You have a dilute sample of a protein of molecular mass 10 kDalton (104 Da) in w

Question

You have a dilute sample of a protein of molecular mass 10 kDalton (104 Da) in water and want to make it more concentrated while retaining a large fraction of the protein. The 1ml sample that you have is in a 10cm tall cylindrical vial and you reason that if you keep the vial on the bench and wait long enough, the protein molecules will sediment to the bottom of the vial. You can then remove the top layer of water and achieve your goal. (one Dalton is 1/12th of the inertia of a C12 atom) (a) Will you achieve your goal? Explain. (Hint: each protein molecule is a particle that has: 1) some internal energy due to the interactions within and between the atoms that make it up, 2) a kinetic energy that is determined on average by the temperature, and 3) a gravitational potential energy that is determined by how high it is above the bottom of the vial. The first two do not change during the course of the experiment since the proteins remain intact and the temperature ‘T’ remains constant. How does the likelihood of finding a protein molecule at a certain height ‘y’ change with y?) (b) Will your answer be the same if the particles were ribosomes of molecular mass ?10 MD (107 Da)? 2 (c) You should have reasoned that the answer to (a) is “No”. Would repeating the experiment on another planet where the gravitational acceleration ‘g’ is much larger than it is on Earth do the job? How big should g be to achieve your goal of having a large fraction of the protein in the bottom 1mm of the vial? (d) Going to another planet to achieve your goal is not an option, but you can achieve your goal on earth by increasing the acceleration of the particles towards the bottom of the vial by spinning it in a centrifuge. Approximately how fast should the centrifuge spin, if you want most of the ribosomes to end up collecting in the bottom millimeter of the vial? Give your answer in RPM (revolutions per minute). Assumption: you place the vial horizontally on its side in the centrifuge with its lid attached to the centrifuge rotation axle. Therefore, the bottom of the vial moves in a circle of radius 10cm.

Explanation / Answer

a)

1ml sample has 104 Da of protein. One Dalton is 1/12th of inertia of the C12 atom.

If the protein is made to sediment at the bottom of the vial, 0.5ml of the top layer of the solvent can be removed. The remaining 0.5ml of the solvent with the same amount of 104Da of protein will definitely be the concentrated solution than the previous one.

As the temperature is kept constant here, there is no change in the kinetic energy of the protein molecules. There is no change in the internal energy of the protein molecules as the interactions of the protein within it will remain the same due to constant temperature and volume of the vial.

If the solvent volume is reduced from 1ml to 0.5ml, then the distribution of the protein molecule all through the solution varies. If there is more solvent, the protein molecules will experience less inertia compared to that when the solvent is less.

Hence, the likelihood of the protein molecule to be found at a certain height will change with that particular height as the buoyancy of the solvent and the inertia of the protein molecules will vary with height.

b)

No.

In the case of ribosomes, along with the change in potential energy of the molecule, the interactions of the ribosome within the molecule will change. So, both these aspects will have an impact on the inertia of the ribosomes. Moreover, the mol. Mass of ribosomes is slightly higher than the previous protein in the experiment.

c)

In the current experiment, 1ml of the solution in the 10cm or 100mm vial will occupy almost 10mm of the vial. So, 1ml of the solution with about 10times the ‘g’ as that on planet earth would keep the protein within the limits of 1mm of the given vial.

d)

Linear velocity (LV) = radius * angular velocity

1 rad/s = 1/2? hertz

2? rad/s = 60 rpm

1 rpm = 2?/60 rad/s

The RPM to Linear Velocity formula is:
  v = r × RPM × 0.10472

LV here is 0.1 * 60 * 0.10472 = 0.628 m/ s

Then angular velocity = 0.628/0.1 = 6.28 revolutions per minute

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