7.03 Problem Set 4 Solutions

PROBLEM SET 4 SOLUTIONS 1. a) The F plasmid has integrated between the Thy A gene and the LysA gene, oriented such that LysA will be transferred early and ThyA late. b) Since the F plasmid has not changed the distance between the ArgA gene and the ThyA gene, their cotransduction frequency will remain 50% in the Hfr. c.) The object here is to find a rare, specific excision event among a population of Hfr cells (see below). This F' can be isolated by mating the above Hfr population (that is streptomycin sensitive) with a streptomycin resistant, ThyA-, LysA-, F- strain. After a brief mating, select for strr, LysA+, ThyA+. Only those cells that have received an F' containing LysA+ ThyA+ will be able to grow and form colonies. When the other Hfr cells (which do not contain an F' ) transfer, they will transfer LysA+ early, but the mating will be disrupted before ThyA+ is transferred. 2.a)Thestructure ofthe F' isseenhere: r_ _i I::' b) Although the LacZ gene is defective on the F' and on the chromosome, the two LacZ- alleles may not be due to the same mutation, If this is the case, then an integration/recombination event could have occurred as diagrammed below. Note, this event happens after the transfer of the F' to the F-. • _-I @ f _ 2. c) If the integration recombination event occurs as diagrammed below, then he 0 ° mutation will be in cis to the LacZ- mutation, while the wildtype operator will be adjacent to the wildtype copy of LacZ. This will result in normal, inducible expression of LacZ. S I I f ! I 0 "2-"" _- . 3. a. The A mutation selectively affects sucrase function without disturbing permease function (line 1). From the data in line 4, we see that the A mutation is recessive to wildtype, since the mero-diploid is normally induced. A simple explanation for these data is that the mutation is a loss of function mutation in the sucrase gene itself. The B mutation is uninducible since both sucrase and permease activities are low even in the absence of glucose. From the data in line 5, we see that the B mutation is dominant, and from the data in lines 7 and 8, we see that the B mutation is trans-acting. This can be explained if the B mutation is in the repressor for the sucrose operon, creating a super-repressor. The C mutation is constitutive, since the sucrase and permease activities are high even in the presence of glucose. From the data in line 6, we see that the C mutation is dominant, and from the data in lines 9-10, we see that the O mutation is cis-acting. (We can only see the constitutive phenotype of the C mutation when it is on the same piece of DNA as a wildtype copy of the A gene). This suggests that the C mutation is in the operator, the binding site for the repressor. The repressor cannot bind the mutant operator. Model: In the presence of glucose, glucose binds the repressor, enabling it to bind the operator and prevent expression of the sucrose operon. In the absence of glucose, the repressor can no longer bind the operator, and expression of the sucrose operon is induced. The B mutation enables the repressor to bind the operator even in the absence of glucose. rCxzs' _ _ __ , , ' b.) In strain with the genotype A÷BC, the operator mutation would prevent binding by the mutant repressor. Therefore, sucrase expression would be constitutive in this strain, and the C mutation is epistatic to B. (Since the phenotype of the double mutant is constitutive, C must be acting downstream of B.