You are characterizing a new DNA polymerase. In a test tube, you incubate the enzyme with all the components needed for catalysis, EXCEPT you don't add any dNTPs. However, you notice that some type of catalysis has occurred. What is the most likely product in this case? O The supercoils in the DNA have been relieved O The sugar has been converted from a deoxyribose to a ribose A DNA strand is one base shorter due to nuclease activity The magnesium ion has been converted to a charge of +2 to +1

You are characterizing a new DNA polymerase. In a test tube, you incubate the enzyme with all the components needed for catalysis, EXCEPT you don't add any dNTPs. However, you notice that some type of catalysis has occurred. What is the most likely product in this case? O The supercoils in the DNA have been relieved O The sugar has been converted from a deoxyribose to a ribose A DNA strand is one base shorter due to nuclease activity The magnesium ion has been converted to a charge of +2 to +1

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 18QP: Replication involves a period of time during which DNA is particularly susceptible to the...

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Ultraviolet light can cause covalent linkages between consecutive pyrimidine bases in DNA (up to 100 per second in a single cell in sunlight!). These bulky lesions (i.e. cyclobutane pyrimidine dimers and 6-4 photoproducts), which inhibit DNA and RNA polymerases, are mostly reversed by CPD photolyase when light >300 nm is available to power the reaction. In the dark, however, which DNA repair system is best able to correct these errors? a) non-homologous end-joining b) mismatch repairc) nucleotide-excision repaird) base-excision repair e) homology-directed repair
DNA is damaged when a base from the DNA chain is removed after an alkylation has occurred. In a depurination reaction, the purine nitrogenous base is displaced from its sugar as shown in this reaction. Draw the mechanism for this reaction and suggest a reason why it occurs so easily. HN' P H2N .N H,O HN° + Pi ОН N- H2N° ОН ОН ОН ОН
Both DNA polymerase (any DNA polymerase) and ligase catalyze the formation of a bond between nucleotides, but these two enzymes do NOT catalyze the same reaction. Briefly describe the differencesbetween the reaction catalyzed by the polymerase activity of DNA polymerase and the one catalyzed by ligase.
Methyl iodide (CH31) can be used to modify DNA in solution. It takes the following mechanism for this reaction: Step 1. (slow) CHз снз* + г Step 2. (fast) DNA + CH3* > DNA-CH3* Which of the following is true about this reaction and its mechanism? O a. lodide ion (I ") is an intermediate in the reaction. Ob.CH3* is a catalyst of this reaction. O. The activation energy of Step 1 must be lower than the activation energy of Step 2. d. The overall reaction order of the reaction predicted by this mechanism is 1st order. By increasing temperature, the rate constant of Step 1 will increase but the rate constant of Step 2 will е. decrease.
Give the name of the enzyme that catalyzes each of the following reactions:(a) Makes a DNA strand from a DNA template. (b) Makes a DNA strand from an RNA template. (c) Makes an RNA strand from a DNA template
Step 1 Lys Lys Ligase-AMP NH2 NH2 NH2 `NH2 PP АТР он он OH Step 2 Lys Lys NH2 NH2 NH2 NH2 DNA- OH OH Adenylate OH OH HO. Step 3 NH2 NH2 NH2 NH2 AMP Phospho- diester OH OH OH OH OH Describe the mechanism shown above for DNA Ligase. Describe the chemistry of each step How the enzyme appears or might facilitate the chemistry How the enzyme increases the reaction rate.
Enediynes are natural products with potent antitumor properties because they are able to cleave DNA (page 288). Their cytotoxic properties are due tothe enediyne undergoing a cyclization to form a highly reactive diradical intermediate. The intermediate abstracts hydrogen atoms from the backbone of DNA, which triggers its damage. Draw the structure of the diradical intermediate.
FIGURE 19.16 Direct repair of dam- aged bases in DNA. (a) The repair of thymine dimers by photolyase. (b) The repair of methylguanine by the transfer of the methyl group to Thymine dimer CH3 SH 06-Methylguanine CH2 ONLINE ANIMATION Н. CH3 Нас. N- н- Н Н N. `NH2 Alkyltransferase alkyltransferase. CONCEPT CHECK: Which of these DNA backbone Alkyltransferase catalyzes the removal of the methyl group onto itself. repair systems is particularly valuable to plants? DNA DNA photolyase cleaves the 2 backbone bonds between the thymine dimer. CH3 Guanine На 'N' CHз Нас. -- н CH2 н Н `NH2 The normal structure of the 2 thymines is restored. The normal structure of guanine is restored. (a) Direct repair of a thymine dimer (b) Direct repair of a methylated base O=
(a) (b) (c) The figure shown below is the mechanism of a DNA nuclease. The DNA molecule is in blue while the protein residues are in black (M1 and M2 are metal ions, they are irrelevant to this question). 3' Base reaction. M² 5' 00 3' Glu 357 M¹ Draw the structure of the final DNA product for this reaction. Draw a square around the nucleophile and draw a circle around the leaving group in this Does this depict the reaction of an endonuclease or an exonuclease?
The anti-viral drug Acyclovir is a nucleotide analog that is lacking the 3’ OH group which is required to form a 3’→5’ phosphodiester bond. This drug is ineffective against DNA polymerases with proofreading abilities, which is why human DNA polymerases are not targeted. Acyclovir can be used to treatsevere cases of Epstein-Barr viral (EBV) infection, but has little to no effect under non-severe infections. Based on this information, EBV will use ________ DNA polymerase during severe infections and __________ DNA polymerase during non-severe infections. Human; Human EBV; Human EBV; EBV Human; EBV
Explain why two metal ions play important role on the palm domain of DNA polymerase.
The restriction endonuclease PstI cuts DNA symmetrically on both strands at the sequence: CTGCAG GACGTC On the resulting DNA fragments are left 3’ overhanging ends of 4 nucleotides. PstI cleavesthe phosphodiester backbone with a-type specificity. In short-hand notation, showing theidentity of phosphates on the ends, draw the two fragments that result from PstI digestion ofthe following double-strand ed oligonucleotide: 5’-pAATTGCTACTGCAGAACCGG-3’ 3’-TTAACGATGACGTCTTGGCCp-5’