🎯⭐ INTERACTIVE LESSON

Biochemistry Foundations

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Biochemistry Foundations - Complete Interactive Lesson

Part 1: Amino Acids & Protein Structure

Biochemistry Foundations

Part 1 of 7 — Amino Acids & Protein Structure

The 20 Amino Acids — Classify by Side Chain

Category	Amino Acids	Key Feature
Nonpolar/Hydrophobic	G, A, V, L, I, P, F, W, M	Interior of proteins
Polar uncharged	S, T, C, Y, N, Q	H-bonding
Positively charged (pH 7)	K, R, H	Basic side chains
Negatively charged (pH 7)	D, E	Acidic side chains

Protein Structure Levels

Level	Held together by	Description
Primary (1°)	Peptide bonds (covalent)	Amino acid sequence
Secondary (2°)	H-bonds (backbone N-H to C=O)	$\alpha$ -helix, $\beta$ -sheet
Tertiary (3°)	Hydrophobic, ionic, disulfide, H-bonds	3D shape of one polypeptide
Quaternary (4°)	Same as tertiary	Multiple subunit assembly

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

<details> <summary>Example 1: Classify amino acid by charge at different pH values</summary>

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:

At pH 7 > pKa (3.9), the carboxylic acid is deprotonated: −COO⁻
The $\alpha$ -carboxylic acid (backbone, pKa ≈ 2) is also deprotonated: −COO⁻
The $\alpha$ -amino group (pKa ≈ 9.6) is protonated: −NH₃⁺
Net charge: (−1 from side chain) + (−1 from backbone) + (+1 from backbone NH₃) = −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.

</details> <details> <summary>Example 2: Determine protein behavior in electrophoresis</summary>

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:

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

</details> <details> <summary>Example 3: Calculate effective charge on a peptide</summary> </details>

Part 2: Enzyme Kinetics

Biochemistry Foundations

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

Michaelis-Menten Equation

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

Part 3: Carbohydrate Metabolism

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

Part 4: Lipids & Membranes

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

Part 5: Nucleic Acids & DNA

Biochemistry Foundations

Part 5 of 7 — Lipids & Fatty Acid Metabolism

Lipid Classification

Type	Structure	Function
Triglycerides	3 fatty acids + glycerol	Energy storage
Phospholipids	2 fatty acids + glycerol + phosphate head	Membranes
Steroids	4 fused rings	Hormones (cholesterol, testosterone, estrogen)
Sphingolipids	Sphingosine backbone	Myelin, cell signaling

$\beta$ -Oxidation (Mitochondrial Matrix)

Each cycle removes 2 carbons and produces:

1 NADH, 1 FADH $_2$ , 1 Acetyl-CoA

Part 6: Bioenergetics & ATP

Biochemistry Foundations

Part 6 of 7 — Nucleic Acids & Molecular Biology

DNA vs RNA

Feature	DNA	RNA
Sugar	Deoxyribose	Ribose ( $2'-OH$ )
Bases	A, T, G, C	A, U, G, C
Structure	Double-stranded	Usually single-stranded
Stability	More stable	Less stable ( makes it prone to hydrolysis)

Part 7: Review & MCAT Practice

Biochemistry Foundations

Part 7 of 7 — Metabolic Regulation & Integration

Fed vs. Fasting State

State	Insulin	Glucagon	Active Pathways
Fed	High	Low	Glycolysis, glycogenesis, lipogenesis, protein synthesis
Fasting	Low	High	Gluconeogenesis, glycogenolysis, $\beta$ -oxidation, ketogenesis

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 -oxidation)

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): −NH₃⁺
N-terminus (pKa ~9): −NH₃⁺
C-terminus (pKa ~3): −COO⁻

What is the net charge?

Asp side chain: pH > pKa, so −COO⁻ → −1
Phe side chain: nonpolar → 0
Lys side chain: pH < pKa, so −NH₃⁺ → +1
N-terminus: pH < pKa, so −NH₃⁺ → +1
C-terminus: pH > pKa, so −COO⁻ → −1
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.

$K_m$ = 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 :

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 = $K_m/V_{max}$

Type	Effect on $V_{max}$	Effect on $K_m$	Overcome by more [S]?
Competitive	No change	Increases (apparent)	Yes
Uncompetitive	Decreases	Decreases	No
Noncompetitive	Decreases	No change	No
Mixed	Decreases	Can increase or decrease	No

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

<details> <summary>Example 1: Calculate enzyme velocity using Michaelis-Menten equation</summary>

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: $v = rac{V_{max}[S]}{K_m + [S]} = rac{120 imes 15}{5 + 15} = rac{1800}{20} = 90 ext{ }µ ext{mol/min}$

Interpretation:

When $[S] = 3 imes K_m$ , the enzyme is at 75% $V_{max}$ (not yet saturated)

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

</details> <details> <summary>Example 2: Distinguish inhibitor types using kinetic parameters</summary>

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

Inhibitor	$K_m$ (mM)	$V_{max}$ ( $µ$ mol/min)

Classify each inhibitor type.

Solution:

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)
Inhibitor B: unchanged (5), decreases (100 → 50) → ✓

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

Inhibitor C: $K_m$ $K_{m}$ increases (5 → 10), $V_{max}$ $V_{ma x}$ decreases (100 → 60) → Mixed ✓
- Binds to both E and ES complex but with different affinities
- Both $K_m$ and are affected (intermediate behavior)

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).

</details> <details> <summary>Example 3: Compare catalytic efficiency of two enzymes at low substrate concentration</summary>

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

Solution:

Catalytic efficiency = $k_{cat}/K_m$ (velocity per unit substrate at low [S])
Enzyme A: $rac{k_{cat}}{K_m} = rac{500}{0.5} = 1000 ext{ s}^{-1} ext{/mM}$
Enzyme B: $rac{k_{cat}}{K_m} = rac{2000}{5} = 400 ext{ s}^{-1} ext{/mM}$
Result: Enzyme A is more efficient (2.5× higher )

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.

</details> <details> <summary>Example 4: Interpret a Lineweaver-Burk (double reciprocal) plot</summary>

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:

Lineweaver-Burk equation: $rac{1}{v} = rac{K_m}{V_{max}} cdot rac{1}{[S]} + rac{1}{V_{max}}$
- Slope = $K_m/V_{max}$
- y-intercept = $1/V_{max}$

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

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

</details>

Key Regulatory Enzymes (HIGH YIELD!)

Enzyme	Step	Activated by	Inhibited by
Hexokinase	Glucose → G6P	—	G6P (product inhibition)
PFK-1	F6P → F-1,6-BP	AMP, fructose-2,6-BP	ATP, citrate
Pyruvate kinase	PEP → Pyruvate	F-1,6-BP	ATP, alanine

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

Condition	Pathway	Product
Aerobic	Pyruvate dehydrogenase	Acetyl-CoA → TCA
Anaerobic (muscle)	Lactate dehydrogenase	Lactate (regenerates NAD $^+$ )
Anaerobic (yeast)	Pyruvate decarboxylase	Ethanol + CO $_2$

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

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

<details> <summary>Example 1: Predict glycolysis flux from metabolite levels</summary>

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

Solution:

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

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

</details> <details> <summary>Example 2: Explain lactate production during exercise</summary>

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

Solution:

Electron transport cannot oxidize NADH fast enough under low oxygen delivery.
Glycolysis still needs NAD+ at glyceraldehyde-3-phosphate dehydrogenase.
Lactate dehydrogenase converts pyruvate to lactate and regenerates NAD+.
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.

</details> <details> <summary>Example 3: Match irreversible glycolysis steps to bypass enzymes</summary>

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

Solution:

Hexokinase/glucokinase bypass: glucose-6-phosphatase.
PFK-1 bypass: fructose-1,6-bisphosphatase.
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.

</details>

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

Electron Transport Chain (Inner Mitochondrial Membrane)

Complex	Accepts from	Pumps H $^+$
I (NADH dehydrogenase)	NADH	Yes (4 H $^+$ )
II (Succinate dehydrogenase)	FADH $_2$	No
III (Cytochrome bc1)	CoQ	Yes (4 H $^+$ )
IV (Cytochrome c oxidase)	Cyt c → O $_2$	Yes (2 H $^+$ )
ATP Synthase (V)	H $^+$ gradient	Makes ATP

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.

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

<details> <summary>Example 1: Compute reducing equivalents from one glucose</summary>

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

Solution:

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

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

</details> <details> <summary>Example 2: Predict effects of cyanide at Complex IV</summary>

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

Solution:

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

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

</details> <details> <summary>Example 3: Distinguish uncouplers from ATP synthase inhibitors</summary>

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

Solution:

DNP carries protons across the inner membrane, dissipating the gradient.
With DNP, ETC can continue and oxygen consumption can increase, but ATP yield falls.
Oligomycin blocks proton flow through ATP synthase directly.
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.

</details>

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

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

<details> <summary>Example 1: Count beta-oxidation cycles for palmitate</summary>

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

Solution:

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

</details> <details> <summary>Example 2: Explain CPT-I regulation by malonyl-CoA</summary>

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

Solution:

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

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

</details> <details> <summary>Example 3: Estimate ATP from one beta-oxidation cycle</summary>

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

Solution:

Each cycle directly yields 1 NADH and 1 FADH $_2$ .
NADH is about 2.5 ATP.
FADH $_2$ is about 1.5 ATP.
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.

</details>

Base Pairing (Chargaff's Rules)

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

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

Process	Key Enzyme	Function
Replication	DNA polymerase III	Synthesizes new DNA strand ( $5' \to 3'$ )
Replication	Helicase	Unwinds double helix
Replication	Primase	Makes RNA primer
Transcription	RNA polymerase	Synthesizes mRNA from DNA template
Translation	Ribosome	Reads mRNA, assembles protein

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

<details> <summary>Example 1: Apply Chargaff's rules quickly</summary>

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

Solution:

In dsDNA, G = C and A = T.
If G is 18%, then C is 18%.
G + C is 36%, so A + T is 64%.
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%.

</details> <details> <summary>Example 2: Determine complement and strand direction</summary>

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

Solution:

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

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

</details> <details> <summary>Example 3: Compare melting temperatures from GC content</summary>

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

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

Solution:

Fragment A has higher GC fraction.
GC pairs have three H-bonds and stronger stacking interactions than AT-rich regions.
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.

</details>

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

<details> <summary>Example 1: Predict pathway dominance after an overnight fast</summary>

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

Solution:

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

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

</details> <details> <summary>Example 2: Connect malonyl-CoA to fat oxidation</summary>

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

Solution:

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

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

</details> <details> <summary>Example 3: Explain diabetic ketoacidosis mechanism</summary>

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

Solution:

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

</details>

V_{max} = k_{cat}[E]_T

As [S] increases further toward infinite concentration, v →

V_{max}

asymptotically

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

Lines converge at a different point on Lineweaver-Burk

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] ✓

x-intercept =

-1/K_m

Plot 1 (Inhibitor A):

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

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

Biochemistry Foundations

Lineweaver-Burk (Double Reciprocal) Plot

Inhibitor Types

Key Takeaways — Part 2

Worked Examples — Enzyme Kinetics

Key Regulatory Enzymes (HIGH YIELD!)

Pyruvate Fates

Key Takeaways — Part 3

Worked Examples — Glycolysis & Gluconeogenesis

Electron Transport Chain (Inner Mitochondrial Membrane)

ATP Yield Per Glucose (approximate)

Key Takeaways — Part 4

Worked Examples — TCA & Oxidative Phosphorylation

Why Fats Store More Energy Than Carbs

Key Takeaways — Part 5

Worked Examples — Lipids & Fatty Acid Metabolism

Base Pairing (Chargaff's Rules)

Central Dogma

Key Enzymes

Key Takeaways — Part 6

Worked Examples — Nucleic Acids & Molecular Biology

Biochemistry Foundations — Complete! ✅

Worked Examples — Metabolic Integration

None (control)	5	100
A	15	100
B	5	50
C	10	60

Biochemistry Foundations

Biochemistry Foundations - Complete Interactive Lesson

Part 1: Amino Acids & Protein Structure

Biochemistry Foundations

The 20 Amino Acids — Classify by Side Chain

Protein Structure Levels

Disulfide Bonds

Isoelectric Point (pI) Logic

Key Takeaways — Part 1

Worked Examples — Amino Acids & Protein Structure

Part 2: Enzyme Kinetics

Biochemistry Foundations

Michaelis-Menten Equation

Part 3: Carbohydrate Metabolism

Biochemistry Foundations

Glycolysis (Cytoplasm, Anaerobic)

Part 4: Lipids & Membranes

Biochemistry Foundations

TCA Cycle (Mitochondrial Matrix)

Part 5: Nucleic Acids & DNA

Biochemistry Foundations

Lipid Classification

β\betaβ-Oxidation (Mitochondrial Matrix)

Part 6: Bioenergetics & ATP

Biochemistry Foundations

DNA vs RNA

Part 7: Review & MCAT Practice

Biochemistry Foundations

Fed vs. Fasting State

Key Regulatory Hormones

Metabolic Integration

Lineweaver-Burk (Double Reciprocal) Plot

Inhibitor Types

Key Takeaways — Part 2

Worked Examples — Enzyme Kinetics

Key Regulatory Enzymes (HIGH YIELD!)

Pyruvate Fates

Key Takeaways — Part 3

Worked Examples — Glycolysis & Gluconeogenesis

Electron Transport Chain (Inner Mitochondrial Membrane)

ATP Yield Per Glucose (approximate)

Key Takeaways — Part 4

Worked Examples — TCA & Oxidative Phosphorylation

Why Fats Store More Energy Than Carbs

Key Takeaways — Part 5

Worked Examples — Lipids & Fatty Acid Metabolism

Base Pairing (Chargaff's Rules)

Central Dogma

Key Enzymes

Key Takeaways — Part 6

Worked Examples — Nucleic Acids & Molecular Biology

Biochemistry Foundations — Complete! ✅

Worked Examples — Metabolic Integration

$\beta$ -Oxidation (Mitochondrial Matrix)