Lab_Report_Eukaryotes_Prokaryotes_and_Viruses (1)

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Apr 3, 2024

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Eukaryotes, Prokaryotes, and Viruses: Structure and Function Ayden Hinojos 3/5/2024
Instructions: 1. Please read all of the introduction and background information within the investigative manual. a. Once you have done so answer the prelab question BEFORE completing any of the lab’s activities. 2. Once you have completed the prelab questions proceed to the activities of the lab within the investigative manual. a. As you read through the instructions for completing each activity make sure you also: i. Complete any instructions (append photos, etc)/ and answer any questions found in the post lab questions for each activity. ii. Take the photos of your experiments in each activity as directed below . IMPORTANT : Don’t clean-up your lab until you know what portion of the experiment you need to take a picture of. 3. Here is a video that will introduce you to the lab and its main concepts. The student is encouraged to watch it. a. Eukaryotes Prokaryotes and Viruses Prelab Questions 1. There are three panels in the figure below (as labeled). Each panel represents two compartments separated by a semi-permeable membrane. Small solid circles represent water and larger hashed circles represent a solute. In each panel label each side (“Side A” and “Side B”) as either hypertonic, hypotonic, or isotonic. After doing this, illustrate (with an arrow) or state which direction water will move (left-to-right, right-to- left, or neither). 1 © 2016 Carolina Biological Supply Company
Panel 1: Side A:Hypotonic Side B:Hypertonic Water: Moving left Panel 2: Side A & B: Isotonic Water: Equilibrium Panel 3 :Side A: Hypertonic Side B: Hypotonic Water : Moving right 2. IKI (also called iodine-potassium iodide) is a reagent that turns black in the presence of starch. Benedict’s regent is a reagent that turns clear blue in the presence of glucose. As a student in BIO111 you are asked to set up an experiment that has a beaker that has been partitioned by a semi- permeable membrane and you have placed a solution contain 20% glucose on one side of the beaker while on the other side you have placed a solution containing 20% starch. See figure below. 2 © 2016 Carolina Biological Supply Company
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Considering this setup answer the following questions: a. After two hours you remove a sample from side A and B and test them for starch and glucose using the IKI solution and benedict’s reagent. Predict, or hypothesize, what you will find for both side A and side B given this scenario. Why did you make that prediction? I think that after two hours, Side B will have turned blue and Side A will have turned black. I think this because benedict’s regent turns clear blue in the presence of glucose and the IKI solution turns black. I also think that it wont take long for the solutions to turn blue and black as I think that glucose and starch react quickly. b. After making your prediction you carry out the test and find that glucose is found on both sides A and B. However, starch is found only on side A. Why do you think this is the case? Answer this question by discussing the molecular difference between glucose and starch. 3 © 2016 Carolina Biological Supply Company
Since glucose molecules are smaller than starch molecules, it will take less time for the glucose molecules to break down and longer for the starch molecules. Because of their small size, they are allowed to pass through the membrane and go through diffusion. The breakdown process might not have been finished within the two hours and the starch’s inability to be diffused through the membrane leaves the presence of starch. How does the scenario described in b. compare to a biological membrane? Biological membrane allows some substances, in this case glucose, to pass through while restricting other substances, in this case starch. The glucose is allowed to diffuse through the membrane but since the starch is restricted it is not diffused and its presence stays there. 3. In Activity 1 of this lab we will be investigating the impact that the surface area-to-volume ratio has on the rate of diffusion. Please read the directions for Activity 1 in the investigative manual. After doing so fill in your purpose and hypothesis statements found under the Activity 1 heading. After completing the lab come back to this section and fill out your evidence/claims and reflection statement. Activity 1 Instructions: 1. Open the investigative manual. Locate all the needed materials supplied in the kit and those you will need to supply yourself. 2. Lay them out in your work area. 3. Read through the entire set of instructions found in the investigative manual for the activity to avoid making mistakes when you go to execute the experiment. 4. Once you have read through the instructions go back to step 1 and begin executing the experiment. 5. Please answer the questions below and/or append appropriate representations of data (photos, graphs, etc). REMEMBER don’t clean 4 © 2016 Carolina Biological Supply Company
up until you have taken the appropriate photos of your experiment as described below. Purpose statement: (This should be the question the experiment is attempting to address. It should be written as a question.) how does surface-to-volume ratio affect the size of a cell and how does it relate to diffusion. Hypothesis statement: (This should be an “if/then” testable prediction that addresses the question/purpose of the lab.) The larger a surface to volume ratio is, the slower the rate of diffusion will be. Evidence/Claim statement: (This should be a statement regarding whether your hypothesis was supported or refuted and what data/evidence allows you to make this claim.) My hypothesis was refuted because in the experiment, the larger surface area had a higher and faster rate of diffusion. Reflection statement: (This should be a statement of what you learned, how your understanding changed, if you have new questions, and what connections can you make between the lab and the content in the book and other assignments.) I learned how different surface-to-volume ratios affect the rate of diffusion. With a larger surface-to-volume ratio, there is a greater surface area for cells to move in and out of the membrane. This means that more cells and molecules will be able to enter the membrane which will allow for a faster rate of diffusion. Photo 1 – Activity 1 Take a picture and insert the image(s) of your potato blocks after step 5 of the “Procedure” section in activity 1 of the investigative manual: 5 © 2016 Carolina Biological Supply Company
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Data Table 1A Length (l) (cm) Width (w) (cm) Height (h) (cm) Size of cross section slice (h x w) (cm) Distance traveled by IKI from potato edge (cm) Area of white region (l × w) (cm 2 ) 2.50 2.50 2.50 6.25cm 0.4cm 4.53 cm^2 2.00 2.00 1.00 2cm .6 cm 2.76cm^2 1.50 1.50 1.50 2.25cm 0.4cm 1.87cm^2 1.00 1.00 1.00 1cm 0.3cm 1.12cm^2 2.00 0.50 0.50 0.25cm 0.5 cm 1 cm^2 0.50 0.50 0.50 0.25cm 0.5 cm 0.87cm^2 6 © 2016 Carolina Biological Supply Company
Data Table 1B Length (l) Width (w) Height (h) Surface area of block (l x w x 2) + (w x h*4) (cm 2 ) Volume (l x w x h) (cm 3 ) Surface area-to- volume ratio (Surface area of block/volum e) Surface area of slice (w x h) (cm 2 ) Surface area of white section (cm 2 ) Surface area of black section (cm 2 ) Percent of potato block saturated with IKI: (Surface area of black section/surface area of slice)*100 2.50 2.50 2.50 37.5cm ^2 15.63 cm^ 3 .8 6.25 cm^ 2 4.53 .43cm 6.88% 2.00 2.00 1.00 16cm^2 4cm ^3 4 2cm ^2 2.76c m^2 .21cm 10.5% 1.50 1.50 1.50 13.5cm ^2 3.375 cm^ 3 4 2.25 cm^ 2 1.87c m^2 .31cm 13.77% 1.00 1.00 1.00 6cm^2 1cm ^3 6 1cm ^2 1.12c m^2 .09cm 9% 0.50 0.50 2.00 4.5cm^ 2 .5cm ^3 9 .25c m^2 1cm^ 2 .2 80% 0.50 0.50 0.50 1.5cm^ 2 .125c m^3 12 .25c m^2 0.87c m^2 .25 100% 1. Go back to the prelab and fill in the “Evidence/Claim” and “Reflection” statement for this lab activity. 2. Make a line graph of the “percent of potato block saturated with IKI” (y-axis) vs. “Surface area-to-volume ratio” (x-axis) and answer the questions below. 7 © 2016 Carolina Biological Supply Company
a. Insert graph here (make sure your graph has a title, labeled axis, and a legend): b. What does this graph, and the results of this experiment, tell us about how the rate of diffusion changes with changing surface area-to- volume ratio? The graph shows that when the surface to area volume increases, the rate of diffusion increases. c. Finally, what can we conclude from these results regarding why biological cells are small rather than large? Biological cells are small rather than large because they need a larger surface area to volume ratio in order to more efficiently move needed materials in and out of the cell. In conclusion, cells are smaller because they need to optimize the surface area-to-volume ratio in order to efficiently move nutrients and other important materials in and out of the cell. 8 © 2016 Carolina Biological Supply Company
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Activity 2 Instructions: 1. Open the investigative manual. Locate all the needed materials supplied in the kit and those you will need to supply yourself. 2. Lay them out in your work area. 3. Read through the entire set of instructions found in the investigative manual for the activity to avoid making mistakes when you go to execute the experiment. 4. Once you have read through the instructions go back to step 1 and begin executing the experiment. 5. Please answer the questions below and/or append appropriate representations of data (photos, graphs, etc). REMEMBER don’t clean up until you have taken the appropriate photos of your experiment as described below. Photo 1 – Activity 2 Take a picture and insert the image(s) of your dialysis tubes after step 14 of the “Preparing the Dialysis Tubing” section in activity 2 of the investigative manual: 9 © 2016 Carolina Biological Supply Company
10 © 2016 Carolina Biological Supply Company
11 © 2016 Carolina Biological Supply Company
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12 © 2016 Carolina Biological Supply Company
Data Table 2 Sample data shown. Treatment Solution in dialysis tubing Solution in cup Initial volume (V i ) (mL) Final volume (V f ) (mL) Change in volume (V f -Vi) (mL) Percent change in volume (Vf- Vi)/V i (mL) The solution inside the tubing was hypotonic, isotonic or hypertonic? A 20% sucrose 20% sucrose 94 94 0 0% isotonic B 40% sucrose 20% sucrose 94 96 2 2.12% increase Hypotonic C 20% sucrose 40% sucrose 94 91 3 3.19% decrease Hypertonic 1. Explain what the change in volume of the dialysis tube indicated. Describe what happened when the volume increased and when the volume decreased. The change in volume indicated that osmosis was taking place. When the volume increased the higher concentration solution inside of the dialysis tubing would be transferred into the solution in the cup and this would result in an increase in volume. The decrease was caused by the solution in the dialysis tubing being released into the solution in the cup which resulted in a decrease. 2. Are the results of your experiment consistent with what you would have expected to happen? Why or why not? 13 © 2016 Carolina Biological Supply Company
yes, because I thought that the more concentrated solution would increase the volume. Although, I did not expect that having a lower solution in the dialysis tube and a higher solution in the cup would decrease the volume. I thought that the higher solution would transfer into the dialysis tube causing the volume to increase or stay the same. 14 © 2016 Carolina Biological Supply Company
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Activity 3 Instructions: 1. Open the investigative manual. Locate all the needed materials supplied in the kit and those you will need to supply yourself. 2. Lay them out in your work area. 3. Read through the entire set of instructions found in the investigative manual for the activity to avoid making mistakes when you go to execute the experiment. 4. Once you have read through the instructions go back to step 1 and begin executing the experiment. 5. Please answer the questions below and/or append appropriate representations of data (photos, graphs, etc). REMEMBER don’t clean up until you have taken the appropriate photos of your experiment as described below. Photo 1 – Activity 3 Insert the photo or scan of your prokaryotic cell drawing from Activity 3. The following should be indicated in this photo: 15 © 2016 Carolina Biological Supply Company
16 © 2016 Carolina Biological Supply Company
1b → 2a→ Prokaryote Dichotomous Key→ 1a→ 2b→ 4a = Listeriamonocytogenes cell membrane type circular DNA ribosomes flagellum (if applicable) Identification of the cell tracing your steps through the Dichotomous key o For example: 1a 🡪 2a 🡪 3a ( Staphylococcus aureus ) Photo 2 – Activity 3 Insert the photo or scan of your eukaryotic cell drawing from Activity 3. The following should be indicated in this photo: 17 © 2016 Carolina Biological Supply Company
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double-stranded DNA inside a nuclear membrane ribosomes mitochondria endoplasmic reticulum lysosomes Golgi apparatus vesicles optional internal organelles means of locomotion if applicable Identification of the cell tracing your steps through the Dichotomous key o For example: 1a 🡪 6a ( plasmodial slime mold ) Photo 3 – Activity 3 Insert the photo or scan of your virus drawing from Activity 3. The following should be indicated in this photo: 18 © 2016 Carolina Biological Supply Company
capsid shape DNA or RNA, if it is single stranded or double stranded, and the replication direction Identification of the cell tracing your steps through the Dichotomous key o For example: 1a 🡪 2b 🡪 9a 🡪 10b ( unidentifed ) 19 © 2016 Carolina Biological Supply Company