Effect of competitive inhibitor on peroxidase enzyme activity

[cmantzioros]

I'm studying for a lab exam tomorrow and I just want to make sure of some things.

One of the experiments was on the effect of inhibitors on enzyme activity. We were studying the reaction 2 H2O2 (hydrogen peroxide) --> 2 H2O + O2. This reaction uses the enzyme peroxidase. Hydroxylamine (HONH2) is structurally similar to hydrogen peroxide so it competes with it for peroxidase's active site thereby preventing peroxidase from binding with hydrogen peroxide and inhibiting the reaction. But a high enough concentration of substrate with a constant concentration of inhibitor will reduce the inhibition. I used guaiacol as an indicator. It turns colourless to brown when it becomes oxidized and the intensity of the brown is proportional to the amount of oxygen produced. I was supposed to have a used a spectrophotometer to measure the absorbance versus time but I ran out of time in the lab period... We had 10 test tubes:

1. water + guaiacol (indicator)
2. water + hydrogen peroxide (substrate)
3. water + guaiacol + hydroxylamine (inhibitor)
4. water + guaiacol + hydrogen peroxide
5. water + guaiacol + peroxidase (enzyme)
6-10. water + guaiacol + increasing volumes of hydrogen peroxide with each + constant volume of peroxidase + constant volume of hydroxlamine

This is what I think: for #6, there is no inhibitor therefore the absorbance should be high because oxygen will be evidently be produced. For #7-10, the increasing amount of substrate at constant amount of inhibitor should reduce inhibition and therefore absorbance should be lowest for 7 and highest for 10. Is this correct? However, for tubes 1-4, I don't know what the absorbance values would look like. Can anyone tell me this?

Thank you.

 

[Kalirren]

Just write out the entire system. Your main reaction is

H2O2 ----> H2O + O2

catalyzed by peroxidase.

But O2 is invisible, so we add an indicator, so that when O2 is produced, it further undergoes the reaction

O2 + guaiacol ---> something brown

and you can measure the brown-ness with a spectrophotometer.

That is to say, you will only see brown-ness when O2 is being produced in the presence of guaiacol. So run through the test tubes one by one.

Spoiler

1: water + guaiacol alone. There is nothing to oxidize the guaiacol.
2. water + hydrogen peroxide. While this will decompose at a baseline rate, there's no indicator to reflect the oxygen being produced.
3. water + guaiacol + hydroxylamine. Again, no chemical change is occurring.
4. water + guaiacol + hydrogen peroxide. As the peroxide decomposes, the released oxygen will oxidize the guaiacol, so you will get a certain baseline rate of appearance of brown.
5. water + guaiacol + peroxidase (enzyme). The enzyme does nothing, because there's no reaction to catalyze.
6-10. water + guaiacol + increasing volumes of hydrogen peroxide with each + constant volume of peroxidase + constant volume of hydroxlamine: Well, all of the components of the above system are present, so the guaiacol correctly reports the amount of oxygen that has been produced. Thus, as the volume of hydrogen peroxide increases, so will the observed brown-ness.