An enzyme (RNA or protein) must (1) increase the rate of a chemical reaction and (2) remain unchanged on completion of a catalytic cycle.

N-3 and N-7.

Within experimental error, the number of purines (A + G) equals the number of pyrimidines (C + T); the fractional amounts of A and T are the same, and the fractional amounts of G and C are the same; the relative ratios of the bases do not vary from one tissue to another.

(a) 5′-TACCAGCCTTAGAATTTAACTAAGGCTGTAATC-3′. Note that nucleic acid sequences are always written in the 5′→3′ direction, and the two strands of DNA are antiparallel. (b) Yes; 5′-CAGCCTTAG-3′ and 5′-CTAAGGCTG-3′ form an inverted repeat, so the strand has the potential to form a hairpin. The duplex can assume a cruciform structure.

Higher; RNA has greater thermal stability than DNA.

The DNA has a backbone with deoxyribose and will take up a B-form helix. The RNA has a backbone with ribose and will take up an A-form helix.

The presence of T rather than U as one of the primary pyrimidine nucleotides in RNA is a likely mechanism by which cells monitor mutations in DNA. Uracil is regularly produced in DNA largely by the slow, nonenzymatic hydrolytic deamination of cytosine; having thymine as the base in DNA allows the efficient detection and repair of C-to-U mutations.

(a) 5′ terminus. (b) GTC. (c) RNA. Note that the identity of a nucleic acid as RNA or DNA depends only on the ribose variant in its backbone. Even though thymine is relatively rare in RNA, an oligo- or polynucleotide containing d-ribose is RNA, and an oligo- or polynucleotide containing deoxy-d-ribose is DNA, regardless of whether thymine or uracil is present.

5′-ATTGCATCCGCGCGTGCGCGCGCGATCCCGTTACTTTCCG-3′

The double helix is the most thermodynamically stable structure. It places the hydrophobic bases in the interior of the molecule, where they interact with one another through base stacking, and the charged phosphate groups on the outside, where they can interact with water and ions.

The phosphate groups between the sugars (deoxyribose or ribose) in the sugar–phosphate backbone are highly acidic, giving the nucleic acids an overall negative charge.

Cytosine deamination to form uracil is a slow but constant reaction in all cells. In many eukaryotes, hundreds of C residues are converted to U residues every day, in every cell, creating G—U base pairs that are “seen” by the repair system as G—T pairs. Because the G is correct and the U is the damaged base, repairing G—T to G≡C restores the correct genetic information. Repair to A=T would cause a mutation.

S-6

The three-dimensional structure of a tDNA would probably be similar to that of a tRNA, in that the base pairing would make it fold into a similar cloverleaf. This is supported by experiments. However, the 2′-hydroxyl groups of ribose contribute significantly to tRNA folding, so subtle structural differences would exist. Enzymes specific for tRNAs modify certain of their bases, and many of these enzymes include the 2′-hydroxyl group in their recognition mechanism. This would prevent many, if not all, of the key base modifications from occurring. A tDNA generated in a cell would thus be unlikely to function as a tRNA substitute.

Both the G and C content and the length of the DNA influence the strength of association between the two strands in the double helix. G≡C pairs contribute more than A=T pairs, due to their stacking properties. The longer the DNA, the greater the number of base pairs and the greater the energy (i.e., the higher the temperature) required to break the hydrogen bonding between them.

An abundance of purines, especially A residues, which play an important role in the three-dimensional folding of RNA; also, multiple short segments capable of base pairing with adjacent or distant regions of the RNA molecule, particularly if these short segments are conserved in related organisms.

The regular repeating properties of the double helix produce characteristic x-ray diffraction patterns for DNA fibers, used in the earliest studies of DNA structure. However, these diffraction patterns result from the averaged properties of the DNA helices in a fiber. Determination of the properties of individual DNA sequences required single crystals containing a single, homogeneous form of the DNA molecule, arranged in a three-dimensional array. The x-ray diffraction patterns produced from single crystals could be used to determine the electron density map of the DNA in the crystal, providing an exact, rather than an averaged, image of the molecular structure.

DMT is a blocking group, preventing unwanted reactions at the 5′-hydroxyl group of the nucleotide.

(a) There is no sulfur in DNA, so proteins are uniquely labeled by ³⁵S. There is little or no phosphate in proteins (at least in bacteria), so DNA is uniquely labeled by ³²P. (b) ¹⁴C or ³H would have labeled both the DNA and the protein, permitting no differentiation. (c) The intact phages, the T2 ghosts, and the DNA are all insoluble in acid, and all are removed from solution by centrifugation. (d) The nucleotides liberated by DNase treatment are soluble in acid. Osmotic shock releases the T2 DNA into solution, where it is degraded by the DNase. The unplasmolyzed T2 phages contain DNA, but it is protected from the DNase, within the phage protein coat. (e) Both the intact viruses and the T2 ghosts adsorb to the bacteria. The components needed for attachment of T2 to the bacteria are located uniquely in the protein coat. (f) The antibodies recognize the T2 protein coat. In both the control and plasmolyzed samples, the protein coats are immunoprecipitated by the antisera, but in the plasmolyzed sample, the DNA is left behind in solution. (g) The material released by osmotic shock is entirely or almost entirely DNA. The T2 ghosts are almost entirely protein. Little or no protein is released from the phages with the DNA. The DNA does not adsorb to phage-susceptible bacteria on its own. The ghosts are protein coats that surround the DNA of the intact phage particles. These coats react with antibodies and protect the DNA within from DNase. They also are responsible for attaching phages to a bacterial host. (h) The centrifugation and resuspension remove unadsorbed phages from the solution, which otherwise would have added to the background signal. (i) About 80% of the ³⁵S-labeled phage heads are stripped from the cells by the blender, with only about 16% found in the supernatant without the blender treatment. The amount in the supernatant without blender treatment increases when the multiplicity of infection increases, as a result of some kind of displacement of phages by other phages attached to the same cells. (j) The bulk of the ³⁵S is removed from the cells by blender treatment, whereas a relatively small amount of the ³²P is removed. The capacity of the cells to survive and continue with the infection process is not affected by the treatment. The results indicate that the bulk of the protein remained in the protein coats at the cell surface during infection, while the bulk of the DNA entered the cells.