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Cell Membranes and Transport: Question 6

Syllabus 4.2

Multiple choice AS 1 mark

Oxygen diffuses across the cell-surface membrane of a cell by simple diffusion, moving from a region of higher oxygen concentration to a region of lower oxygen concentration.

Which change would decrease the rate at which oxygen diffuses across the membrane?

Choose an answer to check it, then compare with the worked solution below.
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Worked solution

The factors that affect the rate of diffusion

The rate of simple diffusion across a membrane depends on the surface area available, the size of the concentration gradient, the distance over which diffusion occurs, and temperature. Broadly:

rate of diffusion∝surface area×concentration gradientdiffusion distance\text{rate of diffusion} \propto \frac{\text{surface area} \times \text{concentration gradient}}{\text{diffusion distance}}

  • A larger surface area gives more room for particles to cross at once, so it increases the rate.
  • A steeper concentration gradient (a bigger difference in concentration between the two sides) means there is a greater net movement of particles down the gradient at any instant, so it increases the rate.
  • A greater diffusion distance means particles must travel further to cross from one side to the other, so it decreases the rate.
  • A higher temperature gives particles more kinetic energy, so they move faster and collide more often, which increases the rate.

Applying this to the options

  • A increases surface area, which would increase, not decrease, the rate.
  • B increases the concentration gradient, which would increase, not decrease, the rate.
  • C increases the diffusion distance, which means oxygen molecules must travel further to cross the same total distance in the same time, so this decreases the rate. This is the correct answer.
  • D raises the temperature, which would increase, not decrease, the rate, since the particles have more kinetic energy.

Final answer

C. Increasing the thickness of the membrane and cytoplasm that oxygen must cross increases the diffusion distance, which decreases the rate of diffusion.