In a Biology experiment, you are charged with finding the average growth rate of
ID: 54041 • Letter: I
Question
In a Biology experiment, you are charged with finding the average growth rate of a bacteria colony. At 8:00 a.m. on day 1, you estimate the colony to be roughly circular with a radius of 1 cm and a depth of 2mm. Due to the irregularity of the colony, you estimate that the radius varies between 0.96 cm to 1.05 cm. The depth of the colony looks fairly uniform but it is rather difficult to probe the depth of the colony in the culture so you estimate that the uncertainity in the depth is about 0.3 mm. On the following day at 2:00 p.m., you measure the size of the colony again and using the same process, decide that the colony has grown to: Radius: 1.5 cm +/- 0.03 cm. Depth: 3.4 mm +/- 0.3 mm.
(a) Calculate the volumes of the colony on day 1 and 2.
(b) Calculate the number of hours between measurements.
(c) What is the rate of growth of the volume of the bacteria colony?
(d) From your estimate of the range of the radii due to the irregularity of the colony, estimate the uncertainty in the radius of the bacteria colony on day 1.
(e) Calculate the uncertainty in the volume of the colony on day 1.
(f) Calculate the uncertainty in the volume of the colony on day 2.
(g) As the measurements on the two different days are independent, you can now calculate the uncertainty in the change in volume using the results in parts (e) and (f). This two step process does not work if the two volumes are not independent they depend on a common variable.
(h) What would you report to your advisor as the uncertainty in your growth rate?
(i) If your advisor is not happy with the accuracy of your results-i.e. she feels the error margin (uncertainty) is too large, which of your two measurements, radius and depth should you pay more attention to when you repeat the experiment? Justify your answer.
I just need (e) through (i) if anybody can help please. I'm very lost on these parts.
Explanation / Answer
V = /6W2L. This is theequation used to calculate cell volumeof a bacteria considering it as a spherical in shape.
Now The change in volume V1 and V2 for the day 1 can be calculated as follows.
V1 = 0.523 X W12 x L1 = 0.523 (.96 x2) x 2 = 1.99 V2 = 0.523 X W22 X L2 = 0.523 (1.05 x2) X 2.3 = 2.55. The difference is +/- 0.53 for day one.
f) In the same manner the difference in the volume for day 2 can also be calculated.
V1 = 0.523x (1.5 x 2) 3.4 = 5.33; V2= 0.523 X (1.53 x2) 3.7 = 5.9 The difference is approximately +/- 0.38.
(g) The difference between two day is +/- 0.1
(h) The uncertainty in the growth rate is permissible since it is very small.
(I) During repetition of experiment care must be taken so that we measure the radius and depth of the cells carefully. So that error does not occur.
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