Nucleotides and Nucleic Acids

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Chapter 10 . Nucleotides and Nucleic Acids9. Based on the information in Table 10.5, describe two different 20-base nucleotidesequences that have restriction sites for BamHI, PstI, SalI, and SmaI. Give the sequencesof the SmaI cleavage products of each.Answer: Table 10.5 lists type II restriction endonucleases, their recognition sequence, where inthe sequence the enzyme cuts in addition to other information. BamHI, PstI, SalI and SmaI allrecognize six-base restriction sites. Since we are asked to produce a 20-base sequence we aregoing to have to have restriction sites overlap. We are not asked to order the sites so this leaveus free to try several combinations.EnzymeBamHIPstISalISmaIRecognitionSequenceGGATCCCTGCAGGTCGACCCCGGGLooking at the recognition sequences it is clear that having BamHI and SmaI adjacent to eachother will allow a two-base overlap (of G’s or C’s, depending on order). So we could start outwith either of these two:1. GGATCCCGGG or 2. CCCGGGATCCTo 1. we could add PstI and SalI sites with single base overhangs as follows:CTGCAGGATCCCGGGTCGACDigestion of this oligonucleotide with SmaI would produce:CTGCAGGATCCC and GGGTCGACTo 2. we could add PstI and SalI sites with single base overhangs as follows:GTCGACCCGGGATCCTGCAGDigestion of this oligonucleotide with SmaI would produce:GTCGACCC and GGGATCCTGCAGAlternatively, we could have started with the BamHI/SmaI overlapped DNA on an end and thenadded in the other sites.For 1. we would get:GGATCCCGGGTCGACTGCAGDigestion of this oligonucleotide with SmaI would produce:GGATCCC and GGGTCGACTGCAGFor 2. we would get:CTGCAGTCGACCCGGGATCCDigestion of this oligonucleotide with SmaI would produce:CTGCAGTCGACCC and GGGATCC10. The synthesis of RNA can be summarized by the reaction:n NTP (NMP)n + n PPiWhat is the •Gº’overall for synthesis of an RNA molecule 100 nucleotides in length,assuming the •Gº’ for transfer of an NMP from an NTP to the 3’-O of polynucleotidechain is the same as the •Gº’ for transfer of an NMP from an NTP to H20? (Use datagiven in Table 3.3.)Answer: From Table 3.3 we are informed that the •Gº’ for hydrolysis of ATP to AMP and PPi is –32.3 kJ/mol. This reaction is in effect a transfer of AMP from ATP to water. We are allowed toassume that transfer of AMP from ATP to the 3’ end of a polynucleotide chain has •Gº’ of +32.3kJ/mol. To make a polymer 100 nucleotides long let us start with hydrolysis of a single NTP toNMP and PP with a very favorable •Gº’ of –32.3 kJ/mol. The NMP would then be used in atransfer reaction with NTP to make the first phosphodiester bond with an unfavorable •Gº’ of151
Chapter 10 . Nucleotides and Nucleic Acids+32.3 kJ/mol. To make 99 phosphodiester bonds, the number in a 100-base-longpolynucleotide, the reaction would have to be repeated 98 times. Thus,•Gº’overall = -32.3 +99 x 32.3 kJ/mol = 3,165.4 kJ/molIf we were to have simply made 99 phosphodiester bonds it would require:•Gº’overall = 99 x 32.3 kJ/mol = 3,198 kJ/molHowever, the RNA would have a triphosphate at its 5’ end.11. Gene expression is controlled through the interaction of proteins with specificnucleotide sequences in double-stranded DNA.a. List the kinds of noncovalent interactions that might take place between a proteinand DNA.b. How do you suppose a particular protein might specifically interact with a particularnucleotide sequence in DNA? That is, how might proteins recognize specific basesequences within the double helix?Answer: a. The weak forces are hydrophobic interaction, van der Waals interactions, hydrogenbonds and ionic bonds. Of these we might not expect excessive hydrophobic interactions tooccur between a protein and dsDNA because dsDNA presents virtually little in the way ofhydrophobic surfaces to the solvent. The bases themselves are hydrophobic but they arestacked in dsDNA. One might expect hydrogen bonding to occur between a protein and bases ina DNA sequence. One might also expect ionic bonding between the negatively-chargedsugar/phosphate backbone of DNA and positively-charged amino acid residues from arginineand lysine.b. Without unwinding dsDNA a protein would have to make base-specific interactions via eitherthe major groove or minor groove. There is more information regarding the identity of base pairsin the major groove. In addition, the dimensions of an α-helix are compatible with the space inthe major groove. Thus, one might expect proteins to bind via the major groove.12. Restriction endonucleases also recognize specific base sequences and then act tocleave the double-stranded DNA at a defined site. Speculate on the mechanism by whichthis sequence recognition and cleavage reaction might occur by listing a set ofrequirements for the process to take place.Answer: Restriction enzymes bind double-stranded DNA at sites that have two-fold rotationalsymmetry and they cleave both strands at identical places on the DNA. For this event to occurefficiently, both strands have to be recognized and cleaved at the same time. An efficient way ofdoing this is to have the enzyme function as a homodimer binding to the major groove.One could envision a restriction endonuclease functioning as a monomer, binding to dsDNA viathe major groove and hydrolyzing a single strand. To simultaneously cleave the second strandwould require a second catalytic site on the enzyme. Alternatively, the enzyme might, uponcleavage of one strand, release from the DNA and rebind to it to cleave the second strand. Nicksin dsDNA can unwind, making it difficult to be recognized by an enzyme that initially bound todsDNA. One could imagine a DNA binding site that recognizes single-stranded DNA sequences.This, however, would complicate binding to dsDNA. So, one could envision a protein with twocatalytic sites or two DNA binding sites, one ssDNA specific and one dsDNA specific.Dimerization, however, seems like a simple way to avoid these complications.13. A carbohydrate is an integral part of a nucleoside.a. What advantage does the carbohydrate provide?Polynucleotides are formed through formation of a sugar-phosphate backbone.b. Why might ribose be preferable for this backbone instead of glucose?c. Why might 2-deoxyribose be preferable to ribose in some situations?Answer: a. A nucleoside is a base in glycosidic linkage to a sugar. Bases are poorly watersolubleand attachment to a sugar will improve their water solubility. Sugars present numeroushydroxyl groups that contribute to water solubility and provide attachment points for othermolecules.152
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