3
$\begingroup$

In learning about how cofactors are essential to proper enzyme function, my textbook mentioned catalase and its relation to the human body. According to my textbook, catalase is similar to hemoglobin in that it has 4 hemes which cradle an Fe (iron) atom; the iron atom is used to neutralize free radicals when it "pulls on the substrate's electrons, which brings on the transition state" after catalase is holding a substrate in its active site. I already know that free radicals have one or more unpaired electrons which make them dangerous as they "seek out" an electron to complete its pair on any molecule, atom, or ion it can find, perhaps causing significant tissue damage in the process. I also know that the transition state is a key point in a reaction in which the activation energy required has been met and the reaction will continue spontaneously until it ends.

My question is, what exactly is happening here that helps to neutralize a free radical, and what is the part after the transition state doing?

If I had to guess, based on that cofactors generally can be affected during a reaction, perhaps Iron somehow binds with the free radical and share its electrons, effectively neutralizing the unpaired electron as well as using up the Iron atom? I imagine the "pulling" in the book could fit a description of a weak chemical bond/attraction, but I'm still unsure.

$\endgroup$
3
$\begingroup$

Catalase (hydrogen peroxide oxidoreductase) does not actually quench free radicals. It catalyzes the conversion of hydrogen peroxide to water; the former is a source of a free radical, not a free radical in itself.

This is the proposed mechanism of action of catalase: First one of the oxygens is deprotonated by at the active site of the enzyme. Then the Iron atom that is bound to heme forms a co-ordination complex with the deprotonated oxygen and pulls it away from the other oxygen. As a result the O-O bond breaks.

Free radical quenchers/antioxidants react with the free radicals and form non-toxic products.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.