BIO 401/501 NUCLEIC ACIDS PROBLEM SET I ANSWERS


  1. a) [T]+[C] =0.46

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    b) [T] = 0.3, [C] = 0.24, [A]+[G]= 0.46
     
     

  2. a) Base stacking is driven by London dispersion forces and the hydrophobic effect. b) The cooperativity occurs because base stacking is a nearest neighbor interaction. Once the unfavorable nucleation parameter (resulting from unfavorable entropy of bringing the two polynucleotide chains together) the bases are already in position to base pair and stack, and these effects are multiplicative.

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  3. 5' nucleotide monophosphate base pair. Each base pair rotates around the helix axis with the phosphate of succeeding base pairs equidistant from it.

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  4. Griffith used heat killed bacterial DNA for his transformation; thus the DNA of these cells was probably denatured. Before transforming, the DNA was cooled allowing it to renature, i.e. reforming the double helical DNA. DNA renatures readily because of the driving forces of base stacking and hydrogen bond formation. These processes are favorable and will occur spontaneously after an unfavorable nucleation event brings the two complimentary strands together.

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  5. The Tm of normal DNA is affected by salt because the DNA phosphate backbone is charged. Salt, monovalent cations in particular, neutralize these negative charges thereby obviating the like-charge base repulsion of the backbone, stabilizing the double helix. Hence if the backbone is uncharged, no charge neutralization is necessary.

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  6. a) In B-DNA the sugars on both strands are in the C2'-endo family of configurations. Upon B->Z transition the sugars of the purine strand flip into the C3'endo (or C1'exo) configuration, concomitant with the conversion of the base from an anti->syn configuration. b) Not all conformations are available to a nucleotide because of the potential of steric clashes between the base on the C1' and the C5' carbon, and moreover, the sugar pucker never is planar.

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  7. The purines in Z-DNA are in the syn configuration, thus the "bulky" part of the base is over the sugar. In order to maximize the distance between the base and the C5' carbon with attached phosphate group, the sugar must be in the C3' endo configuration. This places both the C1'-N and C4'-C5' bonds in an equatorial configuration as opposed to their axial positions in C2' puckering modes.

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  8. No, Z-DNA has a minor groove which is quite narrow and very deep.

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  9. He is probably wrong because the information content of the minor groove is ambiguous, A:T=T:A and G:C=C:G. A specific DNA binding protein picks its particular sequence out of many non-specific sequences, thus requiring the unambiguous information content of the major groove. b) We know that the atomic level structure of B-DNA is very sequence dependent. Therefore, he could be correct if the protein was recognizing a particular 3D array of the minor groove donors and acceptors whose array was specified by a particular DNA sequence.

    10. No. Persistance length is essentially independent of the charges on the phosphates, but is directed much more by base sequence dependent effects.
     
  10. Positive supercoiling resists the unwinding of DNA. Thus the melting temperature of DNA would increase as the DNA became more positively supercoiled. Since this organism lives in an environment that would cause its DNA to denature if it was negatively supercoiled, the positive supercoiling is an adaptation to resist this heat induced denaturation.

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  11. The twist number changes by -2 when a right handed piece of DNA (B-DNA) switches into left-handed Z-form. Hence, since Lk of a circle does not change, Wr changes by +2. Therefore, Lk=100, Tw=102 and Wr=-2 after the transition. Because supercoiled DNA *IS* a high energy form of DNA. Supercoils come from from mechanical (transcription/replication) or chemical (ATP hydrolysis by topoII) work. This energy can be transferred to to the DNA helix to induce Z-DNA formation. That is, the writhe can partition (change) into twist.....