Q.84 Consider a spherical bacterial cell with diameter 2 μm and a eukaryotic cell of spherical
shape with a diameter of 20 μm. The surface area to volume (in μm–1) ratio of the bacterial
cell (round off to 1 decimal place) is ______ times more than that of the eukaryotic cell.
The surface area to volume ratio bacterial cell versus eukaryotic cell explains why smaller prokaryotes like bacteria (diameter 2 μm) thrive with rapid diffusion, while larger eukaryotes (20 μm) rely on organelles. This surface area to volume ratio concept is crucial for CSIR NET questions on cell biology, scaling effects, and metabolic efficiency.
Calculation Steps
For a sphere, surface area SA = 4πr² and volume V = (4/3)πr³, so SA/V = 3/r in μm-1.
- Bacterial cell (diameter 2 μm, r = 1 μm): SA/V = 3/1 = 3.0
- Eukaryotic cell (diameter 20 μm, r = 10 μm): SA/V = 3/10 = 0.3
Ratio: 3.0 / 0.3 = 10.0
Biological Significance
Higher ratios in bacterial cells enable faster nutrient uptake and waste removal via diffusion, supporting high metabolic rates without internal compartments. Eukaryotes compensate with membrane-bound organelles, endocytosis, and cytoskeletal transport, as their low ratio limits passive exchange. This scaling constraint influences cell size evolution and explains prokaryotic simplicity.
CSIR NET Exam Tips
No options provided; focus on deriving SA/V ∝ 1/r for spheres. Round 10 to 1 decimal: 10.0. Practice similar problems on cube or cylinder models for variations.
