[6.1] exchange between organisms and their environment

Question 1

what is the relationship between SA and V?

Answer 1

as the volume increases, the SA:V ratio decreases

Question 2

what does the SA of a cell determine?

Answer 2

the rate at which materials can be transported into or out of the cell eg:

• in - oxygen, glucose, water, ions
• out - carbon dioxide, water, heat energy

Question 3

why is there a point where the SA:V ratio sets a maximum upper limit on the size of a cell for it to still function properly?

Answer 3

mostly due to water heat from production due to volume, but doesn’t have enough SA to deal with the excess, so cell ceases to function

Question 4

how can the problem of SA:V ratio dropping as cell increases be solved?

Answer 4

• by adapting the shape of cells by flattening or extending the membrane
• organisms can also adapt specialised exchange surfaces with large area to increase the SA:V ration eg. lungs in mammals

Question 5

what is the equation for diffusion? (fick’s law)

Answer 5

(SA × difference in concentration gradient) / thickness of exchange surface

Question 6

how can you obtain maximum diffusion with fick’s law?

Answer 6

• high SA
• high concentration gradient
• thin exchange surface

Question 7

what are some features of specialised exchange surfaces? (5)

Answer 7

• large SA relative to volume to increase rate of exchange
• thin surface so diffusion distance is short so materials can cross exchange surface rapidly
• selectively permeable to allow selected materials to cross
• movement of environmental medium eg. air to maintain diffusion gradient
• transport system to ensure the movement of the internal medium eg. blood in order to maintain a diffusion gradient