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TLS Online TPP Program

#Question id: 10132

Which of the following statements would be correct for photophosphorylation ?

#Section 2: General Biology
  1. Water is oxidized and protons are released in the lumen by PSII and reduces NADP+ to NADPH in the stroma both are spontaneous processes becoz their reduction potential is positive
  2. The sequential absorption of four photons (excitons) —known as S states, each absorption causing the loss of one electron from the Mn4Ca cluster, produces an oxidizing agent that can remove four electrons from two molecules of water, producing O2, the electrons lost from the Mn4Ca cluster pass one at a time to an oxidized Tyr residue in a PSII protein, then to P680+
  3. An alternative path of electron is cyclic electron transfer, in which elctron move from ferredoxin back to the cytochrome b6 f complex, the cyclic pathway produces more ATP and Water produces but less NADPH produces than the noncyclic
  4. Transfer of the two electrons to PQB reduces it to PQB 2–, and the reduced PQB 2– takes two protons from the stroma side of the medium, yielding a fully reduced plastohydroquinone (PQH2), the PQH2 then still remain bounded from the reaction center complex
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TLS Online TPP Program

#Question id: 14925

#Section 5: Bioprocess Engineering and Process Biotechnology

Assuming the laminar flow across the filter, the rate of filtration (dVf/dt) can be expressed as a function of pressure drop tap, by the modified D'Arcy's equation as (where A is the area of filtering surface, K is the D'Arcy's filter cake permeability and L is the thickness of the filter cake) 
TLS Online TPP Program

#Question id: 14926

#Section 5: Bioprocess Engineering and Process Biotechnology

The effectiveness factor increases with the 
TLS Online TPP Program

#Question id: 14927

#Section 5: Bioprocess Engineering and Process Biotechnology

Calculate RQ value based on the analysis of the inlet air and the exhaust gas as follows. —Inlet Air: 02, 20.90%; CO2, 0.01% — Outlet Gas: 02, 18.30%; CO2, 2.90% 
TLS Online TPP Program

#Question id: 14928

#Section 5: Bioprocess Engineering and Process Biotechnology

 Escherichia coli grows with a doubling time of 0.5 h in the exponential growth phase. What is the value of the specific growth rate? 
TLS Online TPP Program

#Question id: 14929

#Section 5: Bioprocess Engineering and Process Biotechnology

Escherichia coli grows with a doubling time of 0.5 h in the exponential growth phase. How much time would be required to grow the cell culture from 0.1 to 10 kgdry cell m-3 
TLS Online TPP Program

#Question id: 14930

#Section 5: Bioprocess Engineering and Process Biotechnology

E. coli grows from 0.10 to 0.50 kg dry cell m-3 in 1 h. Assuming exponential growth during this period, evaluate the specific growth rate