1st Unit / Ch 5 Enzymes

Question 1

Enzyme Nomenclature and Properties 5.1

Which one of the six major classes of enzymes is illustrated by
the reaction shown?

Answer 1

Shown is an enzyme that belongs to the class known as
*_oxidoreductases_* (that most commonly function as
dehydrogenases ).

Question 2

Enzyme Nomenclature and Properties 5.1

Why is NAD said to function as a coenzyme– cosubstrate (not
a coenzyme–prosthetic group) in enzymatic reactions such as
the one shown?

Answer 2

NAD functions as a coenzyme–cosubstrate in enzymatic
reactions because it is only loosely bound to the enzyme and
leaves the enzyme in a changed form. [ Note: FAD is an example
of a coenzyme–prosthetic group. It is tightly bound to the
enzyme and is returned to its original form on the enzyme.]

Question 3

Enzyme Nomenclature and Properties 5.1

McArdle disease type V GSD is caused by a defi ciency in
muscle glycogen phosphorylase ( myophosphorylase ), an enzyme
of glycogen degradation. How will a decrease in P i affect the activity
of this enzyme?

Answer 3

Based on its designation as a phosphorylase , myophosphorylase
(deficient in McArdle disease ) uses P i to cleave bonds in
glycogen. Therefore, a decrease in P i will decrease enzymatic
activity. [ Note: The enzyme cleaves the a(1→4) glycosidic bond in
glycogen, thereby generating the phosphorylated product glucose

Question 4

Enzyme Properties 5.2

Enzymes are protein_______ that increase the_____ of a chemical
reaction. As shown, they contain an ,______ which is a small
on the surface of the enzyme to which a specifi c _______binds, forming an________________
complex leading to product formation. Binding may cause a
conformational change in the enzyme, a process known as__________ .

Answer 4

Enzymes are protein catalysts that increase the rate ( velocity ) of a chemicalreaction. As shown, they contain an active site , which is a small pocket ( or cleft)on the surface of the enzyme to which a specifi c substrate binds, forming an enzyme-substrate complex leading to product formation. Binding may cause a
conformational change in the enzyme, a process known as induced fit . [ Note: RNA
catalysts are referred to as ribozymes.]

Question 5

Enzyme Properties 5.2

What is the difference between a holoenzyme and an apoenzyme?

Answer 5

A holoenzyme is an enzyme with its nonprotein component, and an apoenzyme is missing the nonprotein component. The nonprotein component is required for enzymic activity.y.

Question 6

Enzyme Properties 5.2

Elevated blood ALP suggests a pathology. ALP is found primarily in the liver as ALP-1 and in bone as ALP-2. Levels of the two forms can help differentiate between a liver and a bone pathology. What term is used to describe the tissue-specifi c
forms of an enzyme?

Answer 6

Isozyme ( isoenzyme ) is the term used to describe the tissue-specific forms of an enzyme, such as ALP-1 and ALP-2 . Isozymes catalyze the same reaction but differ in
their amino acid composition (primary structure).

Question 7

How Enzymes Work 5.3

1. What name is given to that region of the curve shown at right marked by the asterisk (*)?
2. Which arrow (blue or red) represents the free energy of activation of the uncatalyzed reaction?
3. How do enzymes dramatically increase the reaction rate relative to the uncatalyzed reaction?

Answer 7

1. The asterisk (*) marks the transition state .
2. The blue arrow represents the free energy of activation of the uncatalyzed reaction.
3. The lower the free energy of activation, the faster the reaction rate . Enzymes lower the free energy of activation by (1) providing an alternate, energetically favorable reaction pathway and (2) stabilizing the transition state of this pathway. Stabilization
   increases the concentration of the reactive intermediate that can be converted to product, thereby increasing the reaction rate. [ Note: The turnover number (k cat ), the
   number of substrate molecules converted to product per second, is increased.]

Question 8

How Enzymes Work 5.3

How do enzymes affect the delta G of a reaction??

Answer 8

Enzymes have no effect on the delta G of a reaction. Therefore, the free energies of the reactants and the products are the same in the catalyzed and uncatalyzed reactions.n..

Question 9

Factors Affecting Reaction Velocity 5.4

What processes are shown at right? What effect do they have on the velocity of an enzymecatalyzed reaction?

Answer 9

Phosphorylation and dephosphorylation , covalent modifi cations to proteins, are shown. Depending on the enzyme, these modifications may increase or decrease the velocity of an enzyme-catalyzed reaction. [ Note: The change in enzyme activity is the result
of a conformational change in the enzyme caused by the covalent modifi cation.]

Question 10

Factors Affecting Reaction Velocity 5.4

What general name is given to the enzyme that catalyzes the forward reaction?

Answer 10

Kinases catalyze phosphorylation reactions using ATP as the phosphate source. They are opposed by phosphatases.

Question 11

Factors Affecting Reaction Velocity 5.4

What other environmental factors infl uence the velocity of an enzyme-catalyzed reaction?

Answer 11

Changes in the concentration of the enzyme, coenzyme, and substrate; temperature; and pH are additional factors that infl uence the velocity of an enzyme-catalyzed reaction.

Question 12

Factors Affecting Reaction Velocity 5.4

Tyrosinemia type 1 ( infantile tyrosinemia ) is caused by a deficiency in fumarylacetoacetate hydrolase that catalyzes the last reaction in the degradation of Tyr. It is treated with a drug that inhibits an enzyme earlier in the pathway. What is the biochemical
rationale for this therapy?

Answer 12

Nitisinone is prescribed for infantile tyrosinemia because it decreases production of the substrate for the hydrolase , thereby decreasing the velocity of the reaction. Additionally, by preventing substrate accumulation, this substrate reduction therapy prevents
entry of the substrate into side reactions that produce harmful products..

Question 13

Michaelis–Menten Kinetics 5.5

1. Supply the missing terms in the Michaelis–Menten equation shown?
2. What is the steady state assumption?.
3. True or false: When the [S] is much less than the Km, the V 0 is proportional to [S], and the reaction is said to
   be fi rst order..

Answer 13

1. [S] is the missing term in the Michaelis–Menten equation .
2. The steady state assumption is that the concentration of ES does not change with time. That is, the rate of formation of ES is equal to that of the breakdown of ES to E S and E P.
3. True: When [S] is much less than the K m , the V 0 is proportional to [S], and the reaction is
   said to be f rst order , as shown.

Question 14

Michaelis–Menten Kinetics 5.5

If 1/V 0 and 1/[S] were plotted, what shape would result? What is the X intercept on this plot? The Y intercept?

If a mutation to the gene that codes for an enzyme results in a 12-fold increase in the K m of the enzyme for its physiologic substrate, what effect has the mutation had on the affinity of the enzyme for the substrate?

Answer 14

A straight line would be seen if 1/V 0 and 1/[S] were plotted. The X intercept on this Lineweaver-Burk plot is 1/K m , and the Y intercept is 1/V max .

Increasing the K m of the enzyme for its physiologic substrate decreases the affinity of the enzyme for the substrate.

Question 15

Enzyme Inhibition 5.6

1. What type of inhibition is shown?
2. Which line represents the uninhibited enzyme? Which line represents the highest concentration of inhibitor?

Answer 15

1. Competitive inhibition is shown, in which the inhibitor and the S compete for the same binding site on the enzyme. As a result, the apparent K m increases because a higher [S] is required to
   achieve 1/2 V max .

2.The blue line represents the uninhibited enzyme. The black line represents the highest concentration
of inhibitor.

Question 16

Enzyme Inhibition 5.6

1. What type of inhibition results in a decrease in the apparent V max ? Is K m also affected by the inhibitor?
2. Orlistat, a weight-loss drug, covalently bonds to lipases that hydrolyze dietary fat (TAGs) and inhibits their enzymic activity. Is this an example of reversible or irreversible enzyme inhibition?

Answer 16

1. Noncompetitive inhibition results in a decrease in apparent V max . K m is not affected. [ Note: In
noncompetitive inhibition, the inhibitor does not compete with the S and can bind the E and the ES
complex (as shown).]

2.Covalent bonding of an inhibitor to an enzyme (as seen with orlistat ) irreversibly inhibits the enzyme. [ Note: Covalent modifi cation of an enzyme, such as is seen with the acetylation of COX by
aspirin, also causes irreversible enzyme inhibition.]

Question 17

Allosteric Enzyme Regulation 5.7

1. Which curve shown represents an allosteric enzyme?
2. Will a positive allosteric effector that influences the K 0.5 shift the V 0 versus [S] plot to the left or to the right?
3. Sialuria (sialic acid in the urine) is a rare, AD condition caused by a mutation in the rate-limiting enzyme of sialic acid synthesis. The mutation decreases the enzyme’s ability to bind CMP–sialic acid, the end product of the pathway. Why does this mutation result in increased production (and excretion) of sialic acid?

Answer 17

1. The green curve, with its sigmoidal shape , represents an allosteric enzyme
2. A positive allosteric effector that infl uences K 0.5 will shift the V 0 versus [S] plot to the left (as shown), refl ecting a lower K 0.5 . [Note: K 0.5 is that [S] required to achieve half maximal velocity.]
3. The CMP–sialic acid is a feedback inhibitor of the pathway. Loss of this allosteric inhibition (as the result of decreased binding to the regulated enzyme) results in overproduction of sialic acid and,
   consequently, sialuria .

Question 18

A 66-year-old female is seen in the emergency department in the late evening. She was driven to the hospital by her husband. The woman reports she has had a “pressure” on her chest for the last few hours. She denies overt chest pain; jaw, neck, shoulder, arm, or epigastric pain; shortness of breath ( dyspnea ); and
sweating ( diaphoresis ). The pressure does not seem to increase with exertion. The patient’s history is remarkable for hyperlipidemia that is being treated with diet and drugs. There is no family history of heart disease. The patient’s heart and respiratory rates are elevated. Blood is drawn and sent to the clinical laboratory for measurement of cardiac biomarkers. Sublingual nitroglycerin is given. An EKG is performed, and the results are consistent with a myocardial
infarction( MI ). Cardiac biomarker measurements reveal elevated levels ofCK, theCK- MB(the isoform found predominately in cardiac muscle) to total CK ratio, andcTnI (a cardiac troponin).

Answer 18

CK -MB is one of three isoforms of CK , an intracellular enzyme. CK -MB is the form found virtually exclusively in the heart, with CK -MM found in the brain, and CK -BB found in skeletal muscle.
Troponins ( Tn ) are nonenzymatic, intracellular, regulatory proteins involved in contractility of skeletal and cardiac muscle. Both CK -MB and cTnI become elevated in the blood as a result of tissue necrosis
in an MI, but their patterns are different. Although both cardiac biomarkers rise early in an MI, cTnI remains elevated for up to 10 days, whereas CK -MB remains elevated for up to 3 days.