⚗️ Biochemical Engineering Solved MCQs with Answers

Practice 513 approved ⚗️ Biochemical Engineering questions with verified answers from the 🧬 Medical Science subject for Pakistan exams.

513 solved MCQs

Predominantly uncompetitive inhibition may be called whenAnswer: Competitive inhibition is greater than uncompetitive inhibition An enzyme has a Kₘ of 4.7 x 10-5M. If the Vₘₐₓ of the preparation is 22m moles liter⁻¹ min⁻¹, what velocity would be observed in the presence of 2.0 x 10⁻⁴M substrate and 5.0 x 10⁻⁵M of a competitive inhibitor?Answer: 13.54μ moles liter⁻¹ min⁻¹ The rate equation in non-competitive inhibition based on Michaelis Menten equation is given byAnswer: rₘₐₓS/(Kₘ + S)(1+I/Kᵢ) Which of the following statement(s) regarding enzymes, is/are false?Answer: Enzymes provide activation energy for reactions The slope of Lineweaver Burk plot for Michaelis Menten equation isAnswer: Kₘ/Vₘₐₓ The initial velocity, V₀, of an enzyme catalyzed reaction reaches Vₘₐₓ asAnswer: 1/[S] → 0 The usual method(s) to solve rate equation of simple enzyme kinetics is/areAnswer: All of these Michaelis Menten equation can also be written asAnswer: All of these Which of the following step is assumed to be the slowest step in the Michaelis Menten equation?Answer: The product releasing step When substrate [S] = Kₘ (Michaelis-Menten constant), the velocity of an enzyme catalyzed reaction is aboutAnswer: 0.5 * Vₘₐₓ A classical noncompetitive inhibitor hasAnswer: No effect on substrate binding and vice versa The active site of an enzyme differs from an antibody-antigen binding site in that the enzyme active siteAnswer: Catalyzes a chemical reaction Enzymes are basicallyAnswer: Proteins Which of the following refers to pseudo steady state ?Answer: d(Cₑₛ)/dt = 0 Most enzymes work byAnswer: Decreasing energy of activation Which category of enzymes belongs to class 5 in the international classification?Answer: Isomerases Lock and key theory is based on the compatibility ofAnswer: Enzyme and substrate When an enzyme is functioning at Vmax, the rate of the reaction is limited byAnswer: The rate at which the enzyme can convert substrate to product The equation for the rate of product formation for simple enzyme reaction is given by (Where rₘₐₓ, maximum reaction rate, Cₛ substrate concentration, Cₚ product concentration ES, Cₑₛ enzyme-substrate concentration)Answer: rₚ = rₘₐₓ Cₛ/(Kₘ+Cₛ) When [S] = 0.1 *Kₘ, the velocity of an enzyme catalyzed reaction is about:Answer: 0.1 * Vmax In a glucose-yeast extract medium, Lactococcus lactis has a maximum specific growth rate (um) of 1.23 h⁻¹. What would be the specific growth rate of this organism at steady state in a 4 litre reactor being fed at 2 litres per hour?Answer: 0.5 h⁻¹ Which of the following is not correct with regards to continuous cultures?Answer: The continuous cultures is very useful for the production of pharmaceuticals In a plug flow reactor, the highest concentration of substrates is exposed to the cells thatAnswer: Are located near entrance of feed In an activated sludge process, the biomass is recycled toAnswer: All of these A strain of Escherichia coli has a maximum specific growth rate of 0.8 h⁻¹on a glucose based medium. If this organism is being grown in a chemostat with a dilution rate of 1.2 h⁻¹, then at steady state the concentration of E. coli in the same medium willAnswer: Be zero A fed-batch reactor containing 2 litre of medium and 0.1 g.l⁻¹of biomass is fed with a medium containing 1 g.l⁻¹ of substrate at 1 litre per hour. If after 10 hours, the concentration of biomass in the reactor was 0.2 g.l⁻¹, then the biomass produced during the 10 hour period isAnswer: 2.2 g Which of the following organisms will continue their metabolism at high substrate concentrations?Answer: Pseudomonas aeruginosa Yeast is typically produced in fed batch reactors becauseAnswer: Yeast cells respire at low glucose concentrations Two continuous bioreactors containing the same organisms, fed with the same feed at the same dilution rate were compared. Reactor 1 started with an initial concentration of glucose of 10 g.l-1, while reactor 2 contained 0.1 g.l-1 of glucose at the start of the process then at steady stateAnswer: The concentration of glucose in reactor 1 would be equal to reactor 2 In a fluidized bed reactor mixing is achieved byAnswer: Both (a) and (b)