# Structural Biochemistry/Enzyme/Double-Reciprocal plot: Lineweaver Burk plote
## Double Reciprocal Plot

The double-reciprocal equation is obtained by taking the reciprocal of both sides of the Michaelis-Menten equation. The double-reciprocal (also known as the Lineweaver-Burk) plot is created by plotting the inverse initial velocity (1/V$_0$) as a function of the inverse of the substrate concentration (1/[S]). The V$_{max}$ can be accurately determined and thus K$_M$ can also be determined with accuracy because a straight line is formed. The slope of the resulting line is K$_M$/V$_{max}$, the y-intercept is 1/V$_{max}$, and the x-intercept is -1/K$_M$. Using the Michaelis-Menten equation, the V$_{max}$ is an asymptote and can thus only be approximated and as a result, the K$_M$, which is V$_{max}$/2, can't be determined accurately. This plot is a useful way to determined different inhibitors such as competitive, uncompetitive, and noncompetitive.
For competitive inhibitors, the inhibitor competes with the substrate molecule to bind to the binding site. Consequently, the K$_M$ will increase without changing the V$_{max}$ value. This means that the two graphs will have the same y-intercept as shown below. However the new x-intercept may be quite elusive. For this type of inhibitors, a higher concentration of the substrate is needed to get half of the active sites occupied. Therefore K$_{M2}$ will be larger than K$_{M1}$. This translates to a higher reciprocal value of K$_{M1}$ than that of K$_{M2}$. However the x-intercept has the negative sign in front of it, thus on the graph it has to move to the right relative to the previous intercept. To show this on the double reciprocal plot, the slope will increase to show the strength of the binding competitive inhibitor. While the slope increases with the presence of the inhibitor, the y-intercept remains the same in presence and absence of the inhibitor.
For uncompetitive inhibitors, the inhibitor will only bind to an enzyme-substrate complex; therefore, it does not compete with the substrate for the binding site. Consequently, both K$_M$ and V$_{max}$ values decrease. Consequentially, the reciprocal value of the new V$_{max}$ should be at a higher position on the axis, as a fraction becomes larger when the denominator gets smaller. The new reciprocal value of K$_M$ will move to the left and the explanation should be similar to that of competitive inhibitor. To show this on a double reciprocal plot, the slope will remain the same as if the enzyme was not bound to the inhibitor, but the x-axis intercept will decrease. The double reciprocal plot for enzyme with and without uncompetitive inhibitor will be two parallel lines.
For noncompetitive inhibitors, the inhibitor can bind to the enzyme before the substrate can bind to the binding site. It does not have to wait for the enzyme to become an enzyme-substrate complex in order to bind to the enzyme. The inhibition will cause a decrease in V$_{max}$ value while the K$_M$ is unaffected. This means the value of -1/K$_M$ remains the same for the two lines, while the new value of 1/V$_{max}$ is higher relative to the previous one. To show this on a double reciprocal plot, the decrease in V$_{max}$ will increase the y-intercept with a larger slope.
The black lines represent the reciprocal of velocity when no inhibitor is present and the blue lines correspond to the presence of inhibitors.