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Unformatted text preview: 10.1 Explain why a catalyst must affect the rate of a reaction in both directions. A catalyst speeds up a reaction by lowering the Gibbs energy of the transition state. Therefore, it lowers the activation energy for both the forward and the reverse reaction and affects the rates
of both. W 10.3 The hydrolysis of aeetylcholine is catalyzed by the enzyme acetylcholinesterase, which has a turnover rate of 25,000 3— 1. Calculate how long it takes for the enzyme to cleave one acetylcholine
molecule. The time required for the enzyme to cleave one acetylcholine molecule (one turnover) is the
reciprocal of the turnover rate. .1. = —L__] =4.0 x10_ss=40us
k2 250008—~ M t: 10.5 An enzyme that has a K M value of 3.9 x 10‘5 M is studied at an initial substrate concentration of 0.035 M. After 1 min, it is found that 6.2 ,uM of product has been produced. Calculate the
value of VM and the amount of product formed after 4.5 min. With 6.2 MM of product formed in 1 min, the initial rate, v0 = 6.2 x 10‘6 M min—1. Then solve _ Vmax[S]
U0 — ~——'—'
KM + [S] Vmu(0.035 M) 6.2 x 10—6 M min“l —————————_—
3.9 x 10—5 M + 0.035 M II to ﬁnd Vm : 6.21 x 106 Mminl = 6.2 x 10—6 M min"l. The observed rate is Vm, indi
cating that the substrate concentration is so large that the enzyme is saturated with substrate. Over 4.5 min, the substrate concentration stays essentially constant, and the rate does not change.
Thus, after 4.5 min, (4.5 min)(6.21 x 10‘6 M min”) = 2.8 x 10‘5 M of product has formed. 10.7 The KM value of lysozyme is 6.0 x 10‘6 M with hexa—Nacetylglucosamine as a substrate.
It is assayed at the following substrate concentrations: (a) 1.5 x 10‘7 M, (b) 6.8 x 10‘5 M, (c)
2.4 x 10—4 M, (d) 1.9 x 10”3 M, and (c) 0.061 M. The initial rate measured at 0.061 M was
3.2 ,uM min—1. Calculate the initial rates at the other substrate concentrations. Vmax is calculated from the initial rate given for 0.061 M substrate.
_ Vmax[S]
vo — ————
KM + [S]
Vm(0.061 M) 3.2 x 10—6 Mmin“ = ———————
6.0 x 106 M + 0.061 M Vm = 3.20 x 10—6 M min1
Thus, the substrate concentration is sufﬁcient to have reached Vm. The initial rates at the other substrate concentrations are then found using v _ ths1
0_K +[S]
M _ (3.20 x 10~6 M min—1H8]
6.0 x 106 M + {S} For the ﬁve concentrations given,
(a) v0 = 7.8 x 108 M min1
(b) vo = 2.9 x 106 M min'l
(c) v0 = 3.1 x 106 M min1
(d) 00 = 3.2 x 106 M min—1 (e) 00 = 3.2 x 10‘6 M min“l 10.8 The hydrolysis of urea, (NH2)2CO + H20 —> 2NH3 + C02 has been studied by many researchers. At 100°C, the (pseudo) ﬁrstorder rate constant is 4.2 x
10—5 s”. The reaction is catalyzed by the enzyme urease, which at 21°C has a rate constant
of 3 X 104 s‘l. If the enthalpies of activation for the uncatalyzed and catalyzed reactions
are 134 kJ mol‘1 and 43.9 kJ mol_], respectively, (a) calculate the temperature at which the
nonenzymatic hydrolysis of urea would proceed at the same rate as the enzymatic hydrolysis at
21°C; (b) calculate the lowering of AGi due to urease; and (c) comment on the sign of A51.
Assume that AH i = E a and that AHit and AS1 are independent of temperature. (a) The Arrhenius equation relates reaction rate and activation energy via k = Ae‘Ea/RT. Re—
quiring that the rates of the catalyzed and uncatalyzed reactions be equal at their respective temperatures then means (assuming A to be the same for both the catalyzed and uncatalyzed
reactions), kcat = kuncat Ae—Egm/RT1 : Ae—Egncat/RTZ cat uncat
Ea Ea T1 T2 taking Ea % AHi, 43.9 x 103 Jino11 _ 134 x 103 Jnio11
294.15 K " T2 T2=898K At this temperature the solvent would be vaporized and the urea thermally decomposed, so that
it is in fact impossible to achieve the enzymatic rate without the catalyst. (b) From Equation 9.40, k = "BTTeAGi/RT, or Ac: 2 —RT 111 [cg—g For the uncatalyzed reaction at 373 K, (4.2 x 10*5 5‘1) (6.626 x 1034 J s)
(1.381x 1023 JK“1) (373.15 K) AGi = — (8.314 JK"1mol_1) (373.15 K) 1n =1.234 x 105 Jmol—1 For the catalyzed reaction at 294 K, (3 x 104 SA) (6.626 x 10*34 J 3)
(1.381 x 10—23 JK’I) (294.15 K) A61 = — (8.314JK—1mo1—1) (294.15 K) ln = 4.70 x 104 J mol—1 Thus, A61 is lowered by 1.234 x 105 J mor1 — 4.70 x 104 Jmorl = 7.64 x 104 J nio171,a1
though the comparison is being made at two different temperatures. (0) Since not = AH: — rosi, A51: (Mrt — not) jT. For the uncatalyzed reaction, 134 x 103 Jnior1 — 1.234 x 105 Jinorl
373.15 K A31: = 28.4 J K—1 mot1 There is an increase in entropy upon approaching the transition state as would be expected in a
case where a single molecule is breaking apart in two or more fragments in the transition state. For the catalyzed reaction, 439x1MJnmr4—430x1minmr1 . =—11JK_1mol_1
294.15K ASi= Here there is a decrease in entropy upon approaching the transition state, since the rate deter
mining step now involves the binding of two molecules, enzyme and substrate. ____________————————— 10.11 The initial rates at various substrate concentrations for an enzymecatalyzed reaction are as
follows: [S]/M 001106 M  min1 2.5 x 105 38.0
4.00 x 105 53.4
‘ 6.00 x 105 68.6
8.00 x 105 80.0
16.0 x 105 106.8
20.0 x 105 114.0 (a) Does this reaction follow Michaelis—Menten kinetics? (b) Calculate the value of Vmax of
the reaction. (c) Calculate the KM value of the reaction. (d) Calculate the initial rates at [S] 2
5.00 x 10—5 M and [S] = 3.00 x 10‘1 M. (e) What is the total amount of product formed during
the ﬁrst 3 min at [S] = 7.2 x 10‘5 M ‘? (1‘) How would an increase in the enzyme concentration
by a factor of 2 affect each of the following quantities: K M, VW, and 00 (at [S] = 5.00 x 10‘5
M)? (a) A Lineweaver—Burk plot (1 / 00 versus 1 /[S]) using the following data results in a straight line,
indicating that the reaction does indeed follow Michaelis—Menten kinetics. (l/[S])/103  M1 (1/v0)/103 . Ml . min 40.0 26.32 25.00 18.73 16.67 14.58 12.50 12.50 6.250 9.3633 5.000 8.7719
28
26
.= 24
E 22
g 20
«2'3 18
z 16
g 14
C 12
10
8 0 10 20 30 40 (ll{S])/ 10 3  M" The equation of the best ﬁt line to the data is y : 5.013 x 10—‘x + 6.23 x 103. Thus, since in a
Lineweaver—Burk plot the slope is K M / Vmax and the intercept is 1/ Vmax, 0’) Vmax = ————;——1—— = 1.605 x 10—4 M min—1 = 1.61 x 10—4 M min“
6.23 x 103 M’ min
__1 
(‘9 KM — 5‘0” x 10 mm = 8.047 x 10—5 M = 8.05 x 105 M  6.23 x 103 M" min (d) For Michaelis—Menten kinetics, 00 = ————
KM + [S] Thus at [S] = 5.00 x 10'5 M, (1605 x 104 M min1) (5.00 x 105 M) “5 . _,
v0: W =6.15 x 10 Mmm
8.047 x 10‘5 M + 5.00 x 10—5 M and at [S] = 3.00 x 101 M, _ (1.605 x 104 M min1) (3.00 x 101 M) = 1.60 x 10"4 M min“1
8.047 x 105 M + 3.00 x 101l M 1’0 (e) At [S] = 7.2 x 105 M, (1.605 x 10‘4M minl) (7.2 x 10"5 M)
8.047 X 10‘5M +7.2 X 10—5M 60— :76 x10‘5’Mmin‘l
Although the reaction rate will slow down as substrate is converted into product, the combination of the rapid rate and small concentration of substrate indicates that three minutes will be sufﬁcient
time to convert all of the substrate to product. (1’) KM is independent of [E], but since Vmax = k2[E]O, doubling [E]o will double Vmax. This in
turn leads also to a doubling in no [see equations in parts (d) and (e)]. 10.18 (a) What is the physiological signiﬁcance of cooperative 02 binding by hemoglobin? Why is
02 binding by myoglobin not cooperative? (b) Compare the concerted model with the sequential
model for the binding of oxygen with hemoglobin. (a) Cooperative 02 binding enables hemoglobin to be a more efﬁcient oxygen transporter than
myoglobin. As discussed in detail in Section 10.6 of the text, nearly twice as much oxygen is
delivered to the tissues than would be if 02 binding to hemoglobin were not cooperative. (b) The concerted model, with it’s “allor—none” limitation on the relaxed and tense forms of the
four subunits in hemoglobin does not allow for the existence of mixed forms with some subunits
in one form and the rest in the other. Although not relevant for the binding of oxygen with
hemoglobin, the concerted model is unable to account for negative homotropic cooperativity. Nevertheless, it does allow the characterization of the allotropic behavior of hemoglobin (and
enzymes) in terms of just three equilibrium constants. The sequential model does allow for the exisitence of mixed forms of the subunits comprising
the oligomer, since the binding of substrate (02 in the case of hemoglobin) affects only the
conformation of the subunit bound to substrate. The actual mechanism of oxygen binding to
hemoglobin is more complex than the limiting cases presented by the two models. The sequential
model, however, does have the advantage of being able to account for negative homotropic
cooperativity in enzymes displaying such behavior. /_ on oxygen afﬁnity of adult hemoglobin ’ h f the following actions
10.22 What is the effect of eac o (0 decrease [BPGL (d) (Hb A) in vitro‘? (a) Increase pH, (b) increase partial pressure of C02,
dissociate the tetramer into monomers, and (e) ox1dize Fe(lI) to FeClll). ________________.__ (a) Afﬁnity increases,
(b) afﬁnity decreases,
(c) afﬁnity increases, (d) afﬁnity increases, (e) afﬁnity decreases. 11.2 The threshold frequency for dislodging an electron from a zinc metal surface is 8.54 X
1014 Hz. Calculate the minimum amount of energy required to remove an electron from the
metal. At the threshold frequeHCy, the speed of electron, v, is 0. That is, l
hv=<13+§mevZ=CD Therefore, the minimum amount of energy required to remove an electron from the metal is c1) = (6.626 x 10~34 13) (8.54 x 1014 Hz) = 5.66 x 10719 J 11.4 Calculate the frequency and wavelength associated with the transition from the n = 5 to the
n = 3 level in atomic hydrogen. The wavenumber of the emitted radiation is i_i (i_i)
n? n? 52 32 Therefore, the wavelength and the frequency of this radiation are i? 2 (109737 cm—1) = (109737 cm_l) = 7803.520 ena—1 A = i = —1——— = 1.28147 x 10—4 cm: 1.28147 X103 nm
1) 7803.520 curl
v = ct) = (3.00 x 1010 ems—1) (7803.520 cmT1)= 2.34 x 1014 Hz ...
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 Spring '08
 SEABRA
 Physical chemistry, pH

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