

Center for Disease Control (CDC) West Nile Virus (WNV) IgM ProtocolDocument ID: DQ105 Relevant to following products: Stingray WorkOut 1.5 Document description: How to setup: Center for Disease Control (CDC) West Nile Virus (WNV) IgM Protocol
Center for Disease Control (CDC) West Nile Virus (WNV) IgM Protocol
Typically this assay is run with all controls and patient samples in triplicate. The microplate layout contains 1 blank group, the patient samples (virals and normals) and control groups.
The first transform to setup is the “Blank Correction” this simply removes the average of the blank group from all wells resulting in the “Blank Corrected” data. Use the Blank Correction wizard to setup this transform.
The second transform to setup is the “Viral divided by normal” transform. This simply divides each patient sample by its adjacent normal sample (on the blank corrected data). Create a new userdefined transform with the blank corrected data as the input matrix and tick the option to use automean. The output matrix can be set to “V/Normal” Then for each group to calculate the division, select the wells and enter the expression for the division by an adjacent sample, such as:
sample1/sample2
Then press return to store the settings. Repeat this process across the plate as required.
The third transform is the “Positive divided by negative”. This simply divides each patient sample by its associated control (once again on the blank corrected data). Create a new userdefined transform with the blank corrected data as the input matrix and tick the option to use automean. The output matrix can be set to “P/N". Then for each group to calculate the division, select the wells and enter the expression for the division by the associated control, such as:
sample1/control1
A fourth transform is required to facilitate the interpretation of the patient results. Create a new userdefined transform with the X value set to the “P/N” matrix and Y to the “V/Normal” matrix. The output matrix can be set to “Interp.” Then select all wells and enter the following expression (see below for more details):
x * (((y<2)*2) + 1)
The assay uses the Validity Parameters: P/N for the positive control must be greater than or equal to 3.00 for test to be considered valid. To implement this validation test, select Validations and select the P/N tab and enter the validation expression:
control1>=3
The assay uses the following Patient Result Interpretation:
POSITIVE = P/N greater than or equal to 3.00 with background greater than 2.00
EQUIVOCAL = P/N between 2.00  2.99 with background greater than 2.00
INDETERMINATE = P/N between 2.00  2.99 with background less than 2.00
NEGATIVE = P/N less than 2.00
To implement this, select Cutoff and select the “Interp.” tab. Enter the following cutoff expressions and labels:
POSITIVE: x > = 3
EQUIVOCAL: (x>=2) and (x<2.99)
INDETERMINATE: (x<=2) and (x>=2.99)
NEGATIVE: ((x < 2) and (x > 0)) or ((x >2) and (x < 0))
Now the test can be saved and run.
The following section describes more about the Interpretation step of the protocol.
Since this assay’s patient interpretation is based on data from two matrices (P/N and V/Normal) it is essential to combine the data from these two matrices into one matrix for the qualitative analysis.
This is achieved using the (((y<2)*2) + 1) expression. Here the Boolean logic expression, (y<2) is evaluated to 0 (false) or 1 (true). *2 and +1 is used to convert 0 to 1 and 1 to 1 (so the sign stores the logic result). Multiplying x by this result combines the P/N value (x) with whether the V/Normal is < 2 in one matrix.
So this expression really means:
if y<2 then  if y>=2 then +
Therefore, to do this longhand for background <2 (lets say 1) and P/N of 1.2:
= 1.2 * (((1<2)*2)+1) = 1.2 * (((1)*2)+1) = 1.2 * 1 = 1.2 This example would be a NEGATIVE
Now for background >2 (lets say 3) and a P/N of 1.3:
= 1.3 * (((3<2)*2) + 1) = 1.3 * (((0)*2) + 1) = 1.3 * 1 = 1.3 This example would also be NEGATIVE
More examples:
Background of 3 and a P/N of 2.5:
= 2.5 * (((3<2)*2) + 1) = 2.5 * (((0)*2) + 1) = 2.5 * 1 = 2.5 = EQUIVOCAL
Background of 3 and a P/N of 3.5:
= 3.5 * (((3<2)*2) + 1) = 3.5 * (((0)*2) + 1) = 3.5 * 1 = 3.5 = POSITIVE
Background of 1 and a P/N of 2.5:
= 2.5 * (((1<2)*2) + 1) = 2.5 * (((1)*2) + 1) = 2.5 * 11 = 2.5 = INDETERMINATE
Example files available for download here.
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