Sources: Biology 1……………………………..
Prediction:
Taking the above information into account and looking at the two types of potato –old and new- I predict that the pieces of old potato will take in more water than the new potato. This is because the old potato isn’t as fresh and so therefore will have lost water and become shrivelled and the cells will have become flaccid. This in turn means that because water has been lost, the concentration inside the old potato is higher than that of the new potato. Because of this, I predict that when the old and new potatoes are both tested with the same sucrose concentration, the old potato will gain more weight. This is due to the fact that because the water potential is lower in the old potato, more water is needed to move across the membrane to result in equilibrium.
Method:
- Cut 11 3cm long cores of potato from the first kind of potato to be tested. Remove any skin.
- Set up 11 boiling tubes in two boiling tube racks and label them 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 and 1.0.
- Weigh the potato pieces one at a time and place one in each of the boiling tubes, being sure to record which potato goes in which tube.
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Fill each boiling tube up to halfway with 20cm3 of the corresponding concentration of sucrose solution.
- Leave for 60 minutes. Drain the sucrose solutions from the boiling tubes one at a time and gently dry to remove excess moisture. Reweigh the potato. With these values, calculate the percentage gain of each potato piece. The reason the percentage is calculated is to make fair comparisons between different pieces of potato taking into account the starting weight.
- Repeat procedure for the second type of potato.
- Repeat this whole experiment 3 times so as to ensure experimental errors did not occur and to enable an average to be taken.
Safety:
So as to ensure this experiment is safe, follow these precautions:
- The scalpel and cork borer used will be very sharp. Take care not to cut your fingers and always use a white tile to rest on.
- Keep workbenches clear as glass equipment is being used which could get smashed if knocked.
Equipment:
When it came to carrying out the method, I was unable to keep everything exactly how I had planned. Due to lack of time and resources I was unable to use 11 different sucrose solutions and had to cut down to 6. Lack of resources also meant that I was unable to repeat the experiment 3 times as it would have meant using 33 boiling tubes. Therefore, I decided that to reduce experimental error as much as possible, I would put three pieces of potato in each tube and achieve three results in that way. Also due to lack of time, I was unable to leave the potato pieces in the solutions for 1 hour and I had reduce the time to 30 minutes.
Results:
Table Showing the Amount of Weight Gained by a Piece of Fresh Potato in Various Concentrations of Sucrose
Table Showing the Amount of Weight Gained by a Piece of Fresh Potato in Various Concentrations of Sucrose
Conclusion:
By looking at the graph, it is clear to see that my aim has been achieved because the line for the old potato is generally above the line for the new potato. This means that the old potato has taken up more water and my introductory research tells me that this potato therefore has the lower water potential. By reading the point at which each line meets the x axis, I can estimate the equivalent concentrations of the potatoes as I explained in my introduction. This is because where the line crosses the x axis there was no change in weight. This means that osmosis is not taking place because there is the same level of water potential both inside and outside of the potato. The new potato has an equivalent concentration of 0.21 mol dm-3 while the old potato has a concentration of 0.23 mol dm-3. This shows that the old potato is more concentrated as I predicted.
My results did what I expected from my preliminary experiment as two concentrations, 0 and 0.2 mol dm-3 sucrose, caused weight gain and the other four, 0.4 – 1.0 mol dm-3 sucrose caused increasing weight loss. This is due to osmosis which, as I explained in my introduction, tries to balance the concentration of either side of a membrane by transferring water molecules. The side with the highest water potential (lowest solute concentration) will lose water molecules as they are passed over to the other side in order to even out water potential.
Evaluation:
i was pleased with the data collected by my experimental procedure because the results gathered gave me sufficient evidence to support my prediction. However, I feel that had I been able to carry out the method exactly how I had proposed, I would have achieved a more reliable set of results. For instance, had I been able to use 11 concentrations of smaller intervals than the 6 I used, I would have been able to draw a more accurate graph and easily recognise any anomalies. Also, as I only had time to leave the potato in the solution for 30 minutes and not the 60 I anticipated I feel that it may not have stopped losing or gaining water by the end of the experiment. This could be responsible for reliability errors because my results may not be an indication of how osmosis occurs in potatoes. For example, because I didn’t leave them in the solutions for long enough I do not know that the potato pieces weren’t all going to gain an equal amount of weight but at different speeds.
By looking at the graph I have recognised an anomaly. For both kinds of potato, the results for 0.6 mol dm-3 are lower than expected if they were to follow the trend. To account for this I suggest that because both results for the 0.6 mol dm-3 concentration are wrong, the problem may have been that the solution had been diluted inaccurately or contaminated. By drawing lines of best fit on my graph, I suggest that the accurate concentration was actually 0.7 -0.8 mol dm-3. As we were given the solutions ready made for us to use, I could ensure that this area of unreliability is accounted for by preparing the sucrose solutions myself.
Because I was unable to do as I had planned and use three separate boiling tubes for the three pieces of potato in each concentration, this meant that if one result was wrong, they all were.
Another possible error leading to poor reliability was the drying of the potato before reweighing them. It was impossible for us to consistently dry each piece of potato so as to allow for fair comparisons, so the calculated weight loss or gain may not be exact. This undermines the validity of the statement “The new potato has an equivalent concentration of 0.21 mol dm-3 while the old potato has a concentration of 0.23 mol dm-3” that I made in my conclusion. This is because if, for instance, I had left just 0.01g more water outside of the potato than at the beginning of the experiment, the whole line of best fit would have been higher than it should have been. It would have therefore met the x axis at a different place to give a higher reading of the equivalent concentration.
My measurements were very accurate throughout the experiment as I used an accurate balance for the weighing and always rounded to two decimal places.