clinical immunology lab report (ELISA)

1 EDU30014 Health and Physical Education Teaching Period 1, 2015 EDU30014 Health and Physical Education Assessment 1: Health and physical education fo
June 21, 2020
Ecological History – Catastrophes (Poster & a critical reflection)
June 21, 2020

clinical immunology lab report (ELISA)

clinical immunology lab report (ELISA)

Project description
The aim is to understand the significance of ELISA assay for investigation of disease in biomedical science. This practical has done in two sessions with two different aims. Practical 1 was aimed to find out the optimum dilution for the monoclonal mouse anti-rabbit IgG antibodies and polyclonal goat anti-rabbit IgG antibodies by using different dilutions. This was done to be used in the following practical which was done on (practical 2) that aimed to find out two unknown samples concentrations (X & Y) by drawing a standard curve using known concentrations of Rabbit IgG antibodies.

table before subtracting:
Absorbance
Antigen concentration ng/ml    1/2000.    1/4000.    01/8000.     1/16000    1/32000    1/64000
2000    2.7648    2.7747    2.7177    2.4949    2.006    1.6752
1000    1.9246    1.7403    1.7104    1.7088    1.5611    1.2899
500    1.4618    1.415    1.3463    1.3247    1.2398    0.9302
250    1.0443    1.0552    1.0113    0.9484    0.8535    0.706
125    0.7442    0.7162    0.7102    0.6784    0.6014    0.4944
62    0.4726    0.4468    0.422    0.4004    0.3454    0.2816
31    0.2948    0.2917    0.2816    0.2442    0.2189    0.1823
After subtracting from blank:
Absorbance
Antigen concentration ng/ml    1/2000.    1/4000.    01/8000.     1/16000    1/32000    1/64000
2000    2.6892    2.6957    2.6476    2.4402    1.9545    1.5864
1000    1.849    1.6613    1.6403    1.6541    1.5096    1.2011
500    1.3862    1.336    1.2762    1.27    1.1883    0.8414
250    0.9687    0.9762    0.9412    0.8937    0.802    0.6172
125    0.6686    0.6372    0.6401    0.6237    0.5499    0.4056
62    0.397    0.3678    0.3519    0.3457    0.2939    0.1928
31    0.2192    0.2127    0.2115    0.1895    0.1674    0.0935
0    0    0    0    0    0    0

Absorbance
Antigen concentration ng/ml    1/2000.    1/4000.    01/8000.     1/16000    1/32000    1/64000
2000    4    3.0214    1.8698    1.0056    0.5155    0.2803
1000    3.2376    2.3939    1.4874    0.8459    0.4509    0.2555
500    2.9955    2.145    1.2675    0.7272    0.3639    0.2046
250    2.6414    1.7378    0.9693    0.5602    0.2727    0.1624
125    2.0356    1.3078    0.6753    0.398    0.2205    0.1214
62    1.3528    0.7971    0.4588    0.263    0.1374    0.0825
31    0.8509    0.5246    0.2971    0.1726    0.0944    0.0565
After subtracting from blank:
Absorbance
Antigen concentration ng/ml    1/2000.    1/4000.    01/8000.     1/16000    1/32000    1/64000
2000    3.7442    2.9901    1.8681    0.9853    0.5126    0.2605
1000    2.9818    2.3626    1.4857    0.8256    0.448    0.2357
500    2.7397    2.1137    1.2658    0.7069    0.361    0.1848
250    2.3856    1.7065    0.9676    0.5399    0.2698    0.1426
125    1.7798    1.2765    0.6736    0.3777    0.2176    0.1016
62    1.097    0.7658    0.4571    0.2427    0.1345    0.0627
31    0.5951    0.4933    0.2954    0.1523    0.0915    0.0367
0    0    0    0    0    0    0

2000    0.3899
1000    0.3526
500    0.2643
250    0.2511
125    0.2099
31    0.06965

Calculation:
y = 0.0746ln(x) – 0.1726
Y= Absorbance (X&Y)
X= Concentration (X&Y)
X= 0.2485
Y= 0.111
Applying the equation For X:
0.2485= 0.0746 ln((x) – 0.1726
0.2485 + 0.1726= 0.0746 ln(x)
0.4211= 0.0746 ln(x)
0.4211/0.0746 = 0.0746 ln(x)/ 0.0746
Ln(x)= 5.6447
X= 282.78
Applying the equation for Y:
0.111= 0.0746 ln((x) – 0.1726
0.111 + 0.1726= 0.0746 ln(x)
0.2836= 0.0746 ln(x)
0.2836/0.0746 = 0.0746 ln(x)/ 0.0746
Ln(x)= 3.801
Y= 44.74