so the next type of enzyme Inhibitors we are going to be looking at is the non-competitive Inhibitors non-competitive Inhibitors are molecules that do not compete with the substrates to bind with the active side instead of binding with the active side they will actually attach to something known as the allosteric site of the enzyme which is the site that is not the active site I'm putting it over here and the one where I've highlighted in purple is the allosteric site right there you don't have to know the term allosteric site but it's good to just understand that such parts of the enzymes also exist as well so the substrate is complementary to the active side and the non-competitive inhibitor which is this small little purplish thing that I'm drawing there look at the shape of the non-compact competitive inhibitor is it able to bind to the active site No in fact it doesn't actually need to do so so what happens then is it will actually bind to the allosteric site or the site other than the active site now what's the big deal if it binds with the allosteric site you see when it attaches to the allosteric site of the enzyme it will actually disrupt hydrogen bonds and weak Hy neophobic interactions within the enzyme you see these Bonds were supposed to maintain the 3D structure of the enzymes together with ionic bonds and dulfi bridges but when these two bonds are disrupted by the inhibitor what will actually happen then is the shape of the enzyme is altered now look what happens when the 3D Shape of the enzyme is changed look at what happens to the active side the substate is no longer able to bind to the active side and therefore no es complexes can be formed and in this case over here the enzyme activity is reduced or prevented that is what is known as a non-competitive inhibitor it will bind to the allosteric site or the site other than the active site Cause A disruption of hydrogen bonds and hydrophobic interactions in the enzymes change the 3D structure of the enzymes preventing the substrates from binding with the active site of the enzyme so no es complexes will form so in this case over here it's quite straightforward and for competitive inhibitor it binds with the active side but for non-competitive inhibitor it binds to the allosteric site or the site other than the active site it is important to know that in the exam you do not have to use the word allosteric site you can just say that the the non-competitive inhibitor binds to a site which is not the active site the inhibition is also reversible reversible in the sense where the inhibitor will bind to the the non-competitive inhibitor binds to the enzyme changes its shape but eventually it can also detach and when the non-competitive inhibitor detaches from the enzyme the shape of the enzyme returns to normal so this is what is meant by reversible non-competitive Inhibitors so the next thing we want to know is Will increasing the substrate concentration affect the non-competitive Inhibitors if you remember in the previous video when you increase the concentration of substrates competitive inhibitor effects will reduce because the competitive Inhibitors and the substrates are fighting for the same size site so will the same effect happen with non-competitive Inhibitors let's take a look so I have the enzyme I am I'm doing I'm conducting two experiments over here you can see this orange color thing those are enzymes the green color triangle is a substrate and I'm going to represent the non-competitive inhibitor as a blue dot so the experiment on the left is done without Inhibitors and the experiment on the right is done with the non-competitive Inhibitors now let's do a quick experiment for this so without inhibitor if I have 1% substrate what will be the initial rate of reaction the initial rate of reaction is one if I have 2% substrates obviously the initial rate of reaction is two because they can bind to the enzyme if I have 3% substrates then the initial rate of reaction is three all right and if I have 4% substrates the initial rate of reaction is four as usual and if I were to have 5% substrate what is the initial rate of reaction the initial rate of reaction is still four because of the limited concentration of enzymes and this value is referred to as the value of V Max because you cannot make the initial rate of reaction any higher due to the limited number or concentration of enzymes that you have in this case so that's just a quick experiment now for the non-competitive Inhibitors however I'm going to put two non-competitive Inhibitors the two blue dots I'm going to label it as NCI please in the exam do not say NCI it is noncompetitive Inhibitors I'm just a little lazy okay so cut me some slack um and in this situation here if I have 1% substrate in this case here what will actually happen is the substrate will then bind to the enzyme quite easily if I have 2% substrates it can also bind to the enzyme no problems everything so far so good but here's where it becomes interesting at 3% notice something very interesting earlier actually the non-competitive inhibitor had bound to the allosteric side of those two enzymes which have represented in the blue arrow so what will actually happen is those two enzymes which have highlighted in pink now are no longer occupied so the third substrate particle will not be able to attach to the enzyme right because the 3D structure of the enzyme is changed so only the enzymes highlighted in yellow can be used in this reaction so in this case what has actually happened is the the initial rate of reaction is just two so what has actually happened here is the Vmax has reduced even when I try to put 4% substrates the substrate is no longer able to bind to the enzyme so you see what has actually happened to the Vmax over here the Vmax over here has significantly lowered due to the non-competitive Inhibitors so the presence of more substrates will not be able to fight of the non-competitive Inhibitors at all you will not be able to reach the original Vmax value where I'm just circling over here you will not be able to reach that Vmax value at all no matter how many substrates you add to the reaction so you see here I'm adding as much substrate as possible 10% substrates will they be able to have a a higher chance of binding to the enzyme no because the non-competitive inhibitors can still just easily bind to the allosteric site and when they bind to the allosteric site the initial rate of reaction is only two so this is the effect of non-competitive Inhibitors on the Vmax value on the initial rate of reaction and the Vmax value non-competitive Inhibitors will lower the V-Max value so you see with competitive Inhibitors this is from the previous video right 5% substrate on the left side was able to reach Vmax but with competitive Inhibitors you needed 10% substrate to reach the same Vmax value so if you wanted to reduce the chances of the Inhibitors binding to the enzyme you just had to increase the substrate concentration but with non-competitive Inhibitors no matter how many substrates you put the Vmax value will not be achieved at that high or original value as it was without the Inhibitors because the substrate concentration will not have any effect on the non-competitive Inhibitors at all so as you can see here this is a graph that is showing you the difference between the competitive Inhibitors and also non-competitive Inhibitors with competitive Inhibitors you are still able to reach the original Vmax value however it requires more subst trates but with non-competitive Inhibitors however the V-Max value is always lowered no matter what so then comes the more important question are non-competitive Inhibitors good if you remember when we were talking about competitive Inhibitors I told you that there are some benefits of competitive inhibition as well when I was mentioning the antifreeze and ethanol and non-competitive inhibition it might look dangerous as as well you might think oh it's also reducing the enzyme activity so it must be dangerous but not necessarily as well in fact non-competitive inhibition does happen naturally in our body under normal circumstances so you might be thinking oh okay this is a bit of an interesting snippet now one thing I want you to understand is a lot of chemical reactions in our cells are actually chain reactions you don't have to memorize this fact but I just want to show it to you for example you might have molecule a which is a substrate to convert molecule a into molecule D which is an end product you cannot just instantly make it from a to d it has to go through a series of steps a will become molecule B molecule B is converted to molecule C and molecule c will then be converted into D which is the end product right and each Arrow will represent a chemical reaction catalyzed by a specific enzyme for example A to B is catalized by enzyme 1 B2 C is enzyme 2 which is slightly different and c2d is enzyme 3 which are slightly different as well and I'm going to represent it over here so as you can see a binds to the active side of enzyme 1 gets converted into b b binds into active side of enzyme 2 gets converted into C and C binds to active side of enzyme 3 and gets converted into D and I'm just going to draw molecule d as like this cloudy yellow things now while it is important to form end products too much end products may be harmful in the cell I don't need you to memorize this but I want you to understand this for example let's say the end product is hydrogen ions your cells may need hydrogen ions but too much hydrogen ions can be dangerous because the presence of excess hydrogen ions may cause the cytoplasm to become too acidic and the cell might die so you see it's one of those balancing effect where you must have some of the hydrogen ion but not too much so how does the cell ensure that not too much hydrogen ions are being produced so how does the cell regulate the production of the end product so here is where something very beautiful happens if you notice I'm just looking at enzyme one I'm just redrawing enzyme one and magnifying it notice the active side active side looks pretty normal but I'm also going to draw that allosteric side it looks like that very squiggly shape I'm also going to redraw molecule d as well okay so what do you notice over there by the way you notice something very interesting molecule D which is an end product can bind to the allosteric site of enzyme number one now is this good or is this bad you might be thinking oh this is bad because it is going to inhibit the enzyme but in this case it's fantastic that it does so because when it inhibits enzyme number one look at the shape of the active site is no longer able to be complementary to molecule a anymore so will anym molecule a get converted to molecule B no the reaction will stop and when the reaction stops there will be less molecule B there will be less molecule C and eventually no more molecule D's are being produced so the production of end product in this case is halted in this case it is good to stop the production of the end products because like I told you earlier we don't want too much end products we want some but not too much of it just like everything in life sadly okay like I love oh my God I love ice cream I just do but I I just can't have too much of it okay that's the sad fact now anyway so coming back to so the end product will inhibit the enzymes none competitively and this and this and this is good for two reasons number one you will prevent the end product concentration from being too much but number two also you are not using up all the substrate so the substrates can be kept for the future when the cell needs to use it only when necessary this is an example of something known as end product inhibition so let's try it again now in this example over here your body needs two types of amino acids your body needs tronin or th and isoline which is ISO now here's something very interesting as well well even though your body needs both these amino acids if you only were to eat tronin your liver can convert some of the thionin into isol leucin so your body will have a good amount of thionin and isol leucin as well when you look at this chemical reaction over here now I want you to see this chemical reaction do you want all the threonine to be converted into isoline the answer is no because I told you your body needs both your body needs threein for protein synthesis your body also needs isol leucin for protein synthesis so you must not use up all the tronin and convert it into isoline you must have a good uh you must have a good amount of both amino acids so how does the body number one make enough isoline but also keep thionin for future use here's where non-competitive inhibition helps so I'm going to show you a situation over here so I have the isoline so some of the isoline what exactly happens is it binds with the active site and then it gets converted into the end product which is isoline and the end products over here notice they will bind to the allosteric site of the enzyme what happens to the shape of the enzyme the 3D structure of the enzyme is altered and when the 3D structure of the enzyme is altered will any more of the substrates converted into the end products no it won't you will actually keep them for the future so this is good for two reasons you prevent the concentration of end products from increasing drastically and you also maintain the concentration of the substrates for future use this is a beautiful example of end product inhibition so end product inhibition is what happens when the end product will noncompetitively bind to an enzyme deliberately I might add we want this to happen so that it will stop the chemical reactions from taking place the reason is because we want to prevent the concentration of end products from increasing too much and we also want to maintain the concentration of substrates for future use this is also very important to know for the exam