Enzymes are biological catalysts responsible for supporting almost all of the chemical reactions that maintain animal homeostasis. Because of their role in maintaining life processes, the assay and pharmacological regulation of enzymes have become key elements in clinical diagnosis and therapeutics. The macromolecular components of almost all enzymes are composed of protein, except for a class of RNA-modifying catalysts known as ribozymes. Ribozymes are molecules of ribonucleic acid that catalyze reactions on the phosphodiester bond of other RNAs (see Chapter 35) (Table 9-1).

Enzyme Classifications

Enzymes are also classified on the basis of their composition. Enzymes composed wholly of protein are known as simple enzymes in contrast to complex enzymes, which are composed of protein plus a relatively small organic molecule. Complex enzymes are also known as holoenzymes. In this terminology the protein component is known as the apoenzyme, while the nonprotein component is known as the coenzyme or prosthetic group where prosthetic group describes a complex in which the small organic molecule is bound to the apoenzyme by covalent bonds; when the binding between the apoenzyme and nonprotein components is noncovalent, the small organic molecule is called a coenzyme. Many prosthetic groups and coenzymes are water-soluble derivatives of vitamins (Chapter 8). It should be noted that the main clinical symptoms of dietary vitamin insufficiency generally arise from the malfunction of enzymes, which lack sufficient cofactors derived from vitamins to maintain homeostasis.

The nonprotein component of an enzyme may be as simple as a metal ion or as complex as a small nonprotein organic molecule. Enzymes that require a metal in their composition are known as metalloenzymes if they bind and retain their metal atom(s) under all conditions with very high affinity. Enzymes that have a lower affinity for metal ion, but still require the metal ion for activity, are known as metal-activated enzymes.

Role of Coenzymes

The functional role of coenzymes is to act as transporters of chemical groups from one reactant to another. The chemical groups carried can be as simple as the hydride ion (H⁺ + 2e⁻) carried by NAD or the mole of hydrogen carried by FAD; or they can be even more complex such as the amine (–NH₂) carried by pyridoxal phosphate.

Since coenzymes are chemically changed as a consequence of enzyme action, it is often useful to consider coenzymes to be a special class of substrates, or second substrates, which are common to many different holoenzymes. In all cases, the coenzymes donate the carried chemical grouping to an acceptor molecule and are thus regenerated to their original form. This regeneration of coenzyme and holoenzyme fulfills the definition of an enzyme as a chemical catalyst, since (unlike the usual substrates, which are used up during the course of a reaction) coenzymes are generally regenerated.

Enzyme Relative to Substrate Type

Although enzymes are highly specific for the kind of reaction they catalyze, the same is not always true of substrates upon which they act. For example, while succinate dehydrogenase (SDH) always catalyzes an oxidation-reduction reaction and its substrate is invariably succinic acid, alcohol dehydrogenase (ADH) always catalyzes oxidation-reduction reactions but can act on a number of different alcohols, ranging from methanol to butanol. Generally, enzymes having broad substrate specificity are most active against one particular substrate. In the case of ADH, ethanol is the preferred substrate.

Enzymes also are generally specific for a particular steric configuration (optical isomer) of a substrate. For example, enzymes that hydrolyze D sugars will not recognize the corresponding L isomer. The enzymes known as racemases provide a striking exception to these generalities; in fact, the role of racemases is to convert D isomers to L isomers and vice versa.

As enzymes have a more or less broad range of substrate specificity, it follows that a given substrate may be acted on by a number of different enzymes, each of which uses the same substrate(s) and produces the same product(s). The individual members of a set of enzymes sharing such characteristics are known as isozymes. These are the products of genes that vary only slightly; often, various isozymes of a group are expressed in different tissues of the body.

Enzyme-Substrate Interactions

The favored model of enzyme-substrate interaction is known as the induced-fit model. This model proposes that the initial interaction between enzyme and substrate is relatively weak, but that these weak interactions rapidly induce conformational changes in the enzyme that strengthen binding and bring catalytic sites close to substrate bonds to be altered. After binding takes place, one or more mechanisms of catalysis generate transition-state complexes and reaction products. There are 4 possible mechanisms of enzyme-mediated catalysis.

  1. Catalysis by bond strain: In this form of catalysis, the induced structural rearrangements that take place with the binding of substrate and enzyme ultimately produce strained substrate bonds, which more easily attain the transition state. The new conformation often forces substrate atoms and bulky catalytic groups, such as aspartate and glutamate, into conformations that strain existing substrate bonds.

  2. Catalysis by proximity and orientation: Enzyme-substrate interactions orient reactive groups and bring them into proximity with one another. In addition to inducing strain, groups such as aspartate are frequently chemically reactive as well, and their proximity and orientation toward the substrate thus favors their participation in catalysis.

  3. Catalysis involving proton donors (acids) and acceptors (bases): Other mechanisms also contribute significantly to the completion of catalytic events initiated by a strain mechanism, for example, the use of glutamate as a general acid catalyst (proton donor).

  4. Covalent catalysis: In catalysis that takes place by covalent mechanisms, the substrate is oriented to active sites on the enzymes in such a way that a covalent intermediate forms between the enzyme or coenzyme and the substrate. One of the best-known examples of this mechanism is that involving proteolysis by serine proteases, which include both digestive enzymes (trypsin, chymotrypsin, and elastase) and several enzymes of the blood-clotting cascade. These proteases contain an active site serine whose R group hydroxyl forms a covalent bond with a carbonyl carbon of a peptide bond, thereby causing hydrolysis of the peptide bond.

Chemical Reactions and Rates

According to the conventions of biochemistry, the rate of a chemical reaction is described by the number of molecules of reactant(s) that are converted into product(s) in a specified time period. Reaction rate is always dependent on the concentration of the chemicals involved in the process and on rate constants that are characteristic of the reaction. For example, the reaction in which A is converted to B is written as follows:

The rate of this reaction is expressed algebraically as either a decrease in the concentration of reactant A as

or an increase in the concentration of product B as:

In equation 2, the negative sign signifies a decrease in concentration of A as the reaction progresses, brackets define concentration in molarity, and the k is known as a rate constant. Rate constants are simply proportionality constants that provide a quantitative connection between chemical concentrations and reaction rates. Each chemical reaction has characteristic values for its rate constants; these in turn directly relate to the equilibrium constant for that reaction. Thus, a reaction can be rewritten as an equilibrium expression in order to show the relationship between reaction rates, rate constants, and the equilibrium constant for this simple case. The rate constant for the forward reaction is defined as k₊₁ and the reverse as k₋₁. At equilibrium the rate (v) of the forward reaction (A → B) is, by definition, equal to that of the reverse or back reaction (B → A), a relationship which is algebraically symbolized as

where, for the forward reaction

and for the reverse reaction

In the above equations, k₊₁ and k₋₁ represent rate constants for the forward and reverse reactions, respectively. The negative subscript refers only to a reverse reaction, not to an actual negative value for the constant. To put the relationships of the 2 equations into words, we state that the rate of the forward reaction [v_forward] is equal to the product of the forward rate constant k₊₁ and the molar concentration of A. The rate of the reverse reaction is equal to the product of the reverse rate constant k₋₁ and the molar concentration of B.

This equation demonstrates that the equilibrium constant for a chemical reaction is not only equal to the equilibrium ratio of product and reactant concentrations, but is also equal to the ratio of the characteristic rate constants of the reaction.

Chemical Reaction Order

Empirically, order is easily determined by summing the exponents of each concentration term in the rate equation for a reaction. A reaction characterized by the conversion of one molecule of A to one molecule of B with no influence from any other reactant or solvent is a first-order reaction. The exponent on the substrate concentration in the rate equation for this type of reaction is 1. A reaction with 2 substrates forming 2 products would form a second-order reaction. However, the reactants in second- and higher-order reactions need not be different chemical species. An example of a second-order reaction is the formation of ATP through the condensation of ADP with orthophosphate:

For this reaction the forward reaction rate would be written as:

Enzymes as Biological Catalysts

In cells and organisms most reactions are catalyzed by enzymes, which are regenerated during the course of a reaction. These biological catalysts are physiologically important because they speed up the rates of reactions that would otherwise be too slow to support life. Enzymes increase reaction rates, sometimes by as much as 1-million fold, but more typically by about 1-thousand fold. Catalysts speed up the forward and reverse reactions proportionately so that, although the magnitude of the rate constants of the forward and reverse reactions is increased, the ratio of the rate constants remains the same in the presence or absence of enzymes. Since the equilibrium constant is equal to a ratio of rate constants, it is apparent that enzymes and other catalysts have no effect on the equilibrium constant of the reactions they catalyze.

Enzymes increase reaction rates by decreasing the amount of energy required to form a complex of reactants that are competent to produce reaction products. This complex is known as the activated state or transition state complex for the reaction. Enzymes and other catalysts accelerate reactions by lowering the energy of the transition state. The free energy required to form an activated complex is much lower in the catalyzed reaction. The amount of energy required to achieve the transition state is lowered; consequently, at any instant a greater proportion of the molecules in the population can achieve the transition state. The result is that the reaction rate is increased.

Michaelis-Menten Kinetics

In typical enzyme-catalyzed reactions, reactant and product concentrations are usually hundreds or thousands of times greater than the enzyme concentration. Consequently, each enzyme molecule catalyzes the conversion to product of many reactant molecules. In biochemical reactions, reactants are commonly known as substrates. The catalytic event that converts substrate to product involves the formation of a transition state, and it occurs most easily at a specific binding site on the enzyme. This site, called the catalytic site of the enzyme, has been evolutionarily structured to provide specific, high-affinity binding of substrate(s) and to provide an environment that favors the catalytic events. The complex that forms when substrates and enzymes combine is called the enzyme substrate (ES) complex. Reaction products arise when the ES complex breaks down releasing free enzyme.

Between the binding of substrate to enzyme, and the reappearance of free enzyme and product, a series of complex events must take place. At a minimum an ES complex must be formed; this complex must pass to the transition state (ES*); and the transition state complex must advance to an enzyme product complex (EP). The latter is finally competent to dissociate to product and free enzyme. The series of events can be shown thus:

The kinetics of simple reactions like the above were first characterized by biochemists Michaelis and Menten. The concepts underlying their analysis of enzyme kinetics continue to provide the cornerstone for understanding metabolism today, and for the development and clinical use of drugs aimed at selectively altering rate constants and interfering with the progress of disease states. The Michaelis-Menten equation is a quantitative description of the relationship among the rate of an enzyme-catalyzed reaction [v₁], the concentration of substrate [S] and 2 constants, V_max and K_m (which are set by the particular equation). The symbols used in the Michaelis-Menten equation refer to the reaction rate [v₁], maximum reaction rate (V_max), substrate concentration [S], and the Michaelis-Menten constant (K_m).

This fact provides a simple yet powerful bioanalytical tool that has been used to characterize both normal and altered enzymes, such as those that produce the symptoms of genetic diseases. Rearranging the Michaelis-Menten equation leads to:

From this equation it should be apparent that when the substrate concentration is half that required to support the maximum rate of reaction, the observed rate, v₁, will, be equal to V_max divided by 2; in other words, v₁ = [V_max/2]. At this substrate concentration V_max/v₁ will be exactly equal to 2, with the result that:

The latter is an algebraic statement of the fact that, for enzymes of the Michaelis-Menten type, when the observed reaction rate is half of the maximum possible reaction rate, the substrate concentration is numerically equal to the Michaelis-Menten constant. In this derivation, the units of K_m are those used to specify the concentration of S, usually molarity.

The Michaelis-Menten equation has the same form as the equation for a rectangular hyperbola; graphical analysis of reaction rate (v) versus substrate concentration [S] produces a hyperbolic rate plot (Figure 9-1).

The key features of the plot are marked by points A, B, and C. At high substrate concentrations, the rate of the reaction (the rate represented by point C) is almost equal to V_max, and the difference in rate at nearby concentrations of substrate is almost negligible. If the Michaelis-Menten plot is extrapolated to infinitely high substrate concentrations, the extrapolated rate is equal to V_max. When the reaction rate becomes independent of substrate concentration, or nearly so, the rate is said to be zero order. (Note that the reaction is zero order only with respect to this substrate. If the reaction has 2 substrates, it may or may not be zero order with respect to the second substrate). The very small differences in reaction velocity at substrate concentrations around point C (near V_max) reflect the fact that at these concentrations almost all the enzyme molecules are bound to substrate and the rate is virtually independent of substrate, hence zero order. At lower substrate concentrations, such as at points A and B, the lower reaction velocities indicate that at any moment only a portion of the enzyme molecules are bound to the substrate. In fact, at the substrate concentration denoted by point B, exactly half the enzyme molecules are in an ES complex at any instant and the rate is exactly one half of V_max. At substrate concentrations near point A the rate appears to be directly proportional to substrate concentration, and the reaction rate is said to be first order.

Inhibition of Enzyme-Catalyzed Reactions

To avoid dealing with curvilinear plots of enzyme-catalyzed reactions, biochemists Lineweaver and Burk introduced an analysis of enzyme kinetics based on double reciprocal rearrangement of the Michaelis-Menten equation such that straight line plots of 1/v versus 1/[S] are obtained where the slope reflects K_m/V_max and the y-intercept is 1/V_max. (Figure 9-2)

An alternative linear transformation of the Michaelis-Menten equation is the Eadie-Hofstee transformation

and when v/[S] is plotted on the y-axis versus v on the x-axis, the result is a linear plot with a slope of −1/K_m and the value V_max/K_m as the intercept on the y-axis and V_max as the intercept on the x-axis.

Enzyme inhibitors fall into 2 broad classes: those causing irreversible inactivation of enzymes and those whose inhibitory effects can be reversed. Irreversible inhibitors usually cause an inactivating, covalent modification of enzyme structure. The kinetic effect of irreversible inhibitors is to decrease the concentration of active enzyme, thus decreasing the maximum possible concentration of ES complex. Reversible inhibitors can be divided into 2 main categories: (1) competitive inhibitors and (2) noncompetitive inhibitors—with a third category, uncompetitive inhibitors, rarely encountered (Table 9-2).

The importance of K_I is that in all enzyme reactions where substrate, inhibitor, and enzyme interact, the normal K_m and/or V_max for substrate enzyme interaction appear to be altered. These changes are a consequence of the influence of K_i on the overall rate equation for the reaction. The effects of K_i are best observed in Lineweaver-Burk plots.

Competitive inhibitors always bind at the catalytic or active site of the enzyme. Most drugs that alter enzyme activity are of this type. Competitive inhibitors are especially attractive as clinical modulators of enzyme activity. A decreasing concentration of the inhibitor reverses the equilibrium restoring active-free enzyme. In addition, since substrate and competitive inhibitors both bind at the same site they compete with one another for binding. Raising the concentration of substrate, while holding the concentration of inhibitor constant, results in reversal of competitive inhibition.

Since high concentrations of substrate can displace virtually all competitive inhibitor bound to the active site of an enzyme, it becomes apparent that V_max is unchanged by competitive inhibitors. This characteristic of competitive inhibitors is reflected in the identical vertical-axis intercepts of Lineweaver-Burk plots, with and without inhibitor (Figure 9-3).

Since attaining V_max requires appreciably higher substrate concentrations in the presence of competitive inhibitor, K_m (the substrate concentration at half maximal velocity) is also higher, as demonstrated by the differing negative intercepts on the horizontal axis in panel B.

Analogously, panel C illustrates that noncompetitive inhibitors appear to have no effect on the intercept at the x-axis implying that noncompetitive inhibitors have no effect on the K_m of the enzymes they inhibit. Since noncompetitive inhibitors do not interfere in the equilibration of enzyme, substrate, and ES complexes, the K_m’s of Michaelis-Menten type enzymes are not expected to be affected by noncompetitive inhibitors, as demonstrated by x-axis intercepts in panel C. However, because complexes that contain inhibitor (ESI) are incapable of progressing to reaction products, the effect of a noncompetitive inhibitor is to reduce the concentration of ES complexes that can advance to product. Since V_max = k₂[E_total], and the concentration of competent E_total is diminished by the amount of ESI formed, noncompetitive inhibitors are expected to decrease V_max, as illustrated by the y-axis intercepts in panel C.

A corresponding analysis of uncompetitive inhibition leads to the expectation that these inhibitors should change the apparent values of K_m as well as V_max. Changing both constants leads to double reciprocal plots, in which intercepts on the x and y axes are proportionately changed; this leads to the production of parallel lines in inhibited and uninhibited reactions.

Regulation of Enzyme Activity

Enzyme concentration is continually modulated in response to physiological needs. Three principal mechanisms are known to regulate the concentration of active enzyme in tissues:

  1. Regulation of gene expression controls the quantity and rate of enzyme synthesis.

  2. Proteolytic enzyme activity determines the rate of enzyme degradation.

  3. Covalent modification of preexisting pools of inactive proenzymes produces active enzymes.

Enzyme synthesis and proteolytic degradation are comparatively slow mechanisms for regulating enzyme concentration, with response times of hours, days, or even weeks. Proenzyme activation is a more rapid method of increasing enzyme activity, but, as a regulatory mechanism, it has the disadvantage of not being a reversible process. Proenzymes are generally synthesized in abundance, stored in secretory granules, and covalently activated upon release from their storage sites.

In contrast to regulatory mechanisms that alter enzyme concentration, there is an important group of regulatory mechanisms that do not affect enzyme concentration, are reversible, and are rapid in action, and actually carry out most of the moment-to-moment physiological regulation of enzyme activity. These mechanisms include allosteric regulation, regulation by reversible covalent modification, and regulation by control proteins such as calmodulin. Reversible covalent modification is a major mechanism for the rapid and transient regulation of enzyme activity.

Allosteric Enzymes

In addition to simple enzymes that interact only with substrates and inhibitors, there is a class of enzymes that bind small, physiologically important molecules and modulate activity in ways other than those already discussed. These are known as allosteric enzymes and the small regulatory molecules that bind to these enzymes are known as allosteric effectors.

There are 2 ways that enzymatic activity can be altered by effectors: the V_max can be increased or decreased, or the K_m can be raised or lowered. Enzymes whose K_m is altered by effectors are said to be K-type enzymes and the effector a K-type effector. If V_max is altered, the enzyme and effector are said to be V-type. Many allosteric enzymes respond to multiple effectors with V-type and K-type behavior.

Enzymes in the Diagnosis of Pathology

The measurement of the serum levels of numerous enzymes has been shown to be of diagnostic significance. This is because the presence of these enzymes in the serum indicates that tissue or cellular damage has occurred resulting in the release of intracellular components into the blood. Hence, when a physician indicates that he/she is going to assay for liver enzymes, the purpose is to ascertain the potential for liver cell damage. Commonly assayed enzymes in the blood are the amino transferases: alanine transaminase (ALT; sometimes still referred to as serum glutamate-pyruvate aminotransferase [SGPT]) and aspartate aminotransferase (AST; also referred to as serum glutamate-oxaloacetate aminotransferase [SGOT]). In addition, the enzymes cardiac troponin I (cTnI), lactate dehydrogenase (LDH), and creatine kinase (CK; also called creatine phosphokinase [CPK]) are commonly measured in the diagnosis of cardiac infarct.

The typical liver enzymes measured are AST and ALT. ALT is particularly diagnostic of liver involvement as this enzyme is found predominantly in hepatocytes. When assaying for both ALT and AST, the ratio of the level of these 2 enzymes can also be diagnostic. Normally in liver disease or damage that is not of viral origin the ratio of ALT to AST is less than 1. However, with viral hepatitis the ALT:AST ratio will be greater than 1. Measurement of AST is useful not only for liver involvement but also for heart disease or damage. The level of AST elevation in the serum is directly proportional to the number of cells involved as well as on the time following injury that the AST assay was performed. Following injury, levels of AST rise within 8 hours and peak 24 to 36 hours later. Within 3 to 7 days the level of AST should return to preinjury levels, provided a continuous insult is not present or further injury does not occur. Although measurement of AST is not, in and of itself, diagnostic for myocardial infarction (MI), taken together with LDH and CK measurements (see below) the level of AST is useful for timing of the infarct.

Troponins are complexes composed of 3 regulatory proteins (troponin C, I, and T) attached to tropomyosin and are found in the grooves between actin filaments in muscle tissue. The troponins are found in skeletal and cardiac muscle but not in smooth muscle. Cardiac troponins found in the serum, specifically troponin I and T, are excellent markers for MI as well as for any other type of heart muscle damage. The measurement of plasma troponin I levels is highly diagnostic of necrosis of cardiac muscle. Serum levels of troponin I rise within 4 to 8 hours after the onset of chest pain caused by MI. The levels peak within 12 to 16 hours after the onset of infarction and return to baseline within 5 to 9 days. Although measurement of serum LDH fractions was once considered the ideal marker for onset and severity of a heart attack, the high specificity of troponin I to heart muscle necrosis make this protein the preferred marker to measure in patients suspected of suffering an MI.

The measurement of plasma LDH is useful as a diagnostic indicator of MI because this enzyme exists in 5 closely related, but slightly different forms (isozymes). The 5 types and their normal distribution and levels in nondisease/injury are listed below:

  1. LDH 1: Heart and red blood cells and is 17% to 27% of the normal serum total.

  2. LDH 2: Heart and red blood cells and is 27% to 37% of the normal serum total.

  3. LDH 3: Variety of organs and is 18% to 25% of the normal serum total.

  4. LDH 4: Variety of organs and is 3% to 8% of the normal serum total.

  5. LDH 5: Liver and skeletal muscle and is 0% to 5% of the normal serum total.

Following an MI the serum levels of LDH rise within 24 to 48 hours reaching a peak by 2 to 3 days and return to normal in 5 to 10 days. Especially diagnostic is a comparison of the LDH-1:LDH-2 ratio. Normally, this ratio is less than 1. A reversal of this ratio is referred to as a flipped LDH. Following an acute MI the flipped LDH ratio will appear in 12 to 24 hours and is definitely present by 48 hours in over 80% of patients. Also important is the fact that persons suffering chest pain due to angina only will not likely have altered LDH levels.

CPK is found primarily in heart and skeletal muscle as well as the brain. Therefore, measurement of serum CPK levels is a good diagnostic for injury to these tissues. The levels of CPK will rise within 6 hours of injury and peak by around 18 hours. If the injury is not persistent, the level of CK returns to normal within 2 to 3 days. Like LDH, there are tissue-specific isozymes of CPK and there designations are described below:

     CPK3 (CPK-MM) is the predominant isozyme in muscle and is 100% of the normal serum total.

     CPK2 (CPK-MB) accounts for about 35% of the CPK activity in cardiac muscle, but less than 5% in skeletal muscle and is 0% of the normal serum total.

     CPK1 (CPK-BB) is the characteristic isozyme in brain and is in significant amounts in smooth muscle and is 0% of the normal serum total.

Since most of the released CPK after a myocardial infarction is CPK-MB, an increased ratio of CPK-MB to total CPK may help in diagnosis of an acute infarction, but an increase of total CPK in itself may not. CPK-MB levels rise 3 to 6 hours after an MI and peak 12 to 24 hours later if no further damage occurs and returns to normal 12 to 48 hours after the infarct.

REVIEW QUESTIONS

The interconversion of kJ/mol and kcal/mol in the following questions can be done with the formula: 1kcal/mol = 4.184kJ/mol.

  1. You are studying the effects of a new drug on the activity of your favorite enzyme. The results of your assays indicate that the drug acts as a noncompetitive inhibitor. Which of the following changes to the K_m and V_max allowed that determination to be made regarding the drugs actions?

      A. K_m decreased, V_max increased

      B. K_m decreased, V_max unchanged

      C. K_m increased, V_max decreased

      D. K_m unchanged, V_max decreased

      E. K_m unchanged, V_max unchanged

Answer D: Noncompetitive inhibitors bind to enzyme or ES complex other than at the catalytic site and thus have no effect on substrate binding. However, the resultant ESI complex cannot form products. This type of inhibition cannot be reversed by the addition of more substrate. Under these conditions K_m appears unaltered, while V_max is decreased proportionately to inhibitor concentration.

  2. The drug acetazolamide is used in the treatment of glaucoma since it decreases the production of aqueous fluid, thereby reducing intraocular pressures. This drug is a noncompetitive inhibitor of carbonic anhydrase. Which of the following statements most accurately relates to the actions of this type of inhibitor?

      A. decreases both V_max and apparent K_m

      B. decreases apparent K_m

      C. decreases V_max

      D. increases apparent K_m

      E. increases V_max

Answer C: Noncompetitive inhibitors bind to enzyme or ES complex other than at the catalytic site and thus have no effect on substrate binding. However, the resultant ESI complex cannot form products. This type of inhibition cannot be reversed by the addition of more substrate. Under these conditions K_m appears unaltered while V_max is decreased proportionately to inhibitor concentration.

  3. Drugs that inhibit enzymes in a competitive manner will have which of the following effects on the Michaelis-Menten equation derived parameters: V_max and/or K_m?

      A. decrease both V_max and K_m

      B. decrease K_m but no effect on V_max

      C. decrease V_max but no change on K_m

      D. increase K_m but no change in V_max

      E. increase V_max but no increase in K_m

Answer D: Competitive inhibitors bind enzyme at the catalytic site, thereby competing with substrate for binding in a dynamic equilibrium-like process. This type of inhibition is reversible by the addition of higher concentrations of substrate. With competitive inhibitors the maximum velocity (V_max) of the reaction is unchanged, while the apparent affinity of the substrate to the binding site is decreased. In other words, K_m, as defined by the substrate concentration required for half maximal activity, is increased.

  4. Drugs that inhibit enzymes in a noncompetitive manner will have which of the following effects on the Michaelis-Menten equation derived parameters: V_max and/or K_m?

      A. decrease both V_max and K_m

      B. decrease K_m but no effect on V_max

      C. decrease V_max but no change on K_m

      D. increase K_m but no change in V_max

      E. increase V_max but no increase in K_m

Answer C: Noncompetitive inhibitors bind to enzyme or ES complex other than at the catalytic site and thus have no effect on substrate binding. However, the resultant ESI complex cannot form products. This type of inhibition cannot be reversed by the addition of more substrate. Under these conditions K_m appears unaltered, while V_max is decreased proportionately to inhibitor concentration.

  5. Drugs that inhibit enzymes in an uncompetitive manner will have which of the following effects on the Michaelis-Menten equation–derived parameters: V_max and/or K_m?

      A. decrease both V_max and K_m

      B. decrease K_m but no effect on V_max

      C. decrease V_max but no change on K_m

      D. increase K_m but no change in V_max

      E. increase V_max but no increase in K_m

Answer A: Uncompetitive enzyme inhibition takes place when an enzyme inhibitor binds only to the complex formed between the enzyme and the substrate (the ES complex). This reduction in the effective concentration to the ES complex increases the enzyme’s apparent affinity for the substrate (K_m is lowered) and decreases the maximum enzyme activity (V_max), as it takes longer for the substrate or product to leave the active site. Uncompetitive inhibition works best when substrate concentration is high.

  6. Which of the following best explains why noncompetitive inhibitors of enzymes generally make better pharmaceuticals?

      A. their inhibitory activity is unaffected by substrate concentration

      B. they affect only K_m and not V_max

      C. they bind to the same site as the substrate

      D. they increase the rate of degradation of the target enzyme

      E. they irreversibly inhibit the enzyme

Answer A: Pharmaceuticals that are designed to inhibit an enzyme in a noncompetitive manner are much more effective than inhibitors that function competitively because any increase in substrate will have no effect on the level of inhibition.

  7. The standard free-energy change (ΔG⁰′) for the hydrolysis of phosphoenolpyruvate (PEP) is −14.8 kcal/mol.’ The ΔG⁰’ for the hydrolysis of ATP is −7.3 kcal/mol. What is the ΔG⁰′ for the following reaction?

      Phosphoenolpyruvate + ADP → pyruvate + ATP

      A. −22.1 kcal/mol

      B. −14.8 kcal/mol

      C. −7.5 kcal/mol

      D. −7.3 kcal/mol

      E. +7.3 kcal/mol

Answer C: The reaction, as written, actually represents the coupling of 2 distinct reactions. One is the hydrolysis of PEP to pyruvate and the other is the synthesis of ATP from ADP. Each of the 2 reactions has their own ΔG⁰′. To simplify, a coupled biochemical reaction is one where the free energy of a thermodynamically favorable reaction (PEP to pyruvate) is used to drive a thermodynamically unfavorable one (the formation of ATP from ADP) by coupling the 2 reactions. The coupling in this case is carried out by the enzyme pyruvate kinase. When 2 reactions are coupled, the resultant free-energy change is simply the mathematical sum of the 2 ΔG⁰′ values for each reaction. In the case of the formation of ATP from ADP, the ΔG⁰’ for the reaction in that direction is equal to, but mathematically opposite of, the reciprocal reaction.

  8. Which of the following indicates whether the reaction, A → B, is thermodynamically favorable under standard state conditions at near physiological pH?

      A. ΔG′

      B. ΔG⁰′

      C. K_eq

      D. K_m

      E. V_max

Answer B: The Gibbs free-energy equation (ΔG = ΔT − TΔS) states the relationship between the change in fee energy (ΔG), the change in enthalpy (ΔH), the change in entropy (ΔS), and the temperature (T). When a reaction is carried out under standard state conditions of temperature and pressure, the values are denoted by the superscript 0, and the addition of the prime (ΔG⁰′) indicates that the reaction is carried out at pH 7.0. The spontaneity of a reaction is determined by comparing the free energy of the system before the reaction with the free energy of the system after reaction. If the system, after reaction, has less free energy than before the reaction (ie, the ΔG⁰′ value is negative), the reaction is thermodynamically favorable. One should also note that even if ΔG⁰′ is positive, the reaction can still become spontaneous due to coupling of the thermodynamically unfavorable reaction to a reaction (or reactions) that is favorable.

  9. Which of the following relates to allosteric enzymes?

      A. they are inactive on their substrates in the absence of accessory proteins

      B. they can bind more than one substrate simultaneously

      C. they can convert multiple different substrates to product

      D. they possess binding sites for regulatory molecules in addition to substrate-binding sites

      E. they require additional cofactor-binding sites for subsequent binding of substrate

Answer D: Allosteric enzymes are enzymes that change their conformational orientations upon binding of an effector which results in an apparent change in binding affinity at a different ligand-binding site. This action at a distance through binding of one ligand (or allosteric effector) affecting the binding of another at a distinctly different site is the essence of the allosteric concept.

10. Assessment of enzyme activity under standard condition and plotted as a linear Lineweaver-Burk plot gives the green line (x). The red line (y) was most likely obtained by addition of which of the following to this enzyme assay?

      A. allosteric activator

      B. competitive inhibitor

      C. double the amount of enzyme

      D. half the amount of enzyme

      E. noncompetitive inhibitor

Answer B: Competitive inhibitors bind enzyme at the catalytic site, thereby competing with substrate for binding in a dynamic equilibrium-like process. This type of inhibition is reversible by the addition of higher concentrations of substrate. With competitive inhibitors the maximum velocity (V_max) of the reaction is unchanged, while the apparent affinity of the substrate to the binding site is decreased. In other words, K_m, as defined by the substrate concentration required for half maximal activity, is increased. These changes are visible in Lineweaver-Burk plots where the Y-axis intercept is unchanged, while the X-axis intercept is closer to the Y-axis. This is indicative of a larger value for K_m.

11. Kinases are a class of enzymes that incorporate a phosphate onto their substrates. The catalytic activity of kinases classifies them as members of which of the following enzyme families?

      A. hydrolases

      B. isomerases

      C. ligases

      D. oxidoreductases

      E. transferases

Answer E: Kinases are any of the various enzymes that catalyze the transfer of a phosphate group from a donor, such as ADP or ATP, to an acceptor. As such, kinases represent a subfamily of the transferase class of enzyme.

12. You are studying the effects of the addition of a potential pharmaceutical compound on the activity of your enzyme of interest. You find that the addition of the compound results in an increase in the K_m of the reaction but does not affect the V_max. Which of the following defines the inhibitory action of the compound?

      A. competitive

      B. noncompetitive

      C. suicide

      D. uncompetitive

Answer A: Competitive inhibitors bind enzyme at the catalytic site, thereby competing with substrate for binding in a dynamic equilibrium-like process. This type of inhibition is reversible by the addition of higher concentrations of substrate. With competitive inhibitors the maximum velocity (V_max) of the reaction is unchanged, while the apparent affinity of the substrate to the binding site is decreased. In other words, K_m, as defined by the substrate concentration required for half maximal activity, is increased.

13. If the ΔG⁰′ of the reaction A → B is −40 kJ/mol, under standard conditions which of the following is correct?

      A. the reaction is at equilibrium

      B. ΔG⁰′ will never reach equilibrium

      C. ΔG⁰′ will not occur spontaneously

      D. ΔG⁰′ will proceed at a rapid rate

      E. ΔG⁰′ will proceed spontaneously from left to right

Answer E: ΔG⁰′ is the Gibbs free energy value when a reaction is carried out under standard state conditions of temperature and pressure and at pH 7. The spontaneity of a reaction is determined by comparing the free energy of the system before the reaction with the free energy of the system after the reaction. If the system after reaction has less free energy than before the reaction (ie, the ΔG⁰′ value is negative), the reaction is thermodynamically favorable. Thus, since the reaction as written has a negative free-energy value, it will proceed favorably as written.

14. For the reaction A → B, ΔG⁰′ = −60 kJ/mol. The reaction is started with 10 mmol of A and no B is initially present. After 24 hours, analysis reveals the presence of 2 mmol of B and 8 mmol of A. Which of the following is the most likely explanation?

      A. A and B have reached equilibrium concentrations

      B. an enzyme has shifted the equilibrium toward A

      C. B formation is kinetically slow; equilibrium has not been reached by 24 hours

      D. formation of B is thermodynamically unfavorable

      E. the result described is impossible, given the fact that ΔG⁰′ is −60 kJ/mol

Answer C: When this reaction is at equilibrium, the amount of product (B) would be greater than that of substrate (A) since the reaction proceeds favorably from A to B. Since the amount of B is still less than that of A, the reaction has not yet reached equilibrium even after 24 hours. The reaction is still thermodynamically favorable since the free-energy change for the reaction as written has a large negative value. In addition, since equilibrium has not yet been reached in the time frame of the analysis, it is implied that the reaction proceeds with slow kinetics.

15. When a mixture of 3-phosphoglycerate and 2-phosphoglycerate is incubated at 25°C with phosphoglycerate mutase until equilibrium is reached, the final mixture contains 6 times as much 2-phosphoglycerate as 3-phosphoglycerate. Which one of the following statements is most nearly correct, when applied to the reaction as written? (R = 8.315 J/mol·K; T = 298 K)

3-Phosphoglycerate → 2-phosphoglycerate

      A. ΔG⁰′ is −4.44 kJ/mol

      B. ΔG⁰′ is zero

      C. ΔG⁰′ is +12.7 kJ/mol

      D. ΔG⁰′ is incalculably large and positive

      E. ΔG⁰′ cannot be calculated from the information given

Answer A: Since there is much more product than substrate when the reaction has reached equilibrium, the reaction can be assumed to be favorable when proceeding in the direction as written. Therefore, the free-energy change value for the reaction is most likely to be somewhat negative.

16. When a mixture of glucose 6-phosphate and fructose 6-phosphate is incubated with the enzyme phosphohexose isomerase (which catalyzes the interconversion of these 2 sugars) until equilibrium is reached, the final mixture contains twice as much glucose 6-phosphate as fructose 6-phosphate. Which one of the following statements is best applied to this reaction outlined below? (R = 8.315 J/mol·K; T = 298 K)

      Glucose 6-phosphate → fructose 6-phosphate

      A. ΔG⁰′ is incalculably large and negative

      B. ΔG⁰′ is –1.72 kJ/mol

      C. ΔG⁰′ is zero

      D. ΔG⁰′ is +1.72 kJ/mol

      E. ΔG⁰′ is incalculably large and positive

Answer D: Since there is more substrate than product when the reaction has reached equilibrium, the reaction can be assumed to be less than favorable when proceeding in the direction as written. Therefore, the free-energy change value for the reaction is most likely to be slightly positive.

17. Hydrolysis of 1 M glucose 6-phosphate catalyzed by glucose 6-phosphatase is 99% complete at equilibrium (ie, only 1% of the substrate remains). Which of the following statements is most nearly correct? (R = 8.315 J/mol·K; T = 298 K)

      A. ΔG⁰′ = −11 kJ/mol

      B. ΔG⁰′ = −5 kJ/mol

      C. ΔG⁰′ = 0 kJ/mol

      D. ΔG⁰′ = +11 kJ/mol

      E. ΔG⁰′ cannot be determined from the information given

Answer A: At equilibrium, almost all of the substrate has been converted to product, therefore the reaction can be assumed to be highly favorable when proceeding in the direction of glucose production from glucose 6-phosphate, the direction catalyzed by glucose 6-phosphatase. Therefore, the most likely free-energy change value for the reaction can be expected to be quite negative.

18. The reaction A + B → C has a ΔG⁰′ of −20 kJ/mol at 25°C. Starting under standard conditions, one can predict which of the following?

      A. at equilibrium, the concentration of B will exceed the concentration of A

      B. at equilibrium, the concentration of C will be less than the concentration of A

      C. at equilibrium, the concentration of C will be much greater than the concentration of A or B

      D. C will rapidly break down to A + B

      E. when A and B are mixed, the reaction will proceed rapidly toward formation of C

Answer C: Reactions with large negative free-energy changes are highly favorable. When these types of reactions reach equilibrium, it is expected that the concentration of product(s) would be much greater than that of substrate(s).

19. Which of the following reactions has the largest negative value for the standard free energy-change (ΔG⁰′)?

      A. fructose 1,6-bisphosphate → fructose 6-phosphate

      B. glucose 6-phosphate → fructose 6-phosphate

      C. glycerol 3-phosphate → dihydroxyacetone phosphate

      D. glycerol → glycerol 3-phosphate

      E. phosphoenolpyruvate → pyruvate

Answer E: The standard free-energy change for the reaction catalyzed by pyruvate kinase (PK) is approximately −7.5kcal/mol. This reaction is 1 of the 2 reactions of glycolysis that releases enough energy such that it can be coupled to the synthesis of ATP.

20. For the following reaction, ΔG ⁰′ = +29.7 kJ/mol

L-Malate + NAD⁺ → oxaloacetate + NADH + H⁺

Which of the following correctly describes the characteristics of the reaction as written?

      A. can never occur in a cell

      B. can occur in a cell only if it is coupled to another reaction for which ΔG ⁰′ is positive

      C. can occur only in a cell in which NADH is converted to NAD⁺ by electron transport

      D. cannot occur because of its large activation energy

      E. may occur in cells at some concentrations of substrate and product

Answer E: Although the reaction, as written, has a positive free-energy change and is thus thermodynamically favorable, it is known that within cells this reaction does indeed proceed in the forward direction. The means by which this TCA cycle (tricarboxylic acid cycle) reaction is driven in the forward direction, against the positive free-energy change, is by rapid removal of the product (OAA) via the enzyme catalyzing the next reaction in the cycle. Therefore, when certain reactions are coupled to subsequent favorable reactions that rapidly remove the product of the initial reaction, the less than favorable reaction can occur under these conditions.

21. In glycolysis, fructose 1,6-bisphosphate is converted to 2 products with a standard free-energy change (ΔG⁰′) of 23.8 kJ/mol. Under what conditions, encountered in a normal cell, will the free-energy change be negative, enabling the reaction to proceed spontaneously to the right?

      A. under standard conditions, enough energy is released to drive the reaction to the right

      B. the reaction will not go to the right spontaneously under any conditions because the ΔG⁰′ is positive

      C. the reaction will proceed spontaneously to the right if there is a high concentration of products relative to the concentration of fructose 1,6-bisphosphate

      D. the reaction will proceed spontaneously to the right if there is a high concentration of fructose 1,6-bisphosphate relative to the concentration of products

Answer A: Within the cell, under standard conditions, the 2 products of the hydrolysis of fructose 1,6-bisphosphate (F1,6BP) are rapidly removed by the subsequent reactions of glycolysis. This is, of course, assuming that the energy needs of the cell are high which would likely be the case since F1,6BP has been formed and would only be formed when the cell needs to oxidize glucose for energy.

22. During glycolysis, glucose 1-phosphate is converted to fructose 6-phosphate in 2 successive reactions:

For the overall reaction which of the following correctly identifies the free-energy change?

      A. −8.8 kJ/mol

      B. −7.1 kJ/mol

      C. −5.4 kJ/mol

      D. +5.4 kJ/mol

      E. +8.8 kJ/mol

Answer C: When 2 reactions are coupled, the resultant free-energy change is the arithmetic sum of the ΔG⁰′ values for the coupled reactions. Therefore, the correct free-energy change for these 2 reactions is the sum of their individual free-energy change values.

23. The standard free-energy changes for the reactions below are given.

      A. −73.5 kJ/mol

      B. −12.5 kJ/mol

      C. +12.5 kJ/mol

      D. +73.5 kJ/mol

      E. ΔG⁰′ cannot be calculated without Keq’

Answer B: When 2 reactions are coupled, the resultant free-energy change is the arithmetic sum of the ΔG⁰′ values for the coupled reactions. In the case of the coupled reaction, instead of energy input from the hydrolysis of ATP the opposite reaction is being coupled. The ΔG⁰′ value for a given reaction is its reciprocal value when the reaction is running in the opposite direction to that for which the ΔG⁰′ value is denoted. Thus, for ATP synthesis from ADP, the ΔG⁰′ value would be +30.5kJ/mol. When the 2 reactions are coupled, there is sufficient energy released from the hydrolysis of phosphocreatine to drive the synthesis of ATP from ADP. The ΔG⁰′ value for the coupled reactions is the simple arithmetic sum of the ΔG⁰′ values for the individual reactions.

24. The ΔG⁰′ values for the 2 reactions shown below are given:

      A. −34.1 kJ/mol

      B. −32.2 kJ/mol

      C. −30.3 kJ/mol

      D. +61.9 kJ/mol

      E. +34.1 kJ/mol

Answer C: When 2 reactions are coupled, the resultant free-energy change is the arithmetic sum of the ΔG⁰′ values for the coupled reactions. In the case of the coupled reaction, instead of energy input from the hydrolysis of ATP the opposite reaction is being coupled. The ΔG⁰′ value for a given reaction is its reciprocal value when the reaction is running in the opposite direction to that for which the ΔG⁰′ value is denoted. Thus, for ATP synthesis from ADP, the ΔG⁰′ value would be +30.5kJ/mol. When the 2 reactions are coupled, there is sufficient energy released from the hydrolysis of phosphocreatine to drive the synthesis of ATP from ADP. The ΔG⁰′ value for the coupled reactions is the simple arithmetic sum of the ΔG⁰′ values for the individual reactions.

25. Which one of the following compounds possesses the least negative free energy of hydrolysis?

      A. 1,3-bisphosphoglycerate

      B. 3-phosphoglycerate

      C. ADP

      D. phosphoenolpyruvate

      E. thioesters (eg, acetyl-CoA)

Answer B: Only four of these molecules possess a negative ΔG of hydrolysis. The hydrolysis of 3-phosphoglycerate occurs with a free energy change of +1.1kcal/mol. However, within cells undergoing glycolysis, due to the rapid removal of products of one reaction to substrates of the next reaction the free energy released from 3-phosphoglycerate is an apparent –0.9kcal/mol. The free energy of 1,3-bisphosphoglycerate hydrolysis is –4.5kcal/mol, ADP is –7.3kcal/mol, phosphoenolpyruvate is –14.8kcal/mol, and acetyl-CoA is –8.5kcal/mol.

26. A 57-year-old man has just returned from an overseas trip and reports having had severe substernal chest pain 3 days ago. Which of the following is the most appropriate laboratory test to order for this patient?

      A. aspartate aminotransferase

      B. creatine kinase, MB fraction

      C. creatine kinase, total

      D. lactate dehydrogenase, LD₁ fraction

      E. troponin I

Answer E: Troponin I is now the method of choice for the laboratory diagnosis of MI. There is a detectable increase within 4 to 8 hours of the infarction and the peak level is reached within 12 to 16 hours. Levels do not return to baseline for 3 to 10 days making it an appropriate test for this patient who was 3 days postinfarction. Aspartate aminotransferase was the first serum enzyme marker used for the diagnosis of MI, but it has poor specificity and sensitivity compared to newer markers and is no longer used for this purpose. Creatine kinase total is not used for the diagnosis of MI since the cardiac fraction (MB) can be overwhelmed by the presence of the skeletal muscle fraction. Creatine kinase-MB is still being used in some institutions but it returns to baseline in 2 to 3 days and would not be useful for this patient. Lactate dehydrogenase (LD₁ fraction) returns to baseline later than creatine kinase but has been replaced by troponin I and is seldom used.

27. Enzymes are efficient catalysts because they can do which of the following?

      A. catalyze reactions that otherwise would not occur

      B. decrease the free energy of activation of reactants

      C. decrease the standard free-energy change (ΔG⁰′) of reactions

      D. prevent the conversion of product to substrate

      E. shift the equilibrium of reactions toward more complete conversion to product

Answer B: Like inorganic catalysts, enzymes are efficient catalysts because they are capable of lowering the energy of activation of the reactions in which they are active. A catalyst is a substance that increases the rate of a chemical reaction without itself being changed by the reaction. Enzymes affect the rate but not the equilibrium constant of a reaction. Catalysts, like enzymes, merely reduce the time that a thermodynamically favored reaction requires to reach equilibrium.

28. Which of the following is the mechanism by which allosteric effectors influence enzymatic activity?

      A. binding at the catalytic site

      B. changing the conformation of the enzyme

      C. forming high-energy complexes with substrate

      D. increasing hydration of the active site

      E. protecting against degradation of the enzyme

Answer B: Allosteric enzymes are enzymes whose conformation changes upon binding of an allosteric effector. This binding results in an apparent change in binding affinity in the enzyme at a different ligand-binding site. This action at a distance through binding of one ligand (or allosteric effector) affecting the binding of another at a distinctly different site is the essence of the allosteric concept.

29. Which of the following conditions is most likely if the reaction shown in the diagram is at equilibrium?

      (K₁ = 500/second, K₂ = 5/second)

      A. all of A is consumed

      B. all of B is consumed

      C. both A and B are present, but the concentration of A is greater than that of B

      D. both A and B are present, but the concentration of B is greater than that of A

      E. the concentrations of A and B are equal

Answer D: If the rate constant for the reaction in the direction of the formation of B is 100 times greater than for the reaction in the direction of A formation, the formation of B will be the favored direction. Therefore, when the reaction reaches equilibrium, there will be substantially more B than A present in the reaction vessel.

30. A newborn screening test for galactosemia is positive in your patient. Genetic studies demonstrate the infant harbors a particular mutation in the GALT gene that impairs the activity of galactose-1-phosphate uridyltransferase. The impaired enzyme cannot convert galactose 1-phosphate and UDP-glucose to UDP-galactose and glucose 1-phosphate. Galactose 1-phosphate accumulates and affects the activity of UTP-dependent glucose-1-phosphate pyrophosphorylase (UGP). Which of the following actions by galactose 1-phosphate proves that it is a competitive inhibitor of UGP?

      A. decreases the apparent K_m for glucose 1-phosphate

      B. decreases both the V_max and the apparent K_m for glucose 1-phosphate

      C. decreases the V_max for glucose 1-phosphate

      D. increases the apparent K_m for glucose 1-phosphate

      E. increases the V_max for glucose 1-phosphate

Answer D: Competitive inhibitors bind enzyme at the catalytic site, thereby competing with substrate for binding in a dynamic equilibrium-like process. With competitive inhibitors the maximum velocity (V_max) of the reaction is unchanged, while the apparent affinity of the substrate to the binding site is decreased. In other words, K_m, as defined by the substrate concentration required for half maximal activity, is increased.

31. You are studying the effects of compounds that inhibit the activity of succinate dehydrogenase, SDH. You demonstrate that malonate is a competitive inhibitor of this enzyme. Which of the following sets of findings would most likely provide ideal conditions for maximum inhibition of SDH by malonate?

Answer D: Competitive inhibitors affect only the K_m because they displace the substrate by binding to the same site on the enzyme. Because competitive inhibitors bind to the substrate binding site they exhibit K_i values very similar, if not identical, to the K_m of the substrate. The addition of more substrate can compete with the inhibitor, therefore, in order to effectively inhibit the enzyme one needs to provide a higher concentration of the inhibitor. Given these facts only one of the choices is associated with increased K_m value, a similar K_i value to the substrate K_m, and provides excess inhibitor.

32. Given the table of values for the hydrolysis of some phosphorylated compounds, which of the following best describes the ΔG⁰′ for the following reaction?

      A. −22.1 kcal/mol

      B. −14.8 kcal/mol

      C. −7.8 kcal/mol

      D. −7.0 kcal/mol

      E. −5.0 kcal/mol

Answer C: When 2 reactions are coupled, the resultant free-energy change is the arithmetic sum of the ΔG⁰′ values for the coupled reactions. In the case of the coupled reaction, ATP is not being hydrolyzed but instead is being formed. In this case the ΔG⁰′ value for a given reaction is its reciprocal value when the reaction is running in the opposite direction to that for which the ΔG⁰′ value is denoted. Thus, for ATP synthesis from ADP, the ΔG⁰′ value would be +7.0kcal/mol. When the 2 reactions are coupled, there is sufficient energy released from the hydrolysis of phosphoenolpyruvate to drive the synthesis of ATP from ADP. The ΔG⁰′ value for the coupled reactions is the simple arithmetic sum of the ΔG⁰′ values for the individual reactions.

33. In the absence of an enzyme, the conversion of X → Y exhibits a ΔG⁰′ of +5.0 kJ/mol. Which of the following best describes the ΔG⁰′ of this reaction in the presence of an enzyme that accelerates the reaction 100 fold?

      A. −500 kJ/mol

      B. −50 kJ/mol

      C. −5 kJ/mol

      D. +5 kJ/mol

      E. +50 kJ/mol

Answer D: Enzymes affect the rate but not the equilibrium constant of a chemical reaction. Thus, although the enzyme in this example may indeed accelerate the rate of the reaction 100 fold it has no effect on the equilibrium constant. The equilibrium constant is related to the free-energy change of a reaction via the equation: ΔG = ΔG⁰ + RTlnK_eq.

34. The hydrolysis of ATP to ADP and inorganic phosphate proceeds with a ΔG⁰′ of −7kcal/mol. The hydrolysis of phosphocreatine has a ΔG⁰′ of −10kcal/mol. In the reaction catalyzed by creatine kinase (creatine phosphokinase), in the direction of phosphocreatine formation, which of the following is the overall ΔG⁰′ in kcal/mol?

      A. −17

      B. 17 × TΔS

      C. −3

      D. −3 × TΔS

      E. +3

Answer E: When 2 reactions are coupled, the resultant free-energy change is the arithmetic sum of the ΔG⁰′ values for the coupled reactions. In the case of the coupled reaction phosphocreatine is being formed not being hydrolyzed, therefore the ΔG⁰′ value for the reaction in that direction is the reciprocal value of that for phosphocreatine hydrolysis +10.0 kcal/mol. The ΔG⁰′ value for the coupled reactions is the simple arithmetic sum of the ΔG⁰′ values for the individual reactions.

35. Which of the following is most likely to increase as a result of the binding of an activating effector to the regulatory site of a multi-subunit, allosterically regulated enzyme?

      A. 1/V_max

      B. acetylation of inactive subunits

      C. K_m

      D. proportion of subunits in an active conformation

      E. ribosylation of inactive subunits

Answer D: Allosteric enzymes are enzymes whose conformation changes upon binding of an allosteric effector. This binding results in an apparent change in binding affinity in the enzyme at a different ligand-binding site. This action at a distance through binding of one ligand (or allosteric effector) affecting the binding of another at a distinctly different site is the essence of the allosteric concept.

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