10. The first reaction in glycolysis is the phosphorylation of glucose:\[ \text{Pi + Glucose} \rightarrow \text{Glucose-6-phosphate} \]This is a thermodynamically unfavorable process with \( \Delta G^\circ = +13.8 \, \text{kJ/mol} \). In a liver cell at 37°C, the concentration of both phosphate and glucose are normally maintained at about 5 mM each. The equilibrium concentration of glucose-6-phosphate according to the above data is:A. \( 1.18 \times 10^{-7} \)B. \( 8.50 \times 10^{-4} \)C. \( 4.25 \times 10^{-5} \)D. \( 2.21 \times 10^{-5} \)E. \( 1.38 \times 10^{-8} \)F. \( 3.02 \times 10^{-3} \)G. \( 6.75 \times 10^{-6} \)H. \( 4.08 \times 10^{-5} \)I. \( 9.15 \times 10^{-6} \)

Glycolysis may be described as an energy-producing process that results in a hexose (glucose)…

Answered: 10. The first reaction in glycolysis is…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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1. The first step in the payoff phase of glycolysis is catalyzed by the enzyme glyceraldehyde 3-phosphate dehydrogenase, an enzyme that contains a nucleophilic cysteine playing a central role in the reaction. A) In the direction of gluconeogenesis, what reaction does this enzyme catalyze? AG° = -6.3 kcal/mol for this reaction in the direction of gluconeogenesis. Based on what you know about the substrates involved, provide two chemical reasons as to why the AGO of this reaction is negative.
Consider the graphic representation shown below for the free energy change for the overall reaction of Fructose-6-phosphate and ATP to form the products Fructose-1,6-biphosphate and ADP in the presence (blue line) or absence (red line) of enzyme. Match each of the arrows (labeled A, B, C, or D) indicated in the graph with an apropriate description by dragging the letters into their corresponding box. C D reactants Fructose-6-phosphate + ATP A products Fructose-1,6-biphosphate + ADP Progress of the Reaction Reset Help D BAc В C Energy of activation for the forward reaction in the absence of enzyme Energy of activation for the forward reaction Energy of activation for the reverse reaction in absence AG for the reaction the presence of enzyme of enzyme Energy
The first reaction in glycolysis is the phosphorylation of glucose to form glucose 6-phosphate: P₁+ glucose glucose 6-phosphate + H₂O This is a thermodynamically unfavorable reaction, with AG° = +13.8 kJ/mol. a) In a liver cell at 37°C the concentrations of both phosphate and glucose are normally maintained at about 5 mM each. What is the equilibrium concentration of glucose 6- phosphate, according to the above reaction? b) This very low concentration of glucose 6-phosphate is unfavorable for glycolysis. In fact, the reaction is coupled to the hydrolysis of ATP: ATP + H₂O2 ADP + P₁ +H* AG°¹ = -32.2 kJ/mol Write the expression for the overall reaction and calculate AG°¹. c) In addition to the glucose and phosphate concentrations listed in part (a) above, assume that the liver cell concentrations of ATP and ADP are 3 mM and 1 mM, respectively. Calculate the theoretical equilibrium concentration of glucose 6-phosphate at pH = 7.4 and 37 °C. C ZOOM +
How much fat (in grams) would the body have to burn to produce the daily minimum requirement of 40 kg ATP from ADP and phosphate? Assume that: 1. The fat is metabolized completely to water and carbon dioxide. 2. The energy that is released can be used entirely for ATP production. 3. Complete oxidation of 1 g of fat to water and CO2 releases 9 kcal or 37 kJ. 4. The Delta G for ATP hydrolysis is -30.5 kJ/mol. You will have to look up one more value online to answer this question, but you do not need to know anything about lipid metabolism. A) approx. 16 to 17 g of fat B) approx. 65 to 66 g of fat C) approx 22 to 23 kg of fat D) approx. 267 to 268 g of fat E) approx. 5 to 6 kg of fat
The reaction below occurs in nearly all human cells. It is catalyzed by the enzyme enolase. CO- CO- enolase -OPO3 OPO3 + H20 ČH,OH AGO'=+1.7 kJ/mol CH2 2PGA PEP Under cellular conditions, the AG is -3.3 kJ/mol. Which of the following could account for these changes in cellular AG compared to the standard biological free energy? The concentration of 2PGA is 7 times higher than PEP. The concentration of PEP is 7 times higher than 2PGA Enolase lowers the activation energy of the transition state. The release of water from 2PGA is a form of energy coupling.
The enzyme triosephosphate isomerase (TIM) catalyzes the following reaction in glycolysis, where it converts dihydroxyacetone phosphate (DHAP) to glyceraldehyde-3-phosphate (GAP). CH₂OH C=O CH₂OPO²- DHAP triose phosphate isomerase [DHAP] (MM 1.00 2.00 3.00 6.00 O V. (mM/s) 3.700 6.727 9.250 14.800 = H HCOH CH₂OPO²- Kinetics experiments were performed on TIM. Enzyme activity (initial velocity, V.) was measured at varying concentrations of DHAP. The enzyme kinetics were also measured in the presence of two inhibitors, A and B. The enzyme concentration used in all experiments was 25.00 μM. The data are shown below. GAP V. (mM/s) + A 1.452 2.794 4.038 7.281 V₁ (mM/s) + B 0.755 1.379 1.905 3.077
Assume that in a certain cell, the ratio of products/reactants or Keg = 809.5 (Keq is dimensionless) for the reaction Glucose + 2ATP > Glucose-1,6-diP + 2ADP, at a particular instant, the concentrations of each compound were Glucose =2.4M, ATP =11.1M, ADP -12.8M and G-6-P -28.4M. Calculate the difference (dimensionless) between Keq and the ratio of products/ractants at this instance, in this cell, to five decimal places
8
Assume that the maximum number of ATPs is produced (38) in glycolysis. At pH 7, and in the presence of excess Mg2+, how much of the energy in one mole of glucose is stored as ATP?
During glycolysis, glucose is converted into fructose-6- phosphate in two successive reactions: glucose + ATP glucose 6-phosphate + ADP glucose 6-phosphate fructose 6-phosphate The AG' for the overall reaction is: +15.0 kJ/mol. -18.4 kJ/mol. -16.7 kJ/mol. → -15.0 kJ/mol. +18.4 kJ/mol. AG¹⁰ = -16.7 kJ/mol AG¹⁰ = +1.7 kJ/mol

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