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Virendra Singh is teaching live on Unacademy Plus

Virendra Singh
Gold Medalist (B.Tech and M.Tech.) GATE-BT (AIR-6), CSIR-(AIR-17),DBT-JRF (AIR-70), Admin of YouTube Channel "SMART LEARNING ACADEMY".

U
Unacademy user
AK
Thankyou So Much For Making in Hindi
sir kindly solve ques related to molecular marker, promoter asked in DBT jrf.
Virendra Singh
3 months ago
Okay
sir make a video on specific activity, fold purification of protein etc
Virendra Singh
4 months ago
Yes wait for 1 day
Furqan Choudhary
4 months ago
Thank you sir
sir pls solve more and more numerical as examples..as exam is very near
Virendra Singh
4 months ago
Okqy
and also thankyou sir
  1. Most important Numerical Topics of DBT JRF Special Class Date- 16/3/19 Time 8:00pm VIRENDRA SINGH M.Tech (IT BHU) B.Tech (UPTU) GOLD medalist GATE 6 AIR DBT JRF 70 AIR CSIR 79 AIR unacademy


  2. Follow me On Unacademy unacademy 1544 Follow me on the Unacademy Learning App Virendra Singh Gold Medaist (B. Tech and M.Tech GATE-8T [AIR-6). DBTJRF (AIR-70). CSA(AIR-791 dmin of YouTube ChannelSMART LEARNING CADEMY 861 1 19 MessageR Get updates about new courses Watch all my lessons Download slides and watch offline All courses (19) DBTJRF Preparation Strategy and Important Topics CSIR-UGC NET Q Virendra Singh GATE Life Science 2019, Solution Manual 11 A


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  5. Most important Numerical Topics of DBT JRF Special Class Date- 16/3/19 Time- 8:00pm VIRENDRA SINGH M.Tech (IIT BHU) B.Tech (UPTU) GOLD medalist GATE 6 AIR DBT JRF 70 AIR CSIR 79 AIR unacademy


  6. Mole lo Souuhom use . wote 6023 xid3 mola Sauwhiars


  7. A litre of 18% glucose solution is converted anaerobically into methane and carbon dioxide. The theoretical maximum volume of gases at NTP assuming ideality, produced on complete conversion of the entire glucose would be: Answers: 616 CsH,20% 3CH4 + 3CO2 1. 112 L 2. 67.2 L 3. 134.4L 4. 22.4 L in lobue mole Voluwe CL


  8. 4 ..rnal. 2- 2 masta ch' cm 12


  9. AVOGADRO'S NUMBER R-CH- Protein Size NH OH 1 mole 6.02x1023 atoms (monatomic element) Size measured in kilodaltons (kDa) Dalton mass of hydrogen molecule 1 mole 6.02x1023 molecules (molecular compound or diatomic molecule) -1.66 x 10-24 gram Average amino acid = 110 daltons I mole-6.02% 1023 formula units (ionic compound)


  10. 23 0.24. x 1017


  11. An STE buffer contains 20% sucrose, 100 mM Tris and 10 mM EDTA. Given the stock solutions-50% sucrose, 1 M Tris, and 200 mM EDTA, the volumes of the stock solutions required to make 1 litre of the buffer solution are respectively; Answers 1. 400 ml, 100 ml, 100 ml 2. 200 ml, 50 ml, 100 ml 3. 400 ml, 100 ml, 50 ml 4. 200 ml, 100 ml, 50 ml


  12. lonic strength Calculate the ionic strength of 0.1M solution of Nacl For Na' C 0.1 and Z 1 For CI C 0.1 and Z 1 -(0.1x12 +0.1x12)-0.1 2


  13. Calculate the ionic strength (M) of 50 ml of 0.75 % (w/v) NaCl solution? a. 0.128 b. 0.256 c. 7.8 d. 0.064 0.375 2. wt/no1 -0,375,565 0 OS = 0.0064/0.005 0-12DM


  14. What will be the required volumes of 1N HCl and 4N NaOH to prepare one litre solution of pH7? a. 500 ml; 500 ml b. 800 ml; 200 ml Souuhon c. 600 ml; 400 ml d. 200 ml; 800 ml 1L


  15. Rate Constant orcler of reacton with respect to A ordler of reacion with respect to B rate ab k[A][B] - rate in mol cim-3s-1 rate constant concentations in mol clm3


  16. First Order ratek[A] 1/s In 2 Second Order rate = k[A]2 L/mol s k[Alo [Al [Al Zero Order Rate law Units for k Half-life ratek mol/L s [AJo 2k 1 [Alo Integrated rate law [A kt+ [Alo In[Al, -kt + In[Alo in straight-line form Plot for straight line Slope, y intercept vs. t In[A], vs. t -k, [Alo -k, In[Alo [Alo or display. Copyright O The McGraw-Hill Companies, Inc. Permission required for [Alo In [AJo Al slope =-k In [Al [Al slope - k [Alo


  17. equitbsium dissecakoe.tomslaud kd ka Kd lo


  18. + -,L to sndem lem


  19. WORK DONE


  20. Enzyme Activity Unit of enzyme activity: Used to measure total units of activity in a given volume of solution. Specific activity: Used to follow the increasing purity of an enzyme through several procedural steps. Molecular activity: Used to compare activities of different enzymes. Also called the turn-over number (TON- k cat)


  21. Enzyme Activity New international units: nit of enzyme activity: mol substrate/sec katal Specific activity: mol substrate/sec-kg E katal/kg E Molecular activity: mol substrate/sec-mol E katal/mol E


  22. The protein concentration and enzyme activity in 100 mL of a cell free extract is 5 mg/mL and 2 units/mL, respectively. After multiple steps of purification, the final 10 mL fraction contains 4 mg/mL of protein and 15 units/mL of enzyme activity. The fold purification and percentage recovery, respectively is: 1. 10 and 75 2. 9.4 and 20 3. 9.4 and 75 4. 10 and 20 Ans.3


  23. cahom Volume 100ml Me pol in Comc. bein coen" com as rmkin 3) ISo 3750 O" . E) 40 4 loo


  24. Purification data for an enzyme is given below: Steps Purification step VolumeTotal protein Total activity (ml) 20 4 (mg)(micromoles per min) 100 10 I Cell-free extract 150 120 Il Ni-NTA chromatography hat is the fold-purification? 150-e) 115 hits c. 10.5 12-


  25. Michaelis-Menten Curve 0 max Km Vm LS] (mM) Im


  26. An enzvme reaction follows ichaelis Menten kinetics What will be the reaction S: Km 3 velocity at a substrate concentration Km/3? kr, 3 Km .V max 3 .Vmad2 d. Vmax/4 4 3


  27. Example 2 Answer An enzyme (1.84 gm, MW = 36800) catalyzes a reaction in presence of excess substrate at a rate of 4.2 mol substrate/min. What is the TON ? 1.84 gm -5 x 105 umol E 36800 gm/ mol 4.2 mol/min TON = 84000 min-1 5 x 10-5 umol


  28. Example 3 Ten micrograms of carbonic anhydrase (MW- 30000) in the presence of excess substrate exhibits a reaction rate of 6.82 x 10 3 mol/min. At [S]-0.012 M the rate is 3.41 x 10 3 mol/min. a. What is Vm? b. What is K M? c. What is k 2 (kcat)?


  29. The A26o of a plasmid solution after 100-fold dilution is 0.2. Given that A260 of 1.0 represents 50 g/ml of DNA and the total volume of isolated plasmid solution is 50 Hl, what will be the concentration and amount respectively of the isolated plasmid? a. 1.0 gjul and 50 g b. 1.0 mg/ul and 50 mg C. 10 g/ul and 50 g d. 10 mg/ul and 50 mg


  30. Distribution coefficient .If a solute is in aqueous phase and is extracted into an organic phase: . A solute S will distribute itself between two phases (after shaking/mixing) Ratio of [S] in the two phases will be constant. . KD - distribution coefficient K [Ssolvent 1 eg. organic solvent solvent 2 eg. water


  31. In an organism, the amount of DNA per haploid genome is about 1.6 x 109 nucleotide pairs. Given that the length of DNA helix occupied by one nucleotide pair is 3.4 , approximately how long a double helix could be formed from this DNA? 0 Answers 22 cm 55 cm a. 1.1meter 2.2 mete 4.


  32. A spherical mammalian cell of radius 'R' is infected by a single coccus bacterium having 100 times smaller radius. Given that the host cell will lyse when 1/2 of the cell volume is taken up by the bacterium, approximately how many times will the bacterium divide before the host cell i lysed? 1. 19 2. 3X105 3. 106 4. 40


  33. (vt)/2. se thi al N 0 2x1 mt1 (m 0.633 = (X2.303 1x2.32 = 19.12- 0 843


  34. A 34 Kb linearDNA was digested with Hindill and BamH. The fragments obtained on complete digestion were as follows HindIll BamH Hind BamHl: 4 Kb, 10 Kb, 8 Kb, 12 Kb 14 Kb, 20 Kb 4 Kb, 12 Kb, 18 Kb The appropriate restriction map of the DNA is BamHI Hindll BamHI BamHI Hind BamHI 10 12 4 8 Kb 10 12 Kb Hindlll BamH BamHl BamHl BamHl Hindlll 10 4 12 Kb 4 10 12 8 Kb


  35. A plasmid DNA when digested with EcoRI gave a single band of 16 Kb. When the same plasmid was digested with BamH it gave two bands of 6Kb and 4 Kb. The plasmid has: a. Single site of EcoRl and 2 sites of BamHl b. Single site of EcoRI and 3 sites of BamHl c. Single site of EcoRI and 2 sites of BamHl d. 2 sites of EcoRI and 2 sites of BamHl


  36. ltelf Cmts Biomae ytela 2-6, Pndet yield Cone D- Dituhom rate Volumecv)


  37. A cell suspension (1.5x105 cells per ml) was treated with 1mM HgCl2for 30 min. After treatment, the cell suspension was diluted 10 fold and 100 microliter was plated which gave 5 colonies. Calculate the percentage of cells that survived? b. 3.3 C. 0.033 d. 0.33 -| 0.33 % 3


  38. pen DNA NNo -2 lo N, lo