# lab on cell diffusion

Topics: Diffusion, Osmosis, Surface area Pages: 5 (1137 words) Published: April 11, 2014
﻿Diffusion in Cells
Isabel Zak

Question:
How does the size of a cell affect the distribution of chemicals throughout the cell?

Hypothesis:
The larger the cell is, the more difficult it will be for the chemicals to reach the centre of the cell, and diffuse throughout it. This is because there will be a higher surface-area to volume ratio in the larger cells, making the centre of the cells further away from the surface. Therefore, when dipped in sodium hydroxide, the larger cells will not be dyed pink all the way through like the smaller cells after the ten minutes.

Materials:
1x1x1 cm agar cube
2x2x2 cm agar cube
3x3x3 cm agar cube
100 ml of sodium hydroxide
Goggles
Petri dish
Scissors
Timing device
Tweezers
Beaker

Procedure:
Refer to 'Nelson Science Perspectives 10' pages 38-39, section 2.4; 'What Limits Cell Size?'.

Safety:
All long hair was tied back, and loose clothing was removed. Goggles were worn at all times, and the sodium hydroxide was handled carefully to prevent contact with skin, and was not ingested, as it is poisonous. All of the tests were preformed whilst standing, and in a well-ventilated area. Nothing was eaten or drank during the experiment.

Observations:
Cell
Length of the sides
(mm)
Area of sides
(mm)
Total Surface Area (mm2)
Volume of Cell
(mm3)
Ratio of surface-area to volume
Distance colour extended into cube from surface
Diffusion Rate Per Minute
(depth of dye/time (minutes))
A
10
100
600
1000
3:5
5 mm (all the way through)
.5 mm/minute
B
20
400
2400
8000
3:10
5 mm
.5 mm/minute
C
30
900
5400
27000
1:5
5 mm
.5 mm/minute

Graph:
Ratio of surface-area to volume

Analysis:
The colour change in the cells represents how far into the cells the sodium hydroxide was able to diffuse within ten minutes. The sodium hydroxide diffused into the cells at the same rate, .5 mm per minutes. This meant that only the smallest cube, which was 10 mm long on each side was able to be dyed completely pink. This is the depth to the centre of the small cell was 5 mm form the surface, all of the other cells had a larger distance to the centre, and would need to be soaked for longer for the chemical to diffuse completely. Cell B would need to be soaked for at least 20 minutes, and cell C would need o be soaked for at least 30 minutes.

Along with taking longer for the chemicals to diffuse into them, the larger cells had more extreme surface-area-to-volume ratios. The larger the cell, the longer it takes for the sodium hydroxide to diffuse completely. The smallest cell (cell A) had a ratio of only 3:5, whereas the largest cell (cell C) had a ratio of 1:5. The more extreme the surface-area to volume ratio, the more of the cell was left white. If a real cell was faced with this inability to diffuse the necessary nutrients, it would die from the inside out, as the organelles located in the middle would starve, causing the rest of the cell to die, since their functions would not be completed. Therefore, cells must remain small enough for the nutrients ect., to diffuse through the cell. With this in mind, the only cell which would have been able to survive would have been cell A. Due to its minimal surface-area to volume ratio, the sodium hydroxide was able to completely diffuse, which in a real cell, would have allowed it to carryout its vital functions.

The agar cube is a good model for a cell, as it is very pale, which made the pink colour that the sodium hydroxide dyed them easy to see, also since the agar is non-toxic, they were easy to handle. However, since real cells have constantly undulating membranes to allow for larger surface area, a real cell would have a very different rate of diffusion. Also, this membrane is selectively permeable, which the agar cube could not replicate. A real cell would be able to decide how much or how little of a substance it wants to allow to enter, but the agar cubes could only...