Table 2.

Template in Microsoft Excel format for molar ratio calculation*

	A	B	C	D	E	F	G	H	I
		Experimental data (nmole)			Normalized data (%)			ϓ	(Y−ϓ)2
		X1	X2	Y	X1	X2	Y
1	Asx	2.5886	1.5470	2.2239	0.1999	0.1188	0.1582	0.1579	7.2703 × 10−8
2	Glx	2.0869	2.4461	2.4603	0.1612	0.1879	0.1750	0.1750	3.5468 × 10−9
3	Gly	1.2052	0.5589	0.9185	0.0931	0.0429	0.0653	0.0671	3.1088 × 10−6
4	Ala	0.3728	1.2923	0.9363	0.0288	0.0993	0.0666	0.0653	1.5886 × 10−6
5	Val	1.6232	1.0057	1.4244	0.1254	0.0772	0.1013	0.1004	7.6615 × 10−7
6	Ileu	0.8871	0.3855	0.6789	0.0685	0.0296	0.0483	0.0483	4.3462 × 10−9
7	Leu	0.8676	1.7725	1.4488	0.0670	0.1361	0.1030	0.1029	3.2973 × 10−8
8	Tyr	0.3534	0.5299	0.4975	0.0273	0.0407	0.0354	0.0342	1.2879 × 10−6
9	Phe	0.2359	0.7842	0.5361	0.0182	0.0602	0.0381	0.0400	3.5186 × 10−6
10	His	0.5207	0.3794	0.4873	0.0402	0.0291	0.0347	0.0345	3.2240 × 10−8
11	Lys	2.0096	1.6924	2.0124	0.1552	0.1300	0.1431	0.1421	9.5056 × 10−7
12	Arg	0.1957	0.6254	0.4369	0.0151	0.0480	0.0311	0.0322	1.2362 × 10−6
13	Σ	12.9467	13.0193	14.0613	1.00	1.00	1.00	1.00	1.2603 × 10−5
14	Modified molecular weight of protein 1b			15551.20		Molar portion of protein 1c			3.0967 × 10−5
15	Modified molecular weight of protein 2b			53361.28		Molar portion of protein 2d			9.7156 × 10−6
16	% mass of protein 1a			0.4816		Molar ratio of Protein 1/Protein 2e			3.1873

X1, X2 and Y columns represent the experimentally determined amino acid composition of protein 1, protein2 and conjugate in nmol, respectively; X1, X2 and Y columns represent the normalized amino acid composition of protein 1, protein 2 and conjugate, respectively. Normalization was calculated using this formula (use Asx in protein 1 as example): Normalized Asx in protein 1(cell E1) = 2.5886 (cell B1) ÷ 12.9467 (cell B13) = 0.1999. ϓ column represents the theoretical amino acid composition of a conjugate in % total nmol of protein and calculated as follows (use Asx as example): Asx theoretical conjugates (%) = (normalized Asx in protein 1) × (% mass of protein 1) + ((100% − (% mass of protein 1)) × (normalized Asx in protein 2) = 0.1999 (cell E1) × 0.4816 (cell D16) + (100% − 0.4816 (cell D16)) × 0.1188 (cell F1) = 0.1579 (cell H1). (Y − ϓ)² column represents the squared difference of each amino acid between experimental and theoretical of conjugates and was calculated as follows (use Asx as example): (Y − ϓ)² = (% of Experimentally normalized conjugate − % of theoretical conjugate)² = (0.1582 (cell G1) − 0.1579 (cell H1))² = 7.2703 × 10⁻⁸ (cell I1).

^a% mass of protein 1 is calculated using Microsoft Excel solver.

^bModified molecular weight is the sum of molecular weight of amino acids residues actually used in the calculation. In this example, it is the sum of molecular weight of 14 amino acids (Asx for asparagine and aspartic acid and Glx for glutamine and glutamic acid).

^c $\begin{array}{l}\text{Molar}\text{portion}\text{of}\text{protein}1=(\%\text{mass}\text{of}\text{protein}1)÷(\text{Modified}\text{molecular}\text{weight}\text{of}\text{protein}1)=0.4816\\ (\text{cell}\text{D}16)÷15551.20(\text{cell}\text{D}14)=3.0967\times {10}^{-5}(\text{cell}\text{I}14)\end{array}$.

^d $\begin{array}{l}\text{Molar}\text{portion}\text{of}\text{protein}2=(100\%-\%\text{mass}\text{of}\text{protein}1)÷(\text{Modified}\text{molecular}\text{weight}\text{of}\text{protein}2)\\ =(100\%-0.4816(\text{cell}\text{D}16))÷53361.28(\text{cell}\text{D}15)=9.7156\times {10}^{-6}(\text{cell}\text{I}15)\end{array}$.

^e $\begin{array}{l}\text{Molar}\text{ratio}\text{of}\text{protein}1\text{to}\text{protein}2=\text{Molar}\text{portion}\text{of}\text{protein}1÷\text{Molar}\text{portion}\text{of}\text{protein}2=3.0967\\ \times {10}^{-5}(\text{cell}\text{I}14)÷9.7156\times {10}^{-6}(\text{cell}\text{I}15)=3.1873(\text{cell}\text{I}16)\end{array}$.