Quantitative Microbial Risk Assessment: Main concepts & application in Environmental Health and sanitation Dr Nguyen-Viet Hung Center for Public Health and Ecosystem Research, Hanoi School of Public Health, Hanoi Swiss Tropical and Public Health Institute (Swiss TPH), Basel Sandec/Eawag, Duebendorf, Switzerland International Livestock Research Institute (ILRI) 1 Risk Analysis Training for SEAOHUN June 3 – 7, 2013 Risk Analysis Training for SEAOHUN Hanoi, June 3 – 7, 2013

Exercises: 2 groups of 5 persons Group 1,3,5: conducting QMRA 1 and proposing risk management options Group 2,4: conducting QMRA 2 and proposing risk management options 20mn group work, 5 mn presentation, 10mn discussion

Exercise 1: QMRA of wastewater and management options A peri-urban district of Thailand (10,000 inhabitants) has households equipped with pit latrines or cesspool and septic tanks from which most of the collected sludge is disposed to FS treatment system (not functioning) through the collection from municipality. The rest is disposed by themselves in their land. Most of household greywater are discharged onto the land. The rest about as well as septic tank effluent are discharged into the sewer which then goes into the canal. Farmers utilize canal water for vegetable irrigation. Assess the risk of Salmonella sp. infection (conducting a QMRA) of farmers using canal water for irrigating vegetables for a single exposure and yearly risk. The following information is necessary to do such a QMRA. Concentrations of Salmonella sp. in: –Canal water: 70 organisms/100 mL. –The die-off rate of Salmonella sp. from the canal water to the point of irrigation is 70%. –Water volume ingested per time: 1.5 mL –Frequency of exposure per year (times): 200 Beta-Poisson model is used to assess the dose-response of Salmonella sp. with the values of α= 0.21 and N 50 = 49.8

Exercise 2: QMRA of water treatment plant and management options A water treatment plant uses surface water and treats this water with conventional process comprising filtration, sedimentation and chlorination. The concentration of Campylobacter jejuni in surface water due to the fecal contamination is 150 bacteria per liter. The bacterial removal efficacy of the treatment is 99 %. It is assumed that people consuming directly this water 2L per day from tap water coming from that water treatment plant. 1) Analysis of 10 samples of tap water and found the results as in table 1 below. and define the distribution of the dose of Campylobacter jejuni in the population consuming this water. 2) Assess the risk of C. jejuni infection (conducting a QMRA) of consumer of water mentioned above for a single exposure and yearly risk. The dose response model of Campylobacter spp. is Beta-Poisson model with the values of α= 0.144, N 50 = 890. Based on the result of QMRA, you propose a plan of risk management to reduce health risk to meet the acceptable risk at ppy.

Dose-response models: Beta-Poisson model P: Probability of infection of a single exposure D: Dose ingested with a Poisson distribution α: parameters of the Beta distribution of r N 50 : Dose for 50% of infection k: probability of one organism overcoming host barriers to reach a site for infection. r is constant for the population

Ex 1: Wasterwater in Thailand and health risk Risk of Salmonellosis from surface water for farmers Deterministic Initial concentration (organisms/100mL)70.00 Reduction from canal to irrigation0.30 Water volume ingested1.50 Ingested dose (DOSE)0.32 =B3*(1-B4)*B5/100 Alpha0.21 N Exposures per year200 Probability of infection per exposure =1-((1+B6/B8)*(2^(1/B7)-1))^(-B7) Probability of infection per year =1-(1-B10)^B9 Acceptable risk per year Single risk Detail of calculation DOSE/N (2^1/α - 1) α Risk of single exposure Annual risk0.9984

Ex 2: Set the standard for rotavirus in salad at infection risk at pppy This dose is determined by QMRA as follows: (a) Conversion of the tolerable campy infection risk of 10 −4 pppy PI(A)(d) to the risk of infection per person per exposure event PI(d) − i.e., per consumption of 2L water, which takes place every day throughout the year, so n is 365: Because PI(A)(d) = 1 – [1 – PI(d)] n PI(A)(d) – 1 = [1 – PI(d)] n [PI(A)(d) – 1] 1/n = [1 – PI(d)] PI(d)] = 1- [ PI(A)(d) – 1] 1/n = PI(d) = 1 − (1 − 10 −4 ) (1/365) = × 10 −7 = × 10 −7 N α0.14 Pi(d) × 10 −7

Set the standard for rotavirus in salad at infection risk at pppy The required number of campy is × 10 −7 bacteria per 2 litter in the “treated” water, then × 10 −7 in 1 litter. 150 campy/L raw water, and × 10 −7 /L just after treatment So the required log unit reduction = log (150) – log (1.305 × 10 −7 ) = 2.2 – (-6.9) = 9.1