BIOCHEMISTRY (LOOSELEAF)
BIOCHEMISTRY (LOOSELEAF)
6th Edition
ISBN: 9780190612399
Author: MCKEE
Publisher: Oxford University Press
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Chapter 4, Problem 1Q
Summary Introduction

To review:

The value of ΔG' for the hydrolysis of ATP (adenosine triphosphate), when the concentration (mM) of ATP, ADP (adenosine diphosphate), and Pi (inorganic phosphate) is: 4.0, 1.35, and 4.65 mM, respectively. Temperature is 37°C, pH is 7, and ΔG0' is –3.05 kJ/mol.

Introduction:

In living cells, the concentration of ATP and the product of hydrolysis of ATP (ADP and Pi) islower than that of the standard 1 M concentrations. This is the reason why the actual free energy of ATP hydrolysis differs from the standard free energy. So, it is difficult to obtain an accurate measure of the concentrations of cellular components.

Expert Solution & Answer
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Explanation of Solution

As per the given data, the actual ΔG' for hydrolysis of ATP can be calculated as:

ΔG'=ΔG°'+RTln[ADP][Pi][ATP]

Using the values, as given in the statement. The equation becomes:

ΔG'=30.5+8.134j/mol×310ln[0.00135][0.00465][0.004]

The values of ATP, ADP, and Pi are given in mM. So, to covert, these values in M divide each value by 1000. The resulting values obtained would be: 0.004, 0.00135, and 0.00465 for ATP. ADP, and Pi, respectively. Here, R is the gas constant and its value is 8.134 j/mol. The temperature was given in Celsius and has been converted into kelvin for calculation. ( 37°C+273=310K ). So, the equation becomes:

ΔG'=30.5+8.134j/mol×310ln[0.00135][0.00465][0.004]

ΔG'=30.5+2.577(ln0.00157)

ΔG'=30.516.64

ΔG'=47.14kj/mol

From the above calculation, the actual ΔG' for hydrolysis of ATP at 37° C and neutral pH is calculated as –47.14 kJ/mol.

Conclusion

Thus, it can be concluded that at a certain concentration of ATP, the actual value of ΔG' can bedetermined. From the given data, the actual value of ΔG' is –47.14 kJ/mol.

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Students have asked these similar questions
A total of 30.5 kJ mol-1 of free energy is needed to synthesise ATP from ADP and Pi  when the reactants and products are at 1.0 M concentrations and the temperature is 25oC. Because the actual physiological concentrations of ATP, ADP, and Pi are not 1.0 M, and the temperature is 37oC, the free energy required to synthesise ATP under physiological conditions is actually ~ 46.2 kJ mol-1. A 68 kg adult requires an energy intake of 8,550 kJ of food per day (24 hours). Calculate the mass (in Kg) of ATP synthesised by a human adult in 24 hours, assuming that the percentage efficiency of converting inputted calories in ATP is 50%. What percentage of the body weight does this represent?
ATP Synthase is known to catalyze the synthesis of ATP with a ΔG°’ close to zero, and a Keq' close to 1. Why is the value of ΔG°’ different from the known value which is 30.5 kJ/mol (the energy for the reverse of ATP hydrolysis)? If the Keq' value is close to one, how is it ensured that the reaction is driven to the product side and more ATP is obtained?
Imagine that the concentrations of reactants and products for the coupled reactions above in the cell are at a level that yields a smaller molar ratio of the concentrations than those relevant to the standard state conditions. Would the net free energy of coupled glucose breakdown and ATP synthesis be more or less favorable than the answer for net free energy of coupled glucose breakdown and ATP synthesis in organism in standard conditions. Please explain answer
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