Exam Review Notes: pH, Peptide Sequencing, and Aspirin/Acid-Base Disorders

pH Calculations

Problem 2: Buffer Calculation (Sodium Acetate/Acetic Acid)
- Composition: A solution containing $0.1$ moles of sodium acetate (conjugate base, $A^-$ ) and $0.2$ moles of acetic acid (weak acid, $HA$ ) in enough water to make a $2 ext{ L}$ solution.
- Identification: This mixture constitutes a buffer because it contains a weak acid and its conjugate base.
- Method: The easiest way to calculate the $pH$ of a buffer is to use the Henderson-Hasselbalch equation.
- Henderson-Hasselbalch Equation: $pH = pKa + ext{log} rac{[A^-]}{[HA]}$
- Required Information: The $pKa$ of acetic acid (provided in a previous problem) is given as $4.76$ .
- Concentrations:
  - $[A^-] ext{ (sodium acetate)} = rac{0.1 ext{ moles}}{2 ext{ L}} = 0.05 ext{ M}$
  - $[HA] ext{ (acetic acid)} = rac{0.2 ext{ moles}}{2 ext{ L}} = 0.1 ext{ M}$
- Calculation:
  $pH = 4.76 + ext{log} rac{0.05}{0.1}$
  $pH = 4.76 + ext{log}(0.5)$
  $pH = 4.76 - 0.30$
  $pH = 4.46$
- Intuition Check: Since there is more acid ( $0.1 ext{ M}$ ) than conjugate base ( $0.05 ext{ M}$ ), the $pH$ should be lower than the $pKa$ . The calculated $pH$ of $4.46$ is indeed lower than the $pKa$ of $4.76$ , indicating the calculation is likely correct.
- Exam Information: $Ka$ (or $pKa$ ) values will always be provided on the exam.
- General pKa Estimates: For calculating net charge on a protein, carboxylic acids have $pKa$ values around $1$ or $7$ , and amine groups have $pKa$ values around $7, 8$ , or higher.
Problem 5: Setting up a Buffer Calculation Equation
- Scenario: A $0.5 ext{ M}$ phosphate solution or buffer.
- Goal: Determine the ratio of the conjugate base ( $[A^-]$ ) to the weak acid ( $[HA]$ ) when calculating $pH$ using the Henderson-Hasselbalch equation.
- Relationship: The sum of the concentrations of the weak acid and its conjugate base equals the total buffer concentration: $[HA] + [A^-] = 0.5 ext{ M}$ .
- Variable Assignment:
  - If we arbitrarily define $[A^-] = x$ , then $[HA] = 0.5 - x$ .
  - Alternatively, if $[HA] = x$ , then $[A^-] = 0.5 - x$ . The choice of which species to label $x$ is arbitrary, but it is highly recommended to do it the same way every time (e.g., always letting the conjugate base concentration be $x$ ) to avoid errors. You will obtain a different value for $x$ , but the final answer for $pH$ and concentrations will be the same.
- Equation Setup: $pH = pKa + ext{log} rac{x}{0.5 - x}$
Problem 1a/1b: Strong Acid pH Calculation
- Principle: For a strong acid (e.g., $0.0018 ext{ M}$ ), it fully dissociates, so the concentration of $H^+$ ions is essentially equal to the initial concentration of the strong acid.
- Calculation: $pH = - ext{log}[H^+]$
Problem 1c: Strong Base pH Calculation
- Principle: For a strong base, it fully dissociates, so the concentration of $OH^-$ ions is essentially equal to the initial concentration of the strong base.
- Calculation Steps:
  1. Calculate $pOH = - ext{log}[OH^-]$ .
  2. Calculate $pH = 14 - pOH$ .
Problem 1d: Weak Acid pH Calculation
- Scenario: A weak acid in solution (e.g., initial concentration $0.016 ext{ M}$ ).
- Reaction: $HA ightleftharpoons H^+ + A^-$
- Equilibrium Constant: $Ka = rac{[H^+][A^-]}{[HA]}$
- ICE (Initial, Change, Equilibrium) Chart Setup:
 - Initial: $[HA] = 0.016$ , $[H^+] = 0$ , $[A^-] = 0$
 - Change: $[HA] = -x$ , $[H^+] = +x$ , $[A^-] = +x$
 - Equilibrium: $[HA] = 0.016 - x$ , $[H^+] = x$ , $[A^-] = x$
- Equation: $Ka = rac{x imes x}{0.016 - x} = rac{x^2}{0.016 - x}$
- Approximation: For weak acids, if $Ka$ is very small (e.g., $1.75 imes 10^{-5}$ ), the value of $x$ (the amount of acid that dissociates) is negligible compared to the initial concentration. Therefore, the denominator can be approximated as $0.016 - x hickapprox 0.016$ .
- Simplified Equation: $Ka hickapprox rac{x^2}{0.016}$
- Solving for pH: Solve for $x$ . Since $x = [H^+]$ , you can then calculate $pH = - ext{log}[H^+]$ .
- Note: In this dissociation, $[H^+]$ and $[A^-]$ are produced in a $1:1$ ratio, so their equilibrium concentrations are equal ( $x$ ).
- Weak Bases: In biochemistry, there's typically no focus on $pKb$ values. Instead, the $pKa$ of the conjugate acid is used because $pKa$ correlates nicely with structure and is more useful for determining the net charge on amino acids or peptides.

Protein/Peptide Sequencing

General Approach: Sequencing problems involve using specific chemical or enzymatic cleavages to break a peptide into smaller fragments, identifying N-terminal residues, and then assembling the full sequence from the overlapping information.
Problem 2: Pentapeptide Sequencing
- Composition: Alanine (Ala), Cysteine (Cys), Glycine (Gly), Arginine (Arg), Phenylalanine (Phe).
- Step a: Dansyl Chloride Reaction / Edman Degradation:
  - Reagent Function: Dansyl chloride (or Edman's reagent) reacts with the N-terminal amino acid of a peptide.
  - Result: "…gives you dansyl alanine."
  - Deduction: Alanine (Ala) is the N-terminal amino acid.
- Step b: Trypsin Cleavage:
  - Enzyme Function: Trypsin is an endopeptidase that specifically cleaves peptide bonds after positively charged amino acid residues: Lysine (Lys) and Arginine (Arg).
  - Deduction: The discussion implies Arg is present. If trypsin cleaves, Arg must be internal or at the C-terminus of a fragment. The teacher concludes that Arginine must have been the last residue of a fragment to be cut after it.
- Step c: Chymotrypsin Cleavage:
  - Enzyme Function: Chymotrypsin is an endopeptidase that specifically cleaves peptide bonds after aromatic amino acid residues: Phenylalanine (Phe), Tyrosine (Tyr), and Tryptophan (Trp).
  - Result: "Chymotrypsin does not cleave the peptide."
  - Deduction: Since Phenylalanine (Phe) is an aromatic amino acid present in the composition, and chymotrypsin does not cleave the peptide, Phenylalanine must be the C-terminal amino acid of the entire peptide. If it were internal, chymotrypsin would cleave after it.
- Final Sequence Deduction for Problem 2:
  - From (a), Ala is N-terminal.
  - From (c), Phe is C-terminal.
  - Known composition: Ala, Cys, Gly, Arg, Phe.
  - Combining information (and the teacher's implied solution): Ala-Gly-Cys-Arg-Phe.
    - Verification: Ala is first. Chymotrypsin won't cut after the C-terminal Phe. Trypsin would cut after Arg, yielding Ala-Gly-Cys-Arg and Phe.
Problem 3: Decapeptide from Pentapeptides
- Composition: Double the amino acids from Problem 2 (Two of each: Ala, Cys, Gly, Arg, Phe). This implies the decapeptide is composed of two identical pentapeptides.
- Step a: Edman Degradation: Confirming Alanine as the N-terminus, consistent with the individual pentapeptide sequence.
- Step b: Reduction with Beta-Mercaptoethanol:
  - Reagent Function: Beta-mercaptoethanol is a reducing agent that breaks disulfide bonds ( $-S-S-$ ) between Cysteine residues, converting them to free thiols ( $-SH$ ).
  - Result: "…you get two identical pentapeptides."
  - Deduction: This indicates that the decapeptide is composed of two copies of the pentapeptide (Ala-Gly-Cys-Arg-Phe from Problem 2) connected by a disulfide bond. The Cysteine residues are involved in this bond.
- Step c: Chymotrypsin Cleavage (of intact decapeptide):
  - Result: "Chymotrypsin doesn't cleave either."
  - Deduction: If the two pentapeptides were joined by a peptide bond (e.g., Ala-Gly-Cys-Arg-Phe-Ala-Gly-Cys-Arg-Phe), chymotrypsin would cleave after the first Phenylalanine. Since no cleavage occurs, the connection is not a peptide bond. This further confirms that the connection is a disulfide bond between the Cysteine residues.
- Final Structure for Problem 3: Two molecules of Ala-Gly-Cys-Arg-Phe, connected by a disulfide bond between their Cysteine residues.
Techniques and Considerations in Peptide Sequencing
- Chemical Cleavage: Cyanogen bromide is one chemical cleavage method that students should be aware of (cleaves after methionine).
- Enzymatic Cleavage:
  - Trypsin: Cleaves after Lysine and Arginine.
  - Chymotrypsin: Cleaves after Phenylalanine, Tyrosine, and Tryptophan.
- Quantitative Information: When amino acid composition is given (e.g., "glycine three"), it indicates the number of residues present, not their connectivity. For example, "three glycines" means there are three Gly residues, not necessarily connected.
Limitations of Edman Degradation for Large Proteins:
- Edman degradation is an automated process for sequencing relatively small peptides.
- For very large proteins, the cumulative yield of each step becomes problematic. If each step yields, for example, 90 ext{%}, after many steps, the results become "murky" and conflicting due to the presence of unreacted peptides from previous steps and the decreasing concentration of the desired-length peptide.

Acid-Base Disorders: Aspirin Overdose Case Study

Aspirin (Acetylsalicylic Acid):
- Structure: Contains a carboxylic acid group and an ester linkage (acetyl group).
- Hydrolysis: The ester bond can be hydrolyzed to yield salicylic acid (which then deprotonates to salicylate) and acetic acid.
Drug Solubility and pH in the Stomach:
- Stomach Environment: The stomach has a very acidic $pH$ (typically between $1$ and $2$ ).
- Aspirin/Salicylic Acid in Stomach: In this acidic environment, the carboxylic acid group ( $-COOH$ ) of aspirin and salicylic acid will be largely protonated (uncharged). The uncharged form is less soluble in water.
Lavage (Gastric Lavage) for Overdose Treatment:
- Purpose: To remove ingested substances from the stomach.
- 1. Saline Lavage (Neutral pH):
  - Mechanism: Saline is neutral, so it does not significantly change the acidic $pH$ of the stomach. Therefore, aspirin/salicylic acid remains in its uncharged, less soluble form.
  - Effectiveness: It primarily works by mechanically flushing the stomach. It is "good enough" for smaller overdoses but not optimal for solubilizing the drug.
  - Clinical Use: Often the default or first choice because it doesn't introduce other physiological complications (unlike bicarbonate).
- 2. Bicarbonate Lavage (Basic pH, e.g., > 8):
  - Mechanism: The basic solution significantly increases the $pH$ within the stomach.
  - Effect on Aspirin: This causes the carboxylic acid group to deprotonate, forming the charged carboxylate ion (salicylate, $-COO^-$ ).
  - Effectiveness: The charged (ionized) form is much more soluble in water. Therefore, bicarbonate lavage is more effective at solubilizing and removing aspirin/salicylic acid from the stomach, especially in cases of large overdoses.
  - Considerations: While more effective, bicarbonate can introduce complications like changes in blood $pH$ and kidney reabsorption, which is why it's not always the first choice.
Salicylate Detection in Blood:
- After leaving the acidic stomach and entering the less acidic gastrointestinal tract and then the blood (physiological $pH ext{ of } hickapprox 7.4$ ), salicylic acid will deprotonate to its charged salicylate form. Blood tests therefore measure salicylate levels.
- Normal Levels: In individuals not on aspirin regimens, normal salicylate levels should be $0$ .
Aspirin Overdose and Respiratory Alkalosis (Paradoxical Effect):
- Initial Expectation: Since aspirin is an acid, one might intuitively expect an overdose to cause acidosis (decreased blood $pH$ ).
- Actual Outcome: Aspirin overdose (specifically salicylate poisoning) typically leads to respiratory alkalosis (increased blood $pH$ ).
- Mechanism:
 1. Hyperventilation: Salicylate poisoning induces hyperventilation (rapid and deep breathing).
 2. $CO2$ Loss: Hyperventilation causes the body to lose an excessive amount of carbon dioxide ( $CO2$ ).
 3. Role of $CO2$ : $CO2$ dissolves in blood to form carbonic acid ( $H2CO3$ ), which is an acid. Removing $CO_2$ effectively removes acid from the body.
 4. Result: The excessive loss of acid due to hyperventilation leads to respiratory alkalosis.
- Additional Contributing Factors: Any process that removes stomach acid (e.g., vomiting or effective lavage) also contributes to making the body more basic.
- Specificity: This effect is specific to aspirin and a few other substances; it is not a general response to all drug overdoses.
Impact of pH Changes on Hemoglobin:
- Oxygen Affinity: Changes in blood $pH$ (such as alkalosis or acidosis) directly impact hemoglobin's ability to bind oxygen. This phenomenon is known as the Bohr effect.
- Binding Sites: Hemoglobin binds not only hydrogen ions ( $H^+$ ) but also carbon dioxide ( $CO_2$ ) molecules at sites distinct from the oxygen-binding site.
- Conformational Changes: Binding of $H^+$ (e.g., protonating histidine residues) or $CO_2$ (e.g., derivatizing amino groups to negatively charged carbamines) induces conformational changes in hemoglobin.
- Altered Oxygen Affinity: These conformational changes, even though they occur at distant sites, alter the affinity of hemoglobin for oxygen.
- Hemoglobin vs. Myoglobin: This complex regulatory mechanism of oxygen affinity in response to $pH$ and $CO_2$ occurs in hemoglobin (a tetramer) but not in myoglobin (a monomeric protein, primarily found in muscle tissue, that binds oxygen).
Other Peptide Problem (Decapeptide from a textbook):
- The student's example included oxidation with performic acid, which was noted as a complication not covered in the course. The emphasis for the exam is on the core sequencing methods (Edman degradation, trypsin, chymotrypsin) as taught.
Ionization of Basic Groups and Lavage:
- If a drug has basic groups (e.g., an amino group, $-NH_2$ ) and is ingested:
 - In the acidic stomach ( $pH ext{ } 1-2$ ), the basic group will be protonated ( $-NH_3^+$ ). The charged form is more soluble.
 - Saline Lavage: If saline (neutral $pH$ ) is used, it will not change the stomach environment enough to deprotonate the basic group. The drug will remain in its charged, soluble form, and saline can be effective.
 - High pH Lavage: If a very high $pH$ solution is used, it would deprotonate the basic group ( $-NH3^+ ightarrow -NH2$ ), making it uncharged and less soluble.
- Conclusion: For basic drugs, saline lavage is preferred over high $pH$ lavage to maintain the soluble, ionized form. Generally, you use saline whenever possible unless there's a problem (like an extremely large overdose of an acidic drug that benefits from bicarbonate).
- Universal Principle: All things being equal, the ionized (charged) form of a molecule is more soluble in water than its un-ionized (uncharged) form.