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Chapter 27: Problem 6

(Integrates with Chapters \(18 \text { and } 22 .)\) The reactions catalyzed by PFK and FBPase constitute another substrate cycle. PFK is AMP activated; FBPase is AMP inhibited. In muscle, the maximal activity of PFK (mmol of substrate transformed per minute) is ten times greater than FBPase activity. If the increase in [AMP] described in problem 5 raised PFK activity from \(10 \%\) to \(90 \%\) of its maximal value but lowered FBPase activity from \(90 \%\) to \(10 \%\) of its maximal value, by what factor is the flux of fructose- 6 - \(P\) through the glycolytic pathway changed? (Hint: Let PFK maximal activity = 10, FBPase maximal activity \(=1 ;\) calculate the relative activities of the two enzymes at low \([\mathrm{AMP}]\) and at high \([\mathrm{AMP}] ;\) let \(J,\) the flux of \(\mathrm{F}\) - 6 -P through the substrate cycle under any condition, equal the velocity of the PFK reaction minus the velocity of the FBPase reaction.)

Short Answer

Expert verified
The flux of fructose-6-P through the glycolytic pathway changes by a factor of 89.

Step by step solution

01

Calculate Relative Enzyme Activity at Low AMP

The relative activities of PFK and FBPase at low AMP can be calculated using the given percentages. For PFK, the activity is given as 10% of its maximal activity, which is 10, so the activity at low AMP is \(0.1 \times 10 = 1\). For FBPase, the activity is given as 90% of its maximal activity, which is 1, so the activity at low AMP is \(0.9 \times 1 = 0.9\).
02

Calculate Relative Enzyme Activity at High AMP

Similarly, the relative activities of PFK and FBPase at high AMP can be calculated using the given percentages. For PFK, the activity is given as 90% of its maximal activity, so the activity at high AMP is \(0.9 \times 10 = 9\). For FBPase, the activity is given as 10% of its maximal activity, so the activity at high AMP is \(0.1 \times 1 = 0.1\).
03

Calculate Flux under Low and High AMP

The flux of fructose-6-P through the substrate cycle is given by the equation \(J = \text{PFK activity} - \text{FBPase activity}\). Therefore, the flux at low AMP is \(1 - 0.9 = 0.1\) and the flux at high AMP is \(9 - 0.1 = 8.9\).
04

Calculate Change in Flux

The change in flux due to the increase in AMP can be found by dividing the flux at high AMP by the flux at low AMP. Therefore, the change in flux is \(8.9 / 0.1 = 89\). This means that the flux of fructose-6-P through the glycolytic pathway is increased by a factor of 89 when the concentration of AMP is increased.

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Most popular questions from this chapter

(Integrates with Chapters 3,18 , and \(22 .\) ) The conversion of PEP to pyruvate by pyruvate kinase (glycolysis) and the reverse reaction to form PEP from pyruvate by pyruvate carboxylase and PEP carboxykinase (gluconeogenesis) represent a so-called substrate cycle. The direction of net conversion is determined by the relative concentrations of allosteric regulators that exert kinetic control over pyruvate kinase, pyruvate carboxylase, and PEP carboxykinase. Recall that the last step in glycolysis is catalyzed by pyruvate kinase: \(P E P+A D P \rightleftharpoons\) pyruvate \(+\) ATP The standard free energy change is \(-31.7 \mathrm{kJ} / \mathrm{mol}\). a. Calculate the equilibrium constant for this reaction. b. If \([\mathrm{ATP}]=[\mathrm{ADP}],\) by what factor must [pyruvate] exceed [PEP] for this reaction to proceed in the reverse direction? The reversal of this reaction in eukaryotic cells is essential to gluconeogenesis and proceeds in two steps, each requiring an equivalent of nucleoside triphosphate energy: c. The \(\Delta G^{\circ}\) ' for the overall reaction is \(+0.8 \mathrm{kJ} /\) mol. What is the value of \(K_{\mathrm{eq}} ?\) d. Assuming [ATP] = [ADP], [GTP] = [GDP], and Pi \(=1 \mathrm{m} M\) when this reaction reaches equilibrium, what is the ratio of \([\mathrm{PEP}] /[\text { pyruvate }]\) e. Are both directions in the substrate cycle likely to be strongly favored under physiological conditions?

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The Human Biochemistry box, The Metabolic Effects of Alcohol Consumption, points out that ethanol is metabolized to acetate in the liver by alcohol dehydrogenase and aldehyde dehydrogenase: $$\begin{array}{r} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{NAD}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{CHO}+\mathrm{NADH}+\mathrm{H}^{+} \\\ \mathrm{CH}_{3} \mathrm{CHO}+\mathrm{NAD}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{NADH}+2 \mathrm{H}^{+} \end{array}$$ These reactions alter the NAD \(^{+} /\) NADH ratio in liver cells. From your knowledge of glycolysis, gluconeogenesis, and fatty acid oxidation, what might be the effect of an altered \(\mathrm{NAD}^{+} / \mathrm{NADH}\) ratio on these pathways? What is the basis of this effect?
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