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Current time:0:00Total duration:7:32

Steady states and the Michaelis Menten equation

Video transcript

so today we're going to talk about Michaelis Menten kinetics in a steady-state but first let's review the idea that enzymes make reactions go faster and that we can divide the enzymes catalysis into two steps first The Binding of enzyme to substrate and second the formation of products and each of these reactions has its own rate let's also review the idea that if we keep the concentration of enzyme constant then a really high substrate concentrations will hit the maximum speed for a reaction which we call v-max so first we'll talk about the steady state assumption and what that means so like I said before there are two steps to an enzymes 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 V s 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 in a substrate molecule so I'll call these reverse reactions minus 1 and minus 2 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 1 and rate minus 2 since both of these reactions lead to ES and the rate of loss of es is equal to the sum of rates minus 1 and 2 since both of these lead away from yes now also remember that products very rarely go back to reactants since these reactions are usually thermodynamically stable so rate minus 2 is going to be so small in comparison to rate 1 that we can really just cross it out which means that we can swap out that second double-headed arrow for a single headed arrow so 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 a couple more times if you need to so I'll start out by drawing the same sequence I did before with the three different reactions and also write out that steady state equation I mentioned before where we have rates forming es equal to rates taking away yes now the first thing I'll do is swap out those rate values where there rate constants times the reactants for those reactions so rate 1 will be equal to k1 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 e t or e total is equal to the free enzyme e plus the enzyme bound to substrate or ES and using this equation I'm going to rewrite the EE 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've 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 and now what I'm going to do is I'm going to divide both sides of the equation by k1 so k1 will disappear on our left side and on our right side I've put k1 in with all the other rate constants now since all these rate constants are constant values I'm going to combine them in this expression of K minus 1 plus k2 over k1 into a new term km which I'm going to talk a little bit more about later so 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 and then what I'm going to do is I'm going to divide both sides of the equation by km plus s so I can move that term to the right side so I'll make some more room over here and now what I'm going to 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 to before which is also equal to k2 times es so now using our equation over here I'm going to multiply both sides of the equation by k2 now here's where it gets really tricky remember that if word rmax speed so a reaction speed vo is equal to v-max which happens when our substrate concentration is really high that our total enzyme concentration is going to be equal to es since all of our enzyme is saturated by substrate and there will be any free enzyme left so two times et instead of times es would be equal to v-max instead of being equal to v-0 like you see at the top so I'll make some room here and then sub in k2 es 4 vo and k2 e total 4 v-max 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 so let's take a few steps back and talk about the Michaelis constant so 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 so 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 2's on the bottom the sum of s plus s and then the S will cancel out and we'll be left with v-max over 2 and what this means is that km which we call the Michaelis constant is defined as the concentration of substrate at which our reaction speed is half of the v-max so when vo is equal to one half of v-max and 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 and we can use this km term to quantify and enzymes ability to catalyze reactions which we call catalytic efficiency so I'll rewrite the Michaelis Menten equation remember we defined km as a substrate concentration where vo is 1/2 B max and since it's a concentration it will be in units of molar or moles per liter but now I'm going to throw a new term at you called k-kat which is equal to the maximum speed of a reaction divided by the total enzyme available and we call this the enzymes turnover number and all this term is is how many substrates an enzyme can turn into product in one second at its maximum speed and we measure it in units of seconds to this -1 or per second as in reactions per second so we can define an enzymes catalytic efficiency as a combination of km and k-kat and we do this by saying it's equal to k cat over I am so a higher k-kat or lower km would result in an increase in an enzymes catalytic efficiency and every different enzyme has a different catalytic efficiency in certain conditions and we can use this term to score enzymes on how good they are so we covered a lot of content in this video but the really crucial points to remember are first the idea of the steady state assumption that we make when looking at enzyme kinetics and this is where we assume that the es concentration is constant meaning that the formation and loss of es are equal second we derived the critically important Michaelis Menten equation which you should consider committing to memory and third we talked about how you can score how good an enzyme is at speeding up reactions by looking at that enzymes catalytic efficiency which is a combination of two new terms we learned about k-kat and km