Trapp lecture

Information about Trapp lecture

Published on March 6, 2008

Author: Joshua

Source: authorstream.com

Content

Slide1:  Exposure Modeling: Uptake of Pollutants with Food Stefan Trapp Technical University of Denmark [email protected] Slide2:  Who am I? Slide3:  Stefan Trapp CV * born 1962 – Hassfurt/D - geoecology diploma Bayreuth - PhD botany TU Munich - habil mathematics Osnabrück - Lektor Applied Ecology DTU I am not a chemist Slide4:  Table of contents 1 Current status 2 Equilibrium 3 Roots 4 Stem 5 Fruits 6 Leaves 7 Exposure & risk assessment Soil pollution sources:  Soil pollution sources Slide7:  Status of Soil Risk Assessment in Europe  Soil protection laws passed  Few legal standards raised  Mainly heavy metals, PCB, PAH, PCDD/F and a few others Case by case decisions Slide8:  Soil Risk Assessment Tools CLEA (UK): 1 plant (roots and leaves) CSOIL (NL): 1 plant (roots and leaves) UMS (D): 1 plant (roots and leaves) EUSES: 1 plant (roots and leaves) CALTOX (US): 1 plant (roots and leaves) All consider just roots and leaves! Slide9:  Who eats leaves? Slide10:  We need models for carrots, apple and potato ! Slide11:  Plant-specific models for uptake of (neutral) organic chemicals Leafy vegetables (TGD model, 1995) Carrot model for thick roots (2002) Fruit tree models for apples (2003) Potato model (2004) + Travis & Arms (empirical regression, 1988) Slide12:  How plants function: Roots take up water and solutes Stems transport water and solutes Xylem = water pipe Phloem = sugar pipe Leaves transpire water and take up gas Fruits are sinks for phloem and xylem Slide13:  Advective uptake with water Diffusion Direct soil contact Translocation in xylem Soil – air plant Particle deposition Phloem transport Exchange with air Slide15:  Principles of plant specific models Basic processes identical: Advection, diffusion, sorption, growth dilution, metabolism Chemical data sets needed similar Log Kow, Henry constant, molar mass Slide16:  Test chemicals MTBE = Methyl tertiary butyl ether is a gasoline additive. High production, persistent, water soluble, volatile. TBA Terbutylazine is a herbicide, not very water soluble The vapor pressure is low. BaP Benzo(a)pyrene is a strong carcinogen. It is very lipophilic and has a medium vapor pressure. Slide17:  Chemical data Slide18:  Chemical Equilibrium = endpoint of diffusion, maximum entropy needed: Soil-Water = KSW Roots-Water = KRW Wood-Water KWood Leaves-Air = KLA given: log KOW, KAW (Henry constant) Slide19:  Equilibrium soil water to soil matrix Soil = organic phase (sorption) + water (solution) + gas phase (usually irrelevant) + mineral matter (inert) Sorption to organic phase = Kd = OC x KOC OC = organic matter, KOC = f(KOW) Slide20:  Estimation functions for the KOC ”Standard” (EU 1996): log KOC = 0.81 x log KOW + 0.1 Slide21:  Equilibrium soil water to soil matrix CW is conc. in soil water (mg/L), CSoil in bulk soil (mg/kg), r is density, OC is organic carbon, PW and PA are water and air pores and KWS = 1/KSW is chemical equilibrium. Slide22:  Example calculation CSoil = 1 mg/kg, OC = 0.02 kg kg –1 , rdry = 1.6 kg L-1 PW = 0.35 L L-1 (-> rwet = 1.95 kg L-1), PA = 0.1 L L-1 Chemical BaP, log KOW = 6.13 KOC = 100.81 log Kow +0.1 = 100.81 x 6.13 +0.1 = 116 225 Slide23:  Root Concentration Factor RCF = CRoot / CWater log (RCF - 0.82) = 0.77 log KOW - 1.52   or RCF = 0.82 + 0.03 KOW0.77 Briggs et al. 1982 Slide24:  Equilibrium roots to water ”Root” = Water + Lipids (+ Gas) KRW = WR + LR a KOW b + PA(root) KAW WR is water content (~0.89 L/kg) LR is lipid content (~0.025 kg/kg) a = 1/rOctanol = 1.22 L/kg b = 0.77 for roots (empirical, Briggs) Slide25:  Example calculation MTBE log KOW = 1.14 By Briggs’ equation: log (RCF - 0.82) = 0.77 log KOW - 1.52 log(RCF – 0.82) = 0.77 x 1.14 – 1.52 = -0.6422 RCF – 0.82 = 10-0.6422 = 0.22 RCF = 0.82 + 0.22 = 1.04 mg kg-1 / mg L-1   By Trapp’s equation KRW = WR + LR a KOW b   KRW = 0.89 L kg-1 + 0.025 kg kg-1 x 1.22 L kg-1 x 13.80.77 = 1.12 L kg-1 Slide26:  Similarity of partition coefficients root and soil OC = 2%, WR = 89%, LR = 2.5%, rdry = 1.6 Slide27:  Equilibrium roots to soil KRW / KSW = KRS Roots are similar to organic matter in soil  Chemical equilibrium near 1 Measured data from Trapp et al. 1990 Slide28:  Are all roots in equilibrium with soil? 1 The establishment of the equilibrium depends on a fast diffusive exchange 2 Diffusive exchange is fast for high surface to volume ratio 3 The equilibrium is achieved for fine roots and root hairs 4 The equilibrium is perhaps not achieved for thick roots, including root vegetables Slide29:  Carrot model Change of mass in roots = +uptake with water – transport to shoots dmR/dt = CWQ – CXyQ where Q is water flow [L d-1] Diffusion across the peel is neglected! Slide30:  From mass to concentration m is chemicals’ mass (mg) M is root mass (kg) C = m / M dmR/dt = d(CR MR)/dt The root grows – integration for C and M needed (oh no ...!) Slide31:  How to consider growth For young plants, growth is exponentially. For grasses: while the plant grows, the ratio of mass, leaf area and transpiration remains quite constant. Slide32:  Dilution by growth Chemical mass: m = constant Plant Mass: M(t) = M(0) x e+kt m/M = Concentration in plant: C(t) = C(0) x e-kt Slide33:  Carrot concentration Change of concentration in roots = +uptake with water – transport to shoots – dilution by growth dCR/dt = CWQ/M – CXyQ/M – kCR where k is growth rate [d-1] and CXy is the concentration in xylem = CR/KRW Slide34:  Carrot model steady-state Slide35:  Carrot to Soil Slide36:  Growth k only is important if sorption to roots KRW is high Carrot to Soil - Result Slide37:  Comparison to experimental data Carrotmodel KRW Slide38:  Example calculation for MTBE   What we have from before: KWS = 2.8 kg/L; KRW = 1.12 kg/L  Slide39:  Trapp S (2002): Dynamic root uptake model for neutral lipophilic organics. Environmental Toxicology and Chemistry 21, 203-206. Slide40:  Translocation upwards A ”standard plant” transpires 500 L water for the production of 1 kg dry weight biomass! = approx. 50 L per 1 kg fresh weight = approx. 1 L/day for 1 kg plant mass Slide41:  How do organic chemicals enter the xylem? Discrimination at the ”Casparian strip” Slide42:  Herbs (Briggs et al. 1982) Trees (Burken & Schnoor 1998) Transpiration Stream Concentration Factor TSCF TSCF = CXy/CW Concentration in xylem = ----------------------------------- Concentration in solution Slide43:  Example calculation TSCF MTBE log KOW = 1.14 Slide44:  TSCF from root model: CXy = CR / KRW Slide45:  Example calculation TSCF MTBE log KOW = 1.14 Slide46:  Trees and Fruits Children consume more fruits (as juice) than any other plant part. Some adults consume more than 1000 g vine (from fruits, too) per day. Fruits are very important for nutrition. Slide47:  Sorption to wood KWood = CWood / Cw log KWood = – 0.27 + 0.632 log KOW (oak) log KWood = – 0.28 + 0.668 log KOW (willow) Lignin is a good sorbent for lipophilic chemicals! Slide48:  Tree model   Influx with xylem = Q x CW x TSCF Q is transpired water (m3/a) Loss with xylem = Q  CStem /KWood   Slide49:  Tree model: steady-state solution in stem Slide50:  From tree to fruit For 1 kg of fruit dry weight (dw), about 20 L water are needed QF = dw  20 Then the chemical mass in fruits is   m = QF × CStem /KWood and CF = m / MF or in one step: Slide51:  Example calculation Apple KWood MTBE log KOW = 1.14 log Kwood = –0.266 + 0.632 log KOW = –0.266 + 0.632 x 1.14 = 0.454 KWood = 2.85 kg L-1 Slide52:  Example calculation Apple CStem From before: CW = 2.83 mg L-1, TSCF(Burken & Schnoor) = 0.37 L-1 L-1 Q = 3000 L ha -1 year-1, M = 100 tons ha –1, kG = 0.01 year -1 Slide53:  Example calculation Apple CFruit Water flux into apples: QF = dw  20 = 0.156  20 = 3.12 L/kg Concentration in apples: Slide54:  Uptake from soil into apples Persistent medium-polar compounds may accumulate in fruits! Slide55:  SAR - QSAR Environ. Res. 14, 17-26 Slide56:  Leaves are exposed to air Slide57:  Equilibrium between leaves and air Leaves are plant material, like roots. But they do not hang in soil, and not in water. Leaves hang in air. The concentration ratio between air and water is CAir and CWater are chemicals concentration (mass per volume). The concentration ratio between leaves and air is then Slide58:  Equilibrium between leaves and air L is index for leaves W is for Water and A is for Air Equilibrium leaves to water CL/CW = KLW = WL + LL a KOW b Same as roots but b = 0.95 It follows for equilibrium leaves to air CL/CA = KLW / KAW = KLA [mg/m3: mg/m3] or CL/CA = KLA x rL [mg/kg : mg/m3] Slide59:  Example calculation: Equilibrium leaf-air for benzo(a)pyrene  r = 500 kg m-3 log KOW = 6.13, KAW = 1.35 x 10-5   KLW = W + L a KOW b = 0.8 + 0.02 x 1.22 x 1 348 962 0.95 = 16251 KLA = KLW/KAW = 16251 / 1.35 x 10-5 = 1.2 x 109 (mg/m3 : mg/m3)  KLA/r = 1.2 x 109 (mg/m3 : mg/m3) / 500 kg/m3 = 2.4 x 106 (m3/kg) If CAir = 1 ng m-3 = 10-6 mg m-3 CLeaf = CAir x 2.4 x 106 m3 kg-1 = 2.4 mg kg-1 Slide60:  Leaves – exchange with air Stomata  Cuticle Slide61:  Exchange with Air by gaseous deposition through the cuticle gaseous exchange through the stomata - dry particulate deposition wet particulate deposition A rough estimate for the exchange velocity g is 1 mm/s Slide62:  Exchange with Air Calculation with Fick’s 1st Law of Diffusion Change of mass in leaves = + uptake from air - loss to air A is leaf area [m2], g is conductivity (= exchange velocity 1 mm/s) KLA is needed because the diffusion is between 2 different media Slide63:  The model for leafy vegetables Adapted by the EU in the Technical Guidance Documents for Risk Assessment  ”TGD model” Uptake from soil (via xylem) and from air (or loss to ...) Exponential growth Slide64:  Mass balance for the leafy vegetables The change of mass in leaves = + translocation from roots + uptake from air - loss to air Slide65:  Solution – linear differential equation In steady-state (t oo) : This is identical to with the standard solution where Slide66:  Example calculation Benzo(a)pyrene again CSoil = 1 mg/kg, CAir = 1 ng m-3, CW = 0.5 x 10-3 mg/L; TSCF (Briggs) = 0.00036 = 5 m2 x 10-3 x 86400 m d-1 x 500 kg m-3/(1.2 x 109 m3:m3 x 1 kg) + 0.035 d-1 = 0.035 d-1 = 0.5 x 10-3 mg/L x 0.00036 x 1 L/d / 1kg + 10-6 mg m-3 x 10-3 x 86400 m d-1 x 5 m2 / 1 kg = 0.00043 mg kg-1 d-1 Slide67:  Example calculation Benzo(a)pyrene CSoil = 1 mg/kg, CAir = 1 ng m-3 Uptake from soil = 0.2 x 10-6 mg kg-1 d-1 Uptake from air = 0.00043mg kg-1 d-1 is 2400 times higher! In fact, BaP and many other lipophilic chemicals (PAH, PCB, PCDD/F, DDT, HCB) are mainly taken up from air into leaves. That is a huge problem! Slide68:  Example calculation Benzo(a)pyrene II Steady-state: b/a = 0.00043 mg kg-1 d-1/0.035 d-1 = 0.012 mg/kg (after > 60 days, the solution is always close to steady state for kgrowth is 0.035 d-1) Direct Soil Uptake:  Direct Soil Uptake A ”standard” child eats 200 mg soil a day ”Pica child: 10 grams (acute effects) How much soil do you eat? More than you think ... Slide70:  Soil attached to vegetables Soil on plant surfaces (Li et al. 1994) [mg soil/g plant dw] Lettuce 260 Wheat 4.8 Cabbage 1.1 Default value: 1% attached soil (wet weight) BCF(leafy vegetables to soil) = BCF model + 0.01 Slide71:  Potato Is not a root and not a fruit Is a storage organ of the stem (a ”tuber”) Is not connected to the xylem Is loaded via phloem or by diffusion across the peel Slide72:  For the region 0 < x < r, the initial concentration is zero. There is no flux at x = 0, and x = h is maintained at constant concentration C0 for t > 0. The solution is    (2n + 1) r - x  (2n + 1) r + x C(x,t) = C0 ∑ (-1)n erfc ——————— + C0 ∑ (-1)n erfc ——————— n=0 (4 D t )½ n=0 (4 D t )½ Potato model – diffusion into a sphere Slide73:  Potato model – diffusion into a sphere Parameters needed: C0 = chemical equilibrium D = diffusion coefficient R = Radius Slide74:  Sorption study with potatoes calculation: KPoW = WPo + LPo a KOWb WPo = 79%, LPo = 0.1%, a = 1.22, b = 0.77 Po = Potato  Phenol  Slide75:  x is naphthalene D is benzo(a)pyrene Very lipophilic compounds remain near the peel! Potato model – sensitivity study Slide76:  The regression of Travis and Arms (T&A)   Pesticides, log KOW 1.15 to 9.35, subsurface vegetation log BV (dry wt) = 1.588 – 0.578 log KOW BV is concentration ratio plants (dry wt) to soil (dry wt).   T & A is frequently used in the USA Slide77:  The regression of Travis and Arms (T&A)   log BV (dry wt) = 1.588 – 0.578 log KOW Pesticides log KOW 1.15 to 9.35, above surface vegetation BV is concentration ratio plants (dry wt) to soil (dry wt). Slide78:  Example: BCF of MTBE with the Travis and Arms regression log Kow = 1.14 log BV (dry wt) = 1.588 – 0.578 x 1.14 = 0.92 →BV (dry) = 100.92 = 8.32   BCF (wet) = BV (dry) x (1 - W) x rwet / rdry W water content carrots = 0.89 r is the density of the soil (dry 1.6 kg/L, wet 1.95 kg/L); BCF (wet) = BV (dry) x (1 – 0.89) x 1.95 / 1.6 = 1.11 kg/kg Slide79:  Summary BCFs Slide80:  Summary BCFs   Slide81:  Daily Dietary intake DDI Slide82:  Daily dietary intake   For the calculation of the daily dietary intake (DDI), the consumption of vegetable is multiplied with the calculated concentration in the crop types, i.e. root vegetable, fruits, potatoes and leafy vegetable. DDI [mg/d] = ∑i concentration in crop(i) [mg kg-1] x amount of crop(i) consumed [kg d-1] Slide83:  Average consumption of plants in DK and CZ Slide84:  Example calculation: DDI of BaP (Average DK) BCF carrot to soil was 0.005; BCF apple was 7.8 x 10-8; BCF green vegetables to soil was 0.01 (attached soil particles). CSoil = 1 mg/kg C(carrot) = 0.005 mg BaP/kg consumption 0.027 kg/d C(apple) = 3.1 ng BaP/kg consumption 0.052 kg/d C(leafy vegetable) = 0.01 mg/kg consumption 0.007 kg/d DDI [mg/d] = ∑i concentration in crop(i) [mg kg-1] x amount of crop(i) consumed [kg d-1] = 0.005 x 0.027 + 3.1 x 10-6 x 0.052 + 0.01 x 0.007 [mg kg-1 kg d-1] = 70 ng/d + 0.16 ng/d + 135 ng/d = 205 ng/d Slide85:  Example calculation: Vegetarian C(carrot) = 0.005 mg BaP/kg consumption 0.5 kg/d C(apple) = 3.1 ng BaP/kg consumption 2 kg/d C(leafy vegetable) = 0.01 mg/kg consumption 0.2 kg/d DDI [mg/d] = ∑i concentration in crop(i) [mg kg-1] x amount of crop(i) consumed [kg d-1] = 0.005 x 0.5 + 3.1 x 10-6 x 2 + 0.01 x 0.2 [mg kg-1 kg d-1] = 1000 ng/d + 6.2 ng/d + 2500 ng/d = 3506 ng/d The system can be adapted to sensitive groups ... does this vegetarian have a problem? Slide86:  Dosis facet venunum   “All substances are poisons: there is none which is not a poison. Only the right dose differentiates a poison from a remedy.”     Paracelsus (1493-1541) German physician and father of modern toxicology. Slide87:  Acceptable Daily Intake (ADI) An acceptable daily intake of the chemical under investigation has to be defined by human toxicologists. A "virtually safe dose of BaP as a marker of the mixture of carcinogenic PAH in food would be in the range 0.06 to 0.5 ng BaP kg-1 bw d-1" (EC 2002). “Virtually safe” = 1 dead among 1 000 000 Example: Stefan Trapp, 72 kg bw (= bodyweight) ADI = 72 kg bw x 0.06 ng BaP kg -1 bw d-1 = 4.3 ng BaP d-1 ADI = 72 kg bw x 0.5 ng BaP kg -1 bw d-1 = 36 ng BaP d-1 Slide88:  Acceptable Soil Concentration (ASC) Life-long consumption of food produced from this soil shall not lead to oral intake of chemicals above a “virtually” safe dose! Acceptable soil concentration (ASC) from ADI/DDI(1)  ASC [mg kg-1] = {ADI [mg d-1] / DDI [mg d-1]} x 1 mg kg-1 DDI(1) calculated for a concentration in soil of 1 mg/kg: BCF is linear to concentration in soil Slide89:  Example Acceptable Soil Concentration (ASC) Benzo(a)pyrene   ADI was 4.3 to 36 ng/d DDI (DK) was 205 ng/d (carrots + vegetables)   ASC [mg kg-1] = {ADI [mg d-1] / DDI [mg d-1]} x 1 mg kg-1 = 36 ng/d / 205 ng/d x 1 mg/kg = 0.06 x 1 mg/kg = 0.17 mg/kg (or 0.02 mg/kg with the lower ADI) Slide90:  Benzo(a)pyrene “virtually safe” was 0.02 – 0.17 mg/kg EPA soil standard 0.087 mg/kg CZ, DK quality standard 0.1 mg/kg Intervention value 1 mg/kg Kindergartens CZ 0.12 to 1.65 mg/kg Background D, DK ca. 0.05 mg/kg Highest value Stefan ever worked with: 1600 mg/kg We are at risk! Real Ixisting Soil Konzentration (RISK)

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