Ccea as Biology Coursework: an Investigation to Find the Water Potential of Potato and Carrot Tubers in a Sucrose Solution over a 24 Hour Period
An investigation to find the water potential of potato and carrot tubers in a sucrose solution, of concentration 0.00 – 0.50Mol, over a 24 hour period
Interpretation
Written Communication
C1
From our graph it can be seen that the concentration of sucrose solution is 0.18 M at 0% change in mass for the potato and 0.355 M at 0% change in mass for the carrot. I will use these values to find the solute potential by using the calibration graph. I will work out the water potential by using the equation, Ψ=Ψs +Ψp (Water Potential = Solute Potential + Pressure Potential). The Ψs of the potato at 0% change in mass is -500 kPa and the Ψs of the carrot at 0% change in mass is -1000 kPa. Through the use of our equation, the water potential of the potato and carrot are -500 kPa and -1000 kPa (respectfully) as in this case the solute potential equals the water potential as there is no pressure potential as the solution is open and it isn’t under a membrane so it is not under pressure. The Water Potential (Ψ) of the solution is equal to the Ψ of the tuber as there is no pressure potential.
C2 and C3
As the concentration of the sucrose solution increases, the average percentage change in mass decreases in the potato tubers and this is the same as in the carrot tubers. At low concentrations of sucrose solutions (0.1 M) the mass of the carrot and potato tubers increases due to water moving into the protoplast of the cell from the sucrose solution by osmosis and at high concentrations of sucrose solutions (0.5 M) the mass of the carrot and potato tubers decreases due to water moving out of the protoplast of the cell to the sucrose solution by osmosis. At certain concentrations (0.18 M of the potato and 0.355 M for the carrot) the potato and carrot tubers don’t change in mass due to the water potential inside the cells equalling the water potential of the sucrose solution. My graph displays a distinct negative correlation; the higher the concentration of sucrose
Interpretation
Written Communication
C1
From our graph it can be seen that the concentration of sucrose solution is 0.18 M at 0% change in mass for the potato and 0.355 M at 0% change in mass for the carrot. I will use these values to find the solute potential by using the calibration graph. I will work out the water potential by using the equation, Ψ=Ψs +Ψp (Water Potential = Solute Potential + Pressure Potential). The Ψs of the potato at 0% change in mass is -500 kPa and the Ψs of the carrot at 0% change in mass is -1000 kPa. Through the use of our equation, the water potential of the potato and carrot are -500 kPa and -1000 kPa (respectfully) as in this case the solute potential equals the water potential as there is no pressure potential as the solution is open and it isn’t under a membrane so it is not under pressure. The Water Potential (Ψ) of the solution is equal to the Ψ of the tuber as there is no pressure potential.
C2 and C3
As the concentration of the sucrose solution increases, the average percentage change in mass decreases in the potato tubers and this is the same as in the carrot tubers. At low concentrations of sucrose solutions (0.1 M) the mass of the carrot and potato tubers increases due to water moving into the protoplast of the cell from the sucrose solution by osmosis and at high concentrations of sucrose solutions (0.5 M) the mass of the carrot and potato tubers decreases due to water moving out of the protoplast of the cell to the sucrose solution by osmosis. At certain concentrations (0.18 M of the potato and 0.355 M for the carrot) the potato and carrot tubers don’t change in mass due to the water potential inside the cells equalling the water potential of the sucrose solution. My graph displays a distinct negative correlation; the higher the concentration of sucrose