The shape of the lesson
By the end of the lesson, learners can
- A globular protein with a pocket: Rotate the enzyme in the simulator, then bring in the substrate.
- Lowering the activation energy: Fire 200 collisions with and without the enzyme.
- Following a reaction's progress: Start the reaction in the simulator and watch the substrate curve fall as the product curve rises.
- Active site is complementary to the substrate, not the same shape.
- Enzymes lower activation energy; they do not supply energy.
- Initial rate = gradient of the tangent at t = 0.
- Plateau means substrate used up, not enzyme used up.
One picture every idea returns to
Enzymes are globular proteins that catalyse reactions either inside cells (intracellular, such as catalase) or after being secreted (extracellular, such as amylase in saliva). The substrate binds to the active site, a small region whose shape and charges are complementary to it, forming an enzyme-substrate complex. That makes the enzyme specific. In the lock-and-key hypothesis the site is rigid; in the induced-fit hypothesis it changes shape slightly as the substrate binds, straining its bonds. Either way the enzyme lowers the activation energy, so more collisions succeed at the same temperature, and it leaves the reaction unchanged. Reaction progress is followed by measuring product formed (oxygen from catalase) or substrate lost (starch with amylase), and the initial rate is used to compare conditions.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, A pocket that speeds up chemistry.. Almost every reaction in a cell would be far too slow at 37 °C without help. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: A globular protein with a pocket | Rotate the enzyme in the simulator, then bring in the substrate. Project tab 1 of the lab (Enzymes lab, the active site in 3D) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Lowering the activation energy | Fire 200 collisions with and without the enzyme. Project tab 2 of the lab (Enzymes lab, activation energy) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Following a reaction's progress | Start the reaction in the simulator and watch the substrate curve fall as the product curve rises. Project tab 3 of the lab (Enzymes lab, the course of a reaction) and let a student drive it while the class predicts. | Cooperative Concept Mapping |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Following a reaction's progress, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Say complementary, not the same shape. | The substrate and active site fit together like a key and lock, so their shapes are complementary. Also, the enzyme is not used up: it is released unchanged and can bind another substrate molecule. |
| The curve levels off because the substrate has run out, not because the enzyme has been used up or denatured. | If you add more substrate the reaction carries on at once. |
How enzymes work: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
State what is meant by an intracellular enzyme and an extracellular enzyme, giving one example of each.
Explain how an enzyme catalyses a reaction, using the induced-fit hypothesis.
Suggest why the reactions of a cell need enzymes when the same reactions could be speeded up by heating.
Catalase is added to hydrogen peroxide and the oxygen is collected. Volumes: 0 s, 0 cm3; 20 s, 14 cm3; 40 s, 24 cm3; 60 s, 30 cm3; 80 s, 32 cm3; 100 s, 32 cm3. Calculate the mean rate in the first 20 s and explain why the volume stops rising.
Outline how a colorimeter could be used to follow the hydrolysis of starch by amylase.
The lock-and-key hypothesis was later replaced by the induced-fit hypothesis. Suggest one observation the lock-and-key hypothesis cannot easily explain.
Answer key and marking notes
- Intracellular: catalyses a reaction inside the cell that made it, for example catalase (or DNA polymerase, ATP synthase).
- Extracellular: secreted from the cell and catalyses a reaction outside it, for example amylase (or trypsin, pepsin).
- The substrate collides with and binds to the active site, which has a shape complementary to it.
- As it binds, the active site changes shape slightly to fit more closely around the substrate, forming an enzyme-substrate complex.
- This puts strain on bonds in the substrate / holds groups in the right orientation, lowering the activation energy.
- The products form and are released; the enzyme is unchanged and its active site returns to its original shape.
- Raising the temperature enough would denature the cell's proteins / damage membranes, so cells cannot be heated much.
- Enzymes lower the activation energy so a large proportion of collisions succeed at body temperature.
- Mean rate = 14 ÷ 20 = 0.70 cm3 s−1.
- The rate falls because the substrate concentration falls, so there are fewer successful collisions with active sites.
- The volume stops at 32 cm3 because all the hydrogen peroxide has been broken down (not because the enzyme is used up).
- Add samples of the reaction mixture at intervals to iodine solution (or keep iodine in the mixture); use a red/orange filter, complementary to blue-black.
- Zero the colorimeter with a blank (water / iodine solution without starch) and record the absorbance of each sample.
- Absorbance falls as starch is broken down; convert to starch concentration with a calibration curve made from known starch concentrations.
- Measurements show the shape of the enzyme / active site changes when the substrate binds.
- A rigid active site would not explain how binding strains the substrate's bonds and so lowers the activation energy (or how some enzymes act on a small range of similar substrates).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Trypsin is made by cells in the pancreas and digests proteins in the lumen of the small intestine. Which description of trypsin is correct?
- Aan extracellular enzyme; a globular protein secreted from the cells that make it
- Ban intracellular enzyme; a globular protein made inside pancreas cells
- Can extracellular enzyme; a fibrous protein that is insoluble in water
- Dan intracellular enzyme; it works inside the cells lining the intestine
Which statement correctly describes how an enzyme increases the rate of a reaction?
- Ait provides a pathway with a lower activation energy, so more collisions succeed
- Bit supplies extra energy to the substrate so that its bonds break
- Cit raises the kinetic energy of the substrate molecules at the same temperature
- Dit lowers the energy of the products, so the overall reaction releases more energy
The enzyme maltase hydrolyses maltose but does not hydrolyse sucrose, although both are disaccharides. Which is the best explanation?
- Aonly maltose has a shape and charge pattern complementary to the active site of maltase
- Bsucrose is too large to fit into any active site, because it has more glucose units
- Csucrose binds to maltase but the products cannot leave the active site
- Dmaltose has the same shape as the active site, so it fits exactly into it
Amylase is added to starch at 37 °C. Every 30 s a drop of the mixture is placed on iodine solution on a spotting tile. The drops turn blue-black until 150 s; from 180 s they stay orange-brown. What does the result at 180 s show?
- Athe starch has been hydrolysed, so none is left to give a blue-black colour
- Bmaltose has formed and reacts with iodine to give an orange-brown colour
- Cthe amylase has been denatured, so the reaction has stopped
- Dthe iodine solution on the tile has been used up by the starch
A colorimeter is used to follow the hydrolysis of starch by amylase, with iodine solution in each sample. Which filter should be chosen, and why?
- Aa red filter, because red light is strongly absorbed by the blue-black starch-iodine colour
- Ba blue filter, because the solution is blue-black and blue light matches it
- Cno filter, because white light gives the largest possible absorbance
- Da filter of any colour, because the colorimeter is zeroed with a blank first
Catalase is added to 10 cm3 of hydrogen peroxide solution and the oxygen collected. Curve A shows the result. The experiment is repeated with one change, giving curve B. Which change gives curve B?
- Athe catalase concentration was halved, with the same volume and concentration of H2O2
- Bthe hydrogen peroxide concentration was halved, with the same catalase
- Cthe temperature was raised from 25 to 35 °C, with the same catalase
- Dthe volume of hydrogen peroxide was doubled, with the same catalase
In a catalase experiment the oxygen collected reaches 48 cm3 at 120 s, after which it stops rising. A tangent drawn to the curve at t = 0 passes through the origin and through the point (15 s, 27 cm3). What is the initial rate of reaction?
- A1.8 cm3 s−1
- B0.40 cm3 s−1
- C0.56 cm3 s−1
- D405 cm3 s−1
Hexokinase has two lobes with the active site between them. X-ray studies show that when glucose binds, the lobes close around it, and this change is needed for the reaction to occur. What do these observations most strongly support?
- Athe induced-fit hypothesis, because the active site changes shape as the substrate binds
- Bthe lock-and-key hypothesis, because glucose fits a pre-formed active site exactly
- Cthe idea that glucose is a non-competitive inhibitor of hexokinase
- Dthe idea that binding of glucose denatures hexokinase after each reaction
Quiz key
Trypsin catalyses its reaction outside the cells that made it, after being secreted by exocytosis, so it is an extracellular enzyme, and like all enzymes it is a soluble globular protein. Every enzyme is made inside a cell, so being made in pancreas cells does not make it intracellular. Enzymes are globular, not fibrous; fibrous proteins such as collagen are structural and insoluble. The lumen of the intestine is outside the cells, so trypsin does not work inside the cells lining it.
An enzyme lowers the activation energy, for example by holding the substrate in the right orientation and straining its bonds, so a larger proportion of collisions have enough energy to react. Enzymes do not supply energy; they are catalysts, not fuels. The kinetic energy of molecules is set by temperature, which the enzyme does not change. The energies of reactants and products are unchanged, so the overall energy change of the reaction is the same with or without the enzyme.
The R groups lining the active site of maltase give it a shape and charge distribution complementary to maltose only, so only maltose forms an enzyme-substrate complex. Sucrose is also a disaccharide (glucose plus fructose), so it is not larger in that sense; it simply does not fit this active site. Sucrose does not bind to maltase at all, so there is no complex and no products. The substrate is complementary to the active site, not the same shape as it: a key and its lock fit together because their shapes are complementary.
This method follows the disappearance of the substrate. Iodine gives a blue-black colour with starch; once all the starch has been hydrolysed, the drop stays the orange-brown colour of the iodine itself. Maltose gives no colour with iodine, so the orange-brown is not caused by maltose. At 37 °C amylase is near its optimum and is not denatured; the test stops changing because the substrate is gone. Each drop goes onto fresh iodine solution, so the iodine is not used up.
A blue-black solution absorbs light of the complementary colour, red/orange, so a red filter gives the biggest change in absorbance as starch is broken down. Blue light is transmitted by a blue solution, so a blue filter would show little change. Without a filter, most wavelengths are barely absorbed and the reading is insensitive. Zeroing with a blank sets the starting reading but does not make every filter equally sensitive.
Curve B has a lower initial gradient but reaches the same final volume (40 cm3). The final volume depends only on the amount of substrate, so that must be unchanged; the slower start means fewer active sites are available, so the catalase concentration was lowered. Halving the H2O2 concentration would halve the total oxygen, so B would level off at 20 cm3. Raising the temperature towards the optimum would make the initial gradient steeper, not shallower. Doubling the H2O2 volume would double the final volume to 80 cm3.
Initial rate = gradient of the tangent at t = 0 = 27 ÷ 15 = 1.8 cm3 s−1. The value 0.40 is 48 ÷ 120, the mean rate over the whole reaction, which is lower because the rate falls as the substrate is used up. 0.56 is 15 ÷ 27, the gradient upside down. 405 multiplies 27 by 15 instead of dividing.
In the induced-fit hypothesis the active site is flexible and moulds around the substrate as it binds, putting strain on bonds and lowering the activation energy; the closing lobes are exactly this kind of change. The lock-and-key hypothesis treats the active site as rigid, so a large shape change on binding does not fit it. Glucose is the substrate and binds in the active site, so it is not an inhibitor. The change is temporary and reversible: the enzyme opens again, releases the product and is unchanged, so it is not denatured.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.