MCAT practice · Chemical and Physical Foundations · 5C Separation and purification · 1A Proteins and enzymes
Purifying a bacterial NADH-dependent dehydrogenase
An original practice passage written for this site, not an AAMC item. Exam pace is about 8 minutes for a passage like this.
A soil bacterium grows on lactate as its only carbon source, and researchers suspected that a single NADH-dependent dehydrogenase controls the first step of that pathway. To characterize the enzyme, they set out to purify it from the bacterium itself. Activity was measured in the reverse direction, with pyruvate as substrate, by following the decrease in absorbance at 340 nm as NADH was oxidized to NAD⁺; one unit (U) of activity was defined as the oxidation of 1 µmol of NADH per minute at 25 °C. Total protein in each fraction was measured with a dye-binding assay.
Cells (50 g) were broken open and centrifuged to give a cleared lysate. Protein precipitating between 40% and 65% ammonium sulfate saturation was collected, redissolved and dialyzed. This fraction was loaded onto an anion-exchange (DEAE) column equilibrated in buffer at pH 8.0. The enzyme was retained by the column and came off in a 0–0.5 M NaCl gradient. Active fractions were pooled and applied to an affinity resin carrying an immobilized NAD⁺ analog. After a wash with buffer, the enzyme was eluted with 5 mM NADH. A final size-exclusion step was then run. The results of each step are summarized in Table 1.
To estimate the size of the native enzyme, the size-exclusion column was calibrated with six proteins of known molecular mass (Figure 1). The purified enzyme eluted as a single symmetric peak at 13.2 mL. On SDS-PAGE, the final preparation gave one band near 36 kDa, whether or not β-mercaptoethanol was added to the sample buffer. On a native (nondenaturing) gel, it gave one band, and that band stained for dehydrogenase activity.
| Step | Total protein (mg) | Total activity (U) | Specific activity (U/mg) | Yield (%) | Purification (fold) |
|---|---|---|---|---|---|
| Cleared lysate | 2400 | 4800 | 2 | 100 | 1 |
| Ammonium sulfate, 40–65% | 720 | 4100 | 5.7 | 85 | 2.8 |
| Anion exchange (DEAE) | 96 | 3300 | 34 | 69 | 17 |
| NAD⁺-analog affinity | 4 | 2400 | 600 | 50 | 300 |
| Size exclusion | 3.2 | 2050 | 640 | 43 | 320 |
Data are illustrative, constructed for practice.
1. Assuming the final preparation in Table 1 is homogeneous enzyme, approximately what percentage of the protein in the cleared lysate was the dehydrogenase?
- A. Wrong: 3.2 mg ÷ 2400 mg treats the final protein as all the enzyme the lysate held, ignoring that 57% of the activity was lost along the way.
- B. Right: if the final preparation is pure, the lysate's specific activity (2.0 U/mg) is 1/320 of the pure enzyme's (640 U/mg), so the enzyme was about 0.31% of lysate protein. Equivalently, 3.2 mg ÷ 0.43 yield ≈ 7.5 mg of enzyme in 2400 mg.
- C. Wrong: this is the correct ratio off by a factor of ten, a decimal slip in converting 1/320 to a percentage.
- D. Wrong: 43% is the overall yield, the fraction of starting activity recovered, not the enzyme's share of lysate protein.
2. Based on how the enzyme behaved on the DEAE column, the isoelectric point (pI) of the dehydrogenase is most likely:
- A. Right: DEAE is an anion exchanger with positive groups. The enzyme bound and was displaced by NaCl, so it was net negative at pH 8.0, which means its pI lies below 8.0.
- B. Wrong: this pairs the right pI with the wrong sign; at a pH above its pI a protein loses protons and is net negative, not positive.
- C. Wrong: a protein with no net charge would not be held by an ion-exchange resin, yet the enzyme was retained until salt was applied.
- D. Wrong: a protein at a pH below its pI is net positive and would be repelled by the positively charged DEAE resin, flowing straight through.
3. Which additional experiment would best test whether the enzyme binds the affinity resin through its coenzyme-binding site?
- A. Right: if free NADH already occupies the coenzyme site, the enzyme should no longer be captured by the immobilized NAD⁺ analog. Losing binding under this condition ties binding to that site specifically.
- B. Wrong: a single band speaks to purity, not to which part of the protein contacts the resin.
- C. Wrong: Table 1 already shows this rise; it shows the step enriched the enzyme but says nothing about the mechanism of binding.
- D. Wrong: recovering more activity at higher NADH shows elution is concentration dependent, but the resin may still be binding through some other surface, so it does not isolate the site.
4. Based on Figure 1 and the gel results, the native dehydrogenase is best described as:
- A. Wrong: SDS denatures proteins and coats them with charge but does not break peptide bonds, so a single 145 kDa chain would run near 145 kDa.
- B. Wrong: subunits linked by disulfides would stay together without β-mercaptoethanol and run near 145 kDa, but the band was at 36 kDa either way.
- C. Right: at 13.2 mL, Figure 1 gives log mass ≈ 2.16, about 145 kDa for the native enzyme, and 145 ÷ 36 ≈ 4. The 36 kDa band appears even without a reducing agent, so the subunits are held noncovalently.
- D. Wrong: two 72 kDa subunits would fit the native mass, but SDS-PAGE shows the subunit is 36 kDa, not 72 kDa.
5. The researchers added SDS to the samples for the denaturing gel but not for the native gel. SDS makes migration in a polyacrylamide gel depend mainly on chain length because it:
- A. Wrong: this reverses the gel's sieving; in SDS-PAGE small proteins move fastest because they pass through the pores most easily.
- B. Right: SDS binds at roughly one molecule per two residues, so every chain gets a similar charge-to-mass ratio and the gel sorts them by size.
- C. Wrong: separation by pI is isoelectric focusing; SDS swamps a protein's intrinsic charge instead of preserving it.
- D. Wrong: reducing agents such as β-mercaptoethanol break disulfide bonds; SDS is a detergent that unfolds the chain and binds it.
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