BIOC Problem Set IV ATP Solutions PDF

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Problem Set IV: ATP (Solutions) 1. How long does an ATP molecule last in the cell, before it is hydrolyzed? We can make an estimate of this value by the following order-of-magnitude calculation. 1. a) The steady-state concentration of ATP in a living cell is about 2 mM. The mass of water in an adult human being is about 25 kg. How many moles of ATP do you have in your body? The molar mass (molecular weight) of ATP is about 500. 2. b) A human being requires on the order of 10,000 kJ of energy daily, almost all of which is furnished by ATP hydrolysis. Given that the free energy of hydrolysis of ATP under cellular conditions is on the order of 50 kJ/ mole, how many moles of ATP does a human being use daily? 3. c) Based on your answers to the first two parts of the question, estimate the average lifetime of an ATP molecule in the cell. 1. a) 1 kg water = 1 L -3 -3 2 mM ATP × 25 L = 2 × 10 mol per L × 25 L = 50 × 10 mol ATP -3

-3

2

(In terms of mass: 1 mol ATP = 500 g; so 50 × 10 mol = 50 × 10 mol × 5 × 10 g/ mol = 25 grams ATP; but we don’t actually need this number to answer the question.) 4

-1

2. b) 10,000 kJoules = 10 kJ / (50 kJ mol ) = 200 mol ATP -3

But the body only contains about 50 × 10 mol, so each molecule must be re-used many times daily. -3

3. c) 200 mol / (50 × 10 mol) = 4,000 times. (If you are emptying a swimming pool with a bucket, the number of times you need to use the bucket is simply the volume of the swimming pool divided by the volume of the bucket.) 3

4,000 turnovers per 24 hours (1.4 × 10 minutes) = about 3 turnovers per minute, or an average ATP lifetime of about 20 seconds. 1|Page

Problem Set IV: ATP (Solutions)

2. In the cell, ATP (as well as ADP) exists mainly as a complex with the divalent cation 2+ 2+ Mg (see figure below). The value of ΔG for the hydrolysis of ATP varies with Mg concentration, as shown in the graph at right. Explain these results. The experimental data in the graph show that the absolute value of the free energy change 2+ becomes smaller (i.e., ΔG becomes less negative) as [Mg ] increases (into the millimolar 2+

3. In most cells, the intracellular concentrations of ATP and inorganic phosphate are about the same (see Nelson and Cox, Table 13-5). 32

a) When a small (“tracer”) amount of ATP labelled with radioactive phosphorus ( P) 32

attheterminal(γ)position,[γ- P]ATP,isaddedtoyeastextract,abouthalfofthe radioactivity is found to be in the form of inorganic phosphate within a few minutes, but the concentration of ATP is little changed. Explain. ATP (like most biomolecules) exists in a dynamic steady-state in the cell (or in the extract):ATP is being consumed and produced at about the same rates, so the ATP concentration remains nearly constant. Within a few minutes, about half of the labeled ATP has undergone hydrolysis, releasing the labeled phosphate into the medium as Pi. (Note that regeneration of ATP from ADP 32 and Pi will not restore a significant amount of this radioactivity, because the small amount of Plabeled 32

phosphate that is released by hydrolysis of the [γ- P]ATP is diluted by the relatively large amount of “cold” (unlabeled) phosphate already present in the extract.) 2|Page

-31 2 -33 4 -35 6 -8 -7 - 6 -5 2+-4 -3 -2 log [M g ]

range). This indicates that ATP is stabilized by Mg 2+ becomes less strongly to the right. Why does Mg reduce the free energy change for ATP hydrolysis? The divalent cation binds to the negatively-charged phosphates of ATP and partially neutralizes that charge. ATP hydrolysis is driven largely by the charge repulsion between the products (ADP and Pi). Consequently, the thermodynamic tendency to hydrolysis is reduced by 2+

Mg binding. , i.e., the reaction equilibrium G (kJ/ mol)

Problem Set IV: ATP (Solutions) 32

b) What would be observed if we used [βP] ATP instead? Much less radioactivity would be released as Pi (i.e., most of the radioactivity will remain in ATP), because the most common mode of ATP hydrolysis is to ADP and Pi; relatively few processes involve hydrolysis to AMP and PPi, which would release radiolabel from the β position. 4. The following thermodynamic data are available for the conversion of A to B: -3

(uncoupled reaction) A B Keq = 10 (ATP-coupled reaction) A + ATP B + ADP + Pi Keq = 200 3

In the living cell, the steady state ratio of [ATP] / ([ADP] [Pi]) = 5 ×10 . By what factor will the coupling to ATP hydrolysis increase the ratio of [B] / [A], relative to the value for the uncoupled reaction? For the uncoupled reaction, at equilibrium: [B]/[A] = 10

-3

For the ATP-coupled reaction: ([B][ADP][Pi])/([A][ATP]) = 200 and we are given that: [ATP] / ([ADP] [Pi]) = 5 ×10

3

3

3

therefore, [B]/[A] = (200 [ATP]) / ([ADP] [Pi]) = 200 × 5 ×10 = 200 × 5 ×10 = 10 -3 6 9

So the ratio of [B] / [A] has been increased from 10 to 10 , a 10 -fold (billion-fold)

6

increase. This calculation illustrates the enormous effect of coupling a reaction to ATP hydrolysis. There are two aspects to this coupling. First, ATP coupling dramatically shifts the equilibrium constant for the reaction, because of the large negative free energy change for hydrolysis of ATP. Secondly, living cells use metabolic energy to maintain a high steady- state “energy charge”, i.e., a large excess of ATP relative to ADP and Pi. This high ratio - very far from thermodynamic equilibrium - provides an even greater “push” to an ATP- coupled metabolic process. 3|Page

Problem Set IV: ATP (Solutions) 5. In general, when ATP hydrolysis is coupled to an energy-requiring reaction, the actual reaction often consists of the transfer of a phosphate group from ATP to another substrate, rather than an actual hydrolysis of the ATP. Explain. Hydrolysis of the ATP would result in the loss of most of the free energy as heat. In a transfer reaction, the gamma (third) phosphate of ATP is transferred to the reaction substrate to produce a high-energy phosphorylated intermediate, which can then form the product in an exergonic reaction. 6. Free Energy of Hydrolysis of CTP: Compare the structure of the nucleoside triphosphate CTP with the structure of ATP. 0

0

The ΔG ′ for the reaction ATP ADP + Pi is ~30 kJ/mol. Predict the ΔG ′ for the reaction shown below. CTP CDP + Pi 0

The high ΔG ′ of ATP is related to structural features not of the base or the sugar, but primarily of the anhydride linkages between phosphate groups. In this structural feature, CTP is equivalent to 0 ATP, and thus it most likely has about the same ΔG ′ as ATP. 7. Daily ATP Utilization by Human Adults: a) A total of 30.5 kJ/mol of free energy is needed to synthesize ATP from ADP and Pi when the reactants and products are at 1 M concentrations and the temperature 0

is 25 C (standard state). Because the actual physiological concentrations of ATP, 0

ADP, and Pi are not 1 M, and the temperature is 37 C, the free energy required to

0′ synthesize ATP

under physiological conditions is different from ΔG . Calculate

the free energy required to synthesize ATP in the human hepatocyte when the physiological concentrations of ATP, ADP, and Pi are 3.5, 1.50, and 5.0 mM, respectively. 4|Page

Problem Set IV: ATP (Solutions) 0

ΔG = ΔG ′ + RT ln Q 0′

ADP + Pi ATP + H2O ΔG = 30.5 kJ/mol Mass action ratio, Q = [ATP] [Pi][ADP] -3

= [3.5 x 10 M] -3 -3 [1.5 x 10 M] [5.0 x 10 M] 2

= 4.7 x 10 M

-1

0

ΔG = ΔG ′ + RT ln Q 2 -1 = 30.5 kJ/mol + (2.58 kJ/mol) ln (4.7 x 10 M ) = 46 kJ/mol 2. b) A 68 kg (150 lb) adult requires a caloric intake of 2,000 kcal (8,360 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? The energy going into ATP synthesis in 24 hrs is 8360 kJ x 50% = 4180 kJ. Using the value of ΔG from (a), the amount of ATP synthesized is (4180 kJ)/(46 kJ/mol) = 91 mol

The molecular weight of ATP is 503 (calculated by summing atomic weights). Thus, the weight of ATP synthesized is (91 mol ATP) (503 g/mol) = 46 kg As a % of body weight: 100%(46 kg ATP)/(68 kg body weight) = 68% 3. 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. The concentration of ATP in a healthy body is maintained in a steady state; this is an example of homeostasis, a condition in which the body synthesizes and breaks down ATP as needed. 5|Page

Problem Set IV: ATP (Solutions) 8. Synthesis of the activated form of acetate (acetyl-CoA) is carried out in an ATP- dependent process: 0 Acetate +CoA + ATP acetyl-CoA + AMP + PPi ΔG ′ = 1.7 kJ/mol Almost all cells contain the enzyme inorganic pyrophosphatase, which catalyzes the hydrolysis of PPi to 2 Pi. What effect does the presence of this enzyme have on the synthesis of acetylCoA? Explain. Hydrolysis of PPi would drive the reaction forward, favoring the synthesis of acetyl-CoA....