7.03 Genetics: Solutions to Problem Set #3 (1993)

PROBLEM 2 a)The higherthefrequency oftransductants thecloser thegenesarcinthechromosome. A highercotransduction frequency means thatthechanceoftwo markersbeinginthe same pieceofDNA packagedby thevirus proteins ishighand thatoPe._inside the recipient ccU thechanceofa crossovereventbewccn thetwo markerswithconcomitant Io_5 ofone ofthem islow becausethedistance betweenthemissmall. Accordingtothedatathemap couldbedrawneither as: I) C A B E D B E D OR 2) A C You can't determine whether C is on one side or the other of A. Note: A frequency of 10-8 is the frequency of two independent events, the probability of one marker being transduced rimes the probability of the other marker being transduced, or the chance that two independent molecules of DNA get incorporated independently. b) In order to resolve the ambiguities of this map you could set up the following experiment ( there _,_ more than one way of doing it) You could grow P1 in a strain with the following genotype_ + C + A+ and the lysate used to infect a strain that is B" C" A'. Selection of B + incorporation is done by growing the bacteria in a medium without B and supplemented with C and A. After selection you then screened for incorporation of the other markers. There are four possible classes: A-C + A+C A+C + A'C" Therearctwo possible orders: -__ C+- A+ ........... C" - B+ A+ C+ B+ B-.___ ...................... 1_ A-/_ C- B- orderI order2 When you select forB you arcselecting fora crossover totheright ofB and thedifferent classes aregenerated by different locations ofthesecondcrossing overorby theeventof fourcrossovers. A+C + A'C" A+C- A'C + orderI order2 cros,_vers crossovers I+4 1+4 (2) (2) 1+2 i+2 (2) (2) 1+3 1+2+3+4 (2) (4) 1+2+3+4 1+3 (4) (2) The two orders predict a different least frequent class or the class generated by four crossovers. DctermirdgcJthe frequency ofA+C" and A-C + andcomparingtheir frequency shouldallowyou todetermine theorderofthegenes.IfA+C -istheIcast frequent then order2 isthecorrect orderand ifA-C istheleast frequent thenorderI iscorrect. Problem 3: a) The suc8/F'sucA+ crossshows that the suc8 mutation is recessive,and from the two other F' crosses we can determine that the mutation acts in trans. Since it is constitutive, recessive, and trans-acting, the most likely explanation is that the mutation lies in a _ for the operon. The sucl 1 mutation is dominant, and also acts in trans. Sincewe know that a repressor exists for control of this operon, the sucl 1 mutation is analagousto the I"d mutation in the lac repressor. Therefore, the sucl 1 mutation is a dominant ne0ative mutation in the The sucl 4 mutation is also constitutive and dominant, but only acts when in cis to a functional sucA+ gene. The simplest explanation for this mutation is that it lies in an site that normally binds the repressor. This mutation is analagous to the Oc mutation in the lac operon. b) The suc23 mutaton is uninducible, recessiveand trans-acting. The simplest explanationis that it lies in an _ for the operon. One can postulate that if there is no activator present, as in a suc23- mutant, then only low levelsof sucrasecan be made. The suc21 mutation is uninducibleand recessive,but only acts when in cis to a functional sucA+ gene (as shownin the suc21/F' sucA- rnerodiploid). The most likely explanationis that this mutation liesin an operator site that the activator binds. This site can be different than the site that bindsthe repressor,or it can be the samesite. However, if it is the same site, one must stipulate that the suc21 mutation only affects bindingof the activator, and not bindingof the repressor. Moreover,the suc8 mutation affects repressor binding,but not activator binding. The suc25 mutation hasthe samecharacteristicsas the suc21 mutation, except that no sucraseactivity is detected in the suc25 mutant. There are two possibleexplanationsfor this result. One is that the suc25 mutation is in the promoter for the operon. One would expect no enzyme activity if RNApolymerasecouldnot transcribethe region. Alternatively, the mutation could lie within the sucA gene itself. c) The suc8 suc23 double mutant would be defective in both repressor and activator function. Therefore, the phenotype would be constitutive low levels of enzyme production, or 50 units with or without sucrose. Problem 4 solution: Note that the ptasmid library was made from random DNA fragments from the E coli genome. This has two implications. First, all genes, binding sites and other sequences carried in the plasmid inserts are wild type alleles. Second, each plasmid has only a single continuous DNA insert; in other words the order of genes in a plasmid insert represents their order in the bacterialchromosome. Recall that the suc23 mutation is recessive. The three clones that grow well on sucrose must carry the suc23+ gene. a) Clone three shows high sucrose expression levels because it carries the sucA+ gene (see below) and its regulatory sequences in addition to suc23+. Since the plasmid, and thus suc23+ and sucA+, is present in many copies per cell, sucrase expression is very high under inducing conditions in strains carrying clone 3. This data shows that sucA, and not the activator (suc23+), is limiting in sucrase expression. b) Clones that alleviate the mutant phenotype complemented the recessive mutation in the recipient strain, and thus carry a functional copy of the mutated host gene (suc23, suc8 or sucA). All three clones carry suc23+. Only clone 2 carries suc8+, and only clone 3 carries sucA+ and its regulatory sequences. We know that clone three must carry the regulatory sequences because normal activation and repression of the plasmid sucA ÷ genes is observed. If the repressor binding site were missing there would be several thousand sucrase activity units in the absence of sucrose. Similarly, if the activator binding site were missing there would be only fifty sucrase units from each plasmid in the presence of sucrose, for a total of several hundred sucrase units. c) Remember that the inserts contain continuous DNA inserts derived from the wild type E coli chromosome. There are two possible gene orders based on these three clones: sucA...suc23...suc8 or suc8...suc23...sucA. A schematic map based on the first order is presented Clone# 1 below. 1 2 ' I 3 ) __[abs]_[rbs]__[SucA operon] Gene / re m.tlatory element SucA gene Activator gene Activator binding site (abs) Repressor Repressor binding site (rbs) Other regulation [Activator gene] [Repressor gene]__ Mutation_ suc25 suc23 suc21 suc8 (recessive), suc11 (dominant) suc14 models may be consistent with the data, but this is the simplest.