Nitrate Deficit

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Conversion of organic nitrogen into N2 in

the oceans:where does it happen?

and how?

Yuan-Hui (Telu) Li

Department of OceanographyUniversity of Hawaii at Manoa


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Outline1. Nitrogen cycle in the oceans:

2. Three end-member mixing model and the

aerobic partial nitrification hypothesis.3. Nitrate deficits by the aerobic partial

nitrification and the anoxic denitrification.

4. Conclusions5. Acknowledgement


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+5 NO3- NO3

-

Anammox

+3 NO2-

NO2-

+2 NO NO

+1 N2O N2O

N20 N2 fixation N2

-1 NH2OH NH2OH

-3 NH4+ PON DON NH4+

Uptake &

ammonification

Oxidation

state

[Anoxic]

Dissimilativereduction

(Denitrification)

[Oxic]

Dissimilativeoxidation

(Nitrification)

Air

Gas

exchange

?

AssimilativereductionAir


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1. Nutrient cycle in the ocean:

Redfield Ratios and Nitrification bynitrifying bacteria [oxic]:

138 O2 + (CH2O)106(NH3)16(H3PO4)

H2PO4 + 16 NO3

+ 106 CO2 + 17 H+ + 122 H2O

P\ N\ Corg\- O2 = 1\16\106\138

or

rp = - O2/ P = 138

rn = - O2/ N = 8.63

rc = - O2/ Corg = 1.30

phytoplankton


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Denitrification by denitrifying bacteria [anoxic

and suboxic]phytoplankton

94.4 NO3-+ 93.4 H

+ + (CH2O)106(NH3)16(H3PO4)

H2PO4 + 55.2 N2 + 106 CO2 + 177.2 H2O

P\- N\ Corg\ N2 = 1\94.4\106\55.2

Anaerobic ammonia oxidation (anammox) by

anammox bacteria:

NH4+ + NO2

- N2 + 2 H2O


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2.Three end member mixing model(Li and Peng, 2002)

1 = f1 + f2 + f3 (1)

= f11 + f22 + f33 (2)

S = f1S1 + f2S2 + f3S3 (3)

O2+ rnNO3 = (NO) = f1(NO)1 + f2(NO)2 +f3(NO)3 (4)

O2=0+1+2S- rnNO3 (4a)where, rn = -O2/NO3

Similarly

O2= A0+ A1+ A2SrpPO4 (5a)

where, rp = -O2/PO4

Also:DA =0+1 +2S +3O2 (6a)

where, rc= 1/(30.5/rn); rc = -O2/Corg

DA = (DICAlk/2)


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-70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70

7

8

9

10

11

12

13

14

Latitude

-O2/N

i8

i9

geosecs

-70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70

8

10

12

14

16

18

Latitude

N/P

i8

i9

geosecs

-70 -60 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 70

110

130

150

170

190

Latitude

-O2/P

i8

i9

geosecs

Redfield ratio

Red field ratio

Red field ratio

Indian


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2a. Aerobic Partial nitrification

hypothesis:Unidentified bacteria have evolved in a

low oxygen (but oxic) and high nitrate

environment (such as in oxycline, marine snowand fecal pellets, sediments) to utilize both

oxygen and nitrate as terminal electron

acceptors during oxidation of organic matter,and convert some organic nitrogen into N2,

N2O, and NO.


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N(umol/kg)


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3. Nitrate deficit by partial nitrification (dN) and

denitrification(dN)

Na = 16(P - 0.16)

Nb = -3.223 + 16.772P + 0.574 P2 - 0.465 P3

When Nb N Na

dN = Na - N ;When N < Nb

dN = Na - NbdN = Nb - N ;

N* by Deutsch et al (2001):

N* = (N - Na)

Na = 16(P - 0.181)

-N* dN + dN


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i7n

(mol/kg)


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Additional support for the aerobic partial nitrification hypothesis:

1. Schmidt et al (2004) showed that a wild-type ofNitrosomonaseuropaea in chemostat cell cultures can produce nitrogen gases (N2,

NO, and N2O) during aerobic (O2 ~ 125M) oxidation of ammonia,

using genes encoding reduction enzymes such as nitrite reductase,

nitric oxide reductase etc. For example,NH4+ (ammonia monooxygenase) NH2OH (hydroxylamine

oxidoreductase)NO2(nitrite reductase) NO (nitric oxide

reductase) N2O (not yet identified nitrous oxide reductase) N2.

2. Aerobic and anaerobic ammonia oxidizing bacteria are coupled insuspended organic particles in a low-oxygen (O2 ~ 5 M) CANON

reactor (Nielsen et al., 2005) to produce N2


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3. Codispoti et al. (2001) estimated the excess N2 in the water

column of the Arabian Sea, using the Ar/N2ratio in the water

column and in the air. They found that the excess N2 is

substantially greater than the N2 produced by thedenitrification process.


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2. The aerobic partial nitrification must be performed by yet

unidentified bacteria, which may have evolved in low oxygen and high

nitrate environments (such as oxyclines, marine snows, fecal pellets,

bottom sediments etc) to utilize both oxygen and nitrate as terminal

electron acceptors during oxidation of organic matter. Direct proof is

urgently needed.

Conclusions

1. The dN (nitrate deficit by partial nitrification) maximum coincides

with the P and N maximums, lies within the oxycline below the

oxygen-depleted denitrification zone, and is in contact with the

continental slope sediments. In contrast, dN (nitrate deficit by

denitrification) maximum lies within the denitrification zone, is always

associated with a nitrite maximum in the water column, and intersects

the continental shelf or upper slope sediments.


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Acknowledgement: Ms Lauren Kaupp patiently

showed me how to use the Ocean Data Viewprogram, which was provided by Dr. Reiner

Schlitzer. Discussions with Drs. James Cowen,

David Karl, Marcel Kuypers, FredMackenzie,

Hiroaki Yamagichi and Wajih Naqvi were most

fruitful.

Many thanks to Professor Yoshiki Sohrin for kindly

inviting me here. This work is supported by a NOAAgrant to Y.H. Li and T.H. Peng.


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-dN\O2

=(6 1)\130


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rp= -O2/P; rn= -O2/N; rp/rn = N/P


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V* = Red Sea

IV*w = W Equat. Indian Central

IV*e= E Equat. Indian Central

III* = Equat. Indian P Max.

V = South Indian Central

IV = Sub-Antarctic Oxygen Max

III = Antarctic Intermediate

II = Circumpolar Deep

I = Antarctic Bottom


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Redfield ratios:

P\N\Corg\-O2 = 1\16\106\138; (CH2O)106(NH3)16(H3PO4)

Antarctic Indian Ocean:

P\N\Corg\-O2 = 1\(151)\(832)\(1349)

Deep equatorial Indian Ocean:

P\N\Corg\-O2 = 1\(101)\(945)\(1307)

Average remineralization ratios for the warm water mass:

P\N\Corg\-O2 = 1\(15.60.7)\(1109)\(1598)

Andersons (1995) remineralization ratios and phytoplanktonformula:

P\N\Corg\-O2 = 1\16\106\150; (C106H48)(H2O)38(NH3)16(H3PO4)


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1. The remineralization ratios (P\N\Corg\-O2) of organicmatter in the oxygenated regions of Indian Ocean change

systematically with latitude and depth.

2. The average remineralization ratios for the Indian warm

water masses (potential temperature ~ 10) are

P\N\Corg\-O2 = 1\(15.60.7)\(1109)\(1598).

These are comparable to the traditional Redfield ratios

P\N\Corg\-O2 = 1\16\106\138,

and are in good agreement with Andersons (1995) values ofP\N\Corg\-O2 = 1\16\106\150

within the given uncertainties.

5. Conclusions

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