Chapter 8 Problems Page 1 of 6 11/1/2007

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1 Chapter 8 Problems Page of 6 // he decarboxylation of a beta-eto acid alyzed by an enzyme can be measured by the rate of formation of CO 2. From the initial rates in the table determine the ichaelis- enten constant for the enzyme and the maximal elocity by a graphical method. First the problem should really say determine and V max for this enzyme-substrate combination. he alues for these constants depend on what enzyme and what substrate are being considered. I chose to transform the data for an adie-hofstee plot ( 0 /S s. 0. For this plot the slope is -/ and the ertical axis intercept is V max /. he plot is below. From xcel I get the least-squares alues for the slope ( and intercept (.57 μmol/2min. (I left the alues in the units that were in the table. From the slope we find that - /( From the intercept V max intercept (0.442 (.57 μmol/2min μmol/2min μmol/min..8.6 adie-hofstee plot /S

2 Chapter 8 Problems Page 2 of 6 // n enzymatic reaction has a simple ichaelis-enten mechanism: 2 S S P - and - are ery fast 2 00 /sec and 0-4 at 280 while /sec and at 300. (a For S 0. and what is 0 at 280? (b Calculate the actiation energy for 2. (c What is eq at 280 for binding to S? (d What is ΔH 0 for the formation of S from and S? a We are gien and need to calculate V max 2 0 (00 /sec( /s. icaelis-enten gies us 3 V S s max (0 / ( / s S ln 2 ' 2 R b From the rrhenius relationship e / we can diide 2 at 280 by 2 at 300 (I denote the different s by a prime on the second and then tae the logarithm of the ratio: 00s 5 ln.693 ( mol / J 200s R ' 8.34J / mol so from this we hae (0.693/( mol/j 24.2 J/mol. c eq / - by definition. ( - 2 / - / since - >> 2 from the problem statement that the first reaction is ery fast. hus eq / /( d his problem is asing about how enthalpy of a reaction depends on the change of the equilibrium constant with temperature. his is the an t Hoff relationship: Δ(ln ln(0 ln(.5 0 ( 9.2 ( 8.80 ΔH R R (8.34J / mol 4.2J / mol ( ( Δ (

3 Chapter 8 Problems Page 3 of 6 // Penicillinase inactiates penicillin. One form with W has a single actie site 2000 s - and In response to treatment with 5 μmol of penicillin a ml suspension of bacteria release 0.04 μg of enzyme. (a ssuming rapid mixing and equilibration what fraction of will complex with penicillin early in the reaction? (b How long will it tae for half of the initial amount of penicillin to be inactiated? (c What concentration of penicillin would cause the enzyme to react at half maximal elocity? (d second suspension of bacteria releases 0.06 μg/ml of enzyme. Will this affect the answers to (b and (c and if so by how much? (e odified penicillin acts as a competitie inhibitor. If the affinity of for penicillin and modified penicillin is the same what concentration of the inhibitor will reduce the rate of loss of penicillin fiefold at low penicillin? a he fraction of complexed to S is S S f where S S S S S the second step comes from S/S. Plugging in and S 5 μmol/ ml we get f b We can see that from the - equation when S is still 0.98V max so the system will remain in the approximate zero-order domain. he elocity during this time will be equal to V max to a good approximation. V max 0 (2000 s - (0.04 μg/(30000 g/mol/(0.00 L 2.67 μ/s. he change in the concentration of S is so ΔS/Δt μ/s or Δt ( /(-2.67 μ/s 937 sec 5.6 min. c 0 V max /2 when S so penicillin d Yes this will affect part b because 0 increases by 50% so V max increases by 50% and all elocities will thus increase by this factor. hat means that Δt will decrease by a factor of.5 or Δt 625 sec. Part c won t change since the affinity doesn t depend on how much is around. e If the inhibitor has the same affinity as penicillin then I For a competitie inhibitor we hae Vmax S 0. So when 0 (uninhibited 5 0 (inhibited we hae I S I Vmax S 5Vmax S 5 or. hat is 0 I S S S I I S I I S S 5 so I 4( S 5 thus it depend on how much S is present. If S << then I

4 Chapter 8 Problems Page 4 of 6 // RN polymerase binds rapidly to DN then the complex rearranges slowly. his can be diagrammed as C 2 C D -2 - fast slow t the start of the reaction n and n are present. t final equilibrium the concentrations are 0.6 n C 0.0 n and D 0.39 n. (a Calculate and 2 from the equilibrium concentrations. (b Write the differential equations for dc/dt and dd/dt. (c Setch cures for C and D s. time. (d he alue of 2 was found to be 0.05 sec -. Calculate the alue of -2. a he equilibrium constants are gotten from the equilibrium concentrations: C/ ( 0 - /( D/C ( /( (I ept the units for the first constant in the biochemistry style. b d C C 2 C 2 D 2 D ( 2 C and dt d D 2 C 2 D. dt c he first reaction is ery fast so it will come to equilibrium first. hen the second will slowly come to equilibrium. We now the starting concentrations. and will rapidly drop as the first reaction equilibrates and C increases. hen D will slowly increase to its final alue as C drops. he first reaction will react to the change in C with drops in and to their final alues. he initial plateau when the second reaction hasn t yet iced in will hae n x and C x: C/ x/( n x If we expect x to be small then x hus and C at the intermediate point. setch might loo lie concentration (n and C time (relatie to equlib D d We now 2 which equals 2 / -2. So -2 2 / 2 (0.05 s - /( s -.

5 Chapter 8 Problems Page 5 of 6 // carboxypeptidase was found to hae 2 μ and 50 s - for substrate. (a What is the initial rate of reaction for 5 μ and μ? (b he presence of 5 m of a competitie inhibitor decreased the initial rate by a factor of 2. What is the alue of I? (c competing substrate is added to part (a. Its 0 μ and 00 s -. Calculate /. a We need V max : V max (50 s - (0.0 μ.5 μ/s. So plug into the - equation: V S max (.5 0 / s(5μ / s S 2μ 5μ b competitie inhibitor maes the apparent change so first calculate what the apparent is: Vmax S (.5 0 / s(5μ.07 0 / s 0 or app 9.02 μ. he S app 5μ 2 I apparent in the presence of I is app or after a little algebra I I m (5 (2μ I. 42m. app 9.02μ 2μ c he problem as stated seems to as for two different things: i the relatie rates if the substrates are both present simultaneously at 5 μ and ii the relatie rates of hydrolysis if and are considered separately in separate reaction. he second is easier to deal with so I do it first We already hae 0 for 5 μ from part a. Calculating for 5 μ gies V S s max (.0 0 / (5μ / s. S 0μ 5μ / s herefore / s 0 Notice that this is not what you get if you apply equation 8.25 since that equation is alid only if and are ery small compared to their respectie m alues (if the reactions are in separate pots. Howeer as we will see if both substrates are present together in the same solution then q does hold. 2 If both and are present together in the same solution then simple ichaelis- enten analysis doesn t hold. he reason is that as far as is concerned some of the enzyme that it can interact with is remoed by the complex. We need to account for this in the mathematics that leads to the expression for which gies us the elocity of consumption of substrate. his is done exactly the same way that it is done when a competitie inhibitor I is present in the reaction mix. You can read about that on pages

6 Chapter 8 Problems Page 6 of 6 // in the boo. he whole tric is to note that the total amount of enzyme present equals the sum of and so that Now we now that (gotten in the same way as standard ichaelis- enten setting 0 2 dt d. So If you put this into our deriation of the ichaelis-enten equation you will get Following the same logic for substrate you get. aing the ratio of to gies his is exactly equation 8.25 and it is alid no matter how small or large are the concentrations of and. he alytic efficiencies for substrates and are respectiely / (50 s - /(2 μ 75 s - μ - and / (00 s - /(0 μ 0 s - μ -. So clearly the enzyme has a strong preference for oer. When the two substrates are at equal concentration we hae that the ratio of elocities / equals the ratio of the alytic efficiencies so / (0 s - μ - /(75 s - μ his holds when the two substrates are mixed together with the enzyme.