Understanding the Michaelis-Menten Equation and Its Applications in Enzyme Kinetics

so the infamous equation that is continually used over and over in enzyme kinetics is the miches meal equation and

this is that equation and what this equation does is it builds a relationship between the rate at which

the enzyme operates on a certain substrate and the concentration of that substrate in that surrounding

environment so if we plot the following equation on the XY AIS we get the following red curve and notice an

important point about the red curve the red curve initially increases very rapidly so essentially linearly and then

the slope begins to decrease over time so the slope begins to level off and it approaches a maximum value known as the

Vmax it approaches that value asically and that's a very important mathematical term so what do we mean by a curve

approaching a number ASM totically what that means is every single time we increase the substrate concentration the

red curve gets closer and closer and closer to that Vmax value but what it means so proach asically is it never

quite reaches it so the red curve is never going to actually reach that Vmax quantity now that's problematic because

we want to be able to calculate what Vmax actually is and in fact if we can calculate what Vmax is then we cannot

calculate what the km value is the michaus constant so remember the michis constant basically describes the

Affinity of that substrate for the active side and the km value is the concentration of that substrate at which

that enzyme operates at a velocity that is exact L half of that Vmax value so if we know what this y coordinate is Vmax

divided by two then we can calculate exactly what the km value is we simply draw a horizontal line until it hits the

curve and then we draw a vertical line and that gives us the x coordinate the km value so if we can't calculate what

Vmax is because the slope never actually touches the Vmax value then we can't calculate what Vmax / 2 is and therefore

we cannot calculate what km is so back in the day before computers we really had no way of using this graph to

actually calculate what Vmax and km is so back in the day instead of using this michaus me an equation in this form we

changed the equation into a slightly different form and what we did was we took the reciprocal of the left side and

the right side of that equation now remember in mathematics if we have an algebraic equation and we change the

right side and the left side in the same exact way that uh that does not actually change the information that the equation

provides us with it doesn't change the information but what it does is it Chang the way it changes the way that that

information is actually displayed so the information from this particular equation basically gives us an ASM totic

curve but if we take the reciprocal of both sides we get a linear equation we get a straight line and now the same

pieces of information the km value the V Max and so forth is given to us in the form of a straight line and a straight

line is much more useful than this ASM totic curve because a straight line can be used to actually calculate what km is

and what V Max is so instead of using the above michaa Menan equation that gives us an ASM totic red curve we can

take the reciprocal of the left and the right side to obtain a double reciprocal curve also known as the line weav bur

curve so if we take this equation we take the reciprocal of V KN we get 1 over V KN we we take the reciprocal of

this so the top is Vmax multiplied by this and the bottom is simply km plus the concentration of s so if we

reciprocate this simply goes to the bottom so the bottom becomes Vmax multiplied by the substrate

concentration and the top becomes km plus the concentration of s and now we essentially rearrange the equation we

distribute our denominator to this quantity and this quantity and we get the following result and

notice the concentration of the S cancels out on this second term on the right side of the equation and this

equation has the same exact form as a straight line so this left side 1 over V KN is the Y AIS so here's the Y AIS the

x axis this is the x value of that line so the x coordinate is 1 / the concentration of s the slope of the

straight line is km / V Max so the slope of the red line is KM ID by by Vmax and the Y intercept is 1 / V Max so what

this really tells us is the point where the curve intersects the Y AIS is the quantity 1 / V Max so if we carry out

our experiment we collect the data points and then we plot the double reciprocal curve if we find what this y

value is so let's say it's 20 then we know that is equal to 1 / Vmax and if we solve for VMax we get that Vmax is equal

to 0.05 so that allows us to calculate exactly what the Vmax quantity is unlike

in this case where we had no way of actually determining what that Vmax is because the Curve Cur never actually

touches that Vmax quantity now what about the km well notice that the red curve also touches intersect the x axis

and if we basically let the left side of the equation equal to zero we can solve for what the xaxis is the xaxis is equal

to NE of 1 / km and so if we find exact what the x value is again let's suppose I don't know it's -2 then we said -2

equal to -1 / km and we find that km is equal to 0.5 and in this manner we can calculate exactly what the Vmax value is

and what the km value is in addition if we know any two points on the curve we can calculate what the slope is the

slope is simply km ided by V Max so this double reciprocal plot is a very useful way to basically determine exactly what

these quantities is and these qu or what these quantities are and these quantities can be used to basically

study the way that enzymes increase the rates of different types of chemical reactions now in addition to this

usefulness there's one uh one other important application of the double reciprocal plot we can also actually use

the double reciprocal plot to basically differentiate between the three different types of reversible Inhibitors

remember we have competitive Inhibitors we have uncompetitive and we have non-competitive and we can use the

double reciprocal plot as we'll see in just a moment to basically differentiate between which type of inhibitor is

actually present in our mixture so let's begin by disc discing how a competitive inhibitor will actually affect this

double reciprocal curve so the red curve describes the curve in the absence of that competitive inhibitor and the

purple curve describes that curve in the presence of that inhibitor so notice what happens in the presence of the

inhibitor the slope is greater the Y intercept is the same and the x intercept is less negative it's closer

to that origin the question is why well let's recall how this type of competitive inhibitor actually affects

the kinetics of enzymes so remember a competitive inhibitor binds exactly into the same location the same active side

as the substrate does and because it binds reversibly we can simply increase the concentration of the substrate to

basically replace and kick out that inhibitor and so all ultimately the Vmax value is not changed and if the Vmax

isn't changed because 1 / by Vmax is that y coordinate value it's that location where the curve intersects the

Y AIS both of these curves will intersect the same exact coordinate value because this Vmax does not change

and so one over Vmax also will not change now is affected by a competitive inhibitor well the km value is affected

remember in the presence of a competitive inhibitor the Affinity of that substrate for the active side

decreases and so we have to increase the concentration of s to basically get all those active sides occupied with that

substrate and so km will actually increase now if km increases then because the slope of the line is km /

VMAX Vmax will not change km increases and so the ratio km / by Vmax also increases and that's precisely why the

slope of the purple one uh purple line is greater than the slope of that red line and finally because the km

increases this one ided uh 1 divided by km ratio will essentially decrease and so this x coordinate the x intercept

will be close closer to that origin and so that's exactly what we see in this particular case so if we basically have

some type of unknown inhibitor we take out the uh inhibitor then we plot the red curve and then we place that

inhibitor into our solution and we plot the purple curve if this is how the curve changes we know that it must be a

competitive inhibitor now what about uncompetitive well let's recall what well actually let's first look at the

following plots so again the red describes the absence and the purple describes the presence of that

uncompetitive inhibitor and notice what happens essentially we take the curve and we shift it upward and notice the

slope doesn't change because these lines are parallel what changes is the x coordinate value where the line

intersects the X AIS and the y-coordinate value the Y intercept now why does that actually

take place well recall what an uncompetitive inhibitor does an uncompetitive inhibitor binds onto that

side of the enzyme that is only created when the substrate actually binds onto that particular enzyme so what that does

is it decreases the total number of enzymes that are functional in the mixture and so it brings down the Vmax

value decreases the Vmax value now if the Vmax value is decreased then the ratio 1 / smaller Vmax value means we

have a y intercept that is higher up and that's exactly why the Y intercept is higher for the purple curve than that

red curve now what happens to the km well the km is also affected the km also decreases and that's because when the

inhibitor actually binds onto that enzyme substrate complex to create the enzyme substrate inhibitor complex it

essentially increases the Affinity of that substrate for that enzyme because once the inhibitor binds onto that

enzyme substrate complex that substrate cannot actually leave the active side because its Infinity is higher and if

the infinity is higher the km value is lower so km will also decrease now if km decreases the ratio 1 / by smaller km

value will become more negative it will become larger in the negative Direction and so that means the new km value the

new 1 / km will be farther to the left along the x axis and that's exactly what we see happening here now what about the

slope why does the slope actually not change well the slope is given to us by the ratio km / V Max and in this

particular case both km and Vmax decrease and in fact they decrease by the same exact amount for example if km

decreases by a half this also will be halfed and so what that means is the ratio does not actually change because

the proportion will remain the same and so the slope km / by Vmax will not change and these two lines will be

parallel with respect to one another and so so if we take the case when we have the inhibitor and then we don't have the

inhibitor and we find that the two lines are parallel they have different X and different y intercepts that must mean

the inhibitor is in fact an uncompetitive inhibitor and finally let's see how a non-competitive

inhibitor actually affects that line Weaver Burke curve so if we look at the following uh graph we see that the the

red curve again the absence of that inhibitor the purple curve the presence of that non-competitive inhibitor and

notice that the slopes are different the slope of the purple line is greater we see that the Y intercept is different

this is greater but we see that the x coordinate is actually the same the x intercept is the same and that's because

as we discussed in the non-competitive inhibition case uh that inhibitor non-competitive inhibitor binds onto the

enzyme regardless of whether or not the substrate is actually bound onto that enzyme so the inhibitor can bind onto

the individual enzyme or onto the enzyme substrate mixture and what that basically means is the Vmax will be

smaller and so if the Vmax is smaller the y coordinate the Y intercept will be greater as we see in this particular

case now even though a non-competitive inhibitor binds onto the enzyme and

changes the shape of the active side of the enzyme that substrate