For all rate orders, we will look at a situation where an enzyme (E) converts a substrate (S) into a product (P) at a certain rate (k) (See Equation 1). We can look at the reaction as a change in substrate concentration which decreases at the rate k (Equation 2), or in terms of product concentration which increases at the rate k (Equation 3). The sign in front of k (+ or -) in the equations is very important in this respect. Usually we look at the decrease in substrate concentration.
The two most important equations for Zero Order Reactions are number 4 and number 5.
Equation 4 describes how to determine the rate from a set of data. To calculate the rate, choose an initial time and corresponding substrate concentration from the data set given (usually T = 0 or the earliest time point given), which are represented in the equation as t (initial) and S (initial). Then choose another time point (ideally this should be far away from the initial time point so as to give a good scope) and this time and corresponding substrate concentration are represented by t (time) and S (time) in the equation. Remember: units are very important in kinetic reactions, Zero Order Rates always have units of concentration/time e.g. mM/min.
Equation 5 can only be used when the rate (k) is known (see Equation 4). Once this is known, the substrate concentration at a specified time can be calculated. Choose an initial data set and then input the time at which you want to find the substrate concentration for.
You may be asked to represent this graphically. On the x-axis plot time, on the y-axis, plot concentration of substrate. Plot each of your data sets onto your graph, and the points should be linear (or near linear). If they are not, your data does not follow a Zero Order Reaction scheme! (in which case, see the other Rate Orders). Once you have a linear graph, you can work out the rate of the reaction (k) by taking the gradient.
As in Zero Order Reactions above, First Order Reactions are also used to describe the decrease of substrate concentration over time. However, these reactions are described by a different set of equations.
The rate is now also dependant on the amount of substrate present. As you can imagine, in a abundance of substrate, the reaction is faster than if the substrate is less available. This is described by Equation 6.
Equations 8 and 9 include the function ln, which stands for log natural. To find the rate, use Equation 8.
When finding the substrate concentration at a given time however, getting the answer as a log natural is not useful. So instead we use Equation 9.
To represent this as a graph, as in Zero Order Reactions, time is plotted on the x-axis. However, ln Substrate concentration is plotted on the y-axis. This means you will need to convert the concentrations in your data set to ln, BEFORE plotting them. Once again, the gradient will be the rate.
Second order reactions usually involve dimerization reactions. The reaction is now using up more substrate than in the last two reactions. Equation 10 describes this mathematically.
To find the rate use Equation 11. As the concentrations have been converted to 1/S, the substrate concentrations now have units of concentration-1. This means your rate will have units of concentration-1 time-1 e.g mM-1 min-1.
To find the substrate concentration at a specific time, use Equation 12. NOTE: the rate (k) in this equation is now positive!
This can, again, be represented on a graph, with time on the x-axis. However, 1/S is plotted on the y-axis and the gradient this time is +k.
Three important parameters in a kinetic reaction are Vmax, Km and Vmax/Km.
Vmax: Maximum velocity of the reaction aka how the enzyme behaves when the substrate is in abundance.
Km: The affinity of the enzyme for the substrate, the lower the value, the better the affinity!
Vmax/Km = 1/2 Vmax aka how the enzyme behaves at low substrate concentrations.
These are derived from the rates in Equation 13. This scheme is just a slightly more complicated one than Equation 1, where the enzyme (E) and substrate (S) now form an intermediate (ES) before creating the products (EP). In this scheme the creation of products is also reversible at a certain rate (k-2).
However, for the purposes of constructing the ‘Michealis Menton Equation’ we must make a few assumptions:
This can be assumed to be correct when very little product has been formed. I.e - at the start of the reaction (This is why the initial rates are used.) Under these conditions, Equation 14 is valid.
Catalytic efficiency is how good an enzyme is at catalysing a reaction at low substrate concentrations. When the substrate concentration is much lower than Km, the Michealis Menten equation can be written as in Equation 15.
Looking back to the reaction scheme in Equation 13, and remembering the conditions under which the Michealis Menten Equation is valid, k2 is the slowest and therefore, the rate limiting step. The rate k2 can also be called kcat (the catalytic step that turns ES into E+P).
As Vmax can be calculated by Equation 16 and Km can be calculated by Equation 17, catalytic efficiency can be described as in Equation 18. The equation for catalytic efficiency can be further defined by using the three rates present in the reaction, see Equation 19.
If the creation of P from ES by rate kcat, was much faster than the reaction from ES to E+S (rate k-1), then rate k-1 can be taken out of Equation 19. This would mean that the entire equation would be equal to k1, see Equation 20.
k1 cannot be larger than 10-8 – 10-9M-1 s-1, and so this means that any enzyme with a catalytic efficiency close to k1, has reached ‘Catalytic Perfection’.
A competitive inhibitor is one that competes with the substrate for enzyme binding. It can do this in two ways, either bind to the active site (The same site the substrate binds to), or it can bind to an allosteric site. For a competitive inhibitor, binding to the allosteric site on the free enzyme causes a conformational change of the enzyme. This changes the active site so that the substrate will no longer bind. The reaction scheme in Equation 21 describes this.
Competitive inhibitors work best when there is a lot of free enzyme (or little substrate) but it doesn’t work very well when the enzyme is in very low concentrations compared to the substrate.
Uncompetitve inhibitors don’t bind the active site! They can only bind allosterically. However, an uncompetitive inhibitor can only bind to the enzyme-substrate complex, therefore it does not prevent substrate binding. When the substrate binds the enzyme, there is a conformational change of the enzyme, and this allows the uncompetitive inhibitor to bind a site that previously wasn’t available. The reaction scheme in Equation 22 describes this.
Uncompetitive inhibitors work best when there is a lot of substrate present in comparison to enzyme concentration (as the enzymes will be fully occupied with substrate, meaning the ES complex is abundant).
Compared to the reaction without inhibitor, the reaction with inhibitor:
Vmax stays the same (The inhibitor does not work at high concentrations of substrate)
Km increases (The enzyme seems to have less affinity for the substrate)
Vmax/Km therefore decreases.
Compared to the reaction without inhibitor, the reaction with inhibitor:
Vmax decreases (The inhibitor works best at high concentrations of substrate)
Km decreases (The enzyme seems to have more affinity for the substrate)
Vmax/Km therefore stays the same.
‘Kinetics for Bioscientist’ book, Free to download at: http://bookboon.com/en/textbooks/biology-biochemistry/kinetics-for-bioscientist just give email and institution of study. Friendly and informative style!
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