t /g = (Log Nt – Log N0) /0.301   I introduce a loopful of Escherichia coli cells (say, 1000) into 10 mL of Nutrient Broth at 8 p.m. the night before your lab. The cells were taken from a culture plate (Nutrient Agar) held at 37°C, and inoculated into broth at the same temperature. They were held at 37°C overnight in a shaking water bath. At what time would the culture reach the Stationary Phase? Recall that doubling time under optimal conditions (these are) is 20 minutes.

Human Anatomy & Physiology (11th Edition)
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Author:Elaine N. Marieb, Katja N. Hoehn
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t /g = (Log Nt – Log N0) /0.301

 

  1. I introduce a loopful of Escherichia coli cells (say, 1000) into 10 mL of Nutrient Broth at 8 p.m. the night before your lab. The cells were taken from a culture plate (Nutrient Agar) held at 37°C, and inoculated into broth at the same temperature. They were held at 37°C overnight in a shaking water bath. At what time would the culture reach the Stationary Phase? Recall that doubling time under optimal conditions (these are) is 20 minutes.

 

  1.  A growing bacterial culture has 10,000 CFU/mL at noon and 10,000,000 CFU/mL at 6 p.m. What is the generation time under these conditions? What are your assumptions?

 

  1. At midnight you inoculate 10 mL of a culture of Enterococcus with 103 cells/mL into 990 mL of the same medium, held under the same conditions as the original culture. At what time would the culture reach 107 cells/mL? Assume exponential growth over the period. Assume that g=half an hour. Note: We worked a different variant of this problem in lecture.

 

 

Bacterial Killing Problems

 

Disinfectants, be they heat or radiation or chemicals, usually kill a constant proportion of the cells present per unit time if the cells are all equally susceptible. Hence, the term ‘log kill.’ The probability of death is constant over time given a ‘single hit.’ The more cells present at the start, the greater the number killed per unit time and the longer it will take to kill them ‘all.’ One decimal reduction is one log place (exponent).

 

            Log Nt = Log N0 – t/D.   D is the time required for a log kill.

 

  1. A suspension of 105 CFU/mL of Salmonella is treated with bleach, and 90% of the cells are dead after 10 minutes. How many viable cells per mL would you expect after 30 minutes? Yes, the 30 minutes here implies 3 logs of killing. This is an excellent problem to use with the Death Equation because you can reason it through first, and then solve it using the equation.

 

  1.  A single dose of mouthwash is added to a bacterial culture with 10 million cells/mL and, after 30 minutes, one hundred thousand viable cells/mL remain. How many would survive at the end of 1.5 hours? Based on the equation, can you ever totally decimate the population? In a practical sense, when would you assume that all the cells are dead?

 

 

  1. You are the Water Sanitation Engineer for Waterbury. Levels of bacterial cells in the water supply must be undetectable, which, in terms of the equation, is <1 per mL of drinking water. A single addition of chlorine yields a log kill every 5 minutes. During a storm, levels in the incoming water reach 10,000/mL. You hold the incoming water in a treatment tank and give it a single hit with chlorine. If you add the chemical at 10 a.m., when will it be safe to open the valve and release that water for drinking? (Recall that 10 raised to the 0 power is 1.) 
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