BIOL2010 “Flow Of Genetic Information” Flashcards
Binary fission
Doubling of DNA
How is cell division a fundamental feature of unicellular and multicellular life?
Reproduction in unicellular life
Growth in multicellular life
Bacteriophage genome size
5386bp
Mitochondrial genome size
16569 bp
Name the four DNA nucleotides
Adenosine
Cytidine
Guanosine
Thymidine
Name the four DNA bases
Adenine
Cytosine
Guanine
Thymine
Name the DNA purines
Adenosine (adenine)
Guanosine (guanine)
Name the DNA pyrimidines
Thymidine (thymine)
Cytidine (cytosine)
Describe the structure of DNA
Double helix
Right-handed
Anti-parallel
Phosphodiester backbone
~10nt per turn
Bases on the inside
Hydrogen bonds between bases on opposite strands
How many bonds between DNA bases?
GC = 3
AT = 2
DNA replicated by _____ replication
Semi-conservative
What is required for DNA synthesis?
1 enzyme (Taq - DNA polymerase)
dNTPs (ATP, CTP, GTP, TTP)
Template DNA
Primers
Name the three steps of PCR
Denaturing (heat to 95C)
Annealing (cool to 55C)
Extending (heat to 72C)
What is needed for DNA synthesis in E.Coli, both for one section and the entire chromosome?
DNA polymerase III
dNTPs
Template DNA
Primers
For whole chromosome, also:
Hello case
Topoisomerase
Primase
SSB proteins
DNA pol I
DNA ligase
& more
What are the functions of each of the DNA polymerases?
I: DNA repair and replication
II: repair
III: principal DNA replication enzyme
IV: repair
V: repair
How can we determine which genes (proteins) are important?
Simple knockouts will be lethal
Temperature-sensitive mutants allow proteins to be switched on or off by changing the temperature
Allow cells to begin replication, then deactivate one protein to see the effect
Quick stop mutants
Replication immediately stops
Slow stop mutants
Current round of replication finishes, but a new one can’t start
Name four problems faced in DNA replication
Strands being cooked (topology)
Circular DNA molecules (topology)
Antiparallel strands (polarity & topology)
Mutations/errors (fidelity)
DNA strands are _____ coiled
Plectonemically
How is the problem of strands being coiled overcome?
Helicases separate and unwind the duplex using ATP hydrolysis, creating a replication bubble
Uncoiling at one part of the duplex creates mechanical strain in the rest of the molecule
—> supercoiling occurs
Supercoiling equation
Lk = T + W
Lk = linking number (fixed value in circular DNA)
T = twist (number of duplex turns. Increases with nt)
W = writhe (number of duplex self-crossings)
Twist equation
T = N/h
(Bp/helical repeat)
N can’t change but h can (over- or under-winding)
What does the Lk value mean?
Lk^o = relaxed form where W = 0
Lk > Lk^o = +ive supercoiling (+W)
Lk < Lk^o = -ive supercoiling (-W)
W = +ive
Left
DNA is overwound
W = -ive
Right
DNA is underwound
Superhelical density equation
Sigma = (Lk - Lk^o)/Lk^o = delta (Lk) /Lk^o
Normalised way of expressing how supercoiled a piece of DNA is, removing the effect of chain length
In relaxed DNA, sigma = 0
Sign indicates type of supercoiling as before
Negative supercoiling
Purified cellular DNA
Sigma = -0.06 (euk & pro)
Conversion of -ive writhe to less twist aids unwinding for transcription and replication
Euk DNA -ively supercoiled around histones when forming a nucleosome
Positive supercoiling
Helicase-based unwinding—>overwinding elsewhere
Overwinding will resist replication fork movement
Lk can’t change to relieve the stress without breaking the phosphodiester bond —> +ive supercoils form
Also applies to very long linear DNA
Needs to be removed for replication to continue
Topoisomerases
Cause a change in Lk
Type I: cleave backbone of one strand, allowing duplex rotation and loss of -ive supercoils
Type II: cleave backbone of both strands, using ATP, and introduce a -ive supercoil
Subtypes: IA, IB, IIA, IIB
DNA backbone reseated after manipulation
Topoisomerase type I mechanism of supercoil removal
Phosphodiester bond is transferred to Tyr residue on enzyme, breaking one DNA strand
Unbroken strand passed through the gap
Phosphodiester bond transferred back to DNA, reforming the backbone on the other side
Lk + 1
Topoisomerase type II mechanism of supercoil removal
Horizontal section cut, 5’-P linked to Tyr on A subunit
Vertical section passed through
Introducing a -ive supercoil cancels out a +ive one —> vital for replication fork progression
Lk - 2
How is the problem of circular chromosomes overcome?
Bidirectional synthesis
When circular DNA is replicated, the two daughter rings are interlocked and form a catanene
Topoisomerase IV: cleave, pass through, and re-seal (decatenation) - to separate rings and complete the new DNA
What is a catanene?
Two interlocked daughter rings of circular DNA formed by the bidirectional replication of circular chromosomes
How is the problem of antiparallel strands overcome in DNA replication?
Best solution: 2 polymerases, one for each direction
—> extra genes needed
—> loss of proofreading/editing 3’—>5’ (mutation rate increased by 100x)
All DNA polymerases synthesise DNA by adding to the 3’ end of an existing ds region (primer)
- works but requires complete strand separation before synthesis can start on upper strand
- wasted time (separation time + synthesis time)
—> loop back the lagging strand to correct direction
Synthesis of both strands can occur simultaneously but semi-discontinuously
What is the result of semi-discontinuous replication?
End result of semi-discontinuous replication is two, non-identical dsDNA molecules
Leading strand has 1 primer, lagging strand has multiple primers
Chromosome is replicated, but lagging strand copy is unfinished, with ‘nicks’ in the new backbone, because pol III can’t join fragments
Okazaki fragments (production/discovery)
Short fragments are repeatedly being made on the lagging strand as new sections of DNA are unwound
After a few seconds they are joined to each other and to non-radioactive pieces that had already been synthesised —> size jump
Evidence for Okazaki fragments
Pulse/chase experiments
- After 2s the ^3 H-TTP is removed and chased with normal TTP
—> radioactivity all moves down the tube
DNA ligase mutants
- DNA ligase seals nicks in the backbone and can join Okazaki fragments to each other. ATP consumed
- Short fragments still made, but they remain as fragments
—> lagging strand can’t be finished
Okazaki fragments (structure)
Not completely DNA
Primer is RNA
- We know this because tiny fragments are left over after using DNAase on Okazaki fragments
List three errors that can occur in DNA synthesis/replication (NOT kinds of mutation)
Incorrect nucleotide added
RNA nucleotide instead of DNA nucleotide
Nicks in backbone (fragments)
How/why does incorrect nt addition occur?
Incoming dNTP has to form correct base pair to properly fit into polymerase active site
—> Induced fit
—> Binding + shape discrimination
Incorrect dNTP are excluded by steric collisions, even if there are hydrogen bonds possible
Pol III adds wrong dNTP at a rate of 1 per 100,000 bp
—> drops to 1 per 10,000,000 with proofreading activity of polymerase
Pol III proofreading ability
Pol III has 3’—>5’ exonuclease activity
—> can remove the last nt added if it was incorrect
Mismatched nt at 3’ end won’t be in the right place to add the dNTP
—> polymerase stalls
The end of the strand becomes substrate for the exonuclease active site, which cleaves the terminal Phosphodiester bond, releasing a dNMP
What is produced when DNA pol I is treated with protease?
Small N-terminal fragment (5’—>3’ exonuclease)
Large C-terminal fragment (Klenow fragment) containing polymerase and 3’—>5’ exonuclease
DNA pol I exonucleases
3’—>5’: shortened strand re-engages polymerase active site —> synthesis resumes
5’—>3’: Nick translation—> can remove and replace nt ahead of it
The problem of RNA nt being used instead of DNA nt
rNTPs can’t be directly incorporated into growing DNA strands because the extra OH in ribose causes a steric clash
What are pseudo-Okazaki fragments?
The fragments that need to be joined together in the leading strand of DNA
Leading strand fragments (pseudo-Okazaki fragments)
Pol III will incorporate U instead of T every 1 in 300 times (1 in 1200 nt synthesised)
Since U shouldn’t be present in DNA it must be removed —> nicks in chain —> fragments of ~1200 nt
If enzyme 4 is mutated, more U & shorter fragments
Why is the presence of U in DNA bad?
It suggests damage that has lead to mutation (C can be deaminated to form U)
What removes U from DNA? Describe the process this is involved in
Uracil-N-glycosylase (ung) - removes U
—> baseless nt recognised and phosphodiester backbone cleaved by AP (apyrimidinic) endonuclease (xthA)
—> nicked DNA with incorrect nt
Pol I binds to nick, then removed and replaces baseless nt
ung mutants
No U removal
No fragments/nicks
What seals DNA nicks?
DNA ligase
What is an ori?
Origin of replication
Circular chromosomes and plasmids have a single ori
Region of repetitive dsDNA that is rich in AT
Different in different organisms/plasmids
Initiation of DNA replication
Multiple DnaA products bind to 9bp repeats
Use ATP to separate (melt) the duplex at 13bp repeats
A DnaC protein binds to ssDNA and lids a DnaB helicase onto one strand, facing towards the 3’ end
DnaC detaches and helicase moves to fork
After ~65 nt have been unwound by the helicases, DnaG primase enzymes bind to them, forming a primasome
Primase synthesises a ~10 nt RNA primer and detaches from helicase
Single stranded binding protein (SSB) binds to exposed ssDNA preventing re-annealing
First primer and SSB trigger arrival of the pol III holoenzyme at the 3’ end of the primer
The clamp loser loads a beta clamp onto the DNA
Pol III core binds to the beta clamp
Pol III travels to replication fork, synthesising the leading strand as it goes and displacing SSB
As it catches the helicase, a replisome forms
Primase
An RNA polymerase (but not the one involved in transcription)
Self-priming, adding RNA 5’–>3’ on a 3’–>5’ DNA strand (GTA site preferred)
RNA primers are ~10 nt in length
Has no editing functions (no proof reading)
Activity is increased in the presence of helicase and itself (co-operativity)
Single strand binding protein (SSB)
Encoded by ssb gene
Forms a tetramer
Not sequence specific
Leaves bases exposed when ound
Binds co-operatively to ssDNA
–> Proteins at the ends are bound less well and are easier to dislpace by polymerase
Clamp loader
Binds beta clamp proteins
Transfers the beta clamp onto DNA at primer 3’ end
ATP hydrolysis detaches loader
Beta clamp
Encoded by dnaN gene
Forms a ring dimer
Not sequence specific
Binds Pol III core & imparts processivity
–> Goes from 10s of bp to >50,000
Replisome
Pol III holoenzyme
Primasome
Occupies ~50nm area around replication fork
Elongation (lagging strand synthesis)
1) As helicase unwinds the duplex, primase re-binds and synthesises a new primer
2) A beta clamp is added to the primer by the clamp loader
3) As more ssDNA is created, the lagging strand starts to loop back, reversing the primer polarity
4) A Pol III core binds once enough ssDNA has emerged for the beta clamp to reach it
5) First Okazaki fragment starts. DNA is pulled by helicase and Pol III. The lagging strand loops out, picking up SSB
6) Okazaki fragment lengthens. Lagging strand loop gets longer
7) First Okazaki fragment finished. Primase re-binds helicase, then adds a second primer
8) Pol III core and beta clamp detach from DNA, releasing completed fragment
Fragment removal (lagging strand)
Pol I binds the end of the first Okazaki fragment and replaces the RNA with DNA
DNA ligase seals the nick
Process repeats for each fragment
Termination - DNA forks
Ter and Tus
- To prevent the forks overshooting, there are 23 bp sewuences called Ter, which bind the Tus protein
Tus can be displaced by the fork only in the correct direction, otherwise the helicase stalls
Topoisomerase IV
- As the forks get within 200 bp of each other, there is no longer room for DNA gyrase to bind
–> +ive supercoiling, which is relieved by topoisomerase IV decatenating the molecules
Why does initiation only occur once per cell cycle?
OriC contains GATC sequences, which are substrates for dam (DNA adenosine methylase)
dam methylates N6 of adenine
After replication, only one strand will be methylated
Hemimethylated GATC sequence bind SeqA protein
–> Prevents DnaA binding to OriC
- The GATC sites are methylated by dam very slowly (~13 mins), so newly synthesised dsDNA remains hemimethylated and new initiation is prevented
–> Binds DNA to the membrane
- May assist with separation and moivement of each chromosome into its new cell
In dam strains:
Hemimethylated plasmids are not replicated
Methyated undergo one round of replication
Unmethylated are replicated normally
How does E.Coli replicate so fast? How long does each round of replication take?
Replication of E.coli chromosome takes ~40 mins
Can reproduce every ~20 mins by starting a new round of replication before the last has finished
Similarities between eukaryotic and prokaryotic DNA replication
Use helicases to unwind the diplex and create replication forks
Use SSBs to hold the ssDNA apart
Use RNA primers for the polymerase/clamp
Use various DNA polymerases for synthesis of the new DNA on leading & lagging strands
Broad differences of eukaryotic vs prokaryotic DNA replication
Occurs in nucleus not cytoplasm
More genetic material to replicate (can be 4 orders of magnitude greater)
More than 1 chromosome (46 in humans)
Linear chromosomes (except mitochondria)
Additional packaging (e.g. nucleosomes)
More specific differences in eukaryotic vs prokaryotic DNA replication
Multiple origins per chromosome (10s-1000s)
DNA polymerases are much slower: 50 nt/s vs >1000 nt/s, 40 mins vs 8 hours
Polymerases synthesising the leading and lagging strands aren’t physically linked together
DNA polymerases don’t have 5’–>3’ exonuclease
Okazaki fragments are much shorter: ~165 nt vs ~2000 nt
Eukaryotic DNA replication initiation
Tightly linked to the cell cycle
Initiation must only happen once per cycle
Two step process ensures this
Origins of replication arise from sections of DNA called Autonomously Replicating Sequences (ARS)
Separated by ~30kb, so range from 10 on small chromosomes, to 1000s on the largest
AT-rich consensus sequence (A region) in yeast is: 5’- T/A TTTAYRTTT T/A - 3’ (Y = any pyrimidine, R = any purine)
Eukaryotic DNA replication initiation/licensing (G1 phase)
An Orgigin Recognition Complex of proteins (ORC) binds to the A region
- Less clear how it works in higher eukaryotes
Accessory proteins (licensing factors) accumulate in the nucleus during G1. Cdc-1 & Cdt-1 bind to ORC
2 helicases are loaded by Cdt1. Cdc6/Cdt1 leave
Eukaryotic DNA replication initiation/activation (S phase)
Pre-replication complex must be activated
Pre-replication complex (+ additional proteins + DNA polymerases) –> active initiation complex/replisome progression complex
Eukaryotic DNA replication: elongation
Not every pre-replication complex is used in each round of replication
Replication forks from one origin will pass through another origin (passive replication)
~ Polymerases ~
5 main euk. DNA polymerases: α, β, γ, δ, ε
Polymaerase α has its own primase activity, as well as polymerase, so can make its own primers
- Has no proofreading 3’-5’ exonuclease
- Not v processive since doesn’t associate with the euk. sliding clamp protein called PCNA (proliferating cell nuclear antigen)
Replication factor C (RPC) loads PCNA onto the primer/DNA ready for a different Pol to bind
Pol β: involved in repair
Pol γ: replicates mtDNA
Pol δ: (lagging strand) associates with PCNA and has proofreading
Pol ε: (leading strand) associates with PCNA and has proofreading
–> Pol δ & Pol ε can displace RNA primers, but can’t release them as dNMPs
~ Editing ~
RNA “flap” produced by Pol δ or ε can be mostly digested by RNAse H1, with 1 RNA nt left (like E.coli version)
Flap 1 endonuclease (FEN1) binds PCNA & can remove any incorrect nt, by cutting off a section of ssDNA/RNA
DNA ligase fills in nick
Euk. DNA replication: linear chromosomes
End replication problem:
- Affects the lagging strand
- Last RNA primer may not be at the extreme 3’ end of the DNA –> missed section
- Even if it is at the end, it will be removed because it’s RNA –> ssDNA section
n-10 strand –>shortened strand locked in in next replication –> further end loss (gene shrinkage) after next replication
Euk. DNA replication: telomeres
End replication solution:
-DNA at 3’ ends of chromosomes doesn’t encode genes
Instead it is multiple repeats (100s-1000s) of a simple sequence: TTAGGG
The last 20-200nt (species dependent) at the 3’ end are ssDNA made of these repeats
Once the telomeric DNA is gone (~40 replications) –> gene loss –> cells stop dividing, a point called the Hayflick limit
How do embryonal cells, cancer cells, stem cells, and other types divide more times than the Hayflick limit?
Telomerase is a ribonucleoprotein
Contains an RNA strand (450nt) which has sequence: CUAACCUUAAC
This acts as a template for the synthesis of new telomeric repeats, growing the telomere:
- Telomerase binds and extends ssDNA –> Pol α (primase) binds and adds a primer –> Pol δ binds primer and extends it –> DNA ligase fills in nick
Euk. DNA replication: mitochondrial DNA
2-10 copies of circular mtDNA per mitochondrion
Unidirectional replication (1 fork)
Pol γ synthesises leading strand
Lagging strand is RNA Okazaki fragments
Viral DNA replication: bacteriophages
Small circular genome
Some use a ‘rolling circle’ mechanism to continuously synthesise new DNA
Sigma (σ) replication
Viral DNA replication: RNA genomes
Many animal and plant viruses have genomes composed of RNA (ignores the central dogma)
Have an RNA-dependent RNA polymerase called RNA replicase encoded by the viral genome
The plus strand RNA is copied directly to make the minus strand, which is used as a template for making more plus strand
Can be self-priming –> no primer required
No proof-reading –> highly error prone
Viral DNA replication: retroviruses (e.g. HIV)
RNA genome
Virally encoded reverse transcriptase creates a NDA strand using RNA as template and tRNA Lys as a primer
RNA half is degraded by RNase H
Second NDA strand synthesised using first as template –> incorporated into host genome
