Question

Question 1 - Linked Reactions (10 marks) The reaction catalysed by pyruvate kinase is: Phosphoenolpyruvate+ ADPpyruvateATP Keg 3.63 x 105 a) Calculate the AG" for this reaction. Show your working. 3 marks b) The hydrolysis of ATP has following equation: AG 30.5 kJ/mol Calculate the AG" for the following reaction: Phosphoenolpyruvate pyruvateP Show your working. 2 marks c) At 37 °C, the steady-state concentrations of phosphoenolpyruvate, ATP, and ADP have been measured to be 23 HM, 1.85 mM and 140 HM respectively. Given that the AG for the pyruvate kinase reaction is -23.0 kJ/mol under these conditions, what is the steady-state concentration of pyruvate? Show your working. 3 marks d) Based on your answer to (c) do you think the pyruvate kinase reaction reaches equilibrium in a cell? Explain your answer 2 marks The conversion of G6P to fructose-6-phosphate (F6P), catalysed by phosphohexose isomerase (PI), is thermodynamically unfavourable under standard conditions. Glucose-6-phosphate-) Fructose-6-phosphate ΔGor-1.7 kJ/mol a) Using the data below calculate the mass action ratio (Q) and the AG at 37°C for each pair of G6P and F6P concentrations. Present these values in a table. [G6P] (uM) [F6P] (uM) 0.8 3.3 2.1 1.2 2.6 0.1 1.3 2.2 2 marks b) Plot the natural log (In) of Q against your calculated AG values. Remember to clearly label the axes and give the graph a title. Fit an equation to the data to visualise the relationship between InQ and AG. 4 marks c) How can a cell ensure that the reaction catalysed by Pl is G6P F6P and not F6P G6P? Explain your answer. 4 marks The PFK1 enzyme is regulated by a number of factors in the cell, including the level of ATP. PFK1 is a homo- tetramer (four identical subunits) and ATP binds to multiple sites The binding of ATP can be measured using radio-labelled ATP. The data presented below represents the triplicate determination of the binding of ATP to PFK in the absence and presence of 300 HM citrate. Moles of bound ATP per mole enzyme -citrate +citrate [ATPI (HM) Trial 1 Trial 2 Trial 3 Trial 1 Trial 2 Trial 3 10 9.06 9.08 9.12 11.65 11.84 11.78 5 7.71 7.68 7.67 9.48 9.51 9.58 3.33 6.65 6.66 6.71 8.358.338.32 2.5 5.91 5.86 5.88 7.38 7.45 7.41 1.67 4.38 4.37 4.296.636.686.72 1.25 3.75 3.76 3.81 6.26 6.19 6.31 a) Determine the mean and standard deviation of the data for each concentration of ATP for both citrate and +citrate. Show these in a table. 2 marks b) Create a SINGLE double-reciprocal plot to determine the binding parameters for ATP with and without citrate present. Ensure that each dataset is distinct (use alternate colours and/or shapes for the data points) 4 marks c) For each plot give the fitted equation, the value of Kd, and the value of Bmax. Show these in a table. Ensure you include the units of each value. Show your working for the +citrate experiment. 6 marks d) With reference the data you plotted and Kd values you determined, explain the positive and negative aspects of using a double reciprocal plot. 3 marks e) How many binding sites are there for ATP in a single subunit of PFK1? Explain your answer. 2 marks f) What effect might citrate have on the rate of the reaction catalysed by PFK1? Explain your answer. 3 marks

Answer 1

Write the reaction for the standard change in the free energy for the reaction as follows [Concetration of products Concetration ofreactants AG RT In Here, the ratio of concentration of products to the concentration of reactants raised to power as stochiometric coefficients is called as the equilibrium constant. It is denoted as a Consider the reaction catalysed by pyruvate kinase, Phosphoenolpyruvate ADP pyruvate+ATP (1) The equilibrium constant for the pyruvate kinase reaction is K3.63 x10 Write the standard change in the free energy for the above reaction as follows: [Pyruvate][ATP] Phosphoenolpyruvate] AG-RT In ADP Calculate the AG for the reaction catalysed by pyruvate kinase from the following equation as follows: AG -RT In K (2) Here, AG is the standard change temperature, K is the equilibrium constant on free energy, R is the universal gas constant, T is At standard conditions, the temperature is taking as 25 C 25 C(25 c+ 273.15 K Take temperature, universal gas constant value is as 8.314 Jmol K and equilibrium constant is 3.63 x10 So, substitute all the corresponding values in equation (2).

AG -RT In K. -8314 Jmol- Kx298.15 Kxln(3.63x10 1 kJ =-31734 2 J/mol x 1000 J -31.7342 kJ/mol Thus, the value of AG 31.7342 kJ/mol for the reaction is (b) The hydrolysis of ATP reaction is AG-30.5 kJ/mol (3) ATP+H,O ADP+P Consider the given reaction, Phosphoenolpyruvate ADP+P (4) Calculate AG for the above reaction by adding reactions (1) and (3) The AG for reaction (1) is 31.7342 kJ/mol and for reaction (2) is -30.5 kJ/mol (AG") (AG") (AG")-(AG")+(aG"), Phosphoenolpyruvate ADP> pyruvate+A TP -31.7342 kJ/mo -30.5 kJ/mol ATP+H,OADP+P Phosphoenolpyruvate pyruvate+P So, AG for reaction (4) is, (AG)-(AG),+(AG, kJ =-31.7342 mol kl -30.5 mol =-62.2342 kJ Thus, the AG value is -62.2342 mol

(c) At 37 C(310.15 K) The steady state concentration of phosphoenolpyruvate is 23 uM, ATP is 1.85 mM, ADP is 140 uM The AG value for the pyruvate kinase reaction is -23.0 kJ/mol Calculate the steady state concentration of pyruvate from the above conditions as follows: AG AG RTIn Q at non-standard conditions. Here, Q is the reaction quotient AG AG RTInQ + [Pyruvate][ATP] - AG RT In + Phosphoenolpyruvate][AD So, substitute all corresponding values in the above equation, to calculate the concentration of pyruvate. [Pyruvate]ATP] AG AG +RT In [Phosphoenolpyruvate][ADP] J 1 kJ -x310.15 K 8.314 1000 J mol K kJ -31.7342 kJ/mol -23.0 [Pyruvate]1.85 mM] xln mol [23 HM140 LM kJ +31.7342 kJ/mol mol -23.0 [Pyruvate]1.85 mM] 23 uM140 LM n 1 kJ x310.15 K T 8.314 x 1000 J mol K -3.3872 Then

Pyruvate]1 85 mM] [23 μΜ140 μΜ) 3572 [ 3 μΜ] [140 μ ] Pyruvate 29.58x 1.85 mMx1000 M 1mM 51.49 m Thus, the steady state concentration of pyruvate is 51.49 um (d) Calculate the reaction quotient based on data from question (c) [Pyruvate][ATP] Q= [Phosphoenolpyruvate] [ADP] 1.85 mMx1000 HM mM 51.49 [23 jum][140 um] L -29.58 Thus, the reaction quotient is 29.58. The equilibrium quotient for the pyruvate kinase reaction is 29.58. constant of the pyruvate kinase reaction is 3.63x10 and the reaction To check the reaction reaches equilibrium equilibrium constant or not compare the reaction quotient with the The conditions for equilibrium constant are, 1. At Q-Kg, the system at equilibrium so, the reaction has no shift in either the left or the right 2. At Q<K, the system will reach equilibrium so, the reaction shift towards right. At Q> K, the system not at equilibrium so, the reaction shift towards left. 3. So, in the pyruvate kinase reaction the reaction quotient is less than the equilibrium constant

The reaction shift towards right side means shift towards product side. The reaction will reach at equilibrium Thus, Yes, the pyruvate kinase reaction reaches equilibrum in a cell