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GATE BT-2017 Answer Key | Solutions Discussion
This is post no. 191 under the main topic.
(02-17-2017, 11:52 PM)swarnava Wrote: guys anybody have any idea about the expected highest mark in this years bt paper.
Previous years' highest scores 2011_16    
 
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This is post no. 192 under the main topic.
(02-18-2017, 12:18 AM)RONIT SHARMA Wrote:
(02-17-2017, 11:52 PM)swarnava Wrote: guys anybody have any idea about the expected highest mark in this years bt paper.

This year paper was quite easy compared to previous years, I guess highest marks would be in 80's unlike last year which was 68. Hope for best
I found previous year it was 64 the highest score. And if it reaches 80, the cut off is going to be around 40. Because, when(2011) highest score was 79, cut off was 42.
When (2013) highest score was 78.67, cut off was 38.77.
When (2012) highest score was 76.67, cut off was 35.47.
So, we can "ROUGHLY" conclude as the highest score reaches 80, cut off will be closer to 40.....

(N.B. This is just an observation)
 
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This is post no. 193 under the main topic.
(02-06-2017, 06:10 PM)pragari Wrote: Please answer this.

There are trucks and cars traveling in a single lane. The size of Truck is 10 m while a size of a car is 5 m. There should be 20 m gaps between each truck and 15 m gap in each car. Cars and Trucks can travel alternately at the speed of 36 km/h. How many vehicles can cross a bridge (nothing particular value given aboutto the bridge) in an hour?

very easy. for two vehicle distance occupied is 10+20+5+15=50m. in one hour distance traveled is  36 km = 36000m
so no of vehicle is = 2*36000/50 =1440. opt A
 
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This is post no. 194 under the main topic.
(02-05-2017, 09:50 PM)RONIT SHARMA Wrote: Q. Find the angle between two vectors: i-j+2k and 2i-j-k
Ans: 80.4

actually above question is not the part of GATE 2017.  question was "angle between i-j+2k and 2i-j-1.5k"
and dot product (1*1+(-1*-1)+(2*-1.5))=0. so angle will be 90 degree.
 
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This is post no. 195 under the main topic.
Can anyone help with this question?
Growth of a microbe in a test tube is modeled as dx/dt= rx (1-x/k)..... Where X is biomass....r is growth rate....K is carrying capacity of the environment (r≠0, K≠0)....If the value of the starting biomass is K/100, which one of the following graphs qualitatively represents the growth dynamics?    
 
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This is post no. 196 under the main topic.
(02-16-2017, 10:41 PM)subasuba Wrote:
(02-16-2017, 09:01 PM)CHANDA7 Wrote: I considered the case when the parent female and all the 1 St generation females are carrier.....

Answer with  justification....


Attached Files Image(s)
   
 
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This is post no. 197 under the main topic.
(02-05-2017, 09:45 PM)RONIT SHARMA Wrote: Assertion: Gram -ve bacteria stained by saffranin
Reason: Gram -ve bacteria has LPS.

B option is  correct bcoz lps has rol in endotoxin not in staning....
 
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This is post no. 198 under the main topic.
(02-10-2017, 03:42 AM)Akshay shinde Wrote:
(03-19-2015, 12:21 AM)foodemployment Wrote: GATE BT 2017 Solutions,  Answers, Discussion and Answer keys  shared here....

hi,

In which of the following steps of protien synthesis, tRNA is not Involved.
1 Charging
2 Initiation
3 Elongaltion
4 Termination
Termination.......
(02-15-2017, 03:51 AM)RONIT SHARMA Wrote: Match the plant hormones in column 1 with functions in column 2.....

[Image: c016bb69758998d477074d0d89b21f07.jpg]
 
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This is post no. 199 under the main topic.
In gate 2017  .. Repeated question of  DBT BET JRF 2015Q: 9

And also enzyme concentration question ..of km and vmax ...2012 ,2014,2015,2016....


Attached Files Image(s)
   
 
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This is post no. 200 under the main topic.
(02-19-2017, 04:39 PM)sngm Wrote:
(02-05-2017, 09:45 PM)RONIT SHARMA Wrote: Assertion: Gram -ve bacteria stained by saffranin
Reason: Gram -ve bacteria has LPS.

B option is  correct bcoz lps has rol in endotoxin not in staning....

As you will see in Chapter 4, different kinds of bacteria react
differently to the Gram stain because structural differences in
their cell walls affect the retention or escape of a combination of
crystal violet and iodine, called the crystal violet–iodine (CV–I)
complex. Among other differences, gram-positive bacteria have
a thicker peptidoglycan (disaccharides and amino acids) cell wall
than gram-negative bacteria. In addition, gram-negative bacteria
contain a layer of lipopolysaccharide (lipids and polysaccharides)
as part of their cell wall (see Figure 4.13, page 85). When applied
to both gram-positive and gram-negative cells, crystal violet and
then iodine readily enter the cells. Inside the cells, the crystal violet
and iodine combine to form CV–I. This complex is larger than
the crystal violet molecule that entered the cells, and, because of its
size, it cannot be washed out of the intact peptidoglycan layer of
gram-positive cells by alcohol. Consequently, gram-positive cells
retain the color of the crystal violet dye. In gram-negative cells,
however, the alcohol wash disrupts the outer lipopolysaccharide
layer, and the CV–I complex is washed out through the thin layer
of peptidoglycan. As a result, gram-negative cells are colorless
until counterstained with safranin, after which they are pink. OPTION A IS CORRECT REFER TORTORA MICROBIOLOGY PAGE 69
 
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