Question

32. Work this dilution problem, solve for the number of colony forming units (cfu) in one mL of the E. coli sample. ml 1 ml. 1ml. 1ml. 1m. 1 ml A 2 a. 6.3 x 10 b. 6.3 x 10 c. 4.43 x 106 9 ml. 9 ml. d. 4.43 x 103 9mL Sample of E. coll 443 63 colonies colonles

Answer 1

32) The answer will be 6.3 x 10⁶ (Option a).

Explanation: The countable range for colony is 25-250 CFU/plate. Thus, during calculation we will consider the plate that contains 63 colonies.

Now, dilution factor = Volume of stock solution taken/ Final volume

Now, 1 mL of stock solution is taken from flask to the first tube which contains 9 mL of diluent.

So, dilution factor of first tube = 1/ (1+9) = 1/10 = 10^-1 dilution

Similarly, in each case there will be 10^-1 dilution as similar process is followed in each case.

So, in fifth tube (From which 63 colonies grown on the plate) total dilution factor will be 10^-5.

Now, concentration of fifth tube is 63 CFU/mL (As 63 colonies are grown on the plate by taking 1 mL. Please note that there is no mention that 1 mL or other values of solution taken from this tube for plating. We assume 1 mL & this is only possible from the given answer options).

Now, cell concentration in fifth tube = Initial cell concentration in flask x total dilution factor (TDF) of fifth tube

So, Initial cell concentration in flask = Cell concentration in fifth tube / Total dilution factor (TDF) of fifth tube = (63 / 10^-5) CFU/mL = 63 x 10⁵ CFU/mL = 6.3 x 10⁶ CFU/mL