To find the number of cells in a bacterial culture after 4 hours, we start with two cells, knowing the culture doubles every 30 minutes. We calculate how many doubling periods occur in 4 hours and use this to find the total cell count.
1. **Determine Doubling Periods**: Since 4 hours equals 240 minutes, and each doubling period is 30 minutes, we have:
Number of Doubling Periods = 240 / 30 = 8.
2. **Calculate Cell Count**: The number of cells doubles every period. Therefore, after 8 periods, the number of cells is:
N = N0 × 2n, where N0 is the initial number of cells and n is the number of doubling periods.
Substitute the values:
N = 2 × 28 = 2 × 256 = 512.
3. **Validation**: The calculated number of cells, 512, falls within the given range (512 to 512), confirming its accuracy.
Thus, the number of cells after 4 hours is 512.
List I | List II | ||
|---|---|---|---|
| A | Clostridium butylicum | I | Ethanol |
| B | Saccharomyces cerevisiae | II | Streptokinase |
| C | Trichoderma polysporum | III | Butyric acid |
| D | Streptococcus sp. | IV | Cyclosporin-A |

Identify the taxa that constitute a paraphyletic group in the given phylogenetic tree.
The vector, shown in the figure, has promoter and RBS sequences in the 300 bp region between the restriction sites for enzymes X and Y. There are no other sites for X and Y in the vector. The promoter is directed towards the Y site. The insert containing only an ORF provides 3 fragments after digestion with both enzymes X and Y. The ORF is cloned in the correct orientation in the vector using the single restriction enzyme Y. The size of the largest fragment of the recombinant plasmid expressing the ORF upon digestion with enzyme X is ........... bp. (answer in integer) 