# Km and vmax relationship tips

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Determination of Vmax and Km. Determination of Vmax and Km . Three ways in which the hyperbolic relationship between the initial rate of reaction and the. Skip the theory and go straight to: How to determine Km and Vmax. A simple chemical reaction with a single substrate shows a linear relationship between the rate A number of ways of re-arranging the Michaelis-Menten equation have been. What is the relationship between Km and Kmax to Substrate and Enzyme constant Km is the substrate concentration at which the reaction rate is half of Vmax.

First we'll talk about the Steady-State Assumption and what that means. Like I said before there are two steps to an enzyme's catalysis. Now when we use the term steady-state what we mean is that we're at a point where the concentration of ES or enzyme substrate complex is constant which means that the formation of ES is equal to the loss or dissociation of ES. Now notice that I've used equilibrium arrows between these steps and that was to show the idea that these reactions like any reaction can go forwards or backwards.

Our enzyme substrate complex doesn't have to form products. 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.

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.

## Steady states and the Michaelis Menten equation

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.

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

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. 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. 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. As soon as the catalytic site is empty, more substrate is available to bind and undergo reaction. The rate of formation of product now depends on the activity of the enzyme itself, and adding more substrate will not affect the rate of the reaction to any significant effect. The rate of reaction when the enzyme is saturated with substrate is the maximum rate of reaction, Vmax. The relationship between rate of reaction and concentration of substrate depends on the affinity of the enzyme for its substrate.

This is usually expressed as the Km Michaelis constant of the enzyme, an inverse measure of affinity. For practical purposes, Km is the concentration of substrate which permits the enzyme to achieve half Vmax. An enzyme with a high Km has a low affinity for its substrate, and requires a greater concentration of substrate to achieve Vmax.

An enzyme with a low Km relative to the physiological concentration of substrate, as shown above, is normally saturated with substrate, and will act at a more or less constant rate, regardless of variations in the concentration of substrate within the physiological range. An enzyme with a high Km relative to the physiological concentration of substrate, as shown above, is not normally saturated with substrate, and its activity will vary as the concentration of substrate varies, so that the rate of formation of product will depend on the availability of substrate.

### Determination of Vmax and Km

If two enzymes, in different pathways, compete for the same substrate, then knowing the values of Km and Vmax for both enzymes permits prediction of the metabolic fate of the substrate and the relative amount that will flow through each pathway under various conditions. In order to determine the amount of an enzyme present in a sample of tissue, it is obviously essential to ensure that the limiting factor is the activity of the enzyme itself, and not the amount of substrate available.

This means that the concentration of substrate must be high enough to ensure that the enzyme is acting at Vmax. In practice, it is usual to use a concentration of substrate about 10 - fold higher than the Km in order to determine the activity of an enzyme in a sample. If an enzyme is to be used to determine the concentration of substrate in a sample e.