DNA Polymerase and Replication P2

Question 1

What does DNA polymerase require to carry out DNA polymerization?

Answer 1

DNA polymerase requires a template strand to guide polymerization, following the Watson and Crick base-pairing rules. It reads the single-stranded template in the 3’ → 5’ direction and builds the complementary sequence in the 5’ → 3’ direction.

Question 2

What role does DNA helicase play in DNA replication?

Answer 2

DNA helicase unwinds the parental double-stranded DNA to generate single-stranded templates. This process requires ATP and can be studied through an in vitro helicase assay.

Question 3

Why is single-stranded DNA unstable once unwound, and how is this instability managed during replication?

Answer 3

Single-stranded DNA is unstable because the nitrogen-rich bases are energetically costly to expose, making the DNA prone to re-annealing. To prevent this, single-stranded binding proteins cover the single-stranded DNA, stabilizing it and preventing re-annealing.

Question 4

How accurate is DNA replication in E. coli, and how often do errors occur?

Answer 4

DNA replication in E. coli is highly accurate, with errors occurring once every 1 in 10⁹ to 1 in 10¹⁰ nucleotides added. This equates to an error occurring only once per 1,000 to 10,000 replications.

Question 5

How does the active site of DNA polymerase contribute to replication fidelity?

Answer 5

The active site of DNA polymerase discriminates based on base-pair geometry. It catalyzes the incorporation of correctly matched bases and contorts when incorrect base pairs are introduced, making incorrect pairings less favorable.

Question 6

What is the role of DNA polymerase’s exonuclease proofreading activity?

Answer 6

DNA polymerases (I, II, and III) have intrinsic 3′→5′ exonuclease proofreading activity. This allows the enzyme to remove an incorrectly added nucleotide, preventing the continuation of replication with errors. The incorrect base is removed and the correct nucleotide is added in its place.

Question 7

What happens if a non-canonical base pairing makes it past the proofreading site in DNA polymerase?

Answer 7

If a non-canonical base pairing bypasses the proofreading site, the error cannot be reversed once another phosphodiester bond is formed, making it difficult to correct the mistake through the polymerase’s proofreading mechanism.

Question 8

How much does proofreading activity improve replication accuracy, and how does mismatch repair contribute further?

Answer 8

Proofreading improves replication accuracy by 10²- to 10³-fold. Even after proofreading, one error remains in every 10⁶ to 10⁸ bases. Mismatch repair, which occurs post-replication, further improves accuracy by identifying and correcting distortions between the parent and newly synthesized strands.

Question 9

What are the key steps involved in the formation of the phosphodiester bond during DNA replication?

Answer 9

During DNA replication, the incoming nucleotide forms hydrogen bonds with the template nucleotide and creates a phosphodiester bond with the 3’ OH of the primer or previously incorporated nucleotide. The process occurs only in the 5’ → 3’ direction and is stabilized by magnesium ions.

Question 10

Why is high fidelity in DNA replication important for the organism?

Answer 10

High fidelity in DNA replication is crucial because errors can lead to mutations, which are often detrimental to the organism. Accurate replication ensures genetic stability and minimizes harmful mutations.

Question 11

What 4 mechanisms contribute to the high accuracy of DNA replication?

Answer 11

The high accuracy of DNA replication is achieved through:

1.Base complementarity (correct base incorporation).
2.Shape discrimination in the active site.
3.Exonuclease proofreading activity that removes incorrect bases immediately after incorporation.
4.Mismatch repair mechanisms that correct errors after synthesis.

Question 12

What are the 5 essential components and conditions required for DNA polymerization?

Answer 12

DNA polymerization requires:

A single-stranded template generated by helicase.
Single-stranded binding proteins to stabilize the template.
Free nucleotides (dNTPs).
A 3’ OH from a primer or previously incorporated nucleotide.
Magnesium ions to stabilize the reaction.

Question 13

What is the process by which DNA is faithfully replicated each cell cycle?

Answer 13

DNA replication involves unwinding the two parental strands and synthesizing complementary daughter strands, resulting in two identical DNA molecules. The structure of DNA provides clues to this mechanism, allowing for faithful replication.

Question 14

What were the three models proposed for DNA replication before the mechanism was understood?

Answer 14

The three models of DNA replication were:

Conservative
Semi-conservative
Dispersive

Question 15

What was the purpose of the Meselson-Stahl experiment, and what technique did they use to distinguish between old and new DNA?

Answer 15

The Meselson-Stahl experiment aimed to determine which model of DNA replication was correct. They used nitrogen isotopes (heavy
15𝑁 and light 14𝑁) and separated DNA by density to distinguish between old (heavy) and new (light) DNA strands.

Question 16

Describe the Meselson-Stahl experiment setup and its 3 key steps.

Answer 16

1.E. coli was grown in the presence of heavy nitrogen (15𝑁) to label its DNA.
2.The bacteria were then transferred to light nitrogen (14N) medium and allowed to replicate.
3.DNA was separated by density using a cesium chloride gradient to distinguish between heavy and light DNA.

Question 17

What were the key findings of the Meselson-Stahl experiment after one and two generations of replication?

Answer 17

After one generation, DNA was of intermediate density, ruling out the conservative replication model.
After two generations, the DNA consisted of both light DNA and intermediate DNA, supporting the semi-conservative replication model.

Question 18

How does the semi-conservative model of replication explain the results of the Meselson-Stahl experiment?

Answer 18

In the semi-conservative model, each parental DNA strand serves as a template for a new daughter strand. After one generation, the DNA consists of one old (heavy) and one new (light) strand, creating an intermediate density. After two generations, half of the DNA is fully light and half remains intermediate.

Question 19

How were heavy and light DNA distinguished in the Meselson-Stahl experiment?

Answer 19

Heavy and light DNA were distinguished using a cesium chloride density gradient. Heavy DNA, labeled with 15𝑁, travels further down the gradient than light DNA (14𝑁), allowing clear separation of DNA with different densities.

Question 20

What is bidirectional DNA replication, and how was it validated experimentally?

Answer 20

Bidirectional DNA replication means that replication proceeds in both directions from a single origin. This was validated through experiments showing replication forks at both ends of bacterial chromosomes and electron micrographs capturing expanding loops, which are consistent with bidirectional replication.

Question 21

What is the origin of replication, and how is it identified during DNA replication?

Answer 21

The origin of replication is the specific location on the DNA where replication begins. It is often A-T rich due to the ease of melting. Denaturation mapping, which selectively denatures A=T-rich sequences, is used to identify landmarks like the origin along the DNA molecule.

Question 22

What is denaturation mapping, and what role does it play in understanding replication?

Answer 22

Denaturation mapping involves selectively denaturing A=T-rich sequences, which provides reproducible patterns of single-strand bubbles along the DNA molecule. This technique is used to identify origins of replication, as these are the regions where replication loops are initiated.

Question 23

What does the semi-conservative model of replication indicate about DNA strands during replication?

Answer 23

In the semi-conservative model, each newly replicated DNA molecule contains one parental (old) strand and one newly synthesized (daughter) strand. This mechanism ensures the faithful replication of genetic material in each new DNA molecule.

Question 24

What is the significance of A-T rich regions in DNA replication, particularly at the origin of replication?

Answer 24

A-T rich regions are significant because A-T base pairs have fewer hydrogen bonds than G-C pairs, making them easier to unwind. These regions are often found at the origin of replication, facilitating the initiation of DNA unwinding and replication.

Question 25

Why did the Meselson-Stahl experiment rule out the conservative model of DNA replication after one generation?

Answer 25

The conservative model predicts two distinct DNA bands: one fully heavy (parental) and one fully light (new). However, after one generation, the Meselson-Stahl experiment showed only intermediate DNA, which is inconsistent with the conservative model but supports the semi-conservative model.

Question 26

What would be the expected results if DNA replication followed the conservative model after two generations?

Answer 26

If DNA replication were conservative, after two generations, there would be two distinct DNA bands: one representing fully heavy (parental) DNA and one representing fully light (daughter) DNA. However, the semi-conservative model showed intermediate and light bands, not separate heavy and light bands.

Question 27

What are the 3 universal principles of DNA synthesis?

Answer 27

The universal principles of DNA synthesis are:

1.DNA replication is semi-conservative.
2.Replication begins at an origin and usually proceeds bidirectionally.
3.DNA synthesis occurs in the 5’ to 3’ direction and is semi-discontinuous

Question 28

What does it mean for DNA replication to be semi-discontinuous?

Answer 28

Semi-discontinuous replication means that while one strand (the leading strand) is synthesized continuously in the 5’ to 3’ direction, the other strand (the lagging strand) is synthesized in short fragments (Okazaki fragments), also in the 5’ to 3’ direction, which are later joined together.

Question 29

Before 1967, what was the favored model of DNA replication, and how does it differ from the modern understanding?

Answer 29

Before 1967, the favored model of DNA replication suggested that both strands were replicated continuously: one in the 3’ to 5’ direction and the other in the 5’ to 3’ direction. Modern understanding shows that DNA replication is semi-discontinuous, with the lagging strand synthesized in Okazaki fragments.

Question 30

How did Okazaki and colleagues experimentally distinguish between models of replication?

Answer 30

Okazaki’s team pulsed E. coli with tritiated (radioactive) thymidine (3H-dTTP) to label newly synthesized DNA. They then separated the DNA based on size using an alkaline sucrose density gradient, allowing them to observe the emergence of short DNA fragments over time, supporting the semi-discontinuous model of replication.

Question 31

What is the role of DNA Polymerase III in E. coli replication?

Answer 31

DNA Polymerase III in E. coli is responsible for the faithful replication of the genome. It synthesizes both the leading and lagging strands of DNA during replication, but it requires other proteins to assist in the process, such as helicase, primase, and the clamp loader.

Question 32

What are 4 reasons why DNA polymerase can’t replicate the genome?

Answer 32

DNA polymerase cannot replicate the genome alone because it cannot:

1.Separate DNA strands to generate a single-stranded template.
2.Stabilize the single-stranded DNA.
3.Start replication without a free 3’ OH group.
4.Replicate in the 3’ to 5’ direction.
It is also too slow and not processive enough without the assistance of other proteins like the clamp loader.

Question 33

What is the function of the clamp loader in DNA replication?

Answer 33

The clamp loader is responsible for loading the sliding clamp onto the DNA-RNA hybrid. This clamp holds DNA Polymerase III in place during replication, increasing its processivity. ATP binding and hydrolysis are required to open the clamp and attach it to the DNA substrate.

Question 34

How is lagging strand synthesis different from leading strand synthesis?

Answer 34

Lagging strand synthesis is discontinuous, occurring in short fragments called Okazaki fragments, which are synthesized in the 5’ to 3’ direction away from the replication fork. These fragments are later joined together. Leading strand synthesis is continuous and follows the replication fork in the 5’ to 3’ direction.

Question 35

What proteins are involved in synthesizing the lagging strand, and what are their roles?

Answer 35

Primase: Adds RNA primers to generate free 3’ OH groups required for DNA synthesis.
Clamp loader: Loads the sliding clamp onto the RNA-DNA hybrid.
DNA Polymerase III: Synthesizes Okazaki fragments.
DNA Polymerase I: Removes RNA primers and replaces them with DNA.
DNA Ligase: Seals the remaining nicks between Okazaki fragments, forming a continuous DNA strand.

Question 36

What are Okazaki fragments, and how are they processed during DNA replication?

Answer 36

Okazaki fragments are short DNA fragments synthesized on the lagging strand during replication. After synthesis, the RNA primers are removed by DNA Polymerase I, which replaces them with DNA. The remaining nicks between fragments are sealed by DNA ligase.

Question 37

What happens if the sliding clamp is not loaded properly onto the DNA?

Answer 37

If the sliding clamp is not properly loaded onto the DNA, DNA Polymerase III cannot stay attached to the DNA and will fall off, drastically reducing its processivity and slowing down replication.

Question 38

Why does DNA polymerase require a free 3’ OH group to begin synthesis?

Answer 38

DNA polymerase requires a free 3’ OH group to add nucleotides because it can only extend an existing strand in the 5’ to 3’ direction. Primase generates this free 3’ OH by synthesizing an RNA primer that provides the starting point for DNA synthesis.

Question 39

How are leading and lagging strands coordinated during replication to ensure synchronized progress?

Answer 39

Although the leading strand is synthesized continuously and the lagging strand discontinuously, the overall speed of synthesis is coordinated. The lagging strand forms a loop to allow the replication machinery to move in the same direction as the leading strand. Primase is the rate-limiting step, as it takes time to add primers for Okazaki fragments.

Question 40

What are the two main problems that arise during lagging strand synthesis, and how are they solved?

Answer 40

Two main problems during lagging strand synthesis are the presence of RNA primers and gaps between Okazaki fragments.

Problem: RNA primers need to be removed.
Solution: DNA Polymerase I removes RNA primers and replaces them with DNA.
Problem: Gaps (nicks) remain between Okazaki fragments.
Solution: DNA ligase seals the nicks, creating a continuous DNA strand.

Question 41

What is the role of single-stranded DNA binding proteins (SSBs) during replication?

Answer 41

Single-stranded DNA binding proteins (SSBs) stabilize the single-stranded regions of DNA that are exposed after helicase unwinds the double helix. This prevents the single strands from re-annealing or forming secondary structures before they are replicated.