Lac operon

Solutions to Practice Problems for Molecular Biology, Session 5: Gene Regulation and the Lac Operon ? ? Question 1 ? ? ? a) How does lactose (allolactose) promote transcription of LacZ? 1) Lactose binds to the polymerase and increases efficiency. 2) Lactose binds to a repressor protein, and alters its conformation to prevent it from binding to the DNA and interfering with the binding of RNA polymerase. 3) Lactose binds to an activator protein, which can then help the RNA polymerase bind to the promoter and begin transcription. 4) Lactose prevents premature termination of transcription by directly binding to and bending the DNA. Solution: 2) Lactose binds to a repressor protein, and alters its conformation to prevent it from binding to the DNA

glucose lactose operator (repressor binding site) PromoterZYA lacZ lacY lacAcrp lacI repressor active activator inactive PromoterIPromotercrp + - mRNA mRNA activator binding site polymerase RNA polymerase RNA operator (repressor binding site) PromoterZYA lacZ lacY lacAcrp lacI repressor active repressor lactose inactive activator inactive cAMP activator active PromoterIPromotercrp mRNA mRNA mRNA LacZ (?gal) LacY LacA activator binding site polymerase RNA polymerase RNA repressor active can metabolize lactose!cAMP + - + lactose + cAMP activator active polymerase RNA DNA DNA The lac operon: a model of gene regulation in E. coli MIT OpenCourseWare http://ocw.mit.edu 7.01SC Fundamentals of Biology Fall 2011

Solutions to Molecular Biology Unit Exam Question 1 Consider the following origin of replication that is found on a chromosome. The sequence of region 1 is shown below. Region 1: 5??CTGACTGACA?3? 3??GACTGACTGT?5? 5? 3? 5? 3? Region 1 ori top ? bottom ? a) Within Region 1, which strand will be the template for leading strand synthesis, the top or the bottom? The bottom strand will be the template for leading strand synthesis. b) If we assume that a lagging strand fragment is made from region 1, what will be its sequence? 5? T G T C A G T C A G 3? c) You examine DNA replication in an E. coli mutant, which has a partially defective DNA polymerase. In vitro