2. Under what conditions can an enzyme assay be used to determine the relative amounts of an enzyme present?

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1 Chem 315 In class/homework problems 1. a) For a Michaelis-Menten reaction, k 1 = 7 x 10 7 M -1 sec -1, k -1 = 1 x 10 3 sec -1, k 2 = 2 x 10 4 sec -1. What are the values of K s and K M? K s = k -1 / k 1 = 1 x 10 3 sec -1 / 7 x 10 7 M -1 sec -1 = 1.4 x 10-5 M K m = Michaelis-Menten constant =( k -1 + k 2 )/ k 1 = (1 x 10 3 sec x 10 4 sec -1 ) / 7 x 10 7 M -1 sec -1 K m =3.0 x 10-4 M b) Does substrate binding approach equilibrium or does it behave more like a steady state system? Once the ES complex forms, the substrate may be either converted to product (k 2 ) or released as S (k -1 ). Formation of product is governed by a first order rate constant, k 2, this constant is 20 times larger than the first order rate constant for the dissociation of ES (k -1 ). Thus, the enzyme behaves like a steady state system. Only in the case where k 2 is rate limiting will the substrate binding approach a true equilibrium. 2. Under what conditions can an enzyme assay be used to determine the relative amounts of an enzyme present? To determine the amount of enzyme present you would need to reach saturation. At the early stages of saturation, the concentration of substrate would approximate the concentration of enzyme. At V max the following expression can be used V max = k 2 [E T ], showing that the maximum velocity is proportional to the concentration of enzyme. 3. Liver alcohol dehydrogenase (ADH) is relatively nonspecific and will oxidize ethanol or other alcohols, including methanol. Methanol oxidation, yields formaldehyde, which is toxic, causing blindness in addition to other deleterious effects. Because it has a sweet taste, many animals will drink windshield washer fluid (50% methanol). If the oxidation of methanol to formaldehyde is blocked, it can be excreted through the kidneys without serious harm to the animal. Ethanol is a competitive inhibitor of ADH. (The K M values of alcohol dehydrogenase for ethanol and methanol are 1mM and 10 mm, respectively) Can wine (12% ethanol by volume) effectively be used to lower the activity of the animals ADH towards methanol? Explain. Both methanol and ethanol quickly distribute throughout the entire body as they are structurally similar and both will interact with the enzyme. K m can be used as an estimate of the enzymesubstrate fit. In other words, since K m is a ratio of ES disassociation to ES association, a large value of K M is indicative of a more weakly associate ES complex. Since the K m values of canine ADH for ethanol and methanol are 1mM and 10mM, respectively, ethanol forms a tighter ES complex. The K I for ethanol in its role as a competitive inhibitor of methanol oxidation by ADH is the same as its K M. Since wine is only 12% ethanol by volume, the volume of ethanol would have to be much higher than the volume of pure grain alcohol. However, in sufficient quantity, ethanol could competitively inhibit methanol. (For 50mL of windshield washer fluid, 750 ml of wine would be

2 required; this number can be calculated using the molar mass, density and relative constants of each alcohol. 4. A researcher was studying the kinetic properties of β-galactosidase using an assay in which o-nitrophenol-b-galactoside (ONPG), a colorless substrate, is converted to galactose and o- nitrophenolate, a brightly colored, yellow compound. Upon addition of 0.25mM substrate to a fixed amount of enzyme, o-nitrophenolate (ONP) production was monitored as a function of time by spectrophotometry at λ = 410 nm. The follow data were obtained: Time (s) A 410 nm a) Convert A 410 nm to concentration of o-nitrophenolate, [ONP], using E=3.76mM -1 cm -1 as the extinction coefficient. The relationship between absorbance and concentration is linear as shown by the Beer-Lambert law: A = EbC where A = absorbance, E=molar absorptivity (constant for a particular compound under defined conditions and a specific wavelength; 410 in this case), b is the path length (instrument constant; usually 1 cm) and C is the molar concentration. Solving for C = A/Eb =A/1.0 cm 3.76mM -1 cm -1 Time (s) A410 nm [ONP]

3 b) Plot [ONP] versus time c) Determine the initial velocity of the reaction

4 The initial velocity can be calculated from the slope of a line drawn through the initial points (when [S] is not near saturation). The velocity or rate is by definition change in concentration/change in time or delta y/delta x, the definition of a slope. The slope of the line is mm/s. Answers will vary somewhat depending on the points chosen. d) Explain why the curve is nearly linear initially and later approaches a plateau The enzyme is converting substrate to product (S+E --> ES ---> P); the plot shows the progression of the reaction with time. The initial points are linear because the enzyme is obeying Michaelis- Menten kinetics (i.e., steady state assumption). At later times the concentration of product becomes significant (remember, in M-M kinetics one assumes [P]=0), as the [S] is depleted. As the [P] approaches the initial [S] (all S converted to P), M-M kinetics conditions are no longer appropriate and the curve levels. e) Describe how K M and V max for β-galactosidase can be determined with additional experimentation. To determine K M and V max a Lineweaver-Burke plot should be constructed. This would require running several trials where [S] is varied and the initial velocity is measured. A linear regression analysis of the plot of 1/v vs 1/[S] would yield the required data. 5. The following kinetic data were obtained for an enzyme in the absence of inhibitor (1) and the presence of two different inhibitors (2) and (3) at 5 mm concentration. Assume [E t ] is the same for each experiment. a) Determine V max and K M for the enzyme. b) Determine the type of inhibition and the K I for each inhibitor. [S] (1) (2) (3) mm v(µmol/m L sec) v(µmol/ml sec) v(µmol/ml sec) In order to analyze the data a double reciprocal plot should be constructed. Therefore, you need to calculate 1/vand 1/[S] 1/ v = K M /V max * 1/[S] + 1/V max

5 y = m x + b The y intercept is equal to 1/V max and the x intercept is equal to - 1/K m pay attention to units when comparing data 1/[S] M-1 v1 umol/sml 1/v1 sml/mol v2 umol/sml 1/v2 sml/mol v3 umol/sml 1/v3 sml/mol s m L/ u m ol Linear regressions: uninhibited y = x inhibitor 1 (2) y = x inhibitor 2 (3) y = x Condition Vmax (µmol/mlsec) Km (mm) No inhibitor mm inhibitor (2) mm inhibitor (3) For the first inhibitor (2) the V max value is experiences relatively little change (within experimental error). Graphically, the lack of change is more apparent. This is real data that carries with it real error. [S] (1) (2) (3) mm v(umol/m L sec) v(umol/m L sec) v(umol/m L sec)

6 Inhibitor 2 increases the apparent K M of the enzyme without affecting V max. This is characteristic of a competitive inhibitor. In this case K I is calculated as follows: K M(2) = K M (1 + [I]/K I ) or 10.7 mm = 3.2mM (1 + 5mM/K I ) Solving for K I = 4.04 mm 6. The following graphical patterns obtained from kinetic experiments have several possible interpretations depending on the nature of the experiment ant the variables being plotted. Give at least two possibilities for each.

7 In the top graph, the results may be due to a competitive inhibitor. The line with the steeper slope is from enzyme kinetic measurements in the presence of an inhibitor, whereas the other line is enzyme kinetic data without inhibitor. In competitive inhibition, V max, the reciprocal of the 1/v intercept, is independent of the presence of inhibitor. In other words the y intercept is unchanging with the addition of inhibitor, this is indicative of competitive inhibition. The second graph is most likely pure, noncompetitive inhibition. In noncompetitive inhibition, the inhibitor binds the enzyme and the enzyme substrate complex with equal affinity. The binding of I to E has no effect on the binding of S to E. Thus K M is unaffected because the inhibitor-free enzyme is capable of normal catalysis. In effect, the inhibitor lowers the active enzyme concentration.

8 The bottom graph is most likely an example of mixed noncompetitive inhibition in which the inhibitor I binds to both E and ES but not equally. IN this case, I binds ES with higher affinity than to S. V max is decreased but K M is increased. 7. The presence of inhibitor decreases Vmax; however, it also decreases KM. This seems to suggest that in the presence of an inhibitor, the remaining active enzyme's performance is improved since it saturates at a lower substrate concentration. Explain this paradox. This is an example of mixed competitive inhibition. In this type of inhibition, the inhibitor binds to both E and ES. Decrease in V mzx for an inhibitor is expected as the inhibitor interferes with the natural conversion of ES to P. However, the effect on K M is not as intuitive. Remember, K M can be thought of as a measure of the disassociation of ES (K M = (k 2 + k -1 )/ k1). The inhibitor may not inhibit each of these reactions (reactions governed by the constants that compose K M ) equally and can increase K M (inhibition of ES formation by preferential EI binding) or decrease K M (decreasing k 2 by binding of ES to I which will stop conversion of ES--->P) The following two equilibrium constants apply for a mixed competitive inhibition: K I = [E][I]/[EI] for binding of I to E and K I ' = [ES][I]/[ESI] for binding of I to ES K I is not necessarily equal to K I '. In this example, K M is decreased so the affect on k 2 is most pronounced. In other words, I binds to ES with greater affinity than to E. Therefore, addition of I should affect the E+S ---> ES equilibrium by shifting it to the right since as ES is bound to I, ESI is produced causing depletion of ES. By Le Chatlier's principle, this will cause a corresponding increase the [ES]. As [ES] is increased (with more ESI production, higher [I]), then more ES will be available to be converted to P and the rate of product production increases. This explanation is not something that you may immediately see. Read and study it for a while and it should slowly filter in.