Question

An E. coli culture was inoculated with ${}^{32}P-$ labelled phage which entered lysogenic cycle. A radioactivity measure of 1 ml sample of the culture with a concentration of $1\times {10}^{5}cell/mL$ was taken. The E. coli were then allowed to undergo 3 round of replication. The fraction of the initial radioactivity present in a 1-mL sample containing cells diluted to a final concentration of $1\times {10}^{5}cell/mL$, from the culture containing the final generation of E.coli is

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $\frac{1}{10}$
• D. $\frac{1}{3}$
• E. $\frac{1}{2}$

Answer 1

Answer: (A)

$\frac{1}{8}$

In lysogenic cycle, the virus genome enters the host genome and replicates along with it as a part of a host genome.

The radioactively labelled phage integrates itself into the host genome and starts replication as a part of the bacterial genome. Bacteria undergo replication by 2${}^n$. When the E.coli was allowed to undergo three rounds of replication the number of cells formed will be 2${}^3$= 8. Each replication reduces the amount of radioactivity by 1/2${}^n$. So the reduced radioactivity will be 1/8 times to that of the parent cell.