double transformants. Pair-by-pair testing gives low values whenever B is included but fairly high rates when any drug but B is included. This finding suggests that the gene for B resistance is not close to the other three genes.

  2. To determine the relative order of genes for resistance to A, C, and D, compare the frequencies of double and triple transformants. The frequency of resistance to AC is approximately the same as that of resistance to ACD, which strongly suggests that D is in the middle. Additionally, the frequency of AD coresistance is higher than AC (suggesting that the gene for A resistance is closer to D than to C) and the frequency of CD is higher than AC (suggesting that C is closer to D than to A).

51.  To isolate the specialized transducing particles of phage ϕ80 that carried lac+, the researchers would have had to lysogenize the strain with ϕ80, induce the phage with UV, and then use these lysates to transduce a Lac− strain to Lac+. The Lac+ colonies would then have been used to make a new lysate, which should have been highly enriched for the lac+ transducing phage.