Sensorgram comparision and analysis: similarity score assessment.

Hi Arnoud,
I have a query in understanding the similarity scores obtained during sensorgram comparison.
In an MCK assay (Biacore T200) I have run a sample (n=3) and evaluated in version 3.0 with 1:1 binding fit, I have done sensorgram comparison between those three samples by considering n1 and n3 as standard and n2 as sample, as we should have 2 samples as standard for a comparison. In that case, it is considering n1 and n3 sample as 100% similar (irrespective of variability of the values) and n2 as sample (all user defined) and it is giving similarity score of 92% between set standard and sample n2. And if we take n1 and n2 as standard and n3 as sample during sensorgram comparison the similarity score will change to 96%.
Please note that all the samples run are same.

FYI: values as obtained.

Sample 1

n=1 values ka 4.2E+05, kd 0.0088, KD 2.0E-08 and RMax 25.21, chi2 1.17, U value 3
n=2 values ka 4.2E+05, kd 0.0085, KD 2.0E-08 and RMax 24.03, chi2 1.00, U value 3
n=3 values ka 4.1E+05, kd 0.0094, KD 2.2E-08 and RMax 25.07, chi2 1.02, U value 3
Average ka 4.2E+05, kd 0.0089, KD 2.1E-08 and Rmax 24.77
STD Dev 8.4E+03 0.0005 1.5E-09 0.64
%CV 1.99 5.19 6.95 2.60

Sample 2

n=1 values ka 3.871E+05 kd 0.0094 KD 2.451E-08 RMax 29.22 chi2 1.27 U value 3
n=2 values ka 4.705E+05 kd 0.0085 KD 1.819E-08 RMax 27.18 chi2 1.56 U value 5
n=3 values ka 5.128E+05 kd 0.0093 KD1.819E-08 RMax 26.75 chi2 1.47 U value 3
Average ka 4.568E+05 kd 0.009 KD 2.030E-08 RMax 27.72
STD Dev 6.398E+04 0.000497 3.647E-09 RMax 1.32
%CV 14.01 5.44 17.96 4.77

For the same type of analysis for another sample, Sample 2, I have got a CV of 14.01%, 5.44% and 17.96% for ka, kd and KD respectively. And for this sample the similarity score is 97%.

Could you please explain us how does the values are considered and similarity scores are evaluated, how is the algorithm applied during sensorgram comparision.
Kindly revert if in case of any clarification.

Thanks in advance,
Rajath

Hi Rajath,

I did not use the function myself but found the following paper by GE: www.gelifesciences.co.jp/contact/pdf/29_1519_21aa.pdf

The Similarity Score is based on the number of data points inside and outside the standard deviation corridor. This SD corridor (= Mean ± SD) is calculated from the standard curves. Although you can calculate the SD corridor from two measurements, it is recommended to use more curves that are collected over several days (batches / instruments / operators) to assess the total variation of the assay.

Data points falling inside the SD corridor are always rated 100%. The sum of squared distances (RU² ) to the average curve for outside sample and for SD corridor points are compared for all data points in the sensorgrams. The final score is based on a combined rating of inside vs outside data points.
It is possible to compare either whole sensorgrams or just association or dissociation phases.

The calculated Similarity Score for a sample where 80% of the sample data points are inside the SD corridor: Similarity Score = 80*1 inside + 20* 0.16 = 83.2
0.16 being an example of an obtained ratio between sum of squared distances to the average curve, that is SD corridor point distances/sample distances. Thus the number of data points outside and how far each of them are from the SD corridor both influence the similarity score.
Kind regards
Arnoud