Molecular Biology of the Cell · Part 1 of 2 · 10 MCQs per part · 60 total questions
Part 1 of 2
Transcription and RNA Processing
Every protein in a cell was once a sequence of bases in DNA. The journey from gene to functional protein passes through RNA — a versatile intermediate that is copied, spliced, capped, and finally decoded by the ribosome. Understanding these steps reveals why a single gene can give rise to many different proteins.
6.1 From DNA to RNA: Transcription
Transcription is the process by which RNA polymerase copies one strand of DNA (the template strand) into a complementary RNA molecule. The other DNA strand, which has the same sequence as the RNA (except T instead of U), is the non-template (coding) strand. RNA is synthesised 5′→3′ using ribonucleoside triphosphates (NTPs) as building blocks, releasing pyrophosphate to drive the reaction.
Key term
Promoter
A DNA sequence upstream of a gene where RNA polymerase and associated transcription factors bind to initiate transcription; defines the direction of transcription and the start site.
In bacteria, a single RNA polymerase (with a sigma factor that confers promoter specificity) transcribes all RNA types. In eukaryotes, three RNA polymerases divide the labour: RNA pol I transcribes rRNA genes; RNA pol II transcribes protein-coding genes (mRNAs) and most small nuclear RNAs; RNA pol III transcribes tRNA genes and 5S rRNA. Eukaryotic pol II requires a large set of general transcription factors (GTFs) to assemble at the core promoter (TATA box, initiator element).
6.2 RNA Processing: Capping, Polyadenylation, and Splicing
In eukaryotes, the initial RNA transcript (pre-mRNA) undergoes extensive processing before export to the cytoplasm. Three key modifications occur:
5′ capping: A 7-methylguanosine (m⁷G) cap is added co-transcriptionally to the 5′ end. It protects the mRNA from exonucleolytic degradation and is recognised by the ribosome during translation initiation.
3′ polyadenylation: After a poly(A) signal (AAUAAA), the transcript is cleaved and a poly(A) tail of ~200 adenosines is added by poly(A) polymerase. This stabilises the mRNA and aids export.
Splicing: Non-coding intervening sequences (introns) are removed and the coding sequences (exons) are joined by the spliceosome, a large ribonucleoprotein complex composed of five snRNPs (U1, U2, U4, U5, U6). Splicing occurs via two transesterification reactions involving a branch point adenosine in the intron.
Key term
Spliceosome
A large RNA–protein complex (≥5 snRNPs, ~150 proteins) that catalyses the removal of introns from pre-mRNA via two sequential transesterification reactions, joining the flanking exons.
✒
Pause & Recall
How does alternative splicing allow a single gene to encode multiple protein isoforms?
By including or excluding different combinations of exons during splicing, the same pre-mRNA can be processed into distinct mature mRNAs, each encoding a protein isoform with potentially different structure and function. For example, the human DSCAM gene can in theory generate over 38,000 isoforms via alternative splicing.
6.3 The Genetic Code
The genetic code is the set of rules by which nucleotide triplets (codons) in mRNA specify amino acids. The code is: triplet (3 nucleotides per codon), non-overlapping, degenerate (most amino acids are encoded by more than one codon), and universal (used by virtually all organisms). Of the 64 possible codons, 61 encode the 20 standard amino acids and 3 are stop codons (UAA, UAG, UGA). AUG (methionine) serves as the universal start codon and also defines the reading frame.
6.4 tRNA Structure and Aminoacyl-tRNA Synthetases
Transfer RNAs (tRNAs) are the adaptors that decode mRNA codons into amino acids. Each tRNA has an anticodon loop that base-pairs with the complementary mRNA codon, and a 3′ CCA end that is covalently charged with the appropriate amino acid by aminoacyl-tRNA synthetases (aaRS). There are 20 aaRS enzymes (one per amino acid), each recognising its cognate tRNA via identity elements and catalysing the formation of an aminoacyl-tRNA at the cost of ATP hydrolysis.
Practice Questions — Part 1Score: 0 / 10
1. Which RNA polymerase transcribes protein-coding genes in eukaryotes?
RNA pol II transcribes all protein-coding genes (mRNAs) and most snRNAs/lncRNAs. RNA pol I produces rRNA precursors; RNA pol III produces tRNAs and 5S rRNA.
2. The 5′ cap added to eukaryotic mRNA consists of:
The 5′ cap is a 7-methylguanosine nucleotide added in a unique 5′-to-5′ triphosphate linkage co-transcriptionally. It protects the mRNA from 5′ exonucleases and is required for ribosome recognition during translation initiation.
3. Splicing of pre-mRNA is catalysed by the:
The spliceosome — a dynamic complex of five snRNPs (U1, U2, U4, U5, U6) plus ~150 proteins — removes introns from pre-mRNA via two sequential transesterification reactions, generating a lariat intermediate and joining the flanking exons.
4. The poly(A) signal sequence in eukaryotic mRNA is:
The canonical poly(A) signal AAUAAA (encoded AATAAA in DNA) is recognised by CPSF cleavage factor, which triggers cleavage of the transcript ~10–30 nt downstream, followed by addition of the poly(A) tail by poly(A) polymerase.
5. The genetic code is said to be "degenerate." This means:
Degeneracy means that multiple codons (synonymous codons) can specify the same amino acid. For example, leucine is encoded by 6 codons (UUA, UUG, CUU, CUC, CUA, CUG). This buffering means many mutations in the third codon position are synonymous (silent).
6. Which codon serves as both the universal start codon and encodes methionine?
AUG (methionine) is the universal start codon. It defines the reading frame and is the site where ribosome assembly (in eukaryotes: 43S scanning complex + 60S subunit joining) is completed. UAA and UGA are stop codons.
7. Aminoacyl-tRNA synthetases (aaRS) are essential because they:
aaRS are responsible for "charging" tRNAs — attaching the correct amino acid to the 3′ CCA end of its cognate tRNA using ATP. They are the physical link between the nucleotide language of the codon and the amino acid language of proteins.
8. During splicing, the intron is released as a:
In the first transesterification step of splicing, the 2′-OH of the branch point adenosine attacks the 5′ splice site, forming a 2′-5′ phosphodiester bond that creates the characteristic lariat (loop) structure before the intron is fully excised and debranched.
9. Which strand of DNA is used as the template for RNA synthesis?
RNA polymerase reads the template strand 3′→5′ and synthesises the RNA transcript 5′→3′, complementary to the template. The non-template (coding) strand has the same sequence as the RNA (with U instead of T) and is sometimes called the sense strand.
10. Alternative splicing primarily contributes to proteome diversity by:
Alternative splicing generates distinct mRNAs from the same pre-mRNA by including/excluding different exons. It is estimated that >90% of human multi-exon genes are alternatively spliced, greatly expanding the proteome beyond the ~20,000 protein-coding genes.
Part 1 complete! Score: 0 / 10
Section B · Recall Questions · Part 1
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B1
Describe the basic mechanism of transcription, including the role of the template strand.
Sample answer: RNA polymerase binds to the promoter, unwinds the DNA helix locally, and reads the template strand 3′→5′, synthesising a complementary RNA strand 5′→3′ using ribonucleoside triphosphates. The RNA sequence is identical to the non-template (coding) strand except with U instead of T.
B2
What is the 5′ cap of eukaryotic mRNA, and what are its two main functions?
Sample answer: The 5′ cap is a 7-methylguanosine nucleotide linked 5′-to-5′. Its functions: (1) protects mRNA from 5′ exonucleolytic degradation; (2) is recognised by the cap-binding complex (eIF4E) during translation initiation, recruiting ribosomes.
B3
Outline the steps of pre-mRNA splicing carried out by the spliceosome.
Sample answer: (1) U1 snRNP base-pairs with the 5′ splice site; U2 snRNP binds the branch point. (2) First transesterification: branch point 2′-OH attacks the 5′ splice site, forming a lariat. (3) Second transesterification: the 3′-OH of the upstream exon attacks the 3′ splice site, joining the exons and releasing the lariat intron.
B4
List four properties of the genetic code.
Sample answer: (1) Triplet — each codon is 3 nucleotides. (2) Non-overlapping — adjacent codons are read sequentially, not shared. (3) Degenerate — most amino acids are specified by multiple codons. (4) Universal — virtually all organisms use the same code, reflecting common ancestry.
B5
Explain the adaptor role of tRNA in decoding mRNA.
Sample answer: tRNA bridges two languages: its anticodon loop base-pairs with the mRNA codon (nucleotide language), while its 3′ CCA end carries the corresponding amino acid (protein language). The aminoacyl-tRNA synthetase charges the tRNA, ensuring the correct amino acid is attached to each anticodon.
B6
What is the poly(A) tail and what roles does it play in mRNA metabolism?
Sample answer: The poly(A) tail is a stretch of ~200 adenosines added to the 3′ end after cleavage at the poly(A) signal. It stabilises the mRNA (by protecting from 3′ exonucleases, bound by PABP), facilitates nuclear export, and promotes translation initiation via interaction between poly(A)-binding protein (PABP) and translation initiation factors.
B7
What types of RNA are produced by each of the three eukaryotic RNA polymerases?
Sample answer: RNA pol I: 18S, 5.8S, 28S rRNA (the large pre-rRNA transcript in the nucleolus). RNA pol II: mRNAs, most snRNAs, snoRNAs, lncRNAs. RNA pol III: tRNAs, 5S rRNA, 7SL RNA (SRP), U6 snRNA.
B8
Name the three stop codons and describe how they terminate translation.
Sample answer: The three stop codons are UAA, UAG, and UGA. They are not decoded by tRNAs but instead are recognised by release factors (eRF1 in eukaryotes) that stimulate hydrolysis of the peptidyl-tRNA bond, releasing the completed polypeptide from the ribosome.
B9
What is the role of the sigma (σ) factor in bacterial transcription?
Sample answer: The σ factor associates with the core bacterial RNA polymerase to form the holoenzyme. It recognises specific DNA sequences at the −10 and −35 elements of the promoter, directing the polymerase to the correct start site. Different σ factors (σ70, σ32, etc.) allow transcription of different gene sets under different conditions.
B10
Explain how alternative splicing increases proteome diversity without increasing gene number.
Sample answer: By including or skipping different exons, retaining certain introns, or using alternative 5′ or 3′ splice sites, a single pre-mRNA can be processed into multiple distinct mRNA isoforms. Each isoform encodes a protein with a different domain composition and potentially different function, cellular localisation, or regulatory properties — all from one gene.
Section C · Critical Thinking · Part 1
Develop analytical responses, then compare with the sample.
C1
The "wobble hypothesis" explains why fewer than 61 tRNAs are needed to decode all sense codons. Explain the wobble concept and its consequences for the genetic code.
Sample answer: Wobble (Francis Crick, 1966) proposes that the first two positions of the codon pair strictly (Watson-Crick), but the third position ("wobble position") allows non-standard base pairing. For example, inosine (I) in the tRNA anticodon can pair with U, C, or A in the mRNA. This means one tRNA can recognise multiple synonymous codons differing only at the third position, so ~45 tRNAs suffice for 61 codons.
C2
Mutations in splice sites cause a significant fraction of human genetic diseases. Predict how a point mutation that destroys the 5′ splice site of an intron would affect the mature mRNA and the resulting protein.
Sample answer: If the 5′ splice site of intron N is destroyed, the spliceosome cannot remove that intron, causing intron retention. The retained intron adds extra sequence to the mRNA, usually introducing a premature stop codon or frameshift, producing a truncated or non-functional protein. Alternatively, a nearby cryptic splice site may be activated, causing partial exon skipping or an out-of-frame junction. Either way, normal protein function is disrupted.
C3
In bacteria, transcription and translation are coupled — ribosomes begin translating mRNA while it is still being transcribed. Why is this coupling absent in eukaryotes, and what are the regulatory consequences?
Sample answer: In eukaryotes, transcription occurs in the nucleus and translation in the cytoplasm — they are physically separated. The mRNA must be processed (capped, spliced, polyadenylated) and exported before translation begins. This separation allows more sophisticated gene regulation at each step (e.g., regulated splicing, mRNA export control, cytoplasmic mRNA stability) but sacrifices the rapid translational response to transcription that bacteria exploit (e.g., attenuation).
C4
Why is a frameshift mutation typically more devastating than a point (missense) mutation, even if the frameshift occurs near the 3′ end of the coding sequence?
Sample answer: A frameshift (insertion or deletion not divisible by 3) shifts the reading frame so that all codons downstream are misread, producing a completely wrong amino acid sequence from that point on. It almost always creates a premature stop codon. A missense mutation changes only one amino acid, which may be tolerated if the replacement is conservative or in a non-critical region. Even near the C-terminus, a frameshift eliminates the final protein domain, which may be essential for folding or function.
C5
The ribosome is a ribozyme — its peptidyl transferase activity resides in RNA, not protein. How does this observation support the "RNA world" hypothesis?
Sample answer: The RNA world hypothesis proposes that early life used RNA as both information store and catalyst, before DNA and proteins took over these roles. The fact that the ribosome's peptidyl transferase centre — responsible for all protein synthesis — is catalytic RNA suggests that the translation machinery is a molecular fossil of an ancient RNA-based system. If RNA can catalyse peptide bond formation, early RNA molecules could have synthesised the first proteins, eventually leading to RNA-protein then DNA-RNA-protein systems.
Section D · Interactive Questions · Part 1
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D1
Given the DNA template strand: 3′-TACGGA-5′, write the corresponding mRNA sequence (5′ to 3′).
D2
How many stop codons are there in the standard genetic code? (number)
D3
Which snRNP first recognises the 5′ splice site of an intron in the spliceosome? (format: U#)
D4
The universal start codon is: (three letters, RNA format)
D5
What three-letter nucleotide sequence is found at the 3′ end of all tRNAs, where the amino acid attaches?
Part 2 →
Having explored how mRNA is made and processed, we now follow it into the cytoplasm where the ribosome decodes it into a polypeptide, and then examine how that polypeptide is folded, modified, and quality-controlled into a functional protein.
Part 2 of 2
Translation and Post-Translational Control
6.5 Ribosome Structure and Function
The ribosome is the molecular machine that translates mRNA into protein. Eukaryotic ribosomes are 80S, composed of a large 60S subunit (containing 28S, 5.8S, and 5S rRNA plus ~49 proteins) and a small 40S subunit (18S rRNA plus ~33 proteins). Ribosomes have three tRNA-binding sites: the A (aminoacyl) site accepts incoming aminoacyl-tRNAs; the P (peptidyl) site holds the growing peptide chain; and the E (exit) site from which deacylated tRNAs leave. The peptidyl transferase centre (PTC) in the 23S/28S rRNA catalyses peptide bond formation — a ribozyme activity.
Key term
Reading frame
The particular grouping of nucleotide triplets (codons) used during translation, set by the AUG start codon; a sequence has three possible reading frames, only one of which encodes the correct protein.
6.6 Stages of Translation
Initiation: In eukaryotes, the 43S pre-initiation complex (40S + eIF2-GTP-Met-tRNA + other eIFs) is recruited to the 5′ cap and scans 5′→3′ until an AUG in good Kozak context is encountered. The 60S subunit then joins, forming an 80S ribosome with Met-tRNA in the P site.
Elongation: (1) EF-Tu (prokaryotes)/eEF1A (eukaryotes) delivers aminoacyl-tRNA to the A site; (2) peptide bond is formed by PTC, transferring the growing chain to the A-site amino acid; (3) EF-G/eEF2 catalyses translocation — the ribosome moves 3 nt along the mRNA, shifting tRNAs from A→P→E sites.
Termination: A stop codon (UAA, UAG, UGA) enters the A site; release factors (eRF1/eRF3) trigger hydrolysis of the peptidyl-tRNA bond, releasing the completed polypeptide. The ribosome is recycled by splitting into subunits.
✒
Pause & Recall
Why does puromycin — an antibiotic that mimics aminoacyl-tRNA — cause premature chain termination?
Puromycin structurally resembles an aminoacyl-tRNA and enters the A site. The PTC forms a peptide bond with it, attaching the growing chain to puromycin. However, because puromycin lacks the tRNA body required for normal translocation, the peptidyl-puromycin dissociates from the ribosome, releasing a truncated, puromycin-tagged peptide.
6.7 Post-Translational Modifications
After translation, many proteins undergo covalent modifications that modulate their activity, localisation, or stability. Common modifications include:
Glycosylation (N-linked in ER, O-linked in Golgi) — essential for protein folding, quality control, and cell-surface recognition.
Ubiquitination — marks proteins for proteasomal degradation or alters signalling.
Acetylation (N-terminal or lysine) — affects protein stability, DNA binding (histones), and protein interactions.
6.8 Protein Quality Control
Molecular chaperones (HSP70, HSP90, GroEL/GroES in bacteria, TRiC/CCT in eukaryotes) assist protein folding by transiently binding hydrophobic patches exposed on nascent or misfolded polypeptides. Unfolded or misfolded proteins are recognised by chaperones, given another chance to fold, or directed to the ubiquitin-proteasome system (UPS) for degradation. In the ER, the unfolded protein response (UPR) detects misfolded proteins and expands chaperone capacity or triggers apoptosis under severe stress.
Practice Questions — Part 2Score: 0 / 10
1. Which ribosomal site accepts the incoming aminoacyl-tRNA during elongation?
The A (aminoacyl) site accepts the next aminoacyl-tRNA. The P (peptidyl) site holds the growing chain; the E (exit) site is for departing deacylated tRNAs. There is no standard "T site" in eukaryotic ribosomes (it refers to a bacterial EF-Tu–tRNA complex docking site in some nomenclatures).
2. The peptidyl transferase activity of the ribosome is catalysed by:
The peptidyl transferase centre (PTC) is composed entirely of rRNA (23S rRNA in bacteria, 28S rRNA in eukaryotes). Thomas Steitz, Ada Yonath, and Venkatraman Ramakrishnan confirmed by X-ray crystallography that ribosomal proteins are absent from the active site — the ribosome is a ribozyme.
3. N-linked glycosylation of proteins begins in the:
N-linked glycosylation is initiated co-translationally in the ER lumen, where oligosaccharyltransferase (OST) transfers a pre-assembled 14-sugar oligosaccharide to asparagine residues in the N-X-S/T sequon. Further modification occurs in the Golgi.
4. Which elongation factor delivers aminoacyl-tRNA to the ribosomal A site in eukaryotes?
eEF1A (homologous to bacterial EF-Tu) carries aminoacyl-tRNA as a ternary complex with GTP to the ribosomal A site. Codon-anticodon recognition triggers GTP hydrolysis, releasing eEF1A-GDP and allowing the tRNA to fully accommodate. eEF2 catalyses translocation.
5. The Kozak sequence in eukaryotes is important for:
The Kozak consensus (GCCACCAUGG in ideal form) surrounds the AUG start codon. The 43S scanning complex pauses here; the presence of an A at −3 and G at +4 positions facilitates efficient recognition of AUG as the start codon rather than continuing to scan.
6. Molecular chaperones such as Hsp70 prevent protein aggregation by:
Hsp70 (and co-chaperone Hsp40/DNAJ) binds short hydrophobic peptide segments that are exposed on unfolded or partially folded polypeptides. By shielding these segments, Hsp70 prevents aberrant intermolecular hydrophobic interactions that would otherwise cause aggregation, giving the protein time and opportunity to find its native fold.
7. Ubiquitin-mediated proteasomal degradation requires protein substrates to be tagged with:
Proteasomal degradation is signalled by K48-linked polyubiquitin chains attached to lysine residues of the target protein. K63-linked chains signal other processes (DNA repair, endocytosis). Mono-ubiquitination typically signals endocytic sorting rather than degradation.
8. The Shine-Dalgarno sequence in bacteria:
The Shine-Dalgarno (SD) sequence (consensus AGGAGG) is located ~5–10 nt upstream of the AUG start codon in bacterial mRNAs. It base-pairs with the 3′ end of 16S rRNA in the 30S subunit, precisely positioning the ribosome to initiate translation at the correct AUG.
9. Which cellular compartment is the site of N-linked glycosylation initiation and disulfide bond formation for secretory proteins?
The ER lumen is an oxidising environment containing protein disulfide isomerase (PDI) that catalyses disulfide bond formation. N-linked glycosylation also begins co-translationally in the ER. These modifications are essential for the correct folding of secreted and membrane proteins.
10. Translocation of the ribosome along mRNA during elongation is catalysed by:
eEF2 (bacterial homologue EF-G) is a GTPase that catalyses the translocation step: after peptide bond formation, it uses GTP hydrolysis energy to move the ribosome exactly 3 nt (one codon) in the 3′ direction, shifting the tRNAs from A→P→E sites.
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Section B · Recall Questions · Part 2
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B1
Describe the scanning model of translation initiation in eukaryotes.
Sample answer: The 43S pre-initiation complex (40S subunit + eIF2-GTP-Met-tRNAi + eIF3/eIF1/eIF1A) is recruited to the 5′ cap via eIF4F. It scans 5′→3′ until it encounters an AUG in a favourable Kozak context. GTP hydrolysis by eIF2 fixes the start site, and the 60S subunit joins (triggered by eIF5B-GTP hydrolysis) to form the 80S initiation complex with Met-tRNAi in the P site.
B2
Describe the elongation cycle: how is each amino acid added to the growing polypeptide?
Sample answer: (1) eEF1A-GTP delivers aminoacyl-tRNA to the A site; codon-anticodon matching triggers GTP hydrolysis and accommodation. (2) The PTC (28S rRNA) catalyses peptide bond formation, transferring the peptidyl chain from P-site tRNA to the A-site amino acid. (3) eEF2-GTP catalyses translocation, moving the ribosome 3 nt: the new peptidyl-tRNA moves to the P site, the deacylated tRNA moves to the E site and exits.
B3
How is translation terminated when a stop codon enters the A site?
Sample answer: Release factor eRF1 (which recognises all three stop codons) enters the A site and, together with eRF3-GTP, stimulates hydrolysis of the peptidyl-tRNA bond at the PTC. The completed polypeptide is released. ABCE1 then promotes ribosome splitting into 40S and 60S subunits for recycling.
B4
Why is protein phosphorylation an effective switch for regulating protein activity?
Sample answer: Phosphorylation adds a bulky, negatively charged group to a serine, threonine, or tyrosine side chain, which can alter protein conformation, block or expose binding surfaces, or change enzymatic activity. Crucially, it is reversible — kinases phosphorylate and phosphatases dephosphorylate — allowing rapid on/off switching in response to signals without requiring new protein synthesis or degradation.
B5
What triggers the unfolded protein response (UPR) and what are its main outcomes?
Sample answer: The UPR is triggered by accumulation of misfolded proteins in the ER lumen, which titrate away the chaperone BiP/GRP78 from ER stress sensors (IRE1, PERK, ATF6). Key outcomes: (1) global translation attenuation (via PERK → eIF2α phosphorylation) to reduce ER load; (2) transcriptional upregulation of ER chaperones to increase folding capacity; (3) ER-associated degradation (ERAD) of misfolded proteins. If ER stress is unresolved, apoptosis is triggered.
B6
Describe the enzymatic cascade that attaches ubiquitin to a substrate protein.
Sample answer: (1) E1 (ubiquitin-activating enzyme) activates ubiquitin using ATP, forming a thioester bond. (2) E2 (ubiquitin-conjugating enzyme) receives ubiquitin from E1. (3) E3 (ubiquitin ligase) recognises the substrate protein and transfers ubiquitin from E2 to a lysine on the substrate. Successive rounds build a polyubiquitin chain; K48-linked chains target the substrate to the 26S proteasome for degradation.
B7
What is a polysome and why does it increase translational efficiency?
Sample answer: A polysome (polyribosome) consists of multiple ribosomes simultaneously translating the same mRNA. As one ribosome moves along, another initiates translation at the 5′ end, allowing many copies of the protein to be produced from a single mRNA molecule before it is degraded. This is especially important for highly expressed proteins.
B8
What is nonsense-mediated mRNA decay (NMD) and what is its purpose?
Sample answer: NMD is a surveillance pathway that detects mRNAs with premature termination codons (PTCs) — stop codons more than ~50–55 nt upstream of an exon junction complex (EJC). When the ribosome encounters a PTC and an EJC is still downstream, UPF proteins are recruited, triggering rapid degradation of the aberrant mRNA. This prevents production of truncated, potentially dominant-negative proteins from nonsense or frameshift mutations.
B9
Compare the subunit composition of prokaryotic (70S) and eukaryotic (80S) ribosomes.
Sample answer: Prokaryotic 70S: small 30S subunit (16S rRNA + ~21 proteins) + large 50S subunit (23S + 5S rRNA + ~31 proteins). Eukaryotic 80S: small 40S subunit (18S rRNA + ~33 proteins) + large 60S subunit (28S + 5.8S + 5S rRNA + ~49 proteins). Eukaryotic ribosomes are larger and more complex; differences are exploited by antibiotics that selectively target bacterial 70S ribosomes.
B10
How does an N-terminal signal peptide direct a newly synthesised protein to the ER?
Sample answer: As the signal peptide (typically ~15–30 hydrophobic residues) emerges from the ribosome exit tunnel, it is recognised by the Signal Recognition Particle (SRP). SRP binds the ribosome-nascent chain and halts translation. The SRP-ribosome complex docks onto the SRP receptor on the ER membrane, translation resumes, and the growing polypeptide is co-translationally threaded through the Sec61 translocon channel into the ER lumen, where the signal peptide is cleaved by signal peptidase.
Section C · Critical Thinking · Part 2
Develop analytical responses, then compare with the sample.
C1
Many antibiotics target the bacterial ribosome (e.g., streptomycin binds 30S; erythromycin and chloramphenicol bind 50S). Why are these drugs generally safe for human cells, and what would make a pathogen resistant?
Sample answer: Bacterial ribosomes (70S) differ structurally from eukaryotic 80S ribosomes — the antibiotic binding sites are absent or different in the human cytoplasmic ribosome. (Note: mitochondrial 70S-like ribosomes can sometimes be affected, causing side effects.) Resistance mechanisms include: methylation of the ribosomal RNA target site (e.g., erm genes modify 23S rRNA, blocking erythromycin); ribosomal protein mutations that eliminate the drug-binding site; and active efflux pumps that remove the antibiotic from the cell.
C2
Many neurodegenerative diseases (Alzheimer's, Parkinson's, Huntington's) involve protein misfolding and aggregation. Why does the proteostasis (protein homeostasis) network fail in ageing neurons?
Sample answer: With age, expression of heat-shock proteins and other chaperones declines. Proteasome activity diminishes. Post-mitotic neurons cannot dilute damaged proteins by cell division. Reactive oxygen species accumulate, oxidising and damaging proteins. Aggregation-prone proteins (e.g., α-synuclein, tau, polyglutamine-expanded huntingtin) overwhelm the chaperone and degradation systems, forming toxic oligomers and amyloid fibrils that impair synaptic function and eventually kill neurons.
C3
Many RNA viruses (e.g., HCV, poliovirus) use Internal Ribosome Entry Sites (IRES) for cap-independent translation. What is the biological advantage of this mechanism for the virus?
Sample answer: Viruses like poliovirus cleave eIF4G, disabling cap-dependent translation and shutting down host protein synthesis. IRES elements in viral mRNAs directly recruit ribosomes internally (without the cap), allowing viral proteins to be translated even when host translation is crippled. This diverts the translational machinery to viral use and may also help the virus evade innate immune responses that depend on host protein synthesis.
C4
Although synonymous codons encode the same amino acid, organisms show strong "codon usage bias." What are two consequences of mismatched codon usage when expressing a human gene in bacteria?
Sample answer: (1) Rare codons slow ribosome elongation because the corresponding charged tRNA is scarce in bacteria; this can stall ribosomes and reduce protein yield, or even cause premature drop-off. (2) Slow elongation at certain positions can also cause misincorporation (reduced fidelity) or altered co-translational folding, producing misfolded protein. Biotechnologists routinely codon-optimise human genes (substituting rare bacterial codons with synonymous frequent codons) to maximise expression of recombinant proteins.
C5
MicroRNAs (miRNAs) are ~22 nt RNAs that repress gene expression post-transcriptionally. Explain how a miRNA silences its target mRNA and why partial complementarity (rather than perfect matching) is more common in animals.
Sample answer: The miRNA is loaded into the RISC complex (containing Argonaute 2). RISC base-pairs with complementary sequences in the 3′ UTR of target mRNAs. Partial base-pairing (especially perfect pairing at the ~7-nt "seed" region, positions 2–8) is sufficient for repression by translation inhibition and mRNA deadenylation/decay. Perfect complementarity triggers Ago2-mediated slicing (cleavage). Animal miRNAs typically use partial complementarity, allowing each miRNA to regulate hundreds of targets simultaneously, greatly expanding the regulatory network without requiring exact matches.
Section D · Interactive Questions · Part 2
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D1
Given the mRNA codon 5′-AUG-3′, what amino acid does it encode? (one word)
D2
The ribosomal site that holds the growing polypeptide chain is called the: (letter only, e.g. A)
D3
What polyubiquitin chain linkage (e.g. K48) signals proteasomal degradation?
D4
Which elongation factor catalyses ribosome translocation in eukaryotes? (abbreviation)
D5
The SRP (Signal Recognition Particle) directs ribosomes to which organelle? (one word)