Antimicrobial susceptibility testing was performed to screen a novel compound. The MIC was found to be 10ng/mL. Use the following data to determine the percent recovery for each treatment group. What is the MBC and would you describe the antimicrobial tested as static or cidal? Novel compound concentration CFU/mL Present Recovery
0ng mL 3.4 x 10^5
10ng/mL 2.4 x 10^5
20ng/mL 6.4 x 10^4
40ng/mL 1.4 x 10^4
80n mL 4.4 x 10^3
160n mL 0

a) Percent recovery

100%, 70.58%, 4.11%, 1.29% , 0%

b) MBC = 160 ng mL

c) Static

Explanation:

A) Calculate the percent recovery for each group

i) for 0ng mL

( 3.4*10^5 / 3.4*10^5 ) * 100 = 100%

ii) for 10ng mL

( (2.4 * 10^5) / (3.4 * 10^5) ) * 100 = 70.58%

iii) for 20ng mL

( ( 6.4 * 10^5 ) / ( 3.4 * 10^5 ) ) * 100 = Nil

iv) for 40 ng mL

( ( 1.4 * 10^4 ) / ( 3.4 * 10^5 ) ) * 100 = 4.11 %

v) for 80 ng mL

( ( 4.4 * 10^3 ) / ( 3.4 * 10^5 ) ) * 100 = 1.29%

vi) for 160 n mL

percentage recovery = 0%

B) what is the MBC

The value of MBC = 160 n mL because at this concentration CFU/mL = 0

c) determine if the antimicrobial tested is static or cidal

condition : If, the ratio of MBC : MIC  is  ≤4 then novel compound is Cidal

MBC / MIC =  160ng / 10ng = 16 : 1

hence we can conclude that the antimicrobial tested is STATIC

explanation:

b,bacteria convert nitrates

into nitrogen gas.

d,plants and animals take in

and use nitrogen compounds.

e,bacteria break down waste    

products and return nitrogen  

to the soil.

a,bacteria convert atmospheric

nitrogen to ammonia.

c,bacteria convert ammonia

to nitrites and nitrates.

1. dermis

2. chemical

3. urine

4. viruses

5. antibodies