Relationship between vmax and kcat

relationship between vmax and kcat

However, the Vmax values are inconsistent without a certain pattern (i.e. no people have a tendency to favor the ratio kcat/KM, called the catalytic efficiency. Define the difference between Kcat and Vmax. The book states that Vmax is the maximum velocity (rate) of a reaction at high substrate concentration and that. i) The maximum rate that the enzyme can form product (Vmax) or kcat. .. 5) Summary of the differences between competitive and non-competitive inhibitors.

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.

Steady states and the Michaelis Menten equation

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.

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.

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.

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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.

relationship between vmax and kcat

Let's get to that. Now if you bear with me for a moment and pretend that KM is equal to our substrate concentration then we can sub in that value into our Michaelis-Menten equation which would put two S on the bottom, the sum of S plus S and then the S will cancel out and will be left with Vmax over two. What this means that KM which we call the Michaelis constant is defined as the concentration of substrate at which our reaction speed is half of the Vmax.

If we look at that on a graph from before you'd see that KM is a substrate concentration specific to our circumstances. Where our rate is at half of its max and the lower our KM, the better our enzyme is at working when substrate concentrations are small.

We can use this KM term to quantify an enzyme's ability to catalyze reactions which we call Catalytic Efficiency. I'll rewrite the Michaelis-Menten Equation. Since it's a concentration it will be in units of molar or moles per liter. Now I'm going to throw a new term at you called Kcat which is equal to the maximum speed of a reaction divided by the total enzyme available.

We call this the enzyme's turnover number. All these term is, is how many substrates and enzyme can turn into product in one second at its maximum speed. We measure it in units of seconds minus one or per second as in reactions per second.

We can define an enzyme's catalytic efficiency as a combination of KM and Kcat, and we do this by saying it's equal to Kcat over KM. This small Km will approach Vmax more quickly than high Km value. The enzyme efficiency can be increased as Kcat has high turnover and a small number of Km.

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

Taking the reciprocal of both side of the Michaelis-Menten equation gives: To determined the values of KM and Vmax. The double-reciprocal of Michaels-Menten equation could be used. Lineweaver-Burk graphs are particularly useful for analyzing how enzyme kinematics change in the presence of inhibitors, competitive, non-competitive, or a mixture of the two.

relationship between vmax and kcat

There are three reversible inhibitors: They can be plotted on double reciprocal plot. Competitive inhibitors are molecules that look like substrates and they bind to active site and slow down the reactions. Therefore, competitive inhibitors increase Km value decrease affinity, less chance the substrates can go to active siteand Vmax stays the same.

Uncompetitive inhibitors can bind close to the active site but don't occupy the active site. As a result, uncompetitive inhibitors lower Km increase affinity and lower Vmax. Non-competitive inhibitors are not bind to the active site but somewhere on that enzyme which changes its activity. It has the same Km but lower Vmax to those with no inhibitors. Km value is numerically equal to the substrate concentration at which the half of the enzyme molecules are associated with substrate.