BIO 361 Lecture Notes - Lecture 12: Rheumatoid Arthritis, Hydrolysis, Skeletal Muscle

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BIO361.30
Discussion Clarification Week 3, Lectures 12-16
Lecture 12: Lineweaver-Burk Plot
1. You are trying to calculate the Km and Vmax for an enzyme using the graphical analysis
approach of Lineweaver and Burk.
A. Your determination of the slope of the plot alone is sufficient for you to get separate values
for Km and Vmax.
B. Your determination of the point at which the Lineweaver-Burk plot crosses the X-axis should
give you a good estimate of the value of Km, because the value of that so-called X-intercept is
equal to the value of Km
C. Your determination of the point at which the Lineweaver-Burk plot crosses the Y- axis should
give you a good estimate of the value of Vmax because the value of that so- called Y-intercept is
equal to 1/Vmax.
D. If all of your substrate concentrations are too high, the Lineweaver-Burk plot will intersect the
axes at the origin.
E. The Lineweaver-Burk plot is useful for determining the affinity of an enzyme for its substrate
but it is not useful for estimating the catalytic activity of the enzyme.
The Michaelis constant (KM) can be determined experimentally for any enzyme-substrate pair by
adding the rate constants for ES dissociation (k-1) and product turnover (k2), and dividing the sum
by the rate constant for ES formation (k1). An enzyme’s maximal velocity (Vmax) can be
determined by multiplying k2 by the enzyme concentration when all enzyme has bound substrate
(i.e., all enzyme is in the “ES” form). In terms of the relationship between these two values, KM
is equivalent to the free substrate concentration that results in a reaction velocity, v, that is half of
the value of Vmax for the enzyme (note that KM is NOT half of Vmax); KM is also the free substrate
concentration that would be needed to achieve occupancy of half of all of the enzyme’s active
sites. (Remember that for most enzymes, the total amount of substrate present and the
concentration of free substrate needed to half-saturate the enzyme is much larger than the actual
amount of substrate bound when the enzyme is half-saturated this means that even when the
enzyme is fully saturated, the bulk of the substrate remains free in solution.) KM also serves as a
measure of the enzyme’s affinity for a particular substrate: at lower KM values, a smaller
concentration of substrate is needed to fill half of the enzyme active sites (and therefore the
enzyme has a high affinity for the substrate of interest), while at higher KM values, a larger
concentration of substrate is needed to fill half of the enzyme active sites (and therefore the
enzyme has a low affinity for the substrate). Michaelis-Menten plots are plots of substrate
concentration ([S]) on the x-axis, and the enzyme’s initial velocity (v0) on the y-axis; initial
velocity is plotted to avoid potential issues of substrate depletion during the reaction. While these
hyperbolic plots qualitatively visualize the general effect of [S] on v0, it is often difficult to
accurately predict Vmax or KM from them, as extrapolation of the data can only yield shaky
estimates of Vmax which is asymptotically approached at the highest substrate concentrations.
Without an accurate value for maximum velocity the substrate concentration at which the
velocity is half-maximal is likewise uncertain. Lineweaver-Burk plots, also known as “double
reciprocal” plots, however, remedy this problem. These plots transform a hyperbolic Michaelis-
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Menten plot into a linear plot, allowing Vmax and KM, to be more easily read from the graph: the
y-intercept of the plot becomes equal to 1/Vmax, and the x-intercept becomes equal to -1/KM.
Choice C is therefore the answer. It should be noted that since an enzymes catalytic activity
(kcat) is equal to Vmax divided by the concentration of enzyme with all of its active sites filled with
substrate, Lineweaver-Burke plots can also be used to find kcat.
Lecture 13: Uncompetitive inhibitor
2. Which of the following statements about an uncompetitive enzyme inhibitor is TRUE?
A. Increasing the concentration of the inhibitor will produce an increase in the measured Vmax
of the reaction.
B. Increasing the concentration of the inhibitor will produce an increase in the measured KM of
the reaction.
C. Increasing the concentration of the inhibitor will reduce the concentration of free enzyme
available to bind to substrate.
D. Increasing the concentration of the inhibitor will make the value of the X- intercept on the
Lineweaver-Burk plot more negative.
E. Increasing the concentration of the inhibitor will not alter the measured value of kcat for the
reaction.
Enzyme inhibitors can bind reversibly or irreversibly with an enzyme to affect its activity;
depending on the class of inhibitor, it can alter activity by increasing or decreasing KM, and/or
decreasing Vmax. These qualities make inhibitors eagerly sought after in the pharmaceutical
industry, as they can diminish the activity of enzymes that are involved in various pathologies.
Competitive inhibitors quite literally “compete” with substrate for an enzyme’s substrate-binding
site, and as such, often resemble the structure of the substrate; however, although they are
reversibly bound like the substrate, they cannot be transformed into product (for the purposes of
this course, as long as the competitive inhibitor is bound, no product can be formed). Adding
enough substrate can “out-compete” a competitive inhibitor, as this makes it more likely for
substrate, rather than inhibitor, to encounter free enzyme. In general, competitive inhibitors may
be bound more tightly or less tightly than the substrate. As an important special case, transition
state analogs are almost always strong competitive inhibitors, as the active site of an enzyme has
a structure that binds more tightly to the transition state than the substrate or product in order to
lower the reaction’s energy of activation. Stronger competitive inhibitors like these have a lower
dissociation equilibrium constant, called KI when referring to inhibitors than the actual substrate,
and therefore have a higher affinity for the enzyme than the substrate itself. Because competitive
inhibitors bind to free enzyme only and not to the enzyme-substrate complex, they have the
effect of increasing the apparent KM (the substrate concentration at which the reaction rate is at
half of its Vmax). Uncompetitive inhibitors instead bind only to the enzyme-substrate complex,
distorting the active site of the complex and therefore preventing product turnover. Because they
do not bind to the substrate-binding site (most commonly, the binding site for the first of two
substrates), they do not have to resemble the substrate (or transition state). The prevention of
catalysis decreases the enzyme’s apparent Vmax, as is universally true. Also, because the enzyme-
substrate complex now has three potential decomposition pathways (one towards the product,
one towards the free enzyme and substrate, and another towards the enzyme-substrate-inhibitor
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