How long would it take
e. coli to replicate its entire genome (4.2 x 106 bp), assuming a replication rate of 1,000 nucleotides per second at each fork with no pauses?

Solution:
4.2 x 10^6 bp/10^3 bp/seconds         = 4.2 + 103 s which is 4200 seconds and equivalents to 70 minutes

In addition, assuming a pause of 2 seconds for re initiating after completing every okazaki fragment and assuming the okazaki fragments average 1000 nucleotide long.  
4.2 x 10^6 bp/10^3 bp                        = 4200 okazaki fragments                           4200 * 2 seconds                               = 8400 seconds which is 140 minutes or 2 hours 20 minutes of pauses alone.                                                                        

Therefore, overall time would be pauses plus the 70 minutes so total time of 210 minutes. Assuming that the replisome completely disassociates from the DNA after every okazaki fragment and must spend one-minute rebinding.

4200 okazaki fragments. 60 seconds rebinding time per fragment:                        4200 x 1 minute = 4200 minutes rebinding time plus 70 minutes for actual replication. 4200 minutes is 70 hours which is almost 3 days. 

Answer:

it provides an alternative energy pathway for the reaction to proceed.

explanation:

a catalyst is a substance that speeds up the rate of a chemical reaction but is not consumed during the course of the reaction.

catalysts increase the rate of reaction without being used up. they do this by lowering the activation energy needed. with a catalyst, more collisions result in a reaction, so the rate of reaction increases. different reactions need different catalysts.

a catalyst works by providing an alternative reaction pathway to the reaction product. the rate of the reaction is increased as this alternative route has a lower activation energy than the reaction route not mediated by the catalyst.

Awell breed cat that has no defects or has never been damaged before.hope this ! mark me as brainiest!