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a) The x-intercept shifts further from zero

b) The y-intercept increases

c) The x-intercept shifts closer to zero

d) The y-intercept decreases

**C is correct: The x-intercept shifts closer to zero.**

Lineweaver Burk plots are a graphical method of analyzing the Michaelis–Menten equation and the enzyme-substrate-inhibitor relationship. The x-axis in the graph is 1/[S], and the y-axis is 1/*V*. In competitive inhibition, the inhibitor decreases the binding affinity for the enzyme to the substrate by competing with the substrate for the active site. This competition increases the *K*_{M}, since it will take more substrate to outcompete the inhibitor and move the reaction forward. Increasing the *K*_{M} will cause the x-intercept of the Lineweaver-Burk plot to shift closer to zero (choice C is correct; choice A is incorrect). The y-intercept will not change (choices B and D are incorrect).

V0 = (Vmax x [S]) / (KM + [S])

The reciprocal of this equation is:

1 / *V _{0} = (K_{M} / V_{max} [S]) + (1 / Vmax)*

From this, the Lineweaver Burk equation becomes:

1 / *V _{0} = (K_{M} / V_{max} ) (1 / [S]) + (1 / V_{max})*

**Figure 1**). As a line, interpreting the data points is straightforward. The Michaelis-Menten saturation curve, on the other hand, is more challenging to interpret at times due to the different data points and the hyperbolic nature of the graph.

In Lineweaver Burk plots, the slope is equal to *K _{M} / V_{max}, the x-intercept is equal to -1 / K_{M}*

One crucial point about the Lineweaver Burk plot is that the x-axis goes into negative values. However, there is no such thing as negative substrate concentrations. Instead, researchers can use this graph to extrapolate data from a hypothetical situation in which a negative substrate concentration exists.

**Figure 2** shows Lineweaver Burk plots with the effects of competitive, uncompetitive, and mixed inhibitors.

In terms of competitive inhibitors, recall that they compete directly with the substrate to bind to the active site of the enzyme. In turn, they will decrease the affinity of the enzyme for the substrate, which will increase the *K _{M}. As stated above, the x-intercept is equal to -1 / K_{M}. Increasing the value of the K_{M} will then decrease the value of the x-intercept. Decreasing the value of the x-intercept will shift the line closer to zero. Also, the y-intercept will stay the same because competitive inhibitors do not affect the maximum velocity of the reaction. In competitive inhibitors, the slope of the graph increases, the x-intercept shifts to the right, and the y-intercept is unchanged.*

* *

For uncompetitive inhibitors, both the *K _{M} and the V_{max} will be reduced. Recall that since uncompetitive inhibitors increase the affinity of the enzyme for the substrate, K_{M} decreases. Also, since the x-intercept is equal to -1 / K_{M}, a smaller K_{M} would give a larger value for the x-interceptor. In this way, the x-intercept will shift to the left. Also, since uncompetitive inhibitors also decrease the maximum reaction velocity, the value of the y-intercept, or 1 / V_{max} , will increase. Increasing the y-intercept will shift the line upwards. Therefore, uncompetitive inhibitors will result in a parallel line to the original graph, in which both the x and y-intercepts have changed.*

* *

Lastly, for noncompetitive inhibitors, the *K _{M} is unaffected, and the V_{max} decreases. Recall that *

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