E. coli, to be replicated, is placed in a medium containing radioactive thymidine for five minutes. Then it is made to replicate in a normal medium. Which of the following observation will be correct ?
Both strands of DNA will be radioactive.
One strand of DNA will be radioactive.
Each strand of DNA will be half radioactive.
None of the strands of DNA will be radioactive.
A
Both strands of DNA will be radioactive.
B
None of the strands of DNA will be radioactive.
C
One strand of DNA will be radioactive.
D
Each strand of DNA will be half radioactive.
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Solution
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Matthew Meselson and Franklin Stahl demonstrated the semi-conservative mode of replication in DNA and chromosomes in E.coli cells. After the incorporation of radioactive thymidine 3H, the cells were transferred to growing medium. Radioactive chromosomes (with labelled DNA having 3H) appeared in the form of scattered black dots of silver grains. After replication of DNA and constitution of chromosomes following facts were noticed: (a) It was found in first generation that radioactivity was uniformly distributed in both the chromosomes. In such cases original strand of DNA double helix was labelled with 3H and newly formed strand was non- labelled. (b) During second division only one of two chromosomes represented radioactivity by showing one strand radioactive (original) and other one non-radioactive (newly formed). Hence in E.coli, after the second division, only one strand of DNA will be radioactive.
Therefore, the correct answer is option B.
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