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Running without oxygen, and burning something other than sugar

Take the oxygen away and one stage of respiration keeps going — but only if the cell can solve a supply problem first. This lesson covers what has to be regenerated and why, the two fermentation routes, what lipids and proteins contribute, and the calculation that tells you which substrate an organism is using.

Before this Glycolysis and the link reaction · The electron transport chain and chemiosmosis · Hydrolysis of triglycerides and of proteins

Before you start

Anaerobic respiration is what happens when your muscles run out of energy. It is a tidy story and it has the causation backwards. Nothing runs out of energy — there is glucose and glycogen in the muscle throughout. What runs short is oxygen, and the immediate casualty is not energy but a coenzyme. Glycolysis needs oxidised NAD to keep going, and normally the electron transport chain hands it back. Take oxygen away and NAD stays reduced, so glycolysis would stop within seconds. Lactate production exists to hand that NAD back. The lactate is a by-product of solving the coenzyme problem, not the purpose of the exercise.

What you should be able to do

Explain why glycolysis can continue without oxygen and what must be regenerated for it to do so.
Describe lactate fermentation and explain what happens to the lactate afterwards.
Describe ethanol fermentation in yeast and compare it with the animal route.
Compare anaerobic and aerobic yields per molecule of glucose.
Explain how lipids and proteins enter respiration and why lipid yields more energy per gram.
Calculate a respiratory quotient and interpret values of 0.7, 1.0 and above 1.0.
Describe how a respirometer is used, including its controls.

What actually runs out

A cell holds a tiny pool of NAD, and every molecule of it cycles between oxidised and reduced forms hundreds of times a minute. In aerobic conditions the chain oxidises reduced NAD at the inner membrane and returns it to the matrix and the cytoplasm ready to collect more hydrogen. Remove the oxygen and that return stops.

Glycolysis itself does not need oxygen at any step. What it needs is oxidised NAD, to accept hydrogen when the triose phosphate is oxidised. If none is available, the pathway stalls at that step, and with it goes the cell's last source of ATP. So a cell without oxygen has one problem to solve, and it is a chemical bookkeeping problem: get the hydrogen off the reduced NAD and onto something else.

Both anaerobic routes solve it the same way — by dumping the hydrogen onto pyruvate, or onto something made from pyruvate. The difference between an athlete and a brewer's yeast is only which molecule ends up holding it.

Read the outer arrows first. Both routes exist to return oxidised NAD to glycolysis; lactate and ethanol are simply where the hydrogen was put. Notice the yield is 2 ATP either way, because only glycolysis is running.

Anaerobic respiration: The release of energy from an organic molecule without oxygen, in which glycolysis continues because reduced NAD is oxidised by a reaction other than the electron transport chain.
Fermentation: The anaerobic conversion of pyruvate to lactate or to ethanol and carbon dioxide, regenerating oxidised NAD.

Lactate in animals, and paying it back

In an animal cell, pyruvate itself accepts the hydrogen. Reduced NAD is oxidised and pyruvate is reduced to lactate, catalysed by lactate dehydrogenase. Three carbons in, three carbons out; nothing is released as gas, and no ATP is made in the step itself.

Two features of this reaction matter. It is reversible, because the carbon is all still there — given oxygen again, lactate can be converted back to pyruvate and fed into the link reaction as normal. And it is fast, which is why a sprinter can keep going for thirty seconds at a rate their oxygen supply cannot possibly sustain.

Lactate accumulating in muscle lowers the pH there, which interferes with the enzymes and the contractile proteins and contributes to fatigue. Most of it diffuses into the blood and is carried to the liver. In the Cori cycle, liver cells convert lactate back to pyruvate and then, using ATP, resynthesise glucose from it — which is released into the blood or stored as glycogen. That conversion needs energy and oxygen, and it is a large part of why breathing and heart rate stay elevated after hard exercise has finished. The older name for the raised oxygen consumption afterwards is the oxygen debt.

One correction to the folk version: lactate is not simply waste. Cardiac muscle and slow-twitch fibres take it up from the blood and oxidise it as a fuel, and the muscle soreness felt a day or two after unfamiliar exercise is caused by microscopic damage, not by lactate, which has long since gone.

Ethanol in yeast, and how the two routes differ

Yeast and plant tissues take a different route, and it takes two steps rather than one. Pyruvate is first decarboxylated: a carbon is removed as carbon dioxide, leaving the two-carbon ethanal. Ethanal then accepts hydrogen from reduced NAD and becomes ethanol, regenerating oxidised NAD.

Losing that carbon as gas makes the pathway irreversible in the way the lactate route is not — the carbon has left. It also makes yeast commercially useful twice over: the ethanol is what a brewer wants and the carbon dioxide is what a baker wants, and both come from the same reaction. Yeast eventually poisons itself, since ethanol denatures its own enzymes; most strains stop at around 15% by volume, which is why spirits have to be distilled rather than simply fermented harder.

	In animals	In yeast and plants
Product	Lactate	Ethanol and carbon dioxide
Carbons in the product	3, all of them kept	2 in ethanol, 1 lost as CO₂
Steps	One: pyruvate reduced	Two: decarboxylation, then reduction
Gas released	None	Carbon dioxide
Reversible?	Yes — lactate can become pyruvate again	No — the carbon has gone
ATP per glucose	2 (net, from glycolysis)	2 (net, from glycolysis)

The yield comparison is stark and it is examined often. Anaerobically, a molecule of glucose gives a net 2 ATP, because glycolysis is the only stage running and neither fermentation step makes any ATP at all. Aerobically the same molecule gives around 30. Anaerobic respiration therefore recovers something like seven per cent of what is available — which is why anaerobic organisms consume substrate at a startling rate, and why a muscle working anaerobically empties its glycogen so quickly.

Comparing two flasks of yeast

Two flasks of yeast are given identical amounts of glucose. Flask A is aerated; flask B is sealed. After several hours, flask B has used far more glucose than flask A but has produced less carbon dioxide in total. Explain both observations.

Flask B is respiring anaerobically, gaining 2 ATP per glucose against roughly 30 in flask A. To make a comparable amount of ATP it must respire roughly fifteen times as much glucose, so its glucose disappears far faster.

The carbon dioxide result follows from where the gas comes from. Aerobically, all six carbons of each glucose leave as carbon dioxide, two in the link reaction and four in the Krebs cycle: 6 CO₂ per glucose. Anaerobically in yeast, only the decarboxylation of pyruvate releases gas, so 2 CO₂ per glucose.

So flask A releases three times as much carbon dioxide per glucose respired, and produces more in total even though it consumed less sugar. Any answer that reasons only from 'less glucose used means less gas' has ignored the stage where the gas is made.

Respiring something other than glucose

Glucose is the standard substrate in every diagram, but it is not the only one, and in a person who has not eaten for a few hours it is not even the main one. Lipids and proteins both enter the same pathways, each at its own point.

Lipids are hydrolysed to glycerol and fatty acids. The glycerol is phosphorylated and converted to triose phosphate, joining glycolysis part-way down. The fatty acids are the substantial part: their long hydrocarbon chains are broken down two carbons at a time, and each two-carbon fragment becomes an acetyl group carried by coenzyme A straight into the Krebs cycle. A single 16-carbon fatty acid yields eight acetyl groups and therefore eight turns of the cycle, plus a large haul of reduced coenzyme from the breakdown itself.

Proteins are hydrolysed to amino acids, which are deaminated in the liver — the amino group is removed and converted to urea. What is left of each amino acid enters wherever its carbon skeleton fits: some as pyruvate, some as acetyl CoA, some directly as an intermediate of the Krebs cycle. Protein is respired in quantity only when carbohydrate and lipid are short, since the alternative is to dismantle the body's own structures.

Energy per gram is where lipid pulls away: about 39 kJ from a gram of lipid against roughly 16 from carbohydrate and 17 from protein. The reason is chemical rather than mysterious. A fatty acid chain is almost all carbon and hydrogen, with very little oxygen already attached, so it is highly reduced and carries a great many hydrogen atoms per carbon. Carbohydrate is already part-oxidised — glucose carries an oxygen for every carbon. More hydrogen per gram means more reduced coenzyme, more electrons through the chain, more protons pumped and more ATP. It also means more oxygen is needed to respire it, which is the next section's whole point.

Substrate	Enters at	Energy per gram	RQ
Carbohydrate	Glycolysis, as glucose	≈ 16 kJ	1.0
Lipid	Glycerol at triose phosphate; fatty acids as acetyl CoA	≈ 39 kJ	≈ 0.7
Protein	As pyruvate, acetyl CoA or a Krebs intermediate	≈ 17 kJ	≈ 0.9

The respiratory quotient

If a substrate needs more oxygen per carbon dioxide released, that shows up as a number, and the number has a name. The respiratory quotient is the volume of carbon dioxide given out divided by the volume of oxygen taken in over the same period.

RQ = volume of CO₂ produced ÷ volume of O₂ consumedvolumes measured over the same time, at the same temperature and pressure

Work it out from a balanced equation and the arithmetic is short. For glucose, C₆H₁₂O₆ + 6O₂ gives 6CO₂ + 6H₂O, so RQ = 6 ÷ 6 = 1.0. For a typical lipid such as tripalmitin, C₅₁H₉₈O₆ + 72.5O₂ gives 51CO₂ + 49H₂O, so RQ = 51 ÷ 72.5 = 0.70. The lipid needs far more oxygen because it arrived carrying almost none of its own.

The value falls as the substrate gets more reduced. A reading near 0.7 says lipid is being respired; near 1.0 says carbohydrate; between them, either protein or a mixture — which is why a resting human usually measures around 0.85.

Reading an RQ from data

Germinating seeds in a respirometer take up 21.5 cm³ of oxygen and release 15.1 cm³ of carbon dioxide in the same period. Calculate the RQ and state, with a reason, which substrate the seeds are mainly respiring.

RQ = 15.1 ÷ 21.5 = 0.70. Give it to two decimal places; the ratio has no units, and writing one loses the mark.

A value of 0.70 indicates lipid. These are oil-storing seeds, and a lipid substrate demands much more oxygen per carbon dioxide released because it is highly reduced and carries little oxygen of its own.

A sensible extra sentence: as the seedling's oil reserve runs down and it begins to photosynthesise, the RQ would be expected to rise towards 1.0.

Values above 1.0 need explaining rather than reporting. Carbon dioxide is appearing that no oxygen uptake accounts for, so some of it is coming from a reaction that does not use oxygen — anaerobic respiration alongside the aerobic. Yeast in a sealed flask can show an RQ well above 1.0, and in a completely anaerobic culture, with oxygen uptake at zero, the quotient is undefined rather than merely large. A value slightly above 1.0 in a well-fed animal can also mean carbohydrate is being converted into lipid for storage, which releases carbon dioxide.

Measuring it: the respirometer

The apparatus is simple and the reasoning behind it is what gets examined. Living material — germinating seeds, woodlice, blowfly larvae — sits on a gauze platform in a sealed tube. Beneath the gauze is soda lime or concentrated potassium hydroxide, which absorbs carbon dioxide as fast as it is produced. A capillary tube containing a drop of coloured fluid runs from the tube to the outside air.

With the carbon dioxide absorbed, the only gas change left is the oxygen being used up. The pressure inside falls, and the fluid moves towards the respiring organisms by a volume equal to the oxygen consumed. Multiply the distance moved by the cross-sectional area of the capillary and you have that volume in cubic millimetres.

Two runs, one measurement each. With soda lime the drop reports oxygen uptake alone; without it, the carbon dioxide released pushes back and the drop moves less, and the difference between the two readings is the carbon dioxide.

The controls decide whether the numbers mean anything. An identical tube containing glass beads of the same mass, with the same soda lime, corrects for changes in room temperature and atmospheric pressure — without it, a warm afternoon looks like a burst of respiration. The whole apparatus sits in a water bath so temperature is controlled, and it is left to equilibrate before the first reading, because gas warming up to bath temperature expands and swamps the effect being measured. A syringe lets the drop be returned to the start between readings. And the rate must be expressed per gram of organism per minute, or two results cannot be compared.

TRY IT — Getting an RQ out of two runs

A student runs a respirometer containing woodlice twice at 20 °C. With soda lime present, the fluid moves 40 mm towards the woodlice in ten minutes. With the soda lime removed and the apparatus otherwise unchanged, it moves 12 mm towards the woodlice in ten minutes. The capillary has a cross-sectional area of 1.0 mm². Calculate the RQ and comment on the substrate.

Check your answer

With soda lime, all the carbon dioxide is absorbed, so the movement measures oxygen uptake alone: 40 × 1.0 = 40 mm³ of oxygen in ten minutes.

Without soda lime, carbon dioxide is released back into the tube and partly offsets the oxygen used, so the movement measures the difference between them: 12 mm³. The carbon dioxide produced is therefore 40 − 12 = 28 mm³.

RQ = 28 ÷ 40 = 0.70, which indicates lipid as the main substrate.

The step most often dropped is the second one. The reading without soda lime is not the carbon dioxide; it is oxygen used minus carbon dioxide released, and you have to subtract it from the first reading to get the gas you want.

In the exam

Say what anaerobic respiration is for. The mark is for regenerating oxidised NAD so that glycolysis can continue, not for 'making lactate'.
2 ATP per glucose anaerobically, and it comes from glycolysis. Neither fermentation step produces any ATP, and questions asking how much ATP the lactate pathway makes are checking that you know it makes none.
Name the yeast route in two steps: pyruvate is decarboxylated to ethanal, then ethanal is reduced to ethanol. Answers that go straight from pyruvate to ethanol lose the carbon dioxide.
For 'why does lipid release more energy per gram', the answer is more hydrogen atoms per gram, so more reduced coenzyme, so more protons pumped and more ATP. 'Because fat is more energy dense' restates the question.
RQ has no units and is quoted to two decimal places. State the substrate it indicates, and if a value exceeds 1.0, say that some respiration must be anaerobic.
Respirometer questions almost always carry a mark for the control tube of glass beads and a mark for the water bath. Write both before you write anything about the organisms.

Check yourself

A 100 m sprinter and a marathon runner are both working hard. Explain why the sprinter's muscles produce lactate while the marathon runner's largely do not, and explain why the marathon runner's main respiratory substrate changes during the race.

Answer

The sprinter's muscles demand ATP far faster than the heart and lungs can supply oxygen to them, so the electron transport chain cannot oxidise reduced NAD quickly enough. Without oxidised NAD being returned, glycolysis would stop, so pyruvate is reduced to lactate instead, which hands the hydrogen over and lets glycolysis continue. The yield is only 2 ATP per glucose, but it arrives quickly and without waiting for oxygen.

The marathon runner works at an intensity oxygen delivery can keep up with. Reduced NAD is oxidised at the inner membrane as fast as it is made, oxidised NAD is returned to the cytoplasm and the matrix, and pyruvate goes into the link reaction rather than being reduced. Little lactate accumulates, and what does is taken up and respired by cardiac and slow-twitch muscle.

The substrate changes because the stores run out at different rates. Muscle and liver glycogen last roughly an hour and a half of hard running; as carbohydrate is depleted, an increasing proportion of the ATP comes from fatty acids, hydrolysed from stored triglyceride and fed in as acetyl CoA. Lipid gives about 39 kJ per gram against 16 for carbohydrate, so the reserve is large — but it needs more oxygen per carbon dioxide released, which is why the runner's RQ falls from near 1.0 towards 0.8 or below as the race goes on, and why the pace that can be held drops with it.

Questions

Written to the command words the boards use. Try them on paper before opening a scheme: the marks go to points made, not to length.

Question 14 marks

Explain why glycolysis can continue in a muscle cell that has run short of oxygen, and explain what the reduction of pyruvate to lactate achieves.

Mark scheme

B1 no step of glycolysis uses oxygen; what it needs is oxidised NAD, to accept the hydrogen removed when triose phosphate is oxidised
B1 without oxygen the electron transport chain cannot oxidise reduced NAD, so oxidised NAD is no longer returned to the cytoplasm and glycolysis would stall within seconds
B1 pyruvate accepts the hydrogen from reduced NAD instead, catalysed by lactate dehydrogenase, so pyruvate is reduced to lactate and oxidised NAD is regenerated
B1 glycolysis can therefore keep running and keep supplying 2 ATP per glucose, quickly and without waiting for oxygen, so the lactate is a by-product of solving the coenzyme problem rather than the purpose of the pathway

Question 24 marks

Compare the anaerobic pathway followed in an animal muscle cell with the anaerobic pathway followed in yeast.

Mark scheme

B1 in an animal, pyruvate is reduced to lactate in a single step, whereas in yeast pyruvate is first decarboxylated to ethanal and the ethanal is then reduced to ethanol, which is two steps
B1 the animal route keeps all three carbons in the lactate, whereas the yeast route loses one as carbon dioxide, so only the yeast route releases a gas
B1 the animal route is reversible, so lactate can be converted back to pyruvate and respired once oxygen returns, whereas the yeast route is irreversible because the carbon has left as gas
B1 both routes exist to regenerate oxidised NAD for glycolysis, and both give the same net yield of 2 ATP per molecule of glucose

Question 34 marks

Germinating seeds are placed in a respirometer at 20 °C. With soda lime present the fluid drop moves 50 mm towards the seeds in ten minutes; with the soda lime removed and everything else unchanged it moves 5 mm towards the seeds in ten minutes. The capillary has a cross-sectional area of 1.0 mm². Calculate the respiratory quotient and state which substrate it indicates.

Mark scheme

M1 with soda lime the carbon dioxide is absorbed as fast as it is made, so the movement measures oxygen uptake alone: 50 × 1.0 = 50 mm³
M1 without soda lime the movement measures oxygen consumed minus carbon dioxide released: 5 × 1.0 = 5 mm³, so the carbon dioxide produced is 50 − 5 = 45 mm³
M1 RQ = volume of carbon dioxide produced ÷ volume of oxygen consumed = 45 ÷ 50
A1 RQ = 0.90, quoted with no units, which indicates protein as the main substrate, or a mixture of substrates

Question 43 marks

Explain why a gram of lipid releases roughly 39 kJ when it is respired while a gram of carbohydrate releases only about 16 kJ.

Mark scheme

B1 a fatty acid chain is almost entirely carbon and hydrogen, with very little oxygen already attached, so it is highly reduced and carries many more hydrogen atoms per gram than glucose, which holds an oxygen for every carbon
B1 more hydrogen per gram means more reduced NAD and reduced FAD are produced, so more electrons pass along the chain and more protons are pumped into the intermembrane space
B1 more protons returning through ATP synthase means more ATP is made per gram; it also means more oxygen is needed, which is why lipid gives a respiratory quotient near 0.70

Question 53 marks

Explain why a respirometer investigation includes a second tube holding glass beads of the same mass, and why the whole apparatus is left to equilibrate in a water bath before the first reading is taken.

Mark scheme

B1 the tube of glass beads contains nothing that respires, so any movement of fluid in it is caused by changes in room temperature or atmospheric pressure rather than by oxygen uptake
B1 its reading is used to correct the reading from the experimental tube, so that a warm afternoon is not recorded as a burst of respiration
B1 gas warming up to the temperature of the water bath expands, and that expansion would swamp the small volume change caused by the organisms, so the apparatus is equilibrated before readings begin

Question 63 marks

A sealed flask of yeast, fed on glucose, is found to have a respiratory quotient of 1.6. Suggest what this value shows about the respiration taking place in the flask.

Mark scheme

B1 carbon dioxide is being released that the oxygen uptake cannot account for, so some of it must come from a reaction that does not consume oxygen
B1 the yeast must therefore be respiring anaerobically alongside any aerobic respiration, releasing carbon dioxide when pyruvate is decarboxylated to ethanal
B1 as the oxygen sealed into the flask is used up the quotient would rise further, and once oxygen uptake reaches zero the quotient becomes undefined rather than merely large

Question 72 marks

State the products formed when yeast respires glucose anaerobically, and state the net number of ATP molecules gained per molecule of glucose.

Mark scheme

B1 ethanol and carbon dioxide are the products of the fermentation itself
B1 the net gain is 2 ATP per molecule of glucose, all of it from glycolysis, since neither fermentation step makes any ATP

Worth remembering

Glycolysis needs oxidised NAD, not oxygen. Anaerobic routes exist to hand that NAD back.
Animals: pyruvate to lactate, three carbons kept, reversible. Yeast: pyruvate to ethanal plus CO₂, then ethanal to ethanol, irreversible.
Anaerobic yield is 2 ATP per glucose against about 30 aerobically.
Lipid gives roughly 39 kJ per gram because it is highly reduced: more hydrogen per gram means more reduced coenzyme.
RQ = CO₂ out ÷ O₂ in. About 1.0 for carbohydrate, 0.9 for protein, 0.7 for lipid; above 1.0 means some respiration is anaerobic.

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