Cell Respiration!

Cellular Respiration!

In this lab, we compared the rate of cellular respiration between glass beads, mung beans, and peas. Before we tell you more about this extremely exciting experiment, you have to know some background info on cellular respiration. Cellular respiration takes place in a cell's mitochondria, and produces energy in the form of adenosine triphosphate, commonly known as ATP. Here's the equation:

          C6H12O6 + 6O2 -----> 6CO2 + 6H2O + ATP (energy)

This is a diagram of Cellular Respiration within the cell.

As you can see, cellular respiration requires sugar (glucose) and oxygen in order to take place, and then produces carbon dioxide (CO2), water, and ATP. In this lab, we were able to compare the rate of cellular respiration between glass beads, mung beans, and peas by using sensors that measured the amount of oxygen and carbon dioxide in the chamber where the specimens were placed. We compared the amount of CO2 being produced to the amount of O2 being used up in order to find the rate of respiration. Cooler Than Absolute Zero predicted that the mung beans would have the highest rate of respiration. Let's find out if we were right!

Note: That if the cell does not have oxygen to work with, it will not continue onto the Citric Acid Cycle. It breaks off in to two parts from glycolysis; Lactation or alcoholic fermentation.

Procedure: 

In this lab, we were measuring the respiration rate by focusing on CO2 and O2 production. First, we decided that it was a good idea to have a control group to check our other gathered data with. Our control group in this experiment were solid marbles. We placed these marbles in a clear container, closing it off with a lid. This lid contained two inputs for our CO2 and O2 respirators. We secured these into place and were ready to conduct the experiment. After a minute of waiting, we started calculating the percentage of CO2 and O2 in the container. The system we used was connected to a screen tablet that displayed a graph, and allowed us to see the amounts of these two gases. Shortly after waiting ten minutes, as written in the instructions, we began assessing the data collected. The graph showed that the amount of CO2 and O2 stayed at a constant rate. This data makes sense because the marbles are abiotic and not going through the process of cellular respiration. Next, we cleaned out the container and placed germinated peas to calculate their respiration rate. Following the instructions, we proceeded to do all the steps stated previously to receive our data for the peas. Again, after ten minutes of waiting; we finally got to assess the graph or our data. This time, the graph showed that CO2 progressively increased while,O2 progressively decreased. This data is accurate being the germinated peas are consuming O2, a clear indication that cellular respiration is taking place. Lastly, the group measured the respiration rate in mung beans. This graph showed an increase in CO2 and a decrease in O2; the exact same data in the germinated peas.

Not only did we test the mung beans and peas in normal temperature environments, we cooled the temperature to see if that had any affect on them. Basically, we did the same experiment stated above. However.. Before placing the mung beans and germinated peas into the container, we let each one of them soak in ice water for a minute. After, we simply dried them off and placed them into the container to measure the rate of cellular respiration. The graph for the mung beans showed that CO2 increased and O2 decreased, but at a slower rate. Then, the graph for the germinated peas displayed the same results.

Analysis questions...

1. The effect of germination on the rate of cellular respiration in peas, is that the rate of O2 will become higher. We know this because, when a pea is germinating it means that it is growing. When it keeps growing, the need for O2 consumption grows alongside. Resulting in the rate being higher, as stated previously. 

2. There are many pieces of evidence that cell respiration occurred in peas. One reason that the germinated peas still were undergoing cellular respiration because they were still alive. We know this because the O2 levels increased while the CO2 levels decreased.

3. Germinated peas undergo cellular respiration because they are still producing sugar for themselves

Conclusion:

In this lab, we really started to see how each chapter we learned about starts building on each other. Without knowing vital information from chapters on sugars and proteins, we would not be able to analyze thoroughly on how cellular respiration and photosynthesis works. We see how temperature affects proteins and denaturizes them so these processes are unable to function, and we see how certain molecules behave. Overall, we learned how by changing specific factors, we can change the rate of cellular respiration.

Group Discussion:

We got to observe and accurately measure cell respiration in the lab by using high tech, high quality equipment to measure levels of Oxygen and CO2 within a closed chamber of germinating peas and beans. With dry beans we saw no drastic changing in Oxygen or Carbon Dioxide. When using peas that were germinating, we saw that oxygen was slowly dropping and carbon dioxide levels was rising. This made sense because plants like peas release carbon dioxide during cell respiration. When we soaked the peas in really cold water, the rate at which carbon dioxide was being release drastically slowed down. Is proves that plants have a ideal point of temperature to grow in and hotter/colder climates could slow down their rate of respiration.

Catalase Titration Lab

Catalase Titration Lab

In this lab, we were using enzymes (yeast) to break down hydrogen peroxide. But after a set time (10 seconds, 30, 60, 180, 360), we injected sulfuric acid to stop the catabolic process. We then titrated a 5 ml sample of the solution with potassium permanganate to see how much hydrogen peroxide was left to consume the potassium permanganate in the 5 ml solution. This experiment consisted of three different parts. 

Part B: In this section of the experiment, we determined the amount of hydrogen peroxide (H2O2) initially present in a 1.5% solution without adding a catalase, or enzyme, to the reaction mixture. The amount we find will be our base line, or constant, for the upcoming parts of the experiment. 

                 We started this experiment by measuring out 10 ml of the 1.5% hydrogen peroxide into a plastic cup. (Insert photo) Then, we added a mL of water. For this section, we added water instead of an enzyme solution in order to find constant data that we could compare to the data of the solutions that contained enzymes. After this we added 10 mL of sulfuric acid and mixed the solution. This stopped the reaction by denaturing the enzymes. Next we removed a small sample of the final solution and placed it into the titration cup. As might have guessed, we then titrated the solution. This means that we added small amounts of potassium permanganate until the solution permanently changed to a light pink or brown color. We compared the initial amount of potassium permanganate in the burette to the final amount in order to find out how much potassium permanganate was needed to discover how much hydrogen peroxide was used up. 

Part C: In this portion of the lab, we wanted to calculate the rate at which H2O2 turns into H2O and O2 (or decomposes) in an uncatalyzed reaction. An uncatalyzed reaction simply means that an enzyme was not present. Additionally, this results in the activation energy of the reaction to be higher. After the solution was left to sit uncovered for twenty-four hours, it was then tested to gather this information.

      Natalie made a great video for the paragraph below

To prep for the experiment, we first placed about 15mL of H2O2 in a beaker and let it sit over night. After, we added 1mL of H2O (also known as, water) because we were calculating data for an uncatalyzed reaction. Next, we added 10 mL of H2SO4 (also known as, Sulfuric Acid) to the solution. If the solution we were testing had an enzyme, the acid would denature the enzyme since it is a protein. The enzyme would denature because the acid would lower the pH of the solution and enzymes have a set pH level in which they can function normally. Then, we mixed the contents of the beaker and drew a 5 mL sample of the solution. We took this 5mL sample and added KMnO4 (also known as, the titrate in this experiment) drop by drop, to measure the amount of  H2O2 still remaining in the solution. During this step, we also swirled our sample to insure that the titrate distributed evenly. It only took one drop of KMnO4 to give us the data that we needed for this experiment. Below is the data collected from this portion of the experiment.

                

Part D: In this part of the lab, we wanted to figure out how much H2O2 was eaten up depending on how long we left the H202 to react with a catalase. In this case we used yeast! We noticed a lot from the results when looking at the different amount of times we left the H2O2 with the yeast.       

When we prepared this we had five different beakers that each had a time label. "10 seconds, 30, 90, 180, 360". Each time label represented how long we would let the H2O2 react with the yeast before we applied 10 ml of sulfuric acid in order to stop the process. We added 10 ml of H2O2 and then 1 ml of yeast but the second that we added the yeast, we immediately started timing accord to the label on the beaker. After waiting, we add the acid and then extracted a 5 ml sample from the beaker. Proceeded to place it under the burette and calculated how much potassium permanganate it took to turn the sample a slight pink showing that there wasn't anymore H2O2 to react and "erase potassium permanganate.

 We noticed from our results (after graphing the data) that the longer we left H2O2 to sit with the yeast, the less amount of potassium permanganate was needed to turn the sample a slight pink. Showing that the catalase (yeast) was reacting with the hydrogen peroxide and reacting to make H2O and oxygen.

OUR DISCUSSION

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So in order to determine the course of an enzymatic reaction, we had to measure out how much substrate was disappearing over certain time intervals. So we recorded our data for the amount of hydrogen peroxide used in every test, and made a graph to better depict this information. As you can see below, (insert graph here) the rate of the amount of H2O2 in a certain amount of time is the highest between 180 and 360 seconds. This means that the difference of the amount of hydrogen peroxide used up is greatest between 180 and 360 seconds, most likely because that is the biggest time interval. On the other hand, the smallest rate is between 0 and 10 seconds. This is because it is the shortest gap of time, so hydrogen peroxide does not have nearly the same amount of time to become used up as it does between the 180 and 360 second gap.

CONCLUSION

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Overall, our group experienced a hands-on lab that taught us more about enzyme activity. The rate of the reaction had to do with the substrate concentration. As the concentration went higher, the reaction rate increased as well. At one point, the rate stays the same since the enzymes are always loaded with what they are catalyzing. In order to collect the right data on the reaction rate, we used an acid to denature the enzyme, and then titrate to see how much substrate was left over. In order to better this part, we could have been more precise in our titration. Other than that very tiny error, this lab went very smoothly!

This was our enzyme catalysis lab, hope you enjoyed reading!