3-3. Glucose is actively transported into red blood cells by coupling the transp

Question

3-3. Glucose is actively transported into red blood cells by coupling the transport with the hydrolysis of ATP. The overall reaction can be written as

ATP+H2O+n Glucose(outside)?n Glucose(inside)+ADP+Pi

If the ratio of [ATP]/[ADP] = 1 and [Pi] = 10mM, what is the concentration gradient, [glucose(inside)]/[glucose(outside)], that is established at 298 K? Cal- culate this ratio for n = 1, n = 2, and n = n. Does this suggest a method for determining the value of n? Use the value of ?Go for the hydrolysis of ATP in Appendix 5 for these calculations.

466 APPENDIX 4: STANDARD GIBBS ENERGIES AND ENTHALPIES OF FORMATION Substance Go (k//mol) APPENDIX 5 679.84 -2242.11 -2247.09 Glycerol -670.48 -677.84 -530.62 523.58 Standard Gibbs Energy and Enthalpy Changes for Biochemical Reactions at 298 K, 1 atm, pH 7.0, pMg 3.0, and 0.25 M lonic Strength 5 Succinate 908.68 -688.28 -525.0 -319.29 -1027.12 -1264.31 1038.10 -2042.40 945.46 982.77 -809.11 -1513.66 Glycine Ribulose Ribose -176.08 -328.28 -426.32 331.13 -1219.22 452.10 -372.16 -120.36 Ribose S-phosphate Glutamate Glutamine Reaction G° (kJ/mol) -959.58 -802.12 -682.83 -633.59 -979.06 32.48 32.80 -30.88 -28.86 ATP + H2O = ADP + Pi ADP +H20 AMP +P AMP+H20 adenosine+P ADP ATP+AMP G6P+H20 Glu+P ATP + Glu = ADP + G6P cis-Aconitate -0.31 -11.61 -20.87 +2.02 Glycerol 3-phosphate Fructose 6-phosphate -1077.14 -1315.74 -1311.89 394.36 Data from R. A. Alberty, Arch. Biochem. Biophys. 353, 116 (1998). -393.51 16.40 0.82 99.13 This table is based on the conventions that G'AH. 0 for the species H. adenosine, NAD, and NADP at zero ionic strength. Data are obtained from R. A. Alberty, Arch. Biochem. Biophys. 353, 116 (1998) Physical Chemistry for the Biological Sciences, Second Edition Gordon G. Hammes and Sharon Hammes-Schiffer 2015 John Wiley & Sons, Inc. Published 2015 by John Wiley & Sons, Inc.

Explanation / Answer

ATP+H2O + n Glucose(outside) n Glucose(inside) + ADP + Pi

Go = -RT ln(Keq)

Go = - 32.48 kJ/mol (From table given)

RT ln(Keq) = 32.48 kJ/mol = 32480 J/mol

Keq = [Glucose(inside)]n[ADP][ Pi]/([ ATP][Glucose(outside)]n)

R = 8.314 J/mol-K

T = 298 K

Ln(Keq) = 32480 / (8.314 * 298) = 13.11

Keq = e13.11 = 493663.3

Keq = [Glucose(inside)]n [ Pi]/[Glucose(outside)]n = 493663.3 ( Given that [ATP]/[ADP] = 1 )

([Glucose(inside)]/[Glucose(outside))n = 493663.3 / (10 * 0.001) = 493663.3 x 102

For n = n

([Glucose(inside)]/[Glucose(outside)) = (493863.3 x 102) 1/n

For n = 1

([Glucose(inside)]/[Glucose(outside)) = 493863.3 x 102

For n = 2

([Glucose(inside)]/[Glucose(outside)) = 7026.1

Value of n can be obtained by known concentrations of Glucose(outside) and Glucose(inside)

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