Biochem 432: DNA

Question 1

What experiment is well known for the discovery of semi-conservative DNA replication?

Answer 1

The Meselson-Stahl Experiment

Question 2

How was it proved that DNA replicates in semi-conservative manner?

Answer 2

Meselson and Stahl grew E. coli in 15N medium which contained “heavy” DNA. When the DNA was divided in 14N meduim with “light” DNA, a hybrid DNA was produced which contained one strand of 15N DNA and one strand of 14N DNA. The DNA was separated via density centrifugation, 15N DNA is the bottom, 14N & 15N DNA in the middle and 14N on top.

Question 3

Explain what is meant by origin of replication

Answer 3

The origin of replication is where the formation of a replication fork begins–it allows replication to occur in a bi-directional manner.

Question 4

SUPER IMPORTANT: What direction is DNA synthesized????

Answer 4

The daughter strand of DNA is synthesized in the 5’ to 3’ direction on either side of the origin of replication which leads to one strand being a leading strand (the daughter strand is continuously synthesized) and a lagging strand (the daughter strand is discontinuously synthesized via the formation of Okazaki fragment).

Always attach deoxynucleotides to 3’ hydroxy groups which is why the synthesis is always 5’ to 3’ direction (on the new strand)

Question 5

Does DNA polymerase require a primer?

Answer 5

Yes, it requires a primer which contains a 3’-OH group to begin at the first DNA polymerase-catalyzed reaction and may be produced from DNA or RNA–although RNA primers are used in DNA replication in cells.

Question 6

Explain the oriC in E. coli

Answer 6

Initiation of DNA replication in E. coli occurs at a specific site on the circular genome called oriC (origin).

To initiate replication, DNA at oriC must be separated.

Several steps involve ATP binding or hydrolysis.

Two key elements of oriC include three 13-bp repeats and four 9-bp repeats.

Question 7

What is the use of methylation via DAM during replication?

Answer 7

The OriC contains 11 sites that are recognized by the enzyme Dam methylase. Methylation by Dam is used to distinguish the original strand.

The methylation is used to determine the template strand.

Question 8

What is the use of SeqA during the initiation of replication?

Answer 8

The protein SeqA binds to hemimethylated DNA in the oriC prevents a second round of initiation from occurring before the first round has been completed.

Question 9

What is the function of DnaB helicase?

Answer 9

DnaB (helicase) unwinds the DNA during synthesize

Unwinds the DNA double helix ahead of the replicating complex

hexameric 5’-3’ helicase forming a barrel-like structure

ATP-dependent

Involved in repair and replication

Question 10

Describe the function of DNA Pol III core?

Answer 10

Attached to the helicase (DnaB) and consists of two multisubunit cores which are associated with the β-clamp.

Function: Synthesizes nascent DNA on leading and lagging strands

Mg2+ is a cofactor

NEEDS: Primers, templates, deoxyribonucleotides

The 3’-hydroxyl group is a nucleophile that attacks the α-phosphoryl group of incoming dNTP (5’ position); Hydrolysis of PPi (irreversible).

Have nuclease acitivity that allows them to correct mismatch bases

Question 11

Explain the use of Gyrase (aka Topoisomerase)

Answer 11

Stabilizes the extra coiling from the unwinding of the DNA at the replication fork

Relieves torisonal stress in front of the fork

Question 12

Explain DnaG (aka Primase)

Answer 12

Synthesizes primers for the LAGGING strand of DNA

Synthesizes RNA primers on the lagging strand

Question 13

What is the use of Single-stranded DNA binding proteins (SSBs)

Answer 13

Binds to single-stranded DNA to prevent reannealing of the DNA double helix

Question 14

What is the β-clamp and how does it function?

Answer 14

Prevents DNA polymerase III from dissociating from the DNA

Question 15

What is the complex that is comprised of both primase and helicase?

Answer 15

Primosome

Question 16

What is the use of Mg2+ for DNA polymerase?

Answer 16

Mg2+ ions are there to help stabilize the charges as well as promote the deprotonation of the 3’ OH

The DNA polymerase reaction adds a deoxynucleotide to the 3’ end of the growing DNA chain. The Mg2+ ions stabilize the negative charges on the deoxynucleotide and assist in deprotonation of the 3’-OH by a base. The 3’-oxygen of the growing DNA strand serves as the nucleophile, displacing the pyrophosphate from the deoxynucleoside triphosphate (dNTP) in the active site. The product is a DNA strand that has been extended by one nucleotide in the 3’ direction.

Question 17

What is the Pol I Klenow fragment?

Answer 17

The Pol I Klenow fragment has both polymerizing and editing modes. In the polymerizing mode, nucleotides are added to the end of the growing chain in the polymerase active site. In editing mode, the newly added nucleotide is flipped out of the polymerase active site and into the exonuclease site, where it is excised from the growing DNA strand.

Proteolytic cleavage of DNA polymerase I results in two fragments:
1. 5’ to 3’ exonuclease activity – can both polymerize the DNA and remove it too
2. Klenow fragment which contains polymerizing and editing modes

Question 18

What is Reverse Transcriptase?

Answer 18

An enzyme used to convert RNA to DNA and used in HIV replication because HIV is a virus which turns RNA molecules into DNA molecules. It is highly prone to error, and no proofreading function (the typical error rate 1 out of 2000 nucleotides) and contains a similar structure to DNA pol I.

Question 19

How does the eukaryotic and prokaryotic replication fork differ in relation to the necessary enzymes?

Answer 19

The eukaryotic and prokaryotic replication forks are largely similar, although two different DNA polymerases are required for leading strand (Pol Epsilon) and lagging strand (Pol Delta) synthesisin eukaryotes. A third DNA polymerase (Pol alpha) is associated with primase. Moreover, unlike prokaryotic DNA, eukaryotic DNA is packaged into nucleosomes.

Question 20

How does DNA replication relate to the cell cycle in euykaryotes?

Answer 20

Cyclin-dependent protein kinases (CDKs) control eukaryotic DNA initiation events. CDK activation is required in multiple steps for conversion of the pre-RC to an active replication fork at RPCs. The presence of different cyclins at different times during the cell cycle controls initiation of DNA synthesis at origins.

Cyclin E activates CDK2 present late in G1 and early in S

Question 21

Describe Telomeres

Answer 21

Telomeres are sequences at the ends of eukaryotic chromosomes that help stabilize the chromosome. Telomere shortening is a major factor in aging.

Telomerase adds telomeric DNA sequences to the ends of chromosomal DNA. Length of telomeres decreases with each division without telomerase. Increased activity can cause tumorigenesis (observed in tumor samples).

Limited to undifferentiated embryonic stem cells, male germ cells, and activated lymphocytes in humans.

Regulation of telomere function is an active area of current research.

Question 22

Describe the Telomerase mechanism

Answer 22

Telomerase: complex of RNA and protein

TERC: telomerase RNA component, an ncRNA

TERT: telomerase reverse transcriptase

Telomerase binds to DNA using base pairing between the telomerase RNA template and the DNA

TTAGG is repeated a few thousand times in humans –> Telomerase RNA template: 3’-CCAAUCCC-5’. The template is used to synthesize a complementary strand of DNA in a 5’ to 3’ direction. Primase then adds a new RNA primer to facilitate DNA synthesis by the DNA polymerase complex and the new RNA primer synthesized via primase is removed by exonucleases.

Question 23

Which of the following is true:
1. One of the important function of telomere is to protect the end of linear chromosome.

2. DNA molecules serve as template for telomerase to add telomere at the end of chromosome.

3. DNA synthesis occurs in the mitotic phase of cell cycle.

4. Telomerase is a protein enzyme that is a complex of different proteins with no RNA molecules.

Answer 23

1. One of the important function of telomere is to protect the end of linear chromosome.

FALSE: DNA molecules serve as template for telomerase to add telomere at the end of chromosome.
— WHY? Telomerase uses an RNA template

FALSE: DNA synthesis occurs in the mitotic phase of cell cycle.
—WHY? S-phase and early G1 phase of the cell cycle

FALSE: Telomerase is a protein enzyme that is a complex of different proteins with no RNA molecules.
WHY? TERC is the RNA component.

Question 24

Explain/Describe BRCA1

Answer 24

BRCA1 (Breast cancer type 1 susceptibility protein) is a mutation which promotes DNA double-strand break (DSB) repair

Question 25

Name five causes of DNA damage

Answer 25

1. Nucleotide mismatches can occur during DNA replication (In humans, there are about 0.1-1 mutations per genome per replication) and spontaneous deamination
2. Radiations (patients and interventional radiologists)
3. Exposure to ultraviolet light (sunburn a risk for skin cancer)
4. Biological factors (ROS, virus-transformation, gene editing)
5. Exposure to certain chemicals (tobacco smoking; smoking a pack of cigarettes a day for a year: lung, 150 mutations).

Question 26

Describe the Ames test and its importance

Answer 26

The Ames test is a biochemical test to determine whether or not a substance is mutagenic.

First, rat liver cells are isolated and used to prepare an exymatically active protein extract which is then mixed with the test compound to facilitate chemical modifications. Next, his- bacteria is spread onto an agar plate with minimal histidine because the his- strain of bacteria lacks the ability to produce histidine because of mutations in the genes for histidine synthesis. The test compound and rat liver extract mixture is filtered onto the agar plate and incubated. Then the number of his+ revertant colonies are counted because the presence of his+ colonies give evidence of chemically induced DNA mutations.

Question 27

What is one of the most common types of DNA damage?

Answer 27

Spontaneous deamination of cytosine which produces uracil. Uracil is not present in DNA so it is removed by glycosylase enzymes, generating an abasic site.

There are two different outcomes of cytosine deamination depending on whether replication occurs before or after repair. If replication occurs with the uracil in place, one of the daughter strands will incorporate an adenine base as a complement to the uracil, which will result in conversion of a C-G base pair to T-A. If replication takes place with an abasic site present, a cytosine is inserted opposite the abasic site. This results in a C-G to G-C mutation.

In simple terms: There are two outcomes of cytosine deamination:
1. C-G to T-A mutation (before replication occured)
2. C-G to G-C mutation (after replication occured)

Question 28

What is Glycosylase and why is it important?

Answer 28

Glycoslyase removes uracil in DNA molecules after deamination of cytosine occurs.

Question 29

Explain/Describe Base excision repair mechanism and when it is employed.

Answer 29

Base excision repair mechanism is employed when only ONE base is damamged (Deamination of cytosine)

Responsible for removal and replacement of individual bases that are damaged by various chemical reactions, including damage by reactive oxygen species (ROS)

The removal of the uracil base by Uracil DNA glycosylase to create an abasic site. An endonuclease cuts the DNA strand containing the abasic site on the 5’ side of the lesion, generating an abasic deoxyribose phosphate (dRP)

Question 30

What is one of the most common outcomes of UV radiation?

Answer 30

Pyrimidine dimers which are the primary causes of melanomas in humans which may cause a stalled replication fork and possibly a double-strand break

Question 31

Explain/Describe Nucleotide Excision Repair

Answer 31

Used for large lesions that distort the helical nature of DNA: pyrimidine dimers

Nucleotide excision repair in E. coli proceeds in a similar manner to base excision repair, with recognition of the damage, excision of a string of nucleotides surrounding the damage, filling of the resulting gap, and ligation of the nick.

Specifically: UvrAB complexes scans for errors in DNA and once a lesion is recognized, UvrA dissociates leaving UvrB bound to the lesion. Then UvrC induces UvrB in order to cut the DNA backbone 4-5 nucleotides away on the 3’ side of the lesion. UvrC also cuts the DNA 8 nucleotides away on the 5’ side of the lesion. UvrD helicase removes the resulting 12-13 base section of DNA. Pol I fills the gap left by the damaged DNA and Ligase seals the nick to finish the process.

Pyrimidine Dimers are typically caused by UV exposure (Double Strand Break)

Question 32

Explain/Describe Mismatch Repair in Prokaryotes

Answer 32

Mismatch repair relies on MEHTYLATION

Dam Methylase methylates the parent strand by inserting CH3 at adenines in the GATC sequence in E. Coli

The newly synthesized strand in unmethylated — Replication errors reside in the unmethylated strand

The methylation-directed mismatch repair system will cleave the un-methylated strand in the initial part of the repair process.

Question 32

Explain ROS

Answer 32

UV light can also react with other molecules within the cell, particularly riboflavin and tryptophan. This produces reactive oxygen species (ROS), including hydroxyl radicals and singlet oxygen.

Both of these agents can convert guanine into 8-hydroxyguanine and thymine into thymine glycol.

8-hydroxyguanine and its tautomer 8-oxoguanine can base pair with adenine, and cause a G-C to T-A conversion

Thymine glycols often result in either single-strand or double-strand DNA breaks.

ROS is also byproducts of metabolic pathways.

Question 33

Explain DNA alkylation

What is the impact of DNA alkylation on DNA synthesis?

Answer 33

Alkylating agents can react with a single guanine residue to produce ethenoguanine adducts, which severely alter the normal base pairing and interfere with DNA synthesis. Two adjacent guanine residues can be cross-linked by reaction of the ethenoguanine adduct with the exocyclic amine of the second guanine. This type of lesion changes the structure of DNA, making replication past the lesion difficult or impossible.

The most common site of alkylation is N-7 of guanine.

O-6 and O-4 alkylation yield stable adducts.

Most of the alkylating agents that attack DNA are produced by chemical reactions within cells.

Question 34

What is MGMT?

Answer 34

O6-methylguanine-DNA methyltransferase (MGMT)

The second DNA direct repair mechanism used to removed the CH3 group from the O6-methylguanine

“Suicide enzyme”

Works by suicide inhibition
MGMT is referred to as a “suicide enzyme” because during the reaction with O6MeG, it becomes alkylated itself and cannot be regenerated to its active form. MGMT uses an active site cysteine to remove the methyl group from the guanine, with the cysteine residue being converted to 5-methylcysteine. The only way to convert the 5-methlcysteine residue back to cysteine is by digestion of the protein and metabolism of the free 5-methylcysteine.

Question 35

Describe Single Strand Breaks

Answer 35

Single-strand DNA breaks occur at the rate of tens of thousands per day, largely as a result of reactive oxygen species, alkylating agents, or damage to the deoxyribose.

A single-strand DNA break in eukaryotic DNA is located by poly(ADP-ribose) polymerase 1 (PARP1), which signals XRCC1 to bind to the site.

Question 36

What are the two methods to repair double-strand breaks?

Answer 36

1. Homologous Recombination: only be used during the late S phase or G2 phase, as it requires the presence of an intact sister chromatid as a template for DNA polymerase.
2. Nonhomologous end joining: be used during any portion of the cell cycle, although it is most active during G1. Does not require a homologous template.

Question 37

Which of the following are false?
1. In prokaryotes, the mismatch in the newly synthesized strand is repaired by methylation-directed mismatch repair mechanism.

2. DNA polymerase is not required for mismatch repair in prokaryotes.

3. Spontaneous deamination of cytosine is repaired by base excision repair.
   
4. Pyrimidine dimers are repaired by nucleotide excision repair.

Answer 37

2. DNA polymerase is not required for mismatch repair in prokaryotes.
   —DNA Polymerase III is required for mismatch repair

Question 38

Explain DNA recombination

Answer 38

Segments of DNA can rearrange their location within a singlular chromosome, from one chromosome to another and even between organisms.

Enhances genetic diversity

Repairs DNA, Segregation of chromosomes during meiosis, movement of viral genes to a host gene (e.i HIV) and the rearrangement of immunoglobulin genes

Occurs in the S and G2 phase of the cell cycle

Question 39

Explain Homologous Recombination

Answer 39

Occurs during meiosis

Requires a Double-Stranded break

Holliday junction: region of a quadruplex DNA where four different DNA strands come together and are moved along the now-joined chromosomes by helicase activity (aka branch migration). Revolase resolves Holliday junctions in order to separate the joined chromosomes.

Question 40

Homologous recombination repairs what kind of DNA damage?
- Single Strand Break
- Double Strand Break
- Mismatched Pairs
- Pyrimidine Dimers

Answer 40

Double Strand Break