INSTITUT SENEGALAISDE RECHERCHES AGRICOLES ...
INSTITUT SENEGALAISDE RECHERCHES AGRICOLES
DIRECTION DES RECHERCHES SUR LES PRODUCTIONS FORESTIERES
CENTRE DE RESSOURCES MICROBIOLOGIQUES
(MIRCEN)
~r31~308S6
SOIL AMELIORATION WITH NITROGEN-FIXING
ACACIA SPECIES
First progress report of IAEA contract no. 6375/Rl/RB

FIRST PROGRESS REPORT
CONTRACT NUMBER : 6375/Rl/RB
TITLE OF PROJECT
Soi1 amelioration with nitrogen fixing Acacia albida and Acacia seyal
INSTITUTE WHERE RESEARCH IS BEING CARRIED OUT
Institut Senegalais de Recherches Agricoles (ISRA) : Programme MIRCEN
CHIEF SCIENTIFIC INVESTIGATOR : Mamadou GUEYE
ADDITIONAL SCIENTIFIC STAFF :
Ibrahima NDOYE from Univ. C. A. DIOP, Dakar
Bernard DREYFUS from ORSTOM, Dakar
Simon BADJI from Eaux et Forets, Dakar
Pascal DANTHU from ISRA/DRPF
TIME PERIOD COVERED
: June 1992 -- June 1993
FIRST EXPERIMENT
DESCRIPTION AND CARRYING OUT THE EXPERIMENT
The experiment was carried out from July to December 1992
with seven provenances of Acacia albida from Senegal (5)
and Burkina Faso (2) and one Parka biglobosa
from Senegal.
The isotope dilution technique and the A-value method were
used.
All grounded samples of leaves, stems and roots were sent to
the agency.
ISOTOPE DILUTION TECHNIQUE (ID)
Treatments
- A. albida provenances at 20 Kg N/ha
- Parka biglobosa provenance at 20 Kg N/ha
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1st Report IAEA 6375/Rl/RB

Number of replicates : 8
Number of pots
- (7 pr. A. albida + 1 P. biglobosa) x 8 reps = 64
Calculation of nitrogen requirement
Ammonium sulfate (AS) with 10.09% ’ 5N a.e. was
apllied at 200 mg N/pot.
Total N requirement : 200 mg N x 64 = 12800 mg N
Total AS requirement :12800 x 100/21.2 = 60377.3 mgN
Volume of solution needed : 50(mI/pot) x 64 = 3200 ml
We used 3300 ml because of spillage. Thus, the required
amount of AS for 3300 ml is :
60377.3 x 3300/3200 = 62264 mg AS
Dilution
m l +m2 = 62264 mg AS
Ml = 132.3338 g/mole of AS 10.09% ’ 5N a-e.
M2 = 132 g/ mole of ordinary AS
a’ = 10% 15N a.e. desired in final dilution
a’1 = 10.09% ’ 5N a.e. of AS to be diluted
Then,
62264 x 132.3338 x 10
ml =
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
(132.3338 x10.09) + (132.3338 - 132) x 10
m l = 61554.74 mg,
m2 = 709.26 mg
Summarv
m l = 61.50 g of AS with 10.09% ’ 5N a.e.
m2 = 0.71 g of ordinary AS
Total volume of solution : 3300ml
Number of pots : 64
Application rate : 50 ml
Remaining solution : 100 ml.
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A-VALUE METHOD
Trea tmen ts
Parkia biglobosa at 100 Kg N/ha in comparison
with A. albida treatments described
in ID
section.
Number of replicates : 8
Number of pots : 1 P. biglobosa x 8 reps = 8 pots
Calcula tion of nitrogen requirement
Ammonium sulfate (AS) with 10.09% ’ 5N a.e. was
apllied at 1000 mg N/pot.
Total N requirement : 1000 mg N x 8 = 8000 mg N
Total AS requirement : 8000 x 100/21.2 = 37735.85mgN
volume of solution needed : f>O(mllpot) x 8 = 400 ml
We used 500 ml because of spillage. Thus, the required
amount of As for 500 ml is :
37735.85 x 500/400 = 47169.81 mg AS
Dilution
ml + m2 47169.81
M l = 132.3338 g/mole of AS 10.09% ’ 5N a.e.
M2 = 132 g/ mole of ordinary AS
a’ = 2% ’ 5N a.e. desired in final dilution
a’1 = 10.09% ’ 5N a. e. of AS to be diluted
Then,
47169.81 x 132.3338 x 2
ml =
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
(132.3338 x10.09) + (132.3338 - 132) x 2
m l = 9345.14 mg,
m2 = 37824.67 mg
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1st Report IAEA 6375/Rl/RB

Summary
m l = 9.35 g of AS with 10.09% ’ 5N a.e.
m2 = 37.80 g of ordinary AS
Total volume of solution : 500ml
Number of pots : 8
Application rate : 50 ml
Remaining solution : 100 ml.
SECOND EXPERIMENT : GRAFTING EXPERIMENT
DESCRIPTION AND CARRYING OUT THE EXPERIMENT
The experiment Will be carried out from May to October 1993
with the senegales provenances of A. albida and P. biglobosa
described in the first experience. In addition, one A, seyal
provenance from Senegal Will be used for grafting the A. albida.
TREA TMENTS
The Bradyrhizobium MAO 488 Will be used for inoculating the
five A. albida provenances.
The Rhizobium strain ORS 1088 Will be used for inoculating the
five A. albida grafted on the A. seyal provenance.
The Rhizobium strain ORS 1088 Will be used for inoculating the
A, seyal provenance.
The Fbiglobosa Will serve as reference trop.
NUMBER OF REPLICATES : 8
NUMBER OF POTS : 12 treatments x 8 reps = 96
CALCULATING THE NITROGEN REQUIREMENT
Amount of AS with 10.09% ‘“N a.e. Will be applied at
200 mg Nipots.
Total N requirement : 200 mg N x 96 = 19200 mg N
Total AS requirement : 19200 x 100/21.2 = 90566.0377 mg AS
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Volume of solution needed : 50(ml/pot) x 96 = 4800 ml
We Will use 5000 ml because of spillage. Thus, the required
amount of AS for 5000 ml is :
90566.0377 x 5000/4800 = 94339.62 mg AS
Dilution
ml + m2 = 94339.62
M l = 132.3338 g/mole of AS 10.09% ’ 5N a.e.
M2 = 132 g/mole of ordinary AS
a’ = 10% ’ 5N a.e. desired in final dilution
a’1 = 10.09% ’ 5N a.e. of AS to be diluted
Then,
94339.62 x 132.3338 x10
ml =
- - - - - - - - - - - - - - - - - - - - - - - - - - - - -
(132.3338 x 10.09) + (132.3338 - 132) x 10
m l = 93264.98 mg AS
m2 = 1074.63 mg AS
Summa ry
m l = 9.32 g of AS with 10.09% ’ 5N a.e.
m2 = 1.07 g of ordinary AS
Total volume of solution : 500 ml
Number of pots : 96
Application rate : 50 ml/pot
Remaining solution : 200 ml.
THIRD EXPERIMENT : FIELD EXPERIMENT
DESCRIPTION AND CARRYING OlJT THE EXPERIMENT
The experiment Will be carrying out in the field from May 1993
to April 1994 with one senegalese A. albida provenance and the
P. biglobosa provenance.
TREATEMENTS
The Bradyrhizobium MAO 488 Will be used for inoculating the A.
albida provenance in the nursery before transplanting into the
field. Three, 6, 9 and 12 months after transplantation, the fixed
nitrogen Will estimated.
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The P. biglobosa Will serve as reference tree.
NUMBER OF REPLICATES : 4
NUMBER OF PLOTS :
2 provenances x 4 samplings x 4 reps = 32 plots
Plot size : 8 m x 3 m = 24 m*
Subplot size : 4 m x 2 m = 8 m2
Spacing : 2 m on the row ; lm between rows.
CALCULATING THE NITROGEN REQUIREMENT
Amount of AS with 10.09% ’ 5N a.e. Will be applied at 20 KgN/ha,
Le. 16 gN/subplot.
Total N requirement : 16 g N x 32 = 512 g N
Total AS requirement : 512 x 100/21.2 = 2415.09 g AS
Vol. of solution needed :(500 ml/m’) x 10 x 32 = 160000ml =16Ol
We shall use 170 I because of spillage. Thus, the required amount
of AS for 100 I is :
2415.09 X 170/160 = 2566.03 g AS = 2.6 Kg AS.
Dilution
ml + m2 = 2.6 Kg AS
M l = 132.3338 g/mole of AS 10.09% ’ 5N atom excess
M2 = 132 g/mole of ordinary AS
a’ = 5% l 5N a.e. desired in final dilution
a’1 = 10.09% ’ 5N a.e. of AS to be diluted
Then,
2.6 x 132.3338 x 5
ml =
____-----------------------------
(132.3338 x10.09) + (132.3338 - 132) x 5
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m l = 1.29 Kg AS,
m2 = 1.31 Kg AS
Summary
m l = 1.30 Kg of As with 10.09% 15N a.e.
m2 = 1.30 Kg of ordinary AS
Total volume of solution : 170 I
Number of subplots : 32
Application rate : 500 mlkubplot
Remaining solution : 10 I
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