A mixture holds 120 apoenzyme molecules and 90 molecules of its coenzyme…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A mixture holds apoenzyme molecules and molecules of its coenzyme. Each active holoenzyme needs one of each. At most how many holoenzymes can form? How many apoenzyme molecules are then left without a coenzyme?
Need a reminder? The key ideaShow
Six classes: oxidoreductases (hydrogen, oxidation and reduction), transferases (another group), hydrolases (water), lyases (other breaking, double bond left), isomerases (rearranging), ligases (joining, with ATP). Apoenzyme + cofactor = holoenzyme. Cofactors: prosthetic group (tight, haem), coenzyme (loose, NAD from niacin), metal ion (zinc).
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. The limit. The coenzyme runs out first, at molecules. So at most holoenzymes can form.
Step 2. What is left. apoenzyme molecules have no coenzyme, so they cannot work.
Answer: at most holoenzymes, with apoenzyme molecules left over.
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