IB Biology HLCellular RespirationPaper 1 & 2~12 min read
Glycolysis
The first stage of respiration begins by spending ATP, which sounds like a bad start. It is actually the whole trick: glucose is far too stable to react, so the cell pays two ATP to make it unstable enough to break.
📚 What you need to know
Glycolysis is the first stage of respiration and takes place in the cytoplasm.
It involves trapping glucose in the cell by phosphorylating it, then splitting the molecule in two.
Per glucose it produces two pyruvate (3C), a net gain of two ATP, and two reduced NAD.
Four ATP are made in total, but two were used at the start — hence a net gain of two.
Phosphorylation: glucose (6C) + 2 ATP → fructose-1,6-bisphosphate (6C), which is less stable and therefore more reactive.
Lysis: fructose-1,6-bisphosphate splits into two triose phosphate (3C) molecules.
Oxidation: hydrogen is removed from each triose phosphate by a dehydrogenase and passed to NAD, forming glycerate-3-phosphate and two reduced NAD.
ATP formation: phosphates are transferred from the intermediates to ADP by substrate-linked phosphorylation, giving four ATP.
Every step is catalysed by a different enzyme. You do not need to name the intermediates beyond those above.
Why glucose has to be activated first
Glucose is a very stable molecule. That is exactly what makes it good for storing energy — and exactly what makes it hard to start using.
So the cell attaches two phosphate groups, taken from two ATP molecules. The resulting fructose-1,6-bisphosphate is less stable and therefore more reactive: its activation energy has been lowered, so it will readily split. Phosphorylating glucose also traps it, because the charged, phosphorylated molecule cannot cross the cell membrane and leave.
This is the “phosphorylated intermediates” row from the ATP page, in action. The two ATP are not lost — they are an investment. The stage returns four, so the cell doubles its money.
The four steps
The single most common error on this diagram is forgetting that the bottom half runs twice. Two triose phosphates means two of everything after the split.
Step 1 — Phosphorylation
Phosphorylation
glucose + 2 ATP → fructose-1,6-bisphosphate
Glucose (6C) is activated by receiving two phosphate groups from two ATP. The product is a 6C molecule that is less stable and more reactive.
A dehydrogenase enzyme removes hydrogen from each triose phosphate and transfers it to the coenzyme NAD, forming two reduced NAD. The triose phosphate itself is oxidised to another 3C molecule, glycerate-3-phosphate.
Step 4 — ATP formation
ATP formation
4Pi + 4ADP → 4ATP
Phosphates are transferred straight from the intermediate substrate molecules to ADP. Because the phosphate comes directly from a substrate rather than from a proton gradient, this is called substrate-linked phosphorylation. The end product is two molecules of pyruvate, ready for the next stage.
Two ATP out of a possible thirty-six is a poor return — but it is the only return available when oxygen runs out, which is the whole basis of the next page.
A note on the hydrogens. You may see 4H written instead of 4H+ + 4e−, or 2H written as 2H+ + 2e−. They mean the same thing: a hydrogen atom is a proton plus an electron. Whichever form the question uses, the redox is identical.
Worked examples
WE 1
State the products of glycolysis
State the products of glycolysis from one molecule of glucose, and state where in the cell it occurs. (4 marks)
Location
The cytoplasm of the cell.
Product 1
Two molecules of pyruvate, each with three carbons.
Product 2
A net gain of two ATP — four are produced but two are used in the phosphorylation step.
Product 3
Two molecules of reduced NAD.
2 pyruvate + 2 ATP (net) + 2 reduced NAD, all in the cytoplasmwrite “net” next to the 2 ATP — it shows you know four were made
WE 2
Explain the ATP spent at the start
Explain why ATP is used during a stage of respiration whose purpose is to produce ATP. (3 marks)
Point 1: the problem
Glucose is a very stable molecule, so it will not react readily on its own.
Point 2: what the ATP does
Two ATP phosphorylate glucose to form fructose-1,6-bisphosphate, making it less stable and more reactive by lowering the activation energy.
Point 3: why it is worth it
The stage goes on to produce four ATP, so there is still a net gain of two, and phosphorylation also traps the glucose inside the cell.
Spend 2 to make 4 — an investment, not a loss“lowers the activation energy” is the phrase that secures the second mark
WE 3
Scale the yield up
A cell respires three molecules of glucose. Calculate the number of pyruvate molecules, net ATP molecules and reduced NAD produced by glycolysis. (3 marks)
Pyruvate
3 × 2 = 6 pyruvateNet ATP
3 × 2 = 6 ATP net (12 made, 6 used)
Reduced NAD
3 × 2 = 6 reduced NAD6 pyruvate, 6 net ATP, 6 reduced NADif a question asks for total ATP rather than net, the answer here would be 12
💡 Exam tips
Learn the four step names in order: phosphorylation, lysis, oxidation, ATP formation.
Always distinguish 4 ATP produced from a net gain of 2.
State the location as the cytoplasm — not the mitochondrion.
Say substrate-linked phosphorylation for the ATP made here; it contrasts with oxidative phosphorylation later.
Note that each step has a different enzyme, and that a dehydrogenase does the oxidation.
⚠ Common mistakes
Writing that glycolysis produces 2 ATP in total. It produces 4 and uses 2.
Forgetting that the pathway doubles after lysis, so halving the NAD and ATP counts.
Saying glycolysis needs oxygen. It runs identically with or without it.
Placing glycolysis in the mitochondrial matrix. That is the link reaction.
Saying pyruvate is a waste product. It still holds most of the energy of the glucose.
Claiming the ATP used at the start is wasted. It lowers the activation energy and traps the glucose.
Up next: Anaerobic Respiration. Glycolysis has left the cell with two pyruvate and two reduced NAD. What happens to them when there is no oxygen to hand them on to?
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