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Small recombination frequencies should be approximately additive. Note that recombination distances are twice the value of wildtype plaques since the double mutant recombinants were not counted. Thus, the data table should be:
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19. The order is a c b, and c is close to a. Genes c and a are cotransformed 76% of the time, suggesting that these two genes are very close and b is far away (a-b cotransduction is 10%, or 0.1). Two orders are possible: a-c b and c-a b. If the rst order is correct, a b c results from a double crossover; this class should be the least frequent. If the second order is correct, a single exchange between a and c would yield a b c , but this frequency should be similar to a b c , and it is not. 21. thr leu pro his. We see that cells that are thr are the most frequent. The chance of interruption in the conjugation increases with the length of time for the mating. Therefore, genes farther from the origin of transfer appear less frequently. We can order the genes based merely upon the frequency of genotypes seen. The order must be thr leu pro his. Since we see no his , and since we stopped the mating at 25 minutes, his must be after minute 25 on the map of this Hfr strain. 23. a and c are close; b is farther away; c is probably in the middle. The numbers of the rst three transformant classes indicate that each gene, by itself, is readily transformed. We notice that classes with b and any other gene are quite rare, a situation that indicates b is far from a and c. We notice that a and c are cotransformed about 13% of the time, b and c about 3% of the time, and a and b about 2% of the time. 25. a. lys his val lys his val lys his val lys his val Since there is no lysine in the medium, lys must be present to allow growth. b. lys val his lys val his Both lys and val must be present to allow growth.
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Percent Wild-Type Plaques 0.3 1.0 0.4 0.7 0.1 0.6
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Percent Recombinants 0.6 2.0 0.8 1.4 0.2 1.2
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a a a b b c
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b c d c d d
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The largest distance is between a and c; therefore a and c must be at opposite ends. Since a-b 0.6, b must be 0.6 units to the right of a. This position gives b-c as 1.4, the observed distance. We now have the following map 0.6 a b 1.4 c
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If d is to the left of b, then d-c should be greater than 1.4, a result not seen. Therefore, d is 0.2 units to the right of b. 31. 2 0.04 1 0.02 3
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To calculate map distance, you must have the number of recombinations and the total number of progeny. Since all phages grow on strain B, this number must equal the total number of progeny; this is 250 107. Since only wild-type phages grow on K12, and since wild-types result from recombination between two genes, a a b X b
a b and a
the number that grow on K12 must be recombinants. But this number represents only half of the recombinants, for the double mutant will not grow on K12. Total recombinants are:
Tamarin: Principles of Genetics, Seventh Edition
Back Matter
App. A: Brief Ans. to Selected Exercises, Problems, and Critical Thinking Ques.
The McGraw Hill Companies, 2001
Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions
1 1 2
2 3 3
(2 (2 (2
50)(104) 25)(104) 75)(104) 104) 107) (0.4 10
106 5 105 1.5 106 100
33. Use replica-plating on selective media with arabinose as the sole carbon source, thus selecting for ara cells. Although all three loci can be cotransduced, the rarity of ara leu ilvH indicates leu is the middle locus (ara leu ilvH). Cotransductance frequencies: ara to leu 4 10
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