REM 610 - Exercise 4 February 10, 2010

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# REM 610 - Exercise 4 February 10, 2010 - PowerPoint PPT Presentation

REM 610 - Exercise 4 February 10, 2010. Q1. Relationship between internal concentration of dioxin and rainbow trout mortality. Q1. Conclusions/Observations.    - Cumulative mortality increases with increasing internal concentration

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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)

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

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

Q3. What has happened in the one Compartment Model?

Dioxin concentration increases over time ~ linear

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)

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
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.
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

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

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.
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.

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.