The determination of D-dimer by dual particle size latex immunoturbidimetry

doi:10.3969/j.issn.1673-8640.2016.09.015

Abstract

Abstract:

Objective To improve the analytical sensitivity and linear range of latex immunological reagent,and to compare the sensitivities and linear ranges of dual and single particle size latex immunoturbidimetry in the determination of D-dimer（DD）. Methods Carboxylic latex particles with diameters of 49 and 150 nm were prepared,and a covalent coupling method was used to link DD monoclonal antibody to the particles. The 2 sensitized latex particles were mixed at various proportions,and the sensitivity and linear range were determined,which were compared with those of imported reagents. Results If 4 parts of 49 nm sensitized latex particles were mixed with 1 part of 150 nm sensitized latex particles,the optimal linearity of calibration curve was obtained,with the sensitivity of 0.03 and linear range of 61.88 μg/mL,and it correlated well with imported reagents. Conclusions Dual particle size latex immunoturbidimetry is better than traditional DD latex diagnostic assay,and is of potential for clinical application.

Key words: D-dimer, Immunoturbidimetry, Dual particle size latex, Sensitivity, Linear range, Automatic blood coagulation analyzer

CLC Number:

R446.1

FANG Qing. The determination of D-dimer by dual particle size latex immunoturbidimetry[J]. Laboratory Medicine, 2016, 31(9): 801-806.

Figures/Tables 10

References 7

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胶乳浓度（mg/mL）	回归方程	R²值
3	Y = 0.000 6X + 0.000 4	0.997 1
4	Y = 0.001 3X + 0.000 7	0.997 8
5	Y = 0.001 9X + 0.000 7	0.998 3
6	Y = 0.002 5X + 0.000 8	0.998 9
7	Y = 0.002 6X + 0.000 7	0.999 3
8	Y = 0.002 9X + 0.000 6	0.999 6
9	Y = 0.003 2X + 0.003 5	0.976 2

胶乳浓度（mg/mL）	回归方程	R²值
3	Y = 0.000 6X + 0.000 4	0.997 1
4	Y = 0.001 3X + 0.000 7	0.997 8
5	Y = 0.001 9X + 0.000 7	0.998 3
6	Y = 0.002 5X + 0.000 8	0.998 9
7	Y = 0.002 6X + 0.000 7	0.999 3
8	Y = 0.002 9X + 0.000 6	0.999 6
9	Y = 0.003 2X + 0.003 5	0.976 2

胶乳浓度（mg/mL）	回归方程	R²值
1	Y = 0.006 6X -0.001 7	0.999 4
2	Y = 0.007 8X -0.002 8	0.997 7
3	Y = 0.008 6X -0.003 1	0.996 9
4	Y = 0.007 4X + 0.004 3	0.987 9

胶乳浓度（mg/mL）	回归方程	R²值
1	Y = 0.006 6X -0.001 7	0.999 4
2	Y = 0.007 8X -0.002 8	0.997 7
3	Y = 0.008 6X -0.003 1	0.996 9
4	Y = 0.007 4X + 0.004 3	0.987 9

49 nm:150 nm	回归方程	R²值
1:1	Y = 0.008 1X + 0.006 1	0.989 3
1:2	Y = 0.008 6X + 0.016 4	0.931 2
1:3	Y = 0.007 6X + 0.027 6	0.810 4
1:4	Y = 0.008 2X + 0.028 9	0.797 6
2:1	Y = 0.005 4X -0.003 4	0.992 0
3:1	Y = 0.006 6X -0.001 2	0.999 6
4:1	Y = 0.007 2X + 0.001 6	0.999 7