8.1.1 Introduction to exchange and transport

Substances move into and out of organisms

1. Cells need a continuous supply of useful substances and must remove substances that would otherwise build up.
2. Oxygen enters animals for aerobic respiration and leaves photosynthesising tissues when it is produced faster than it is used.
3. Carbon dioxide leaves respiring cells as a waste product and enters photosynthesising cells as a raw material.
4. Water is needed as a solvent, as a reactant and for transport, so it moves into, through and out of organisms.
5. Dissolved food molecules, including glucose and amino acids, move from exchange surfaces to cells for respiration, growth and repair.
6. Mineral ions enter plants through their roots and are transported to tissues where they are used to make essential substances.
7. Urea is carried from the liver to the kidneys so that it can be removed from the blood and excreted.

Size changes exchange needs

1. A small organism has a large surface area to volume ratio, so it has a large exchange surface compared with the amount of living tissue it must supply.
2. Its cells are close to the outside, giving a short diffusion distance for oxygen, nutrients and waste products.
3. As length increases, surface area rises with the square of length while volume rises with the cube of length.
4. A large organism therefore has a smaller surface area to volume ratio and many cells that are far from the body surface.
5. Diffusion across the outer surface cannot move enough material over these longer distances to meet the demands of all the cells.

Calculate surface area to volume ratio

1. Calculate the ratio using $surface\; area\; to\; volume\; ratio=\frac{surface\; area}{volume}$.
2. For a cube with side length $l$, use $surface\; area=6l^2$ and $volume=l^3$.
3. Write the result as $surface\; area:volume$ and simplify both parts to the lowest convenient whole-number ratio when possible.

Exchange surfaces speed transfer

1. An exchange surface provides a specialised boundary where substances enter or leave an organism.
2. A large surface area allows more particles to cross at the same time, while a thin barrier gives a short diffusion distance.
3. A blood supply, water flow or air flow carries substances away and replaces them, which maintains a concentration gradient.
4. Examples include alveoli in lungs, gills in fish, root hair cells in roots and the small intestine in mammals.

Transport systems connect surfaces and cells

1. A transport system moves substances rapidly between exchange surfaces and cells, reducing the distance over which diffusion must act.
2. In mammals, the circulatory system carries oxygen and dissolved food molecules to cells and carries carbon dioxide and urea away.
3. In plants, xylem carries water and mineral ions from roots, while phloem transports dissolved sugars and other organic substances.
4. Transport maintains concentration gradients at exchange surfaces because arriving fluid supplies materials and departing fluid removes them.

8.1.2 Alveoli adapted for gas exchange

Alveoli exchange two respiratory gases

1. Air reaching the alveoli contains a higher concentration of oxygen than the deoxygenated blood arriving in nearby capillaries.
2. Oxygen dissolves in the moist lining and diffuses through the alveolar and capillary walls into the blood.
3. The arriving blood contains a higher concentration of carbon dioxide than the air in the alveoli.
4. Carbon dioxide diffuses from the blood into the alveoli and is removed from the lungs when you breathe out.

Large area increases gas exchange

1. Each lung contains millions of tiny alveoli, which together create a very large surface area.
2. A larger area allows more oxygen and carbon dioxide molecules to cross the exchange surface at the same time.
3. The result is a higher rate of diffusion than a single smooth lung surface could provide.

Thin moist walls shorten the path

1. The wall of each alveolus is one cell thick, and the wall of the neighbouring capillary is also one cell thick.
2. These thin walls give oxygen and carbon dioxide a very short diffusion distance.
3. The alveolar surface is moist so oxygen and carbon dioxide dissolve before crossing the cell membranes.
4. The walls are permeable to the respiratory gases, so dissolved oxygen and carbon dioxide can pass through them.

Airflow and blood flow maintain gradients

1. Ventilation replaces alveolar air, keeping oxygen concentration high and carbon dioxide concentration low inside the alveoli.
2. The capillary network brings deoxygenated blood with little oxygen and more carbon dioxide close to every alveolus.
3. Blood flow carries absorbed oxygen away and continually brings more carbon dioxide to the lungs.
4. Oxygen binding to haemoglobin helps keep the concentration of dissolved oxygen low in the blood beside the alveolus.
5. Together, ventilation and blood flow maintain steep concentration gradients for oxygen into the blood and carbon dioxide into the alveoli.

8.1.3 Factors affecting rate of diffusion

Diffusion is net particle movement

1. Particles move randomly in gases and liquids, crossing an exchange surface in both directions.
2. If more particles cross from the more concentrated side to the less concentrated side, there is a net movement down the concentration gradient.
3. Diffusion is passive, so it does not require energy released by respiration.

Four factors change diffusion rate

1. A steeper concentration gradient increases the net movement of particles because the difference between the two sides is greater.
2. A higher temperature increases particle kinetic energy, so particles move faster and cross the surface more frequently.
3. A larger surface area allows more particles to cross at the same time.
4. A shorter diffusion distance reduces the distance particles must travel, so they cross the barrier sooner.

A gradient compares two regions

1. A concentration gradient describes the difference in concentration between two places, not the concentration in one place alone.
2. Increasing the difference makes the gradient steeper and increases the rate of diffusion.
3. As diffusion continues in a closed system, the difference becomes smaller, so the net rate falls until there is no concentration gradient.

Area and distance control the route

1. Folding or dividing a surface increases its area without requiring the whole structure to become much larger.
2. A thin membrane gives a short diffusion distance because particles cross less material.
3. A thick membrane gives a longer diffusion distance and therefore a lower rate of diffusion when the other factors stay constant.

Apply each factor to the context

1. In lungs, gills and intestines, folded surfaces increase area, thin walls shorten the path and fluid flow maintains concentration gradients.
2. When temperature rises, explain the faster rate through increased kinetic energy and faster particle movement.
3. Use comparative words such as larger, steeper, shorter and faster so the direction of each change is unambiguous.

8.1.4 Rate of diffusion using Fick's law

Fick's law links three factors

1. The relationship is $rate\; of\; diffusion\propto \frac{surface\; area\times concentration\; difference}{membrane\; thickness}$.
2. The symbol $\propto$ means is proportional to, so the expression shows how the variables change together rather than giving a universal constant.
3. The concentration difference is the size of the concentration gradient across the membrane.
4. The membrane thickness represents the diffusion distance through the exchange surface.

Numerator factors increase the rate

1. Surface area and concentration difference are in the numerator, so increasing either one increases the diffusion rate when the other variables stay constant.
2. If surface area doubles, diffusion rate doubles, provided concentration difference and membrane thickness do not change.
3. If concentration difference triples, diffusion rate triples, provided surface area and membrane thickness do not change.

Thickness has an inverse effect

1. Membrane thickness is in the denominator, so increasing thickness decreases diffusion rate when surface area and concentration difference stay constant.
2. Halving membrane thickness halves the diffusion distance and doubles the diffusion rate.
3. Multiplying thickness by four divides the diffusion rate by four.

Calculate a relative diffusion value

1. Substitute the values into $\frac{surface\; area\times concentration\; difference}{membrane\; thickness}$.
2. Calculate the numerator first, then divide by the thickness and keep the units supplied in the question.
3. When comparing two exchange surfaces, calculate the expression for each one and divide one value by the other to find the fold change.

Apply the law to exchange surfaces

1. Alveoli and villi have large surface areas, so more molecules can diffuse at the same time.
2. Their thin walls reduce membrane thickness, which raises the rate because thickness is in the denominator.
3. Ventilation, blood flow and the removal of absorbed substances maintain concentration differences across exchange surfaces.