I am aware that certain organs in the body have specific pH environments to increase the activity of their enzymes. I'm also aware that enzymes only work at certain pHs due to the configuration of the amino acid in their active site and the way this is altered as the pH is changed. What I want to know, however, is why their amino acids are as they are to begin with - has it got to do with the substrate molecules they react with? Particular reference to catalase would be helpful.

  • $\begingroup$ Are you asking about how the amino acids of an enzyme interact with the substrate molecule with particular reference to catalase? $\endgroup$
    – Tyto alba
    Jul 19 '16 at 14:37
  • $\begingroup$ It would be helpful if you could answer this for me but, in particular, I was asking whether there is any reasoning behind why the amino acids of an enzyme's active site are arranged in a specific way. $\endgroup$ Jul 21 '16 at 8:32

The optimum pH for catalase can be found be a simple search. According to the Wikipedia entry for catalase it is approximately 7.

With the exception of extreme pHs, which can cause denaturation, the pH optimum of enzymes is often related to the ionization of the amino acid residues that participate in the reaction (and hence their pKa). Often what happens is that the reaction mechanism requires a residue to alternate between an ionized and non-ionized form, so that the optimum pH approximates to the pH of half-ionization. (This may not be the same as the pKa of the amino acid in solution, as the environment at the active site can alter this greatly.)

You need to understand ionization chemistry to appreciate this. If you do not, you could try this section from Berg et al. first. If you do, look at the mechanism of action of well-established enzymes like lysozyme and the serine proteases.

I cannot help you further with catalase. Apparently it is not well understood.

  • $\begingroup$ Do you have any recommendations for what to read to understand the mechanisms as how reactants reach the Transition state without a catalyst? I'm finding all sorts of information but all are unrefrenced. $\endgroup$
    – Tyto alba
    May 3 '17 at 17:40
  • $\begingroup$ @Tytoalba — Heat. But you could post on SE Chemistry (or check previous questions there) for a more comprehensive answer. $\endgroup$
    – David
    May 3 '17 at 17:52
  • $\begingroup$ I don't prefer asking there the answers are often not backed with references. (This isn't the right place to ask unrelated questions, I just don't know how we may discuss if you are interested). $\endgroup$
    – Tyto alba
    May 3 '17 at 18:15
  • 1
    $\begingroup$ @Tytoalba — I make a point of not doing discussions on this list. In any case my chemistry is rusty. However the Wikipedia entry on activation energy — en.wikipedia.org/wiki/Activation_energy seems reasonable to me. In general, if you want some of the reacting molecules to be able to adopt the transition state without a catalyst, you need to put in the energy in the form of heat. (For particular reactions there may be other factors such as pH that would help.) $\endgroup$
    – David
    May 3 '17 at 19:57

From your comment:

I was asking whether there is any reasoning behind why the amino acids of an enzyme's active site are arranged in a specific way.

After the binding of a substrate to the active site of an enzyme a series of chemical reactions occur at the active site. These reactions in turn convert the substrate into product. So for these reactions to occur a specific arrangement of amino acids at the active site is necessary.

Here's an example:

In the $6th$ step of Glycolysis Glyceraldehyde-3-phosphate is oxidised and phosphorylated to form 1,3-Bisphosphoglycerate.


This reaction is cataysed by the enzyme Glycealdehyde-3-phosphate dehydrogenase. The active site of this enzyme has two amino acids; Cysteine and Histidine. The side chains of these two amino acids participate in the conversion of Glyceraldehyde-3-phosphate into 1,3-Bisphosphoglycerate.


  1. The thiol group(-SH) of the Cysteine residue reacts with the aldehyde group of the Glycealdehyde-3-phosphate to form a Hemithioacetal.Its a nucleophilic addition reaction.
  2. Then a H+ (proton) and a H- (hydride) from the Hemithioacetal moves to the imidazole group of His and the benzene ring of NAD+ respectively oxidising the Hemithioacetal to a Thioester.
  3. The reduced NAD+ (NADH) then leaves the active site and is replaced by another NAD+ molecule.
  4. Finally, a molecule of inorganic phosphate attacks the thioester and forms a tetrahedral intermediate, which then collapses to release 1,3-bisphosphoglycerate.

Thus after the completion of 1,3-bisphosphoglycerate formation the active site of the enzyme gets back to its original state. By state I mean that the H+ from the imidazole group returns to the Cys forming the thiol group of Cys.


So you can see that the specific arrangement of amino acids (Cys and His) at the active site was necessary for the conversion of Glyceraldehyde-3-phosphate to 1,3-Bisphosphoglycerate, had there been some other amino acids the reaction wouldn't be possible.Even if the amino acids had a different arrangement than what is present in the diagram, the enzyme would not be able to catalyse the reaction.


  1. Wikipedia

  2. Biochemistry, 5th edition-Jeremy M Berg

  • $\begingroup$ Interesting answer, but you leave out catalase completely. $\endgroup$
    – Chris
    Jul 22 '16 at 19:13
  • $\begingroup$ Yeah actually after seeing the question I was overjoyed as I found something on this site for the first time that I could answer and cited the enzyme from which I had understood the concept. I'll add catalase to the answer soon. $\endgroup$
    – Tyto alba
    Jul 23 '16 at 7:23
  • $\begingroup$ @Biology Student I found two links that explain the chemical reactions taking place at the active site of catalase. I couldn't fully understand the mechanism so could not add it to my answer. Expecting you can comprehend here are the links: [1]: en.wikipedia.org/wiki/Catalase#Molecular_mechanism [2]: researchgate.net/figure/… $\endgroup$
    – Tyto alba
    Jul 23 '16 at 8:09

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