![Q.64 The product of complete digestion of the plasmid shown below with EcoRI and HaeIII was purified and used as a template in a reaction containing Klenow fragment of DNA polymerase, dNTPs and [α-32P]-dATP in a suitable reaction buffer. The product thus obtained was purified and subjected to gel electrophoresis followed by autoradiography. EcoRI 5′–GAATTC–3′ 3′–CTTAAG–5′ HaeIII 5′–GGCC–3′ 3′–CCGG–5′ The number of bands that will appear on the X-ray film is ________](data:image/svg+xml;nitro-empty-id=MTE5MzoyNzA0-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTkyMCAxMDgwIiB3aWR0aD0iMTkyMCIgaGVpZ2h0PSIxMDgwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==)
![Q.41 Determine the correctness or otherwise of the following Assertion [a] and the Reason [r] Assertion: The association constant in water for the G-C base pair is three times lower than that for the A-T base pair. Reason: There are three hydrogen bonds in the G-C base pair and two in the A-T base pair. (A) Both [a] and [r] are true and [r] is the correct reason for [a] (B) [a] is false but [r] is true (C) Both [a] and [r] are false (D) Both [a] and [r] are true and [r] is not the correct reason for [a]](data:image/svg+xml;nitro-empty-id=MTE5MzozMzUzNA==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTkyMCAxMDgwIiB3aWR0aD0iMTkyMCIgaGVpZ2h0PSIxMDgwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==)
![Q.39 Determine the correctness or otherwise of the following Assertion [a] and the Reason [r] Assertion: Ab initio gene finding algorithms that predict protein coding genes in eukaryotic genomes are not completely accurate. Reason: Eukaryotic splice sites are difficult to predict. (A) Both [a] and [r] are false (B) [a] is true but [r] is false (C) Both [a] and [r] are true and [r] is the correct reason for [a] (D) Both [a] and [r] are true but [r] is not the correct reason for [a]](data:image/svg+xml;nitro-empty-id=MTE5MzozNjIwNQ==-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTkyMCAxMDgwIiB3aWR0aD0iMTkyMCIgaGVpZ2h0PSIxMDgwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==)
Q.65 A rod shaped bacterium has a length of 2 µm, diameter of 1 µm and density the same as that of water. If proteins constitute 15% of the cell […]
Category: GATE Biotechnology 2018 Previous Year Papers with Solutions
GATE Biotechnology 2018 Previous Year Papers with Solutions
![Q.64 The product of complete digestion of the plasmid shown below with EcoRI and HaeIII was purified and used as a template in a reaction containing Klenow fragment of DNA polymerase, dNTPs and [α-32P]-dATP in a suitable reaction buffer. The product thus obtained was purified and subjected to gel electrophoresis followed by autoradiography. EcoRI 5′–GAATTC–3′ 3′–CTTAAG–5′ HaeIII 5′–GGCC–3′ 3′–CCGG–5′ The number of bands that will appear on the X-ray film is ________](data:image/svg+xml;nitro-empty-id=MTI0MjoxMjgw-1;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTkyMCAxMDgwIiB3aWR0aD0iMTkyMCIgaGVpZ2h0PSIxMDgwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==)
GATE Biotechnology 2018 Previous Year Papers with Solutions
EcoRI + HaeIII Digestion and Klenow Labeling
Q.64 The product of complete digestion of the plasmid shown below with EcoRI and HaeIII was purified and used as a template in a reaction containing Klenow fragment of DNA […]
GATE Biotechnology 2018 Previous Year Papers with Solutions
Unveiling Enzyme Inhibition
Q.63 Consider a simple uni-substrate enzyme that follows Michaelis-Menten kinetics. When the enzyme catalyzed reaction was carried out in the presence of 10 nM concentration of an inhibitor, there was […]
GATE Biotechnology 2018 Previous Year Papers with Solutions
Enzyme Half-Life Ratio
Q.62 First order deactivation rate constants for soluble and immobilized amyloglucosidase enzyme are 0.03 min-1 and 0.005 min-1, respectively. The ratio of half-life of the immobilized enzyme to that of […]
GATE Biotechnology 2018 Previous Year Papers with Solutions
Calculate DNA Alignment Score
Q.61 Consider the following alignment of two DNA sequences: AGTAAC AA—-AC Assuming an affine gap scoring scheme of an identity matrix for substitution, a gap initiation penalty of 1 and […]
GATE Biotechnology 2018 Previous Year Papers with Solutions
Calculate Initial Culture Volume
Q.60 At the end of a batch culture, glucose solution is added at a flow rate of 200 ml/h. If the culture volume after 2 h of glucose addition is […]
GATE Biotechnology 2018 Previous Year Papers with Solutions
Ethanol from Glucose
Q.59 Yeast converts glucose to ethanol and carbon dioxide by glycolysis as per the following reaction: C6H12O6 → 2C2H5OH + 2CO2 Assuming complete conversion, the amount of ethanol produced (in […]
GATE Biotechnology 2018 Previous Year Papers with Solutions
Mammalian Cell Population Doubling Time
Q.58 Mammalian cells in active growth phase were seeded at a density of 1×105 cells/ml. After 72 hours, 1×106 cells/ml were obtained. The population doubling time of the cells in […]
GATE Biotechnology 2018 Previous Year Papers with Solutions
Chemostat Effluent Substrate Concentration
Q.57 In a chemostat, the feed flow rate and culture volume are 100 mL/h and 1.0 L, respectively. With glucose as substrate, the values of μmax and Ks are 0.2 […]
GATE Biotechnology 2018 Previous Year Papers with Solutions
Fermentation KLa Calculation
Q.56 A microbial strain is cultured in a 100 L stirred fermenter for secondary metabolite production. If the specific rate of oxygen uptake is 0.4 h-1 and the oxygen solubility […]
