In a sedimentation experiment using a centrifuge rotating at 20,000 rpm at 25°C the position of the boundary between the settled phase and the supernatant for a protein was found to be at 6.153 cm at t = 300 s, 6.179 at 600 s, and 6.206 at 900 s. If D for the protein is 7.62 × 10–11 m2s–1 at 20°C and it has a density of 1.373 g cm–3, what is its molecular weight?

To calculate the molecular weight of the protein, we need to make use of the sedimentation coefficient formula:

s = (M * D * g) / (6 * pi * eta * r)

Where:
- s is the sedimentation coefficient
- M is the molecular weight of the protein
- D is the diffusion coefficient
- g is the acceleration due to gravity
- pi is a mathematical constant (approximately equal to 3.14159)
- eta is the viscosity of the medium
- r is the distance from the center of rotation to the boundary

First, we need to convert the temperature from °C to Kelvin (K) because the diffusion coefficient is given at 20°C.

20°C + 273.15 = 293.15 K

Next, we need to convert the density from g cm–3 to kg m–3.

1.373 g cm–3 * (1000 g / 1 kg) * (1 cm / 0.01 m)^3 = 1373000 kg m–3

Now, let's calculate the diffusion coefficient at 25°C using the Arrhenius equation:

D2 = D1 * exp((Ea / R) * [(1 / T1) - (1 / T2)])

Where:
- D2 is the diffusion coefficient at 25°C
- D1 is the diffusion coefficient at 20°C
- Ea is the activation energy (assumed to be 0 for proteins)
- R is the ideal gas constant (approximately 8.314 J/mol K)
- T1 is the initial temperature in Kelvin (293.15 K)
- T2 is the final temperature in Kelvin (298.15 K)

D2 = 7.62 × 10–11 m2s–1 * exp((0 / 8.314 J/mol K) * [(1 / 293.15 K) - (1 / 298.15 K)])

D2 = 7.62 × 10–11 m2s–1 * exp(0 * [(1 / 293.15 K) - (1 / 298.15 K)])

D2 = 7.62 × 10–11 m2s–1 * exp(0)

D2 = 7.62 × 10–11 m2s–1

Now we can calculate the sedimentation coefficient (s) at each time point by rearranging the sedimentation coefficient formula:

s = (6 * pi * eta * r) / (M * D * g)

Let's calculate the sedimentation coefficient (s) at each time point:

s1 = (6 * pi * eta * r1) / (M * D2 * g)

s2 = (6 * pi * eta * r2) / (M * D2 * g)

s3 = (6 * pi * eta * r3) / (M * D2 * g)

Since the sedimentation coefficient is constant throughout the experiment, we can set up the following equations:

s1 = s2 = s3

(6 * pi * eta * r1) / (M * D2 * g) = (6 * pi * eta * r2) / (M * D2 * g) = (6 * pi * eta * r3) / (M * D2 * g)

r1 / M = r2 / M = r3 / M

Now, we can calculate the molecular weight (M) by taking the average of the ratios:

M = (r1 + r2 + r3) / 3

M = (6.153 + 6.179 + 6.206) / 3

M = 18.538 / 3

M = 6.179 g cm–3

Therefore, the molecular weight of the protein is 6.179 g cm–3.