Pharmacology Exam Prep Notes: pKa/pH, Degradation, MRSA, and Beta-Lactamase Resistance

pKa, pH, and solubility in drug molecules

• Key idea: many drug instability issues arise when exposed to acids or bases; pKa–pH relationships affect ionization, solubility, and activity.
• Major relationship: pKa and pH determine the ionization state of ionizable groups in drugs; this changes solubility, permeability, and stability.
• Practical exam mnemonic used in class: if pKa > pH, the molecule tends to be predominantly neutral; if pH = pKa, you have ~50% neutral and ~50% charged.
• Example setup from slides:
  • Given pH = 6.5 and pKa = 7.4, the compound will be predominantly neutral (since pKa > pH).
  • In the multiple-choice example, all other structures were charged ions (positive, negative, or zwitterions) except the neutral one, which is the correct answer.
• Key quantitative relation (Henderson–Hasselbalch):
  • For an acid HA ⇌ A⁻ + H⁺, \mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[A^-]}{[HA]}\right)
  • Therefore, \frac{[A^-]}{[HA]} = 10^{\mathrm{pH} - \mathrm{p}K_a}
  • Fraction protonated (neutral HA) for an acid:
    f{HA} = \frac{1}{1 + 10^{(\mathrm{pH} - \mathrm{p}Ka)}}
  • Fraction deprotonated (A⁻) for an acid:
    f{A^-} = \frac{1}{1 + 10^{(\mathrm{p}Ka - \mathrm{pH})}}
• Practical takeaway: as pH approaches pKa, ionization shifts from predominantly neutral to more charged; a small change in pH near pKa can greatly alter charge state and behavior.
• If the pH is higher than the pKa, the compound is predominantly charged (ionized).
• Charge state is also media-dependent: in acidic media, basic groups tend to be protonated; in basic media, acids tend to be deprotonated.
• Specific example: for a pair of nearby pKa values (e.g., pKa ~10 and pKa ~5 for two functional groups on a molecule), protonation/deprotonation depends on the surrounding medium and the specific pKa values.

Tetracycline degradation and Fanconi-like syndrome

• Tetracycline can degrade to toxic products under certain conditions; one degradation pathway can lead to Fanconi-like syndrome if an anhydrous species is formed.
• Key structural requirement for this degradation pathway: presence of a hydroxy group at C6 is involved; the discussion explicitly states C6 hydroxy is necessary (the example mentions a C6 hydroxy group and labels the correct structure accordingly).
• Conditions leading to different degradation products:
  • Acidic conditions: formation of an anhydrous species that is linked to Fanconi-like toxicity.
  • Basic conditions: formation of a lactone, which inactivates the antibiotic (reduces efficacy).
• Mechanistic note: hydrolysis/degradation depends on whether the medium is acidic or basic; base attack on the carbonyl can open a ring and drive decarboxylation to form a penicillin-like product (illustrative naming in lecture material).
• About alpha vs beta, and ring fusion in tetracyclines:
  • The hydroxy at C6 is described as alpha in the discussion; there is also a beta substituent elsewhere.
  • The instructor notes that tetracyclines are generally described as having beta ring fusion in this context; however, the important point is that cis (syn) ring junctions (the two rings joining at A/B) are required for potency because they constrain the molecule into a conformation that binds its target.
• Practical implications: if the drug degrades in basic solution, potency drops because the active form is lost; if the drug degrades in acidic solution, a toxic species may form.
• Example names (as given in lecture):
  • Acid-catalyzed degradation product discussed as a penaldehyde species (penaldehyde).
  • Base-catalyzed/dehydrocyclization product discussed as penaloic acid and related species (linked to penicillin G mapping in the lecture example).
• Takeaway: structure + media determine the degradation product and its pharmacologic consequence; familiarity with the specific structures shown in the slides is essential for identifying which product corresponds to which condition.

Penicillin/β-lactam degradation and products under different conditions

• Basic hydrolysis product (base):
  • Hydroxide attacks the carbonyl, opens the four-membered β-lactam ring, generating a carboxylate at the attack site, which can decarboxylate to form a penaloic acid-type product (penicillin G derivative).
  • In the example, the product shown for base conditions is penaloic acid (Pen-aloic acid).
• Acidic hydrolysis product (acid):
  • The most reactive site in the side chain/benzylic region can undergo acid-catalyzed transformations producing penaldehyde species (penaldehyde) in the example.
  • The lecturer emphasizes being able to recognize these hydrolysis products from structures shown in the exam, noting the exact product depends on which penicillin analogue is present (e.g., methicillin, ampicillin, or others).
• Co-administered inhibitor to prevent degradation by renal enzymes (imipenem context):
  • The drug mentioned with a co-administered inhibitor to prevent hydrolysis by renal dehydropeptidase I (DHP-1) is effectively cilastatin (noted in the talk as liselstatine).
  • Real-world pairing: Imipenem + cilastatin, where cilastatin inhibits renal dehydropeptidase I, increasing imipenem’s urinary half-life.
• Drug structure examples discussed:
  • Ampicillin: described as having a phenylalanine side chain.
  • Foxicillin: described as containing an isoxazole ring (isoxazole-containing penicillin).
  • A penicillin that decarboxylates to penicillin G (benzylpenicillin) was mentioned as part of the degradation discussion.
• Practical exam note: be able to map structural changes to the products under acid vs base conditions and recognize which product corresponds to which condition.

Fluoroquinolones: side effects and structure–activity hints

• Which is NOT a common side effect? The talk mentions three concern categories for fluoroquinolones:
  • CNS side effects can be reduced by having a fluorine at the C6 position.
  • Other common side effects include toxicity (general toxicity) and joint erosion (arthropathy).
• The slide/teacher’s point: the presence of a fluorine at C6 reduces CNS adverse effects; the other two listed side effects (toxicity and joint erosion) are known concerns with fluoroquinolones.
• Practical implication: structural modifications (like C6 fluorination) can mitigate some CNS adverse effects, but other toxicities remain relevant for clinical decisions.

Methicillin resistance and beta-lactamase resistance concepts

• Mechanism of MRSA resistance (methicillin-resistant Staphylococcus aureus):
  • Resistance is conferred by modification of the biological target, specifically penicillin-binding proteins (PBPs).
  • The altered PBP binds β-lactam antibiotics poorly, reducing drug efficacy.
• Beta-lactamase resistance factors (two main contributors):
  • Steric hindrance around the β-lactam ring (bulky groups block β-lactamase access).
  • Electron-withdrawing groups on the drug structure (stabilize the amide bond against hydrolysis and reduce electrophilicity of the β-lactam carbonyl).
• Example ranking (as discussed in the class):
  • Methicillin is the most resistant to β-lactamase among the listed penicillins due to its bulky, electron-withdrawing side chain and steric hindrance.
  • The instructor described a ranking exercise: 1 = greatest resistance (methicillin), with subsequent positions involving other penicillins (Ampicillin, etc.) and ultimately Penicillin G having the least resistance among the set. The exact order beyond methicillin was discussed as a reasoning exercise rather than memorization of a fixed list.
• Why structure matters: Both steric bulk (hindrance) and electronic effects (withdrawal) influence how readily β-lactamases can attack the β-lactam carbonyl; drugs with bulky, electron-withdrawing substituents are less susceptible to hydrolysis.
• Practical outcome: When selecting a penicillin in a resistant setting, consider whether the antibiotic’s structure affords β-lactamase resistance; methicillin-like structures show the strongest resistance in the examples discussed.

Penicillins and their canonical structural notes from the lecture

• Methicillin: highest β-lactamase resistance; bulky, electron-withdrawing substituents contribute to resistance.
• Ampicillin: contains an amine/side chain that is electron-withdrawing in certain environments; protonation state around physiological pH affects electrophilicity and β-lactamase interaction.
• “Foxicillin” (isoxazole-containing penicillin): demonstrates how heterocyclic ring substitutions affect properties.
• Benzylpenicillin (Penicillin G): discussed as a reference or baseline with comparatively lower β-lactamase resistance.
• Instructor emphasized being able to identify products of hydrolysis and degradation across different penicillins; exam questions may require selecting the correct degradation product given a specific penicillin and reaction condition.

Specific question-and-answer highlights from the lecture

• Which drug exhibits the greatest beta-lactamase resistance? Methicillin (structure 1 in the example).
• How to compare others? The two key concepts are steric hindrance and electron-withdrawing groups; a ranking can be constructed by evaluating these features, though exact ordering beyond methicillin was presented as an on-the-fly exercise.
• What is co-administered with liselstatine to prevent renal hydrolysis? Cilastatin (inhibitor of renal dehydropeptidase I).
• Which agent has the lowest minimum inhibitory concentration (MIC) against E. coli at pH 7? Sulfonamide antibiotics; the PK window around 6.7–7.4 is highlighted as relevant to potency and PK/PD alignment.
• What structural modification reduces CNS side effects in fluoroquinolones? A fluorine at C6 reduces CNS toxicity; other adverse effects include general toxicity and joint erosion.
• Cis/trans ring fusion in tetracyclines: what does cis AV ring fusion mean? The junctions A and B must be on the same face (syn) for a cis ring junction; a trans junction would reduce potency by altering the molecule’s preferred conformation for target binding.
• Why is ring fusion important for activity? The preferred conformation stabilizes binding to the drug’s target; a trans arrangement can reduce or abolish activity.

Ring numbering and structural labeling (practice note from the lecture)

• The instructor referenced a specific figure with ring numbering (A, B, C, D) and a sequence to count carbons (OC67). Students were advised to refer to the labeled picture in their notes for accuracy.
• When asked about counting, the instructor noted that the notes contain a labeled figure; students were reminded to review that figure to understand how the rings are labeled and how carbons are counted in the scaffolds discussed (A, B, C, D rings).

Exam and study strategy notes

• The class mentioned 36 exam questions; it may feel large but was framed as manageable for capable students.
• Advice to students:
  • Spend time studying the material rather than relying on last-minute cramming.
  • Practice identifying ionization states, knowing which forms are neutral vs charged in given pH contexts.
  • Be able to predict degradation products under acid vs base conditions for penicillins and tetracyclines, given the structure.
  • Understand how structural features (bulky, electron-withdrawing substituents) contribute to β-lactamase resistance.
• Administrative reminders from the lecture:
  • Attendance/check-in on Pointsolutions is open; fill out the card with group members present and your group number for test identification.
  • Decide on an official “scratch-off” presenter during the session.
  • Bring nonprogrammable calculators for the exam.
• Communication:
  • The instructor offered to reply to emails and to return phone calls if a number is left; students are encouraged to reach out with questions.

Quick reference formulas and numbers to remember

• Henderson–Hasselbalch for acids: \mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[A^-]}{[HA]}\right)
• Ratio of ionized to unionized: \frac{[A^-]}{[HA]} = 10^{\mathrm{pH} - \mathrm{p}K_a}
• Fraction ionized for acids: f{A^-} = \frac{1}{1 + 10^{(\mathrm{p}Ka - \mathrm{pH})}}
• Fraction ionized for bases: f{BH^+} = \frac{1}{1 + 10^{(\mathrm{pH} - \mathrm{p}Ka)}}
• Example reference values mentioned: pH = 6.5, pKa = 7.4 -> ratio [A^-]/[HA] = 10^{-0.9} ≈ 0.126, so ~11% ionized, ~89% neutral (approximate).

Note on terminology in lecture materials

• The lecture uses some specific product names that may appear slightly different in written texts (e.g., penaldehyde, penaloic acid, penaldehyde species). The core idea is to recognize that acid-catalyzed hydrolysis tends to produce one product, while base-catalyzed hydrolysis produces a different product, and that these labeling conventions depend on the exact penicillin analogue.
• The inhibitor name mentioned in the talk for renal dehydro-peptidase I is commonly known as cilastatin in clinical practice; the transcript uses liselstatine, which appears to be a mispronunciation or misnaming in the lecture.

Summary takeaway

• Understanding pKa–pH relationships is critical for predicting ionization, solubility, and drug stability.
• Drug instability often arises from acid/base exposure; predicting degradation products requires knowing the drug’s structure and the reaction conditions.
• Beta-lactamase resistance hinges on steric hindrance and electron-withdrawing substituents; methicillin is a canonical example of high resistance.
• Co-administration strategies (cilastatin with imipenem) illustrate how pharmacokinetic interactions can preserve drug activity by preventing enzymatic degradation.
• Ring fusion and stereochemical orientation (cis vs trans) in tetracyclines directly influence binding and potency.
• The exam will test recognition of degradation products and the ability to infer ionization states and potential clinical implications from structure and pH context; expect a mix of mechanism-based questions and structure-identification questions.

References to the lecture organization and support

• All slides and figures referenced in these notes were shown in class; students should review the Blackboard resources for the exact structures and names used in the exam examples.
• If you have questions, email the instructor or leave a voicemail; the instructor indicated willingness to respond.