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20 Aug 2024

The number of α-helical turns permeating the membrane Common Data: The width of the lipid bilayer membrane is 30 Å. It is permeated by a protein which is a right-handed α-helix. (A) 5.6 turns (B) 3.5 turns (C) 6.5 turns (D) 5.0 turns

The number of α-helical turns permeating the membrane Common Data: The width of the lipid bilayer membrane is 30 Å. It is permeated by a protein which is a right-handed α-helix. (A) 5.6 turns (B) 3.5 turns (C) 6.5 turns (D) 5.0 turns

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20 Aug 2024

The ΔG° for the oxidation of NADH by FAD Statement: The standard redox potential values for two half-reactions are given. NAD⁺ + H⁺ + 2e⁻ ↔ NADH -0.315 V FAD + 2H⁺ + 2e⁻ ↔ FADH₂ -0.219 V (A) -9.25 kJ mol⁻¹ (B) -103.04 kJ mol⁻¹ (C) +51.52 kJ mol⁻¹ (D) -18.5 kJ mol⁻¹

The ΔG° for the oxidation of NADH by FAD Statement: The standard redox potential values for two half-reactions are given. NAD⁺ + H⁺ + 2e⁻ ↔ NADH -0.315 V FAD + 2H⁺ + 2e⁻ ↔ FADH₂ -0.219 V (A) -9.25 kJ mol⁻¹ (B) -103.04 kJ mol⁻¹ (C) +51.52 kJ mol⁻¹ (D) -18.5 kJ mol⁻¹   … Read more

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20 Aug 2024

The number of amino acid residues present in the protein Common Data: The width of the lipid bilayer membrane is 30 Å. It is permeated by a protein which is a right-handed α-helix. (A) 15 (B) 18 (C) 20 (D) 17

The number of amino acid residues present in the protein Common Data: The width of the lipid bilayer membrane is 30 Å. It is permeated by a protein which is a right-handed α-helix. (A) 15 (B) 18 (C) 20 (D) 17

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20 Aug 2024

The flow rate required to result in a steady-state concentration of sucrose as 1.5 g/L in the bioreactor Common Data: A culture of Rhizobium is grown in a chemostat (100 m³) bioreactor. The feed contains 12 g/L sucrose, Ks for the organism is 0.2 g/L and μm = 0.3 h⁻¹. (A) 15 m³/h (B) 26 m³/h (C) 2.6 m³/h (D) 150 m³/h

The flow rate required to result in a steady-state concentration of sucrose as 1.5 g/L in the bioreactor Common Data: A culture of Rhizobium is grown in a chemostat (100 m³) bioreactor. The feed contains 12 g/L sucrose, Ks for the organism is 0.2 g/L and μm = 0.3 h⁻¹. (A) 15 m³/h (B) 26 … Read more

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20 Aug 2024

Determine the correctness or otherwise of the following Assertion (a) and the Reason (r) Assertion: Cytoplasmic male sterility (cms) is invariably due to defect(s) in mitochondrial function. Reason: cms can be overcome by pollinating a fertility restoring (Rf) plant with pollen from a non cms plant. (A) both (a) and (r) are true and (r) is the correct reason for (a) (B) both (a) and (r) are true and (r) is not the correct reason for (a) (C) (a) is false but (r) is true (D) (a) is true but (r) is false

Determine the correctness or otherwise of the following Assertion (a) and the Reason (r) Assertion: Cytoplasmic male sterility (cms) is invariably due to defect(s) in mitochondrial function. Reason: cms can be overcome by pollinating a fertility restoring (Rf) plant with pollen from a non cms plant. (A) both (a) and (r) are true and (r) … Read more

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20 Aug 2024

Thermal death of microorganisms in the liquid medium follows first order kinetics. If the initial cell concentration in the fermentation medium is 10⁸ cells / ml and the final acceptable contamination level is 10⁻³ cells, for how long should 1m³ medium be treated at temperature of 120° (thermal deactivation rate constant = 0.23 / min) to achieve acceptable load? (A) 48 min (B) 11 min (C) 110 min (D) 20 min

Thermal death of microorganisms in the liquid medium follows first order kinetics. If the initial cell concentration in the fermentation medium is 10⁸ cells / ml and the final acceptable contamination level is 10⁻³ cells, for how long should 1m³ medium be treated at temperature of 120° (thermal deactivation rate constant = 0.23 / min) … Read more

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20 Aug 2024

True breeding Drosophila flies with curved wings and dark bodies were mated with true breeding short wings and tan body Drosophila. The F1 progeny was observed to be with curved wings and tan body. The F1 progeny was again allowed to breed and produced flies of the following phenotype, 45 curved wings tan body, 15 short wings tan body, 16 curved wings dark body and, 6 short wings dark body. The mode of inheritance is (A) Typical Mendelian with curved wings and tan body being dominant (B) Typical non-Mendelian with curved wings and tan body not following any pattern (C) Mendelian with suppression of phenotypes (D) Mendelian with single crossover

True breeding Drosophila flies with curved wings and dark bodies were mated with true breeding short wings and tan body Drosophila. The F1 progeny was observed to be with curved wings and tan body. The F1 progeny was again allowed to breed and produced flies of the following phenotype, 45 curved wings tan body, 15 … Read more

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