A is Yes.
B is No.
C is No.
D is Yes.
E is No.
E is No because as we can see one glucose give rise to 2 pyruvates with gain of 2 ATP. Then 2 pyruvate will give rise to 2 Acetyl CoA and then These 2 Acetyl CoA will form 2 butanol So 2 butanol will form not one.
1- Consider the fermentation pathways of Clostridium as diagrammed. A common goal of metabolic engineering is...
Consider the degradation of glucose to pyruvate by the glycolytic pathway: glucose+2ADP+2Pi+2NAD+→2pyruvate+2ATP+2H2O+2NADH+2H+ Part A Calculate ΔG for this reaction at pH=7.4 and 37 ∘C when the reactants and products are at the concentrations given below. ATP=4.00mM Pi=5.90mM ADP=220μM glucose=5.10 mM pyruvate=64.0 μM NAD+=350μM NADH=14.5μM CO2=15.0torr half reaction E∘′(V) NAD++H++2e− ⟶ NADH −0.315 2pyruvate+6H++4e− ⟶ glucose −0.590 pyruvate+NADH+2H+ ⟶ ethanol+NAD++CO2 ΔG∘′=−64.4kJ/mol ATP+H2O ⟶ ADP+Pi+H+ ΔG∘′=−32.2kJ/mol