Biotechnology Forums

(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

(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.

(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"

Sunil sir, what is your opinion on this question ?

Q: Transcription factor X binds a 10bp DNA. X was found to bind at 20 distinct sites, distribution is as following. What is the Consensus sequence pattern.

(02-15-2017, 07:00 PM)Manoj_Shastry Wrote: [ -> ]Q: Transcription factor X binds a 10bp DNA. X was found to bind at 20 distinct sites, distribution is as following. What is the Consensus sequence pattern.

Option (A) NGTCNNNTNN seems most acceptable consensus (assuming a >=95% confidence).

Q: A bacteria has genome of size 6 million bp. If DNA synthesis rate is 1000 bp/ second. How long does it take (in minutes) for replication of genome? 

Q: A proto-oncogene is suspected to have undergone duplication in a certain type of cancer. Which technique can verify gene duplication?

A. Northern blotting
B. Southern blotting
C. South western blotting
D. Western blotting

Q: During anerobic growth, an organism converts Glucose (P) into biomass (Q), Ethanol ®, Acetaldehyde (S) and Glycerol (T). Every mole of carbon in Glucose gets distributed as follows:

1(C-mole P) -> 0.14(C-mole Q) + 0.25(C-mole R) + 0.3(C-mole S) + 0.31(C-mole T) 

From 1800 grams of Glucose fed to organism, amount of Ethanol produced(in grams) is?

A protein expressed in Ecoli under lac promoter and operator. Leaky expression seen in absence of IPTG. 
Which of these methods will minimize leaky expression? 

Which of the following graph qualitatively represents the growth dynamics?