I can see a few options:

1. Assuming you follow growth as OD - are you sure its growth? Many times when you move bacteria to a high salt media the cell shape change, resulting in a fast increase in OD. This increase will be short time only and will stop. Follow the OD for a few hours, at least, and make sure it continue to go up or use CFU instead.

2. Are you adding the same VOLUME of cells? Adding more cells from the same cell concentration would mean more dilution of the salts. Add spent media (remove cells by centrifugation+filtration) to the 10^6 to get to the same volume.

3. If the above is not the case one could assume that 100 fold more cells will mean more diversity of mutants to select from.

Dear Gerchman,

thanks for your answer.

1- Yes, I am sure about growth due to observation of byproducts resulted by microbial growth.

2- the bacteria concentration was adjusted somehow the only difference between two media is the number of cells. everything else are the same (I modified the question description to make it more clear).

3- could you please explain the third one more?

This is a far shot - but assuming a certain mutation rate the more bacteria the more mutants you will have. E.g., if we assume 0.003 mutations per genome per generation and you have 10^6 cells (or genomes) --> 10^6*0.03=30,000 'mutants' to choose from. If you start from 10^8 you will 100 fold more...

Another option to consider is "the bacteria concentration was adjusted somehow". Is the starting point the same? Many times when cells are under mild stress they express stress related proteins and thus can handle better bigger stress (that how Heat Shock Proteins were discovered). You have to start from exactly similar starting culture - i.e. grow them in the same media to the same OD. If the 10^8 grew longer they were probably exposed to some stress, either due to lack of nutrients or accumulating waste. Assuming you dilute your cells to the high ionic strength media I would guess you start at at least 10^9~10^10 cell/ml, definitely stress conditions.

So I suggest you take the 10^8 intended culture, dilute part of it 100 fold, and add THE SAME VOLUME to your final broth.

Hope this helps

Yoram

Ms. Alavi

Thank you for the answer.

-Yes, the bacteria is halophilic.

- I mean the cell concentration in solution at T=0 is adjusted to 10^6 or 10^8 cell/ml.

- Assuming the destination solution is mixture of solution A and B. Salinity of the destination solution is ~ 130 mS/cm. in order to adjust the bacteria concentration to 10^6, 10^7 and 10^8 cell/ml in this solution at T=0, I followed the following steps:

for 10^8: I took 1ml of origin bacterial solution (10^9 cell/ml, salinity of the fresh solution without bacterial was ~ 25 mS/cm). it was centrifuged and the supernatant was replaced with solution A (salinity= ~65 mS/cm). The destination solution (9ml) was then inoculated with the whole 1ml solution.

for 10^7: the same as above but the destination solution (9.9 ml) was inoculated with 0.1ml of that 1ml centrifuged solution.

for 10^6: I took 0.1ml of origin bacterial solution (10^9 cell/ml, salinity of the fresh solution without bacterial was ~ 25 mS/cm). then it was diluted with 0.9 ml of solution A (salinity = ~ 50 mS/cm). it was centrifuged and the supernatant was replaced with solution A (salinity= ~65 mS/cm). The destination solution (9.9 ml) was then inoculated with 0.1ml of the 1ml centrifuged solution.

The proportion of A and B in destination solutions for all concentrations is the same.

The bacteria were taken from the same origin bacterial solution.

THERE IS NO GROWTH FOR 10^6, SMALL GROWTH FOR 10^7, HIGH GROWTH FOR 10^8!

Looking forward your helpful answers,

Thanks,

Here is a big part of your problem. You assume that centrifugation will be effective in collecting ALL the cells but this is simply wrong, especially with 10^6 cells. The way you do it you probably end up with no cells.

Centrifuge only once: Take 1.5 ml of 10^9 cells, centrifuge and resuspend in 1.4 ml solution A (it is very hard to remove all the liquid). Now take from THIS suspension 1 ml, 0.1 ml + 0.9 ml A and 0.1 ml from the previous + 0.9 ml (after through mixing). Mix each with with solution B. Keep in mind the real cell concentration will be lower then 10^8/7/6. You can test for the real numbers using dilutions and plating for CFU counting.

Yoram

Dear Mr. Gerchman

I did colony counting using serial dilution method. there was a negligible change before and after centrifuging (4000rpm, 15min).

Dear Ms. Alavi

Centrifuging and replacing the supernatant (metabolic wastes produced by bacteria in the culture medium during growth) with a fresh saline solution increases the activity of my bacteria (according to the literature), that's why I do centrifugation and mixing with fresh saline solution.

my destination solution is a mixture of urea, CaCl2, and nutrient broth for bacterial growth.

Urea (NH4+)+ CO2

CO2 + H2O HCO3-

(Ca2+) + (CO3-) CaCO3 (higher pH (more NH4+) and CO3- concentration, more CaCO3 precipitation)

the bacteria that I am using produce an enzyme catalyzing the first reaction and consequently the CaCO3 precipitation process which is the aim of my study.

thanks again,

I must admit I am surprised centrifugation worked, from our experience, in low cell count efficiency of retrieval is ~2% (from water). But as you said, if you tested it then its probably OK...

I would guess that the higher cell density has modified the medium sufficiently to encourage growth. One of the factors could be the pH for instance which if made lower by sufficient metabolism could provide more protons to encourage more cell metabolism via co-transport of whatever nutrients are available in your medium. You could try to find the nutrient responsible or you could try adjusting the pH with a pH-stat which might counteract the growth stimulation of the metabolic protons.

As a first stage try adding bis-tris-propane or MES to your medium. 50mM should be OK.