Telomerase solves the end-replication problem of linear chromosomes because it

is a reverse transcriptase that carries its own RNA template and extends the 3′ G-rich overhang

is a DNA-dependent DNA polymerase that fills the gap left after removal of the last lagging-strand primer

is a ligase that joins the two strands at each chromosome end into a covalently closed hairpin

is a primase that lays a final RNA primer at the very end of the template so the strand is completed

A temperature-sensitive *E. coli* mutant is shifted to the restrictive temperature. It keeps synthesising DNA, but pulse-labelled DNA stays as fragments of 1 to 2 kb that carry no RNA. Which gene product is most likely defective?

DNA ligase, because the nicks between finished Okazaki fragments cannot be sealed

DNA polymerase I, because its 5′→3′ exonuclease is needed to remove the primer from each fragment

DnaB helicase, because without unwinding the lagging-strand template is never exposed

DNA gyrase, because positive supercoils ahead of the fork prevent the fragments from being joined

DNA ligase, because the nicks between finished Okazaki fragments cannot be sealed

Rifampicin is added to an *E. coli* culture in which a gene is being actively transcribed. What happens to transcription of that gene over the next few minutes?

Chains already elongating are completed, but no new chains start, because rifampicin blocks early RNA extension

All RNA chains stop at once, because rifampicin intercalates into the template DNA ahead of each polymerase

Chains already elongating are completed, but no new chains start, because rifampicin blocks early RNA extension

Only Rho-dependent termination fails, so the polymerases read through into the neighbouring genes downstream

Transcription continues normally, because rifampicin binds the 30S ribosomal subunit and blocks only translation

Which of the following statements about transcription and RNA processing are correct?

The 5′ cap of eukaryotic mRNA is attached to the first nucleotide through a 5′–5′ triphosphate linkage

In the first step of splicing, the 2′-OH of the branch-point adenosine attacks the 5′ splice site

An intrinsic bacterial terminator is a GC-rich hairpin in the RNA followed by a run of U residues

The 5′ cap of eukaryotic mRNA is attached to the first nucleotide through a 5′–5′ triphosphate linkage

In the first step of splicing, the 2′-OH of the branch-point adenosine attacks the 5′ splice site

An intrinsic bacterial terminator is a GC-rich hairpin in the RNA followed by a run of U residues

RNA polymerase I is the eukaryotic nuclear polymerase most sensitive to inhibition by α-amanitin

A tRNA has the anticodon 5′-IGC-3′, where I is inosine. Which mRNA codons can it read?

GCU, GCC and GCA, because inosine at the wobble position pairs with U, C or A

GCU, GCC, GCA and GCG, because inosine pairs with any of the four bases

UCG, CCG and ACG, because inosine pairs with U, C or A at the first codon position

GCU, GCC and GCA, because inosine at the wobble position pairs with U, C or A

GCC only, because every anticodon base pairs strictly by Watson–Crick rules

Why are methylated CpG sites hotspots for C·G → T·A transitions in mammalian DNA?

Deaminating 5-methylcytosine gives thymine, a normal DNA base that uracil-DNA glycosylase ignores

The methyl group of 5-methylcytosine lets it pair directly with adenine during replication

Deaminating 5-methylcytosine gives uracil, which is excised so quickly that a gap is left behind

5-Methylcytosine stalls the replicative DNA polymerase, which then inserts adenine opposite it by default

A heteroduplex plasmid carrying a single G·T mismatch is made so that the strand containing G is fully methylated at its GATC sites and the strand containing T is unmethylated. It is introduced into wild-type *E. coli* and repaired before replication. Which base pair do most repaired molecules carry at the mismatch?

G·C, because MutH nicks the unmethylated strand, so the T-containing strand is excised and resynthesised

A·T, because MutH nicks the methylated strand, marking the G-containing strand as the one to be replaced

G·C, because MutH nicks the unmethylated strand, so the T-containing strand is excised and resynthesised

G·C and A·T in equal numbers, because mismatch repair has no means of telling the two strands apart

Neither, because MutS cuts out the mismatched pair and the gap is sealed without replacing the bases

Which of the following statements about DNA repair pathways are correct?

Photolyase uses light energy to split cyclobutane pyrimidine dimers without removing any nucleotides

O⁶-Methylguanine-DNA methyltransferase transfers the methyl group to its own cysteine and is inactivated

Photolyase uses light energy to split cyclobutane pyrimidine dimers without removing any nucleotides

O⁶-Methylguanine-DNA methyltransferase transfers the methyl group to its own cysteine and is inactivated

Non-homologous end joining copies the missing sequence from the sister chromatid, so it is error-free

In base excision repair, the glycosylase cuts out an oligonucleotide of 12 to 13 nucleotides

A 250-bp DNA fragment labelled at one 5′ end is incubated with increasing amounts of a purified protein and then lightly digested with DNase I. On a denaturing gel the control lane shows a band at every position, but with protein the bands for positions 81 to 100 disappear and the band at position 78 becomes much stronger. Which is the best interpretation?

The protein covers positions 81 to 100 and distorts the DNA beside them, making position 78 hypersensitive

The protein is itself a nuclease that cuts at position 78 and destroys the DNA between 81 and 100

DNase I was used at too high a concentration, so molecules cut many times lost the bands from 81 to 100

The protein binds only at position 78, and positions 81 to 100 are a sequence that DNase I cannot cut

The protein covers positions 81 to 100 and distorts the DNA beside them, making position 78 hypersensitive

Which of the following statements about methods that probe chromatin accessibility are correct?

Light micrococcal nuclease digestion of chromatin gives DNA fragments that are multiples of the nucleosome repeat length

Active promoters and enhancers commonly appear as DNase I hypersensitive sites in nuclei

Light micrococcal nuclease digestion of chromatin gives DNA fragments that are multiples of the nucleosome repeat length

Active promoters and enhancers commonly appear as DNase I hypersensitive sites in nuclei

In DNase I footprinting, the fragment should be labelled at both ends so that both strands are read in one lane

ATAC-seq needs an antibody against the protein of interest to direct the Tn5 insertions

An *E. coli* partial diploid has the genotype I⁺ Oᶜ Z⁻ / I⁻ O⁺ Z⁺ (I, repressor gene; O, operator; Z, β-galactosidase gene). How is the synthesis of functional β-galactosidase regulated in this strain?

It is inducible, because the I⁺ repressor acts in trans on the O⁺ copy that carries Z⁺

It is constitutive, because the Oᶜ operator on one chromosome cannot bind repressor

It is never made, because the I⁻ allele makes no repressor that could respond to inducer

It is made only without inducer, because the I⁻ repressor responds to inducer in reverse

It is inducible, because the I⁺ repressor acts in trans on the O⁺ copy that carries Z⁺

In the trp operon leader of *E. coli*, what happens when tryptophan is plentiful?

The ribosome translates past the Trp codons and covers region 2, so the 3:4 terminator hairpin forms

The ribosome stalls at the two Trp codons, so region 2 pairs with region 3 and transcription continues

The ribosome translates past the Trp codons and covers region 2, so the 3:4 terminator hairpin forms

Tryptophan binds the leader RNA directly, so region 1 pairs with region 4 and blocks the ribosome

The TrpR repressor binds the leader RNA, so the polymerase pauses until region 2 pairs with region 4

When cellular iron falls, iron regulatory proteins (IRPs) bind iron-responsive elements (IREs). What is the effect on ferritin and transferrin receptor synthesis?

Ferritin translation is blocked through the IRE in its 5′ UTR, and transferrin receptor mRNA is stabilised through IREs in its 3′ UTR

Ferritin mRNA is stabilised through IREs in its 3′ UTR, while transferrin receptor translation is blocked through an IRE in its 5′ UTR

Both mRNAs are degraded, because IRP binding recruits the RNA-induced silencing complex to each IRE

Both mRNAs are translated faster, because IRP binding unfolds the secondary structure of their untranslated regions

Gene Expression and Its Regulation, and DNA Repair: Replication, Transcription, Translation, Repair and Chromatin Accessibility

This chapter answers Section XL1.6 of XL1 Biochemistry, one of the five optional sections of the Life Sciences (XL) paper, of which a candidate answers any two. The heading reads: “DNA replication, transcription, and translation; DNA damage and repair; Nuclease accessibility methods; Biochemical regulation of gene expression.” It follows information from DNA to protein, with the enzymes, the energy and the inhibitors of each step; the lesions DNA suffers and the pathways that remove them; how DNase I, micrococcal nuclease and the Tn5 transposase reveal which DNA is bare and which is protected; and how bacteria and eukaryotes switch genes on and off. The numericals are replication time and origin counts from fork speed, Meselson–Stahl band fractions, codon and amino acid counts, the energy cost of protein synthesis and nucleosome linker lengths.

1. DNA replication

Replication is semiconservative. Meselson and Stahl grew E. coli on ¹⁵N, shifted it to ¹⁴N and separated DNA by density in a CsCl gradient: after one generation all DNA was of hybrid density; after two, half was hybrid and half light. A conservative model predicts separate heavy and light bands after one generation, and a dispersive model a single band at every generation. After n generations the hybrid fraction is 2/2ⁿ.

In E. coli replication starts at one origin, oriC, and runs in both directions. DnaA binds repeated sites in oriC and melts the AT-rich region; DnaC loads the DnaB helicase; single-strand binding protein coats the separated strands; DNA gyrase (a type II topoisomerase, the target of quinolones) removes the positive supercoils that build up ahead of the fork. Primase (DnaG) lays RNA primers. DNA polymerase III, held on the template by the β sliding clamp, makes the leading strand continuously and the lagging strand as Okazaki fragments (1,000–2,000 nucleotides in bacteria, 100–200 in eukaryotes). DNA polymerase I removes primers with its 5′→3′ exonuclease and fills the gaps, and DNA ligase (NAD⁺-dependent in E. coli, ATP-dependent in eukaryotes) seals the nicks.

Every DNA polymerase extends 5′→3′ from a primer with a free 3′-OH; none can start a chain.
Fidelity comes from base selection, 3′→5′ exonuclease proofreading (the ε subunit of Pol III) and mismatch repair, leaving about one error per 10⁹–10¹⁰ nucleotides.
Eukaryotes license many origins in G1: ORC, Cdc6 and Cdt1 load the MCM2–7 helicase, which kinases activate only in S phase, so each origin fires once per cycle. Pol α-primase starts chains, Pol ε extends the leading and Pol δ the lagging strand, and the clamp is PCNA.
Linear chromosomes lose sequence at their ends because the last primer cannot be replaced. Telomerase, a reverse transcriptase carrying its own RNA template, extends the 3′ G-rich overhang in germ cells, stem cells and most cancers.

ℹ️ Worked example: replication time

With one origin and two forks each moving at 1,000 nucleotides per second, the 4.6 Mb E. coli chromosome needs 4.6 × 10⁶ / (2 × 1,000) = 2,300 s, about 38 minutes. Cells dividing every 20 minutes start new rounds before the previous round ends. Eukaryotic forks are far slower, so a long chromosome needs many origins: origins = genome length ÷ (2 × fork speed × S-phase time).

2. Transcription and RNA processing

Bacterial RNA polymerase core (α₂ββ′ω) becomes the holoenzyme with a σ factor; σ⁷⁰ recognises the −35 (TTGACA) and −10 (TATAAT) promoter elements, the duplex opens near −10, and after some abortive starts σ is released and the core elongates 5′→3′ without a primer. Termination is either intrinsic, a GC-rich hairpin followed by a run of U in the RNA, or Rho-dependent, where the hexameric Rho helicase binds a C-rich site on the RNA and uses ATP to catch the polymerase. Rifampicin binds the β subunit and blocks the RNA path once a few nucleotides are made, so it stops initiation but not chains already elongating; actinomycin D intercalates into DNA and blocks elongation in all cells.

Eukaryotic nuclear RNA polymerases
Polymerase	Transcribes	α-Amanitin
RNA polymerase I	28S, 18S and 5.8S rRNA (in the nucleolus)	Insensitive
RNA polymerase II	mRNA, most snRNAs, miRNA precursors	Very sensitive
RNA polymerase III	tRNA, 5S rRNA, U6 snRNA	Moderately sensitive

Pol II needs general transcription factors: TFIID, whose TATA-binding protein binds the TATA box about 25–30 bp upstream, TFIIB, TFIIF, TFIIE and TFIIH, whose helicase opens the DNA and whose kinase phosphorylates serine 5 of the heptad repeats of Pol II's C-terminal domain (CTD), releasing the polymerase into elongation. The phosphorylated CTD recruits the processing machinery, so pre-mRNA is processed as it is made:

5′ cap: 7-methylguanosine joined by a 5′–5′ triphosphate bridge; it protects against 5′ exonucleases and helps recruit the ribosome.
Splicing by the spliceosome (U1, U2, U4, U5, U6 snRNPs): introns begin with GU and end with AG. The 2′-OH of a branch-point adenosine attacks the 5′ splice site, forming a lariat, and the freed exon's 3′-OH then attacks the 3′ splice site. Alternative splicing lets one gene make several proteins.
3′ end: cleavage 10–30 nucleotides downstream of AAUAAA and addition of a poly(A) tail of about 200 A by poly(A) polymerase, without a template.
Editing changes the coded message: in the human intestine a C→U deamination creates a stop codon in apolipoprotein B mRNA, giving apoB-48 instead of the liver's apoB-100.

3. Translation

The genetic code is read in non-overlapping triplets from a fixed start: 64 codons, 61 for the 20 amino acids and 3 stops (UAA, UAG, UGA); AUG encodes methionine and starts nearly all proteins. The code is degenerate and nearly universal (mitochondria have exceptions, such as UGA read as tryptophan). Wobble: the first (5′) anticodon base pairs loosely with the third codon base, so G pairs with C or U, U with A or G, and inosine with U, C or A; one tRNA can therefore read several codons.

Each amino acid is attached to its tRNA by an aminoacyl-tRNA synthetase in two steps (amino acid + ATP → aminoacyl-AMP + PPi, then transfer to the 3′ end of the tRNA), costing two high-energy bonds. Synthetases proofread: isoleucyl-tRNA synthetase hydrolyses valine wrongly attached to tRNA(Ile). The ribosome checks only codon–anticodon pairing, not the amino acid. Bacterial 70S ribosomes have 30S (16S rRNA) and 50S (23S and 5S rRNA) subunits; eukaryotic 80S ribosomes have 40S (18S) and 60S (28S, 5.8S and 5S). The peptidyl transferase centre is 23S rRNA: the ribosome is a ribozyme.

Initiation (bacteria): the 30S subunit with IF1, IF3 and IF2·GTP binds mRNA by base pairing between the Shine–Dalgarno sequence and the 3′ end of 16S rRNA, and fMet-tRNA enters the P site. In eukaryotes eIF4E binds the cap and the small subunit, carrying Met-tRNAi with eIF2·GTP, scans to the first AUG in a good Kozak context.
Elongation: EF-Tu·GTP delivers each aminoacyl-tRNA to the A site (EF-Ts recycles EF-Tu); peptidyl transferase moves the growing chain onto the A-site amino acid; EF-G·GTP translocates the ribosome by one codon. The eukaryotic equivalents are eEF1A, eEF1B and eEF2.
Termination: RF1 reads UAA and UAG, RF2 reads UAA and UGA, and RF3·GTP assists release; eukaryotic eRF1 reads all three stops.

⚠️ Energy and inhibitors

Counting 2 bonds for charging and one GTP each for EF-Tu and EF-G, each peptide bond costs at least 4 high-energy bonds. Inhibitors act at named steps: streptomycin (30S, misreading), tetracycline (blocks aminoacyl-tRNA binding to the A site), chloramphenicol (bacterial peptidyl transferase), erythromycin (50S exit tunnel), cycloheximide (eukaryotic translocation), diphtheria toxin (ADP-ribosylates eEF2) and puromycin, an aminoacyl-tRNA mimic accepted in the A site of all ribosomes, which releases the chain prematurely with puromycin attached.

4. DNA damage and repair

Common lesions and their consequences
Damage	Cause	Consequence if unrepaired
Depurination	Spontaneous hydrolysis of the glycosidic bond	Abasic (AP) site; any base may be inserted opposite
Deamination	Spontaneous; nitrous acid	C → U (pairs with A); 5-methylcytosine → T, a normal base, so methylated CpG sites are mutation hotspots
Oxidation	Reactive oxygen species	8-oxoguanine pairs with A: G·C → T·A transversion
Alkylation	EMS, nitrosamines	O⁶-methylguanine pairs with T: G·C → A·T transition
Pyrimidine dimers	UV light	Cyclobutane dimers and 6-4 photoproducts distort the helix and block polymerases
Strand breaks, cross-links	Ionising radiation; cisplatin	Double-strand breaks lose chromosome segments; cross-links block replication
Intercalation	Acridines, ethidium	Insertion or deletion of one base: frameshift

Direct reversal: photolyase (FADH⁻ cofactor; absent in placental mammals) uses visible light to split pyrimidine dimers; O⁶-methylguanine-DNA methyltransferase (Ada in E. coli, MGMT in humans) moves the methyl group onto its own cysteine and is used up, a suicide enzyme.
Base excision repair (BER) for small base lesions: a DNA glycosylase (for example uracil-DNA glycosylase) cuts the glycosidic bond, AP endonuclease nicks the backbone, a polymerase (Pol β in mammals, Pol I in E. coli) fills one or a few nucleotides, and ligase seals. Using thymine instead of uracil in DNA lets every U be recognised as damage.
Nucleotide excision repair (NER) for bulky, helix-distorting lesions such as UV dimers: E. coli UvrA₂B finds the lesion, UvrC cuts on both sides, and UvrD removes 12–13 nucleotides that Pol I and ligase replace. Humans excise 24–32 nucleotides using XP proteins and TFIIH, globally or coupled to transcription. Defects cause xeroderma pigmentosum: extreme sun sensitivity and skin cancers.
Mismatch repair (MMR) for replication errors: MutS binds the mismatch, MutL joins, and MutH nicks the unmethylated strand at a hemimethylated GATC, because Dam methylase marks the new strand only after a delay. The new strand is excised and resynthesised. Inherited defects in the human homologues (MSH2, MLH1 and others) cause Lynch syndrome (hereditary non-polyposis colorectal cancer) with microsatellite instability.
Double-strand breaks: non-homologous end joining (Ku70/80, DNA-PKcs, ligase IV) works throughout the cycle but often adds or loses a few bases; homologous recombination (RecA in bacteria; Rad51 with BRCA1 and BRCA2 in humans) resects the ends, invades the sister chromatid and copies it accurately, so it operates in S and G2.
SOS response in E. coli: RecA bound to single-stranded DNA stimulates self-cleavage of the LexA repressor, inducing repair genes and the error-prone translesion polymerase V, which copies past lesions at the cost of mutations.

5. Nuclease accessibility methods

A bound protein or a nucleosome shields DNA from enzymes that cut it, while bare DNA is cut easily. Mapping where a nuclease or a transposase cuts therefore maps protein-binding sites and open chromatin. Every such method depends on limited digestion: enough to sample the cutting sites, not enough to destroy every molecule.

DNase I footprinting maps where a purified protein binds a DNA fragment. The fragment is labelled at one end of one strand, incubated with and without the protein, and digested so lightly that each molecule is cut about once, at random. On a denaturing sequencing gel the control lane gives a continuous ladder, one band per cut position measured from the labelled end. Where the protein sits, cutting is blocked and the ladder shows a gap, the footprint; bands that strengthen at its edges mark DNA the protein bends, which becomes hypersensitive. A sequencing ladder run alongside fixes the positions, and titrating the protein gives its affinity.

In nuclei, DNase I hypersensitive sites (DHS) are stretches of a few hundred base pairs that DNase I cuts at far lower concentrations than bulk chromatin. They are nucleosome-depleted regions at active or poised promoters, enhancers, insulators and origins, and often appear before a gene is expressed; DNase-seq sequences the cut ends genome-wide. Micrococcal nuclease (MNase), a Ca²⁺-dependent nuclease, cuts the linker DNA between nucleosomes first. Light digestion gives a ladder of fragments in multiples of the nucleosome repeat length (often about 180–200 bp); further digestion trims them to the 147 bp of the core particle wrapped around the histone octamer. MNase-seq maps nucleosome positions, and the linker length is the repeat length minus 147 bp.

ATAC-seq (assay for transposase-accessible chromatin) uses a hyperactive Tn5 transposase preloaded with sequencing adapters. In one step, 'tagmentation', it cuts DNA and inserts the adapters, preferentially into open chromatin; the tagged fragments are amplified by PCR and sequenced. It works with thousands of cells rather than the millions DNase-seq uses, and its fragment sizes report both nucleosome-free regions (short fragments) and nucleosome positions (a periodicity of about one nucleosome).

⚠️ Accessibility is not identity

Accessibility shows that DNA is open, not which protein binds there. To name the protein use chromatin immunoprecipitation (ChIP) with an antibody, or footprint the purified protein. A missing band means protection only if the control lane, digested to the same extent, shows it.

6. Biochemical regulation of gene expression

The lac operon of E. coli (lacZ, β-galactosidase; lacY, permease; lacA, transacetylase) is under dual control. Negative: the LacI repressor binds the operator and blocks transcription; allolactose, made from lactose by β-galactosidase, binds LacI and releases it (IPTG is a gratuitous inducer, not metabolised). Positive: when glucose is low, cAMP rises and the cAMP–CAP complex binds beside the promoter and recruits RNA polymerase, so the operon is fully on only with lactose present and glucose absent (catabolite repression). Partial diploids show the logic: the repressor gene acts in trans (I⁺ is dominant over I⁻, and the non-inducible super-repressor Iˢ over I⁺), whereas an operator-constitutive mutation Oᶜ acts only in cis, on the genes of its own DNA molecule.

The trp operon is repressed when the TrpR repressor binds tryptophan (a corepressor) and attaches to the operator. A second control, attenuation, uses coupled transcription and translation in the leader. The leader RNA has four regions: 3:4 pairing forms a terminator, 2:3 pairing an antiterminator, and region 1 encodes a leader peptide with two adjacent Trp codons. With plenty of Trp-tRNA the ribosome passes them and covers region 2, so 3:4 forms and transcription stops before the structural genes. When Trp-tRNA is scarce the ribosome stalls on region 1, region 2 pairs with 3 and the operon is transcribed. Riboswitches achieve similar control by binding a metabolite directly in the 5′ untranslated region.

In eukaryotes chromatin sets the baseline. Histone acetylation by HATs neutralises lysine charges and recruits bromodomain proteins, opening chromatin; deacetylases (HDACs) reverse it. H3K4me3 marks active promoters, whereas H3K9me3 (bound by HP1) and Polycomb-deposited H3K27me3 mark silent chromatin; ATP-dependent remodellers such as SWI/SNF slide or evict nucleosomes. DNA methylation at CpG (DNMT3A/B de novo, DNMT1 maintenance) silences promoters and underlies imprinting and X-inactivation. Transcription factors are modular: a DNA-binding domain (helix-turn-helix, zinc finger, leucine zipper, helix-loop-helix) plus an activation domain that recruits Mediator, TFIID and coactivators. Enhancers act in either orientation and at long distances by looping, limited by insulators bound by CTCF. Steroid hormone receptors are ligand-activated transcription factors that bind hormone response elements.

miRNAs are processed by Drosha and Dicer and loaded onto Argonaute in RISC; the seed (nucleotides 2–8) pairs with the 3′ UTR of target mRNAs, repressing translation and promoting decay.
Iron: when iron is low, iron regulatory proteins (IRPs) bind iron-responsive elements (IREs). The IRE in the 5′ UTR of ferritin mRNA then blocks its translation, and the IREs in the 3′ UTR of transferrin receptor mRNA protect it from degradation; iron reverses both.
Global translation control: phosphorylation of eIF2α under stress (for example by the haem-regulated inhibitor in reticulocytes when haem is short) prevents recycling of eIF2·GTP and slows the initiation of most mRNAs.

Key takeaways

Replication is semiconservative (hybrid fraction 2/2ⁿ), bidirectional and primed; Pol III with its β clamp extends, Pol I removes primers and ligase seals; origins needed = genome length ÷ (2 × fork speed × time).
σ⁷⁰ reads the −35 and −10 elements; termination is intrinsic or Rho-dependent; Pol I, II and III make rRNA, mRNA and tRNA/5S RNA and differ in α-amanitin sensitivity; capping, splicing (GU…AG, lariat) and polyadenylation accompany Pol II transcription.
There are 61 sense codons and 3 stops; wobble lets one tRNA read several codons; charging costs 2 high-energy bonds and EF-Tu and EF-G one GTP each, about 4 per peptide bond; antibiotics block named steps.
Match lesion to pathway: dimers to photolyase or NER, O⁶-methylguanine to MGMT, small base damage to BER, mismatches to MMR (MutH nicks the unmethylated strand), breaks to NHEJ or HR; SOS induces error-prone bypass.
Nucleases read accessibility: DNase I footprints show protected bases on singly end-labelled DNA, DHS mark regulatory elements, MNase ladders give the nucleosome repeat (linker = repeat − 147 bp), and ATAC-seq tags open chromatin with Tn5.
lac: LacI acts in trans and the operator in cis, with cAMP–CAP as positive control; trp: repressor plus attenuation; eukaryotes add histone marks, DNA methylation, modular activators and enhancers, miRNAs and IRE–IRP control.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Telomerase solves the end-replication problem of linear chromosomes because it
1. is a reverse transcriptase that carries its own RNA template and extends the 3′ G-rich overhang
2. is a DNA-dependent DNA polymerase that fills the gap left after removal of the last lagging-strand primer
3. is a ligase that joins the two strands at each chromosome end into a covalently closed hairpin
4. is a primase that lays a final RNA primer at the very end of the template so the strand is completed
Show answer
Answer: A — is a reverse transcriptase that carries its own RNA template and extends the 3′ G-rich overhang
Telomerase is a ribonucleoprotein whose protein subunit is a reverse transcriptase; it copies a short segment of its own RNA into repeats added to the 3′ G-rich overhang, which ordinary lagging-strand synthesis can then partly fill in. It does not copy a DNA template, so it cannot simply fill the gap left by the last primer, which is precisely the gap no DNA polymerase can fill.
A temperature-sensitive E. coli mutant is shifted to the restrictive temperature. It keeps synthesising DNA, but pulse-labelled DNA stays as fragments of 1 to 2 kb that carry no RNA. Which gene product is most likely defective?
1. DNA polymerase I, because its 5′→3′ exonuclease is needed to remove the primer from each fragment
2. DnaB helicase, because without unwinding the lagging-strand template is never exposed
3. DNA gyrase, because positive supercoils ahead of the fork prevent the fragments from being joined
4. DNA ligase, because the nicks between finished Okazaki fragments cannot be sealed
Show answer
Answer: D — DNA ligase, because the nicks between finished Okazaki fragments cannot be sealed
Okazaki-sized fragments that have already lost their RNA primers have been processed by DNA polymerase I and await only the final phosphodiester bond, which ligase makes. A Pol I defect would leave RNA on the fragments; a DnaB or gyrase defect would halt fork movement and DNA synthesis rather than leave a pool of unjoined, primer-free fragments.
E. coli grown for many generations on ¹⁵NH₄Cl is transferred to a medium containing only ¹⁴NH₄Cl. After exactly three generations of synchronous growth, DNA is centrifuged to equilibrium in CsCl. What percentage of the double-stranded DNA molecules have hybrid (¹⁵N/¹⁴N) density?
Numerical answer — type the value.
Show answer
Answer: 25
Semiconservative replication keeps the two original heavy strands intact, each ending up in a separate hybrid duplex. After 3 generations there are 2³ = 8 duplexes per original duplex, of which 2 are hybrid and 6 light: 2/8 = 25%. A conservative model would give 1 heavy and 7 light duplexes and no hybrid, and a dispersive model a single band of intermediate density.
A eukaryotic chromosome contains 144 Mb of DNA. Each replication fork moves at 50 nucleotides per second; every origin fires once at the start of S phase and sends two forks in opposite directions; origins are evenly spaced, and forks keep moving until they meet. If S phase lasts 8 hours, what is the minimum number of origins needed to replicate the chromosome?
Numerical answer — type the value.
Show answer
Answer: 50
One origin's two forks copy 2 × 50 × (8 × 3,600) = 2 × 50 × 28,800 = 2.88 × 10⁶ bp, that is 2.88 Mb, during S phase. Origins needed = 144 / 2.88 = 50. Forgetting that each origin sends out two forks gives 100.
Rifampicin is added to an E. coli culture in which a gene is being actively transcribed. What happens to transcription of that gene over the next few minutes?
1. All RNA chains stop at once, because rifampicin intercalates into the template DNA ahead of each polymerase
2. Chains already elongating are completed, but no new chains start, because rifampicin blocks early RNA extension
3. Only Rho-dependent termination fails, so the polymerases read through into the neighbouring genes downstream
4. Transcription continues normally, because rifampicin binds the 30S ribosomal subunit and blocks only translation
Show answer
Answer: B — Chains already elongating are completed, but no new chains start, because rifampicin blocks early RNA extension
Rifampicin binds the β subunit of bacterial RNA polymerase in the channel that the RNA occupies once it is a few nucleotides long, so a newly initiated complex cannot extend its RNA, whereas a polymerase already elongating is unaffected. Transcripts in progress therefore finish and no new ones start. Intercalation into DNA is the action of actinomycin D, which stops elongating polymerases too.
Which of the following statements about transcription and RNA processing are correct?
1. The 5′ cap of eukaryotic mRNA is attached to the first nucleotide through a 5′–5′ triphosphate linkage
2. In the first step of splicing, the 2′-OH of the branch-point adenosine attacks the 5′ splice site
3. An intrinsic bacterial terminator is a GC-rich hairpin in the RNA followed by a run of U residues
4. RNA polymerase I is the eukaryotic nuclear polymerase most sensitive to inhibition by α-amanitin
Show answer
Answer: A — The 5′ cap of eukaryotic mRNA is attached to the first nucleotide through a 5′–5′ triphosphate linkage; B — In the first step of splicing, the 2′-OH of the branch-point adenosine attacks the 5′ splice site; C — An intrinsic bacterial terminator is a GC-rich hairpin in the RNA followed by a run of U residues
The cap is 7-methylguanosine joined 5′ to 5′ through three phosphates. Splicing begins when the branch-point adenosine's 2′-OH attacks the phosphate at the 5′ splice site, forming the lariat; the freed exon's 3′-OH then attacks the 3′ splice site. An intrinsic terminator is a stable GC-rich hairpin followed by U residues, whose weak rU·dA pairing lets the transcript dissociate. Pol II is the most α-amanitin-sensitive polymerase; Pol I is insensitive.
A tRNA has the anticodon 5′-IGC-3′, where I is inosine. Which mRNA codons can it read?
1. GCU, GCC, GCA and GCG, because inosine pairs with any of the four bases
2. UCG, CCG and ACG, because inosine pairs with U, C or A at the first codon position
3. GCU, GCC and GCA, because inosine at the wobble position pairs with U, C or A
4. GCC only, because every anticodon base pairs strictly by Watson–Crick rules
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Answer: C — GCU, GCC and GCA, because inosine at the wobble position pairs with U, C or A
Codon and anticodon pair antiparallel, so the anticodon's 5′ base (I) pairs with the codon's 3′ base, the wobble position. The anticodon 5′-IGC-3′ therefore reads 5′-GCN-3′, and inosine pairs with U, C or A but not G: GCU, GCC and GCA, all alanine codons. Reading the anticodon in the wrong orientation gives UCG, CCG and ACG; because inosine does not pair with G, GCG needs another tRNA.
In an mRNA, the A of the initiating AUG is nucleotide 61, and the first in-frame stop codon, UGA, occupies nucleotides 961–963. How many amino acids does the primary translation product contain, counting the initiating methionine?
Numerical answer — type the value.
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Answer: 300
Sense codons run from nucleotide 61 to 960: (960 − 61 + 1)/3 = 900/3 = 300 codons, each adding one amino acid; the stop codon adds none. Counting the stop codon as well gives 301, and dividing 963 − 61 by 3 without adding 1 gives a non-integer, a sign of an off-by-one error.
A bacterial protein of 200 amino acids is synthesised. Count 2 high-energy phosphate bonds for attaching each amino acid (including the initiator) to its tRNA, 1 GTP for each EF-Tu delivery of an aminoacyl-tRNA and 1 GTP for each EF-G translocation. Ignore initiation factors, release factors, proofreading and mRNA synthesis. How many high-energy phosphate bonds are consumed?
Numerical answer — type the value.
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Answer: 798
Charging: 200 × 2 = 400. The initiator fMet-tRNA enters the P site through IF2, not EF-Tu, so EF-Tu delivers the other 199 aminoacyl-tRNAs: 199. Each of the 199 peptide bonds is followed by one translocation: 199. Total 400 + 199 + 199 = 798. The shortcut '4 per amino acid' gives 800 and '4 per peptide bond' gives 796; the stem's counting rules give 798.
Why are methylated CpG sites hotspots for C·G → T·A transitions in mammalian DNA?
1. Deaminating 5-methylcytosine gives thymine, a normal DNA base that uracil-DNA glycosylase ignores
2. The methyl group of 5-methylcytosine lets it pair directly with adenine during replication
3. Deaminating 5-methylcytosine gives uracil, which is excised so quickly that a gap is left behind
4. 5-Methylcytosine stalls the replicative DNA polymerase, which then inserts adenine opposite it by default
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Answer: A — Deaminating 5-methylcytosine gives thymine, a normal DNA base that uracil-DNA glycosylase ignores
Deamination of cytosine gives uracil, which uracil-DNA glycosylase recognises as foreign to DNA. Deamination of 5-methylcytosine gives thymine instead, producing a T·G mismatch in which both bases are normal; unless it is repaired correctly first, replication fixes it as T·A in one daughter. 5-Methylcytosine still pairs with guanine and does not stall polymerases, and uracil arises only from unmethylated cytosine.
A heteroduplex plasmid carrying a single G·T mismatch is made so that the strand containing G is fully methylated at its GATC sites and the strand containing T is unmethylated. It is introduced into wild-type E. coli and repaired before replication. Which base pair do most repaired molecules carry at the mismatch?
1. A·T, because MutH nicks the methylated strand, marking the G-containing strand as the one to be replaced
2. G·C, because MutH nicks the unmethylated strand, so the T-containing strand is excised and resynthesised
3. G·C and A·T in equal numbers, because mismatch repair has no means of telling the two strands apart
4. Neither, because MutS cuts out the mismatched pair and the gap is sealed without replacing the bases
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Answer: B — G·C, because MutH nicks the unmethylated strand, so the T-containing strand is excised and resynthesised
Dam methylation of the A in GATC lags behind the fork, so for a while after replication the new strand is unmethylated. MutS binds the mismatch, MutL links it to MutH, and MutH nicks the unmethylated strand at a hemimethylated GATC. The T-containing strand is therefore removed and resynthesised against G, giving G·C. Nicking the methylated strand would destroy the parental information the system exists to protect, and MutS recognises mismatches but does not cut DNA.
Which of the following statements about DNA repair pathways are correct?
1. Photolyase uses light energy to split cyclobutane pyrimidine dimers without removing any nucleotides
2. O⁶-Methylguanine-DNA methyltransferase transfers the methyl group to its own cysteine and is inactivated
3. Non-homologous end joining copies the missing sequence from the sister chromatid, so it is error-free
4. In base excision repair, the glycosylase cuts out an oligonucleotide of 12 to 13 nucleotides
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Answer: A — Photolyase uses light energy to split cyclobutane pyrimidine dimers without removing any nucleotides; B — O⁶-Methylguanine-DNA methyltransferase transfers the methyl group to its own cysteine and is inactivated
Photolyase (with FADH⁻) absorbs visible light and reverses the dimer directly. The alkyltransferase moves the methyl group from O⁶ of guanine onto an active-site cysteine and cannot be regenerated, so each molecule acts once. NHEJ joins ends without a template and often adds or loses a few nucleotides; copying the sister chromatid is homologous recombination. A glycosylase removes a single base by cutting the glycosidic bond; excising a 12–13-nucleotide oligomer describes bacterial nucleotide excision repair.
A 250-bp DNA fragment labelled at one 5′ end is incubated with increasing amounts of a purified protein and then lightly digested with DNase I. On a denaturing gel the control lane shows a band at every position, but with protein the bands for positions 81 to 100 disappear and the band at position 78 becomes much stronger. Which is the best interpretation?
1. The protein is itself a nuclease that cuts at position 78 and destroys the DNA between 81 and 100
2. DNase I was used at too high a concentration, so molecules cut many times lost the bands from 81 to 100
3. The protein binds only at position 78, and positions 81 to 100 are a sequence that DNase I cannot cut
4. The protein covers positions 81 to 100 and distorts the DNA beside them, making position 78 hypersensitive
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Answer: D — The protein covers positions 81 to 100 and distorts the DNA beside them, making position 78 hypersensitive
A protein-dependent gap in an otherwise continuous ladder is a footprint: positions 81–100 (20 bp) are covered. An enhanced band beside it is a hypersensitive site, where binding bends the DNA or widens a groove so that DNase I cuts better. The control lane is cut normally at 81–100, so the sequence is cuttable; over-digestion would shorten the products in both lanes rather than remove one block only when protein is present; and a nuclease would create new bands, not a protected gap.
Which of the following statements about methods that probe chromatin accessibility are correct?
1. Light micrococcal nuclease digestion of chromatin gives DNA fragments that are multiples of the nucleosome repeat length
2. Active promoters and enhancers commonly appear as DNase I hypersensitive sites in nuclei
3. In DNase I footprinting, the fragment should be labelled at both ends so that both strands are read in one lane
4. ATAC-seq needs an antibody against the protein of interest to direct the Tn5 insertions
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Answer: A — Light micrococcal nuclease digestion of chromatin gives DNA fragments that are multiples of the nucleosome repeat length; B — Active promoters and enhancers commonly appear as DNase I hypersensitive sites in nuclei
MNase cuts linkers first, so partial digests give a ladder of mono-, di- and trinucleosome DNA. Nucleosome-depleted regulatory elements are cut at low DNase I concentrations and are the classic hypersensitive sites. Footprinting needs a single labelled end so that each band reports its distance from that end; two labelled ends would superimpose two ladders. ATAC-seq uses no antibody: Tn5 inserts wherever DNA is accessible, and an antibody is needed for ChIP.
Partial micrococcal nuclease digestion of nuclei gives a ladder with bands at 195, 390 and 585 bp. Taking the DNA of the nucleosome core particle as 147 bp, what is the average linker DNA length, in bp?
Numerical answer — type the value.
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Answer: 48
The ladder spacing is the nucleosome repeat length: 390 − 195 = 195 bp, and 585 − 390 is the same. Linker = 195 − 147 = 48 bp. Taking the core as 146 bp, or as the 166 bp chromatosome, would give other values; the stem fixes it at 147 bp.
An E. coli partial diploid has the genotype I⁺ Oᶜ Z⁻ / I⁻ O⁺ Z⁺ (I, repressor gene; O, operator; Z, β-galactosidase gene). How is the synthesis of functional β-galactosidase regulated in this strain?
1. It is constitutive, because the Oᶜ operator on one chromosome cannot bind repressor
2. It is never made, because the I⁻ allele makes no repressor that could respond to inducer
3. It is made only without inducer, because the I⁻ repressor responds to inducer in reverse
4. It is inducible, because the I⁺ repressor acts in trans on the O⁺ copy that carries Z⁺
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Answer: D — It is inducible, because the I⁺ repressor acts in trans on the O⁺ copy that carries Z⁺
Only the Z⁺ gene makes active enzyme, and it lies next to a normal O⁺ operator. The I⁺ repressor is diffusible, so it binds that operator and keeps Z⁺ off until inducer is added: the strain is inducible. Oᶜ is cis-acting, so it makes only the Z⁻ gene on its own DNA constitutive, and that gene yields inactive protein. I⁻ is recessive to I⁺, and nothing in the genotype reverses the response to inducer.
In the trp operon leader of E. coli, what happens when tryptophan is plentiful?
1. The ribosome stalls at the two Trp codons, so region 2 pairs with region 3 and transcription continues
2. The ribosome translates past the Trp codons and covers region 2, so the 3:4 terminator hairpin forms
3. Tryptophan binds the leader RNA directly, so region 1 pairs with region 4 and blocks the ribosome
4. The TrpR repressor binds the leader RNA, so the polymerase pauses until region 2 pairs with region 4
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Answer: B — The ribosome translates past the Trp codons and covers region 2, so the 3:4 terminator hairpin forms
Attenuation relies on translation of the leader being coupled to transcription. With abundant Trp-tRNA the ribosome reads through the Trp codons in region 1 to the leader stop codon and overlaps region 2, leaving region 3 free to pair with region 4 and form an intrinsic terminator, so transcription stops before trpE. A stall at the Trp codons is the response to scarce tryptophan and leads to read-through. TrpR acts at the DNA operator, not on the leader RNA.
When cellular iron falls, iron regulatory proteins (IRPs) bind iron-responsive elements (IREs). What is the effect on ferritin and transferrin receptor synthesis?
1. Ferritin translation is blocked through the IRE in its 5′ UTR, and transferrin receptor mRNA is stabilised through IREs in its 3′ UTR
2. Ferritin mRNA is stabilised through IREs in its 3′ UTR, while transferrin receptor translation is blocked through an IRE in its 5′ UTR
3. Both mRNAs are degraded, because IRP binding recruits the RNA-induced silencing complex to each IRE
4. Both mRNAs are translated faster, because IRP binding unfolds the secondary structure of their untranslated regions
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Answer: A — Ferritin translation is blocked through the IRE in its 5′ UTR, and transferrin receptor mRNA is stabilised through IREs in its 3′ UTR
When iron is low the cell needs to take up more iron and store less. IRP bound to the 5′ IRE of ferritin mRNA prevents the small ribosomal subunit from loading, so ferritin falls; IRP bound to the 3′ IREs of transferrin receptor mRNA hides it from endonucleolytic cleavage, so the mRNA lasts longer and more receptor is made. Swapping the two locations gives the opposite, wrong physiology, and IRPs do not act through RISC.