Lab Module Three Biology

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Southern New Hampshire University *

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120L

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Biology

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Feb 20, 2024

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Name: _Paul Sangeavur Nomsule_____________________________ Date: _01/28/2024_____________________________ BIO 120L Module Three Lab Report: Cell Structure and Function Data Activity 1 Data Table 1a Length ( l) (cm) Width ( w ) (cm) Height ( h ) (cm) Size of cross section slice (h x w) (cm) Distance traveled by I 2 KI from potato edge (cm) Area of white region (l × w) (cm 2 ) 2.50 2.50 2.50 2.50 x 2.50 = 6.25 .68 2.50 x 2.50 =6.25 2.00 2.00 1.00 2.00 x 1.00 = 2.00 .51 2.00 x 2.00 =4.00 1.50 1.50 1.50 1.50 x 1.50 = 2.25 .42 1.50 x 1.50 =2.25 1.00 1.00 1.00 1.00 x 1.00 = 1.00 .22 1.00 x 1.00 =1.00 2.00 0.50 0.50 0.50 x 0.50 = 0.25 .21 2.00 x .50 =1.00 0.50 0.50 0.50 0.50 x 0.50 = 0.25 .22 .50 x .50 =.25

1 Data Table 1b Lengt h ( l ) Widt h ( w ) Heig ht ( h ) Surface area of block ( l x w x 2) + ( w x h *4) (cm 2 ) Volum e (l x w x h ) (cm 3 ) Surface area/volum e Surfac e area of slice ( w x h ) (cm 2 ) Surface area of white section (cm 2 ) Surface area of black section (cm 2 ) Surface area of black section/surfa ce area of slice 2.50 2.50 2.50 2.50x2. 5… =15.6 0 37.50/15 .60 = 6.25 4.00 2.3 2.3/6.25= 0.358 2.00 2.00 1.00 = 16.00 = 4.00 16.00/4. 00 = 2.0 3.00 15.8 7.9 1.50 1.50 1.50 = 13.50 = 3.4 13.50/3. 4 =4.00 = 2.25 2.40 .50 0.125 1.00 1.00 1.00 = 6.00 = 1 6.00/1.0 0 =6.00 1.00 = 4.40 .42 0.42 0.50 0.50 2.00 = 4.50 = .50 4.50/.50 = 9.00 = .2 5 1.30 .60 2.4 0.50 0.50 0.50 1.5 .13 1.5/.13 = 11.54 .25 .2 .1 0.4 1. The potato blocks used in this experiment are rich in starch molecules. Iodine potassium iodide is a starch indicator that forms a blue-black complex in the presence of starch. What does the width of the black of the potato cube represent in the context of diffusion? The dark coloration of the potatoes demonstrate the degree to which the chemical permeated into the potatoes. It also demonstrated the depth to which the chemical penetrated the cells of the potatoes. 2. How does the width of the black iodide starch complex of each potato block compared to the others? Why? Density of the potatoes matters in this experiment. The smaller the density of the potatoes, the more pigmentation the cells were able to absolve. © 2016 Carolina Biological Supply Company

2 Activity 2 Data Table 2 Treatme nt Solution in dialysis tubing Solutio n in cup Initial volume (V i ) (mL) Final volume (V f ) (mL) Chang e in volume (V f -Vi) (mL) Percent change in volume (Vf- Vi)/V i (mL) Hypotonic, isotonic or hypertonic? A 20% sucrose 20% sucros e 90 ml 90 ml 90-90 =0 0.00 Isotonic solution B 40% sucrose 20% sucros e 93 ml 95 ml 95-93 = 2 0.02 Hypertonic Solution C 20% sucrose 40% sucros e 91 ml 89 ml -2 -0.02 Hypotonic Solution 3. Explain what the change in volume of the dialysis tube indicated. Describe what happened when the volume increased and when the volume decreased. The exchange in volume is the difference between the final volume against the initial volume. When the volume increased. It means that there has been an increase in solution absorption into the mass. When the volume decrease, there is a decrease in the absorption of the solvent by the solute. © 2016 Carolina Biological Supply Company

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