REM 610 - Exercise 4 February 10, 2010

1. REM 610 - Exercise 4February 10, 2010

2. Q1. Relationship between internal concentration of dioxin and rainbow trout mortality

3. Q1. Conclusions/Observations    - Cumulative mortality increases with increasing internal concentration   - Appears to increase more slowly at higher internal concentrations - Mean internal concentration causing death ~64 ng/kg (i.e., ~50% of fish die)

4. Q2. Observed vs. Estimated Mortality • Solver used to minimize sum of squared deviations • a = 4.54; b = 62.11 • Appears to be a typical dose-response relationship

5. Q2. Calculate LC50 (based on internal concentration) M (%) = 100*Cfa / (ba + Cf a) 50/100 = Cfa / (ba + Cf a) Cfa = ½ *(ba + Cf a) Cf a = ½ ba + ½ Cf a ½ Cf a = ½ ba Therefore Cf = b (at 50% mortality) = 62.11 ng/kg

6. Q3. What has happened in the one Compartment Model? Dioxin concentration increases over time ~ linear

7. Q4. What does One-Compartment model tell us about rainbow trout toxicity test? • Concentration in rainbow trout after 14 days (toxicity test period) has not reached steady state – still increasing • Fish concentration after 14 days is only 4.2 ng/kg, which is below NOAEL of 34 ng/kg and LC50 of 62 ng/kg • Results demonstrate the problem of extrapolating from lab to the field (i.e., from acute exposure to chronic exposure) • If we were to predict a “safe” water concentration based on results of acute toxicity tests, we would be underestimating the potential occurrence of toxic effects in organisms that are chronically exposed to contaminant • After 1 yr the concentration is ~102 ng/kg = ~90%Mortality (see Table 1 or % mortality equation)

8. Q5. Mortality at Steady State Steady state: Inputs = Outputs or dC/dt = 0 • dCf /dt = k1*Cw – k2*Cf • 0 = k1*Cw – k2*Cf • k2*Cf = k1*Cw • Cf = (k1*Cw)/ k2 = [(300 L/kg.day)*(1pg/L)]/0.0004 day-1 = 750000 pg/kg = 750 ng/kg Given a = 4.54; b = 62.11 Calculate Mortality (%) M(%) = 100*Cfa /(ba + Cfa) = 100*(750)4.54/(62.114.54 + 7504.54) = 99.98% or ~100% • In the field we would expect 100%mortality of rainbow trout at steady state

9. Q6. Two Compartment Model

10. Q6. Explain the Model • 2 compartment model  fish are composed of 2 compartments • C1 exchanges chemical with both the water & with C2 • Comp 2  exchanges chemical only with Comp 1 • All compartments initially increase because the chemical in the environment is much greater than inside the fish (uptake phase of curve, steady state has not been achieved) • Why does compartment 1 increase faster than compartment 2? • -  Compartment 1 has two intake sources: water & Comp 2. Direct uptake from water occurs rapidly & still in uptake phase of curve • - Compartment 2 only has the one source of chemical uptake: Comp 1. Chemical moving from compartment 1 to 2 & vice versa occurs relatively slowly. Compartment 2 receives only 0.2% of the chemical in compartment 1, per day.

11. Q6. Explain Two Compartment Model Movement into clean water: • Compartment 1 • 2 intake routes: water & C2. Initially during elimination, Cw = 0 (so k01*Cw =0), and C2 << C1 therefore uptake rates are much lower than elimination rates (i.e., k12*C1 + k10*C1 >> k01*Cw + k21*C2) • k10 is relatively large; thus elimination into the water is the main sink of the chemical from compartment 1 - Both (a) and (b) lead to a rapid elimination rate & corresponding drop in chemical concentration in compartment 1 • As C1 decreases, the rate of elimination decreases (rate(t-1) = 1.1%C1) – as chemical leaves, C1 decreases • Eventually, C1< C2 such that uptake of the chemical from C2 becomes a more important process (i.e., “k21*C2” becomes important relative to “(k12+k10)*C1”). - Both factors (c) and (d) lead to a slower and slower rate of decline

12. Q6. Compartment 2 • After movement to a clean tank, compartment 2 still receives chemical from compartment 1 (i.e., C1>>C2, so k12*C1>>k21*C2), thus C2 continues to increase. Since k12 is relatively low, the rate of increase is also slow. • Compartment 2 reaches steady state when C1 = ½ C2. At this point C2 is constant over time steady state: dC2/dt = 0 K12*C1 = k21*C2 C1/C2 = k21/k12 = 0.001/0.002 C1 = ½ C2 • However, compartment 1 is continually losing chemical to the ambient water, so C1 is continually decreasing. When C1 < ½ C2 then elimination term is greater and C2 declines.

13. Q6. Overall Fish a) The overall fish concentration combines chemical uptake/elimination rates in both compartments (i.e., weighted average). The initial drop is relatively rapid. The second phase of the decline in concentration is slower due to the slow rate-limiting chemical transport process from compartment 2 to 1.

14. Q7.     Elimination Rates vs. Exposure Period • e.g., Cave on day 5 = maximum, Cave day 82  ½ max; thereore ½ life = 82-5 days = 77 days (see spreadsheet) • Result: the longer the exposure period, the slower the rate of elimination. This is because with a longer exposure period, the chemical in compartment 2 will have time to “fill-up”. Thus compartment 2 is more significant in the overall depuration of the chemical and compartment 2’s relatively slow elimination rate will now slow the overall loss of chemical from the fish.