Genes in action

Question 1

What is the central dogma? What is the key biochemical property of DNA that gave rise to the central dogma?

Answer 1

Central dogma- flow of genetic information
DNA->RNA-> Protein ,DNA-> DNA
Also
RNA-> DNA( reverse transcription by reverse transcriptase, telomere maintenance)
RNA->RNA( virus replication)
Transmissible proteins/protein-> protein(BSE/mad cow disease)
Specific base pairing of AT and GC, suggested transmission of information by copying into complementary DNA/RNA strands

Question 2

How is eukaryotic DNA packaged?

Answer 2

cellular DNA-> tightly packed as chromatin, hierarchical and highly ordered.
Nucleosome- negatively charged( phosphodiester backbone) DNA wrapped around 8 positively charged histones (histones have arg and lys residues-> + charge)
Histone octamers consist of 2 copies of 4 types of histone , nucleosome closed by 9th histone, H1, acts as a clamp.
Nucleosomes form extended, then condensed chromatin , becomes scaffold associated (scaffold proteins bind A-T rich regions of DNA), then condensed scaffold associated, then forms the metaphase chromosome. Stabilised/maintained by condensin proteins (fluorescence microscopy). Sequence specific recognition only possible in extended/condensed chromatin.

Question 3

How was the amount of DNA per nucleosome determined?

Answer 3

Micrococcal nuclease digests DNA linking nucleosomes, not that wrapped around histones. Samples-> full/partial nuclease digest, remaining DNA removed from histones-> gel electrophoresis. Around 147 bp/nucleosome.

Question 4

What is the function of histone tails ?

Answer 4

Histone tails- flexible( aren’t visible in X-ray crystallography studies ), site of post translational modifications-> regulate transcription through level of DNA packing+ accessibility of sequence

Question 5

What were the 4 main experiments contributing to understanding of DNA replication?

Answer 5

Chargaff- %A=%T and %G=%T for any given organism
Watson and Crick-resolved DNA double helix with specific base pairing.
Meselson and Stahl- Semi conservative nature of replication. E.coli- grown on 15-nitrogen labelled medium until all bases had heavy nitrogen. E.coli transferred to 14-nitrogen labelled growth medium, samples removed at each generation, DNA extracted-> density gradient centrifugation, positions visualised. Gen 0- single heavy band, Gen 1-single lighter band (not conservative ) Gen 2- 2 discrete bands, 1 the same as gen 1 and 1 lighter (not dispersive-only 1 band), Gens progressed - gen 1 band remained but the lighter band (14/14 DNA) signal continuously stronger. To confirm results, denatured gen 1 DNA pre centrifugation, 2 single stranded bands observed (again not dispersive ).
Numerous- identification of replication fork( eu and prokaryotes), bidirectionality of replication
E.coli grown in tritium (H3) labelled thymidine-visualisation of replication fork by autoradiography in cells arrested mid replication
E.coli grown in low activity H3 labelled thymidine, exposed to pulse of high specific activity H3 thymidine, only picked up by recently replicated DNA-> showed 2 active replication forks.

Question 6

What are the essential initiation elements found in the prokaryotic DNA replication origin ?

Answer 6

Prokaryotes-single origin-OriC (E.coli) , minimal length 245bp
DNA unwinding elements- 3 A-T rich 13bp sequences, 2 HB-separate more readily than GC rich
DnaA recognition sites-5 9bp sequences
Dam methylase target sites- 11 GATC repeats, palindromic , A is methylated. Only oriC methylated on both strands can initiate replication-newly synthesised DNA is hemi-methylated-> replication only occurs once per cell cycle

Question 7

What are the key events in the prokaryotic initiation of DNA replication?

Answer 7

1. DnaAs complex with ATP, bind to 9bp sequences-> partial unwinding of 13bp sequences , can only bind if DNA fully methylated.
2. DnaC/helicase loader loads helicase /DnaB onto each strand of ssDNA.
3. Each helicase moves 5’ to 3’ toward replication forks
4. Primase/DnaG recruited by helicase interaction-> primisome complex.
5. Primase synthesise 10nt RNA primers, moving in 3’ to 5’ direction. RNA sequence synthesised in 5’ to 3’.
6. DNA polymerase III holoenzyme recruited onto initial RNA primer on leading strand- initiation complete
7. Primase continues to lay down primers for lagging strand synthesis, helicase unwinds DNA and displaces DnaA-ATP. Primase continues to lay down primers periodically for lagging strand synthesis

Question 8

What is the structure of DNA polymerase III?

Answer 8

17 subunits.
alpha subunit- synthesise DNA in 5’ to 3’ direction (move in 3’ to 5’). 2 alpha subunits, one synthesises leading, other synthesises lagging at same rep fork
beta subunit- dimer associated with each alpha subunit , acts as sliding clamp, contributes to some unwinding
gamma subunit- clamp loading/unloading function
tau subunit-dimerisation of 2 alpha subunits
Epsilon and theta subunit- 3’ to 2’ exonuclease activity

Question 9

How is the lagging strand synthesised ? Describe the experiments which led to this conclusion

Answer 9

Polymerase- only has 5’ to 3’ synthesising capability, antiparallel lagging strand needs to be synthesised in opposite direction, strand still synthesised with copy stand being formed 5’ to 3’-> discontinuous synthesis. Strand synthesised in fragments beginning at primers formed at top of replication fork as DNA continuously unwound .Okazaki fragments- discovered in pulse chase experiments- tritium labelled thymidine added to E.coli in a pulse, some cells immediately quenched+ harvested, others allowed to continue replicating( “chased” with unlabelled thymidine). DNA denatured-ALWAYS REMEMBER, OR STRANDS COULDN’T BE RECOGNISED-and ultra centrifuged, visualised by autoradiography. No chase->small and large fragments. Small and large fragments must be synthesised at same time.
Chase-> large fragments, larger proportion large fragments as period of chase increased. Could only result from joining together of small fragments.
(radiolabelling still used , fluorescent tagging of small molecules can interrupt function/ may not be recognise by enzymes)

Question 10

What is the proposed architecture of the replication fork ?

Answer 10

Semi-discontinuous replication requires conformational flexibility- replication of each strand occurring in opposite direction. Trombone model-> lagging strand template and primer looped, effectively synthesised in same direction as leading strand. Trombone-loop increases in size as helicase continues to unwind but polymerase synthesises complementary strand to section leading on from previous primer , then is released as polymerase alpha subunit reaches new primer. Cyclic extension and contraction of ssDNA loop.

Question 11

What are the other key proteins involved in DNA replication ?

Answer 11

single strand binding proteins- prevent pre-mature re-annealing or damage to ssDNA
topoisomerase- relieves torsional stress caused by unwinding of strand, in both pro and eukaryotes, induces single strand breaks then re-anneals
Remember exonuclease 3’ to 5’ proofreading activity of DNA pol III subunits also

Question 12

How are the okazaki fragments joined together ?

Answer 12

DNA synthesis stops when RNA primer of previous fragment reached, “nick” produced due to single lacking phospodiester bond
DNA polymerase I, recognises nick, binds to DNA-RNA hybrid, synthesises new DNA from while displacing primer using 5’ to 3’ exonuclease activity. Detaches- new nick recognised by DNA ligase, forms new phosphodiester bond.

Question 13

How is replication terminated ?

Answer 13

Ter sites, termination sequences opposite oriC, interact with replication fork-> replication fork pauses. Fully replicated chromosomes-> topologically linked, topoisomerase IV indices double stranded break-> separation of circular DNAs

Question 14

What are the common principles behind eukaryotic and prokaryotic DNA replication?(6)

Answer 14

origins of replication (though multiple in eukaryotes )
bi-directional
5' to 3' directionality 
semi conservative 
semi discontinuous
multiprotein replication complex

Question 15

What are the main differences between the 2 processes ?

Answer 15

Eukaryotic chromosomes have multiple origins of replication- much larger and eukaryotic elongation is slower. Each origin-> must only be used once per cell cycle
DNA replication occurs in S phase , defined point in cell cycle, in bacteria just initiated when environmental and nutritional conditions favour replication.
Multiple polymerases
RNA primer removal different
End replication problem-chromosomes linear

Question 16

How does initiation take place in eukaryotes? How is it linked to the cell cycle?

Answer 16

Origin recognition complex (ORC) binds to ATP, ORC-ATP complex binds to AT rich sequences in origin , allows binding of CDC6 and CDT1 (helicase loaders ).
2 of the helicase MCM2-7 loaded onto pre replication complex (inactive-unlike in prokaryotes). Helicases activated , DNA polymerases recruited +other proteins-> replisome formed
ORC, can bind at any point, but helicase loading can only occur in G1 phase when CDK activity low , and helicase activation can only occur in S phase when CDK activity is high (CDK promotes activation but prevents loading-reciprocal regulation ).

Question 17

What are the 3 different nuclear replication eukaryotic DNA polymerases ? Function?

Answer 17

DNA polymerase alpha-initiates synthesis of DNA on both leading and lagging strands. 2 small subunits-create 10 nt RNA primer, rest of DNA pol alpha then elongates with up to 30 DNA nucleotides, then detaches-polymerase switching .
DNA polymerase delta- lagging strand synthesis (okazaki fragments) , interacts with PCNA (sliding clamp protein)- increases processivity. Has 3’ to 5’ exonuclease activity
DNA polymerase epsilon- leading strand synthesis, interacts with PCNA and also has P domain, even more processive.

Question 18

How does RNA primer removal occur in eukaryotes?

Answer 18

DNA pol delta continues to synthesise, displaces primer. Detaches, leaving 5’ flap. Flap cleaved by FEN-1 (Flap endonuclease 1). Remaining nick joined by DNA ligase.

Question 19

What is the end replication problem ?

Answer 19

Lagging strand- synthesis always requires a primer to be upstream-> lagging chromosome never gets fully replicated. Telomeres, repetitive sequences (TTAGGG) cap ends of chromosomes. Shortened every round of replication until so short-> cell senescence
In certain cells-> Telomerase , ribonucleoprotein enzyme , protein reverse transcriptase subunit and integral RNA template, adds telomere repeat sequences to ends of chromosome.( telomerase activation occurs in cancer cells)

Question 20

How do the 4 DNA repair mechanisms
1.mismatch repair
2. photoactivation
3. nucleotide excision repair
4. base excision repair
work?

Answer 20

Mismatch repair-exonuclease (first defence), E.coli, mismatched base pairs recognised by MutS, working with MutL. MutH, recognises methylated GATC sequences-> ensures daughter strand repaired instead of parent. MutH, activated by MutL, nick newly synthesised strand. 3’ to 5’ exonuclease removes segment including mismatch. DNA pol III replaces removed segment, DNA ligase seals nick. Eukaryotes don’t use methylation to discern strands.
UV light exposure- UV-induces pyrimidine dimers can form , joined by cyclobutane ring between C5 and C6 (T-T dimers form most quickly, but other combinations do form)-> misincorperation during transcription/replication, arrest of replication. Photoreactivation (not mammals)- photolyase, flavin cofactor absorbs photon-> singlet excited state. Donates electron to pyrimidine dimer-> flavin free radical. Electron returned-> ground state flavin and separated pyrimidines. Direct repair of damaged bases.. Mammals, use nucleotide excision repair.
Nucleotide excision repair-distortion of DNA helix recognised by UvrA dimer , in complex with single UvrB molecule-> “scans” DNA molecule and stops at sites of damage . UvrB-> unwinds DNA, UvrA released , UvrC recruited. ss strand site excised , encompassing damage site. Replaced by DNA pol I, nick repaired by DNA ligase (like replacement of primers). Eukaryotes-> many more proteins involved
Base excision repair- repair DNA with bases damaged , e.g by oxidative stress/ionising radiation, no distortion of DNA ocurring. DNA glycolyase enzyme , recognises and indices base flipping , exposes damaged base. Glycosyl bond cleaved, base removed. Remainder of nucleotide removed by endonuclease, replaced and sealed by DNA polymerase and DNA ligase.

Question 21

What are the main features of the prokaryotic RNA polymerases?

Answer 21

Don’t require primers
Single RNA polymerase for all types of RNA
Exists in 2 forms, core and holoenzyme.
Core form , 2 alpha subunits(required for assembly) , 2 beta subunits (form a pincer clamp with each other, one has conserved motif essential for catalysis, coordinates Mg2+ and required for phosphodiester bond formation)
Can transcribe efficiently but lacks any specificity
Holoenzyme form- has additional sigma factor, DNA binding specificity. Porkaryotes have multiple sigma factors , often initiate transcription of groups of genes with related functions. sigma 70 -housekeeping genes.

Question 22

Outline the process of transcription initiation in prokaryotes (incl promoter structure)

Answer 22

Region of DNA with coding sequence-> Open reading frame. Transcription initiated upstream of ORF at transcription start site (TSS). TSS recognised by promoter.
Promoter
TSS-(+1) has purine
-10 box, 6bp sequence upstream of TSS, consensus sequence TATAAT
-35 box,6bp sequence with consensus sequence TTGACA.
Deviation from consensus sequences/distance between promoter sequences(should be 16-18nt) -> reduce efficiency of transcription
Holoenzyme initially binds non specifically, scans DNA until sigma subunit recognises -35 sequence. Holoenzyme then unwinds 14-17nt of DNA-> open promoter , also involves recognition of -10 promoter.
8-9 RNA nts pair with open template strand , then sigma dissociates-> elongation.

Question 23

Describe the processes of elongation and termination

Answer 23

Elongation- RNA nt covalently added on 3’ end of growing mRNA, high processivity-> Beta clamps 20nt of downstream DNA. Continual unwinding and processive movement.
Termination-extended pausing->termination. Can also involve dissociation of RNA polymerase.
Rho independant- GC rich RNA stem loop, upstream of run of Us, stem loop-> pausing of RNA polymerase. RNA:DNA duplex in section is weak-> transcription complex can dissociate.
Rho dependant- hexameric helicase protein (Rho factor), aids dissociation

Question 24

What are the eukaryotic RNA polymerases and how do they differ in function?

Answer 24

RNA polymerase for each type of RNA. I-rRNA, II-mRNA, III-tRNA
Each have 10-12 subunits. RNA pol II and prokaryotic core RNA polymerase , structurally v similar despite limited sequence similarity-largest 2 subunits form pincer clamp enclosing 20nt of DNA in both.
RNA pol II-> unable to specifically initiate transcription, requires general transcription factors, have sigma like properties.

Question 25

What is the structure of the RNA pol II promoter?

Answer 25

TATA box- usually 25nt upstream, recognised by TATA binding protein, part of general transcription factor D.
pyrimidine rich sequence around +1 position, not always
Many additional upstream elements, required for full activity, bind to activating transcription factors. e.g CCAAT box, -50 to -70, GC boxes, -90 position.
Enhancers, strong activating elements , can be 10-50kb from stream, can be within gene. often change structure of DNA region( DNA flexible).

Question 26

What is the pre-mRNA processing which has to occur?

Answer 26

Product of RNA polymerase transcription-> precursor
Addition of 5’ cap-RNA polymerase, pauses around 30nt into transcription-> addition of 7-methylguanine group by guanylyl transferase by 5’ 5’ triphosphate linkage. Cap-> important in splicing, translation and prevents mRNA degradation by 5’ to 3’ exonucleases (effectively adds 3’ OH onto 5’ end)
Poly(A) tail- RNA pol II continues to transcribe well beyond end of mature mRNA. AAUAAA sequence in pre -mRNA recognised as signal for cleavage, multiprotein complex cleaves and adds poly(A) tail 15-30nt downstream. Stabilises mRNA , inhibits 3’ exonucleases and required for translation
Splicing- Eukaryotic DNA has introns and exons( prokaryotic DNA couldn’t-co transcriptional translation) introns must be removed and exons spliced together. Spliceosome- large ribonucleoprotein. snRNPs recognise GU (5’) and AG (3’) at ends of intron , assemble spliceosome across intron. Spliceosome (similar in size +complexity , also has RNA based catalytic centre- U2 and U6 snRNAs ). 2 sequential chemical reactions, exons spliced together, intron removed in looped lariat form. Must be v accurate or shifts reading frame.

Question 27

What are the key features of the genetic code?

Answer 27

Colinearity- 5’ to 3’ and N to C terminus
Triplet code
Non overlapping
Redundant ( 3rd base of a codon can often be changed without consequence)
Start codon AUG, methionine
Stop codons UAA, UAG, UGA
Nearly universal( species prefer certain codons for each amino acids, vertebrate mitochondria have slightly different codons for some amino acids compared to standard code-AUA can also be a start codon )

Question 28

What were the key experiments behind these key features?

Answer 28

Already knew it couldn’t be 1/2 base codons, didn’t know if 3 or more- carried out deletion experiments ,+1/-1-misfunctional protein (shift rf), +1 and -1 functional protein if close together, +3/-3, functional protein (only if mutations close together)
Specific amino acids codon relate to investigated by synthesising RNA with specific codes and analysing protein product, e.g polyU
Non overlapping- sequencing mutated RNA , determine how many amino acids had changed

Question 29

What is the open reading frame and how many reading frames are there in a code? How is the correct reading frame chosen?

Answer 29

Proteins encoded by open reading frame that start with AUG and end in stop codon, have 3 potential reading frames in sequence( 4th frame shift would just return to first reading frame ), only 1 has long ORF without premature stop codon. For double stranded DNA segemnt , 3 reading frames on each strand, effectively 6 reading frames

Question 30

Describe the structure of rRNA/ the ribosome

Answer 30

rRNAs->2/3 mass of ribosome. High levels of primary and secondary level conservation between species, provide key function of ribosome and determine structure. Ribosomal proteins-> auxillary roles. Bacteria-> 23s rRNA provides core petidyl transferase activity.
Ribosome, large and small subunit, assembles on mRNA(80s eukaryotic, 70s prokaryotic). Has 3 binding sites, A-aminoacyl-tRNA site, incoming tRNAs bind here,
P-Petidyl-tRNA site
E-exit site

Question 31

Describe the structure of the tRNA molecules

Answer 31

tRNA molecules-> function as adaptors, short, 70nt, cloverleaf secondary structure and L shape tertiary structure. 3’ end->CCA , post transcriptionally added, amino acid binding site (ester bond via hydroxy group on 3’ nucleotide, carboxyl group of amino acid). Attached by aminoacyl-tRNA synthetase enzymes, high specificity, uses ATP, releases AMP and pyrophosphate. Have many unusual bases(modified bases normally pyrimidines, more easily modififed) e.g inosine, more flexible at hydrogen bonding than G, often in wobble position of anticodon-> can recognise wider range of codons.

Question 32

How is translation initiated in prokaryotes? How does this relate to ability of prokaryotes to have polycistronic mRNA?

Answer 32

Initiation-> always methionine tRNA
Prokaryotes- purine rich Shine-Dalgarno sequence, 6-8nt upstream of initiator AUG, recruits ribosome by base pairing with 16S RNA of ribosome.(ensure primers never correspond to ). Binding positions 30s subunit adjacent to AUG initiator codon, helped by initiation factor 3. Initiator tRNA (formylated met tRNA) and other IFs recruited. GTP hydrolysis by IF2, release of IFs-> recruitment of 50S subunit.
Prokaryotic mRNAs are polycistronic- more than one ORF, each ORF has own SD sequence. Also due to co-linearity of transcription/ translation(both 5’ to 3’ )- translation can occur while transcription still occurring, coupled.

Question 33

Eukaryotic translation initiation?

Answer 33

40s ribosome subunit , interacts with eukaryotic initiation factors, recruit to 5’ cap structure. Complex scans mRNA until first AUG-> recognised by initiator fMet-tRNA. Initiation factors released when large subunit binds, GTP hydrolysis by eIF2. Initiation-> first AUG from m7G 5’ cap, monocistronic mRNA. Transcription and translation spatially and temporally separated. In both eukaryotic and prokaryotic initiator tRNA directly enters the ribosome P site.

Question 34

Describe the elongation process

Answer 34

Cyclical process. Ternary complex of EF-Tu:GTP:aminoacyl-tRNA enters A site.If anticodon-codon interaction occurs, GTP hydrolysis occurs-> great conformational change in EF-Tu, dissociates. free aminoacyl end of A site tRNA-> petptidyl transferase centre, positioned by A loop of 23s subunit of large subunit . Transfer of peptide from P site tRNA to amino group of aminoacyl tRNA, new peptide bond formed. EF-G: GTP recruited-> translocation of ribosome, uncharged P-site tRNA into E site, peptidyl-tRNA from A to P site as GTP hydrolysed.

Question 35

Describe the termination process

Answer 35

Termination codon entry into A site. No corresponding tRNA, instead release factors 1 and 2 recognise. Hydrolysis of GTP bound to RF3 triggers ribosome dissociation
Every step of translation process has involved GTP hydrolysis followed by conformational change.

Question 36

What are the main gene expression regulation mechanisms in bacteria ?

Answer 36

Modulation of RNA polymerase specificity via the lac operon, e.g PbrA sigma factor-> iron-stress response genes
Transcriptional activators/ repressors- e.g lac operon
(operons-tend to be functionally related genes) Lacl repressor protein, binds to operator sequence(overlaps TSS) under normal conditions. Lactose-> allolactose, binds to Lacl , stops binding to operator sequence. Some transcription. Only if glucose low->cAMP high->efficient binding of CAP( activator protein- cAMP binds+ causes conformational shape change) to promoter sequence-> transcription. CAP also stabilises binding of polymerase
Attenuation- formation of transcription termination structure in mRNA. Ribosome mediated-> can only occur in prokaryotes( co transcriptional translation, found in same compartment). e.g Trp operon. Trp abundant- can transcribe segments 1/2 (have trp), segemnts 3 and 4 form termination hairpin. Trp low- ribosome pauses on segemnt 1. segment 2 and 3 form structure-> termination loop cannot form.

Question 37

What are the main pre-translational mechanisms of gene expression regulation in eukaryotes?

Answer 37

Chromatin remodelling- DNA packaged into nucleosomes-> tight packing (30nm fibre or above) of chromatin represses transcription. Modification of histone tails -> modification of chromatin compaction. Acetylation of lysine residues, histone acetyltransferases, reduces positive charges of histones-> chromatin opens up. Methylation (and acetylation) allows binding of transcription regulatory/chromatin remodelling proteins( acetylation-bromodomain , methylation-chromodomain). For example-transcription of HO gene in s.cerevisiae-presence of SW15 and SBF required.SW15- binds to distant sites, recruits histone acetyltransferase and remodelling enzymes. Post acetylation, SBF binds to promoter, recruits RNA pol II. Deacetylation represses GAL1 transcription in yeast, recruited by transcription factor Mig1.
Methylation of cytosines-> recognised by protein which switches gene off. X chromosome inactivation, imprinting of genes , e.g IGF2, only paternal copy expressed , imprinting control region methylated so repressor protein can’t bind and IGF2 expressed.
Transcription factors-activating transcription factors required for transcription, often dimers, function by recruiting RNA pol II by GTFs , or recruiting histone acetyl transferases( e.gSW15). Level of transcription factor can change in response to stimuli, e.g steroid hormone receptors, have ligand binding domain , retained in cytoplasm by Hsp90 interaction , when hormone binds interaction broken, TF-> nucleus.
Attenuation- can also occur in eukaryotes, just protein mediated/ other pathway. e.g proto-oncogene myc-c shown to undergo attenuation.

Question 38

What is alternative splicing? How is it involved in regulation?

Answer 38

In eukaryotes, transcripts of a gene can often be spliced in multiple ways. AS regulated by ell type / developmental specificity. e.g Sex determination in drosophila. Can lead to rapid evolution of gene function
TRPV1 heat sensitive ion channel in vampire bats, novel penultimate exon spliced into bats nose- shorter form of TRPV1 activated at 31 degrees rather than normal 40

Question 39

What is the role of small RNAs in transcriptional regulation?

Answer 39

micro RNAs, processed from longer precursors, bind to target mRNAS , specific but imperfect base pairing, binding can reduce translation efficiency or reduce mRNA stability. e.g MYC targeted
siRNAS- normally base pair perfectly, cause rapid degradation of target mRNAs by RNAi (RNA interference). Can also induce histone modifications and chromatin packing. Produced from long dsRNA precursors, like viral RNAs. Can alsobe produced from transposable elements.