Energy and enzymes | Biology | Science | Khan Academy
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The transition state is at the top of the energy "hill" in the diagram above. Active sites and substrate specificity To catalyze a reaction, an enzyme will grab on bind to one or more reactant molecules. These molecules are the enzyme's substrates. In some reactions, one substrate is broken down into multiple products. In others, two substrates come together to create one larger molecule or to swap pieces. In fact, whatever type of biological reaction you can think of, there is probably an enzyme to speed it up!
A substrate enters the active site of the enzyme.
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- What is the relationship between an enzyme and its substrate?
This forms the enzyme-substrate complex. The reaction then occurs, converting the substrate into products and forming an enzyme products complex. The products then leave the active site of the enzyme.
Basics of enzyme kinetics graphs
Image modified from " Enzymes: Proteins are made of units called amino acidsand in enzymes that are proteins, the active site gets its properties from the amino acids it's built out of. These amino acids may have side chains that are large or small, acidic or basic, hydrophilic or hydrophobic.
The set of amino acids found in the active site, along with their positions in 3D space, give the active site a very specific size, shape, and chemical behavior. Thanks to these amino acids, an enzyme's active site is uniquely suited to bind to a particular target—the enzyme's substrate or substrates—and help them undergo a chemical reaction.
How specific is the matching between enzyme and substrate? Different types of enzymes have different degrees of specificity, or "pickiness" about which molecules can be used as substrates. If the starting conditions are set such that the total substrate concentration exceeds the total enzyme concentration by orders of magnitude, the amount of substrate getting into the ES complex will be negligible compared to the total amount of substrate. Consequently, the free substrate concentration at the beginning of the reaction will practically equal the total substrate concentration.
This way, both the total enzyme concentration, [E]T, and the free substrate concentration, [S], will be experimentally-set known parameters. In the next step, based on Equation 9. In this case, the substrate saturates the enzyme molecules.
Enzymes and the active site
According to this, the kcat [E]T product in Equation 9. By taking this into account, we can formulate Equation 9. P1 is also the horizontal asymptote of the hyperbola, the maximal value of Y that the graph of the function approaches as X tends to infinity. The P2 parameter is KS. Note that this equation is in a perfect accordance with the experimental observations regarding the [S]-V0 relationship illustrated in Figure 9.
Accordingly, in the [S] range where the substrate concentration is orders of magnitude lower than the value of Ks, the initial reaction rate will be linearly proportional to substrate concentration, exactly as the experiments show. In other words, in this substrate concentration range the reaction is a pseudo -first order reaction in respect of the substrate. Accordingly, if we consider only this [S] range, we get equation 9.
Accordingly, in this substrate concentration range, the initial reaction rate is practically independent of the substrate concentration, i. This is why the [S]-V0 plot illustrated in Figure 9. When the substrate concentration equals the value of KS, the initial reaction rate is the half of the Vmax value. In a simple descriptive way, the deduced Equation 9. Yet, it leads to serious theoretical contradictions. The more efficient the enzyme, the less rational the initial assumptions of the above model.
If the enzyme is highly efficient, the rate of ES conversion into product should be very high. If so, that process should interfere with the presumed quasi-equilibrium between the enzyme, the substrate and the ES complex. Therefore, the assumption of equilibrium for the first step of the reaction renders the model ill-suited for describing the action of genuinely efficient enzymes. The second—and equally significant—problem is that this first model also contradicts the thermodynamic bases of catalysis.
KS is a dissociation constant and, as such, it defines the affinity, i.B.7.3 Describe the relationship between substrate concentration and enzyme activity.
In the first model, KS describes how strongly the enzyme binds the substrate or, in other words, how stable the ES complex is. The lower the KS, the more stable the complex.
Moreover, based on the model, the lower the KS, the more effective the enzyme. This is because a low KS means that the enzyme reaches half-maximal reaction rate at low substrate concentration. But there is a discrepancy here. By increasing the stability of the interaction between the enzyme and the substrate, the reaction rate should decrease because the enzyme would stabilise the substrate in the ground state.
Basics of enzyme kinetics graphs (article) | Khan Academy
Naturally, the enzyme must bind the substrate, but it should not bind it too tightly. Instead, the enzyme should bind tightly the transition state, thereby decreasing the activation free enthalpy of the reaction. Due to these contradictions, the first kinetic model had to be developed further.
The improved model accounts for the two additional kinetic rate constants that were left out from the first model. One of these, denoted as k1, corresponds to the formation of the ES complex from free enzyme and substrate. The other, denoted as k-1, corresponds to the reverse reaction, the dissociation of the ES complex towards enzyme and substrate.
The improved scheme is illustrated in Equation 9. Depending on the exact initial conditions, the steady-state may last long, i. The steady-state requires equal rates for ES generation and ES decomposition.
This is illustrated in Figure 9. The reaction is triggered by adding the substrate. Proportions of the figure are not realistic, as otherwise various parts of the figure could not be shown on the same page. In reality, the steady-state can be reached in milliseconds and it can last for minutes. The steady-state Decomposition of the ES complex can happen on two different routes: The development of the steady-state is illustrated in Figure 9. In this equation, the ES complex forms at a rate k1[E][S]; it decays back towards the substrate at a rate k-1[ES] and decomposes towards the product at a rate k2[ES].
In the steady-state, the rate of ES formation and the sum of the two types of ES decomposition rates are equal in magnitude and, thus, the concentration of the ES complex does not significantly change.
In the next section, Equation 9. The improved model has the same initial requirement: The rate of the reaction can be easily formulated, as shown in Equation 9. By rearranging Equation 9.
This quotient has been defined as the Michaelis constant, with the abbreviation KM. Replacing the quotient for KM yields Equation 9.
In other words, it quantifies the instability of the ES complex. Therefore, the k2[E]T product in Equation 9. Taking this into consideration, we get Equation 9. Consequently, just like Equation 9. When the numerical value of [S] equals that of KM, the rate of the reaction is exactly half of the maximal one.
In spite of the many formal similarities, there are principal differences in the interpretations of the two models. Note that the KS constant of the simple model and the KM constant of the improved model have different meanings.
It is readily apparent from the comparison of Equations 9.