1) (Define…)
R1: CH2O (glucose) → pyruvate (CH4/3O) + 1/3 NADH2 + 1/3 ATP (EMP pathway)
R2: pyruvate (CH4/3O) → 2/3 acetyl-CoA (CHO0.5) + 1/3 NADH2 + 1/3 CO2
R3: CH4/3O + 1/3NADH2 → 2/3Ethanol (CH3O0.5) + 1/3CO2
R4: 3/4CH4/3O + 1/4CO2 + 1/2NADH2 → Succinate(CH1.5O) + 1/4H2O
R5: 1.27CHO0.5 + 0.2NH3 + K ATP + 0.405H2O → CH1.8O0.5N0.2 + 0.27 CO2 + 0.44NADH2
2) (Relate…)
𝑟
= −𝑟
𝑔𝑙𝑢𝑐𝑜𝑠𝑒
1
𝑟
= −0.2 𝑟
𝑁𝐻3
5
𝑟 = 𝑟
𝑋
5
2
𝑟
= 𝑟
𝑒𝑡ℎ𝑎𝑛𝑜𝑙
3 3
𝑟
= 𝑟
𝑠𝑢𝑐𝑐𝑖𝑛𝑎𝑡𝑒
4
1
1
1
𝑟
= 𝑟 + 𝑟 – 𝑟 + 0.27 𝑟
𝐶𝑂2
5
3 2 3 3 4 4
𝑟
𝑟
𝐻2𝑂
1
= 𝑟 – 0.405 𝑟
5
4 4
3
= 𝑟 –𝑟 –𝑟 – 𝑟
𝑝𝑦𝑟𝑢𝑣𝑎𝑡𝑒
1
2
3 4 4
2
= 𝑟 – 1.27 𝑟
𝑎𝑐𝑒𝑡𝑦𝑙 𝐶𝑜𝐴
5
3 2
−
1
1
1
1
𝑟
= 𝑟 + 𝑟 – 𝑟 – 𝑟 + 0.44 𝑟
𝑁𝐴𝐷𝐻
5
3 1 3 2 3 3 2 4
𝑟
𝑟
1
= 𝑟 – 𝐾𝑟
𝐴𝑇𝑃
5
3 1
3) (Assume Steady State)
𝑟
𝐴𝑇𝑃
= 𝑟
𝑁𝐴𝐷𝐻
= 𝑟
𝑝𝑦𝑟𝑢𝑣𝑎𝑡𝑒
= 𝑟
𝑎𝑐𝑒𝑡𝑦𝑙
−
𝐶𝑜𝐴
= 0
4) (Find DOF)
5 (𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠) − 4 (𝑢𝑛𝑘𝑛𝑜𝑤𝑛𝑠) = 1
5) (The system is solved)
0 = 𝑟 – 𝑟 – 𝑟 – 3/4𝑟
1
2
3
4
2
0 = 𝑟 – 1.27 𝑟
5
3 2
0 =
1
1
1
1
𝑟 + 𝑟 – 𝑟 – 𝑟 + 0.44 𝑟
5
3 1 3 2 3 3 2 4
1
0 = 𝑟 – 5𝑟
5
3 1
𝑟
𝑟
𝑟
𝑟
5
2
3
4
=
1
𝑟
15 1
= 0.13 𝑟
= 0.52 𝑟
= 0.46 𝑟
1
1
1
𝑟
= −𝑟
𝑔𝑙𝑢𝑐𝑜𝑠𝑒
1
𝑟
= −0.013 𝑟
𝑁𝐻3
1
𝑟 = 0.067𝑟
𝑋
1
𝑟
= 0.35 𝑟
𝑒𝑡ℎ𝑎𝑛𝑜𝑙
1
𝑟
= 0.46 𝑟
𝑠𝑢𝑐𝑐𝑖𝑛𝑎𝑡𝑒
1
𝑟
= 0.12 𝑟
𝐶𝑂2
1
𝑟
= 0.088 𝑟
𝐻2𝑂
1
Overall:
−𝐶𝐻2 𝑂 − 0.01𝑁𝐻3 + 0.07𝐶𝐻1.8 𝑂0.5 𝑁0.2 + 0.35𝐶𝐻3 𝑂0.5 + 0.46𝐶𝐻1.5 𝑂 + 0.12𝐶𝑂2 + 0.09𝐻2 𝑂 = 0
When assuming Monod kinetics, 𝜇 = 𝜇𝑚𝑎𝑥 when [𝑆] ≫ 𝐾𝑚 (i.e. during exponential growth)
ln(X) vs time
2
ln(X) = 0.2971(t) - 1.2076
R² = 0.9985
1.5
ln(X)
1
0.5
0
-0.5
0
2
4
6
8
-1
-1.5
Time( (h)
ln(X)
Linear (ln(X))
10
12