The gene that we studied, TAS2R38, has three possible point mutations (SNPs), the functions of which are crucial to the taste receptors. In order to identify these point mutations, we had to procure two DNA probes for each set. One of the probes can bind to DNA when no mutation is present (wild type), and the other when a mutation (mutant) is present. Each of the probes is labeled with a different fluorescent color; in our case the dyes FAM and HEX were used. Using these dyes, we could determine which probe would bind to our DNA sample. Apart from this dye, the probes also carry a so-called quencher, which would suppress the fluorescence of the dye as long as they are bound to the DNA probes. However, if the dye is released by the probe, the quencher can no longer suppress the fluorescence and it emits a - for us measurable – fluorescence.

The release of the dye is caused by the PCR (PCR = Polymerase Chain Reaction, a technique of amplifying DNA), which can only happen when the DNA probe can bind completely to a target sequence (= our area of the particular point mutation). Generally, DNA fragments are copied through PCR. The exact size and position is determined through so-called primers. To do so, a PCR requires two primers that flank the desired DNA segment, which, in our case, are the segments that can carry a mutation. Only these DNA sections are amplified through PCR. The reproduction occurs exponentially, meaning that during each PCR cycle, the previous number of DNA fragments doubles (after 50 PCR cycles we have 250 = ca. 1015 times as much DNA as at the beginning). The formation of the new DNA strands happens because of the DNA polymerase. In our system, the DNA polymerase has a further task: it also ensures the release of our fluorescent dye in the probes (Fig. 1). But only the probes that can bind to DNA one hundred percent of the time lose their dye. When the dye splits off from the DNA, the fluorescence is no longer suppressed by the quencher, which is still attached to the probe. This fluorescence is what we can measure. Using the qPCR, we could measure the fluorescence generated by each PCR cycle. With the resulting curve (example Fig. 2), we could determine whether the wild type variant or mutant was present.

At this point, we can determine which genotype is present based on whether the dye that marks the wild type sequence, the dye of the mutant probes or both dyes are measurable. If only one dye can be determined, the homozygote genotype is present, which means that both chromosomes carry the same type of gene variation. This can be either the non-mutated wild type or the mutation. If we can measure both dyes, the wild type gene is present on one chromosome and on the other the mutation. We cannot determine on which chromosome which variety is present. For a better evaluation and representation of the results, we felt that we needed to plot the value of the fluorescence of the single dye on a coordinate system (Fig. 3). In the graphic, three groups of measured values are represented. The squares on the far left stand for the DNA, which are homozygous of the wild type. The circles to the far right show the homozygous mutation. The largest group of triangles in the middle of the graphic show the DNA trials that are heterozygous, where one chromosome carries the point mutation and the other one does not.

Heterozygote and homozygote samples can be easily recognized by the fluorescence intensity curves of both of the dyes that were used. It shows striking differences between the homozygote wild type gene and heterozygote and homozygote mutations (Fig. 4). In homozygous samples, only one of the two dyes showed a measurable increase in fluorescence, while the other dye did not show any notable increase in fluorescence (Fig. 5). Whether it has to do with a homozygous wild type or a homozygous mutation can easily be ascertained by finding out which dye shows an increase in fluorescence and which does not. In a heterozygous genotype, both dyes show a strong increase in fluorescence of roughly the same amount (Fig. 6). Therefore, it is the case that the wild type, as well as the mutant gene, are present.