Life with Oxygen Lab

10.19.2011

TITLE: The Importance of Oxygen for Life
PURPOSE: To determine whether test tubes filled with bacteria, a few salts, distilled water, and glucose would grow in the presence or absence of oxygen. By adding oxygen to certain tubes and withholding it from other tubes, one can determine which bacteria grow the most efficiently.
HYPOTHESIS: The test tubes of bacteria with oxygen will grow and thrive over the test tubes without oxygen.
METHODS: Using the data shown in Table 7-A-1, construct two graph lines on the same graph. First plot “concentration of glucose” against the “number of bacteria” in series A (tubes without air). Label the vertical axis “millions of cells per ml” and the horizontal axis as “glucose (mg/100ml).” For the second graph plot the “concentration of glucose” against the “number of bacteria” in series B (tubes with air). Label the first graph line: “Growth without air.” Label the second line “Growth with air.” RESULTS:
1.) Bacteria numbers in tubes without air only grew to a certain level and then growth stopped/leveled off.
2.) Bacteria numbers in tubes with air seemed to grow and multiply exponentially. **** See Attached Data Graph ****
ANALYSIS:
1. What are the most obvious differences between the two graph curves? The curve for the tubes with oxygen go up and the the curve for the tubes with no oxygen level off and peak quickly due to no more growth. 2. Look at Table 7-A-1 and compare tubes 4A and 4B. How many times greater was the growth when air was present? The growth was 5 times greater in tube 4B. 3. Compare the other tubes in A and B series at various glucose concentrations. How much greater is the growth in air for each pair of tubes from 1A and B through 5A and B? It starts at 4 times greater with air, peaks at 5 times greater, and ultimately decreases back down to 4.67 times greater than with air. 4. Notice that the number of bacteria not given for tubes 6B,7B,8B, and 9B. How many bacteria would you predict in tubes 6B? 3450 million bacteria per ml. In 7B? 4175 million bacteria per ml. These numbers were omitted from the table because they are so large. 5. After each tube reached maximum growth, the solution was tested for the presence of glucose. In all the tubes from 1A to 6A and from 1B to 9B there was no glucose. In contrast, tubes 7A, 8A, and 9A contained some glucose even after maximum growth had been reached. Now compare tubes 4A and 4B. How many bacteria were produced per milligram of glucose in each case? 4A produced 3.06 million bacteria/ml of glucose and 4B produced 15.28 million bacteria/ml of glucose. 6. Now offer a hypothesis that can explain the numbers you calculated from question 5. As more oxygen is available, the bacteria will utilize the glucose more and as a result will produce higher levels of bacteria. Why are there so many cells per milligram of glucose in the B tubes? Aerobic respiration in the “B” test tubes yields a higher amount of ATP (36) than the glycolysis in the “A” test tubes which yields a lower amount of ATP (2). Glycolysis is very inefficient, but it’s the only available source in the beginning. 7. Each milligram of glucose has an equal amount of energy available to do work. The B tubes produces more cells per milligram of glucose than did the A tubes. Assuming that each cell produced requires a certain amount of energy. Which tubes should contain some “unused” energy? Tubes 7A, 8A, and 9A would be expected to have some unused energy. 8. In additional tests it was also determined that the alcohol accumulates in the A tubes. How does this information relate to your answer concerning “unused energy?” The energy that was “unused” was actually converted to ethanol.

CONCLUSION: The results show that the “B” test tubes of bacteria that had unlimited access to oxygen thrived and multiplied to higher maximum level than that of the “A” test tubes.