Biology › Enzymes and metabolic control › Four factors, four curves, and one honest way to measure a rate
Four factors, four curves, and one honest way to measure a rate
Temperature, pH, substrate concentration and enzyme concentration each bend the rate in a shape of their own. Getting full marks means drawing each shape correctly and, before any of that, measuring a rate that refuses to stay constant while you measure it.
Before this Enzyme action and the active site · Denaturation as loss of tertiary structure
Before you start
A denatured enzyme is one that has been switched off, so cooling it down should switch it back on. Every class contains someone who believes this, usually because they have been told that enzymes 'work best at 40 °C' and have reasoned sensibly from there. Cooling really does reverse the slowing you get below the optimum. It does not reverse denaturation, because denaturation is not a slowing — it is the tertiary structure coming apart. Once the hydrogen and ionic bonds have gone and the chain has unravelled, a cooled protein reassembles into some folded state, but not reliably into the one shape that made an active site. A cooled fried egg does not turn back into a raw one.
What you should be able to do
- Explain why the rate of an enzyme-catalysed reaction falls during a run, and why initial rate is the only fair comparison between conditions.
- Account for the shape of the temperature curve as the sum of two opposing effects, and calculate a Q10.
- Explain the effect of pH in terms of charge on R groups and the bonds that depend on it.
- Interpret a substrate-concentration curve, naming what is limiting the rate in each part of it.
- Draw all four curves with the right shape, including the asymmetry of the temperature curve.
The rate depends on when you measured it
Set an enzyme going and take readings every thirty seconds — oxygen collected from catalase, reducing sugar produced by amylase — and you get a curve that starts steep and flattens out. It is tempting to read the rate off the middle of it. Do not.
Three things eat the rate as a run proceeds. Substrate is used up, so collisions with active sites get rarer. Product piles up, and for many enzymes it competes for the site or binds elsewhere and slows things down. And over a long warm run a fraction of the enzyme denatures. By minute ten you are no longer measuring the experiment you set up.
The fix is the initial rate: the gradient of the tangent at time zero, when the substrate concentration is still the one you chose and no product exists. In practice you either draw that tangent, or you time a fixed small change — the seconds until an iodine test stops giving blue-black — and use 1/time. The second method only works if the change happens near the start.
- Initial rate
- The rate of reaction at time zero, found from the gradient of the tangent to a progress curve at the origin.
- Limiting factor
- The factor that, of all those affecting a rate, is the one restricting it at that moment.
- Optimum
- The value of a variable at which an enzyme's rate is highest.
Temperature: two effects pulling in opposite directions
Warm a reaction mixture and two things start happening at once, and the curve you are asked to draw is the sum of them.
The helpful one is kinetic energy. Molecules move faster and collide harder, so more collisions between substrate and active site carry enough energy to react. Below about 40 °C this dominates and the rate roughly doubles for every 10 °C rise — the temperature coefficient, Q10, the rate at a temperature divided by the rate ten degrees lower.
The unhelpful one is vibration inside the protein itself. The heat that is shaking substrate molecules is also shaking the enzyme, and the hydrogen and ionic bonds holding its tertiary structure are weak. Past a certain point they start to break faster than they re-form, the chain unfolds, and the active site is no longer complementary to anything.
The optimum is therefore not a temperature at which something special happens. It is the crossing point of two trends — the last temperature at which the gain in kinetic energy still outweighs the loss to denaturation. Most human enzymes have it near 37 to 40 °C, which is not a coincidence. Taq polymerase, taken from a bacterium living in hot springs, has it near 72 °C and survives brief spells at 95 °C, which is exactly why PCR uses it.
Keep the two directions distinct, because questions test the difference. Take an enzyme from 37 °C down to 4 °C and it slows to a crawl: fewer successful collisions, tertiary structure intact, fully reversible on rewarming. Take it from 37 °C up to 70 °C and it stops: bonds broken, structure lost, nothing to recover. Refrigerating food and boiling food are not the same intervention.
Getting a Q10, and knowing when to stop using it
An enzyme gives an initial rate of 12 arbitrary units at 15 °C and 25 units at 25 °C. Calculate Q10, predict the rate at 35 °C, and say why the same method should not be used to predict the rate at 45 °C.
Q10 is the ratio of the rates ten degrees apart: 25 ÷ 12 = 2.08, which is the near-doubling you should expect well below the optimum.
Predicting 35 °C means multiplying by that ratio once more: 25 × 2.08 = 52 units. Acceptable, because 35 °C is still on the rising part of the curve where kinetic energy is the only thing changing much.
At 45 °C the prediction would be 108 units, and it would be nonsense. By then a significant fraction of the enzyme has denatured, so the second effect is no longer negligible and the curve has turned downwards. Q10 describes the rising limb only.
pH: charge first, then bonds, then shape
The pH curve on the right of the figure above is sharper than students expect, and the reason is chemical rather than thermal.
Several of the twenty amino acids have side chains that carry a charge, and whether they carry it depends on the concentration of hydrogen ions around them. Change the pH and you protonate or deprotonate those side chains; the ionic bonds between them fail, hydrogen bonds are disrupted too, and the tertiary structure shifts. The active site is a product of that structure, so its shape changes and the substrate stops being complementary. Charged groups taking part in the catalysis directly are affected as well.
Small deviations are reversible — return the enzyme to its optimum and the bonds re-form. Large ones denature it permanently, in the same way and for much the same reason as excessive heat.
Learn the optima as a set, because they are the examples questions reach for. Pepsin works in the stomach at around pH 2. Salivary amylase works in the mouth at around pH 7 and is destroyed within minutes of being swallowed. Trypsin, secreted by the pancreas into the small intestine, works at around pH 8 — which is why the pancreas also secretes hydrogencarbonate to neutralise the acid arriving from the stomach.
| Enzyme | Where it works | Optimum pH | What it acts on |
|---|---|---|---|
| Pepsin | Stomach | about 2 | Proteins |
| Salivary amylase | Mouth | about 7 | Starch |
| Trypsin | Small intestine | about 8 | Proteins |
| Catalase | Peroxisomes of most cells | about 7 | Hydrogen peroxide |
Substrate and enzyme: work out which one has run out
The last two factors are the ones that reward careful reading of a graph, because their curves have different shapes for a reason you can state.
Raise the substrate concentration from zero and the rate climbs steeply. Active sites are sitting empty, so every extra substrate molecule finds one quickly and substrate concentration is the limiting factor. Keep going and the climb eases, because more and more of the sites are already occupied when the next substrate molecule arrives. Eventually the curve levels off: every active site is working flat out, and the only thing that limits the rate now is how many sites there are and how fast each one turns over. The enzyme is saturated, and the plateau height is called Vmax. Adding more substrate past that point achieves precisely nothing.
Raise the enzyme concentration instead, with plenty of substrate present, and you get a straight line through the origin. Twice the enzyme means twice the active sites means twice the rate, and there is no ceiling for as long as substrate remains in excess. Let the substrate become limiting and the line bends over too — which is the small print that separates a full-mark answer from a three-quarters one.
TRY IT — Reading a plateau correctly
A student measures the initial rate of a catalase-catalysed reaction at five hydrogen peroxide concentrations. The rate rises with concentration up to 0.4 mol dm⁻³ and is then the same at 0.5 and 0.6 mol dm⁻³. She concludes that the enzyme has been denatured at the higher concentrations. Explain why that conclusion is wrong, and say what the plateau does show.
Check your answer
Nothing in the experiment could denature the enzyme. Denaturation needs heat, extreme pH or something similar; adding more substrate does none of those. A denatured enzyme would also have given a falling rate, not a constant one.
The plateau shows saturation. Above about 0.4 mol dm⁻³ every active site is occupied as soon as it becomes free, so substrate concentration has stopped being the limiting factor. What limits the rate now is the number of active sites available, which is set by the enzyme concentration, and the speed at which each site works.
The test of that explanation is a prediction: adding more catalase should raise the rate again, while adding more peroxide should not.
In the exam
- Write the temperature curve's fall as denaturation described in bonds: hydrogen and ionic bonds break, tertiary structure is lost, the active site is no longer complementary, no enzyme-substrate complex forms. 'The enzyme is killed' scores nothing and irritates the examiner, since a protein was never alive.
- Below the optimum an enzyme is slowed and unharmed; above it, it is damaged. Any question that cools something after heating it is testing exactly that distinction.
- On a substrate-concentration graph, name the limiting factor for each region separately — substrate concentration on the rising part, number of active sites or enzyme concentration at the plateau.
- When you sketch the temperature curve, make the fall steeper than the rise and bring it right down to the axis. Marks are given for the shape.
- 'Initial rate' means the tangent at time zero, or 1/time to a fixed early end point. A rate read from the middle of a run is a different quantity and cannot be compared across conditions.
- Quote a number when you have one: 37 to 40 °C for most human enzymes, pH 2 for pepsin, pH 8 for trypsin. Specific beats vague on every mark scheme in circulation.
Check yourself
Two students investigate how temperature affects amylase. One times how long a starch solution takes to stop giving a blue-black colour with iodine and calculates 1/time. The other measures the mass of reducing sugar formed after ten minutes. They disagree about whether 45 °C or 35 °C is better. Explain how the same enzyme can give them opposite answers, and say which method you would trust.
Answer
The two methods are measuring different things, and the difference matters most at exactly the temperature they disagree about.
Timing to an early end point is close to an initial-rate measurement. It samples the first part of the run, before much enzyme has denatured, so at 45 °C it records the genuinely high rate that the extra kinetic energy delivers. That method will report 45 °C as the better temperature.
Measuring product after ten minutes gives a mean rate over the whole period. At 45 °C the enzyme denatures steadily during those ten minutes, so the reaction that began quickly slows and may stop altogether, while at 35 °C the enzyme survives and keeps working. The mean rate over ten minutes can therefore be higher at 35 °C even though the initial rate is higher at 45 °C. Both students have measured correctly; neither has measured the same quantity.
Which to trust depends on what is being asked. For how temperature affects the enzyme's activity, the initial-rate method is the fair comparison. For which temperature to run a ten-minute process at, the second student's answer is the useful one. The full response is that stability over time is a second variable, and that a proper investigation would take a progress curve at each temperature.
Questions
Question 14 marks
Explain why the rate of an enzyme-catalysed reaction climbs gradually as the temperature is raised towards the optimum, and then falls steeply above it.
Mark scheme
- B1 raising the temperature gives the molecules more kinetic energy, so substrate and active site collide more often and with more energy
- B1 more of those collisions are successful, so more enzyme-substrate complexes form and the rate roughly doubles for each 10 °C rise
- B1 above the optimum the increased vibration breaks the hydrogen bonds and ionic bonds holding the tertiary structure, so the enzyme denatures
- B1 the active site is then no longer complementary to the substrate, and denaturation proceeds far faster than a doubling per 10 °C, which is why the fall is steeper than the climb
Question 24 marks
A student collects the oxygen released by catalase every thirty seconds for ten minutes and finds that the rate falls steadily throughout the run. Suggest three reasons for the fall, and suggest what she should measure instead if she wants to compare different temperatures fairly.
Mark scheme
- B1 substrate is being used up, so collisions between hydrogen peroxide and active sites become less frequent
- B1 product accumulates, and for many enzymes it competes for the active site or binds elsewhere and slows the reaction
- B1 a fraction of the enzyme denatures over a long warm run
- B1 she should compare initial rates, taken from the gradient of the tangent at time zero or from 1/time to a fixed early end point
Question 33 marks
Explain how a change in pH away from the optimum reduces the rate of an enzyme-catalysed reaction.
Mark scheme
- B1 a change in hydrogen ion concentration protonates or deprotonates the side chains, altering the charge they carry
- B1 ionic bonds between those side chains fail and hydrogen bonds are disrupted, so the tertiary structure shifts
- B1 the active site is a product of that structure, so its shape changes and the substrate is no longer complementary to it
Question 43 marks
An enzyme has an initial rate of 0.8 µmol per minute at 10 °C and 1.8 µmol per minute at 20 °C. Calculate the Q10 for this enzyme, and calculate the rate you would predict for it at 30 °C.
Mark scheme
- M1 Q10 is the rate at one temperature divided by the rate ten degrees lower, so 1.8 divided by 0.8
- A1 a Q10 of 2.25
- A1 a predicted rate of 1.8 × 2.25 = 4.05 µmol per minute at 30 °C
Question 53 marks
Describe the shape of a graph of initial rate against substrate concentration, and describe what is limiting the rate in each part of that graph.
Mark scheme
- B1 the rate climbs steeply from the origin and the climb then eases, before the curve levels off at a plateau
- B1 on the rising part the limiting factor is substrate concentration, because active sites are sitting empty and each extra substrate molecule finds one
- B1 at the plateau every active site is occupied, so the limiting factor is the number of active sites, set by the enzyme concentration
Question 62 marks
Give the optimum pH of pepsin and the optimum pH of trypsin, naming where in the gut each of these enzymes works.
Mark scheme
- B1 pepsin has an optimum of about pH 2 and works in the stomach
- B1 trypsin has an optimum of about pH 8 and works in the small intestine
Worth remembering
- Compare initial rates, never rates read part-way through a run.
- The temperature optimum is a balance between rising kinetic energy and rising denaturation, so the curve is lopsided: gentle up, steep down.
- Q10 is roughly 2 below the optimum, and meaningless above it.
- pH acts by changing charge on R groups, which breaks ionic and hydrogen bonds and alters the shape of the active site.
- Substrate concentration gives a curve that plateaus at saturation; enzyme concentration gives a straight line, for as long as substrate stays in excess.