To confirm that your restriction digests were complete (or nearly so), you analyzed a sample of each reaction by agarose gel electrophoresis. In addition to the restriction enzyme digested DNAs, you also included a sample of each uncut DNA and DNA size markers (which contains lambda virus DNA digested with the restriction enzyme BstEII to produce fragments of known size). Each uncut DNA serves as a negative control; the DNA size markers allow you to determine the size of the fragments produced by the restriction digests.
Since each uncut DNA is circular, you might expect to see a single band that comigrates with a similarly sized fragment of the DNA size markers. However, because linear and circular DNA molecules have different conformations (shapes), they migrate at different rates. For example, a 5,000 base pair (bp) linear fragment will not comigrate with a 5,000 bp circular plasmid. Furthermore, a circular plasmid can actually exist in two conformations. In the bacterial cell, the plasmid is wound around proteins making it supercoiled. When isolated from the cell, nicks may be introduced into the sugar-phosphate backbone of the circular plasmid DNA causing it to become relaxed. As a result, the isolated plasmid DNA then consists of two populations: supercoiled circles and relaxed circles. Because the supercoiled form of the plasmid has a more compact structure, it migrates faster than the relaxed form. (The linear form of the plasmid usually runs at an intermediate rate.)
To make things even more complicated, you may see more than two bands because E. coli recombination enzymes may join two plasmid circles together to make a double length circle. In summary, you should expect to see several bands in each lane of uncut DNA; the fastest two bands should correspond to the supercoiled and relaxed circles.
First, to confirm that digestion has indeed occurred, you should compare the lanes with the digested DNAs to the corresponding uncut DNA lanes. Then to confirm that the digestion products are the expected size, you need to compare their migration with those of the known fragments in the DNA size markers lane (see Figure 1).
Note that the pattern of bands in your DNA size markers lane may vary slightly from Figure 1. First, the 702 base pair (bp) and smaller DNA fragments may not be visible (because of their lower concentration). Second, the bands at the top (containing the largest fragments) may blur and consequently be difficult to distinguish individually. Third, the 8.5 kb and 5.7 kb fragments can link to each other to form a larger fragment of ~14 kb (kilobases = 1000 bases). Consequently, you'll see an additional band at the top of this lane and the 8.5 kb and 5.7 kb bands may be barely visible.
To identify each band unambiguously, line up bands from your photo with those shown in Figure 1 by looking for similar patterns. For example, note that near the bottom there are two closely spaced bands of 1,264 and 1,371 bp. (Remember that the single band of 0.7 kb may not be visible.) Starting from this pair on your photograph, you can move upward to identify the larger fragments (realizing, as noted above, that the 8.5 kb and 5.7 kb bands may be less intense or missing because they can join together to make the top, ~14 kb band, which is not shown in the figure).
Once you have identified the size of each band in your DNA size markers, you can then determine the size of the DNA fragments produced by the restriction digests. To accurately determine the size of the restriction fragments, you should prepare a graph plotting the migration distance of each DNA size marker fragment against its length. However, for our purposes, an "eyeball" estimate is sufficient: therefore simply confirm that each of your restriction fragments comigrates with the DNA size marker fragment(s) that are approximately the expected size of your restriction fragment.
You may have noticed that the bands vary in their staining intensity. Band intensity reflects the amount of ethidium bromide bound to each fragment, which is directly related to the mass of the DNA present in a band. Because the DNA size markers were produced by digesting a single DNA molecule (the lambda virus chromosome) with a restriction enzyme, the molar amount (or number of molecules) of each fragment is identical. However, the same amount of a larger fragment (e.g., the 8,454 bp fragment) will have a greater mass and will be able to bind more ethidium bromide than a smaller fragment (e.g., the 702 bp fragment). The mass of DNA in a given band can be calculated using the following equation:
DNA mass = fragment length (in bp) x original DNA concentration x
volume loaded
÷ total length of the DNA fragments (= length of lambda DNA
chromosome = 48, 502 bp)
The original DNA concentration of the DNA size markers was 0.4 µg / 6 µl and you loaded 10 µl. So if we want to determine how much DNA is in the 4,822 bp band, we simply plug in the numbers:
DNA mass = 4,822 bp x (0.4 µg / 6 µl) x (10 µl) / 48, 502 bp) = 0.07 µg = 70 ng
So if one of your bands has an intensity similar to the 4,822 bp DNA fragment, it has about 70 ng of the fragment present.
To determine the amount of DNA of your fragments, first find a DNA size marker fragment that has a similar intensity. Then determine the amount of DNA present in the size marker band (and hence yours) using the above equation. Determine the mass amount of each of your restriction fragments.
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