Cells: Unit Diagnostic (Biology Unit 2)

What this diagnostic is for

Use this page before or right after revising Unit 2 to find which topics to revisit. There is one question per topic, and two for the largest (cell structure and tonicity).

These are original Marlbridge practice questions, not past exam questions. They are not calibrated, so your number right does not predict an exam score.

Calculator and data. A calculator is allowed on both exam sections; a scientific one is enough. For solute potential use Ψs = −iCRT with R = 0.0831 L bar mol⁻¹ K⁻¹ and T in kelvin (°C + 273). All data, drugs and cells are fictional.

How to work. Answer everything first, writing a reason for each choice.

Question 1 (multiple choice · 2.1)

The Golgi complex normally adds sugar chains to a membrane protein. In a mutant cell line it cannot. The protein is still made and folded, but most of it ends up in the wrong place. Which is the best explanation?

(A) The sugar chains help label the protein so it is sorted to the right destination.
(B) Without the sugar chains, ribosomes cannot make the protein.
(C) The sugar chains supply the energy that moves the vesicle.
(D) The rough ER now sends the protein straight to lysosomes.

Answer and explanation

Answer: (A). The Golgi chemically modifies new products, for example by glycosylation, and these changes can decide where a protein goes.

(B) is ruled out: the protein is still made.
(C) confuses a label with a fuel.
(D) invents a route that skips the Golgi.

If you missed this: read “Golgi complex: finishing and dispatch” in the cell structure and function study guide.

Question 2 (multiple choice · 2.1)

Which statement about mitochondria and chloroplasts in a flowering plant is correct?

(A) Plant cells have chloroplasts instead of mitochondria.
(B) Leaf cells contain both; cells deep inside a root typically have mitochondria but no chloroplasts.
(C) Both organelles have a single membrane.
(D) Chloroplasts make all of a leaf cell’s ATP, so leaf cells need no mitochondria.

Answer and explanation

Answer: (B). Plant cells respire aerobically all the time, so they need mitochondria. Chloroplasts are found mainly in cells that receive light.

(A) and (D) forget that making sugar does not remove the need to break it down for ATP.
(C) is wrong: both have a double membrane.

If you missed this: read “Mitochondria and chloroplasts: two membranes, two energy roles” in the cell structure and function study guide.

Question 3 (multiple choice · 2.2)

A cell is modelled as a cube with sides of 10 µm. Microvilli on its top face double the area of that face. Ignoring their small volume, what is the new surface area-to-volume ratio?

(A) 0.60 µm⁻¹
(B) 0.70 µm⁻¹
(C) 0.80 µm⁻¹
(D) 1.20 µm⁻¹

Answer and explanation

Answer: (B). SA = 6 × 10² = 600 µm²; V = 10³ = 1000 µm³. Doubling the 100 µm² top face adds 100 µm², so SA/V = 700 ÷ 1000 = 0.70 µm⁻¹.

(A) is the ratio before the microvilli.
(C) adds 200 µm², the face’s whole new area, not the extra 100 µm².
(D) doubles all six faces.

If you missed this: read “Ways to keep SA/V high” in the cell size study guide.

Question 4 (multiple choice · 2.3)

A transmembrane protein forms a water-filled pore that lets ions cross. Which description of its R groups is most likely?

(A) Nonpolar R groups line the pore; charged R groups face the fatty acid tails.
(B) Charged R groups line the pore and face the tails.
(C) Nonpolar R groups line the pore and face the tails.
(D) Polar and charged R groups line the pore; nonpolar R groups face the fatty acid tails.

Answer and explanation

Answer: (D). Hydrophilic regions sit inside the protein (the pore) or in the fluid on each side; hydrophobic regions touch the tails.

(A) swaps the regions.
(B) puts charges among the nonpolar tails, which is unstable.
(C) would anchor the protein but block hydrated ions.

If you missed this: read “Membrane proteins: R groups decide where each part sits” in the plasma membrane study guide.

Question 5 (multiple choice · 2.4)

Vesicles of pure phospholipid are placed in a solution of ammonia (NH₃: small, polar, uncharged) and ammonium ions (NH₄⁺). Which prediction is best?

(A) Both cross freely, because both are smaller than glucose.
(B) Neither crosses, because both contain nitrogen.
(C) NH₃ crosses in small amounts; NH₄⁺ hardly crosses.
(D) NH₄⁺ crosses faster, because the phosphate heads attract it.

Answer and explanation

Answer: (C). Small polar, uncharged molecules pass through a bilayer in small amounts. An ion, held in its shell of water, cannot enter the nonpolar layer of tails.

(A) ignores charge; (B) invents a rule about elements.
(D) forgets that crossing means getting through the tails.

If you missed this: read “Why the hydrophobic interior decides” in the membrane permeability study guide.

Question 6 (multiple choice · 2.5)

A white blood cell takes in a whole bacterium. Which description is correct?

(A) Exocytosis: vesicles fuse with the membrane, with no energy needed
(B) Facilitated diffusion: it passes through a large channel protein
(C) Passive transport: it moves down its own concentration gradient
(D) Endocytosis: the membrane encloses it in a new vesicle, using energy

Answer and explanation

Answer: (D). A whole cell is far too large for any membrane protein, so the membrane encloses it in a vesicle, using energy.

(A) names the reverse process, which moves material out.
(B) and (C) describe real routes for small solutes, not for particles.

If you missed this: read “Bulk transport: endocytosis and exocytosis” in the membrane transport study guide.

Question 7 (multiple choice · 2.6)

Glucose enters a cell only through carrier proteins. Outside it is at 5 mmol L⁻¹, inside at 1 mmol L⁻¹. A poison suddenly stops ATP production. What happens to glucose entry in the next minute?

(A) It stops, because carriers need ATP to change shape.
(B) It reverses.
(C) It continues, because glucose still moves down its concentration gradient.
(D) It speeds up, because the carriers no longer spend ATP.

Answer and explanation

Answer: (C). Facilitated diffusion is passive: the carrier provides a path, the gradient provides the drive.

(A) and (D) confuse these carriers with ATP-using pumps.
(B) would need the inside to be more concentrated.

If you missed this: read “Large polar molecules: carriers and saturation” in the facilitated diffusion study guide.

Question 8 (multiple choice · 2.7)

All four solutions are at 25 °C in open beakers. Which has the lowest water potential?

(A) 0.40 M sucrose
(B) 0.25 M NaCl
(C) 0.15 M CaCl₂
(D) 0.30 M glucose

Answer and explanation

Answer: (B). R and T are the same, so compare i × C: sucrose 0.40, NaCl 2 × 0.25 = 0.50, CaCl₂ 3 × 0.15 = 0.45, glucose 0.30. NaCl gives the most particles: Ψ = −2 × 0.25 × 0.0831 × 298 = −12.4 bar.

(A) has the highest molarity, but sucrose does not ionize (i = 1).
(C) has the largest i, but 0.45 is less than 0.50.
(D) has the highest water potential (−7.43 bar).

If you missed this: read “Calculating solute potential” in the tonicity and osmoregulation study guide.

Question 9 (multiple choice · 2.8)

How does the sodium–potassium pump directly help make the inside of a resting cell negative?

(A) Each cycle moves 3 Na⁺ out and 2 K⁺ in, so one more positive charge leaves than enters.
(B) Each cycle moves 2 Na⁺ out and 3 K⁺ in.
(C) It pumps Cl⁻ ions into the cell.
(D) It has no direct effect, because it swaps ions one for one.

Answer and explanation

Answer: (A). Per ATP, 3 Na⁺ leave and 2 K⁺ enter: a net loss of one positive charge. The pump also keeps up the gradients the membrane potential depends on.

(B) reverses the numbers, which would make the inside more positive.
(C) names the wrong ion.
(D) forgets that the swap is unequal.

If you missed this: read “The sodium–potassium pump” in the mechanisms of transport study guide.

Question 10 (multiple choice · 2.9)

A eukaryotic cell builds compound F in one membrane-bound compartment and breaks F down in another. Which is the best explanation?

(A) Mixing all the enzymes together makes F faster.
(B) Separating the opposing reactions stops F being broken down as soon as it is made.
(C) The cell needs two copies of one enzyme.
(D) Every reaction needs a membrane to supply ATP.

Answer and explanation

Answer: (B). Internal membranes reduce competing interactions; otherwise energy spent making F would be wasted.

(A) is the opposite of what compartments do.
(C) misreads the question: building and breaking need different enzymes.
(D) overgeneralizes from the membranes that do make ATP.

If you missed this: read “Idea 1: compartments keep competing reactions apart” in the cell compartmentalization study guide.

Question 11 (short answer · 2.10)

(a) A prokaryote has no membrane-bound organelles. Name two regions or structures in it that have specialized jobs, with each job. (b) Give two observations that support the claim that mitochondria evolved from once free-living prokaryotes, and explain each.

Answer and explanation

(a) Any two: the nucleoid, holding the main DNA without a membrane; ribosomes, which make proteins; the plasma membrane, which carries the proteins of aerobic respiration; thylakoid membranes in cyanobacteria, where light is captured.

(b) Any two, each with its reason: two membranes (a prokaryote enclosed by a host membrane); own circular DNA (like a bacterial chromosome); bacteria-like ribosomes (leftover machinery of the ancestor); division by splitting (a cell cannot build one from scratch); DNA sequences closest to one bacterial group (shared ancestry).

If you missed this: read “Inside a prokaryotic cell” and “The evidence” in the origins of cell compartmentalization study guide.

Question 12 (short answer · 2.7)

A fictional animal cell at 37 °C has cytoplasm with the same solute potential as 0.15 M NaCl, and zero pressure potential. It is placed in solution X (0.30 M sucrose) or solution Y (0.10 M NaCl). Neither solute crosses the membrane.

(a) Calculate the water potential of the cell, X and Y. (b) Describe X and Y as hypotonic, isotonic or hypertonic to the cell, and predict what happens in each. (c) Explain why a plant cell placed in Y does not burst.

Answer and explanation

(a) T = 310 K. Cell: −2 × 0.15 × 0.0831 × 310 = −7.73 bar. X: −1 × 0.30 × 0.0831 × 310 = −7.73 bar. Y: −2 × 0.10 × 0.0831 × 310 = −5.15 bar.

(b) X is isotonic (i × C = 0.30 for both): no net water movement. Y is hypotonic (higher Ψ): water moves into the cell by osmosis, so it swells and may burst.

(c) The cell presses on its wall, which resists. Pressure potential rises until the cell’s Ψ reaches −5.15 bar, and net entry stops: the cell is turgid.

Watch for: i = 1 for NaCl gives −3.86 bar for the cell and makes X look hypertonic.

If you missed this: read “Water potential: one number that predicts the direction” in the tonicity and osmoregulation study guide.

Your next step

A short answer missed in one part counts as a miss.

Topic	Question(s)	If you missed it, read
2.1 Cell structure and function	1, 2	Study guide 2.1
2.2 Cell size	3	Study guide 2.2
2.3 Plasma membrane	4	Study guide 2.3
2.4 Membrane permeability	5, 12(c)	Study guide 2.4
2.5 Membrane transport	6	Study guide 2.5
2.6 Facilitated diffusion	7	Study guide 2.6
2.7 Tonicity and osmoregulation	8, 12	Study guide 2.7
2.8 Mechanisms of transport	9	Study guide 2.8
2.9 Cell compartmentalization	10	Study guide 2.9
2.10 Origins of compartmentalization	11	Study guide 2.10

How to use your result

Wrong answers first. For each topic you missed, read the named section, then do that topic’s practice questions.
Lucky guesses count as misses if your reason did not match the explanation.
Missed 6, 7 and 9? The thread is energy: which routes need ATP. Start with Topic 2.5.
No misses? Go straight to the mixed unit review.