Recent estimates conclude that macromolecules occupy approximately 25% of the vo
ID: 51656 • Letter: R
Question
Recent estimates conclude that macromolecules occupy approximately 25% of the volume of a cell. A rod-shaped prokaryotic E. coli cell has dimensions 2 µm long x 0.5 µm in diameter. Assume, for the sake of simplicity, that the macromolecules occupying the cell are all globular proteins of 4 nm diameter. How many of these protein molecules would be found in 25% of the bacterial volume -- Calculations showed there were 2918288 protiens able to be found in this volume.
A) Now assume that half of the crowded volume is occupied by 70S ribosomes with a spherical shape and diameter of 25 nm. How many ribosomes would you find in this crowded space? Show your calculations!
Explanation / Answer
1)
Based on the given data,
The shape of E. coli is: Rod
Length (l) = 2 µm
Diameter (w) = 0.5 µm
Radius = (r) = 0.5/2 = 0.25 µm
The Volume of cylinder (V) = r2h
= 3.14 × (0.25)2 × 2
= 0.39 µm3
Thus, the volume of one E. coli cell is about 0.39 µm3.
Here, the given proteins are globular (sphere) with 4 nm diameter, so radius = 2 nm.
The Volume sphere V= (4/3)r3
= (4/3) × (3.14) × (2)3
= 33.51nm3
Thus, the volume of one globular protein is about 33.51nm3.
The 25% of E. coli volume is: 0.39 µm3 × 0.25 = 0.0975 µm3
Thus, the 25% of bacterial cell volume = 0.0975 µm3 ×1000 = 97.5 nm3
Thus, the number of proteins for this cell volume is: 97.5 nm3 / 33.51nm3
= 2.9
Hence, the number of proteins that can fit in 25% of the E. coli volume is: 2.9 molecules.
2)
Here, the given ribosome is spherical and has 25 nm diameter, so radius = 12.5 nm.
The Volume sphere V= (4/3)r3
= (4/3) × (3.14) × (12.5)3
= 8177 nm3
Thus, the volume of one Ribosome is about 8177 nm3.
The 50% (half of the crowded volume) of E. coli volume is: 0.39 µm3 × 0.50 = 0.195 µm3 or 195 nm3
Thus, the number of ribosomes for this cell volume is: 195 nm3 / 8177 nm3
= 0.023
Hence, the number of ribosomes that can fit in 50% (half of the crowded volume) of the E. coli volume is: 0.023.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.