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? Byhow much?

Given Information: [Fructose-6-phosphate] = 1 mM [Fructose-1,6-bisphosphate] = 10 mM [AMP] = 0.1 mM…

Answered: Intracellular concentrations in resting…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter22: Gluconeogenesis, Glycogen Metabolism, And The Pentose Phosphate Pathway

Section: Chapter Questions

Problem 5P

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Intracellular concentrations in resting muscle are as follows: fructose6-phosphate, 1.0 mM; fructose-1,6-bisphosphate, 10 mM; AMP, 0.1 mM;ADP, 0.5 mM; ATP, 5 mM; and Pi, 10 mM. Is the phosphofructokinasereaction in muscle more or less exergonic than under standard conditions?By how much?
is this stement false? Intracellular concentrations in resting muscle are as follows: Fructose-6-phosphate (1.0 mM)Fructose-(1-6)-bisphosphate (10.0 mM)AMP (0.1 mM)ADP (0.5 mM)ATP (5.0 mM)Pi (10.0 mM)Under the above conditions the Phosphofructokinase reaction in muscle is more exergonic than under standard conditions.
What is the rationale for the inhibition of muscle glycogen phosphorylase by glucose 6- phosphate when glucose 1- phosphate is the product of the phosphorylase reaction?
In muscle tissue, the rate of conversion of glycogen to glucose 6-phosphate is determined by the ratio of phosphorylase a (active) to phosphorylase b (less active). Determine what happens to the rate of glycogen breakdown if a muscle preparation containing glycogen phosphorylase is treated with: (a) phosphorylase kinase and ATP; (b) PP1; (c) epinephrine
Calculate glucose concentration. Na (sodium) and glucose secondary active transport. Na transport (which drive glucose import) G=R*T*In(Na in/ Na out)+Z*F*Y(psi symbol) Na in=14mM Na out=145MM Z=+1 F=96.5 KJ/V*mol Y(psi)=-0.05V What is the glucose in and out concentration? Please be very through when explaining this calculation. (I am stuck at why the energy sign changes from negative to positive when using the calculated energy from sodium to glucose)
Kt of glucose for GLUT1=3 mM, GLUT2=17 mM, GLUT 3=1.3 mM, and GLUT11=0.3 mM. Sketch a graph with [glucose] on the x-axis and rate of transport on the y-axis. Sketch a line for each GLUT transporter above. Which of these GLUT transporters will be working at its maximum rate at [glucose]=5 mM? Which of these GLUT transporters will be working closest to ½ Vmaxwhen [glucose]=5 mM? What will be the approximate rate of glucose transport for each of these transporters if [glucose]=17 mM?
The Vmax of muscle glycogen phosphorylase is much larger than that of the liver enzyme. Discuss the functional signifi cance of this phenomenon.
Heart and muscle cells, where myoglobin resides, maintains an intracellular pO2 of about 2.5 torr. Calculate the fractional saturation (for human myoglobin) if a small change (1 torr) in oxygen partial pressure occurs in either direction (ie what happens at 1.5 torr, what happens at 3.5 torr?) and explain how a small change in oxygen pressure dramatically changes the myoglobin oxygen binding.
The interconverison of DHAP and GAP greatly favors the formation of DHAP at equilibirum. Yet the conversion of DHAP by triose phosphate isomerase proceeds readily. Why?
Intramitochondrial ATP concentrations are about 5 mM, and phosphate concentration is about 10 mM. If ADP is five times more abundant than AMP,calculate the molar concentrations of ADP and AMP at an energy charge of 0.85. Calculate ∆G for ATP hydrolysis at 37 °C under these conditions.The energy charge is the concentration of ATP plus half the concentration of ADP divided by the total adenine nucleotide concentration.
In active muscle cells, the pO2 is about 10 torr at the cell surface and 1 torr at the mitochondria(the organelles where oxidative metabolism occurs). Calculate the percentage of bound oxygentransported to the mitochondria of muscle cells by myoglobin (KD = 2 torr).
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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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?