Abstract
The ELISA test on the “Ebola virus” was performed. In general, 25 students’ “blood” samples were tested. Initially, only two of them contained “the infection”. However, after samples mixing “Ebola virus” became spread among half of the population. For the determination of “infected” samples, the standard indirect ELISA protocol was used. Positive and negative controls confirmed the veracity of the obtained results. The purpose of the analysis was to estimate the initially “infected” samples. Accomplishing this, the analysis of the samples’ mixing scheme was conducted.
Methods
As a material, “blood” samples were used. Each student has a sample. Only two of them were infected with the “Ebola virus”, while 23 were free from viral infection. Each student randomly mixed his “blood” sample with three other students’ samples, which resulted in the “Ebola virus” spreading among the population. Table 1 shows the scheme of sample mixing. Afterward, ELISA analysis was performed for testing all the samples on “the infection”.
Table 1.
The Scheme of Students’ Samples Mixing
The standard indirect ELISA protocol was used for the analysis. One 12-well strip was used for two students: three wells for the positive control, three wells for the negative control, and three wells for each student’s sample. In the first stage, 50µl of controls and samples were added to the appropriate wells. After 5 min of incubation, the strip was washed two times with 100µl of wash buffer. Before adding the buffer, liquid from the wells was removed from the paper. After washing, 50µl of the first antibody solution from the green tube was added to each well. After 5 min of the incubation and washing two times, 50 µl of the second antibody solution from the orange tube was added. After 5 min of incubation, the strip was carefully washed three times with the purpose to remove unbounded antibodies. Finally, 50 µl of enzyme-substrate was added. The results of the analysis could be observed after 5 min. Positive results were blue, negative ones were colorless.
Results
Figure 1 shows the results of the ELISA test of one team’s samples. According to Figure 1, both students’ “blood” was “infected” with the “Ebola virus”. The liquid in wells from the 7th to the 12th was colored, which means that the analysis was positive. The liquid in positive controls wells was blue, while the negative controls wells remained colorless. Positive and negative controls confirm the correctness of the ELISA analysis procedure and obtained results.
Table 2 demonstrates the results of the ELISA for the whole group’s samples. According to it, a part of students’ “blood” became “infected” after the procedure of mixing, while the rest of them remained free of “virus”. Wells with “infected” samples were blue after the reaction, and wells with “non-infected” samples were colorless. Positive and negative controls also were correct. According to Table 2, twelve samples were colored and contained “Ebola virus”, and thirteen samples were free from the “viral infection”.
Table 2.
Results of the ELISA Analysis
Discussion
The ELISA test shows the presence of the “viral particles” in the sample. If “blood” was “infected”, “viral particles” bounded with the specialized antibodies to them. The second antibodies solution contained the antibodies to the anti-viral antibodies. In the final stage, the enzyme solution was added. This solution changed its color in the case of the second antibody’s presence in the well. Thus, this analysis allows detecting the “infected” samples.
Initially, only two “blood” samples contained the “Ebola virus”. After samples were mixed, the “virus” was spread, and half of the population became “infected”. For the determination of the primary infected samples, connections between participants of the experiment should be studied. First of all, all the samples with negative results should be excluded, as well as the samples, they had contacts with. After this procedure, only three numbers of possible “infected” samples remained: No 1, 7, 13. However, sample No 7 could be excluded because it was mixed with No 24 and 5 and did not spread “the infection” to them. It could be assumed that sample No 7 received “the infection” after mixing with sample No13. Therefore, it could be concluded that samples No 1 and 13 were initially “infected”, and they contaminated all other samples.