The shape of the lesson
By the end of the lesson, learners can
- Temperature: faster, then denatured: Drag the temperature slider for the human enzyme (optimum 37 °C), then switch to the Antarctic fish enzyme and the thermophile enzyme (optimum 75 °C).
- pH and the charges in the active site: Choose pepsin, salivary amylase, trypsin or arginase and move the pH slider.
- Concentration, V max and inhibitors: With no inhibitor, slide [S] up and watch the rate level off at V max; read K m = 4 mmol.
- Immobilised in alginate: Run several batches through the free enzyme and the alginate beads at 37 °C, then raise the temperature above 55 °C and compare the yields.
- Temperature and pH: explain via collisions and active site shape.
- Km = [S] at ½Vmax; low Km = high affinity.
- Competitive: Vmax same, Km up. Non-competitive: Vmax down.
- Immobilised: reusable, pure product, more stable.
One picture every idea returns to
Raising the temperature increases kinetic energy and successful collisions until the optimum; above it, hydrogen and ionic bonds break, the active site loses its shape and the enzyme is denatured. Each enzyme has an optimum pH, kept constant in experiments with buffer solutions; other pH values alter the charges on R groups in the active site. Rate rises with enzyme concentration while substrate is in excess, and with substrate concentration until every active site is occupied and the rate reaches Vmax. The Michaelis-Menten constant Km is the substrate concentration giving ½Vmax: a low Km means high affinity. A competitive inhibitor competes for the active site (Vmax unchanged, Km raised); a non-competitive inhibitor binds elsewhere and changes the active site (Vmax lowered). Enzymes immobilised in alginate can be reused, do not contaminate the product and are more stable.
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, Turn the dials, watch the rate.. Temperature, pH, the amounts of enzyme and substrate, and inhibitors all change how fast an enzyme works. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 11 min | Teach and check: Temperature: faster, then denatured | Drag the temperature slider for the human enzyme (optimum 37 °C), then switch to the Antarctic fish enzyme and the thermophile enzyme (optimum 75 °C). Project tab 4 of the lab (Enzymes lab, temperature) and let a student drive it while the class predicts. | Whole class, then pairs |
| 11 to 17 min | Teach and check: pH and the charges in the active site | Choose pepsin, salivary amylase, trypsin or arginase and move the pH slider. Project tab 5 of the lab (Enzymes lab, pH) and let a student drive it while the class predicts. | Pairs |
| 17 to 23 min | Teach and check: Concentration, V max and inhibitors | With no inhibitor, slide [S] up and watch the rate level off at V max; read K m = 4 mmol. Project tab 6 of the lab (Enzymes lab, concentration and inhibitors) and let a student drive it while the class predicts. | Pairs |
| 23 to 30 min | Teach and check: Immobilised in alginate | Run several batches through the free enzyme and the alginate beads at 37 °C, then raise the temperature above 55 °C and compare the yields. Project tab 7 of the lab (Enzymes lab, immobilised enzymes) and let a student drive it while the class predicts. | Think-Pair-Share |
| 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, 19 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 Immobilised in alginate, 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.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
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 |
|---|---|
| Do not write that enzymes are killed: they are not alive. | Say the enzyme is denatured and explain it: bonds in the tertiary structure break, the active site loses its complementary shape, and fewer enzyme-substrate complexes form. |
| A competitive inhibitor does not lower V max: at a very high substrate concentration the substrate out-competes it. | Only a non-competitive inhibitor lowers V max. And K m is a substrate concentration, read on the x-axis, not a rate. |
Factors that affect enzyme action: worksheet
6 questions, 19 marks. Written in the style of the papers; no past paper question is reproduced.
Explain the effect of increasing temperature from 20 °C to 60 °C on the rate of a reaction catalysed by a human enzyme.
Describe how you would investigate the effect of pH on the activity of catalase, identifying the independent variable and two variables to control.
For an enzyme, Vmax is 24 μmol min−1. At a substrate concentration of 3.0 mmol dm−3 the rate is 12 μmol min−1. State Km and explain how a second enzyme with Km 0.8 mmol dm−3 for the same substrate compares.
Compare the effects of a competitive and a non-competitive reversible inhibitor on enzyme activity.
Explain why increasing the substrate concentration has no further effect on the rate once Vmax is reached, and state what would raise the rate.
A student compares lactase immobilised in alginate beads with free lactase. Describe how the beads are made and give two advantages of immobilisation.
Answer key and marking notes
- From 20 °C to the optimum (about 37-40 °C) the rate increases, because the molecules have more kinetic energy.
- So there are more frequent collisions between substrate and active site, and more enzyme-substrate complexes form per second.
- Above the optimum, hydrogen and ionic bonds in the tertiary structure break, so the active site changes shape.
- The substrate no longer fits / is no longer complementary; the enzyme is denatured and the rate falls towards zero.
- Independent variable: pH, set with a range of buffer solutions (for example pH 4, 5, 6, 7, 8, 9) added to the hydrogen peroxide.
- Measure the volume of oxygen released in a fixed time (or the initial rate) with a gas syringe.
- Control two of: temperature (water bath), concentration and volume of hydrogen peroxide, concentration and volume of catalase.
- 12 is half of 24, so Km = 3.0 mmol dm−3.
- The second enzyme has a lower Km, so it reaches half of its Vmax at a lower substrate concentration.
- It therefore has a higher affinity for the substrate.
- Competitive: similar shape to the substrate, binds to the active site, so blocks substrate binding.
- Its effect is reduced by increasing substrate concentration; Vmax is unchanged but Km increases.
- Non-competitive: binds to a site other than the active site, changing the shape of the active site.
- Increasing substrate does not overcome it; Vmax decreases (Km unchanged). Both bind reversibly by weak bonds.
- At Vmax all the active sites are occupied at any moment (saturated), so enzyme concentration is the limiting factor.
- Increasing the enzyme concentration (or the temperature, up to the optimum) would raise the rate.
- Mix lactase with sodium alginate solution and drip it into calcium chloride solution; calcium alginate beads form, trapping the enzyme.
- Advantage: the enzyme can be recovered and reused / allows a continuous process.
- Advantage: the product is not contaminated with enzyme; or the enzyme is more stable at higher temperatures and over a wider pH range.
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.
What is the Michaelis-Menten constant, Km, of an enzyme?
- Athe substrate concentration at which the rate of reaction is half of Vmax
- Bthe rate of reaction when the substrate concentration is half its maximum
- Cthe substrate concentration at which the rate of reaction first reaches Vmax
- Dhalf of the maximum rate of reaction that the enzyme can achieve
Trypsin has an optimum pH of about 8. A trypsin solution is placed in a buffer at pH 3 and its rate of protein digestion is almost zero. Which is the best explanation?
- AH+ ions change the charges on R groups in the active site, so the substrate cannot bind
- BH+ ions raise the activation energy, so no collisions have enough energy
- Cthe acid hydrolyses the peptide bonds in the primary structure of trypsin
- DH+ ions act as a competitive inhibitor and fill every active site
Samples of a human enzyme are kept for 30 minutes at 5 °C or at 70 °C, then all are tested at 37 °C. Which result is expected?
- Athe 5 °C samples are active at 37 °C; the 70 °C samples show little or no activity
- Bneither set of samples is active, because both were kept far from the optimum
- Cboth sets of samples are fully active, because they are tested at the optimum
- Dthe 70 °C samples are more active, because they gained kinetic energy
How are enzymes immobilised in alginate beads in a school laboratory?
- Amix the enzyme with sodium alginate solution and drip it into calcium chloride solution
- Bmix the enzyme with calcium chloride solution and drip it into sodium alginate solution
- Cmix the enzyme with sodium alginate solution and drip it into hot water to set it
- Dmix the enzyme with sodium chloride solution and drip it into calcium alginate solution
The graph shows the rate of reaction against substrate concentration for two enzymes, P and Q, that act on the same substrate. Both have a Vmax of 40 μmol min−1. Which statement is correct?
- AP has a Km of 2 mmol dm−3 and a higher affinity for the substrate than Q
- BQ has a Km of 6 mmol dm−3 and a higher affinity for the substrate than P
- CP has a Km of 20 μmol min−1 and the same affinity as Q
- DP has a Km of 5 mmol dm−3, half of the highest substrate concentration tested
The initial rate of an enzyme-catalysed reaction is 0.80 cm3 s−1 at 25 °C and 1.36 cm3 s−1 at 35 °C. What is the percentage increase in rate?
- A70%
- B41%
- C170%
- D56%
The table shows the initial rate of an enzyme reaction at a low and a high substrate concentration, with no inhibitor and with the same concentration of inhibitor X or inhibitor Y. Which conclusion is supported?
- AX is a competitive inhibitor and Y is a non-competitive inhibitor
- BX is a non-competitive inhibitor and Y is a competitive inhibitor
- Cboth are competitive, because both halve the rate at low [S]
- Dboth are non-competitive, because both lower the rate at low [S]
Equal masses of lactase are immobilised in alginate beads of two sizes, 2 mm and 5 mm diameter. Milk is run through columns of each at the same flow rate. More glucose is detected in the milk from the 2 mm beads. Which is the best explanation?
- Asmaller beads have a larger surface area to volume ratio, so lactose diffuses a shorter distance to the active sites
- Bsmaller beads contain more lactase in total, so there are more active sites in the column
- Cthe lactase in the larger beads has been denatured by the calcium chloride solution
- Dsmaller beads lower the activation energy of the reaction more than larger beads do
Quiz key
Km is a substrate concentration: find Vmax, halve it, and read the substrate concentration that gives that rate. It is not a rate, so it is neither the rate at some substrate concentration nor half of Vmax itself (that is the y-value used to find it). Vmax is approached gradually as the curve levels off, so there is no clear concentration where it is first reached, and that would not be Km anyway.
At a pH far from the optimum, the extra H+ ions alter the ionisation of R groups in and around the active site, so the substrate no longer binds well, and ionic and hydrogen bonds holding the tertiary structure are disrupted. The activation energy is a property of the reaction pathway; the problem is that enzyme-substrate complexes do not form. A pH change disrupts ionic and hydrogen bonds, not the strong covalent peptide bonds of the primary structure. H+ ions are not shaped like the protein substrate and do not occupy the active site as a competitive inhibitor; they change the charges of the enzyme itself.
At 5 °C the enzyme molecules simply have little kinetic energy; their structure is intact, so they work normally once warmed to 37 °C. At 70 °C hydrogen and ionic bonds in the tertiary structure break, the active site loses its complementary shape and the enzyme is denatured, which is permanent, so cooling back to 37 °C does not restore activity. Low temperature does not denature, so the samples from 5 °C are not inactive. Denaturation is not reversed by testing at the optimum. Kinetic energy depends on the current temperature, not on past heating.
The enzyme is mixed with soluble sodium alginate, and drops of this mixture fall into calcium chloride solution, where calcium ions replace sodium ions to form insoluble calcium alginate beads that trap the enzyme. Reversing the solutions would leave the enzyme in the calcium chloride, not trapped through the alginate gel. Hot water does not set alginate and could denature the enzyme. Calcium alginate is insoluble, so there is no calcium alginate solution to drip into.
Half Vmax = 20 μmol min−1. P reaches 20 at 2 mmol dm−3 and Q at 6 mmol dm−3, so Km is 2 mmol dm−3 for P and 6 mmol dm−3 for Q. A lower Km means half-maximal rate at a lower substrate concentration, so P has the higher affinity. Q's Km is correctly 6, but a higher Km means a lower affinity. 20 μmol min−1 is half Vmax, a rate, not Km, which is a concentration. Km is found from half of Vmax, not from half of the substrate axis.
Increase = 1.36 − 0.80 = 0.56 cm3 s−1; percentage increase = 0.56 ÷ 0.80 × 100 = 70%. 41% divides the increase by the new value (1.36) instead of the original. 170% is the new rate as a percentage of the old one (1.36 ÷ 0.80 × 100), not the increase. 56% is the raw difference multiplied by 100, without dividing by the original rate.
At high substrate concentration the rate with X (9.3) is almost back to the uninhibited rate (9.5): substrate molecules out-compete X for the active site, so Vmax is unchanged and X is competitive. With Y the rate stays at about half (4.8) even at high [S], so Vmax is lowered: Y binds away from the active site and changes its shape, which extra substrate cannot overcome, so Y is non-competitive. Swapping them reverses this evidence. Both inhibitors reduce the rate at low [S], so the low [S] column alone cannot show the type of inhibition.
In immobilised enzymes the substrate has to diffuse into the bead to reach the active sites. Smaller beads give a larger total surface area for the same volume of gel and a shorter diffusion distance to the enzyme at the centre, so more lactose is hydrolysed in the time the milk spends in the column. The total mass of lactase was the same, so the number of active sites is not the reason. Calcium chloride is used to make all the beads and does not denature lactase in one size only. The activation energy depends on the enzyme, not on the size of the bead it is trapped in.
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.