1) Calculate the ratio of optic density versus viable cell. 2) Your teacher has
ID: 252937 • Letter: 1
Question
1) Calculate the ratio of optic density versus viable cell. 2) Your teacher has told you that this ratio is specific for this bacterial type only. Explain why? 3) How much is the CFU of the same bacterial type if the absorbance is found to be 0.108?
During the construction of the growth curve of a certain bacterial type you have calc CFU in parallel with measuring the optical density. At absorbance 0.315 the CFU was 25.105CFU/ml. ulated the a. Calculate the ratio of optic density versus viable cells Your teacher has told you that this ratio is specific for this bacterial type only. Explain why b. c. How much is the CFU of the same bacterial type if the absorbance is found to be 0.108?
Explanation / Answer
Answer a.
CFU refers to the colony forming units present. These are the total number of viable cells. Given that CFU at 0.315 absorbance is 25 * 10^5 CFU/ml.
So, ratio of optical density and viable cells is 0.315 / 25* 10^5 = 0.0126* 10-5
Answer b.
Optical density values can be directly related to bacterial growth. At the maximum optical density, cell growth will be reduced; now bacteria will start growing slowly.
Every bacterial species has a particular value of optical density at which exponential growth occurs in them. There is also a maximum value, at which bacterial growth will be start depleting due to depletion of resources.
Same is the case with these bacterial cells.
Answer c.
Given that absorbance is 0.108
So, 0.0126* 10-5 = 0.108/ CFU count;
CFU/ml = (0.108/0.0126* 10-5) = 8.57 * 105
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