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. 2016 May 11;4:41. doi: 10.3389/fcell.2016.00041

Table 4.

Steps of steady-state determination for a model with six states and eight reactions.

Loop	Steady-state graph	Steady-state equations	Simplification
1		A: $0 = k_{0} + k_{2} B - k_{1} A - k_{3} A^{2}$	Cycle: [A, A]
		B: 0 = k₁A − k₂B − k₅BC − k₇BF	Type 3
		C: $0 = k_{3} A^{2} - k_{5} B C - k_{4} C$	$k_{0} = \frac{A^{2} k_{3} + A k_{1}}{r_{1} + 1}$
		D: 0 = k₅BC − k₆DG
		F: 0 = k₆DG − k₈F	$k_{2} = r_{1} \frac{k_{0}}{B} = r_{1} \frac{A^{2} k_{3} + A k_{1}}{B (r_{1} + 1)}$
		G: 0 = k₈F − k₆DG
2		B: $0 = k_{1} A \frac{1}{r_{1} + 1} - k_{3} A^{2} \frac{r_{1}}{r_{1} + 1}$	Cycle: [B, B]
		− k₅BC − k₇BF	Type 1
		C: $0 = k_{3} A^{2} - k_{5} B C - k_{4} C$
		D: 0 = k₅BC − k₆DG	$k_{1} = \frac{r_{1} + 1}{A} k_{3} A^{2} \frac{r_{1}}{r_{1} + 1}$
		F: 0 = k₆DG − k₈F
		G: 0 = k₈F − k₆DG	+k₅BC + k₇BF]
3		C: $0 = k_{3} A^{2} - k_{5} B C - k_{4} C$	Cycle: [D, G, D]
		D: 0 = k₅BC − k₆DG	Type 0
		F: 0 = k₆DG − k₈F	G is part of CQ
		G: 0 = k₈F − k₆DG
4		C: $0 = k_{3} A^{2} - k_{5} B C - k_{4} C$	Tree-like
		D: 0 = k₅BC − k₆DG
		F: 0 = k₆DG − k₈F	Solve for C, D and F