An enzyme has a maximum rate of 60 µmol/min…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
An enzyme has a maximum rate of µmol/min. Its rate at some substrate concentrations is: µmol/min at mmol/L, at mmol/L, at mmol/L, at mmol/L, at mmol/L and at mmol/L. Find the Km. At mmol/L, what percentage of the maximum rate has been reached?
Need a reminder? The key ideaShow
Each enzyme has an optimum temperature and pH. Cold makes an enzyme inactive; heat denatures it. The rate rises with substrate and levels off at Vmax; Km is the substrate concentration at half of Vmax. A competitive inhibitor looks like the substrate and blocks the active site; more substrate reduces its effect.
The worked solution
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. Half the maximum. µmol/min.
Step 2. The Km. The rate is µmol/min when the substrate is at mmol/L. So the Km is mmol/L.
Step 3. The rate at 20 mmol/L. , which is of the maximum.
Answer: Km is mmol/L, and of the maximum rate is reached at mmol/L.
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