Effect of Surface Area on Rate of Diffusion in Plant Cells
Published: Last Edited:
Disclaimer: This essay has been submitted by a student. This is not an example of the work written by our professional essay writers. You can view samples of our professional work here.
Any opinions, findings, conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of UK Essays.
The effect of SURFACE AREA : VOLUME RATIO on the rate of DIFFUSION IN PLANT cells
 INTRODUCTION
Diffusion is movement of the particles from a place higher concentration to a place of low concentration. In the experiment, agar cubes were used to resemble plant cells. The surface area to volume ratio of the agar cubes was investigated by cutting out different volumes agar and then immersing them in sulphuric acid to investigate how the rate of diffusion is affected.
Simple passive diffusion occurs when molecules pass through the lipid bilayer of a cell membrane. A plasma membrane is an outer membrane of a cell, which consists of two layers of phospholipids and entrenched with proteins. It is a thin selectively permeable membrane layer, which surrounds the cytoplasm and other elements of the cell. Membranes are in motion with the fast drifting lipids and slower drifting proteins. The labelled diagram below shows a plasma membrane.
(image courtesy of http://micro.magnet.fsu.edu/cells/plasmamembrane/plasmamembrane.html)
The different parts of the cell membrane play different significant roles to enable it work as a whole. The proteins in a plasma membrane serve diverse functions such as transport, enzymatic activity and cell to cell recognition amongst many other functions. The carbohydrates are only found on the outside face and attach to lipids or protein to enable cells to distinguish/ recognize one another. The membrane hence allows nutrients and other essential elements to enter the cell and waste materials to leave the cell. Small molecules, such as oxygen, carbon dioxide, and water, are able to pass freely across the membrane
Temperature, size of the molecules, concentration gradient, pH sensitivity, and surface area to volume ratio are some factors that affect the diffusion rate. With this experiment the effect of surface area to volume ratio on the diffusion rate will be examined.
The aim of this investigation was to examine the relation between surface area and volume ratio and how it affects the rate of diffusion in plant cells.
The hypothesis was that the greater the SA:V, the greater the diffusion rate
VARIABLES
INDEPENDENT VARIABLES: surface area volume ratio of the agar cubes.
DEPENDENT VARIABLES: rate of diffusion
CONTROLLED VARIABLES:
 Temperature
 Concentration of sulphuric acid
 Volume of sulphuric acid
 Pressure
 Time taken
 MATERIALS
 1 30  cm long ruler
 A knife
 4  200ml large beakers
 Forceps
 Paper towel
 800 ml of sulphuric acid
 Stop watch
 2 large tabs of prepared agar blocks
 METHOD
 Four agar cubes were cut as accurately as possible using a knife on a chopping board: 1, 2, 3, and 4 cm cubes.
 The same was done for 2 more repeats. The cubes were remeasured a couple of times with a ruler to get accurate measurements. (See below diagrams)
 200 ml of sulphuric acid was put in four beakers.
 The four cubes of different sizes were immersed in in the solution noting the time. Each size was put in the same beaker i.e. the 1cm cubes were in one beaker and so were the rest. Then each beaker was labelled.
 They were then let to soak for approximately 10 minutes.
 Periodically, the solution was gently stirred, or cubes turned over.
 After 10 minutes, the acid was poured out in the sink making it easier to remove the cubes out of the beaker without damaging them.
 Using the pair of forceps the blocks were removed from the beakers and blotted with a paper towel.
 Each cube was then cut into half and each the depth to which the pink colour had penetrated was measured.
 The measurements were then recorded and put in a table.
 RESULTS
Calculating % diffusion in each cube:
 Calculate total volume of each cube (volume = L x W x H)
 Calculate volume that did not turn pink. (That is, calculate total volume of the small portion of the cube that did not turn pink – use the same formula L x W x H)
 Calculate volume diffused = total volume – volume not pink.
 Calculate % diffusion = Volume diffused /total volume x 100
Calculate the surface area of each cube and the surface area to volume ratio:
 Calculate the surface area of a cube = L x W x # of sides
 Calculate surface area/volume ratio.
FIGURE 
TOTAL NO. OF CUBES 
SURFACE AREA OF FIGURE(=6) 
VOLUME OF FIGURE (=) 
SURFACE AREA TO VOLUME RATIO (= SA/V) 
TOTAL SURFACE OF INDIVIDUAL CUBES* (=6* #CUBES) 
A 
1 
6 
1 
6 
1 
B 
8 
24 
8 
3 
48 
C 
27 
54 
27 
2 
162 
D 
64 
96 
64 
1.5 
384 * because surface area of one cube is 6 
CUBE 
A 
B 
C 
D 


NO. OF REAPEATS 

1 
8 
27 
64 


 
0.01 
0.012 
0.01 
0.97 
0.94 
0.86 
9.26 
8.12 
8 

1 
7.99 
7.98 
7.99 
26.03 
26.06 
26.14 
54.74 
55.88 
56 

100% 
99.83% 
96.57% 
86.77% 
 DISCUSSION
Surface Area is the measure of how much exposed area a solid object has, expressed in square units (x^{2}). It determines the exchange of materials between the cells and its environment. Volume is how much threedimensional space a substance (solid, liquid, gas, or plasma) or shape occupies and is expressed in cubed units (x^{3}) and it determines the amount metabolism in the cytoplasm.
From table 1.1 above, the surface area to volume ratio decreases as the surface area and volume get bigger and this is graphed against the length in graph 1. When the length increases the ratio gets smaller; this shows as organisms get bigger it becomes more difficult for them to exchange materials with their surroundings. This is evident in the calculations in table 1.2; as the cubes got bigger, the percentage of how much acid had diffused became lesser. In cube A, which was the smallest, the sulphuric acid diffused in 100% but in cube D, which was the biggest, only 86.77% diffused in. The images below show how the acid was diffusing into the agar when immersed in sulphuric acid.
To meet its metabolic needs, a cell must have sufficient surface area for its volume and this relationship is described as a “large” SA: V ratio. Thus the bigger cells have greater metabolism than smaller cells, but bigger cells have proportionally less surface for exchange. The ratio of the two is critical. Cells surface area and volume increases at different rates. The volume (V=needs) increases faster than surface area (SA= supply)
Therefore SA/V ratio actually decreases with increasing cell size. The decreasing SA/V ratio limits the cells from growing larger than 100 mm. Without enough surface area for its size, such a large cell will not be able to supply its own needs as a cell grows larger, while both the supply and demand increase, the ‘need’ quickly outpace the ‘supply’. At that point, the cell must stop growing or divide into two smaller cells (cell division) or it will die due to suffocation of waste unable to diffuse out fast enough or get nutrients fast enough so it will either be poisoned or starve.
Eukaryotic cells are typically larger than 10 mm because you need at least 10 mm to hold the minimum structures for survival. Eukaryotic cells are smaller than 100mm because a decreasing SA/V limits a growing cell’s ability to supply its own needs.
 CONCLUSION
The experiment showed that if the surface area volume ratio gets bigger the diffusion rate increases. It means that when the surface area volume ratio increased the amount of Sulphuric acid that entered the agar block decreased hence supported the hypothesis. So the bigger surface area volume ratio causes increasing diffusion rate. That’s why, it was concluded that increasing surface area and volume ratio, increases the rate of diffusion. Thus, surface area and volume ratio is proportional to rate of diffusion.
Limitations:
 As the time taken for diffusion to be completed is measured by an observer, errors due to observer are possible.
 As a stopwatch is used to measure time, thinking time and reaction time of the observer cause random errors due to observer.
 Errors could be made while observing the exact time of colour change.
 As the experiment is done in an open beaker, the entry of possible impurities may not be realized.
 Inaccurately cut cubes i.e. by minimal units/ measures.
Improvements:
 The experiment can be improved by:
 Repeating the experiment more than 2 times.
 Insulating the beaker while the diffusion is taking place.
 Using a colorimeter in order to measure the colour change.
 First of all, wash all the equipment in order to ensure that no errors are caused due to dust and other impurities.
Cite This Essay
To export a reference to this article please select a referencing stye below: