Hi Shivagi, mixed inhibition and substrate inhibition are not synonyms, but in this case they may coincide. I mean, if GDP changes both kcat and Km (not by the same extent), then it is a mixed inhibitor by definition. As GDP is also the product of your reaction, then you also have product inhibition.

In principle, I would have expected GDP to be a (mostly) competitive inhibitor, since it is so similar to the substrate GTP, but you never know...

Dear Peracchi,

Thank you for your response.

I also had the same opinion that GDP should be a competitive inhibitor of GTPase but in competitive inhibition only km increases and there should be no change in kcat and vmax.

But the kinetic parameters in this case is more favourable to mixed inhibition theory.

So as you mentioned in your answer,

GDP here act both as mixed and product inhibitor?

Can both these inhibition happen at the same time?

As long as you're sure that your kcat is changing significantly (and this is not due to some trivial artifacts, such as pH changes due to the addition of GDP from an unbuffered stock or chelation of Mg by the excess GDP, for example), the answer is yes to both of your questions.

Dear Peracchi,

Thank you so much for your clarification over this doubt.

Yes GDP significantly changes the kcat and it is not due to any artifacts.

Once again thank you.

It may be relevant to this discussion that the GTPase reaction has 2 products: GDP and inorganic phosphate. The pattern of product inhibition by GDP may differ depending upon whether it is the first or the 2nd product to dissociate from the enzyme, if dissociation is ordered. In addition, GDP will very likely act as a competitive inhibitor, since it is so close to GTP, as already mentioned by others.

Dear Shapiro,

Thank you for your response.

The hydrolysis of GTP gives GDP and Pi at the same time.

I also thought that the inhibition should be competitive but the kinetic parameters were not in same favour.

The Km increases and vmax and kcat decreases.

Also GDP and GTP are differ by only one phosphate and GDP is also the product then why it can't be product inhibition

If you have a copy of the book by Irwin H. Segel, Enzyme Kinetics, Behavior and Analysis of Rapid Equilibrium and Steady-State Enzyme Systems (1993), this topic is covered starting on p. 544, but especially on pp. 553-555.

For a Uni Bi system (one substrate A, 2 products P and Q), in which there is an ordered release of products P and Q, product P acts as a mixed-type inhibitor and product Q acts as a competitive inhibitor.

If you are seeing mixed-type inhibition by GDP, then it is probably product P, the first one to dissociate.

This assumes that you are looking at a steady-state system and measuring initial rates of GTP hydrolysis.

If so, you can learn some additional things about the enzyme from doing product inhibition studies, such as the equilibrium dissociation constants of the two products (Kip and Kiq) and the ratio of the Kms of the two products (KmQ/KmP) for the reverse reaction of GTP synthesis (see equations IX-73 and IX-74).

• In competitive inhibition, the inhibitor binds only to the free enzyme E, but not to the enzyme substrate complex ES. Thus, increasing [S] can compete out I, Vmax is constant and in the Lineweaver-Burk plot all lines go through a common point on the y-axis (1/Vmax).
• In uncompetitive inhibition, it is the other way round. IN an LB-plot you get parallel lines for different [I]
• In non-competitive inhibition, the inhibitor can react with both E and ES. In the LB-plot you get lines intersecting left of the y-axis. The intersection is on the x-axis if the dissociation constant for the inhibitor of E and ES are identical, if they are different, the intersection will be above or below the x-axis, depending on which is larger

In all these cases, EIS has no enzymatic activity. Mixed inhibition occurs, if EIS can form EI + P. It looks like non-competitive in LB-plots, but the secondary plot (slope and x-intercept of the LB-plot against [I]) is non-linear.

Bisswanger's "Enzyme kinetics" is a very readable introduction.

Engelbert Buxbaum

Thank you for your response.

In my case: the enzyme, inhibitor and substrate are GTPase, GDP and GTP respectively.

As you mention in mixed inhibition: GTPase, GDP and GTP can form GDP (but with lower extent as in comparison to normal case).

This inhibition increases the Km and decreases Vmax and Kcat.

GDP is more pronounced to be as competetive inhibitor. But in this case the kinetic parameters are in accordance with the product inhibition (a type of mixed inhibiton).

Is it possible?

Adam B Shapiro

Thank you so much.

I dont have the book for now But I am looking for it.

In the example you have mentioned the inhibition by product P.

In my case I have added different concentratins of P as an inhibitor and measured the kintetic parameters of inhibiton by measured the initial rate of GTP hydrolysis.

If the product binds only to the substrate binding (= product release) site, you get competitive inhibition, as the enzyme can bind only either substrate or product. If the product can bind to an additional site, you can get a PES-complex, which would result in either non-competitive or mixed inhibition, depending on the enzymatic activity of that ternary complex. Remember also that you have two products: GDP and Pi. In addition, you have to distinguish between those and their complexes with Mg2+, which may behave differently (for example, in Mdr1 substrate inhibition by ATP is due to free ATP, but not the Mg-ATP complex, doi:10.1046/j.1432-1327.1999.00643.x).

Engelbert Buxbaum

Yes the product binds to the other site.

Thank you so much for this valueable suggestion. I will definitely check this.

Sourav Roy

What do you mean. Do you mean a physical interpretation of phenomena or a mathematical description. Start by specifying the stoichiometric patterns,

Mirosław Grzesik : I don't understand your question. A mathematical description should follow from a physical interpretation of a phenomenon.

The books cited give a good introduction to enzyme kinetics, perhaps somewhat easier to read is my own in ISBN 978-3-319-19919-1.

Dear Engelbert Buxbaum, The question was directed to the author of the question, and not to you. For me the question was not clear and therefore I asked for a further explanation. However, it just so happens that I know perfectly well how to describe an enzymatic reaction with the accompanying mixed inhibition and product inhibition - and more. I would like to add that the times when graphical methods were used are gone. Best regards,