Biochemistry Foundations

Biochemistry Foundations

Part 1 of 7 — Amino Acids & Protein Structure

The 20 Amino Acids — Classify by Side Chain

Protein Structure Levels

Disulfide Bonds

Cysteine + Cysteine → Cystine (C-S-S-C), covalent bond stabilizing tertiary structure.

Isoelectric Point (pI) Logic

• At pH < pI, amino acids/proteins tend to carry net positive charge.
• At pH > pI, they tend to carry net negative charge.

This is heavily tested in electrophoresis and separation contexts.

Amino Acids & Protein Structure 🎯

Key Takeaways — Part 1

• Know amino acid categories cold (nonpolar, polar, positive, negative)
• Proline = helix breaker (rigid ring); Glycine = most flexible (no side chain)
• Primary: sequence. Secondary: $\alpha$-helix/$\beta$-sheet. Tertiary: 3D fold. Quaternary: subunits.
• Disulfide bonds (Cys-Cys) = only COVALENT bond in tertiary structure

Worked Examples — Amino Acids & Protein Structure

Question: Aspartic acid (Asp, D) has a side chain with a carboxylic acid (pKa ≈ 3.9). Is aspartic acid positively, negatively, or neutrally charged at pH 7?

Solution:

1. At pH 7 > pKa (3.9), the carboxylic acid is deprotonated: $-COO^{-}$
2. The $\alpha$-carboxylic acid (backbone, pKa ≈ 2) is also deprotonated: $-COO^{-}$
3. The $\alpha$-amino group (pKa ≈ 9.6) is protonated: $-NH_{3}^{+}$
4. Net charge: (−1 from side chain) + (−1 from backbone) + (+1 from backbone $NH_{3}$) = −1 (negatively charged)

MCAT Strategy: At physiological pH (~7), acidic amino acids (D, E) are negative; basic amino acids (K, R, H) are positive because their pKa values are far from 7.

Question: A protein has an isoelectric point (pI) of 5.5. At pH 7, will it migrate toward the positive or negative electrode during electrophoresis?

Solution:

1. At pH 7 > pI (5.5), the protein is above its isoelectric point
2. Above pI, the protein carries a net negative charge (more deprotonation than protonation)
3. Negative proteins migrate toward the positive electrode (anode)

MCAT Strategy: pH > pI → negative charge → migrates to (+) electrode. This is a one-liner on the MCAT.

Question: A tripeptide is Asp-Phe-Lys. At pH 7:

• Asp side chain (pKa 3.9): $-COO^{-}$
• Phe side chain: nonpolar (no charge)
• Lys side chain (pKa 10.5): $-NH_{3}^{+}$
• N-terminus (pKa ~9): $-NH_{3}^{+}$
• C-terminus (pKa ~3): $-COO^{-}$

What is the net charge?

Solution:

1. Asp side chain: pH > pKa, so $-COO^{-}$ → −1
2. Phe side chain: nonpolar → 0
3. Lys side chain: pH < pKa, so $-NH_{3}^{+}$ → +1
4. N-terminus: pH < pKa, so $-NH_{3}^{+}$ → +1
5. C-terminus: pH > pKa, so $-COO^{-}$ → −1
6. Net charge = −1 + 0 + 1 + 1 − 1 = 0

MCAT Strategy: For charges near physiological pH, check each ionizable group against its pKa. If pH >> pKa, deprotonate; if pH << pKa, protonate.

Biochemistry Foundations

Part 2 of 7 — Enzyme Kinetics (ULTRA HIGH YIELD)

Michaelis-Menten Equation

$v = \frac{V_{max}[S]}{K_m + [S]}$

• $K_m$ = Michaelis constant = [S] at which $v = V_{max}/2$
• Low $K_m$ → high affinity (enzyme binds substrate tightly at low [S])
• $V_{max}$ depends on $[E]_{total}$ and $k_{cat}$: $V_{max} = k_{cat}[E]_T$

Lineweaver-Burk (Double Reciprocal) Plot

$\frac{1}{v} = \frac{K_m}{V_{max}} \cdot \frac{1}{[S]} + \frac{1}{V_{max}}$

• y-intercept = $1/V_{max}$
• x-intercept = $-1/K_m$
• Slope = $K_m/V_{max}$

Inhibitor Types

Catalytic efficiency is often summarized by $k_{cat}/K_m$, useful for comparing enzymes at low substrate concentrations.

Enzyme Kinetics 🎯

Key Takeaways — Part 2

• $K_m$ = [S] at half $V_{max}$. Low $K_m$ = high affinity.
• Competitive: same $V_{max}$, higher apparent $K_m$
• Noncompetitive: lower $V_{max}$, same $K_m$
• Lineweaver-Burk: know how each inhibitor changes the plot

Worked Examples — Enzyme Kinetics

Question: An enzyme has $K_m = 5$ mM and $V_{max} = 120$ $µ$mol/min. What is the reaction velocity when $[S] = 15$ mM?

Solution: $v = \frac{V_{max}[S]}{K_m + [S]} = \frac{120 \times 15}{5 + 15} = \frac{1800}{20} = 90 \text{ }µ\text{mol/min}$

Interpretation:

• When $[S] = 3 \times K_m$, the enzyme is at 75% $V_{max}$ (not yet saturated)
• As [S] increases further toward infinite concentration, v → $V_{max}$ asymptotically

MCAT Strategy: You don't memorize $v = 0.75 V_{max}$ when $[S] = 3K_m$. Instead, use the equation. At low [S] << $K_m$, velocity is nearly first-order in [S]. At high [S] >> $K_m$, velocity is nearly zero-order.

Question: Three inhibitors (A, B, C) are tested. The data (without inhibitor vs. with inhibitor):

Classify each inhibitor type.

Solution:

1. Inhibitor A: $K_m$ increases (5 → 15), $V_{max}$ unchanged → Competitive ✓

   • Competes with substrate for the active site
   • Increasing [S] can overcome inhibition
   • Can be distinguished on Lineweaver-Burk (intersects on y-axis)

2. Inhibitor B: $K_m$ unchanged (5), $V_{max}$ decreases (100 → 50) → Noncompetitive ✓

   • Binds to both E and ES complex with equal affinity
   • Cannot be overcome by increasing [S]

• On Lineweaver-Burk, x-intercept stays fixed while y-intercept increases

1. Inhibitor C: $K_m$ increases (5 → 10), $V_{max}$ decreases (100 → 60) → Mixed ✓
   • Binds to both E and ES complex but with different affinities
   • Both $K_m$ and $V_{max}$ are affected (intermediate behavior)
   • Lines converge at a different point on Lineweaver-Burk

MCAT Strategy: Rapid classification: if only $V_{max}$ changes → noncompetitive; if only $K_m$ changes → competitive; if both change → mixed or uncompetitive (uncompetitive is rare but decreases both proportionally).

Question: Enzyme A: $K_m = 0.5$ mM, $k_{cat} = 500$ $s^{-1}$. Enzyme B: $K_m = 5$ mM, $k_{cat} = 2000$ $s^{-1}$. Which enzyme is more efficient when cellular [S] ≈ 0.1 mM?

Solution:

1. Catalytic efficiency = $k_{cat}/K_m$ (velocity per unit substrate at low [S])

2. Enzyme A: $\frac{k_{cat}}{K_m} = \frac{500}{0.5} = 1000 \text{ s}^{-1}\text{/mM}$

3. Enzyme B: $\frac{k_{cat}}{K_m} = \frac{2000}{5} = 400 \text{ s}^{-1}\text{/mM}$

4. Result: Enzyme A is more efficient (2.5× higher $k_{cat}/K_m$)

5. At [S] = 0.1 mM:

   • Enzyme A: $v_A = 500(0.1)/(0.5 + 0.1) ≈ 83$ $s^{-1}$ (relative to $[E]$)
   • Enzyme B: $v_B = 2000(0.1)/(5 + 0.1) ≈ 40$ $s^{-1}$
   • Enzyme A wins at low [S] ✓

Why this matters: Under cellular conditions where substrate is limiting, high $k_{cat}/K_m$ predicts enzyme performance better than either parameter alone.

MCAT Strategy: When comparing enzymes at low [S], use $k_{cat}/K_m$ as a single metric. The enzyme with higher efficiency wins "at the racetrack" when substrate is scarce.

Question: Two Lineweaver-Burk plots are shown:

Plot 1 (with inhibitor A):

• y-intercept = 0.01 (same as control)
• x-intercept = -0.2 (control: -0.25)
• Lines converge on y-axis

Plot 2 (with inhibitor B):

• y-intercept = 0.02 (control: 0.01)
• x-intercept = -0.25 (same as control)
• Lines intersect on the x-axis

Identify the inhibitor types.

Solution:

1. Lineweaver-Burk equation: $\frac{1}{v} = \frac{K_m}{V_{max}} \cdot \frac{1}{[S]} + \frac{1}{V_{max}}$

   • Slope = $K_m/V_{max}$
   • y-intercept = $1/V_{max}$
   • x-intercept = $-1/K_m$

2. Plot 1 (Inhibitor A):

   • y-intercept unchanged → $V_{max}$ unchanged
   • x-intercept changes (-0.25 → -0.2) → $K_m$ increases (from $-1/(-0.25) = 4$ mM to $-1/(-0.2) = 5$ mM)
   • Diagnosis: Competitive inhibition ✓
   • Lines converge on y-axis (characteristic pattern)

3. Plot 2 (Inhibitor B):

   • y-intercept changes → $V_{max}$ decreases (0.01 → 0.02 = $1/V_{max}$, so $V_{max}$ goes from 100 to 50)
   • x-intercept unchanged → $K_m$ unchanged

• Pattern matches noncompetitive inhibition (x-intercept conserved, y-intercept increased)

1. Answer: Inhibitor A = competitive; Inhibitor B = noncompetitive

MCAT Strategy: Memorize the Lineweaver-Burk signatures:

• Competitive: Conserved y-intercept, changed x-intercept, lines meet on y-axis
• Noncompetitive: Changed y-intercept, conserved x-intercept, lines meet on x-axis
• Uncompetitive: Both intercepts change, lines parallel

Biochemistry Foundations

Part 3 of 7 — Glycolysis & Gluconeogenesis

Glycolysis (Cytoplasm, Anaerobic)

$\text{Glucose} \xrightarrow{10\text{ steps}} 2\text{ Pyruvate}$

Net yield per glucose: 2 ATP, 2 NADH, 2 Pyruvate

Key Regulatory Enzymes (HIGH YIELD!)

PFK-1 is the PRIMARY rate-limiting step of glycolysis!

Pyruvate Fates

Gluconeogenesis bypasses the three irreversible glycolysis steps using pyruvate carboxylase/PEPCK, fructose-1,6-bisphosphatase, and glucose-6-phosphatase.

Glycolysis 🎯

Key Takeaways — Part 3

• Glycolysis: glucose → 2 pyruvate + 2 ATP + 2 NADH (cytoplasm)
• PFK-1 is the key regulatory enzyme — know its activators and inhibitors
• Lactate production regenerates NAD$^+$ for anaerobic glycolysis to continue
• Gluconeogenesis bypasses the 3 irreversible steps with different enzymes

Worked Examples — Glycolysis & Gluconeogenesis

Question: A hepatocyte has high ATP, high citrate, and low AMP. What happens to glycolysis?

Solution:

1. ATP and citrate both inhibit PFK-1.
2. Low AMP removes a major activator of PFK-1.
3. PFK-1 activity falls, so F6P to F-1,6-BP slows.
4. Net effect: glycolytic flux decreases and glucose is conserved.

MCAT tip: If ATP and citrate are both high, think "energy abundant" and downregulate glycolysis.

Question: Why do muscle cells convert pyruvate to lactate during sprinting?

Solution:

1. Electron transport cannot oxidize NADH fast enough under low oxygen delivery.
2. Glycolysis still needs NAD+ at glyceraldehyde-3-phosphate dehydrogenase.
3. Lactate dehydrogenase converts pyruvate to lactate and regenerates NAD+.
4. This allows ATP production from glycolysis to continue short term.

MCAT tip: Lactate formation is about NAD+ regeneration, not ATP gain from that step itself.

Question: Which enzymes bypass the three irreversible glycolysis steps in gluconeogenesis?

Solution:

1. Hexokinase/glucokinase bypass: glucose-6-phosphatase.
2. PFK-1 bypass: fructose-1,6-bisphosphatase.
3. Pyruvate kinase bypass: pyruvate carboxylase then PEP carboxykinase (PEPCK).

MCAT tip: The PFK-1 and fructose-1,6-bisphosphatase pair is a common regulation target in passage questions.

Biochemistry Foundations

Part 4 of 7 — TCA Cycle & Oxidative Phosphorylation

TCA Cycle (Mitochondrial Matrix)

$\text{Acetyl-CoA} + \text{OAA} \to \text{Citrate} \to \cdots \to \text{OAA (regenerated)}$

Per acetyl-CoA: 3 NADH, 1 FADH$_2$, 1 GTP

Per glucose (2 acetyl-CoA): 6 NADH, 2 FADH$_2$, 2 GTP from TCA

Electron Transport Chain (Inner Mitochondrial Membrane)

ATP Yield Per Glucose (approximate)

• Glycolysis: 2 ATP + 2 NADH (~3-5 ATP)
• PDH: 2 NADH (~5 ATP)
• TCA: 6 NADH (~15 ATP) + 2 FADH$_2$ (~3 ATP) + 2 GTP
• Total: ~30-32 ATP per glucose

Chemiosmosis links ETC to ATP synthase: proton-motive force (electrochemical gradient) drives ATP production.

TCA & ETC 🎯

Key Takeaways — Part 4

• TCA: 3 NADH + 1 FADH$_2$ + 1 GTP per acetyl-CoA
• ETC: NADH → Complex I; FADH$_2$ → Complex II (fewer ATP)
• Poisons: Rotenone (I), Antimycin A (III), Cyanide/CO (IV), Oligomycin (ATP synthase)
• ~30-32 ATP per glucose total (aerobic metabolism)

Worked Examples — TCA & Oxidative Phosphorylation

Question: How many NADH and FADH$_2$ come from the TCA cycle per glucose?

Solution:

1. One acetyl-CoA yields 3 NADH and 1 FADH$_2$ in TCA.
2. One glucose gives 2 acetyl-CoA.
3. Multiply by 2: 6 NADH and 2 FADH$_2$.

MCAT tip: Distinguish TCA output from total cellular output (which also includes glycolysis and PDH).

Question: Cyanide blocks cytochrome c oxidase. What are the immediate biochemical consequences?

Solution:

1. Electrons cannot transfer to oxygen.
2. Upstream carriers remain reduced and electron flow stalls.
3. Proton pumping collapses at I, III, and IV as flow stops.
4. ATP synthase loses proton-motive force, so oxidative ATP production drops sharply.
5. NADH accumulates, pushing metabolism toward anaerobic pathways.

MCAT tip: A block at Complex IV functionally backs up the entire chain.

Question: Compare DNP (uncoupler) with oligomycin (ATP synthase inhibitor).

Solution:

1. DNP carries protons across the inner membrane, dissipating the gradient.
2. With DNP, ETC can continue and oxygen consumption can increase, but ATP yield falls.
3. Oligomycin blocks proton flow through ATP synthase directly.
4. With oligomycin, proton backpressure slows ETC and oxygen use decreases.

MCAT tip: Uncoupler = high heat, low ATP; ATP synthase blocker = low ATP and reduced ETC throughput.

Biochemistry Foundations

Part 5 of 7 — Lipids & Fatty Acid Metabolism

Lipid Classification

$\beta$-Oxidation (Mitochondrial Matrix)

Each cycle removes 2 carbons and produces:

• 1 NADH, 1 FADH$_2$, 1 Acetyl-CoA

$\text{Palmitate (C16)} \to 7\text{ cycles} \to 8\text{ Acetyl-CoA} + 7\text{ NADH} + 7\text{ FADH}_2$

Total ATP from palmitate: $8(10) + 7(2.5) + 7(1.5) - 2 = 106$ ATP

Why Fats Store More Energy Than Carbs

Fats are more reduced (more C-H bonds) → more electrons to donate to ETC → more ATP per gram (~9 kcal/g vs ~4 kcal/g for carbs).

In fasting, lipolysis and beta-oxidation increase while fatty acid synthesis decreases due to hormonal regulation (low insulin, high glucagon).

Lipid Metabolism 🎯

Key Takeaways — Part 5

• $\beta$-oxidation: each cycle yields 1 NADH + 1 FADH$_2$ + 1 acetyl-CoA
• Carnitine shuttle: required for long-chain FA entry into mitochondria
• CPT-I is rate-limiting, inhibited by malonyl-CoA (fed state)
• Fats yield ~9 kcal/g vs ~4 kcal/g for carbs/protein

Worked Examples — Lipids & Fatty Acid Metabolism

Question: How many beta-oxidation cycles and acetyl-CoA molecules are produced from palmitate (C16:0)?

Solution:

1. Number of acetyl-CoA units for an even-chain fatty acid = $n/2$.
2. For C16: $16/2 = 8$ acetyl-CoA.
3. Number of beta-oxidation cycles = acetyl-CoA count minus 1.
4. Cycles = $8 - 1 = 7$.

So palmitate gives 7 cycles and 8 acetyl-CoA.

MCAT tip: Cycles generate NADH and FADH$_2$; acetyl-CoA then feeds TCA.

Question: In a fed state with high insulin, why is mitochondrial fatty acid oxidation suppressed?

Solution:

1. Insulin promotes fatty acid synthesis.
2. Fatty acid synthesis increases malonyl-CoA.
3. Malonyl-CoA inhibits CPT-I on the outer mitochondrial membrane.
4. Long-chain acyl groups cannot enter mitochondria efficiently.
5. Beta-oxidation decreases, avoiding futile cycling.

MCAT tip: High malonyl-CoA means synthesis mode, not oxidation mode.

Question: What is the approximate ATP equivalent from one beta-oxidation cycle before counting TCA from acetyl-CoA?

Solution:

1. Each cycle directly yields 1 NADH and 1 FADH$_2$.
2. NADH is about 2.5 ATP.
3. FADH$_2$ is about 1.5 ATP.
4. Direct total per cycle is about 4 ATP equivalents.

This does not include ATP from oxidizing the produced acetyl-CoA in TCA.

MCAT tip: Keep direct cycle yield separate from downstream TCA yield to avoid double counting.

Biochemistry Foundations

Part 6 of 7 — Nucleic Acids & Molecular Biology

DNA vs RNA

Base Pairing (Chargaff's Rules)

• A-T: 2 hydrogen bonds
• G-C: 3 hydrogen bonds → higher GC content = higher melting temp ($T_m$)

Central Dogma

$\text{DNA} \xrightarrow{\text{Transcription}} \text{mRNA} \xrightarrow{\text{Translation}} \text{Protein}$

Key Enzymes

Directionality is crucial: polymerases synthesize nucleic acid in the 5' to 3' direction by adding to a free 3'-OH.

Nucleic Acids 🎯

Key Takeaways — Part 6

• GC content correlates with $T_m$ (3 H-bonds vs 2 for AT)
• DNA polymerase: $5' \to 3'$ synthesis, needs primer, has proofreading
• RNA polymerase: no primer needed, reads template $3' \to 5'$
• Chargaff: A=T, G=C in double-stranded DNA

Worked Examples — Nucleic Acids & Molecular Biology

Question: A double-stranded DNA sample has 18% guanine. What are the percentages of all four bases?

Solution:

1. In dsDNA, G = C and A = T.
2. If G is 18%, then C is 18%.
3. G + C is 36%, so A + T is 64%.
4. Therefore A is 32% and T is 32%.

Final composition: A 32%, T 32%, G 18%, C 18%.

MCAT tip: Solve these in two moves: pair equalities first, then total to 100%.

Question: If the template DNA strand is 3'-TAC GGA TTT-5', what mRNA sequence is transcribed?

Solution:

1. RNA polymerase reads template 3' to 5'.
2. mRNA is synthesized 5' to 3'.
3. Apply base pairing with U instead of T in RNA:
   • T to A
   • A to U
   • C to G
   • G to C
4. Result: 5'-AUG CCU AAA-3'.

MCAT tip: Always write orientation marks first to avoid reversing the final answer.

Question: Which fragment has higher melting temperature, and why?

Fragment A: 5'-GCGCGCAAAT-3' Fragment B: 5'-ATATATATGC-3'

Solution:

1. Fragment A has higher GC fraction.
2. GC pairs have three H-bonds and stronger stacking interactions than AT-rich regions.
3. More thermal energy is needed to separate the duplex.

So Fragment A has higher $T_m$.

MCAT tip: GC content is a fast proxy for duplex stability in PCR and denaturation questions.

Biochemistry Foundations

Part 7 of 7 — Metabolic Regulation & Integration

Fed vs. Fasting State

Key Regulatory Hormones

• Insulin: promotes anabolism (storage). Activates PFK-1, pyruvate kinase, glycogen synthase.
• Glucagon: promotes catabolism (mobilization). Activates glycogen phosphorylase, lipase.
• Epinephrine: fight-or-flight. Similar to glucagon + increases heart rate.

Metabolic Integration

• High ATP/NADH → inhibits TCA, glycolysis (feedback)
• Acetyl-CoA activates pyruvate carboxylase (gluconeogenesis) and inhibits PDH
• Malonyl-CoA (from fatty acid synthesis) inhibits CPT-I (blocks $\beta$-oxidation)
• You can't do fatty acid synthesis AND $\beta$-oxidation simultaneously!

Integrated MCAT passages often ask you to infer dominant pathways from hormone state plus one metabolite signal (ATP, acetyl-CoA, or malonyl-CoA).

Metabolic Integration 🎯

Biochemistry Foundations — Complete! ✅

Key themes: Fed state = insulin = anabolism. Fasting = glucagon = catabolism. Know how pathways are coordinated and regulated — the MCAT loves integrative metabolic questions.

Worked Examples — Metabolic Integration

Question: A healthy person has fasted for 14 hours. Which liver pathways are upregulated?

Solution:

1. Hormone state shifts to low insulin and higher glucagon.
2. Glycogenolysis is active early in fasting.
3. Gluconeogenesis rises as fasting continues.
4. Lipolysis in adipose tissue increases fatty acid delivery to liver.
5. Hepatic beta-oxidation rises to provide ATP for gluconeogenesis.

MCAT tip: Pair fasting with glucagon, glucose output, and increased fatty acid use.

Question: Why is beta-oxidation low in the fed state even when fatty acids are available?

Solution:

1. Insulin promotes acetyl-CoA carboxylase activity.
2. Malonyl-CoA concentration rises.
3. Malonyl-CoA inhibits CPT-I.
4. Long-chain fatty acids cannot enter mitochondria efficiently.
5. Fat oxidation drops while synthesis/storage pathways dominate.

MCAT tip: High malonyl-CoA is a direct signal that mitochondrial fatty acid entry is blocked.

Question: In untreated Type 1 diabetes, why do ketone bodies rise despite hyperglycemia?

Solution:

1. Absolute insulin deficiency makes cells unable to use glucose effectively.
2. Counterregulatory hormones rise, especially glucagon.
3. Adipose lipolysis releases large amounts of fatty acids.
4. Liver increases beta-oxidation, generating excess acetyl-CoA.
5. Oxaloacetate is diverted to gluconeogenesis, limiting TCA flux.
6. Excess acetyl-CoA is shunted to ketogenesis.

Result: elevated acetoacetate and beta-hydroxybutyrate, causing metabolic acidosis.

MCAT tip: DKA is a hormone-signaling problem first, glucose concentration problem second.