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.
Hello there,
Your answer would be "Dissolved Oxygen"
Hope this helps
~HotTwizzlers
Answer:
The answer is B. from east to west
Answer;
B. Water purification programs
Explanation;
The goal of hygiene promotion is to help people to understand and develop good hygiene practices, so as to prevent disease and promote positive attitudes towards cleanliness.
Hygiene promotion encompasses a systematic attempt to adequately promote personal, domestic, environmental and food hygiene practices that prevent or mitigate the transmission of diseases. Water purification programs is among the government programs that encourages hygiene practices.
In cells with a nucleus, as in eukaryotes, the cell cycle is also divided into three periods: interphase, the mitotic (M) phase, and cytokinesis. During interphase, the cell grows, accumulating nutrients needed for mitosis, preparing it for cell division and duplicating its DNA.
