If YX/S = 0.4 g / g for the above culture and steady state

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Direction: A culture of Rhizobium is grown in a chemostat (100 m³ bioreactor). The feed contains 12 g/L sucrose, K_S for the organism is 0.2 g/L and µ_m = 0.3 h^–1.

If Y_X/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

1. 2 g/L
2. 8 g/L
3. 4 g/L
4. 6 g/L

Correct Option: A

We know, Yield, Y_X/S = – ΔX / ΔS
Y_X/S = 0.4 g/L
Therefore, ΔX = X_final – X_initial = 4 g/L
From the equation, Y_X/S = – ΔX/ΔS
0.4= –[4/( S_final – 12)] 0.4 S_final – 4.8
= –4, 0.4 S_final
= 0.8, S_final = 2
Final substrate concentration, S_final = 2 g/L