Question

The monoanion of adenosine monophosphate (AMP) is an intermediate in phosphate metabolism: where \(\mathrm{A}=\) adenosine. If the \(\mathrm{p} K_{\mathrm{a}}\) for this anion is \(7.21\), what is the ratio of \(\left[\mathrm{AMP}-\mathrm{OH}^{-}\right]\)to \(\left[\mathrm{AMP}-\mathrm{O}^{2-}\right]\) in blood at \(\mathrm{pH}\) 7.4?

Short answer

In blood with a pH of 7.4, the ratio of [AMP-OH⁻] to [AMP-O²⁻] is approximately 1.549. This value was determined using the Henderson-Hasselbalch equation, plugging in the given pH and pKa values, and solving for the ratio.

Step-by-step solution

Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \[pH = pKa + \log \frac{[A^-]}{[HA]}\] In this problem, HA is AMP-OH⁻ and A⁻ is AMP-O²⁻.

Plug in the given values

We have the values for pH (7.4) and pKa (7.21). We will plug these values into the Henderson-Hasselbalch equation: \[7.4 = 7.21 + \log \frac{[AMP-O^2-]}{[AMP-OH^-]}\]

Solve for the ratio

Now, we need to solve for the ratio of [AMP-O²⁻] to [AMP-OH⁻]: \[0.19 = \log \frac{[AMP-O^2-]}{[AMP-OH^-]}\] To undo the logarithm, raise both sides as powers of 10: \[\frac{[AMP-O^2-]}{[AMP-OH^-]} = 10^{0.19}\] Now, calculate the value of 10 raised to the power of 0.19: \[\frac{[AMP-O^2-]}{[AMP-OH^-]} \approx 1.549\]

Write the final answer

The ratio of [AMP-OH⁻] to [AMP-O²⁻] in blood at pH 7.4 is approximately 1.549.

Key concepts

pH Calculation

Understanding how to calculate the pH is crucial in many fields of science, including chemistry and biology. The pH is a logarithmic measure of the hydrogen ion concentration in a solution, and it indicates the level of acidity or alkalinity. To grasp this concept, consider that a pH of 7 is neutral, as found in pure water. Values below 7 indicate acidity while those above 7 indicate alkalinity.

To perform a pH calculation, we often use the Henderson-Hasselbalch equation, which relates the pH to the pKa (the acid dissociation constant) and the ratio of the concentrations of the deprotonated and protonated forms of an acid. From our exercise:
\[pH = pKa + \log \frac{[A^-]}{[HA]}\]
By rearranging and inserting the known pH and pKa values, we can solve for the concentration ratio of the different forms of a compound in a solution, such as in the buffer systems of our bodies.

pKa Value

The pKa value is a fundamental concept in chemistry which indicates the strength of an acid. pKa is the negative base-10 logarithm of the acid dissociation constant (Ka) of a solution. The lower the pKa value, the stronger the acid is because it means the acid more readily donates protons (H+) to the solution.

In biochemical systems, knowing the pKa of a compound can be pivotal for understanding how the compound will behave under various pH conditions. For instance, the pKa of 7.21 for the AMP anion given in the exercise tells us that at a pH of 7.21, the concentrations of AMP-OH⁻ and AMP-O²⁻ are equal. When the pH is above this pKa, the deprotonated form (AMP-O²⁻) will be more prevalent. The difference between the pH of the environment and the pKa of the compound determines the ratio of the protonated and deprotonated forms, as seen through the Henderson-Hasselbalch equation.

Biochemical Metabolism

Biochemical metabolism encompasses all the chemical reactions taking place in living organisms. It involves the transformation of substrates through metabolic pathways, often resulting in the production of energy and new molecules necessary for life. One aspect of metabolism is the interconversion between different molecular forms, which can be influenced by pH levels.

For example, adenosine monophosphate (AMP), mentioned in our exercise, participates in energy transfer through its conversion to ATP and ADP. The balance between different ionic forms of such molecules can be critical for their proper functioning in metabolic reactions. By applying principles like the Henderson-Hasselbalch equation, we can predict and control the ionic forms of these biomolecules, ensuring that biochemical pathways proceed correctly within the intricate environment of the cell.