GATE BT-2017 Answer Key | Solutions Discussion - Printable Version +- Biotechnology Forums (https://www.biotechnologyforums.com) +-- Forum: Education and Careers (https://www.biotechnologyforums.com/forum-26.html) +--- Forum: Competitive Exams (https://www.biotechnologyforums.com/forum-38.html) +--- Thread: GATE BT-2017 Answer Key | Solutions Discussion (/thread-7776.html) |
RE: GATE BT-2017 Answer Key | Solutions Discussion - CHANDA7 - 02-18-2017 (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[attachment=368] RE: GATE BT-2017 Answer Key | Solutions Discussion - CHANDA7 - 02-18-2017 (02-18-2017, 12:18 AM)RONIT SHARMA Wrote: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.(02-17-2017, 11:52 PM)swarnava Wrote: guys anybody have any idea about the expected highest mark in this years bt paper. 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) RE: There are trucks and cars traveling in a single lane.. - sadanandkumar469@gmail.com - 02-18-2017 (02-06-2017, 06:10 PM)pragari Wrote: Please answer this. 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 RE: GATE BT-2017 Answer Key | Solutions Discussion - sadanandkumar469@gmail.com - 02-18-2017 (02-05-2017, 09:50 PM)RONIT SHARMA Wrote: Q. Find the angle between two vectors: i-j+2k and 2i-j-k 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. GATE BT-2017 Answer Key | Solutions Discussion - CHANDA7 - 02-18-2017 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?[attachment=371] RE: GATE BT-2017 Answer Key | Solutions Discussion - sngm - 02-19-2017 (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.... RE: Gram -ve Gram +ve | GATE BT-2017 Answer Key | Solutions Discussion - sngm - 02-19-2017 (02-05-2017, 09:45 PM)RONIT SHARMA Wrote: Assertion: Gram -ve bacteria stained by saffranin B option is correct bcoz lps has rol in endotoxin not in staning.... RE: GATE BT-2017 Answer Key | Solutions Discussion - sngm - 02-19-2017 (02-10-2017, 03:42 AM)Akshay shinde Wrote:Termination.......(03-19-2015, 12:21 AM)foodemployment Wrote: GATE BT 2017 Solutions, Answers, Discussion and Answer keys shared here.... (02-15-2017, 03:51 AM)RONIT SHARMA Wrote: Match the plant hormones in column 1 with functions in column 2..... RE: GATE BT-2017 Answer Key | Solutions Discussion - sngm - 02-19-2017 In gate 2017 .. Repeated question of DBT BET JRF 2015Q: 9 And also enzyme concentration question ..of km and vmax ...2012 ,2014,2015,2016.... RE: Gram -ve Gram +ve | GATE BT-2017 Answer Key | Solutions Discussion - swarnava - 02-20-2017 (02-19-2017, 04:39 PM)sngm Wrote:(02-05-2017, 09:45 PM)RONIT SHARMA Wrote: Assertion: Gram -ve bacteria stained by saffranin 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 |