Question 1 · 2.1 Cell Structure and Function
Agar cubes containing a pH indicator were soaked in the same acid bath for 10 minutes, then cut open and measured for the fraction of the interior that had changed color. The data most strongly support which explanation of the pattern?
- A The volume needing acid grows faster than the surface available for uptake
- B The surface available for uptake shrinks as cube edge length increases
- C Agar becomes denser toward the center of a cube, slowing acid movement
- D The acid concentration gradient across the surface is smaller for large cubes
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A — Acid entered every cube for the same 10 minutes and penetrated roughly 1 mm inward from each face, so the stained shell is about the same thickness in all four. What changes is how much interior that shell has to cover. Surface area rises with the square of edge length while volume rises with the cube, so the ratio falls from 3.0 mm⁻¹ at 2 mm to 0.5 mm⁻¹ at 12 mm. The 12 mm cube has 36 times the surface of the 2 mm cube but 216 times the interior to supply. Real cells face the same geometric limit, which is why large cells stay thin, fold their surfaces, or divide.
Question 2 · 2.2 Cell Size
A single-celled organism takes up twice as much dissolved nutrient per minute as a close relative of the same volume. Which difference in its form best accounts for this?
- A A thicker cell wall surrounding the same amount of cytoplasm
- B A rounder shape that brings the cell closer to a sphere
- C A flattened shape with many folds along the outer surface
- D A cytoplasm holding a higher concentration of the nutrient
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C — Uptake happens across the boundary, so the rate depends on how much boundary the organism has, not on how much interior. Holding volume fixed, flattening the cell and folding its boundary raises the exchange area without adding cytoplasm. The same solution appears wherever exchange rates matter: microvilli on intestinal cells, cristae inside mitochondria, and the flattened discs of a red blood cell all buy area at constant volume.
Question 3 · 2.3 Plasma Membrane
A mouse cell and a human cell had their surface proteins labeled with two different fluorescent dyes, and the two cells were then fused into one. Within an hour the two dyes were evenly intermixed across the entire surface of the fused cell. Which property of the membrane does this outcome demonstrate?
- A Proteins are held in fixed positions by a rigid sheet lining the inner face of the bilayer
- B Proteins are free to drift laterally within a bilayer that behaves as a fluid
- C Proteins are exchanged between the two faces of the bilayer at a rapid rate
- D Proteins are digested and rebuilt from shared subunits within a single hour
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B — The dyes started segregated on two halves of the fused cell and ended up uniform, so something carried the labeled proteins over the whole surface. Lateral drift through a fluid bilayer does exactly that, and it needs no energy input, which is why the fluid mosaic model treats the membrane as a two-dimensional liquid with proteins floating in it. The experiment measures movement in the plane of the membrane; it says nothing about movement between the two faces, which for a protein is far slower.