Michaelis menten km and vmax relationship

Steady states and the Michaelis Menten equation (video) | Khan Academy

michaelis menten km and vmax relationship

How to read enzyme kinetics graphs (and how they're made). Km and Vmax. If you want a more detailed look at the Michaelis-Menten equation and the model. In biochemistry, Michaelis–Menten kinetics is one of the best-known models of enzyme kinetics. It is named after German biochemist Leonor Michaelis and Canadian physician Maud Menten. The model takes the form of an equation describing the rate of enzymatic "Km & Vmax". jingle-bells.info Retrieved. The Michaelis-Menten equation arises from the general equation for an enzymatic Vmax is equal to the product of the catalyst rate constant (kcat) and the.

Steady states and the Michaelis Menten equation

Vmax is the maximal rate of the reaction. Km is the Michaelis-Menten constant which shows the concentration of the substrate when the reaction velocity is equal to one half of the maximal velocity for the reaction.

It can also be thought of as a measure of how well a substrate complexes with a given enzyme, otherwise known as its binding affinity.

michaelis menten km and vmax relationship

An equation with a low Km value indicates a large binding affinity, as the reaction will approach Vmax more rapidly. An equation with a high Km indicates that the enzyme does not bind as efficiently with the substrate, and Vmax will only be reached if the substrate concentration is high enough to saturate the enzyme.

Michaelis–Menten kinetics

As the concentration of substrates increases at constant enzyme concentration, the active sites on the protein will be occupied as the reaction is proceeding. When all the active sites have been occupied, the reaction is complete, which means that the enzyme is at its maximum capacity and increasing the concentration of substrate will not increase the rate of turnover.

Here is an analogy which helps to understand this concept easier.

michaelis menten km and vmax relationship

Vmax is equal to the product of the catalyst rate constant kcat and the concentration of the enzyme. Kcat is equal to K2, and it measures the number of substrate molecules "turned over" by enzyme per second. The reciprocal of Kcat is then the time required by an enzyme to "turn over" a substrate molecule. The higher the Kcat is, the more substrates get turned over in one second. Km is the concentration of substrates when the reaction reaches half of Vmax.

A small Km indicates high affinity since it means the reaction can reach half of Vmax in a small number of substrate concentration. This small Km will approach Vmax more quickly than high Km value.

michaelis menten km and vmax relationship

It could just as easily dissociate back to an enzyme and a substrate molecule. I'll call these reverse reactions minus one and minus two. If we look at that in terms of our rates we can say that the rate of formation of ES would be the sum of rate one and rate minus two since both of these reactions lead to ES and the rate of loss of ES is equal to the sum of rates minus one and two since both of these lead away from ES. Now I also remember that products very rarely go back to reactants since these reactions are usually thermodynamically stable.

Rate minus two is going to be so small in comparison to rate one that we can really just cross it out. Which means that we can swap out that second double headed for a single headed arrow.

Using this information let's do some math.

Enzyme kinetics vmax and km

Now I'm going to be deriving a new equation. This can get a bit confusing so don't worry if you have a little trouble with this. Just rewind the video and try watching it a couple more times if you need to.

I'll start up by drawing the same sequence I did before with the three different reactions, and I'll also write out that steady-state equation I mentioned before where we have rates forming ES equal to rates taking away ES.

Now first thing I'll do is swap out those rate values for their rate constants times the reactants for those reactions. Rate one will be equal to K one times E times S and so on for the other two. Next I'll introduce a new idea and say that the total amount of enzyme available which we'll call ET or E total is equal to the free enzyme E plus the enzyme bound to substrate or ES.

Using this equation I'm going to rewrite the E on the left side of our equation as the total E minus the ES which would be equal to the E we had there before. On the right side of the equation I just factored out the common term ES. Next I'm just going to expand the left side of the equation so take a moment to look at that. Now what I'm going to do is I'm going to divide both sides of the equation by K one. K one will disappear on our left side and on our right side I've put K one in with all the other rate constants.

Now since all these rate constant are constant values I'm going to combine them in this expression of K minus one plus K two over K one into a new term KM which I'm going to talk a little bit more about later. In this next line I've done two things.

  • Basics of enzyme kinetics graphs
  • Structural Biochemistry/Enzyme/Michaelis and Menten Equation

First I've thrown in that KM value that I just mentioned, but I've also added ES times S to both sides of the equations and thus moved it from the left side to the right. In the next line I've done two things. First I switched the left sides and right sides of the equation just to keep things clear, but I've also factored out the common term ES on our new left side.

Then what I'm gonna do is I'm gonna divide both sides of the equation by KM plus S so I can move that term to the right side.

Michaelis–Menten kinetics - Wikipedia

I'll make some more room over here and now what I'm gonna do is remind you that the speed of our whole process which I'll call Vo is equal to the rate of formation of our product which we called rate two before which is also equal to K two times ES. Now using our equation over here I'm gonna multiply both sides of the equation by K two. Here's where it gets really tricky. Remember that if we're if at our max speed so our reaction speed Vo is equal to Vmax which happens when our substrate concentration is really high, then our total enzyme concentration is going to be equal to ES since all of our enzyme is saturated by substrate and there won't be any free enzyme left.

I'll make some room here and then sub in K two ES for Vo and K two E total for Vmax and then we finally get to our end equation which is called the Michaelis-Menten Equation and is super important when we talk about enzyme kinetics. Let's take a few steps back and talk about the Michaelis constant.

First I'll write out the Michaelis-Menten equation and if you remember we created this new term which I called KM, but we never really talked about what it meant.