Please solve 2-4 You are given a sample thought to contain about 2.4 times 109 v
ID: 44745 • Letter: P
Question
Please solve 2-4 You are given a sample thought to contain about 2.4 times 109 viable cells/ml. Devise a dilution scheme (showing or describing required steps) so that 0.1 ml plated from last dilution would yield the countable plate. From the 3rd serial 10-fold dilution of sample you plate 0.1 ml and count 200 colonies. If the total count of the original sample was 4 times 107 cells/ml - what is viable count of original?. What is percent viability? A sample with total count of 8 times 1010 cells/ml is estimated to be 75% viable. Devise a dilution scheme so that 0.1 ml plated from the last dilution would yield the countable plate. (Show or describe required steps)Explanation / Answer
2) Serial dilution is an usual technique. Take 1 ml from the original sample and add it to 9ml of medium to make it 10 fold diluted. Take 1 ml from this already diluted sample and add it to 9ml of medium to further dilute it. Now, the current sample is 100 fold diluted to the original sample. Continue the process till 6 dilutions and plate it by taking 0.1 ml of sample. The original count of cells is 2.4 x 109 cells/ml. After diluting it to 6 times (10 fold each time), final concentration would be 2.4 x 103 cells/ml. So, by plating 0.1ml you would get a countable plate forming around 240 colonies.
3) Viable count = colony count x dilution used x volume used
viable count = 200 x 10-3 x 0.1
viable count = 2 x 106.
Percent viability = viable cell count/total cell count
percent viability = 2 x 106 / 4 x 107
= 0.5 x 10 -1
4) 75% viable cells of 8 x 10 10 cells would be 6 x 10 10
Serial dilute the sample by taking 1 ml from original sample and add it to 9ml of medium (10 fold). Take 1 ml from this diluted sample and add it to 9ml of medium (100 fold to original sample). Continue this procedure 6 times. Whne you plate 0.1 ml from the sixth dilution and plate it you would get around 60 colonies.