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 × 10^59.9×105 cell divisions.
Now, the expected number of mutants can be calculated using the formula:
Expected mutants=Mutation rate×Number of cell divisions×Initial cell density\text{Expected mutants} = \text{Mutation rate} × \text{Number of cell divisions} × \text{Initial cell density}Expected mutants=Mutation rate×Number of cell divisions ×Initial cell density
Expected mutants=(1×104)×(9.9×105)×(1×102)=99 mutants.
Thus, the number of lacZ mutants expected in the population is approximately 100.
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 9.9×1059.9 × 10^59.9×105
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:
Expected mutants=Mutation rate×Number of cell divisions×Initial cell density\text{Expected mutants} = \text{Mutation rate} \times \text{Number of cell divisions} \times \text{Initial cell density}Expected mutants=Mutation rate×Number of cell divisions ×Initial cell densityThe expected result is a value near 100, meaning approximately 100 mutants are expected in the population.
