1. Taq polymerase is the DNA polymerase from the thermophilic bacterium that lives in hot springs. Consequently, it is heat stable and can withstand the high temperatures required for PCR without denaturing.

2. Ovalbumin cDNA should be used. E. coli lacks the machinery to splice the primary transcript arising from genomic DNA.

3. Consistent with its planar, aromatic structure, ethidium bromide is a DNA intercalator: it aligns itself between the paired bases in a DNA duplex.

4. The presence of the AluI sequence would, on average, be (1/4)⁴, or 1/256, because the likelihood of any base being at any position is one-fourth and there are four positions. By the same reasoning, the presence of the NotI sequence would be (1/4)⁸, or 1/65,536. Thus, the average product of digestion by AluI would be 250 base pairs (0.25 kb) in length, whereas that by NotI would be 66,000 base pairs (66 kb) in length.

5. No, because most human genes are much longer than 4 kb. A fragment would contain only a small part of a complete gene.

6. Southern blotting of an MstII digest would distinguish between the normal and the mutant genes. The loss of a restriction site would lead to the replacement of two fragments on the Southern blot by a single longer fragment. Such a finding would not prove that GTG replaced GAG; other sequence changes at the restriction site could yield the same result.

7. Although the two enzymes cleave the same recognition site, they each break different bonds within the 6-bp sequence. Cleavage by KpnI yields an overhang on the 3′ strand, whereas cleavage by Acc65I produces an overhang on the 5′ strand. These sticky ends do not overlap.

8. A simple strategy for generating many mutants is to synthesize a degenerate set of cassettes by using a mixture of activated nucleosides in particular rounds of oligonucleotide synthesis. Suppose that the 30-bp coding region begins with GTT, which encodes valine. If a mixture of all four nucleotides is used in the first and second rounds of synthesis, the resulting oligonucleotides will begin with the sequence XYT (where X and Y denote A, C, G, or T). These 16 different versions of the cassette will encode proteins containing either Phe, Leu, Ile, Val, Ser, Pro, Thr, Ala, Tyr, His, Asn, Asp, Cys, Arg, or Gly at the first position. Likewise, degenerate cassettes can be made in which two or more codons are simultaneously varied.

9. Because PCR can amplify as little as one molecule of DNA, statements claiming the isolation of ancient DNA need to be greeted with some skepticism. The DNA would need to be sequenced. Is it similar to human, bacterial, or fungal DNA? If so, contamination is the likely source of the amplified DNA. Is it similar to that of birds or crocodiles? This sequence similarity would strengthen the case that it is dinosaur DNA because these species are evolutionarily close to dinosaurs.

10. PCR amplification is greatly hindered by the presence of G–C-rich regions within the template. Owing to their high melting temperatures, these templates do not denature easily, preventing the initiation of an amplification cycle. In addition, rigid secondary structures prevent the progress of DNA polymerase along the template strand during elongation.

11. At high temperatures of hybridization, only very close matches between primer and target would be stable because all (or most) of the bases would need to find partners to stabilize the primer–target helix. As the temperature is lowered, more mismatches would be tolerated; so the amplification is likely to yield genes with less sequence similarity. In regard to the yeast gene, synthesize primers corresponding to the ends of the gene, and then use these primers and human DNA as the target. If nothing is amplified at 54°C, the human gene differs from the yeast gene, but a counterpart may still be present. Repeat the experiment at a lower temperature of hybridization.

12. Digest genomic DNA with a restriction enzyme, and select the fragment that contains the known sequence. Circularize this fragment. Then carry out PCR with the use of a pair of primers that serve as templates for the synthesis of DNA away from the known sequence.

13. The encoded protein contains four repeats of a specific sequence.

14. Use chemical synthesis or the polymerase chain reaction to prepare hybridization probes that are complementary to both ends of the known (previously isolated) DNA fragment. Challenge clones representing the library of DNA fragments with both of the hybridization probes. Select clones that hybridize to one of the probes but not the other; such clones are likely to represent DNA fragments that contain one end of the known fragment along with the adjacent region of the particular chromosome.

15. The codon(s) for each amino acid can be used to determine the number of possible nucleotide sequences that encode each peptide sequence (Table 4.5):

Ala–Met–Ser–Leu–Pro–Trp:

4 × 1 × 6 × 6 × 4 × 1 = 576 total sequences

Gly–Trp–Asp–Met–His–Lys:

4 × 1 × 2 × 1 × 2 × 2 = 32 total sequences

Cys–Val–Trp–Asn–Lys–Ile:

2 × 4 × 1 × 2 × 2 × 3 = 96 total sequences

Arg–Ser–Met–Leu–Gln–Asn:

6 × 6 × 1 × 6 × 2 × 2 = 864 total sequences

The set of DNA sequences encoding the peptide Gly-Trp-Asp-Met-His-Lys would be most ideal for probe design because it encompasses only 32 total oligonucleotides.

16. Within a single species, individual dogs show enormous variation in body size and substantial diversity in other physical characteristics. Therefore, genomic analysis of individual dogs would provide valuable clues concerning the genes responsible for the diversity within the species.

17. On the basis of the comparative genome map shown in Figure 5.28, the region of greatest overlap with human chromosome 20 can be found on mouse chromosome 2.

18. T_m is the melting temperature of a double-stranded nucleic acid. If the melting temperatures of the primers are too different, the extent of hybridization with the target DNA will differ during the annealing phase, which would result in differential replications of the strands.

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19. Careful comparison of the sequences reveals that there is a 7-bp region of complementarity at the 3′ ends of these two primers:

In a PCR experiment, these primers would likely anneal to one another, preventing their interaction with the template DNA. During DNA synthesis by the polymerase, each primer would act as a template for the other primer, leading to the amplification of a 25-bp sequence corresponding to the overlapped primers.

20. A mutation in person B has altered one of the alleles for gene X, leaving the other intact. The fact that the mutated allele is smaller suggests that a deletion has occurred in one copy of the gene. The one functioning copy is transcribed and translated and apparently produces enough protein to render the person asymptomatic.

Person C has only the smaller version of the gene. This gene is neither transcribed (negative northern blot) nor translated (negative western blot).

Person D has a normal-size copy of the gene but no corresponding RNA or protein. There may be a mutation in the promoter region of the gene that prevents transcription.

Person E has a normal-size copy of the gene that is transcribed, but no protein is made, which suggests that a mutation prevents translation. There are a number of possible explanations, including a mutation that introduced a premature stop codon in the mRNA.

Person F has a normal amount of protein but still displays the metabolic problem. This finding suggests that the mutation affects the activity of the protein—for instance, a mutation that compromises the active site of enzyme Y.

21. Chongqing: residue 2, L → R, CTG → CGG

Karachi: residue 5, A → P, GCC → CCC

Swan River: residue 6, D → G, GAC → GGC

22. This particular person is heterozygous for this particular mutation: one allele is wild type, whereas the other carries a point mutation at this position. Both alleles are PCR amplified in this experiment, yielding the “dual peak” appearance on the sequencing chromatogram.