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The cleavage of ATP (adenosine triphosphate) to ADP (adenosine diphosphate) and \(\mathrm{H}_{3} \mathrm{PO}_{4}\) may be written as follows: It is interesting to determine which bond (the \(\mathrm{P}-\mathrm{O}\) bond marked a or the \(\mathrm{O}\) - \(\mathrm{P}\) bond marked \(\mathbf{b}\) ) is cleaved by hydrolysis (reaction with water). (a) Outline an experiment (using radioactivity) that can be used to determine where the cleavage results. (b) Describe the results that would lead you to conclude that cleavage results at a, (c) Describe the results that would lead you to conclude that cleavage results at \(\mathbf{b}\). Results show that the cleavage occurs at \(\mathrm{b}\).

Short Answer

Expert verified
Answer: If the cleavage occurs at the 'b' bond, the radioactive label (\({}^{32}\mathrm{P}\)) would be found in the \(\mathrm{H}_{3}\mathrm{PO}_{4}\) product after hydrolysis.

Step by step solution

01

(a) Experiment Outline

To determine the cleavage position, we can use radioactively labeled phosphorus (P) in ATP. Specifically, we can use two separate samples of ATP with either the "a" or "b" phosphorus atom labeled. This can be achieved by using \({}^{32}\mathrm{P}\) instead of the standard \({}^{31}\mathrm{P}\) in the phosphate groups. 1. Synthesize two ATP samples by incorporating labeled phosphorus (\({}^{32}\mathrm{P}\)) selectively in either the \(\mathrm{P}-\mathrm{O}\) bond (a) or the \(\mathrm{O}-\mathrm{P}\) bond (b). 2. Carry out the hydrolysis reaction for both samples separately, mixing each ATP sample with water. 3. Analyze the products formed (ADP and \(\mathrm{H}_{3}\mathrm{PO}_{4}\)) using appropriate techniques like paper chromatography or electrophoresis to separate them. 4. Measure the presence of radioactivity in the separated products using a radiation detector. 5. Observe whether the radioactive label (the \({}^{32}\mathrm{P}\)) stays on the ADP or it goes with the released phosphate. By tracing the fate of the radioactive label, we can pinpoint the bond being cleaved during hydrolysis.
02

(b) Results for Cleavage at 'a'

If the cleavage occurs at the \(\mathrm{P}-\mathrm{O}\) bond marked as 'a', we would expect the radioactive label (\({}^{32}\mathrm{P}\)) to remain with the ADP molecule. This is because the labeled phosphorus is still part of the ADP structure. In this scenario, our results would show significant radioactivity in the ADP product and no radioactivity in the \(\mathrm{H}_{3}\mathrm{PO}_{4}\) product.
03

(c) Results for Cleavage at 'b'

If the cleavage occurs at the \(\mathrm{O}-\mathrm{P}\) bond marked as 'b', we would expect the radioactive label (\({}^{32}\mathrm{P}\)) to go with the released phosphate. In this case, the \({}^{32}\mathrm{P}\) would be included in the \(\mathrm{H}_{3}\mathrm{PO}_{4}\) molecule. Our results would show significant radioactivity in the \(\mathrm{H}_{3}\mathrm{PO}_{4}\) product and no radioactivity in the ADP product. As the exercise states, the results show that the cleavage occurs at 'b', which implies that the radioactive label (\({}^{32}\mathrm{P}\)) would be found in the \(\mathrm{H}_{3}\mathrm{PO}_{4}\) product after hydrolysis.

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Key Concepts

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

ATP to ADP and H3PO4
Adenosine triphosphate, commonly known as ATP, is a crucial energy carrier in biological systems. The transformation of ATP to adenosine diphosphate (ADP) and inorganic phosphate (\(\mathrm{H}_{3}\mathrm{PO}_{4}\)) occurs through a biochemical process called hydrolysis. This reaction releases energy that is used by cells to perform various functions, from muscle contraction to nerve impulse propagation.

During hydrolysis, one of the phosphate-oxygen bonds in the ATP molecule is broken, leading to the formation of ADP and \(\mathrm{H}_{3}\mathrm{PO}_{4}\). The specific bond that is cleaved can be at different positions, labeled 'a' or 'b', with each one producing ADP and \(\mathrm{H}_{3}\mathrm{PO}_{4}\) through slightly different paths. By understanding this fundamental conversion, we can better appreciate the controlled manner in which energy is released and used within our cells.
Radioactively Labeled Phosphorus
The use of radioactively labeled phosphorus, particularly \({}^{32}\mathrm{P}\) as opposed to \({}^{31}\mathrm{P}\), provides critical information in biochemistry experiments. By introducing a detectable radioactive marker into one of the phosphate groups of ATP, scientists can track the pathway through which chemical bonds are broken during hydrolysis.

\({}^{32}\mathrm{P}\) emits radiation that can be picked up by detectors, allowing researchers to ascertain which molecular fragment the radioactive phosphorus ends up in after the ATP has been hydrolyzed to ADP and \(\mathrm{H}_{3}\mathrm{PO}_{4}\). This proves invaluable in identifying the mechanisms behind biochemical reactions and understanding how energy is harnessed within living organisms.
Analysis of ATP Hydrolysis Products
To analyze the products of ATP hydrolysis, researchers employ various techniques such as paper chromatography or electrophoresis. These methods separate the resulting compounds, ADP and \(\mathrm{H}_{3}\mathrm{PO}_{4}\), based on their physical and chemical properties.

Following separation, a radiation detector is used to determine the presence of radioactivity in the products. If the experiment included radioactively labeled phosphorus, one would expect to find radiation in the compound that retains the labeled phosphate group. This method provides a clear-cut means to deduce which bond, 'a' or 'b', is cleaved during ATP hydrolysis, adding a layer of insight into the intricate workings of cellular energy management.
Biochemistry Education
Biochemistry education serves as a foundation for students to grasp essential life processes at the molecular level. By integrating experiments like ATP hydrolysis with the use of radioactively labeled phosphorus, students can visually and practically understand core biochemical concepts. Learning about ATP's role in energy transfer and the meticulous investigation techniques used to study its behavior lays the groundwork for future scientific inquiry and innovation.

Providing a hands-on approach to teaching, such as carrying out laboratory experiments, enables students to directly observe the processes they learn about, thereby reinforcing their understanding and retention of complex biochemical principles.

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Most popular questions from this chapter

Bromine-82 has a half-life of 36 hours. A sample containing Br-82 was found to have an activity of \(1.2 \times 10^{5}\) disintegrations \(/ \mathrm{min}\). How many grams of Br-82 were present in the sample? Assume that there were no other radioactive nuclides in the sample.

Classify the following statements as true or false. If false, correct the statement to make it true. (a) The mass number increases in beta emission. (b) A radioactive species with a large rate constant, \(k\), decays very slowly. (c) Fusion gives off less energy per gram of fuel than fission.

A source for gamma rays has an activity of 3175 Ci. How many disintegrations are there for this source per minute?

The amount of oxygen dissolved in a sample of water can be determined by using thallium metal containing a small amount of the isotope Tl- 204\. When excess thallium is added to oxygen-containing water, the following reaction occurs. $$2 \mathrm{Tl}(s)+\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{Tl}^{+}(a q)+2 \mathrm{OH}^{-}(a q)$$ After reaction, the activity of a 25.0-mL water sample is 745 counts per minute (cpm), caused by the presence of \(\mathrm{Tl}^{+}-204\) ions. The activity of Tl-204 is \(5.53 \times 10^{5} \mathrm{cpm}\) per gram of thallium metal. Assuming that \(\mathrm{O}_{2}\) is the limiting reactant in the above equation, calculate its concentration in moles per liter.

Krypton-87 has a rate constant of \(1.5 \times 10^{-4} \mathrm{~s}^{-1}\). What is the activity of a \(2.00-\mathrm{mg}\) sample?

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