DNA Replication

Question 1

Prokaryotic Replication Overview

Answer 1

Creating a new DNA strand from an existing DNA strand
Phosphodiester bonds attach the sugars
N-glycosidic bond connects base to the sugar
Next phosphate attaches to the 3’OH
During replication, DNA polymerase recognises the DNA bases + adds a complementary nucleotide (A+T, C+G) of dNTP (sugar, base and phosphate group) -> removes two phosphates from dNTP + leaves one to form a phosphodiester bond.
DNA polymerase needs a 3’OH to start replication -> RNA primer is laid down first.
dNTP added in 5’ to 3’ direction

Question 2

Semi-conservative Replication - Meselson Stahl Experiment

Answer 2

Grow E.coli in 15 nitrogen isotope (heavy) and then transfer to a 14 nitrogen (lighter) isotope environment
DNA then isolated from cell and mixed with CsCl solution -> separated DNA by density
Rings of bases absorp UV light -> able to identify where the DNA had settled
1st generation: original DNA so both strand had 15 nitrogen isotope -> sunk to the bottom
2nd generation: both bits of DNA had one original strand (15N) and one new strand (14N) so settled in the middle.
3rd generation: had a 2 DNA molecules with one original strand (15N) and one new strand (14N) which settled around the middle and 2 DNA molecules with two new strands (14N) which settled at the top.

Question 3

DNA Replication in E.coli

Answer 3

Circular Genome
Origin of Replication (oriC) = made up of a 13-mer motif sequence (3 sequences of the same 13 nucleotides -> AT rich -> only 2 hydrogen bonds -> easier to split) and a 9-mer motif (5 sequences of the same 9 nucleotides).
Origin recognition protein binds to the 9-mer motif sequence -> change the structure of the DNA -> strands split apart at the 13-mer motif
Helicase unzips the DNA strands bidirectionally
Single stranded DNA binding protein (SSB) binds to DNA and keeps the strands apart whilst it is being copied.
RNA primer laid down (once on leading strand + multiple times on lagging strand)
DNA polymerase III : alpha subunit adds nucleotides, beta subunit keeps polymerase attached to the DNA (high processivity -> able to catalyse consecutive reactions).
To join Okazaki fragments: RNA primer removed by exonuclease -> DNA polymerase I removes last RNA nucleotide + adds dNTP -> DNA ligase joins the fragments.

Question 4

Ensuring DNA is Identical to Parent

Answer 4

DNA polymerase must select the correct base
DNA Pol I + III proofread -> remove incorrect nucleotides 3’->5’ exonuclease activity (end of strand)
DNA repair mechanism: remove incorrect bases at any point along the DNA strand

Question 5

Challenge of Supercoiling

Answer 5

Supercoiling = twist in DNA strands -> condenses genome
Type I topisomerase; cuts one strand of DNA -> twists DNA -> seals back up.
Type II topoisomerase: cuts both strands -> allows another strand to pass through the gap + then seals back up. Allows decatination (spits the replicated rings into two seperate E.coli genomes)

Question 6

5’ –> 3’ polymerase

Answer 6

Both pol I + II

Question 7

5’ –> 3’ exonuclease

Answer 7

Remove last bit of primer

Pol I

Question 8

3’ –> 5’ exonuclease

Answer 8

Proof reading

Both Pol I + II

Question 9

Polymerase Chain Reaction

Answer 9

Copies a particular section of DNA
Three steps
1. DNA double strand seperated (heat = 95°C)
2. Attach primers (55°C)
3. Synthesise new DNA (72° for Taq pol)
Taq polymerase does not denature at high temperatures -> can do PCR multiple times without the polymerase denaturing + having to add more.
Primer = short single strands of DNA -> the section of DNA to be amplified to specified by the primer.

Question 10

Polymerase Chain Reaction - Advantages

Answer 10

Quickly synthesise many copies
Completely in vitro
Can use low amount of low quality DNA

Question 11

Polymerase Chain Reaction - Disadvantages

Answer 11

Some information of target sequence needs to be known

Susceptible to corruption

Question 12

DNA Sequencing

Answer 12

Uses ddNTPs -> dont have an OH group on the 3’ carbon -> only H.
ddNTPs also have 32 phosphate isotope on alpha isotope (this is the phosphate that will be used to create the phosphodiester bond).
Add DNA, DNA polymerase + DNTPs
If we want to find the position of A bases -> add ddATP
Replication occurs -> stops when a ddNTP is added (no 3’ OH)
Different lengths of DNA strands produced -> run on gel -> separates ragments on basis of size -> able to determine nucleotides position,

Question 13

Fkuorescence Based Dideoxy Sequencing

Answer 13

Uses 4 different fluorecence markers on ddNTPs rather than 32 phosphate isotope
Strands separated by column chromatography
Klenow fragment of DNA pol I used -> wasn’t very good at incorporating ddNTP’s -> use altered Taq polymerase instea.

Question 14

Probe Labeling

Answer 14

Highlight small sequence of DNA
Design complementary sequence of DNA that matches with desired sequence -> needs to be 32 phosphate isotope labelled -> able to identify where it is located
Making probe (2 ways)
1. Nick translation: DNase I breaks DNA -> DNA Pol I cuts back more DNA -> gap filled with 32P dNTPs.
2. Random primed labeling by Klenow: separate strands -> add random hexamers (sequence of 6 nucleotides) -> act as primers -> 32P labeled dNTPs used to fill in gaps.

Question 15

The Eukaryotic Genome

Answer 15

Only 2% of the genome codes for proteins
The rest of the genome: regulatory sequences, introns, pseudogenes (mutated version of functional genes), non-coding RNA
40% of the genome is composed of repetitive sequences -> code for high demand proteins.

Question 16

Compaction of DNA

Answer 16

2 copies of the four core histones (H2A, H2B, H3, H4) bind together to form an octame
DNA wound around octame forms a nucleosome (octame +vely charged and DNA -vely charged -> attracts each other).
A single DNA strand to wound around multiple octames with linker DNA in between
Histone 1 (H1) interacts with nucleosome + linker DNA -> pulls together the multiple nucleosomes to form a 30nm fibre
30nm fibres looped further -> held together with by a nuclear scaffold.

Question 17

Eukaryotic DNA Replication + Cell Cycle

Answer 17

G0 = no dividing cells
G1 = growth
S = synthesis (where replication occurs)
G2 = prepare for chromosome division 
M = chromosome segragation (mitosis) -> 2 cells 
Interphase = G1, S, G2 -> time between one mitosis + the start of the next.

Question 18

Initiation of Eukaryotic DNA Replication - STAGE ONE

Answer 18

Origins of Replication (OOR) needs to be selected - G PHASE
Multiple OOR -> many need to be activated to copy the whole genome
One OOR cannont be activated more than once -> mutation
Origin recognition complex (group of proteins) identify the OOR and bind to it.
Helicase attaches
NOTE: these ASSEMBLE at the OOR but do not become active until the S phase.

Question 19

Initiation of Eukaryotic DNA Replication - Stage Two

Answer 19

OOR activation (S PHASE)
Pre replicaive complexes are attracted -> phosphorilation activates them -> promotes attraction of DNA polymerase.

Question 20

Elongation - Eukaryotic DNA Replication

Answer 20

Replication if biodirectional from OOR

1. DNA polymerase α/primase: lays down an RNA primer -> low processivity so only about 20 nucleotides.
2. DNA polδ: high processivity -> bulk of replication
3. Other DNA polymerases involved in specilized processes e.g repai

Question 21

Finishing Eukaryotic DNA Replication

Answer 21

Challenging because genome is linear (end replication problem)
The end of the strands are not able to be syntheisised after the RNA primer is removed (no 3’ OH) -> this results in a 3’ overhang
Telomers solve this problem

Question 22

Telomers + Telomerase - Eukaryotic DNA Replication

Answer 22

1. RNA part of the telomerase contains 1 1/2 copies of the complement of the telomere sequence -> binds to it and adds more DNA to the original strand (translocates along strand)
2. RNA primer laid down by telomerase
3. DNA polymerase completes strand
   NOTE: most somatic cells have little telomerase activity -> only able to replicate a finite number of times (DNA strands become too short + important coding information is lost) -> senescence.

Question 23

Replication + Chromosome Segregation

Answer 23

2 copies of DNA together = sister chromatids -> held together by a protein complex called cohesin.
M PHASE:
1. Prophase: chromosomes become condensed
2. Metaphase: chromosomes line up on spindle
3. Anaphase: cohesin complexes break + spindle fibres separate sister chromatids.
4. Telophase: nuclear envelopes reform around separate chormosomes.
5. Cytokinesis: cytoplams separated + chromosomes decondense

Question 24

Transcriptional Regulation E.g lactose metabolism in E.coli

Answer 24

Enzyme β-galactatosidase breaks down lactose -> expressed in an inducible fashion (lactose is the inducer).
lac 1 produces repressor protein which binds to the operator.
The regulatory region contains the promotor (where RNA polymerase binds) and the operator
The operon contains the lac Z, lac Y and lac A genes.
How it works:
- Lac 1 gene always switched on -> always producing the repressor protein.
- The repressor protein has 2 binding sites -> one for lactose and one for the operator site. Only able to bind one at at time.
- When lactose is not present the repressor protein is unable to bind to lactose and thus binds to the operator. This stops RNA polymerase from binding to the promotor -> transcription does not occur.
- When lactose is present the repressor protein binds to lactose and therefore CANT bind to the operator. RNA polymerase is then able to bind to the promotor and transcription occurs.

Question 25

Operon

Answer 25

Collection of genes under a single transcriptional control.

Question 26

Post Transcriptional Processing

Answer 26

DNA -> pre mRNA -> mRNA -> protein
Pre mRNA has both introns + exons -> splicing cuts + removes introns + remaining exon pieces are joined -> now mature mRNA -> leaves nucleus for translation

Question 27

Exon

Answer 27

Wanted sections of DNA

Question 28

Intron

Answer 28

Unwanted sections of DNA

Question 29

Pre mRNA

Answer 29

Poly A tail = lots of A bases stuck to the 3’ end by polA polymerase.
Modified G cap at 5’ end

Question 30

Splicing

Answer 30

Splicing machinery attaches to the tail of the RNA polymerase.
RNA Pol A produces a bit more RNA than the DNA template (overshoots).
CPSF and CstF protein recognises unwanted RNA and cuts it.
Poly A polymerase adds A’s to the 3’ end of RNA.
Activators and repressors can bind to splicing site on pre mRNA -> encourage or prevent splicing of pre mRNA.
Proteins are bound to mRNA at splicing junctions -> goes to ribosomes.
Ribosome identifies 5’ cap and scans the mRNA for the start sequence (AUG) -> saves off proteins at junctions as it copies.
Stop sequence before the string of As at 3’ end of mRNA.

Question 31

Alternative Splicing

Answer 31

Different sequence of exons e.g 1, 3, 4 and 1, 2, 4,

Different proteins can be produced from the same section of DNA.