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GATE BT-2017 Answer Key | Solutions Discussion
(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?

What must Ben the solution for this numerical ? Kindly answer..
(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
Sunil Nagpal
MS(Research) Scholar, IIT Delhi (Alumnus)
Advisor for the Biotech Students portal (BiotechStudents.com)
Computational Researcher in BioSciences at a leading MNC


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(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 .
(02-15-2017, 04:00 AM)RONIT SHARMA Wrote:
(02-15-2017, 03:59 AM)Akshay shinde Wrote: Which one of the following graphs represents kinetics of protein precipitation...
"A"

Ohh! I had marked "C" as I was bit confused with those two options.
  

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