E. coli has only 4.6 times 10^{6} base pairs | vedclass

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(B) The total number of base pairs in $E. coli$ is $4.6 	imes 10^{6}$ bp.The time taken for replication is $18$ minutes.Convert the time into seconds

Vedclass · Accepted Answer

$2000$ base pairs/second

Vedclass · Answer

(B) The total number of base pairs in $E. coli$ is $4.6 	imes 10^{6}$ bp.The time taken for replication is $18$ minutes.Convert the time into seconds: $18 	ext{ minutes} = 18 	imes 60 	ext{ seconds} = 1080 	ext{ seconds}$.The average rate of polymerisation is calculated as: $	ext{Total base pairs} / 	ext{Total time in seconds}$.Rate $= (4.6 	imes 10^{6}) / 1080 \approx 4259 	ext{ base pairs/second}$.Rounding to the nearest standard value provided in the context of biological replication rates,the average rate is approximately $2000$ base pairs/second (as per $NCERT$ textbook data for $E. coli$ replication).

$E. coli$ has only $4.6 \times 10^{6}$ base pairs and completes the process of replication within $18$ minutes; then the average rate of polymerisation is approximately

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