Bio-366-Key-Test 3-Fall-2009

Bio-366-Key-Test 3-Fall-2009 - Bio-366 Test-3 October 1...

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1 Bio-366 Test-3 October 1, 2009 Allowed time: 11.00 am to 12.10pm (I hr and 10 min) Name: UTEID: Please be brief and to the point I . A . Briefly indicate the steps involved in the initiation of replication at the E. coli origin and the proteins involved in these steps. During movement of the replication fork, what type of supercoiled domain in built up behind the fork, and in front of it? How is the torsional stress relieved to help forward movement of the fork? DnaA protein binds to the protein A boxes present at the E. coli replication origin region. This opens the origin, and allows recruitment of the DnaB helicase with assistance of DnaC protein. DnaB helicase unwinds DNA by hydrolyzing ATP. The single stranded regions are stabilized by the binding of the single strand binding protein (SSB). The DNA primase and the DNA polymerase complex can be loaded at the origin to start DNA synthesis .
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2 A negatively supercoiled domain is built up behind the replication fork. A compensatory positively supercoiled domain is built up ahead of the fork. The torsional stress of positive supercoiling is relieved by the action of gyrase associated with the replication fork. Gyrase cuts the DNA strands and ligates them back. B . Is the E. coli chromosome replicated by a single replication fork or two replication forks? How many polymerase holoenzyme complexes are present at a replication fork? Why does an advancing replication fork synthesize one strand continuously and the other discontinuously? How are the RNA primers in the discontinuous strand removed? What is the enzyme that carries out this function, and what are the two activities required for this purpose? Replication of E. coli chromosome is bidirectional, carried out by two replication forks. Each fork comprises of two holoenzyme complexes, one to replicate the leading strand and the other to replicate the lagging strand.
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3 The continuous and discontinuous modes of DNA synthesis is the result of the opposite polarity of the template strands, one 5’ to 3’ and the other 3’ to 5’. DNA polymerase can synthesize the new strands only in the 5’ to 3’ direction. The fork moves in one direction. Thus, the 3’-5’ template will be copied in a continuous fashion. The 5’-3’ template is synthesized in pieces (Okazaki fragments) with the help of RNA primers laid down to initiate them. The RNA primers are removed by the Pol I DNA polymerase. It has a 5’ to 3’ exonuclease activity, and also a 5’-3’ DNA polymerase activity. As the enzyme removes the ribonucleotides, it replaces them with deoxyribonucleotides. C . How does the replisome manage to synthesize the two DNA strands in opposite directions while maintaining the movement of the replication fork in the same direction? How is the misincorporation of a non-complementary base in the nascent DNA chain corrected?
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This note was uploaded on 02/21/2010 for the course BIO 5 taught by Professor Mcclelland during the Spring '08 term at University of Texas.

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Bio-366-Key-Test 3-Fall-2009 - Bio-366 Test-3 October 1...

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