Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
An enzyme was tested free, and with each of two drugs, A and B. The table gives the rates in µmol/min at five substrate concentrations in mmol/L.
| [S] | 1 | 2 | 4 | 8 | 16 |
|---|---|---|---|---|---|
| free | 10.0 | 16.7 | 25.0 | 33.3 | 40.0 |
| A | 3.85 | 7.14 | 12.5 | 20.0 | 28.6 |
| B | 5.00 | 8.33 | 12.5 | 16.7 | 20.0 |
Competitive inhibitor: , unchanged, with . Non-competitive inhibitor: , unchanged. Feedback from an end product usually acts on the first committed step, and a regulated enzyme such as phosphofructokinase-1 has activators and inhibitors that bind away from its active site.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. The fits. The enzyme alone gives µmol/min and mmol/L. Drug A gives and . Drug B gives and .
Step 2. The kinds. Drug A raises three times and leaves alone, so it is competitive, with . Drug B halves and leaves alone, so it is non-competitive, with .
Step 3. The crossing. Drug A shares with the enzyme alone, so their lines share the y-intercept . They cross on the y-axis.
Answer: with A is 12 mmol/L and with B is 25 µmol/min. A is competitive, B is non-competitive, and the lines for A and the enzyme alone cross on the y-axis.
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