Wednesday 5 February 2014

Practical #1- Beetroot experiment

Aim: To investigate the factors affecting the cellular homeostasis of beetroots cells

Apparatus:
Apparatus

 
Procedure
  • Set up a boiling water-bath with the bigger beaker. When the water starts to boil, turn off the Bunsen burner.
  • Use a ruler and scalpel to cut the cylinder of beetroot into 25 discs of 2mm each.
  • Take 5 discs of beetroot and cut them further into smaller pieces.
  • Rinse the beetroot discs and pieces until the water is colourless.


Label and prepare 5 test tubes as follows:

Tube
Content
A
4ml of water
B
4ml of 25% alcohol
C
4ml of 50% alcohol
D
4ml of hot water (90 degree C – 100 degree C)
E
4ml of water with chopped beetroot

  • Place 5 discs of betroot in A-D and all the chopped beetroot in tube E using the forceps.
  • Leave the tubes to stand in your test tube rack for 15 minutes.
  • Shake the tubes gently after 15 minutes and hold it agasint the white tile to note the colour. Record your observations in a table.
  • Use a colouring meter to record the percentage of transmission of light.
  • Decant a small amount of liquid from each tube into a cuvette to measure the percentage of transmission.
  • Dispose the content of the tubes after the exeriment. Do not throw the beetroot into the sink.


Results:
Tube
Contents of tubes
% Of transmission of light
Description of liquid in tube
A
4ml of water
86.7
The liquid was colourless at first but turned a hue of pink after 15 minutes.
B
4ml of 25% alcohol
80.03
The liquid was colourless at first but turned pale pink after 15 minutes.
C
4ml of 50% alcohol
33.6
The liquid was colourless at first but turned dark pink after 15 minutes.
D
4ml of hot water (90 degree C – 100 degree C)
5.43
The liquid was colourless at first but turned very dark pink after 15 minutes.
E
4ml of water with chopped beetroot
20.73 (very low-supposed to be the second highest) (72.0)
The liquid was colourless at first but turned a dark shade of pink after 15 minutes. (turned light pink)
*Note: Use the same colour but to vary the range, use workds like ‘darker’, ‘dark’, ‘light’ etc






Discussion Questions
1.     Why was it necessary to wash the beetroot slices thoroughly befure using them in this experiment?
- The cell wall and cell membranes of the beetrot had been punctured causing the betalain to leak out. Hence to ensure a fair experiment, it has to be rinsed.
2.     Identify the independent and dependent variables in this experiment
-        Independent variable: Concentration of alcohol, temperature, surface area
-        Control set-up: Test tube A
-        Dependent Varable: Percentage of light transmission (colour of liquid)
3.     Construct a suitable table with appropriate headings and units to tabulate data
- See above for results
4.     Explain with reference to the tabulated data the effect of different solutions in tubes A-C on the readings obtained in the experiment. Make reference to the knowledge you acquired from the lessons on cellular homeostasis.
-        Temperature --> denature proteins
-        Cell membrane and Tonoplast --> becomes more porous
-        Alcohol: interaction with the lipid layer (phospholipid bilayer) and hence the cell membrane and tonoplast becomes more porous.
5.     Suggest an explanation for the observations of tube D and E.
- Beetroot in D was submerged in hot water bath. When expoded to heat the beetroots cell is disrupted. The molecules start to move faster causing the membrane and pigment insde to become denatured. The lipids in the membrane becomes more fluid, as the temperature is high, ths the membranes become more gragile. Coils of protein will break aparat and start vibrating. Holeswill tart to form and the cell membrame will be destroyed. The betalina pigments will then leak out. Thsus the percentage of light transmissio of tube D is 5.43%. For tube E there was a larger surface area for diffusion to occur among the membranes, thus more betalain pigments leaked out. The rate of diffusion from internal to ecternal environment increased as a result to a larger surface area exposed. Thus E has a reading of 20.73%.



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