Topic 6 — Recycling of Proteins and Other Molecules
How the cell disposes of and recycles its own components through three degradation systems — the lysosome (bulk digestion at low pH), the peroxisome (oxidative breakdown), and the ubiquitin–proteasome pathway (selective, tagged destruction of individual proteins). Verified against Alberts, Molecular Biology of the Cell, 7th ed., Chapters 3, 6 & 13.
Source: Ch. 3 & 6 (ubiquitin/proteasome) + Ch. 13 (lysosomes/autophagy)Format: Scantron MCQStudy mode: click any option for full reasoning
correct answer + mechanism distractor = named misconception▲ amber = lecture diverges from textbook
§Introduction
Everything the cell builds — proteins from the rough ER, lipids from the smooth ER, whole organelles — eventually wears out and must be destroyed and replaced. Degradation is not an afterthought: it is as tightly controlled as synthesis, and the balance between the two sets the level of every protein in the cell. This is the disposal-and-recycling half of the story that began with protein sorting (Topic 5b).
The cell runs three degradation systems, and the exam expects you to know which does what. The lysosome is bulk digestion — a membrane sac of ~50 acid hydrolases that breaks down engulfed material and worn organelles, kept safe by a low-pH activation switch. The peroxisome is oxidative breakdown — a single-membrane compartment that oxidizes fatty acids and detoxifies, abundant in the liver. The ubiquitin–proteasome pathway is the precision tool — a non-membrane protein machine that destroys individual, specifically tagged proteins, controlling the cell cycle, immune presentation, DNA repair, and more.
This guide follows the lecture through all three, and does three things with the source material. It flags one arithmetic slip (ubiquitin is 96% conserved across 76 residues — that means about three residues differ, not "one amino acid"). It sharpens a point students routinely miss (the proteasome cuts proteins to peptides, not to amino acids). And it makes explicit the payoff connection the lecture leaves implicit: I-cell disease is a direct consequence of the mannose-6-phosphate sorting failure from Topic 5b.
◎Learning Objectives
What the exam will hold you responsible for
Distinguish the three degradation systems (lysosome, peroxisome, proteasome) by mechanism, contents, and what each destroys.
Explain lysosomal low-pH activation (~5.5) and why it is a safety mechanism against leakage.
Contrast phagocytosis, autophagy, and heterophagy, and describe the autophagosome route to the lysosome.
Connect I-cell disease to M6P sorting (Topic 5b) and describe Tay-Sachs as an enzyme-deficiency storage disease.
Characterize peroxisomes: single membrane, no genome but self-replicating, oxidative breakdown (β-oxidation, catalase/H₂O₂), and biosynthetic roles.
Describe the proteasome: the 20S core (four stacked rings) plus 19S cap, ATP-driven unfolding, and degradation to peptides.
Characterize ubiquitin: 76 residues, highly conserved, the Gly76 isopeptide linkage, and the ubiquitin code (Lys48 vs Lys63 vs mono).
Trace the E1–E2–E3 cascade and explain why E3 diversity provides substrate specificity.
1The Recycling Problem & Three Degradation Systems
Bottom line: synthesis and degradation must be balanced precisely — the steady-state level of any protein is set by both. The cell runs three degradation systems that differ on two axes: membrane-bound vs not, and bulk vs selective.
Two directions of digestion. Phagocytosis engulfs external particles into a phagosome; autophagy packages the cell's own worn organelles into a double-membrane autophagosome. Both then fuse with the lysosome for degradation. Textbook Fig. 13–41.
Why degrade at all — first principles
Replace worn components. Proteins, lipids, and organelles accumulate damage; they must be cleared and rebuilt.
Regulate levels. Controlled destruction is how the cell turns proteins off — driving the cell cycle, ending signals, and clearing misfolds.
Recycle building blocks. Breakdown products (amino acids, nucleotides, sugars) are reused for new synthesis.
Different proteins have very different lifetimes: some cytosolic regulators turn over in minutes, while structural proteins (actin, myosin) and red-cell hemoglobin persist for weeks to months. The machinery must be selective enough to destroy the right protein at the right time.
Three degradation systems. Lysosome (membrane, bulk, from Golgi, acid hydrolases); peroxisome (membrane, oxidative, from ER); proteasome (no membrane, selective, ubiquitin-tagged single proteins). Textbook Figs. 13–36, 3–64.
The three systems
Lysosome — a membrane-bound sac of ~50 acid hydrolases; bulk digestion of engulfed material and worn organelles. Buds from the Golgi.
Peroxisome — a membrane-bound single-membrane compartment of oxidative enzymes; oxidative breakdown (and some synthesis). Buds from the ER.
Proteasome — a non-membrane protein complex (like the ribosome, it's not an organelle); selective destruction of individually ubiquitin-tagged proteins.
2Lysosomes: Structure & Low-pH Activation
Bottom line: the lysosome is a membrane bag of ~50 digestive enzymes that only work at acidic pH (~5.5) — and that pH requirement is a deliberate safety mechanism. If the enzymes escape into the neutral cytosol, they simply don't work.
Contents and origin
Buds from the Golgi (its hydrolases are made in the ER, processed in the Golgi, and delivered via the M6P route from Topic 5b).
Contains ~50 hydrolytic enzymes ("acid hydrolases") capable of digesting essentially every class of macromolecule: proteins, nucleic acids, lipids/phospholipids, and carbohydrates.
The low-pH switch — a fail-safe, not just a preference
The lysosomal interior is kept acidic (~pH 5.5) by proton pumps, and the hydrolases are activated by that low pH. This does double duty:
The pH safety switch. Proton pumps keep the lysosome at ~pH 5.5, where the hydrolases are active. In the neutral cytosol (~7.2), the same enzymes are inactive — so leakage or accidental secretion causes little damage. Textbook Fig. 13–37.
It gives the enzymes their optimal working conditions inside the lysosome.
More importantly, it's a containment safeguard: the cytosol is at pH ~7.2, where these enzymes are inactive. So if a lysosome leaks — or if hydrolases are accidentally secreted — they do little damage at neutral pH. The cell doesn't rely on the membrane alone; it relies on the enzymes being switched off everywhere except inside the lysosome.
3Lysosome Functions: Phagocytosis & Autophagy
Bottom line: the lysosome digests material from outside the cell (engulfed particles) and from inside the cell (worn organelles). The same machinery serves immune defense, cellular housekeeping, and programmed cell death.
Phagocytosis — digesting external material
Large particles (bacteria, debris, whole cells) are engulfed into a vesicle called a phagosome, which then fuses with a lysosome; the acid hydrolases degrade the contents. In animals this is immune defense by macrophages and neutrophils; in single-celled eukaryotes the identical process is used for feeding. Strikingly, the machinery is nearly the same across ~1 billion years of divergence.
Cell death and development
Apoptosis (programmed cell death). Lysosomes participate in the controlled cell death that sculpts tissues during development — e.g. removing the cells between developing digits so fingers separate.
Necrosis. A different, disordered death from sudden injury such as oxygen deprivation (e.g. myocardial infarction) — less regulated than apoptosis, but still involving lysosomal enzymes and often causing much of the damage after the initial insult.
4Lysosomal Storage Diseases
Bottom line: if a lysosomal enzyme is missing or mislocalized, its substrate accumulates — the cell fills with undigested material. Two canonical examples: I-cell disease (a sorting failure) and Tay-Sachs (a single enzyme deficiency).
I-cell disease — the Topic 5b payoff
This is where the mannose-6-phosphate pathway you learned in Topic 5b comes back with real consequences. In I-cell (inclusion-cell) disease, the enzyme that adds the M6P tag is defective. Without the tag, lysosomal hydrolases can't be recognized by the M6P receptor, so instead of being delivered to lysosomes they are secreted from the cell by default. The result: lysosomes are missing nearly all their hydrolases, undigested substrates pile up as inclusion bodies, and the cell malfunctions. (The secreted enzymes themselves do little harm — they're inactive at the neutral pH outside.)
◆ Connect it — no M6P tag ⇒ default secretion
This is the exact logic from Topic 5b §9: a protein reaching the TGN without a diversion signal is secreted by default. I-cell disease is that principle failing in the body — knock out the M6P-adding enzyme, and hydrolases that should be diverted to lysosomes instead follow the default secretory route out of the cell. The lysosome is starved of enzymes not because the enzymes are broken, but because they were never delivered. (§4; cf. Topic 5b §9)
Tay-Sachs — a single missing enzyme
Tay-Sachs disease is an autosomal recessive disorder caused by the near-total absence of a specific lysosomal enzyme (hexosaminidase). Its substrate (a lipid, GM2 ganglioside) accumulates in neurons, causing progressive neurodegeneration — loss of neurological function, blindness, and death typically between ages 2 and 4. Here the sorting works; the enzyme itself is absent.
I-cell disease. A defect in M6P tagging sends lysosomal hydrolases out of the cell by default instead of to lysosomes; the enzyme-starved lysosomes accumulate undigested material. A direct clinical consequence of the M6P sorting logic. Textbook (I-cell disease); cf. Fig. 13–24.
5Peroxisomes
Bottom line: the peroxisome is a single-membrane compartment for oxidative chemistry — breaking down fatty acids and detoxifying byproducts — and it also builds key lipids. It has no genome but still self-replicates, budding from the ER and dividing.
Structure and biogenesis
Single membrane enclosing a matrix packed with oxidative enzymes; no DNA of its own.
Self-replicating from the ER. Peroxisomes bud from the ER; precursor vesicles fuse, the peroxisome grows by importing proteins and lipids, and when large enough it divides into two — adding phospholipids from the cytosol.
Imports folded proteins. Peroxisomal proteins are made on free ribosomes, carry a peroxisomal targeting signal, and are imported as complete (folded) polypeptides by a receptor that delivers them and recycles — analogous to mitochondrial import in being post-translational, but accepting folded protein.
Breakdown functions
Detoxification by oxidation — abundant in the liver, where toxic byproducts from food accumulate; the peroxisome is where most such oxidation happens.
β-oxidation of long fatty acids — peroxisomes break down the longer fatty acids (roughly >20 carbons); mitochondria handle the shorter ones.
H₂O₂ handling by catalase — oxidations generate hydrogen peroxide; catalase decomposes it (to water + O₂) or uses it to oxidize other substrates.
Purine breakdown — degradation of AMP/GMP toward uric acid.
Biosynthetic functions
Not only breakdown — peroxisomes contribute to synthesizing cholesterol, steroid hormones (estrogen, testosterone), bile acids (in the liver, from cholesterol), and lipids used to make myelin (nerve insulation).
Peroxisome: breakdown and synthesis. A single-membrane, DNA-free organelle that oxidizes long fatty acids, decomposes H₂O₂ via catalase, and processes purines — while also helping build cholesterol, steroids, bile acids, and myelin lipids. Textbook Fig. 12–30.
6Protein Turnover & the Ubiquitin–Proteasome Pathway
Bottom line: for individual, selective protein destruction — not bulk digestion — the cell uses the ubiquitin–proteasome pathway (UPP): tag a specific protein with ubiquitin, then feed it to the proteasome. This is the principal route for protein degradation in the cytosol and nucleus, and it controls an enormous range of processes.
The logic: tag, then destroy
The UPP is a two-step system. First, a doomed protein is covalently marked with a chain of ubiquitin (§8–9). Second, the proteasome recognizes that mark, unfolds the protein, and chops it up (§7). Selectivity lives in the tagging step — the cell decides which protein to destroy by controlling what gets ubiquitinated.
Why it matters — the reach of the UPP
Because it can remove any specific protein on cue, the UPP regulates a huge span of biology:
Cell cycle and division — timed destruction of regulators drives progression.
Antigen presentation — the peptides the proteasome produces are displayed to helper T cells, a core of adaptive immunity.
Apoptosis, DNA transcription and repair, differentiation and development, and modulation of receptors/channels.
Cancer — the UPP degrades tumor-suppressor proteins (e.g. p53); dysregulation can drive a cancer phenotype.
Its importance was recognized with the Nobel Prize in Chemistry (2004) for the discovery of ubiquitin-mediated degradation.
The two-step UPP. A target protein is tagged with a polyubiquitin chain (via E1→E2→E3), then recognized, unfolded, and cut into peptides by the proteasome. Selectivity is set at the tagging step. Textbook Figs. 3–64, 3–65.
7The Proteasome
Bottom line: the proteasome is a barrel-shaped protein machine that recognizes ubiquitin-tagged proteins, unfolds them using ATP, and threads them into an enclosed core to be chopped into peptides — not all the way to amino acids. Degradation happens inside a sequestered cavity so it can't attack the wrong proteins.
The degradation cycle
The cap recognizes the polyubiquitin chain on the doomed protein.
Ubiquitin is cleaved off early and recycled — it was only needed for recognition, so a deubiquitinating enzyme frees it for reuse.
ATPases unfold the protein; it enters the core pore as an extended chain (its folded structure is lost).
Inside the cavity it is cut into short peptides, which exit; cytosolic peptidases later break most peptides down to amino acids, and some peptides are used for antigen presentation.
◆ Misconception — the proteasome makes peptides, not amino acids
A common error is that the proteasome digests proteins straight to amino acids. Fix: the proteasome cuts proteins into short peptides (typically several residues each). Those peptides are then further degraded to amino acids by separate cytosolic peptidases — and, importantly, some intact peptides are diverted to antigen presentation (displayed to helper T cells). The "peptide, not amino acid" endpoint is exactly why the proteasome feeds the immune system. (§7; Ch. 6)
Textbook note. Notice the structural echo of Topic 5b: a substrate is recognized, unfolded, and threaded through a channel — just like the ribosome→SRP→translocon route into the ER. The contrast worth remembering: passing through the ER translocon begins a protein's functional life (it's delivered to fold and work), whereas passing through the proteasome ends it (it's destroyed). Same mechanical motif, opposite biological meaning. (§7; cf. Topic 5b §4)Flag this for the exam.
8Ubiquitin & the Ubiquitin Code
Bottom line: ubiquitin is a tiny, extraordinarily conserved protein whose C-terminal glycine attaches to a target's lysine through an isopeptide bond. And the type of chain matters: a Lys48-linked chain means "destroy," while other linkages mean entirely different things.
The molecule
Small and conserved. Ubiquitin is 76 amino acids (~8.5 kDa) and one of the most conserved proteins known.
Key residues. The C-terminal tail — especially glycine 76 — is the attachment point; ubiquitin is also rich in lysines (seven of them), which matters for building chains.
Textbook note. The lecture states that human and yeast ubiquitin share 96% sequence identity, then says this "means one amino acid is different." Correction — check the arithmetic: ubiquitin is 76 residues long. 96% identity means ~96% × 76 ≈ 73 residues are identical, so about 3 residues differ, not one. (A single difference in 76 residues would be ~99% identity, not 96%.) The headline point — ubiquitin is extraordinarily conserved — is right; the "one amino acid" gloss is not. Expect roughly 3 differences. (arithmetic on 76 residues; Ch. 3, Fig. 3–65)Flag this for the exam.
The isopeptide linkage
A family of enzymes couples ubiquitin's Gly76 carboxyl group to the ε-amino group of a lysine side chain on the target. This is an isopeptide bond — it looks like a peptide bond (a C–N linkage) but is not a standard one, because it uses the lysine's side-chain amino group, not the backbone. The target protein's main-chain sequence is untouched; ubiquitin hangs off a lysine side chain.
The ubiquitin code — chain type is the message
Ubiquitin can attach to a lysine of another ubiquitin, building chains — and which lysine is used encodes different instructions:
Lys48-linked chain (Gly76 → Lys48 of the next ubiquitin) → proteasomal degradation. A chain of about five ubiquitins is enough for the proteasome to recognize.
Lys63-linked chain → non-degradative signals such as DNA repair.
Monoubiquitination (single or multiple single tags) → signals like endocytosis and histone regulation.
These marks are "read" by proteins that recognize each specific linkage — the same tag molecule, different meanings depending on how it's assembled.
Isopeptide bond and the ubiquitin code. Gly76 links to a lysine side chain (isopeptide). Chain type carries the message: Lys48 → proteasomal degradation, Lys63 → DNA repair, mono → endocytosis. Textbook Figs. 3–65, 3–66.
9The E1–E2–E3 Cascade & Specificity
Bottom line: attaching ubiquitin takes three enzymes acting in sequence — E1 activates, E2 carries, E3 ligates onto the target — and it's E3 diversity that decides which protein gets destroyed. This is an ATP-powered relay, and the last enzyme is the specificity gatekeeper.
The three-step relay (ATP-dependent)
E1 — ubiquitin-activating enzyme. Activates ubiquitin's C-terminus, attaching it to E1's active-site cysteine via a thioester bond, powered by ATP hydrolysis.
E2 — ubiquitin-conjugating (carrier) enzyme. Ubiquitin is handed from E1 to E2's active-site cysteine.
E3 — ubiquitin ligase. Brings the E2–ubiquitin and the substrate together and catalyzes transfer of ubiquitin onto a lysine of the target. Repeat to build the chain; about five ubiquitins → the 26S proteasome recognizes and degrades it.
Why E3 is the specificity gatekeeper
The numbers tell the story: cells have one (or few) E1, roughly 30 E2s, but hundreds of E3s (~500 in humans). Because each E3 recognizes specific substrate proteins, the huge diversity of E3s — combined with E2 pairings (potentially tens of thousands of combinations) — is what lets the system target one particular protein at one particular time. E3 is where "which protein dies" is decided.
The E1→E2→E3 relay. E1 activates ubiquitin (ATP); E2 carries it; E3 — the specificity gatekeeper, with hundreds of variants — transfers it onto the target's lysine. A ~5-ubiquitin chain sends the protein to the proteasome. Textbook Figs. 3–65, 3–66.
E3 classes.RING E3s bind E2–ubiquitin and substrate and stimulate direct transfer to the substrate lysine. HECT E3s have their own active-site cysteine and receive ubiquitin onto the E3 first, then pass it to the substrate.
Extra specificity layers. Beyond E3 identity, degradation is tuned by signals: an E3 can be switched on by phosphorylation or ligand binding, or a substrate can expose a degradation signal (degron) — often when it is phosphorylated — that the E3 then recognizes.
✦Core Concepts — Rapid Review
Everything most likely to be tested, compressed. If any line isn't instantly familiar, reread that section.
The Recycling Problem & Three Degradation Systems
Know. Three systems: the lysosome (bulk, organelles, extracellular material), the proteasome (short-lived/misfolded cytosolic proteins, ubiquitin-tagged), and autophagy.
Memorize. Lysosome vs proteasome vs autophagy · Ubiquitin tag routes to the proteasome
Lysosomes: Structure & Low-pH Activation
Know. A V-ATPase pumps H⁺ INTO the lumen to keep it at ~pH 5. Acid hydrolases work best at low pH — so a leak into the neutral cytosol does little damage (built-in safety).
Know. Lysosomes degrade material delivered by phagocytosis (engulfing particles) and by endocytosis; the vesicle fuses with a lysosome and acid hydrolases digest the contents.
Memorize. Phagosome fuses with lysosome · Acid hydrolases do the digestion
Lysosomal Storage Diseases
Know. A missing hydrolase lets its substrate accumulate (e.g., Tay-Sachs, Gaucher). I-cell disease is a trafficking failure — enzymes are secreted instead of delivered (M6P tagging defect).
Protein Turnover & the Ubiquitin–Proteasome Pathway
Know. The pathway targets short-lived, misfolded, and regulatory proteins; a K48-linked polyubiquitin chain is the degradation signal; the process is ATP-dependent.
Know. The proteasome degrades ubiquitin-tagged proteins in an ATP-dependent way, chopping them into short peptides (not free amino acids).
Memorize. ATP-dependent, ubiquitin-targeted · Products are short peptides (NOT free amino acids)
Ubiquitin & the Ubiquitin Code
Know. Mono- vs poly-ubiquitin and the linkage type set the fate (K48 → degradation, K63 → signaling/trafficking). Ubiquitin is one of the most conserved proteins known.
Memorize. ~96% identity across 76 residues (~3 differ — NOT “one amino acid”) · K48 vs K63 linkages · The ubiquitin code
The E1–E2–E3 Cascade & Specificity
Know. E1 activates ubiquitin (ATP), E2 conjugates it, and E3 ligase confers substrate specificity — hundreds of E3s give the system its selectivity.
Memorize. E1 (activating) / E2 (conjugating) / E3 (ligase, specificity) · ATP is consumed at E1 · Degron = the recognition signal
✎End-of-Topic Problems
Every Alberts end-of-chapter problem for this topic, one multiple-choice item each. Click any option to check it instantly — the correct choice is marked Answer as lectured. No submit step; feedback is immediate.
3–1verdict + reasonClaim: "Each strand in a β sheet is a helix with two amino acids per turn." This is —
A strand qualifies as a 2-fold helix by definition. (Ch. 3/6)
Answer as lectured Formally an extended strand is a 2-fold helix; the α helix is ~3.6 residues per turn. (Ch. 3/6)
An extended strand is formally a twofold helix. (Ch. 3/6)
That is the α-helix value; a strand is ~2 per turn. (Ch. 3/6)
3–2verdict + reasonClaim: "Loops of polypeptide that protrude from a protein's surface often form binding sites for other molecules." This is —
Loops occur in all fold types and can bind. (Ch. 3/6)
Loops are flexible and are common binding elements. (Ch. 3/6)
Many binding sites are surface loops, not buried. (Ch. 3/6)
Answer as lectured Protruding loops are conformationally flexible and exposed, so they often form binding sites. (Ch. 3/6)
3–3verdict + reasonClaim: "An enzyme reaches a maximum rate at high substrate concentration because it has a fixed number of active sites where substrate binds." This is —
Answer as lectured A finite number of sites, all busy at high substrate, sets the maximum rate. (Ch. 3/6)
Substrate is in excess; the sites are the limit. (Ch. 3/6)
Enzymes do saturate at Vmax. (Ch. 3/6)
It is site saturation, not product inhibition, that sets Vmax. (Ch. 3/6)
3–4verdict + reasonClaim: "Higher concentrations of an enzyme give rise to a higher turnover number for that enzyme." This is —
kcat is unchanged by enzyme amount, not decreased. (Ch. 3/6)
Answer as lectured Turnover number is intrinsic to each molecule; adding enzyme scales Vmax. (Ch. 3/6)
Total rate (Vmax) rises, but per-molecule kcat is unchanged. (Ch. 3/6)
kcat is per active site, not the total. (Ch. 3/6)
3–5verdict + reasonClaim: "Enzymes that undergo cooperative allosteric transitions invariably consist of symmetrical assemblies of multiple subunits." This is —
Cooperativity is common in multimers but not strictly required. (Ch. 3/6)
The 'invariably' claim is too strong. (Ch. 3/6)
Answer as lectured Cooperativity is typical of symmetric multimers but is not an absolute rule. (Ch. 3/6)
Most cooperative enzymes are multimeric, just not invariably symmetric. (Ch. 3/6)
3–6verdict + reasonClaim: "Continual addition/removal of phosphates by kinases and phosphatases is wasteful of ATP, but is a necessary consequence of effective regulation by phosphorylation." This is —
Reversibility inherently costs energy here. (Ch. 3/6)
Phosphatases work normally; the cost is inherent to the cycle. (Ch. 3/6)
Answer as lectured The futile cycle is the price of fast, reversible regulation. (Ch. 3/6)
Kinases consume ATP; that is the cost. (Ch. 3/6)
3–7concept · figure→conceptPulling on titin gives a sawtooth force–extension curve. In 8 M urea the peaks vanish and extension lengthens; after glutaraldehyde crosslinking the peaks vanish and extension shortens. The sawtooth is best explained by —
Answer as lectured Urea pre-unfolds them and cross-linking blocks unfolding — both abolish the peaks, confirming each peak is one domain popping open. (Ch. 3/6)
The backbone stays intact; domains unfold non-covalently. (Ch. 3/6)
The peaks track domain unfolding, not slippage. (Ch. 3/6)
Each peak is an unfolding event, not refolding. (Ch. 3/6)
3–8conceptCould you synthesize one molecule of every possible 300-residue polypeptide? Considering mass, the answer is —
Answer as lectured Even one copy each of ~10³⁹⁰ sequences vastly exceeds the mass of the universe. (Ch. 3/6)
The barrier is mass and number, not cost. (Ch. 3/6)
20³⁰⁰ is astronomically large — not manageable. (Ch. 3/6)
Each sequence is distinct; there are 20³⁰⁰ of them. (Ch. 3/6)
3–9concept · figure→conceptDomains can be "in-line" (threaded through the chain) or "plug-in" (inserted into a loop of another domain). What structural feature marks a domain as a good "plug-in" module?
Secondary-structure content is not what defines plug-in. (Ch. 3/6)
Plug-in domains are folded; it is about termini position. (Ch. 3/6)
Far-apart termini suit in-line, not plug-in, domains. (Ch. 3/6)
Answer as lectured Close termini let a domain be spliced into a surface loop without disrupting the host. (Ch. 3/6)
3–10concept · SVGTwo identical subunits can associate "head-to-head" (2-fold symmetric) or "head-to-tail." Which arrangement is far more common, and why?
Dimer geometries. A symmetric (head-to-head) dimer forms when one surface on each monomer binds the identical surface on its partner; a head-to-tail dimer needs two distinct complementary surfaces.
Answer as lectured A symmetric interface closes the assembly at two subunits; head-to-tail tends to polymerize. (Ch. 3/6)
Symmetric head-to-head is more common. (Ch. 3/6)
The contact is non-covalent; symmetry is the reason. (Ch. 3/6)
Head-to-tail tends to form open polymers; symmetric dimers are more common. (Ch. 3/6)
3–11conceptAn antibody binds protein 1 with K = 5×10⁹ M⁻¹. Against protein 2 it forms 3 fewer H-bonds, weakening binding by 11.9 kJ/mol. Given that ΔG° = −2.3RT·log K and 2.3RT ≈ 5.9 kJ/mol at 310 K, K for protein 2 is —
Fewer H-bonds weaken, not strengthen, binding. (Ch. 3/6)
Answer as lectured 11.9 kJ/mol ≈ RT·ln(100), so K drops ~100×: 5×10⁹ → 5×10⁷. (Ch. 3/6)
11.9 kJ/mol corresponds to ~100×, not 10×. (Ch. 3/6)
3–12conceptPlotting fraction of ligand bound vs ligand concentration (fraction = [Pr]/([Pr]+Kd)) gives an S-shaped curve on a log scale. At what free-ligand concentration is the protein exactly half-saturated?
At 2·Kd the fraction is ⅔, not ½. (Ch. 3/6)
That gives full saturation, not half. (Ch. 3/6)
Zero ligand gives zero binding. (Ch. 3/6)
Answer as lectured At [ligand] = Kd, fraction bound = Kd/(Kd+Kd) = ½. (Ch. 3/6)
3–13conceptNormal hexokinase (built from L-amino acids) phosphorylates D-glucose but ignores L-glucose. A hexokinase synthesized entirely from D-amino acids would —
The mirror-image enzyme flips specificity to L-glucose. (Ch. 3/6)
Answer as lectured An all-D enzyme is the enantiomer, so it recognizes the mirror-image substrate. (Ch. 3/6)
It is stereospecific for the mirror substrate, L-glucose. (Ch. 3/6)
It folds fine — into the mirror-image structure. (Ch. 3/6)
3–14conceptFor Michaelis–Menten kinetics, rate = Vmax[S]/([S]+Km). The rates at [S] = 0, [S] = Km, and [S] = ∞ are, respectively —
At [S]=Km the rate is Vmax/2, not Vmax. (Ch. 3/6)
The middle value is Vmax/2, not Km. (Ch. 3/6)
At [S]=0 the rate is 0 and at ∞ it is Vmax — reversed. (Ch. 3/6)
Answer as lectured At Km the rate is Vmax/2; at zero, 0; at infinity, Vmax. (Ch. 3/6)
3–15concept · SVGAMP and GMP are made from ribose-5-phosphate (R5P) through a common trunk that branches at IMP. What feedback-inhibition strategy ensures balanced supply of both while minimizing intermediate buildup?
Branched purine pathway. A common trunk (R5P→IMP) splits to AMP and GMP. Balanced regulation requires each end product to inhibit both the shared first step and the first committed step of its own branch.
It is feedback inhibition, not activation. (Ch. 3/6)
Answer as lectured Branch-specific plus shared feedback keeps AMP and GMP balanced and prevents overproduction. (Ch. 3/6)
Feedback inhibition is required to balance the branches. (Ch. 3/6)
Both products regulate, and each also controls its own branch. (Ch. 3/6)
3–16conceptHemoglobin binds O₂ efficiently in the lungs yet releases it efficiently in tissues. The property that makes this possible is —
Fixed high affinity would not release O₂ in tissues; cooperativity does. (Ch. 3/6)
Hemoglobin is a cooperative tetramer, not single-site. (Ch. 3/6)
O₂ binds reversibly; cooperativity is the key property. (Ch. 3/6)
Answer as lectured The steep sigmoid gives high saturation at lung pO₂ and sharp release at tissue pO₂. (Ch. 3/6)
3–17conceptNormal Src carries a myristoyl group that anchors it to the plasma membrane; a non-myristoylated mutant stays cytosolic. Both have equal kinase activity, but only membrane-bound Src drives proliferation. Why?
The effect is local concentration, not degradation. (Ch. 3/6)
kcat is equal; the difference is localization. (Ch. 3/6)
Both bind ATP; localization is the difference. (Ch. 3/6)
Answer as lectured Confining Src to the thin membrane layer where its target sits boosts effective activity without changing kcat. (Ch. 3/6)
6–1verdict + reasonClaim: "Errors in transcription are less dangerous to an organism than errors in DNA replication." This is —
Protein, not RNA, is usually the product; mRNA is transient. (Ch. 3/6)
Replication errors are permanent; transcription errors are not. (Ch. 3/6)
The reason is transience and heritability, not relative accuracy. (Ch. 3/6)
Answer as lectured A bad mRNA is one of many transient copies; a replication error is permanent and inherited. (Ch. 3/6)
6–2verdict + reasonClaim: "Because introns are largely genetic 'junk,' they do not have to be removed precisely during RNA splicing." This is —
The ribosome cannot fix a mis-spliced mRNA. (Ch. 3/6)
Answer as lectured A one-nucleotide error shifts the reading frame or alters the protein — precision is essential. (Ch. 3/6)
Imprecise splicing corrupts the coding sequence. (Ch. 3/6)
Introns are removed; the point is precise junction removal. (Ch. 3/6)
6–3verdict + reasonClaim: "Wobble pairing occurs between the first position in the codon and the third position in the anticodon." This is —
They pair antiparallel; wobble is codon-3 / anticodon-1. (Ch. 3/6)
Answer as lectured Reading antiparallel, the flexible pair is the third codon base with the first anticodon base. (Ch. 3/6)
The second position is strict; wobble is at the third. (Ch. 3/6)
It is the third codon base, not the first. (Ch. 3/6)
6–4verdict + reasonClaim: "During protein synthesis, the thermodynamics of base-pairing between tRNAs and mRNAs sets the upper limit for the accuracy with which proteins are made." This is —
Proofreading exceeds the thermodynamic limit at an energy cost. (Ch. 3/6)
Base-pairing sets a baseline; proofreading improves on it. (Ch. 3/6)
Proofreading acts during selection and uses GTP; it is not post-synthesis repair. (Ch. 3/6)
Answer as lectured Proofreading uses extra GTP hydrolysis to reject wrong tRNAs beyond what binding energy alone allows. (Ch. 3/6)
6–5verdict + reasonClaim: "Protein enzymes outnumber ribozymes in modern cells because they can catalyze a greater variety of reactions, and all of them have faster rates than any ribozyme." This is —
Not universally faster — the ribosome is a fast ribozyme. (Ch. 3/6)
Answer as lectured Proteins catalyze more reaction types, but some ribozymes (the ribosome) rival or beat protein rates. (Ch. 3/6)
Variety and speed are distinct; the speed claim fails. (Ch. 3/6)
Protein enzymes are more common; the error is 'all faster.' (Ch. 3/6)
6–6concept · figure→conceptAn RNA polymerase is fixed to a slide; the DNA (both strands tethered to a bead) is transcribed. As RNAP advances, why must the bead rotate?
DNA is helical, so tracking it requires rotation. (Ch. 3/6)
Rotation comes from helical tracking, not the RNA product. (Ch. 3/6)
The rotation is directional, driven by transcription. (Ch. 3/6)
Answer as lectured Reading bases along a helix requires relative rotation; if RNAP cannot spin, the DNA does. (Ch. 3/6)
6–7concept · figure→conceptA moving RNA polymerase generates supercoils in DNA that cannot freely rotate. What is the pattern, and what happens if the polymerase can rotate freely?
They form when rotation is blocked, not when free. (Ch. 3/6)
Blocked rotation builds the twin supercoil domains. (Ch. 3/6)
Answer as lectured Tracking the helix overwinds DNA ahead and underwinds it behind when rotation is blocked. (Ch. 3/6)
6–8concept · SVGIn the α-tropomyosin gene, exons 2 and 3 are mutually exclusive alternatives (every mRNA has one or the other). Which constraint on exons 2 and 3 is most accurate — and does it also hold for the mutually exclusive pair 7 and 8?
α-tropomyosin exon structure. Exons 2/3 and 7/8 are mutually exclusive cassettes spliced into a shared frame downstream. Whichever is chosen must not shift the reading frame of exons 4–onward.
Answer as lectured Mutually exclusive exons must keep the downstream reading frame identical, so their lengths must be congruent mod 3. (Ch. 3/6)
The frame must be preserved, imposing a mod-3 constraint. (Ch. 3/6)
They need the same remainder mod 3, not necessarily divisible by 3. (Ch. 3/6)
Only the length mod 3 must match, not the exact length. (Ch. 3/6)
6–9conceptBy single-nucleotide changes, Val mutates to Ala or Met, and both then mutate to Thr. Given Val=GUN, Ala=GCN, Met=AUG, Thr=ACN, could you get Val→Thr in a single step?
Many pairs need two or more changes; Val→Thr is one. (Ch. 3/6)
Answer as lectured Thr is reached only via the Ala or Met intermediates, not directly from Val. (Ch. 3/6)
They share position 3; the block is two differing positions, reachable via intermediates. (Ch. 3/6)
Chemistry aside, their codons differ at two positions. (Ch. 3/6)
6–10most deleteriousWhich mutation is predicted to be the most deleterious to gene function?
A late frameshift affects only a short C-terminal stretch. (Ch. 3/6)
A point substitution changes at most one residue. (Ch. 3/6)
That removes one codon but preserves the frame. (Ch. 3/6)
Answer as lectured An early frameshift garbles the entire downstream protein — the most damaging. (Ch. 3/6)
6–11conceptBoth prokaryotes and eukaryotes have systems to deal with broken/partial mRNAs. What danger do truncated mRNAs pose that justifies such systems?
They pose real risks, which is why surveillance exists. (Ch. 3/6)
The danger is bad protein and stalled ribosomes, not nucleotide supply. (Ch. 3/6)
Answer as lectured Truncated products can be dominant-negative, and ribosomes stall on broken mRNA — hence surveillance systems. (Ch. 3/6)
mRNA problems do not alter genomic DNA. (Ch. 3/6)
6–12concepthsp60- and hsp70-type chaperones both recognize exposed hydrophobic patches on proteins. Why is an exposed hydrophobic patch a reliable signal of incomplete folding?
They indicate the opposite — misfolding. (Ch. 3/6)
Folded soluble proteins bury hydrophobics; exposure is abnormal. (Ch. 3/6)
Answer as lectured Correctly folded soluble proteins hide hydrophobics in the core, so an exposed patch marks an unfolded or aggregating protein. (Ch. 3/6)
Soluble proteins have them too, buried in the core. (Ch. 3/6)
6–13conceptIf chaperones are needed to fold many proteins, how do the chaperones themselves fold correctly — the apparent chicken-and-egg problem?
They fold inside the cell, often with help. (Ch. 3/6)
Answer as lectured Spontaneous folding plus mutual chaperone assistance resolves the apparent paradox. (Ch. 3/6)
Chaperones fold many clients and assist each other. (Ch. 3/6)
Many proteins fold spontaneously; the paradox is resolvable. (Ch. 3/6)
6–14concept · figure→conceptAn RNA can fold into a hairpin with a symmetric internal loop. Can the complementary RNA form a similar structure, and would any regions be identical?
The complement forms an analogous paired hairpin. (Ch. 3/6)
Only the paired stems correspond; single-stranded loops differ. (Ch. 3/6)
The paired stem regions do correspond between the two. (Ch. 3/6)
Answer as lectured Stems pair as complements in both, so the double-stranded stem regions correspond; the loops differ. (Ch. 3/6)
6–15conceptWhy is RNA hypothesized to be an evolutionary precursor to both DNA and protein, yet DNA is the better material for information storage?
RNA does carry information, e.g. RNA viruses. (Ch. 3/6)
Answer as lectured RNA's catalytic and coding versatility could start life; DNA's stability makes it a safer store. (Ch. 3/6)
DNA is the more stable archive; RNA is reactive. (Ch. 3/6)
RNA is the catalytic one; DNA's edge is stability. (Ch. 3/6)
6–16conceptA single copy of a self-catalytic RNA (a ribozyme replicase) assembles by chance. Can that one molecule use itself as a template to catalyze its own replication?
Answer as lectured Copying requires an extended template while catalysis requires the folded enzyme — one molecule cannot be both simultaneously. (Ch. 3/6)
Copying needs a template strand. (Ch. 3/6)
It cannot be template and enzyme at the same time. (Ch. 3/6)
RNA can catalyze; the issue is the template/enzyme conflict. (Ch. 3/6)
Topic 6 — Recycling of Proteins & Other Molecules. Content verified against Alberts, Molecular Biology of the Cell, 7th ed. Textbook-only concepts (formerly badged 📖) have been removed for lecture-scope focus; textbook corrections are inlined as notes and flagged for the exam. Midterm 1 scope.