• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Investigation to find the Water Potential of Apple Tissue

Extracts from this document...


Investigation to find the Water Potential of Apple Tissue Introduction The water molecules move in and out of the cell through the partially permeable cell membrane by the process of Osmosis. The definition of osmosis as given in "Understanding Biology For A-level, Fourth Edition" by "Susan and Glenn Toole" is as follows, "the passage of water from a region where it is highly concentrated to a region where its concentration is lower, through a partially permeable membrane." Diagram of Osmosis This investigation is to establish the exact water potential of apple tissue. The definition of water potential, as written in "Biology 1" endorsed by OCR "Water potential is the tendency of a solution to lose water; water moves from a solution with high water potential to one with low water potential. Water potential is decreased by the addition of a solute, and increased by the application of pressure. Symbol is ?" This is called moving down a water potential gradient. When the water potential in both regions is equal both areas are in equilibrium and there is not further net movement of molecules. The water potential of a cell is determined by two factors: the solute potential in the cell (?s), and the pressure potential (?p). * The solute potential (?s) is a measure of the reduction in water potential due to the presence of solute molecules. It is the negative component of water potential, sometimes referred to as the osmotic potential or osmotic pressure. ...read more.


5. I also need to find the percentage change by dividing the change in by the initial and multiplying by 100 for both length and mass. Results for Preliminary Test 1 Concentration Mass before/g Mass after/g Change in mass/g % Change in mass /Mol 1 2 Mean 1 2 Mean 1 2 Mean 1 2 Mean 0.0 0.87 0.87 0.87 0.85 0.82 0.835 -0.020 -0.050 -0.035 -2.30 -5.75 -4.02 0.2 0.92 0.92 0.92 0.87 0.90 0.885 -0.050 -0.020 -0.035 -5.43 -2.17 -3.80 0.4 0.92 0.90 0.91 0.90 0.94 0.92 -0.020 0.040 0.010 -2.17 4.44 1.10 0.6 0.86 0.83 0.845 0.90 0.85 0.875 0.040 0.020 0.030 4.65 2.41 3.55 0.8 0.91 0.88 0.895 0.87 0.83 0.85 -0.040 -0.050 -0.045 -4.40 -5.68 -5.03 1.0 0.93 0.90 0.915 0.78 0.75 0.765 -1.150 -0.150 -0.150 -16.13 -16.67 -16.39 Concentration Length before/mm Length after/mm Change in Length/mm % Change in Length /Mol 1 2 Mean 1 2 Mean 1 2 Mean 1 2 Mean 0.0 50 50 50 50 51 50.5 0 1 0.5 0 2 1 0.2 50 50 50 51 50 50.5 1 0 0.5 2 0 1 0.4 50 50 50 51 52 51.5 1 2 1.5 2 4 3 0.6 50 50 50 51 51 51 1 1 1 2 2 2 0.8 50 50 50 48 49 48.5 -2 -1 -1.5 -4 -2 -3 1.0 50 50 50 47 46 46.5 -3 -4 -3.5 -6 -8 -7 Preliminary Test 2: to find what shape to use I also needed ...read more.


It was also very difficult to cut the strips into exactly the right thickness and the masses were much less than the cores. By using a cork borer, it ensures the diameter of each piece of apple is always the same and the only measurement I have to cut by hand is the length, this reduces risk of inaccuracy within the results. From the preliminary tests I can tell that apple tissue has a very high sugar content so the solute concentration will be higher than the one demonstrated. I can also conclude that I will use a range of concentrations around 0.2M and 0.8M concentrations. This is because I am trying to find the exact point at which the molarity of the solution causes the percentage change = 0. If the results were drawn on the graph, this would be the point at which the curve crossed the x-axis, thus showing the concentration of sucrose within the apple tissue. I will also use four pieces of apple instead of two to allow a greater range of results to get a much more accurate mean average. This would also make it easier to see any anomalies. Prediction Using my preliminary tests I can estimate the sucrose concentration of apple tissue to be between 0.4M and 0.8M, and therefore the water potential will also be between 0.4M and 0.8M. The results of the main experiment will conclude whether this hypothesis is accurate or not. ?? ?? ?? ?? ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE Green Plants as Organisms section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Green Plants as Organisms essays

  1. Marked by a teacher


    4 star(s)

    - 52 2 U1 = 5 U2 = 6 x 6 + 6(6 + 1) - 26 2 U2 = 31 Critical Value = 5 (No. of samples: 6) "If the smallest U value is less than or equal to the critical value we reject the null hypothesis and accept the alternative hypothesis" As the smallest U value (5)

  2. Peer reviewed

    Osmosis in potato and apple

    3 star(s)

    7 Time in solution: all the 'chips' will be kept in the solutions for the same period of time. The timing will be staggered, each chip will be placed in the solution two minutes after the previous so each have been in the solution for an equal period of time, and measured.

  1. Determine the water potential of potato tuber cell with the varying affect of solute ...

    Lower water concentration so water moves out However even though I predict that there will be a loss of mass, I also predict that the loss will only be a little since the cylinder is almost at equilibrium. This means that I predict that the rate of water entering the

  2. Determine the Water Potential of Potato tissue.

    Water potential equals to solute potential because pressure potential is zero. This is because there is no pressure added or taken away in this experiment, pressure does not vary in this experiment we are leaving it natural. Therefore we now don't need to use the formula above for this reaction

  1. Investigate and find the water potential of baking potatoes and sweet potatoes in (N/mm2) ...

    If you know the water potential of the solution then you can find the water potential of the cell. When the external water potential of the cell is less negative than the cell, more dilute, water flows into the cell.

  2. The aim of the investigation is to find the exact concentration of the cell ...

    This is the variable I am interested in for my results. In this experiment there are other variables such as room temperature, the initial mass of the potato may effect it as well and the type of potato or size of the potato.

  1. An experiment to determine the osmotic values ofChinese radish and Potato cores.

    Molority M cm3 of water cm3 of 1.0M Glucose 0.2 8 2 0.4 6 4 0.6 4 6 0.8 2 8 1.0 - 10 Aim: To find and prove the molarities of Chinese Radish and Potato, along with scientific backing, and also to see how they 'perform' when exposed to differing molarity solutions.

  2. Osmosis investigation. My prediction is that as the concentration of the solute increases, ...

    To make my solutions of 0, 0.25, 0.5, 0.75 and 1 mole sucrose, I will use ratios- e.g. for 0.5: 1:1, so 10ml of each. Next I will prepare the potato pieces for their "soak". 2. I will aim to attain all the pieces from one potato.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work