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Lab.7 determination of different organic matter
1.
2.
3. Subject objective: Each student should be able to
• Being able to determine which kind of organic compounds (carbohydrates, proteins,
and lipids) are more decomposed by soil microorganisms under different :
1. Temperatures (20,25 and 37°C)
2. Humidity (40%, 60% and 80% of field capacity)
• Effect of incubation time (7, 14, 21 and 28 days) on organic compound
decomposition
Materials per Group of Students:
• 1 kg of garden soil.
• 0.5gm of each (cellulose, starch, glucose, peptone, lipids, HCl (1N), NaOH (1N),
phenophthalene, (BacL2 )
• 5 beakers with 5 test tubes,(wax pencil, Bunsen burner, Oven, pipette with pipetter)
4. CARBON CYCLE```
• The concentration of carbon in living matter (18%)is
almost 100 times greater than its concentration in
the earth (0.19%).
• So living things extract carbon from their non-living
environment.
• For life to continue, this carbon must be recycled.
5. Strictly speaking the “total carbon” of the soil comes from two principal
sources:
• Inorganic carbon Carbon dioxide in the atmosphere and dissolved in water
(forming bicarbonate - HCO3, Carbonate rocks(lime stone and coral - Ca
CO3,
• Organic carbon (only slightly processed organic residues of plant and
animal origin, humus, charcoal, petroleum, fossil organic matter, Dead
organic matter, e.g., humus in the soil, microorganisms). In the majority of
methods, the gas phases present in the atmosphere of the soil (CO2 linked
with biological activity, CH4).Soil organic matter (SOM) can be of plant,
animal, or microbial origin and the terms “soil organic matter” and “humus”
are considered synonyms.
Organic matter is anything that contains carbon compounds that were
formed by living organisms. Four main components are:
• 1-dead forms of organic material - mostly dead plant parts (85%)
• 2-living parts of plants - mostly roots (10%)
• 3-living microbes and soil animals
• 4-Partly decayed organic matter is called humus
6. Organic matter is the vast array of carbon compounds in soil. Originally created by
plants, microbes, and other organisms, these compounds play a variety of roles in
nutrient, water, and biological cycles. For simplicity, organic matter can be divided
into two major categories: stabilized organic matter which is highly decomposed
and stable, and the active fraction which is being actively used and transformed by
living plants, animals, and microbes. Two other categories of organic compounds are
living organisms and fresh organic residue. These may or may not be included in
some definitions of soil organic matter.
Organic matter plays a determining role in pedogenesis and can drastically modify the
physical, chemical, and biological properties of soil (structure, plasticity, color, water
retention). The fundamental processes of evolution include phenomena of
mineralization and immobilization and, in particular, of carbon and nitrogen.
• Mineralization: allows the transformation of organic residues into inorganic
compounds in the soil, the atmosphere, and the hydrosphere, these are usable by
flora and by micro-organisms.
Carbon returns to the atmosphere by
1. respiration (as CO2)
2. burning
3. Decay (producing CO2 if oxygen is present, methane (CH4) if O2 is absent.
Immobilization: is the transformation of organic matter into more stable organic and
organomineral compounds with high molecular weights that are fixed in the interlayer
spaces of clays. These processes are summarized by the following diagram
12. • Major steps in the degradation of organic matter and
their types:
1. The dead organic matter is colonized by microbes and
degraded with help of microbial enzymes
2. Macromolecules are broken down into simpler units and
further degraded into constituent elements.
– Breakdown of compounds that are easy to decompose (e.g. sugars,
starches and proteins)
– Breakdown of compounds that may take several years to decompose
(cellulose and lignin)
– Breakdown of compounds that may take 10 years (e.g. waxes and
phenols)
– Compounds that may take 100-1000’s of years (e.g. humus like
substances, which are very complex)
13. Atmospheric
CO2
CO2 from Assimilation of
degradation CO2 by plants
of lignin CO2 from plant and
animal respiration
Glucose from degradation of
Fungal mycelia cellulose transferred to fungivores
like insect larvae, ants, and squirrels
16. Decomposition
• When organisms die and decay, the carbon
molecules in them enter the soil.
• Microorganisms break down the molecules,
releasing CO2
• Oxygenic photosynthesis:
CO2 + H2O (CH2O) + O2
• Respiration:
(CH2O) + O2 CO2 + H2O
17. Procedure:
1. After knowing the volume of water that need for obtaining 60% of soil
humidity, we add 0.5gm of different organic compound (Cellulose, glucose,
starch, peptone) to each beaker respectively, with remaining 5th beaker
without addition of organic compound it act as a control.
2. Vertically fix or put test tube containing (15ml) of NaOH (1N) in each soil
sample, then put cover on each beaker to avoid reaction of NaOH with air
CO2.
3. Incubate the samples at 25°C for 3 weeks (interval= 1 week)
4. At the end of each week we estimate volume of released CO2 form organic
compound decomposition by titrating NaCH (1N) test tube with HCl (1N)
after addition of BaCl2 and some drops of phenolphthalein as an indicator
for determination end point of reaction between HCl and NaOH by changing
their color from pink to colorless.
18. After titration calculation is done by the following steps:
we designate the letter (X) for the (ml) of NaOH that reacted with CO2 in controlled
test tube.
X=15 ml of NaOH- (?)ml of NaOH reacted with HCl= (?)
we designate the letter (Y) for the (ml) of NaOH that reacted with CO2 in a different
test tube
Y=15 ml of NaOH- (?)ml of NaOH reacted with HCl= (?)
We designate the letter (Z) for the volume of NaOH that reacted with released CO2
form decomposed of organic compounds.
Z = Y – X = (? ) ml of NaOH purely reacted with released CO2 from decomposition of
studied organic compound
Amount of CO2 released from = Volume of NaOH that reacted with CO2 = CO2 (mg)
organic compound decomposition
19. Amount of CO2 (mg) = Equivalent weight × Z(1?) = (2?)
Equivalent weight (CO2)= Molecular weight / equivalent= 12+ 2×16 / 2=
=44/ 2= 22
Amount of CO2 (mg) = Equivalent weight × Z = ?
22 × (2?) = (3?)
CO2 C
M.wt. 44 12
Mg (3?) X
X= (3?)×12 / 44= (4?) mg of C that released from the 1st week and so on for the next
week.
Then at the end of three weeks carbon (C) measurements draw a diagram showing C
mg and time as follow:
20. Starch
Glucose
Peptone
Cellulose
1 2 3
Time by week
24. 80
different amino acids and proteins
70
mg of mineralized C/ 2gm of 60
50
40
30
20
10
Incubation time (days)
0
3 6 9 12 15 18 21 24 27 30
Alanine 48 30.6 28.8 27 22 18.8 16.2 12 8.2 4.8
Lysine 46.8 31.8 27.2 22 18.6 13.8 8 4 3.1 0.9
Albomin 19.8 72.9 31.2 21.6 18 16.2 13 5.7 4.8 2.2
Peptone 69 35.4 27 24.6 18 9 5 3.8 1.4 1.2
Casein 44 25 21 15 6.3 4.9 3.8 2.9 2.1 1.2
CO2 efflux by soil microorganisms, mean (mean=3) respiration among different
polypeptides and amino acids different in different time intervals (30 days).
25. 200
100% of FC
180
80% of FC
160 60% of FC
mg of C / 2.5 gm of plant residue
40% of FC
140
120
100
80
60
40
20
Incubaction time ( weeks )
0
1 2 3 4 5 6 7
Cumulative C mineralized (mean; n = 3) in different humidity conditions of soils, at
10°C and in different durations.
26. 250
100% of FC
80% of FC
60% of FC
200
mg of C / 2.5 gm of plant residue
40% of FC
150
100
50
Incubation time ( weeks )
0
1 2 3 4 5 6 7
Cumulative C mineralized (mean; n = 3) in different humidity conditions of soils,
at 15°C and in different durations.
