Semester 1 Flow of Genetic Information Flashcards

1
Q

What is the genetic material?

A

Nucleic acids are the genetic material

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2
Q

What is cell division called in different types of cells?

A

In unicellular life its called reproduction

In multicellular life its called growth

Creating new cells means synthesising new DNA

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3
Q

What is binary fission?

A

The doubling of DNA

2 –> 4 –> 8 –> 16

This is the most common type of cell division but doesn’t occur in yeast

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4
Q

What are some generic genome sizes?

A

NON CELLULAR:
- Bacteriophage = 5380
- Mitochondrial = 16,569

CELLULAR:
- E.Coli = 4.6x10^6
- Fruit Fly = 1.8x10^8
- Mouse = 2.7x10^9
- Human = 3.1x10^9

Exceptions are present;
- Amoeba Dubia = 6.7x10^11
Even though it is non cellular

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5
Q

DNA polynucleotide chain composition?

A

A polynucleotide DNA chain is composed of three components:
- Nitrogenous base
- Pentose sugar (deoxyribose in DNA)
- Phosphate group.

There are two types of nitrogenous bases:

  • Purines (adenine and guanine)
  • Pyrimidines (cytosine and thymine)

A nitrogenous base is connected to the 1′ carbon of the pentose sugar through an N-glycosidic linkage to form a nucleoside such as:

  • Adenosine
  • Guanosine
  • Cytidine
  • Thymidine
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6
Q

What is the structure of DNA?

A

Double-helix

Right-handed

Antiparallel (one strand runs in opposite direction of the other, 5’ -> 3’, 3’ -> 5’)

Phosphodiester backbone

~10 nucleotides per turn

Bases are on the inside

Hydrogen bonds between bases on opposite strands

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7
Q

What is complementary base-pairing?

A

The idea that Guanine and Cytosine will always bond together and Adenine and Thymine will always bond together

G – C = 3 Hydrogen bonds
(stronger bond, more stable)

A – T = 2 Hydrogen bonds
(weaker bond, less stable)

Chargaffs rule states that the amount of A and T, and the amount of G and C will always be the same as each other

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8
Q

What is required for DNA synthesis?

A

In lab:
- Taq (DNA Polymerase)
- dNTPs
- Template DNA
- Primers

In cell:
- DNA Polymerase III
- dNTPs
- Template DNA
- Primers
- But also… many many other proteins are involved in a cell

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9
Q

What are the types of DNA Polymerases and what do they do?

A

Named in order of discovery not importance

Pol I:
- DNA repair and replication

Pol II:
- DNA repair

Pol III:
- Principal DNA replication enzyme

Pol IV:
- DNA repair

Pol V:
- DNA repair

All synthesize 5’ –> 3’

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10
Q

Comparison of Pol I and Pol III?

A

Pol 1:
- One gene
- 109kDa
- 400 copies per cell
- 10 nucleotides/s
- 20-100 nucleotides at a time (before falling off)
- 100 hours per genome

Pol III:
- 22 genes
- 10^6kDa
- 10 copies per cell
- 1600 nucleotides/s
- >50,000 nucleotides at a time
- 40 minutes per genome

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11
Q

How can we determine which genes (proteins) are important in DNA replication?

A

Simple knock-outs will be lethal

Temperature-sensitive mutants allow proteins to be switched on or off by changing the temperature
- e.g. protein works at 20, but not 37°C

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

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12
Q

What are some issues that arise with DNA replication?

A

Strands being coiled (topology)

Circular DNA molecules (topology)

Antiparallel strands (polarity & topology)

Mutations/errors (fidelity)

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13
Q

What is the issue with strands being coiled and what is the solution?

A

PROBLEM:
- DNA strands are plectonemically coiled so they will have to be unwound to separate them without getting them tangled

RESOLUTION:
- Helicases can separate and unwind the duplex using ATP hydrolysis

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14
Q

What are helicases and what is their structure??

A

Helicases separate and unwind doulbe stranded DNA using ATP hydrolysis (3 bp/ATP)

Hexamer ring surrounds a single DNA strand

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15
Q

Helicase Function & Action

A

Conformational changes in the helicase pull on the DNA strand, separating it from its partner

Helicases move towards the 3’ end of the strand they are clamped to

One helicase on each strand of DNA

Helicases separate and unwind the two DNA strands, creating a replication bubble

Uncoiling at one part of the duplex, creates mechanical strain in the rest of the molecule

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16
Q

DNA Replication Speed & Mechanics

A

Pol III can synthesize ~1600 nucleotides/s

There are ~10 nucleotides per turn

Helicases must rotate DNA at ~10,000rpm!

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17
Q

What is an issue with circular chromosomes?

A

When DNA helicases unwind and seperate DNA, it can create torsional strain elsewhere in the duplex which results in supercoiling

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18
Q

What equation can DNA topology be described by?

A

Lk = T + W

Lk = linking number (fixed value in circular DNA, number of times the DNA crosses over itself)

T = twist - number of turns of the duplex
Twist = N/h (number of BP / helical repeats)
N cant change but h can (over/under-winding)

W = writhe - number of duplex self crossings

The relaxed form, where W=0, is called Lk0
If Lk > Lk0 then there is +ve supercoiling (+W)
If Lk < Lk0 then there is –ve supercoiling (-W)

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19
Q

Different amount of writhe results?

A
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20
Q

What is the equation used to determine superhelical density?

A

This is a normalised way of expressing how supercoiled a piece of DNA is, removing the effect of chain length.

In relaxed DNA, σ = 0

Sign indicates type of supercoiling

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21
Q

What is the biological significance of negatively supercoiled DNA?

A

Purified cellular DNA is always slightly negatively supercoiled

σ = -0.06 (eukaryotes and prokaryotes)

Conversion of –ve Writhe to less Twist aids unwinding for transcription and replication

Allows easier separation of DNA

Also, Eukaryotic DNA is negatively supercoiled around histones when forming a nucleosome

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22
Q

What is the biological significance of positively supercoiled DNA?

A

Helicase-based unwinding results in overwinding elsewhere so will always produce positive supercoiling

Overwinding will resist replication fork movement

Lk can’t change to relieve the stress without breaking the phosphodiester bonds

So… Positive supercoils form

Also applies to very long linear DNA

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23
Q

What is the result of positive supercoiling?

A

Positive supercoiling prevents DNA replication

It will need to be removed for replication to continue

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24
Q

How can we remove or add supercoiling?

A

Specific enzymes that introduce / remove supercoils

TOPOISOMERASES:
Type I:
- Cleave backbone of one strand, allowing duplex rotation and loss of negative supercoils

Type II:
- Cleave backbone of both strands, using ATP and introduces a negative supercoil

DNA backbone is sealed after manipulation

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25
Q

Supercoil removal via Topoisomerase Type I mechanism

A

Phosphodiester bond is transferred to Tyr residue on enzyme, breaking one DNA strand

The unbroken strand is passed through the gap

The phosphodiester bond is transferred back to DNA, reforming the backbone on the other side

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26
Q

Supercoil removal via Topoisomerase Type II mechanism

A
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27
Q

What is special about supercoiled dna in gel?

A

Supercoiled DNA runs faster than linear DNA in an agarose gel

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28
Q

What issues occur when chromosomes are circular

A

DNA replication is bidirectional, so with circular DNA the DNA Polymerases can collide with each other

Also, as the DNA is seperated and expanded, two duplexes will be created that when finished, will be
entwined with one another

When circular DNA molecules are replicated, the two daughter rings are interlocked and form a CATANENE

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29
Q

How are catanenes dealt with?

A

Topoisomerase IV can be used

Cleaves 1 duplex, the second duplex passes through and then the cleavage is sealed

Process called decatenation

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30
Q

How do antiparallel DNA strands manage synchronous replication?

A

DNA strands are antiparallel: one 5’→3’ and the other 3’→5’

DNA polymerase can only add nucleotides in the 5’→3’ direction

The leading strand is continuously synthesized in the direction of the replication fork

The lagging strand is synthesized discontinuously in short segments called Okazaki fragments

To facilitate this, the lagging strand loops out, forming a structure that allows the DNA polymerase to synthesize the strand in the 5’→3’ direction, but effectively in the opposite direction of replication fork movement

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31
Q

What is the end result of antiparallel strand replication?

A

End result of semi-discontinuous replication is two, non-identical dsDNA molecules

Chromosome is replicated, but lagging strand copy is unfinished, with ‘nicks’ in the new backbone, because Pol III can’t join fragments

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32
Q

What did Okazaki investigate?

A

Okazaki investigated the synthesis of DNA over time using radioactivity

1) E.coli culture infected with T4 bacteriophage

2) Add 3H-TTP (tritiated thymidine) so that the new DNA strands created will be radioactive

4) Mapped out sizes of radioactive ssDNA over time using density centrifugation

5) Small new strands of radioactive DNA were always present, as pictured

6) Even after full size strand had been created, new smaller strands were still present, proving size of radioactive ssDNA doesn’t simply increase smoothly over time. Short pieces are always present

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33
Q

Okazaki fragment explanation?

A

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 synthesized

Results in a size jump

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34
Q

Further evidence of Okazaki fragments using chase experiment

A

Chase/Pulse experiments carried out

After 2 seconds the 3H-TTP
is removed and “chased” with normal TTP

Radioactivity all moves down the tube as pictured

The observations supported the hypothesis that the lagging strand is synthesised discontinuously in short fragments, which are later connected to form a continuous strand.

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35
Q

Further evidence of Okazaki fragments using DNA Ligase mutant experiment

A

DNA Ligase consumes ATP to join Okazaki fragments of DNA together

When DNA Ligase mutant is used instead of DNA Ligase, the fragments can not be joined together

This results in density centrifugation only showing short, small okazaki fragments and no fully formed DNA fragments, as pictured

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36
Q

Are okazaki fragments completely DNA?

A

Not completely DNA

Primer is RNA

We know this because tiny fragments are left over after using DNAse on Okazaki fragments

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37
Q

What are some mutations / errors that occur during DNA synthesis?

A

1) Incorrect nucleotide added

2) RNA nucleotide instead of DNA nucleotide

3) Nicks in backbone (fragments)

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38
Q

How is adding the wrong nucleotide avoided?

A

The incoming dNTP has to form the
correct base pair to fit properly into the
polymerase active site

DNA Polymerase has an induced fit mechanism

Has binding and shape discrimination

Incorrect dNTP are excluded by steric collisions, even if there are hydrogen bonds possible

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39
Q

How often does the addition of the wrong nucleotide occur?

A

Pol III adds the wrong dNTP at a rate of 1 per 100,000 bp (error rate of 10-5)

Would be ~ 46 point mutations per E.coli replication

Error rate drops to 10-7 with proofreading activity of polymerases

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40
Q

What is proofreading in dna polymerase?

A

Proofreading in DNA polymerase is the enzyme’s ability to detect and correct mistakes during DNA replication

When DNA polymerase inserts an incorrect nucleotide opposite the template strand, it can recognise this error and correct it

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41
Q

Proofreading in Pol III?

A

Pol III has 3’-5’ exonuclease activity
- Can remove the last nucleotide added, if it was incorrect

A mismatched nt at the 3’ end won’t be in the right place to add the next dNTP so:
- The polymerase stalls

The end of the strand become substrate
for the exonuclease active site, which cleaves the terminal phosphodiester bond, releasing a dNMP

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42
Q

What are the activities in Pol I?

A

An enzyme with multiple activities:
- 5’-3’ polymerase (synthesis)
- 3’-5’ exonuclease (proofreading)
- 5’-3’ exonuclease (nick translation)

When pol I is treated with protease it produces 2 main fragments:

Small N-Terminal Fragment
- Contains 5’-3’ exonuclease

Large C-Terminal Fragment (Klenow Fragment)
- Contains polymerase
- Contains 3’-5’ exonuclease

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43
Q

DNA Pol I 5’-3’ exonuclease?

A

Nick translation

Binds to nick in synthesised dna fragment

Will then remove nucleotides ahead of it and replace with correct nucleotide

Doesn’t remove nick, just moves it further along

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44
Q

Is there RNA in ozaki fragments? What removes the RNA?

A

Yes, there is RNA at the 5’ end, and a nick at their 3’ end

Pol I binds to the nicks in in the fragments, but Pol III does not

RNA primers are removed by Pol I 5’-3’ exonuclease

Pol I detaches after 100bp and a new nick is left behind

Pol I starts joining together the fragments while Pol III is still expanding the strand

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45
Q

What is RNAse H?

A

It is a protein that removes RNA in DNA strands

It cant cut between RNA and DNA nucleotides, so leaves a gap between the end 2

Pol I comes along and fills the gap

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46
Q

What are pseudo-Okazaki fragments?

A

Leading strand also consists of fragments that need to be joined together

Strand is grown continuously and nicks are added in afterwards

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47
Q

Why are pseudo-Okazaki fragments created?

A

The synthetic pathway for making dTTP, includes dUTP

Enzyme 7 can sometimes add a dUTP instead of a dTTP

During synthesis, Pol III will incorporate U instead of T, 1 out every 300 times (once every 1200 nt synthesized) by accident

Since U shouldn’t be present in DNA it must be removed:
- Nicks in chain
- Fragments of ~1200nt

If enzyme 4 is mutated:
- More U & shorter fragments

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48
Q

Why is U in the place of T a problem?

A

When U is accidentally added in place of a G there is no problem

When U is formed by deamination of C, mutations are caused

Presence of U in DNA is therefore offensive as it suggest damage and therefore it must be removed

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49
Q

How is U removed from DNA?

A

Base is removed by:
- Uracil-N-Glycosylase (gene = ung)

Baseless nucleotide is recognised and phosphodiester backbone cleaved by:
- AP(apyrimidinic) endonuclease (xthA)

Results in nicked DNA with incorrect nucleotide

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50
Q

How are nicks in dna sealed?

A

DNA Ligase seals nicks in DNA

Can’t do RNA - DNA nicks (wont accidentally join in the RNA primer)

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51
Q

What is the origin of DNA replication?

A

Circular chromosomes and plasmids have a single origin of replication (ori)

The origin is a region of repetitive dsDNA that is rich in A-T

Different in different organisms/plasmids e.g. oriC in E.coli chromosome

oriC is about 245bp long, split into 2 main sections:
- 3, 13-bp repeats
- 4, 9-bp repeats

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52
Q

What is DnaA?

A

It is a protein that multiple of bind to 9bp repeats in the ori, and cause it to coil which in turn induces unwinding at the 13-bp repeats

It uses ATP to separate (melt) the duplex at the 13bp repeats

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53
Q

What is DnaC and DnaB helicase?

A

DnaC is a protein that binds to the ssDNA created by DnaA and loads a DnaB helicase onto one strand, facing the 3’ end

The DnaC then detaches and the helicase moves to fork

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54
Q

What is DnaG primase?

A

After 65 nucleotides have been unwound by the DnaB helicases, DnaG primase enzymes bind to the helicases, forming a ‘primosome’

Primosomes are the functional units that create Primase

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55
Q

What is the primase enzyme?

A

Creates a primer for DNA polymerase to bind to

Primase is an RNA polymerase (but not the one involved in transcription)

It is self priming, adding RNA 5’-3’ on a 3’-5’ DNA strand (GTA site preferred)

RNA primers are -10 nucleotides in length

It has no editing functions (no proof reading)

Activity is increased in the presence of helicase and itself (co-operativity)

Primase synthesises a 10 nucleotide RNA primer and detaches from helicase

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56
Q

What is single stranded binding protein? (SSB)

A

Single stranded binding protein (SSB) binds to exposed ssDNA preventing re-annealing

  • Encoded by ssb gene
  • Forms a tetramer
  • Not sequence specific
  • Leaves bases exposed when bound
  • Binds co-operatively to ssDNA
  • Therefore, proteins at the ends are bound less well and are easier to displace by polymerases
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57
Q

What does the primer strand and SSB trigger?

A

First primer and SSB trigger arrival of the Pol III holoenzyme at the 3’ end of the primer

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58
Q

What is the structure of Pol III holoenzyme?

A
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59
Q

How does Pol III bind to the primer?

A

The clamp loader on Pol III loads a beta-clamp onto the DNA

Pol III core then binds to the beta-clamp

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60
Q

What does the clamp loader do?

A

Binds β clamp proteins

Transfers the β clamp onto DNA at primer 3’ end

ATP Hydrolysis is required to detach the clamp loader from the beta-clamp

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61
Q

Beta-clamp structure?

A

Encoded by dnaN gene

Forms a ring dimer

Not sequence specific

Binds to the Pol III core and imparts processivity

Changes ability of Pol III from doing 10s of basepairs at a time to >50,000

Pol III on its own isnt any better than Pol I at staying on DNA, beta-clamp improves this

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62
Q

What happens as Pol III catches the helicase?

A

Pol III travels to replication fork, synthesising the leading strand as it goes and displacing SSB

As it catches the helicase, a replisome forms

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63
Q

What is a replisome?

A

It is a region around the replication fork that contains these proteins:
- Pol III Holoenzyme
- Primosome

It occupies around 50nm area around the replication fork

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64
Q

Initiation of Lagging Strand Synthesis

A

As helicase unwinds the duplex, primase re-binds and synthesises a new primer

A β clamp is added to the primer by the clamp loader

A Pol III core binds once enough ssDNA has emerged for the β clamp to reach it

Primer 1 bound, with correct polarity

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65
Q

First Okazaki Fragment Formation

A

First Okazaki fragment starts

DNA is pulled by helicase and Pol III

The lagging strand loops out, picking up SSB

The first Okazaki fragment is finished

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66
Q

Cycle Repeat for Subsequent Fragments after okazaki fragment formation

A

Primase re-binds helicase, then adds a second primer

Pol III core and β clamp detach from DNA, releasing completed fragment

Primer 2 gets a β clamp

Looping process repeats for primer 2

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67
Q

How are the lagging strands okazaki fragments edited?

A

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

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68
Q

How are the replication forks prevented from overshooting?

A

Ter and Tus

To prevent the forks overshooting there are 23 bp sequences called Ter

The Ter sequences bind the Tus protein

Tus can be displaced by the fork only in the correct direction, otherwise the helicase stalls

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69
Q

What is topoisomerase IV and what is it used for?

A

As the forks get within 200bp of each other, there is no longer room for DNA gyrase to bind

This results in positive supercoiling, which is relieved by
topoisomerase IV decatenating the molecules

The topoisomerase IV binds to one of the strands of dsDNA, cuts it, passes the other dsDNA strand through it, and then rejoins the original strand

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70
Q

How does initiation only occur once per cell cycle?

A

OriC contains GATC sequences (within the 13bp repeats), which are substrates (binding sites) for dam (DNA adenosine methylase)

The ‘A’ within the GATC sequences in the 13bp repeats will have a methyl group attached to it

dam methylates N6 of adenosine

After replication, only one strand will be methylated (template strand) (referred to as hemimethylated, one strand methylated, one strand not methylated)

Hemimethylated GATC sequence bind SeqA protein

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71
Q

What does the SeqA protein do?

A

Prevents DnaA binding to OriC

The GATC sites are methylated by dam very slowly (~13 mins)

So newly synthesized dsDNA remains hemimethylated and new initiation is prevented

SeqA also binds DNA to the membrane, localising it to the membrane (assists with separating two chromosomes during cell division)

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72
Q

What are the similarities in eukaryotic and prokaryotic DNA replication?

A

Both use helicases to unwind the duplex and create replication forks

Both use SSBs to hold the ssDNA apart

Both use RNA primers for the polymerase/clamp

Both use various DNA polymerases for synthesis of the new DNA on leading and lagging strands

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73
Q

What are the differences in eukaryotic and prokaryotic DNA replication?

A

Eukaryotic occurs in the nucleus not the cytoplasm

Eukaryotes have much more genetic material to replicate (can be 4 orders of magnitude greater)

More than 1 chromosome in eukaryotes

Linear chromosomes (except mitochondria) in eukaryotes

Eukaryotes have more packaging (nucleosomes, histones)

Eukaryotic replication has multiple origin points (10s-1000s), replication starts from multiple points

DNA polymerases are much slower in eukaryotes (50nt/s vs 1000nt/s)

Polymerases synthesising the leading and lagging strands aren’t physically linked together in eukaryotes

Eukaryotes do NOT have any polymerases with a 5’-3’ exonuclease

Okazaki fragments are much shorter in eukaryotes (165nt vs 2000nt)

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74
Q

DNA replication initiation in eukaryotes

A

Tightly linked to the cell cycle

Initiation must only happen once per cycle

A two step process ensures this ^

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75
Q

What is M-Phase, S-Phase, G1-Phase, G2-Phase?

A

G1 Phase (Gap 1):
- This is the first phase of the cell cycle
- During G1, the cell grows in size, carries out its normal functions, and prepares for DNA replication.
- At the end of G1, the cell decides whether to continue the cell cycle and divide or enter a non-dividing state called G0.

S Phase (Synthesis):
- In the S phase, DNA synthesis occurs
- The cell replicates its genetic material by duplicating the chromosomes

G2 Phase (Gap 2):
- After DNA replication in the S phase, the cell enters the G2 phase.
- During this phase, the cell continues to grow, synthesizes proteins, and prepares for mitosis (or meiosis if it’s a germ cell).

M Phase (Mitosis or Meiosis):
- The M phase is when the actual cell division occurs
- It includes mitosis or meiosis

76
Q

At what part of DNA does DNA replication initation in eukaryotes occur?

A

Origins of replication arise from sections of DNA called Autonomously Replicating Sequences (ARS)

Autonomously Replicating Sequences (ARS) are separated by ~30kb, so range from 10 on small chromosomes, to 1000s on the largest

ARS contain an AT-rich consensus sequence

In yeast (A region):

5’- (T/A)TTTAYRTTT(T/A) -3’

Where:
- Y = any pyramidine
- R = any purine

77
Q

What binds to autonomously replicating sequences?

A

In yeast, an Origin Recognition Complex of proteins (ORC) binds to the A region of the ARS

It is unclear how this works in higher eukaryotes

78
Q

What occurs after ORC binds to the ARS?

A

Accessory proteins (licensing factors) accumulate in the nucleus during G1

Cdc6 and Cdt1 (examples of licensing factors) bind to ORC

79
Q

What happens after licensing factors bind the ORC?

A

Two helicases are loaded by the licensing factors

The licensing factors then leave

Pre-replication complex is formed

80
Q

How is the pre-replication complex activated?

A

During the S phase, additional proteins + DNA polymerases are added, and the pre-replication complex is activated

This results in a active initiation complex/replisome progression complex

81
Q

Is every pre-replication complex used every time?

A

Not every pre-replication complex is used in each round
of replication

Replication forks from one origin will pass through another origin (passive replication)

82
Q

What DNA polymerases are in eukaryotes?

A

5 main eukaryotic DNA polymerases: α, β, γ, δ, ε

Polymerase α is involved in making primers

Polymerase β is involved in repair

Polymerase γ replicates mtDNA

Polymerase δ (lagging strand) associates with PCNA and has proofreading

Polymerase ε (leading strand) associates with PCNA and has proofreading

Pol δ & Pol ε can displace RNA primers, but can’t break phosphodiester bonds, and cant release them as nucleotides

83
Q

DNA polymerase α mechanism

A

Polymerase α has its own primase activity, as well as polymerase, so can make its own primers

(main role is making primers)

It has no proofreading 3’-5’ exonuclease

It is not very processive since it does not associate with the eukaryotic sliding clamp protein called PCNA (proliferating cell nuclear antigen)

84
Q

What are th

A
85
Q

What is the proliferating cell nuclear antigen?

A

It is the eukaryotic sliding clamp protein

PCNA

Replication factor C (RPC) loads the PCNA onto the primer/DNA ready for a different Pol to bind

86
Q

How are RNA primers removed in eukaryotic dna replication?

A

Pol δ & Pol ε displace the primers and push them to the side generating an RNA ‘flap’

The RNA “flap” produced by Pol δ or ε can be mostly digested by RNAse H1, with 1 RNA nucleotide left at the end

Flap 1 endonuclease (FEN1) binds PCNA and can remove any incorrect nt, by cutting off a section of ssDNA/RNA

DNA ligase fills in nick generated by FEN1

87
Q

What is the end replication problem in linear chromosomes?

A

Affects the lagging strand

Last RNA primer may not be at the extreme 3’ end of the DNA which results in a missed section

88
Q

What are telomeres?

A

They are the end replication solution

The DNA at the 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

These repeated ends are called telomeres

The more the chromosomes are replicated, the shorter the telomeres become

89
Q

What is the hayflick limit?

A

The Hayflick limit is the maximum number of times normal human somatic cells can divide

As the chromosome is replicated, the telomere shortens, until there is no telomeric DNA

Once telomeric DNA is gone, gene loss occurs and the cell stops dividing

89
Q

So how do embryonal cells, cancer cells, stem cells
and other types divide more times than the Hayflick limit?

A

They can synthesise new telomeric DNA

Telomerase is a ribonucleoprotein

It contains an RNA strand (450 nt) which has the sequence CUAACCUAAC

This acts as a template for the synthesis of new telomeric repeats, growing the telomere

89
Q

Telomerase mechanism with Pol alpha nad Pol Delta

A
90
Q

How is mitochondrial DNA replicated in eukaryotes?

A

There are 2-10 copies of circular mitochondrial DNA per mitochondrion

Mitochondrial DNA undergoes unidirectional replication (1 replication fork)

Pol γ synthesises the leading strand

Lagging strand is unknown but thought to be RNA Okazaki fragments

90
Q

How does bacteriophage DNA replication occur?

A

They have a small circular genome

Uses a ‘rolling circle’ mechanism to continuously synthesise new DNA

Called Sigma (σ) replication

  • Start off with circular DNA molecule
  • Nick is deliberately created in DNA
  • Nick acts as binding site for polymerase
  • Pol displaces existing strand and replaces with newly synthesised one
91
Q

How are RNA genomes replicated?

A

Many animal and plant viruses have genomes composed of RNA

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 meaning it is highly error prone

92
Q

How do retroviruses undergo replication?

A

Contain RNA genome

Virally encoded reverse transcriptase creates a DNA strand using RNA as template and tRNALys as a prime RNA half is degraded by RNase H

Second DNA strand synthesized using first as template and is then incorporated into host genome

93
Q

What is DNA damage?

A

Something wrong with DNA that may lead to:
- Mutations
- Unable to replicate

94
Q

What are some types of DNA damage?

A

Abnormal base pairs

Chemical adducts

Base pair mismatches

Double-stranded breaks

Single-stranded breaks

Abasic site

Thymine dimers

DNA cross-links

Nucleotide insertions

Nucleotide deletions

95
Q

What are some spontaneous sources of DNA damage?

A

Replication errors

Tautomerisation

Deamination

Depurination

96
Q

What are some induced sources of DNA damage?

A

Intercalating agents

Base analogues

Deaminating agents

Alkylating agents

Oxidising agents

Radiation, U/V

97
Q

What are replication errors?

A

Normal replication introduces the wrong base once every 10^7 bp, even with proofreading

There is a good chance this error will be repaired

Repair systems reduce error rate to 10^-10

Some repetitive regions cause slippage of the growing strand insertion of more repeats which can’t be repaired

Slippage is particularly prevalent with trinucleotide repeats

98
Q

What are tautomerisation errors in dna replication?

A

Tautomerization is a reversible chemical process in which a DNA base temporarily changes its structure, leading to the incorrect pairing of nucleotides during DNA replication

In the example, the shifted enol form of T binds G instead of A

The mutations are consolidated on next replication

99
Q

What are deamination errors in dna replication?

A

It is the loss of a base amino group

C –> Uracil (pairs A)
A –> Inosine/Hypoxanthine (pairs C)
G –> Xanthine (pairs C less strongly)

C, A and G deaminations can be repaired

  • Around 100 C are converted to U per day
  • Around 1 a day per cell for A and G
100
Q

What is 5-Methyl-C?

A

It is a regular Cytosine base that has had a methyl group attached to it

When deaminated, 5-Methyl-C turns into a T

This T generated, can not be distinguished from a normal T, and therefore cannot be repaired

101
Q

What are depurination errors in dna replication?

A

Occurs mainly in A or G

It is the cleavage of base-sugar bond

Forms an abasic site (apurinic)

Around 10,000 purine glycosidic bonds
hydrolyse/cell/day

Around 600 pyrimidine glycosidic bonds
hydrolyse/cell/day

4x more likely in ssDNA

Mutation is consolidated on next replication

102
Q

What are the 5 chemical mutagens?

A

Intercalating agents

Base analogues

Deaminating agents

Alkylating agents

Oxidising agents

103
Q

What are intercalating agents?

A

They are chemicals that insert themselves between bases, and distort the DNA

They look similar to bases, so are able to insert themselves

e.g. ethidium bromide

Intercalating agents cause frameshift mutations
- Insertions
- Deletions

104
Q

What are base analogues?

A

These are chemicals that are very similar in structure to bases

For example, bromouracil (T analogue) (pictured)

Because the structure is so similar, they are incorporated into the DNA

These base analogues are more prone to tautomeric shifts

105
Q

What are alkylating agents?

A

Alkylating agents add carbon groups (alkyl) to nucleotides

e.g:
- Nitrosamines
- Methyl bromide

Most common is G –> O6-Methyl-G, which is G with an additional methyl group

O6-Methyl-G pairs T

Alkylation can also speed up depurination

106
Q

What are deaminating agents?

A

They are chemicals that remove amino groups from the bases

e.g:
- Nitrous acid
- Nitrosamines
- Nitrite
- Nitrate

Far quicker than spontaneous deamination

107
Q

What are oxidising agents?

A

Cause of most mutations

Any chemical that makes bonds or breaks bonds

e.g:
- Free radicals
- Superoxide ions (O2-)
- H2O2

Many possible nucleotide alterations

e.g. (OH- free radical) + G —> 8-oxo G

8-oxo G pairs A

108
Q

How does UV light cause DNA errors?

A

It can cause UV-induced formation of dimers between adjacent thymines

Stops correct base-pairing in polymerases

109
Q

How does Radiation, γ & X rays & UV rays cause DNA errors?

A

Breaks bonds & creates free radicals cause:
- Single or double-strand breaks
- Bases chemically altered, linked or detached

This is the leading source of mutation

Number of mutations is in direction proportion to the radiation dose

110
Q

What are the two options of dna repair?

A

Damage can be directly repaired

Damage that can’t be directly repaired:
→ Remove and replace the affected DNA

111
Q

What are the repair mechanisms?

A

Sequence repair
- Direct reversal/repair
- Base Excision Repair (BER)
- Nucleotide Excision Repair (NER)
- Mismatch Repair (MMR)

Molecule repair
- Homologous recombination
- Non-homologous end repairH

112
Q

How does sequence repair occur?

A

Errors in sequence must first be detected by scanning the DNA

Checking if it’s:
→ Not a DNA nt
→ The wrong DNA nt for the base pair

Specialised proteins either directly reverse the error, or edit the DNA to reset that section of sequence

113
Q

When does direct repair occur?

A

Where the damage has converted A,C,G,T to something else, it may be possible to convert back to the original nt

e.g. Demethylation following alkylation
e.g. Removal of crosslinks following UV light damageH

114
Q

How does direct reversal of demethylation occur?

A

Occurs via specific reversal proteins
e.g. O6-methylguanine-DNA methyltransferase (MGMT)

Transfers methyl/ethyl group from G to a Cys
residue on itself:
→ G restored

115
Q

How does direct reversal of dimers created via ultraviolet light occur?

A

Via DNA photolyase

Absorbs blue light and breaks T-T internucleotide bonds, using FADH → 2 Ts restored

Mammals have to use another system for repairing thymine dimers,as they don’t have this enzyme

116
Q

What is base excision repair?

A

It is a removal of individual bases (local correction)

Group of >6 DNA glycosylases which recognise abnormal bases and cleave them from the deoxyribose, creating an abasic site

Glycosylases flip out bases for closer checking

If the base is found to be incorrect, the sugar-base bond is cleaved, but UDGase remains attached to DNA

117
Q

How does UDGase discriminate between U and T?

A

T has a methyl group and the U doesn’t

That methyl group clashes with the tyrisine residue on the active site

There is a separate TDGase for GT pairs

If Tyr is mutated to Ala, both T and U will be removed by the enzyme

118
Q

What happens if there is a U in DNA? Base excision repair…

A

U in DNA –>

Base removed by Uracil-N-glycosylase (ung) –>

Baseless nt recognised and phosphodiester backbone cleaved by AP(apyrimidinic) endonuclease (xthA) –>

Results in nicked DNA –>

Pol I nick translation restores T + DNA ligase seals nick (pol beta in eukaryotes)

119
Q

What is nucleotide excision repair?

A

Removal of oligonucleotide fragments from one strand (bulk correction)

Triggered by changes in the physical structure of the duplex as a result of damage

Corrects any of the types of sequence damage e.g. thymine dimer removal

Achieved by a protein complex called UvrABC exinuclease in E.coli (many more proteins in eukaryotes)

120
Q

What is mismatch repair?

A

Detection and removal of incorrect base pairs

Principle:
- A mismatched pair will distort the helix
- This can be detected by specialist proteins
- Incorrect base removed

It must somehow know which one of the two is the wrong one: strand-directed mismatch repair

121
Q

How does methyl-directed mismatch repair work?

A

Hemi-methylation provides the information on which strand is parent (correct) and daughter

  1. MutH binds to unmethylated GATC at OriC,
    identifying the daughter strand.
  2. MutS binds to a distorted site on the duplex
  3. MutL binds to MutS
  4. MutL/MutS complex travels back to the origin and activates MutH
  5. MutH cleaves daughter strand (nicked)
  6. Specialized helicase and exonucleases remove nt until past the distortion
  7. Pol III fills in missing nt. DNA ligase seals nick
122
Q

How does mismatch repair occur in eukaryotes?

A

Eukaryotes have several homologues of MutL and MutS e.g. MSH1-6

No homologues of MutH, but they don’t use hemimethylation replication tags either

123
Q

What is translesion synthesis?

A

Sometimes a risk of mutation is better than the alternative, so there are polymerases that can add nt
where processive, proof-reading polymerases cannot

e.g. Pol IV&V in E.coli and at least 8 Pols in eukaryotes
→ Blockage is bypassed, but the nt added may not be correct

124
Q

What are the repair methods in double-stranded breaks?

A

1) Homologous recombination
- Following replication, while the sister chromatids are still joined one can be used as a template to repair the other.

2) Non-Homologous end-joining (NHEJ)
- A protein complex binds the naked ends of duplex fragments and recruits DNA ligase IV, which can ligate both strands – but it does it blindly, to any two pieces of DNA, with loss of some nt

125
Q

What is xeroderma pigmentosum?

A

Individuals show dry, parchment-like skin (xeroderma) and many freckles (pigmentosum)

Increased sensitivity to UV light

1000-fold increased risk of skin cancer

–> Due to inherited defects in one of eight distinct genes responsible for components of the NER complex

126
Q

What is hereditary non-polyposis colon cancer?

A

Individuals exhibit a predisposition to colon cancer (2-3% of all colon cancer cases)

Due to defects in the human equivalents of the MutS/L MMR system (MSH2 and MLH1)

Leads to the accumulation of mutations throughout the genome

127
Q

How does DNA encode for proteins?

A

DNA contains the code for proteins

DNA is in the nucleus

Proteins are made in the cytoplasm

Requires mRNA

128
Q

Where does replication, transcription and translation occur in prokaryotes?

A

Replication + transcription + translation all happens in the cytoplasm in prokaryotes

129
Q

Why is the intermediate mRNA required?

A
  • Better regulation of the protein production process
  • Fidelity of DNA replication is very important. Having an intermediate step prevents interference and potential additional mistakes
  • Proteins can still be made when DNA is being replicated.
130
Q

What is RNA?

A

RNA is a linear polymer like DNA

Residues linked by phosphodiester bonds

Contains ribose not deoxyribose

RNA is single stranded

Contains:
A (adenine)
G (guanine)
C (cytosine)
but not T (thymine); uses U (uracil) instead

U base-pairs with A (just like A:T base pair)

131
Q

can RNA form structures?

A

Although RNA is single stranded, it can fold into specific structures

This involves base-paring and covalent bonds

This allows some RNA molecules to have structural and catalytic functions

132
Q

What types of RNA are there?

A

rRNA: Ribosomal RNA, form the basic structure of the ribosome and catalyse protein synthesis

tRNA: Transfer RNAs, central to protein synthesis as adaptors between mRNA and amino acids

Non-coding RNA, e.g. microRNAs, long non-coding RNAs

mRNA: Messenger RNA, codes for proteins (3-5%)

132
Q

What are all of the types of transcription control?

A
  1. Transcription control

Regulation of transcription is the most common form of control of protein production

  1. RNA processing control
  2. RNA transport and localization control
  3. Translation control
  4. mRNA degradation control
  5. Protein activity control

(Long non-coding RNA)
- (>200 nt, not transcribed. Involved in regulation through binding RNA, proteins and DNA thus influencing important interactions)

133
Q

DNA transcription labelling

A
134
Q

What does RNA polymerase do?

A

Separates the 2 strands of DNA

Uses one of the DNA strands as a template for RNA synthesis

Does not require a primer

It is very accurate 1/10,000 bases

Moves along the gene in a 5’ to 3’ direction

Synthesises a complementary RNA copy of the DNA template strand

135
Q

How many RNA copies are made per gene?

A

Many RNA copies are made at the same time

Multiple RNA polymerases per gene

Multiple transcripts per gene
- As soon as the first RNA polymerase starts to move down the gene, the next polymerase can bind and initiate RNA transcription

136
Q

What is the mechanism of transcription?

A

RNA Polymerase binds, has helicase activity to begin with, breaking hydrogen bonds and unwinding DNA

Does this as it moves in 5’ to 3’ direction, forming a transcription bubble, which opens up so that the template strand can be copied

Free nucleotides join and are added to complement the template strand and generate a copy of the coding strand

Length of the bubble is 12-14bp

Reaction rate approx 40 bases per sec

137
Q

What are the three main steps of transcription? (important to know the differences)

A
  1. Initiation - Template recognition
    - Polymerase locates promoter
    - Polymerase unwinds the DNA
    - Transcription begins
  2. Elongation
    - Polymerase places RNA in exit hole
    - Once RNA is 10 or more base pairs long
    - Polymerase conformation change - tightens grip
  3. Termination
    - Termination sequence
    - Polymerase separates
    - RNA released
138
Q

What are the differences in eukaryotic and prokaryotic transcription?

A

Differences in initiation

Differences in promotor

Different RNA Polymerases

139
Q

What are the two types of genes involved in transcriptional control in bacteria?

A

Constitutive Genes:
– Housekeeping genes

Regulated genes:
e.g. Changes in food sources

Switches on genes that encode enzymes which are needed to metabolise that sugar

e.g. Changes in environmental stresses (pH and temp)

Switches on genes that encode proteins which help the
bacterium survive

140
Q

What is an operon?

A

Single unit that controls multiple genes

Genes encoding for proteins in the same pathway are located adjacent to one other and controlled as a single unit that is transcribed into a polycistronic RNA - No introns

141
Q

What is the lac operon?

A

Deals with changes in lactose

Switches on genes that encode enzymes which are needed to metabolise that sugar

LacZ –> LacY –> LacA

Contains three genes:
- B-Galactosidase
- Permease
- Transacetylase

142
Q

What are bacterial promoters?

A

Control of transcription most commonly occurs at transcription initiation

Promoters are the sites of transcription initiation in the DNA

Promoters are recognised by RNA polymerase by having a consensus (common pattern of) DNA sequence

The first is a hexamer (6bp) at -35
Second is a TATAAT sequence at -10 (Pribnow box)

  • Asymmetric (only in 1 DNA strand), hence RNA polymerase knows which way to go!
143
Q

RNA Polymerase Structure?

A

Holoenzyme (complete enzyme) consists of 5 types of subunit (6 units)

  • 2 alpha subunits (40 kD) – enzyme assembly
  • beta and beta’ = form catalytic centre
  • Omega subunit – enzyme assembly and stability
  • Sigma factor (32- 90 kD) – binds promoter

Core enzyme = 2 alpha, beta, beta’, omega

The core enzyme has a general affinity for DNA- this is known as loose binding

positively charged (Mg2+ and Zn2+ bound ions which has affinity for the negatively charged DNA

The σ (sigma) unit ensures RNA polymerase only binds at promoter sequences

It increase 1000 X binding strength

There is enough σ unit for 1/3 of all of the RNA polymerases

144
Q

What are alternative sigma factors?

A

e.g., Heat Shock

σ32 is induced by high temperatures

σ32 is induced by the accumulation of unfolded proteins

σ32 recognises a different -35 and -10 sequence

145
Q

What are the stages of the bacterial growth curve?

A

Lag Phase
- No increase in number of living cells

Log phase
- Exponential increase in number of living bacterial cells

Stationary phase
- Plateau in number of living cells; rate of cell division and death roughly equal

Death or decline phase
- Exponential decrease in number of living bacterial cells

sigma70 in log phase
sigma38 replaces it near stationary phase

146
Q

What is a regulon?

A

A regulon is a group of genes that are regulated as a unit

This is different from an operon.

Commonly studied regulons in bacteria are those involved in response to stress such as heat shock.

The heat shock response in E. coli is regulated by the sigma factor σ32 (RpoH), whose regulon has been characterized as containing at least 89 open reading frames.

Hence, a regulon involves a more diverse set of genes that are regulated at the same time.

147
Q

Transcription summary

A

Initiation:
RNA polymerase binds to the promoter
- Consensus DNA sequence .
- Regulates transcription.

  1. Elongation:
    RNA polymerase makes an RNA copy complementary to the template strand.
    - Bacterial polymerase: only 1, with 5 units (2x a, b, b’, w ,σ).
    - Operons, no introns, no post-transcriptional modification of RNA.
  2. Termination:
    RNA polymerase stops when it recognises termination sequences (prokaryotes).
148
Q

Initiation in bacteria

A

RNA polymerase binds to the promoter

Unwinds the DNA

Transcription begins

Regulates transcription

149
Q

Elongation in bacteria

A

RNA polymerase makes an RNA copy complementary to the template strand

Bacterial RNA polymerase: only 1, with 5 units (sigma)

150
Q

Termination in bacteria

A

RNA polymerase stops at a termination sequence

RNA polymerase separates

RNA released

151
Q

What are the types of transcriptional regulators?

A

Transcription is controlled by regulatory proteins (transcription factors)

a) Negative regulation - transcriptional repressors
- Repressor binds to sites known as Operator sites
- Stops RNA polymerase binding

b) Positive regulation - Transcriptional activators
- Activator binds to specific site
- Helps RNA polymerase bind

152
Q

Lac Operon regulation

A

Regulated by both negative and positive regulators

Negative = lac repressor (lacI)
Positive = Catabolite activating protein (CAP)

LacI, the lac repressor, is said to be constitutively expressed hence it is always on hence lacI is always present

The lacI gene encodes the lac repressor

The lac repressor is the protein that binds down stream and its job is to control transcription of the lactose operon

153
Q

What happens to the lac repressor when lactose is absent?

A

Lac repressor is produced

Lac repressor binds to the operator

Lac operon transcription blocked

154
Q

What happens to the lac repressor when lactose is present?

A

Allolactose (derived from lactose metabolism) is produced

Allolactose binds to the lac repressor

This induces a conformational change in LacI

Repressor cannot bind to operator

155
Q

How does lac repressor regulate

A

Lac repressor binds to the operator site which overlaps the start site of transcription

The sequence where lac repressor binds and the first A is where transcription starts and this is labelled plus 1

156
Q

How many operator sites does the lac operon have?

A

3

157
Q

Lac repressor binding to operator sites

A

Lac repressor binds to the operator site which overlaps the start site of transcription

The lac repressor must bind 2 out of 3 sites

  • Repressor can bind to O1 and O2 or O1 and O3
  • But not to O2 and O3
158
Q

CAP function

A

When glucose is available it is used in preference to other sugars (e.g., lactose)

Low Glucose –> cAMP High –> CAP Active

High Glucose –> cAMP Low –> CAP Inactive

When cAMP is high it binds to CAP and causes conformational change allowing it to bind upstream of the -35 sequence and stabilizes RNA polymerase

159
Q

Termination of transcription

A

Termination sequences in DNA tell RNA polymerase when to stop transcription

Also known as intrinsic terminators

These are run of A-Ts in the template strand in around 50% genes

160
Q

What is the stem-loop structure?

A

Stem loops formed by inverted repeats

E.g. CCCCXXXGGGG

Inverse sequences result in folding of strand

You find these in the 3 prime control region of a gene

The stem-loop structure causes RNA polymerase to pause

When the Polymerase pauses the RNA:DNA hybrid unravels from the weakly bonded A: U terminal region

The sequence of the hairpin and length of the U rich region determine the efficiency of termination (2 -90%)

161
Q

Rho dependant termination

A

Rho is a prokaryotic transcription protein
~275kD hexamer

Each subunit has a RNA binding domain & an ATP hydrolysis domain: moves along the RNA

Also requires an inverted repeat
Stalls the RNA polymerase

Rho is a helicase
Unwinds DNA:RNA hybrids

162
Q

From the perspective of transcription, why is regulation important?

A

Expressing all genes at once is a waste of resources. Biology prefers not to waste energy as organisms are always in competition. Those organisms that can save energy while still carrying out their essential functions have an advantage over other competing organisms.

Overproduction of some proteins are likely to be toxic. A simple example is ion channels or transporters.

Knowing how systems are regulated allows us the ability to manipulate the system. Protein over-expression systems are used extensively in the Biotech/Pharma industries.

Development and growth requires certain groups of proteins to be present at certain times.
This can define how cells ultimately become tissue, how tissues become organs and how organisms are formed.

Responding to external stimuli can require changes in the proteome. For example, we all produce more melanin when exposed to UV light, apart from the congenital absence of melanin in an animal (albinism) which is known to reduce the survivability of animals.

163
Q

How do prokaryotes regulate their gene expression?

A

Some genes are constitutively expressed - on all the time e.g., lacl.

Other genes are regulated in a positive and negative way e.g., see details of the lac operon.

164
Q

Which strand is the helicase on in DNA replication?

A

Prokaryotes:
- Lagging strand

Eukaryotes:
- Leading strand

165
Q

What is the difference between eukaryotes and prokaryotes (transcription)

A

Eukaryotes seperate transcription (nucleus) and translation (cytoplasm) which allows for more control and complexity

166
Q

Eukaryote RNA polymerases

A

Eukaryotes have 3 types of RNA polymerases

RNA Pol 1
- Structurally important

RNA Pol 2
- All genes that encode proteins
- snoRNA (small nucleolar - non-coding)
- snRNA (small nuclear - pre-mRNA splicing)

RNA Pol 3
- Structurally important

167
Q

What is the Typical arrangement of a eukaryotic promoter region

A

Core promoter elements:
- TATA Box (-25)
- Inr (+1)
- DPE (+25)

Regulatory elements:
Proximal
- CAAT Box
- GC Box

Distal
- Enhancer Elements

168
Q

Core promoters in eukaryote region

A

Only 30% of genes have TATA Box

Other 70% lack it and are known as TATA-less

These genes have an Inr (initiator) and a DPE (downstream promoter element) usually located downstream (generally +28 and +32) and contain sequence AGAC

Recognised by TF2I

169
Q

Typical arrangement of a TATA core promoter

A

TATA Box and INR (initiator) make up the ‘Core Promoter Region’

INR:
- Simplest functional promoter
- Can initiate basal transcription in absence of TATA box
- Conserved Y is pyrimidine
- Consensus sequence YYANWYY in humans, where, Y = C/T, W = A/ T, N=A/C/G/T, and +1 is underlined.
- The INR element facilitates binding of TFIID

TATA Box:
- Conserved sequence TATAAAA
- Approx 25 bp upstream of initiator
- Found in ~30% of promoters
- Core promoter similar concept in Pribnow box (prokaryotes) and TATA box (eukaryotes).

170
Q

What is an RNA Pol II Pre initiation complex?

A

Eukaryotic RNA Pol II does not bind directly to the TATA box region of the DNA, it requires proteins called “General Transcription Factors (GTF)” to position it

Each transcription factor is a complex of polypetides

TFIID begins whole process

Need to assemble an RNApol II pre-initiation complex

This positions the RNA pol II over transcription start sites.

171
Q

How is the RNA Pol II Pre initiation complex formed?

A

General transcription factors for RNA Pol II (TFII) bind to the TATA box

The first transcription factor added to the TATA box is TFIID

The TFIID complex consists of TBP (TATA binding protein) and TAF (TATA associated factors) and acts as a scaffold for remainder of the preinitiation complex

Next, TFIIA binds followed by TFIIB

TFIIB is able to then recruit RNA Pol II to the TATA Box

TFIIF accompanies the RNAPol II and stabilises it

TFIIE and TFIIH bind and make up the initiation complex

Pre- initiation complex (PIC) is now assembled

172
Q

What is TFIIH and TFIIE Function?

A

The function of TFIIH is a helicase, it melts the DNA and separates the strands

A short sequence of pre-mRNA is added to the gap

TFIIH and TFIIE also have kinase activity

Phosphorylation of the C terminal domain of RNApol II by TFIIH then occurs

After phosphorylation, TFs are released from the complex

The phosphorylation also causes a conformational change in RNApol II, which tightens the grip it has, also allows binding of new factors (elongation factors) that increase efficiency

173
Q

What determines cell fate?

A

Cell fate is determined by:
- Gene expression
- Cell-cell/cell-matrix interactions
- External factors e.g. hormones

WHICH CAUSES:
- Unique transcriptional factor (TFs) environment for each cell type

174
Q

What are upstream sequence elements (USE)?

A

Transcription can be enhanced by the binding of transcription factors to sites upstream of the PIC

Upstream sequence elements (USE)

Examples:

GC Box:
- GGGCGG
- Binds SP1 TF

CAAT Box:
- GG(T/C)CAATCT
- Binds CAAT box TF

These must be in same orientation as the RNA pol II initiation site

Upstream position is important.

175
Q

What are enhancers (eukaryotes)

A

Regulatory sequences that act at a distance

Cis acting (up to 1Mb away) with reversible orientation

Bound by activator proteins

These interact with the mediator complex

Encourage binding of RNApol II

Enhancers work because DNA structure is complex and twisted around itself, so regions distant in sequence may be physically close to each other

176
Q

What are some properties of enhancer elements?

A
  • They can activate transcription when placed thousands of bp away from the TATA box
  • They act in either orientation
  • Can act when placed upstream or downstream of the TATA box, or when placed within an intron
177
Q

What are transcription factors?

A

Determine whether transcription occurs

Determine cell specificity

Confer response to specific timed stimuli

Structure is central, determined by the exact sequence of amino acids

Mechanisms:
- How they initiate transcription can be difficult to determine due to TFs being expressed at low levels in cells

178
Q

What are transcription factors structure

A

TFs are proteins made up of amino acids hence their 3D structure is important to their function

Modular structure:
- One region binds DNA
- All the amino acids for DNA binding in 1 region
- Another region binds to other components
- Also contains activation or inhibitory domains

179
Q

What are the TFs DNA binding domains

A
  • Zinc fingers
  • Helix turn helix
  • Basic binding domains

Binding mostly based on a-helices fitting into DNA grooves

180
Q

What are zinc fingers?

A

Contains a loop of 23 aa

Usually have multiple zinc fingers per TF

The linker between the fingers is 7-8 aa

a-helix contacts the major groove of DNA

Often multiple zinc fingers involved in binding the specific DNA sequence.

Zn2+ ion does not directly interact with the DNA but is essential for the folding of the finger.

Zinc fingers bind both to the major and minor grooves.

181
Q

What is a helix turn helix

A

Two helices held at a fixed angle

Recognition helix binds major groove of DNA

Bind DNA as dimers, so the 2 recognition helices are separated by one turn of the DNA helix

182
Q

What are basic +ve charged binding domains

A

Transcription factors with basic binding domains cannot bind to DNA alone

Transcription factors with basic binding domains must dimerise