Diffusion, active transport and bulk transport

Carbon dioxide is a small, non-polar molecule, so it dissolves in the hydrophobic core and diffuses straight through the bilayer. Glucose is polar, so it is repelled by the fatty acid tails and crosses by facilitated diffusion through carrier proteins. Sodium ions are charged and cross through channel proteins (facilitated diffusion) or are moved by carrier proteins in active transport. Amino acids are polar (and often charged), so they also need carrier proteins.

Surface area = 6 × 10² = 600 μm²; volume = 10³ = 1000 μm³; SA : V = 600 ÷ 1000 = 0.6 : 1. 6 : 1 is the ratio for a cube of side 1 μm, so it ignores the actual size. 1.67 : 1 is volume divided by surface area, the ratio upside down. 0.06 : 1 comes from using only one face (100 μm²) over the volume instead of all six faces.

In endocytosis the membrane folds around a particle or droplet and pinches off as a vesicle inside the cell; moving the membrane and vesicles needs energy from ATP. A vesicle fusing with the membrane and releasing its contents is exocytosis, the reverse process. Passage through a carrier protein down a gradient is facilitated diffusion, which only works for single small molecules or ions. Diffusion through the bilayer is for small non-polar molecules, and endocytosis is an active process that does use energy.

K⁺ is 140 mmol dm⁻³ inside and only 5 outside, so moving K⁺ into the cell is against its concentration gradient and needs carrier proteins and ATP (the Na⁺/K⁺ pump). Na⁺ is more concentrated outside (145) than inside (12), so Na⁺ entering the cell is down its gradient and can be facilitated diffusion through channels. Cl⁻ is also more concentrated outside, so its entry is down the gradient. K⁺ leaving the cell goes from 140 to 5, down its gradient, so it can be by facilitated diffusion.

A thicker surface means a longer diffusion distance, so the rate of diffusion falls, not rises. A higher temperature gives the molecules more kinetic energy, so they move faster and diffuse more quickly. A steeper concentration gradient increases the net movement per second. A larger surface area allows more molecules to cross at the same time.

SA : V for each block: 4 × 1 × 0.5: SA = 2(4 + 2 + 0.5) = 13 cm², V = 2 cm³, ratio 6.5 : 1, and no point is more than 0.25 cm from the surface, so it clears first. The 1 cm cube has the smallest volume but a ratio of only 6 : 1 and its centre is 0.5 cm in; smallest volume is not the same as largest ratio. The 3 × 2 × 1 block has SA = 22 cm², V = 6 cm³, ratio 3.67 : 1. The 2 cm cube has the largest surface area (24 cm²) but the lowest ratio, 3 : 1, so it clears last; a large surface area alone does not make diffusion to the centre quick.

The rate levels off, so a limited number of proteins is involved and they become saturated; the inhibitor has no effect and X is never taken up against its gradient, so no ATP is involved. That is facilitated diffusion through carrier proteins. Active transport also saturates, but it depends on ATP from respiration, so the inhibitor would have reduced the rate, and it can accumulate X above the outside concentration. Simple diffusion through the bilayer does not level off; its rate keeps rising with the concentration gradient. Endocytosis needs ATP, so it would also be stopped by the inhibitor.

Number of cycles = 9.0 × 10⁶ ÷ 3 = 3.0 × 10⁶ per second (one ATP each), and each cycle moves 2 K⁺ in, so K⁺ = 2 × 3.0 × 10⁶ = 6.0 × 10⁶. 9.0 × 10⁶ assumes one K⁺ for every Na⁺, but the pump exchanges them 3 : 2. 3.0 × 10⁶ is the number of ATP molecules used, not the number of K⁺ ions. 1.35 × 10⁷ multiplies by 3/2 instead of 2/3, which would mean more K⁺ in than Na⁺ out.