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BCHM-3050 Lecture Notes - Lecture 5: Dna Replication, S Phase, Dogma

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scarletmouse537
23 May 2018
School
Clemson University
Department
Biochemistry
Course
BCHM-3050
Professor
Geoffrey Ford

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OVERVIEW OF CENTRAL DOGMA
DNA Replication: template directed duplication of DNA genome prior to
cell division (Copy the DNA)
1.
Transcription: DNA template directed biosynthesis of RNA (create
RNA from copied DNA)
Reverse transcription: RNA template directed biosynthesis of DNA
a.
2.
Translation: mRNA template directed biosynthesis of proteins (Create
protein from mRNA)
3.
Brief overview of Replication process
DNA strands separated at regions called forks
1.
Replication begins at one or more fixed site called replication origins
2.
After replication each new strand receives one of the parent strands
3.
Replication in bacteria
Replication initiates from a fixed origin but proceeds in both directions
(bi-directional)
•
Unidirectional: termination and initiation occur at the same site
•
Bidirectional: occurs at different sites
•
Eukaryotic replication
Bidirectional and proceeds from several fixed origins
•
Forks advance till they meet another fork travelling in the opposite
direction
Coming from the replication origin
○
•
Origin programmed to initiate replication at fixed time in S phase
•
Prokaryotes v Eukaryotes: DNA Replication
Similarities
Bi directional replication
○
Origin sites ( 3 may vary)
○
Replication forks
Replisomes but number or type of protein varies
§
○
Leading and lagging strands
Plays into bidirectional replication
§
○
DNA polymerase synths
5' to 3' direction
§
○
Primers are required
For free 3' -OH end
§
○
•
Differences:
Prokaryotes:
Simple packaging of DNA
§
Circular chromosome
No ends
□
§
Small genome
§
Low number of DNA polymerase copies
§
○
Eukaryotes:
Bigger genome
§
Complex packaging
Histones, proteins etc
□
§
Linear chromosome
§
○
•
DNA polymerase
DNA polymerase synthesizes DNA in 5' to 3' direction
•
DNA polymerase catalyzes the nucleophilic attack of 3' OH side of the
primer to the 5' end of the dNTP strand
dNTP: general template for base pair
Ex. dGTP, dATP, dCTP, dTTP
§
○
Attacks the alpha phosphate of the dNTP
○
•
Requires:
DNA template
Need to make a copy of this
§
○
RNA or DNA primer
In nature DNA uses a RNA primer
§
In the lab, DNA primer is used
§
○
dNTPs
○
•
DNA polymerase is capable of proofreading
3' to 5' reverse
○
5' to 3' forward
○
•
Structure and function of polymerase
Polymerase III, I, IV: polymerase in bacteria
Multimer: collection of different protein subunits
○
•
Polymerase alpha, delta and epsilon: polymerase in eukaryotes
Only one protein subunit
Depending on the protein domain the subunit has different
function
§
○
Polymerase delta has similar function as polymerase I
○
•
Polymerase clamps onto DNA
"Palm, finger and thumb"
○
•
DNA Polymerase I
Function:
Polymerase
○
Exonuclease activity (3' to 5')
○
Exonuclease activity (5' to 3')
○
•
Lab used polymerase I
Has klenow fragment
○
Klenow fragment can not remove primers
○
•
2 metal mechanisms of Polymerase
Mg2+ ions stabilize the attack of the primer to the incoming dNTP
Help with coordination
○
•
Sanger sequencing
•
DNA synthesis is discontinuous at lagging strand
polymerase can not continuously synthesize DNA in the 3' to 5' direction
Leaves fragments and tiny gaps
○
•
Okazaki fragments: small discontinuous fragments in the 3' to 5' direction
Link DNA together using ligase
○
•
DNA ligase rxn
5' phosphate attacks the AMP of the ligase 1.
3' OH attacks 5' end of the phosphate
Releases AMP a.
2.
The attack of AMP gives the whole rxn enough energy
Allow 3' OH end to stretch and attack the 5' end of the phosphate
○
•
Removal of RNA primers
RNA primer: help with initiation of replication
•
Primer has to be removed
Cant have RNA in the already existing DNA chain
○
•
Primer is removed and ligation of its hole allows for the replacement with
dNTP
Process facilitated by DNA pol I (bacteria) and delta polymerase
(eukaryotes)
○
•
Replication Process (in review)
Helicase unwinds helix 1.
Single strand binding proteins stabilize parental DNA2.
Leading strand synthesis begins in 5' to 3' direction 3.
The lagging strand synthesis is discontinuous, forming okazaki fragments 4.
RNA primers are removed and gaps are replaced by DNA 5.
Chapter 22: REPLICATION
Sunday, May 20, 2018
7:21 PM
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Document Summary

Dna replication: template directed duplication of dna genome prior to cell division (copy the dna) Transcription: dna template directed biosynthesis of rna (create. Reverse transcription: rna template directed biosynthesis of dna. Translation: mrna template directed biosynthesis of proteins (create protein from mrna) Replication begins at one or more fixed site called replication origins. After replication each new strand receives one of the parent strands. Replication initiates from a fixed origin but proceeds in both directions (bi-directional) Unidirectional: termination and initiation occur at the same site. Forks advance till they meet another fork travelling in the opposite direction. Origin programmed to initiate replication at fixed time in s phase. Replisomes but number or type of protein varies. Dna polymerase synthesizes dna in 5" to 3" direction. Dna polymerase catalyzes the nucleophilic attack of 3" oh side of the primer to the 5" end of the dntp strand dntp: general template for base pair. In the lab, dna primer is used dntps.

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Related Questions

1. In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that
DNA is a double helix.
DNA may be taken up by bacterial cells and be active.
DNA is present in bacteriophage T2.
eukaryotic cells may be genetically transformed.
DNA binds to the cell wall of the bacterium.
2. In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.
temperatures high enough to kill cells
enzymes that hydrolyze proteins
enzymes that hydrolyze DNA
enzymes that hydrolyze polysaccharides
enzymes that hydrolyze RNA
3. Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?
Pure RNA from S cells
Cell-membrane extracts from S cells
A mixture of pure RNA and protein from S cells
Cell-free extracts from S cells
Pure protein from S cells
4. A-T base pairs in a DNA double helix
form three hydrogen bonds with each other.
are more heat-stable than G-C base pairs.
are of a substantially different length than G-C base pairs.
are not accessible to DNA binding proteins.
are chemically distinct from G-C base pairs.
5. If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?
27
50
0
73
23
6. The uniform diameter of the DNA structure provides evidence for
antiparallel DNA strands.
a right-handed helix.
a double-stranded helix.
3′ end phosphorylation.
purine–pyrimidine pairing.
7. If a sequence of one strand of DNA is 5′-TGACTATC-3′, what is the complementary strand?
5′-ACTGATAG-3′
3′-TGACTATC-5′
3′-ACTGATAG-5′
3′-ACTGATAG-3′
5′-TGACTATC-3′
8. What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?
3′ end phosphorylation
Purine–pyrimidine pairing
Antiparallel DNA strands
Covalent bonds between sugars and phosphate groups
Hydrogen bonding between paired bases
9. If the Meselson–Stahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only 15N and the second only 14N, this result would have indicated that replication was
unidirectional.
dispersive.
conservative.
bidirectional.

semiconservative.
10. The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3′ hydroxyl group (–OH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.
The replication complex will no longer bind to origins of replication.
The DNA polymerase will only incorporate acyclovir molecules.
Acyclovir will bind single-strand binding proteins.
DNA polymerase will not be able to link a successive nucleotide.
Acyclovir will block the activity of the DNA helicase.
11. Which of the following does not demonstrate the stability of the DNA double helix?
Single-strand binding proteins are required to keep the strands apart.
Paired bases form hydrogen bonds.
High temperatures are required to denature it.
DNA synthesis requires energy.
DNA helicases require ATP.
12. What effect would a primase inhibitor have on DNA replication?
DNA polymerase would not be able to add bases to the DNA.
Primase would be blocked from joining the DNA replication complex.
Primase would not be able to provide primers for DNA polymerases.
DNA polymerase would not be able to use DNA as a template.
There would be no effect on DNA replication.
13. To replicate their DNA in a timely manner, most eukaryotic chromosomes
normally have uncoiled DNA.
have multiple origins of replication.
contain linear molecules of single-stranded DNA.
have two replication forks that move in the same direction.
are composed entirely of DNA.
14. Which statement about DNA replication is false?
Okazaki fragments are synthesized as parts of the leading strand.
Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
The sliding clamp increases the rate of DNA synthesis.
Replication forks represent areas of active DNA synthesis on the chromosomes.
Ligases and polymerases function in the vicinity of replication forks.
15. In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.
leading; telomerase
lagging; telomerase
lagging; DNA polymerase I
leading; DNA polymerase I
leading; replicase
16. A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?
DNA polymerase is constantly active.
Primase is able to extend the telomeres.
The mismatch repair system is extra efficient.
The telomerase gene is being expressed.
The cells do not need the ends of their chromosomes.
17. If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?
The replication complex excises the incorrect nucleotide.
The excision repair proteins remove the incorrect nucleotide.
The mismatch repair system scans for base-pairing mismatches.
There is no correction, because nucleotide errors by DNA polymerases are not repaired.
DNA ligase connects the correct nucleotide.
18. Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.
1, 2, 3, 4
2, 1, 3, 4
3, 4, 2, 1
2, 4, 1, 3
4, 2, 3, 1
19. Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.
64
32
12
128
6
20. When double-stranded DNA is heated to temperatures above 90°C, it denatures. Denaturation is a process that
breaks covalent bonds.
hydrolyzes DNA.
separates double-stranded DNA into two single strands.
causes new hydrogen bonds to form.
irreversibly destroys the double helical structure.

BIOCHEM QUESTIONS:

Question 1) (1.5 points) True or false: The initial native DNA is usually in an overwound state, making the initial strand separation during replication energetically favorable, with continued strand separation at the replication fork resulting in underwound DNA that is then fixed by DNA gyrase.

Question 2) (1.5 points) True or false: UV based DNA damage largely causes dimerization of C and T bases that are fixed by the NER DNA repair pathway.

Question 3) (1.5 points) True or false: There is no enzyme within humans that has a reverse transcriptase activity.

Question 4) (1.5 points) True or false: One reason why uracil is not observed in DNA is that deamination of cytosine would result in Uracil and therefore potentially changing a GC base bp to an AT bp after a couple of rounds of replication.

Question #5: (1 point) Which of these statements is correct about the RNA polymerase? a) It makes tRNA, mRNA and rRNA b) It uses an RNA template c) It uses ATP, CTP, GTP and TTP d) It requires a primer

Question #6: (1 point) RNA polymerase grows the RNA chain in the 3’ to 5’ direction. a) True b) False

Question #7: (1 point) The mRNA is copied from the template strand. The mRNA has the exact same nucleotide sequence of the non-template strand except for U instead of T. a) True b) False

Question #8: (1 point) Only a limited portion of the DNA strand is copied by RNA polymerase, and there are initiation and termination signals for transcription. a) True b) False

Question #9: (1 point) The hairpin loop is the only way to conduct transcription termination in Prokaryotes. a) True b) False

5. Alpha amanitin is a drug that binds RNA polymerase andprevents it from functioning in the cell (this is the compound thatis found in poison mushrooms). The reason people die when theyconsume alpha amanitin is:

a) Alpha amanitin prevents gene expression from occurring in thecell

b) DNA replication is directly inhibited by alpha amantin

c) Alpha amanitin binds tightly to the messenger RNAs that aremade in the cell, preventing translation

d) Alpha amanitin binds to nuclear pores preventing thetransport of messenger RNA into the cytoplasm

6. In order for transcriptional initiation to begin in E. colithe following elements are needed:

a) an origin and RNA polymerase

b) initiator protein and a promoter

c) the sigma subunit of RNA polymerase and an RNA template

d) RNA polymerase and a promoter

e) helicase and primase

Questions 7 and 8. You are studying two species of closelyrelated snakes, one species lives in sandy desert area and has paletan skin and the species lives in a volcanic rock area and has darkbrown skin. You a sequence the DNA from each type of snake anddiscover a gene encoding a protein that transports pigment into theskin cells. The pigment transporter protein is responsible for skincolor. When you analyze the DNA coding sequence of the pigmenttransporter from each snake you find these two DNA sequences (hashmarks have been added to indicate the reading frame):

pale tan snake: GCG/TTG/GAC/TGC/ACC (DNA template strand)CGC/AAC/CTG/ACG/TGG (complementary strand) dark brown snake:ACG/TTG/GAC/TGC/ACC (DNA template strand) TGC/AAC/CTG/ACG/TGG(complementary strand)

7. From these sequences you conclude:

a) that the amino acid sequence of the pigment transporters forthe two proteins are almost identical

b) that the amino acid sequence of the pigment transporters forthe two proteins are very dissimilar

c) that the amino acid sequence of the pigment transporters forthe two proteins are identical

8. Since the skin color is different in each snake, you concludefrom the sequence data above that:

a) the genetic code in the skin cells in each of these snakes isdifferent

b) the DNA sequence encoding the entire gene for pigmenttransporters in these two snakes is different

c) that one of the codons in the gene for the pigmenttransporter in these two snakes is different

9. You are analyzing eukaryotic cells that have been exposed tothe radioactive uracil. In most cells the radioactive uracil isinitially localized in the nucleus and then moves from the nucleusto the cytoplasm. However, you observe that in some cells theradioactive molecules stay in the nucleus and never arrive in thecytoplasm. This is most likely a defect in:

a) the transcription of the DNA

b) the processing of the mRNA

c) the translation of the RNA

d) the labeling of the RNA

e) the replication of the DNA

10. Which of the following is true of both eukaryotic andbacterial transcription? (select all that are true)
a) Transcription factors must bind the promoter before RNApolymerase initiates transcription

b) Messenger RNAs must be spliced before translation begins inorder to produce the appropriate gene product

c) During transcription only one DNA strand is used as atemplate to make the messenger RNA

d) Termination of transcription occurs when the RNA polymerasetranscribes a hairpin structure in the RNA followed by a string ofU