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IB Biology B2.1: Membranes and membrane transport worked solutions

Worked solutions to every practice question in B2.1: Membranes and membrane transport of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
The diagram shows part of a plasma membrane with four proteins, P, Q, R and S. Which statement is correct? [Graph: Part of a membrane: P is a small protein lying on the outer surface; Q is a large protein spanning both halves of the bilayer; R is embedded in the inner half only; S lies on the inner surface.] (A) P is a peripheral protein and R is an integral protein. (B) Q is an integral protein and S is an integral protein. (C) P is an integral protein and Q is a peripheral protein. (D) R is a peripheral protein because it does not span the bilayer.
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P rests on the outer surface, so it is peripheral. R is embedded in the inner half of the bilayer, so it is integral even though it does not span the membrane (A). S rests on the inner surface, so it is peripheral, which rules out B. C reverses the types of P and Q. D is wrong because embedded means integral, not peripheral.

Final answer

A

Question A2

[1 mark]
Potassium ions move into a cell down their concentration gradient. Which statement is correct? (A) They diffuse between the phospholipids because they are small. (B) They pass through a channel protein by facilitated diffusion. (C) They are pumped through a protein using ATP. (D) They cross the membrane by osmosis.
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Ions are charged, so they cannot cross the hydrophobic core between the phospholipids (A is wrong). Moving down the gradient needs no energy, so the route is a specific channel by facilitated diffusion (B). ATP pumping (C) is for movement against a gradient. Osmosis (D) is the movement of water.

Final answer

B

Question A3

[1 mark]
The concentration of ion X is 0.40\,\mathrm{mol\,dm^{-3}} inside a root cell and 0.02\,\mathrm{mol\,dm^{-3}} in the soil water. Uptake of X stops when the root is placed in nitrogen gas, which excludes oxygen. What does this show? (A) X enters by simple diffusion. (B) X enters through channel proteins by facilitated diffusion. (C) X enters by active transport using ATP from aerobic respiration. (D) X enters by osmosis.
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X moves into a region where it is already more concentrated, so it is moving against its gradient. Uptake stops when oxygen is removed because aerobic respiration, and so ATP, stops (C). Simple and facilitated diffusion are passive and would not depend on oxygen. Osmosis is the movement of water.

Final answer

C

Question A4

[1 mark]
Cells P and Q have the same size and the same solute concentration inside. P has many aquaporins and Q has few. Both are placed in the same dilute solution. What is the expected result? (A) Water enters P faster, and both reach the same final volume. (B) Water enters P more slowly because aquaporins use ATP. (C) Water enters P faster and P reaches a larger final volume. (D) Water enters P only, because Q is impermeable to water.
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Aquaporins raise the rate of water movement without changing its direction or the balance point, which depend on the solute concentrations (A). They do not use ATP (B). The final volume is set by the solutions, not by the number of channels (C). Water still crosses the bilayer of Q slowly (D).

Final answer

A

Question A5

[1 mark]
Phospholipids placed in water form a bilayer. Which statement explains the arrangement? (A) The hydrophilic heads face the water and the hydrophobic tails are shielded in the middle. (B) The hydrophobic tails face the water on both sides and the heads meet in the middle. (C) The heads face the water on one side and the tails face the water on the other. (D) The hydrophobic tails form hydrogen bonds with the water on both sides.
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The heads are polar and interact with water, while the tails are non-polar and are kept away from it, so the tails meet in the centre and the heads face the water on both sides (A). B and C place tails against water. D is wrong because hydrophobic tails cannot form hydrogen bonds with water.

Final answer

A

Question A6

HL[1 mark]
Four membranes have the same amount of phospholipid. Which will stay fluid at the lowest temperature? (A) Tails that are all saturated and short. (B) Tails with many cis double bonds. (C) Tails that are all saturated and long. (D) Tails that are all saturated, with extra peripheral proteins.
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Cis double bonds kink the tails, so they cannot pack closely and the membrane stays fluid when it is cold (B). Saturated tails are straight and pack tightly, which makes the membrane stiff at lower temperature. Peripheral proteins do not unpack the tails.

Final answer

B

Question A7

HL[1 mark]
Which row correctly pairs each channel in a neuron with the event that opens it? (A) Neurotransmitter-gated channel: the neurotransmitter binds. Voltage-gated channel: the membrane potential changes. (B) Neurotransmitter-gated channel: the membrane potential changes. Voltage-gated channel: the neurotransmitter binds. (C) Both channels open when ATP binds. (D) Both channels open when a neurotransmitter binds.
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A neurotransmitter-gated channel has a binding site for the transmitter, and a voltage-gated channel has a voltage sensor that responds to the membrane potential (A). B swaps the signals. In C and D neither channel needs ATP, and a voltage-gated channel does not respond to a transmitter.

Final answer

A

Question A8

HL[1 mark]
Cells lining the gut stay attached to one another as food passes. Which description best fits the molecules that hold the cells together? (A) Integral membrane proteins whose outer parts bind to matching proteins on the neighbouring cell. (B) Peripheral proteins that pump ions from one cell to the next. (C) Phospholipids that form a bridge between the two cells. (D) Vesicles that fuse the two plasma membranes into one.
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Cell-adhesion molecules are membrane proteins, usually integral, with outer regions that bind a matching molecule on the neighbouring cell or the matrix (A). They do not pump (B), are not lipid bridges (C) and do not fuse the membranes (D).

Final answer

A

Tutorial practice

Question B1

[4 marks]
Sketch and label a plasma membrane to show the fluid mosaic model, marking its outside and inside.
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Any four of these (1 each): a bilayer of two rows of phospholipids with the heads outward and the tails inward, with the heads and tails labelled; an integral protein drawn through both rows (or in one half) and labelled; a peripheral protein drawn on a surface, outside the bilayer, and labelled; a glycoprotein or glycolipid with a carbohydrate chain on the outer face; cholesterol drawn among the tails near the heads; the outside and inside of the cell labelled.

Final answer

Two rows of phospholipids, heads outward and tails inward; integral protein through both rows; peripheral protein on a surface; glycoprotein with carbohydrate outside; cholesterol among the tails; labels.

Question B2

[5 marks]
Facilitated diffusion and active transport both move particles across the plasma membrane. (a) State two features that the two processes share. [2] (b) Distinguish between the two processes in terms of the direction of movement and the source of energy. [2] (c) Name one particle moved by each process. [1]
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(a) Any two: both need a membrane protein (1); both are specific to particular particles, so both are selective (1); both move particles that cannot cross the lipid core easily (1). (b) Facilitated diffusion moves particles down their concentration gradient with no energy input (1); active transport moves particles against the gradient using energy from ATP (1). (c) For example potassium ions through a channel protein, and sodium ions pumped out by the sodium-potassium pump (1, both needed).

Final answer

(a) Both use specific membrane proteins; both are selective. (b) Facilitated: down the gradient, no ATP. Active: against the gradient, uses ATP. (c) \ce{K+} through a channel; \ce{Na+} pumped out.

Question B3

[6 marks]
Cells were placed in solutions of two substances, P and Q. The rate of uptake was measured at different concentrations.\begin{array}{lcccccc}\hline \text{Concentration / mmol\,dm}^{-3}&\text{0}&\text{2}&\text{4}&\text{8}&\text{16}&\text{32}\\ \hline \text{Uptake of P / units}&\text{0}&\text{3}&\text{6}&\text{12}&\text{24}&\text{48}\\ \hline \text{Uptake of Q / units}&\text{0}&\text{14}&\text{25}&\text{38}&\text{46}&\text{47}\\ \hline\end{array} (a) Describe the relationship between concentration and uptake of P. [1] (b) Calculate the percentage increase in the uptake of Q when the concentration doubles from 16 to 32 mmol dm^{-3}. [1] (c) Explain the plateau in the uptake of Q. [2] (d) Predict, with a reason, the effect on the uptake of Q of doubling the number of channel proteins. [1] (e) Suggest why P shows no plateau. [1]
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(a) Uptake of P rises in direct proportion to concentration: each doubling doubles the uptake (1). (b) (47-46)/46\times100=2.2\% (1). (c) Q crosses through a limited number of channel proteins (1); at high concentration all the channels are in use, so a higher concentration cannot raise the rate (1). (d) The plateau would be higher, about twice as high, because more channels can carry particles at once (1). (e) P crosses between the phospholipids by simple diffusion, which has no protein to become saturated (1).

Final answer

(a) Directly proportional. (b) 2.2 %. (c) Limited number of channels, all busy. (d) Higher plateau. (e) P crosses the lipid, no protein to saturate.

Question B4

[6 marks]
Protein T may carry substance Y into root cells. Uptake of Y was measured (mean \pm SD, six samples).\begin{array}{lc}\hline \text{Treatment}&\text{Uptake of Y / }\mu\text{mol\,g}^{-1}\text{\,h}^{-1}\\ \hline \text{Control}&\text{8.0 }\pm\text{ 0.5}\\ \hline \text{Compound Z, which binds protein T}&\text{2.0 }\pm\text{ 0.4}\\ \hline \text{Compound Z}'\text{, similar shape, does not bind T}&\text{7.6 }\pm\text{ 0.6}\\ \hline \text{Cyanide, which stops respiration}&\text{1.5 }\pm\text{ 0.3}\\ \hline\end{array} (a) Calculate the percentage reduction in uptake caused by compound Z. [1] (b) Using the standard deviations, state whether Z' differs from the control. [1] (c) Deduce what the cyanide result suggests about the way Y enters. [1] (d) Evaluate the claim that these results prove that protein T carries Y, including the purpose of treatment Z'. [3]
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(a) (8.0-2.0)/8.0\times100=75\% (1). (b) The ranges 7.5 to 8.5 and 7.0 to 8.2 overlap, so Z' probably does not differ from the control (1). (c) Uptake falls when respiration stops, so Y needs ATP and enters by active transport (1). (d) Z' shows that the fall with Z is caused by binding to T and not by the chemical in general or by damage to the cells (1). Z lowers uptake by 75% while Z' does not, which supports a role for T (1); the data cannot prove T is the carrier, because Z could block other proteins, and a further test such as cells with less T is needed (1).

Final answer

(a) 75 %. (b) No: ranges overlap. (c) Needs ATP: active transport. (d) Z' is a control for non-specific effects; supports T but does not prove it.

Question B5

[6 marks]
Frog egg cells were placed in a solution much more dilute than their cytoplasm. Some cells had been given the gene for an aquaporin and control cells had not. Volumes were measured at the start and after 10 minutes.\begin{array}{lcc}\hline \text{Group}&\text{Start volume / nl}&\text{Volume after 10 min / nl}\\ \hline \text{Control}&\text{940}&\text{970}\\ \hline \text{Aquaporin}&\text{1010}&\text{1210}\\ \hline\end{array} (a) Calculate the percentage increase in volume for each group. [2] (b) Explain why percentage change is a better comparison than final volume. [1] (c) Explain the result for the aquaporin cells. [2] (d) Predict the change in volume if the aquaporin cells were placed in a solution with the same solute concentration as their cytoplasm. [1]
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(a) Control: 30/940\times100=3.2\% (1). Aquaporin: 200/1010\times100=20\% (1). (b) The two groups started at different volumes, so a percentage change measures the change relative to the starting size (1). (c) The solute concentration inside is higher, so water enters by osmosis (1); aquaporins give water a faster route through the membrane, so more enters in the same time (1). (d) There would be no net change in volume, because there is no difference in solute concentration to drive osmosis (1).

Final answer

(a) Control 3.2 %; aquaporin 20 %. (b) Start volumes differ. (c) Aquaporins let water in faster. (d) No net change.

Question B6

[5 marks]
Dialysis tubing was filled with a mixture of starch and glucose solutions, sealed and placed in water. After 30 minutes the water outside was tested. Glucose was present and starch was absent. (a) Explain why starch was not found in the water. [1] (b) Explain how glucose reached the water. [2] (c) Identify two ways in which dialysis tubing is a limited model of a plasma membrane. [2]
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(a) Starch molecules are too large to pass through the pores of the tubing (1). (b) Glucose molecules are small enough to pass through the pores (1) and moved from the higher concentration inside to the lower concentration outside by diffusion (1). (c) Any two: it has no phospholipid bilayer; it has no channel or pump proteins, so no active transport; it sorts particles by size only, not by charge or polarity; it has no cholesterol or glycoproteins (1 each, up to 2).

Final answer

(a) Starch molecules too large for the pores. (b) Small enough; diffuses down its gradient through the pores. (c) No transport proteins or pumps; sorts by size only.

Question B7

[6 marks]
Equal cubes of beetroot were placed in water at different temperatures for 10 minutes. The absorbance of the water, which measures the red pigment that leaked out, was recorded. Each value is the mean of three tubes.\begin{array}{lccccc}\hline \text{Temperature / }^\circ\text{C}&\text{20}&\text{30}&\text{40}&\text{50}&\text{60}\\ \hline \text{Absorbance}&\text{0.04}&\text{0.05}&\text{0.09}&\text{0.31}&\text{0.78}\\ \hline\end{array} (a) State two variables that should be kept constant. [2] (b) Describe the trend in the results, using values. [1] (c) Explain the result at 60 ^\circC. [2] (d) Suggest one way to make the conclusion more reliable. [1]
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(a) Any two: size and number of cubes, volume of water, time in the water, source of the beetroot (1 each). (b) Absorbance stays low up to 40 ^\circC (0.04 to 0.09), then rises steeply to 0.78 at 60 ^\circC (1). (c) At 60 ^\circC the heat disrupts the phospholipid bilayer and denatures membrane proteins (1), so the membrane becomes permeable and more pigment leaks out (1). (d) Repeat with more tubes, or test more temperatures between 40 and 60 ^\circC (1).

Final answer

(a) Cube size and number, volume of water, time. (b) Low to 40, then rises to 0.78. (c) Heat disrupts the bilayer and denatures proteins. (d) More replicates or more temperatures.

Question B8

HL[5 marks]
Part of a transmembrane protein forms a channel for ions. The table shows the main amino acids in three regions of the chain. Regions X and Y lie within the thickness of the bilayer.\begin{array}{ll}\hline \text{Region}&\text{Main R groups}\\ \hline \text{X}&\text{leucine, valine, phenylalanine}\\ \hline \text{Y}&\text{serine, glutamate, asparagine}\\ \hline\end{array} (a) Deduce which of X and Y faces the phospholipid tails and which lines the pore. [2] (b) Explain why the lining of the pore must be polar. [2] (c) Predict the effect of replacing the lining of the pore with leucine. [1]
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(a) X has non-polar R groups, so it faces the hydrophobic tails (1); Y has polar or charged R groups, so it lines the pore (1). (b) The ions are charged and cannot enter a non-polar environment (1); polar groups lining the pore interact with the ions and with water, so the ions can pass through (1). (c) The lining would be non-polar and hydrophobic, so ions would not enter the pore and transport would stop (1).

Final answer

(a) X faces the tails; Y lines the pore. (b) Polar groups attract and hold ions in water-like surroundings. (c) Ions blocked.

Question B9

[4 marks]
Oxygen crosses the plasma membrane without a protein, but sodium ions need one. (a) Explain why oxygen can cross the phospholipid bilayer unaided. [2] (b) Explain why sodium ions cannot. [2]
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(a) Oxygen is a small non-polar molecule, so it can dissolve in the hydrophobic tails of the core (1) and diffuses between the phospholipids down its concentration gradient (1). (b) Sodium ions are charged, so they are repelled by the hydrophobic core and held by water around them (1); they need a channel protein for facilitated diffusion or a pump for active transport (1).

Final answer

(a) Small and non-polar, so it dissolves in the hydrophobic core and diffuses down its gradient. (b) Charged, so repelled by the core; needs a channel or pump.

Advanced practice

Question C1

HL[8 marks]
A secretory cell has a surface membrane area of 1200\,\mu\mathrm{m}^2. Secretory vesicles fuse with the surface at 60 per minute, and each has a membrane area of 0.50\,\mu\mathrm{m}^2. Endocytic vesicles each have a membrane area of 0.20\,\mu\mathrm{m}^2. (a) Calculate the membrane area added to the surface per minute by exocytosis. [1] (b) Express this as a percentage of the surface area. [1] (c) The surface area of the cell stays constant. Calculate the number of endocytic vesicles that must form per minute. [2] (d) Describe how endocytosis forms a vesicle and why fluidity is needed. [2] (e) Explain why the contents of a new endocytic vesicle are not yet in the cytoplasm. [1] (f) State one use of ATP in exocytosis. [1]
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(a) 60\times0.50=30\,\mu\mathrm{m}^2 per minute (1). (b) 30/1200\times100=2.5\% (1). (c) The area removed must equal 30\,\mu\mathrm{m}^2 (1); 30/0.20=150 vesicles per minute (1). (d) The plasma membrane infolds around material outside the cell and pinches off to form a vesicle inside (1); the bilayer is fluid, so it can bend and its edges can fuse again without leaving a gap (1). (e) The contents are still enclosed by the membrane of the vesicle, which separates them from the cytoplasm (1). (f) Any one: ATP powers motor proteins that move vesicles along the cytoskeleton; ATP is used to bring about fusion of the vesicle with the membrane (1).

Final answer

(a) 30 \mum^2. (b) 2.5 %. (c) 150. (d) Membrane infolds and pinches off; fluid bilayer can bend and reseal. (e) Enclosed by the vesicle membrane. (f) Movement of vesicles on the cytoskeleton.

Question C2

HL[8 marks]
An intestinal epithelial cell has 2.0\times10^{5} sodium-potassium pumps in its basal membrane. Each pump completes 50 cycles per second. (a) Calculate the number of sodium ions pumped out of the cell per second. [2] (b) Calculate the net number of positive charges removed from the cell per second. [2] (c) Explain how the pump allows the cell to take up glucose from the gut against its concentration gradient. [3] (d) The drug ouabain blocks the pump. Predict the effect on glucose uptake over the next hour, with a reason. [1]
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(a) Sodium ions out per second =3\times50\times2.0\times10^{5}=3.0\times10^{7} (1 for the method, 1 for the value). (b) Each cycle removes 3-2=1 net charge, so 1\times50\times2.0\times10^{5}=1.0\times10^{7} charges per second (1 for the method, 1 for the value). (c) The pump uses ATP to keep the sodium concentration low inside the cell (1); this makes sodium tend to flow into the cell down its gradient through a cotransporter (1); the cotransporter carries glucose in with the sodium, even though glucose is moving against its gradient (1). (d) Uptake falls gradually (1) because sodium builds up inside the cell, the sodium gradient is lost and the cotransporter no longer has a driving force.

Final answer

(a) 3.0\times10^7. (b) 1.0\times10^7. (c) Pump keeps \ce{Na+} low inside; cotransporter uses the Na gradient to bring glucose in. (d) Falls gradually as \ce{Na+} builds up.

Question C3

HL[7 marks]
Fluidity was measured in two artificial membranes at three temperatures. Membrane M contains phospholipids only. Membrane N contains the same phospholipids with cholesterol. Higher values mean more fluid.\begin{array}{lccc}\hline \text{Temperature / }^\circ\text{C}&\text{10}&\text{25}&\text{40}\\ \hline \text{Fluidity of M}&\text{12}&\text{45}&\text{90}\\ \hline \text{Fluidity of N}&\text{30}&\text{42}&\text{55}\\ \hline\end{array} (a) Calculate the change in fluidity between 10 and 40 ^\circC for each membrane. [2] (b) Compare the effect of cholesterol at 10 ^\circC and at 40 ^\circC. [2] (c) Explain these two effects. [2] (d) Suggest one change in the phospholipids that helps a cold-water fish keep its membranes fluid. [1]
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(a) M: 90-12=+78 (1). N: 55-30=+25 (1). (b) At 10 ^\circC cholesterol raises fluidity, from 12 to 30 (1); at 40 ^\circC it lowers fluidity, from 90 to 55 (1). (c) At low temperature the rings of cholesterol keep the tails from packing tightly, so the membrane does not stiffen (1); at high temperature they restrain the movement of the tails, so the membrane is less fluid (1). (d) A higher proportion of unsaturated fatty acids, whose kinked tails pack loosely (1).

Final answer

(a) M +78; N +25. (b) Raises fluidity at 10; lowers at 40. (c) Cholesterol prevents tight packing when cold, restrains tail movement when hot. (d) More unsaturated tails.

Question C4

[6 marks]
Lipids were extracted from red blood cells and spread in a single layer on water. The lipid from 1.0\times10^{8} cells formed a film of area 2.0\times10^{10}\,\mu\mathrm{m}^2. The mean surface area of one cell is 100\,\mu\mathrm{m}^2. (a) Calculate the ratio of the film area to the total surface area of the cells. [2] (b) One model says a membrane has a single layer of lipid and another says it has two layers. State the ratio each predicts. [1] (c) State which model the data support. [1] (d) Explain why this evidence cannot show where the proteins are. [1] (e) Suggest one source of error in the investigation. [1]
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(a) Total cell surface area =1.0\times10^{8}\times100=1.0\times10^{10}\,\mu\mathrm{m}^2 (1); ratio =2.0\times10^{10}/1.0\times10^{10}=2.0 (1). (b) One layer predicts a ratio of 1 and two layers predict a ratio of 2 (1). (c) The ratio is 2, so the data support two layers of lipid, a bilayer (1). (d) Only the area of lipid was measured, so nothing is shown about the position of proteins (1). (e) Any one: some lipid was lost during extraction; the film was not fully compressed; the cell surface area was estimated (1).

Final answer

(a) 2.0. (b) 1 and 2. (c) Two layers. (d) Only lipid area measured. (e) Some lipid lost, or film not compressed.

Connected questions

Question X1

Connected questions · HL[6 marks]
B2.1 + C2.2. In a resting neuron the concentration of \ce{K+} is 140 mmol dm^{-3} inside and 5 outside. The concentration of \ce{Na+} is 12 inside and 145 outside. (a) State the direction of net passive movement of each ion if its channels open. [2] (b) Explain how the sodium-potassium pump maintains these concentrations. [2] (c) A toxin blocks voltage-gated sodium channels. Predict the effect on an action potential and explain it. [2]
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(a) Potassium ions move out of the neuron and sodium ions move in, each down its concentration gradient (1 each). (b) The pump uses ATP to move 3 sodium ions out and 2 potassium ions in each cycle (1), both against their gradients, which restores the gradients lost by leakage through channels (1). (c) The action potential would not occur or would be much smaller (1), because sodium cannot enter to depolarise the membrane when the potential reaches the threshold (1).

Final answer

(a) \ce{K+} out; \ce{Na+} in. (b) Pumps 3 Na out, 2 K in per ATP, against gradients. (c) No action potential: no Na entry, no depolarisation.

Question X2

Connected questions · HL[6 marks]
B2.1 + D3.3. Glucose is filtered from blood into the nephron and is normally reabsorbed completely by the cells of the proximal convoluted tubule, even where the glucose concentration in the filtrate is lower than in the blood. (a) Explain how the tubule cells reabsorb glucose against its concentration gradient. [3] (b) In diabetes mellitus the blood glucose is very high and glucose appears in the urine. Suggest why reabsorption is incomplete. [2] (c) Explain why the tubule cells contain many mitochondria. [1]
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(a) Sodium-potassium pumps in the membrane next to the blood use ATP to keep the sodium concentration low inside the cell (1); a cotransporter in the tubule membrane brings in sodium and glucose together, the sodium moving down its gradient (1); glucose then leaves to the blood through a channel by facilitated diffusion (1). (b) There is a fixed number of cotransporters (1); when the filtrate glucose is very high they are all in use, so the extra glucose stays in the tubule and is excreted (1). (c) Mitochondria make the ATP that the sodium-potassium pumps use (1).

Final answer

(a) Pump keeps Na low inside; cotransporter couples glucose to Na entering; channel to blood. (b) Fixed number of cotransporters saturated. (c) ATP for the pumps.

Question X3

Connected questions[6 marks]
B2.1 + B2.2 + B1.2. A pancreatic cell makes a digestive enzyme, which is a protein, and secretes it into a duct. (a) Outline the pathway of the enzyme from synthesis to release. [3] (b) Explain what happens to the membrane of the secretory vesicle during release. [2] (c) Suggest why the cell contains many mitochondria. [1]
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(a) The enzyme is made by ribosomes on the rough endoplasmic reticulum (1); it is carried in a vesicle to the Golgi apparatus, where it is processed (1); a secretory vesicle carries it to the plasma membrane, where it is released by exocytosis (1). (b) The vesicle membrane fuses with the plasma membrane (1) and so becomes part of the surface; the inside of the vesicle is now continuous with the outside of the cell, so the enzyme is released (1). (c) Mitochondria supply ATP for protein synthesis and for moving the vesicles (1).

Final answer

(a) Ribosomes on rough ER; vesicle to Golgi; vesicle to plasma membrane; exocytosis. (b) Fuses with the plasma membrane and becomes part of it. (c) ATP for synthesis and vesicle movement.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Membranes and membrane transport from the core ideas to full exam questions, with question types, worked examples and graded practice.

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