Biochemistry PDF

31 Dua’ Al-Shrouf Leen Hawamdeh Nafez Abutarboush Kinetics of enzymatic reactions ➢ In biochemistry we consider kinetics the science that studies rates of chemical reactions which means studying the pathway between the reactants and products ➢ An example is the reaction (A —> P), The velocity, v,...

31 Dua’ Al-Shrouf Leen Hawamdeh Nafez Abutarboush Kinetics of enzymatic reactions ➢ In biochemistry we consider kinetics the science that studies rates of chemical reactions which means studying the pathway between the reactants and products ➢ An example is the reaction (A —> P), The velocity, v, or rate, of the reaction A —> P is the amount of P (product) formed or the amount of A (reactant) consumed per unit time, t. That is: ➢ The rate of the reactants consumption (the change in their concentrations/time) =the rate of products formation =V ➢ At a fixed concentration of the enzyme, V is almost linearly proportional to [S], when it’s a small value ➢ However, V is nearly independent of the [S] when it has a large value ✓ Enzyme kinetics ➢ Enzymatic reactions may either have a simple behavior or complex (allosteric) behavior ➢ Simple behavior of enzymes: as the concentration of the substrate rises, the velocity rises until it reaches a limit ➢ Thus, enzyme-catalyzed reactions have hyperbolic (saturation) plots ➢ The maximal rate, Vmax, is achieved when the catalytic sites on the enzyme are saturated with substrate ➢ Vmax reveals the turnover number of an enzyme ➢ Turnover number:The number of substrate molecules converted into product by an enzyme molecule in a unit of time when the enzyme is fully saturated with substrate ➢ At Vmax, the reaction is in zero-order rate since the substrate has no influence on the rate of the reaction ➢ The limiting factor here is the number of active sites and according to the active sites 1) if you have more active sites compared to the substrate then you’re talking about a linear relationship 2) if the substrate molecules are equal to the number of active sites or slightly higher we’re going to see a curve 3) if the number of substrate molecules is much higher than the active sites then the velocity won’t be affected ➢ We can see the concentration of the substrate is on the x axis while the velocity is on the y axis (but we don’t have any numbers) ➢ If you want to measure the velocity when [S]=5 nanomolar, if it’s in the linear range then you can easily measure it ➢ But when [S]=15 for example, you have to integrate the area under the curve, you do this when there is an increasing curvature, you have to measure the area under the curve and divide it into periods ➢ Accordingly in any reaction we have a substrate, a certain period of time and a resulting product ➢ This reaction is moving in a constant behavior, which means there is a constant controlling it ➢ So instead of measuring the velocity by the integration of the area under the curve, you can multiply the [S](substrate) that we want to calculate the velocity at by the rate constant (how the rate of a reaction depends on the reactant concentration) of that reaction ✓ Reaction Rate Law ➢ The rate is a term of change over time ➢ The rate will be proportional to the conc. of the reactants ➢ It is the mathematical relationship between reaction rate ➢ and concentration of reactant(s) ➢ For the reaction (A + B—>P), the rate law is ➢ From this expression, the rate is proportional to the concentration of A, and k is the rate constant ➢ It’s universally applied that if you want the velocity at a specific concentration multiply this [S] by the rate constant ➢ But if we have two substrates, D and E for example,then V=k[D][E] ➢ So in general, V is equal to the concentrations of the substrates multiplied by each other and by the rate constant ✓ The order of the reaction and the rate constant ➢ A multistep reaction can go no faster than the slowest step v = k(A)n1(B)n2(C)n3 ➢ k is the rate constant: the higher the activation energy (energy barrier), the smaller the value of k ➢ (n1+n2+n3) is the overall order of the reaction ➢ In these two pictures we can see the substrate concentration on the x axis and the initial V on the y axis ➢ Usually as you’re increasing the substrate concentration you’re increasing the chance of binding it to the active site (probability of binding), accordingly you’re increasing the actual binding, so you’re increasing the reactions and as a result the velocity (so increasing the substrate concentration leads to increasing the velocity), which means it’s a linear relationship like in the second graph, if the enzyme isn’t limiting it, which is logical ➢ The second graph is called the first order reaction which means that the velocity here depends only on one substrate (variable) ➢ The non logical thing is when we’re increasing the substrate without effecting the velocity, which means that the velocity is independent of the substrate concentration. But this makes sense and actually happens when the substrate concentration is very high, like in the first graph ➢ The first graph is called the zero order reaction which means that the velocity of the reaction doesn’t depend on any substrate concentration ➢ If you have one variable in the equation you can easily solve it but having two variables will make it harder, same thing here if you have two materials (substrates) that define the velocity of the reaction ➢ Having two substrates that affect the velocity of the reaction means that the reaction is a second order reaction and so on, but we won’t be able to solve this type of reactions ➢ For example, the glucokinase enzyme that transfers a phosphate group from the (ATP) to the (glucose) to produce (glucose 6-phosphate and ADP), has an active site that actually binds two substrates Any transferase will bind two (glucose and ATP) ➢ So the velocity of the glucosekinase enzyme substrates and produce two depends on the concentrations of both substrates substances (variables) ➢ In order to determine the velocity, we have to make it independent of one of the substrates and find the velocity with regard to the second one ➢ We can make it independent of the ATP for example by increasing the ATP concentration to a very high level ➢ After making the glucokinase independent of one substrate we start adding glucose gradually to find the effect of the glucose concentration on the velocity of the reaction(how many glucose 6- phosphate molecules are being produced) ➢ Then in a different experiment we do the opposite, we make the enzyme independent of the glucose and see the effect of the ATP concentration on its velocity ➢ So, because the glucokinase enzyme has a second order reaction that cannot be solved, we solve it by making one of the reactions a zero order reaction (making it independent of this substrate) and the other one a first order reaction (the velocity depends only on it) (we call it a pseudo-first order reaction) ➢ Dimensions of k ➢ In the zero order reaction the velocity is independent on the substrate so V=k, which the unit of the constant here should equal the unit of the velocity which is ( conc. time-1) ➢ While in the first order reaction V=k[A], so the unit of k should equals (unit of V/unit of [A]= conc.time-1/conc= time-1), which means reactions per time (second,minute,hour…) ➢ So if the constant of a first order reaction= 1000 /sec that means there’s a 1000 reaction happening per second ✓ Expressions of enzyme kinetic reactions “Steady State Assumption” ➢ Michaelis and Menten are two scientists that tested a lot of enzymes such as glucokinase, and when they drew the graphs of each substrate (substrate concentration on the x axis and the velocity on the y axis regardless of the used concentrations), they noticed that both graphs (of glucose and ATP) have the same hyperbolic shaped plot ➢ A group of these enzymes showed the same shape of this hyperbolic plot, so they concluded that these enzymes are catalyzing their reactions in the same manner ➢ Another group of enzymes resulted in a sigmoidal graph (S shaped graph) ➢ Going back to the first group of enzymes, they said that if all of these enzymes adapt, belong and adopt the same plot that means there’s a principle that control their catalysis for them to behave in the same manner ( a repetitive behavior) ➢ They decided to find this shared principle/ law that controlled these enzymes and we need laws 1) To save money, effort and time because if you find a shared principle between these enzymes then you won’t have to experiment each and every one of them 2) Dealing with numbers is easier because you can compare them to each other ➢ In science to come up with a law you have to assume things such as • neglecting very small numbers in addition and subtraction because they won’t have a big effect on the resulting values but you don’t neglect them in division and multiplication because they have a great effect (100+0.5=100.5 but 100/0.5=200) E:enzyme S:substrate P:product Both are decreasing the [ES] ➢ Each reaction has a constant so • The formation of ES from E and S has a constant called k1 • The degradation of ES to E and S has a constant called k-1 • The formation of P and E from ES has a constant called k2 • And the reversible reaction of forming ES from E and P has a constant k-2 ➢ So, they assumed two things 1)That the reaction is irreversible which means once you formed the product you cannot reform ES, so we don’t have the k-2 constant: 2)Steady State Assumption, the [ES] is in a steady state which means it doesn’t change throughout the reaction, so the rate of the change in the concentration (dES) /time (dt)=zero which means The rate of ES formation= the rate of ES degradation V of formation=k1 [E][S], V of degradation=k-1[ES]+ k2 [ES] Vd=ES ((k-1) + k2)—> ES= v/((k-1) + k2) So k1 [E][S]= [ES ]((k-1) + k2) ➢ 1) 2) ➢ Any enzyme in a solution has two forms The bound enzyme (ES) The free enzyme (E ) So the total enzyme concentration (Et)= [E] + [ES] So know k1 ([Et]-[ES])[S]= [ES ]((k-1) + k2) k1[Et] [S]- k1[ES] [S]= [ES ]((k-1) + k2) k1[Et] [S]= [ES ]((k-1) + k2)+ k1[ES] [S] k1[Et] [S]= [ES] ((k-1) + k2) + k1[S] [ES]= k1[Et] [S]/ ((k-1) + k2) + k1[S] to determine the rate of product formation (d[P]/dt = k2[ES]) v= d[P]/dt ➢ The Vmax is reached when all the enzymes are in the bound form, Et=ES, that’s why we can replace the (Et.k2) by Vmax in the equation (Vmax=k2 [ES]) ➢ Michaelis and Menten equation ➢ You can measure the velocity at any substrate using this equation, and the results will show a hyperbolic plot ➢ V max is the plateau and it’s the inherent property of the enzyme ✓ The Michaelis constant (Km) ➢ Km, called the Michaelis constant is ➢ In other words, Km is related to the rate of dissociation of substrate from the enzyme to the enzyme-substrate complex ➢ Km describes the affinity of enzyme for the substrate ➢ As we can see Km=[E][S]/[ES] and the unit of Km=conc.conc/conc=conc ➢ Km=(k-1)+k2/k1 (degradation/formation) so it indicates unbinding to binding this means it indicates affinity ➢ It indicates the affinity of the enzyme toward the substrate but it’s not an actual measurement for the affinity because in the actual affinity we’re measuring the chemistry between S and E, which means we’re measuring the rate constant of the dissociation of ES to E and S / rate constant of associating E and S to produce ES (the rate of unbinding S to the active site/ rate of binding S)= k-1/k1 ➢ But in the Km we’re also talking about the unbinding of the P which is controlled by the rate constant k2 that’s why it’s not an actual measurement because we have a third-rate constant that doesn’t express the unbinding of S but the production of P ➢ The actual affinity = k-1/k1 and that’s a different constant called kd (dissociation constant) ➢ The Km is easier to measure because it depends on the product formation so it’s easily measured using central machines while measuring kd is very expensive and need sophisticated machines, so we usually deal with the km ✓ Expressions of enzyme kinetic reactions (MichaelisMenten equation) ➢ A quantitative description of the relationship between the rate of an enzyme catalyzed reaction (V0) & substrate concentration [S] • The rate constant (Km) and maximal velocity (Vmax) ➢ Looking at the graph the km is going to be on the x axis because it’s a concentration ➢ At the beginning of the reaction (at a very low substrate concentration, [S] is much lower than the km) the initial velocity is measured by the following equation : V0= Vmax [S]/ km. (We neglect small values in addition and subtraction, that’s why we neglected the [S] in the denominator) ➢ Now we proved mathematically that the equation at the beginning of the reaction is linear (both Vmax and km are constants and we have one variable which is [S]) ➢ At very high substrate concentrations, we remove the km from the equation because the [S] is much higher than the km V0= Vmax [S]/[S]——> V0= Vmax ➢ This proves that the velocity here is independent of the substrate concentration ➢ When km=[S], then the following equation will be used: V0= Vmax [S]/2*[S] ——> V0= Vmax/2 ➢ Accordingly, km: is the substrate concentration at which V0 is half maximal (like the half life) • km is the concentration needed to reach 50% of the Vmax ➢ The lower the Km of an enzyme towards its substrate, the higher the affinity because the association constant is in the denominator ➢ When more than one substrate is involved? Each will have a unique Km & Vmax (if they are producing the same product they’ll have the same Vmax) ➢ To explain this we’re going to talk about the glucokinase enzyme that binds two substrates, as we know the affinity is controlled by the bonds between the chemical structures, this is why this enzyme active site will have two different values and likenesses of binding (affinities) ➢ But because both of these substrates are involved in producing (glucose 6phosphate) during the same reaction, both of them will have the same maximum velocity ➢ The maximum velocity here is the highest amount that can be produced of glucose 6-phosphate from ATP and glucose per second ➢ If the maximum velocity is 100 molecule/second and the concentration (km) at v=50 equals 2 what does this number indicates the affinity of ATP or that of glucose ➢ The results of having two affinities(substrates) are shown in the upper graph, each substrate has its own km so each has a plot that represents it, since both of them have the same Vmax they will eventually meet because we’re talking about the same reaction that produces the same product ➢ Each km will describe how the velocity is changing with respect to the substrate, and it describes what concentration of this substrate will give us half the Vmax ➢ That’s why the lower the Km the higher the affinity because you’ll need a lower concentration of this substrate to achieve half of the Vmax ➢ And this is what we talked about in the pseudo-first order reaction you focus on how the velocity is changing with respect to each substrate and they’ll have the same general hyperbolic plot but with different dimensions ➢ As we said the Vmax is an inherent property of the enzyme which means it’s the highest velocity that can be achieved by the enzyme and it’s a special property that cannot be changed, so if you’re at V0=0 or 50 you know that the Vmax of this enzyme is 100 for example ➢ this velocity can be increased only if you increased the number of active sites by increasing the enzyme concentration and you will go back to the linear phase (exam question) and then you’ll reach the new maximum velocity ➢ For example if the Vmax of an enzyme is 9 and then: 1)you decreased the enzyme concentration by 1/3 then the new Vmax is 3 2)you repeated the experiment by increasing the enzyme concentration up to 3 folds the new Vmax is 27 ➢ So the only thing that effect the velocity is the enzyme concentration ➢ Km values have a wide range. Mostly between (10-1 & 10-7 M) End of sheet 31