Filter Qualification

Technical report no. 26 from the Parenteral Drug Association [63] identifies the following factors that should be part of selecting and qualifying a filter for use as a product sterilizing filter:

1. Particle-shedding characteristics 2. Extractables

3. Chemical compatibility 4. Adsorption

5. Thermal stress resistance 6. Hydraulic stress resistance 7. Toxicity testing

8. Bacterial challenge testing 9. Physical integrity testing

Physical integrity testing has already been discussed. Subsequent discussion will focus on extractables and bacterial challenge testing.

E. Bacterial Challenge Test

Microbiological challenging of a filter is the only true means of determining the bacterial retention properties of the system. Such a test is sensitive because of the large number of organisms used and because the organism self-replicate and allow even low numbers of bacteria that might pass through a filter system to make themselves known.

Filter media are not repetitive-use items, and although used for more than one lot in production, the media are usually discarded after some predetermined number of uses or time. Therefore, it is impossible to test every filter medium individually, since the challenge test is a destructive test. The nondestructive tests, therefore, require a high degree of correlation with a retention test. When such correlated tests are established and controls maintained, filtration users can depend on filtration to produce a sterile parenteral product.

The level of sensitivity of the challenged test is dependent on the chal- lenge organism, culture environment of the organism, challenge level of the organism, test volume filtered, challenge rate or the duration of the challenge test, and pressure used during the challenge test [63,64].

In 1987, FDA published its guideline on validation of aseptic processing [43] specifying requirements for challenging filters with 107cells of Pseudomo- nas (now Brevundimonas) diminuta per cm2

of filter surface and for validating aseptic processes using sterile media fills.

The challenge organism utilized in filter testing is Brevundimonas dimi- nuta (ATCC 19146). The rationale for using B. diminuta follows the same logic as used in choosing B. stearothermophilus for steam sterilization testing. Bacil-

lus stearothermophilus is resistant to heat and therefore severely challenges the lethality given by an autoclave. Because filtration is a removal process, the most resistant organism to filtration would be the smallest known bacterium. Brevundimonas diminuta has been adopted for several reasons. First, the organ- ism is quite small. The gram-negative rod-shaped cell has a mean diameter of 0.3µm. The bacteria were first isolated when found to consistently pass through 0.45-µm filter membranes to contaminate filtered protein solutions. The organ- ism can be grown to high cell densities in a short period of time, and with proper culturing the cells are small and arranged singly. In addition, B. diminuta shows only limited biochemical activity. A growth curve for B. diminuta in saline-lactose broth (SLB) at 30°C is shown in Figure 12. The initial lag time lasts about 3 hr. In the exponential growth phase, the organism has a population doubling time (generation time) of 2.6 hr and an instantaneous growth rate con- stant (µ) of 0.27 hr−1_{. The growth curve levels off in the stationary phase at}

approximately 107cells/mL [65].

For reproducible challenge tests, care must be taken in culturing and han- dling the bacteria to maintain bacterial cells of equal morphology. Studies have shown that differences in cell morphology can be produced by using different growth media or by the use or nonuse of agitation during culturing. Brevundimo- nas diminuta grown in trypticase-soy broth (TSB) without agitation produces a cell that is distinctly rod-shaped, having a length-to-diameter ratio of 2 to 5. Grown in the same TSB medium but with 200 rpm agitation, B. diminuta was more dense and had longer cells with a length-to-diameter ratio of about 4. In

Figure 12 Growth curve of P. diminutia (ATCC 19146) in saline-lactose broth incu- bated without agitation at 30°C. (From Ref. 72.)

addition, the cells tended to form clusters of from 3 to 8 cells each. Brevundimo- nas diminuta grown in SLB without agitation are found to have a length-to- diameter ratio of 1 to 2.5 and are arranged singly [65].

The growth state of a B. diminuta culture is also important in obtaining the smallest cell size on a reproducible basis. Brevundimonas diminuta cells are observed to increase in cell size during the lag phase and become smaller during the declining growth period. Therefore, challenge cells for retention testing are most appropriate when in the early stationary phase of growth. Early stationary phase rather than late stationary phase is taken to reduce the chance of the challenge culture containing nonviable cells and cellular debris, which could prematurely clog the test filter medium.

Maintenance of a pure culture of B. diminuta must be done in such a manner as to keep the probability of mutational changes that might alter cellular characteristics to a minimum.

The microbial challenge test can be performed on a particular filtration medium, whether disk or cartridge type, by following these general steps:

1. Sterilize the filter system. Figure 13 shows a hypothetical test system for a disk filter medium.

2. Integrity test the filter medium using a sterile 0.1% peptone solution or saline solution to wet the medium. The wetting solution also serves as a negative control sterility check. The entire wetting solution is

forced through a sterility control filter, incubated, and checked for sterility.

3. The bacterial challenge suspension is placed in the appropriate con- tainer and the test filter medium is challenged.

The challenge suspension should have a microbial concentration of 107 B. diminuta per square cm of effective filter area (EFA). Many challenge levels have appeared in the literature: 107per 100 ml for 1400 liters, 105

–107

per ml, 2–4× 105

per liter per min, 1.2× 1012

–1.9× 1013

per liter, and 108

per cm2

EFA [44,50,64–68]. Much discussion has also appeared in the literature concerning the challenge level and the potential adverse effects of excessive levels of challenge bacteria [66]. The rationale for the 107 B. diminuta per cm2EFA challenge is that while this level of bacteria might not challenge every membrane pore (approximately 108 pores per membrane medium), it is enough to challenge any oversized pore. Since the flow through pores varies as the fourth power of the radius of the pore, a larger fraction of the total flow is carried by the larger pores. Therefore, it is felt that at the 107challenge level enough increased flow will pass through any oversized pores that challenge bacteria will inevitably encounter an oversized pore, pass through, and indicate a negative test. The 107 level is also under the filter-clogging concentration [10].

The challenge suspension should be forced through the test medium at a pressure differential greater than 2 kg/cm2

(approximately 30 psi) for disk filters and at fluxes of greater than 2 liters per 0.1 m2

up to around 3.86 liters per 0.1 m2

for cartridge-type filters [64,65]. A pres- sure relief valve on the downstream side of the filter should be provided to allow maximum pressure differentials. The suggestion has also been made that the pressure be applied full strength immediately rather than a gradual buildup in order to stress the filter system further [65]. 4. The entire volume of the challenge filtrate is subsequently forced

through a sterility test filter system and incubated in the same manner as the negative control filtrate.

5. A postchallenge integrity test is performed.

6. The challenge test results are then observed. The challenge tests are considered invalid if the negative control contains any organisms. The filter system is considered to have failed the test if the filtrate contains any test organisms.

F. Extractables

Filter validation now includes tests to prove that sterilizing filters do not gener- ate extractable materials when exposed both to water and to the drug product

formulation. Tests for filter extractables may be found in the USP, Section<87> Biological Reactivity Tests, in Vitro and Section <88> Biological Reactivity Tests, in Vivo. These tests involve soaking filter material in different solvents, then evaluating them in two animal models and in cell culture. USP Section <661> also describes testing of filters to ensure that no extraneous contaminants are found in the filter material. Filter extracts have been identified as surfactants, wetting agents, additives used in filter manufacture, higher molecular weight polymers of the filter polymer, and general particulates [69–71]. Extraction pro- cedures with actual drug product may include immersing the filter into the drug product solution, then exposing it to high temperatures and mechanical agitation before taking samples and assaying by various analytical techniques [71].

G. Retention Efficiency

In the past, several terms have been coined to describe the retention efficiency of the filter system: beta value, microbiological safety index, reduction ratio, and titer reduction ratio [64,68,72]. The log reduction value (LRV) is a filter retention efficiency term that is the logarithm to the base of 10 of the ratio of the number of organisms in the challenge suspension to the number of organisms in the filtrate.

LRV= logN0

N (26)

LRV= log N0− log N (27)

where

LRV= log reduction value

N0 = number of organisms in the challenge

N = number of organisms in the filtrate

With a sterile filtrate, the term log N becomes log 0, which is undefined and is eliminated from the expression. The LRV is then expressed as being equal to or greater than N0.

LRV≥ log N0 (28)

For example, if a 293-mm-diameter disk filter system having an EFA of 530 cm2 is challenged and the 107/cm2level is used, the total challenge to the filter is 5.3× 109 organisms. If a sterile filtrate is assumed the LRV would be calcu- lated and reported as follows:

LRV= log 5.3 × 109− log 0 LRV≥ 9.72

The probability of passing a single organism through this filter system, or in other words, the probability of nonsterility (PNS) can be calculated by the following equation:

PNS= N N0

(29) For the above example, the PNS is calculated as

PNS= 1

5.3× 109= 1.89 × 10

−10

Equations (27) and (29) can be used to calculate the total PNS of replicated experimental filter challenges. For example, five filter membrane media are to be challenged at the following P. diminuta levels.

18× 109 9× 109 5× 109 14× 109 12× 10

The total challenge (N0) is 5.8× 10 10

, and the assumption is made that the filtrate for each is sterile. The LRV then is

LRV= log N0− log N = log 5.8 × 10

10_{= 10.76}

The PNS then is calculated to be PNS= N

N0

= 1.72 × 10−11_{, N= 1}

These equations may be used to calculate an estimate of the degree of nonsteril- ity associated with a particular filtration process. In order to determine such a sterility assurance associated with the process, some knowledge of the initial microbiological bioburden of the product to be sterilized must be known. If it is assumed that the microbiological bioburden of a product is 104

organisms and the product is to be sterilized by filtration through filters from the example above, the PNS is the sum of all the probabilities of all of the combinations of the 104organisms passing through the filter. The expression for this is

PNS=

∑

N₀ n=1

Pi (30)

where

N0= the bioburden of the product

Pi = probability of i organisms passing the filter medium n = 1, 2, . . . i

In other words, the PNS is equal to the probability of one organism passing the filter plus the probability of two organisms passing the filter, and so on. With the bioburden level greater than one organism, however, there result many com- binations of sets or organisms having a probability of passing the filter. The probability of all combinations of one organism passing a filter with a given retention efficiency from a bioburden level N0can be written as

P1= N0! (N0− 1)!1 1× RV (31) where RV= reduction value or in logarithmic form logP1= log N0! (N0− 1)!1 − LRV (32)

Similarly, the probability of all combinations of i organisms passing the filter is log Pi= log

N0!

(N0− i)i

− i(LRV) (33)

When Eq. (30) is expanded into the format of Eqs. (32) and (33), the following expression results: log PNS= Σ

冋冉

log N0! (N0− 1)!1 − LRV

冊

+ ⴢ ⴢ ⴢ +

冉

log N0! (N0− i)!i − iLRV

冊册

(34) In a convergent series, as in Eq. (34), the bracketed quantity representing P1Ⰷ

P2Ⰷ ⴢ ⴢ ⴢ Pican be approximated by using P1 only. Therefore, Eq. (34) can be

simplified to

log PNS= log N0! (N0− 1)!1

− LRV (35)

This simplifies further to

log PNS= log N0− LRV (36)

A sterility assurance (SA) can be calculated from

SA= 1 − PNS (37)

By way of example, the LRV for the previous five-filter example was 10.76. If a product having a bioburden, N0, of 10

4

organisms is to be filtered, the PNS can be calculated using Eq. (36).

log PNS= log N0− LRV = 4 − 10.76 = −6.76

PNS= 1.74 × 10−7

The SA can be calculated then by Eq. (37) where SA= 1 − PNS = 1 − 1.74 × 10−7

SA= 0.9999998 or a 99.99998% assurance of sterility

Terminally sterilized parenteral products have a level of SA in the range of 0.999999. If we assume that our example solution has 100 organisms per liter, how much could we filter before the SA dropped below 0.999999? By using Eq. (37)

SA= 1 − PNS

PNS= 1 − .999999 = 10−6

When we substitute into Eq. (36) and rearrange log N0= log PNS + LRV = −6 + 10.76 = 4.76

N0= 57,544 organisms

When we use the following relationship

N0= C × V (38)

where

N0= the bioburden of the product

C = the bioburden concentration per unit volume V = the volume of the product

then V=N0 C= 57,544 organisms 100 organisms/liter V= 575.44 liters

Therefore, 575.44 liters can be filtered before going below an SA of 0.999999. Data generated by the Millipore Corporation show that a mixed cellulosic ester membrane filter medium with an average bubble point of 3.44 kg/cm2, when challenged with an average of 2.78× 107organisms/cm2, had an average LRV of 9.96. Millipore claims that 20 years of quality control testing has con- firmed that mixed esters of cellulose filter media having a minimum bubble test of 3.3 kg/cm2

or greater will quantitatively retain 107

P. diminuta/cm2

EFA at a differential pressure of 2.6 kg/cm2

. Such is the type of correlative data that are needed to validate each product for filtration sterilization [69].

In document Pharmaceutical Process Validation.pdf (Page 171-179)