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# PS3A KEY - BICD100 Winter 2008 Chisholm Problem Set 3A...

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Unformatted text preview: BICD100 Winter 2008 Chisholm Problem Set 3A Phage genetics 1. Five mutants a-e in phage π are temperature sensitive for growth. All the mutants grow normally (i.e. make plaques) at 30° but do not grow at all at 40°. As your ﬁrst step in characterizing these mutants you perform double infections of the host E. coli strain at 30°. You allow the doubly infected bacteria to lyse then titrate the phage by plating on E. coli at 40°. You obtain the following data (the number is the burst size for each infection). a 0 b 90 0 c 0 0 0 d 85 100 0 0 e 0 100 0 85 0 WT 105 95 100 98 110 a b c d e In a second experiment you perform double infections as before, but then separately titrate the resultant bursts at 30° and at 40°. Below are your data, plotted as the ratio (burst size at 40°/burst size at 30°). a 0 b 0.003 10-7 c 0 0.001 0 d 0.009 0.03 0 10-7 e 0.001 0.014 0 0.007 10-7 a b c d e (a) How many genes do the mutants deﬁne? Draw a genetic map showing the distances between these mutations and the boundaries between the genes. The ﬁrst experiments are complementation tests. All the mutations are recessive to WT, so complementation tests are meaningful. a and e are in the same complementation group (gene). b and d are in separate complementation groups. c fails to complement all other mutations suggesting it may be a deletion. So, the short answer is THREE genes. The second experiments allow you to measure recombination, however note that frequencies of the order of 10-7 are reversion not recombination. a and c do not revert when coinfected with themselves, so they are likely deletions. The only mutation that recombines with c is b, so b is outside the c deletion; all others are within the c deletion. BICD100 Winter 2008 Chisholm Multiply the RFs by 200 to get map units (cM). The map is constructed by pairwise comparisons. Start with the mutations that map furthest apart, b and d. a and e must map between them, consistent with complementation data. The d-a distance is greater than the d-e distance, so the most likely order is: d, e, a, b. c is a deletion spanning d and e/a. (b) In a double infection with d and e, what fraction of plaques ts for growth would be expected to be double mutant? To make the d e double mutant, assume that the double mutant will have the same phenotype as the single mutant, ie be ts for growth. Double infect bacteria (using a high m.o.i. such that most bacteria should be double infected). Plate the infected bacteria at the permissive temperature. Phage from resulting plaques can be transferred to new lawns and tested for growth at the restrictive temperature. Take plaques that are ts for growth and do separate double infections with the parental strains d and e. Since d and e are in different complementation groups, a double mutant could be recognized either by complementation testing (failure to complement both d and e) or by mapping (failure to recombine with d or e). The d-e map distance = 1.4 cM. The percentage of ts plaques that are recombinant will be 0.7% since half the recombinants will be the double wild type. Thus on average about 1 in 140 plaques would be double mutant. (c) Another mutation f maps close to d. How could you use the d e double mutant to determine which side of d the f mutation maps? BICD100 Winter 2008 Chisholm This is a three factor cross. Either f is to the right of d (between d and e) or it is to the left. If f is between d and e, the fraction of WT recombinants from the cross d e x f should be much less than the fraction from the cross d x f. If f is to the left, the fractions should be equal. (d) In a rush to get to class, you accidentally put unlabeled tubes of the precious d e double mutant phage in the same rack as unlabeled tubes of c. What crosses could you do to distinguish your d e double mutant from c? Cross the unknowns to the a mutant and look for wt recombinants. The d e double mutant will recombine with a, the c deletion will not. (You could also cross with the b mutant; d e will recombine more often with b than c will. However again this involves more work for no reason; the cross with a gives you a yes/no answer.) ...
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