To solve the problem of finding the number of cells in the 5th generation from a single bacterium population where the 20th generation has 1,048,576 cells, we need to understand the concept of exponential growth. In bacterial growth, each generation typically doubles the population. Thus, the number of cells after n generations can be described by the equation: \( N = N_0 \times 2^n \) where \( N \) is the number of cells after n generations, and \( N_0 \) is the initial number of cells. Given the final population (after the 20th generation) is 1,048,576, we know:
\( N = 1, \! 048, \! 576 = 1 \times 2^{20} \)
This confirms that the population size indeed doubled 20 times starting from a single bacterium.
To find the number of cells in the 5th generation, we apply the same formula:
\( N = 1 \times 2^5 = 32 \)
Thus, after the 5th generation, there are 32 cells.
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) 