/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 103 The monoanion of adenosine monop... [FREE SOLUTION] | 91影视

91影视

The monoanion of adenosine monophosphate (AMP) is an intermediate in phosphate metabolism: where \(\mathrm{A}=\) 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

Expert verified
The ratio of \([\mathrm{AMP}^{2-}]\) to \([\mathrm{AMP}^{-}]\) in blood at a pH of 7.4 is approximately 1.54.

Step by step solution

01

Understand the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is used to find the pH or pKa of a solution when the ratio of the concentrations of the acidic and basic forms of the solute is known, or vice versa. The equation is: \(\mathrm{pH} = \mathrm{p}K_{a} + \log_{10} \left(\dfrac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\right)\) where pH is the pH of the solution, pKa is the pKa of the acidic form of the solute (protonated form), [A鈦籡 is the concentration of the deprotonated form, and [HA] is the concentration of the protonated form. In this exercise, we are given the pKa and asked to find the ratio of two forms of AMP at a known pH.
02

Substitute the given values into the equation

We are given the pKa of AMP (\(7.21\)) and the pH of blood (\(7.4\)). We want to find the ratio of the concentrations of two forms of AMP, [\(\mathrm{AMP}^-\)] and [\(\mathrm{AMP}^{2-}\)], also represented as [\(\mathrm{A}^{-}\)] and [\(\mathrm{HA}\)] respectively in the equation. We substitute the given values into the Henderson-Hasselbalch equation: \(7.4 = 7.21 + \log_{10} \left(\dfrac{[\mathrm{AMP}^{2-}]}{[\mathrm{AMP}^{-}]}\right)\)
03

Solve for the ratio of [\(\mathrm{AMP}^{2-}\)] to [\(\mathrm{AMP}^-\)]

To find the ratio of [\(\mathrm{AMP}^{2-}\)] to [\(\mathrm{AMP}^{-}\)], we first need to isolate the fraction inside the logarithm. \(7.4 - 7.21 = \log_{10} \left(\dfrac{[\mathrm{AMP}^{2-}]}{[\mathrm{AMP}^{-}]}\right)\) \(0.19 = \log_{10} \left(\dfrac{[\mathrm{AMP}^{2-}]}{[\mathrm{AMP}^{-}]}\right)\) Now, we take the antilog (inverse log) of both sides: \(10^{0.19} = \dfrac{[\mathrm{AMP}^{2-}]}{[\mathrm{AMP}^{-}]}\)
04

Calculate the ratio and the final answer

Now that we have the equation isolated, we can calculate the ratio of [\(\mathrm{AMP}^{2-}\)] to [\(\mathrm{AMP}^{-}\)]: \(\dfrac{[\mathrm{AMP}^{2-}]}{[\mathrm{AMP}^{-}]} = 10^{0.19} \approx 1.54\) This means that in blood at a pH of 7.4, the ratio of [\(\mathrm{AMP}^{2-}\)] to [\(\mathrm{AMP}^{-}\)] is approximately 1.54.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

pH Calculation
Understanding how to calculate pH is essential in chemistry, particularly in acid-base chemistry. The pH scale measures how acidic or basic a solution is and ranges from 0 (strongly acidic) to 14 (strongly basic), with 7 being neutral.

pH can be determined using the equation \( \mathrm{pH} = -\log_{10}([H^{+}]) \) where \( [H^{+}] \) is the concentration of hydrogen ions in moles per liter. Calculations can become more complex when dealing with weak acids or bases, since they only partially dissociate in solution. In these cases, the Henderson-Hasselbalch equation provides a more suitable method for determining the pH, especially when information on the protonated and deprotonated forms of a compound is available.
pKa Values
In the context of acid-base reactions, the pKa value is a crucial concept. It represents the strength of an acid by indicating the equilibrium constant for its dissociation into a proton and its conjugate base.

The lower the pKa, the stronger the acid, as it more readily donates protons. For example, a strong acid like hydrochloric acid (HCl) has a very low pKa, whereas acetic acid, a weak acid, has a higher pKa.

Understanding pKa values helps predict reaction outcomes and is also vital in biochemistry for understanding enzyme activity and the behavior of amino acids, nucleotides, and other biomolecules at different pH levels.
Acid-Base Equilibrium
Acid-base equilibrium refers to the balance established in chemical reactions between acids and bases. Many biological and chemical processes depend on maintaining a specific pH, which is the result of this equilibrium.

For weak acids and bases, which only partially dissociate in solution, the equilibrium can be expressed by the Henderson-Hasselbalch equation, which helps calculate the ratio of protonated and deprotonated species. In an equilibrium situation, knowing the concentrations of these species and the pKa allows one to use this equation to determine the pH of the solution or vice versa.
Biochemistry
Biochemistry is the study of chemical processes within and related to living organisms. The field is fundamental for understanding cellular processes, including metabolism, enzyme function, genetic coding, and molecular biology.

In the given exercise, understanding the Henderson-Hasselbalch equation allows biochemists to predict the ionization state of AMP, a nucleotide that plays a central role in energy transfer within cells. This understanding is crucial for researchers because the state of AMP can affect its interaction with other biomolecules, influencing various biochemical pathways.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An organic compound is analyzed and found to contain \(66.7 \%\) carbon, \(11.2 \%\) hydrogen, and \(22.1 \%\) oxygen by mass. The compound boils at \(79.6^{\circ} \mathrm{C}\). At \(100{ }^{\circ} \mathrm{C}\) and \(0.970 \mathrm{~atm}\), the vapor has a density of \(2.28 \mathrm{~g} / \mathrm{L}\). The compound has a carbonyl group and cannot be oxidized to a carboxylic acid. Suggest a structure for the compound.

Give the molecular formula of a hydrocarbon containing five carbon atoms that is (a) an alkane, (b) a cycloalkane, (c) an alkene, (d) an alkyne. Which are saturated and which are unsaturated hydrocarbons?

Describe two ways in which the octane number of a gasoline consisting of alkanes can be increased.

Indole smells rather terrible in high concentrations but has a pleasant floral-like odor when highly diluted. It has the following structure: Indole is a planar molecule. The nitrogen is a very weak base, with a \(K_{b}\) of \(2 \times 10^{-12}\). Explain how this information indicates that the indole molecule is aromatic in character.

(a) Describe the primary, secondary, and tertiary structures of proteins. (b) Quaternary structures of proteins arise if two or more smaller polypeptides or proteins associate with each other to make an overall much larger protein structure. The association is due to the same hydrogen bonding, electrostatic, and dispersion forces we have seen before. Hemoglobin, the protein used to transport oxygen molecules in our blood, is an example of a protein that has quaternary structure. Hemoglobin is a tetramer; it is made of four smaller polypeptides, two "alphas" and two "betas." (These names do not imply anything about the number of alpha helices or beta sheets in the individual polypeptides.) What kind of experiments would provide sound evidence that hemoglobin exists as a tetramer and not as one enormous polypeptide chain? You may need to look into the chemical literature to discover techniques that chemists and biochemists use to make these decisions.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.