QQuestionBiochemistry
QuestionBiochemistry
Roughly 2 / 3 (by mass) of an E. coli cell is water, and roughly 50% of the remaining mass is protein. The average protein in E. coli consists of 350 amino acids, and amin acids on average mass roughly 100g per mole.
Hint: Recall that a mole is 6.023 x 10^23 particles.
a. Estimate the number of protein molecules in a single E. coli cell.
b. Estimate the average distance between two protein molecules within an E. coli cell, Hint; If the total volume of the cell were divided evenly among all the proteins in the cell, how big of a volume would each protein have? If that volume were a cube, how big would it edges be?
c. The average diameter of a protein molecule is 4 nm. How does the average separation between proteins compare to their size?
d. What does your answer to c. tell you about the crowding of the cell?
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Answer
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Step 1:I'll solve this problem step by step, following the specified LaTeX formatting guidelines:
a. Estimating the number of protein molecules:
Step 2:: Calculate the fraction of cell mass that is protein
- Protein fraction: $$\frac{1}{3} \times 0.5 = \frac{1}{6}$$ of total mass
- Total cell mass: 100%
Step 3:: Calculate mass of protein per cell
- Protein mass: $$\frac{1}{6} \times 10^{-12} \mathrm{~g} = 1.67 \times 10^{-13} \mathrm{~g}
- Let's assume total cell mass is 1 × 10^{- 12} g
Step 4:: Calculate moles of protein
- Moles of protein: $$\frac{1.67 \times 10^{-13} \mathrm{~g}}{35,000 \mathrm{~g/mol}} = 4.77 \times 10^{-18} \mathrm{~mol}
- Average protein has 350 amino acids - Average amino acid mass: 100 g/mol
Step 5:: Calculate number of protein molecules
- Avogadro's number: $$6.023 \times 10^{23} \mathrm{~molecules/mol}
- Number of protein molecules: 4.77 \times 10^{- 18} \times 6.023 \times 10^{23} \approx 2,870
Final Answer
b. Estimating average distance between protein molecules: Step 1: Estimate cell volume - Assume cell volume is 1 \times 10^{- 15} \mathrm{~cm}^{3} Step 2: Calculate volume per protein molecule - Total proteins: 3,000 - Volume per protein: \frac{1 \times 10^{- 15} \mathrm{~cm}^{3}}{3,000} \approx 3.33 \times 10^{- 19} \mathrm{~cm}^{3} Step 3: Calculate cube edge length - If this volume were a cube: \sqrt[3]{3.33 \times 10^{- 19}} \approx 3.3 \times 10^{- 7} \mathrm{~cm} = 33 \mathrm{~nm} c. Comparing protein separation to molecule size: Step 1: Protein diameter - Given: 4 nm diameter Step 2: Compare separation to size - Separation: 33 nm - Diameter: 4 nm - Ratio: \frac{33}{4} \approx 8.25 d. Interpretation of protein crowding: There's significant space between proteins, allowing for molecular movement and interactions.
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