/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 A direct measurement of the stan... [FREE SOLUTION] | 91Ó°ÊÓ

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A direct measurement of the standard free-energy change associated with the hydrolysis of ATP is technically demanding because the minute amount of ATP remaining at equilibrium is difficult to measure accurately. The value of \(\Delta G^{\prime \circ}\) can be calculated indirectly, however, from the equilibrium constants of two other enzymatic reactions having less favorable equilibrium constants: \(\begin{array}{ll}\text { Glucose } 6 \text { -phosphate }+\mathrm{H}_{2} \mathrm{O} \longrightarrow \text { glucose }+\mathrm{P}_{\mathrm{i}} & K_{\mathrm{eq}}^{\prime}=270 \\ \text { ATP }+\text { glucose } \longrightarrow \mathrm{ADP}+\text { glucose } 6 \text { -phosphate } & K_{\mathrm{eq}}^{\prime}=890\end{array}\) Using this information for equilibrium constants determined at \(25^{\circ} \mathrm{C},\) calculate the standard free energy of hydrolysis of ATP.

Short Answer

Expert verified
The standard free energy change for ATP hydrolysis is approximately 7.55 kJ/mol.

Step by step solution

01

Understanding the Problem

We need to calculate the standard free energy change (\(\Delta G^{\prime \circ}\)) of ATP hydrolysis using the equilibrium constants from two other reactions. These reactions are given with their equilibrium constants.
02

Write the Equations and Relationships

We are given two reactions and their equilibrium constants: \[ \text{Glucose 6-phosphate} + \mathrm{H}_2\mathrm{O} \rightarrow \text{glucose} + \mathrm{P}_i \quad K_1^{\prime} = 270 \] and \[ \text{ATP} + \text{glucose} \rightarrow \mathrm{ADP} + \text{glucose 6-phosphate} \quad K_2^{\prime} = 890 \]. The overall reaction of interest, ATP hydrolysis, is: \[ \text{ATP} + \mathrm{H}_2\mathrm{O} \rightarrow \text{ADP} + \mathrm{P}_i \].
03

Combining the Reactions

The conversion of the given reactions to the desired reaction involves combining them in a way to end with the ATP hydrolysis. This means adding the equations such that the intermediates (glucose and glucose 6-phosphate) cancel out. Effectively, we need to add equation 1 and reverse equation 2.
04

Calculating the Equilibrium Constant for ATP Hydrolysis

Since we are summing the reactions and reversing one, the equilibrium constant for the resultant reaction (ATP hydrolysis) is given by \( K_3^{\prime} = \frac{K_1^{\prime}}{K_2^{\prime}} \). Substituting the values, \[ K_3^{\prime} = \frac{270}{890} \approx 0.303 \].
05

Calculating \( \Delta G^{\prime \circ} \) for ATP Hydrolysis

The standard free energy change is calculated using the relationship \( \Delta G^{\prime \circ} = -RT \ln K^{\prime} \). Substituting \(R = 8.314 \ J/mol \cdot K\), \(T = 298 \ K\) (which corresponds to \(25^{\circ}C\)), and \(K_3^{\prime} = 0.303\), we find: \[ \Delta G^{\prime \circ} = -8.314 \times 298 \times \ln(0.303) \approx 7.55 \ kJ/mol \].

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Key Concepts

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

ATP Hydrolysis
ATP hydrolysis is a vital biochemical reaction that involves the breakdown of adenosine triphosphate (ATP) into adenosine diphosphate (ADP) and an inorganic phosphate group (\( \text{P}_i \)). This process releases energy that cells use to perform various functions, such as muscle contraction, nerve impulse propagation, and active transport across membranes.

The hydrolysis of ATP is considered highly exergonic, meaning it releases a significant amount of energy. The standard free energy change (\( \Delta G^{\prime \circ} \)) is commonly negative, reflecting the spontaneous nature of the reaction under standard conditions (1 M concentrations, 25°C, and 1 atm pressure). ATP itself is remarkable in its capacity to donate a phosphate group thanks to the high-energy phosphate bonds in its structure. These bonds are not inherently unstable but release a lot of energy upon breaking, which is harnessed for cellular tasks.

Understanding ATP hydrolysis is fundamental in biochemistry because it is so closely linked to energy transfer in biological systems.
Equilibrium Constant
The equilibrium constant (\( K_{eq}^{\prime} \)) is a critical concept in understanding chemical reactions, indicating the ratio of product concentrations to reactant concentrations at equilibrium. In biochemical terms, it helps understand how much of any given reactant will be transformed into product at equilibrium.

For an ATP hydrolysis reaction, the equilibrium constant informs us about the balance between ATP and its hydrolysis products (ADP and \( \text{P}_i \)).
The value of \( K_{eq}^{\prime} \) is dependent on temperature and ionic strength, but at 25°C, biochemical standard conditions include 1 M concentration of products and reactants.

Using equilibrium constants from other known reactions can help calculate values for more challenging reactions, like ATP hydrolysis, using manipulation of these constants. The key relationship for standard free energy change \( \Delta G^{\prime \circ} \) is \( \Delta G^{\prime \circ} = -RT \ln K^{\prime} \), where \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
Thermodynamics
Thermodynamics in biochemistry refers to the study of energy transformation in biological processes. The focus is often on how organisms extract energy from their environment to drive metabolic processes. One cornerstone concept is the `free energy change`, which determines the spontaneity of a reaction.

For a process such as ATP hydrolysis, the standard free energy change (\( \Delta G^{\prime \circ} \)) signifies energy released for use by a cell.
This parameter would indicate a negative value in exergonic processes, which provide the energy that powers cellular work.
Total energy exchanges and availability are encapsulated in the principles of free energy, enthalpy (\( \Delta H \), the total heat content), and entropy (\( \Delta S \), the degree of disorder).

Thermodynamics in biology considers the efficiency and regulation of these energy transformations, pivotal for understanding how cells maintain order and function. ATP is central to these processes, often referred to as the "energy currency" of the cell. It serves as a way for cells to store and release energy efficiently.
Biochemistry Education
Biochemistry education plays an essential role in helping students understand the molecular mechanisms underpinning life. It bridges biology and chemistry, providing insights into how biological systems operate at a chemical level.

A knowledge of ATP hydrolysis, equilibrium constants, and thermodynamic principles forms the backbone of biochemistry curriculums. These concepts are crucial for comprehending metabolic pathways, enzymatic activity, and energy transformations.

Educational approaches should include more than just textbook problems; they should incorporate interactive learning experiences, laboratory work, and real-world examples. This way, abstract concepts become tangible, and students can visualize how pathways interconnect within cells.

Biochemistry education also often tackles the challenge of integrating these complex systems in a manner accessible to all learning stages, ensuring a strong foundation for future study or careers in biological sciences.

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Most popular questions from this chapter

(a) A total of \(30.5 \mathrm{kJ} / \mathrm{mol}\) of free energy is needed to synthesize ATP from ADP and \(\mathrm{P}_{\mathrm{i}}\) when the reactants and products are at \(1 \mathrm{M}\) concentrations and the temperature is \(25^{\circ} \mathrm{C}\) (standard state \() .\) Because the actual physiological concentrations of \(\mathrm{ATP}, \mathrm{ADP},\) and \(\mathrm{P}_{\mathrm{i}}\) are not \(1 \mathrm{M},\) and the temperature is \(37^{\circ} \mathrm{C},\) the free energy required to synthesize ATP under physiological conditions is different from \(\Delta G^{\prime \circ}\). Calculate the free energy required to synthesize ATP in the human hepatocyte when the physiological concentrations of ATP, ADP, and \(\mathrm{P}_{\mathrm{i}}\) are 3.5,1.50 and \(5.0 \mathrm{mM},\) respectively. (b) \(\mathrm{A} 68 \mathrm{kg}(150 \mathrm{lb})\) adult requires a caloric intake of 2,000 kcal \((8,360 \mathrm{kJ})\) of food per day \((24\) hours). The food is metabolized and the free energy is used to synthesize ATP, which then provides energy for the body's daily chemical and mechanical work. Assuming that the efficiency of converting food energy into ATP is \(50 \%\), calculate the weight of ATP used by a human adult in 24 hours. What percentage of the body weight does this represent? (c) Although adults synthesize large amounts of ATP daily, their body weight, structure, and composition do not change significantly during this period. Explain this apparent contradiction.

Biochemical reactions often look more complex than they really are. In the pentose phosphate pathway (Chapter 14 ), sedoheptulose 7 -phosphate and glyceraldehyde 3 -phosphate react to form erythrose 4 -phosphate and fructose 6 -phosphate in a reaction catalyzed by transaldolase. Draw a mechanism for this reaction (show electron-pushing arrows). (Hint: Take another look at aldol condensations, then consider the name of this enzyme.)

The standard reduction potential, \(E^{\prime \circ}\), of any redox pair is defined for the half-cell reaction: Oxidizing agent \(+n\) electrons \(\longrightarrow\) reducing agent The \(E^{\prime \circ}\) values for the \(\mathrm{NAD}^{+} / \mathrm{NADH}\) and pyruvate/actate conjugate redox pairs are \(-0.32 \mathrm{V}\) and \(-0.19 \mathrm{V},\) respectively. (a) Which redox pair has the greater tendency to lose electrons? Explain. (b) Which pair is the stronger oxidizing agent? Explain. (c) Beginning with 1 M concentrations of each reactant and product at \(\mathrm{pH} 7\) and \(25^{\circ} \mathrm{C}\), in which direction will the following reaction proceed? $$\text { Pyruvate }+\mathrm{NADH}+\mathrm{H}^{+} \rightleftharpoons \text { lactate }+\mathrm{NAD}^{+}$$ (d) What is the standard free-energy change \(\left(\Delta G^{\prime \circ}\right)\) for the conversion of pyruvate to lactate? (e) What is the equilibrium constant \(\left(K_{\mathrm{eq}}^{\prime}\right)\) for this reaction?

The phosphorylation of glucose to glucose 6 -phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by \(\mathrm{P}_{\mathrm{i}}\) is described by the equation Glucose \(+\mathrm{P}_{\mathrm{i}} \longrightarrow\) glucose 6 -phosphate \(+\mathrm{H}_{2} \mathrm{O}\) $$ \Delta G^{\prime \circ}=13.8 \mathrm{kJ} / \mathrm{mol} $$ (a) Calculate the equilibrium constant for the above reaction at \(37^{\circ} \mathrm{C}\). In the rat hepatocyte the physiological concentrations of glucose and \(P_{1}\) are maintained at approximately \(4.8 \mathrm{mm} .\) What is the equilibrium concentration of glucose 6 -phosphate obtained by the direct phosphorylation of glucose by \(\mathrm{P}_{\mathrm{i}} ?\) Does this reaction represent a reasonable metabolic step for the catabolism of glucose? Explain. (b) In principle, at least, one way to increase the concentration of glucose 6 -phosphate is to drive the equilibrium reaction to the right by increasing the intracellular concentrations of glucose and \(\mathrm{P}_{\mathrm{i}}\). Assuming a fixed concentration of \(\mathrm{P}_{\mathrm{i}}\) at \(4.8 \mathrm{mM}\), how high would the intracellular concentration of glucose have to be to give an equilibrium concentration of glucose 6 -phosphate of \(250 \mu \mathrm{M}\) (the normal physiological concentration)? Would this route be physiologically reasonable, given that the maximum solubility of glucose is less than 1 M? (c) The phosphorylation of glucose in the cell is coupled to the hydrolysis of ATP; that is, part of the free energy of ATP hydrolysis is used to phosphorylate glucose: (1) \(\quad\) Glucose \(+\mathrm{P}_{\mathrm{i}} \longrightarrow\) glucose 6 -phosphate \(+\mathrm{H}_{2} \mathrm{O}\) \\[ \Delta G^{\prime \circ}=13.8 \mathrm{kJ} / \mathrm{mol} \\] (2) \(\quad \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ADP}+\mathrm{P}_{\mathrm{i}} \quad \Delta G^{\prime \circ}=-30.5 \mathrm{kJ} / \mathrm{mol}\) Sum: Glucose + ATP \(\longrightarrow\) glucose 6 -phosphate \(+\) ADP Calculate \(K_{\mathrm{eq}}^{\prime}\) at \(37^{\circ} \mathrm{C}\) for the overall reaction. For the ATPdependent phosphorylation of glucose, what concentration of glucose is needed to achieve a \(250 \mu\). intracellular concentration of glucose 6 -phosphate when the concentrations of ATP and ADP are \(3.38 \mathrm{mM}\) and \(1.32 \mathrm{mM}\), respectively? Does this coupling process provide a feasible route, at least in principle, for the phosphorylation of glucose in the cell? Explain. (d) Although coupling ATP hydrolysis to glucose phosphorylation makes thermodynamic sense, we have not yet specified how this coupling is to take place. Given that coupling requires a common intermediate, one conceivable route is to use ATP hydrolysis to raise the intracellular concentration of \(\mathrm{P}_{1}\) and thus drive the unfavorable phosphorylation of glucose by \(\mathrm{P}_{\mathrm{i}}\). Is this a reasonable route? (Think about the solubility products of metabolic intermediates.) (e) The ATP-coupled phosphorylation of glucose is catalyzed in hepatocytes by the enzyme glucokinase. This enzyme binds ATP and glucose to form a glucose- ATP-enzyme complex, and the phosphoryl group is transferred directly from ATP to glucose. Explain the advantages of this route.

Consider a system consisting of an egg in an incubator. The white and yolk of the egg contain proteins, carbohydrates, and lipids. If fertilized, the egg is transformed from a single cell to a complex organism. Discuss this irreversible process in terms of the entropy changes in the system, surroundings, and universe. Be sure that you first clearly define the system and surroundings.

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