Biochemical Miscellaneous

Biochemical Miscellaneous

Easy Questions

Biochemical

Biochemical Miscellaneous
  1. ​A roller bottle culture vessel perfectly cylindrical in shape having inner radius (r) = 10 cm and length (l) = 20 cm was fitted with a spiral film of length (L) = 30 cm and width (W) = 20 cm. If the film can support 105 anchorage dependent cells per cm2, the increase in the surface area after fitting the spiral film and the additional number of cells that can be grown respectively are ​​








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    Given, ​​​​Inner radius (r) = 10 cm ​​​​
    Length (l) = 20 cm ​​​
    Length of spiral film (L)​= 30 cm ​​​​
    Width (W) = 20 cm ​​
    Thus, the increase in the surface area after fitting of surface film = 30 * 20 = 600 cm2 ​​
    Also, the film can support 105 anchorage dependent cells per cm2
    ​​Thus, the film will add​ = 105 cells/cm2 * 600 cm2 ​​​​​= 6 * 107 cells.

    Correct Option: B

    Given, ​​​​Inner radius (r) = 10 cm ​​​​
    Length (l) = 20 cm ​​​
    Length of spiral film (L)​= 30 cm ​​​​
    Width (W) = 20 cm ​​
    Thus, the increase in the surface area after fitting of surface film = 30 * 20 = 600 cm2 ​​
    Also, the film can support 105 anchorage dependent cells per cm2
    ​​Thus, the film will add​ = 105 cells/cm2 * 600 cm2 ​​​​​= 6 * 107 cells.

  1. 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? ​








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    Initial number of cells = 108 cells/ml * 1m3 ​​
    But, we know, ​​​​​  1 m3 = 106 cm3
    Also, ​103 cm3 ​= 1000 ml
    Therefore, total initial number of cells = 108 cells * 106 = 1014 cells
    Final acceptance of contamination = 10–3 cells/ml or 103 cells/m3
    Thus, using the first order rate constant equation :  
    t = 2.303 / K log10 [1014] /103
    = 2.303/ 0.23 per min log10 (1014/103)
    = 10 * 11 log10 (10)
    t = 110 min.

    Correct Option: C

    Initial number of cells = 108 cells/ml * 1m3 ​​
    But, we know, ​​​​​  1 m3 = 106 cm3
    Also, ​103 cm3 ​= 1000 ml
    Therefore, total initial number of cells = 108 cells * 106 = 1014 cells
    Final acceptance of contamination = 10–3 cells/ml or 103 cells/m3
    Thus, using the first order rate constant equation :  
    t = 2.303 / K log10 [1014] /103
    = 2.303/ 0.23 per min log10 (1014/103)
    = 10 * 11 log10 (10)
    t = 110 min.

Direction: 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. ​

  1. The flow rate required to result in steady state concentration of sucrose as 1.5 g/L in the bioreactor will be ​








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    By using the Monod Equation under steady state,
    We have, µ = µmax [S]/[S] + KS = 0.3 hr–1 * 1.5g/l/ [1.5g/l] + [0.2g/l]
    µ = 0.2648 hr–1
    D = µ (by using the Monod model)
    Also, D = F / V
    Therefore, equating the two, we get, F = µV = 0.2648 * 10
    F= 26.5 m3hr–1

    Correct Option: B

    By using the Monod Equation under steady state,
    We have, µ = µmax [S]/[S] + KS = 0.3 hr–1 * 1.5g/l/ [1.5g/l] + [0.2g/l]
    µ = 0.2648 hr–1
    D = µ (by using the Monod model)
    Also, D = F / V
    Therefore, equating the two, we get, F = µV = 0.2648 * 10
    F= 26.5 m3hr–1

  1. If YX/S = 0.4 g / g for the above culture and steady state cell concentration in the bioreactor is 4 g/ L the resulting substrate concentration will be ​








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    We know, Yield, YX/S = – ΔX / ΔS
    YX/S = 0.4 g/L
    Therefore, ΔX = Xfinal – Xinitial = 4 g/L
    From the equation, YX/S = – ΔX/ΔS
    0.4= –[4/( Sfinal – 12)] 0.4 Sfinal – 4.8
    = –4, 0.4 Sfinal
    = 0.8, Sfinal = 2
    Final substrate concentration, Sfinal = 2 g/L

    Correct Option: A

    We know, Yield, YX/S = – ΔX / ΔS
    YX/S = 0.4 g/L
    Therefore, ΔX = Xfinal – Xinitial = 4 g/L
    From the equation, YX/S = – ΔX/ΔS
    0.4= –[4/( Sfinal – 12)] 0.4 Sfinal – 4.8
    = –4, 0.4 Sfinal
    = 0.8, Sfinal = 2
    Final substrate concentration, Sfinal = 2 g/L

  1. A culture vessels in which physical, physicochemical and physiological conditions, as well as cell concentration, are kept constant is known as ​








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    ​Biostat keeps all the given parameters constant.

    Correct Option: B

    ​Biostat keeps all the given parameters constant.