1. Lecture 12 Primary Production – Nutrient Stoichiometry Topics Stoichiometry Biolimiting Elements

2. Surface NItrate Q. Why high NO3 ocean areas?

3. Surface Phosphate

4. Surface Silicate

5. Chemical Composition of Biological Particulate Material Hard Parts - Shells NameMineralSize (mm) Plants CoccolithsCaCO 3 Calcite5 DiatomsSiO 2 Opal10-15 Silicoflagellates SiO 2 Opal30 Animals ForaminiferaCaCO 3 Calcite~100 and Aragonite RadiolariaSiO 2 Opal~100 PteropodsCaCO 3 Aragonite~1000 AcanthariaSrSO 4 Celestite~100

6. Soft Parts - protoplasm from Redfield, Ketchum and Richards (1963) The Sea Vol. 2 Also for particles caught by sediment traps.

7. The Redfield or "RKR" Equation (A Model) The mean elemental ratio of marine organic particles is given as: P : N : C = 1 : 16 : 106 The average ocean photosynthesis (forward) and aerobic ( O 2 ) respiration (reverse) is written as: 106 CO 2 + 16 HNO 3 + H 3 PO 4 + 122 H 2 O + trace elements (e.g. Fe) light (h )  ( C 106 H 263 O 110 N 16 P ) + 138 O 2 or (CH 2 O) 106 (NH 3 ) 16 (H 3 PO 4 ) Algal Protoplasm The actual chemical species assimilated during this reaction are: HCO 3 - NO 3 - PO 4 3- NO 2 - NH 4 +

8. 1.This is an organic oxidation-reduction reaction - during photosynthesis C and N are reduced and O is oxidized. During respiration the reverse occurs. There are no changes in the oxidation state of P. We assume C has an oxidation state of 0 which is the value of C in formaldehyde (CH 2 O), that N has an oxidation state of -III and that H and P do not change oxidation states. 2. Photosynthesis is endothermic. This means it requires energy from an outside source. In this case the energy source is the sun. Essentially plants convert the photo energy from the sun into high energy C - C bonds. This conversion happens in the plants photosystems. Respiration is exothermic. This means it could occur spontaneously and release energy. In actuality it is always mediated by bacteria which use the reactions to obtain their energy for life. 3. Stoichiometry breakdown of oxygen production CO 2 + H 2 O  (CH 2 O) + O 2 C : O 2  1 : 1 H + + NO 3 - + H 2 O  (NH 3 ) + 2O 2 N : O 2  1 : 2 4. Total oxygen production: 106 C + 16 N x 2 = 138 O 2 Oxidation State of C?

9. 5. If ammonia is available it is preferentially taken up by phytoplankton. If NH 3 is used as the N source then less O 2 is produced during photosynthesis 106 CO 2 + 16 NH 3 + H 3 PO 4 + 122 H 2 O + trace elements light (h )  (CH 2 O) 106 (NH 3 ) 16 (H 3 PO 4 ) + 106 O 2 The relationship between O 2 and NO 3 /NH 4 is 2:1 (as shown in point #3) 16 HNO 3 + 16 H 2 O = 16 NH 3 + 32 O 2

10. Dissolved seawater data from Atlantic GEOSECS Program (Broecker and Peng, 1982) small deficit in NO3 Remarkable congruence between ratios in the ocean and plankton composition.

11. Nutrient stoichiometry from US JGOFS EqPac Suspended Matter and sediment trap particles Lines show Redfield slopes Murray et al, 1992, unpublished EqPac

12. Why the Redfield Ratios?? From a chemistry point-of-view. Each class of organic compounds has its own unique stoichiometry carbohydrates are C rich but N and P poor (e.g. (CH 2 O) n ) proteins are C and N rich but P poor (e.g. amino acids) nucleotides are rich in both N and P (e.g. 4 bases) lipids are C rich Questions: Why 16:1? Why not 6:1 or 60:1? See Arrigo 2005 (to be read later) How does an organism end up with a certain composition? What happens if one constituent is not available in adequate amounts?

13. Stoichiometry based on organic composition Average Plankton 65% protein 19% lipid 16% carbohydrate Average formula for plankton biomass C 106 H 177 O 37 N 17 PS 0.4 Oxidation consumes 154 moles of O 2 Hedges et al (2002) Marine Chemistry 106 CO 2 + 17 HNO 3 + H 3 PO 4 + 122 H 2 O + trace elements light (h )  ( C 106 H 177 O 37 N 17 PS 0.4 ) + 154 O 2

14. Controls on Atmospheric CO 2 Remarkable consistency for glacial/interglacial concentrations of CO 2. A main Control on atm CO 2 is the B flux! We need to understand B… PI CO2 = 280 ppm PI CO2 w/o B = 970 ppm!

15. How do we get from the marine food web to a global assessment of CO 2 flux??? With great difficulty!

16. Broecker two-box model (Broecker, 1971) v is in m y -1 then flux is mol m -2 y -1 Flux = V mix C surf = m yr -1 mol m -3 = mol m -2 y -1 see Fig. 2 of Broecker (1971) B fB (1-f)B B = VmCd + VrCr - VmCs

17. Mass balance for surface box V s dC s /dC t = V r C r + V m C d – V mix C s – B At steady state: B = V r C r + V mix C d – V mix C s and fB= V riv C riv

18. Broecker (1971) defines some parameters for the 2-box model g = B / input = (V mix C D + V r C r – V mix C s ) / V mix C d + V r C r f = V r C r / B = V r C r / (V mix C d + V r C r - V mix C s ) f x g In this model V r = 10 cm y -1 V mix = 200 cm y -1 so V mix / V r = 20 From 14 C mass balance (next slide) Here are some values: gff x g N0.950.010.01 P0.950.010.01 C0.200.020.004 Si1.00.010.01 Ba0.750.120.09 Ca0.010.120.001 Q. Explain these values and why they vary the way they do. because fB = V r C r fraction of element removed to sediment per visit to the surface fraction of input to surface removed as B See Broecker (1971) Table 3

19. How large is the transport term: If the residence time of the deep ocean is 1850 yrs (from 14 C) and  = Vol d / V mix then: V mix = (3700m/3800m)(1.37 x 10 18 m 3 ) / 1850 y = 0.72 x 10 15 m 3 y -1 If River Inflow = 3.7 x 10 13 m 3 y -1 Then River Inflow / Deep Box Exchange = 3.7 x 10 13 /72 x 10 13 = 1 / 19.5 This means water circulates on average about 19.5 times through the ocean (surface to deep exchange) before it evaporates and returns as river flow. volume fraction of total depth that is deep ocean

20. Example – 14 C Deep Ocean Residence Time substitute for B v mix in cm yr -1 ; vC in cm yr -1 x mol cm -3

21. Rearrange and Solve for V mix Use pre-nuclear 14 C data when surface 14 C > deep 14 C ( 14 C/C)deep = 0.81 ( 14 C/C)surf V mix = (200 cm y -1 ) A A = ocean area for h = 3200m thus age of deep ocean box (t) t = 3700m / 2 my -1 = 1850 years

22. Why is this important for chemical oceanography? What controls ocean C, N, P? g ≈ 1.0 Mass Balance for whole ocean:  C/  t = V R C R – f B C S = 0; C D = C D V U = V D = V MIX Negative Feedback Control: if V MIX ↑ V U C D ↑ B ↑ f B ↑ ( assumes f will be constant! ) assume V R C R  then C D ↓ ( because total ocean balance V U C D ↓ has changed; sink > source ) B ↓ CSCS CDCD if V MIX = m y -1 and C = mol m -3 flux = mol m -2 y -1 The nutrient concentration of the deep ocean will adjust so that the fraction of B preserved in the sediments equals river input!

23. Example: Perturbation analysis – Mass Balance Control Double Upwelling Rate sequence of events Paleo record Double rate of ocean mixing V r C r = fB at the beginning and at the end! The deep concentration (C d ) is cut in half

25. Example: Perturbation Analysis 1. Double River Input 2. CaCO 3 burial increases and Carbonate Compensation Depth (CCD) deepens

26. R = RKR P = Protein L = Lipid C = Carbohydrate E = Equatorial Pacific A = Arabian Sea 1-3 = Southern Ocean 1a = Anderson et al From Hedges et al (2002)