Question

CHEM 351 Biochemistry I - Fall 2018 Study Problems for Metabolism and Glucose Catabolism For the reaction: aA + bB =cC + dD Gibbs free energy: AG = AG" + RT IN THE glucose + glucose 6-phosphate + H20 1. The equilibrium constant for phosphorylation of glucose is: Ke = 3.97 x 10- a. Calculate the standard Gibbs free energy for the uncatalyzed reaction. b. Use the physiological concentrations for glucose (2.0 mm) and inorganic phosphate (20 mM) to determine what would be the equilibrium concentration of glucose 6-phosphate in the absence of hexokinase at 298 K. glucose + ATP glucose 6-phosphate + ADP + H+ 2. Hexokinase catalyzes the phosphorylation of glucose by ATP, above, with the equilibrium constant: Kg - 1.32 x 10 a. Calculate the standard Gibbs free energy for the catalyzed reaction. b. Use the intracellular concentration of glucose 6-phosphate (2.0 mm) and ratio of ATPIADP = 7.0, to find the equilibrium concentration of free glucose at 298 K and pH 7.0. Note, under standard biochemical conditions the activity of (H') equals 1. ATP + H20 = ADP+P 3. The free energy released by hydrolysis of ATP is sometimes misleadingly attributed to breaking the "high energy phosphoanhydride bond. a. Which bonds are actually broken and which new bonds are formed in the hydrolysis of ATP? b. What is the net enthalpy change, AH, based on bond enthalples? Given the bond enthalphies for O-P 502 kJ/mol and H-O 460 kJ/mol. c. What is the source of the large negative free energy change for ATP hydrolysis? AG" = -30.5 kg-mol

Answer 1

AG ZAG' + R1 ly terroid и і ві z Rt luk + RT ln teic EDTA [+]" в glease + Pi ② glucose 6-phosphale thao 2. О м омм Taj Кі, ја ја 0 -ху 2 3. 43 х 3 14- х 1/2 - 1 . 14-х јте - x][зят моз) 2) до о. % x 20 - 2) 44" - Rт и 1 41 103 2 1.1 kJ / ил | glease + ATP ~ glucose 6-phosphate + ADP + Ht ke 1.32 xio Д - Кт и к 1 - 314 х 21с ҳ, 3.32 x 3 -11, ся 1 м a Glut ATP - ghn-6-p + ADP + H² х 2 . v IM 2. м

2mm 31.3.2.8103 Yequ? Tx-27 X 7. 2 1:32X10377 - 12-2 a 2. 2.0002 1 3) In the hydrolysie of ATP P-0 bond. breaks down to generate free phosphat ion 6) AH 2 tlp - HR Exot 2 (460 - 502) kil mol 2-42 KT/mol
The source of the large negative free energy for the hydrolysis of ATP is due to three phosphate units destabilize the molecule and makes it of higher energy , due to acidic , basic property of the molecules they react considerably giving large negative value for free energy .