Question

Consider the scale-up of a chromatography column for purification of a protein. A column of \(40 \mathrm{~cm}\) length is used with a superficial velocity of \(40 \mathrm{~cm} / \mathrm{h}\). The peak concentration of the target protein exits at a time of \(100 \mathrm{~min}\). The standard deviation of the peak is \(14 \mathrm{~min}\). a. How long must you wait to collect \(90 \%\) of the protein? b. If the same column is used, but velocity is increased to \(60 \mathrm{~cm} / \mathrm{h}\) and external or Taylor dispersion controls, what will be the value of \(\sigma\) ? c. If the column is lengthened to \(60 \mathrm{~cm}\) while the velocity is at \(60 \mathrm{~cm} / \mathrm{h}\), how will \(\sigma\) change? Will the peak be sharper or broader than at \(40 \mathrm{~cm} / \mathrm{h}\) with a \(40-\mathrm{cm}\) column?

Short answer

To collect 90% of the protein, wait approximately 1.97 hours. The new standard deviation if velocity increases is approximately 0.286 hours. For longer column, new \( \sigma \) will be approximately 0.351 hours, making the peak broader.

Step-by-step solution

- Convert Time to Hours

Since the original values are given in minutes, convert the time to hours for consistent units. Given time = 100 min, standard deviation \(\sigma\) = 14 min. Convert 100 min to hours: 100 min = 100/60 hours = 5/3 hours Convert 14 min to hours: 14 min = 14/60 hours ≈ 0.233 hours

- Calculate Time for 90% Protein Collection

To collect 90% of the protein, use the property of normal distribution where approximately 90% of the data lies within the limits \( \mu \pm 1.28 \sigma \). Peak concentration time \( \mu \) = 5/3 hours Standard deviation \( \sigma \) ≈ 0.233 hours Time to collect 90%: \[ \mu + 1.28 \sigma = \frac{5}{3} + 1.28 \times 0.233 \approx 1.67 + 0.298 = 1.97 \text{ hours } \]

- Find New Standard Deviation if Velocity Increases

Under Taylor dispersion, the standard deviation scales with the square-root of the velocity: \[ \sigma_2 = \sigma_1 \times \sqrt{\frac{v_2}{v_1}} \] Initially, \sigma_1 = 0.233 hours Initially, velocity \( v_1 = 40 \text{ cm/h} \) New velocity \ v_2 = 60 \text{ cm/h} \ Calculate new standard deviation: \[ \sigma_2 = 0.233 \times \sqrt{\frac{60}{40}} = 0.233 \times \sqrt{1.5} \approx 0.286 \text { hours } \]

- Calculate Standard Deviation for Longer Column

The standard deviation scales with the square root of the column length if external dispersion controls: Original length = 40 cm, \(\sigma_1 = 0.286 \text{ hours } \) New length = 60 cm Calculate new standard deviation: \[ \sigma_3 = \sigma_2 \times \sqrt{\frac{L_3}{L_2}} \] \[ \sigma_3 = 0.286 \times \sqrt{\frac{60}{40}} = 0.286 \times \sqrt{1.5} \approx 0.351 \text { hours } \]

- Compare Peak Sharpness

To determine if the peak will be sharper or broader, compare the standard deviations. Original \( \sigma_1 \) at 40 cm column with 40 cm/h was 0.233 hours. The recalculated \( \sigma_3 \) with 60 cm column and 60 cm/h is approximately 0.351 hours. Since \( \sigma \) increased, the peak will be broader.

Key concepts

Protein Purification

Protein purification is essential in biotechnology and biochemical labs. It involves separating a target protein from a complex mixture, often using chromatography. Chromatography relies on different interactions between the components in a mixture and the stationary phase within a column. The goal is to obtain a high yield of the pure target protein, which often involves scaling up the process from small laboratory columns to larger ones for industrial use.
Scaling up requires careful adjustments to maintain the quality and efficiency of purification. When scaling up a chromatography column, knowing how various parameters like superficial velocity, column length and dispersion factors interact is crucial. This understanding helps ensure that the protein elutes in a predictable, concentrated peak without unnecessary broadening, which could compromise the purity and yield.

Normal Distribution

Normal distribution is a probability function that describes how values are spread over a range. Most data points will cluster around the mean (average) and create a bell-shaped curve. In chromatography, this distribution helps describe the elution profile of a protein.
The peak concentration of the protein at a particular time is the mean (

Superficial Velocity

Superficial velocity is the linear flow rate of the mobile phase in the chromatography column. It is critical for protein purification as it affects the time it takes for a protein to travel through the column. It is calculated as the mobile phase flow rate divided by the column's cross-sectional area.
In the exercise, the superficial velocity changed from 40 cm/h to 60 cm/h. This alteration impacts the standard deviation of protein elution profiles, which tells us about the spread of the protein peak across the column. By adjusting the superficial velocity, one can optimize the protein elution, balancing between run time and peak broadness.

Taylor Dispersion

Taylor dispersion refers to the spreading of solute (such as protein) in the mobile phase as it travels through the column. It combines molecular diffusion and convective distribution effects. In chromatography, understanding Taylor dispersion is essential because it influences the broadening of the elution peak.
When the velocity increases, Taylor dispersion generally increases, leading to broader peaks. This happens due to enhanced variations in the flow velocity across the column's cross-section, which causes more mixing of the protein. By controlling Taylor dispersion (for example, by adjusting column length and flow rate), the elution profile can be fine-tuned to achieve sharper peaks.