Binary Fission & Generation Time - Microbiology

1. If a single bacterial cell undergoes binary fission, how many generations are required to produce a population of at least 25,000 cells?
   Answer: 15 We are looking for the smallest integer n (number of generations) such that $1 \times 2^n \ge 25,000$. We can solve this by testing powers of 2. $2^{10} = 1024$. $2^{12} = 4096$. $2^{13} = 8192$. $2^{14} = 16384$. This is not enough. $2^{15} = 32768$. This is the first power of 2 that exceeds 25,000. Therefore, 15 generations are required. Using logarithms gives $n = \log_2(25000) \approx 14.6$, which means 15 full generations are needed to surpass the target.
2. The mathematical model for exponential growth, $N_t = N_0 \times 2^n$, is most accurate during the log phase. Which of the following factors is the primary reason this model becomes invalid as the culture enters the stationary phase?
   Answer: The rate of cell division equals the rate of cell death due to nutrient limitation and waste accumulation. When you encounter questions about bacterial growth models, focus on understanding what changes as cultures transition between growth phases. The exponential growth equation $N_t = N_0 \times 2^n$ assumes unlimited resources and constant doubling, which only occurs during log phase. The correct answer is D because the stationary phase is defined by the equilibrium between cell division and cell death. As nutrients become depleted and toxic waste products accumulate, the culture can no longer sustain exponential growth. The birth rate equals the death rate, resulting in a stable population size rather than exponential increase. This fundamentally breaks the mathematical model, which assumes continuous doubling. Let's examine why the other options are incorrect. Option A is wrong because bacterial cells continue dividing by binary fission throughout all growth phases - the mechanism doesn't change. Option B incorrectly suggests that synchronous division is required for the exponential model. In reality, even asynchronous division can produce exponential growth as long as the average generation time remains constant. Option C describes sporulation, which is a specialized stress response that doesn't occur in all bacterial species and isn't the primary factor limiting growth in typical laboratory cultures. For microbiology exams, remember that growth phase transitions are driven by environmental factors, not changes in cellular mechanisms. When you see questions about mathematical models failing, think about what environmental conditions would violate the model's assumptions - usually resource limitation or waste accumulation that prevents the assumed constant growth rate.
3. A strain of Listeria monocytogenes has a generation time of 90 minutes at refrigeration temperature (4°C). If a food product is contaminated with 100 cells of this bacterium, approximately how many cells will be present after one week (7 days) of storage?
   Answer: $5.2 \times 10^{35}$ cells When you encounter bacterial growth problems, you're dealing with exponential growth calculations. The key is understanding that bacteria multiply by binary fission, doubling their population each generation. To solve this, you need three pieces of information: initial cell count (100 cells), generation time (90 minutes), and total time (7 days = 10,080 minutes). First, calculate the number of generations: 10,080 ÷ 90 = 112 generations. Then apply the exponential growth formula: Final count = Initial count × $2^{number of generations}$. So: 100 × $2^{112}$ = 100 × 5.2 × $10^{33}$ = 5.2 × $10^{35}$ cells. This confirms answer A is correct. Looking at the wrong answers: Answer B (1.2 × $10^32$) represents a calculation error, likely from miscounting generations or making an arithmetic mistake with the exponents. Answer C (1.9 × $10^18$) suggests a significant underestimation of either the time period or generation count - perhaps confusing hours with days or using an incorrect generation time. Answer D (5.0 × $10^12$) indicates a major calculation error, possibly treating this as linear rather than exponential growth. This problem highlights why Listeria monocytogenes is particularly dangerous - it's one of the few pathogens that continues growing at refrigeration temperatures. For microbiology exams, always remember that exponential growth problems require careful attention to units (convert everything to the same time unit) and that even small initial contamination can lead to massive populations given enough time and favorable conditions.
4. A sample of pasteurized milk is found to contain $1.0 \times 10^3$ CFU/mL of a contaminating psychrotrophic bacterium. The milk is stored at a temperature that permits a generation time of 2 hours for this organism. How much time must elapse before the bacterial concentration reaches $4.096 \times 10^6$ CFU/mL, a level considered unsafe?
   Answer: 24 hours First, determine the number of generations (n) required to reach the final concentration. Use the formula $N_t = N_0 \times 2^n$, which can be rearranged to $2^n = N_t / N_0$. $2^n = (4.096 \times 10^6) / (1.0 \times 10^3) = 4096$. To solve for n, take the base-2 logarithm: $n = \log_2(4096)$. Since $2^{10} = 1024$ and $4096 = 4 \times 1024 = 2^2 \times 2^{10}$, then $n = 12$. The total time (t) is the number of generations multiplied by the generation time (g): $t = n \times g = 12 \text{ generations} \times 2 \text{ hours/generation} = 24$ hours.
5. A bacterial culture growing exponentially increases in population from $5.0 \times 10^4$ cells/mL to $3.2 \times 10^6$ cells/mL over a period of 180 minutes. What is the mean generation time (g) of this bacterium under these conditions?
   Answer: 30 minutes First, determine the number of doublings (generations, n). The population increased by a factor of $(3.2 \times 10^6) / (5.0 \times 10^4) = 64$. We need to solve $2^n = 64$. By inspection or logarithms, $n=6$, since $2^6 = 64$. Second, the generation time (g) is the total time (t) divided by the number of generations (n). Here, $t = 180$ minutes. Therefore, $g = t / n = 180 \text{ minutes} / 6 \text{ generations} = 30$ minutes/generation.
6. The equation for calculating the number of generations (n) of microbial growth is $n = (\log_{10}N_t - \log_{10}N_0) / \log_{10}2$. Why is the term $\log_{10}2$ present in the denominator of this equation?
   Answer: It accounts for the fact that binary fission produces two daughter cells from one parent cell. The fundamental model of bacterial growth is $N_t = N_0 \times 2^n$, where the base '2' signifies that the population doubles with each generation due to binary fission. To solve for n, we take the logarithm of both sides: $\log(N_t) = \log(N_0) + n \log(2)$. Rearranging gives $n = (\log N_t - \log N_0) / \log 2$. The $\log 2$ term is a direct consequence of the doubling nature of the growth process (base 2) and is used to convert the change in population magnitude (expressed in base 10) into the number of doubling events (generations).
7. A microbiologist is studying two different bacterial species. Species A has a generation time of 30 minutes. Species B has a generation time of 45 minutes. If two separate, identical flasks are inoculated with the same initial number of cells of each species, what will be the ratio of the population of Species A to Species B ($N_A/N_B$) after 3 hours of exponential growth?
   Answer: 4.0 to 1 First, calculate the number of generations for each species over the 3-hour (180-minute) period. For Species A, $n_A = 180 \text{ min} / 30 \text{ min} = 6$ generations. For Species B, $n_B = 180 \text{ min} / 45 \text{ min} = 4$ generations. Let the initial number of cells be $N_0$. The final populations will be $N_A = N_0 \times 2^{n_A} = N_0 \times 2^6$ and $N_B = N_0 \times 2^{n_B} = N_0 \times 2^4$. The ratio is $N_A / N_B = (N_0 \times 2^6) / (N_0 \times 2^4) = 2^6 / 2^4 = 2^{(6-4)} = 2^2 = 4$. So the ratio is 4 to 1.
8. After 4 hours of incubation in a rich medium, a bacterial culture reached a density of $1.28 \times 10^7$ cells/mL. Assuming the culture was in exponential phase for the entire duration and the generation time was 30 minutes, what was the initial cell density ($N_0$) in cells/mL?
   Answer: $5.0 \times 10^4$ cells/mL First, calculate the number of generations (n). The total time is 4 hours = 240 minutes. The generation time (g) is 30 minutes. So, $n = 240 / 30 = 8$ generations. The growth formula is $N_t = N_0 \times 2^n$. To find the initial density, rearrange the formula to $N_0 = N_t / 2^n$. Substitute the known values: $N_0 = (1.28 \times 10^7) / 2^8$. Since $2^8 = 256$, the calculation is $N_0 = (1.28 \times 10^7) / 256 = (128 \times 10^5) / 256 = 0.5 \times 10^5 = 5.0 \times 10^4$ cells/mL.
9. A pure culture of Staphylococcus aureus is grown for 2 hours, reaching a density of $2.0 \times 10^8$ CFU/mL. A dose of penicillin is then added that causes rapid lysis of 99.9% of the population. The surviving cells, which are penicillin-resistant, immediately resume growth with a generation time of 40 minutes. What is the approximate CFU/mL two hours after the addition of penicillin?
   Answer: $1.6 \times 10^6$ CFU/mL This is a multi-step problem. First, calculate the number of surviving cells after penicillin addition. A 99.9% kill rate means 0.1% (or 0.001) of the cells survive. Survivor population = $(2.0 \times 10^8) \times 0.001 = 2.0 \times 10^5$ CFU/mL. This is the new starting population ($N_0$). Second, calculate the number of generations in the next phase of growth. The time is 2 hours = 120 minutes, and the generation time is 40 minutes. Number of generations $n = 120 \text{ min} / 40 \text{ min} = 3$. Third, calculate the final population: $N_t = N_0 \times 2^n = (2.0 \times 10^5) \times 2^3 = (2.0 \times 10^5) \times 8 = 1.6 \times 10^6$ CFU/mL.
10. A microbiologist inoculates a flask of sterile broth with 100 cells of Escherichia coli. The culture is then incubated at 37°C. If the generation time of this E. coli strain under these conditions is 20 minutes, what will be the approximate cell population after 1.5 hours of exponential growth?
    Answer: 2,260 cells The calculation requires three steps. First, convert the total time to be in the same units as the generation time: 1.5 hours = 90 minutes. Second, calculate the number of generations (n) by dividing the total time by the generation time (g): n = 90 min / 20 min/generation = 4.5 generations. Third, use the formula for exponential growth, $N_t = N_0 \times 2^n$, where $N_t$ is the final cell number and $N_0$ is the initial cell number. $N_t = 100 \times 2^{4.5} = 100 \times (16 \times \sqrt{2}) \approx 100 \times 22.62 \approx 2262$ cells.
11. A bacterium has a generation time of 20 minutes in nutrient broth and 60 minutes in a minimal salts medium. A flask containing $10^3$ cells in minimal medium is incubated for one hour. The cells are then transferred to nutrient broth and incubated for another hour. What is the final number of cells?
    Answer: $1.6 \times 10^4$ When you encounter bacterial growth calculations, you're dealing with exponential growth where cells double at regular intervals called generation times. The key is tracking how many doublings occur in each phase and applying them sequentially. Let's work through this step-by-step. You start with $10^3$ cells in minimal medium (60-minute generation time) for one hour. Since the generation time equals the incubation time, exactly one doubling occurs: $10^3 \times 2^1 = 2 \times 10^3$ cells. Next, these $2 \times 10^3$ cells are transferred to nutrient broth (20-minute generation time) for another hour. In 60 minutes with 20-minute generations, you get $60 ÷ 20 = 3$ doublings. So: $2 \times 10^3 \times 2^3 = 2 \times 10^3 \times 8 = 1.6 \times 10^4$ cells. Answer A ($1.6 \times 10^4$) is correct. Answer B ($8.0 \times 10^3$) represents a common error of forgetting the first doubling in minimal medium—calculating only the nutrient broth phase from the original $10^3$ cells. Answer C ($4.0 \times 10^3$) occurs if you miscalculate the doublings in nutrient broth as just two instead of three. Answer D ($3.2 \times 10^4$) results from incorrectly treating both phases as having the faster 20-minute generation time. Remember: always calculate growth phases sequentially, and double-check your division when determining the number of generations. The output of one phase becomes the input for the next.
12. Using a spectrophotometer, a student observes that a bacterial culture in log phase takes exactly 66 minutes to increase its optical density (OD) from 0.2 to 0.8. Assuming OD is directly proportional to cell density, what is the generation time for this culture?
    Answer: 33 minutes When you encounter bacterial growth problems involving optical density changes, you're dealing with exponential growth kinetics. The key insight is that during log phase, bacterial populations double at regular intervals called generation times. Since OD is proportional to cell density, you can treat OD values as direct measures of population size. The culture grew from OD 0.2 to OD 0.8 in 66 minutes. Notice that 0.8 ÷ 0.2 = 4, meaning the population quadrupled during this time period. In exponential growth, quadrupling represents exactly two doublings: the first doubling takes the population from 0.2 to 0.4, and the second doubling takes it from 0.4 to 0.8. Since two generation times occurred in 66 minutes, each generation time must be 66 ÷ 2 = 33 minutes. Looking at the wrong answers: Answer B (22 minutes) would result from incorrectly dividing 66 by 3, perhaps misunderstanding how many doublings occurred. Answer C (16.5 minutes) comes from dividing by 4 - this might happen if you confused the fold-increase (4×) with the number of doublings. Answer D (66 minutes) represents the total time elapsed, not the generation time, which would be the trap for students who don't recognize that multiple doublings occurred. Remember this pattern: always determine how many doublings occurred by finding the ratio of final to initial values, then expressing that ratio as a power of 2. The generation time equals total time divided by the number of doublings.
13. Which of the following statements most accurately describes the fate of the original parent cell after binary fission is complete?
    Answer: The parent cell enlarges and then divides equally to form two new daughter cells, ceasing to exist as a separate entity. Binary fission is a process of asexual reproduction where a single cell divides into two identical daughter cells. The parent cell elongates, replicates its DNA, and then the cytoplasm divides, resulting in two cells. The original parent cell does not remain; it is effectively transformed into the two daughter cells. This process makes the concept of a 'parent' and 'offspring' different from multicellular organisms and leads to the 'immortality' of the cell line under ideal conditions.
14. A single cell of a pathogenic bacterium divides by binary fission every 40 minutes inside a host. If the infectious dose for this pathogen is approximately $10^6$ cells, how long after the initial infection will this threshold be reached, assuming ideal growth conditions and no host immune response?
    Answer: Approximately 13 hours When you encounter bacterial growth problems, you're dealing with exponential growth through binary fission, where one cell becomes two, two become four, and so on. The key is using the exponential growth formula: $N = N_0 \times 2^{n}$, where N is the final number of cells, N₀ is the initial number, and n is the number of generations. Starting with 1 cell and needing to reach $10^6$ cells, we solve: $10^6 = 1 \times 2^{n}$. Taking log₂ of both sides: $n = \log_2(10^6) ≈ 19.93$ generations. Since we can't have partial generations in this context, we round to 20 generations. With each generation taking 40 minutes, the total time is: $20 × 40 = 800$ minutes, which equals 13.33 hours, or approximately 13 hours. Answer D (approximately 13 hours) is correct. Answer A (8 hours) represents only 12 generations, yielding about 4,000 cells—far short of the infectious dose. Answer B (27 hours) calculates to about 40 generations, producing over $10^{12}$ cells—vastly exceeding the threshold. Answer C (20 hours) represents 30 generations, yielding about $10^9$ cells, which overshoots by a factor of 1,000. For microbiology exams, remember that bacterial growth problems always involve powers of 2. Practice converting between logarithms and exponents, and always double-check your time unit conversions. These calculation errors are common traps in growth kinetics questions.

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