SANS 5550-2009 Potency of antibiotics - Microbiological agar diffusion method of assay (two-point technique)《抗生素药效 微生物琼脂扩散化验法(2点法)》.pdf

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1、 Collection of SANS standards in electronic format (PDF) 1. Copyright This standard is available to staff members of companies that have subscribed to the complete collection of SANS standards in accordance with a formal copyright agreement. This document may reside on a CENTRAL FILE SERVER or INTRA

2、NET SYSTEM only. Unless specific permission has been granted, this document MAY NOT be sent or given to staff members from other companies or organizations. Doing so would constitute a VIOLATION of SABS copyright rules. 2. Indemnity The South African Bureau of Standards accepts no liability for any

3、damage whatsoever than may result from the use of this material or the information contain therein, irrespective of the cause and quantum thereof. ISBN 978-0-626-22716-6 SANS 5550:2009Edition 3SOUTH AFRICAN NATIONAL STANDARD Potency of antibiotics Microbiological agar diffusion method of assay (two-

4、point technique) Published by SABS Standards Division 1 Dr Lategan Road Groenkloof Private Bag X191 Pretoria 0001Tel: +27 12 428 7911 Fax: +27 12 344 1568 www.sabs.co.za SABS SANS 5550:2009 Edition 3 Table of changes Change No. Date Scope Foreword This South African standard was approved by National

5、 Committee SABS TC 1023, Microbiological evaluation of pharmaceutical products, in accordance with procedures of the SABS Standards Division, in compliance with annex 3 of the WTO/TBT agreement. This document was published in November 2009. This document supersedes SABS SM 550:1993 (first revision).

6、 SANS 5550:2009 Edition 3 1 Contents Page Foreword 1 Scope . 3 2 Normative references . 3 3 Apparatus and equipment 3 4 Reagents, buffer solutions, media and test organisms 4 4.1 General 4 4.2 Reagents . 4 4.3 Buffer solutions . 5 4.4 Media . 6 4.5 Test organisms 8 5 Procedure . 9 5.1 Preparation of

7、 suspensions and solutions 9 5.2 Test procedure 10 6 Calculations and interpretation 13 SANS 5550:2009 Edition 3 2 This page is intentionally left blank SANS 5550:2009 Edition 3 3 Potency of antibiotics Microbiological agar diffusion method of assay (two-point technique) 1 Scope This standard specif

8、ies a microbiological method for the determination of the potency of various antibiotics in pharmaceutical and other products by means of an agar diffusion two-point assay technique. 2 Normative references The following referenced documents are indispensable for the application of this document. For

9、 dated references, only the edition cited applies. For undated references, the latest edition of the referenced document (including any amendments) applies. Information on currently valid national and international standards can be obtained from the SABS Standards Division. 2.1 Standards SANS 171, G

10、lassware and equipment for microbiological tests. 2.2 Other publications British Pharmacopoeia (BP). Code of Federal Regulations Food b) the concentrations of the doses are increased (see 5.1.3 and 5.1.4); or c) the volume administered to each dose site is increased (see 5.2.8). Table 5 Concentratio

11、ns of standard and sample solutions 1 2 3 4 5 6 7 8 9 Concentration of standard stock solutionaMaximum storage at 4 C Concentrations of dose solutionsaDose volume per site Antibiotic name Solvent g/mL or units/mL or IU/mL d Diluent g/mL or units/mL or IU/mL Test organism suspension Test mediumL Amox

12、ycillin 0,1 M B8 1 000 7 0,1 M B8 1 and 0,25 M. luteus Mcr 10 M11 75 Bacitracin Zinc 0,01 N HCl 100 14 1 % B6 5 and 1,25 M. luteus Mcr 25 M1 or M2 50 to 75 Doxycycline 0,1 N HCl 1 000 1 0,1 M B 4,5 2 and 0,5 B. cereus Bac 97 M1 25 to 50 Erythromycin Methanol 1 000 14 0,05 M B8 or 0,1 M B8 10 and 2,5

13、 B. pumilus Bac 42 M11 50 to 75 Gentamycin Distilled water 1 000 14 0,05 M B8 2 and 0,5 B. pumilus Bac 42 M11 75 to 100 Kanamycin Distilled water 1 000 14 0,05 M B8 10 and 2,5 B. subtilis Bac 2 M11 50 to 75 Mitomycin 5 mL distilled water plus 45 mL 1 % B6 1 000 14 1 % B6 5 and 1,25 B. subtilis Bac 2

14、 M1 75 to 100 Neomycin Distilled water 1 000 30 0,05 M B8 10 and 2,5 or 40 and 10 B. pumilus Bac 42 Med E or M11 50 to 100 Nystatin Dimethylform-amide (DMF) 1 000 max. 2 h 10 % B6 200 and 50 S. cerevisiae 23 M19 75 to 100 Oxytetracycline 0,1 N HCl 1 000 1 0,1 M B4,5 5 and 1,25 B. cereus Bac 97 M1 50

15、 Polymyxin B 2 mL distilled water plus 98 mL10 % B6 1 000 14 10 % B6 200 and 50 B. bronchiseptica M10 50 Streptomycin Distilled water 1 000 14 0,05 M B8 20 and 5 B. subtilis Bac 2 M11 50 to 75 Tetracycline 0,1 N HCl 1 000 1 0,1 M B4,5 2,5 and 0,625 B. cereus Bac 97 M1 50 to 75 Tobramycin 0,05 M B8 1

16、 000 14 0,05 M B8 20 and 5 B. subtilis Bac 2 M11 50 to 75 Vancomycin Distilled water 1 000 7 0,05 M B8 80 and 20 B. subtilis Bac 2 M11 50 to 75 aPrepare the standard stock solutions and the dose solutions in concentrations with the same units. SANS 5550:2009 Edition 3 12SANS 5550:2009 Edition 3 13 6

17、 Calculations and interpretation 6.1 Calculations 6.1.1 Calculate the average diameter, in millimetres, of the inhibition zones produced by each of the four dose solutions. 6.1.2 Calculate the potency ratio (M) using the following equation: M = IEF22= SLSHTLTHSLSHTLTH+ 0,602 06 and the difference in

18、 effect due to difference in slope (G) using the following equation: G = TH TL SH + SL where M is the potency ratio (the logarithm of the ratio between the potencies of the sample dose solutions and those of the standard); F is the difference in effect due to the sample; E is the difference in effec

19、t due to the dose; I is equal to 0,602 06; NOTE is the logarithm of the fixed ratio 4:1 of the concentrations of the two dose solutions in column 6 of table 5. TH is the average diameter of the zones produced by the dose solution of the sample at high concentration, in millimetres; TL is the average

20、 diameter of the zones produced by the dose solution of the sample at low concentration, in millimetres; SH is the average diameter of the zones produced by the dose solution of the standard at high concentration, in millimetres; SL is the average diameter of the zones produced by the dose solution

21、of the standard at low concentration, in millimetres; G is the difference in effect due to difference in the slope. 6.1.3 For each of the four dose solutions, calculate the difference (d) between the diameter of each inhibition zone (see 6.1.1) and the average diameter calculated for each relevant d

22、ose solution. Calculate the square of each difference so found and summate these. The sum of these squares is denoted by d2. 6.1.4 Calculate the variance of the assay (V) using the following equation: V = )1(2nndSANS 5550:2009 Edition 3 14 where d is the difference between the diameter of each inhib

23、ition zone and the average diameter calculated for each relevant dose solution (see 6.1.3); n is the number of doses of each dose solution in the assay (see table 6). Table 6 Values of n and t 1 2 3 4 n Degrees of freedom (n 1) n(n 1) ta6 20 120 2,09 8 28 224 2,05 9 32 288 2,04 10 36 360 2,03 at is

24、a constant value for the Students index (tc) (see 6.1.5) for a given value of n. 6.1.5 Calculate the Students index (tc) using the following equation: tc = VG4where G is the difference in effect due to difference in the slope; V is the variance of the assay (see 6.1.4). If the numerical value of tcs

25、o calculated is smaller than the corresponding value of t tabulated in table 6, the assay is valid. 6.1.6 Calculate the index of the significance of the slope (g) using the following equation: g = 222)1( Etnndwhere d is the difference between the diameter of each inhibition zone and the average diam

26、eter calculated for each relevant dose solution (see 6.1.3); n is the number of doses of each dose solution in the assay (see table 6); t is a constant value for the Students index for a given value of n (see table 6); E is the difference in effect due to the dose. SANS 5550:2009 Edition 3 15 6.1.7

27、Calculate the potency factor (PF) using the following equation: PF = antilog M NOTE M is the potency ratio (the logarithm of the ratio between the potencies of the sample dose solutions and those of the standard). 6.1.8 Calculate the dilution factor (DF) using the following equation: DF = HS DS wher

28、e HS is the highest sample dose solution, in micrograms or units per millilitre (see column 6 of table 5); DS is the dilutions to highest sample dose solution, in micrograms or units per millilitre. 6.1.9 Calculate the potency (P) of the sample using the following equation: P = PF DF where PF is ant

29、ilog M (see 6.1.7); DF is the dilution factor (see 6.1.8). 6.1.10 Calculate the fiducial limits (L) of the estimate of the potency of the sample resulting from the assay, expressed as a percentage of the result itself, using the following equation: L = antilog()()+221112IVMgVgEtIggMwhere g is the in

30、dex of the significance of the slope (see 6.1.6); M is the potency ratio (the logarithm of the ratio between the potencies of the sample dose solutions and those of the standard); t is a constant value for the Students index for a given value of n (see table 6); I is equal to 0,602 06; E is the diff

31、erence in effect due to the dose; V is the variance of the assay (see 6.1.4). SANS 5550:2009 Edition 3 16 6.2 Interpretation Repeat the assay if a) the numerical value of M exceeds 0,0415, b) the Students index exceeds the prescribed value, or c) the fiducial limits exceed the prescribed value of (9

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