Question

#1) What should be the concentration of kanamycin if a serial 2-fold dilution was done starting with tube 1 and ending in tube 9 with an initial concentration of 100 ug/mL

tube #	1	2	3	4	5	6	7	8	9	10 [control]
Kanamycin concentration (ug/ml)	100	?	?	?	?	?	?	?	?	0
growth	no	no	no	no	no	yes	yes	yes	yes	yes

#2) Culture C has a titer of 2.3 x 107 bacteria/mL. Draw a dilution scheme showing 1 method in which this culture could be diluted to give plates with between 30 and 300 colonies

#3) Culture F has a titer of 1,555,233 bacteria/mL. Draw a dilution scheme showing 1 method in which this culture could be diluted to give plates with between 30 and 300 colonies

Answer 1

#1) What should be the concentration of kanamycin if a serial 2-fold dilution was done starting with tube 1 and ending in tube 9 with an initial concentration of 100 ug/mL

tube #	1	2	3	4	5	6	7	8	9	10 [control]
Kanamycin concentration (ug/ml)	100	?	?	?	?	?	?	?	?	0
growth	no	no	no	no	no	yes	yes	yes	yes	yes

Initial concentration of Kanamycin: Tube#1 = 100 ug / mL.

This is serially dilulted two-fold, which means a 1:1 or a 1 / 2 dilution. This means the kanamycin concentration becomes half at each dilution. Now, let us find the Kanamycin concentration for each tube. This is determined by multipying the Kanamycin concentration with the dilution:

Tube#2: 1 /2 * 100 = 0.5 * 100 = 50 ug / mL

Tube#3: 1 /2 * 50 = 0.5 * 50 = 25 ug / mL

Tube#4: 1 / 2 * 25 = 0.5 * 25 = 12.5 ug / mL

Tube#5: 1 / 2 * 12.5 = 0.5 * 12.5 = 6.25 ug / mL

Tube#6: 1 / 2 * 6.25 = 0.5 * 6.25 = 3.125 ug / mL

Tube#7: 1 / 2 * 3.125 = 0.5 * 3.125 = 1.56 ug / mL

Tube#8: 1 / 2 * 1.56 = 0.5 * 1.56 = 0.78 ug / mL

Tube#9: 1 / 2 * 0.78 = 0.5 * 0.78 = 0.39 ug / mL

#2) Given, the titer of culture C = 2.3 x 107 bacteria/mL

Usually, the volume used for plating is 0.1 mL

We need to find the dilution so that the number of colonies obtianed for this culture is between 30 and 300.

A simplistic way of looking at it would be:

Given concentration = 2.3 * 107 bacteria / mL

For a 0.1 mL plating volume, the concenration would become:

2.3 * 0.1 * 107 = 0.23 * 107 = 2.3 *106 bacteria / 0.1 mL

Which actually means that if we make a 10-6 dilution and plate 0.1 mL, then we would get 2.3 colonies.

A 10-5 dilution would give us 2.3 * 10 = 23 colonies and a 10-4 dilution would give us 23 * 10 = 230 colonies. So the dilution scheme would be:

Tube#	1	2	3	4	5
Culture (mL)	1	1	1	1	1
Source tube#	culture C	1	2	3	4
Water (mL)	9	9	9	9	9
Dilution	10-1	10-2	10-3	10-4	10-5
Plating				0.1 mL	0.1 mL

#3) Culture F concentration = 1,555,233 bacteria/mL.

Number of bacteria in 0.1 mL = 1,555,233 * 0.1 = 15, 55, 233 bacteria / mL = 1.5 * 10^6 bacteria / 0.1 mL

Number of bacterial colonies expected when we plate 0.1 mL of 10^-6 dilution = 1.5

No. of colonies expected with 10^-5 dilution = 1.5 * 10 = 15

No. of colonies expected with 10^-4 dilution = 15 * 10 = 150.

So, the dilution to be used here is 10^-4, as the next dilution (10^-3) would give us 1500 colonies and it would be out of range.

The dilution scheme for culture F would be:

Tube#	1	2	3	4
Culture (mL)	1	1	1	1
Source tube#	culture F	1	2	3
Water (mL)	9	9	9	9
Dilution	10-1	10-2	10-3	10-4
Plating				0.1 mL