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Measurement of Acetylcholine Levels by pH Changes You have a \(15 \mathrm{~mL}\) sample of acetylcholine (a neurotransmitter) with an unknown concentration and a \(\mathrm{pH}\) of \(7.65\). You incubate this sample with the enzyme acetylcholinesterase to convert all of the acetylcholine to choline and acetic acid. The acetic acid dissociates to yield acetate and hydrogen ions. At the end of the incubation period, you measure the \(\mathrm{pH}\) again and find that it has decreased to \(6.87\). Assuming there was no buffer in the assay mixture, determine the number of nanomoles of acetylcholine in the original \(15 \mathrm{~mL}\) sample.

Short Answer

Expert verified
1687.5 nanomoles of acetylcholine present in the sample.

Step by step solution

01

Understand the Problem

We need to find the amount of acetylcholine originally present in the solution by analyzing the change in pH after reaction completion. The pH change is due to the formation of acetic acid which dissociates to produce hydrogen ions (H+).
02

Calculate the Change in Hydrogen Ion Concentration

The initial pH is 7.65, translating to an initial \[\text{{[H+]}} = 10^{-7.65} \] moles/L. The final pH is 6.87, which translates to \[\text{{[H+]}} = 10^{-6.87} \] moles/L. Calculate the change in [H+]: \[\Delta[\text{H+}] = 10^{-6.87} - 10^{-7.65} = 1.349 \times 10^{-7} - 2.239 \times 10^{-8} \].
03

Calculate the Increase in Hydrogen Ion Moles

The change in [H+] indicates the moles of acetic acid produced, which corresponds to the moles of acetylcholine initially present. Multiply the change in concentration \([-1.125 \times 10^{-7} \text{ moles/L} - 2.239 \times 10^{-8} \text{moles/L} ]\) by the total volume (15 mL or 0.015 L) to find the total moles: \[-1.125 \times 10^{-7} \times 0.015 = -1.6875 \times 10^{-9} \text{moles}\].
04

Convert Moles to Nanomoles

Convert the moles of hydrogen ions, which corresponds to acetylcholine, to nanomoles: \[1.6875 \times 10^{-9} \text{moles} = 1.6875 \times 10^{-9} \times 10^{9} \text{nanomoles} = 1687.5 \text{nanomoles} \].
05

Verify the Results

Check each step to ensure there are no errors. The decrease in pH indicates an increase in [H+], which is caused by the conversion of acetylcholine to acetic acid and is consistent with the calculated moles/nanomoles.

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Key Concepts

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

pH Change
When working with solutions, the pH is a measure of its acidity or alkalinity. It is calculated as the negative logarithm base 10 of the hydrogen ion concentration ([H+]). For our exercise, the initial pH of the acetylcholine sample is 7.65, and it drops to 6.87 after the enzyme reaction. This signifies a shift towards greater acidity. A decrease in pH implies an increase in hydrogen ion concentration within the solution. Calculating the exact change in [H+] is crucial to understanding how much acetic acid was produced, as this directly correlates to the amount of acetylcholine originally present. The formula for calculating the hydrogen ion concentration is \[ \text{[H+]} = 10^{-\text{pH}} \]Tracking the change in pH thus provides direct insights into the amount of substrate converted during a biochemical reaction.
Enzyme Reaction
Enzymes act as catalysts, speeding up chemical reactions without being consumed in the process. In this exercise, the enzyme acetylcholinesterase specifically targets acetylcholine, converting it to choline and acetic acid. This reaction can be simplified into two main categories:
  • Substrate: Acetylcholine, which is the initial molecule that the enzyme acts on.
  • Products: Choline and acetic acid formed as a result of the reaction.
The enzyme enables this conversion by lowering the activation energy required for the reaction to proceed. By monitoring the resulting acetic acid, we can deduce how much acetylcholine was processed by the enzyme, because the quantity of acetic acid produced directly corresponds to the acetylcholine initially present.
Concentration Calculation
Concentration changes are directly tied to the pH shift observed in the exercise. By calculating the difference in hydrogen ion concentration, we can deduce the amount of acetic acid produced. Using the observed pH drop from 7.65 to 6.87:The initial hydrogen ion concentration can be calculated from the initial pH (7.65) as \[ \text{[H+]}_\text{initial} = 10^{-7.65} \]The final hydrogen ion concentration, from the final pH (6.87), is \[ \text{[H+]}_\text{final} = 10^{-6.87} \]The change in hydrogen ion concentration \[ \Delta[\text{H+}] = \text{[H+]}_\text{final} - \text{[H+]}_\text{initial} \]This difference represents the amount of acetic acid produced per liter, which is directly proportional to the original acetylcholine concentration.
Biochemical Assay
A biochemical assay is a vital method used to quantify chemical components in a biological sample. In this case, it measures the acetylcholine levels using the enzyme reaction that results in a pH change. Key steps in a biochemical assay process include:
  • Sample Preparation: Ensuring a controlled environment, with specific volumes and concentrations of reactants, is crucial.
  • Reaction Initiation: Adding the enzyme to catalyze the specific biochemical conversion.
  • Measurement: Monitoring changes, such as pH, to quantify the levels of the desired molecule, acetylcholine in this instance.
This structured approach allows for a precise and meticulous measurement process, ensuring that calculated results reflect accurate concentrations of biochemical entities within a given sample.

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Most popular questions from this chapter

a. In what pH range can glycine be used as an effective buffer due to its amino group? b. In a \(0.1 \mathrm{~m}\) solution of glycine at pH \(9.0\), what fraction of glycine has its amino group in the \(-\mathrm{NH}_{3}^{4}\) form? c. How much \(5 \mathrm{M}\) KOH must be added to \(1.0 \mathrm{~L}\) of \(0.1 \mathrm{M}\) glycine at pH \(9.0\) to bring its pII to exactly \(10.0 ?\) d. When 9996 of the glycine is in ?ts \(-\mathrm{NH}_{3}^{+}\)form, what is the numerical relation between the pH of the solution and the p \(K_{\mathrm{n}}\) of the amino group? Properties of a Buffer The amino acid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a \(\mathrm{p} K_{\mathrm{n}}\) of \(9.6\), can exist either in the protonated form \(\left(-\mathrm{NH}_{3}^{+}\right)\)or as the free base \(\left(-\mathrm{NH}_{2}\right)\), because of the reversible equilibrium $$ \mathrm{F}-\mathrm{NH}_{3}^{+} \rightleftharpoons \mathrm{H}-\mathrm{NH}_{2}+\mathrm{H}^{+} $$

Preparation of a Phosphate Buffer Phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\), a triprotic acid, has three \(\mathrm{p} K_{\mathrm{a}}\) values: \(2.14,6.86\), and 12.4. What molar ratio of \(\mathrm{HPO}_{4}^{2-}\) to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)in solution would produce a \(\mathrm{pH}\) of \(7.0 ?\) Hint: Only one of the \(\mathrm{p} K_{\mathrm{a}}\) values is relevant here.

Electronegativity and Hydrogen Bonding The Pauling electronegativity is a measure of the affinity of an atom for the electron in a covalent bond. The larger the electronegativity value, the greater the affinity of the atom for an electron shared with another atom. $$ \begin{aligned} &\begin{array}{cc} \text { Abem } & \text { Electrenegativity } \\ \mathrm{H} & 2.1 \\ \mathrm{C} & 2.55 \\ \mathrm{~s} & 2.58 \\ \mathrm{~N} & 3.04 \end{array}\\\ &349 \end{aligned} $$ote that \(\mathrm{S}\) is directly beneath \(\mathrm{O}\) in the periodic table. a. Do you expect \(\mathrm{H}_{2} \mathrm{~S}\) to form hydrogen bonds with itself? With \(\mathrm{H}_{2} \mathrm{O}\) ? b. Water boils at \(100^{\circ} \mathrm{C}\). Is the boiling point for \(\mathrm{H}_{2} \mathrm{~S}\) higher or lower than for \(\mathrm{H}_{2} \mathrm{O}\) ? c. Is \(\mathrm{H}_{2} \mathrm{~S}\) a more polar solvent than \(\mathrm{H}_{2} \mathrm{O}\) ?

Calculation of \(\mathrm{p} K_{\mathrm{a}}\) An unknown compound, \(\mathrm{X}\), is thought to have a carboxyl group with a \(\mathrm{p} K_{\mathrm{n}}\) of \(2.0\) and another ionizable group with a \(\mathrm{p} K_{\mathrm{n}}\) between 5 and 8 . When \(75 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) is added to \(100 \mathrm{~mL}\) of a \(0.1 \mathrm{~m}\) solution of \(\mathrm{X}\) at \(\mathrm{pH} 2.0\), the \(\mathrm{pH}\) increases to \(6.72\). Calculate the \(\mathrm{p} K_{\mathrm{a}}\) of the second ionizable group of \(X\).

Biological Advantage of Weak Interactions The associations between biomolecules are often stabilized by hydrogen bonds, electrostatic interactions, the hydrophobic effect, and van der Waals interactions. How are weak interactions such as these advantageous to an organism?

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