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# Determining MIC

MIC

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### #1 Gamewizard

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Posted 15 July 2015 - 08:16 AM

Hi all

I basically need some help with interpreting my MIC results.

I have carried out a MIC test on some raw materials against my bug. Bacterial inoculum was at 0.6 OD. Used a 96 well A to H microtire plate.

Obviously made serial dilutions of the raw materials, now some of my raw materials were liquid so i used them straight away (100%) but some were solids so i had to dilute them first with with sterile water.

I made a 1 in 2 dilution so put in 10g of solid raw material in to 20 g water. So thier starting percentage was 50%, and then the serial dilutions were made from Column 2 (total vol of column 2 was 200ul as i had 100ul sterile water and 100ul of raw material in it) by taking 100ul out and transfering to next one and so on. (1 in 2 dilution so the percentage was halved each time like so):-

Raw material A

100%, 50% 25% 12.5%, 6.25%, 3.125%, 1.5625%, 0.78174%, 0.390625%, 0.1953125%, 0.09765625%, and 0.0244140625% for all wells.

Now i need to determine at what concentrations the raw materials inhibited the bacterial growth and I dont know how to figure out the concentration, and i believe the concentration has to be in ug/ml as well ?

Can some one please help me figure this out, I have looked on the internet but it has confused me and I do not understand it.

### #2 phage434

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Posted 15 July 2015 - 09:24 AM

Well, skip the calculations in terms  of %, which will just confuse you. For your solids, your initial concentration is 10 grams in 20 ml, or 0.5 grams/ml. This is 500,000 ug/ml. You will halve this number with each stage of the serial dilution.

For the liquids, you need to know the density of the liquid, so that you can calculate the grams/ml number. If you can't look it up, you can determine this by weighing a measured volume.

The concentration will again halve at each stage of the serial dilution.

### #3 Gamewizard

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Posted 16 July 2015 - 12:39 PM

Thank you so much for your reply. You have explained it very well.   So, starting from the first original well, where the initial concentration is 500,000 ug/ml, then 250, 000 ug/ml,

125000 ug/ml, 62500 ug/ml, 31,250 ug/ml, 15625 ug/ml, 7812.5 ug/ml, 3906.25 ug/ml, 1953.125 ug/ml, 976.5625 ug/ml, 488.28125 ug/ml, 244.140625 ug/ml.  Is that correct?

For the liquids, you determine the density by weighing out the liquid in question and then dividing its mass (g) by volume to get the density right ?   but I don't know what volume you use when weighing it out  ?

And when you have your density how do you calculate grams/ml ?

Thanks

### #4 phage434

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Posted 16 July 2015 - 12:59 PM

Density *is* grams per milliliter. You can measure it by taring a tube, adding 1 ml of liquid, and observing the weight. That weight is the density. It is also the number you need for your serial dilution calculation. Say you measure 0.8 grams, so your density is 0.8 grams/ml. If you add pure liquid to your first well, then it will have a concentration of 800,000 ug/ml. The second well will have a concentration of 400,000 ug/ml, etc.

Note that if you have a pure substance, the density can often be looked up in tables (or it may be listed on the label).

### #5 Gamewizard

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Posted 19 July 2015 - 11:31 AM

Sorry my bad,  yes  it is g/ml.  I have just found out that the density for my liquids is 1.26g/ml,(it was listed on a paper I had, I should have looked harder before)  but I still don't know how I will use this to calculate concentration of the raw materials in each of the well ?   is there any formula I can use

thank you

### #6 phage434

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Posted 20 July 2015 - 06:49 PM

Well, if your first well is pure liquid, then it will be 1260000 ug/ml. The second, after a 2:1 dilution, will be 630000 ug/ml. The third, 315000 ug/ml etc.

### #7 Gamewizard

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Posted 23 July 2015 - 03:16 AM

thank you so much,  can you check if i hav calculated the concentrations for the solids correctly?   they are in my previous post

thanks

### #8 pito

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Posted 24 August 2015 - 08:34 AM

Maybe better to repeat it..

You just dilute it 1/2 each time, so you divide by 2 each time.

If your startconcentration is 500000 µg/ml, than the second one is 500000/2 µg/ml (250000 µg/ml)... and so on...

thank you so much,  can you check if i hav calculated the concentrations for the solids correctly?   they are in my previous post

thanks

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