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Question

The doubling time for a bacterial population is 60 minutes. Given a density of 35 cells/ml in a population in its exponential growth phase and assuming unlimited resources, the number of hours that the population will take to grow to 560 cells/ml is ________.

Bacterial Population Growth Calculation

This problem involves calculating the time required for a bacterial population to grow from an initial density to a final density, given its doubling time during exponential growth.

Identify Given Information

  • Initial Density ($N_0$): 35 cells/ml
  • Final Density ($N$): 560 cells/ml
  • Doubling Time ($t_d$): 60 minutes
  • Growth Phase: Exponential
  • Resources: Unlimited

Growth Formula

The formula for exponential growth based on doubling time is:

$ N = N_0 \times 2^{(t/t_d)} $

Where:

  • $N$ is the final population density.
  • $N_0$ is the initial population density.
  • $t$ is the time elapsed.
  • $t_d$ is the doubling time.

Calculate Time (t)

  1. Determine the number of doublings: First, find the ratio of the final density to the initial density. $ \frac{N}{N_0} = \frac{560 \text{ cells/ml}}{35 \text{ cells/ml}} = 16 $ This ratio represents the factor by which the population has increased.
  2. Relate ratio to doublings: Since the population doubles each cycle, we need to find how many doublings result in a factor of 16. $ 2^x = 16 $ Where $x$ is the number of doublings. We know that $2^4 = 16$, so $x = 4$.
  3. Calculate total time: The population has doubled 4 times. Since each doubling takes 60 minutes: $ t = \text{Number of doublings} \times t_d $ $ t = 4 \times 60 \text{ minutes} $ $ t = 240 \text{ minutes} $
  4. Convert to hours: To express the time in hours, divide by 60. $ t (\text{hours}) = \frac{240 \text{ minutes}}{60 \text{ minutes/hour}} = 4 \text{ hours} $

Conclusion

The calculated time is exactly 4 hours. This value falls within the specified range of 3.9 to 4.1 hours, confirming the calculation.

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Important Questions from Population growth curves

  1. The population of whirligig beetles in a lake grows or declines exponentially i.e. 

    $N(t) = N(0)e^{rt}$ 

    where $N(t)$ is the population size at time $t$, $N(0)$ is the initial population size and $r$ is the per capita rate of population change, occurring only due to birth and death. 
    A researcher tracks population sizes for a year and finds the following:

    Time intervalNumber of beetles at startNumber of beetles at end
    January - March1000150
    April – June1503013
    July – September3013100
    October - December1002009

    Assuming that the individual birth rates remain constant throughout the year and only death rates are affected, which one or more of the following statements is/are true? 
    (In your calculations, round off the birth and date rates to two decimal places)

  2. The population size at which net recruitment is the highest is also when the greatest amount can be harvested, while ensuring the long-term survival of the population. The amount harvested at this population size is known as
  3. The graphs shown represent the relationship between population size ($N$) and population growth rate ($\frac{dN}{dt}$). Which one of the following growth curves represents a density-dependent population that experiences a strong Allee effect?

  4. Overfishing reduced food availability for sea lions in California, causing a decline in their population size. In 1972, under the US Endangered Species Act, fishing was banned from sea lion foraging areas. Subsequently, the population of sea lions increased in a logistic form as shown in the figure.

    The per capita growth rate is highest in the interval __________ and the population growth rate is highest in the interval __________

  5. Consider the logistic population growth model, given by $$ \frac{dn}{dt} = rn \left(1 - \frac{n}{k}\right) $$ where $r$ is the intrinsic growth rate, $n$ is the population size and $k$ is the carrying capacity. Which one or more of the following is/are assumption(s) of the model?

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