BIOLOGY INTERNAL ASSESMENT

DATA COLLECTION AND PROCESSING

CONCLUSION AND EVALUATION

RELATIONSHIP BETWEEN SURFACE AREA AND THE VOLUME OF A CELL, HOW IT AFFECTS THE RATE OF DIFFUSION

Nazirul Ibrahim

04.10.2011

DATA COLLECTION AND PROCESSING

We have determined the following variables:

* dV- the time for the color indicator phenolphthalein to reach the center of the cube (diffusion) * iV- the size of the cube

* cV- the amount of Sodium Hydroxide the perfect dimensions of the cube, how the cube is fully submerged in the solution, the temperature of the cube and the constant surrounding temperature, the concentration of the NaOH solution

During this experiment of investigating the relationship between the surface area and the volume of a cell, the following was determined:

* The amount of time (sec) taken for the pink color to reach the center of the cube * How ratio of the cube (the volume and surface area) affects the time of diffusion * The ratio of the

With the results obtained accurately, calculations were made to find out the following:

* The mean of each group’s data

* The standard deviation of each group’s data

As each trial consist for 5 different sizes of the agar cubes (1cm, 1.5cm, 2cm, 2.5cm and 3cm) and that the trials are repeated 5 times, we expect that results will be as follows:

* As the size increases, it will take a longer time for the pink color (phenolphthalein indicator) to reach the center of the cube while it is submerged in the solution * That the experiment can produce different results in each due to many affecting factors of the controlled variables

The next set of tables of raw data include the data with their mean and standard deviation have been recorded and calculated throughout the experiment:

Length of each side of a cube (cm) (0.1 cm) | Time (sec) (0.01 sec)| | Group 1 (My group)| Group 2| Group 3| Group 4 | Group 5| Mean time of each group to a given size of the cube (0.01sec)| Standard deviation| 1.0| 175.00| 1180.00| 479.28| 360.00| 498.87| 538.63| 380.98| 1.5| 874.84| 1508.60| 794.18| 914.62| 807.37| 979.92| 299.62| 2.0| 1215.38| 1692.60| 954.03| 1126.00| 1262.10| 1250.02| 273.96| 2.5| 1479.84| 1782.60| 1369.90| 1605.00| 1458.15| 1539.10| 159.94| 3.0| 2363.66| 1650.00| 1539.49| 2101.00| 1591.78| 1849.19| 363.98|

Calculations:

Mean of each group’s data: ()

1.0cm: 2693.15 5 = 538.63 (sec)

1.5cm: 4899.61 5 = 979.92 (sec)

2.0cm: 6250.01 5 = 1250.00 (sec)

2.5cm: 7695.49 5 = 1539.10 (sec)

3.0cm: 9245.93 5 = 9245.93 (sec)

Standard deviation

= Standard Deviation

= Sum of

= Value of each group’s data

m= Mean value of that

n= Number of trials

1.0cm:

, ,

1.5cm:

, ,

2.0cm:

, ,

2.5cm:

, ,

3.0cm:

, ,

Standard error of the mean

1.0: = 170.08

1.5: = 133.76

2.0: = 122.30

2.5: = 71.40

3.0: =162.49

Calculating the surface area to volume ratio to determine the relationship between them:

1cm cube Surface area: 6 sides of 1cm x 1cm = 6.00cm2 , Volume: 1.00cm3

SA:V ratio= 6:1 – 6.00

1.5cm cube Surface area: 6 sides of 1.5cm x 1.5cm = 13.50cm2 , Volume: 3.375cm3

SA:V ratio= 13.50:3.38

2cm cube Surface area: 24 sides of 2cm x 2cm = 6.00cm2 , Volume: 8.00cm3

SA:V ratio= 24:8 – 3.00

2.5cm cube Surface area: 6 sides of 2.5cm x 2.5cm = 37.50cm2 , Volume: 15.625cm3

SA:V ratio= 37.50:15.63

3cm cube Surface area: 6 sides of 3cm x 3cm = 54.00cm2 , Volume: 27.00cm3

SA:V ratio= 54:27 – 2.00

FIGURE 1

FIGURE 2

CONCLUSION AND EVALUATION

CONCLUSION

The main aim of the experiment was to investigate the relationship between the surface area and the volume, and how it greatly affects the rate of diffusion. Hypothesis was given that as the surface area and volume ratio (size of the agar cube) increases, the rate of...

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