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Determination of MIC & MBC

Minimum inhibitory concentrations (MICs) are defined as the lowest concentration of anantimicrobial that will inhibit the visible growth of a microorganism after overnight incubation, while minimum bactericidal concentrations (MBCs) known as the lowest concentration of antimicrobial that allow less than 0.1% of the inoculated bacteria to grow this mean that more than 99.9% of the bacteria should be killed.

MIC and MBC can be determined by several techniques:

  1. Tube dilution method ( macro or microtube)
  1. Plate dilution method
  1. Using spectrophotometer
  1. Using epsilometer to determine MIC

How to calculate MIC and MBC using tube dilution method ?

  1. Obtain antimicrobial powders directly from the manufacturer or from commercial sources. The agent must be supplied with a stated potency (mg or International Units per g powder, or as percentage potency).
  1. Store powders in sealed containers in the dark at 4 °C with a desiccant unless otherwise recommended by the manufacturer.

3. Prepare / antibiotic / stock / solutions / byusing / the / following / formula:
Weight / of powder / (mg) = / Volume of / solvent (ml) / X / Concentration / (µg/ml)

Potency of powder (µg /mg).

  1. Prepare serial dilution from the stock solution of the antimicrobial agent as it shown in the figure below.

Each tube contain 1 ml of different concentrations of the antimicrobial

  1. From broth inoculated with the bacteria to be tested and overnight incubated, transfer 1 ml of the bacterial broth to each tube containing antimicrobial agent. ( the broth used is Muller Hinton broth).

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Note: the volume in the tubes duplicated, this mean that both the antimicrobial agent and bacterial concentration have been decreased to the half of the initial concentrations.

  1. After an overnight incubation the tubes noticed to have decreased turbidity should be plated. The volume to be plated is important for calculation.

  1. The antimicrobial agent concentration _in the first tube contain decreasing in turbidity ( inhibition of growth) or no growth_ considered as MIC.

Note: For some drugs, MIC can be MBC as well, so we need to do calculations to find out the MBC.

  1. After overnight incubation for the inoculated plates, the number of colonies in each plate should be counted. Thus the bacterial concentration of each tube can be determines using the following formula : Bacterial concentration= no. of colonies/ ( volume plated * dilution).

By applying the concentration formula the bacterial concentration in the tube containing 8 µg of the drug contained 120 CFU/ml. Hereby, we need to know the percentage of grown cells from the initial bacterial concentration used.

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  1. The initial concentration of bacteria can be obtained or confirmed from plating the control tube containing 0 µg of drug.

  1. The concentration of drug in the tube giving percentage of growth less than .1% considered as MBC.

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Answer the following questions:

A

  1. If the initial bacterial concentration was 6.2×105 CFU/ml , calculate MIC, MBC and number of colonies should be obtained in the control plate

Initial bacterial concentration 6.2×105

CFU/ml.

0.5mlofbacterial

broth transferred to

1.5mlbroth

containing tube

.001 ml

Volume plated in each tube .001 ml

How many colonies??

0512

5

Solution:

  1. MIC is the minimum concentration of drug that cause bacterial growth inhibition which is 8 µg/ml.
  1. To find out MBC: we need to calculate the bacterial concentration of the last three tubes as well as the bacterial percentage from the initial concentration in each of them. The tube that established less than 0.1% bacterial growth should be considered MBC.

Firstly: the e initial bacterial concentration is 6.2 ×105 CFU/ml

  1. In the tube of drug concentration 8 µg/ml

C = no. of colonies / ( volume plated*dilution) C = 12/ ( 0.001 * 1) = 12000 CFU/ml

Bacterial content percentage = (concentration at tube 8 µg/ml / initial

bacterial concentration) * 100%

Bacterial content percentage = (12000 / 3.1×105) * 100% = 3.8%

  1. In the tube of drug concentration 16 µg/ml

C = no. of colonies / ( volume plated*dilution) C = 5/ ( 0.001 * 1) = 5000 CFU/ml

Bacterial content percentage = (concentration at tube 16 µg/ml / initial

bacterial concentration) * 100%

Bacterial content percentage = (5000 / 3.1×105) * 100% = 1.6% So the drug concentration 32 µg/ml should be considered MBC.

Note: If the dilution considered is ½ in the concentration formula, the initial bacterial concentration should be considered 6.2×105 CFU/ml in the percentage formula.

  1. To find out the no. of colonies should be obtained in the control plate Total dilution in the control tube = 0.5 / (0.5 + 1.5) = 0.25

C = no. of colonies / ( volume plated*dilution) 3.1×105 = ?? / ( 0.001 * 0.25)

So , no. of colonies = 77.5 ~ 78 CFU

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  1. Using 2.5 × 105 CFU/ml was initial concentration of bacteria calculate MIC, MBC, and volume plated in the control plate.

0.5 ml of the bacterial
broth transferred to 1.5
Volume plated
ml broth tube
???

Volume plated in each tube 0.1 ml

31 colonies

512

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Solution:

  1. MIC is the minimum concentration of drug that cause bacterial growth inhibition which is 8 µg/ml.
  1. To find out MBC: we need to calculate the bacterial concentration of the last three tubes as well as the bacterial percentage from the initial concentration in each of them. The tube that establish less than 0.1% bacterial growth should be considered MBC.
  1. In the tube of drug concentration 8 µg/ml

C = no. of colonies / ( volume plated*dilution) C = 12/ ( 0.1 * 1) = 120 CFU/ml

Bacterial content percentage = (concentration at tube 8 µg/ml / initial

bacterial concentration) * 100%

Bacterial content percentage = (120/1.25×105) * 100% = 0.096%

So the drug concentration 8 µg/ml can be considered as both MIC & MBC.

Note: If the dilution considered is ½ in the concentration formula, the initial bacterial concentration should be considered 2.5×105 CFU/ml in the percentage formula.

  1. To find out the volume plated should be obtained in the control plate Total dilution in the control tube = 0.5 / (0.5 + 1.5) = 0.25

C = no. of colonies / ( volume plated*dilution) 1.25×105 = 31 / ( ?? * 0.25)

So , volume plated = 9.9×10-4 ml

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  1. Using 1×106 CFU/ml as initial bacterial concentration, Find MIC and MBC

Day 1

Day 2

Day 3

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Solution:

  1. MIC is the minimum concentration of drug that cause bacterial growth inhibition which is 16 µg/ml.
  1. To find out MBC: we need to calculate the bacterial concentration of the last three tubes as well as the bacterial percentage from the initial concentration in each of them. The tube that establish less than 0.1% bacterial growth should be considered MBC.
  1. In the tube of drug concentration 16 µg/ml C = no. of colonies / ( volume plated*dilution) C = 7/ ( 0.01 * 1) = 700 CFU/ml

Bacterial content percentage = (concentration at tube 16 µg/ml / initial

bacterial concentration) * 100%

Bacterial content percentage = (700/ 0.5×106) * 100% = 0.14%

So the drug concentration 32 µg/ml can be considered as both MIC & MBC.

Note: If the dilution considered is ½ in the concentration formula, the initial bacterial concentration should be considered 1×106 CFU/ml in the percentage formula.

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4.Using data in the figure calculate the initial bacterial concentration, MIC and MBC

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Solution:

1.To find out the initial bacterial concentration

Total dilution in the control tube = 0.5 / (0.5 + 1) = 1/3

C = no. of colonies / ( volume plated*dilution)

?? CFU/ml = 80 / ( 0.001 * 1/3)

Bacterial concentration = 2.4 × 105 CFU/ml

So, the initial bacterial concentration = 4.8 × 105 CFU/ml

  1. MIC is the minimum concentration of drug that cause bacterial growth inhibition which is 8 µg/ml.
  1. To find out MBC: we need to calculate the bacterial concentration of the last three tubes as well as the bacterial percentage from the initial concentration in each of them. The tube that establish less than 0.1% bacterial growth should be considered MBC.
  1. In the tube of drug concentration 8 µg/ml

C = no. of colonies / ( volume plated*dilution) C = 30/ ( 0.1 * 1) = 300 CFU/ml

Bacterial content percentage = (concentration at tube 8 µg/ml / initial

bacterial concentration) * 100%

Bacterial content percentage = (300 / 2.4 × 105) * 100% = 0.125%

  1. In the tube of drug concentration 16 µg/ml

C = no. of colonies / ( volume plated*dilution) C = 5/ ( 0.1 * 1) = 50 CFU/ml

Bacterial content percentage = (concentration at tube 16 µg/ml / initial

bacterial concentration) * 100%

Bacterial content percentage = (50 / 2.4 × 105) * 100% = 0.020% So the drug concentration 16 µg/ml should be considered as MBC.

Note: If the dilution considered is ½ in the concentration formula the initial bacterial concentration should be considered 4.8 × 105 CFU/ml in the percentage formula.

Ms. Noor Abu Tayyem