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GATE BT-2017 Answer Key | Solutions Discussion
(02-15-2017, 04:40 PM)SunilNagpal Wrote:
(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
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 .
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Messages In This Thread
RE: GATE BT-2017 Answer Key | Solutions Discussion - by Akshay shinde - 02-15-2017, 04:45 PM
A GATE BT Marks Survey Page - by reviewer21 - 02-28-2017, 05:31 AM
RE: A GATE BT Marks Survey Page - by SunilNagpal - 03-01-2017, 02:09 AM
RE: A GATE BT Marks Survey Page - by CHANDA7 - 02-28-2017, 04:37 PM
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