(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?

Short Answer

Expert verified
a. \(K_{eq} = 2188293465\), b. \([pyruvate]/[PEP] = 2188293465\), c. \(K_{eq} = 2.1\), d. \([PEP]/[pyruvate] = 2.1\), e. Both directions are not likely to be highly favored under physiological conditions.

Step by step solution

01

Calculation of equilibrium constant for the reaction

The standard free energy change (\(\Delta G^{\circ}\)) and the equilibrium constant (\(K_{eq}\)) are related by the formula \(\Delta G^{\circ}= -RT \ln K_{eq}\), where \(R = 8.3145 \, J \, mol^{-1}K^{-1}\) is the gas constant and \(T = 298 \, K\) is the temperature. Rewriting for \(K_{eq}\), we get \(K_{eq}= e^{-\Delta G^{0} / RT}\). Substituting for \(\Delta G^{\circ}\), \(R\), and \(T\), we find that \(K_{eq}= e^{(-31.7 \times 10^{3} J/mol)/ (8.3145 \, J \, mol^{-1}K^{-1} \times 298 \, K)} = 2188293465\).
02

Calculation of the stoichiometric ratio required for the reaction to proceed reversibly

In the situation where the concentrations of ATP and ADP are equal, the reaction equation simplifies to \(PEP \rightleftharpoons\) pyruvate. From the calculated equilibrium constant, the concentration of products over reactants at equilibrium is given by \(K_{eq} = [pyruvate]/[PEP]\). Rewriting gives the necessary ratio as \([pyruvate]/[PEP] = K_{eq} = 2188293465\).
03

Calculation of equilibrium constant for the overall reaction

Following a similar process as in Step 1, using the given \(\Delta G^{\circ}\') for the overall reaction, we find that \(K_{eq}= e^{0.8 \times 10^{3} J / (8.3145 \, J \, mol^{-1}K^{-1} \times 298 \, K)} = 2.1\).
04

Calculation of the concentration ratio at equilibrium

The equilibrium concentrations of ATP, ADP, GTP, and GDP are all given as equal, simplifying the complex reaction to \(PEP \rightleftharpoons pyruvate\). Substituting the equilibrium constant from Step 3, we get the equilibrium concentration ratio as \([PEP]/[pyruvate] = K_{eq} = 2.1\).
05

Final analysis of the reactions

From calculations in Steps 2 and 4, we see that for the reaction to proceed from PEP to pyruvate (glycolysis), the concentration of pyruvate must vastly exceed that of PEP. Similarly, for the reverse reaction (gluconeogenesis), the concentration of PEP must be higher than that of pyruvate. Such significant differences in concentration might not be physiologically probable, therefore both directions of the cycle are not likely to be highly favored under physiological conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Glycolysis
Glycolysis is a fundamental metabolic pathway that converts glucose into pyruvate, generating a small amount of energy in the form of adenosine triphosphate (ATP). This ten-step process occurs in the cytoplasm of cells and is vital for cellular respiration.

During glycolysis, glucose undergoes a series of enzymatic reactions. It is first phosphorylated and eventually split into two three-carbon molecules of glyceraldehyde-3-phosphate. After a series of redox reactions, the final product, pyruvate, is formed alongside a net gain of two ATP molecules and two nicotinamide adenine dinucleotide (NADH) molecules. Pyruvate can then be further metabolized in the mitochondria in the presence of oxygen (aerobic respiration) or convert to lactate in the absence of oxygen (anaerobic respiration).

Glycolysis is tightly regulated by key enzymes such as hexokinase, phosphofructokinase-1, and pyruvate kinase to ensure that the rate of glucose catabolism matches the energy needs of the cell.
Gluconeogenesis
Gluconeogenesis is the metabolic process of generating glucose from non-carbohydrate carbon substrates, essentially the reverse of glycolysis. It occurs primarily in the liver and to a lesser extent in the kidneys.

The process involves the conversion of pyruvate into phosphoenolpyruvate (PEP) through reactions catalyzed by pyruvate carboxylase and PEP carboxykinase, then reversing the glycolytic pathway with some bypasses that overcome the irreversible steps in glycolysis. These bypasses are essential because they help maintain glucose levels during fasting or intense exercise.

Gluconeogenesis is significant because it ensures a continual supply of glucose to tissues that depend on it as their energy source, such as the brain and red blood cells.
Equilibrium Constant
The equilibrium constant, denoted as \( K_{eq} \), is a numerical value that represents the ratio of concentrations of products to reactants at equilibrium for a given chemical reaction. It's derived from the standard free energy change \( \Delta G^{\circ} \) using the equation: \[ \Delta G^{\circ} = -RT \ln{K_{eq}} \]where \( R \) is the ideal gas constant, \( T \) the temperature in Kelvin, and \( \Delta G^{\circ} \) the standard free energy change.

For the reaction PEP to pyruvate, the high equilibrium constant calculated suggests that under standard conditions, the reaction will proceed towards pyruvate formation. However, in the cell where conditions are not standard, other factors such as the concentrations of ATP and ADP will affect the actual direction of the reaction.
Allosteric Regulation
Allosteric regulation is a mechanism by which the activity of an enzyme is modulated through the binding of an effector molecule at a site other than the enzyme's active site, known as the allosteric site. The binding of effectors can either inhibit or activate the enzyme's function, causing conformational changes that affect the enzyme's activity.

Enzymes in the glycolytic pathway and gluconeogenesis, like pyruvate kinase, are subject to allosteric regulation. In the context of the substrate cycle between PEP and pyruvate, regulators like ATP and alanine inhibit pyruvate kinase, slowing down glycolysis, while acetyl-CoA activates pyruvate carboxylase, stimulating gluconeogenesis when energy is abundant.
Free Energy Change
Free energy change, denoted as \( \Delta G \), is an expression of the amount of work a system can perform. When \( \Delta G \) is negative, the process occurs spontaneously. Conversely, a positive \( \Delta G \) indicates the process is non-spontaneous and requires energy.

In the context of biochemical reactions, such as the interconversion of PEP and pyruvate, the standard free energy change (\( \Delta G^{\circ} \)) determines the spontaneity under standard conditions. However, it’s important to note that intra-cellular conditions deviate from standard, and hence actual free energy changes (\( \Delta G \)) can differ, influencing the direction of metabolic processes.
Pyruvate Kinase
Pyruvate kinase is a key glycolytic enzyme catalyzing the final step of glycolysis, in which phosphoenolpyruvate (PEP) is converted to pyruvate with the concomitant production of ATP. This reaction is one of the three irreversible steps in glycolysis.

The activity of pyruvate kinase is regulated by allosteric effectors and covalent modification. Allosteric inhibitors, such as ATP and alanine, can downregulate the enzyme's activity, while fructose-1,6-bisphosphate, a glycolytic intermediate, acts as an allosteric activator to enhance its activity. This regulation ensures the proper balance between energy production and consumption within the cell.
Pyruvate Carboxylase
Pyruvate carboxylase is an enzyme involved in gluconeogenesis that converts pyruvate to oxaloacetate using the energy from ATP hydrolysis. This reaction is also the first step in reversing the final reaction of glycolysis.

Oxaloacetate is then converted to PEP, continuing the gluconeogenic pathway. Pyruvate carboxylase is activated by acetyl-CoA, signaling a surplus of energy and the need for gluconeogenesis to produce glucose for maintaining blood sugar levels or storing energy as glycogen.
PEP Carboxykinase
PEP carboxykinase (PEPCK) is the next enzyme in gluconeogenesis, catalyzing the conversion of oxaloacetate to phosphoenolpyruvate (PEP), augmented by the hydrolysis of GTP.

PEPCK effectively bypasses the irreversible step of pyruvate kinase in glycolysis, providing a pathway for the production of glucose from pyruvate. This enzyme plays a crucial role in maintaining blood glucose levels and is also subject to regulation by various hormones and dietary conditions, reflecting the body's metabolic state.

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Most popular questions from this chapter

(Integrates with Chapter \(23 .\) ) Assuming each NADH is worth 3 ATP, each \(\mathrm{FADH}_{2}\) is worth \(2 \mathrm{ATP}\), and each NADPH is worth \(4 \mathrm{ATP}\) How many ATP equivalents are produced when one molecule of palmitoyl-CoA is oxidized to 8 molecules of acetyl-CoA by the fatty acid \(\beta\) -oxidation pathway? How many ATP equivalents are consumed when 8 molecules of acetyl-CoA are transformed into one molecule of palmitoyl-CoA by the fatty acid biosynthetic pathway? Can both of these metabolic sequences be metabolically favorable at the same time if \(\Delta G\) for ATP synthesis is \(+50 \mathrm{kJ} / \mathrm{mol}\) ?

(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.)

Strenuous muscle exertion (as in the 100 -meter dash) rapidly depletes ATP levels. How long will 8 m \(M\) ATP last if 1 gram of muscle consumes \(300 \mu\) mol of ATP per minute? (Assume muscle is \(70 \%\) water.) Muscle contains phosphocreatine as a reserve of phosphorylation potential. Assuming [phosphocreatine] \(=40 \mathrm{m} M,[\text { creatine }]=4 \mathrm{m} M,\) and \(\left.\Delta G^{\circ \prime} \text { (phosphocreatine }+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \text { creatine }+\mathrm{P}_{\mathrm{i}}\right)=-43.3 \mathrm{kJ} / \mathrm{mol}\) how low must [ATP] become before it can be replenished by the reaction: phosphocreatine \(+\mathrm{ADP} \rightleftharpoons \mathrm{ATP}+\) creatine? [Remember \(\Delta G^{\circ \prime}\) (ATP hydrolysis) \(=-30.5 \mathrm{kJ} / \mathrm{mol} .\)]

Would it be appropriate to call neuropeptide \(Y\) (NPY) the obesitypromoting hormone? What would be the phenotype of a mouse whose melanocortin-producing neurons failed to produce melanocortin? What would be the phenotype of a mouse lacking a functional MC3R gene? What would be the phenotype of a mouse lacking a functional leptin receptor gene?

a. Leptin was discovered when a congenitally obese strain of mice \((o b / o b \text { mice })\) was found to lack both copies of a gene encoding a peptide hormone produced mainly by adipose tissue. The peptide hormone was named leptin. Leptin is an anorexic (appetitesuppressing agent; its absence leads to obesity. Propose an experiment to test these ideas. b. A second strain of obese mice \((d b / d b\) mice ) produces leptin in abundance but fails to respond to it. Assuming the \(d b\) mutation leads to loss of function in a protein, what protein is likely to be nonfunctional or absent? How might you test your idea?

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