BIOENERGETICS, ENZYMES AND METABOLISM

OBJECTIVES

 Define and explain thermodynamic terms.

 Clarify the concepts of potential and kinetic energy.

 Explain the significance to Cell Biology of the First and Second Laws of Thermodynamics.

 Define free energy and its relationship to the spontaneity of chemical processes.

 Clarify the distinction between steady state and equilibrium.

 Acquaint students with the operation of coupled reactions.

 Point out the similarities in and differences between enzymes and inorganic catalysts.

 Explain the concepts of activation energy and transition state.

 Describe the principles of enzyme kinetics and define important terms (KM, Vmax, etc.).

 Describe the action of competitive and noncompetitive enzyme inhibitors.

 Explain different mechanisms of metabolic regulation (feedback inhibition, activators, etc.).

 Define oxidation - reduction (redox) reactions.

HUMAN PERSPECTIVES QUESTIONS

1. Why has the rate of development of antibiotic resistance accelerated since the late 1970s and early 1980s? One reason may be the AIDS epidemic. Since victims of AIDS suffer from opportunistic infections that healthy humans normally don't get, the antibiotics are being used with more frequency and perhaps in higher doses. Furthermore, many people use antibiotics incorrectly; they either do not take the full course of medication or they use bacterial antibiotics for viral infections. This increases the selection pressure for microbial immunity to antibiotics. Furthermore, many of the genes conferring antibiotic immunity are located on the same bacterial plasmid. Consequently, more than one antibiotic immunity gene can be passed simultaneously from one bacterium to another.

2. A number of antibiotics attack prokaryotic protein synthesis, but not eukaryotic protein synthesis. Name two common antibiotics that work this way. The antibiotics are streptomycin and the tetracyclines. What is their site of action and why don't they affect eukaryotic protein synthesis? They act by binding to prokaryotic ribosomes. Eukaryotic ribosomes are sufficiently different from those in prokaryotes to prevent binding of these substances. Thus, eukaryotic protein synthesis is unaffected.

3. Penicillin is an irreversible inhibitor of the transpeptidases, enzymes that cross-link components of the bacterial cell wall. The cell wall is thus fragile and the bacteria die. Penicillin and its derivatives are structural analogs of the natural substrates of these enzymes. Why doesn't penicillin normally kill humans, unless a severe allergic reaction (anaphylaxis) develops? Humans don't have cell walls and transpeptidases. Thus, penicillin cannot inhibit an enzyme that humans do not possess and it will not normally harm humans. How does penicillin inhibit transpeptidase? Since penicillin occupies the transpeptidase active site, it acts somewhat like a competitive inhibitor, but since it binds irreversibly, it is not truly competitive.

4. If an antibiotic were found to bind to a site on an essential bacterial enzyme other than the active site, what would its most likely mode of action be? It would probably be a noncompetitive inhibitor of the affected enzyme.

5. You isolate the enzyme that synthesizes folic acid in bacteria and conduct some enzyme kinetics experiments. You find, not surprisingly, that sulfa drugs inhibit the enzyme's activity. What happens to the Vmax and KM of this enzyme when it is treated with sulfa drugs? Since the sulfa drugs are competitive inhibitors, the Vmax should stay the same and the KM should increase. Why do sulfa drugs have no effect on human metabolism? Sulfa drugs work by acting as a competitive inhibitor of an enzyme that converts p-aminobenzoic acid (PABA) to the essential coenzyme folic acid. Since humans lack a folic-acid synthesizing enzyme, they must obtain this essential coenzyme in their diet and, consequently, sulfa drugs have no effect on human metabolism.

6. What is unusual about the mechanism by which vancomycin inhibits transpeptidation of the bacterial cell wall? Vancomycin binds to the peptide substrate of the transpeptidase, rather than to the enzyme itself. Why is it more difficult for bacteria to develop resistance to vancomycin than to other antibiotics? Normally, the transpeptidase substrate terminates in a D-alanine—D-alanine dipeptide. To become resistant to vancomycin, a bacterial cell must synthesize an alternate terminus that does not bind the drug. This is a roundabout process that requires the acquisition of several new enzymatic activities. Consequently, vancomycin is the antibiotic to which bacteria have been least able to develop resistance and thus it is usually given as a last resort when other antibiotics have failed.

7. What is the reason for treating a patient simultaneously with two antibiotics against the same bacterium? The hope is that while a bacterium may have the gene for resistance to one antibiotic, it will not have a gene conferring resistance to the second antibiotic. Also, in the case of penicillin, treatment with penicillin and an agent that inhibits -lactamase may protect penicillin against -lactamase and allow it to work.

8. The development of bacterial resistance to antibiotics is an example of what engine of evolution? Natural selection.

9. Which antibiotic inhibits the enzyme DNA gyrase, which is required for bacterial DNA replication? Quinolones.

10. Penicillin fits into the active site of transpeptidases and thus acts as what kind of inhibitor? Penicillin acts as a competitive inhibitor since it occupies the active site. How is the effect of such an inhibitor usually able to be reversed? Such inhibition is usually reversible by increasing substrate concentration, which would tend to displace the inhibitor from the active site. Why would this approach not work with penicillin? An increase in substrate concentration does not reverse penicillin's effect because penicillin forges a covalent bond between itself and the enzyme active site, thus occupying the active site permanently and irreversibly inactivating the enzyme.

11. What are some ways that antibiotic resistance genes can be passed from bacterium to bacterium? DNA can pass from bacterium to bacterium through conjugation in which bacteria pass DNA through tubelike connections joining two of them together. Bacteria may exchange genes by the process of transduction in which a virus carries bacterial DNA from one bacterium to another. DNA may also be passed by transformation in which a bacterium can pick up naked DNA from its surrounding medium.

12. How does the AIDS virus manage to avoid the effects of drugs that attack its enzymes so effectively? The replicating enzyme of HIV is reverse transcriptase. This enzyme has a much higher error rate than DNA polymerase and makes a large number of mistakes leading to a higher mutation rate. When combined with the high rate of virus production, the likelihood of producing drug-resistant variants in a single infected individual is relatively high.

13. Many bacteria have acquired resistance to penicillin by picking up the gene for -lactamase. However, some have developed resistance without acquiring this gene. How do these bacteria escape the fatal effects of penicillin? Some are resistant because they possess modifications in their cell walls that block the entry of the antibiotic. Others are resistant because they are able to selectively export the antibiotic once it has entered the cell. Still others are resistant because they possess modified transpeptidases that fail to bind the antibiotic.

14. What strategies are employed to combat the ability of the AIDS (HIV) virus to develop drug-resistant variants? One strategy employed is having patients take several drugs at once (cocktails), each of which is targeted at different viral enzymes. This strategy greatly reduces the likelihood that a variant will emerge that is resistant to all of the drugs. Second, drugs have been designed that interact with the most highly conserved portions of each targeted enzyme, those portions within which mutations are most likely to produce a defective enzyme.

EXPERIMENTAL PATHWAYS QUESTIONS

1. Two important amino acids project into the active site of lysozyme: a glutamic acid and an aspartic acid. Both contain carboxyl groups, but when the enzyme's active site is unoccupied, the carboxyl group on the glutamic acid residue is not ionized while the carboxyl group on the aspartic acid residue is ionized. These two residues are only 0.3 nm apart. Why is one ionized and the other not ionized? Despite their proximity, they are far enough apart to exist in different chemical environments. With the active site unoccupied, the glutamic acid is in a nonpolar environment and is, therefore, not ionized. On the other hand, the aspartic acid residue is in a hydrophilic environment, which facilitates the loss of its proton.

2. The glutamic acid residue eventually does lose its proton. What precipitates this event and what is the result? When the substrate enters the active site, the strained glycosidic linkage that is to be broken approaches the glutamic acid very closely. The linkage is strained and contains an oxygen atom that is sufficiently electronegative to draw the proton from the undissociated carboxyl group of the nearby glutamic acid. This leads to the ionization of the glutamic acid carboxyl group. The released proton attacks the strained linkage and breaks it.

3. Why is the glycosidic linkage between the fourth and fifth peptidoglycan residues in the active site strained? The active site is not big enough to accommodate all six sugar residues of the substrate. Like fitting a size 6 foot into a size 5 shoe, strain is introduced, specifically between the fourth and fifth sugar residues. The fourth or D sugar does not fit easily into the available space in the active site. In order to fit, it is forced out of its normal chair conformation and flattened into a shape approaching a half-chair or sofa. It is this change in shape that introduces the strain in the bond between the fourth and fifth (D and E) sugars in the substrate.

4. Why is lysozyme considered to be a hydrolytic enzyme? The enzyme splits the substrate by introducing water across the bond. How do you know? This is the definition of a hydrolysis reaction.

5. What feature of the fourth (D) sugar of the lysozyme substrate causes the distortion of the substrate leading to the catalytic ability of lysozyme? The distortion results from the presence of a bulky hydroxymethyl (—CH2OH) group on carbon 6 of the substrate's fourth sugar. What happens if this feature is removed prior to the substrate's binding to the enzyme active site? If the hydroxymethyl group is removed from the substrate, the substrate will bind to the enzyme with greater ease. In fact, it binds 40 to 50 times more strongly than the unaltered substrate.

6. Recently, another amino acid residue (asparagine at residue 46) has been shown to be essential for lysozyme activity. What is its role? This residue forms an H bond with the aspartic acid at position 52. The H bond stabilizes the negative charge on Asp52, which promotes the latter’s ability to stabilize the oxocarbonium intermediate. Mutation of Asp52 and/or Asn46 disrupts the enzyme’s catalytic activity.

7. In a study of an enzyme called bustozyme that is similar to lysozyme, the active site is found to accommodate 8 sugar residues. When the enzyme is exposed to substrates of different lengths, there is no reaction when the substrate contains one, two, three, four or five sugar residues. However, when substrates with six, seven or eight residues are used, a reaction occurs. Where do you think the normal substrate is cleaved? The normal substrate of bustozyme is cleaved between the fifth and sixth residues.

8. In a continuation of the study mentioned above in Question 7, the normal substrate of bustozyme is chemically altered and the effect of the alteration on the binding assessed. When a hydroxymethyl group on carbon-6 of the fifth residue is removed, the altered substrate binds 60 to 70 times more tightly than the normal substrate. Why? The hydroxymethyl group is relatively big and it probably hinders tight binding of the normal substrate and also introduces strain into the bond between the fifth and sixth residues. Its removal, therefore, allows tighter binding to occur and prevents the straining of the bond. Does the reaction still occur? It does not occur. Why or why not? The reaction will not occur, since a strained bond between the fifth and sixth residues is essential to the reaction.

9. You discover two inhibitors of bustozyme. One binds significantly more tightly to the enzyme than the other. What is a possible explanation? The inhibitor that binds most tightly is probably a transition state analog.

10. You suspect that the R groups of two amino acid residues are important for the operation of bustozyme. How can you determine whether this is true? Chemically modify the amino acid residues and determine if the activity of the enzyme is affected. If the enzyme's activity is affected by the modification, the amino acid is involved in and probably essential for the reaction.

11. What is site-directed mutagenesis (SDM) and how can it be used to prove that the glutamic acid and aspartic acid residues mentioned in Question 1 are essential for lysozyme function? SDM is a relatively new technique in DNA technology that allows investigators to alter a specific codon within a gene so that the amino acid for which it codes is changed when the gene directs the synthesis of the corresponding protein. If either of the two codons that code for the glutamic acid and aspartic acid residues respectively is changed so that the residues in the resultant enzyme change, the activity of the enzyme should also change. This experiment has been performed and the two residues have been shown to be essential.

12. What technique has been used to demonstrate that the aspartic acid residue of the unoccupied lysozyme active site is ionized, while the R group of the unoccupied active site's glutamic acid remains protonated at the enzyme's optimal pH? Circular dichroism (CD) was used to confirm the ionization states of these amino acid residues.

LECTURE OUTLINE

Bioenergetics: Definitions

I. Bioenergetics – the study of the various types of energy transformations that occur in living organisms; to maintain its high level of activity, a cell must acquire & expend energy

II. Energy - capacity to do work (the capacity to change or move something)

A. Energy exists in two alternate states

1. Potential energy - energy of object by virtue of its position; rock perched on edge of cliff; has potential to do work since it exists in a field of force (a gravitational field)

2. Kinetic energy - energy of motion; push rock over cliff edge -> falls (can perform work); e. g., directed Na+ ion movement into cell is kinetic energy; can be used to do work (nerve impulse)

B. Two factors must be considered when you are involved in the measurement of energy at work

1. Potential factor - proportional to the intensity of the field of force (distance rock will fall; for the movement of charged ions, potential factor is voltage)

2. Capacity factor - provides some measure of the "size" of the subject being considered (rock’s mass; for the movement of charged ions, capacity factor is combined charge of particles)

III. Work (energy released during such events) - multiple of potential & capacity factors; as either one increases so too does the amount of energy

The Laws of Thermodynamics and the Concept of Entropy

I. Thermodynamics - study of the changes in energy that accompany events in the Universe

A. Use it to predict direction events take & whether or not energy input is needed for them to happen

B. Gives no help in defining how rapidly specific process occurs or specific mechanism cell uses to do it

II. The First Law of Thermodynamics - energy can be neither created nor destroyed (Law of Conservation of Energy); total energy in Universe remains constant (regardless of transduction process)

A. Energy can, however, be transduced - burning fuel, polysaccharide breakdown, photosynthesis

1. Several organism communities are independent of photosynthesis – communities residing in hydrothermal vents on ocean floor; depends on energy obtained by bacterial chemosynthesis

2. Some animals (fireflies, luminous fish) convert chemical energy back into light

B. Universe divided into 2 parts to discuss energy transformations involving matter - system under study (certain space in Universe or amount of matter, e.g., living cell) & surroundings (rest of Universe)

1. Usually stipulate that system does not exchange matter with environment, but may exchange energy with surroundings - closed system

C. Change in a system's energy during an event is manifested by a change in system heat content (positive [endothermic], negative [exothermic] or no change) or in the performance of work

1. Gain/loss in energy balanced by corresponding gain/loss in surroundings (Universe is constant)

2. The energy of a system is termed internal energy (E) & its change during a transformation is DE

D. An equation describing 1st Law - DE = Q – W, where Q = heat energy & W = work energy

E. Depending on process, system internal energy at end can be greater than, equal to or less than internal energy at start (depending on relationship to its surroundings); DE - positive, zero, or negative

F. As long as there is no change in pressure or volume of the contents of a reaction vessel, there is no work being done by the system on its surroundings, or vice versa; thus…..

1. If heat is absorbed from surroundings during the event, E is greater than it was at the start; if heat is released to surroundings during the event, E is lower than it was at the start

2. Reactions that result in heat lost to the environment are called exothermic; those that result in heat gained from the environment are called endothermic

3. There are many reactions of both types

G. Since E for a particular reaction can be positive or negative, it gives us no information as to the likelihood that a given event will occur; need other concepts to do this

III. The Second Law of Thermodynamics – expresses concept that events in Universe have direction

A. Events tend to proceed "downhill" from higher to lower energy state; in any energy transformation, there is a decreasing availability of energy for doing additional work

1. Such events are said to be spontaneous

2. They are thermodynamically favorable & can occur without the input of external energy

B. Concept of 2nd law originally formulated for heat engines; carried with it the idea that perpetual motion machines were impossible (no machine is 100% efficient which would be necessary)

1. Some energy is inevitably lost as machine works (same is true of living organism)

C. Energy unavailable for additional work after an event has both a potential & capacity factor

1. The potential factor is temperature (in degrees); capacity factor is entropy (S; with dimensions of energy/degree [calories/degree])

2. The unavailable energy term is TDS, where DS is change in entropy between initial & final states

3. Loss of available energy during process results from tendency for randomness (disorder) of Universe to rise with every transfer of energy; entropy provides measure of this disorder

4. Entropy is also associated with random movement of particles of matter, which, because they are random cannot accomplish a directed work process

5. 2nd law - every event accompanied by increase in entropy of universe (dissolving sugar cube)

D. Heat (thermal energy) release by live organisms increases rate of random movement of atoms & molecules; cannot be redirected to accomplish additional work – so entropy rises with temperature

1. Examples: glucose oxidation, friction generated as blood flows through vessels

2. Only at absolute 0 (0°K) do all movements cease; at that point, entropy is 0

E. Must distinguish between the system & its surroundings

1. System can increase order (lower entropy) as long as Universe entropy (disorder) rises overall; surroundings must increase in entropy more than the system decreased

2. Example: heat sugar H2O solution —> sugar recrystallizes (more order) by H2O evaporation, but gaseous H2O molecules cause surroundings to gain even more entropy —> entropy rises overall

3. Living organisms work on this principle (lower their entropy while raising that of environment)

4. Maintaining low entropy (high structural order; macromolecule information content) requires energy input (example – proteins patrol DNA constantly looking for & repairing damage

The Concept of Free Energy Arises from a Combination of the First & Second Laws

I. Free energy (G) - energy available to do work; defined by J. Willard Gibbs, American chemist (1878)

A. Together the 1st & 2nd laws of thermodynamics show that the energy of the universe is constant, but that entropy continues to increase toward a maximum

B. Gibbs combined concepts inherent in 1st & 2nd Laws to get equation:DH = DG + TDS where:

1. DG is the change in free energy (the change during a process in energy available to do work)

2. DH - change in enthalpy (total energy content of system; equivalent to DE for our purposes)

3. T - absolute temperature (°K; °K = °C + 273)

4. DS - change in entropy of system

C. Total energy change = sum of changes in useful energy available (DG) & unavailable (TDS) for work

D. Rearrange to DG = DH - TDS - can predict direction in which process will proceed & the extent to which the process will occur

1. DG size shows the maximum amount of energy that can be passed on for use in another process

2. Spontaneous process has -DG (exergonic) & proceeds toward state of lower free energy; such a process is thermodynamically favored

3. Non-spontaneous process, +DG (endergonic); cannot occur spontaneously; it is thermodynamically unfavorable; make it go by coupling to high -G (energy-releasing) reaction

E. The signs of DH & DS can be "+" or "-", depending on relation between system & surroundings

1. DH is "+" if heat is gained by the system & "-" if heat is lost during process

2. DS is "+" if system becomes more disordered & "-" if it becomes more ordered during process

F. The counterplay of DH & DS determines if DG is "+" or "-"; water-ice transformation is example

1. Conversion of water from liquid to solid is accompanied by decreases in entropy & enthalpy

2. For ice to form, DH must be more "-" than TDS, a condition that only occurs below 0°C

3. At 10°C, 0°C & -10°C, regardless of temperature, ice energy level is less than that of liquid H2O

4. At higher temperatures, the entropy term of the equation is more "-" than enthalpy term so that DG is "+" & the process is not spontaneous; at 0°C, the system is in equilibrium

5. At –10°C, solidification process is favored (DG is "-")

II. Free energy changes in chemical reactions - all cell chemical reactions are reversible, so one must consider forward & reverse reactions at the same time

A. Law of Mass Action - reaction rate proportional to reactant concentration (Ex.: A + B <—> C + D)

1. Forward reaction rate - proportional to A & B concentrations (k1[A][B]); k1 - forward rate constant

2. Reverse reaction rate - proportional to C & D concentrations (k2[C][D]); k2 – reverse rate constant

B. All reactions proceed toward equilibrium (can be slow), where forward & reverse rates are equal

1. At equilibrium, the same number of A & B molecules are converted into C & D molecules per unit time as are formed from them

2. k1[A][B] = k2[C][D] at equilibrium - allows calculation of equilibrium constant (Keq) by combining k1 & k2 into Keq (Keq = k1/k2 = [C][D]/[A][B])

3. Gives predictable ratio of product to reactant concentrations at equilibrium & allows prediction of reaction direction (forward or reverse) under given set of conditions

4. Ratio can be determined experimentally by measuring product & reactant concentrations at equilibrium

5. If Keq > 1 & initial concentration ratios = 1 —> products increase at expense of reactants

6. If Keq < 1 & initial concentration ratios = 1 —> reactants increase at expense of products

C. Net direction in which reaction is proceeding at any moment depends on relative concentrations of all participating molecules & can be predicted if Keq is known

III. Free energy changes in chemical reactions - -DG favors product production; +DG favors reactant production; a greater -DG reaction is farther from equilibrium & the system can perform more work

A. -DG when total free energy of reactants > total free energy of products (+DG - vice versa)

1. As reaction proceeds, difference in free energy content between reactants & products decreases (DG gets less negative)

2. At equilibrium, no further work can be obtained & the free energy difference is 0

B. G for a given reaction depends on the reaction mixture present at a given time, thus it is not useful for comparing the energetics of various reactions

1. To allow such comparisons & various types of calculations, scientists began to consider reactions under standard conditions

2. DG at standard conditions (DG0’ - standard free energy change; 25°C [298°K], 1 atm pressure, reactant/product concentrations - 1 M [water - 55.6 M], pH 7 [H+ ion concentration at 10-7 M])

3. DG0’ describes the free energy released when reactants are converted to products under these standard conditions; must be kept in mind that such conditions do not prevail in cell……

4. Therefore, one must be cautious in the use of values for standard free-energy differences in cell energetics calculations

C. But there is an equation that describes relationship between Keq & DG0’: DG0’ = -RT ln K’eq = 2.303 RT log K’eq

1. R = gas constant (1.987 cal/mol-°K); T = absolute temperature (°K)

2. DG0’ & K'eq indicate that standard conditions include pH7, while DG0 & Keq indicate standard conditions in 1 M H+ (pH = 0.0)

3. If Keq > 1, then DG0’ negative (spontaneous) & if Keq < 1, then DG0’ positive (nonspontaneous) under standard conditions

4. The right-hand side of this equation is equivalent to the amount of free energy lost as reaction proceeds from standard conditions to equilibrium

D. If DG0' is negative, reaction goes to right when reactants & products are all present at 1.0 M concentrations at pH 7

1. The greater the negative value, the farther to right reaction will go before equilibrium is reached

2. If DG0' is positive, reaction proceeds to left; reverse reaction is favored

E. If reaction spontaneous, this says nothing about how fast reaction occurs

IV. Free energy changes in metabolic reactions - ATP hydrolysis is important cell chemical reaction

A. ATP + H2O <—> ADP + Pi - DG0’ = -7.3 kcal/mole; highly favorable, so under standard conditions, ADP builds up

1. Electrostatic repulsion of negative phosphate groups partially explains its favorable nature

B. Cell, however, does not have standard conditions, so to know actual DG in cell compartment, you must know reactant & product concentrations that are present in the cell at the time

1. Plug in typical cell values & calculate the resultant G to reveal the reaction's direction in cell

2. G in cell is much more negative (about -12 kcal/mole) than with standard conditions due to high cell [ATP]/[ADP] ratio

V. How do cells do +DG reactions?- high reactant:product ratio & coupling (stored chemical energy input)

A. Some cell reactions maintain reactant/product ratio above that defined by Keq -> keeps DG “-” (ex.: dihydroxyacetone phosphate conversion to glyceraldehyde-3-phosphate, DG0' = +1.8 kcal/mole)

1. Works because of metabolic pathways where product of one reaction is substrate in next one

2. So, product is removed rapidly; maintains favorable ratio; keeps G "-" & reaction spontaneous

3. Specific reactions cannot be considered independently as if they were occurring in isolation; each reaction affects others in cell

B. Coupling endergonic & exergonic reactions - drives reactions with large +G by input of energy (ex.: glutamic acid + NH3 <—> glutamine; DG0’ = +3.4 kcal/mole); ATP hydrolyzed in 2 steps

1. Reaction catalyzed by glutamine synthetase

2. Occurs in 2 sequential reactions: glutamic acid + ATP <—> glutamyl phosphate + ADP, then glutamyl phosphate + NH3 <—> glutamine + Pi; overall DG0’ = -3.9 kcal/mole

3. One downhill reaction (ATP hydrolysis) is used to drive the uphill synthesis of glutamine; glutamic acid accepts phosphate group from ATP, then phosphate is displaced by NH3

4. Add DG of two reactions - if overall DG is negative, overall reaction is spontaneous; get overall reaction below: glutamic acid + ATP + NH3 <—> glutamine + ADP + Pi

5. Coupled reactions must have common intermediate (1st reaction product is reactant in 2nd; glutamyl phosphate above)

6. In example, higher [ATP]/[ADP] ratio than at equilibrium maintained & responsible for overall -DG; used in most highly endergonic cell reactions since phosphate is easily transferred

7. At equilibrium, [ADP] should be >107 times greater than [ATP]; in fact [ATP] in cells is 10 – 100 times greater than [ADP]

8. If cell had equilibrium concentration of ATP, ADP & PI, it would not matter how much ATP was preset, cell would have no capacity to perform work

VI. ATP hydrolysis is used to drive most cellular endergonic processes

A. ATP is used for diverse processes because its terminal phosphate group can be transferred to a variety of different types of molecules (amino acids, lipids, sugars, & proteins)

B. In most coupled reactions, phosphate group is transferred in initial step from ATP to one of above acceptors & is subsequently removed in second step

VII. Equilibrium vs. steady-state metabolism - if you stay far from equilibrium, you preserve capacity to do work & lose less energy to increased entropy

A. As reactions tend toward equilibrium, the free energy available for work decreases toward a minimum & entropy increases toward a maximum

1. Cell metabolism is essentially nonequilibrium metabolism (nonequilibrium ratios of products to reactants); some reactions do occur at or near equilibrium

2. Usually, at least one reaction or several reactions in a pathway are poised far from equilibrium, making them essentially irreversible; they keep the pathway going in a single direction

3. They are also subject to cell regulation, since flow of material through pathway can be greatly increased or decreased by stimulating or inhibiting activity of enzymes catalyzing these reactions

B. Thermodynamic principles developed for closed systems (no exchange of matter between system & its surroundings) are under reversible equilibrium conditions

1. But cells are open systems (exchange both matter & energy with surroundings); they are often under irreversible nonequilibrium conditions (nonequilibrium product:reactant ratios maintained)

2. Materials & energy are continually flowing into cell from blood or culture medium & other substances (wastes) leave the cell

3. Thus, cells are not at equilibrium, instead, they are in state that resemble it called steady state

C. Steady-state resembles equilibrium (reactant & product concentrations are essentially constant) even though individual reactions are not necessarily at equilibrium

1. Products of one reaction are acted on as substrates of next reaction

2. Concentrations of each metabolic intermediate stays essentially the same as long as new substrates are brought in from outside & terminal products are removed

Enzymes as Biological Catalysts

I. Brief history of enzymes (biological catalysts)

A. Justus von Liebig vs. Pasteur - reactions in living organisms not special vs. they are special (can only occur in living cell); based on ethanol formation in yeast (must yeast be alive to do it?)

B. Hans (bacteriologist) & Eduard (chemist) Büchner (1897) – found yeast extracts (with enzymes) can do fermentation; intact cells not needed (serendipitous discovery)

1. They tried to preserve extract by adding sugar as was done with jams & jellies

2. Instead, yeast juice produced gas from sugar & it bubbled for days —> bubbles were CO2; ethanol also made —> so fermentation did not require intact, living cells

C. Others soon found that fermentation was different from reactions usually done by organic chemists

1. Process required catalysts others in non-living world (called enzymes - Greek for “in yeast”)

2. Speed reactions up in living organisms; without them, rates imperceptible & life impossible

3. Enzymes are the mediators of metabolism, responsible for nearly every reaction in living cell

D. James Sumner (1926) - crystallized urease from jack beans; it & others soon found to be protein

1. It was soon accepted that all biological catalysts (enzymes) were proteins, but…..

E. RNA "enzymes" eventually discovered (called ribozymes; enzyme still used for protein catalysts)

II. Enzymes can be conjugated proteins with nonprotein components that carry out activities for which amino acids not suited - cofactors may be inorganic (metals) or organic (coenzymes)

A. When present, they are important participants in enzyme function

B. Myoglobin – heme group iron atom is site where O2 is bound & stored until needed for metabolism

III. The properties of enzymes that are like those of other catalysts; they:

A. Are required only in small amounts

B. Are not altered irreversibly during course of reaction (used over and over again)

C. They have no effect on reaction thermodynamics - can only increase rate of favorable reaction

1. Do not supply energy for reaction or determine whether reaction is exergonic or endergonic

2. Do not determine what the ratio of products to reactants is at equilibrium; these are inherent properties of reacting chemicals

IV. There is no necessary relationship between the size of DG & the rate of a particular reaction – a substance that is thermodynamically unstable (like glucose) can be kinetically stable

A. DG magnitude gives only the difference in free energy between beginning state & equilibrium dependent of pathway it takes to reach equilibrium & how long it takes to get there

V. How do enzymes differ from inorganic catalysts? - they are remarkably adept catalysts

A. Much more effective than non-enzyme catalysts; do it under very mild temperature & pH conditions

1. Non-enzyme catalysts (acid, metallic platinum, magnesium) speed up reactions 100 - 1000 times

2. Enzymes often speed reactions 108 - 1013 times; in 1 sec, enzymes accomplish what would normally take 3 – 300,000 years if the enzyme were absent

B. Most are highly specific in reactants (substrates) they bind & the reaction they catalyze

1. Such specificity in enzymes (and other proteins & ligands) is essential for maintenance of order needed to sustain life

C. Also act as metabolic traffic directors; their reactions are very orderly & only make the appropriate products —> no side reactions or unwanted products that would gum up the works

D. Unlike other catalysts, they can be regulated to meet particular cellular needs at particular time

VI. Overcoming the activation energy barrier - why do we need enzymes?

A. Reactants must have enough energy to get over activation energy barrier that slows up reactions even if spontaneous - sufficient kinetic energy can do it, but usually don’t have enough

1. Chemical transformations require that certain covalent bonds in reactants be broken

2. Molecule energies have normal distribution, some with little energy, some with much more; only those in rightmost tail of graph below have enough energy to react without a catalyst

3. Activated (high energy) molecules stay that way a short time; they either undergo a reaction to produce product or lose energy by colliding with others

4. Can speed up by heat, which increases all molecule energies (dangerous for living organisms); applying heat to enzyme-mediated reactions leads to enzyme denaturation & rapid inactivation

B. At energy barrier crest, reactants in transition state, activated complex forms (bonds formed & broken)

1. Example: bacterial enzyme proline racemase – reaction proceeds in either direction through the loss of a proton from -carbon of proline

2. Transition state structure contains negatively charged carbanion in which the 3 bonds formed by carbon atom are all in same plane

C. Unlike standard free energy difference, activation energy required to reach transition state is not fixed value but varies with the reaction mechanism utilized

D. Enzymes speed reactions by decreasing activation energy barrier under relatively mild conditions

1. Enzyme cause their substrates to be very reactive without having to be raised to particularly high energy levels

2. Enzymes lower activation energy by binding more tightly to transition state than to reactants (stabilizes activated complex & decreases its energy)

3. As transition state is converted to products, the enzyme's affinity for bound molecules decreases & the products are expelled

E. Transition state important as demonstrated by following findings:

1. Compounds resembling transition state are often very effective inhibitors since they bind active site (catalytic region) so tightly

2. Antibodies (usually simply bind to molecules with high affinity & do not catalyze reaction) against transition-state-like compounds can sometimes catalyze their breakdown

I. Enzyme kinetics is study of reaction rates under various experimental conditions – enzymes vary greatly in their ability to catalyze reactions

II. Leonor Michaelis & Maud Menten (1913) - worked out mathematical relationship between substrate concentration & enzyme reaction velocity (rate of product formed/substrate consumed in a given time)

A. Set up incubation mixture at desired temperature with all ingredients required except one

1. When add missing ingredient, reaction starts

2. If when reaction begins, no product is present in mixture, the amount of product that appears over time provides a measure of reaction velocity

3. If incubation time is too great, substrate concentration becomes measurably reduced

4. Also, as product appears, it can be converted back to substrate by reverse reaction (also catalyzed by enzyme)

5. Thus, researchers want to know initial velocity (velocity at instant when no product has yet formed)

6. To get accurate initial velocity measurements, need to use brief incubation times & sensitive measuring techniques

B. Curve is hyperbolic if assess initial reaction velocity in series of mixtures containing same amount of enzyme & rising substrate concentration; initial reaction velocity varies markedly with [substrate]

1. At low[S], substrate concentration is rate limiting; collisions of enzymes & substrates are relatively infrequent so enzyme has idle time

2. At high [S], enzyme concentration limits rate since enzyme is saturated; enzyme & substrate collide faster than conversion to product; works as fast as possible, reach maximal velocity (Vmax)

C. If know MW & enzyme concentration —> can calculate turnover number or catalytic constant, kcat (maximum number of product molecules formed/unit time by 1 enzyme molecule) from Vmax

1. Relatively few enzymes can rapidly convert a fairly high number of substrate molecules to product

2. 1 - 1000/sec turnover is typical but can get up to as much as 106 (for carbonic anhydrase)

D. Michaelis-Menten equation –

1. When [S] is set at value equivalent to KM, then reaction velocity (v) is equal to Vmax/2; thus KM = [S], when v = Vmax/2

2. When [S] is low relative to KM, [S] can be ignored in denominator of the equation so that v = Vmax x [S]/KM; thus, at low [S], velocity (v) is directly proportional to [S]

3. When [S] is high relative to KM, the KM term in denominator can be ignored so that v = Vmax; in other words, reaction velocity is independent of [S]

E. One can also determine Michaelis constant (KM) - equal to [S] when reaction velocity is 1/2 Vmax

1. KM usually serves as a measure of enzyme affinity for substrate

2. If higher value, lower affinity, since a greater substrate concentration is needed to reach 1/2 Vmax

3. If lower value, higher affinity, since a lower substrate concentration is needed to reach 1/2 Vmax; most enzymes have KM range from 10-1 - 10-7 M, 10-4 M is a typical value

4. KM is constant for a given enzyme & thus independent of substrate & enzyme concentration

III. Hans Lineweaver & Dean Burk - graph Michaelis-Menten data as 1/[S] vs. 1/velocity instead of [S] vs. velocity —> transforms Michaelis-Menten equation from hyperbolic into linear function

A. Generation of hyperbolic curve & accurate determination of KM & Vmax requires the plotting of a considerable number of points

B. With a Lineweaver-Burk plot, one can determine KM & Vmax by extrapolation of the straight line drawn from relatively few points (x-intercept = -1/KM; y-intercept = 1/Vmax; slope = KM/Vmax)

1. Easier & more accurate description of data & determination of KM & Vmax

IV. pH optimum and temperature optimum also affect enzyme activity – each enzyme has an optimum pH & a temperature at which it operates at maximal activity

A. Influence of temperature on enzyme activity is illustrated by the striking effect of refrigeration in slowing the growth of microorganisms

Enzyme Inhibitors

I. Enzyme inhibitors - molecules that are able to bind to an enzyme and decrease its activity

A. Cell depends on inhibitors to regulate the activity of many of its enzymes

B. Biochemists use them to study enzyme properties; many are used as drugs, antibiotics & pesticides

II. Inhibitors can be divided into two types: reversible or irreversible; reversible inhibitors can, in turn, be considered as competitive or noncompetitive

A. Irreversible inhibitors bind very tightly to an enzyme, often by forming a covalent bond to one of the enzyme's amino acid residues; they are often effective poisons, antibiotics & pesticides

1. A number of nerve gases (diisopropylphosphofluoridate & the organophosphate pesticides) act as irreversible inhibitors of acetylcholinesterase (ACHase)

2. Example: ACHase plays crucial role in destroying acetylcholine (neurotransmitter that causes muscle contraction); if ACHase inhibited, muscle stimulated continuously & contracts permanently

3. The antibiotic penicillin is irreversible inhibitor of a key enzyme in bacterial cell wall formation

B. Reversible inhibitors bind only loosely to an enzyme & thus are readily displaced – there are 2 types of reversible inhibitor: competitive & noncompetitive inhibitors

III. Types of reversible inhibitors – competitive inhibitors

A. They are reversible inhibitors that compete with substrate for access to enzyme active site

1. Since substrates have complementary structure to active site to which they bind, competitive inhibitors must resemble substrate to compete for the same binding site

2. However, competitive inhibitors must differ in a way that prevents them from being transformed into product

3. Analysis of types of molecules that can compete with substrate for enzyme active site provides insight into structure of active site & nature of interaction between natural substrate & enzyme

B. Competitive inhibition forms basis of action of a wide variety of common drugs; example - angiotensin converting enzyme (ACE)

1. ACE is a proteolytic enzyme that acts on 10-residue peptide (angiotensin I) to produce 8-residue peptide (angiotensin II)

2. High angiotensin II levels are big risk factor in high blood pressure (hypertension) development

3. John Vane et al., Eli Lilly Co. (1960s) – searched for compounds that could inhibit ACE

4. Earlier studies had found that Brazilian pit viper venom contained proteolytic enzyme inhibitors & that one venom component (peptide called teprotide) was potent ACE competitive inhibitor

5. Teprotide did lower blood pressure in hypertensives but was not very useful since it had peptide structure & was rapidly degraded if taken orally

6. Attempts to develop non-peptide ACE inhibitors led to synthesis of captopril, the first useful antihypertensive drug that acted by binding to ACE

C. Effectiveness of competitive inhibitor depends on its relative affinity for enzyme

1. Competitive inhibition can be overcome if substrate/inhibitor ratio is great enough

2. If number of collisions between enzyme & inhibitor becomes insignificant relative to those between enzyme & substrate, the inhibitor effect becomes minimal

3. With sufficient substrate concentration, it is theoretically possible to achieve the enzyme's Vmax even in presence of competitive inhibitor

IV. Types of reversible inhibitors – noncompetitive inhibitors

A. They are inhibitors that do not compete with substrate for binding to enzyme active site; generally, the noncompetitive inhibitor acts at a site other than the active site

B. The level of inhibition depends only on the inhibitor concentration; increasing substrate concentration cannot overcome it

C. In the presence of a noncompetitive inhibitor, a certain fraction of enzyme molecules is necessarily inactive at a given instant, thus Vmax of the population of enzyme molecules cannot be reached

V. Effects of different types of inhibitors on Michaelis-Menten and Lineweaver - Burk plots

A. In presence of noncompetitive inhibitor, Vmax, is lowered; in presence of competitive inhibitor, KM is increased

B. In both types of inhibitor, the slope of the Lineweaver-Burk plot (KM/Vmax) is increased relative to the uninhibited reaction

C. Figures below show the effect of both inhibitor types on the kinetics of enzymes in both Michaelis-Menten & Lineweaver-Burk plots

Metabolism

I. Metabolism - collection of biochemical reactions that occur in cell; they are often interconnected into metabolic pathways; include a tremendous diversity of molecular conversions

A. Metabolic pathways contain a sequence of chemical reactions in which each is catalyzed by a specific enzyme & in which the product of one reaction is the substrate for the next

B. Enzymes of a particular metabolic pathway are usually confined to a particular cell region (cytosol, mitochondria, etc.)

1. Evidence suggests that they are often physically linked so that the product of one passes directly to the active site of the next one in sequence

C. Metabolic intermediates (metabolites) form at each step leading to end product (play cellular roles)

1. Compound made in one pathway may be shuttled to a number of others depending on cell needs

II. An overview of metabolism – metabolic pathways divided into 2 broad types

A. Catabolic pathways – serve 2 functions: make available raw materials from which other molecules can be synthesized & provide chemical energy required for many cell activities

1. Lead to disassembly of more complex molecules to form simpler products

2. Energy produced by catabolic pathways is stored temporarily in 2 forms: high-energy phosphates (mostly ATP) & high-energy electrons (mostly in NADPre & NADre)

B. Anabolic pathways – lead to synthesis of more complex compounds from simpler starting materials; they require energy & use stored chemical energy from the exergonic catabolic pathways

III. Greatly simplified profile of ways in which major anabolic & catabolic pathways are interconnected

A. Macromolecules are first broken down (hydrolyzed) into the building blocks of which they are made

B. Building blocks are reutilized directly to:

1. Form other macromolecules of same class (stage I),

2. Convert them into different compounds to make other products

3. Degrade them further to extract some measure of their free energy content (stages II & III)

C. Degradation pathways of diverse building blocks vary according to the particular compound being catabolized; however, all of them are converted into a small variety of compounds (stage II)

1. This small variety of compounds can be metabolized similarly

2. Thus, although their initial macromolecular structures are quite different, they are converted into the same low MW metabolites; therefore, catabolic pathways are thus said to be convergent

D. Most of these reactions & pathways are found in virtually every living cell; this is evidence that they appeared very early in, & have been retained throughout, biological evolution

Oxidation and Reduction – A Matter of Electrons

I. Oxidation - reduction (redox) reactions must occur simultaneously; involve change in the electronic state of reactants; both oxidation & reduction are reversible; both anabolic & catabolic pathways include them

A. Oxidation - reaction involving loss of ≥1 electrons; ex.: metallic iron (Fe0) to ferrous state (Fe2+) via loss of electron pair; since an electron acceptor is needed, it is accompanied by reduction

1. The molecule oxidized is called a reducing agent & becomes more positive

2. Reaction is reversible; oxidized molecule can pick up electrons & return to earlier state (Fe0 -> Fe2+)

B. Reduction - a reaction involving a gain of ≥1 electrons; ex.: Fe2+ to Fe0; molecule reduced is called an oxidizing agent; since electron donor is needed, it is accompanied by oxidation

1. Example: Fe0 + Cu2+ <—> Fe2+ + Cu0; Cu2+ is reducing agent, Fe0 is oxidizing agent

II. Oxidation or reduction of metals involves complete loss or gain of electrons; this cannot happen with most organic compounds

A. Organic compound oxidation & reduction during cellular metabolism involve carbon atoms that are covalently bonded to other atoms

B. When a pair of electrons is shared by 2 different atoms, the electrons are more strongly attracted to one of the 2 atoms of the polarized bond

1. In C—H bonds, C pulls electron more strongly & is in reduced state

2. In C—O & C—N bonds, electrons are attracted by electronegative atom, thus C is in oxidized state

C. Carbon atom has a number of oxidation states, since it can share a number of valence electrons (4)

1. Range - CH4 (fully reduced) to CO2 (fully oxidized); oxidation state of a carbon atom in an organic molecule is a measure of the molecule's free energy content

2. One can roughly determine oxidation state by counting number of H vs. O & N atoms per C atom

III. Capture & utilization of energy – chemical fuels are highly reduced organic compounds; energy is released when they are burned in presence of O2 converting them to more oxidized states (CO, CO2)

A. As carbon reduction increases, can do more chemical work in cell - fats (H—C—H) have more energy per unit weight than carbohydrates (H—C—OH)

1. The more H atoms that can be stripped from a "fuel" molecule, the more ATP can be made

2. Carbohydrates are rich in chemical energy because they contain strings of H—C—OH units

3. Fats have even more energy/unit weight due to their strings of more reduced units (H—C—H)

B. Carbohydrates broken down in small, energy-releasing steps; some are coupled to reactions leading to ATP formation (free energy difference between reactants & products relatively large)

1. Glucose monomer of glycogen & starch is key molecule in animal & plant energy metabolism

2. Free energy released by complete glucose oxidation is very large

3. C6H12O6 + 6 O2 —> 6CO2 + 6 H2O DG0' = -686 kcal/mol

4. Free energy needed to convert ADP to ATP is relatively small: ADP + Pi —> ATP + H2O, DG0' = +7.3 kcal/mol; complete glucose oxidation can drive production of lots of ATP

5. Up to ~36 ATP molecules are formed/glucose molecule oxidized under conditions existing within most cells; to make so much ATPs, sugar must be disassembled in many small steps

C. Glucose catabolism divided into 2 stages - glycolysis in cytoplasm & Krebs (TCA) cycle in eukaryote mitochondria or prokaryote cytoplasm (energy stored as high energy electrons)

1. The 2 steps are virtually identical in all aerobic organisms

2. Glycolysis leads to the formation of pyruvate; tricarboxylic acid (TCA) cycle leads to final oxidation of carbon atoms to carbon dioxide

3. Some of these steps release enough free energy to drive ATP production by coupled reactions

4. Most of glucose chemical energy is stored in high energy electrons removed from glucose as substrate molecules are oxidized during glycolysis & TCA cycle

5. Energy from these electrons is ultimately used to synthesize ATP

D. Glycolysis is presumably the energy capture pathway used by our early anaerobic ancestors & is the major anabolic pathway used by anaerobic organisms today

1. DG values measured for glycolysis reactions (using metabolite concentrations from cell)

2. All but 3 reactions have DG values of nearly 0; they are near equilibrium

3. 3 reactions are far from equilibrium (essentially irreversible in cell); they provide driving force that moves metabolites through glycolysis in directed manner

IV. Arthur Harden & William Young, British chemists (1905) – studied glucose breakdown by yeast cells, a process that generates bubbles of CO2 gas

A. Noted that bubbling eventually slowed & stopped, even though much glucose was left to metabolize

1. Some other essential component exhausted – adding inorganic phosphates started it up again

2. First clue that phosphate groups played a role in metabolic pathways

B. First few reactions in glycolysis illustrate importance of phosphate groups

V. ATP formed 2 ways in cell: oxidative phosphorylation & substrate-level phosphorylation

A. Oxidative phosphorylation - dehydrogenases move 2 electrons & proton to NAD+ to make NADH

1. High energy NADH donates electrons to other molecules at electron transport (ET) chain

2. Because NADH transfers electrons so readily, it is said to have high electron transfer potential

3. As electron travels down ET system, it loses energy used to make ATP & is added to O2 to make H2O

B. Substrate-level phosphorylation - phosphate group moved from a substrate to ADP —> ATP

1. ATP formation is not that endergonic, formation of other molecules is more endergonic

2. Such molecules can donate their phosphates to ADP to make ATP

Glycolysis: The Steps

I. Activation (pump priming) steps with addition of high-energy phosphates to 6-carbon chain to make 6-carbon diphosphates; use up 2 ATPs to do this

A. Sugar linked to first phosphate group (making glucose-6-phosphate) from ATP (an energy investment); glucose-6-phosphate is then converted to fructose-6-phosphate

1. Phosphorylation activates sugar; makes it capable of participating in subsequent reactions

2. Phosphorylation also lowers the cytoplasmic concentration of glucose, which promotes continued diffusion of the sugar from the blood into the cell

B. Fructose-6-phosphate is converted to fructose-1,6-bisphosphate by using up a second ATP

C. 6-carbon bisphosphates break into two 3-carbon monophosphates (follow both to count up ATPs)

1. Sets the stage for the first exergonic reaction to which ATP formation can be coupled

II. Two 3-carbon monophosphates lose electrons that bind to NAD+ (NADox) to make NADH (NADre), while a new high-energy phosphate group is added

A. This is the conversion of glyceraldehyde-3-phosphate to 3-phosphoglycerate

B. The overall reaction is the oxidation of an aldehyde to a carboxylic acid; occurs in 2 steps catalyzed by 2 different enzymes

C. First enzyme requires nonprotein cofactor (coenzyme), nicotinamide adenine dinucleotide (NAD) to catalyze the reaction; NAD plays a key metabolic role by accepting & donating electrons

1. First reaction is oxidation-reduction; 2 electrons & proton (equal to a hydride ion; :H-) are transferred from glyceraldehyde-3-phosphate (oxidized) to NAD+ (reduced) forming NADH

2. Enzyme is glyceraldehyde-3-phosphate dehydrogenase; NAD+ (derived from vitamin niacin) acts as a loosely associated coenzyme to dehydrogenase which is able to accept hydride ion

3. NADH is released from enzyme in exchange for fresh NAD+; NADH is thought to be high-energy compound due to ease with which it transfers electrons to other molecules that attract electrons

4. NADH has high electron-transfer potential relative to other electron acceptors; usually transfers electrons to series of membrane-embedded electron carriers (electron-transport system or chain)

5. As electrons move along chain, they move to lower & lower free-energy state & are ultimately passed to molecular oxygen, reducing it to water

6. Energy released during electron transport is used to form ATP by oxidative phosphorylation

D. Glyceraldehyde-3-phosphate to 3-phosphoglycerate conversion includes direct ATP formation route

1. In 2nd step of overall reaction, phosphate group is moved from 1,3-bisphosphoglycerate to ADP to form ATP molecule; catalyzed by phosphoglycerate kinase by substrate-level phosphorylation

2. Formation of ATP is not that endergonic, not so energetic that it cannot be readily formed by metabolic reactions

3. There are many phosphorylated molecules whose hydrolysis has more negative G0' than ATP

4. Any donor molecule with higher negative G0' on scale can phosphorylate any molecule lower on scale, including phosphorylation of ADP to ATP

5. Molecules higher on scale (with higher free energy or larger –G0') have higher transfer potential

6. G0' of such a reaction will be equal to the difference between the two values

7. Such donor molecules have lower affinity for the phosphate group than the molecule accepting it; the lower the affinity, the better the donor; the greater the affinity, the better the acceptor

E. Neither phosphate group donation to ADP by 1.3-bisphosphoglycerate or by phosphoenolpyruvate requires molecular oxygen; thus, glycolytic production of ATP is anaerobic

1. 2 ATPs are produced by substrate-level phosphorylation from each molecule of glyceraldehyde-3-phosphate oxidized to pyruvate; since there are 2 of them, 4 ATPs are made

2. Since 2 ATPs are used up to start pathway, there is a net gain of 2 ATPs

3. Overall equation: Glucose + 2ADP + 2Pi + 2NAD+ —> 2Pyruvate + 2ATP + 2NADH + 2H+ + 2H2O

4. Pyruvate (glycolysis end product) is key compound; sits at junction between anaerobic & aerobic pathways: without O2, pyruvate undergoes fermentation; in its presence, aerobic respiration occurs

III. Fermentation - recycle NAD+ for use in glycolysis (stops without NAD+); occurs when O2 low because NAD+ cannot be recycled by ET system

A. Glycolytic reactions occur at rapid rates so cell can produce a significant amount of ATP

1. Some cells (yeast, tumor & muscle cells) rely heavily on glycolysis to make ATP

2. NADH formation in glycolysis occurs at expense of NAD+, which is in short supply in cells

3. NAD+ is required reactant in important glycolysis step & must be regenerated from NADH

4. If it is not, neither glyceraldehyde-3-phosphate oxidation nor any of the succeeding glycolytic reactions can take place

5. Without oxygen, NADH cannot be oxidized to NAD+ by means of the electron transport chain since oxygen is the final electron acceptor of the chain

B. Cells can regenerate NAD+ by fermentation, transfer of electrons from NADH to pyruvate or a compound derived from pyruvate

1. Like glycolysis, fermentation occurs in cytosol of eukaryotic & prokaryotic cells

C. Pyruvate is converted to various substances (CO2 & ethanol in plants, yeast; lactate in animals)

1. When muscles contract repeatedly, O2 levels drop too low to keep pace with cell metabolic demands so the muscle cells regenerate NAD+ by converting pyruvate to lactate (fermentation)

2. When enough O2 is again available, lactate can be converted back to pyruvate for further oxidation (aerobic respiration) & much more efficient harnessing of energy in the form of ATP production

D. Fermentation is a necessary adjunct to metabolism in many organisms & the sole source of metabolic energy in some anaerobes

1. It is usually a stopgap measure to regenerate NAD+ when O2 levels are low, continuing glycolysis & maintaining ATP production

2. Energy gained from glycolysis alone is meager when compared to the complete oxidation of glucose to CO2 & H2O

3. Complete glucose oxidation – 686 kcal/mole; under standard conditions, conversion of glucose to ethanol releases 57 kcal/mole, while conversion of glucose to lactate releases only 47 kcal/mole

4. Very inefficient use of glucose energy (>90% of energy discarded); that’s why ETOH burns

E. Before O2 in atmosphere, glycolysis & fermentation were primary metabolic pathways for energy extraction from sugars by primitive prokaryotic cells

1. After the appearance of cyanobacteria, O2 in atmosphere rose dramatically

2. When O2 appeared in higher levels, aerobic metabolic strategies became possible, more efficient mechanisms evolved (glycolysis products were more completely oxidized —> more ATP)

IV. Reducing power- NADH moves electrons about in catabolic reactions; NADPH in anabolic ones; a cell's NADPH reservoir is an important measure of a cell's reducing power (its usable energy content) A. Many materials, like fats & lipids, are more reduced than the metabolites from which they are built

1. Fat synthesis requires reduction of metabolites via the transfer of high-energy electrons from NADPH (similar in structure to NADH but with an additional phosphate group)

B. Example of use of NADPH – 1,3-bisphosphoglycerate —> glyceraldehyde-3-phosphate reaction of photosynthesis

1. Pair of electrons (together with a proton) is transferred from NADPH to 1,3-bisphosphoglycerate, reducing a carbon atom

C. NADP+ is formed from NAD+ by the following reaction: NAD+ + ATP <—> NADP+ + ADP

1. NADPH then formed by reduction of NADP+; NADPH is a high-energy compound due to its high electron-transfer potential

2. The transfer of free energy in the form of these electrons raises the acceptor to a more reduced, more energetic state

D. Separation of reducing power into 2 distinct & related molecules (NADPH & NADH) reflects separation of their primary metabolic roles

1. NADH & NADPH are recognized as coenzymes by different enzymes

2. Enzymes with reductive role in anabolic pathways recognize NADPH as coenzyme

3. Enzymes acting as dehydrogenases in catabolic pathways recognize NAD+ as coenzyme

E. Even though they are used differently, one coenzyme can reduce the other in this reaction:

1. NADH + NADP+ <—> NAD+ + NADPH; reaction is catalyzed by transhydrogenase (they are interconverted); the ability to interconvert the two coenzymes facilitates regulation

2. If energy is abundant, NADPH formation is favored, providing supply of electrons needed for biosynthesis of new macromolecules that are essential for growth (more anabolism)

3. If energy is scarce, NADH is used to make ATP (more catabolism); only enough NADPH made to meet cell's minimal biosynthetic requirements

Metabolic Regulation

I. Amount of ATP in cell at any given time is small (bacteria - ~1 million molecules); half-life is very brief (a second or two); with so limited a supply, not much free energy is stored in this form

A. Cell energy reserves are stored instead as carbohydrates (polysaccharides) & fats - when ATP needed or too plentiful, reactions leading to ATP enhanced or inhibited, respectively

1. If ATP levels start to fall, energy-rich storage forms (CHOs, fats) broken down to make ATP 2. If ATP levels are too high, reactions that would normally lead to ATP production are inhibited

B. Cells regulate these important energy-releasing reactions by controlling certain key enzymes in a number of metabolic pathways; do this in a couple ways

1. Modify enzyme's activity by altering enzyme active site & changing activity in 2 ways: covalent modification & allosteric modulation (both play key roles in regulating glucose oxidation)

2. Can also regulate metabolism by controlling concentrations of enzymes by regulating the relative rates at which enzymes are synthesized or degraded

II. Altering enzyme activity by covalent modification - Edmond Fischer & Edwin Krebs, U. of Washington (mid-1950s) – glycogen phosphorylase (muscle cell enzyme that breaks glycogen down into glucose)

A. Glycogen phosphorylase can exist in an active or inactive state – Fischer & Krebs found that if they added ATP to crude extract of muscle cells, inactive phosphorylase was converted to active form

1. Second enzyme found in extract; transferred phosphate group from ATP to 1 of 841 glycogen phosphorylase amino acids; called it converting enzyme (later named phosphorylase kinase)

2. Presence of phosphate group altered enzyme active site shape & increased its catalytic activity

B. It has since been discovered that a general mechanism for altering enzyme activity is the addition or removal of phosphate groups (can increase or decrease enzyme activity)

1. Protein kinases are enzymes that transfer phosphate groups to many different enzymes; they regulate such diverse activities as hormone action, cell division & gene expression

2. There are 2 basically different types of protein kinases: one adds phosphates to specific tyrosine residues in substrate protein; the other adds phosphates to specific serine or threonine residues

3. Importance of protein kinases illustrated by fact that ~2% of all yeast cell genes (113 out of ~6200) encode members of this class of enzymes

III. Altering enzyme activity by allosteric modulation - compound binds to site spatially distinct from the enzyme active site (allosteric site) & either inhibits or stimulates enzyme activity

A. Substances bind to allosteric sites; binding causes defined change in active site shape activating or inhibiting its activity

1. Allosteric site may be located on opposite side of enzyme from the allosteric site or even on a different polypeptide within protein

2. Allosteric modulation illustrates intimate relationship between molecular structure & function

B. Feedback inhibition - type of allosteric modulation where end product of pathway allosterically inhibits first committed step in metabolic pathway making it; shuts down anabolic assembly lines

1. Keeps cells from wasting energy & materials by producing compounds that are not utilized

2. Pathway is temporarily inactivated when its end product's concentration reaches certain level

3. Feedback inhibition exerts immediate, sensitive control over a cell's anabolic activity

IV. Separating catabolic and anabolic pathways

A. Glycolysis & gluconeogenesis (synthesis of glucose) overlap in places running reverse reactions; in other places, reactions are very different using different enzymes

1. Most cells can synthesize glucose from pyruvate, while they are oxidizing glucose for energy

2. Some reactions in these pathways are identical although they run in opposite directions

3. Enzymes can catalyze reactions in both directions, but reactions of gluconeogenesis cannot proceed simply by reversal of reactions of glycolysis

4. The glycolytic pathway contains 3 thermodynamically irreversible reactions that must be bypassed in gluconeogenesis

5. If all glycolytic reactions could be reversed, it would be an undesirable way to handle cell metabolic activities, since it is hard to control glycolysis & gluconeogenesis independent of each other

6. Cell could not shut down glucose synthesis & crank up glucose breakdown, since the same enzymes would be active in both directions

B. Regulation of the reactions that are different in glycolysis & gluconeogenesis & their enzymes allows separate regulation of these opposing pathways

1. Regulating these different key enzymes in 2 opposing pathways solves both the thermodynamic & regulatory problems inherent in being able to manufacture & degrade the same molecules

V. Example: phosphofructokinase (PFK; adds phosphate group to fructose-6-phosphate) & fructose 1,6-bisphosphatase (FBP; cuts phosphate group off of fructose-1,6-biphosphate; simple hydrolytic reaction)

A. These enzymes are regulated, in part, by AMP & ATP; AMP concentration in cell is inversely related ATP concentration; when ATP levels are low, AMP levels are high & vice versa

1. Thus, elevated AMP concentrations signal to cell that its ATP fuel reserved are becoming depleted

B. PFK - ATP is both substrate & allosteric inhibitor of PFK; AMP is allosteric activator; runs essentially irreversible reaction: Fructose-6-phosphate + ATP <—> Fructose-1,6-bisphosphate + ADP

1. It is essentially irreversible largely because it is coupled to ATP hydrolysis (G0' = -3.4 kcal/mol)

2. If ATP concentration is high —> PFK activity decreases (no new ATP is formed by glycolysis)

3. If ADP & AMP concentrations are high relative to ATP —> PFK activity goes up, get more ATP

C. FBP - inhibited by high AMP, when ATP needs to be made through glucose breakdown; thus it helps to halt glucose production, while it is being broken down to produce ATP

VI. Importance of allosteric modulation in metabolic control is illustrated by recent studies on enzyme AMP-activated protein kinase (AMPK)

A. AMPK controls activity of other enzymes by adding phosphate groups to specific serine or threonine residues within their structure; it is regulated allosterically by AMP

1. As AMP concentrations rise, AMPK is activated, causing it to phosphorylate & inhibit key enzymes involved in anabolic pathways

2. At the same time, it phosphorylates & activates key enzymes in catabolic pathways

3. End result of AMPK activation is decrease in activity of pathways that consume ATP & increase in activity of pathways that produce ATP

4. Leads to elevation in concentration of ATP within cell

B. AMPK is involved not only in cellular energy regulation but, at least in mammals, in regulation of energy balance within the whole body

1. In mammals, appetite is controlled by centers within the hypothalamus of the brain that respond to the levels of certain nutrients & hormones within the blood

2. A drop in blood glucose levels, acts on the hypothalamus to stimulate appetite, which appears to be mediated through activation of AMPK in hypothalamic nerve cells

3. Conversely, feeling of being "full" experienced after eating meal is thought to be triggered by certain hormones (leptin secreted by fat cells) that inhibit AMPK activity in same brain cells

Experimental Pathways

Determining the Mechanism of Lysozyme Action

I. Alexander Fleming, Scottish bacteriologist (1922) – discovered that nasal mucus, a product of his cold, added to a bacterial culture caused the lysis of the cells; cause was the enzyme lysozyme in mucus

A. It had no clinical value for killing bacterial infections

B. It has been important in the study of enzyme mechanisms

II. David Phillips, et al., Oxford Univ. (1965) – published detailed model of lysozyme tertiary structure, the first enzyme whose 3-D structure had been determined using X-ray crystallography

A. The lysozyme was purified from egg white (it protects developing embryo from bacterial infection)

1. Lysozyme kills bacteria by hydrolyzing glycosidic bonds in polysaccharide of their cell walls

2. Cell walls of sensitive bacteria (gram-positive bacteria) are made of a copolymer of 2 alternating amino sugars (N-acetylglucosamine & N-acetylmuramic acid

3. Normal substrate is large polymeric polysaccharide & unsuitable for investigations

B. Phillips. et al. found a low MW inhibitor that resembled substrate but was not hydrolyzed

1. The longest chain of sugars that would fit into lysozyme active site when packed into crystal consisted of 3 linked units of N-acetylglucosamine, (NAG)3

2. When crystals were made in presence of (NAG)3 & subjected to X-ray diffraction, the (NAG)3 was found to sit in cleft of egg-shaped enzyme, filling about one-half of cleft length

3. Fragment was bound to enzyme by H bonds & van der Waals forces

4. Extrapolated findings to larger molecules & proposed that active site of lysozyme contains 6 subsites (A-F), each of which bound a single sugar along polysaccharide

C. So active site is occupied by 6 adjacent cell wall sugar units; analysis of complex suggested mechanism

1. When 6 adjacent sugars were bound in cleft, the fourth (D) sugar could not be readily accommodated in available space

2. For the sugar to fit in space adjacent to tryptophan side chain,, it must be forced out of its normal chair conformation & flattened into a shape approaching half chair (sofa)

3. Since this part of substrate was subjected to physical strain & for other reasons, Phillips proposed that glycosidic bond linking D & E sugars would be the bond that was hydrolyzed

D. Vicinity of the D-E bond possesses 2 amino acid residues approaching within ~0.3 nm of bond from either side

1. One residue was aspartic acid, the other glutamic acid, both with carboxyl-containing side chains

2. Environment of glutamic acid is nonpolar, which should suppress dissociation of proton

3. Environment of aspartic acid is polar which should promote proton dissociation, leaving aspartyl residue with a negative charge

III. The proposed mechanism of lysozyme action

A. Splitting of glycosidic bond is accomplished by interaction of bond with nearby glutamic acid

1. The oxygen atom connecting sugars D & E is sufficiently electronegative to draw proton from undissociated glutamic acid carboxyl group

2. Causes acid-catalyzed hydrolysis of substrate bond

3. Breaking of bond by proton leaves both the carbon & oxygen atoms of substrate (C1 & O5 atoms of sugar D with partial positive charge

4. Such a shared positive charge between an O & C atom is called an oxocarbonium ion

5. Formation of oxocarbonium ion is facilitated by distortion of sugar after its binding to enzyme

6. This distorted, positively charged molecule corresponds to substrate molecule as it exists in its transition state, thus favoring product formation

B. The positive oxocarbonium ion is stabilized by the very close presence of the negatively charged aspartic acid of the enzyme active site

C. Oxocarbonium ion reaction with hydroxyl ion from surrounding solvent completes hydrolysis

IV. Support for mechanism

A. Ability of lysozyme to hydrolyze substrates of various lengths provided support - oligosaccharides consisting of 2, 3 or 4 NAG units were not hydrolyzed by enzyme

1. When 5-unit substrate was used, a low rate of hydrolysis was observed

2. A 6-unit oligosaccharide (hexamer) was hydrolyzed as efficiently as cell wall polysaccharide prep

3. A 6-unit oligosaccharide is just big enough to fill active site; as predicted, it was split between the fourth & fifth sugar, generating products consisting of 4 & 2 sugar units

B. Distortion of substrate results from presence of bulky hydroxymethyl (—CH2OH) group on carbon 6 of fourth sugar of the substrate

1. If the hydroxymethyl were to be removed, the substrate should be able to bind the enzyme more easily —> this happened when enzyme was incubated with modified version of (NAG)4

2. Fourth NAG was altered by removal of CH2OH (sugar was N-acetylxylosamine instead of NAG)

3. As predicted, the modified oligosaccharide bound 40 – 50 times more strongly to enzyme than (NAG)4

C. If model is correct & only fourth of 6 sugars in active site experiences strain, then binding of this sugar to enzyme subsite should have different thermodynamic properties than binding of other sugars

1. Studied kinetics of different sugar compounds binding to their respective subsites, their favorability or unfavorability

2. Found that only binding of sugar residue to subsite D was found to be unfavorable (at least +2.9 kcal/mole) —> supported model

3. Energy needed to bind fourth sugar is supplied by energy released as other sugars bind enzyme

D. Distortion of fourth sugar residue in substrate placing strain on glycosidic bond is important element in structure of substrate as it exists in transition state

1. Linus Pauling (1948) suggested enzyme is most complementary to high-energy transition state that forms as reactants are ready to be converted to products

2. If this is true, then inhibitors resembling transition state molecule should bind more strongly than inhibitors resembling original substrate(s); called transition state analogues (TSAs)

3. A TSA for lysozyme called TACL (tetra-N-acetylchitotetrose) was tested in 1972

4. It resembles (NAG)4 except that its terminal sugar residue is oxidized to a beta-lactone (mimics distorted conformation of D sugar proposed by Phillips as transition state structure)

5. Binding of TACL to lysozyme measured by its ability to inhibit lysozyme-catalyzed lysis of a culture of sensitive bacterial cells

6. When tested near optimum pH of enzyme, it required more than 100x the concentration of (NAG)4, an inhibitor that resembles the substrate, to cause the same inhibition as TACL

7. This supports suggestion that distorted substrate corresponds to transition state; TACL binds ~3600 times more strongly to enzyme than does the unmodified tetrasaccharide, (NAG)4

8. Concluded that strain caused by substrate binding could increase rate of catalysis by 103 to 104

E. Different ionization states of Glu 35 & Asp 52 because of polarity differences in their environments

1. At pH 5 (pH optimum), model proposes that Glu 35 carboxyl group should be holding onto its proton; also states that Asp 52 should be negatively charged & able to stabilize oxocarbonium

2. Circular dichroism (CD) gives best measurements of this; it depends on absorption of polarized light & is very sensitive to changes in state of protein

3. CD spectrum of lysozyme shows strong absorption band at 305 nm traced to significant tryptophan residue (Trp 108); Trp 108 is very close to Glu 35 which is close to Asp 52

4. Due to proximities, pH changes that alter ionization states of Glu 35 or Asp 52 have effect on CD spectrum derived from Trp 108

5. CD data at varying pHs indicate that the Glu 35 pK (pH at which half of groups are protonated & half are ionized) is unusually high at 6.1 (typical value would be 4.4); pK of Asp 52 is about 3.4

6. As predicted by model, glutamyl carboxyl group would retain its proton at pH 5, while aspartyl carboxyl would be negatively charged

F. Can also selectively modify key amino acid residues & measure the effect on enzyme catalytic activity

1. Enzyme can then be digested into fragments & the precise modification verified; the first such report of a chemically modified lysozyme appeared in 1969

2. Treat lysozyme with agent that removed Asp 52 COOH group (triethyloxonium fluoroborate; converts COOH into ethyl ester) —> all catalytic activity lost but substrate bound with high affinity

3. Subsequent experiments in which either Glu 35 or Asp 52 was modified chemically confirmed expectation that both residues must remain in their native state for enzyme to remain active

G. With DNA technology, it is possible to make deletions, additions or substitutions to genes that have been isolated; called site-directed mutagenesis (SDM)

1. DNA nucleotide sequence is specifically altered so that amino acid in polypeptide is replaced by another one of investigator's choosing; can replace any amino & all proteins will have alteration

2. First use of SDM with lysozyme was published in 1989; mutant lysozyme in which Asp 52 or Glu 35 were altered with SDM had very little or no catalytic activity

H. More recent crystallographic studies of lysozyme at higher resolution (1.5 Å) have confirmed prediction that sugar in site D adopts a strained conformation (allows it to fit into available space)

1. All the atoms that make up the ring of the bound sugar lie in essentially the same plane, which is very different from their arrangement in the normal chair conformation

2. This study also found evidence that the Asp 52 residue was H bonded to an asparagine residue (Asn 46); this H bond would be expected to stabilize the negative charge on Asp 52

3. This would, in turn, promote the latter's ability to stabilize the oxocarbonium intermediate

4. Subsequent studies in which the Asp 52 and/or the Asn 46 residues were mutated has provided support for this interpretation

5. When the Asn 46 residue is mutated to an alanine, the ability of the Asp 52 residue to bind to the substrate is disrupted & so too is the catalytic activity of the enzyme

V. Evidence suggests that the mechanism suggested by Phillips 30 years ago has stood the test of time very well

Human Perspectives

The Growing Problem of Antibiotic Resistance

I. Not long ago, it was thought that antibiotics had defeated serious bacterial infections forever

A. Antibiotics selectively kill bacteria without harming the human host in which they grow

B. State of affairs has changed dramatically in last couple of decades; pathogenic bacteria have become increasingly resistant to these "wonder drugs"

1. Has led to death of many who would once have been successfully treated

C. To make things worse, pharmaceutical industry has drastically cut resources devoted to development of new antibiotics

1. This is generally attributed to a lack of financial incentives

2. Antibiotics are taken only for short period of time (as opposed to drugs prescribed for chronic conditions (diabetes, depression)

3. New antibiotics run risk of having relatively short lifetime in market place as bacteria become resistant to each successive product

4. The most effective antibiotics may be held back from widespread use to be kept instead as a weapon of last resort when other drugs have failed

D. Ideas for forming nonprofit institutions for development of new antibiotics widely discussed

II. Most antibiotics are derived from natural products made by microorganisms to kill other microorganisms

A. They work because they interfere with certain bacterial activities without affecting those of eukaryotic cells

B. Several types of targets in bacterial cells have proven most vulnerable

1. Enzymes involved in the synthesis of the bacterial cell wall

2. Components of the system by which bacteria duplicate, transcribe & translate their genetic information

3. Enzymes that catalyze metabolic reactions specifically in bacteria

III. Enzymes involved in the synthesis of the bacterial cell wall

A. Penicillin & its derivatives (e.g., methicillin) are structural analogues of the substrates of a family of transpeptidases that catalyze the final cross-linking reactions that give the cell wall its rigidity

1. If these reactions do not occur, a rigid cell wall does not develop

2. Penicillin is an irreversible inhibitor of transpeptidases; it fits into the active sites of these enzymes, forming a covalently bound complex that cannot be dislodged

B. Vancomycin (originally derived from microorganism living in soil samples taken from Borneo)

1. It inhibits transpeptidation by binding to the peptide substrate of the transpeptidase, rather than to the enzyme itself

2. Normally, the transpeptidase substrate terminates in a D-alanine—D-alanine dipeptide

3. To become resistant to vancomycin, a bacterial cell must synthesize an alternate terminus that does not bind the drug

4. This is a roundabout process that requires the acquisition of several new enzymatic activities

5. Vancomycin is thus the antibiotic to which bacteria have proven least able to develop resistance & is thus usually given as a last resort when other antibiotics have failed

C. However, vancomycin-resistant strains of several pathogenic bacteria (like Staphylococcus aureus) have emerged in recent years

1. S. aureus is common inhabitant of human body surface

2. Usually relatively harmless, but it is most frequent cause of life-threatening infections that develop in hospitalized patients recovering from surgery or those with compromised immune systems

3. For years, methicillin-resistant S. aureus (MRSA) strains have reeked havoc in hospital wards, killing tens of thousands of patients; vancomycin is often only drug that can stop its infections

4. Alarmingly, MRSA found to get vancomycin-resistant by getting cluster of vancomycin-resistance genes from another bacterium (E. faecium), also a common cause of hospital-based infections

5. So far, no known vancomycin-resistant MRSA has gained foothold in either a hospital or community setting, but it may only be a matter of time

D. Thus, infectious disease specialists have urged hospitals to institute better hygiene programs, which have proven to reduce incidence of lethal infections

IV. Components of the system by which bacteria duplicate, transcribe & translate their genetic information

A. Prokaryotes & eukaryotes have similar systems for storing & using genetic information, but there are basic differences that pharmacologists can take advantage of

B. Streptomycin & tetracyclines bind to prokaryotic ribosomes, but not eukaryotic ribosomes

C. Quinolones, a rare example of wholly synthetic antibiotics (not based on natural products), inhibit the enzyme DNA gyrase, which is required for bacterial DNA replication

1. Most widely prescribed quinolone is ciprofloxacin (brand name Cipro)

D. Nearly all new antibiotics are derivatives of existing compounds that have been modified chemically in the lab; new compounds are typically screened in one of two ways

1. Compound is tested for its ability to bind & inhibit a particular target protein that has been purified from bacterial cells or

2. It is tested for its ability to kill bacterial cells that are growing in culture dish or lab animal

E. Remarkably, there have been only 2 new classes of antibiotics developed since 1963

1. One includes antibiotic linezolid (brand name Zyvox); acts specifically on bacterial ribosomes & interferes with protein synthesis; introduced in 2000, within ~1 year resistance seen in S. aureus

2. The other includes daptomycin (brand name Cubicin); introduced in 2003, they are cyclic lipopeptides, which disrupt bacterial membrane function

F. Researchers hope that Cubicin & Zyvox are used so sparingly that resistance can be kept to minimum

V. Enzymes that catalyze metabolic reactions specifically in bacteria - sulfa drugs are effective since they closely resemble p-aminobenzoic acid (PABA)

A. Bacteria convert PABA enzymatically to the essential coenzyme folic acid

B. Humans lack a folic-acid synthesizing enzyme & thus must get this essential coenzyme in the diet

1. Thus, sulfa drugs have no effect on human metabolism

VI. Bacteria become resistant to antibiotics through a number of distinct mechanisms – example: penicillin

A. Penicillin is a -lactam; it contains a characteristic 4-membered -lactam ring

1. As early as 1940, researchers discovered that certain bacteria possess an enzyme called -lactamase or penicillinase that can open the lactam ring, rendering the compound harmless to the bacterium

2. In World War II, when penicillin was introduced, none of the major disease-causing bacteria had a gene for -lactamase

3. We know this by examining DNA of bacteria descended from cultures established before antibiotics appeared

4. Today the -lactamase gene is found in a wide variety of infectious bacteria; it is the main cause of penicillin resistance

B. The widespread occurrence of the -lactamase gene demonstrates how easily genes can spread from one bacterium to another, not only in a given species, but between species

1. Can happen by conjugation in which DNA is passed from one bacterial cell to another

2. Can happen by transduction, in which a bacterial gene is carried from cell to cell by a virus

3. Can happen by transformation, in which a bacterial cell is able to pick up naked DNA from its surrounding medium

C. Pharmacologists have attempted to counter the spread of -lactamase by synthesizing penicillin derivatives (cefuroxime) that are more resistant to hydrolytic enzyme

1. Natural selection leads rapidly to the evolution of bacteria whose -lactamase can split the new forms of the antibiotic

D. Some penicillin-resistant bacteria do not have the -lactamase gene; they have used other strategies

1. Some possess modifications in their cell walls that block entry of the antibiotics

2. Others are able to selectively export the antibiotic once it has entered the cell

3. Others possess modified transpeptidases that fail to bind the antibiotic

E. Bacterial meningitis is caused by bacterium Neisseria meningitidis, which has yet to display it has acquired -lactamase; their resistance comes from transpeptidases losing affinity for antibiotics

VII. Drug resistance is not restricted to bacterial diseases; it has become a major issue in AIDS treatment

A. Replicating enzyme of AIDS virus (HIV), reverse transcriptase, makes a large number of mistakes unlike bacteria whose DNA replicating enzymes operate at a very high level of accuracy

1. This leads to a high rate of mutation (error rate), ~1 mistake for every 10,000 bases duplicated

2. This combined with the high virus production rate (>108 virus particles produced in a person/day) makes it very likely that drug-resistant variants will emerge in individual as infection progresses

B. This problem is being combated in 2 ways:

1. Patients take several different drugs targeted at different vital enzymes; this greatly reduces the likelihood that a variant will emerge that is resistant to all of the drugs

2. By designing drugs that interact with the most highly conserved parts of each targeted enzyme (where mutations are most likely to produce a defective enzyme)

CRITICAL THINKING QUESTIONS

1. You are observing a reaction and discover that the reaction vessel is warm to the touch. The reaction also results in an increase in entropy. Is the reaction spontaneous? The reaction has a negative DG and is, therefore, spontaneous. How do you know? The reaction is exothermic (-DH) as demonstrated by the warm reaction vessel and entropy is increased (DS). When these values are plugged into the equation DG = DH - TDS, the only possible result is that DG is negative.

2. If a reaction vessel is cold to the touch and the reaction results in an increase in order in the reaction vessel, is the reaction spontaneous or nonspontaneous? The reaction is nonspontaneous. Explain your answer. The DH is positive and the DS negative. When these values are plugged into the equation, DG = DH - TDS, the result is a positive DG and, therefore, a nonspontaneous reaction.

5. In the reaction A + B <—> C + D, how might the reaction take place in the cell if the DG is very positive? If DG is very positive, the reaction would probably take place by coupling it to a reaction with a larger -DG so that when the DG values are added up the sum is negative. How might the reaction occur if the DG is slightly positive? If the reaction is slightly positive, it may be coupled as well, although it may also be driven forward by increasing the amounts of the reactants and/or decreasing the amounts of products. The Law of Mass Action would be likely to allow the reaction to run under these conditions.

6. You are observing an enzyme driven reaction. To the reaction mixture you add a chemical X which inhibits the reaction. If you add more substrate, the reaction rate approaches the Vmax of the uninhibited reaction. Furthermore, the structure of X is similar to the natural substrate. What kind of inhibitor is X? X is a competitive inhibitor. First, competitive inhibitors resemble the substrates of the reactions they inhibit. Second, their effects can be reversed by increasing substrate concentration, since both the substrate and the competitive inhibitor are capable of binding to the enzyme active site. As the amount of substrate increases relative to the competitive inhibitor, the active site is more likely to pick up substrate and thus reaction rate increases and the inhibition is reversed.

7. If ATP is present in relatively high amounts, what is likely to happen to the rate of glycolytic activity in that cell? The glycolytic pathway will probably slow down. ATP will act like a noncompetitive inhibitor and will inhibit one of the early reactions in the glycolytic pathway.

8. An enzyme has a KM of 20 µM and a Vmax of 50 mmoles of product/minute/µg of enzyme. After exposure to an inhibitor and analysis on a Lineweaver - Burk plot the following values are obtained: -1/KM = - 0.05 liters/µmole and 1/Vmax = 0.04 (mmoles of product/minute/µg of enzyme)-1. What kind

of inhibitor was used in the experiment? Since the KM has remained the same and the Vmax has decreased, the inhibitor was noncompetitive.

9. Why are alcoholic beverages often made in airtight containers? The alcohol in these beverages is produced by fermentation, which only occurs in the absence of oxygen.

10. Below is a segment of a cell's collection of biochemical pathways. M is a product of one series of these reactions. It is also a regulatory molecule. Look at the pathway below and indicate the position(s) at which M is most likely to act as a feedback inhibitor when its concentration gets too high.

M will be most likely to act as a feedback inhibitor at position 1, 2 or maybe 3. Usually, a feedback inhibitor will work on an enzyme near the beginning of the pathway that leads to its production. Therefore, positions 1 and 2 are the most likely with 1 being slightly favored.

11. You are studying metabolic pathways and discover that two pathways intersect so that the enzyme basinase participates in both of the intersecting pathways, in one case using substrate K and in the other using substrate M. When presented with substrate K in amounts significantly larger than M, basinase converts K to L which leads eventually to the production of the end product R. The activity of the second pathway is depressed simultaneously. In the presence of large amounts of substrate M and lower amounts of substrate K, the second pathway is activated and culminates in the production of that pathway's end product Y. The activity of the first pathway is depressed simultaneously. What are the alternative substrates K and M acting like? K and M are acting like competitive inhibitors. If one is present in excess, the pathway involving the other is inhibited.

12. 4 ATPs per glucose molecule are made during glycolysis. Why then is there a net production of only 2 ATPs for each glucose molecule in the pathway? Two ATPs are used up early in glycolysis before the 4 ATPs are produced. Consequently, there is a net gain of only 2 ATPs per glucose.