BIOCHEMISTRY AND MOLECULAR BIOLOGY Problem Unit Two ...

SIU School o f Medic ine BIOCHEMISTRY

Enzymes/Membrane Transport

t

BIOCHEMISTRYAND

MOLECULAR BIOLOGY

Problem Unit Two1999/2000

Enzymes/Membrane TransporCopyright 1999, E.C. Niederhoffer. All Rights Reserved.All trademarks and copyrights are the property of their respectiveowners.

Modu le 1: Enzyme Kinetics

Modu le 2: Clinical Enzymology

Modu le 3: Membrane Transport

Faculty: P.M.D. Hardwicke Problem Unit 2 - Page 1



Faculty: Dr. Peter M.D. HardwickeBiochemistry & Molecular Biology

Office: 210 Neckers Bldg.email: [email protected]:453-6469

Learning Resources: ESTIMATED WORK TIME: 40 hours.

A. This study guide is provided in two forms: printed and elec-tronic (cowritten and produced by Dr. E.C. Niederhoffer, Biochemis-try and Molecular Biology). It is best viewed in electronic form as apdf file which can be read on your computer using Adobe AcrobatReader. See Appendix I for an introduction on how to view a pdffile. The pdf file can be downloaded from the biochemistry server(http://www.siu.edu/departments/biochem) and Acrobat Readercan be downloaded free from Adobe’s web page (http://www.adobe.com/acrobat). They should also be installed on the stu-dent computers. There are a number of advantages to using the elec-tronic version including color, a hypertext index, and hypertext linkswithin the text. Hypertext links in the text body are in blue under-lined characters (such as this). Clicking on these will lead to a jumpto the linked material for further details. The destination material isindicated by red underlined characters (such as this). (Clicking onthe black double arrows in the menu bar will allow you to “hyper-jump” back and forth.)

This and other study guides are provided to help you focus on thetopics that are important in the biochemistry curriculum. These aredesigned to guide your studying and provide information that maynot be readily available in other resources. They are not designed toreplace textbooks, and are not intended to be complete. They areguides for starting your reading and reviewing the material at a laterdate.

B. Textbooks:

1. Devlin, Textbook of Biochemistry with Clinical Correla-tions, 4th ed. ('97), Wiley-Liss. Core text for Biochemistry& Molecular Biology.

2. Champ & Harvey, Lippincotts Illustrated Reviews: Bio-chemistry, 2nd ed. (‘94), Lippincott. Efficient presentationof basic principles.

3. Murray et al., Harper's Biochemistry, (24th ed.) ('96),Appleton & Lange. An excellent review text for examina-


http://www.siu.edu/departments/biochem

http://www.adobe.com/acrobat




tions.

4. Marks, Marks, and Smith, Basic Medical Biochemistry: AClinical Approach, (‘96), Williams & Wilkins. Good basicpresentation with clinical relevance.

5. Cohn and Roth, Biochemistry and Disease, (‘96), Williams& Wilkins. A good bridge between the basic sciences andclinical medicine.

6. Garrett and Grishham, Biochemistry, 1st ed., (‘95), Saun-ders College Publishing.

7. Garrett and Grishham, Molecular Aspects of Cell Biology,1st ed., (‘95), Saunders College Publishing.

Most texts of biochemistry have sections on enzymes and membranetransport. The content of the subject is much the same from text totext; the differences are basically in style and rigor. The Study Guide,Pretest, and Post Test in the Problem Unit will set the level of rigorexpected of you. Read the sections on enzymes and membrane trans-port in several texts. What differences there will be between thesetexts and the Study Guide will be helpful to you in gaining perspec-tive on the subject. Additional material can be found on the web atthe National Institutes of Health (http://www.nih.gov), theNational Library of Medicine (http://www.nlm.nih.gov), and thefree MEDLINE PubMED Search system at the National Library ofMedicine (http://www3.ncbi.nlm.nih.gov/PubMed/).

C. Journals/Reviews.

You may find worthwhile reading in some of the more popular jour-nals and review series (see also the searchable SIU-SOM database).These resources typically contain specific articles involving enzymesand membrane transport. Suggestions for journals include AmericanFamily Physician, Journal of Biological Chemistry, Nature, Science, andScientific American (and SA’s Science and Medicine). Excellent reviewsmay be found in the Annual Review of Biochemistry, Cell and Develop-mental Biology, Genetics, Medicine, and Microbiology.

D. Lecture/Discussions

Especially recommended for those who have not had biochemistryand for those who have questions.

Evaluation Criteria: A written examination will be scheduled. Answers to questions andthe solving of problems will be judged against the learning resources.Examples of exam questions are given in the Problem Sets. The passlevel is 70%.


http://www.siu.edu/departments/biochem/MLdatabase.html

http://www.nih.gov/

http://www.nlm.nih.gov/

http://www3.ncbi.nlm.nih.gov



Module 1: Enzyme KineticsIntroduction: Biochemical reactions that occur in living cells are in most respects

ordinary chemical reactions. What makes these reactions unique isthat they proceed very rapidly at relatively low temperatures (physio-logical temperature, 37°C or 98.6°F, is low when compared to thechemical reactions used in industrial processes).

These low temperature reactions have accelerated rates because of theaction of very efficient catalysts which we refer to as enzymes. Dif-ferent cells have different numbers and types of enzymes. The exactnumber of enzymes in any cell remains unknown, but there are cer-tainly thousands. There may be as many as 100,000 in some unicel-lular organisms.

Enzymes, then, are necessary for the normal functioning of cells.Disease states may be caused by the absence or alteration of anenzyme (i.e., a genetic disease), the introduction of an enzymeinhibitor (i.e., a bacterial toxin, a chemical, etc.), the overproductionof an enzyme, or the introduction of a foreign enzyme (i.e., viralinfection). Enzymes are popular therapeutic targets. Pharmaceuti-cals are frequently enzyme inhibitors.

Enzyme assays are important in diagnosis. A diminution or increasein enzyme activities in tissues and fluids is indicative of the variouscauses of disease listed above. In addition, injury may release tissueenzymes to the blood and thus, their concentrations in the blood canbe used to locate the site of injury. Temperature (fever) affects therate of enzyme-catalyzed reactions, as do other physical factors.

For these reasons, it is necessary to have an understanding ofenzymes, how they act, how they are made, and how their activity iscontrolled.

Objectives: 1. After reading a passage from a medical journal, monograph, ortextbook that describes a clinical investigation of changes in enzymeactivities, or a clinical assay procedure, or which gives a moleculardescription of an enzyme, answer questions about the passage (whichmay involve drawing inferences or conclusions) or use the informa-tion given for the solving of a problem.

Examples of the kinds of passages about which understanding isrequired are found within this Study Guide. Examples of the types ofquestions and problems to be solved are included in the Problem




Sets.

In order to accomplish objective 1, you will need to be able to do thefollowing, which are also objectives:

2. Define the terms in the Nomenclature and Vocabulary list anduse them properly in answering questions concerning this module.

Nomenclature and Vocabulary:

Activation energy Activator

Active site Allosteric enzyme

Binding site Catalyst

Catalytic site Coenzyme

Cofactor Competitive inhibition

Dissociation constant Enzyme assay

Enzyme Enzyme inhibitor

Enzyme-substrate complex First-order reaction

Free energy of reaction Induced fit

Isoenzyme Isozyme

kcat Km

Lineweaver-Burke plot Maximal velocity (Vmax)

Mechanism-based inhibitor Michaelis constant (Km)

Michaelis-Menten equation Microsomal enzyme

Monomeric enzyme Negative allosteric effector

Noncompetitive inhibition Oligomeric enzyme

Ordered Positive allosteric effector

Product inhibition Prosthetic group

Protomer Random

Saturation velocity Sequential mechanism

Specific activity Steady state

Substrate Subunit

Suicide substrates Turnover number

Unit of enzyme activity Vmax

Vitamin Zero-order reaction

Zymogen

3. Enzymes have systematic and trivial names that must be learned.Describe the action of given enzymes from their systematic namesand the action of those enzymes that have the more obvious trivial




names (i.e., alkaline phosphatase, urease, ribonuclease). [The actionof those enzymes with less obvious trivial names (i.e., aldolase, eno-lase, trypsin) will have to be picked up as your studies proceed.]

4. Explain the concept of the active site of an enzyme.

5. Discuss the effects of the following on enzyme activity andunderstand the origin of the relationships in terms of enzyme cata-lyzed reactions:

ActivatorsCoenzymesCompetitive inhibitorsEnzyme concentrationIrreversible inhibitorsNoncompetitive inhibitorspHProductsSubstrate concentrationTemperature

6. Discuss the criteria that must be met for a valid enzyme assayand for using an enzyme as a reagent.

7. Describe the construction and interpretation of a linearized formof the Michaelis-Menton equation (Lineweaver-Burk plot) and beable to evaluate the pertinent constants from it.

8. By plotting v against [S], show the relationship of brain hexoki-nase and liver glucokinase to the normal blood glucose concentra-tion. For this plot assume the Vmax values are identical for the twoenzymes.

9. Discuss the rationale for design of mechanism-base inhibitorsand their therapeutic applications.

10. Use the material covered on the above objectives and previousobjectives to solve new, related problems such as those given in thePrestest and Post Test.

Key Words: Binding sites BiochemistryCoenzymes Enzyme activationEnzyme inhibitors EnzymesKinetics




Pretest:A Pretest is available for those interested in assessing their currentknowledge base and conceptual understanding. You may hyper-jumpto Study Guide-1.

1. Serum alkaline phosphatase is a phosphomonoesterase. The pHoptimum of its activity is at pH 9.0. Free -SH groups inhibit theactivity of this enzyme, while magnesium ions enhance it. Thereare two possible ways of measuring the activity of phosphatase:first, determination of phosphate liberated by the enzyme perunit of time, and second, determination of the organic part of aphosphoric acid ester used as substrate, which is liberated underthe same conditions.

Sample questions:

a. What can be inferred about this enzyme from the terms alka-line phosphatase and phosphomonoesterase? answer

b. The anticoagulants EDTA and citrate may not be used inthis assay. Why? answer

c. Must the reaction be done at exactly pH 9.0? Why? answer

2. Skeletal muscle aldolase, which catalyzes the reverse aldol con-densation of D-fructose-1,6-diphosphate into dihydroxyacetonephosphate and D-glyceraldehyde-3-phosphate and vice versa, hasa molecular weight of 150,000 and contains four major subunits.Acidification of the enzyme causes it to dissociate into the sub-units, which are inactive; upon neutralization, the subunits reas-semble quickly and spontaneously to reform the active enzyme.Aldolase contains free -SH groups, some of which are essentialfor catalytic activity.

Sample questions:

a. Enzymes with multiple subunits are _____________enzymes. answer

b. With this enzyme, an intact ___________ structure is neces-sary for activity. answer

c. From the information given, what can be inferred about theactive site of the enzyme? answer

d. The assay procedure for this enzyme uses the enzymes triosephosphate isomerase and glyceraldehyde-3-phosphate dehy-drogenase which catalyze the reactions respectively:

dihydroxyacetone phosphate → glyceraldehyde-3-phosphate




glyceraldehyde-3-phosphate + NAD+ →3-phosphoglycerate + H+ + NADH

These enzymes are used because the appearance of NADH can beeasily measured spectrophotometrically. How much of the twoenzymes must be added? answer

3. Enolase, which catalyzes the dehydration of 2-phosphoglycerateto convert it into phosphoenolypruvate, has an absolute require-

ment for a divalent cation (Mg2+ or Mn2+), which complexeswith the enzyme before the substrate is bound. The enzyme isstrongly inhibited by fluoride.

Sample questions:

a. It is likely that F- is what kind of inhibitor? answer

b. Enolase catalyzes a critical reaction in the energy producingscheme. Fluoride is essential in small amounts for bones andteeth but has been used as a rat poison. Explain. answer

c. What is the inferred role of divalent cations in this reaction?answer

4. "The conversion of glycine to glyoxylate and NH3 is catalyzed byan enzyme purified by Ratner et al. This enzyme, which ispresent in liver and kidney, is the same enzyme as D-amino acidoxidase. It has a high Km and is therefore thought not to play amajor role in glycine degradation." (From "Non-ketotic Hyper-glycinemia" by W. L. Nyhan in The Metabolic and MolecularBases of Inherited Disease.)

Sample questions:

a. What is an enzyme? answer

b. What does it mean that glycine oxidase is "the same enzymeas D-amino acid oxidase?" answer

c. Why does high Km indicate that the enzyme does not "play amajor role in glycine degradation?" answer

5. "Trypsin is an example of a large class of enzymes with a reactiveserine at the catalytic site. Trypsin is most active in the pH range7 to 9. Calcium ion increases the stability and the activity oftrypsin solutions and is a necessary component for the completeconversion of trypsinogen to trypsin by autocatalytic activation."

The conversion of bovine trypsinogen into trypsin is effected byrelease of a single peptide, Val-Asp-Asp-Asp-Asp-Lys, from theamino terminal of the zymogen, accompanied by a conforma-tional change. Trypsinogen is stable at pH 3, but when it is dis-




solved in neutral or slightly alkaline solution, conversion totrypsin occurs. The rate of this activation increases rapidly as thetrypsin formed begins to catalyze the activation reaction."[From "Zymogens of Proteolytic Enzymes" by B. Kassell and J.Kay, Science, 180, 1022-1027 (1973)].

Sample questions:

a. Explain the meaning of the phrase "reactive serine at the cat-alytic site". answer

b. What is calcium ion called with respect to trypsin? answer

c. What is meant by the phrase "from the amino terminal ofthe zymogen"? answer

d. What is the significance of the fact that when trypsinogen "isdissolved in neutral or slightly alkaline solution conversionto trypsin occurs"? answer

e. What is the significance of the fact that "the rate of this acti-vation increases rapidly"? answer

f. The small intestine, into which trypsinogen is released andin which trypsin acts, is slightly alkaline; but what is the pHoptimum of trypsin? answer

g. What is a zymogen? answer

6. The following statements are taken from the article "The Isola-tion and Properties of Phenylalanine Hydroxylase from HumanLiver" by S.L.C. Woo, S.S. Gillman, and L.I. Woolf, Biochem.J., 139, 741-749 (1974), but not in the order in which they orig-inally appeared. "Phenylketonuria is a genetically determineddisease in which the enzyme phenylalanine 4-hydroxylase (EC1.14, 16.1) is absent or has very low activity . . . The phenylala-nine hydroxylase activity of the liver was confined to a singleprotein of molecular weight of approximately 108,000 . . . Itseems to consist of two polypeptide chains, each with a molecu-lar weight of approximately 54,000 . . . By using the double-reciprocal plots, the apparent Km values for phenylalanine were

3.5 x 10-4 M for the full-term infant and 3.8 x 10-4 M for theadult preparations, respectively. For the synthetic cofactor,

apparent Km values were 6.8 x 10-5 M and 6.6 x 10-5 M, respec-tively . . . The activity of the enzyme was assayed, by using vari-ous concentrations of phenylalanine, in the presence and absenceof 0.5 mM p-chlorophenylalanine . . . Iron-chelating and cop-per-chelating agents inhibited human phenylalanine hydroxy-lase. Thiol-binding agents inhibited the enzyme but, as with the




rat enzyme, phenylalanine both stabilized the human enzymeand offered some protection against these inhibitors."

Sample questions:

a. The Km value of the enzyme from human fetal liver for

phenylalanine was 3.18 x 10-4 M. How does the affinity ofthe enzyme for the substrate change with development?answer

b. What is the reaction catalyzed by phenylalanine 4-hydroxy-lase? answer

c. Is this a systematic (I.U.B.) or trivial name? How can youtell? answer

d. How can a synthetic cofactor have a Km value? answer

e. The apparent Km of the enzyme from human fetal liver for

the synthetic cofactor was 7.15 x 10-5 M. What changeswith development can be inferred from this information?answer

f. Is this a monomeric enzyme? Explain. answer

g. What kind of inhibitor would you guess they found p-chlo-rophenylalanine to be? Why? answer

h. What does the next to the last sentence tell you? answer

i. What does the last sentence tell you? answer

Answers to Pretest Questions 1. a. The enzyme catalyses the hydrolysis of monoesters of phos-

phoric acid and has a pH optimum above 7.

b. Magnesium ions are required as cofactors, and chelators suchas EDTA and citrate would tie them up, making themunavailable to the enzyme.

c. No. As a matter of fact, the Bessey-Lowry method for alka-line phosphatase uses a pH 10.25 buffer. Since the enzymeactivities are small, the assay should be done close to theoptimum pH to keep the reaction time reasonably short, andthe assays must always be done at the same and a constantpH; but the pH chosen does not have to be exactly at theoptimum.

2. a. Oligomeric

b. Quaternary

c. It can be inferred that the active site contains a sulfhydryl




group, i.e., the amino acid cysteine.

d. They must be in large excess so that the two reactions theycatalyze are essentially instantaneous, making the reactioncatalyzed by aldolase the only rate-limiting step. Neither can

NAD+ be limiting.

3. a. Since it is so unlike the substrate, it must be either a non-competitive inhibitor or perhaps an irreversible inhibitor.

b. Simply, fluoride strongly inhibits a key enzyme for energyproduction.

c. The negatively charged phosphate group of the substrate

probably coordinates with the Mg2+ or Mn2+ ions, i.e., elec-trostatic interaction of substrate to the active site.

4. a. A catalyst for a biological reaction which is usually a protein.

b. It means that the enzyme has both activities. [Actually, sinceglycine does not have an asymmetric carbon atom, it is nei-ther D nor L (or both, depending on how you look at it); soit could be considered a D amino acid where R = H.]

c. A high Km means that it takes a high substrate concentrationto achieve one-half maximum velocity (a low affinity of theenzyme for the substrate). Although we are not given the Kmor the physiological concentration of glycine, we can assumethat the latter is far below the Km and that there is really a D-amino acid oxidase that also happens to work on glycine, butvery poorly.

5. a. The R groups of serine is one of the groups at the site wherereaction takes place and is somehow more reactive than the Rgroups of other serines in the protein.

b. Cofactor

c. This means "from the end of the protein chain that is termi-nated with a free alpha-amino group".

d. Trypsinogen by itself must have some proteolytic activity.Conversion must come from splitting off of a hexapeptide(proteolysis), not by ionization alone.

e. Trypsin is a better enzyme for this conversion, so as trypsin ismade from trypsinogen, the conversion goes faster and faster.Trypsin acts at the carboxyl end of lysine.

f. pH 7-9

g. A zymogen is an inactive precursor of an enzyme (a proen-zyme).


SIU School o f Medic ine BIOCHEMISTRY Enzymes/Membrane Transport

6. a. Decreases

b.

c. Systematic. It has an Enzyme Commission number after it.

d. One can determine the concentration that gives 1/2 of theVmax for any substrate, coenzyme, or synthetic coenzymethat binds with the enzyme and is required for reaction totake place.

e. The affinity of the enzyme for the cofactor increases withdevelopment.

f. No. It is a dimeric enzyme, i.e., it contains 2 subunits.

g. Competitive. It is a derivative of both its substrate and theproduct. Hence, it would likely bind at the active site butnot reaction would take place since it cannot be hydroxylatedat the 4 position.

h. The enzyme requires iron and copper ions for its activity.When they are tied up and unavailable to the enzyme, noreaction occurs.

i. There is a -SH group at the active site because reagents thatreact with -SH (sulfhydryl or thiol) groups inactivate theenzyme; but if the active site is occupied by a substrate mole-cule, the thiol-binding agents cannot get to the -SH group.

CH2

CH

CO2H

H2N

CH2

CH

CO2H

H2N

OH

NADPH + H+ NADP+

O2 H2O

phenylalanine tyrosine



STUDY GUIDE-1I. Where are enzymes located?

Enzymes are present in the cytosol, in all organelles (e.g., the nucleus,mitochondria, ribosomes, etc.), and in membranes. They are foundsingly and in multienzyme complexes.

II. What are catalysts and how do they work?

Enzymes are proteins (or ribonucleic acids, ribozymes) that are cata-lysts. As catalysts:

1. They are unchanged in the overall reaction, but may be tempo-rarily modified during intermediate steps.

2. They are effective in small amounts.

A unit of enzyme activity is the amount of enzyme that will catalyzethe transformation of 1 µmole of substrate/minute at a stated temper-ature and pH (usually optimal for both). One katal (or kat) = con-version of 1 mole substrate per second.

Specific activity is the number of enzyme units/mg protein = µmolesof substrate reacted/minute/mg protein.

3. They do not affect the equilibrium of a reversible chemical reac-tion. The function of a catalyst is to speed up the process ineither direction, i.e., enzymes affect only the kinetic and not thethermodynamic properties of a reaction. Therefore, the samechemical equilibrium will be reached with or without theenzyme, although it may not be reached in an reasonable timewithout the enzyme.

4. They exhibit specificity in their ability to accelerate chemicalreactions, although the degree of specificity varies greatly.Enzymes involved in digestion are generally rather nonspecific.Many others are very specific in that they act only with a singlesubstrate or with a very limited number of chemically similarcompounds. (The substrate is the reactant in an enzyme-cata-lyzed reaction.) For example, pancreatic lipase is a digestiveenzyme, an esterase, which will catalyze the hydrolysis of glycer-ides without much specificity in terms of the fatty acid, whileacetylcholine esterase is an enzyme that rather specifically cata-lyzes the hydrolysis of acetylcholine. Some proteases are specificfor peptide bonds involving certain amino acids; others are non-discriminating about the nature of the amino acids forming thepeptide bond.



Enzymes, like other catalysts, lower the activation energy of a reac-tion.

Consider the general reaction:

Substrate (S) → Product (P)

In any given population of molecules, there is a range of energies in aBoltzmann distribution (Fig. 1, curve A). If temperature is increased,a new distribution is set up (Fig 1, curve B). (The areas under thetwo curves are identical.)

Molecules must have a certain energy level before they can react, i.e.,before S can be converted into P, they must be activated. Theamount of energy required is called the activation energy. (∆E, Fig-ure 2; energies exceeding that given by the solid vertical line in Figure1.) It is seen that the distribution that pertains at the higher temper-ature has a larger number of molecules with the necessary activationenergy than does the low temperature distribution. Consequently,there will be a proportionately larger number of molecules undergo-ing reaction per unit time in the case of the high temperature distri-bution.



∆E = Eact = energy of activation for S P

∆G = the free energy of reaction, i.e., the difference in freeenergy of S and P

S* = activation of S (or P)Eact determines the rate of the reaction. ∆G (sometimes

termed ∆F) determines the position of equilibrium

There are two ways to accelerate a chemical reaction. The reactioncan be heated, which increases the energy of the molecules and hence,increases the percentage of molecules with the required energy ofactivation (compare dark stippled area of Figure 1 with lightly stip-pled area). However, living cells of homeothermic animals such asman operate in a very limited temperature range so that applicationof heat energy cannot be used as a means of speeding up chemicalreactions.

Catalysts provide the second way to accelerate a chemical reactions;they lower the activation energy, i.e., the energy barrier that sub-strates must overcome before they can be converted into products(dotted vertical line, Figure 1 versus the dashed vertical line).Enzymes do this by providing the reaction a different route. Theenzyme (E) reacts with the substrate to form an enzyme-substratecomplex or compound (ES); in a second step, ES (which now canalso be considered to be an enzyme product complex or compound)dissociates to regenerate the enzyme and release the product (P).

Each of these reactions has its own activation energy, which is much

E + S ES EP E + Pbinding

release

catalysis release

binding



lower than that of the uncatalyzed reaction (Figure 3).

Because catalysis is very fast compared to the other steps, it isassumed that the rate of reaction is determined by the rate of bindingand the rate of release and the reaction is usually written:

The ∆G and the equilibrium constant of the reaction is unchanged.However, the Eact is lowered, which also has the effect of increasingthe proportion of molecules that are in an activated state.

For example, Eact for the decomposition of H2O2 (2H2O2 →2H2O + O2 ) is about 18,000 cal.mol-1 for an uncatalyzed reaction

and = 2,000 cal.mol-1 in the presence of the enzyme catalase. Ureasecatalyzes the following reaction:

Urease lowers the activation energy for the reaction to less than

10,000 cal.mol-1. Eact for the acid-catalyzed reaction is = 25,000

cal.mol-1. What this means for the reaction can be seen from the fol-

E + S ES E + P

H2N-C-NH2 + H2O → CO2 + 2NH3

O

Urea



lowing table:

As will be discussed in more detail later, kinetic analysis is undoubt-edly the most useful tool for the determination (assay) of enzymeactivity and for investigation of the specific means of reaction.Therefore, we will review kinetics briefly.

III. What are zero-order and first-order reactions?

Consider the reaction S → P. The reaction rate is the amount of Sreacted in a unit of time or the amount of P formed in a unit of time.

n = order of reaction

Enzymes have either first-order, fractional order or zero-order kineticsdepending on the substrate concentration.

x = the amount of S reacting in time ta = the initial amount of Sa - x = the amount of S remaining unchanged at time t.c = concentration of S undergoing change at time t

A first-order reaction depends only on the substrate concentration(for example, radioactive decay).

(c is to the first power and, hence, the reaction is first order)

Table 1:

Eact, cal.mol-1k(first-order rate

constant)(sec-1)half-time

10,000 7.7 x 105 9 X 10-6 s

15,000 1.7 x 102 0.004 s

25,000 8.0 x 10-4 145 min

time S P

zero a zero

t a-x x

rate v( ) d S[ ]dt

-----------– d P[ ]dt

----------- k S[ ] n= = =

dcdt------– k1 c[ ]=



If integrated within limits of a - x and a, then

All units except 1/t cancel. Hence, the unit for k1 is 1/t. For exam-

ple, if for a certain reaction with first-order behavior k1 = 0.001 sec-1,this means that each succeeding second finds 1/1000th of the remain-ing reactant [S] converted into product (Figure 4).

A zero-order reaction is independent of substrate concentration. Ifthe reactant is present in sufficient excess that, for all practical pur-poses, its concentration remains constant during the measured courseof a reaction, than that reaction rate is constant and the amount ofproduct formed depends only on the time elapsed (Figure 5).

(c is to the zero power, i.e., [c]i = 1 and the reaction is, therefore,zero-order)

If this equation is integrated within limits of a-x and a, then ko = x/t

and the unit for ko = moles x sec-1.

k11t--- a

a x–-----------ln=

dcdt------– k0=



IV. What factors affect enzyme-catalyzed reac-tions?

The following factors affect enzyme-catalyzed reactions:

1. Concentration of substrate2. Concentration of enzyme3. Concentration of cofactors (activators and coenzymes)4. Concentration of inhibitors5. Temperature6. pH7. Concentration of allosteric effectors

V. What is the effect of substrate concentra-tion?

When doing enzyme assays, i.e., determining the amount of enzymeactivity, enzyme solutions are added to substrate solutions and theinitial rate is measured. Under these conditions, the concentration ofproduct (P) is zero and the equation becomes.

In general, one enzyme, molecule (polypeptide chain) can combinewith one molecule of substrate, although there are exceptions witholigomeric enzymes. Thus, as the ratio of substrate molecules toenzyme molecules increases, the initial rate (velocity) increases in pro-portion to the substrate concentration ([S]). This proportionalitydecreases with increasing substrate concentration, i.e., the reactionorder becomes less than unity and continues to decrease to zero orderin the limit of very high substrate concentration. Under conditionsof a large excess of substrate, essentially all enzyme molecules are tiedup in an enzyme-substrate complex or compound. At this point,there are, for all practical purposes, no free enzyme molecules thatcan combine with more substrate molecules, so the velocity will notincrease no matter how many more substrate molecules are added.The curve depicting the effect of substrate concentration on reactionrate is given in Figure 6.

E + S ES EP E + P

Init

ial

Rat

e

[S]

Vmax/2

v =

dc/d

t

First order v = k1[S]

Zero order v = k0

Km

Maximum rate (Vmax) (Saturation velocity)

Figure 6



This effect can be likened to a crew rowing a boat. Two crew mem-bers can row the boat faster than one, and three faster than two.However, a point is reached when all crew positions are filled, thereare no more available oars, and no more crew can be put in the boat.At this point the boat is going at its maximum speed, and no matterhow many additional crew are standing on the shore, watching, theboat cannot go faster because those not in the boat cannot increase itsspeed.

VI. What is the importance of Km? What is the basis of enzyme assays?

The enzyme concentration ([E]) also enters into the rate equationbut, when it is held constant as above, the numerical value of [E]merges into the rate constant (k). When [S] is held constant and [E]is varied, the substrate concentration becomes part of the propor-tionality constant and the initial rate is proportional to the enzymeconcentration, i.e.

Therefore, if the substrate concentration is in saturating amounts(large excess) and [E] is doubled, the rate is doubled, etc. (Figure 7).This fact is the basis for enzyme assays (determination of the amountof enzyme activity).

Michaelis and Menten defined a constant that is a dissociation con-stant for the enzyme-substrate complex (ES). It is a rapid equilib-rium constant and can be defined mathematically as follows:

v dcdt------– k E[ ]= =

Km

k 1–

k1------- E[ ] S[ ]

ES[ ]-----------------= =



Where

Km, termed the Michaelis constant, is characteristic of the enzymejust as melting point, boiling point, etc. are characteristic of organiccompounds. It was defined by Michaelis and Menten as a measure ofthe affinity of the enzyme for the substrate (KS). However, this istrue only when k-1 >> k2 which is not always the case.

It can be shown that (see Appendix)

which is the equation for any surface catalysis. If one solves for Km,then

and Km = [S] when Vmax/v = 2,

i.e., v = 0.5Vmax (Figure 6)

Thus, Km is the substrate concentration that gives an initial rate thatis 1/2 of the maximum rate of a given enzyme concentration.

The Michaelis-Menten equation relates the experimentally deter-mined initial velocity (v) of an enzyme-catalyzed reaction to the satu-ration, limiting or maximal velocity at infinite substrateconcentration (Vmax ) and to the Michaelis constant (Km).

To find Km, which has the units of concentration, several plots can beused. The most common is the Lineweaver-Burke plot (Figure 8)which is based on a reciprocal form of the Michaelis-Menten equa-tion.

E + S ES (EP) E + P

k1

k-1

k2

vVmax S[ ]Km S[ ]+----------------------

Vmax

1Km

S[ ]--------+

-----------------= =

Km S[ ]Vmax

v------------ 1–

=

1v---

Km S[ ]+

Vmax S[ ]----------------------

Km

Vmax------------ 1

S[ ]--------× 1

Vmax------------+= =



It should be noted here that these derivations and equations are forsingle substrate reactions and that biochemical reactions often involvetwo and three substrates reacting together.

Multisubstrate, multiproduct enzyme-catalyzed reactions are morecomplicated to treat mathematically. For our purposes we shouldsimply understand that these more complex reactions are character-ized by sequential mechanisms. Substrates and products may add toor be released from the enzyme in a random (phosphorylation of glu-

cose by ATP as catalyzed by hexokinase) or ordered (NAD+ and

NADP+ requiring dehydrogenases) manner, respectively. One specialcase, where the first substrate adds and the first product releasesbefore the second substrate adds and the second product releases, isdenoted as a Ping-Pong reaction and is characteristic of serine pro-teases and transaminases.

As stated earlier, Km is often an indicator of the dissociation constantfor the ES complex. The importance of Km to metabolic controlwhich is important in the etiology of several diseases and in treatmentof some conditions can be seen in the following examples.

Consider the situation with brain hexokinase and liver glucokinase(also denoted hexokinase D or hexokinase IV), two enzymes that cat-alyze the phosphorylation of D-glucose by ATP to yield D-glucose-6-phosphate, a process involved in the transport of glucose into cells.Brain hexokinase has a Km for glucose of <0.1 mM and liver glucoki-nase reaches 1/2 of its maximal velocity at [glucose] of ~5 mM.[Please note that glucokinase (a monomeric enzyme) displays sigmoi-dal kinetics (it does not follow Michaelis-Menten kinetics) with aHill coefficient of 1.5 and therefore we denote a K0.5 instead.]Because the Km is the substrate concentration that gives 1/2 Vmaxand the normal concentration of glucose in the blood is about 4.5mM (80 mg/100 ml), the normal concentration in the blood is about

Figure 8

1/v

1/[S]

-1/Km

slope = Km/Vmax

1/Vmax



45 times the Km for brain hexokinase. Thus, it is saturated (zero-order) at all normal blood glucose concentrations. Because the brainis almost totally dependent on blood glucose for its energy supply, itis very efficient in glucose utilization, hexokinase making glucosealways available at an essentially constant rate, a very necessary condi-tion.

On the other hand, 4.5 mM approximates the K0.5 for liver glucoki-nase. It is acting within the region of the v versus [S] curve in whichthe rate would change with any change in glucose concentration; asituation in keeping with the physiological role of the liver in control-ling blood glucose levels.

However, this says nothing about the total amount of glucose con-verted to glucose-6-phosphate and thus taken up respectively by thetwo tissues. What determines that is the relative Vmax's, the relativerates of the conversions catalyzed by the two enzymes at 4.5 mM glu-cose, and the relative concentrations of the two enzymes in theirrespective tissues.

To cite another example, drinking methanol can be fatal. Methanolitself is not toxic, but it is converted by alcohol dehydrogenase intoformaldehyde, which is fatal. (Alcohol dehydrogenase of the liver, anenzyme requiring NAD as a coenzyme, normally oxidizes ethanol toacetaldehyde.) One treatment for methanol poisoning is to give largeamounts of ethanol. If one were to ask, "Why get the patientdrunk?", the answer would be "because the Km of alcohol dehydroge-nase for ethanol is lower than the Km for methanol". Thus, floodingthe system with ethanol can competitively inhibit oxidation of meth-anol until it is excreted, an example of the use of an alternative sub-strate as a competitive inhibitor.

VII. What is the effect of the concentration of coenzymes and activa-tors?

As will be discussed later, many enzymes require inorganic ions foractivity, coenzymes that are cosubstrates and act as acceptor or donormolecules, and/or coenzymes that are cocatalysts. The concentrationof these cofactors, if limiting, will affect the rate of the enzyme-cata-lyzed reaction. Hence in any determination of enzyme activity, theconcentrations of required cofactors must be in amounts that are inexcess of the stoichiometric requirements.

There are four terms used to refer to the nonprotein components ofan enzyme that influence the rate of the catalyzed reaction. They areas follows:

Cofactor--nonprotein component of an enzyme (which can be ametal ion or a small organic molecule) required for catalytic activity



by the enzymes. The component can be loosely or tightly linked tothe enzyme.

Prosthetic group--small organic compound more or less tightlylinked to the protein and required for efficient performance of cata-lytic function.

Coenzyme--low molecular weight, dialyzable, organic compoundrequired for the action of some enzymes. A hazy distinction is drawnbetween coenzyme and prosthetic group with those "tightly bound"to enzyme termed a prosthetic group. (Vitamins are coenzymes thatmust be obtained through the diet.)

Coenzymes may be structurally altered in the course of reaction butin subsequent reactions (with other enzymes) the original structure isregenerated. This distinguishes coenzyme from substrates.

Activators are not required for enzyme activity but can increaseenzyme activity by putting the enzyme molecule in the proper stateto combine with the substrate or remove inhibitors.

VIII. What are com-petitive and noncom-petitive inhibitors? What is their effect on enzyme-catalyzed reac-tions?

Studies of the effects of enzyme inhibitors on the kinetics of enzyme-catalyzed reactions are a most important part of biochemistry andpharmacology.

As has been discussed above, enzymes are catalysts. Therefore, as thereaction proceeds, the concentration of products continuously andstoichiometrically increases at the expense of the disappearance of thecorresponding substrates, while, ideally, the total concentration ofenzyme (in all its possible forms) remains fixed and invariant. Thus,a consideration of the rate behavior or kinetics of such reactions fur-nishes a powerful tool, not only for their determination, but also for adefinition of their properties and possible modes of action. On themolecular level, we are always dealing with interrelations betweenstructure and function. Kinetics provides us with the most important(and with impure preparations, the only means of coming to gripswith enzyme function).

Selective inhibitions by natural or synthetic compounds (antimetabo-lites) forms the base of a broad approach to pharmacology and che-motherapy. Antimetabolites will be considered in the module onmetabolic integration and regulation.

Before we discuss inhibitors and their action, we must first discuss inmore detail the mechanism of enzyme action. Each enzyme moleculehas an active site. This active site is the part of the enzyme moleculethat interacts directly with the substrate, determining both the speci-



ficity (high affinity for specific compounds) and the catalytic functionof the active protein molecule. The active site (or active center) is fre-quently spoken of in terms of binding sites that hold the substrate ina specific conformation and the catalytic site where reaction takesplace. Active site = binding sites + catalytic site.

It is reasonable to consider the active site as composed of severalamino acid residues (at least 3 to explain stereospecificity) broughttogether in the specific, active, geometrical pattern by the folding ofthe polypeptide chain, and thus being determined by weak (second-ary and tertiary) bonds as well as the primary amino acid sequence.

Enzymes are often much, much larger than their substrates; at leastthey are much larger than the part of the substrate with which theyinteract. Therefore, only a small portion of the enzyme can be incontact with the substrate. Sometimes part of the protein molecule isdispensable, sometimes not. The bulk of the enzyme may be requiredto bring the active site in contact with the substrate. It certainly isrequired in most cases to put the reactive groups into proper juxtapo-sition to each other.

Emil Fischer pictured a "lock and key fit" between the enzyme andthe substrate. Koshland and others have suggested that someenzymes change their shape on combination with their substrate, i.e.,the substrate induces the enzyme to conform to the binding geometry("induced fit" hypothesis).

Reactive groups frequently found in the catalytic site are the -CH2OH group of serine, the -CO2H group of aspartic and glutamicacids, the imidazole group of histidine, and the -CH2SH group ofcysteine. The groups involved can be determined by kinetic analysis(use of multiple substrates and inhibitors and investigation of pHdependence), chemical modification (reaction with group specificreagents and specific labeling of the enzyme with substrate or sub-strate analogues), and X-ray analysis of crystalline ES complexes orcompounds.

There are two general types of reversible enzyme inhibition; competi-tive and non competitive.

In competitive inhibition, the inhibitor competes with the substratefor the active site of the enzyme. Thus, competitive inhibitors areusually structurally similar to the substrate. Product inhibition isquite common.

In competitive inhibition, inhibition depends on the relative concen-trations of the inhibitor and substrate. One such example is the use



of sulfa drugs to block the synthesis of folic acid.

Sulfa drugs, such as sulfanilamide, which are structurally similar tothe B vitamin, p-aminobenzoic acid, are effective competitive inhibi-tors in the synthesis of 7,8-dihydropteroate, and because some micro-organisms make their own folic acid, another B vitamin, provide aneffective means to prevent their growth.

In this case, the initial reactions are

(no further reaction)

The Michaelis-Menten equation becomes

where

from which it can be seen that, if [S] >> [I], the effect of the inhibitorcan be overcome.

To test for competitive inhibition, Km is determined in the absence ofthe inhibitor, then increasing concentrations of inhibitor are addedand a Lineweaver-Burke plot is made (Figure 9).

H2N CO2-

p-aminobenzoate

sulfanilamide

H2N S-NH2

O

O

E + S ES EP E + P

E + I EI

vVmax S[ ]

Km 1 I[ ]Ki------ S[ ]+ +

--------------------------------------------=

KiE[ ] I[ ]EI[ ]

---------------=



An increasing slope without a change in the y-intercept (without achange in Vmax which remains constant because high concentrationsof S can overcome the effect of the inhibitor) is evidence for compet-itive inhibition.

In noncompetitive inhibition, the initial reactions are

Both Ki’s are the same.

where in the formation of ES and EI, respectively, S and I are reacting

with independent sites. is also possible. In the

Lineweaver-Burke plot, as [I] increases, both the slope and the y-intercept increase (Figure 10).

Figure 9

1/v

1/Vmax

1/[S]

[I] = 0

Increasing [I]

-1/{Km(1 + [I]/KI)}

E + S ES EP E + P

E + I EI

ES + I ESI

EI + S ESI

vVmax S[ ]

Km S[ ]+( ) 1 I[ ]Ki------+

-------------------------------------------------=



Some inhibitors are true competitive inhibitors, but many others areneither true competitive nor noncompetitive inhibitors. Hence,other kinds of inhibition (uncompetitive) have been hypothesizedand treated mathematically in attempts to explain the observed Lin-eweaver-Burke plots.

IX. What is the effect of irreversible inhibi-tion?

There are at least two classes of inhibitors that form covalent or verytight linkages with the enzyme and effectively prevent catalysis.

In many enzymes, sulfhydryl (-SH) groups are required for activityand the reaction of heavy metal ions (e.g. silver, mercury, lead, etc.)with sulfhydryl groups on these enzymes render them irreversiblyinactive. The effect of this type of inhibitor on enzyme kinetics is todecrease the apparent Vmax without affecting the Km. Consequently,the kinetic plots that occur for this class of irreversible inhibitors areidentical to those observed for noncompetitive reversible inhibitors.This has led to confusion as to the differences between reversiblenoncompetitive and irreversible inhibition. Reversible noncompeti-tive inhibitors can be distinguished from irreversible inhibitors bydialyzing the enzyme preparation containing the inhibitor. Fullenzyme activity will be regenerated on dialysis of the enzyme prepara-tion containing a non-competitive inhibitor but not with an enzymepreparation containing an irreversible inhibitor.

Reversible noncompetitive inhibition:

Figure 10

1/v

1/[S]

[I] = 0

Increasing [I]

-1/Km

1/Vmax(1 + [I]/KI)

slope = (Km/Vmax)(1 + [I]/KI)

E + S ES E + P

EI + S EIS

I I



Irreversible inhibition:

In a number of diseased states, the activity of one or more enzymes isoften greater than needed to maintain normal metabolic balance. Inthe past few years, a strategy for drug design has developed in whichirreversible inhibitors are used to maintain control over the metabolicimbalance. These irreversible inhibitors are referred to as "suicide"inhibitors or "mechanism-based" inhibitors. The rationale for thistherapeutic strategy is to design a molecule that looks very much likethe substrate, but when the enzyme binds this molecule and beginsthe catalytic process, it creates a species that is chemically more reac-tive than the normal reaction intermediate. This intermediate mustbe reactive enough to covalently modify a residue in the active site ora bound coenzyme rather than form a product metabolite molecule asdoes the normal substrate molecule.

Let us consider the action of the suicide substrate gabaculine on thepyridoxal phosphate requiring bacteria enzyme, GABA transaminase.

The carbon skeleton of GABA forms part of the structure of gabacu-line and the enzyme will accept gabaculine into its active site. Theenzyme tries to transaminate gabaculine and in the process it irrevers-ibly modifies its pyridoxal moiety. Gabaculine renders inactive everyenzyme molecule that takes part in this reaction.

Recent examples of target enzymes for which suicide substrates havebeen designed include: (a) aromatase, the key enzyme in the inter-conversion of androgenic and estrogenic steroid sex hormones;(b) GABA transaminase, an enzyme target for antiepileptic drugs;

E + S ES E + P

EI

I

CO2-

NH3+

CO2-

NH3+

GABA gabaculine



and (c) DOPA decarboxylase, an enzyme involved in regulation ofblood pressure.

Knowledge of the mechanistic details of the enzyme-catalyzed reac-tion is used to design the irreversible inhibitor (mechanism-basedsubstrate) and cause the enzyme to self destruct (thus the name sui-cide inhibitor). In principle, if the suicide substrate is designed sothat only the targeted enzyme can begin catalysis to produce the reac-tive species, then a drug with maximum specificity and minimizedside effects should result. Any class of enzymes including proteasesand phosphatases, in which a normal intermediate can be re-routedto mechanism-based auto destruction, should be amenable to target-ing. In point of fact, the action of several classes of therapeutic agentsin current use can be explained in molecular terms by the suicide sub-strate analysis.

X. What is the effect of temperature?

As with other chemical reactions, the rates of enzyme-catalyzed reac-tions are increased by increases in temperature. However, becauseenzymes are proteins, thermal denaturation of the protein withincreasing temperature will decrease the effective concentration of theenzyme. The result on an enzyme-catalyzed reaction is depicted inFigure 11. The upper limit for most human enzymes is 40 to 50˚C(normal body temperature 37˚C).

CO2-

NH3+

CO2-

NH+

NH+

2-3OPO

OH

CH3

HH

CO2-

NH2+

NH+

2-3OPO

O-

CH3

Enzymepyridoxal phosphate

:B



XI. What is the effect of pH on enzyme-cata-lyzed reactions?

Changes in pH will change the ionization of the enzyme that is, ofcourse, amphoteric; for example,

If the substrate possesses ionizable groups, as many do, its ionic statewill change with changes in pH; for example, focusing on the sub-strate ionization.

Thus, as the ionic nature of either the substrate or the enzyme ischanged from that form that combines with the other, the reactionswill be slowed and a curve such as that in Figure 12 will result.

To illustrate one example of how a bell-shaped pH profile mightoccur, let's consider an example in which the enzyme has a group atthe active site which must be positively charged in order for substrate

H+

E + S ES (EP) E + P

EH+ EHS+

+ +

H+

H+

HE + S- HES- (HEP-) HE + P-

E- + H+ HESH+

+ +H+

SH

H+

PH+E- + H+

+



to bind, and the pK of the group is, e.g., 8.0. Furthermore, let us saythat the substrate has to be negatively charged in order for it to beattracted to the enzyme active site and the pK of the group on thesubstrate is, e.g., 4.0.

At low pH the substrate is protonated (SH) and will not want to bind

to the enzyme that also is in its protonated form (EH+). As the pH is

increased we remove the proton from substrate to form S-, which isthe form that will bind to enzyme. Thus, we will see an increase inenzymatic activity proportional to the fraction of substrate molecules

in the S- form. At pH 6 we should have essentially 100% of the sub-

strate as S- and maximum catalytic activity will be observed. As wego to pH values higher than 6, we begin to remove the proton fromthe group at the enzyme active site. There will be a decrease in activ-ity proportional to the amount of unprotonated enzyme present

because the unprotonated form will not bind substrate (S-). Theresult is the bell-shaped pH profile seen in Figure 12.

The pH optimum may be either narrow or broad. As the pHincreases or decreases above or below the optimum pH, the activity ofthe enzyme (the velocity of the enzyme-catalyzed reaction) willdecrease because:

a. There will be decrease in enzyme substrate binding due to repul-sion or to loss of charge on either the substrate or the enzyme.

b. the catalytic functional groups will be in the wrong ionizationstate.

c. The enzyme may become denatured.

XII. What are allos-teric enzymes and allosteric effectors?

Allosteric enzymes are enzymes whose kinetic properties cannotbe accounted for by the Michaelis-Menten model. The activity ofallosteric enzymes is altered by regulatory molecules, and there is acooperative binding of substrates.



With allosteric enzymes, an inhibitor or an activator (termed nega-tive and positive effectors, respectively) interacts at a site other thanthe active site on a multi-subunit protein. These effector moleculesare quite different in shape and structure in comparison with the sub-strate molecules. The sites at which the allosteric effectors bind arecalled regulatory or allosteric sites.

Allosteric control is dependent on binding sites for the effectors andon the existence of at least two conformational states of the enzyme,one in which the binding sites have a high affinity for the substrate,and another in which they bind weakly, if at all, with the substrate.

One model of the action of allosteric enzymes is that binding of theeffector molecules stabilizes the enzyme in one of its possible confor-mation. Binding of a negative effector stabilizes the enzyme in aninactive state, and binding of a positive effector stabilizes the enzymein its active state. This can be seen diagrammatically in Figure 13.

Sometimes allosteric enzymes have regulatory subunits and catalyticsubunits. This kind of inhibition is of the greatest physiological sig-nificance because of its importance in regulation of biochemical reac-tions. The effector molecules can be hormones, the immediateresults of the action of a hormone (with polypeptide hormones theresult is usually cyclic AMP), or intermediates or the end-product ofa pathway. Several examples of the different kinds of control involv-ing allosteric enzymes will be covered as you progress through basic



metabolism.

XIII. Why do enzyme assays?

The thing you as a physician will be most interested in is the enzymeconcentration. This parameter varies from patient to patient and isimportant in diagnosis. In some cases, a low concentration is signifi-cant (e.g., erythrocyte glucose-6-phosphate dehydrogenase isdecreased in some cases of non-spherocytic hemolytic anemia), andin some cases a high concentration is important (e.g., serum creatinephosphokinase is increased following a myocardial infarction).

Enzyme concentration is not reported in milligrams per liter or mil-liequivalents per liter, for this type of measurement cannot be doneon blood samples, or other impure fluids. Therefore, enzymes aremeasured by the work they do, i.e., the amount of substrate con-verted to product (the activity).

Enzyme results are reported in activity units and it is extremelyimportant to know the methodology used and the normal values forthat method. Because there are several methods for assaying mostenzymes, the activity of the enzyme will depend on the methodologychosen by the particular hospital; and results from one hospital toanother cannot necessarily be compared unless identical methods areused. This problem is evident at hospitals into which patients aretransferred from other hospitals.

It is hoped that, in the future, some uniformity can be achieved inreporting enzyme activities so that the problem of comparing resultsfrom one institution to another can be overcome. Two attempts toachieve uniformity were by the use of international units and thekatal. An international unit (IU) is what was defined earlier as a unit,i.e., the amount of enzyme activity that will catalyze the conversionof one micromole of substrate to product in one minute, while onekatal equals the conversion of one mole of substrate per second. Ashas been discussed, the pH, temperature, and other factors affectingenzyme activity must be specified and carefully controlled in deter-mining units; their control is the responsibility of the laboratory. Itmay be many years before a uniform system of reporting enzymeresults can be implemented; until then remember that you cannotnecessarily compare results from one institution to another. Youmust compare the results from any hospital with the normals estab-lished for that hospital.

Now, let us consider how enzymes get into the serum. The concen-tration of an enzyme in serum is a result of (a) its rate of release fromthe tissue and (b) the rate at which it is removed from the circulation.There is some leakage of enzymes into serum from normal cells.



When a cell is damaged, it releases enzymes into the serum at a ratefaster than it can be removed, and therefore, an increase in the serumconcentration is noted. The nature and the amount of the enzymereleased will depend on the type of cell injury that has occurred.

Some mild degree of cell injury, such as is caused by progressive mus-cular dystrophy, will increase plasma membrane permeability. Thisincreased permeability will result in a partial loss of soluble enzymesfrom the cytoplasm to the serum, in turn resulting in a moderateserum elevation.

Cell necrosis, such as is caused by myocardial infarction, acute hepa-titis and acute pancreatitis, will result in a complete loss of all cellularenzymes, including those in organelles such as mitochondria. Thiscomplete release of cellular enzymes results in high serum levels.

An increased number of cells with an accompanying increased pro-duction of enzymes, such as occurs in neoplastic disease, can givemoderate to high serum levels of enzymes.

To take a brief look at how enzyme assays are used in diagnosis, wecan consider the following. Some enzymes are fairly localized; that is,only specialized tissues have a high concentration. Amylase is anexample of this type of enzyme. It is found in high concentrationonly in the pancreas, and therefore, a high serum amylase concentra-tion reflects specific pancreatic damage. Other enzymes such as lacticdehydrogenase are present in almost all tissues, and thus, total LDHanalysis will provide less information about specific tissue damage.However, as we will see in a later section, determination of the LDHisoenzyme pattern, although more difficult to do than an ordinaryenzyme assay, can be an important diagnostic tool.

XIV. How does one assay an enzyme?

Specific activity can be determined in one of two ways. In the "one-point method," the reaction is allowed to proceed for a time and thenthe enzyme is denatured by acid, base, or heat. The reaction is thenanalyzed by a chemical or physical method to see how far it has gone.

In the other method, data are obtained continuously without stop-ping the reaction. This is usually done through physical methodssuch as spectrophotometry and polarimetry, although titration mayalso be used. The rate is then calculated and converted to specificactivity.

Before a procedure can be used as an assay, it must be shown that theenzymic reaction is linear over the time period studied and that thereaction rate is proportional to the amount of enzyme present.



An example of an enzyme assay is the determination of alkaline phos-phatase in human serum. The phosphatases are generally located inthe cytoplasm of the cell; alkaline phosphatase is a microsomalenzyme. (Microsomes are fragments of endoplasmic reticulum iso-lated after tissue homogenization and differential centrifugation.)Phosphatases catalyze the hydrolysis of phosphate esters with libera-tion of inorganic phosphate.

Clinical assay procedures make use of synthetic substrates. For exam-ple, the Bessey-Lowry method used p-nitrophenyl phosphate which,in the presence of the enzyme, undergoes hydrolysis to yield p-nitro-phenol and inorganic phosphate. The rate of reaction is determinedby measuring the change in absorbance at 405 nm, the absorbancemaximum for the product.

Product is measured because an excess of substrate is used. Smallchanges in concentration can be determined with much, much lesserror when the initial concentration is zero than when the initial con-centration is some large value as it would be if disappearance of sub-strate were to be measured. also, measurement of product formationis more desirable than measurement of substrate disappearancebecause the fluids assayed contain many enzymes and, hence, sub-strate disappearance may occur via other reactions.

The normal values vary according to the method used. With theBodansky method the normal adult serum activity is 1.5 to 4.0Bodansky units. With the King-Armstrong method, the normalrange is 1.35 to 4.0. These are both arbitrary units. In InternationalUnits (Bessey-Lowry method), the normal range is 20 to 85 mU/ml.

It is essential that the serum sample be properly processed. Greatvariation may occur in the assay results if certain precautions are notobserved. Things that may alter activity are storage (even underrefrigeration), losses or increases in dissolved carbon dioxide, andaddition of excessive oxalate or fluoride to the specimen.

The level of alkaline phosphatase is higher in growing children thanin adults. There are many other physiologic and pathologic condi-tions in which there is an increase in the activity of serum alkalinephosphatase and still others that lead to lower than normal adult lev-els.

The use of glucose oxidase to determine glucose in the blood andurine is an example of the use of an enzyme for assay and of the use ofcoupled enzyme reactions. The reactions are as follows:

glucose oxidase



D-glucose + O2 → D-gluconolactone + H2O2

peroxidase

H2O2 + reduced dye → oxidized dye + 2H2O(colorless) (green)

The intensity of the green color is proportional to the original glu-cose concentration.

XV. How are enzymes named?

There are four ways of naming enzymes. Three give the enzyme atrivial name. The first enzymes discovered were given names thathave no relationship to their activity; these are largely digestiveenzymes (e.g., pepsin, trypsin, chymotrypsin). Another way to nameenzymes is to name the substrate and add the suffix :"ase" (e.g., mal-tase, fructose-1,6-disphosphatase, fumarase, ribonuclease, urease) orthe reaction alone can be named and the suffix "ase" added (e.g.,invertase). A third way names the substrate and the reaction cata-lyzed and adds the suffix "ase" (e.g., malic dehydrogenase, glutamic-oxaloacetic transaminase, hexokinase). (Kinases are enzymes thattransfer a phosphate group from ATP to a substrate.)

There is now an international system of enzyme classification that hasspecific and systematic rules for naming enzymes. This methodassigns to each enzyme a number that is an indication of its activity.The six main groupings are as follows:

1. oxidoreductases (oxidation-reduction reactions)2. transferases (transfer of functional groups)3. hydrolases (hydrolysis reactions)4. lyases (addition to double bonds)5. isomerases (isomerization reactions)6. ligases (formation of bonds with ATP cleavage)

The international systematic name describes the reaction, substrates,and products in question. The trivial name is used in general refer-ence where the longer systematic name would be cumbersome.

As an example of the difficulty, the commonly used names for thetwo clinically significant transaminases described above areglutamate-oxaloacetate transaminase (glutamic-oxaloacetic transami-nase) (SGOT) and glutamate-pyruvate transaminase (glutamic-pyru-vic transaminase) (SGPT). (S stands for serum). More meaningfulcommon names for these particular enzymes (to be discussed in alater domain) would be aspartate aminotransferase (AST) and alanineaminotransferase (ALT), respectively.



Abbreviations are another matter. In many cases two or more abbre-viations are used for the same enzyme. The names and abbreviationsused in this module are ones that are, more or less, in general usage.However, we will have to live with the confusion in nomenclatureand abbreviations, with the fact that biochemists and pathologistsmay use different names for the same enzyme, and with the fact thatnames and abbreviations may change from one hospital laboratory toanother.

XVI. Do enzymes have prosthetic groups?

Sometimes an organic molecule is covalently linked to the proteinmolecule (the apoenzyme) forming a conjugated protein (the holoen-zyme). In this case, the nonpeptide part of the protein is called aprosthetic group. The addition of a prosthetic group occurs after ini-tiation of protein synthesis and usually considered as one form ofpost-translational modification.

The prosthetic group of an enzyme generally has the same function asthat of a coenzyme. Popular examples of prosthetic groups includeFAD of the flavoenzymes and TPQ (trihydroxyphenylalaninequinone) of serum amine oxidase. However, proteins other thanenzymes have prosthetic groups, for example, the heme group foundin hemoglobin.

XVII. What are isoen-zymes?

Some proteins are associations of several polypeptide chains. Eachindividual chain is termed a subunit; thus, a subunit is any polypep-tide chain in the completed functioning protein that is not covalentlybound via a peptide linkage to other peptide units and can thus bereadily separated from other subunits. A protomer is a subunit in aprotein containing a finite number of identical subunits. Oligomersare combinations of similar or different subunits to form the totallyfunctioning protein. Allosteric enzymes have such a structure. Thoseenzymes consisting of a single polypeptide chain are termed mono-meric enzymes. Those enzymes consisting of two or more polypep-tide chains are termed oligomeric enzymes.

Some enzymes occur in multiple molecular forms that catalyze thesame reaction in the same organism. These forms are known asisoenzymes, or more simply as isozymes, i.e. isoenzymes are differ-ent proteins within an organism with similar enzymic activity. Themost thoroughly studied of these enzymes is lactate dehydrogenase(LDH), which can occur in five hybrid forms (LDH1 to LDH5), eachof which catalyzes the reaction:



When serum of a tissue extract is subjected to electrophoresis on anyusual electrophoretic medium (agar, starch acrylamide gel, celluloseacetate), five zones of LDH activity are seen.

These LDH isoenzymes are tetramers (M.W. about 135,000 daltons)that can be split into four monomeric subunits (H and M subunits)(M.W. about 32,000).

There are other enzymes of clinical importance that exist as isoen-zymes (for example, creatine phosphokinase, CPK1, CPK2, andCPK3).

Appendix: Derivation of Km and Relation of Km to [E]

The simplified equation for an enzyme-catalyzed reaction is:

However, in an enzyme assay, only the initial rate is measured so thatin the limit of t = 0, the concentration of product is zero. Therefore,there is no back reaction and the equation can be simply written.

Michaelis and Menten and Henri defined a dissociation constant for

CO2

C

CH3

O + NADH + H+

CO2

C

CH3

H + NAD+HO

pyruvate lactate

E + S ES E + Pk1

k-1

k2

k-2

E + S ES E + Pk1

k-1

k2



ES

and when k2 << k-1

then

However, k2 is not always negligibly small in comparison with k-1.An exact derivation (made by Briggs and Haldane), in which we donot have to assume k2 << k-1, involves the assumption that the reac-tion is in steady state; i.e., we assume that, under the conditionsunder which the rate of reaction is measured, the concentrations of Sand ES are constant. In other words, rate of formation of ES = rate ofdisappearance of ES. Thus, for the mechanism given:

v1 = v-1 + v2k1 [E] [S] = k-1 [ES] + k2 [ES]

[E]t = total enzyme concentration[S]t = total substrate concentration

but [S]t >>> [E] by experimental designtherefore [S]t >>> [ES]

and since [S]t = [S] + [ES]then [S]t = [S] = concentration of free substrate[E] = [E]t - [ES] = concentration of free enzyme

With these equations and definitions, it is possible to derive the rateequation for any single-substrate system where only initial rates (v)are measured.

k1 [E] [S] = k-1 [ES] + k2 [ES]

k1 ([E]t - [ES]) [S] = (k-1 + k2) [ES]

It is convenient to rearrange the equation as follows

(solving for [ES]):

Km

k 1– k2+

k1-------------------=

Km

k 1–

k1------- E[ ] S[ ]

ES[ ]-----------------= =

k 1– k2+

k1-------------------

E[ ] t ES[ ]–( ) S[ ]ES[ ]

----------------------------------------- E[ ] S[ ]ES[ ]

----------------- Km= = =

E[ ] t

ES[ ]------------ 1–

S[ ] Km=



or

The measured rate (v) is the rate of formation of product, i.e., v2 =k2 [ES] = v

Therefore,

This is the same as the Michaelis-Menten equation except that Km isa steady state constant (that is, it has a different definition here). Youmay wish to think of Km as a commitment constant. As a dissocia-tion constant (k-1/k1), it would be a measure of enzyme-substrateaffinity. However, it is a true affinity constant only when k-1 >> k2.

Km is a characteristic constant for the enzyme and independent of theconcentration of enzyme used in its determination. Why this is socan be seen from the following graph.

Post Test1. Definitions

a. What is meant by the order of a reaction? answer

b. Define Km, Ks, kcat. answer

c. Km is all too frequently equated with Ks. In fact, in mostreactions there is an appreciable disparity between the values

E[ ] t S[ ] S[ ]–

ES[ ]-------------------------------- Km=

ES[ ]E[ ] t S[ ]

Km S[ ]+----------------------=

vk2 E[ ] t S[ ]Km S[ ]+------------------------=



for Km and Ks. For the reaction A → B define conditionsunder which Km = Ks. Describe conditions under which thisis not true. answer

d. What is the steady-state approximation, and under whatconditions is it valid? answer

2. True or False. If False, explain why they are false.

a. At saturating levels of substrate, the rate of an enzyme cata-lyzed reaction is proportional to the enzyme concentration.answer

b. The Michaelis constant Km equals the substrate concentra-tion at which v = 1/2 Vmax. answer

c. The Km for a regulatory enzyme varies with enzyme concen-tration. answer

d. If enough substrate is added, the normal Vmax of an enzyme-catalyzed reaction can be attained even in the presence of anoncompetitive inhibitor. answer

e. The Km of some enzymes may be altered by the presence ofmetabolites structurally unrelated to the substrate. answer

f. The rate of an enzyme-catalyzed reaction in the presence of arate-limiting concentration of substrate decreases with time.answer

g. The sigmoidal shape of the v versus (S) curve for some regu-latory enzymes indicates that the affinity of the enzyme forsubstrate decreases as the substrate concentration isincreased. answer

3. a. The _____________ of a reaction is the numerical relation-ship between substrates and products. answer

b. The rate constant ____________ of an enzyme-catalyzedreaction is a measure of the catalytic efficiency at saturatinglevels of substrate. answer

c. _______________ inhibitors do not alter the Vmax of anenzyme-catalyzed reaction. answer

d. The sigmoidal shape of the v versus [S] curve for some regu-latory enzymes results from a _______________ effect ofsubstrate on the substrate binding sites. answer

e. For an enzyme whose Km can be regulated, the presence of a_____________ effector increases the level of substraterequired to attain a given reaction rate. answer

4. Assume that an enzyme catalyzed reaction follows Michaelis-



Menten kinetics with a Km of 1 x 10-6 M. If the initial reaction

rate is 0.1 µmol/min at 0.1 M, what would it be at 0.01 M, 10-3

M, and 10-6 M? answer

5. Consider the reaction

glucose-6-phosphate → fructose-6-phosphate

which is catalyzed by phosphoglucose isomerase.

a. What is its stoichiometry? answer

b. What is the simplest representation of this reaction in termsof S,E, and P? answer

c. What are S,E, and P in this reaction? answer

6. A more general form of an equation for an enzyme catalyzedreaction is:

Consider the essentially irreversible reaction represented by the free-energy diagram below.

HCO

CH OH

C

C

OH H

C

H OH

H OH

CH2OPO32-

CH2OH

C O

C

C

OH H

C

H OH

H OH

CH2OPO32-

E + S ES EP E + Pk1

k-1

k2

k-2

k3

k-3



a. Using the letters indicated in the diagram, relate each of therate constants in the Equation above to the energy-level dif-ference that determines it. answer

b. Which rate constant limits the rate of formation of product?answer

c. Does Km approximately equal Ks for this enzyme? answer

7. To study the dependence of the rate of an enzyme-catalyzed reac-tion on the substrate concentration, a constant amount ofenzyme is added to a series of reaction mixtures containing dif-ferent concentrations of substrate (usually expressed in mol/L).The initial reaction rates are determined by measuring the num-ber of moles (or µmoles) of substrate consumed (or product pro-duced) per minute. Consider such an experiment in which theinitial rates in Table 2 were obtained at the indicated substrateconcentrations.

a. What is Vmax for this reaction? answer

Table 2: Initial rates at various substrate concentrations for a hypothetical enzyme-catalyzed reaction

[S] (mol/L) v (µmol/min)

2.0 X 10-1 60

2.0 X 10-2 60

2.0 X 10-3 60

2.0 X 10-4 48

1.5 X 10-4 45

1.3 X 10-5 12



b. Why is v constant above substrate concentrations of 2.0 x

10-3 M? answer

c. What is the concentration of free enzyme at 2.0 x 10-2 Msubstrate concentration? answer

8. Using the experimental procedure described in Problem 7, thedata in Table 3 were obtained for an enzyme in 10-ml reactionmixtures. Use numerical (not graphical) calculations in answer-ing the following questions.

a. What is Vmax for this concentration of enzyme? answer

b. What is the Km of this enzyme? answer

c. Show that this reaction does or does not follow simpleMichaelis-Menten kinetics. answer

d. What are the initial rates at [S] = 1.0 x 10-6 M and at [S] =

1.0 x 10-1 M? answer

e. Calculate the total amount of product made during the first

five minutes when [S] = 2.0 x 10-3 M. Could you make the

same calculation at [S] = 2.0 X 10-6 M? answer

f. Suppose that the enzyme concentration in each reaction mix-ture were increased by a factor of 4. What would be thevalue of Km? of Vmax? What would be the value of v at [S]

= 5.0 x 10-6 M? answer

9. The Km of a certain enzyme is 1.0 X 10-5 M in a reaction that isdescribed by Michaelis-Menten kinetics. At a substrate concen-

Table 3: Initial rates at various substrate concentrations for a hypothetical enzyme-catalyzed reaction.


5.0 X 10-2 0.25

5.0 X 10-3 0.25

5.0 X 10-4 0.25

5.0 X 10-5 0.20

5.0 X 10-6 0.071

5.0 X 10-7 0.0096



tration of 0.10 M, the initial rate of the reaction is 37 µmol/minfor a certain concentration of enzyme. However, you observethat at a lower substrate concentration of 0.010 M the initialreaction rate remains 37 µmoles/min.

a. Using numerical calculations, show why this tenfold reduc-tion in substrate concentration does not alter the initial reac-tion rate. answer

b. Calculate v as a fraction of Vmax for [S] = 0.20 Km, 0.50 Km,1.0 Km, 2.0 Km, 4.0 Km, and 10 Km. answer

c. From the results in (b), sketch the curve relating v/Vmax to[S]/Km. What is the best range of [S] to use in determiningKm or investigating the dependence of v on [S]? answer

10. The hydrolysis of pyrophosphate to orthophosphate is importantin driving forward biosynthetic reactions such as the synthesis ofDNA. This hydrolytic reaction is catalyzed in E. coli by a pyro-phosphatase that has a mass of 120 kDa and consists of six iden-tical subunits. Purified enzyme has a Vmax of 2800 units permilligram of enzyme. For this enzyme, a unit of activity isdefined as the amount of enzyme that hydrolyzes 10 µmol ofpyrophosphate in 15 minutes at 37°C under standard assay con-ditions.

a. How many moles of substrate are hydrolyzed per second permilligram of enzyme when the substrate concentration ismuch greater than Km? answer

b. How many moles of active site are there in 1 mg of enzyme?Assume that each subunit has one active site. answer

c. What is the turnover number of the enzyme? answer

Answers to Post Test1. a. Reactions can be independent of the concentration of sub-

strate (0-order), directly dependent on substrate (1st-order)or dependent on substrate concentration raised to somehigher power (2, etc.).

b. Km = Michaelis constant. It is the substrate concentrationthat gives 1/2 Vmax. That is, it is the concentration of sub-strate at which half the active sites are filled. Km is alsorelated to rate constants of the individual steps in the reac-tion.



Ks = dissociation constant of the ES complex for

At equilibrium the rate of formation of ES must equal itsrate of dissociation to E + S. Therefore you should be able toshow that

In the case where k2 is much smaller than k-1, Kmapproaches k-1/k1 = Ks. Under these conditions, the Kmdetermined kinetically is equal to the dissociation constant,Ks.

kcat = the turnover number and is equal to k2 in theMichaelis Menten mechanism.

The turnover number of an enzyme is equal to the numberof moles of substrate converted to product per minute permole of enzyme present when the enzyme is fully complexedwith substrate.

c.

If k2 << k-1, Km = Ks

Conditions where Km not equal Ks

If k2 is approximately equal to k-1

If k2>>k-1

d. The steady state approximation assumes that the concentra-tions of the intermediates in a reaction do not change whilethe rate of product formation is being measured. This holdsfor the early stages of a reaction, after the ES complex hasformed and before appreciable changes have occurred ineither the substrate or product concentrations.

2. a. True. Vmax = k2[E]t

b. True

c. False. The value of Km is independent of enzyme concentra-

Km

k 1– k2+

k1-------------------=

E + S ES

k1

k-1

KS

k 1–

k1-------=

A + E EA E + B



tion for almost all enzymes.

d. False. A non-competitive inhibitor cannot be overcome bysubstrate concentration.

e. True. This occurs in regulatory enzymes.

f. True.

g. False. The initial increasing slope of the curve shows thatbinding of the first substrate molecule increases the affinityof the enzyme for subsequent substrate molecules.

3. a. Stoichiometry

b. kcat

c. Competitive

d. homotropic

e. negative

4.

5. a. glc-6-P fru-6-P

or S P

b.

Notice that k-2 is included for a reversible Rx.

c. S = glucose-6-P

E = phosphoglucose isomerase

P = fructose-6-phosphate

6. a. The rate constant for each step is inversely related to the dif-

[S], M v(µmol/min

0.1 0.1 Vmax

0.01 0.09999

0.001 0.09990

0.000001 0.050 Km = 1 x 10-6 M

E + S ES E + Pk1

k-1

k2

k-2



ference between the energy level of the reactants and thehighest energy barrier between the reactants and the prod-ucts of that step. In terms of the letters in the Figure,

k1 is determined by b-ak-1 is determined by b-ck2 is determined by d-ck-2 is determined by d-ek3 is determined by f-e, andk-3 is determined by f-g.

b. k2 corresponds to a much higher energy barrier than theother forward rate constants and therefore must limit therate of product formation.

c. Since k2 is small relative to k-1 , Km approximates Ks for thisenzyme. Therefore, Km is a measure of affinity for substrate.

7. a. Vmax = 60µmol/min

b. v is constant because it has reached Vmax; the enzyme is satu-rated with substrate.

c. The concentration of free enzyme is negligible because all ofthe enzyme is in the ES form.

8. a. Vmax = 0.25 µmol/min

b. For a reaction obeying Michaelis-Menten kinetics, Vmax andKm are simply constants relating v to [S]. Km can be calcu-lated by substituting Vmax and any pair of v and [S] values atv < Vmax. For example, at [S] = 5.0 x 10-5 M and v = 0.20µmol/min the equation becomes

Solve: Km= 1.25 X 10-5 M

c. If the reaction follows simple Michaelis-Menten kinetics,then the Michaelis-Menton equation should relate v to [S]over a wide range of [S]. This can be tested by determiningwhether the equation yields the same value of Km at severaldifferent values of [S] and v < Vmax. Under the conditions of

this problem, the same value, Km = 1.3 x 10-5 M, is obtained

at [S] = 5.0 x 10-6 M, v = 0.071 µmol/min and at [S] = 5.0 x

0.20 0.25

1Km

5.0x105–

----------------------+

-------------------------------=



10-7 M, v= 0.0096 µmol/min. Therefore, Michaelis-Mentenkinetics are obeyed.

d. At [S] = 1.0 x 10-6 M

µmol/min

e. At [S] = 2.0 x 10-3 M, v = Vmax = 0.25 µmol/min. Since0.25 µmole is much less than the amount of substrate

present (2.0 x 10-3 mole/liter x 10-2 L x 106 µmol/mol = 20µmol) the reaction can proceed for five minutes without sig-nificantly changing the substrate concentration. Thus,

0.25 µmol/min x 5 min = 1.25 µmol

At [S] = 2.0 x 10-6 M,

µmol/min

During 5 minutes at this rate, 0.033 µmol/min x 5 min =0.17 µmol of product would be produced. However, this

value exceeds the total amount of substrate present (2.0 x 10-

6 mol/L x 10-2 L x 106 µmol/mol = 0.020 µmol). Clearly,during the-5 min reaction, [S] and therefore v woulddecrease significantly. Calculation of the exact amount ofproduct made would require integration of a differentialequation; this amount obviously cannot exceed 0.020 µmole.

f. Km is independent of enzyme concentration, since a changein [E] does not affect the three rate constants, k1, k2, and k3.

Hence Km would remain equal to 1.25 x 10-5 M. SinceVmax = k3[E]o, increasing the enzyme concentration by a fac-tor of 4 increases Vmax by a factor of 4. Therefore, Vmax =1.0 µmole/min.

At [S] = 5.0 x 10-6 M,

µmol/min

9. a. Since both substrate concentrations are well above Km, youcan assume that Vmax = 37 µmol/min. Then

v 0.25

1 1.3x105–

1.0x106–

----------------------+

------------------------------- 0.251 13+--------------- 0.018= = =

v 0.25

1 1.3x105–

2.0x106–

----------------------+

------------------------------- 0.251 6.5+----------------

0.257.5---------- 0.033== = =

v 1.0

1 1.3x105–

5.0x106–

----------------------+

------------------------------- 11 2.6+---------------- 1

3.6------- 0.28= = = =



µmol/min.

Therefore, at [S] 1.0 x 10-2 M, v still is equal to Vmax.

b. From the Michaelis-Menten equation, the following rela-tionships can be calculated:

c. When you plot these values you will be able to see that thebest range of [S] for studying the dependence of v on [S] is inthe neighborhood of Km or below it, since changes in [S]below Km cause greater changes in v than do changes in [S]above Km. Therefore, when using graphic methods to deter-mine Km and Vmax, several measurements should be made at[S] well below Km.

10. a.1 'unit' here = 10 µmole/15 min at 37ºC

= 10/15 µmole min-1

[S], Km v, Vmax

0.20 0.17

0.50 0.33

1.0 0.50

2.0 0.67

4.0 0.80

10.0 0.91

v 37

1 1.0x105–

1.0x102–

----------------------+

-------------------------------37

1 1.0x103–

+------------------------------- 37≈= =



= 10/15.60 µmole s-1

When [S] >> Km,

Vo = Vmax

Here Vmax = 2800 units mg-1

= 2800 x 10/15.60 = 31.3 µmol s-1 mg-1.

NOTE. THIS IS BY DEFINITION THE SP. ACT.

b. 1mg of enzyme = 10-3/MW moles = 103/MW µmoles

= 103/120000 = 1/120 µmoles.

But each mole of enzyme has 6 active sites. Therefore 1 mgof enzyme ≡ 1/120 x 6 = 1/20 = 0.05 µmoles active site.

c. The units of sp. act. are,

µmoles (unit time)-1 (mg enzyme)-1.

Here µmoles s-1 mg-1 is used, and the value is 31.3 µmole s-

1 mg-1.

The turnover number, kcat, is effectively the activity in terms

of (µmole active site)-1, instead of (mg enzyme)-1 as in sp.act., and its units are,

µmoles s-1 (µmole active site)-1 = s-1.

We know from '(b)' that there are 0.05 µmoles active site permg of enzyme, i.e.,

µmole of active site = 1/0.05 = 20 mg of enzyme,

and that the sp. act. (activity per mg of enzyme) = 31.3

µmoles s-1 mg-1. Therefore the kcat (activity per mmole ofactive site, i.e. its activity per 20 mg of enzyme) is

= 31.3 x 20 = 626 s-1.

Problem Set1. In the cases of severe liver damage, an enzyme EL is released into

the blood. After severe exercise, an isozyme from muscle, EM, isfound in the blood. EL and EM can be differentiated since theyhave different kinetic constants. The Km of the liver enzyme is 3

x 10-4 M; the Km of the muscle enzyme is 7 x 10-5 M.

Data from assays on an unconscious patient's blood are givenbelow. Ten microliters of blood was used in each assay.



a. Is the patient likely to be suffering from a liver disease or hadshe been exercising too strenuously? answer

b. Explain your reasoning for your answer to part a. answer

2. Define the following:

a. a unit of enzyme activity answer

b. Km answer

c. steady-state conditions answer

d. oligomeric enzyme answer

3. Under what experimental conditions does an enzyme-catalyzedreaction follow zero-order kinetics? answer

4. Indicate the effects of substrate concentration, enzyme concen-tration, temperature, inhibitors or activators on enzyme activityby labeling correctly both axes of the graphs given shown below.

S (M) v (µmol/min/10 µl

3 x 10-2 990

3 x 10-3 909

1 x 10-3 769

7 x 10-4 700

3 x10-4 500

1 x 10-4 250

7 x 10-5 190

3 x 10-5 91

1 x 10-5 32



Your choice for axes are: Energy, [E], Temperature, [S], 1/[S], 1/v, v.

(You may use the same label on more than one graph.) answer

5. An enzyme-catalyzed reaction was assayed at several substrateconcentrations. Two data points which fell on the Lineweaver-

Burk plot are v = 41.7 µmol S/min when [S] = 5 x 10-4 M and v

= 16.7 µmol S/min when [S] = 5 x 10-6 M. Place the two pointson a line on the accompanying graph.

a. Determine the value of Km and Vmax in the correct units.answer

When an inhibitor was added, the velocities fell to 1/2 theiruninhibited values.

b. Plot the inhibited line on the graph. answer

c. Is the inhibitor competitive or non-competitive? On whatevidence did you base your decision? answer



6. Fumarase (L-malate hydrolase) catalyzes the reversible hydrationof fumarate to L-malate. The fumarase from pig heart has beencrystallized (M.W. 197,000 Da) and consists of four subunitsthat can be dissociated into inactive monomers under relativelymild conditions. Substrate can induce reformation of tetramerswith complete recovery of the activity. The subunits have amolecular weight of 48,500 Da and each contains three free -SHgroups. Fumarase requires no cofactors. Kinetic studies impli-cate the participation of a pair of groups on the enzyme (oneacidic, one basic) with pKa's of 6.2 and 6.8. These groups havebeen postulated to be two imidazole groups of histidine residues,one in the imidazole form and one in the imidazolium form.The reverse reaction is stereospecific for L-malate and only L-malate is produced from fumarate.

a. For fumarase, an intact ___________ structure is necessary.answer

b. Predict graphically the effect of pH on the activity of fuma-

C

C-O2C

CO2-H

H

+ H2O

CO2-

CHO H

CO2-

H H



rase. At what pH would you expect maximal enzyme activ-ity? answer

c. Malonate, -OOC-CH2-COO- is an inhibitor of fumarase.What type of inhibitor would you expect malonate to be?answer

d. Using a Lineweaver-Burk plot, graphically illustrate theeffect of malonate on the kinetics of fumarase. Be sure tolabel all parts of the graph as well as the inhibitor data.answer

7. a. What is meant by an allosteric enzyme? answer

b. What is the difference between the active site and the regula-tory site of an allosteric enzyme? answer

c. How does an allosteric inhibitor produce its effect on anenzyme? answer

8. a. The activation energy for a non enzyme-catalyzed reaction is_____________ than the activation energy for the samereaction catalyzed by an enzyme. answer

b. Why would increasing the temperature of an enzyme-cata-lyzed reaction from 25°C to 37°C increase the rate of thereaction? answer

c. Why would increasing the temperature to 60°C probablycause a decrease in the rate of the enzyme-catalyzed reaction?answer

9. a. If the concentration of substrate is 10-3 M and the concen-

tration of enzyme is 10-8 M, what would be the effect of theobserved rate of production of product if the enzyme con-

centration were doubled? (Assume Km is 10-6 M.) answer

b. What would be the effect on the observed rate if the sub-

strate concentration were doubled to 2 x 10-3 M but the

enzyme concentration remained at 10-8 M? answer

10. Answer the following with true or false; justify your answer ineach case.

a. The initial rate of an enzyme-catalyzed reaction is indepen-dent of substrate concentration. answer

b. If enough substrate is added, the normal Vmax of an enzyme-catalyzed reaction can be attained even in the presence of anoncompetitive inhibitor. answer

c. The rate of an enzyme-catalyzed reaction in the presence of arate-limiting concentration of substrate decreases with time.



answer

d. The sigmoid shape of the v-versus-[S] curve for some regula-tory enzymes indicates that the affinity of the enzyme forsubstrate decreases as [S] is increased. answer

11. Two forms of isocitric dehydrogenase exist in mammals. One

which is NAD+ specific and found only in mitochondria and a

NADP+ specific enzyme found in both cytosol and mitochon-dria

(isocitric acid; pK1 = 3.2, pK2 = 4.8, pK3 = 6.4)

NADP+ specific isocitric dehydrogenase catalyzes the decarboxylation

by formation of an unstable enzyme bound chelate of Mn2+ and a α-keto acid intermediate. The free energy change for formation of a-keto glutaric acid under physiological conditions is: ∆G˚ = -5 kcal/mol. AMP regulates the enzymatic activity by reducing Km for isoci-

trate by 10 fold. Only {isocitrate2-} was found to be the substrateform that binds to the enzyme.

a. Mn2+ is an example of a(an) answer

b. NAD+ is an example of a(an) answer

The reaction velocity was found to be 4th order with respectto isocitric acid indicating high cooperativity.

c. The fact that "high cooperativity"is found indicates that thisenzyme is a(an) ______________ enzyme. answer

The NAD+ specific enzyme has a molecular weight of330,000 Daltons and is made up of 8 identical subunits.

d. The NAD+ specific enzyme is an example of a(an)______________ protein and the level of structural organi-zation for the 8 identical subunits is termed the structure ofthe protein. answer

e. AMP is said to act as a(an) answer

Reaction velocity is decreased in presence of ATP which acts

by binding at the same site that NAD+ binds.

CH2-COOH

HC

C

COOH

CH2-COOH

C

C-COOH

HH

COOHHO

H O

+ NAD+

or NADP+

+ H+ + NADH

or NADPH

isocitric acid α-ketoglutaric acid



f ATP is said to act as a(an) ____________ answer

g.Production of NADH during the course of the reaction wouldbe expected to ____________ the reaction velocity and NADHwould be an example of a(an) ____________. answer

h. What is the significance of ∆Go = -5 kcal/mole in terms of:

1. The desire of the reaction to go as written? (One sentenceanswer).

2. The activation energy of the reaction?

answers

i. If isocitrate ionization was the only important controllingfactor for the pH-activity profile of the reaction. (See char-acteristics at the beginning of this exam). Using the infor-mation given, construct an accurate pH-activity profile forthe enzyme catalyzed reaction. Use graph paper. answer

12. Zero order kinetics in an enzyme catalyzed reaction only occurswhen we have:

a. a high specific activity

b. an isozyme present

c. high substrate concentration

d. a high Km

e. high enzyme concentration

f. a high turnover number

answer

13. The redox pairs NAD+ NADH and NADP+

NADPH are well suited for use in coupled clinical enzyme assaysystems:

a. because of their acid-base properties.

b. because of their distinctly different absorbance properties oftheir oxidized and reduced species.

c. because of their occurrence in cells.

d. because of their ability to take the place of enzyme reactions.

e. because they are coenzymes.

answer

14. The success of a measurement of an enzyme activity using a cou-pled enzyme assay depends on having

a. The NAD+ NADH dehydrogenase reaction.



b. the second reaction as non-limiting.

c. A non-buffered reaction medium.

d. a colorimeter adequate in the visible region of the spectrum.

e. a trained M.D. to supervise.

answer

15. In measuring the rate of a coupled reaction one must know:

a. how to control the humidity in the sample chamber.

b. where the lag phase ends.

c. the maximum absorbance of the unknown enzyme.

d. the rate of absorbance change during the preincubationperiod.

e. the total absorbance change throughout the assay.

answer

16. The levels of LDH isozymes in the blood are indicative of certaindisease states: Some of the isozymes present will react with aparticular substrate while others will not. Thus, one can experi-mentally measure the activity of these particular isozymes whileother isozymes are also present in the sample. In particular, thisassay makes particular use of:

a. the colligative properties of the solution.

b. the substrate specificity of the isozyme of interest.

c. the preincubation phase of the isozyme of interest.

d. the tertiary structure of the isozymes of interest.

e. the total activity of all enzymes species.

answer

17. Let's say the following assay was established to measure E1 levelsin serum.

(1)

(2)

a. List the solution conditions you would have to control tomake this a valid assay. answer

b. List the species to equation 1 and 2 whose concentrations

AE1

B

B + NAD+ NADH + H+ + CE2



you would have to manipulate in order to make a valid assay.answer

c. One of your technicians ran the assay under conditionswhich, with normal serum levels of E1, the Vmax for equa-tion (1) turned out to be about equal to Vmax for equation(2). State what is wrong with the assay. answer

d. In one sentence state what you would do to rectify the prob-lem. answer

18. Clinical data for enzymes are often reported in terms of interna-tional units. What is an enzyme unit of activity and what rela-tionship does it bear to specific (enzyme) activity? answer

19. What is meant by the term "zero order" reaction as it pertains toenzymes? answer

20. Why is the preincubation phase of a coupled enzyme assay neces-sary? answer

21. Which one of the following relationships best describes howreaction velocity can be used to indicate the level of enzyme inblood serum as developed for clinical labs?

a.

b.

c.

d. Vmax = k [E0]

e. when Km >> [S],

f. [S0] - [Sf ] + [ES]

answer

ANSWERS TO PROBLEM SET1. a. Liver

b. The [S]Vmax/2 is close to the Km of the liver enzyme

Km

k 1– k2+

k1-------------------=

S[ ]S0[ ]

----------ln kt–=

1v---

Km

Vmax------------ 1

S[ ]-------- 1

Vmax------------+=

vVmax

Km------------ S[ ]=



2. a. The amount of enzyme which converts 1 µmole of substrate

to product per min (at 25/37oC)

b.

c. Under steady state conditions, S is converted into P at a con-stant rate, the [S] and [P] vs time plots are linear, and [ES] isconstant

d. An enzyme with a quaternary structure, i.e., made up ofmore than one subunit (protomer, monomer). The subunitsmay be the same (e.g., a homodimer), or different (e.g., aheterodimer).

3. Saturating substrate ([S] >> Km)

4.

Upper L: vo vs [S]

Upper R: Vmax vs [E]tot

Lower L: vo vs Temperature

Lower R: vo vs [S] for a Km-type allosteric enzyme showing +vehomotropy.

5. a. Km = 8.3 x 10-6 M, Vmax = 45.5 µmoles min-1

Km S[ ] Vmax 2⁄

kcat k 1–+

k1-----------------------

E[ ] SS S[ ] SS

ES[ ] SS----------------------------= = =



b.

c. Non-competitive (Km same, Vmax smaller)

6. a. Quaternary (oligomeric)

b. pH 6.5

c. Competes with fumarate because of similar structure, there-fore a competitive inhibitor

d.

7. a. They are enzymes whose kinetic properties cannot beaccounted for by the Michaelis-Menton model.

b The active site is where the substrate binds to the regulatoryenzyme; the regulatory site is where the effector moleculesbind. These two sites are different.

c. An allosteric inhibitor typically binds and stabilizes theenzyme in an inactive or less active (conformation) state.

8. a. Greater

b. Yes

c. Yes

9. a. [S] >> KmTherefore, vo = VmaxTherefore, the velocity is doubled when [E]tot is doubled

1/v

1/[S]

Km/Vmax

KmI/Vmax

Islope =

1/VmaxI

1/Vmax-1/Km

I = -1/Km

1/v

1/[S]

Km/Vmax

KmI/Vmax

Islope =

1/Vmax = 1/VmaxI

-1/KmI

-1/Km

+malonate

-malonate



b. None, because the enzyme is already saturated with substrate

10. a. F

b. F

c. T

d. F (positive Km-type)

11. a. Cofactor

b. Coenzyme

c. Allosteric (positive Km-type)

d. Oligomeric

e. Quaternary

f. +ve heterotropic effector

g. -ve heterotropic effector

h. Equal, coenzyme

i.

1. If NAD+/NADP+ and NADH /NADPH and isocitrate anda-ketoglutarate all at 1 M concentration are mixed at 1 atmpressure, since the free energy change under standard condi-

tions for the forward (L R) reaction is negative,

NAD+/NADP+ and isocitrate will be converted intoNADH/NADPH and α-glutaric acid. (Also, at equilibrium,the ratio of products to reactants (Keq), will be greater than1, i.e., the equilibrium lies to the right.)

2. The standard free energy change in the L R direc-

tion is the difference between the standard free energies ofactivation in the forward and reverse directions (∆G0' =∆G*for - ∆G*rev)

j.

12. c

13. b

14. b

15. b

16. b



17. a. The standard conditions of assay for E1 must be met (pH,ionic strength, cofactors, temperature, etc), so the resultsmay be compared to the range of normal values.

b. [A] must be at a sufficiently high concentration to saturateE1.

[NAD+] must be at a sufficiently high concentration to satu-rate E2.

E2 must be present at sufficiently high levels of activity for itnot to be a limiting factor in the assay

c. The rate of the second reaction must be much greater thanthat of the first reaction (E2 at high activity), so that only theamount of E1 limits the rate of the reaction, i.e., Vmax2 mustbe >> Vmax1.

d. Increase the activity of E2 present in the assay medium, sothat it is much greater than that expected for E1.

18. The IU is the amount of enzyme converting 1 µmole of S into Pper min (at a given temperature, pH etc).

The specific activity = IU per mg total protein (it increases as theenzyme is purified)

19. When the enzyme is saturated with substrate,

v0 = Vmax = kcat Et = a constant for that assay,

i.e., the rate = a constant = k, therefore it is not dependent on[S], and zero order kinetics in [S} are observed.

20. To take into account:

Breakdown of NADH/NADPH due to

a. Its intrinsic chemical instability

b. The effect of other factors in our serum sample on NADHlevels besides the enzyme we want to measure.

21. d. Vmax = k [Eo] (i.e., Vmax = kcat [E]tot)



Module 2: Clinical EnzymologyObjectives: 1. Be able to define or identify the correct from the incorrect use of

each of the following terms as they relate to enzymes and cataly-sis.

VmaxcoenzymeKmzero order reactionsaturation velocityspecific activityactive sitesubstrate specificityisoenzymeenzyme assay

2. Be able to describe the two general ways in which enzymes areused clinically and be able to supply a specific example of each.

3. Be able to explain why the conditions of pH, temperature, sub-strate concentration and inhibitors must be controlled in enzymeassays.

4. One of the most useful enzyme assay procedures makes use of

the redox pair: NAD+/NADH. Describe those features of the

NAD+/NADH interconversion that make it well suited for clini-cal measurements.

5. Explain the concept of coupled reactions. In your answer be sureto comment on each of the following:

a. Why is it useful to couple reactions?

b. What is meant by preincubation and lag phase?

c. What aspects of the coupled reactions one must be concernedwith in order to assure an accurate measurement?

6. Give an example of an organ specific enzyme which can appearin blood plasma and describe a clinical procedure that uses the

NAD+/NADH (or NADP+/NADPH) couple in its assay.

7. Several enzymes found in blood plasma originate from specifictissues or organs. Give examples of 4 different enzymes thatbelong to this category. State the conditions leading to their ele-



vated concentration in blood plasma.

8. Elevation of concentrations.

a. Describe the relationships between enzyme and substrateconcentrations which provide the best evaluation of enzymeactivity (see goal 3).

b. Differentiate between enzyme concentration and enzymeactivity.

c. Depending on the spectrophotometric equipment available,a number of approaches are possible for evaluating reactionrate from a given assay system. Describe three ways thatreaction rate may clinically be evaluated and give the advan-tages and disadvantages of each.

9. Given a passage from a journal of text concerning the principlesand uses of enzymes in clinical measurements (which may beeither a clinical investigation or enzymology description), answerquestions about the passage. (Answers may involve drawinginferences or conclusions based on principles mastered in BIO-CHEM PU02-1 and/or in this domain.)

10. Be able to answer questions such as those given in the ClinicalProblem and the Self Test. In addition, be able to define andcorrectly use the words in the NOMENCLATURE andVOCABULARY and KEY WORD lists (if abbreviation, knowwhat the letters stand for). Be able to answer questions such asthose in the Problem Sets and the Practice Exam.

CLINICAL PROBLEM:

Consider the following clinical problem with a focus on the biochem-istry of clinically important enzymes. (Glucose-6-phosphate Dehy-drogenase Deficiency. p. 298, Biochemistry: A Case OrientedApproach, Montgomery, Dryer, Conway, Spector, 1974])

A 29-year-old previously healthy Iranian physician, S.I., was admit-ted to the hospital because of generalized myalgia of 4 days duration.His major complaint was preceded, in sequence, by a temperature of102 degrees Fahrenheit (38.9 ˚C), dark urine, anorexia, nausea, andultimately scleral icterus. He had been in contact with jaundicedpatients in the course of his hospital work but he denied exposure tohepatotoxins in any form, including injections or blood transfusions.

The significant findings of the physical examination were deep jaun-dice, a moderately tender liver palpable 2 cm below the right costalmargin, and no splenomegaly.



The immediately relevant hematologic and liver function studies aresummarized in Table 1 Please note that a nonstandard unit (U/dl)has been provided by this particular clinical laboratory. A red bloodcell (RBC) survival study with sodium chromate-51 Cr showed aninitial precipitous drop in counts over a 3-day period leading to anextrapolated Cr half-life of 6 days. Findings from paper electro-phoresis of the serum protein were normal except for a moderatedecrease in albumin.

Biopsy of the liver demonstrated a marked inflammatory reactioninvolving the portal areas, with occasional areas of focal necrosis inthe lobule. Myeloid metaplasia was also noted in the biopsy speci-men, most probably the result of intermittent, lifelong hemolysis.After careful questioning, the patient described repeated episodes ofmalaise, myalgia, and severe fatigability that had accompanied rela-tively minor respiratory infections and that probably representedundiagnosed episodes.

The color generated during the methemoglobin reductase test forglucose-6-phosphate dehydrogenase estimated visually was compati-ble with complete absence of the enzyme. This suspicion was sup-ported later by quantitative assay.Further discussion of the case is contained on pp. 299-302 in Mont-gomery et al., Biochemistry, 1st edition.

Table 4: Laboratory Data for Patient S.I.

Test Values S.I.* Normal

Hemoglobin 15.1 (9.2) (g/100 ml)

14-18 (g/100 ml)

Hematocrit 47.0 (29) (%) 40-54 (%)

Reticulocytes 12.4 (%) 0.5-1.5 (%)

Alkaline phos-phatase

8.0 (U/dl) 5-13 King Armstrong units/dl

AST (or SGOT) 620.0 (U/dl) 15-40 Karmen units/dl

ALT (or SGPT) 1220.0 (U/dl) 5-40 Karmen units/dl

Direct bilirubin 35.5 (mg/100 ml)

0.0-0.1 (mg/100 ml)

Total bilirubin 57.0 (mg/100 ml)

0.2-1.0 (mg/100 ml)



* Figures in parentheses indicate lowest value measured during hos-pitalization.


Acid Phosphatase AlbuminAlkaline Phosphatase ALT (or SGPT) Assayαααα-Amylase ββββ-AmylaseAntibiotics AsparaginaseAST (or SGOT) Assay Blood Urea Nitrogen (BUN)Ceruloplasmin CholinesteraseCoupled Assays CPK AssayCreatinine levels Functional Plasma EnzymesGalactosemia Glucose-6-phosphatase deficiencyγγγγ-Glutamyl Transferase (GGT)αααα-Hydroxybutyrate Dehydrogenase (αααα-HBD)Hyperlipoproteinemias Inborn errors of metabolismLag Phase LDHLipoprotein Lipase (LPL) LovostatinLower Limit Normal (LLN) Non-functional plasma enzymesPancreatic Lipase PhenylketonuriaPlasminogen PreincubationScoline Sorbitol DehydrogenaseSphingolipidoses StreptokinaseUpper Limit Normal (ULN)

Key Words: Binding SitesBiochemistryCoenzymesEnzyme InhibitorsEnzyme TestsEnzymes

Serum iron 247.0 (µg/100 ml)

75-175 (µg/100 ml)

LDH 985.0 (U/dl) 90-200 IU/dl

Glucose-6-phos-phatedehydroge-nase activity

Absent

Table 4: Laboratory Data for Patient S.I.

Test Values S.I.* Normal



Self-Test 1. Describe the function of a catalyst.

2. How do enzymes differ from other chemical catalysts?

3. Why are enzymes necessary in the biological system?

4. What are isoenzymes and how are they useful clinically?

5. How may enzymes be used to provide useful clinical informa-tion?

6. What is the function of the coenzyme NAD+ and NADP+ inenzyme reactions?

7. Distinguish between lag phase and preincubation.

8. What reaction parameters affect the enzyme-catalyzed reaction?

9. What is a coupled enzyme reaction?

10. How can hemolysis affect the enzyme assay?

11. What is the significance of a-hydroxybutyrate dehydrogenase?

12. In what clinical conditions would you expect AST to be greaterthan ALT and vice versa?

13. How would the clinical laboratory's use of a nonstandard unit(U/dl) impact on your interpretation?

14. What is (are) the true physiological substrate(s) for alkalinephosphatase?



STUDY GUIDE-2I Diagnostic Enzyme levels are often altered in disease. There are two basic situa-

tions:

1. Where the abnormal level of enzyme is in fact the fundamentalcause of the disease. Many inherited diseases are due directly toan insufficiency or complete absence of an enzyme activity. Suchinborn errors of metabolism derive from a fault at the geneticlevel, i.e., the condition is passed from one generation to thenext through transmission of the parental DNA. For example, amutation at a codon coding for an amino acid critical for thefunction of the enzyme may affect its Km or kcat, reducing itsefficiency, or completely inactivating it.

There are many examples of such conditions. E.g.,

Glucose-6-phosphatase deficiency

Ordinarily, this enzyme is only expressed in liver and kidney calls. Itallows glucose-6-phosphate, produced by breakdown of liver or kid-ney glycogen under conditions where blood glucose is low, to be con-verted into glucose that can then be released into the bloodstream forby other tissues, particularly the brain and muscles. If its activity istoo low, a severe hypoglycemia results.

Phenylketonuria Here, the enzyme activity (phenylalanine hydroxylase) which ordi-narily converts the amino acid phenylalanine to tyrosine is missing,and phenylalanine accumulates, with damaging consequences.

The Sphingolipidoses Sphingolipids are catabolized in lysosomes. The enzymes carryingthis out have to work in a specific sequence, so that if only one ofthem is absent, its substrate accumulates with pathological conse-quences, even if all the prior and subsequent enzymes in the pathwayare functional. (See Module 4, PU6).

Galactosemia In this condition, the enzyme activity galactosyl-1-phosphate uridyltransferase is absent. This enzyme is important for the conversion ofgalactose into glucose, which is then metabolized by the glycolyticpathway, pentose phosphate pathway, and glycogen synthase. Thus,galactose cannot be utilized, and accumulates as galactose-1-phos-phate, producing many symptoms.

Enzyme assays allow diagnosis of these genetically determined dis-eases. Such diagnosis often allows intervention to ameliorate the con-



dition. E.g., in galactosemia, galactose intake can be restricted.Lactose contains galactose, so that removal of dairy foods from thediet is useful. Again, in phenylketonuria, restriction of phenylalanineintake is useful, for example, by reducing dairy food consumption,particularly cheese.

2) Where the abnormal enzyme activity is a consequence of dis-ease or trauma. For example, a permeability barrier may bedamaged secondary to some disease process, allowing high levelsof some enzyme activity to leak into the blood; or, a tissue whichnormally produces an enzyme may be damaged or destroyed,thereby reducing the enzyme activity.

Determination of Whether an Enzyme Activity Level is Abnormal

This is done by assaying under a fixed, well-defined standard set ofconditions, that are used in all hospitals, so that we can COMPAREour measured activity directly with the normal activity as assayedunder the same conditions. Thus, our assay must be carried out atexactly the same pH, temperature, and ionic strength as the standardassay, and all the necessary cofactors and ingredients must be presentat their standard concentrations. We determine whether our assayvalue lies within the range defined by the Upper Limit Normal(ULN) and the Lower Limit Normal (LLN) which depend on theparticular enzyme.

Sampling 1. Sometimes a specific tissue biopsy has to be taken; e.g., skeletalmuscle in the case of a suspected muscular dystrophy.

2. Blood is the most convenient tissue to sample, if possible. Ifblood is used, certain precautions have to be taken. If the sampleis being used for diagnosis of a suspected disease associated withblood clotting, the blood plasma obtained by centrifuging downthe blood cells has to be used. This requires the use of anti-coag-

ulants agents such as EDTA or citrate (which chelate Ca2+

needed for clotting) or heparin. These compounds interferewith many enzyme assays, and are avoided if at all possible.Thus, serum is the most convenient sample form. This lacks theclotting factors, and is obtained by allowing clot formation andretraction to take place in the sample, when the straw-coloredsupernatant is collected. Additionally, because the RBC's con-tain enzymes which are sometimes the same as those we are try-ing to assay, it is most important to avoid hemolysis whenhandling blood samples. It is necessary to distinguish between'Functional Plasma Enzymes' such as lipoprotein lipase, whichare present under normal circumstances in the blood, which havea real physiologic role, and 'Non-functional plasma enzymes',



which only appear in the blood in pathological conditions.

Measurements Units The International Enzyme Unit (I.U.): typically, the amount of

enzyme converting 1 mmole of substrate into produce min-1, usuallyat 37 or 25ºC. (However, if the enzyme is of intrinsically low activ-

ity, the units might be, say, nmoles hr-1.)

The Specific Activity: effectively, enzyme units mg-1. Typically,

µmoles min-1 mg-1.

Note that if we give the activity in terms of µmole-1 instead of per

mg-1, we obtain a quantity identical with kcat, i.e.,

µmole min-1 µmole-1 = min-1, the units of a first order rate constant.Thus, if we know the molecular weight of our enzyme, we can inter-convert kcat and the specific activity,

( )

In practice, we always work under conditions where the assayedenzyme should be completely saturated with its substrate ([S]>>Km),so that Vmax is determined, and the measured activity is directly pro-portional to the amount of enzyme added.

Students are sometimes confused by the fact that when they are intro-duced to Vmax, they see the equation,

Vmax = kcat [E]T,

where Vmax is typically in units of mM min-1, kcat is in min-1, and[E]T is in mM (i.e., the units are concentration per unit time), while

in practice Vmax is often given in, say, µmoles min-1, i.e., in absoluteamounts of substrate converted or product formed.

This is because if we multiply the above equation for Vmax by the vol-ume of the system, we obtain

Vmax = kcat ET,

where Vmax is typically in µmoles min-1, kcat is in min-1, and ET is inµmoles.

kcatSpAct pureEnzyme MW×,

1000---------------------------------------------------------------------= 1µmole

MW1000------------mg=



Thus, our measured activity is directly proportional to the amount ofenzyme added, i.e., the plot of activity against [E]T or ET is linear. Inthis way, the specific activity and kcat are given simply by the slope ofthe plot.

It is important to remember that serum is a very complex mixture ofenzymes, proenzymes, transport proteins, immunoglobulins, cofac-tors etc, and many trace molecules. Thus, in general, activities basedon serum samples are given in units per unit volume of blood for thepurposes of comparison with normal enzyme levels, say per ml or per100 ml, and the amount of sample is quantitated in terms of the vol-ume added to the assay mixture. If 10 µl, of serum give a Vmax of,

say, 2 µmoles min-1, then 1 ml of blood will give 200 µmoles min-1,which we can compare with the range of normal values based onactivity per ml blood.

Since ET or [E]T is directly proportional to Vmax, provided we knowthere is nothing intrinsically wrong with the enzyme (no geneticdefect), we ought to be able to calculate the absolute amount ofenzyme present in our sample,

, or

However, in general, unless we have other data, we cannot distinguishbetween low total amounts of good (functional) enzyme, and a lowactivity due to the poor intrinsic activity of a defective enzyme.

Thus, we make our measurements under steady-state conditions (S→ P at a constant rate), with sufficient substrate to keep the enzymesaturated, so that our measured velocity is the maximal velocity(Vmax), and our progress curve is linear, with a slope proportional tothe specific activity/kcat of the enzyme. In practice, we can either fol-

Vmaxµmoles min-1

or mM min-1

ET or [E]T (usually µl of serum)

ET

Vmax

SpAct----------------mg= ET

Vmax

kcat------------µmoles=



low the decrease in substrate concentration or the increase in productconcentration. The assay has to be carried out under exactly the cor-rect conditions -pH, ionic strength, temperature, correct levels ofcoenzymes etc, so as to allow comparison with tabulated data.

A very common and convenient method used to determine the con-centration of a chemical species is near ultraviolet or visible spectro-photometry. If radiation of a wavelength corresponding to the energyneeded to excite the molecules is absorbed, the intensity of the radia-tion is reduced by its passage through the sample. The amount ofradiation absorbed at a particular wavelength λ (units nm) is mea-sured by a quantity called the Absorbance, which is on a logarithmicscale.

Often there is a very simple relationship between the absorbance, andthe concentration of the species, which can be summarized in theBeer-Lambert Law,

Aλ = ελcl,

where ελ is the extinction coefficient for the molecule at wavelengthλ, c is its concentration, and l is the pathlength. Usually, l = 1 cm.

Thus, the rate of change of the absorbace is directly proportional tothe rate of change in concentration.

i.e., the rate of the reaction is effectively given by the rate of changeof the absorbance.

Sometimes a substrate or product itself may absorb light, and we can

I0

I

I0 = initial intensity

I = intensity afterpassage through sample

Absorption bands

Aλ

λ nm

Aλ = log(I/I0)λ = -log(I0/I)λ

∆Aλ∆t

-----------ελ∆c

∆t------------=



follow the reaction directly. This can be done in particular if NAD+/

NADH or NADP+/NADPH are involved. (These are cosubstrates inmany oxidation-reduction reactions.) This is because the reducedforms absorb light at 340 nm, while the oxidized forms do not.

Thus, in an enzyme catalyzed reaction,

AH2 + NAD+ → B + NADH + H+

we will see an increase in A340 as the absorbing species NADH is

formed from the non-absorbing species NAD+, while in

A + NADH → BH2 + NAD+

we will observe a decrease in A340.

Our sample is usually serum, so that many enzymes and other extra-neous molecules including enzymes that use NADH/NADPH will bepresent during the assay, which may cause oxidation/reduction of the

NADH or NAD+ that we add to the assay medium. Also, our moni-tored molecule may be unstable, and also contribute to any change inA340 - this is particularly true of NADH or NADPH. Suppose we

have oxidation of NADH to NAD+,

Aλ

λ nm

NAD+/NADP+

NADH/NADPH

200 250 300 350 400



We can circumvent these potential problems by allowing the assay toproceed at first without having any substrate (A) added to the assaymixture. This is called ‘preincubation’. After we have determinedour background rate of change of A340, we then add our specific sub-strate. Only the enzyme that uses that substrate will bind it and con-vert it into product, so that any increase in the reaction rate after weadd the specific substrate can only be due to the enzyme we are inter-ested in. I.e., we ‘isolate’ our particular enzyme by making use of thefact that only that enzyme can use the specific substrate we add toform the specific product. We then substrate the ‘background’ ratefrom the total rate seen after adding the specific substrate.

True rate = Rate after adding specific substrate - Preincubation back-ground rate

Synthetic Chromoge-nic Substrates

For a lot of enzyme activities we may wish to measure, NAD+/

NADP+ may not be a cosubstrate, but many synthetic (chromogenic)substrates that either absorb light or give products that absorb lightare now available, e.g., p-nitrophenolphosphate (pNPP). This is usedin the determination of phosphatase activities.

A yellow color develops as the pNPP is hydrolyzed, the increase in theintensity of which is followed at 400 nm. In this case, we simply

A340

Time

preincubation

Rates ~ slopes

serum added excess ‘A’ added

phase

O2N P

O-

O-

O

O2N OH

O2N O-

+ H2O + PO43

+ H+

p-nitrophenolphosphate

phosphatase

p-nitrophenol

p-nitrophenolate anionabsorbs maximally at 400 nm



measure the rate of change in A400 before and after adding our sam-ple (serum), and substract the background rate. I.e. although we cor-rect for instability of the artificial substrate by following backgroundbreakdown of the pNPP, we do not strictly actually do a ‘preincuba-tion’ of the serum in this situation.

True rate = Rate after adding serum - Rate before adding serum

Coupled Assays Sometimes there is no appropriate synthetic substrate available forthe enzyme we wish to assay, and the enzyme does not itself useNADH/NADPH. However, sometimes the product of the enzyme(E1) we wish to assay is a substrate of an enzyme (E2) using NADH/

NAD+.

or

Provided the two sets of assay media are compatible, we can combinethe two assays into a single system, so that as B/BH2 is formed fromA by E1, it is converted into BH2/B by E2 with the concomitant oxi-

dation/reduction of NADH/NAD+, i.e., we can indirectly follow theE1 activity using the rate of change of A340. We have coupled thetwo reactions. We must ensure:

1. As usual, the pH, temperature, etc, correspond to standard assayconditions.

but in addition,

2. The only limiting factor in the overall reaction is the amount ofE1.

Thus,

A BE1

B + NADH + H+E2

BH2 + NAD+

A BH2

E1

BH2 + NAD+E2

B + NADH + H+



(i) the concentration of the initial substrate A must be saturating forE1.

(ii) the concentration of NADH/NAD+ must be saturating for E2.

(iii) the second enzyme, E2, must be present at sufficiently high levelsthat the overall rate will not depend on the second reaction, i.e.,the second reaction is not limiting.

The assay medium initially contains all of the components EXCEPT

(1) the specific substrate (A), and

(2) the serum sample containing the enzyme whose activity we wantto measure (E1).

Suppose we have a reaction where in the second step NADH is oxi-

dized to NAD+. As with all NADH/NADPH assays using serum asthe sample, we first add the serum and allow the system to preincu-bate to determine any background breakdown of NADH due toinstability of the NADH and extraneous enzymes in the serum. Afterwe have determined this rate, we add the specific substrate for E1 (A),and measure the increase in rate of change of A340 over the back-ground preincubation rate. Because only E1 can use A with the cou-pled oxidation of NADH, this increase is specific for E1. We thensubtract the background preincubation rate from the rate after A hasbeen added.

Note that after we have added A, the reaction at first accelerates overa period of time called the ‘lag phase’. The part of the progress curvewhose slope we use for the determination of activity is the linearphase after the lag phase has ended. The lag phase arises because afterwe have first added A, although E1 is now saturated with its substrate,and that reaction is going at its maximum rate, the steady state levelof B, the product of E1 and substrate for E2 has not yet been attained.This takes a finite period of time to be generated. The second reac-tion (catalyzed by E2), which is the reduction of B with the concom-itant oxidation of NADH, will increase in rate until theconcentration of its substrate B has achieved its steady state level.

In such a coupled reaction we might see the following progress curve-



er on

We can divide the time course into the following sections:

I. Substrate A and serum both not yet added. Low backgroundbreakdown of NADH due to its chemical instability.

II. Preincubation. Serum sample added. Preincubation begins.Background breakdown of NADH due to non-specific factors inthe serum as well as intrinsic instability of NADH.

III. Lag Phase. A saturating amount of A, the specific substrate forE1, has been added. Only E1 can use this, so any increase in ratemust be due to the activity of E1. The reaction accelerates dur-ing the lag phase until the steady state concentration of B hasbeen reached.

IV. Steady-State, Linear phase. For every molecule of A convertedinto B by E1, a molecule of NADH is oxidized by E2. Underthese conditions, the rate of the reaction catalyzed by E1 is thesame as the rate of the reaction catalyzed by E2, so we can mea-sure the activity of E1 by the rate of fall of A340 during this phaseafter we have subtracted the rate observed during the preincuba-tion phase.

V. Depletion of NADH occurs.

Some of the Commonly Determined Enzyme Activities and OthBiochemical Quantities: the Meaning of Some of the Acronymsa Lab Report.

Albumin This transport protein is made by the liver, and serum levels of albu-min are one of the indications of liver dysfunction. Since gammaglobulin is synthesized by the immune system, while albumin is made

A340

Time

I IIIII

IV

V

Add serum

preincubation

Add excess ‘A’

lag

Steady-State linear

Depletionof NADH



by the liver, the albumin/globulin ratio is sometimes measured, beingdecreased in liver disease.

BUN, Blood Urea Nitrogen

Used to monitor renal function. It may be elevated in renal disease,and in states of dehydration.

Creatinine levels Used along with BUN to monitor renal function. Creatinine isformed by spontaneous (non-enzymatic) breakdown of creatine phos-phate in the muscles,

and is normally excreted only by the kidney into the urine. Serumlevels of creatinine are raised in renal insufficiency. The rate of for-mation is proportional to muscle mass, and is normally at a constantrate, which allows the creatinine level in a urine sample to be used asa measure of the time over which the urine was produced.

Ceruloplasmin This is a serum protein (an α2-globulin) that has two functions.Firstly, it carries 90% of the Cu in the blood. Secondly, it acts as a

ferroxidase, oxidizing Fe2+ to Fe3+. This is important in the blood,because the major Fe carrier, transferrin, can only bind Fe in its ferric

(Fe3+) state. Ceruloplasmin thus acts to link Cu and Fe metabolism.Its main diagnostic use is as an indicated of Wilson's disease (a genet-ically determined hepatoreticular degeneration), when there is a fail-ure of the liver to excrete Cu into the bile, and Cu accumulates in theorgans of the body. Although an abnormality in ceruloplasmin is notthe primary cause of Wilson's disease, its levels are decreased in thatcondition.

ACP, Acid Phosphatase A non-specific phosphatase with an acid (~pH 4) pH optimum.High levels are present in the prostate gland. Serum levels are raisedin ~80% of cases where prostate cancer has metastasized. Prostateacid phosphatase is inhibited by tartrate, unlike most other acidphosphatases, which are inhibited by formaldehyde (e.g., those ofRBC's, bone, liver and spleen). Often assayed using the chromogenicsynthetic substrate p-nitrophenyl phosphate, which gives a yellowcolor absorbing at 400 nm as it is hydrolyzed. The physiological

-OC

CH2

NC

NH

PO-

O CH3

NH2+ O

O-

HN

C

N CH2

COHN

H3C

+ PO43-spontaneous

creatine phosphate creatinine



function of this enzyme is unclear.

ALP, Alkaline Phos-phatase

Another non-specific phosphatase, but with an alkaline pH optimum(~10). Also assayed with p-nitrophenylphosphate. It has less specificlocations than acid phosphatase. It is present in high levels in liver,bone (osteoblasts), and the intestine. The hepatic content of ALP isincreased in cholestasis (obstruction of the bile duct), and can giveincreased serum concentrations. It is released into the blood by activeosteoblasts, and can therefore indicate active bone formation. Thus,it is at high levels in infants of 1-2 years and in the early teens. It isan important indicator of bone pathology, and is a valuable guide tothe activity of Paget's Disease. It indicates bone remodelling due tobone resorption. The serum activity is often elevated in bone cancer.The physiological function of this enzyme is unclear.

The Amylases These are digestive enzymes produced in two tissues, namely the sali-vary glands and the exocrine pancreas. There are tissue-specificisozymes, which can be distinguished by the effect of inhibitors, andby mobility on electrophoresis.

ββββ-amylase is an exoglycosidase, cleaving the α (1 → 4) glycosidiclinks in starch by hydrolysis, working from the non-reducing end ofthe glucose chain.

αααα-amylase is an endoglycosidase, cleaving the internal α (1 → 4) gly-cosidic linkages randomly.

The most important clinical use is in detection of acute pancreatitis(where serum levels of >10 x ULN may be seen). Increased serumlevels are seen in mumps, and some other conditions, such as pepticulcer, gallstones and intestinal obstruction.

Pancreatic Lipase This is a digestive enzyme of the exocrine pancreas released into theduodenum that hydrolyses the ester links at the 1 and 3 positions of

triglycerides in the presence of colipase, bile and Ca2+ to give 2-monoglyceride and the fatty acids from the 1 and 3 positions (seeModule 3, PU6).

H2C

CH

H2C

O

O CR1

O

R2C

O

O CR3

O

CH2OH

CH

CH2OH

OR2C

O

triglyceride

2H2O+ R1COOH + R3COOH

2-monoglyceride



This may be a better indicator for diagnosing pancreatitis than amy-lase.

LPL, Lipoprotein Lipase

This enzyme is bound to the lumen of the capillaries, and catalysesthe breakdown of triglyceride to 2-monoglyceride and R1COOH andR3COOH, (like pancreatic lipase). (It is discussed in detail in PU 6,Modules 3 and 4.)

Its levels are low in Type I diabetes (juvenile onset), and lipoproteinlipase activity may be greatly reduced in Type I hyperlipoproteinemia,an autosomally recessive inherited disease.

CK, Creatine Kinase/CPK, Creatine Phos-phokinase

This enzyme of the muscle sarcoplasm catalyzes the transfer of aphosphoryl group from creatine phosphate to ADP to regenerateATP used up in muscle contraction (creatine phosphate is the imme-diate energy reserve in muscle.)

The active enzyme consists of a dimer. The two subunits (protomers,monomers) may be of two types: M and B. Therefore, there are 3types of active isozyme.,

MM in skeletal muscle

MB 30% of the CK in the heart

BB Brain and Thyroid

Usually, most of the CK in serum is from skeletal muscle (MM), andraised levels of MM are seen after strenuous exercise. High levels ofthe MB isozyme are associated with myocardial infarction (MI). Ifseen in young boys, high levels of MM are consistent with Duchenne

H2C

CH

H2C

O

O CR1

O

R2C

O

O CR3

O

CH2OH

CH

CH2OH

OR2C

O

triglyceride

2H2O+ R1COOH + R3COOH

2-monoglyceride

-OC

H2C

NC

NH

PO-

O

NH2+ O

O-

-OC

H2C

NC

NH2

O

NH2+

+ ADP + ATP

creatine phosphate creatine



muscular dystrophy (DMD).

LDH, Lactate Dehy-drogenase

This catalyzes the interconversion of ketopyruvate and L-lactate,

The active enzyme is a tetramer, which is made up of two types ofsubunit, M and H. There are thus 5 types of isozyme, which are dis-tributed differently among the tissues,

Heart, RBC, kidney and brain (inhibited by pyru-vate, use α-hydroxybutyrate as a substrate)

LDH3 H2M2 Pancreas, lung, spleen, lymph, adrenal and thyroidglands

Skeletal muscle, liver (not inhibited by pyruvate,cannot use α-hydroxybutyrate as a substrate)

The different isozymes represented by the different combinations ofH and M have evolved to suit the requirements of a particular tissue.The inhibition of the H4 and H3M isozymes by the substrate pyru-vate is an excellent example of a negative homotropic effect. In skele-tal muscle, where isozymes insensitive to pyruvate are present, lactatemay accumulate without disastrous results, apart from a leg cramp.However, in heart muscle, any form of cramp would be fatal, and sothe isozymes in that tissue cannot in practice exceed a certain activityto prevent the lactate levels rising too high.

α-HBD, αααα-Hydroxy-butyrate Dehydroge-nase

An important difference between the LDH isozymes is that H4 andH3M show higher activity with α-hydroxybutyrate as the substratethan with the normal substrate, lactate, while the other isozymes donot.

C

CH

CH3

HO

O-O

lactate

+ NAD+

C

C

CH3

O-O

O + NADH + H+

ketopyruvate

LDH1 H4LDH2 H3M }

LDH4 HM3LDH5 M4

}



Thus, if we use α-hydroxybutyrate as the substrate, we are effectivelyonly assaying heart or RBC LDH - we can distinguish between theisozymes both by the degree of inhibition by pyruvate, and theiractivity towards α-hydroxybutyrate dehydrogenase. This is veryimportant in practice, as elevated serum levels of LDH occur undermany situations, so that total serum levels of LDH are not very use-ful. Raised levels of the heart and RBC isozymes (H4 and H3M)occur in myocardial infarction and sickle cell hemolytic crises. Sincethe H4 and H3M isozymes are present in RBC's, care has to be takenwhen sampling for diagnosis of myocardial infarction that hemolysisdoes not occur.

The Transaminases These enzymes catalyze the transfer of an α-amino group from an α-amino acid to an α-keto acid.

Frequently, α-ketogluarate is the acceptor, when glutamate is formed,

C

CH

CH2

HO

O-O

L-α-hydroxybutyrate

α-HBD

+ NAD+

C

C

CH2

O-O

O + NADH + H+

α-ketobutyrate

CH3 CH3

NH3+

C

R1

H CO2- C

R2

CO2- C

R1

CO2-

NH3+

C

R2

H CO2-

O O

+ +



AST, Aspartate Tran-saminase/GOT, Glutamate Oxaloace-tate Transaminase (SGOT = Serum GOT)

This catalyzes the transfer of the amino group of apartate to α-keto-glutarate,

AST/GOT is found in many tissues, both in the cytosol and mito-chondria of the cells. In myocardial infarction and liver damage/dis-ease, it leaks out of the damaged cells so that the SGOT rises.

ALT, Alanine Transam-inase/GPT, Glutamate Pyruvate Transami-nase (SGPT = Serum GPT)

This catalyzes the transfer of the amino group of alanine to α-ketogl-utarate,

NH3+

C

R1

H CO2-

C

CH2

CO2-

C

R1

CO2-

NH3+

C

CH2

H CO2-

O

O

CH2

CO2-

CH2

+

CO2-

+

α-amino acid α-KG α-keto acid glutamate

NH3+

C

CH2

H CO2-

C

CH2

CO2-

C

CH2

CO2-

NH3+

C

CH2

H CO2-

O

O

CH2

CO2-

CH2

+

CO2-

+

L-Asp α-KG oxaloacetate L-glutamate

CO2- CO2

-



ALT/GPT is a cytosolic enzyme particularly concentrated in the livercells, but is less abundant in the liver than AST/GOT. In situationswhere the entire liver is affected, such as cirrhosis, cancer, or hypoxia,SGOT levels are raised more than SGPT levels; but if the liver cellplasma membrane is mainly affected, as in viral hepatitis, SGPT ishigher than SGOT (because it is mainly cytosolic enzymes which leakout of the damaged liver cells).

The levels of CK, AST (GOP), and α-HBD have characteristic timecourses in the serum after a myocardial infarction (MI).

GGT, γγγγ-Glutamyl Transferase

This is located in the ER of liver cells (where its function is not clear),and in the plasma membrane of renal tubular cells, where it may beinvolved in the transport of certain amino acids in the kidney. It isused as an indicator of some hepatobiliary diseases, when it is releasedinto the blood.

Sorbitol Dehydroge-nase

Increased serum levels are seen after liver damage.

NH3+

C

CH3

H CO2-

C

CH2

CO2-

C

CH3

CO2-

NH3+

C

CH2

H CO2-

O

O

CH2

CO2-

CH2

+

CO2-

+

L-Ala α-KG ketopyruvate L-glutamate

RelativeEnzymeActivity

Time, Days5 10

CK

ASTα-HBD



Cholinesterase A non-specific pseudocholinesterase (sometimes called butyrylcho-linesterase or serum cholinesterase) of unknown physiological func-tion is secreted by the liver into the blood. (This must bedistinguished from the specific cholinesterase, acetylcholinesterase,which mediates the termination of the acetylcholine signal at bothnicotinic and muscarinic cholinergic synapses by hydrolyzing acetyl-choline into acetic acid and choline.) The liver enzyme is importantclinically because we rely on it to hydrolyze succinylcholine('Scoline'), a depolarizing cholinergic blocker used to obtain relax-ation during surgery, into succinic acid and choline. In some patients(~0.05%), there is a genetic defect, and the liver enzyme is inactive,so that recovery from scoline is greatly delayed, giving rise to a condi-tion called 'scoline apnea'. Thus, it is important to screen patientsbefore surgery, to make sure this condition does not exist.

Other Enzyme Activi-ties

Clearly, if a specific condition is suspected, then other assays may becarried out, e.g., for glucose-6-phosphatase if glucose-6-phosphatasedeficiency is likely, etc, etc. Click on the hypertext link for a listingof some popular enzyme-based diseases and syndromes.

II. Therapeutic The major use of our knowledge of enzymes is in the treatment ofdisease by enzyme inhibitors. There are many examples of this, e.g.,the use of lovostatin, an inhibitor of cholesterol synthesis in thetreatment of certain hyperlipoproteinemias. Occasionally, anenzyme may be used directly, e.g., the use of asparaginase to destroyasparagine,

in the treatment of adult leukemias.

Note that streptokinase, used to remove bloodclots in the legs, is notonly not a kinase, but is in fact not an enzyme at all. It is a small pro-tein which binds to plasminogen in the blood, and is able to activatethe zymogen to its plasmin activity without cleaving it by proteolysis.

C

CH2

CH

O

H3N+ CO2-

NH2

+ NH4+

L-Asn L-Asp

H2O C

CH2

CH

O

H3N+ CO2-

O-



Many antibiotics are effectively enzyme inhibitors, e.g., penicillin isa suicide substrate for the enzyme involved in last step in bacterial cellwall biosynthesis.

Post Test1. All of the following enzymes are markers for liver disease

EXCEPT

a. alanine transaminase (ALT)

b. aspartate transaminase (AST)

c. creatine phosphokinase (CPK)

d. sorbitol dehydrogenase

e. lactate dehydrogenase (LDH)

answer

2. The isozymes of lactate dehydrogenase

a. demonstrate the evolutionary development of this enzyme

b. range from monomers to tetramers

c. differ only in a single amino acid

d. exist in five forms, depending upon the content of M and Hmonomers

e. are forms of the enzyme that differ in activity but not in elec-trophoretic mobility

answer

3. An individual suffering chest pain went to the local hospitalemergency room for treatment. Analysis of the patient’s bloodshowed elevated levels of the enzyme lactate dehydrogenase(LDH) and predominance of the H4 isozyme. Give one possibleexplanation for the increased levels of LDH. Explain the bio-chemistry behind your medical diagnosis. answer

Answers to Post Test1. c This enzyme catalyzes the transfer of a phosphoryl group from

creatine phosphate to ADP to regenerate ATP used up in musclecontraction (creatine phosphate is the immediate energy reservein muscle). The enzyme comprises two subunits, M and B,



which forms three isozymes. MM is found in skeletal muscle,MB found in the heart, and BB found in brain and thyroid.

2. d The LDH tetramer may be composed of subunits characteris-tic of heart or muscle, in any possible proportion, making fivepossible tetramers.

3. The patient may have had a heart attack, causing damage toheart tissue. LDH molecules of the H4 variety leaked out fromthe damaged heart tissue into the blood.



Module 3: Membrane TransportObjectives: 1. Describe the permeability properties of lipid bilayer membranes.

2. Be able to describe the difference between simple diffusion andfacilitated passive diffusion through biological membranes.

3. Be able to distinguish between facilitated passive diffusion andactive transport

4. Be able to distinguish between primary and secondary activetransport..

5. Describe symport and antiport transport. Be familiar with howcoupled transport processes are used a) at the plasma membraneto move sugars and amino acids with Na+ and b) at the mito-chondrial membrane to shuttle metabolites in and out of themitochondrion.

6. After reading a short passage from a medical journal or textbookbe able to interpret the data and draw conclusions about the sig-nificance of the data.

7. Understand the terms in the Nomenclature and VocabularyYlist as well as the Key Words list. Be able to answer questionsand problems similar to those on the Practice Exam.


Active transport Antiport

Ca2+-ATPase CotransportCoupled cotransport DigitalisElectrochemical gradient Facilitated diffusion

Flux (J) H+-pumpIon channel IonophoreLigand-gated channel Membrane potential

Mobile carriers Na+,K+-ATPaseNet flux Passive diffusionPassive mediated diffusion Passive facilitated diffusionPermeable PermeabilityPrimary active transport Saturation kineticsSecondary active transport Simple diffusionSymport Voltage-gated channel



Key Words: Adenosine Triphosphatase, Sodium, PotassiumBiochemistryBiological TransportCell Membrane PermeabilityIon ChannelsMembrane PotentialsMembranesPermeability



STUDY GUIDE-3I. Membrane Trans-port

Biological membranes are constructed around the phospholipidbilayer. This is a 2-dimensional micelle, with the non-polar acylchains of the phospholipids forming an oily internal film sandwichedbetween two polar hydrophilic surfaces formed by the head groups ofthe phospholipids, together with the solvating medium of water andcounterions. The phospholipid bilayer represents a barrier to the freediffusion of solutes, acting to separate the internal contents of the cellfrom the external medium, and to form internal compartmentsbetween different functional parts of the cell. Because of the highelectrical resistance of the hydrocarbon chains in its interior, thebilayer also acts as a capacitor, separating electrical charges across themembrane. Operation of three phenomena, the Donnan effect, the

differential permeability of the membrane to ions and the Na+, K+-pump combine to give the interior of the cell a negative potential rel-ative to the outside. There is an electrical potential gradient acrossthe plasma membrane tending to pull positively charged solutes intothe cell, and push negatively charged solutes out.

II. The Electrochemi-cal gradient

When there is a difference in concentration of an uncharged solutebetween adjacent regions with no barrier between them, diffusionallows net transport (flux) of the solute to occur from the regionwhere it is initially at high concentration to the region where it is ini-tially at low concentration, until the difference in concentration hasdisappeared. This process is thermodynamically spontaneous, and its∆G is negative. Free diffusion is described by Fick's Laws, the detailsof which do not concern us here, except to note that an unchargedsolute will flow down its concentration gradient, and that the rate oftransport depends on the size of the concentration gradient of the sol-ute between the two regions, i.e., the difference in concentration. Ifthere is a barrier between the regions, the permeabilty of that barrierto the solute will determine the rate of flow of molecules from thehigh concentration compartment to the low concentration compart-ment. (Strictly, we should use the term 'chemical potential' insteadof 'concentration', but for us the difference is not important. Theterm 'net' is used because diffusion is a random (stochastic) process,with molecules moving in both directions. Some will be movingfrom the low to the high concentration region, but more will be mov-ing from the high to the low. It is this excess which represents the nettransport from high to low.)

If the molecule possesses an electric charge, the presence of the mem-



brane potential has to be taken into account, since the negativecharge on the internal surface of the cell tends attract positive ionsinto the cell, and repel negative ones, whilst the relatively positiveouter surface attracts negatively charged solutes and repels positivelycharged solutes. Thus, for a charged molecule there is a combinationof two effects to be taken into account in determining its directionand rate of net movement across a biological membrane: its concen-tration gradient across the membrane; and its electrical potential gra-dient across the membrane. These may act in the same direction andreinforce one another, or oppose one another. The net driving forcecan be easily calculated by converting the concentration driving forceinto electrical units (using the Nernst equation), and adding or sub-tracting it from the electrical driving force depending on its sign. Acharged molecule moves spontaneously down its electrochemical gra-dient, until its electrochemical potential is the same on both sides ofthe membrane (∆G < 0).

(We will usually use the general term 'electrochemical gradient',remembering that for uncharged molecules this simply means con-centration gradient.)

III. Simple Passive Diffusion Across Phos-pholipid Bilayers

Fig. 1 Permeability of a Phospholipid Bilayer to a Selection of Solutes

In this case, the direction and rate of net transport is determinedsolely by the electrochemical gradient of the particular solute acrossthe bilayer, and the permeability of the bilayer towards that solute.Net transport of solute occurs spontaneously down its electrochemi-

very non-polar O2N2

benzenesmall uncharged polar CO2

H2Oureaglycerol

small charged polar acetate

large polar sucrose

inorganic ions K+

Ca2+

Na+

Cl-



cal gradient until its electrochemical potential is the same on bothsides of the membrane, with a decrease in the free energy of the sys-tem (∆G < 0). The molecule diffuses across enclosed in a cavityformed by kinks and defects in the packing of the acyl chains. Thepermeability of the membrane to a solute depends on a number offactors, one of the most important of which is its solubility in thenon-polar hydrocarbon interior of the bilayer. Thus, very non-polaruncharged molecules, which dissolve well in the oily interior of themembrane, move easily down their concentration gradient: i.e., theactivation free energy for the process is low. The membrane is quitepermeable to small uncharged polar molecules, such as urea, glycerolor water itself. The presence of an actual electrical charge on a polarmolecule greatly lowers its permeability, but if the molecule is small,and the formal charge can be reversibly lost and regained on the twosurfaces of the membrane, quite high permeabilites can be seen (e.g.,acetate anion->acetic acid-> acetate anion). Phospholipid bilayers arevery impermeable to large polar molecules (e.g., sucrose). The mem-brane is most impermeable towards inorganic ions; these cannot losetheir charge, and the activation free energy for their movement is veryhigh.

Movement of a solute across membranes is quantitated in terms of itsnet flux, J, which is the rate of net transport per unit area of mem-brane. As with rates in enzyme kinetics, we always measure the initialrate, Jo, since obviously the rate falls off as the electrochemical gradi-ent is dissipated. For an uncharged solute moving across a givenmembrane, the electrochemical gradient is just the concentration gra-dient, which for a membrane of a defined thickness we can estimateby the concentration difference of the solute between the two sides ofthe membrane. As Fig. 2 shows, the initial flux does not reach a sat-urating (constant) value as the concentration difference across themembrane is increased, but shows a linear increase with ∆G.

Fig. 2 Simple and Facilitated Passive Diffusion across a membrane

JM

JM/2

KM ∆c

hyperbolic

linear

facilitated

simple

J0



Facilitated (Mediated) Passive Diffusion and Active Transport

Clearly, since the cell needs to admit and expel an enormous numberof different solutes in a highly selective way for many purposes, spe-cial mechanisms must have evolved to control movement of mole-cules and ions across biological membranes. These all use proteins,which, since they all act to lower the activation free energy for move-ment, have the characteristics of enzymes. (Only O2 and CO2 moveacross cell membranes without special proteins being involved-thereare even specific channels for water.) A fixed, limited, number of pro-tein molecules, often located in special regions of the membrane areinvolved. These are usually very specific for the solute whose trans-port they are catalyzing, and must be regulatable to control the flux.

We can divide the transport proteins into two broad classes: facili-tated passive diffusion and active transport.

Facilitated (Mediated) Passive Diffusion

Here the driving force for the transport process is just the electro-chemical gradient for the solute, which moves spontaneously downits electrochemical gradient. ∆G is negative, and the process is pas-sive just as with the simple case, because we do not have to do any-thing to the system to make the net transport occur. Because thereare only a limited number of transporter protein molecules, the fluxwill reach a limiting value as the concentration difference of the sol-ute across the membrane is increased and all the protein moleculesbecome occupied in transporting the solute; i.e., we have saturationkinetics, as shown in Fig. 2. The protein effectively catalyzes thetransport of the solute and we get equations for J0 and ∆c very similarto the Michaelis -Menten equation for v0 and [S].

and the J0 vs. ∆c curve looks just like the hyperbolic curve for anordinary Michaelis-Menten enzyme.

Thus, Lineweaver-Burke plots can be made, just as with solubleenzymes

Since the binding sites on the protein molecules which catalyze trans-port can only accomodate the particular solute or structurally similarmolecules, facilitated (mediated) passive diffusion thus differs fromfrom simple (passive) diffusion by:

1) Speed and specificity

2) Saturation kinetics

J0 JM∆c

∆c KM+---------------------=



3) It can be chemically inactivated (by reagents that react withthe transporter)

4) It can be competitively inhibited.

Fig. 4 Lineweaver-Burke plots can be used to identify competitiveinhibitors of facilitated diffusion.

Sub-Types of Facili-tated Diffusion

Uniport:-A single type of solute moves down its electrochemical gra-dient

Symport and Antiport Coupled Cotransport

In symport, two different types of solute are transported in the samedirection across the membrane

In antiport, two different types of solute are transported in oppositedirectons

+I -I

1/J0

1/∆c

1/JM

-1/KM -1/KIM

Uniport

Gap Junctions (Allows ions and small molecules to move from one cell to another)

The GLUT class of glucose transporters e.g. in red blood cells, muscle, adi-pocytes

Ion Channels

Ligand Gated

Voltage Gated

X



Transport of 'X' depends on thesimultaneous or sequential trans-port of 'Y' in both symport andantiport mechanisms

E.g., the redblood cell anion

exchanger (antiport) of HCO3- for Cl-. (Symport and Antiport also

occur in active transport, see below).

Ionophores provide good examples of facil-itated diffusion.

These are organic molecules, often from microorganisms, that medi-ate facilitated diffusion of ions across membranes. They sometimeshave antibiotic properties. The driving force for net solute flow isagain the electrochemical gradient of the solute. There are two classesof ionophore:

1. Mobile Carriers. These are characterized by having: (1) A non-polar outer surface that dissolves well in the hydrocarbon interior ofthe membarne bilayer; (2) A hydrophilic interior with a polar surfacethat can solvate the transported ion in its unhydrated form. The cav-ity is often very specific for a given type of ion. Both the complexedand uncomplexed forms of this type of ionophore can diffuse backand forth across the membrane bilayer. Some, like valinomycin (spe-

cific for K+) and the nactins (also mainly K+), catalyze a uniport typeof process which is therefore electrogenic (since net charge is trans-

ported across the membrane). Others, such as nigericin (mainly a K+

carrier) and monensin (mainly Na+), lose a proton on binding the ion

on one side of the membrane, and bind a H+ when they release theion on the other side: thus, they catalyze an electroneutral exchange

(antiport) of K+ or Na+ for H+. Mobile carriers can only functionwhen the interior of the membrane is in its liquid crystalline state, i.e., the membrane is fluid and above its transition temperature.

2. Channel Formers. These do not depend on the membrane beingin a fluid state. They often show only poor selectivity, with very hightransport rates compared to mobile carriers. Hydrophobic groups onthe outside of the channel allow it to be inserted into the bilayer,whilst its interior is lined with polar groups which let ions passthrough the channel in their hydrated forms, so that a pore is formedin the membrane. E. g., gramicidin A.

Active Transport Here the electrochemical gradient for the process is unfavorable -inthe absence of the transporter, the ∆G is positive (> 0), and transportwill not occur spontaneously in the direction needed. The electro-chemical gradient opposes the desired movement. To get movement

Symport Antiport

xy

xy



against the electrochemical gradient, we have to couple the transportprocess to a second process with a sufficiently negative ∆G to makethe ∆G for the total coupled process negative.

∆Gtot(< 0)) = ∆Gtrans (> 0) + ∆Gcoupled process (more negative than∆Gtrans is positive)

In order for two processes to be coupled, there must be an obligatoryrelationship between them, i.e., one cannot happen with out theother also happening. The free energy of hydrolysis of ATP is oftencoupled to active transport, either directly or indirectly. Thus, inactive transport, the cell has to invest energy to accomplish the move-ment of the solute in the desired direction.

There are two forms of active transport:

Primary (Direct) Active Transport

In this, an energy source is used directly to push the solute against itselectrochemical gradient. There are three main forms:

1. Where the free energy of hydrolysis of ATP is coupled to thetransport process. Examples are the P-type ATPases, in which a

phosphorylated intermediate is formed, such as the Na+, K+-

ATPase in the plasma membrane, and the Ca2+, H+-ATPases in

the plasma membrane and SER. The V-type H+ transportingATPases of endosomes and lysosomes represent another class,where a phosphorylated intermediate is not formed.

2. Where the transport of H+ against their concentration gradientout of the mitochondria is coupled to passage of electronsthrough Complexes I, III, and IV of the electron transport chainin the inner mitochondrial membrane.

3. Where light energy is captured and used to move H+ acrossmembranes, as in bacteriorhodopsin.

Secondary (Indirect) Active Transport

In this, an electrochemical gradient is first established by some energy

requiring primary active transport process, such as the Na+ gradient

generated across the plasma membrane by the Na+, K+-pump, andthis is then coupled to movement of a second solute against its elec-trochemical gradient through a coupled cotransporter working ineither the Symport or Antiport mode, i.e. the movement of one sol-ute down its electrochemical gradient is coupled to the movement ofa second solute against its electrochemical gradient. An example ofsecondary active transport across the plasma membrane is the uptakeof glucose or amino acids from the gut lumen against their electro-



chemical gradients in symport with Na+ moving back down its elec-trochemical gradient into the intestinal epithelial cell. Another

example is the movement of Ca2+ out of a heart muscle cell against its

electrochemical gradient in antiport with Na+ moving into the cell

down its electrochemical gradient. This is catalyzed by the Na+-Ca2+

exchanger,

(Since digitalis inhibits the Na+, K+ pump, which thus reduces the

Na+ gradient across the sarcolemma of the muscle cell, less Ca2+ are

pumped out, and the concentration of Ca2+ in the muscle cell rises.This increases the contractile force of the heart muscle cell, and is thebasis for the positive inotropic effect of the cardiac glycosides.)

The H+ gradient produced by the H+ pumping complexes of theelectron transfer chain is coupled to the secondary active transport ofmany solutes across the inner mitochondrial membrane against their

electrochemical gradient in the symport or antiport modes, as H+

move down their electrochemical into the mitochondrial matrix.

(Movement of OH- out of the mitochondria in symport or antiport

also is used, when we remember that movement of OH- out is equiv-

alent to movement of H+ in.)

Both primary active transport, and therefore secondary active trans-port as well, are inhibited by metabolic poisons interfering with ATP

production or H+ pumping in the mitochondria. Secondary active

transport is blocked by ionophores which dissipate the driving Na+ or

H+ electrochemical gradient.

Post-Test1. Selective permeability of cell membranes is achieved, in part, by

active transport systems. Active transport differs from passivetransport in that it

a. requires energy but no transport carrier

b. depends primarily on diffusion and osmosis

c. requires a carrier but no energy

d. requires energy, usually in the form of phosphate anhydridebonds

3Na+out + Ca2+

in 3Na+in + Ca2+

out



e. is necessarily associated with pinocytosis

answer

2. The best described cotransport systems involving glucose andsodium are

a. uniports

b. two oppositely oriented uniports

c. two uniports oriented in the same direction

d. antiports

e. symports

answer

3. Ionophores include

a. pore-forming antibiotics that act by passive transport

b. carrier antibiotics that act by active transport mechanisms

c. pore-forming and carrier antibiotics that act by active trans-port and passive transport respectively

d. pore-forming and carrier antibiotics that act by passive trans-port and active transport, respectively

answer

4. All of the following substances freely diffure across biologicallipid bilayer membranes EXCEPT

a. carbon dioxide

b. malate

c. nitric oxide

d. oxygen

e. urea

answer

5. The sodium dependent transport of glucose in the kidney is anexample of

a. antiport

b. group translocation

c. passive transport

d. primary active transport

e. secondary active transport

answer

6. What type of plot represents substrate concentration versus rate



of transport by simple diffusion?

a. bell-shaped curve

b. rectangular hyperbola

c. signoidal curve

d. straight line

answer

7. All of the following are true statements regarding the sodium/potassium-ATPase EXCEPT

a. It consists of two α-subunits and two β-subunits

b. It forms an aspartyl-phosphate high-energy bond duringtranslocation

c. It forms an aspartyl-phosphate low-energy bond duringtranslocation

d. It translocates three sodium ions per ATP

e. It translocates three potassium ions per ATP

answer

8. Which of the following is a true statement regarding cardiac gly-cosides such as digitalis, which are used in the treatment of heartfailure?

a. They inhibit calcium-ATPase

b. They activate calcium-ATPase

c. They inhibit sodium/potassium-ATPase

d. They activate sodium/potassium-ATPase

e. They inhibit translocation by the plasma membrane sodium-calcium antiport protein

answer

9. The enzyme responsible for maintaining the pH inside of lysos-omes is a class

a. P ATPase

b. V ATPase

c. F ATPase

d. M ATPase

answer

Answers to Post Test


1. d Active transport is the passage of material against a concentra-tion gradient and requires the expenditure of energy, often thehydrolysis of ATP or PEP.

2. e

3. a

4. b Malate is a polar compound that does not cross biologicallipid bilayers by simple diffusion. The inner mitochondrialmembrane contains several specific transport proteins that trans-locate this substance.

5. e Secondary active transport uses a source of chemical energysuch as a sodium gradient that is produced by primary activetransport. The cotransport of sodium and glucose in the samedirection is called symport; this is an example of secondary activetransport.

6. d The rate of simple diffusion is a linear function of the sub-strate concentration.

7. e Three potassium ion are NOT translocated per ATP, only twopotassium ions are translocated per ATP.

8. c Cardiac glycosides bind to the exterior surface of the sodium/potassium-ATPase and inhibit the enzyme. As a result, the intra-cellular concentration of sodium is increased somewhat. Thisintracellular sodium exchanges for extracellular calcium by a pro-cess that is mediated by an antiport protein. The higher intracel-lular calcium ion concentration resulting from this process isthought to augment cardiac muscle contraction.

9. b This class of enzymes is responsible for maintaining the pHinside of lysosomes and secretory vessicles. These enzymes lack aphosphorylated enzyme intermediate. The class V enzymes arecomposed of from three to five different polypeptide subunits.



Practice ExamThe two hour exam for Biochemistry PU02 will be composed ofmultiple choice questions. The pretest, post test, and problem sets inthis Problem Unit provide examples of short answer questions tocheck your knowledge base while the (25) questions below are actualquestions of the "Board Type" given to a previous MSI class. Youmay wish to time your use of this Practice Exam to one hour or less inorder to pace yourself for the exam.

For each of the following multiple choice questions, choose the mostappropriate answer.

1. Which one of the following statements about the maximumvelocity (Vmax) of an enzyme catalyzed reaction is true?

a. Vmax is unaltered in the presence of a noncompetitive inhib-itor.

b. Vmax is directly proportional to enzyme concentration.

c. Vmax is directly proportional to substrate concentration.

d. Vmax is inversely proportional to the measured rate (v).

e. Vmax is directly proportional to Km, being higher forenzymes with higher Km values for their substrates.

answer

2. Ingested glucose is phosphorylated in the liver by both hexoki-nase and glucokinase. Hexokinase has a Km for glucose of about

1 x 10-5 M and a glucokinase has a Km for glucose of about 1.5 x

10-2 M. The glucose concentration available to liver cells isapproximately equal to the concentration in the blood, and this

value is about 135 mg% - or 0.75 x 10-2 M. When the glucoselevel falls from 135 mg% to 80 mg%, the in vivo rate of phos-phorylation of glucose in the liver may change because:

a. the rate of the reaction catalyzed by hexokinase will fall butthat of glucokinase will be little affected.

b. the rate of the reaction catalyzed by glucokinase will fall butthat of hexokinase will be essentially unaffected.

c. the rates of both enzymes will decrease.

d. the rates of both enzymes will not be affected much.

e. the rate of glucose will increase but that of hexokinase willremain about the same.



answer

3. For any enzymic reaction, when the ratio [ES]/[E] increases:

a. the velocity of the reaction will increase.

b. the velocity of the reaction will decrease.

c. the Km is increased.

d. the turnover number decreases.

e. an allosteric effector must be present.

answer

4. In the presence of effector A an enzyme displays sigmoidalbehavior when initial velocity is plotted against substrate con-centration. With increasing concentrations of A the curve is dis-placed to the left (i.e., towards the ordinate). Effector A is thus:

a. a positive allosteric effector.

b. a competitive inhibitor.

c. a negative allosteric effector.

d. a non-competitive inhibitor.

e. an uncompetitive inhibitor.

answer

5. The activity of enzyme 1 (E1) is to be measured in a coupledassay system utilizing enzyme 2 (E2) according to the scheme:

(measured product)

Where A and B are substrates for enzyme 1, P1 is the product ofthe reaction catalyzed by enzyme 1, and P2 is the product of theaction of enzyme 2 on P1. Which of the following conditionsmust be satisfied to have a valid assay for enzyme 1?

a. The concentration of substrate A must be much greater thansubstrate B in order to have the rate of the reaction depen-dent on substrate B.

b. The activity of enzyme 2 must be much greater than enzyme1 in the assay system.

c. The rate of reactions of enzyme 1 and enzyme 2 must beadjusted so that a steady state concentration of P1 will bemaintained during the reaction.

d. The formation of P1 must be detectable.

e. None of the above.

answer

A + B P1 P2E1 E2



6. The initial velocity of a reaction at various initial substrate con-centrations was determined with an enzyme of connective tissuemetabolism from a HEALTHY patient. The results obtainedwere as follows:

The best estimates from this data with respect to this enzyme are:

a. Km = about 5 x 10-5 M, Vmax about 63 µmol/min.

b. active sites are unoccupied on the great majority of enzyme

molecules when the substrate concentration equals 3 x 10-5

M.

c. Km = about 3 x 10-5 M, Vmax about 65 µmol/min.

d. Km = about 2 x 10-5 M, Vmax about 130 µmol/min.

e. the enzyme's active sites are approximately half-saturated at

1 x 10-4 M.

answer

7. The role of most vitamins in metabolism is:

a. to serve as coenzymes or precursors of coenzyme.

b. to serve as precursors of essential amino acids.

c. to serve as key intermediates in the citric acid cycle.

d. to build up resistance to bacterial and viral infections.


1.0 x 10-3 65

5 x 10-4 63

1 x 10-4 51

5 x 10-5 42

3 x 10-5 33

2 x 10-5 27

1 x 10-5 17

5 x 10-6 9.5

1 x 10-6 2.2

5 x 10-7 1.1



e. to "spark" metabolism.

answer

8. All of the following statements regarding diagnostic enzymologyare true EXCEPT:

a. serum enzymes are usually measured by their activity ratherthan by the absolute concentration of enzyme molecules.

b. enzyme activity should be measured with the substrate con-centration at less than 10 times the Michaelis constant toensure reaction rate linearity.

c. multiple point determinations yield superior accuracy andprecision over two-point determinations in kinetic analyses.

d. the presence of cofactors in the reagent may increase theenzyme activity.

e. enzymes generally show higher activity at 37°C than at30°C.

answer

9. If in a Lineweaver-Burk plot the 1/v intercept is 2 x 10-3 min per

mole and the 1/[S] intercept is -2 x 104 M-1, what is the value ofKm?

a. 5 x 10-3 M

b. 5 x 10-4 M

c. 5 x 10-5 M

d. 0.5 x 104 M

e. None of these.

answer

10. What is the reaction rate in a simple enzymatic system, if thesubstrate concentration(s) is much less than Km?

a. It is maximal.

b. It would be too slow to measure.

c. It is virtually proportional to substrate concentration.

d. It would be reduced by addition of more substrate.

e. It would be uninfluenced by the addition of more substrate.

answer

11. What is one good reason why initial velocities are employed inenzyme assays?



a. Substrate inhibition is minimized.

b. Substrate activation is minimized.

c. Errors in technique are minimized.

d. Product inhibition is minimized.

e. The sensitivity of the assay is increased.

answer

12. Which one of the following applies to competitive enzyme inhi-bition?

a. Vmax is decreased from the uninhibited value.

b. Km is unchanged from the uninhibited value.

c. Vmax is the same.as the uninhibited value.

d. Velocity is independent of substrate concentration.

e. Km is decreased from the uninhibited value.

answer

13. Enzymes affect the rate of a chemical reaction by:

a. decreasing the free energy of the reaction.

b. increasing the free energy of the reaction.

c. lowering the energy of activation of the reaction.

d. raising the energy of activation of the reaction.

e. displacing the equilibrium constant.

answer

14. Ingested glucose is phosphorylated in the liver by both hexoki-nase and glucokinase. Hexokinase has a Km for glucose of about

1 x 10-5 M and glucokinase has a Km for glucose of about 1.5 x

10-2 M. The glucose concentration available to liver cells isapproximately equal to the concentration in the blood, and this

value is about 135 mg% or 0.75 x 10-2 M. Assuming that theATP concentration and other factors are optimal, the velocity ofthe glucokinase reaction will be about:

a. 75% of Vmax.

b. 66% of Vmax.

c. 50% of Vmax.

d. 33% of Vmax.

e. 16% of Vmax.



answer

15. The fact that an enzyme-catalyzed reaction is first order withrespect to substrate at low substrate concentration but becomeszero order with respect to substrate at high substrate concentra-tion is explained by:

a. the catalytic constant.

b. reversibility of formation of the enzyme-substrate complex.

c. three point attachment.

d. specificity.

e. saturation.

answer

16. If a competitive inhibitor of an enzyme is added to the assaymixture, the effect on a Lineweaver-Burk plot is to:

a. decrease the slope.

b. increase the slope.

c. increase the slope and decrease the Y intercept.

d. increase the Y intercept and decrease the slope.

e. decrease both slope and the Y intercept.

answer

17. The effect of pH on enzyme-catalyzed reactions can be related toall of the following EXCEPT:

a. the influence of pH on the equilibrium.

b. the need for certain forms of ionizable groups on the sub-strate.

c. the need for certain forms of ionizable groups in the enzyme.

d. the general effect of pH on protein structure.

e. the influence of pH on the interaction of hydrophobic resi-dues in the protein.

answer

18. All of the following statements about enzymes are true EXCEPT:

a. enzymes reduce the activation energy needed for a reactionto occur.

b. pH affects enzyme activity by determining the degree of ion-ization of catalytic groups.

c. turnover number is a measure of the maximal activity on amolecular level.



d. the Michaelis constant Km is equal to the maximal velocityVmax/2.

e. induced fit may occur in nonallosteric enzymes.

answer

Answer the following questions using the key outlined below:

A. If 1, 2, and 3 are correct

B. If 1 and 3 are correct

C. If 2 and 4 are correct

D. If only 4 is correct

E. If all four are correct

19. A sigmoidal substrate saturation curve implies:

1. that the enzyme must possess more than one subunit.

2. that the enzyme does not obey the Michaelis-Menten Equa-tion.

3. a slower reaction velocity than hyperbolic kinetics.

4. that over some concentration range the change in enzymeactivity will be greater than the change in substrate concen-tration.

answer

20. For a certain enzyme, an irreversible, non-protein inhibitor isfound that completely blocks enzymatic activity. The enzyme isreacted with the inhibitor until the enzymatic activity is exactly1/2 its starting value, and the excess inhibitor is removed.Which of the following statements concerning the kinetic prop-erties of this modified enzyme preparation are true?

1. Km remains unchanged.

2. Vmax remains unchanged.

3. The plot of v vs [S] is identical to the plot for the unmodi-fied enzyme at 1/2 the enzyme concentration.

4. The kinetic behavior cannot be predicted from the informa-tion provided.

answer

21. A homotropic allosteric effector:

1. is usually a substrate for the enzyme.

2. always gives positive cooperativity.

3. binds to an allosteric site which is also an active site.



4. binds to an allosteric site which is not an active site.

answer

22. A non-competitive inhibitor causes:

1. an apparent decrease in Vmax.

2. no apparent change in Km.

3. the formation of an EIS complex.

4. a family of parallel lines in 1/V vs 1/[S] plots.

answer

23. The Km for glucokinase is approximately 5 mM, while the Kmfor hexokinase is about 0.01 mM. At normal blood glucose(M.W. = 180) concentrations (80-100 mg/100 ml):

1. the activity of hepatic glucokinase is equal to one-half Vmax,while the hexokinase activity is near Vmax.

2. the activity of hepatic glucokinase is approximately 0.01Vmax, while that of hexokinase activity is 100 Vmax.

3. glucokinase activity will be unaffected by glucose 6-phos-phate levels, while hexokinase activity may be reduced.

4. the activity of extrahepatic glucokinase will be variable,depending upon the intracellular glucose concentration.

answer

24. In enzyme assays, it is preferable to measure initial velocities inorder to:

1. avoid substrate inhibition.

2. avoid product inhibition.

3. increase the sensitivity of the assay.

4. avoid the reverse reaction.

answer

25. In clinical isozyme determinations on plasma, which quan-tity(ies) is/are significant?

1. Enzyme levels (concentration).

2. Relative amounts of the isozymes in the plasma.

3. Isozyme distribution pattern of various tissues.

4. Number of subunits of the enzyme.

answer



-
Answers to Practice Exam Questions
1. B 2.B

3. A 4.A

5. B 6.C

7. A 8.B

9. C 10.C

11. D 12.C

13. C 14.D

15. E 16.B

17. E 18.D

19. C 20.B

21. B 22.A

23. B 24.C

25. A

APPENDIX I: Using Acrobat Reader with pdf Files

Portable Document Format (PDF) files can be read by AcrobatReader, a free program which can be downloaded from the AdobeWeb site (http://www.adobe.com/acrobat). If Acrobat Reader isinstalled on your system, it will automatically open simply by double-clicking on the pdf file that you wish to read.

Acorbat Window The document will be displayed in the center of your window and anindex will appear at the left side of the screen. Each entry in theindex is a hypertext link to the associated topic in the text.

Using hypertext links in a pdf document is exactly like that in a webpage or html document. When you place the cursor over a hypertextlink, it changes to a hand with the index finger pointing to the under-lying text. Clicking the mouse causes the text window to jump tothat location. The index does not change. Magnification may needto be adjusted using the menu option in the lower part of the screen




to optimize the view and readability. The best magnification is usu-ally around 125%.

Subheadings in the index can be viewed by clicking on the open dia-monds to the left of appropriate entries to cause them to point down-wards. Clicking again will close the subheadings lists.

Hypertext links Hypertext links in the text (not in the index) are indicated by blueunderlined text. The cursor should change to a hand with the indexfinger pointing to this text when it passes over it. Clicking will causethe text page to move to the associated or linked text which will behighlighted in red underlined text. Red underlined text is not ahyperlink, only a destination.

How to back up to a previous window:

If you wish to return to a previous text window after following ahypertext link, use the black double solid arrow key at the top of theAcrobat window (or use the key equivalent “command - “). Acrobatkeeps a record of your last 20 or so windows so that multiple stepsback can be made by repeating the command.

Links to web sites A number of url links to web sites are located in the pdf file andappear in blue underlined type starting with http:// (e.g. http://www.som.siu.edu). Clicking on these should open a web browsersuch as Netscape and take you to those web sites. You may need toresize the Acrobat Window to view the web browser window dis-played underneath it.

COMMENTSI hope that you find this pdf file useful. Comments on how to makeit better would be greatly appreciated. Please notify me in person orby email ([email protected]) of any errors so that they can beremoved. The online version on the Biochem server can be easilyupdated.


http://www.som.siu.edu

http://www.som.siu.edu

BIOCHEMISTRY AND MOLECULAR BIOLOGY Problem Unit Two ...

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