Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Question
Chapter 27, Problem 3P
Interpretation Introduction
Interpretation:
The time required to last 8 mM of ATP when ATP consumed by muscle is
Concept introduction:
The relation between
Here, R is Universal gas constant, T is temperature,
The reaction is spontaneous if the value of change in Gibbs free energy is negative.
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Intracellular concentrations in resting muscle are as follows: fructose-
6-phosphate, 1.0 mM; fructose-1,6-bisphosphate, 10 mM; AMP, 0.1 mM;
ADP, 0.5 mM; ATP, 5 mM; and P, 10 mM. Is the phosphofructokinase reac-
tion in muscle more or less exergonic than under standard conditions? By
how much?
The sarcoplasmic reticulum Ca2+-ATPase, pumps 2 mol Ca2+ out of sarcomeres per mol ATP hydrolyzed.(a) Given the following steady-state concentrations and a membrane potential of 65 mV (inside negative), calculate ΔG for the following activetransport process at 37 °C and pH = 7.4:2Ca2+(in) + ATP + H2O → 2Ca2+(out) + ADP + Pi + H+ATP = 2.6 mM, ADP = 210 μM, Pi = 5.1 mM, Ca2+(in) = 32 mM,Ca2+(out) = 2.2 mM(b) In active muscle the pH can drop below 7.4. Is the reaction above moreor less favorable under these conditions?(c) The activity of the Ca2+-ATPase is regulated reversibly under normalconditions to maintain homeostatic concentrations of Ca2+ inside thesarcomere. However, in a rare genetic disorder, irreversible activation ofthe Ca2+ -ATPase can occur. Assuming 37 °C, pH = 7.4, and the steadystate concentrations for ATP, ADP Pi, and Ca2 +(out) given in part (a),calculate the minimum [Ca2 +] inside a sarcomere that has irreversibly activated Ca2 +-ATPase (i.e., the Ca2 +-ATPase…
a) (1)
Calculate the physiological AG (not AG.) for the reaction:
Phosphocreatine + ADP - creatine + ATP
Given;
Phosphocreatine + H;0 - creatine + Pi
ADP + Pi → ATP + H;0
AG.' -43 kJ/mol
AG.- +30.5 kJ/mol
at 25°C as it occurs in the cytosol of neurons, in which phosphocreatine is present at
4.7 mM, creatine at 1.0 mM, ADP at 0.20 mM, and ATP at 2.6 mM.
(R = 8.315 JK-' mol-)
(ii)
Caleulate the free energy change at standard conditions for the following reaction:
Acetaldehyde + NADH + H* + Ethanol + NAD*
The half- reactions are:
Acetaldehyde + 2H + 2e +
Ethanol
E°- - 0.20V
NAD-+ 2H- + 2e ++
NADH + H-
E=-0.32V
(F= 96.485 kJ/V/mol)
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