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Bases cause the litmus to turn blue. The strength of a base is dependent on the concentration of OH- ions produced. On the pH scale, the basicity of a substance increases from 7 to 14: 7 being neutral and 14 being the strongest base (one that generates maximum OH- ions when dissolved). Name two acids and bases that you often use. Acids: Milk, Orange Juice Bases: Soap, Toothpaste Cube Dimensions Surface Area (cm2) Volume (cm3) Surface Area: Volume Time Required for Complete Color Change Distance of Diffusion 1 cm X 1 cm X 1 cm 6 1 6:1 3 min 0.
50 cm 1 cm X 2 cm X 2 cm 16 4 4:1 2 min 50 sec 0.30 cm 1 cm X 1 cm X 6 cm 36 6 6:1 2 min 42 sec 0.48 cm How did the surface area affect the diffusion of the cube? What about the volume? What about the surface area to volume ratio? Which of these had the greatest effect on the diffusion of the cube? The time required for complete colour change is found to be approximately the same for all three cubes (average time: 2 minutes 50 seconds) which conclude that the rate of diffusion of vinegar into the agar is the same irrespective of shape and size.
Further, volume had no impact on the time required for colour change. No regular trend has been observed for the distance of diffusion of vinegar inside the agar with respect to surface area or volume when each is considered alone. Ideally, a greater distance of diffusion should have been found in Cube B ( 1 cm X 2 cm X 2 cm) as compared to Cube A ( 1 cm X 1 cm X 1 cm) as it has the greater surface area (larger surface area enhances the amount of diffusion); however, the irregular pattern reveals that there is a third factor that has a direct bearing on the distance of diffusion.
Studying the results of the experiment, it becomes evident that surface area to volume ratio is the single parameter that majorly affects the amount of diffusion of vinegar into the agar. Cube A ( 1 cm X 1 cm X 1 cm) and Cube C (1 cm X 1 cm X 6 cm) having the same surface area to volume ratio yield the same distance of diffusion. On the other hand, cube B ( 1 cm x 2 cm x 2cm) tails at 0.3 cm. Even though a greater amount of vinegar diffuses into cube B than in cube A due to greater surface area, a larger time period is required for it to reach the centre.
In conclusion, the experiment demonstrates that depth of diffusion is lesser for materials having smaller surface areas to volume ratio than that with higher surface area to volume ratio. How does this experiment demonstrate the need for larger cells to divide? A cell meets its requirements for growth, repair and removal of waste materials through its cell membrane, the amount of transfer directly relating to its surface area. More so, the extent of movement of substances into a cell-towards its centre- depends upon its size and volume.
An increase in cell size is accompanied by an increase in the requirements of the cell. More nutrients need to be diffused in and a greater amount of waste materials require to be diffused out. Consequently, a mechanism needs to be established that would allow a rapid transfer of substances through the cell membrane. As the cell grows uncontrollably without division, the surface area to volume ratio decreases. This limits the distance to which substances diffuse into the cell, posing a restriction to the functionality of various components of the cell-especially those located towards the centre. To counter this, it is necessary that cells divide as they grow to keep the surface area to volume ratio within the limits where effective transfer with the surroundings can take place.
Cube B (0.5 cm x 0.5 cm x 6 cm) would serve best as a cell because of its flat, elongated shape that gives it a higher surface area to volume ratio. This ensures complete diffusion of substances even to the centre of the cell resulting in good functionality of all parts of the cell. On the contrary, cube A and C have a large volume that would inhibit the substances to diffuse into the centre of the cell at a relatively fast pace to fulfil the cell’s needs. This renders them an inadequate choice.
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