Water Relationships in Two Plant Tissues Results: Table 5 Table to show the sucrose concentration and water potential of each tissue. The sucrose solution was extrapolated from graph 1, which shows the percentage change in tissue mass when immersed in different sucrose solutions. A line of best fit was drawn, where the line of best fit intercepts the x-axis (concentration of the sucrose solution) which is the sucrose concentration of the tissue because at this point there is no loss of gain mass (reading on the y axis). the water potential of each tissue was read from graph 2. The solute potential is equal to the water potential because the pressure potential = 0. Tissue Sucrose concentration in the tissue based on data extrapolated from the graph 1 /mol dm-3 Solute potential read from graph 2 /Kpa (equal to water potential/Kpa)PotatoSwedishTable 6Table to show the results of the iodine test, the Benedict test and the non-reducing sugars test for potato and swede tissues.TissueIodine test for starchBenedict's test for reducing sugarsTest for non-reducing sugarsPotatoTurned black/blueTurned greenTurned yellow/pale orangeSwedeGrey/blackTurned orangeTurned dark orangeAnalysis: The aim of the experiment was to find out the water potential of swede and potato tissues. First, 12 potato strips and 12 turnip strips were cut to 7 cm long, 0.5 cm wide and 0.5 cm deep. The strips were weighed (initial mass in the results tables). Two potato strips were placed in each Petri dish, each containing a different sucrose solution (0.0, 0.2, 0.4, 0.6, 0.8, 1.0 mol dm-3). This was repeated with the Swedish strips. So in total 12 Petri dishes were used (6 for each tissue). Diagram 1 shows how the survey was conducted. The next day the strips were extracted from the solutions, dried and weighed (final mass).