A bacterial sample has a cell density of 1.85 x106 CFU/mL. What plate dilution should yield a countable plate? Which two dilutio

Question

3 years ago

15

n tubes could produce this plate dilution? How

1 answer:

insens350 [35] · Answer 1 · 2021-12-25T01:30:23+03:00

Answer:

Explanation:

Bacterial count in stock- 1.85x10^6 cfu/ml

Dilution methods

Take 100 uL or (0.1ml) from stock and add to 900ul (0.9ml) saline and mixed it- this makes 10^1dilution.

Now take 100ul from 10^1 dilution and add to next 900ul saline this is 10^2 dilution, similarly do upto 10^5 dilution.

Then take 100ul from 10^ 4 and 10^5 dilution seperately and plate on LB agar plate seperetely and count the colonies.

Cfu/ml formula= (No.of colonies x dilution factor)/0.1 ml

So suppose, 18 colonies formed on 10^4 dilution then total no. Of cells in stock will be 18x10^4/ 0.1= 18x10^5 cfu/ml.

If we dilute 10^4 or 10^5 that's leads to colony count of 18-19 colonies on 10^4 dilution while 2 colonies should come on plate of 10^5 dilution.