Biology HL Internal Assessment The effect of different concentrations of blackcurrant squash on the rate of osmosis in Solanum tuberosum.
School name: ********************** Candidate name: Agnes ********* Candidate number: *********
Candidate number: *******-006 Research question:
What is the effect of different concentrations of blackcurrant squash solution on the rate of osmosis and therefore on the mass of equally sized pieces of potatoes?
Background information: In biology there are numerous important processes and one of them is osmosis. It is a passive movement of water molecules from a region of low concentration to the region of high concentration through partially permeable membrane.
1
This results in minimizing
osmotic gradient (the difference of concentration on both sides of partially permeable membrane) and therefore is essential in maintaining homeostasis. Let assume that we put a cell into a solution, depending on the concentration of those mediums, there are three possible outcomes : 1) The solution in which a cell is placed is hypotonic (the concentration of water is
higher in medium than in the cell) and therefore the cell will gain water. 2) The medium is hypertonic, the concentration of water outside the cell is lower and the
cell shrinks and lose water. 3) The water concentrations on both sides of partially permeable membrane are equal.
There is no flow of water molecules and such a solution is called isotonic. Picture 1.2 The graphical representation of the effect of osmosis on cell.
1
Roberts M., Reiss M., 2000. Advanced Biology, UK, Cheltenham, Nelson
2
Picture taken from http://antranik.org/movement-of-substances-across-cellmembranes/ 2
Candidate number: *******-006
Hypothesis:
According to the background information, placing potato chips in a solutions with different blackcurrant squash concentration, three different results may be obtained. When the concentration of blackcurrant in a solution is high, the water molecules will move from the cell to dilute the medium, potato will lose mass and shrink. In the case, where water concentration will be lower in the cell, potato will absorb water and get bigger. Finally, when both potato and medium have the same water concentration, no change will be observed.
Variables:
Independent: The concentration of the solution. There will be 5 different mediums with 0%,
25%, 50%, 75% and 100% concentration. Dependent: The final mass of potato chips. The mass will be measured in g with accuracy
g. Controlled: 3
Candidate number: *******-006 One type of potatoes. To minimise the source of errors only the middle part of
vegetable will be used, because in different places there may be different water content. Once prepared medium for all trials. This will eliminate the error, because in
all six trials, the same concentration will be used. Moreover, medium will be kept in closed containers to avoid evaporation. The same source of blackcurrant squash. For all experiments squash will be
used from the same bottle. Precise timing. All the test tubes will be left for the same amount of time -1h. The same temperature of distilled water, blackcurrant squash and potatoes. All
these ingredients will be left in lab to get the room temperature, which will prevent from errors of different rate of osmosis. 3
Size of potato chips will be the same. All chips will be cut into cubes 1cm . The same laboratory conditions.
Apparatus:
1litr of blackcurrant squash.
15 the same type potatoes.
Balance with accuracy
5 beakers with lids and with volume at least 300cm3.
Sharp knife and cutting board.
Ruler with accuracy
1litr of distilled water.
Paper kitchen towel.
2 times 50ml volumetric pipette with accuracy
30 the same type of test tubes with radius at lest
Watch.
Marker.
30 test tubes lids.
Tweezers.
.
.
. .
4
Candidate number: *******-006
Rubber gloves.
Methodology: 1. Label test tubes. 2. Prepare solutions with 0%, 25%, 50%, 75% and 100% concentration by adding
adequate amount and distilled water and blackcurrant squash.
Table 1. Ratio of volume of blackcurrant squash and volume of distilled water needed to produce different concentration of medium. The volume of The volume of Concentration The volume of medium blackcurrant of blackcurrant distilled water prepared squash used squash / % used / / 100% 0 40 40 75%
10
30
40
50%
20
20
40
25%
30
10
40
0%
40
0
40
3. Cut 3 potatoes into half and from each half, from the middle cut
chip. In this
way only 6 cubes will be obtained, what will prevent the water evaporation from other chips. To cut potatoes use gloves, this will prevent leaving oil from human hands, which may affect the rate of osmosis. Touch cut chips only through gloves. 4. Put the chips on paper kitchen towel. The excreted juice from the potatoes might
change the weight on the chips. 5. Weight chips and record the results in a table. 6. Put the potato chips into labelled test tubes. 7. To each test tube add
of 100% concentrated solution.
8. Cover test tubes to prevent evaporation of water from test tubes, which might be a
source of error. 9. Leave test tubes for an hour to allow osmosis to occur. 10. Remove potato chips from test tubes and place them on paper kitchen towel for 5
minutes. In this way we will get rid of water which was not absorbed by chips and might be a source of error. 5
Candidate number: *******-006 11. Weight chips and record results. 12. Repeat the experiment using 0%, 25%, 50% and 75% concentrated solutions.
Safety rules:
Use knife carefully.
Wear lab coat and goggles.
Respect laboratory rules.
Data collection:
Qualitative data: After taking potato chips from test tubes, I could easily notice that some of them became shrivelled, especially those taken from high blackcurrant concentration medium. This is perhaps the result of water loss. However I could also observe that in the solution consisting just only of distilled water, chips became bigger, which means that medium was hypotonic and Solanum tuberosum gained water. Quantitative data: Table 2. Initial and final masses, difference in masses and % difference in weight of Solanum tuberosum in 100% blackcurrant concentration.
Sample
Initial weight / g
Final weight / g
Difference between initial and final masses / g
% Difference in weight /%
A
2.26
1.48
0.78
34.51
B
2.15
1.40
0.75
34.88
C
2.09
1.34
0.75
35.89
D
2.18
1.48
0.70
32.11 6
Candidate number: *******-006 E
2.21
1.49
0.72
32.58
F
2.14
1.42
0.72
33.64
G
2.21
1.48
0.73
33.03
H
2.24
1.53
0.71
31.70
I
2.12
1.41
0.71
33.49
J
2.19
1.45
0.74
33.79
Table 3. Initial and final masses, difference in masses and % difference in weight of Solanum tuberosum in 75% blackcurrant concentration.
Sample
Initial weight / g
Difference between % Final weight initial and Difference in / g final masses weight /% / g
A
2.23
1.61
0.62
27.80
B
2.12
1.50
0.62
29.25
C
2.20
1.61
0.59
26.82
D
2.09
1.46
0.63
30.14
E
2.12
1.51
0.61
28.77
F
2.15
1.52
0.63
29.30
G
2.23
1.57
0.66
29.60
H
2.15
1.51
0.64
29.77
I
2.18
1.57
0.61
27.98
J
2.11
1.48
0.63
29.86
7
Candidate number: *******-006
Table 3. Initial and final masses, difference in masses and % difference in weight of Solanum tuberosum in 50% blackcurrant concentration.
Sample
Initial weight / g
Difference between % Final weight initial and Difference in / g final masses weight /% / g
A
2.14
1.72
0.42
19.63
B
2.21
1.75
0.46
20.81
C
2.20
1.81
0.39
17.73
D
2.11
1.72
0.39
18.48
E
2.12
1.72
0.40
18.87
F
2.25
1.88
0.37
16.44
G
2.14
1.72
0.42
19.63
H
2.15
1.71
0.44
20.47
I
2.09
1.70
0.39
18.66
J
2.18
1.77
0.41
18.81
8
Candidate number: *******-006 Table 5.
Initial and final masses, difference in masses and % difference in weight of Solanum tuberosum in 25% blackcurrant concentration.
Final weight / g
Difference between initial and final masses / g
2.12
1.96
0.16
7.55
B
2.08
1.93
0.15
7.21
C
2.26
2.11
0.15
6.64
D
2.22
2.05
0.17
7.66
E
2.23
2.07
0.16
7.17
F
2.12
1.97
0.15
7.08
G
2.12
1.98
0.14
6.60
H
2.10
1.97
0.13
6.19
I
2.21
2.08
0.13
5.88
J
2.17
2.02
0.15
6.91
Sample
A
Initial weight / g
% Difference in weight /%
Table 6. Initial and final masses, difference in masses and % difference in weight of Solanum tuberosum in 0% blackcurrant concentration.
Sample
Initial weight / g
Final weight / g
Difference between initial and final masses
% Difference in weight /%
9
Candidate number: *******-006 /
g
A
2.08
1.8
-0.28
-13.46
B
2.11
1.81
-0.3
-14.22
C
2.23
1.91
-0.32
-14.35
D
2.25
1.94
-0.31
-13.78
E
2.05
1.76
-0.29
-14.15
F
2.12
2.42
-0.30
-14.15
G
2.15
2.46
-0.31
-14.42
H
2.21
2.54
-0.33
-14.93
I
2.17
2.48
-0.31
-14.29
J
2.24
2.58
-0.34
-15.18
Data processing:
1) Mean:
To make a conclusion from the experiment, average values of percentage weight change of Solanum tuberosum in different blackcurrant squash concentrations will be essential. Therefore mean has to be calculated. The formula is:
, where
are the respective values of % difference in weight of
potato chips from one kind of blackcurrant squash concentration.
10
Candidate number: *******-006
Table 7. The mean values for the % difference in mass of Solanum tuberosum in different medium concentrations. Blackcurrant squash concentration /%
Mean value of % difference in weight / %
100%
33.56
75%
28.93
50%
18.95
25%
6.89
0%
-14.29
2) Standard deviation:
To present results most accurately, standard deviation must be calculated. Standard deviation is ’’a measure of variability (dispersion or spread) of any set of numerical values about their arithmetic mean’’.3 The formula is:
, where
is the value of % difference in weight of potato chips from one kind of
blackcurrant squash concentration.
3
http://www.britannica.com/EBchecked/topic/562938/standard-deviation 11
Candidate number: *******-006 Table 8. Standard deviation of %% difference in mass of Solanum tuberosum in different medium concentrations Blackcurrant squash concentration /%
Standard deviation of % difference in weight / %
100
1.29
75
1.07
50
1.28
25
0.57
0
0.50
3) ANOVA test
ANOVA test is a statistical tool which compares mean values obtained for each independent variable in the experiment. In this case, mean values of percentage difference mass change in solutions with different concentrations of blackcurrant squash are independent. The ANOVA test has a null hypothesis, which states that all means of independent variable are the same. Conducting ANOVA test will verify the hypothesis.
12
Candidate number: *******-006
Data presenting the data obtained with critical value 0.05 using Excel, clearly shows that F value is much bigger than F critical value. This with very small
contradict
the null hypothesis. It means that mean values of percentage change in mass of Solanum tuberosum in different concentrations of blackcurrant in medium are significantly different.
4) Uncertainties:
Uncertainties are very important part of data analysis and conclusion. We already have calculated standard deviation, however we also have to calculate uncertainties connected with solution.
a) Uncertainty of preparing solution.
b) Total uncertainty.
c) Absolute uncertainty.
13
Candidate number: *******-006 Table 9. Percentage and absolute uncertainty of volume used for different blackcurrant concentrations. Uncertainty Blackcurrant for squash preparing concentration / solution % /%
(Uncertainty of pipette /v of medium *100%)
Total uncertainty /%
Concentration of blackcurrant with absolute uncertainty / %
/%
100
0.24
(0.1/5)*100
2.24
75
0.24
(0.1/5)*100
2.24
50
0.24
(0.1/5)*100
2.24
25
0.24
(0.1/5)*100
2.24
0
0.24
(0.1/5)*100
2.24
14
Candidate number: *******-006 Data presentation:
From the data obtained above we can present the results with uncertainties.
Table 10. The total results of the experiment presenting relationship between the concentration of blackcurrant with absolute uncertainty and the mean value of % difference in weight ± standard deviation Concentration of blackcurrant with absolute uncertainty / %
Mean value of % difference in weight ± standard deviation / %
15
Candidate number: *******-006
16
Candidate number: *******-006
Now, as the graph was produced and best-fit line drawn, we can see that despite of high R2 value, there is some uncertainty. The gradient is a measurement of how much the valume on y axis change with the increase on x axis by one unit. In our example, for every 1% of blackcurrant squash concentration increase, there is 0,4710% increase in mean value of percentage difference in weight. To calculate the uncertainty of the slope, we have to draw the maximum and minimum gradient lines. The values obtained –gradient of maximum and minimum gradient lines determine the boundaries of the slope. We would consider just two points, for 0% and 100% concentrated solutions.
Table 11. The values of points to draw the maximum and minimum gradient lines. maximum gradient
minimum gradient
Y value / %
X value / %
Y value / %
X value / %
-14.79
0
-13.79
0
34.85
102.24
32.27
97.74
In the Table 11, no uncertainties are calculated, because the table represent the maximum and minimum data with the uncertainties.
17
Candidate number: *******-006
onclusion and evaluation:
From the results obtained I am able to confirm that my hypothesis was correct. In a hypertonic medium Solanum tuberosum shrivelled, especially those samples taken from solutions with high concentration of blackcurrant squash. The mass of cells decreased, which means that potato chips lost water. However in samples taken from 0% blackcurrant concentration, there was a water gain by cells and therefore the increase in mass was recorded. Uncertainty of the gradient is relatively small, it varies from 0.4788 to 0.5203, so the difference between the highest possible gradient and the lowest is just only 0.0415. This proves that the results obtained are reliable. In my experiment I decided to cut potato into equally sized chips. I decided to not to consider the weight of the chips as a controlled variable –it was not the same for every sample. My decision was based on the fact that surface area of cell has a great influence on the rate of the reaction and in my experiment I focused on the same surface area for all the samples. To get the most accurate results I was trying to control as many variables as possible. To eliminate error, the solution on one concentration was prepared at once for all trials. This ensured me that in all trials I had the same percentage concentration of blackcurrant squash. Unfortunately it did not prevented me from systematic error during the preparation of mediums –I had to read the values from pipette with eye and this may be a source of error. It could be improved by using electrical pipette. What is more, to prevent the loss of water, I covered beakers and test tubes. Potato chips were cut just a while before they were put into test tubes, so again no water was lost. To obtain the most accurate results, I was wearing gloves and using tweezers, to not to leave oil from hands, which might have influence on the rate of reaction.
18
Candidate number: *******-006
19
Candidate number: *******-006 Temperature also affects the rate of reaction, therefore all substrates were left in laboratory for night. All laboratory conditions, such as light and temperature were kept constant. The results obtained were reasonable and justifiable, what is proven by the small value of standard deviation (see Table 8.). Another proof is high
value of the trend line on the
Graph 1. However as we can see in the Graph 1, the regression line does not pass through the points, which means that both systematic and random errors were present. In my opinion one of the biggest sources of error was immeasurable water content of potato chips4. I was trying to minimise that by cutting chips from the middle of potatoes. Very interesting would be conducting an experiment with chips cut from the whole potatoes to check if that would give different results. Another systematic error was caused by the balance. Dealing with such small pieces of Solanum tuberosum, even equipment with high accuracies as 0.05 cm3 in case of pipette, may cause big uncertainties in results, therefore more accurate balance would be essential. What is more, using ruler to cut chips also require eye reading and even if 1mm of accuracy seems a little, in this experiment, but might cause o big error. This human mistake could be improved by using special machine to cut potatoes. To eliminate another source of error, after taking Solanum tuberosum from mediums, I placed them for 5 minutes on the kitchen towel to get rid of excess water present on the surface, which was not absorbed by potatoes. However 5 minutes is a small amount of time and there is no assurance that all chips dried to the same extent and did not lose water from insight of the cells. Another way to deal with this problem is smoothly touching chips with paper kitchen towel. Water will be quickly absorbed by the paper and the probability of water loss from the cells will minimised. If we consider to repeat experiment, interesting would be taking samples with bigger surface area. Dealing with bigger samples would minimise errors, however the results would
4
It could be measured but with advanced method and for this lab I had a time limitation. 20
Candidate number: *******-006 not be proportional to the change of surface area, because the ratio volume to surface area would increase and it would highly affects the rate of osmosis.
Bibliography:
Books: 1. Greenwood T. , Allan R., 2009. Senior Biology 1, UK, Edinburgh, BIOZONE. 2. Greenwood T. , Allan R., 2009. Senior Biology 2, UK, Edinburgh, BIOZONE. 3. Roberts M., Reiss M., 2000. Advanced Biology, UK, Cheltenham, Nelson.
Webs: 1. Movement of substances across cell membranes. Available at:
http://antranik.org/movement-of-substances-across-cell-membranes/. [Accessed 10.01.2014]. 2. Encyclopaedia Britannica, Standard deviation. Available at:
http://www.britannica.com/EBchecked/topic/562938/standard-deviation. [Accessed 15.10.2013].
21
