2. In the frog muscle of the rectus abdominous concentrations of ATP, ADP, and phosphate are M, M, and M, respectively.

Size: px

Start display at page:

Download "2. In the frog muscle of the rectus abdominous concentrations of ATP, ADP, and phosphate are M, M, and M, respectively."

Winfred Dylan Mathews
5 years ago
Views:

1 Homework Assignment #3 Solutions Chapter 4, Problems: 2, 4 *, 8 &, 14a #, 2, and S1-S4 below. Due Tuesday, October 5th, in class. Note that G for hydrolysis of ATP is given in the footnote of Table In the frog muscle of the rectus abdominous concentrations of ATP, ADP, and phosphate are M, M, and M, respectively. a. Calculate the Gibbs free-energy change Δ r G 298 of the muscle as 25 C and 7, respectively. for the hydrolysis of ATP in muscle. Take the temperature and ph ATP + H 2 O ADP + P i Δ r G = Δ r G + RT ln ( [ADP][P i] ) Eqn 4.15 [ATP] = 31. kj + (8.314 kj K ) (298 K) ln ([ ][ ] [ ) ] Δ r G = 48.1 kj b. For the muscle described, what is the maximum amount of mechanical work it can do per e of ATP hydrolyzed? w max,by the system = w max,on the surroundings = Δ r G = 48.1 kj w rev,max = 48.1 kj c. In muscle, the enzyme creatine phosphokinase catalyzes the following reaction: phosphocreatine + ADP creatine + ATP The standard Gibbs free energy of hydrolysis of phosphocreatine at 25 C and ph 7 is kj -1 : phosphocreatine + H2O creatine + phosphate Calculate the equilibrium constant of the creatine phosphokinase reaction. ΔG phosphocreatine + H 2 O creatine + P i 43.1 kj P i + ADP ATP + H 2 O 31. kj phosphocreatine + ADP creatine + ATP 12.1 kj K = e ΔG RT = e 12.1 kj ( kj Page 1 of 12 K )(298 K) = In the red blood cell, glucose is transported into the cell against its concentration gradient. The energy for this

2 transport is supplied by the hydrolysis of ATP: ATP + H 2 O + 2glucose(out) 2glucose(in) + ADP + P i a. At 25 C, under conditions where [ATP], [ADP], and [P i ] are held constant at M by cell metabolism, find the maximum value of [glucose(in)] [glucose(out)] Assume all activity coefficients are equal to 1. At equilibrium ΔG = Therefore Q eq = K ΔG = ΔG + RT ln Q Q K = e ΔG RT ΔG = ΔG ATP ADP + 2ΔG glucose out glucose in = 31. kj + ΔG = 31 kj K = [ADP][P i][glu(in)] 2 [ATP][glu(out)] 2 K = [1 1 2 M][1 1 2 M] [1 1 2 M] 1 2 [glu(in)] 2 [glu(out)] 2 = e [glu(in)] 2 [glu(out)] 2 = 12 e [glu(in)] 2 [glu(out)] 2 31 kj ( kj )(298 K) K 31 kj ( kj )(298 K) K [glu(in)] 2 = [glu(out)] 2 [glu(in)] = [glu(out)] b. If the stoichiometry of transport were 1 of glucose transported per e of ATP hydrolyzed, what would be the maximum concentration gradient of glucose under the same conditions as part (a)? [glu(in)] = [glu(out)] c. In an actual cell, the glucose inside the cell may have an activity coefficient much less than 1 due to nonideal Page 2 of 12

3 behavior. Would this increase or decrease the maximum concentration gradient obtainable? (Assume that all other activity coefficients are equal to 1.) (Find the maximum concentration ratio of glucose(in)/glucose(out) using the stoichiometry of 4(b) if the activity coefficient for glucose(in) =.72.) γ[glu(in)] = [glu(out)].72[glu(in)] [glu(out)] = [glu(in)] = [glu(out)].72 [glu(in)] = [glu(out)] 8. An important metabolic step is the conversion of fumarate to malate. In aqueous solution, an enzyme (fumarase) allows equilibrium to be attained: fumarate + H 2 O malate Page 3 of 12

4 at 25 C; the equilibrium constant K = (a M /a F ) = 5.. The activity of malate is a M, and the activity of fumarate is a F, defined on the arity concentration scale (a = c in dilute solution). (Do this for compounds X and Y, i.e., replace the word fumarate with X, and replace the word malate with Y. & 8a K eq = 5. at 25 C & 8e K eq = 15. at 4 C) a. What is the standard Gibbs free-energy change for the reaction at 25 C? Δ r G = RT ln K = ( kj ) (298 K) ln(5.) Δ r G = 3.99 kj b. What is the Gibbs free-energy change for the reaction at equilibrium? equilibrium = c. What is the Gibbs free-energy change when 1 of.1 M X is converted to 1 of.1 M Y? Δ r G = Δ r G + RT ln (.1.1 ) = Δ rg Δ r G = 3.99 kj 1 = 3.99 kj d. What is the Gibbs free-energy change when 2 of.1 M X are converted to 2 of.1 M Y? Δ r G = Δ r G + RT ln (.1.1 ) = Δ rg Δ r G = 3.99 kj 2 = 7.98 kj e. If K = 15. at 4 C, calculate the standard enthalpy change for the reaction; assume that the enthalpy is independent of temperature. Δ r H = RT 1T 2 ΔT ln (K 2 K 1 ) = (8.314 kj 1 3 ) (313 K)(298 K) K ln ( K 5 ) Δ r H = 56.8 kj f. Calculate the standard entropy change for the reaction; assume that Δ r S is independent of temperature. Δ r S = Δ rh Δ r G T Δ r G = Δ r H TΔ r S kj kj ( = ) 298 K Δ r S J = 24 K 14 a. You made a ph 6. buffer solution at 25 C by mixing NaOH and histidine (HisH) to give a solution that is.3 M in total concentration of histidine. Calculate the concentrations of all species at 25 C. If this cannot be done using NaOH, use HCl instead. Page 4 of 12

5 [H + ] = 1 ph = M [H + ] = M [OH ] = K w [H + = ] M = M [OH ] = M HisH 3 2+ HisH H + pka = 1.82 K a = HisH 2 + HisH + H + pka = 6. K a = HisH His + H + pka = 9.16 K a = Therefore, [HisH + 2 ][H + ] [HisH 2+ = ] [HisH 2 + ] [HisH 3 2+ ] = = M [HisH][H + ] [HisH 2 + ] [HisH] [HisH 2 + ] [His ][H + ] [HisH] = = M = 1 = [His ] = [HisH] = M [HisH 2 + ] + [HisH] =.3 M [HisH 2 + ] = [HisH] 1 [HisH] + [HisH] =.3 M [HisH] =.15 M [HisH 2 + ] = [HisH] =.15 M [HisH 2 + ] =.15 M [HisH + 2 ] [HisH 2+ = ] Page 5 of 12

6 [HisH 3 2+ ] = M [HisH 3 2+ ] = M [His ] = [HisH] [His ] = M [His ] = M Lastly we need the concentration of Na + We can determine this using charge balance: [Na + ] + ( 1)[His ] + [HisH] + (1)[HisH 2 + ] + (2)[HisH 3 2+ ] + (1)[H + ] + ( 1)[OH ] = [Na + ] = [His ] [HisH 2 + ] 2[HisH 3 2+ ] [H + ] + [OH ] [Na + ] = ( M) (.15 M) 2(1 1 5 ) (1 1 6 M) + ( M) =.15 M This would mean that it cannot be done using NaOH alone. Therefore,.15 of HCl would need to be added. 2. An elastomer, such as a rubber band, maintains a constant volume when it is stretched since the band things, to a first approximation. What thermodynamic quantity represents the maximum non-pv work when a rubber band is stretched at constant temperature? Recall that A = U TS Page 6 of 12

7 Therefore, da = du TdS SdT From the first law of thermodynamics du = dq rev + dw rev Therefore, da = dq rev + dw rev TdS SdT Recall that work does not only have to be pv work but it can also be fdl for a rubber band, where f is the force applied and l is the length. Therefore, dw rev = fdl pdv da = TdS + fdl pdv TdS SdT In a constant T and constant V process da = fdl pdv SdT da = fdl Which indicates that Helmholtz Free Energy represents the maximum non-pv work in an isothermal, isochoric stretching of an elastomer. Supplemental Problems S1. Find the ΔG hydrolysis of fructose-1,6-diphosphate using Table 4.3. Note that G for hydrolysis of ATP is given in the footnote of Table 4.3 fructose-1,6-diphosphate=fdp D-Fructose-6-phosphate=DFP Page 7 of 12

8 DFP + ATP FDP + ADP Δ r G = 14.2 kj FDP + ADP DFP + ATP Δ r G = 14.2 kj ATP + H 2 O ADP + P i Δ r G = 31. kj FDP + H 2 O DFP + P i Δ r G = 23.8 kj kj kj 31. = 16.8 S2. Consider the equilibrium constant for dissociation (K d) ("melting") of the double stranded probe (P) + target DNA (P) at 37 o C: TP --> P + T Page 8 of 12 TP T P +

9 where P (probe) is a small piece of DNA, complementary to a small part of the target (T). (a) Write down the equilibrium constant, K d, for this reaction in terms of [P] and the ratio, r, of free target, [T], and complexed target, [TP]. (r = [T]/[TP] ) This is as trivial as it sounds. K d = [Probe] r (b) Using the sequence 5 - CTCAA-3 for the probe, and Table 4.4, calculate G and K d for the reaction in (a) at 37 o C. (Note: Table 4.4 is for association not dissociation. You will have to convert the numbers to values for dissociation.) 5 C T C A A 3 3 G A G T T 5 C T T C C A A A Δ r G = Δ r G (initiation) + Δ r G ( ) + Δ r G ( ) + Δ r G ( ) + Δ r G ( ) G A A G G T T T Using Table 4.4, The following applies for association: Δ r G association = 8.1 kj 5.4 kj 5.4 kj 6. kj 4.2 kj = 12.9 kj Therefore, for dissociation Also, Δ r G dissociation = Δ r G association = 12.9 K dissociation = e Δ rg dissociation RT kj Or K dissociation = e 12.9 kj kj K dissociation = 1 Δ rg dissociation 2.3RT = 1 K 31 K = kj kj K 31 K = K d = (c) Find the temperature (T m) for which 5% of the target is bound to probe, given that: C probe = [P] + [TP] = 3 x 1-4 M and C target = [T] + [TP] = 2 x 1-7 M. (T m is called the melting temperature ). (Note that [P] C probe because probe is in great excess and only a small fraction of it can bind). We know that the melting temperature is defined as when [T] = [TP]. Notice that r = [T] [TP] so when these two species are at equal concentrations, r = 1. Page 9 of 12

10 ΔH association ΔH association [T] + [TP] = M [T] = [TP] = M 2 [P] + [TP] = M = M [P] = M M M K melt = [P] = M ln K melt K 31 K = ΔH R ( 1 T m 1 31 K ) C T T C C A A A = ΔH (initiation) + ΔH ( ) + ΔH ( ) + ΔH ( ) + ΔH ( ) G A A G G T T T =.8 kj 32.7 kj 34.3 kj 35.6 kj 33.1 kj = kj ΔH dissociation = ΔH association ln = kj kj K = kj 3.11 = K ( 1 T m 1 31 K ) K = ( 1 1 T m 31 K ) K K = 1 T m ( K K ) 1 = T m T m = 293 K ( 1 T m 1 31 K ) S3. The atmosphere closely obeys the Boltzmann distribution. (This math is exactly the same as K=exp(- G /RT) except we use U instead of G ). For this problem K =p(altitude 2)/p(altitude 1), i.e., the ratio of pressures at any two altitudes. Using the Boltzmann distribution, find the atmospheric pressure (bars) in a place whose elevation = 12,345 feet above sea level), assuming T=298 K. Assume the degeneracy (which is the W in S=k bln W) is effectively a constant for altitudes below 1, ft. Use the average ecular mass of the air ecules, which is.2(32) +.8(28) = Page 1 of 12

11 28.8 g/. (remember that the ratio of the pressures is the ratio of concentrations) ft 1 m 3.28 ft = 3763 m p high p low = e mgh RT = e (.288 kg)(9.8 m2 )(3763 m) s J (8.314 p high =.65 bar 1 bar p high =.65 bar K )(298 K) =.65 S4. From the Boltzmann distribution, find the ratio of probabilities to be in the first excited vibrational quantum state compared to being in the lowest (zero-point) state for N 2 and I 2 gases, given that for N 2, the vibrational energy level spacing is about 28.5 kj/e, but for I 2 the energy of the vibrational energy level spacing is only 2.55 kj/. How do your results show that it is reasonable that the vibrational degrees of freedom do not contribute significantly to heat capacity for N 2 but they do for I 2. (The degeneracy of vibrational states = 1). Page 11 of 12

12 (ρ = Probability) N 2 : I 2 : ρ 2 ΔE = e RT = e ρ 1 ρ 2 ΔE = e RT = e ρ kj ( kj 2.55 kj ( kj K )(298 K) = K )(298 K) =.36 Since the spacing is larger for the N 2 energy states than in I 2 there is not sufficient energy at room temperature to promote an electron to the next energy state. For I 2 however, the energy spacing is small enough that we see a more probable chance that an electron is on a higher energy state. Page 12 of 12

The Second Law of Thermodynamics (Chapter 4)

The Second Law of Thermodynamics (Chapter 4) First Law: Energy of universe is constant: ΔE system = - ΔE surroundings Second Law: New variable, S, entropy. Changes in S, ΔS, tell us which processes made