ESum OV-Discussion Worksheet for Students.MCB 250.Fall 2017 .Worksheet 5.Blanke.10-02-27 to 10-06-17 copy PDF

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Zil Patel Discussion ADY

MCB 250 Fall 2017

Discussion Worksheet Week 6 – October 2-6

1. The initiation of DNA replication is tightly controlled in all organisms. Discuss how this is accomplished in E. coli. Consider the following: a) What protein is responsible for initiating DNA replication? DnaA is the specific initiator protein that oriC recognizes in E. coli.

b) What ligands must be bound to this protein for it to work? DnaA-ATP recruits other proteins to form the pre-replication complex, which unzips dsDNA. DnaA recruites DnaC (helicase loader) which loads hexameric DnaB (helicase) onto the single strand DNA. ssDNA-binding proteins stabilize the ssDNA to maintain the replication bubble. DnaB is 5’-3’ helicase that travels on the lagging strand and associates with Primase to generate the RNA primer for the leading strand on the lagging strand

c) Describe 3 ways that the activity of this protein as an initiator is regulated. 1) Control of DnaA-ATP levels - Concentration of Dna-ATP is tightly controlled 2) Control of access to oriC by methylation - oriC is inactive until fully methylated. DnaA-ATP binds fully methylated DNA 3) Control of access to oriC: SeqA binding - The protein SeqA binds hemimethylated oriC, preventing DnaA binding and slowing Dam methylation

d) What would be the consequence of over-initiation of DNA replication? It might lead to extensive - DNA damage, potentially due to head-to-tail replication fork collisions - replication fork collapse, growth, inhibition of cell division and cell proliferation - cell death due to untimely extra initiations and asynchrony in initiation of multiple origin

e) With what cellular process does the initiation of DNA replication have to be tightly coordinated? Cell division - must have an adequate cell size and synthesis of all the proteins needed to form two complete cells

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2. Discuss the role of telomerase in eukaryotic DNA replication considering the following issues: a) What is the “end replication problem” in eukaryotic DNA replication?

b) What is a telomere and what is its function believed to be? Telomeres are the ends of eukaryotic chromosomes. They contain tandem repeats of TG-rich sequences. The sequence TTAGGG provides scaffolding. The structure protects the ends of linear chromosomes from degradation.

c) What is the structure of telomerase and how does it work to compensate for the end replication problem? They contains an RNA that contains sequences, which pair with the repeats in the telomere. It can serve as the template for DNA polymerase activity that extends the 3’ ends by adding new repeats. The extended 3’ end can maintain the length of the chromosome despite the inability to fully replicate it.

d) It has been proposed that loss of telomeric DNA as a result of replication might be responsible for cellular (and organismal) senescence. Since telomerase should be able to solve the problem, why is this proposal a reasonable explanation? most cells contain low levels of telomerase - too low to fix chromosome ends.

e) How might telomerase be involved in cancer? - Senescence = aging due to steady shortening of telomeres - somatic cells lack telomerase and a predetermined number of cell divisions - germ cells contain telomerase - when telomeres get too short, they trigger programmed cell death (apoptosis) - it has been found that telomerase is activated in cancer cells, and these cells are immortal in lab culture

f) Telomerase is a ribonucleoprotein. Is it also a ribozyme? Why or why not? Telomerase is an enzyme; its protein component catalyzes the chemical reaction. Its RNA component only merely serves as a template that influences the order in which dNTPs are added.

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3. E. coli DNA polymerase III is both highly accurate and highly processive. a) Describe how Pol III avoids the incorporation of improperly paired bases. Mistake can be minimized either by not making them in the first place or by correcting them through proofreading. Pol III is designed to have a very low frequency of misincorporations of bases of bases in the first place, and furthermore, to fix misincorporations if they occur.

b) Describe how Pol III avoids incorporating ribonucleotides. the 2’ OH of the ribonucleotide doesn’t fit in the active site

c) Describe the role of protein confor mational changes in contributing to the accuracy of Pol III.

d) What is processivity and how is it achieved in E. coli DNA replication? Processivity is an enzyme’s ability to catalyze “consecutive reactions without releasing its substrate” in DNA replication it is the average number of nucleotides added by DNA polymerase per association event with the template strand. Pol I is much less processive than Pol III because its primary function in DNA replication is to create many short DNA regions rather than a few very long regions.

4. A promoter for an E. coli gene that is transcribed by a s-70 RNA polymerase has the following sequence: -30 -20 -10 +1 | | | | 5’GGCTTTACACTTTATGCTTCCGGCTCGTATGTTGTGTGGA 3’CCGAAATGTGAAATACGAAGGCCGAGCATACAACACACCT 3

The transcription start site (+1) is identified. a. Identify the –10 and –35 sequences. consensus –10 and –35 sequences?

How close are they to the

TATGTT at -12: 4/6 match to consensus TATAAT TTTACA at -36: 5/6 match to consensus TTGACA

b. What is the spacing between the –10 and the –35 sequences? How does this compare with the consensus spacing? The spacer is 18 nucleotides. Consensus is 17, but usually 17 plus/minus 2 also works

c. The sequence of bases in a transcribed RNA is identical (except for U's instead of T's) to the non-template strand. Explain. RNA strand that is synthesized is the complement of the template strand RNA strand is the same as the DNA of non-templated strand

d. Define promoter strength. A strong promoter gives a high rate of transcription and produces lots of product A weak promoter has a low rate of transcription and produces relatively little product

e. Predict the effect of the following mutations on the strength of this promoter (stronger, weaker, no change). The changes indicated show the sequence as top strand/complementary strand. 1) 2) 3) 4) 5)

Replacement of CTT/GAA at –23 to –20 with AAG/TTC. Replacement of G/C at –9 with A/T. Replacement of T/A at –12 with C/G. Replacement of G/C at –38 with C/G. Insertion of AT/TA between –18 and –17.

1) no change; spacer region changed without changing its length 2) stronger because its closer to concensus 3) weaker because its farther from the concensus 4) no change because its outside the -35 region 5) much weaker, probably wont work at all. the spacer goes from 18 to 21

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