Enzyme Rate Graphs: Why the Plateau
Why the rate of an enzyme-controlled reaction rises with substrate concentration then levels off, and which factor is limiting at the plateau. Includes the 4-mark exam question, a worked model answer and the examiner trap on saying the substrate has run out.
Chapters (8)
Exam question [4 marks]
A student added increasing amounts of substrate to a fixed amount of enzyme and measured the rate of reaction. The rate rose steeply at first, then levelled off. Explain the shape of the graph, and state what would happen to the plateau if the enzyme concentration were doubled.
Try it on paper first, then check the mark scheme below. The question is original and was written for this lesson.
Mark scheme
| Mark | Mark point | Also accept | Do not accept |
|---|---|---|---|
| 1 | At low substrate concentration, rate increases because more substrate gives more collisions with active sites / more enzyme-substrate complexes formed | "more frequent collisions", "more complexes per second", "substrate is the limiting factor" | "more substrate so faster" with no collisions or complexes |
| 2 | At the plateau, all the active sites are occupied / enzymes are saturated | "all enzymes working at maximum rate", "no free active sites" | "the enzymes are full" with no mention of active sites; "the substrate has run out" |
| 3 | Adding more substrate does not increase the rate, because enzyme concentration (number of active sites) is now the limiting factor | "enzyme concentration limits the rate" | "substrate concentration is limiting" at the plateau; "enzymes are used up" |
| 4 | Increasing enzyme concentration gives more active sites, so the rate increases / the plateau is higher (if substrate is in excess) | "more enzyme, more complexes", "higher plateau" | "more enzyme so the enzymes work faster" |
Each point is a separate cause-and-effect step. "The substrate has run out" is the classic wrong answer and scores nothing; the substrate is in excess at the plateau. Enzymes are not used up. Temperature and pH are assumed constant.
Model answer (4/4)
At low substrate concentration the rate increases because more substrate molecules means more collisions with active sites, so more enzyme-substrate complexes form each second. At the plateau all the active sites are occupied, so the enzymes are saturated. Adding more substrate does not increase the rate because enzyme concentration is now the limiting factor. Increasing the enzyme concentration would provide more active sites, so the rate would increase and the plateau would be higher.
Examiner trap
- Saying the rate levels off because the 'substrate has run out' (enzymes are saturated)
- Not naming enzyme concentration as the limiting factor at the plateau
- Mixing up what the axes show
Key terms
- limiting factor
- enzyme concentration
- substrate concentration
- active sites
- saturated
- plateau
- collisions
Lesson notes
A written version of the video, chapter by chapter, for revision. Times link to the video.
0:19Reading the axes
First, read the axes. Along the bottom is substrate concentration. Up the side is the rate of reaction. So we are asking what happens to the speed of the reaction when we keep the enzyme the same, and keep adding substrate. Enzyme amount, temperature and pH are all held constant, so the only thing changing is the substrate.
0:42Low substrate: rate rises
Start at low substrate concentration. There are lots of empty active sites, and few substrate molecules. Add more substrate, and there are more collisions between substrate molecules and active sites. So more enzyme-substrate complexes form every second, and the rate goes up. Notice the line is steep here. Double the substrate, and you roughly double the rate.
At this stage, substrate concentration is the limiting factor. That is a factor that is in short supply, so it limits how fast the reaction can go.
1:16Saturation: all active sites full
Keep adding substrate, and eventually virtually every active site is occupied at the same moment. We say the enzymes are saturated. Extra substrate molecules have to queue. They cannot react until an active site is free. Remember, an enzyme is not used up. It releases the product, and then takes the next substrate.
Each enzyme can only work at its own top speed, so the rate stops rising and the graph plateaus.
And here is the key point. The substrate has not run out. There is plenty of it, queuing up.
1:53The limiting factor
So at the plateau, what is limiting the rate? The enzyme concentration. There are only so many active sites, and they are all busy. Adding more substrate makes no difference. Only more enzyme can raise the rate, while temperature and pH stay fixed.
Other factors such as temperature or pH could also limit it, but in this question they are fixed.
2:17Adding more enzyme
Now, change the other variable. Add more enzyme, and keep plenty of substrate. More enzyme means more active sites, so more collisions between substrate and active sites, so more complexes form, and the rate goes up. If substrate is in excess, the rate is directly proportional to enzyme concentration. Double the enzyme, double the rate. On the substrate graph, doubling the enzyme gives a higher plateau.
3:23Examiner trap and recap
The trap. Never say the substrate has run out. It has not. The active sites are all occupied.
Recap. At low substrate, substrate is limiting. At the plateau, the enzyme is limiting.
Next time, we add inhibitors, molecules that interfere with enzymes, and see what they do to this graph.
Watch first
This lesson builds on:
- How Enzymes Work: Active Site and Induced Fit · AQA 7402 · 3.1.4.2
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Independent revision resource, not affiliated with or endorsed by AQA. The exam question is original and was written for this lesson; it is not an AQA question. Spotted an error? Email info@medservatory.com with the lesson code AB-012.