7.03 Genetics: Problem Set 4 Solutions (2000)

7.03 Problem Set 41 Solutions Fall 2000 Problem 1: a) The transductants that havenormalregulated Lac geneexpression are result of Tn5 and Lac+cotransduction. Therefore, the c.f. = 70/100 = 70%. Tn5 Lac÷ ... X X ... LacI s b) Here are four possible genotypeswith their phenotypes: A Tn5 1sO÷ uninducible 648 B Tn5 1sOC constitutive 52 C Tn5 Z÷Oc constitutive 298 b Tn5 1÷O÷ inducible (wildtype) 2 We are giventhat A = 648, B + C = 350, and D = 2. We_knowthat the c.f. betweenTn5 and i is 170%, SO we expect A + B = 700. Fromthis we cando somealgebra and saythat B = 52 and C= 298. Therefore the rarest class is b which is the result of four crossovers,giving us that Lac! is between LacO and Tn5. " ... Tn5 i s O+ X X X X ...... i÷ O C X Tn5 O+ i s X X X ... O _ i+ _Tn51÷O÷_ Tn5i _Os c) In this experiment, the donor/host genotype is different, howeverthe gent order/distance is the same,so we expect the samefrequency for each type of transduction. A' Tn5 I ÷Os constitutive 648 B' C' b' Tn5 1_O+ inducible (wildtype) Tn5I _0 + uninducible Tn5 TSO_ constitutive 52 298 2 Therefore, we expect 298 uninducible,650 constitutive and 52 inducible transductants. % d) Since Lac÷never cotransduced with TnS, we can say that the distance between the mutation and Tn5 is more than one phage head length (10kb). So the new mutation is unlinked and likely ¢o be trans-acting. e) A possible reason for the uninducible expression is over-expression of Laci. If there are more repressors present than inducers, there will always be some repressor thatcan bindtothe operatorto preventtranscription. Here are two possible mechanisms: 1) LacI is positively regulated (as in Mal operon), and activator _ mutation is causing constitutive (over-expression) of LacI 2) LacI is negatively regulated (as in Lac operon), and repressormutation is causing constitutive (over-expression)of LacI. To test between these two mechanisms, we can transform a plasmid containing a wildtype version of Lac-. Since activator _ is dominant, and repressor is recessive, if the resulting merodiploid is uninducible we take Model 1, otherwise Model 2. Problem 2: a) urel could be an operator mutation which prevents repressor binding (at). It could also be a repressor mutation (R- or R-d),or a super activator (A_) b) This shows that urel is a dominant mutation. It could still be all of the above with the exception of R-. c) This shows that urel needs to be cis to show the constitutively active phenotype. Therefore it is likely to be a mutation in the Operator/Promoter (at). d) ure2 is a recessive and constitutive mutation, so it is most likely a mutation in a repressor of Urea transcription (R-). Inactivation is usually associated with recessive mutations. _ e) ure3 is uninducible, therefore wildtype ure3 activates transcription The two possible mechanisms are: urea urea -> -> ure3 ure2 --I --l ure2 --I ure3 -> urel/UreA urel/UreA of UreA. ure2-/ure3-=const tutive ure2-/ure3-=uninducible- The top modelisconsistent withthe epistasis test,where ure2isepistatic to ure3. urea r_® ure3 _5_ ure2 __ urel UreA Also: ure3 urea _X_ where ure3transports ureaintothecell. ure2 UreA urel (i: f) ure2 is tightly linked to ure3, so they are likely in the same gene. There are two possible molecular mechanisms suggested by this information: 1- ure3 is a mutation element, therefore ure2 transcript. 2- ure3 is a mutation protein (R), which in the ure2 operator which destroys the urea response rendering it constitutively active, always making the which makes a superrepressor (R_), while ure2 kills the makes them different alleles of the same gene 1. 2. urea__ urea ure3ure2 - _urea urel k "_- ure2/3_mSurea urel