Essential Cell Biology 5th edition

How Proteins Work
143
rate of reaction
V max
½V max
K M substrate concentration
ECB5 04.35
of substrate, the amount of enzyme−substrate complex—and the rate at
which product is formed—will depend solely on the concentration of the
substrate. If the concentration of substrate added is large enough, however,
all of the enzyme molecules will be filled with substrate. When this
happens, the rate of product formation depends on how rapidly the substrate
molecule can undergo the reaction that will convert it to product.
At this point, the enzymes are working as fast as they can, a value termed
V max . For many enzymes operating at V max , the number of substrate molecules
converted to product is in the vicinity of 1000 per second, although
turnover numbers ranging from 1 to 100,000 molecules per second have
been measured for different enzymes. Enzymes can speed up the rate of
a chemical reaction by a factor of a million or more.
The same type of experiment can be used to gauge how tightly an enzyme
interacts with its substrate, a value that is related to how much substrate
it takes to fully saturate a sample of enzyme. Because it is difficult to
determine at what point an enzyme sample is “fully occupied,” biochemists
instead determine the concentration of substrate at which an enzyme
works at half its maximum speed. This value, called the Michaelis constant,
K M , was named after one of the biochemists who worked out
the relationship (Figure 4−35). In general, a small K M indicates that a
substrate binds very tightly to the enzyme—due to a large number of
noncovalent interactions (see Figure 4−31A); a large K M , on the other
hand, indicates weak binding. We describe the methods used to analyze
enzyme performance in How We Know, pp. 144–145.
Lysozyme Illustrates How an Enzyme Works
We have discussed how enzymes recognize their substrates. But how do
they catalyze the chemical conversion of these substrates into products?
To find out, we take a closer look at lysozyme—an enzyme that acts
as a natural antibiotic in egg white, saliva, tears, and other secretions.
Lysozyme severs the polysaccharide chains that form the cell walls of
bacteria. Because the bacterial cell is under pressure due to intracellular
osmotic forces, cutting even a small number of polysaccharide chains
causes the cell wall to rupture and the bacterium to burst, or lyse—hence
the enzyme’s name. Because lysozyme is a relatively small and stable
protein, and can be isolated easily in large quantities, it has been studied
intensively. It was the first enzyme to have its structure worked out at
the atomic level by x-ray crystallography, and its mechanism of action is
understood in great detail.
The reaction catalyzed by lysozyme is a hydrolysis: the enzyme adds a
molecule of water to a single bond between two adjacent sugar groups in
the polysaccharide chain, thereby causing the bond to break (see Figure
2−19). This reaction is energetically favorable because the free energy of
the severed polysaccharide chains is lower than the free energy of the
intact chain. However, the pure polysaccharide can sit for years in water
Figure 4−35 An enzyme’s performance
depends on how rapidly it can process its
substrate. The rate of an enzyme reaction
(V ) increases as the substrate concentration
increases, until a maximum value (V max ) is
reached. At this point, all substrate-binding
sites on the enzyme molecules are fully
occupied, and the rate of the reaction is
limited by the rate of the catalytic process
on the enzyme surface. For most enzymes,
the concentration of substrate at which
the reaction rate is half-maximal (K M ) is a
direct measure of how tightly the substrate
is bound, with a large value of K M (a large
amount of substrate needed) corresponding
to weak binding.
QUESTION 4–5
Use drawings to explain how
an enzyme (such as hexokinase,
mentioned in the text) can
distinguish its normal substrate
(here, D-glucose) from the optical
isomer L-glucose, which is not a
substrate. (Hint: remembering
that a carbon atom forms four
single bonds that are tetrahedrally
arranged and that the optical
isomers are mirror images of each
other around such a bond, draw the
substrate as a simple tetrahedron
with four different corners and
then draw its mirror image. Using
this drawing, indicate why only
one optical isomer might bind to a
schematic active site of an enzyme.)