Questions on surface area to volume ratio: why size limits diffusion

Question 1

A cube-shaped organism has sides of 0.5 mm. A second organism of the same shape has sides of 2.0 mm. Calculate the surface area to volume ratio of each, giving a unit, and calculate how many times greater the smaller organism's ratio is.

Accepted Answer

M1 ratio is surface area divided by volume, which for a cube simplifies to 6 divided by the length of a side A1 smaller organism: 6 ÷ 0.5 = 12 mm⁻¹, with the unit given as one over length M1 larger organism: 6 ÷ 2.0, worked in the same length unit as the first A1 larger organism gives 3 mm⁻¹, so the smaller ratio is four times greater

Question 2

A flatworm 0.3 mm thick supplies all of its cells with oxygen by diffusion across its body surface. A block of tissue one centimetre thick cannot. Explain the difference.

Accepted Answer

B1 surface area rises with the square of length while volume rises with the cube, so the surface area to volume ratio falls as an organism gets thicker B1 demand for oxygen goes with the volume of respiring tissue, so each unit of surface has more tissue to supply in the larger body B1 diffusion time rises with the square of the distance, so ten times further takes about a hundred times as long A1 over a centimetre diffusion would take hours and could never meet demand, whereas the flatworm's flattened shape keeps every cell a fraction of a millimetre from the outside

Question 3

Diffusion of a substance over 10 micrometres takes about 5 milliseconds. Using the relationship that diffusion time rises with the square of distance, calculate the time diffusion would take over 100 micrometres, and calculate the time it would take over 1000 micrometres.

Accepted Answer

M1 increasing the distance by a factor of ten increases the time by a factor of ten squared, which is 100, since time rises with the square of distance A1 5 × 100 = 500 milliseconds, or 0.5 seconds, for 100 micrometres M1 increasing the distance from 100 to 1000 micrometres is another factor of ten, so the time increases by another factor of 100 A1 500 × 100 = 50 000 milliseconds, or 50 seconds, for 1000 micrometres, one millimetre

Question 4

A tapeworm can grow to several metres in length while remaining under a millimetre thick, and survives with no gas exchange system, gills or lungs. Explain how its shape allows this, and explain why its length is not limited by diffusion in the same way its thickness would be.

Accepted Answer

B1 the tapeworm's flattened, ribbon-like shape keeps every cell within a short diffusion distance of its outer surface, since the animal is thin however long it grows B1 diffusion distance depends on thickness rather than length, because oxygen crosses inwards from the surrounding surface rather than travelling the length of the worm B1 as long as the worm stays thin, adding length adds more surface and more volume in the same fixed ratio, so the surface area to volume ratio of a cross-section does not fall as the animal gets longer B1 growing thicker, by contrast, would increase the diffusion distance to the centre and reduce the surface area to volume ratio

Question 5

A single sphere has a radius of 6 mm. Its total volume is divided equally among eight smaller spheres, each with a radius of 3 mm. Compare the total surface area of the eight small spheres with the surface area of the original single sphere, and compare their surface area to volume ratios.

Accepted Answer

M1 surface area of the original sphere = 4π × 6² = 144π mm², and its ratio = 3 ÷ 6 = 0.5 mm⁻¹ M1 surface area of one small sphere = 4π × 3² = 36π mm², so the total surface area of the eight small spheres = 8 × 36π = 288π mm², twice that of the single original sphere A1 the ratio of each small sphere = 3 ÷ 3 = 1 mm⁻¹, also double the original sphere's ratio B1 dividing a fixed volume into many smaller units always raises the total surface area and the surface area to volume ratio, the same principle behind subdividing an exchange surface into alveoli or gill lamellae

Question 6

A sphere and a flattened sheet contain identical volumes of tissue. The sphere has a surface area to volume ratio of 2.4 cm⁻¹ and the sheet one of 8.8 cm⁻¹. Compare the two shapes as exchange surfaces.

Accepted Answer

B1 the sheet's ratio is about 3.7 times the sphere's, even though the two contain the same volume of material B1 the sheet therefore has far more surface serving each unit of volume, while the sphere is the shape with the least surface possible for a given volume B1 the centre of the sphere is also much further from the nearest surface than any point in the sheet, so the diffusion distance is longer in the sphere as well

Question 7

A hibernating dormouse curls its body into a tight ball. Suggest how this reduces the rate at which it loses heat to its surroundings.

Accepted Answer

B1 curling brings the body closer to a sphere, which is the shape with the smallest surface area for a given volume B1 the surface area to volume ratio is therefore lowered without any change in the volume of the animal A1 heat is lost across the body surface, so less surface per unit of body volume means a slower rate of heat loss

Question 8

A shrew must eat almost constantly, and respires far faster per gram of body tissue than an elephant does. Explain why, using the surface area to volume ratio.

Accepted Answer

B1 a small mammal has a much higher surface area to volume ratio, because the ratio falls as size rises B1 heat is lost across the body surface, so each gram of a shrew has far more surface losing heat for it than each gram of an elephant B1 to stay warm the shrew must release heat faster per gram, so it must respire faster per gram, which raises its oxygen and food demand per gram

Question 9

Describe three features shared by specialised exchange surfaces, pairing each with the term of Fick's law it serves.

Accepted Answer

B1 a large surface area, usually made by folding, as in alveoli or gill lamellae, which raises the area term on top of the fraction B1 a wall one cell thick, which cuts the thickness of the surface underneath the fraction B1 a blood supply, ventilation or another way of renewing the medium on each side, which carries the substance away or renews it and so holds up the difference in concentration

Question 10

The time a substance takes to diffuse across a barrier rises with the square of the barrier's thickness. In pulmonary fibrosis, scar tissue thickens the barrier between the air in the alveoli and the blood. Calculate the factor by which the diffusion time for oxygen increases if the barrier becomes three times as thick, and suggest why this lowers the concentration of oxygen in the blood leaving the lungs.

Accepted Answer

M1 diffusion time is proportional to thickness squared, so the factor is 3² A1 the diffusion time becomes 9 times as long B1 blood spends only a short time passing through an alveolar capillary, so with slower diffusion less oxygen crosses into it before it leaves the lung

Question 11

Explain why an Amoeba, a single-celled organism, needs no specialised exchange surface and no transport system, referring to its surface area to volume ratio and to the distance oxygen must diffuse.

Accepted Answer

B1 an Amoeba is very small, so it has a high surface area to volume ratio, meaning a large area of cell surface membrane serves a small volume of cytoplasm B1 no part of its cytoplasm is far from the cell surface membrane, so the diffusion distance from outside to any point inside is short B1 the whole cell surface membrane acts as the exchange surface, so oxygen entering and carbon dioxide leaving by diffusion across it are fast enough to meet the cell's demand

Question 12

A land animal's gas exchange surface must be both thin and moist to allow gases to diffuse across it. Suggest why this requirement creates a problem for a land animal that a water-dwelling animal does not face, and suggest how a mammal's lungs help to solve it.

Accepted Answer

B1 a surface thin and moist enough for gases to cross easily is also thin and moist enough for water vapour to diffuse out readily, a much greater risk in air than in water, where the surrounding medium is already saturated B1 a water-dwelling animal's gas exchange surface is surrounded by water, so it faces no equivalent drying risk B1 a mammal's lungs are folded away inside the body, in a cavity that stays warm and humid, rather than being exposed directly to the drying effect of the outside air, which limits water loss from the moist exchange surface

Question 13

Explain why an exchange surface across which a gas is exchanged must be kept moist, referring to how a gas actually crosses a cell membrane.

Accepted Answer

B1 a gas must first dissolve in the thin film of moisture covering the exchange surface before it can diffuse across the cell membrane B1 a dry surface would not allow this dissolving step to happen, so the gas could not cross even if the surface were thin and well supplied with blood B1 this is why the alveolar lining, the moist walls of a leaf's mesophyll cells, and the surface of a gill lamella are all kept wet

Question 14

State the relationship known as Fick's law, naming all three of the quantities on which the rate of diffusion across a surface depends.

Accepted Answer

B1 rate of diffusion is proportional to the surface area multiplied by the difference in concentration B1 and inversely proportional to the thickness of the surface, so a thinner barrier gives a faster rate

Question 15

A spherical cell has a radius of 30 µm. Calculate its surface area to volume ratio, giving a unit.

Accepted Answer

M1 for a sphere the ratio is 3 ÷ radius, so 3 ÷ 30 A1 0.1 µm⁻¹, and the unit is one over length because the ratio divides an area by a volume