Team:Yonsei Korea/Results
In the Laboratory:
The Science
Behind It All
In this section, we plan to analyze the most important results that we obtained from our experiments. This is divided into two sections largely: part 1 is about DNAzyme cleavage assay results and part 2 is about nanoparticles. In part 1, we have analyzed PAGE and urea PAGE results. First, we have interpreted and analyzed each band on the gel. Second, to quantify the amount of DNA present in each band, we measured the net band intensity with Fiji, an open source program. In part 2, we have been expecting to yield results from our nanoparticles experiments; however, we are still in the process of them.
PART 1: DNAzyme CLEAVAGE ASSAYSamples DNAzyme-gene (substrate) A were reacted with 1x, 2x, and 5x reaction buffers. For each concentration of reaction buffer, 4 samples were tested: A-1, A-2, A-3, and A-4. Details about each sample can be found below:
| 1 | DNAzyme only |
|---|---|
| 2 | Gene (substrate) only |
| 3 | DNAzyme + gene, but without catalyzing the cleavage reaction |
| 4 | DNAzyme + gene, with catalyzing the cleavage reaction |
After the cleavage reaction, it was visualized with native PAGE (12% and 15%). The gel was run at 90V for 120 minutes. The figure below represents the native PAGE results.
1. Interpretation & analysis of bands
It was possible to observe bands at 4 distinct locations, which are (a), (b), (c), and (d) in Figure 1. (c) and (d) each represents DNAzyme A and Gene A. (a) is observed in A-3 and A-4, but not in A-1 and A-2. Therefore, (a) would represent DNAzyme-Gene A hybridized, but not reacted or cleaved. This is because DNAzyme requires specific ions such as Cu2+ and Mn2+ to exercise its cleavage activity. However, since none of these are present in A-3, DNAzyme would not be catalyzed, and therefore would not cleave the gene (substrate) even though the two stands are hybridized with each other. (b) is only observed in A-4, and this represents that the cleavage activity occurred. Since A-4 contains important ions such as Cu2+ and Mn2+, activated DNAzyme could cleave the gene (substrate). As gene (substrate) is cleaved at CTGC by DNAzyme, this 4bp would be released and the structure of the DNAzyme-gene complex would change. In other words, the cleaved gene except for CTGC is still hybridized with the DNAzyme binding arms, but has a different secondary structure compared to unreacted DNAzyme-gene hybridized structure. Since migration in PAGE is also affected by structure as well as size, the resulting product would be present at (b).
The diagram below represents possible structures that might have formed:
- ● Green: DNAzyme
- ● Yellow: Left-binding arm of the target gene
- ● Red: CTGC sequence - sequence where DNAzyme specifically recognizes the target gene
- ● Orange: Right-binding arm of target gene
2. Quantification of the band intensity
Using Fiji, an image analysis software, we measured the light intensity of each of the bands (Figure 2a). Fiji measures the mean of the light intensity of the pixels within the selected area. The area of measurement was kept constant as 925. The result is shown as in Figure 2b (15% PAGE) and 2c (12% PAGE). (a), (b), (c), (d) represent the same thing as above.
The quantification process was conducted as:
- 1) Since the background is blue, adjust the picture into an 8-bit image.
- 2) Measure the mean light intensity of the background and each band.
- 3) Subtract the background value from the value of each band to calculate the net band intensity.
3. Analysis
With the band intensity measurement results, we compared and analyzed the following:
- ● Comparison between A-1 and A-4 at each concentration of reaction buffer: this would allow us to see whether or not our designed DNAzyme participates in the cleavage reaction.
- ● Comparison between A-3 and A-4 at each concentration of reaction buffer: this would allow us to see whether or not our designed DNAzyme exercised the cleavage activity.
As shown in Figure 4, the net band intensity of A-4 at (c) is decreased compared to the net band intensity of A-1 at (c). Since (c) represents solely the DNAzyme, this clearly shows DNAzyme participated in cleavage reaction. In 15% PAGE, the net band intensity of A-1 is almost the same; however, the net intensity of A-4 at (c) decreases more, as the reaction buffer concentration increases. This is because as the concentration of the reaction buffer increases, the amount of important ions (Cu2+ and Mn2+) increases, which means that more DNAzyme will be activated, since they require these ions for catalytic activity. However, this trend is not quite observable in 12% PAGE: the net band intensity of A-1 is low in 5x reaction buffer condition. As these samples were run at the end, it was hard to obtain a high quality picture of them, since the light intensity of the picture at the end is diminished compared to the other lanes.
As shown in Figure 3, the net band intensity of A-4 at (b) is higher than the net band intensity of A-3 at (b). Since (b) represents the cleaved DNAzyme-gene (substrate), this clearly shows that DNAzyme exercised cleavage activity in the presence of the ions (Cu2+ and Mn2+) in the reaction buffer.
4. Summary of the results
- ● It is clear that the DNAzyme participates in the cleavage reaction, and this amount is proportional to the concentration of the ions that catalyze its cleavage activity.
- ● DNAzyme cleavage activity is present only in the presence of certain ions. Without it, DNAzyme and substrate only hybridize: no cleavage activity is observable.
Sample DNAzyme-gene (substrate) A was reacted with a 2x reaction buffer. Here, a total 7 samples were tested: A-1 to A-7. Unlike in native PAGE, only the 2x buffer was used. Details about each sample can be found below:
| 1 | DNAzyme only |
|---|---|
| 2 | Gene (substrate) only |
| 3 |
DNAzyme + gene + 2x reaction buffer without CuCl2 (DNAzyme : gene = 1 : 0.1) |
| 4 |
DNAzyme + gene + 2x reaction buffer with CuCl2 (DNAzyme : gene = 1 : 0.1) |
| 5 |
DNAzyme + gene + 2x reaction buffer with CuCl2 (DNAzyme : gene = 1 : 0.25) |
| 6 |
DNAzyme + gene + 2x reaction buffer with CuCl2 (DNAzyme : gene = 1 : 0.5) |
| 7 |
DNAzyme + gene + 2x reaction buffer with CuCl2 (DNAzyme : gene = 1 : 1) |
Samples A-4, 5, 6, 7 all contain 2x reaction buffers including CuCl2, but the ratio between DNAzyme: gene concentration varies. However, the concentration of DNAzyme is kept constant. A-4 has the minimum concentration and A-7 has the maximum concentration of gene A.
After the cleavage reaction, it was visualized with urea PAGE (8%). The gel was run at 200V for 40 minutes. The figure below represents the urea PAGE results.
1. Interpretation & analysis of bands
Neglecting side bands and noises, it was possible to observe meaningful bands at 4 distinct locations, which are (a), (b), (c), and (d) in Figure 5.
Some things to note is that although urea PAGE was used to separate possible hybridization complexes into single stranded DNA, the results showed that denaturing did not occur properly. In other words, the urea within the gel mesh would not be useful because the denaturation was not complete during the sample preparation step. To get rid of this, we could extend the denaturation time for 30 minutes or more. Also, since our target gene substrate and DNAzyme is single-stranded as its own, it could have formed additional secondary structures on its own. This led us to observe a similar band pattern as in native PAGE results.
In addition, a lot of side bands and noises were detected, as seen at the upper and lower part of the urea PAGE gel. We ordered our DNA products without HPLC purification, meaning that other truncated synthesis products were not removed. Ensuring the high purity of our samples was not possible due to funding constraints, so this could have affected our results. This was evident from the appearance of several unspecific bands (contributing to a high background noise) in all lanes including the control groups: ssDNAzyme and single-stranded target gene.
(c) and (d) each represent DNAzyme A and Gene A. (a) is observed in A-3 and A-4, but not in A-1 and A-2. Therefore, (a) would represent DNAzyme-Gene A hybridized, but not reacted or cleaved. This is due to the same reasons as in the native PAGE section. However, (b) appeared in urea PAGE, which was not present in native PAGE results. Unlike the thin band (b) in native PAGE, 2 bands were overlapping in urea PAGE (b). It is difficult to observe the 2 overlapping bands when the gene (substrate) concentration is low, but it becomes clear as the concentration increases, as shown in A-7 sample. (b) represents the cleaved products: the intensity is very weak in A-3, whereas it is clearly visible in A-4 to A-7. In fact, the weak intensity of (b) at A-3 might be just noise, as the intensity is very dim. The cleaved products (b) in urea PAGE would be analogous to (b) in native PAGE, although there is a difference in the appearance and the location of the bands. Although denaturation by urea was not 100% efficient, urea definitely would have affected the structure of the cleaved DNAzyme-gene structure, resulting in a different structure from (b) in native PAGE.
The diagram below represents possible structures that might have formed:
- ● Green: DNAzyme
- ● Yellow: Left-binding arm of the target gene
- ● Red: CTGC sequence - sequence where DNAzyme specifically recognizes the target gene
- ● Orange: Right-binding arm of target gene
2. Quantification of the band intensity
As we did in native PAGE result analysis, we used Fiji to measure the light intensity of each of the bands (Figure 6a). The area of measurement was kept constant as 940. The result is shown in Figure 6b. (a), (b), (c), (d) represent the same thing as above.
The quantification process was conducted as:
- 1) Measure the mean light intensity of the background and each band.
- 2) Subtract the background value from the value of each band to calculate the net band intensity.
3. Analysis
With the band intensity measurement results, we compared and analyzed the following:
- Comparing (b) net band intensities for A-3 to A-7 to check the following:
- ● Whether or not our DNAzyme is specifically catalyzed by Cu2+
- ● Whether or not DNAzyme cleavage activity of proportional to the amount of DNA (substrate) present
Cu2+ requirement for DNAzyme catalytic activity
The difference between A-3 and A-4~A-7 was the presence of Cu2+ ions in the reaction buffer: in A-3, all the other ions such as Mg2+, Na+, K+, Mn2+ were present, except for Cu2+. When we compare the net band intensity of A-3 to others (A-4~A-7), the net band intensity of A-3 is significantly low. However, the catalytic activity of A-3 is not exactly zero: due to Mn2+, it was possible for the DNAzyme to cleave the gene (substrate), but with a very low efficiency. Therefore, in order for our DNAzyme to be efficient in cleaving DNA, Cu2+ is essential.
DNAzyme cleavage activity is proportional to the amount of DNA (substrate)
Samples from A-4 to A-7 had an increasing concentration of the target substrate, while that of DNAzyme remained unchanged (10μM). The 1:1 ratio between DNAzyme and substrate (10μM:10μM) was the most effective reaction since we observed a shrunken band size at (b) (represents DNAzyme) compared to other samples. This indicated that DNAzyme cleaved the available substrate fully, leaving only unreacted DNAzymes. Therefore, it is possible to conclude that our detection system based on the DNAzyme cleavage assay is effective and efficient.
4. Summary of the results
- ● Cu2+ is specifically required for the DNAzyme cleavage activity, since it highly increases the efficiency of cleavage. Without Cu2+, cleavage activity is hardly observable.
- ● DNAzyme cleavage activity is proportional to the amount of DNA (substrate present), which makes the detection system based on the DNAzyme cleavage assay effective and efficient.
Phase 1 of using gold nanoparticles for our detection system is to functionalize them with our 20 base pair complementary sequences to our target gene. Here we began the salt aging method to functionalize both thiolated DNA sequences TS1: 3’-tgccggtcacggccgctgtc-5’ and TS2: 3’-atcgttggggtgaccgagcc-5’ to two samples of gold nanoparticles. In total we have four samples (2 for TS1 and 1 for TS2). This enables better chances of proceeding with better functionalized gold nanoparticles as some samples could aggregate early on in the experiments.
- 1. Sample A1: 3’-tgccggtcacggccgctgtc-5’ functionalized
- 2. Sample A2: 3’-atcgttggggtgaccgagcc-5’ functonalized
- 3. Sample A3: 3’-tgccggtcacggccgctgtc-5’ functonalized
- 4. Sample A4: 3’-atcgttggggtgaccgagcc-5’ functonalized
To functionalize our gold nanoparticles, first the oligonucleotide probes were prepared with PBS buffer (pH 8) and purified water. In order to functionalize them to our GNPs we needed to determine which samples had the DNA present in them; therefore, we used a nanodrop UV-Vis machine to detect the DNA absorbance at 260nm. The results were obtained as followed for all of our samples:
Some samples did not contain DNA while others contained it, so we employed those with the DNA further in the experiment.
In the next steps, phosphate adjustment buffer (final 9mM), surfactant solution (final ~0.1% (wt/v)) and salting buffer (final 0.3M NaCl) were added with the DNA and gold nanoparticles. The salting buffer was added six times over the course of two days. Our gold nanoparticles did not have any visible aggregates except for one sample, meaning that they have been properly functionalized.
Due to the wiki freeze deadline, we are still in the process of the next phase to hybradize our functionalized gold nanoparticles with our target gene and observe a color change from red to purple. We shall present these results in time for the Giant Jamboree.
Our DNAzyme needs to be checked against another substrate sequence, to confirm that exact sequence specificity is required for cleavage activity. Moreover, our gold nanoparticles should be further hybridized with our target gene to induce a color change from red to purple.
In conducting the DNAzyme cleavage assay, it is important that all DNA samples are kept on ice while performing the experiments. Otherwise, it would cause DNA degradation, leading to inexact results. In addition, the denaturing step in urea PAGE should be done under a sufficient time. Since DNAzyme and target gene DNA are single stranded, there is a higher possibility of forming secondary structure on its own.
Yonsei_Korea is a convergent team of students based in Seoul, South Korea striving to create meaningful change utilizing synthtic biology in the iGEM 2021 Competition.