02-15-2017, 04:45 PM
(02-15-2017, 04:40 PM)SunilNagpal Wrote:Thank you sir for the solution . I believe it is a simple numerical as I guess there is no other possible approach for this numerical .(02-15-2017, 04:23 AM)Akshay shinde Wrote: One hundred E coli cells are infected by a single lambda phage particle. the ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80 , the number of free phage particles released after on round of infection is?
I am sorry if I'm missing some crucial information regarding lytic and lysogenic cycles (which should have been used for approaching this question), I am rather assuming it as a simplistic question....
Given the 4:1 for lysogenic : lytic phages, it is clear than 20 would be lytic phages.
Now each E.coli will burst once 80 particles have been accumulated.
So total free particles should be 20x80 = 1600