Lecture 6- DNA replication

Question 1

Learning Outcomes

Answer 1

*Define the key terms used in molecular biology located in the glossary.

*Explain why there is a need to maintain high sequence similarity during replication.

*Identify key enzymes needed for replication in prokaryotes and explain their function.

*Describe the steps of replication andidentify which oftheenzymesaboveareinvolved at each step.

*Describetheproperties of the genome that are needed for replication to occur effectively.

Question 2

recap from when I saw you last….

Answer 2

*Prokaryotic genomes usually have 1000-6000 genes.

*Prokaryotes have a high density of coding DNAand that mRNA is polycistronic.

*ProteinshavetointeractwiththeDNAtohaveanyeffectonthegenome,thereforethenon-codingDNA in these genomes are often promoters which provide an interface for that to happen and influence gene expression.

*In prokaryotes genes are organised in operons where genes controlling a process may be located under the control of a single promoter.

Question 3

What is the function of DNA?

Answer 3

Storage of genetic information.

➢Replication
➢Gene expression

The Central Dogma of Molecular Biology
DNA(4 bases)–>mRNA(4 bases)–>Proteins (20 amino acids)

Question 4

what is the key outcome?

Answer 4

Generating protein is the “key outcome”

Question 5

why is Replication accuracy is essential

Answer 5

Changes in nucleic acid sequence causes Changes in amino acid sequence
which in turn changes
-Changes protein shape
-Changes protein function
-Changes protein expression

The mutation rate in bacteria is approximately three nucleotide changes per 1010 nucleotides per cell generation

Question 6

how is Replication based upon the formation of base-pairs

Answer 6

*A key property of nucleic acids which enable them to store information is their ability to form complementarybase pairs.

*The 4 bases can interact by hydrogen bonding in a specific manner:-
-T forms 2 bonds with A
-C forms 3 bonds with G

*Note that each base pair contains one large purine and one small pyrimidine

Question 7

how is DNA replication is semi-conservative

Answer 7

To function as the hereditary molecule, DNA must be replicated accurately.

*During semi-conservative replicationeach new DNA molecule contains half (one strand) of the original DNA.

*Each strand serves as a template for a new complementary strand.

*Each of the two resulting double strands is exactly the same as the parent

Question 8

Replication forks form at Origins of Replication

Answer 8

In prokaryotic origins of replication, where replication forks form, there are specific sequences to attract initiator proteins and contain sequences that are easy to pry apart.

Question 9

Binding and opening of DNA structure at OriC

Answer 9

*OriCcontains easy to separate AT rich regions

*DnaAcan then bind to several sites (some strongly and some weakly) to begin strand separation and formation of the replication fork.

*TheDnaAoligomerrecruitsthe DnaB-DnaCheterodimer

Question 10

Steps of DNA replication

Answer 10

The original copy is a template used to generate a complimentary copy of genomic DNA. This occurs in a stepwise manner:

1.The double helix must be separated into 2 strands by DNA helicase, generating the replication fork (Separation).

2.DNA Primase makes short RNAs (primers) acting as a platform for DNA polymerase (Initiation –leading strand only).

3.The template strand bases are recognised by free bases and then joined together by DNA polymerase (Elongation)

4.Termination is the process by which DNA polymerase dissociates from the DNA template

(at the same time on the lagging strand DNA primase creates primers every 100-200 bp, which are extended by DNA polymerase and joined together by DNA ligase)

Question 11

Step 1a: Separation

Answer 11

Before replication can begin, the helix needs to be separated into 2 template strands by DNA helicase.

DNA helicase binds to a single strand of DNA and spins around it, propelled by the hydrolysis of ATP, physically breaking the hydrogen bonds between bases.

Question 12

Step 1b: Separation

Answer 12

Once separated the single stranded DNA is bound by single stranded binding (SSB) proteins to prevent secondary structures forming

This is especially important on the lagging strand..

Question 13

Step 2: Initiation

Answer 13

Once the replication fork is formed DNA Primasemakes short RNA primers of around 10nt in length which recognise bases at the replication fork

Question 14

Step 2: Elongation

Answer 14

This presents DNA Polymerase with the template that is needed to initiate its action, nucleotides are added by phosphodiester bonds in a 5’ to 3’ direction

Question 15

Meanwhile…on the lagging strand

Answer 15

DNA Polymerasecannot generate a long polymer in the 3’ to 5’ direction so instead DNA primase has tohelp out…

Short Okazaki Fragments are generated by DNA polymerase using the primersas a starting point. This creates short sections of complementary, double stranded (ds) DNA which are then extended to replace the next primerin a 5’ to 3’ direction.

The sections of DNA are stuck together by DNA ligaseto form the second complimentary copy

Question 16

how are Replication forks are bidirectional

Answer 16

*Each replication fork has an asymmetric structure:
*At the leading strand, the daughter copy of DNA is synthesised continuously.
*On the lagging strand, the daughter strand that is synthesized discontinuously, using the Okazaki fragments.

*The structure of these are maintained and created by accessory proteins which include a sliding clamp and DNA Gyrase

Question 17

The sliding clamp

Answer 17

*DNA polymerase cannot maintain contact with the DNA for more than around 50-200bp.

*A sliding clamp is used to keep DNA polymerase attached to the DNA template until it reaches some double stranded DNA

The clamp complex is made of 3 parts:
*One of two parts of the sliding ring binds to the back of DNA polymerase
*The clamp loader binds to both parts of the sliding ring and controls its opening/closing

Question 18

Unwinding can cause tension elsewhere too!

Answer 18

The separation of 2 wound strands, from each other, during replication leads to torsional stress as replication forks move along the DNA

Gyrase (a type II topoisomerase) that formscovalent bonds with both strands, forming a brief double strand break, active where 2 double helixes cross one another. This causes negative supercoiling and relieves torsional stress.

Question 19

Step 4: Termination

Answer 19

*Terminationoccurs when the two replication forks essentially run into each other, at the difsite (opposite side of the genome to Ori)

*DNA is unwound by helicase, which then dissociates dromthe DNA

*Anygaps are filled in using polymerase or okasakifragments andthenligated together by DNA ligase

Covalentbondsthatarelinkingthe2daughtercopiesarebrokenOtherreplication enzymes are unloaded

Question 20

DNA replication: Summary

Answer 20

*The steps are separation, initiation, elongation, termination.

*The enzymeswork together and assemble at origins of replication to establish replication forks at Ori.

*DNA helicase “unzips” the DNA then DNA polymerase is “loaded” onto the DNA by the clamp and loader.

*On the lagging strand DNA primase makes primers that are used by DNA polymerase to generate okasakifragments (in 5’ to 3’ direction, which are then joined together by DNA ligase.

*Termination occurs at the difsite and occurs where replication forks “collide”and enzymes disassemble

Question 21

Test your knowledge

Answer 21

*Which bp are easier to pry apart and why?

*What are the 4 steps of replication? (extension summarise what happens at each step)

*What are the functions of the following enzymes: DNA gyrase, DNA polymerase, DNA ligase, DNA primase, DNA helicase, single stranded binding proteins.

*What do we call it when there is a single base change in sequence?

*What happens when there is an error during replication?

*When might changes in DNA sequences be considered good