Contact:
sales@biotechnologyforums.com to feature here

  •  Previous
  • 1
  • ...
  • 24
  • 25
  • 26(current)
  • 27
  • 28
  • ...
  • 122
  • Next 
Thread Rating:
  • 2 Vote(s) - 3 Average
  • 1
  • 2
  • 3
  • 4
  • 5
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


Suggested Reads:
Top Biotech Companies | Top places to work
Indian Biotech Companies and Job Openings
Aiming a PhD in Top Grad School? | These are the Important Points to Consider
Careers in Biotechnology | A list of various Options
Biotechnology Competitive Exams in India
(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.
  

Possibly Related Threads...
Thread
Author
  /  
Last Post
Replies: 1
Views: 108
1 hour ago
Last PostLavkeshsharma
Replies: 1
Views: 45
1 hour ago
Last PostLavkeshsharma
Replies: 1
Views: 154
08-09-2017, 12:36 PM
Last PostLavkeshsharma
Replies: 9
Views: 1,015
08-07-2017, 02:07 PM
Last PostNisha96
Replies: 3
Views: 493
07-26-2017, 12:57 AM
Last PostBETSY ANN



Users browsing this thread:
8 Guest(s)

GATE BT-2017 Answer Key | Solutions Discussion32