Chapter 5. Exploring Genes and Genomes

Question 1

Question 5.1

It’s not the heat … Why is Taq polymerase especially useful for PCR?

Answer 1

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.

Question 2

Question 5.2

The right template. Ovalbumin is the major protein of egg white. The chicken ovalbumin gene contains eight exons separated by seven introns. Should ovalbumin cDNA or ovalbumin genomic DNA be used to form the protein in E. coli? Why?

Answer 2

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

Question 3

Question 5.3

Handle with care. Ethidium bromide is a commonly used stain for DNA molecules after separation by gel electrophoresis. The chemical structure of ethidium bromide is shown here. Based on this structure, suggest how this stain binds to DNA.

Answer 3

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

Question 4

Question 5.4

Cleavage frequency. The restriction enzyme AluI cleaves at the sequence 5′-AGCT-3′, and NotI cleaves at 5′-GCGGCCGC-3′. What would be the average distance between cleavage sites for each enzyme on digestion of double-stranded DNA? Assume that the DNA contains equal proportions of A, G, C, and T.

Answer 4

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.

Question 5

Question 5.5

The right cuts. Suppose that a human genomic library is prepared by exhaustive digestion of human DNA with the EcoRI restriction enzyme. Fragments averaging about 4 kb in length would be generated. Is this procedure suitable for cloning large genes? Why or why not?

Answer 5

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

Question 6

Question 5.6

A revealing cleavage. Sickle-cell anemia arises from a mutation in the gene for the β chain of human hemoglobin. The change from GAG to GTG in the mutant eliminates a cleavage site for the restriction enzyme MstII, which recognizes the target sequence CCTGAGG. These findings form the basis of a diagnostic test for the sickle-cell gene. Propose a rapid procedure for distinguishing between the normal and the mutant gene. Would a positive result prove that the mutant contains GTG in place of GAG?

Answer 6

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.

Question 7

Question 5.7

Sticky ends? The restriction enzymes KpnI and Acc65I recognize and cleave the same 6-bp sequence. However, the sticky end formed from KpnI cleavage cannot be ligated directly to the sticky end formed from Acc 65I cleavage. Explain why.

Answer 7

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.

Question 8

Question 5.8

Many melodies from one cassette. Suppose that you have isolated an enzyme that digests paper pulp and have obtained its cDNA. The goal is to produce a mutant that is effective at high temperature. You have engineered a pair of unique restriction sites in the cDNA that flank a 30-bp coding region. Propose a rapid technique for generating many different mutations in this region.

Answer 8

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.

Question 9

Question 5.9

A blessing and a curse. The power of PCR can also create problems. Suppose someone claims to have isolated dinosaur DNA by using PCR. What questions might you ask to determine if it is indeed dinosaur DNA?

Answer 9

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.

Question 10

Question 5.10

Rich or poor? DNA sequences that are highly enriched in G–C base pairs typically have high melting temperatures. Moreover, once separated, single strands containing these regions can form rigid secondary structures. How might the presence of G–C-rich regions in a DNA template affect PCR amplification?

Answer 10

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.

Question 11

Question 5.11

Questions of accuracy. The stringency of PCR amplification can be controlled by altering the temperature at which the primers and the target DNA undergo hybridization. How would altering the temperature of hybridization affect the amplification? Suppose that you have a particular yeast gene A and that you wish to see if it has a counterpart in humans. How would controlling the stringency of the hybridization help you?

Answer 11

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.

Question 12

Question 5.12

Terra incognita. PCR is typically used to amplify DNA that lies between two known sequences. Suppose that you want to explore DNA on both sides of a single known sequence. Devise a variation of the usual PCR protocol that would enable you to amplify entirely new genomic terrain.

Answer 12

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.

Question 13

Question 5.13

A puzzling ladder. A gel pattern displaying PCR products shows four strong bands. The four pieces of DNA have lengths that are approximately in the ratio of 1 : 2 : 3 : 4. The largest band is cut out of the gel, and PCR is repeated with the same primers. Again, a ladder of four bands is evident in the gel. What does this result reveal about the structure of the encoded protein?

Answer 13

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

Question 14

Question 5.14

Chromosome walking. Propose a method for isolating a DNA fragment that is adjacent in the genome to a previously isolated DNA fragment. Assume that you have access to a complete library of DNA fragments in a BAC vector but that the sequence of the genome under study has not yet been determined.

Answer 14

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.

Question 15

Question 5.15

Probe design. Which of the following amino acid sequences would yield the most optimal oligonucleotide probe?

Answer 15

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.

Question 16

Question 5.16

Man’s best friend. Why might the genomic analysis of dogs be particularly useful for investigating the genes responsible for body size and other physical characteristics?

Answer 16

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.

Question 17

Question 5.17

Of mice and men. You have identified a gene that is located on human chromosome 20 and wish to identify its location within the mouse genome. On which chromosome would you be most likely to find the mouse counterpart of this gene?

Answer 17

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.

Question 18

Question 5.18

Designing primers I. A successful PCR experiment often depends on designing the correct primers. In particular, the Tm for each primer should be approximately the same. What is the basis of this requirement?

Answer 18

18. Tm 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.

Question 19

Question 5.19

Designing primers II. You wish to amplify a segment of DNA from a plasmid template by PCR with the use of the following primers: 5′-GGATCGATGCTCGCGA-3′ and 5′-AGGATCGGGTCGCGAG-3′. Despite repeated attempts, you fail to observe a PCR product of the expected length after electrophoresis on an agarose gel. Instead, you observe a bright smear on the gel with an approximate length of 25 to 30 base pairs. Explain these results.

Answer 19

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.

Question 20

Question 5.20

Any direction but east. A series of people are found to have difficulty eliminating certain types of drugs from their bloodstreams. The problem has been linked to a gene X, which encodes an enzyme Y. Six people were tested with the use of various techniques of molecular biology. Person A is a normal control, person B is asymptomatic but some of his children have the metabolic problem, and persons C through F display the trait. Tissue samples from each person were obtained. Southern analysis was performed on the DNA after digestion with the restriction enzyme HindIII. Northern analysis of mRNA also was done. In both types of analysis, the gels were probed with labeled X cDNA. Finally, a western blot with an enzyme-linked monoclonal antibody was used to test for the presence of protein Y. The results are shown here. Why is person B without symptoms? Suggest possible defects in the other people.

Answer 20

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.

Question 21

Question 5.21

DNA diagnostics. Representations of sequencing chromatograms for variants of the α chain of human hemoglobin are shown here. What is the nature of the amino acid change in each of the variants? The first triplet encodes valine.

Answer 21

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

Karachi: residue 5, A → P, GCC → CCC

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

Question 22

Question 5.22

Two peaks. In the course of studying a gene and its possible mutation in humans, you obtain genomic DNA samples from a collection of persons and PCR amplify a region of interest within this gene. For one of the samples, you obtain the sequencing chromatogram shown here. Provide an explanation for the appearance of these data at position 49 (indicated by the arrow):

Answer 22

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.