Book Viewer

Short Contents | Full Contents | Other books @ NCBI

Modern Genetic Analysis Answers to Selected Problems
Chapter 9*

6. | The regions with the bars above or below are identical in sequence. The order on the circular map is See question

7. | An F strain will respond differently to an F+ (L) or an Hfr (M) strain, whereas Hfr × Hfr, Hfr × F+, F+ × F+, and F × F will give 0. Thus strains 2, 3, and 7 are F. Strains 1 and 8 are F+, and strains 4, 5, and 6 are Hfr. See question

11. | a. The gene order is arg bio leu. b. arg-bio: RF = 12.76 m.u. bio-leu: RF = 2.12 m.u. See question

12. | The most straightforward way would be to put an Hfr at both ends of the same sequence and measure the time of transfer between two specific genes. For example, See question

13. | The best explanation is that the integrated pro+ was incorporated onto an F factor that was transferred into recipients early in the mating process. These cells now carry the F factor and are able to transmit F+ in the second cross as part of the F factor, which still carries pro+. See question

18.

a. Notice that each gene was transferred into about one-tenth of the cells (single drugs tested). Also notice that pairwise testing gives low values whenever B is involved but fairly high rates when any drug but B is involved. This suggests that the gene for B resistance is not close to the other three genes and that the low rates come from double crossovers.
b. To determine the relative order of genes for resistance to A, C, and D, notice that the frequency of resistance to AC is approximately the frequency of resistance to ACD. Also notice that AD resistance is roughly 50% higher and CD resistance is about 25% higher. This suggests that the gene for D resistance is between the other two genes.
See question

20.

a. Specialized transduction is at work here. It is characterized by the transduction of one to a few markers.
b. The prophage is located in the cys-leu region, which is the only region that gave rise to colonies when tested against the six nutrient markers.
See question

22. | Recognize that, if a compound is not added and growth occurs, the E. coli has received the genes for it by transduction. Thus, the BCE culture must have received a+ and d+. The BCD culture received a+ and e+. The ABD culture received c+ and e+. The order is thus d a e c. Notice that b is never cotransduced and is therefore distant from this group of genes. See question

24.

a. The colonies are all cys+ and either + or for the other two genes.
b. (1)cys+ leu+ thr+ and cys+ leu+ thr

(2)cys+ leu+ thr+ and cys+ leu thr+

(3)cys+ leu+ thr+

c. Because none grew on minimal medium, no colony was leu+ thr+. Therefore, medium (1) had cys+ leu+ thr, and medium (2) had cys+ leu thr+. The remaining cultures were cys+ leu thr, and this genotype occurred in 39 percent of the colonies.
d.
See question

© 1999 by W. H. Freeman and Company. All rights reserved.