UB Chapter 3: Enzymes (Exercises 3.6) p.44
1. a. Factor: Temperature / pH / Enzyme concentration should be kept constant.
b.i. The rate of reaction is higher in the absence of the inhibitor.
The competitive inhibitor molecules bind to some active sites of the enzyme. It prevents the
substrate from occupying that site. This would reduce the rate of reaction.
ii. Substrate molecules compete with the inhibitor for the active site of enzyme. The higher the
substrate concentration the greater chance of it binding to the active site. At very high
concentration, e.g. 30umol cm-3, the substrate overcomes the inhibition effect.
c.i. Negative feedback mechanism. (End-product inhibition)
ii. Z binds to a point on the surface of enzyme e1 far away from the active site. It changes the
active site that it no longer fits substrate V. Therefore, the rate of V changing to W is reduced.
So are the production rates of X, Y and Z. The presence of more Z results in greater inhibition.
R O
R
2. a.i. H2N-CH-C-NH-CH-COOH
R
+ H2O ----------- > H2N-CH-COOH +
R
N2H-CH-COOH
ii. hydrolytic enzyme
b. Once the reaction starts the substrate molecules disappear at a high rate. Increase of products
would have end-product inhibition on the enzyme. Both factors reduce the reaction rate.
c.i. Phase 1: The reaction rate rises as more substrate molecules collide with the enzyme.
(Phase 2: The acceleration rate slows down (but the rate is still rising). At any one moment
almost all the active sites are occupied by substrate.)
Phase 3: The reaction rate stops to rise and the reaction goes on at the maximum rate. At any
moment all the active sites are occupied by substrate.
ii. Curve B: The reaction goes on at maximum rate as though there is no inhibitor. Substrate
competes with inhibitor for the active site. The inhibition effect can be overcome by the
increase of substrate concentration.
Curve C: The maximum reaction rate is much reduced. The non-competitive inhibitor attaches
on the enzyme molecule other than the active site. This alters the shape of the active site and
the enzyme loses its function. The number of active sites becomes less and the maximum rate is
lowered that curve A.
d.i. The substrate is attracted to the enzyme. It binds to and modifies the active site so that the
substrate can fit in. After that, the change of substrate will take place readily.
ii. A non-competitive inhibitor binds to a point far away from the active site. It alters the shape of
the active site rendering it non-functional. The induced fit theory suggests that the active site
can change shape while the lock and key model does not.
e. Enzymes are globular proteins that have special 3-D structure. The active site keeps its shape
by H-bonds, ionic bonds, disulphide bonds and hydrophobic interaction. High temperature
supplies heat energy, and subsequently kinetic energy to the enzyme molecule. It vibrates so
much that chemical bonds break up and active sites are damaged.
f. At temperature as high as 80oC, immobilized enzyme still has 50% efficiency while that of the
soluble one has only 10%.
g. Place enzyme pellets into a packed reactor column.
h. 1. Label 6 pairs of test tubes. 2. Add 2ml of amylase and starch solution into a tube of each
pair respectively. 3. Place each pair into water bath with the following temperature (oC):
30,40,50,60,70 and 80. 4. 3 minutes after, mix the contents of each pair. Wait for 5 minutes.
5. Perform Benedict's test for the amount of reducing sugar formed. Compare and find out the
differences.
3. a.i. The reaction rate rises as the substrate concentration increases. The acceleration slows down
and almost stops increasing.
ii. The reaction rate rises slowly at first. It rises faster at very high substrate concentration.
b.i. Increase in substrate concentration can increase the chance of collision between enzymes and
Chapter 3 Enzymes
p.2
substrates. Thus the rate of the reaction increases. When the active sites of the enzymes are
nearly occupied by the high concentration of substrate, the acceleration slows down..
ii. Compound A is a competitive inhibitor that reduces the number of active sites for the substrate.
The inhibition can be overcome by increasing the substrate concentration.
c. The inhibition doubles as there is double amount of inhibitor that takes up double number of
active sites.
d. Suitable inhibitors are used to combat bacterial infections by inhibiting the growth of bacteria.
4. a.
Amylase digests starch into maltose. Iodine solution stains starch blue-black but not maltose.
The clear zone around the well indicates the disappearance of starch.
b.
Amylase digests starch most efficiently at pH 7. The clear zone around the well indicates the
absence of starch. The larger is the clear zone, the higher activity is amylase.
c.
At pH 11 amylase loses its H+. Some chemical bonds break. The amylase peptide chain alters
the shape of the active site. Amylase becomes non-functional.
d.
Repeat the experiment but replace amylase with distilled water to show the activity of the
enzyme.
5. a. Enzymes are polypeptide chains made of amino acids. They are linked by peptide bonds.
Disulphide bonds are formed between the amino acids, cystine.
b. Mercaptoethanol breaks all the disulphide bonds that maintain the 3-dimensional structure of
ribonuclease. Since the active site of ribonuclease has been damaged, the enzyme is inactivated.
c. i. Competitive inhibition is the result of adding a molecule that has similar structure to the
substrate to slow down an enzyme-catalyzed reaction. The inhibitor reduces the chance of
the substrate fitting into the active site.
ii. Non-competitive inhibition is the result of adding a molecule that binds to a region remote
from the active site on the surface of an enzyme to slow down an enzyme-catalyzed reaction.
The inhibitor changes the shape of the active site.
6. a. The 3-dimensional structure of an enzyme is maintained by chemical bonds like ionic bond and
hydrogen bond. Any changes in pH will break these bonds and form new linkages. Usually the
active site will be damaged.
b. The optimum pH for trypsin is 8. The reaction rate drops as the pH falls below or rises above 8.
The reaction stops at pH 2 and pH 11.
The optimum pH for pepsin is 2. The rate falls as the pH rises. At pH 5.7 the reaction stops.
Activity of papain is not affected by changes within pH 4 to 10. The rate remains high.
c. Papain. Products of the reaction are amino acids that lower the pH. This does not affect the
activity of papain.
d. The activity of Maxiren® may not be affected by pH changes, while rennin needs a low pH.
The cost of producing Maxiren® may be cheaper. Extraction of rennin from calves may be
more expensive.
7. a.i. High temperature provides more kinetic energy to the enzyme molecule that will vibrate more
vigorously. Most chemical bonds that maintain the active site break. The rate drops.
ii. Certain amino acids in the polypeptide chain forms an active site that specifically allows the
substrate to bind to. Molecules that do not have the substrate structure cannot fit into it.
iii. Inhibitors reduce the enzyme-catalyzed reaction rate. This is done either by competitive
inhibitors or non-competitive inhibitors. The former ones have similar structure to the substrate.
They compete with the substrate for active sites. The latter ones do not fit into the active site.
They bind to somewhere else on the enzyme surface and change the shape of active site.
b. Measure the initial reaction rate of an enzyme-catalyzed reaction with: an inhibitor and 0.2 %
of a substrate. Repeat the experiment by changing the substrate concentration, such as 0.4, 0.6,
0.8…%. If the inhibition can be overcome by adding more substrate, it is the effect of a
competitive inhibitor. If not, a non-competitive inhibitor.
8. a.i. The rate of production of D will be reduced. Lower concentration of A produces less B or C.
ii. The rate of production of D will be increased at first, until the deficiency of either enzyme 2 or
Chapter 3 Enzymes
p.3
3 occurs. Then the rate stops to increase. It is because the amount of active sites is all saturated.
b. Excess of D inhibits either enzyme 1, 2 or 3. As the formation of B or C is reduced, the further
production of D will be reduced.
–End-