The following data were obtained from pour plates used to test the effectiveness of a food preservative. Two samples of cottage cheese were inoculated with bacteria; the preservative was added to one sample. After incubation, samples of the cottage cheese (control) and samples treated with the preservative (experimental) were diluted and plated on nutrient agar. Calculate the number of bacteria. Was the preservative effective

The colony-forming unit includes the fungal or the bacterial cells and can be used for the cell counting. The colony forming unit is used for the viable cells.

The cfu/ ml can be calculated as follows:

cfu/ml=(no. of colonies × dilution factor)/volume of culture plate

In control:

We have given, the no. of colonies=160

, dilution factor=400  and volume=1ml

The cfu of the control is

cfu/ml(control)=(160 × 400)/1=64000 cfu/ml

For the experimental data,

We have given the no. of colonies=32

, dilution factor=200  and volume=0.1 ml

.

cfu/ml=32 × 200/0.1=64000 cfu/ml

.

Since, the cfu of the control and the experiment is equal, the preservative is of no use.

an active site is a region that is present on an enzyme where the reactants or the substrate molecules we can say bind and experience a chemical reaction. the substrate is basically a reactant whose concentration is changing and that is converting into a product after binding at the active site of an enzyme

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explanation:

ok