Cell Structure and Function: Organelles, Membranes, Transport, Protein Targeting, Signalling and Ion Channels

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.

1. Which organelle carries out the first steps in the oxidation of very-long-chain fatty acids and contains catalase to destroy the hydrogen peroxide made by its oxidases?

   1. Peroxisome, a single-membrane organelle that imports its folded enzymes
   2. Mitochondrion, the main site of β-oxidation of most dietary fatty acids
   3. Lysosome, which digests lipids with acid hydrolases at a pH near 5
   4. Smooth endoplasmic reticulum, the main site of membrane lipid synthesis
   Show answer

   Answer: A — Peroxisome, a single-membrane organelle that imports its folded enzymes

   Peroxisomes begin the β-oxidation of very-long-chain fatty acids with an oxidase that hands electrons straight to O₂, making H₂O₂, which their catalase destroys; in Zellweger syndrome these fatty acids accumulate. Mitochondria oxidise most ordinary fatty acids but have no catalase-based H₂O₂ system for this; lysosomes hydrolyse rather than oxidise lipids; the smooth ER synthesises lipids.

2. A rat liver homogenate is centrifuged in steps at 1,000 g for 10 min, 20,000 g for 20 min and 100,000 g for 60 min, collecting the pellet after each step. In which fraction is glucose 6-phosphatase activity most enriched?

   1. The 1,000 g pellet, which contains nuclei and unbroken cells
   2. The 20,000 g pellet, which contains mitochondria, lysosomes and peroxisomes
   3. The final supernatant, which contains the soluble cytosolic proteins
   4. The 100,000 g pellet, which contains microsomes formed from the ER
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   Answer: D — The 100,000 g pellet, which contains microsomes formed from the ER

   Glucose 6-phosphatase is an integral ER membrane enzyme, the classic microsomal marker, so it sediments with the small ER vesicles at 100,000 g. Mitochondria (marker succinate dehydrogenase) pellet at 20,000 g, and the cytosol (marker lactate dehydrogenase) stays in the supernatant. The enzyme's role in releasing glucose from the liver tempts some to place it in the cytosol, but its active site faces the ER lumen.

3. Cultured fibroblasts from a child secrete many different lysosomal acid hydrolases into the medium, while their lysosomes fill with undigested material. Which single defect best explains this?

   1. Loss of the lysosomal V-type H⁺-ATPase, so the lumen is not acidified and the hydrolases cannot work
   2. Loss of the cis-Golgi phosphotransferase that starts the mannose 6-phosphate tag on lysosomal enzymes
   3. Loss of the SRP receptor, so lysosomal enzymes are made on free ribosomes and stay in the cytosol
   4. Loss of one lysosomal hydrolase, so that its own particular substrate builds up inside lysosomes
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   Answer: B — Loss of the cis-Golgi phosphotransferase that starts the mannose 6-phosphate tag on lysosomal enzymes

   This is I-cell disease. Without the phosphotransferase, lysosomal enzymes carry no mannose 6-phosphate, are not captured by M6P receptors in the trans-Golgi, and follow the default route out of the cell, so many hydrolases appear in the medium and lysosomes lack them. A V-ATPase defect would leave enzymes inside, merely inactive; an SRP receptor defect would block entry into the ER, so the enzymes could not be secreted at all; a single-enzyme defect such as Tay–Sachs affects one substrate and does not divert many enzymes outward.

4. In a healthy cell, phosphatidylserine is confined almost entirely to the cytosolic leaflet of the plasma membrane. What does its appearance on the outer leaflet usually signal?

   1. The cell is preparing to divide and is adding new membrane to its surface
   2. Cholesterol has been removed from the membrane, so the leaflets have mixed
   3. The cell is undergoing apoptosis and should be engulfed by phagocytes
   4. A hormone has bound and activated a receptor tyrosine kinase on the cell
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   Answer: C — The cell is undergoing apoptosis and should be engulfed by phagocytes

   ATP-dependent flippases keep phosphatidylserine in the inner leaflet. In apoptosis they are switched off and scramblases are activated, so phosphatidylserine reaches the surface, where macrophage receptors recognise it as an “eat me” signal (and annexin V detects it in the laboratory). Cholesterol depletion does not mix the leaflets, and neither division nor receptor activation is signalled this way.

5. Which of the following statements about transport across the plasma membrane are correct?

   1. GLUT1 of erythrocytes shows saturation with a characteristic Km but cannot move glucose against its concentration gradient
   2. The intestinal Na⁺–glucose symporter SGLT1 hydrolyses ATP directly to pump glucose into the epithelial cell
   3. The Na⁺/K⁺-ATPase moves 3 Na⁺ out and 2 K⁺ in for each ATP hydrolysed, so it is electrogenic
   4. Ion channels saturate like carriers and move ions more slowly than carriers move their solutes
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   Answer: A — GLUT1 of erythrocytes shows saturation with a characteristic Km but cannot move glucose against its concentration gradient; C — The Na⁺/K⁺-ATPase moves 3 Na⁺ out and 2 K⁺ in for each ATP hydrolysed, so it is electrogenic

   GLUT1 is a uniporter for facilitated diffusion: saturable and specific, but only downhill. The Na⁺/K⁺-ATPase exchanges 3 Na⁺ for 2 K⁺ per ATP, moving one net charge out per cycle. SGLT1 is secondary active: it uses the Na⁺ gradient made by the pump, not ATP directly. Channels conduct up to 10⁷–10⁸ ions per second, far faster than carriers, because no conformational cycle is needed per ion.

6. The hydropathy plot of a newly cloned plasma-membrane protein shows seven peaks, each a run of about 20 consecutive hydrophobic residues. What is the most likely structure of the protein?

   1. A peripheral protein held on the surface by ionic bonds to the lipid head groups
   2. A β-barrel porin whose short alternating polar and nonpolar strands line a pore
   3. A soluble cytosolic enzyme whose hydrophobic core contains seven buried segments
   4. An integral protein with seven membrane-spanning α-helices, like a GPCR
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   Answer: D — An integral protein with seven membrane-spanning α-helices, like a GPCR

   About 20 residues in an α-helix span roughly 3 nm, the thickness of the hydrophobic core, so each peak marks one transmembrane helix and seven peaks fit a seven-helix receptor such as a G protein-coupled receptor. β-barrel strands are shorter and alternate polar and nonpolar residues, so they do not give long hydrophobic peaks; peripheral proteins and soluble enzymes lack such long uninterrupted hydrophobic runs.

7. A normally cytosolic protein is engineered to carry an N-terminal ER signal sequence and the C-terminal sequence Lys-Asp-Glu-Leu (KDEL). Where will most of the protein be found at steady state?

   1. Secreted into the medium, since proteins entering the ER follow the default pathway
   2. In the mitochondrial matrix, since the N-terminal signal also acts as a presequence
   3. In the ER lumen, since KDEL receptors return escaped copies from the Golgi by COPI
   4. In the cytosol, since a protein that normally folds there cannot use a signal sequence
   Show answer

   Answer: C — In the ER lumen, since KDEL receptors return escaped copies from the Golgi by COPI

   The signal sequence sends the protein into the ER co-translationally via SRP and Sec61, and KDEL is the ER-retrieval signal: copies that reach the cis-Golgi bind KDEL receptors and return in COPI vesicles. Without KDEL the protein would indeed be secreted by default, which is the tempting answer. ER signal sequences are hydrophobic and differ from the amphipathic, basic mitochondrial presequences, and sorting signals work when fused to proteins that are normally cytosolic.

8. Which of the following statements about the ubiquitin–proteasome pathway are correct?

   1. The proteasome degrades ubiquitinated proteins inside the lysosomal lumen at acidic pH
   2. Proteins tagged with Lys48-linked polyubiquitin are unfolded and degraded by the 26S proteasome in an ATP-dependent process
   3. Ubiquitin is attached through its N-terminal methionine to aspartate side chains of the target protein
   4. E3 ubiquitin ligases recognise particular substrates and so give the system its specificity
   Show answer

   Answer: B — Proteins tagged with Lys48-linked polyubiquitin are unfolded and degraded by the 26S proteasome in an ATP-dependent process; D — E3 ubiquitin ligases recognise particular substrates and so give the system its specificity

   Lys48-linked chains are the degradation signal read by the 19S caps of the 26S proteasome, which unfold the substrate and feed it into the 20S core using ATP; the many E3 ligases, each recognising particular degrons, provide specificity. The proteasome is a cytosolic and nuclear particle, not a lysosomal one. Ubiquitin is joined through the carboxyl group of its C-terminal glycine to lysine ε-amino groups by an isopeptide bond.

9. Proteins with a nuclear localisation signal accumulate in the nucleus even though nuclear pores are aqueous channels open in both directions. What makes the import directional?

   1. A Ran-GTP gradient, high in the nucleus where its exchange factor binds chromatin, makes importins release cargo only inside
   2. The localisation signal is cut off inside the nucleus, so the imported protein can no longer bind the export machinery
   3. Pore proteins hydrolyse ATP and pump each cargo inward through the central channel against its concentration gradient
   4. The signal recognition particle binds the localisation signal in the cytosol and then docks only on the nuclear face of the pore
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   Answer: A — A Ran-GTP gradient, high in the nucleus where its exchange factor binds chromatin, makes importins release cargo only inside

   The Ran exchange factor (on chromatin) keeps nuclear Ran in the GTP form, while the Ran GTPase-activating protein in the cytosol keeps cytosolic Ran as GDP. Importin binds cargo in the cytosol, crosses the pore, and releases it when Ran-GTP binds in the nucleus; the importin–Ran-GTP complex returns and GTP is hydrolysed in the cytosol. This asymmetry, paid for by GTP, sets the direction. Nuclear localisation signals are not cleaved (proteins are re-imported after every mitosis), the pore is not an ATP pump, and SRP serves the ER.

10. Which of the following statements about toxins are correct?

    1. Cholera toxin ADP-ribosylates Gsα and blocks its GTPase, so adenylyl cyclase stays active in intestinal cells
    2. α-Bungarotoxin blocks voltage-gated Na⁺ channels along the axon and so prevents action potentials
    3. Ouabain activates the Na⁺/K⁺-ATPase and so hyperpolarises heart muscle cells
    4. Botulinum toxin is a protease that cleaves SNARE proteins and so stops acetylcholine release
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    Answer: A — Cholera toxin ADP-ribosylates Gsα and blocks its GTPase, so adenylyl cyclase stays active in intestinal cells; D — Botulinum toxin is a protease that cleaves SNARE proteins and so stops acetylcholine release

    Cholera toxin locks Gsα in its GTP-bound active state, so cAMP stays high and CFTR drives Cl⁻ and water into the gut; botulinum toxin cleaves SNAREs at motor nerve terminals, causing flaccid paralysis. α-Bungarotoxin binds the nicotinic acetylcholine receptor, not the voltage-gated Na⁺ channel (that is tetrodotoxin's target). Ouabain inhibits the Na⁺/K⁺-ATPase; the resulting rise in cell Na⁺ slows Ca²⁺ extrusion and strengthens heart contraction.

11. In a bacterial two-component system, the sensor kinase autophosphorylates on a conserved histidine when it detects its stimulus and then transfers the phosphoryl group to a conserved aspartate in the receiver domain of the response regulator. If that aspartate is replaced by asparagine, what is expected?

    1. The sensor can no longer autophosphorylate, because its histidine needs that aspartate as a catalytic partner
    2. The sensor still autophosphorylates, but the regulator cannot accept the phosphate, so the response is lost
    3. The regulator is phosphorylated on a nearby serine instead, so the response continues at a lower rate
    4. The regulator becomes permanently active, so its target genes are expressed even without the stimulus
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    Answer: B — The sensor still autophosphorylates, but the regulator cannot accept the phosphate, so the response is lost

    Asparagine has an amide instead of the carboxylate that forms the acyl-phosphate, so the regulator cannot be phosphorylated and cannot switch on its targets. Autophosphorylation of the sensor histidine happens within the kinase (using ATP) and does not need the regulator, so the sensor side still works. Histidine kinases do not phosphorylate serine, and an Asp→Asn change removes, rather than mimics, the phosphorylated state, so constitutive activation is not expected.

12. A hormone binds a single class of independent receptor sites with Kd = 3.0 nM. At what free hormone concentration, in nM, are 80% of the receptors occupied?

    Numerical answer — type the value.

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    Answer: 12

    θ = [L]/(Kd + [L]) = 0.80 gives [L] = 0.80Kd/0.20 = 4Kd = 4 × 3.0 = 12 nM. Using 0.8 × Kd (2.4 nM) confuses occupancy with concentration.

13. An agonist binds its receptor with Kd = 2.0 nM, and a competitive antagonist binds the same site with Ki = 5.0 nM. With free agonist at 10 nM and free antagonist at 40 nM, what percentage of the receptors are occupied by the agonist at equilibrium? Give the answer to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 35.7

    The antagonist raises the apparent Kd to Kd(1 + [I]/Ki) = 2.0 × (1 + 40/5.0) = 2.0 × 9 = 18 nM. Agonist occupancy = 10/(18 + 10) = 0.357, i.e. 35.7%. Without antagonist it would be 10/12 = 83.3%.

14. A hormone acting through a Gq-coupled receptor raises cytosolic Ca²⁺ within seconds. Which sequence explains this?

    1. Gq activates adenylyl cyclase, and cAMP then opens Ca²⁺ channels in the plasma membrane
    2. Gq activates phospholipase Cβ, and diacylglycerol then opens Ca²⁺ channels in the ER membrane
    3. Gq activates a receptor tyrosine kinase, whose phosphotyrosines then release bound Ca²⁺
    4. Gq activates phospholipase Cβ, and IP₃ then opens Ca²⁺ channels in the ER membrane
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    Answer: D — Gq activates phospholipase Cβ, and IP₃ then opens Ca²⁺ channels in the ER membrane

    Gqα activates phospholipase Cβ, which splits PIP₂ into IP₃ and diacylglycerol. Water-soluble IP₃ diffuses to IP₃-gated Ca²⁺ channels in the ER membrane; diacylglycerol stays in the membrane and, with Ca²⁺, activates protein kinase C. The distractor that gives diacylglycerol the channel-opening role swaps the two messengers; adenylyl cyclase is the Gs effector.

15. Which of the following statements about hormones and neurotransmitters are correct?

    1. The insulin receptor is a G protein-coupled receptor that activates adenylyl cyclase through Gs
    2. Steroid hormones such as cortisol cross the plasma membrane and bind intracellular receptors that act as transcription factors
    3. GABA acting on GABA_A receptors opens Cl⁻ channels and usually inhibits the postsynaptic neuron
    4. Acetylcholine's action at the neuromuscular junction ends mainly by reuptake of intact acetylcholine into the nerve terminal
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    Answer: B — Steroid hormones such as cortisol cross the plasma membrane and bind intracellular receptors that act as transcription factors; C — GABA acting on GABA_A receptors opens Cl⁻ channels and usually inhibits the postsynaptic neuron

    Lipid-soluble steroids act through nuclear receptors that bind DNA, and GABA_A receptors are ligand-gated Cl⁻ channels whose opening usually hyperpolarises or clamps the neuron. The insulin receptor is a receptor tyrosine kinase, not a GPCR. Acetylcholine is hydrolysed in the synaptic cleft by acetylcholinesterase, and only the choline is taken back up; reuptake of intact transmitter is how monoamines such as serotonin are cleared.

16. At 37 °C (310 K), [K⁺] is 5.0 mM outside and 140 mM inside a neuron. Using R = 8.314 J mol⁻¹ K⁻¹ and F = 96485 C mol⁻¹, what is the K⁺ equilibrium (Nernst) potential, inside relative to outside, in mV? Give the nearest integer, with its sign.

    Numerical answer — type the value.

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    Answer: -89

    E_K = (RT/zF) ln([K⁺]out/[K⁺]in). RT/F = 8.314 × 310/96485 = 0.02671 V = 26.71 mV; z = +1. ln(5.0/140) = ln(0.03571) = −3.332. E_K = 26.71 × (−3.332) = −89.0 mV, so −89. Inverting the ratio gives +89, the wrong sign.

17. A cell has [K⁺] 140 mM inside and 5.0 mM outside, and [Na⁺] 15 mM inside and 145 mM outside. Its resting permeabilities are in the ratio PK : PNa = 1 : 0.05, and Cl⁻ can be ignored. Using the Goldman–Hodgkin–Katz equation with RT/F = 26.7 mV, what is the resting membrane potential (inside relative to outside) in mV? Give the answer with its sign, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: -65.2

    Vm = (RT/F) ln[(PK[K⁺]o + PNa[Na⁺]o)/(PK[K⁺]i + PNa[Na⁺]i)]. Numerator: 1 × 5.0 + 0.05 × 145 = 5.0 + 7.25 = 12.25. Denominator: 1 × 140 + 0.05 × 15 = 140.75. ln(12.25/140.75) = ln(0.08703) = −2.441. Vm = 26.7 × (−2.441) = −65.2 mV, between E_K (−89 mV) and E_Na (+61 mV) and closer to E_K because K⁺ is far more permeant.

18. Tetrodotoxin applied to an isolated nerve abolishes action potentials within minutes but leaves the resting potential almost unchanged. Ouabain applied to the same nerve leaves action potentials intact at first, but the resting potential declines slowly over hours. Which explanation fits both observations?

    1. Tetrodotoxin blocks the K⁺ leak channels, while ouabain blocks the voltage-gated Na⁺ channels that set the resting potential
    2. Both inhibit the Na⁺/K⁺-ATPase, but tetrodotoxin acts faster because it is a much smaller molecule that diffuses in quickly
    3. Tetrodotoxin blocks the voltage-gated Na⁺ channels of the upstroke; ouabain stops the pump, so ion gradients run down slowly
    4. Tetrodotoxin stops acetylcholine release from the nerve terminal, while ouabain blocks the nicotinic receptors on the axon
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    Answer: C — Tetrodotoxin blocks the voltage-gated Na⁺ channels of the upstroke; ouabain stops the pump, so ion gradients run down slowly

    The resting potential depends on K⁺ leak through open channels and on the existing gradients, not on voltage-gated Na⁺ channels, which are closed at rest; so tetrodotoxin abolishes only the regenerative upstroke. The pump's direct electrogenic contribution is small, so blocking it with ouabain lets the large ion stores run down only gradually, and the potential falls over hours. Tetrodotoxin does not act on the pump or on synaptic release, and ouabain is not a channel blocker.