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ikadub [295]
3 years ago
11

2. Suppose that a broth culture has 52 million bacteria per milliliter. A serial dilution is made using three bottles of 99 mill

iliters each. One (1.0) mL is transferred from the sample to the first bottle (A); then 1 mL is transferred from the first bottle to the second (B); and then 1 mL is transferred from the second bottle to the third (C). From the third bottle 0.1 mL is transferred to a spread plate. How many colonies would you expect to find on this plate
Biology
1 answer:
sdas [7]3 years ago
8 0

Answer:

Approximately 5.2 colonies

Explanation:

Given that the original broth culture has 52 million bacteria/ml. = 52 * 10^6 bacteria/ml

It is diluted 100 times in bottle A = 52 * 10^4 bacteria/ml

It is diluted 100 times again in bottle B = 52 * 10^2 bacteria/ml

It is diluted 100 times again in bottle C = 52 bacteria/ml

From bottle C, 0.1 ml is plated = 5.2 bacteria

Since a single colony on plate represents a single bacteria in solution, approximately 5.2 colonies will be observed on plate.

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Answer:

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Explanation:

Reaction 1 would be the most suitable in this case. This is so because in reference to the question you are wanting to keep the food warm over long periods. So in reference to the data in the chart reaction 1 continues to heat up over long periods of time. However, with the other reactions (2, 3, 4) temperatures vary and some even decrease constantly. Therefore showing that the other reactions are invalid to help reach your goal referencing back to the question.

7 0
3 years ago
if there are 500 people in a population, and 150 are homozygous HbA/hba, 150 are homozygous hbs/hbs, and 200 are heterozygous hb
sineoko [7]

Answer:

0.547

Explanation:

Given -

Total Population - 500

Individuals with homozygous HbA/hbA genotype - 150

Individuals with homozygous hbs/hbs genotype -150

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Let us assume the given population is in Hardy Weinberg' equilibrium

The frequency of individuals in the given population with homozygous HbA/hba genotype is equal to number of individuals with homozygous HbA/hba genotype divided by total population.

\frac{150}{500} \\= 0.3

The frequency of hba allele is equal to

\sqrt{0.3} \\= 0.5477

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4 years ago
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