Two samples of the same enzyme are prepared…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Two samples of the same enzyme are prepared. Sample A is kept in ice for ten minutes. Sample B is boiled for ten minutes. Each is then cooled or warmed to its optimum temperature and mixed with the substrate. Predict the activity of each sample and explain the difference.
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. Sample A. Cold only slows the enzyme down and makes it inactive. The fold is not lost. At the optimum temperature the enzyme works again, and sample A shows normal activity.
Step 2. Sample B. Boiling breaks the weak bonds that hold the fold. The protein is denatured, and the active site is lost. At the optimum temperature there is little or no activity.
Answer: sample A recovers fully, because cold only inactivates. Sample B stays inactive, because heat denatures.
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