Question

The monoanion of adenosine monophosphate (AMP) is an intermediate in phosphate metabolism:where \(A=\) adenosine. If the \(\mathrm{p} K_{a}\) 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

The ratio of [AMP-OH⁻] to [AMP-O²⁻] in blood at pH 7.4 is approximately 1:1.54.

Step-by-step solution

Write down the Henderson-Hasselbalch equation.

The equation we will use is: pH = pKa + log([A⁻]/[HA]) Where pH is the blood pH, pKa is the pKa of the monoanion AMP, [A⁻] is the concentration of AMP-O²⁻, and [HA] is the concentration of AMP-OH⁻.

Plug in the given values.

We are given the following values: pH = 7.4 pKa = 7.21 We will now plug these values into the Henderson-Hasselbalch equation: 7.4 = 7.21 + log([AMP-O²⁻]/[AMP-OH⁻])

Solve for the ratio of [AMP-OH⁻] to [AMP-O²⁻].

Rearrange the equation to isolate the concentration ratio on one side: log([AMP-O²⁻]/[AMP-OH⁻]) = 7.4 - 7.21 Now, calculate the difference: log([AMP-O²⁻]/[AMP-OH⁻]) = 0.19 To remove the logarithm, use the inverse function, which is the exponential function: [AMP-O²⁻]/[AMP-OH⁻] = 10^0.19 Calculate the exponent: [AMP-O²⁻]/[AMP-OH⁻] ≈ 1.54

Write down the final answer.

The ratio of [AMP-OH⁻] to [AMP-O²⁻] in blood at pH 7.4 is approximately 1:1.54.

Key concepts

Acid-Base Chemistry

Acid-base chemistry is a branch of chemistry that explores the behavior of acids and bases. It focuses on how they react with one another and the impact they have on a solution's pH level. The pH scale, ranging from 0 to 14, is a measure of how acidic or basic a solution is:

• A pH below 7 indicates an acidic solution.
• A pH above 7 represents a basic or alkaline solution.
• A pH of 7 is neutral, meaning it is neither acidic nor basic.

In the context of biological systems, maintaining a balanced pH is vital for proper cellular functions. Enzymatic reactions, for instance, often depend on a specific pH range to proceed efficiently. In blood, the typical pH range is tightly regulated around 7.4. This ensures optimal conditions for various biochemical processes.

Understanding acid-base interactions in the body, such as those involving phosphate groups in ATP or AMP, reveals how pH changes can influence metabolism and cellular activities. Acids donate protons (H⁺ ions) to solutions, while bases accept them, creating a dynamic balance crucial for life.

Phosphate Metabolism

Phosphate metabolism is essential for energy transformation and storage within cells. Phosphates are a key part of molecules like ATP (adenosine triphosphate), which cells use as the primary energy currency. During metabolism, phosphate groups are transferred to different molecules in reactions facilitated by enzymes.
Phosphates also play a role in signaling pathways. They act as switches that turn on or off different cellular functions by being added or removed from proteins in a process called phosphorylation and dephosphorylation.
In the given exercise, adenosine monophosphate (AMP) is part of phosphate metabolism. AMP is one of the products generated when ATP is broken down to release energy. The balance between AMP and its phosphate derivatives governs many regulatory pathways that control cellular energy levels.

• AMP can exist in forms with different numbers of phosphate groups.
• These forms interconvert depending on the cell’s energy needs and conditions.

By being intermediates in these transformations, phosphates ensure that nutrients and energy are efficiently managed and utilized in biological systems.

pKa and pH Relationship

The relationship between pKa and pH is foundational to understanding acid-base equilibria, especially in biological contexts. The pKa of a substance is the pH at which it equilibrates between its acidic and basic forms. It provides insight into the strength of an acid or base; the lower the pKa, the stronger the acid.
The Henderson-Hasselbalch equation clarifies this relationship:\[ \text{pH} = \text{pKa} + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]Where:

• \([\text{A}^-]\) is the concentration of the deprotonated (base) form.
• \([\text{HA}]\) is the concentration of the protonated (acid) form.

In the given problem, we apply this equation to find the ratio of the deprotonated AMP form (AMP-O²⁻) to its less deprotonated form (AMP-OH⁻) at a specific pH. By comparing the blood's pH to AMP’s pKa, we can assess how the AMP species distribute between their available forms. A higher pH than the pKa means the deprotonated form is favored.
This relationship is crucial for predicting how molecules behave in different pH environments. Understanding these dynamics allows scientists and healthcare professionals to make informed decisions about biochemical and physiological processes.