Question

Assume that a chemical of concern has a concentration of 12 mg/L in drinking water. Determine the dose of the chemical for adults and children. What is the cancer risk associated with this exposure if the slope factor is determined to be 0.065 (mg/kg-d)-1 for oral exposure? If 10,000 individual consumed this water over their lifetime what number would likely get cancer?

Answer 1

Answer:

Concentration of chemical in water = 12mg/L
For Adult, assuming an average weight of 70 kg, a water intake rate of 2 L/day, and daily exposure:

Dose of the chemical for adult = 12mg*2/70 = 0.343 mg/kg/day

For children, assuming an average weight of 10 kg, a water intake rate of 1 L/day, and daily exposure:

Dose of the chemical for children = 12mg* 1/10= 1.2 mg/kg/day

Slope factor for oral exposure = 0.065 (mg/kg-d)-1

ER = CSF x dose

• Where, ER = Estimated theoretical risk
• CSF = Cancer slope factor [(mg/kg/day)-1]
• Dose = Estimated exposure dose (mg/kg/day

Cancer Risk = (Exposure dose x risk factor x years of exposure) / 70 years (lifetime)

For Children, ER = [1.2 mg/kg/day * 0.065 (mg/kg-d)-1] / 70 = 0.001

For Adult, ER = [0.343 mg/kg/day * 0.065 (mg/kg-d)-1] / 70 = 0.0003

If 10,000 individual consumed this water over their lifetime what number would likely get cancer?

For Children = 0.001*10000 = 10

For Adult = 0.0003 *10000 = 3