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Concept explainers
(a)
To draw: The structure of each end of a linear DNA fragment produced by an EcoRI restriction digest.
Introduction:
EcoRI is a restriction endonuclease enzyme isolated from species E. coli. The restriction site of EcoRI consists of six palindromic
(a)
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Explanation of Solution
Explanation:
Recognition sequence of EcoRI is GAATTC.
So, the fragments produced at both the ends are:
Ad
(b)
To draw: The structure resulting from the reaction of this end sequence with DNA polymerase I and the four deoxynucleoside triphosphates.
Introduction:
DNA polymerase I is an important enzyme used in the replication of DNA. It binds to specific sequence called initiator sequence. DNA polymerase I has multiple functions which include nick translation during DNA repairing, proofreading and DNA dependent DNA polymerase activity.
(b)
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Explanation of Solution
Explanation:
The strands formed after DNA polymerase I activity are:
DNA polymerase enzyme adds the complementary nucleotide base pairs to the sequence, which makes the sticky end turn into blunt ends.
(c)
To draw: The sequence produced at the junction that arises if two ends with the structure derived in (b) are ligated.
Introduction:
Ligation is the process of joining two or more sequences with the help of an enzyme called DNA ligases. DNA ligase forms phosphodiester bonds between the nucleotides.
(c)
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Explanation of Solution
Explanation:
The sequence obtained after ligation of the two ends is:
DNA ligase forms the phosphodiester bonds between two nucleotides which end up joining two DNA fragments.
(d)
To draw: The structure produced if the structure derived in (a) is treated with a nuclease that degrades only single stranded DNA.
Introduction:
Nuclease enzyme is important in breaking the phosphodiester bond between the nucleotide. There are two types of nuclease enzymes endonuclease and exonuclease. Exonuclease attacks the DNA from the terminals and endonuclease attacks the DNA from anywhere between the DNA.
(d)
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Explanation of Solution
Explanation:
The sequence obtained after treatment with nuclease that degrades only single stranded DNA is:
And
Nuclease enzyme removed the nucleotide sequences present at the end turning the sticky ends into blunt ends.
(e)
To draw: The sequence of the junction produced if an end with structure (b) is ligated to an end with structure (d).
Introduction:
Ligation is the process of joining two or more sequences with the help of an enzyme called DNA ligases. DNA ligase forms phosphodiester bonds between the nucleotides.
(e)
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Explanation of Solution
Explanation:
The resulting sequence obtained after ligation structure (b) with structure (d) is:
The ligase enzyme joined the ends of the two separate DNA fragments and formed a continuous sequence.
(f)
To draw: The structure of the end of a linear DNA fragment that was produced by a PvuII restriction digest.
Introduction:
PvuII restriction enzyme is isolated from a bacterium known as Proteus vulgaris. The recognition sequence of PvuII is made up of six nucleotide sequence and the ends produced are blunt ends
(f)
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Explanation of Solution
Explanation:
The sequences obtained after digestion with restriction enzyme PvuII are:
PvuIIproduces blunt ends on digesting the DNA fragment and the recognition sequence for PvuII is 5ʹ CAGCTG 3ʹ PvuII is CAGCTG
(g)
To determine: The sequence of the junction produced if an end with structure (b) is ligated to an end with structure (f).
Introduction:
Ligation is the process of joining two or more sequences with the help of an enzyme called DNA ligases. This enzyme is important in repairing single stranded breaks in the DNA. DNA ligase forms phosphodiester bonds between the nucleotides.
(g)
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Explanation of Solution
Explanation:
The resulting sequence obtained after ligation structure (b) with structure (f) is:
The ligase enzyme joined the ends of the two separate DNA fragments by catalyzing the formation of phosphodiester bonds between them.
(h)
To determine: A protocol for removing a EcoRI restriction site from DNA and incorporate BamHI restriction site at the same location.
Introduction:
BamHI restriction enzyme is isolated from a bacterium known as Bacillus amyloliquefaciens. BamHI is a type II restriction enzyme and the recognition sequence is made up of six nucleotide sequence and the ends produced are four nucleotide long sticky ends.
(h)
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Explanation of Solution
Explanation:
First, the DNA is digested with EcoRI. It generates staggered ends these ends are treated with polymerase I enzyme and four nucleotide base pair. DNA polymerase binds the NTPs to the complementary sequence of the DNA, and converts sticky ends to blunt end.
After conversion of sticky ends produced by EcoRI into blunt ends the DNA fragment is ligated with a synthetic fragment which has recognition sequence for BamHI enzyme this will result into conversion of EcoRI restriction site into BamHI.
The sticky ends produced by restriction digestion can be converted to blunt ends by using polymerase enzyme and the four nucleotide tri phosphate containing solution. Similarly a restriction site can be added by ligation with the ligase enzyme and synthetic nucleotide sequence.
(i)
To design: Four different short synthetic double-stranded DNA fragments that would permit ligation of structure (a) with a DNA fragment produced by a PstI restriction digest.
Introduction:
EcoRI is a restriction endonuclease enzyme isolated from species E. coli. The restriction site of EcoRI consists of six palindromic nucleotide sequences and the restriction digestion produce sticky ends. PstI is a restriction endonuclease enzyme isolated from species Providencia stuartii. It is a type II restriction enzyme and produces sticky ends.
(i)
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Explanation of Solution
Explanation:
The sequence in which the final junction contains the recognition sequences for both EcoRI and PstI is:
The sequence that contains only the EcoRI restriction segment is:
The sequence that contains the recognition site for only PstI is:
The sequence that has neither of the two recognition sites present is:
The recognition site for specific enzyme is specifically based on the nucleotide sequence. The enzyme binds to a specific nucleotide sequence present inside the DNA and cleaves it producing sticky or blunt ends.
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Chapter 9 Solutions
Lehninger Principles of Biochemistry
- Problem 15 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:33 CO Problem 4 of 15 4G 54% Done On the following Lineweaver-Burk -1 plot, identify the by dragging the Km point to the appropriate value. 1/V 40 35- 30- 25 20 15 10- T Км -15 10 -5 0 5 ||| 10 15 №20 25 25 30 1/[S] Г powered by desmosarrow_forward1:30 5G 47% Problem 10 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without a competitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' s mM¹ with 10 mg pe 20 V' 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
- Problem 14 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:36 CO Problem 9 of 15 4G. 53% Submit Using the following reaction data points, construct a Lineweaver-Burk plot by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Based on the plot, determine the value of the catalytic efficiency (specificity constant) given that the enzyme concentration in this experiment is 5.0 μ.Μ. 1 [S] ¨‚ μM-1 1 V sμM-1 100.0 0.100 75.0 0.080 50.0 0.060 15.0 0.030 10.0 0.025 5.0 0.020 Answer: ||| O Гarrow_forwardProblem 11 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without a noncompetitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' 20 V' s mM¹ with 10 μg per 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
- Problem 13 of 15 Submit Using the following reaction data points, construct Lineweaver-Burk plots for an enzyme with and without an inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Using the information from this plot, determine the type of inhibitor present. 1 mM-1 1 s mM -1 [S]' V' with 10 μg per 20 54 10 36 20 5 27 2.5 23 1.25 20 Answer: |||arrow_forward12:33 CO Problem 8 of 15 4G. 53% Submit Using the following reaction data points, construct a Lineweaver-Burk plot by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. Based on the plot, determine the value of kcat given that the enzyme concentration in this experiment is 5.0 μM. 1 [S] , мм -1 1 V₁ s μM 1 100.0 0.100 75.0 0.080 50.0 0.060 15.0 0.030 10.0 0.025 5.0 0.020 Answer: ||| Гarrow_forward1:33 5G. 46% Problem 12 of 15 Submit Using the following reaction data points, construct a Lineweaver-Burk plot for an enzyme with and without an uncompetitive inhibitor by dragging the points to their relevant coordinates on the graph and drawing a line of best fit. 1 -1 1 mM [S]' 20 V' s mM¹ with 10 μg per 54 10 36 > ст 5 27 2.5 23 1.25 20 Answer: |||arrow_forward
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