Lab.7 determination of different organic matterPresentation Transcript
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 decompositionMaterials 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)
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.
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
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 by1. respiration (as CO2)2. burning3. 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
• 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 enzymes2. 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)
Atmospheric CO2CO2 from Assimilation ofdegradation CO2 by plantsof lignin CO2 from plant and animal respiration Glucose from degradation of Fungal mycelia cellulose transferred to fungivores like insect larvae, ants, and squirrels
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
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.
After titration calculation is done by the following steps:we designate the letter (X) for the (ml) of NaOH that reacted with CO2 in controlledtest 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 differenttest 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 CO2form decomposed of organic compounds.Z = Y – X = (? ) ml of NaOH purely reacted with released CO2 from decomposition ofstudied organic compoundAmount of CO2 released from = Volume of NaOH that reacted with CO2 = CO2 (mg)organic compound decomposition
Amount of CO2 (mg) = Equivalent weight × Z(1?) = (2?)Equivalent weight (CO2)= Molecular weight / equivalent= 12+ 2×16 / 2= =44/ 2= 22Amount of CO2 (mg) = Equivalent weight × Z = ? 22 × (2?) = (3?) CO2 CM.wt. 44 12Mg (3?) XX= (3?)×12 / 44= (4?) mg of C that released from the 1st week and so on for the nextweek.Then at the end of three weeks carbon (C) measurements draw a diagram showing Cmg and time as follow:
Starch Glucose Peptone Cellulose1 2 3 Time by week
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 1.2CO2 efflux by soil microorganisms, mean (mean=3) respiration among different polypeptides and amino acids different in different time intervals (30 days).
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 7Cumulative C mineralized (mean; n = 3) in different humidity conditions of soils, at10°C and in different durations.
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 7Cumulative C mineralized (mean; n = 3) in different humidity conditions of soils,at 15°C and in different durations.