Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Chapter 5, Problem 19P
Interpretation Introduction
Interpretation:
The basis for the given requirement of having approximately the same
Concept introduction:
The reaction that is used to make the duplicates of a particular DNA segment is known polymerase chain reaction (PCR).
A small single strand of DNA or RNA containing approximately
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Pstl.
EcoRI
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(ori)
Ampicillin Tetracycline
resistance resistance
(Amp) (Tet")
pBR322
(4,361 bp)
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Pvull
Sall
Recombinant
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Bacterial cell contains...
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pBR322 (no insert)
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00,000.
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of E. coli cells
+AMP plate
pBR322
Figure 7-5
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Based on the recombinant plasmid growth pattern (bottom row of blue table), which of
the depicted plasmid's restriction sites was used to prepare this sample? Explain how
you can tell.
Question. What would the forward primer sequence look like if it were intended to
bind the area of the DNA template?
gene. If the JM109 strain is transformed by the PBKSK plasmid, the strain will produce the B-galactosidase (from the lac gene)
and will hydrolyze X-gal to produce the blue compound. Therefore, colonies that were transformed and contain the pBSKS wil
you
appear blue.
IPTG &
X-Gal &
NO colonies
Amp
E. coli JM109
E. coli JM109
50 mM calcium
chloride-15% glycerol
lac
lac
lac
IPTG &
I Recovery
X-Gal
solution at -702C
PBSKS
White colonies
E. coli JM109
E. coli JM109
ampR
amp I
amp
lac
lac
Heat Shock
Non-transformed
42°C
E. coli JM109
E. coli JM109
amps
amps
lac
lac
IPTG &
X-Gal
lac I
Recovery
lac
PBSKS
BLUE colonies
PBSKS
ampRI
(amp
Transformed
IPTG &
X-Gal &
BLUE colonies
Amp
Hypotheses: Circle the correct answer
1. If PBSKS is transformed into JM109 cells, colonies will be (able/not able) to grow in the presence of ampicillin.
a. Why?
_
2. If PBSKS is transformed into JM109 cells, colonies in media with IPTG (will/will not) induce the production the B-
galactosidase enzyme.
a. Why?_
3. If…
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- Pick a plasmid . What was its approximate transformation? Express it in # colonies per microgram of DNA transformed. Assume the original DNA was about .001 ug/ul . Count how many colonies you got on one plate (or estimate that number) and figure out how much of the total solution you plated on that plate. Multiply by all the plates, if you plated all of it. OR, if you only plated some of it, figure out how many colonies you would have gotten had you plated all of it. Divide by the number of ug used.arrow_forwardThis is DWA. You hope to clone an extinct animal species by taking the easy route - using museum bones or tissues. You extract the DNA and use PCR to amplify, then sequence all segments of the genome You are unsuccessful. Later you discover that the museum specimens have been treated with formaldehyde, which forms covalent bonds in DNA, where once there existed H-bonds. What step in your PCR reaction would be inhibited? g -S Knowing this, if a human was exposed to formaldehyde, what enzyme(s) of DNA replication in vivo would be prevented from doing their job?arrow_forwardNeed help ASAP. How do you use FISH(fluorescence in situ hybridization) to detect gene rearrangement? Describe an example with the detail of the disease and the procedure.arrow_forward
- Primer designing: A single-stranded DNA sequence (963 nucleotides) that codes for a hypothetical protein are shown below (lower case shaded blue). 1. Design a pair of forward and reverse primers (~18 nucleotides long each) with EcoRI and BamHI added at 5' and 3' ends, respectively, for the amplification and cloning of this a plasmid with the same restriction sites. gene into GTATCGATAAGCTTGATATCGAATTCatggctaaaggcggagct cccgggttca aagtcgcaat acttggcgct gccggtggcattggccagccccttgcgatgttgatgaagatgaatcctctggtttctgttctacatctatatgatgtagtcaatgcccctggtgtcaccgctgatatta gccacatggacacgggtgctgtggtgcgtggattcttggggcagcagcagctggaggctgcgcttactggcatggatcttattatagtccctgcaggtgttcctcg aaaaccaggaatgacgagggatgatctgttcaaaataaacgcaggaattgtcaagactctgtgtgaagggattgcaaagtgttgtccaagagccattgtcaacctg atcagtaatcctgtgaactccaccgtgcccatcgcagctgaagttttcaagaaggctggaacttatgatccaaagcgacttctgggagttacaatgctcgacgtagt cagagccaatacctttgtggcagaagtattgggtcttgatcctcgggatgttgatgttccagttgttggcggtcatgetggtgtaaccatttgccccttctatctcagg…arrow_forwardIn DNA extracting. What is the purpose of clear shampoo in the DNA extraction buffer?arrow_forwardCalculation problem on genetics. What would be the formula?arrow_forward
- Please answer this asap. Thanks, You have discovered a new plasmid RK21 in a unique bacterial community. As a first step towardunderstanding this plasmid, you digest the plasmid with three restriction enzymes: SspI, XhoI andSmaI. You run the digested plasmid DNA on an agarose gel, along with an uncut sample of theRK21 plasmid DNA as a control.Unfortunately you forget to load a DNA ladder, and obtain the following results. Assumecomplete digestion of all samples or all the digests worked completelyarrow_forwardRestriction sites of Lambda (A) DNA - In base pairs (bp) The sites at which each of the 3 different enzymes will cut the same strand of lambda DNA are shown in the maps (see figure 3 B-D), each vertical line on the map is where the respective enzymes will cut. A DNA A (bp) 48502 10 000 20 000 30 000 40 000 9162 17 198 B Sal I 7059 14 885 28 338 35 603 42 900 (bp) Hae III 11 826 21 935 29 341 38 016 (bp) 11648 29,624 Eco R1 (bp) 10 592 16 246 28 915 41 864 Figure 3: Restrictrion site map showing the following A) inear DNA that is not cut as reference B) DNA CLt with Sal L C) DNA cut with Hae , D) DNA cut with Eco RI 1. Calculate the size of the resulting fragments as they will occur after digestion and write the sizes on the maps below. Note that linear DNA has a total size of 48 502 bp (see figure 3A). Page 3 of 7 9162 17 198 Sal i (bp) 7059 14 885 28 338 35 603 42 900 Hae I (bp) 11 826 21 935 29 341 38 016 11648 29,624 Eco R1 (bp) 10 592 16 246 28 915 41 864arrow_forwardPreparing plasmid DNA (double stranded, circular) for Sanger sequencing involves annealing a complementary, single-stranded oligonucleotide DNA primer to one strand of the plasmid template. This is routinely accomplished by heating the plasmid DNA and primer to 90°C and then slowly bringing the temperature down to 25°C. Why does this protocol work? What enzyme is used and what other components are required in the sequencing reaction? How does the Sanger method determine the sequence?arrow_forward
- You have another circular plasmid. Complete and effective digestion of this plasmid with a restriction enzyme yields three bands: 4kb, 2kb, and 1 kb. In comparing the band intensity on an ethidium bromide-stained gel, you notice that the 4 kb and the 2 kb bands have the exact same brightness. The 1 kb band is exactly one fourth as bright as each of these. (Assume there is uniform staining with ethidium bromide throughout the gel.) How many times did the enzyme cut the plasmid? What is the size of the plasmid? Justify your answers to a and b above using a clearly labeled diagram showing the relative location of the cut-sites on the plasmid.arrow_forwardorganisms ha varieties. Through this process, several genetically (10) been produced. What I Can Do Activity 7: MATCHY MATCHY Direction: Match the purpose to the components found in the box below. Antibiotic Resistance Gene Multiple Cloning Site Promoter DNA Inserted Gene Sequence Multiple Cloning Site 1. Allows the controlled expression of the desired gene in the presence of an inducing agent (e.g., beta-galactosidase; heat treatment (~65°"C) 2. DNA sequence or portion for the insertion of the desired gene. This section may contain sequences that will be cut by specific restriction endonucleases ( cuts within the molecule) 3. Successful insertion of a gene allows the expression of its protein product. This usually provides a specific trait to the "transformed" bacteria. 4. Provides a way to screen a population of bacteria for those that took up the plasmid. For example, if an ampicillin resistance gene is encoded in the plasmid, then only bacteria 10 t4arrow_forwardRestriction mapping sample question You have a 5.3 kb PstI fragment cloned into the PstI site of the vector pUC19, which is 2.7 kb in size. This vector has unique sites for the following enzymes in a multiple cloning site: PstI, HincII, Xbal, BamHI, SmaI, EcoRI A restriction map of the 5.3 kb insert is prepared. The recombinant plasmid is digested with the enzymes listed above in single digests, and then several combinations of enzymes are tested in double digests. The following bands are observed when the digests are run on a gel: Enzyme(s) used PstI ECORI HincII Band sizes observed (kb) 5.3, 2.7 5.4, 2.6 4.5, 3.5 6.7, 1.3 | 4.0 (high intensity band) 3.9, 3.7, 0.4 4.0, 3.5, 0.5 3.5, 2.6, 1.9 3.7, 3.6, 0.4, 0.3 3.7, 2.2, 1.7, 0.4 3.7, 3.0, 0.9, 0.4 3.9, 3.5, 0.4, 0.2 Smal Xbal ВатHI HinclI + Xbal HincII + ECORI XbaI + BamHI ECORI + BamHI Smal + BamHI HincII + BamHI Use the data above to construct a map of the cloned insert. Note that fragments smaller than 100 bp will not usually be…arrow_forward
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