Given the protein concentration in a cell of 300 mg/ml. Let’s assume e a typical

Question

Given the protein concentration in a cell of 300 mg/ml. Let’s assume e a typical protein is 20 kDa in mass. Find the average volume occupied by each protein molecule and the resulting mean spacing between molecules. How does this compare with our assumed size of the typical protein molecule. (Hint: The volume of a protein molecule can be approximated very simply and reliably from the molecular weight of the protein and an average protein partial specific volume. (Partial specific volume = volume / molecular weight.) The average of experimentally determined partial specific volumes for soluble, globular proteins is ~0.73 cm**3/g. This value varies from protein to protein, but the range is rather narrow.)

Explanation / Answer

Mass of a single protein molecule = 20 kDa = 3.32*10-20 grams

So, molecular weight of protein = 3.32*10-20 * 6.02*1023 = 19.98*103 grams

Mass of protein per mL = 300 mg = 0.3 grams

So, number of molecules of protein present in this solution = Mass of protein in sample/Mass of single protein molecule = 0.3/(3.32*10-20) = 9.036*1018 molecules

So, volume occupied by one molecule = 1cm3 / 9.036*1018 = 0.11*10-18 cm3

Now, using the average of experimentally determined partial specific volume, we get:

Total volume occupied by single protein molecule = 0.73 (cm3/g) * Mass of sinlgle protein molecule = 0.73*3.32*10-20 = 0.0242*10-18 cm3

This value is 4.5 times smaller than the one calculated above.

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.