DSE Biology | Cellular Respiration: Tutor Lesson Plan

2026-11-03

Tutor lesson-prep guides · Secondary Biology

Students can recite that glycolysis is in the cytoplasm and the Krebs cycle in the mitochondrion, but go quiet when asked where the 30 ATP come from. This lesson builds one conveyor belt, locks the location and products of each stage, and uses an energy balance to explain why oxygen shortage leaves only 2 ATP.

Objectives — Students can: (1) name the cytoplasm for glycolysis and the mitochondrion for the Krebs cycle and oxidative phosphorylation; (2) find 6 carbon dioxide and about 30 ATP per glucose; (3) use 2870 kJ and 30.5 kJ per ATP to get 31.9 percent efficiency against 2.1 percent anaerobically; (4) explain why fermentation still runs and name the yeast and muscle products.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Enzyme action	What happens to respiratory enzymes at 60 degrees?	They denature and the rate collapses
Gas exchange	Which way does carbon dioxide move?	Down its gradient, out of the cell
Energy currency	What happens when ATP becomes ADP?	Energy is released for cell work
Cell structure	Which organelle runs aerobic respiration?	The mitochondrion

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Opening contrast	Ask why breathing is faster after a sprint	Link breathing rate to energy demand
8-20 min	Glycolysis	Split glucose into two pyruvate in the cytoplasm	State the site, products and ATP
20-32 min	Krebs cycle	Pyruvate enters the mitochondrion and carbon dioxide leaves	Mark where carbon dioxide appears
32-44 min	Oxidative phosphorylation	The chain passes electrons to oxygen	Name the final electron acceptor
44-52 min	Energy balance	2870 kJ against 30 ATP	Compute efficiency and compare
52-60 min	Independent practice	Three questions in eight minutes	Mark the errors and name the cause

03 Teaching script (read it out)

Scene 1: Three stages form one conveyor belt, not three stories

Glycolysis sits in the cytoplasm, cuts glucose into two pyruvate and nets 2 ATP without oxygen. Pyruvate enters the mitochondrion, the Krebs cycle breaks it fully and releases 6 carbon dioxide while loading electron carriers. Oxidative phosphorylation passes those electrons down the chain to oxygen and supplies more than twenty ATP. Students should string it in one sentence: cut in the cytoplasm, dismantle in the matrix, collect at the membrane. Wrong locations poison every later question, so slow down and make them draw the belt once.

Ask the student to name the site, input and product of each stage without notes.

Scene 2: Oxygen is the final electron acceptor, not a raw material

Many students write that oxygen joins glucose to make carbon dioxide. That is wrong: the oxygen in carbon dioxide comes from glucose and water. Oxygen receives electrons and hydrogen at the end of the chain and becomes water. Without oxygen the chain backs up, ATP production stops and lactate or ethanol accumulates. Show the contradiction on the board and let them find it.

Ask where the oxygen atoms in the carbon dioxide come from.

Scene 3: An energy balance makes the 2 against 30 gap visible

One mole of glucose releases 2870 kJ, one mole of ATP releases 30.5 kJ on hydrolysis. Thirty ATP hold 915 kJ, an efficiency of 31.9 percent, and the rest leaves as heat. Anaerobically only 2 ATP form, 61 kJ and 2.1 percent. Once students do that division they see why sprinting cannot last: the fuel is there but the yield is tiny. It also sets up the follow-up on why breathing stays fast after the race.

Ask for the percentage of 2870 kJ captured in 30 ATP and where the rest goes.

04 Question types and worked steps

Type	Example	Steps	Answer
Locate the stages	Where does each of the three stages happen?	Cytoplasm, matrix, inner membrane in order	Three named sites
Product count	Carbon dioxide and water from one glucose?	Write the equation, balance it, read the coefficients	6 carbon dioxide, 6 water
Efficiency	30 ATP against 2870 kJ	Multiply by 30.5, divide by 2870, convert to percent	915 kJ, 31.9 percent
Anaerobic case	Why does lactate build up in a sprint?	Oxygen runs short, the chain stops, pyruvate becomes lactate to regenerate NAD	Lactate, 2 ATP only

Sanity check (teach this): Order check: the ATP count must sit between 2 and 38, so 300 means a lost decimal point, and aerobic must far exceed anaerobic. Coefficient check: 6 carbon dioxide against 6 oxygen gives a quotient of 1.0, so 12 carbon dioxide means the six carbons of glucose were forgotten. An efficiency near 32 percent is plausible for a cell; anything above 90 percent denies heat loss.

05 Independent practice (3 questions)

Question	Answer	What to watch
Net ATP from anaerobic fermentation of one glucose	2	Is the Krebs cycle wrongly included
Moles of carbon dioxide from 180 g of glucose	6	Was mass converted to moles first
Aerobic efficiency using 30 ATP	31.9 percent	Is ATP multiplied before dividing

06 Common mistakes and what to say

Mistake	Student writes	Say this
Treating oxygen as a raw material	“Glucose plus oxygen just becomes carbon dioxide and water”	“Your oxygen atoms do not balance: the carbon dioxide oxygen comes from glucose and water, while inhaled oxygen takes electrons at the end of the chain.”
Wrong ATP totals	“Aerobic respiration gives 2 ATP”	“Two is only the net gain of glycolysis; the mitochondrion has not been counted yet.”
Fermentation described as aerobic	“The mitochondrion keeps making ATP without oxygen”	“With no final electron acceptor the chain jams, and a jammed chain cannot keep making ATP.”
Confused locations	“The Krebs cycle runs in the cytoplasm”	“Glycolysis is cytoplasmic; the Krebs cycle sits in the matrix, so draw them in separate boxes.”

07 Homework (paper, section and question numbers)

Biology chapter index — Revise enzymes and metabolism, then questions 1 and 2
Cell cycle and mitosis — Question 3 compares the energy demand of division and respiration
Enzymes and factors — Temperature and pH only, to prepare the rate experiment

08 Exit check (last 5 minutes)

Question	Answer	Decision
Site of glycolysis	Cytoplasm	Error: written as the mitochondrion
Carbon dioxide per glucose	6	Error: written as 12
Percentage of 2870 kJ held in 30 ATP	31.9 percent	Error: ATP count left out
Product of anaerobic muscle respiration	Lactate	Error: written as ethanol

09 Tutor FAQ

Why does aerobic respiration capture only 30 ATP?

Part of the energy leaves as heat and part is lost along the electron chain; the cell trades efficiency for control and simply burns more glucose.

How do yeast and muscle fermentation differ?

Yeast turns pyruvate into ethanol and carbon dioxide, muscle turns it into lactate; both only regenerate NAD so glycolysis can continue, and both yield 2 ATP.

What is the respiratory quotient for?

It is carbon dioxide released over oxygen used; about 1.0 for carbohydrate and 0.7 for fat, so it hints at the fuel being burned.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.