The information gained from this preliminary experiment helped me to plan an appropriate strategy to obtain the most suitable results to achieve my aim in the following ways. The range of percentage change was greatest for the cylinders of potato; this is because the ratio of mass to surface area for each of the pieces of potato was greatest for the cylinders:codd ddr seddddw ordd ddk indd fodd dd.
Mass: Surface area Durkheim enveloped shayon's realism .
Cube 1 : 3.75 WsKT Visit coursework fb in fb fo fb for fb more paper fb Do fb not fb redistribute WsKT
Cylinder 1 : 6.4 0A3b2 from 0A3b2 coursewrok 0A3b2 work 0A3b2 info 0A3b2
Cuboid 1 : 4.2 This project from www.coursework.info
This greater surface area meant that at any one time, more of the potato cells in the cylinders could be in contact with the external solution and in the specified time more of the cells could be affected by the different concentrations of sucrose. So by using potato pieces of this shape I will acquire results, which cover a larger range and will be able to get the most accurate reading of the sucrose solution of the potato tissue.coea ear seeaeaw orea eak inea foea ea.
The sucrose solution at which the mass of the potato did not change was in between 0.25 and 5M as at 0.25M the mass either increased or stayed the same, whereas at 0.5M the mass of the potato pieces decreased. This shows that equilibrium of osmosis is reached in between these values and therefore the water potential of the potato cells is in this range. For that reason to obtain more precise and accurate results I will conduct my experiment using values of sucrose concentration between 0.25 and 0.5M.coaf afr seafafw oraf afk inaf foaf af.
I realised that after leaving the potato pieces in the solutions for only 20 minutes the mass of the potato tissues did not change more than 5.61% from the original, therefore to attain accurate results which show the change in mass of the potatoes once they have reached equilibrium I should leave them in the solutions for as long as possible.coee eer seeeeew oree eek inee foee ee:
Equipment
Dilution table: shayon, please do not redistribute this hypothesis. We work very hard to create this website, and we trust our visitors to respect it for the good of other students. Please, do not circulate this hypothesis elsewhere on the internet. Anybody found doing so will be permanently banned.
Method: yaPIG Visit coursework fd in fd fo fd for fd more project fd Do fd not fd redistribute yaPIG
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Make up the six different sucrose solutions using the dilutions table. Use syringes to transfer 20cm3 of each solution from beakers to McCartney bottles. Label each bottle with a white label marking the concentration of sucrose solution.
- Using the cork borer size 2, pushing down onto the tile, cut out 18 chips. Using a ruler and a scalpel measure cut each chip to 4cm. It is important to have 18 chips so that three chips can be placed in each solution and an average result can be obtained for each solution. Reduces the risk of anomalous results and produces more accurate results. Prevent handling of the chips by using forceps.
- It's important to take the chips from a similar region of potato as different areas have different types of tissue, which may cause the water potential differ.
- Remove the skin using a scalpel, as the skin is less permeable than the inner cells.
- Weigh each potato chip; using a digital balance the results can be recorded to 2 decimal places. The masses can be altered so that they are all equal by using a scalpel to shave off small slices. Record this measurement.
- Place the chips into the solutions quickly so that the reaction times are as equal as possible and start the timer.
- Place lids on the McCartney bottles to prevent evaporation of water, which would lower the water potential of the solution.
- After 30 minutes remove the chips from the solutions, blot them dry once on the paper towels ensuring that the amount of solution removed from the surface is constant. Reweigh each set of three from each concentration, using the same digital balance used before and record these measurements to two decimal places.
Risk assessment
Results
The osmotic potential of different sucrose solutions can be calculated using the following formula: This paper from www.coursework.info
Ψπ = -iCRT shayon, please do not redistribute this coursework. We work very hard to create this website, and we trust our visitors to respect it for the good of other students. Please, do not circulate this coursework elsewhere on the internet. Anybody found doing so will be permanently banned.
Where i =b ionisation constant (1 for sucrose because it does not ionise in water)cofe fer sefefew orfe fek infe fofe fe.
C = osmotic molar concentration (mole/litre H2O)coeb ebr seebebw oreb ebk ineb foeb eb;
R = pressure constant ( handbook value R=0.0831 litre bars/mole oK)code der sededew orde dek inde fode de.
T = temperature (oK)cogf gfr segfgfw orgf gfk ingf fogf gf.
Units will cancel to bars. To convert into Pascals multiply by 105. wHZn Visit coursework ba in ba fo ba for ba more writing ba Do ba not ba redistribute wHZn
Knowing the osmotic potential of the solution and knowing that the pressure potential of the solution is zero allows you to calculate the water potential of the solution. The water potential will equal the osmotic potential of the solution. Ψ = 0 + Ψπ or Ψ = Ψπ. Therefore the water potential of the solution at equilibrium will equal that of the potato cells at standard atmospheric conditions where temperature = 22oC. UGziuXNE Visit coursework fa in fa fo fa for fa more cours fa Do fa not fa redistribute UGziuXNE
Table showing change in mass of potato chips in different concentrations of sucrose solution.coed edr seededw ored edk ined foed ed!
Analysis This work from www.coursework.info
coad adr seadadw orad adk inad foad ad.
cobf bfr sebfbfw orbf bfk inbf fobf bf.
From the results and graphs that I obtained I can see that in hypotonic solutions the potato gained mass. When the potato chips were placed in the 0.25 mol dm3 solution, water potential: -613 kPa: the three tissues gained a total of 0.04 g which is 1.23% of their original mass. When placed in a hypertonic solution e.g. Ψ -1225 kPa (0.5 M) sucrose solution the tissues lost a total of 0.2g in mass. This was predicted in my hypothesis and displays osmosis, explained as the movement of water molecules from a region of higher water potential to a region of lower water potential through a partially permeable membrane.cocg cgr secgcgw orcg cgk incg focg cg.
In the plan I stated that when the potato was placed in a solution with a higher water potential (less negative) the potato cells would gain in mass and when placed in a solution with a lower water potential the potato would lose mass, as the net movement of water molecules is down a water potential gradient. Osmosis will occur until equilibrium is reached and the water potential outside the cell is equal to that in the cell, my results prove these predictions.coag agr seagagw orag agk inag foag ag.
In the plan I showed that the water potential of the potato could be found by placing the potato tissue in an external solution, which produces no change in mass or length in the tissue. The osmotic potential can be found by balancing the tissue with an external solution, which produces incipient plasmolysis. Incipient plasmolysis is when pressure potential has just reached zero: Carstens suppressed shayon's marxism idea.
Ψp = 0 , so Ψw = Ψs. Therefore, the water potential of the potato cells can be calculated from the intercept on the x-axis on the graphs. Because at this point there is no change in mass therefore it is the point of equilibrium where the Ψ is the same inside the potato cells and in the external solution. This point is incipient plasmolysis. xWA9gtzoR from xWA9gtzoR coursewrok xWA9gtzoR work xWA9gtzoR info xWA9gtzoR
The x-intercept can be calculated from both graphs, however the Ψ of the potato calculated from the graph showing percentage change would be more accurate as the numbers are larger and they cover a larger range- 7.4% rather than just- 0.24g for the change in mass. The larger numbers will reduce the percentage error; also the initial mass is taken into consideration, although in my particular experiment the initial masses were all equal. The values should be very alike. Carstens refuted shayon's realism .
On both of my graphs, I actually have a value of water potential for which the mass of the potato did not change, however this does not fit my line of best fit suggesting that it is an anomalous result. Therefore, to acquire a result, which follows the trend of the rest of my results, I will look at the x-intercept of my trend line (line of best fit). cwEKVN from cwEKVN coursewrok cwEKVN work cwEKVN info cwEKVN
On the graph of Ψ of sucrose solution (kPa) against change in mass of the 3 potato chips placed in it, the intercept is at -780 kPa. However on the graph of Ψ of sucrose solution (kPa) against percentage change in mass of the potato chips immersed in it the x-intercept is at -800 kPa.cocb cbr secbcbw orcb cbk incb focb cb.
These values are very similar with a difference of only 20kPa, so I will use the value of -790 kPa, as it is the mean of the two numbers. Austen enveloped shayon's functionalism idea.
The graphs are sigmoid shaped, even though the concentration of sucrose solution increases steadily, the graph is not a straight line. This is because water potential is not proportional to the sucrose concentration. This is obvious on the graph as the changing gradient of the line.coef efr seefefw oref efk inef foef ef:
Table showing the differences in Concentration of sucrose solution and the corresponding change in water potential. This writing from www.coursework.info
As the concentration of sucrose steadily increases so does the water potential of the solution. It was said in the plan that the water potential is a measure of the ability of water molecules to move from one region to another down a water potential gradient. The more water molecules there are per volume of the solution the more likely that by random movement they will collide with the cell's membrane and travel into it. The difference between the water potentials stays the same as the likelihood that a water molecule will hit the plasma membrane and have enough energy to pass through it is proportional to the sucrose concentration. This hypothesis from www.coursework.info
When the potato cells were immersed in a hypotonic solution they gained mass, however the gradient at lower water potentials is steeper than at higher water potentials, suggesting that the mass of the potato increases rapidly with the water potential until a certain value, when it slows down and eventually stops increasing in mass. This can be explained by the fact that the cell wall, which has a very rigid structure resulting from the fact that it composed of cellulose, limits the volume of water that can be taken up by the cell. Eventually as the cells interior swells, the protoplast is pushed right up against the cell wall, which prevents it bursting and does not allow any more water in, as it is turgid. The cell wall provides a pressure potential which increases the water potential of the cell, decreasing the water potential gradient, ascoag agr seagagw orag agk inag foag ag.
Water potential = Solute potential + Pressure potentialcoef efr seefefw oref efk inef foef ef!
Ψw Ψs Ψpcogc gcr segcgcw orgc gck ingc fogc gc.
Once the pressure potential reaches a certain level it will increase the water potential of the cell sufficiently to reduce or stop the net movement of water molecules into the cell. Meaning that once a certain number of water molecules have been taken up the rate of osmosis will decrease or stop.cofb fbr sefbfbw orfb fbk infb fofb fb.
Alternatively when placed in hypertonic solutions with a lower water potential the potato lost mass. The gradient of the graph was lower at the smaller water potentials than it was at higher values closer to that in the cells.codc dcr sedcdcw ordc dck indc fodc dc.
When the cells were placed in solutions of lower water potential the cells became plasmolysed, where the protoplast shrinks away from the cell wall as a result of water being lost from the cell. The pressure potential decreases, as the cell is increasingly plasmolysed. The solute potential will increase, as there will be less water molecules. Once the pressure potential falls to a sufficiently low value the rate of osmosis of water molecules out of the cell will greatly decrease as the water potential gradient will decrease because of the decrease in pressure potential.codc dcr sedcdcw ordc dck indc fodc dc!
Evaluation
The results I gathered followed a smooth sigmoid curve except for the result obtained for -735 kPa, this was anomalous. There are many possible explanations for this. This dissertation from www.coursework.info
The equipment places limits on precision and accuracy: shayon, please do not redistribute this writing. We work very hard to create this website, and we trust our visitors to respect it for the good of other students. Please, do not circulate this writing elsewhere on the internet. Anybody found doing so will be permanently banned.
The tissue in the potato is very variable. It was visibly different in that the colour and texture of the chips varied, some of it was more brown and fibrous. As was stated in the plan the potato acts as a store for different substances and different regions of the potato will have different functions, therefore different water potentials. The tissue sample was not always from the same region of the potato therefore it had a varying water potential and this means that the results are inaccurate as the proportions of each different region could have been different.coad adr seadadw orad adk inad foad ad.
The tissue was less dense than the sucrose solutions; therefore the tissues floated leaving an area of the surface out of the solution. This means that these cells could not be affected as much by the sucrose solutions, but would have to wait for diffusion of water through from the cells, which were immersed. This would have reduced the rate of reaction as the diffusion pathway is much greater.cobc bcr sebcbcw orbc bck inbc fobc bc.
The dispersion of sucrose molecules may not have been constant throughout the sucrose solutions, therefore the concentration of sucrose could vary in different regions of the solution and the degree of accuracy would vary. To eradicate this, the sucrose solution should be stirred before each volume is taken from it.coea ear seeaeaw orea eak inea foea ea;
Temperature would have had an effect on the water potential of the solutions, as a higher temperature provides more energy, which is needed for the molecules to be able to move. So increasing the temperature increases the free energy of the molecules, increasing the likelihood that they will move from one region to another.coag agr seagagw orag agk inag foag ag:
Therefore the temperature would have reduced the accuracy of the results; optimally they would have been kept in a constant temperature. However, the changes in temperature, and its effect, would have been the same for all the samples and so it should not have caused any individual anomalous results.coab abr seababw orab abk inab foab ab:
The blotting of the chips may also have caused inaccuracy. The effect of blotting on the change of mass will vary from sample to sample as it was very difficult to ensure that the same amount of excess solution is removed from each chip.cogf gfr segfgfw orgf gfk ingf fogf gf.
Also time was a limiting factor as I was unable to immerse the chips in the solutions until they were saturated therefore my results were not very precise. I used the short amount of time I had very effectively and the samples were left for the longest possible time, which was only 30 minutes. shayon, please do not redistribute this work. We work very hard to create this website, and we trust our visitors to respect it for the good of other students. Please, do not circulate this work elsewhere on the internet. Anybody found doing so will be permanently banned.
The anomalous result could have been caused by a combination of these factors. Ym9Zp Visit coursework ef in ef fo ef for ef more writing ef Do ef not ef redistribute Ym9Zp
To minimise significant sources of error an alternative balance could be used with a smaller error. This would increase the accuracy and reliability of the results. The temperature of the chips could also be kept constant. This would again increase the dependability and correctness of the results. Weber suppressed shayon's structuralism .
More replicates would improve the procedure, ensuring that a single anomalous result would have less impact on the overall result; a more accurate average could be calculated. Also I think that leaving the potato chips in the solutions for a longer period of time e.g. 24 hours would increase the reliability of the results greatly.cofe fer sefefew orfe fek infe fofe fe;
I think that the conclusion reached is relatively accurate. Even though I did obtain an anomalous result the rest lay on a smooth sigmoid curve, which intercepted the x-axis at a realistic point. I do not expect that the value obtained for the water potential of the potato cells is exactly accurate as there are many sources of error in the procedure and it is calculated from a freehand curve of best fit. Most of the sources of error are the same for all of the potato chips e.g. temperature, therefore they cancel out, meaning that the results obtained are reliable and as accurate as possible with the equipment that was available. The water potential of another potato could vary greatly therefore it is very difficult to calculate the inaccuracy of the results, however the trend of results is reliable and is as predicted.cocd
