-Abstract-Introduction-Materials and Methods-I still have toReplace all instances of Micrograms and microliters with the correct symbol.
change "Log of MW" in the standard curve to "Log of (# of Base Pairs)".
Add the KS lane bands into the table of the Excel graph (Uncut:4.6, 8.8, 10.4 Pst+, Xho,
and Pst+/Xho: 8.8)
Perhaps increase accuracy of present Excel graph by splitting it into two best-fit lines.
Mark the lanes of Both Gel Images (Numerically, 1-20) and number the marker bands.
Transfer "efficiency of transformation" data into an excel graph.
-ResultsThe following is the efficiency of the transformation according to the counts of the
successful bacterial colonies.
The number of colonies that resulted on the different plates were:
Plate
# of microL of plasmid
# of colonies
-Amp+XGal+IPTG no plasmid
Many many colonies
+Amp+XGal+IPTG no plasmid
Zero colonies
+Amp+XGal+IPTG 2 microL of KS+
152
+Amp+XGal+IPTG 10 microL of KS+
334
+Amp+XGal+IPTG 2 microL of Chit
81
+Amp+XGal+IPTG 10 microL of Chit
378
The equation for the calculation of success of transformation per bacterium is as follows:
1mL of competent cells * 0.5*10^8 cells/mL * 0.1mL/2mL * dilution factor of 9 =
2.25*10^7 cells
so # of colonies/2.25*10^7 cells = X successful colonies/bacteria
The equation for the calculation of success of transformation per plasmid is as follows:
0.1 micrograms of plasmid dna * 1 g/10^7 microg * 1 plasmid/2961 Base Pairs * 1 Base
Pair/600 Daltons * 1 Dalton/(1g/mole) * 6.02*10^23 molecules/mole * dilution factor of
9 = 3.05*10^10 plasmids
so # of colonies/(3.05*10^10) plasmids = # of successful colonies/plasmid
The equation for the calculation of success of transformation per microgram of plasmid
DNA is as follows:
2 microL of plasmid * 50 micrograms of plasmid/1 microL * 1 something/.1 something *
dilution factor of 9 = 9000 microg of plasmid
so # of colonies/9000 microg of plasmid = X successful colonies/microg of plasmid
[Insert Excel efficiency table here]
Quantification of Purified DNA
The purity of the DNA sample with respect to RNA contamination was determined using
the ratio of absorbance at optical densities of DNA and RNA.
The ratio of OD260 to OD280 was 0.047/0.027, or 1.74, which is equivalent to 2.35
Micrograms/microLiter.
The DNA sample Concentration was calculated using the following equation:
(0.047 OD260 * 50 microg/mL per 1.0 OD260) * 500 times dilution factor = 1175
microg/mL = 1mL/1175microg = .000851 mL/microg * 1000 microL/mL = 0.851
microL/microg * 3 microg = 2.55 microL of DNA
Results of the SYBR Gel
[Insert labelled SYBR Gel Image here]
The SYBR Gel was not used due to the lack of clarity of the bands on the gel. Instead,
the more distinct Ethidium bromide gel was used, since the bands on the gel were of the
same type in the same locations.
Results of the Ethidium Bromide Gel
[Insert labelled Ethidium Bromide Image here]
The distance moved,
[Insert Table of Marker base pairs, Distance moved, and Log of Base Pairs, Standard
curve, and]
Standard Curve from the Marker
[Insert Standard Curve Graph here]
Using the equation taken from the best fit line of the standard curve, the distances moved
by the unknown length sequences were used in the equation "# of base pairs=(.1277*distance travelled by the band)+4.6915" and then 10 was taken to the power of the
resulting number in order to get the number of Base Pairs that the DNA in the band
contained according to the standard curve.
[Insert Table for Unknown Sequence lengths here]
The second lane directly after the marker is the undigested Chit plasmid, and shows the
presence of three bands. The first band moved 4.8 cm, indicating 12,00 Base Pairs, the
second moved 6.8 cm, indicating 6,650 Base Pairs, and the third band moved 9.4 cm,
indicating 3,100 Base Pairs.
In both lane 3 and 4, the Pst+ digest of Chit and the Xho digest of Chit, respectively, one
band that moved 7.9 cm is visible, indicating the presence of 4,800 Base Pairs.
In lane 5, the double digested Chit plasmid, there was a slower band that moved 8.7 cm,
indicating 3,800 Base Pairs, and a faster band that moved 13.8 cm, indicating 850 Base
Pairs.
In lane 6, the undigested KS plasmid showed three bands which moved 4.6, 7.9, and 10.4
cm, indicating 24,000, 9000, and 4300 Base Pairs, respectively.
In lanes 7, 8, and 9, only one band, which moved 8.8 cm, is visible, indicating the
presence of 6,900 Base Pairs.
In lane 19, three is a single band which moved 15 cm, indicating a sequence of 600 Base
Pairs.
There are no bands in lane 20.
-DiscussionThe first band in lane 2 with 12,000 base pairs is probably the result of multiple DNA
strands binding to eachother and forming a conglomerate DNA cluster. The second band
in lane 2 with 6,654 apparent Base Pairs seems to be the nicked circular plasmid DNA,
which would have moved slower than it should have for its true weight and the third band
with an apparent 3,100 Base Pairs is probably the supercoiled plasmid DNA, which
would have moved faster than average for its weight.
Lanes 3 and 4 both contain only one band at 4,800 Base Pairs, which is indicative of a
linearized strand of DNA that was unable to recircularize after being fully cut by the
restriction enzymes.
In Lane 5, the presence of a 3,800 Base Pair band and a 850 BP band as a result of two
restriction enzyme digestions can be used to map the distance of the Pst+ digest site from
the Xho digest site on the Chit plasmid, indicating a distance of 850 BP between the two
restriction sites.
The difference of 150 base pairs between the sum of the lane five sequences and the
4,800 base pair sequences of lanes 3 and 4 may be attributable to the generalization the
trendline makes on the standard curve. Seperate trend-lines for different areas of the
curve, or ignoring certain data points at the smaller extreme entirely might allow for a
more consistent estimation of the restriction site locations on the Chit plasmid.
[Insert a drawing of the the resulting restriction enzyme map of the 4,800 bp Chit
plasmid, including Pst+ site, and Xho site]
Lane 6 repeats the trend of the three bands, with a conglomerate DNA cluster of 12,000
BP, a bande for the nicked circular DNA at 9,000 apparent base pairs, and a band for the
Supercoiled DNA at 4,300 apparent base pairs.
Lanes 7 and 8 show the single band, of 6,900 base pairs that indicate a linearized strand
of DNA resulting from the restriction digest.
Lane 9 also shows a single band of about 6,900 base pairs, despite being cut by both
restriction enzymes, implying that the restriction sites for the two enzymes are at
essentially the same location on the plasmid.
[Insert a drawing of the resulting restriction enzyme map of the 6,900 BP KS Plasmid,
including Pst+ and Xho]
Lane 19 shows the single band that corresponds to a very small sequence of about 600
base pairs, which therefore must be the distance between the two primers used in the PCR
of the chit plasmid.
The explanation for the presence of the PCR band in the Chit lane (19) but not in the KS+
lane (20) is the fact that Chit primers were used for the PCR process, and since those
primers would only be present in the Chit plasmid insert, only that Chit sequence
between those primers would have been amplified. This 600 base pair sequence must be
part of the chit sequence that is inserted into the Multiple Cloning Site.
-Literature Cited(I kept these in only for examples of the citation structure.)
Garret R. H., & Grisham, C. M.. (2005). Biochemistry Third Edition. California:
Thomson Brooks/Cole.
Ream, W., & Feild, K. G.. (1999). Molecular Biology Techniques An Intensive
Laboratory Manual. New York: Academic Press.
Weaver, Robert F. (2005). Molecular Biology Third Edition. New York: McGraw Hill.
Grdadolnik, J., Marechal, Y., (2001).
Bovine Serum Albumin observed by infrared spectrometry. I. Methodology, structural
investigation, and water uptake, Biopolymers, 62, 40-53.
Retrieved February 10, 2005, from www.ncbi.nlm.nih.gov