Genetics Lab 2

1. Below is a map for a plasmid called ‘pGFP’: The sequence (5’ to 3’) of pGFP containing the promoter region, the 5’ multiple cloning site (with a BamHI site, underlined), and the 5’ end of the GFP coding sequence is as follows: 1 AGCGCCCAAT ACGCAAACCG CCTCTCCCCG CGCGTTGGCC GATTCATTAA TGCAGCTGGC 61 ACGACAGGTT TCCCGACTGG AAAGCGGGCA GTGAGCGCAA CGCAATTAAT GTGAGTTAGC (Lac P) 121 TCAC TCATTA GGCACCCCAG GCTTTACACT TTATGCTTCC GGCTCGTATG TTGTGTGG (+1A)A (LacO) (SD) (LacZ #1) 181 TTGTGAGCGG ATAACAATTT CACACAGGAA ACAGCTATGA  CCATGATTAC GCCAAGCTTG (GFP #1) 241 CATGCCTGCA GGTCGACTCT AGAGGATCCC CGGGTACCGG TAGAAAAAAT GAGTAAAGGA 301 GAAGAACTTT TCACTGGAGT TGTCCCAATT CTTGTTGAAT TAGATGGTGA TGTTAATGGG 361 CACAAATTTT CTGTCAGTGG AGAGGGTGAA GGTGATGCAA CATACGGAAA ACTTACCCTT The GFP open reading frame was cloned as a BamHI (5’ MCS)/ XhoI (3’MCS) fragment. In this plasmid, GFP is expressed as a fusion protein with LacZ. Clearly indicate on the above sequence the following elements: (you may highlight in different colors –blue, red, green, etc. and indicate the color for each element below) lac Promoter : lacP The start codon for LacZ: LacZ#1 The first codon of the GFP sequence : GFP#1 (here are the first ten amino acids for GFP: MSKGEELFTG) Lac Operator : lacO The transcription start (+1) of the mRNA : circle the nucleotide and indicate +1 The Shine Dalgarno sequence : SD ● Is the plasmid-encoded fusion protein subject to regulation by cAMP (catabolite repression)? Explain your answer. ○ Yes, the plasmid-encoded fusion protein is subject to regulation by cAMP because there is a CAP binding site in the sequence and the cAMP and CAP can form a complex that attaches to a binding sequence upstream of the lac promoter. This then allows the RNA polymerase to be recruited to the promoter. HINT: To do this, you will need to consult figures 12.5 and 9.7 in your textbook! 2. Here is a simple depiction of the pSK+ vector: The vector allows for protein expression in E. coli. Briefly explain the role of the following elements in the context of gene expression and protein production: LacP: LacP is the promoter region and it has regulatory sequences with -10 and -35 consensus sequence sites that are critical for RNA polymerase binding. LacO: Lac O is the operator region that helps regulate the lac operon. It binds near the repressor protein produced by LacI and overlaps LacP near the start of transcription. When bound to the repressor protein, transcription cant take place because it blocks RNA polymerase from binding to LacP and prevents transcription initiation LacZa: LacZa helps determine if the protein was inserted in the correct orientation. It also produces an enzyme called beta galactosidase which results in the breaking of the lactose into glucose and galactose components. The vector also contains other important elements. Explain their role: AmpR: AmpR stands for ampicillin or antibiotic resistance gene which allows for the bacteria to grow in an ampicillin plate Ori: This is the sequences that indicates the origin of replication in the bacteria 3. You plated your transformations on agar containing Ampicillin, IPTG and X-Gal. - explain the use of IPTG: IPTG acts like allolactose which acts as an inducer, so in this case, it binds to the lac repressor and helps start lac transcription. When IPTG binds to the lac repressor, it prevents the repressor from binding to the operator. This allows for the promoter to be bound by RNA polymerase and allows transcription - explain the use of X-Gal: When broken down by beta-galactosidase, X-gal allows for the emission of blue light. This blue light emission through X-gal allows for us to see which plasmids have a functional lacZ gene 4.You are attempting to clone an EcoRI fragment into an EcoRI-restricted vector treated with CIP, and you obtain the following results on your transformation plates (with Amp): #1 vector only 107 colonies +ligase #2 vector+insert+ligase 663 colonies ● Would you expect more, fewer or about the same number of colonies if no ligase were added in #2? Explain. ○ I would expect to see fewer colonies in #2 if there was no ligase added due to the fact that ligase helps join the 5’ phosphate end of the insert with the 3’ -OH end. With no ligase, there would be less transformed bacteria and less colonies ● Would you expect more, fewer or about the same number of colonies with #1 with a vector not treated with CIP? Explain ○ I would expect more colonies with #1 with a vector not treated with CIP because CIP removes the 5’ phosphate. If ligase were then added to a CIP untreated vector, the vector ends would be able to re-ligate and produce more colonies PART II (Lab data provided). 1. You’ve grown miniprep cultures to isolate plasmids from your transformants. Paste the picture of your gel below (see pp.92-97, picture provided for Spring 2020) : Please refer to p. 92 for the content of the lanes. Answer the following three questions: 1) Why did we use different concentrations of the native plasmid (i.e., pBS-SK) on lanes 2-6? ● We use different concentrations of the native plasmid on lanes 2-6 so we can analyze the exact concentration of DNA. This would allow us to see if the four mini-preps in lanes 7-10 were able to isolate different levels of plasmid. The serial dilution allows us to see the visible band intensities as concentration increases and comparing the intensities helps us estimate the DNA concentration of our minipreps. 2) Based on visual observation of the gel's picture, can your plasmids be used for restriction analysis in lab 10? ● The plasmids can be used for restriction analysis in Lab 10 based on the visual observation of the gel’s picture. For the concentration and amount of the DNA, we would need at least 114 µg/ml and 2.5 µg respectively. Lane 4 has a concentration of 100 µg/ml and Lane 3 has a concentration of 50 µg/ml. When looking at lanes 7, 8 and 9, they appear to have the same DNA concentration as Lane 4, while lane 9 has the same DNA concentration as lane 3. Since all four DNA concentrations are under the minimum value, we would need the maximum volume for all of these in order to do restriction analysis. 3) Just by inspection of the gel above, is there a clone that appears to have an insert? Justify your answer. ● The clone that appears to have an insert would be lane 8 due to the fact that it did not travel as far as the other lanes. A heavier clone would run slower which could also indicate the presence of an insert, as this would increase the weight of the plasmid. You have then digested your plasmids with XhoI, AlwNI and both enzymes. Below is the photograph of this gel (see p. 105): Refer to the manual p.99 for the content of the lanes [note: the lanes are loaded slightly differently here] ● Why did you digest the plasmid with XhoI? What is your expectation if the plasmid is a reclosed vector without an insert? What is your expectation if your plasmid has an insert (how many bands, what are their sizes)? ○ We digested the plasmid with Xho1 because after we insert a gene, this allows us to make sure we have the correct plasmid size. The plasmid would have one band if the vector did not have an insert in the vector, as there would only be one cut from the Xho1 at ~3000 bp. If Xho1 was used on a plasmid with an insert, there would be two bands, as there one be a band for the vector, and one for the insert. The vector would show up around 3000 bp and the insert would show up at 700 bp. ● Describe your result from your gel obtained with the XhoI digest, for clone #1 and clone #2 ○ Clone #1 was in Lane 5 while clone #2 was in Lane 9, and they would both have two bands when digested by Xho1. For clone #1 in Lane 5, we would see two bands. Because clone #1 has a vector and an insert, one appears at around 2,000 bp and the other shows up at 600 bp, and the Xho1 separates the two. For clone #2, this also contains a vector and an insert, so it makes sense that there would also be two bands after it has been digested, and they also appear at the same base pair length. ● Why did you digest the plasmid with AlwNI? What is your expectation if the plasmid is a reclosed vector without an insert? What is your expectation if your plasmid has an insert in the correct orientation (how many bands, what are their sizes)? Refer to the gel in your answerexplain the bands you see for the correct orientation. ○ We digested the plasmids with AlwNI because we wanted to see what the plasmids would look like if the insert’s orientation was correct or incorrect. A plasmid that has an insert with a correct orientation would have two bands that are close in size, while a plasmid that has an insert with an incorrect orientation would have two bands that are farther in size. I would expect there to be one band if our plasmid is a reclosed vector without an insert. The expected bp size would be at ~2000-3000 bp, and we are able to see a band at ~2,500 bp in Lane 3. Since there would be two bands for an incorrect insert orientation, Lane 6 shows two bands, as one is at ~1700 bp and the other is ~950 bp. We get two bands in this instance because AlwNI cuts the plasmid twice. If there was an insert with a correct orientation, the restriction sites would be farther from each other and the bands would be similar sizes while an insert that is incorrectly oriented in the plasma would not have similar sized bands. ● What is your expectation if your plasmid has an insert in the incorrect orientation (how many bands, what are their sizes)? Again, refer to the gel- explain the bands you see for the incorrect orientation. ○ If the plasmid has an insert in the incorrect orientation, there would be two bands visible that have a large difference between them because the incorrect insert orientation allows the restriction sites from the AlwN1 to be closer together. This would equate to a large and small DNA fragment. Looking at our gel, Lane 10 shows that for one of the bands, we would see its appearance around 1900 bp and the other around 550 bp. ● Describe your result from gel with the XhoI/AlwNI double digest, for clone #1 and clone #2. Do you see a difference between the two? Why or why not? ○ We do not see a difference between clone #1 and clone #2 because either of the insert orientations left the four restriction sites create four bands. Lane 7 correlates to clone #1 and Lane 11 correlates to clone #2, so when we look at the four bands on both of them, each of clone #1’s bands (Lane 7) has a corresponding band size to clone #2’s bands (Lane 11). We were able to see that the total number of basepairs was similar to the size of the vector and insert, the values we observed are reasonable. The top band is ~1700 for both lanes, the next band is ~700 bp for both lanes, the third band is ~500 bp for both lanes, and the smallest band for both lanes is ~50 bp.