Soil viable count, an anaerobe that will not grow, and a medium that supports nothing
In a Class 11 practical examination a student is given 1 g of garden soil and asked to estimate the total viable bacterial count. She suspends the soil in 9 mL of sterile water, makes ten-fold dilutions up to 10⁻⁶, and spreads 0.1 mL of the 10⁻⁴, 10⁻⁵ and 10⁻⁶ dilutions on nutrient agar in duplicate. After 48 hours at 37 °C the mean colony counts are 480, 48 and 4 respectively. She reports the count obtained from the 10⁻⁴ plate. In a second experiment she tries to isolate Clostridium from the same soil by streaking on nutrient agar and incubating in air at 37 °C; no colonies appear. In a third, she prepares a synthetic medium containing only glucose, ammonium chloride, potassium phosphate and trace salts, inoculates it with the same soil suspension, and again gets no growth, even though the same suspension grew well on nutrient agar.
1. Which plate should be used for the calculation? Calculate the viable count per gram of soil and explain why the other two plates are rejected.
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Choice of plate. The countable range for a spread plate is roughly 30–300 colonies. The 10⁻⁴ plate (480 colonies) is too crowded: colonies merge, compete for nutrients and suppress slow growers, so it undercounts. The 10⁻⁶ plate (4 colonies) is far too sparse for statistical reliability - a single pipetting or plating error changes the answer several-fold. The 10⁻⁵ plate with 48 colonies is the only one properly placed.
Calculation. 0.1 mL of the 10⁻⁵ dilution was plated. The 10⁻¹ suspension is 1 g in 10 mL, that is 0.1 g per mL, so 1 mL of the 10⁻⁵ dilution contains 0.1 × 10⁻⁴ = 10⁻⁵ g of soil. Therefore 0.1 mL contains 10⁻⁶ g.
CFU per gram = 48 ÷ 10⁻⁶ = 4.8 × 10⁷ CFU/g of soil.
State the result as colony-forming units, not cells, because clumps of bacteria and spores each give a single colony; the true cell number is higher.
2. Why did Clostridium fail to appear on nutrient agar, and how should the experiment be redesigned?
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The medium is not the problem; the atmosphere is. Clostridium species are obligate anaerobes. Their spores survive in soil, but germination and growth require a low oxidation-reduction potential, and oxygen is toxic to them because they lack the enzymes to detoxify its reduction products. Nutrient agar incubated in ordinary air is therefore the wrong environment, not the wrong food.
Redesign.
- Incubate in an anaerobic jar with a gas-generating sachet and indicator, or use freshly boiled, cooled thioglycollate broth, or a deep agar shake tube in which the organisms grow below the surface.
- Give the soil suspension a heat shock (about 80 °C for 10 minutes) before plating. This kills vegetative cells, activates Clostridium spores and greatly improves selectivity.
- Keep nutrient agar or use reinforced clostridial medium; the colonies are typically irregular and spreading.
A control jar incubated aerobically alongside is worth including, so that growth under anaerobic conditions can be confidently attributed to an anaerobe.
3. Why did the glucose–ammonium chloride medium support no growth while nutrient agar did? What does this prove about the soil population?
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Nutrient agar is a complex medium: peptone and beef extract supply amino acids, peptides, vitamins, purines and growth factors in addition to a carbon source. The synthetic medium supplies only glucose as carbon and ammonium chloride as nitrogen.
An organism that grows on the complex medium but not on the defined medium is fastidious or auxotrophic - it cannot synthesise some essential growth factor, commonly a vitamin such as biotin or nicotinic acid, an amino acid, or a growth factor, and must obtain it preformed. In a mixed soil population, only a minority of species can grow on a minimal salts medium; those that can are prototrophs, and they may still be slow growers easily outcompeted in liquid culture.
What this proves. The soil suspension grows on nutrient agar because the agar supplies the missing nutrients; failure on the synthetic medium is a statement about the nutritional requirements of the organisms, not about their absence. Adding yeast extract, soil extract or a vitamin mixture should restore growth, and that test would confirm the explanation.