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+--- Thread: GATE BT-2017 Answer Key | Solutions Discussion (/thread-7776.html)
RE: GATE BT-2017 Answer Key | Solutions Discussion - Akshay shinde - 02-15-2017
(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..
RE: GATE BT-2017 Answer Key | Solutions Discussion - SunilNagpal - 02-15-2017
(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
RE: GATE BT-2017 Answer Key | Solutions Discussion - Akshay shinde - 02-15-2017
(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
RE: GATE BT-2017 Answer Key | Solutions Discussion - Akshay shinde - 02-15-2017
(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.
RE: GATE BT-2017 Answer Key | Solutions Discussion - Akshay shinde - 02-15-2017
(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 ?
RE: GATE BT-2017 Answer Key | Solutions Discussion - Manoj_Shastry - 02-15-2017
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.
![[Image: 585_217875_0_587434_r_bt_q23.jpg]](https://www.digialm.com//per/g01/pub/585/touchstone/TempQPImagesStoreMode1/adcimages/1486612285314/2178754///585_217875_0_587434_r_bt_q23.jpg)
RE: GATE BT-2017 Answer Key | Solutions Discussion - SunilNagpal - 02-15-2017
(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).
RE: GATE BT-2017 Answer Key | Solutions Discussion - Manoj_Shastry - 02-16-2017
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?
RE: GATE BT-2017 Answer Key | Solutions Discussion - Manoj_Shastry - 02-16-2017
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?
![[Image: 585_217875_0_587434_r_bt_q42.jpg]](https://www.digialm.com//per/g01/pub/585/touchstone/TempQPImagesStoreMode1/adcimages/1486612285314/2178754///585_217875_0_587434_r_bt_q42.jpg)
RE: GATE BT-2017 Answer Key | Solutions Discussion - CHANDA7 - 02-16-2017
Which of the following graph qualitatively represents the growth dynamics?
![[Image: b1513aebf5cd2992f27e3be4f431127f.jpg]](https://uploads.tapatalk-cdn.com/20170215/b1513aebf5cd2992f27e3be4f431127f.jpg)