How to Calculate the Concentration Ratio of Myoglobin in a Test Tube?

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Myoglobin has a molar mass of approximately 17,000 g/mole, and buoyant correction reduces this to an effective mass of about 0.25m. The discussion involves calculating the concentration ratio of myoglobin at the top versus the bottom of a 4 cm test tube, with the conclusion that myoglobin sinks due to its density being greater than that of water. The initial calculation suggests a concentration ratio of 0 at the top and M at the bottom, resulting in a ratio of 0, which raises concerns about its reasonableness. The problem's mention of molar mass indicates its relevance in understanding the behavior of myoglobin in solution.
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Myoglobin is a globular protein, with molar mass m ≈ 17,000g/mole. The buoyant correction typically reduces m to m' ≈ 0.25m. Calculate the concentration ratio of the protein at the top of a 4 cm test tube and that at the bottom.

Here, m' stands for effective mass.

I tried as below.

I used the equation of buyoancy so that the effective mass reduce into 0.25m.

m'g = (ρ-ρ')Vg = 0.25mg = 0.25ρVg
(ρ is the density of myoglobin, and ρ' is the density of water)
→ 0.75ρ = ρ'

So since ρ > ρ', I think the myoglobin should sink to the bottom of the test tube.

As a result, the concentration of myoglobin at the top is 0 and let the concentration of the bottom is M.

Then, concentration ratio of bottom and the top becomes 0 / M which gives me 0.

However, I don't think my answer is reasonable.
Why does the problem give me the molar mass?


I hope anyone could ever help me.
 
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Greg Bernhardt said:
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

sorry, I still couldn't find out anything more, wish I knew anything new.

deeply sorry.
 
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