IB Biology C1.2: Cell respiration worked solutions

B. Cell respiration is the controlled release of energy from organic compounds to make ATP. A describes gas exchange, C describes photosynthesis, and D describes a process that uses ATP.

B. Hydrolysis of ATP to ADP and phosphate releases energy. A and C describe making ATP, which needs energy, and D is wrong because active transport uses ATP and does not make it.

A. Without oxygen, glucose is broken down in the cytoplasm to lactate with a small ATP yield. B and D put the process in the mitochondrion, and C describes yeast, not a human cell.

B. Oxygen supply cannot meet the demand for ATP, so more ATP is made by anaerobic respiration, which makes lactate. Aerobic respiration does not make lactate (A), and lactate rising shows it is made faster than it is removed, not used up (C, D).

B. The seeds use oxygen and release carbon dioxide, which the alkali absorbs. Gas is removed, so the pressure falls. In A the carbon dioxide is not absorbed, in C the absorber is ignored, and D invents a heating effect.

A. 0.60\div4.0=0.15\,\mathrm{cm^3\,min^{-1}}, and 0.15\div5.0=0.030. B forgets to divide by mass, C divides by mass but not time, and D multiplies instead of dividing.

B. Without mitochondria only glycolysis can go on, and in a human cell pyruvate then becomes lactate. Carbon dioxide and water need the mitochondrial stages (A), ethanol is made by yeast (C), and ATP can still be made in the cytoplasm (D).

B. In alcoholic fermentation pyruvate becomes ethanol and carbon dioxide, and the carbon dioxide forms bubbles in the dough. Oxygen is absent (A), and yeast does not make methane (C) or hydrogen (D).

C. Glycolysis converts one glucose to two pyruvate in the cytoplasm with a net gain of 2 ATP and 2 reduced NAD. A reverses used and made ATP, B gives the wrong site, and D is wrong because no carbon dioxide is released.

B. Carbon dioxide leaves in the link reaction (pyruvate loses a carbon) and in the Krebs cycle (two decarboxylations). Glycolysis and the chain release none.

(a) Any two properties with their benefit (2): ATP is small and soluble, so it moves quickly through the cytoplasm to where it is needed; hydrolysis of one phosphate releases a useful but modest amount of energy, so little is lost as heat and one reaction can be supplied; ATP is recycled from ADP and phosphate, so it is not used up; it is stable until hydrolysed, so energy is not lost while it waits. A property without a benefit scores no mark. (b) Any two (2): active transport across a membrane; synthesis of macromolecules such as proteins or DNA; movement of the whole cell or of parts such as chromosomes or muscle filaments. Diffusion and osmosis do not use ATP.

(a) Glucose \to lactate, with no oxygen on the left and no carbon dioxide on the right (1). (b) Any three distinctions, each with both halves in one sentence (3): only aerobic respiration needs oxygen; aerobic respiration gives carbon dioxide and water while anaerobic respiration gives lactate only; aerobic respiration gives a large yield of ATP, anaerobic a small one (2 ATP per glucose); aerobic respiration needs mitochondria, anaerobic respiration is in the cytoplasm only. Two separate lists of facts do not earn the marks.

(a) Lactate rises from 1.2 to 8.2\,\mathrm{mmol\,dm^{-3}} between 10 and 16\,\mathrm{km\,h^{-1}} (1), a rise of 7.0, and the rise is small up to 12\,\mathrm{km\,h^{-1}} and steep above it (1). (b) (3.1-2.0)\div2.0\times100=55\% (1). (c) Oxygen uptake levels off near 3.2\,\mathrm{dm^3\,min^{-1}} (1), so aerobic respiration cannot supply all the ATP the faster running needs (1); the muscles make the extra ATP by anaerobic respiration, whose product is lactate (1).

(a) Volume =40\times0.50=20\,\mathrm{mm^3} (1); rate =20\div10=2.0\,\mathrm{mm^3\,min^{-1}} (1). (b) At 15^\circ\mathrm{C} the volume is 18\times0.50=9.0\,\mathrm{mm^3}, a rate of 0.90\,\mathrm{mm^3\,min^{-1}}, and 2.0\div0.90=2.2 times (1). (c) The enzymes of respiration have more kinetic energy at the higher temperature (1), so collisions between enzyme and substrate are more frequent and the reactions, and the use of oxygen, are faster (1). (d) The alkali absorbs the carbon dioxide released, so the volume change equals oxygen use (1); the control tube shows changes in the gas volume caused by temperature or pressure alone, and these are subtracted (1).

(a) Glucose \to ethanol + carbon dioxide (1). (b) Glycolysis needs NAD (1); with no oxygen the electron transport chain cannot oxidise reduced NAD (1); making ethanol from pyruvate oxidises reduced NAD back to NAD, so glycolysis can go on (1). (c) The net yield is 2 ATP per glucose, all from glycolysis (1).

(a) In the cytoplasm (1). (b) Any three in order (3): glucose is phosphorylated using ATP; the 6C compound is split (lysis) into two 3C compounds; the 3C compounds are oxidised and NAD is reduced to reduced NAD; ATP is formed, giving a net gain of 2 ATP and two pyruvate. (c) The substrate that loses hydrogen (the 3C sugar phosphate) is oxidised, and NAD that gains it is reduced (1).

(a) X is oxaloacetate (1) and Y is citrate (1). (b) The link reaction releases one carbon dioxide per pyruvate (1) and one turn of the Krebs cycle releases two (decarboxylations from 6C to 5C and 5C to 4C) (1). (c) The hydrogen removed in the oxidation steps is accepted by NAD, which becomes reduced NAD (1).

(a) The inner mitochondrial membrane, the cristae (1). (b) Protons are pumped from the matrix into the intermembrane space (1). (c) Oxygen accepts the electrons at the end of the chain (1) and combines with protons to form water (1).

Reduced NAD gives electrons to the chain in the inner membrane (1). Energy released as the electrons pass between carriers is used to pump protons into the intermembrane space (1), which builds a proton gradient (1). Protons flow back into the matrix through ATP synthase, down the gradient (1), and the energy of the flow drives phosphorylation of ADP to ATP (1). Oxygen as the final acceptor keeps electrons flowing, and an answer that mentions it may replace the first mark.

(a) Fatty acids are broken down into 2C acetyl groups (1), which are carried by coenzyme A as acetyl CoA into the Krebs cycle (1). (b) Lipid molecules contain less oxygen and more oxidisable hydrogen and carbon per gram than carbohydrate (1), so more energy is released per gram (1). Glycolysis, and so anaerobic respiration, needs a carbohydrate and cannot use fat.

(a) The parasite uses glycolysis in the cytoplasm followed by lactate formation, as in anaerobic respiration (1); the yield is 2 ATP per glucose (1). (b) A cell using aerobic respiration makes about 30 ATP per glucose, so the parasite needs 30\div2=15 times as much glucose for the same ATP (1). (c) Fatty acids need the aerobic stages, which are in mitochondria (1), so a parasite without mitochondria cannot use fat as a fuel and depends on the host's glucose (1).

(a) Oxygen is not reduced to water, so oxygen consumption falls (1). (b) Reduced NAD cannot give up its electrons, so reduced NAD accumulates (1) and NAD becomes scarce (1). (c) Electron flow stops, so energy is not released to pump protons (1); the proton gradient is not maintained (1); little flows through ATP synthase, so far less ATP is made by chemiosmosis (1). (d) The cells can still make a small amount of ATP by glycolysis if NAD is regenerated (1) by converting pyruvate to lactate, so lactate concentration rises (1).

(a) 0.65\div5.0\div2.5=0.052 (1) and 1.85\div5.0\div2.5=0.148\,\mathrm{cm^3\,g^{-1}\,min^{-1}} (1). (b) The rate rises as temperature rises from 10 to 40^\circ\mathrm{C} (1) and falls sharply at 50^\circ\mathrm{C}, to below the value at 20^\circ\mathrm{C} (1). (c) At 50^\circ\mathrm{C} the enzymes of respiration are denatured (1), so the active sites no longer fit the substrates and ATP production falls, and the maggots may die (1). (d) The ratios for successive 10^\circ\mathrm{C} steps are 0.65\div0.30=2.2, 1.20\div0.65=1.8 and 1.85\div1.20=1.5 (1); the claim holds only from 10 to 20^\circ\mathrm{C}, and the data come from one species and one run, so the rule cannot be generalised (1).

(a) Contraction, active transport and synthesis need energy supplied by ATP (1), and respiration releases energy from organic fuels to make ATP (1). (b) Any five: glucose is converted to pyruvate in the cytoplasm by glycolysis (1), with a net gain of 2 ATP and reduced NAD (1); pyruvate enters the mitochondrion if oxygen is present, and the link reaction and Krebs cycle release carbon dioxide and make reduced NAD (1); fatty acids are broken into acetyl groups that join the cycle as acetyl CoA (1); reduced NAD passes electrons to the chain, which pumps protons (1); flow back through ATP synthase makes most of the ATP (1); oxygen is the final acceptor and forms water (1). (c) Aerobic respiration gives a much larger yield than anaerobic respiration, which stops at lactate with 2 ATP per glucose and cannot use fatty acids (1).

(a) (18-120)\div120\times100=-85\%, a fall of 85\% (1). (b) Protons cross the membrane through the leak, not through ATP synthase (1), so the proton gradient is smaller (1) and less of the energy drives phosphorylation of ADP, so less ATP is made (1). (c) With a smaller gradient there is less resistance to proton pumping, so electrons pass along the chain faster (1) and more oxygen is used as the final acceptor (1). (d) The energy released by electron transport is not captured in ATP and is released as heat (1).

(a) Length =60\div25=2.4\,\mu\mathrm{m} (2: one for the ratio, one for the unit). (b) The tail moves by using ATP (1), which the mitochondria produce by aerobic respiration (1). (c) The cristae fold the inner membrane (1), which increases the area for electron carriers and ATP synthase, so more ATP can be made (1).

(a) The leaf releases 1.5\,\mu\mathrm{mol\,m^{-2}\,s^{-1}}, the rate of respiration, since there is no photosynthesis in the dark (1). (b) Gross photosynthesis =3.5+1.5=5.0 (1). (c) At light intensity 10 respiration releases carbon dioxide faster than photosynthesis takes it up (1), so the net value is negative (1). (d) Respiration equals photosynthesis where net uptake is zero, between 10 and 20 units (1). (e) The leaf cells need ATP by aerobic respiration in the light too, for active transport and synthesis (1).

(a) Ventilation rate at rest =12\times0.50=6.0\,\mathrm{dm^3\,min^{-1}} (1); during exercise =24\times1.5=36\,\mathrm{dm^3\,min^{-1}} (1). (b) Working muscle cells need much more ATP, so aerobic respiration uses more oxygen and makes more carbon dioxide (1); faster ventilation keeps the concentration gradients steep, so oxygen supply and carbon dioxide removal keep up (1). (c) Gas exchange is passive diffusion of gases across a surface (1), while cell respiration is the chemical release of energy from organic compounds to make ATP in the cells (1).