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Chapter 15: Problem 1

Measuring the concentrations of metabolic inter-mediates in a living cell presents great experimental difficulties-usually a cell must be destroyed before metabolite concentrations can be measured. Yet enzymes catalyze metabolic interconversions very rapidly, so a common problem associated with these types of measurements is that the findings reflect not the physiological concentrations of metabolites but the equilibrium concentrations. A reliable experimental technique requires all enzyme-catalyzed reactions to be instantaneously stopped in the intact tissue so that the metabolic intermediates do not undergo change. This objective is accomplished by rapidly compressing the tissue between large aluminum plates cooled with liquid nitrogen \(\left(-190^{\circ} \mathrm{C}\right),\) a process called freeze- clamping. After freezing, which stops enzyme action instantly, the tissue is powdered and the enzymes are inactivated by precipitation with perchloric acid. The precipitate is removed by centrifu-gation, and the clear supernatant extract is analyzed for metabolites. To calculate intracellular concentrations, the intracellular volume is determined from the total water content of the tissue and a measurement of the extracellular volume. The intracellular concentrations of the substrates and products of the phosphofructokinase- 1 reaction in isolated rat heart tissue are given in the table below. $$\begin{array}{lc} \text { Metabolite } & \text { Concentration }(\mu \mathrm{M})^{*} \\ \hline \text { Fructose } 6 \text { -phosphate } & 87.0 \\ \text { Fructose } 1,6 \text { -bisphosphate } & 22.0 \\ \mathrm{ATP} & 11,400 \\ \mathrm{ADP} & 1,320 \\ \hline \end{array}$$ (a) Calculate \(Q, \text { [fructose } 1,6 \text { -bisphosphate }][\mathrm{ADP}] /\) [fructose \(6 \text { -phosphate }][\mathrm{ATP}],\) for the PFK-1 reaction under physiological conditions. (b) Given a \(\Delta G^{\prime \circ}\) for the PFK-1 reaction of \(-14.2 \mathrm{kJ} / \mathrm{mol}\) calculate the equilibrium constant for this reaction. (c) Compare the values of \(Q\) and \(K_{\mathrm{eq}^{\prime}}^{\prime}\) Is the physiological reaction near or far from equilibrium? Explain. What does this experiment suggest about the role of PFK-1 as a regulatory enzyme?

Short Answer

Expert verified
The physiological reaction is far from equilibrium, indicating PFK-1 as a regulatory enzyme.

Step by step solution

01

Calculate the Reaction Quotient, Q

The reaction quotient \( Q \) for the phosphofructokinase-1 (PFK-1) reaction can be calculated using the concentrations provided. The formula for \( Q \) is: \[Q = \frac{[\text{Fructose 1,6-bisphosphate}][\text{ADP}]}{[\text{Fructose 6-phosphate}][\text{ATP}]}.\] Substituting the given concentrations, we have: \[Q = \frac{(22.0 \ \mu M)(1320 \ \mu M)}{(87.0 \ \mu M)(11400 \ \mu M)}.\] This simplifies to: \[Q = \frac{29040}{991800}.\] Finally, we calculate \( Q \) to get approximately \( 0.029 \).
02

Calculate the Equilibrium Constant, K_eq'

Using the formula for the equilibrium constant \( K_\mathrm{eq}^{\prime} \) which is related to the standard Gibbs free energy change \( \Delta G^{\prime \circ} \): \[ \Delta G^{\prime \circ} = -RT \ln K_{\mathrm{eq}^{\prime}}, \] where \( R \) is the gas constant \( 8.314 \ \mathrm{J} / \mathrm{mol} \cdot \mathrm{K} \) and \( T \) is the temperature in Kelvin (assuming physiological temperature of 298 K). Rearranging the equation: \[ K_{\mathrm{eq}^{\prime}} = e^{-\Delta G^{\prime \circ} / RT}. \] Substituting the values, \[ K_{\mathrm{eq}^{\prime}} = e^{-(-14200 \ \mathrm{J/mol})/(8.314 \ \times 298)}. \] This yields \( K_{\mathrm{eq}^{\prime}} \approx 165.59 \).
03

Compare Q and K_eq'

With \( Q \approx 0.029 \) and \( K_{\mathrm{eq}^{\prime}} \approx 165.59 \), we compare these values. Since \( Q \) is much less than \( K_{\mathrm{eq}^{\prime}} \), it indicates that the reaction is far from equilibrium under physiological conditions. This large difference suggests that PFK-1 acts as a regulatory enzyme that controls the rate of the glycolytic pathway.

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Key Concepts

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

Enzyme-Catalyzed Reactions
Understanding enzyme-catalyzed reactions is crucial in studying cellular metabolism. Enzymes are biological catalysts that speed up reactions by lowering the activation energy needed. A unique attribute of enzymes is their specificity鈥攖hey typically catalyze only one type of chemical reaction. Here's why enzyme-catalyzed reactions are significant:
  • Speed: Enzymes can accelerate reactions up to millions of times faster than if the reaction occurred without them.
  • Control: Enzymes are regulated by the cell to ensure that reactions occur as needed, maintaining metabolic balance.
  • Energy Efficiency: By lowering activation energy requirements, enzymes make reactions more energy-efficient, crucial for maintaining life processes.

In the context of metabolism, enzymes like phosphofructokinase-1 (PFK-1) play a pivotal role in regulating pathways such as glycolysis. They ensure that cells efficiently convert substrates into necessary products, supporting vital cellular functions. Without enzyme-catalyzed reactions, life as we know it would not be feasible due to the slow pace of natural chemical processes.
Reaction Equilibrium Constant
The reaction equilibrium constant, denoted as \( K_{eq}' \), is an essential concept in understanding chemical reactions within a biological context. It indicates the ratio of concentrations of products to reactants at equilibrium. Here鈥檚 how it functions:
  • Direction: A high \( K_{eq}' \) suggests that a reaction strongly favors the formation of products, while a low \( K_{eq}' \) means reactants are favored.
  • Gibbs Free Energy: The relationship between \( K_{eq}' \) and Gibbs free energy \( \Delta G^{\prime \circ} \) is vital as it predicts reaction spontaneity. A negative \( \Delta G^{\prime \circ} \) indicates that the reaction is spontaneous under standard conditions.

For the PFK-1 reaction, an equilibrium constant calculation reveals insights into how the reaction behaves under physiological conditions. Comparing the equilibrium constant to the reaction quotient \( Q \) can further uncover whether a reaction is at, near, or far from equilibrium. This comparison plays a critical role in understanding the regulatory mechanisms within cells.
Phosphofructokinase-1 (PFK-1)
Phosphofructokinase-1 (PFK-1) is a key regulatory enzyme in the glycolysis pathway, which is the process that breaks down glucose to release energy. Here are some important aspects of PFK-1:
  • Regulation: PFK-1 is allosterically regulated by various molecules. High levels of ATP can inhibit PFK-1, signaling that the cell has sufficient energy, whereas AMP can activate it, indicating a need for energy production.
  • Rate Limiting Step: PFK-1 catalyzes one of the rate-limiting steps of glycolysis鈥攃onverting fructose 6-phosphate into fructose 1,6-bisphosphate.
  • Physiological Role: Due to its strategic position in the glycolytic pathway, PFK-1 serves as a metabolic checkpoint, where the cell decides to either proceed with further glucose breakdown or divert it to other pathways.

The action of PFK-1 is crucial, particularly because its activity determines the flux of metabolites through the glycolysis pathway. Moreover, its regulation affects how cells respond to the energy demands of various physiological conditions. Understanding PFK-1's role helps explain its importance in energy metabolism within cells.

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Most popular questions from this chapter

Muscle cells need rapid access to large amounts of glucose during heavy exercise. This glucose is stored in liver and skeletal muscle in polymeric form as particles of \(g\) lycogen. The typical glycogen particle contains about 55,000 glucose residues (see Fig. \(15-35\) b) Mel茅ndez-Hevia, Waddell, and Shelton (1993) explored some theoretical aspects of the structure of glycogen, as described in this problem. (a) The cellular concentration of glycogen in liver is about \(0.01 \mu\) M. What cellular concentration of free glucose would be required to store an equivalent amount of glucose? Why would this concentration of free glucose present a problem for the cell? Glucose is released from glycogen by glycogen phosphorylase, an enzyme that can remove glucose molecules, one at a time, from one end of a glycogen chain. Glycogen chains are branched (see Figs \(15-28\) and \(15-35 \mathrm{b}\) ), and the degree of branching - the number of branches per chain-has a powerful influence on the rate at which glycogen phosphorylase can release glucose. (b) Why would a degree of branching that was too low (i.e., below an optimum level) reduce the rate of glucose release? (Hint: Consider the extreme case of no branches in a chain of 55,000 glucose residues.) (c) Why would a degree of branching that was too high also reduce the rate of glucose release? (Hint: Think of the physical constraints.) Mel茅ndez-Hevia and colleagues did a series of calculations and found that two branches per chain (see Fig. \(15-35\) b) was optimal for the constraints described above. This is what is found in glycogen stored in muscle and liver. To determine the optimum number of glucose residues per chain, Mel茅ndez-Hevia and coauthors considered two key parameters that define the structure of a glycogen particle: \(t=\) the number of tiers of glucose chains in a particle (the molecule in Fig. \(15-35 \text { b has five tiers }) ; g_{\mathrm{c}}=\) the number of glucose residues in each chain. They set out to find the values of \(t\) and \(g_{\mathrm{c}}\) that would maximize three quantities: (1) the amount of glucose stored in the particle \(\left(G_{\mathrm{T}}\right)\) per unit volume; (2) the number of unbranched glucose chains \(\left(C_{\mathrm{N}}\right)\) per unit volume (i.e., number of chains in the outermost tier, readily accessible to glycogen phosphorylase); and (3) the amount of glucose available to phosphorylase in these unbranched chains \(\left(G_{\mathrm{PT}}\right)\). (d) Show that \(C_{A}=2^{l-1}\). This is the number of chains available to glycogen phosphorylase before the action of the debranching enzyme. (e) Show that \(C_{\mathrm{T}},\) the total number of chains in the particle, is given by \(C_{\mathrm{T}}=2^{t}-1 .\) Thus \(G_{\mathrm{T}}=g_{\mathrm{c}}\left(\mathrm{C}_{\mathrm{T}}\right)=g_{\mathrm{c}}\left(2^{t}-1\right),\) the total number of glucose residues in the particle. (f) Glycogen phosphorylase cannot remove glucose from glycogen chains that are shorter than five glucose residues. Show that \(G_{\mathrm{PT}}=\left(g_{\mathrm{c}}-4\right)\left(2^{t-1}\right) .\) This is the amount of glucose readily available to glycogen phosphorylase. (g) Based on the size of a glucose residue and the location of branches, the thickness of one tier of glycogen is \(0.12 g_{\mathrm{c}} \mathrm{nm}+0.35 \mathrm{nm} .\) Show that the volume of a particle, \(V_{\mathrm{s}}\) is given by the equation \(V_{\mathrm{s}}=\frac{4}{3} \pi t^{3}\left(0.12 g_{\mathrm{c}}+0.35\right)^{3} \mathrm{nm}^{3}\) Mel茅ndez-Hevia and coauthors then determined the optimum values of \(t\) and \(g_{\mathrm{c}}-\) those that gave the maximum value of a quality function, \(f,\) that maximizes \(G_{\mathrm{T}}, C_{\mathrm{A}},\) and \(G_{\mathrm{PT}},\) while minimizing \(V_{\mathrm{s}}: f=\frac{G_{\mathrm{T}} C_{\mathrm{A}} G_{\mathrm{PT}}}{V_{\mathrm{S}}} .\) They found that the optimum value of \(g_{\mathrm{e}}\) is independent of \(t\) (h) Choose a value of \(t\) between 5 and 15 and find the optimum value of \(g_{c}\). How does this compare with the \(g_{c}\) found in liver glycogen (see Fig. \(15-35\) b) \(?\) (Hint: You may find it useful to use a spreadsheet program.)

The \(V_{\max }\) of the glycogen phosphorylase from skeletal muscle is much greater than the \(V_{\max }\) of the same enzyme from liver tissue. (a) What is the physiological function of glycogen phosphorylase in skeletal muscle? In liver tissue? (b) Why does the \(V_{\max }\) of the muscle enzyme need to be greater than that of the liver enzyme?

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) \(\mathrm{PP} 1\) (c) epinephrine.

Between your evening meal and breakfast, your blood glucose drops and your liver becomes a net producer rather than consumer of glucose. Describe the hormonal basis for this switch, and explain how the hormonal change triggers glucose production by the liver.

A man with insulin-dependent diabetes is brought to the emergency room in a near-comatose state. While vacationing in an isolated place, he lost his insulin medication and has not taken any insulin for two days. (a) For each tissue listed below, is each pathway faster, slower, or unchanged in this patient, compared with the normal level when he is getting appropriate amounts of insulin? (b) For each pathway, describe at least one control mechanism responsible for the change you predict. Tissue and Pathways 1\. Adipose: fatty acid synthesis 2\. Muscle: glycolysis; fatty acid synthesis; glycogen synthesis 3\. Liver: glycolysis; gluconeogenesis; glycogen synthesis; fatty acid synthesis; pentose phosphate pathway
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