Investigating water relations in two different plant tissues

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A-Level Biology Coursework:

Investigating Water Relations In Two Different Plant Tissues

Ying-Jun Ng

Equipment used in practical

  • 0.1mol dm-3 Hydrochloric acid
  • 2 plant tissues (swede & potato)
  • 4 test tubes
  • 5x100ml beakers containing 6 different sucrose solutions
  • 6” ruler
  • 6 toothpicks
  • 12 boiling tubes & boiling tube covers
  • 50cm3 measuring cylinder
  • 80oC Water bath
  • Benedict’s solution
  • Blotting paper/paper tile
  • Chinagraph pencil
  • Distilled water
  • Dropping pipette
  • Filter paper
  • Iodine solution
  • Litmus paper
  • Pestle and mortar
  • pH indicator chart
  • Sand
  • Scalpel
  • Scissors
  • Sharp knife
  • Sodium hydrogen carbonate
  • Spatula
  • Stop clock
  • Top balance (2dp)
  • White tiles

Table I:

Results table to show the appearance of potato and swede chips after being submerged in six different sucrose solution concentrations for 24 hours

Table II:

A table to summarise the food tests results on samples of potato and swede

Analysis

Graph I shows the average percentage change in potato and swede tissue samples in six different concentrations of sucrose solutions after being submerged for 24 hours. The average changes in mass are taken from table’s III & IV and plotted as a graph where a line of best fit is drawn.

On graph I, there is a clear trend to show that the average percentage mass of the tissue samples decreases as the sucrose solution concentration increases. This is because in a more concentrated sucrose solution (e.g. 0.6 mol dm-3, 0.8 mol dm-3, and 1.0 mol dm-3) there are more sugar molecules present compared to the number of water molecules. This is a hypertonic solution. It is this small amount of water molecules in the solution, which gives the solution a low water potential. When plant tissue is placed into this hypertonic solution, water molecules from within the tissue (vacuole) will pass through the cell membrane to the surroundings, which results in the decrease in mass of the tissue sample.

This passive process is known as osmosis. Osmosis is where water moves from an area of high water potential to an area of low water potential (down its concentration gradient) through a partially permeable membrane. The difference in solute concentration between the tissue samples and surroundings caused the potato and swede chips samples decreased in mass (g) when submerged in hypotonic solutions. The potato and swede chips increased in mass when the tissues were submerged into weak sucrose solutions (0.0 mol dm-3, 0.2 mol dm-3 and 0.4 mol dm-3.). These solutions are hypotonic and more water molecules then sugar molecules. This has a high water potential and water will move from the solution and into the cell causing it to increase in mass.

When a tissue sample was placed in a concentrated sucrose solution it decreased in mass and also became flaccid (refer to Table II). This is because plasmolysis occurred. When a plant cell is placed in a hypertonic solution, it will lose its water and pressure potential causing the cell to become flaccid. The point on the graph where the line of best fit intercepts the x-axis can be used to calculate the water potential of the two tissue samples with use of Reference graph II. The point where the line of best fit intercepts the x-axis is the point of incipient plasmolysis. This is where pressure potential had reached 0 and plasmolysis is about to occur. The pressure potential is 0 because the protoplast has shrunk and is no longer in contact with the cell wall.

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From Graph I, the point of incipient plasmolysis for the potato tissue sample is 0.26 mol dm-3 and the point of incipient plasmolysis for the swede sample is 0.40 mol dm-3. Using Reference graph II the solute potential of each tissue can be calculated. The solute potential for the potato tissue sample at the point of incipient plasmolysis was 725 kPa. The solute potential for the swede sample at the point of incipient plasmolysis was 1125kPa.

 

In a plant cell the equation, Water potential = Solute potential + pressure potential however, at the point of incipient plasmolysis Pressure potential ...

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