​UV mutagenesis was performed to isolate mutants of the lacZ gene. The mutation rate of this gene is 1 × 10⁻⁴ per cell division. Assuming that an E. coli culture was initiated from a culture density of 1 × 10² cells and grown to a density of 1 × 10⁶ cells, how many lacZ mutants are expected in this population?​

Correct option is C

The mutation rate of the lacZ gene is 1 × 10⁻⁴ per cell division. Given that the culture started with 1 × 10² cells and grew to a density of 1 × 10⁶ cells, the number of cell divisions is:

​1×10⁶ cells - 1×1021 × 10^21×10² cells = 9.9×1059.9 × 10^59.9×10^$5$ cell divisions.​

Now, the expected number of mutants can be calculated using the formula:

$\text{Expected mutants}=\text{Mutation rate}\times \text{Number of cell divisions }\times \text{Initial cell density}$

Expected mutants=(1×10⁴)×(9.9×10⁵)×(1×10^$2$$)=99$
Thus, the number of lacZ mutants expected in the population is approximately 100.

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Information Booster:

• The mutation rate of a gene is the likelihood that a mutation will occur during each cell division.

• In this case, the mutation rate for the lacZ gene is 1 × 10⁻⁴ per cell division.

• The initial number of cells was 1 × 10², and after growing to 1 × 10⁶ cells, the cell population increased by $9.9\times 10$^$5$

• The total number of mutants is calculated by multiplying the mutation rate by the number of cell divisions and the initial cell density.

• The formula to calculate the expected number of mutants is:
  $\text{Expected mutants}=\text{Mutation rate}\times \text{Number of cell divisions }\times \text{Initial cell density}$

• The expected result is a value near 100, meaning approximately 100 mutants are expected in the population.