The difference between autotrophic and heterotrophic bacteria lies in the ability of autotrophes to make use of a source of energy other than that of the potential of organic compounds, with which to build these from simple inorganic substances. It is not unusual to find bacteria capable of existing as both autotrophe and heterotrophe according to environmental conditions. WE shall take a look now at photo-autotrophic bacteria.
Photo-Autotrophic Bacteria: Light energy can be made to perform chemical work by living things only in the presence of the necessary pigments. These pigments are possessed by three groups of bacteria only. Photo-autotrophic bacteria are the green sulphur bacteria(chlorobacteriaceae) , the purple bacteria( Thiorhodaceae) and the purple and brown bacteria(Athiorhodaceae).
The green sulphir bacteria can assimilate carbon(IV) oxide in essentially the same way as can higher plants, but instead of using water as reducing agent or hydrogen donor for the reduction of carbon(IV) oxide, hydrogen sulphide is used. The overall synthesizing reactions of green plants and the green sulphur bacteria can be compared thus:
Green plants: 2H2O + CO2 → (CH2O) + O2 + H2O
Green Bacterium: 2H2O + CO2 → (CH2O) + 2S + H2O
The strictly anaerobic bacterium Chlorobium, abundant in the soil, mud and sulphuretted waters, is able to perform this synthesis. In the equations as written, (CH2O) merely represents a starting point for further carbohydrate synthesis. Nitrogen requirements are filled by ammonia and nitrate.
The purple sulphur bacteria perform the same reaction as the green bacteria but in this case the sulphur appears as visible granules within the cells. Should the organisms be deprived of H2S, they can use the elemental sulphur stored in the cells to reduce more carbon(IV) oxide.
2S + 8H2O + 3CO2 → 2H2SO4 + 3(CH2O) + 3H2O
The above reaction is an oxidation-reduction process which may be split into two parts as shown below.
2S + 8H2O – 12H → 2H2SO4
3CO2 + 12H → 3(CH2O) + 3H2O
Starting from H2S as the reducing agent, the carbon(IV) oxide reduction can be written as:
2H2S + 8H2O + 4CO2 → 2H2SO4 + 4(CH2O) + 4H2O
These sulphur bactreria can use other sulphur compounds such as thiosulphate for a similar purpose. An example of such a bacterium is Chromatium.
The purple and brown bacteria do not use sulphur compounds as hydrogen donors in the carbon(IV) oxide reduction process. They are not ‘sulphur bacteria.’ Instead they use organic acids and may be confused with the heterotrophic bacteria because of this. However, they can be distinguished because they grow only anaerobically and in the presence of light.