Laboratory Methods & Separations

Laboratory Methods & Separations

Part 1 of 5 — Chromatography (Separating by Affinity)

Chromatography separates a mixture distributed between a stationary phase (fixed) and a mobile phase (moving). Molecules that interact MORE with the stationary phase move SLOWER.

$\text{Separation} \;\propto\; \frac{\text{affinity for stationary phase}}{\text{affinity for mobile phase}}$

Major Chromatography Types

Retention Factor ($R_f$) for TLC

$R_f = \frac{\text{distance traveled by spot}}{\text{distance traveled by solvent front}}$

• $R_f$ is always between 0 and 1.
• Normal phase (polar plate): nonpolar compounds travel farther → higher $R_f$.
• A compound that "streaks" or sticks at the baseline ($R_f \approx 0$) is too polar for the chosen solvent.

Chromatography 🎯

Worked Examples — Chromatography

Question: On a TLC plate, a spot travels 3.0 cm while the solvent front travels 7.5 cm. Find $R_f$. If a second compound has $R_f = 0.80$, which is more polar (normal phase)?

Solution: $R_f = \frac{3.0}{7.5} = 0.40$ The first compound ($R_f = 0.40$) traveled less far than the second ($R_f = 0.80$). On a polar normal-phase plate, lower $R_f$ = stronger binding = MORE polar. So compound 1 is more polar. ✓

MCAT note: To raise a stuck polar compound's $R_f$, use a MORE polar mobile phase (it competes the analyte off the plate).

Question: A lysate contains proteins differing mainly in net charge at pH 7. Which method best separates them, and how are they eluted?

Solution:

1. Differences in CHARGE → ion-exchange chromatography. ✓
2. A cation-exchange resin (negatively charged) binds positively charged proteins; an anion-exchange resin binds negative ones.
3. Elution: apply a salt gradient (competing ions displace the protein) or change pH (toward the protein's pI to neutralize it). Weakly bound proteins elute first.

Key idea: Match the molecular property that differs (size, charge, polarity, specific binding) to the mechanism of the column.

Question: A mixture of pentane (bp 36 °C), hexane (bp 69 °C), and octane (bp 126 °C) is run on a nonpolar GC column. Predict the elution order.

Solution:

1. GC separates largely by volatility (boiling point) on a nonpolar column.
2. The most volatile (lowest bp) spends the most time in the gas/mobile phase → elutes FIRST.
3. Order: pentane → hexane → octane. ✓

Connection: On GC, low boiling point = short retention time. Adding polarity to the column would further retain polar analytes regardless of bp.

Key Takeaways — Part 1

• Chromatography = partition between stationary and mobile phases; stronger stationary affinity = slower migration.
• $R_f$ = (spot distance)/(solvent front); normal phase → nonpolar travels farther.
• Size-exclusion: LARGE elutes first. Ion-exchange: separates by charge, elute with salt/pH. Affinity: most specific (e.g., His-tag/Ni).
• GC separates volatile analytes; lowest boiling point elutes first.

Laboratory Methods & Separations

Part 2 of 5 — Electrophoresis (Separating by Size & Charge)

Electrophoresis drives charged molecules through a gel matrix in an electric field. Migration depends on charge, size, and the gel's sieving.

$\text{Migration velocity} \;\propto\; \frac{q \, E}{f}$

where $q$ = net charge, $E$ = field strength, and $f$ = frictional drag (rises with size and gel density).

Key Variants

Direction of Migration

• DNA and SDS-coated proteins carry NET NEGATIVE charge → migrate toward the anode (+).
• Smaller molecules thread through the gel faster → travel FARTHER.

Isoelectric Point (pI)

$\text{At pH} = \text{pI}, \quad \text{net charge} = 0 \Rightarrow \text{molecule stops migrating}$

• pH > pI → molecule is net negative → moves to anode (+).
• pH < pI → molecule is net positive → moves to cathode (−).

Electrophoresis 🎯

Worked Examples — Electrophoresis

Question: On an SDS-PAGE gel, a ladder shows bands at 75, 50, 37, and 25 kDa from top to bottom. An unknown protein runs slightly below the 37 kDa marker. Estimate its mass and explain the logic.

Solution:

1. In SDS-PAGE, SMALLER proteins migrate FARTHER (lower on the gel). ✓
2. The unknown sits between 37 and 25 kDa, closer to 37 → roughly ~33 kDa. ✓
3. Migration distance varies linearly with log(MW), so the ladder calibrates the estimate.

MCAT trap: "Lower on the gel = larger" is WRONG. Lower = traveled farther = smaller.

Question: Three proteins (pI = 4, 7, 9) are loaded into native gels buffered at pH 7. Predict each protein's migration.

Solution:

• pI 4 (pH 7 > pI): net negative → migrates to anode (+). ✓
• pI 7 (pH 7 = pI): net charge ≈ 0 → essentially does not migrate. ✓
• pI 9 (pH 7 < pI): net positive → migrates to cathode (−). ✓

Key rule: Compare buffer pH to pI. Above pI → negative → anode. Below pI → positive → cathode.

Question: A researcher runs a 2D gel (IEF horizontally, SDS-PAGE vertically). Two spots have the same vertical position but different horizontal positions. What does this tell you?

Solution:

1. Same VERTICAL position (SDS-PAGE dimension) → same molecular weight/size. ✓
2. Different HORIZONTAL position (IEF dimension) → different pI. ✓
3. Conclusion: the two species are the same size but differ in charge — consistent with post-translational modifications (e.g., phosphorylation adds negative charge, lowering pI) producing protein isoforms.

Why 2D matters: It resolves proteins that co-migrate in a single dimension by separating on two independent properties.

Key Takeaways — Part 2

• Migration ∝ charge × field / friction; smaller molecules travel farther through the gel.
• SDS-PAGE: size only (SDS normalizes charge). Agarose: DNA by size. Native PAGE: size + shape + charge.
• IEF separates by pI; molecule stops where pH = pI (net charge 0).
• pH > pI → net negative → anode (+); pH < pI → net positive → cathode (−).

Laboratory Methods & Separations

Part 3 of 5 — Centrifugation (Separating by Density & Size)

Centrifugation spins samples to generate a force that drives denser/larger particles outward (to the bottom = pellet). The rest stays in the supernatant.

$\text{RCF} = 1.118 \times 10^{-5} \; r \; (\text{RPM})^2$

RCF (relative centrifugal force, in × g) depends on rotor radius $r$ (cm) and rotation speed.

Sedimentation Logic

$v_{sed} \;\propto\; (\rho_{particle} - \rho_{medium}) \, r_{particle}^2 \, \omega^2$

• Larger and DENSER particles sediment faster.
• A particle stops moving when its density equals the surrounding medium's density (isopycnic point).

Two Major Techniques

Differential Centrifugation of a Cell Lysate (Order of Pelleting)

$\text{Low speed} \to \text{nuclei} \to \text{mitochondria} \to \text{microsomes/ER} \to \text{ribosomes (ultracentrifuge)}$

• Increase speed in steps; pellet the largest organelles first, re-spin the supernatant faster.

Svedberg Units (S)

• S is a sedimentation coefficient — it is NOT additive because it depends on shape and density, not mass alone.
• Example: the 70S prokaryotic ribosome = 50S + 30S subunits (70 ≠ 50 + 30).

Centrifugation 🎯

Worked Examples — Centrifugation

Question: A rotor has radius $r = 8$ cm and spins at 10,000 RPM. Find the RCF (in × g).

Solution: $\text{RCF} = 1.118 \times 10^{-5} \times r \times (\text{RPM})^2 = 1.118 \times 10^{-5} \times 8 \times (10000)^2$ $= 1.118 \times 10^{-5} \times 8 \times 10^{8} = 8944 \times g \approx 8.9 \times 10^{3} \; g \checkmark$

MCAT note: RCF scales with the SQUARE of RPM — doubling RPM quadruples the force. Always report g-force, not RPM, because RPM alone is meaningless without the radius.

Question: You want to isolate mitochondria from liver cells without nuclear contamination. Outline the spin scheme.

Solution:

1. Homogenize cells in isotonic buffer (preserve organelles).
2. Low-speed spin (~1000 g): pellets NUCLEI and debris → discard the pellet, keep the supernatant. ✓
3. Medium-speed spin (~10,000 g) of that supernatant: pellets MITOCHONDRIA → keep the pellet. ✓
4. The remaining supernatant holds microsomes and cytosol.

Key logic: Remove the biggest contaminant (nuclei) first at low force, then collect your target at the next-higher force.

Question: DNA labeled with heavy ${}^{15}N$ is shifted to ${}^{14}N$ medium. After one round of replication, a CsCl gradient shows a single band of intermediate density. What does this rule out?

Solution:

1. CsCl gradient bands DNA by buoyant density (isopycnic). ✓
2. A single INTERMEDIATE band means every duplex is half-heavy/half-light → each daughter has one old and one new strand.
3. This rules out conservative replication (which would give two bands: one fully heavy, one fully light) and supports semiconservative replication. ✓

Why it matters: Density-gradient centrifugation provided the classic experimental proof of semiconservative DNA replication.

Key Takeaways — Part 3

• RCF = $1.118\times 10^{-5}$ · r · $RPM^{2}$; force scales with $RPM^{2}$. Report g, not RPM.
• Sedimentation rises with particle size and (density − medium density).
• Differential centrifugation: increasing speeds pellet nuclei → mitochondria → microsomes → ribosomes.
• Isopycnic/density-gradient: particle bands where its buoyant density = medium density (size-independent). Svedberg units are NOT additive (50S + 30S = 70S).

Laboratory Methods & Separations

Part 4 of 5 — Spectroscopy & Beer's Law (ULTRA HIGH YIELD)

Spectroscopy measures how molecules absorb/emit electromagnetic radiation. Different energies probe different transitions.

What Each Method Probes

Beer–Lambert Law

$A = \varepsilon \, l \, c$

• $A$ = absorbance (unitless, log scale)
• $\varepsilon$ = molar absorptivity $(M^{-1}cm^{-1})$ — an intrinsic property at a given wavelength
• $l$ = path length (cm)
• $c$ = concentration (M)

Linearity: absorbance is DIRECTLY proportional to concentration (and to path length) — the basis of quantitative assays. The relationship breaks down at high absorbance (A > ~1) due to stray light and deviations.

Absorbance vs. Transmittance

$A = -\log_{10}(T) = -\log_{10}\!\left(\frac{I}{I_0}\right)$

• $T = 1$ (100% transmitted) → $A = 0$.
• $T = 0.1$ (10%) → $A = 1$. $T = 0.01$ → $A = 2$.

Biochemistry Hooks

• $A_{260}$: nucleic acids (aromatic bases). $A_{280}$: proteins (Trp, Tyr).
• $A_{260}/A_{280}$ ratio ≈ 1.8 → pure DNA; ≈ 2.0 → pure RNA; lower → protein contamination.

Spectroscopy & Beer's Law 🎯

Worked Examples — Spectroscopy

Question: A dye has $\varepsilon = 5000$ $M^{-1}cm^{-1}$ at 500 nm. In a 1 cm cuvette its absorbance is 0.75. Find the concentration.

Solution: $c = \frac{A}{\varepsilon l} = \frac{0.75}{5000 \times 1} = 1.5 \times 10^{-4} \text{ M} \checkmark$

MCAT note: Always check path length (often 1 cm but not always) and that A is in the linear range (≈ 0.1–1.0). At A > 1, dilute and remeasure.

Question: A sample transmits 10% of incident light. What is its absorbance? What about 1% transmittance?

Solution: $A = -\log_{10}(T)$

• $T = 0.10 \Rightarrow A = -\log_{10}(0.10) = 1.0$ ✓
• $T = 0.01 \Rightarrow A = -\log_{10}(0.01) = 2.0$ ✓

Key idea: Absorbance is logarithmic — each unit of A means a 10× drop in transmitted light. A = 2 lets only 1% through.

Question: Standards give A = 0.10, 0.20, 0.30 at c = 2, 4, 6 µM (path 1 cm). An unknown reads A = 0.25. Find its concentration and $\varepsilon$.

Solution:

1. The data are linear: A/c = 0.10/2 = 0.05 per µM → slope = $\varepsilon l$. With $l = 1$ cm, $\varepsilon = 0.05$ µ$M^{-1}cm^{-1}$ = $5\times 10^{4}$ $M^{-1}cm^{-1}$. ✓
2. Unknown: $c = A/\text{slope} = 0.25/0.05 = 5$ µM. ✓

Why a standard curve: It empirically captures $\varepsilon l$ for your exact instrument and conditions, so you read concentration directly off the line.

Key Takeaways — Part 4

• Beer–Lambert: $A = \varepsilon l c$. Absorbance ∝ concentration (linear, valid ~A 0.1–1).
• $A = -\log_{10}(T)$: T = 10% → A = 1; T = 1% → A = 2 (logarithmic).
• UV-Vis = electronic/conjugation & quantitation; IR = functional groups (C=O ~1700 $cm^{-1}$); NMR = H/C environments; MS = molecular weight.
• $A_{260}$ = nucleic acids, $A_{280}$ = protein; $A_{260}/A_{280}$ ≈ 1.8 (DNA), 2.0 (RNA); low = protein contamination.

Laboratory Methods & Separations

Part 5 of 5 — Molecular & Immunological Assays + Review

This part covers the detection/amplification methods that follow separations, plus a method-selection review.

Nucleic Acid & Protein Detection (Know the Blots)

Mnemonic — SNoW DRoP: Southern = DNA, Northern = RNA, Western = Protein.

PCR Logic

$\text{Copies} = N_0 \times 2^{n}$

after $n$ cycles (ideal doubling). Steps per cycle:

1. Denature (~95 °C): separate strands.
2. Anneal (~50–65 °C): primers bind.
3. Extend (~72 °C): Taq polymerase synthesizes.

qPCR quantifies in real time; reverse-transcription (RT-PCR) starts from RNA (via cDNA).

ELISA & Antibody Specificity

• Direct ELISA: antigen bound, enzyme-linked primary antibody → color.
• Sandwich ELISA: capture antibody traps antigen; detection antibody reports it (high specificity).
• Signal (color/absorbance via Beer's law) ∝ amount of antigen.

Method-Selection Cheat Sheet

Molecular Assays Review 🎯

Worked Examples — Molecular Assays

Question: A lab wants to know whether a gene is being TRANSCRIBED (made into mRNA) in a tumor sample, and to measure how much. Which method, and why not a Western blot?

Solution:

1. Transcription produces mRNA → measure RNA → RT-qPCR (reverse-transcribe mRNA to cDNA, then quantitative PCR). A Northern blot also detects RNA but qPCR quantifies it precisely. ✓
2. A Western blot measures PROTEIN, not mRNA — a gene could be transcribed but the protein degraded, so Western answers a different question.

MCAT skill: Map the molecule of interest (DNA / RNA / protein) to the assay before anything else.

Question: How many cycles are needed to amplify a single template molecule to over one million copies (assume ideal doubling)?

Solution: $2^n > 10^6 \Rightarrow n > \log_2(10^6) = \frac{6}{\log_{10}2} \approx \frac{6}{0.301} \approx 20$ So about 20 cycles exceed one million copies ($2^{20} \approx 1.05 \times 10^6$). ✓

Note: Real PCR plateaus as reagents deplete and efficiency drops below 2×, so practical yields fall short of the ideal — a common experimental caveat.

Question: A sandwich ELISA gives absorbances of 0.2, 0.4, 0.6 for antigen standards of 10, 20, 30 ng/mL. A patient sample reads 0.5. Estimate the antigen concentration.

Solution:

1. The standards are linear: 0.02 absorbance units per ng/mL (0.2/10). ✓
2. Patient: $c = 0.5 / 0.02 = 25$ ng/mL. ✓

Connection: This mirrors the Beer's-law standard curve from Part 4 — ELISA converts antigen amount into a colorimetric signal read by a spectrophotometer.

Key Takeaways — Part 5 (and Suite Review)

• Blots: Southern = DNA, Northern = RNA, Western = protein (SNoW DRoP); ELISA = antibody-based quantitation.
• PCR amplifies DNA exponentially: copies = $N_{0}$ × $2^{n}$; steps = denature/anneal/extend; RT-PCR starts from RNA.
• Match the molecule to the method: MW → SDS-PAGE; specific protein → Western/ELISA; amplify DNA → PCR; mRNA level → RT-qPCR; concentration → UV-Vis.
• Detection assays (ELISA, UV-Vis) report concentration via signal proportional to amount (Beer's law).