homework 10

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Dec 6, 2023

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Version Beta Assignment #10 (50 pts) Name: Phoebe Lin Work that is done inside Lab 10 – DNA I Names: Phoebe Lin, Julian Zhang Without belaboring a repetitive comment, this assignment will SIGNIFICANTLY challenge you. We’ve gone from doing ‘ABC’ in Lab 8, to writing a book in Lab 9, to now making a 7-book series that a movie series gets based on…the theme park happens in weeks 11 and 12. It is a lot. You have to understand the terminology and how they apply from Lab 8 through Lab 10. NOTE: #1 may be the same image from both partners. While there will be significant discussion between partners and other students in lab on 2-5, the answers should be individually created. 1. Take a picture of the gel as it is running. Annotate it with arrows pointing to the wells and indicate the direction that the bands are moving through the gel . [__ / 1 pts]

Version Beta 2. The image below is the MCS that you read about for pUC19. The capital letters are the MCS, and restriction sites are indicated. Below the numbers are codon designations. A. Using BamH I and Kpn I, if the MCS were to be cut with those two restriction enzymes, what would be the result? Draw the small fragment that would be cut out of the MCS. Show both strands. For reference, here are the recognition sequences: You can draw on paper and upload the image, draw on a screen and upload that. Or type the letters in sequence (remember that there are two strands) : [ 2 pts] Answer: 5’ CCGGG 3’ 3’ CATGGGCCCCTAG 5’ B. Below is the map of the entire pUC 19 plasmid. What is the size of the entire circular pUC19 plasmid in bp? What will be the size (in bp) of the linear, cut vector after the small fragment (the part you did in 1A) was removed? (Base your answer on the reference numbers given on the map and ignore the issue of single-stranded overhangs.) What kinds of sticky ends will the vector fragment have? (Sticky ends are referred to by the name of the restriction enzyme that cut them.) [ 2 pts] Full restriction map of the pUC19 cloning vector. Total length: 2686 bp. Note: recognition sites shown in bold occur only at that location. Unbolded sites occur at multiple places on the plasmid. (Image from New England Biological technical information)

Version Beta Answer: Total length: 2686 bp Small fragment = 417 – 408 = 9 bp size of cut vector = 2677 bp The vector fragments will have sticky ends will be Kpn I and BamH I 3. As you have seen this semester, working with DNA is more of a mental exercise than a visual. It is too small to see. We must use clever tricks to cut them where we want, and to fuse the pieces back together the way we want. You have already seen how specific restriction endonucleases are used to cut plasmid vectors in particular places. A. How does directional cloning control the orientation of an insert in the cut MCS of the cloning vector plasmid? [ __/3 pts Answer: In the process of directional cloning, it can control the orientation of an insert in the cut MCS of the cloning vector plasmid by having the vector undergo cleavage using a pair of restriction enzymes featuring recognition sites on the vector. This results in the formation of different sticky ends. Only sticky ends cleaved by identical restriction enzymes can form bonds. As a result, when the foreign DNA fragment is cleaved using the matching restriction enzyme, the resultant sticky ends exclusively join with their complementary counterparts, also cleaved by the same restriction enzyme on the vector. B. This question is about the importance of the fact that restriction recognition sites engineered to be in the MCS should be found nowhere else in the plasmid. If a particular type of restriction recognition site were found in the MCS and one other place in the plasmid, how many pieces would be formed if the plasmid were cut with that restriction enzyme? After those pieces were cut apart and were floating around in the buffer, how many possible ways would there be to ligate those pieces back together? Why would it cause problems if we tried do directional cloning if one of the restriction enzymes used cut the plasmid in more than one place? [ __/4 pts] Answer: Two different pieces would be formed if plasmid were cut with restriction enzyme. The pieces can ligate back together in two different ways. If the restriction enzyme cuts the plasmid in more than one place, this would produce two pairs of sticky ends and thus two different ways of ligation, which would prevent directional cloning. This is because in one pair of the sticky ends, it can orient themselves because they are different. However, if there are two pairs, it is possible that they can recognize the sticky end of the other pair, thus changing the directionality of the vector. 4A. For each of the four inserts listed below and referring to the plasmid shown (below in 4B), indicate which end(s) would ligate with the cut circular plasmid (left end only, right end only, both ends, neither end). [ __/4 pts] Insert A: neither Insert B: neither Insert C: right only Insert D: both ends

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Version Beta 4B. Which of the inserts would re-form a circular plasmid, and also what recognition sites would form after the ligation? You will probably need to refer to Fig. 4 of ‘Restriction Endonucleases’ in Experiment 9 to answer this. [ __/3 pts] Answer: BamH1, ECOR1

Version Beta 5. Below is a simplified version of the pUC19 plasmid that you used in question 2. It shows only the restriction recognition sites in the MCS. DNA was extracted from a piece of frozen Siberian mammoth flesh and double digested with Bam H I and Eco R I restriction enzymes. pUC19 vector was also digested with Bam H I and Eco R I. One of the mammoth DNA fragments was ligated into the cut pUC19 vector. A. What will the recombinant plasmid look like? Draw a diagram showing this recombinant plasmid. Label the locations of: - the Bam H I and Eco R I restriction sites - identify the sections of the recombinant plasmid that are the vector and the insert -On the vector part of the diagram, show the positions of the ApR and lacZ α genes and the ori . Note: The map that you have created here will form the starting point of your work on next week's assignment . [__ / 6 pts] Answer:

Version Beta B. What we know at this point: -We know that we made a recombinant plasmid. -We know how much of the original plasmid was removed. What we do NOT know is how big the insert of mammoth DNA is, it could be a few bases to several hundred bases. After making the recombinant plasmid mix, it was transformed into bacteria, cloned and then purified to obtain a sample containing a large number of identical recombinant plasmids. The recombinant plasmid was then extracted from the bacteria and then was next digested with a combination of three different restriction enzymes: Acc I, Bci V I, and Kpn I. Because the sequence of the mammoth DNA insert is not known, it is not possible to know where these three restriction enzymes will cut in the insert part of the recombinant plasmid. However, it is possible to predict the number and size of fragments that will be produced from the vector part of the plasmid (refer to the full pUC19 map above in 2B ). On your diagram from part A, mark and label the positions where the vector portion of the plasmid will be cut by each of the three restriction enzymes ( Acc I, Bci V I, and Kpn I). Calculate the number and size (in basepairs) of the fragments that will form entirely from parts of the vector (do not include the parts of the vector that are attached to part of the insert). [__ / 8 pts] Answer:

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Version Beta Postlab: Recombinant DNA I NOTE: 6-8 is to be individually completed. #6 cannot be attempted until your gel is posted. #6 will take several minutes to get A set-up. B and C should be fairly quick IF you have kept up with the material. 6. Insert the image of your gel below. Label/state the lanes which contain your sample (remember, this is the PCR sample from week 9 and successful ones were split to perform two different digests). Note: If your gel is not usable in general, you will be instructed to use the sample gel on Brightspace. [__ / 1 pts (put image here)

Version Beta A. Using the instructions in part C of the experimental procedure, create a standard curve using the molecular mass standard bands on the gel you are analyzing (#6). Include the equation of the trendline. Upload the Excel spreadsheet to Brightspace but also show your work in this document by including a screen capture of the best-fit line and a table containing the measured lengths and the calculated logs of the molecular masses. NOTE: Do NOT just paste from Excel into this document. Insert a JPEG or PNG image acquired using screen capture (e.g. using Windows Snipping Tool). [__ / 9 pts] Learning objective 10.11: construct a standard curve from measurements of mass standards, determine the equation of the best-fit line (trendline) through the data, and use that equation to determine the masses of an unknown fragment. Answer:

Version Beta B. Find a lane (or two adjacent lanes) on your gel that has bands for all three sizes of fragments. Use the trendline equation to estimate the molecular mass (in bp) of each of the three kinds of fragments in the Kpn I and Bst E II digests, and the uncut PCR product. Report these masses and compare them to what you calculated in question 2 of the homework last week . [__ / 4 pts] Answer: Fragment frag distance log10 mass calc mass Lane 2 4.8cm 2.6699 467.627 Lane 2 6.1cm 2.2786 189.9328 Lane 3 3.8cm 2.9709 935.1903 C. Based on your gel and your answers to question 3.A. of last week's assignment, what is the genotype at the "type O" SNP locus of ALL of the samples on the gel? Explain. If you believe that one or both of the restriction digests did not cut in a particular sample, explain why you think that. [__ / 3 pts] Learning objective 10.12: infer genotypes based on the result of restriction digests of PCR-amplified DNA fragments containing the "type O" SNP locus. Answer: The genotype at the “type O” SNP locus of the samples on the gel is most likely O, A/B heterozygous. In lane 2, there are two bands with fragment masses of approximately 468bp and 190bp. This implies that there were two restriction sites where cuts were made by restriction enzymes. Because two cleavages occur, we know that the genotype is O,O homozygote. In lane 3, BstEII is used, but only one band appears. This cleavage produces a fragment mass of roughly 935bp signifying the presence of an A/B, A/B homozygote. Lanes 4 and 5 have all three bands present indicating the genotype is O, A/B heterozygote.

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Version Beta Prelab: Recombinant DNA II BE SURE TO READ NEXT WEEK’S LAB prior to coming to lab and to sometime between this lab next week’s lab to log into Brightspace to do your Reading Assessment Questions. Five multiple choice questions, randomly selected from a large question bank. AS A NOTICE, the difficulty level in Lab 11 reading is higher than the last few lab weeks. YOU REALLY NEED TO UNDERSTAND what is happening to allow for easier understanding in Lab 11 (and homework)

homework 10

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