Understanding Enzyme Inhibition: The Impact of Competitive, Uncompetitive, and Non-Competitive Inhibitors

previously we discussed the concept of enzyme inhibition and we said there are two categories of enzyme inhibitors we

have irreversible enzyme inhibitors that bind unset the enzymes and don't let go don't dissociate very easily and we also

have the reversible enzyme inhibitors that bind unto the enzymes but they can dissociate quite easily under specific

conditions now we also said we can subdivide reversible inhibitors into three types we have competitive

inhibitors we have uncompetitive and we have non-competitive reversible inhibitors and what I want to focus in

this lecture is how exactly do these three types of reversible inhibitors actually affect the kinetics of enzymes

how do they affect things like the turnover number the Michaelis constant and v-max the maximum velocity of that

enzyme so let's begin by focusing on competitive inhibition so this is the equation that describes competitive

inhibition so in the absence of an inhibitor that substrate is going to collide into and bind to the active site

of the enzyme forming the functional enzyme substrate complex and then that complex will catalyze and transform the

substrate into the product which will then dissociate and be released from that active site now in the absence of

an inhibitor the curve that describes the substrate concentration to the velocity is this black curve here and

notice that black curve eventually reaches a maximum velocity that's the point when all the active sites of all

the enzymes are filled by that substrate now in the presence of an inhibitor what happens is because the inhibitor

resembles that substrate it's going to bind to that same active site and once it bonds it forms this enzyme inhibitor

complex and because that inhibitor is found in that active site the substrate is not

found in that active side and so that enzyme is inhibited now the thing about competitive inhibition is because the

substrate binds to the same exact region as the inhibitor if we increase the concentration of that substrate there is

a greater likelihood that the substrate is going to collide into that active site and that can displace and replace

that inhibitor in the active site to form back this enzyme substrate complex and so all we have to do to basically

overcome competitive inhibitors is to increase the concentration of the substrate and what that ultimately means

is if we examine the red curve which ascribes the presence of the inhibitor if we increase the concentration of s if

we move along to the right side along the x-axis eventually the red curve is going to reach the same v-max value as

the black curve and what that means is in the presence of a competitive inhibitor that v-max does not actually

change and once again this is because that inhibitor binds to the same exact section as that substrate and so that

inhibitor can be overcome by increasing the concentration of that substrate so even though we have to increase the

concentration of that substrate eventually all those inhibitors in the active sites will be replaced with that

substrate and all the same active sites are going to be filled by that substrate in the inhibition case as in the absence

of the inhibitor and so the same v-max will actually be reached now let's move on to fact number two about competitive

inhibition and enzyme kinetics the turnover number K CAD does not actually change is not affected by a competitive

inhibitor inhibitor and that can be seen from the following equation so in our lecture in our discussion on the turn

over number we said that the turnover number basically describes the efficiency of that active site though so

it's basically the number of substrate molecules that can be transformed into the product molecules over some amount

of time per a single personal active site so per enzyme and because the turnover number is not changed what that

means is the efficiency of that active site in the presence of the inhibitor does not actually change and this can be

seen from this equation if v-max does not actually change and the total concentration of that enzyme that is

functional does not change then k-kat also does not change if remains constant and so because these two don't change

the turnover number also does not change so v-max doesn't change k-kat doesn't change but

what does change is the km the km value essentially increases and all that means is to reach the same rate in the

inhibition case as in the absence of the inhibitor to reach the same rate of activity all we have to do is increase

the substrate concentration to basically reach that same velocity of that enzyme and that means our K and value will

increase because remember the km value basically describes the V Max divided by two so when our mixture reaches a

concentration of substrate that is equal to km the velocity of that enzyme will be exactly midway between the v-max and

the zero point so V Max divided by 2 because v-max doesn't change V Max divided by 2 doesn't change and so if we

look at the corresponding y coordinate point for the at the corresponding x coordinate point for this y coordinate

in the case of no inhibitor present it's here in the case of the inhibitor present it's farther along that

axis and so the km is greater so competitive inhibitors increase the apparent K and value so since most of

the active sites are occupied by NIT by the inhibitor a larger amount of substrate needs to be present to

actually overcome and displace that inhibitor to reach that same enzyme rate and this means a higher amount of

substrate is actually needed to reach that rate of V Max divided by two now what this also means is because the

km increases the affinity of that substrate for the active site decreases and that's because now we have an

inhibitor that has a higher affinity for that active site and so we have to increase the number of substrate

molecules to increase the likelihood of collision with that active side to actually displace and replace that

inhibitor in the active site now let's move on to uncompetitive inhibition in this type of inhibition the only time

the inhibitor can bind on to that enzyme is when the substrate is bound to that enzyme so we have the substrate collides

and binds into the active site of the enzyme and once we form the enzyme substrate complex that creates a

conformational change that creates this brand-new Park in the allosteric site that the inhibitor can now bind to now

in the absence of the inhibitor the enzyme substrate complex will simply form the product and then the product

will dissociate but in the presence of the inhibitor that inhibitor will bind onto that brand-new allosteric site

found on the enzyme substrate complex and that will form the enzyme substrate inhibitor complex and once this complex

is formed no reaction will take place and that's because once the inhibitor binds on to

the enzyme substrate complex it will keep that substrate inside that active site and that active site will not

be able to catalyze the transformation of that substrate into that particular product now this leads us directly into

point number one uncompetitive inhibitors actually decrease the v-max value and that's because they decrease

the number of enzyme substrate complexes that are efficient that are functional that can actually convert the substrate

into that particular product so uncompetitive inhibitors bind to the enzyme substrate complex and decrease

the total number of functional enzymes and we can see how that affects v-max by looking at this particular equation so

if we rearrange this equation we get this equation here and what it basically tells us is the v-max is reached when

all the active sites are filled with that particular substrate and in this particular case because we decrease the

total number of functional enzymes we decrease this quantity because some of them are transformed into this quantity

we decrease the v-max value now what about the cake at what about the turn over number well the turnover number

basically describes the ability of that active side to actually transform the substrate molecules into the product

molecules per unit time now because when the end when the inhibitor is not balanced to that particular enzyme

substrate complex that active sites ability or efficiency to change that substrate to the product doesn't

actually change we see that the cake add value in uncompetitive inhibition also doesn't actually change it remains the

same so the cake add value in this equation remains unchanged but because this change is decreases the v-max also

decreases and finally what about what that KN Valley what happens to the michaelis constant well to answer that

question let's take a look at the following equation so when that inhibit so remember km the Michaelis constant

scribe's the affinity of that particular substrate to the active site if km increases the Finiti decreases if km

decreases the affinity increases so what happens in the presence of the inhibitor well when the inhibitor binds on to the

enzyme substrate complex what happens is once we form the enzyme substrate inhibitor complex that basically that

inhibitor by binding unto that complex it prevents that substrate from actually leaving that active site and what that

means is because that active site does not release that substrate the affinity of that substrate for the enzyme

increases and if the affinity of that enzyme and if the affinity of the substrate for the active site of the

enzyme increases when the inhibitor binds unto the complex the km value decreases and that's one way that we can

explain how the km actually decreases another way of explaining it is in the following manner so since the total

number of functional enzymes decreases and the v-max decreases that means we need to add a lower concentration of

that substrate to actually reach v-max / - and that's why the km decreases so if we compare no inhibitor to inhibitor

present we see that the v-max is lower in the presence of the inhibitor and the km value is closer it's more to the left

side along the x-axis so the apparent km decreases compared to this km so let's move on to the final one non-competitive

inhibition so in non-competitive inhibition on that enzyme there is a permanent

allosteric side batters present and what that means is that inhibitor can bind onto that enzyme regardless of whether

or not that substrate is actually bound onto that active site and so this is the equation that describes this process so

in the absence of the inhibitor the substrate is going to bind to the active side of the enzyme forming the enzyme

substrate complex and that will transform the substance of the product and then the product will be released

now in the presence of the inhibitor that inhibitor can either bind onto that individual enzyme forming that enzyme

inhibitor complex or a combined unto the enzyme substrate complex to form that particular enzyme substrate inhibitor

complex and notice that if we form the enzyme inhibitor complex valdes enchain j-- the likelihood that the substrate is

going to bind to that enzyme and that is important in in part 3 as we'll see in just a moment so let's begin with 1 the

v-max is lowered and the v-max is lowered for the same reason that the v-max is lowered in this uncompetitive

inhibition case so we simply decrease the number of active sites number of enzymes that are functional and if we

decrease that value we essentially decrease the v-max so since some number of inhibitors are balanced at the enzyme

at any given moment in time and that inhibits the functionality of the enzyme less functional enzymes will be present

and so v-max will decrease now let's move on to to the k-kat is lowered the cake at the turnover number is lowered

and what that basically means is the efficiency of that active site in transforming the substrate into that

particular product is decreased and this makes sense why well because as it turns out

when that enzyme binds on to that inhibitor that inhibitor changes the shape of the active side and so the

active site is essentially normal it's it's no longer complementary to that particular substrate and although the

substrate can bind to that active site just as likely as in the case of the absence of the inhibitor once the

substrate binds on to this enzyme inhibitor complex that fit will not be perfect and what that means is the

active sites ability to actually