14 Genomes and Genomics WORKING WITH THE FIGURES 1. Based on Figure 14-2, why must the DNA fragments sequenced overlap in order to obtain a genome sequence? Answer: Sequence overlap is required to align the sequenced segments relative to each other. 2. Filling gaps in draft genome sequences is a major challenge. Based on Figure 14-6, can paired-end reads from a library of 2-kb fragments fill a 10-kb gap? Answer: No, because typically these gaps consist of repetitive sequences. The sequencing reads in these regions cannot be aligned with certainty. Alternatively, gaps arise because the sequences in the region of the gap cannot be easily cloned in E. coli and would therefore not be represented in the libraries. 3. In Figure 14-9, how are the positions of codons determined? Answer: Several methods for identifying protein-coding sequences are discussed in the companion text. Computer analysis of the nucleotide sequence to look for open reading frames can identify at least some of the exons. These searches can be refined to include the detection of codon bias, binding site predictions, and conservation of predicted amino acid sequences. However, comparison of cdna sequences to the genomic sequence will directly reveal the sequences comprising the coding portion of the gene. 4. In Figure 14-10, expressed sequence tags (ESTs) are aligned with genomic sequence. How are ESTs helpful in genome annotation?
Chapter Fourteen 347 Answer: The ESTs delineate the 5 and 3 ends of a transcription unit, indicating that the exons must lie between these boundaries. 5. In Figure 14-10, cdna sequences are aligned with genomic sequence. How are cdna sequences helpful in genome annotation? Are cdnas more important for bacterial or eukaryotic genome annotations? Answer: The cdna defines the exons of a particular gene. The sequences in between, therefore, represent the introns. Consequently, cdnas are more important for eukaryotic genome annotation because bacterial genes lack introns. 6. Figure 14-16 shows syntenic regions of mouse chromosome 11 and human chromosome 17. What do these syntenic regions reveal about the genome of the last common ancestor of mice and humans? Answer: Because these genes in these large conserved blocks are found in the same order in mice and in humans, they were found in the same order in the last common ancestor of these two species. 7. Based on Figure 14-17 and the features of ultraconserved elements, what would you predict you d observe if you injected a reporter-gene construct of the rat ortholog of the ISL1 ultraconserved element into fertilized mouse oocytes and examined reporter gene expression in the developing embryo? Answer: Because these regulatory elements are so highly conserved that the expression pattern conferred by the human element mimics the expression of the native mouse ISL1 locus, you would expect the pattern conferred by the rat element to look like the image seen in Figure 14-17a. 8. The genomes of two E. coli strains are compared in Figure 14-18. Would you expect any third strain to contain more of the blue, tan, or red regions shown in Figure 14-18? Explain. Answer: The blue regions represent the genetic sequences common to both E. coli strains, K-12 and O157:H7. These common sequences contain those that define these strains as E. coli, and therefore, would be more likely to be shared with a third E. coli strain. 9. Figure 14-21 depicts the Gal4-based two-hybrid system. Why don t the bait proteins fused to the Gal4 DNA-binding protein activate reporter-gene expression?
348 Chapter Fourteen Answer: The bait proteins do not activate transcription because they do not contain activation domains. If a bait activated transcription on its own, the assay would not work. BASIC PROBLEMS 10. The word contig is derived from the word contiguous. Explain the derivation. Answer: Contig describes a set of adjacent DNA sequences or clones assembled using overlapping sequences or restriction fragments. Because the pieces are assembled into a continuous whole, it makes sense that contig is derived from the word contiguous. 11. Explain the approach that you would apply to sequencing the genome of a newly discovered bacterial species. Answer: Because bacteria have relatively small genomes (roughly three megabase pairs) and essentially no repeating sequences, the whole genome shotgun approach would be used. 12. Terminal-sequencing reads of clone inserts are a routine part of genome sequencing. How is the central part of the clone insert ever obtained? Answer: Terminal sequence reads of cloned inserts, such as those generated during whole genome shotgun sequencing, are assembled into a scaffold by matching homologous sequences shared by reads from overlapping clones. In essence, the central sequence of any single clone will be generated from the terminal sequences of overlapping clones. 13. What is the difference between a contig and a scaffold? Answer: A scaffold is also called a supercontig. Contigs are sequences of overlapping reads assembled into units, and a scaffold is a collection of joinedtogether contigs. 14. Two particular contigs are suspected to be adjacent, possibly separated by repetitive DNA. In an attempt to link them, end sequences are used as primers to try to bridge the gap. Is this approach reasonable? In what situation will it not work? Answer: Yes. If the gap is short, PCR fragments can be generated from primers based on the ends of your contigs, and these fragments can be directly
Chapter Fourteen 349 sequenced without a cloning step. The process does not work if the gap is too long. 15. A segment of cloned DNA containing a protein-encoding gene is radioactively labeled and used in an in situ hybridization to chromosomes. Radioactivity was observed over five regions on different chromosomes. How is this result possible? Answer: The clone may contain DNA that hybridizes to a small family of repetitive DNA that is adjacent to the gene being studied or within one of its introns. Alternatively, the clone may share enough homology with members of a small gene family to hybridize to all or there may be four pseudogenes that are descendants of the unique gene being studied. 16. In an in situ hybridization experiment, a certain clone bound to only the X chromosome in a boy with no disease symptoms. However, in a boy with Duchenne muscular dystrophy (X-linked recessive disease), it bound to the X chromosome and to an autosome. Explain. Could this clone be useful in isolating the gene for Duchenne muscular dystrophy? Answer: Yes. The clone hybridizes to and spans a translocation breakpoint that involves the X chromosome and an autosome. Because the DMD gene normally maps to the X chromosome, it is of interest that a translocation of the X is found in a patient with DMD. If the translocation is the cause of DMD mutation, the clone identifies a least a portion and/or location of the gene. 17. In a genomic analysis looking for a specific disease gene, one candidate gene was found to have a single-base-pair substitution resulting in a nonsynonymous amino acid change. What would you have to check before concluding that you had identified the disease-causing gene? Answer: You still need to verify that the gene with the amino acid change is the correct candidate for the phenotype under study. For example, you might transform mutants with a wild-type copy of the candidate gene and see if the transgene reverts the mutant phenotype (functional complementation). Alternatively, you can attempt to compare the phenotype of the mutant with the known function of the candidate gene. A third alternative solution is to determine if the presence of the amino acid change (or homozygosity for the amino acid change if the phenotype is recessive) correlates with the phenotype, and whether the change is absent in the general population that do not have the phenotype (or transmit the phenotype if recessive). 18. Is a bacterial operator a binding site?
350 Chapter Fourteen Answer: Yes. The operator is the location at which repressor functionally binds through interactions between the DNA sequence and the repressor protein. 19. A certain cdna of size 2 kb hybridized to eight genomic fragments of total size 30 kb and contained two short ESTs. The ESTs were also found in two of the genomic fragments each of size 2 kb. Sketch a possible explanation for these results. Answer: There are numerous possible drawings that can be made. The goal is to indicate that at least one exon be present in all eight genomic fragments and that the ESTs define the 5 and 3 ends of the transcript. EST cdna 2 kb EST 2 kb 4 kb 3 kb 6 kb 1 kb 7 kb 5 kb 2 kb genomic DNA exon 20. A sequenced fragment of DNA in Drosophila was used in a BLAST search. The best (closest) match was to a kinase gene from Neurospora. Does this match mean that the Drosophila sequence contains a kinase gene? Answer: A level of 35 percent or more amino acid identity at comparable positions in two polypeptides is indicative of a common three-dimensional structure. It also suggests that the two polypeptides are likely to have at least some aspect of their function in common. However, it does not prove that your particular sequence does in fact encode a kinase. 21. In a two-hybrid test, a certain gene A gave positive results with two clones, M and N. When M was used, it gave positives with three clones, A, S, and Q. Clone N gave only one positive (with A). Develop a tentative interpretation of these results. Answer: The yeast two-hybrid test detects possible physical interactions between two proteins. These results indicate that gene A codes for a protein that interacts with proteins encoded by clones M and N, and further, that clone M encodes a protein that also interacts with proteins encoded by clones S and Q. For example:
Chapter Fourteen 351 S N A M Q 22. You have the following sequence reads from a genomic clone of the Drosophila melanogaster genome: Read 1: TGGCCGTGATGGGCAGTTCCGGTG Read 2: TTCCGGTGCCGGAAAGA Read 3: CTATCCGGGCGAACTTTTGGCCG Read 4: CGTGATGGGCAGTTCCGGTG Read 5: TTGGCCGTGATGGGCAGTT Read 6: CGAACTTTTGGCCGTGATGGGCAGTTCC Use these six sequence reads to create a sequence contig of this part of the D. melanogaster genome. Answer: This is just a matter of aligning the sequences to determine their overlap. Read 1: TGGCCGTGATGGGCAGTTCCGGTG Read 2: TTCCGGTGCCGGAAAGA Read 3: CTATCCGGGCGAACTTTTGGCCG Read 4: CGTGATGGGCAGTTCCGGTG Read 5: TTGGCCGTGATGGGCAGTT Read 6: CGAACTTTTGGCCGTGATGGGCAGTTCC And this creates the contig: CTATCCGGGCGAACTTTTGGCCGTGATGGGCAGTTCCGGTGCCGGAAAGA 23. Sometimes, cdnas turn out to be monsters ; that is, fusions of DNA copies of two different mrnas accidentally inserted adjacent to each other in the same clone. You suspect that a cdna clone from the nematode Caenorhabditis elegans is such a monster because the sequence of the cdna insert predicts a protein with two structural domains not normally observed in the same protein. How would you use the availability of the entire genomic sequence to assess if this cdna clone is a monster or not? Answer: You can determine whether the cdna clone was a monster or not by alignment of the cdna sequence against the genomic sequence. (There are
352 Chapter Fourteen computer programs available to do this.) Is it derived from two different sites? Does the cdna map within one [gene-sized] region in the genome or to two different regions? Of course, introns may complicate the issue. 24. In browsing through the human genome sequence, you identify a gene that has an apparently long coding region, but there is a two-base-pair deletion that disrupts the reading frame. a. How would you determine whether the deletion was correct or an error in the sequencing? b. You find that the exact same deletion exists in the chimpanzee homolog of the gene but that the gorilla gene reading frame is intact. Given the phylogeny of great apes below, what can you conclude about when in ape evolution the mutation occurred? Answer: a. To determine whether the 2-bp deletion was a sequencing error, you would simply resequence that region of the genome, or a cdna copy of the gene. b. You would conclude that the 2-bp deletion occurred in the common ancestor of both chimpanzee and human after the divergence from the ancestor of gorillas. 25. In browsing through the chimpanzee genome, you find that it has three homologs of a particular gene, whereas humans have only two. a. What are two alternative explanations for this observation? b. How could you distinguish between these two possibilities? Answer: a. It may be that the ancestor of both chimpanzee and humans had three copies of the gene and one was lost after the human ancestors diverged from
Chapter Fourteen 353 chimpanzee, or conversely that the common ancestor had two and a duplication has occurred in the chimpanzee line. b. One would determine how many copies are present in the other great apes. This would allow one to determine when in the evolution of apes the change occurred. 26. The platypus is one of the few venomous mammals. The male platypus has a spur on the hind foot through which it can deliver a mixture of venom proteins. Looking at the phylogeny in Figure 14-15, how would you go about determining whether these venom proteins are unique to the platypus? Answer: One would perform an analysis similar to the one described for vitellogenin in the companion text. One would search the genome sequences of other mammals and at least one evolutionary outgroup for orthologs of the venom protein genes. 27. You have sequenced the genome of the bacterium Salmonella typhimurium, and you are using BLAST analysis to identify similarities within the S. typhimurium genome to known proteins. You find a protein that is 100 percent identical in the bacterium Escherichia coli. When you compare nucleotide sequences of the S. typhimurium and E. coli genes, you find that their nucleotide sequences are only 87 percent identical. a. Explain this observation. b. What do these observations tell you about the merits of nucleotide- versus protein-similarity searches in identifying related genes? Answer: a. Because the triplet code is redundant, changes in the DNA nucleotide sequence, (especially at those nucleotides coding for the third position of a codon) can occur without change to its encoded protein. b. Protein sequences are expected to evolve and diverge more slowly than the genes that encode them. Due to the flexibility in the third positions of most codons, the DNA sequence can accumulate changes without affecting protein structure. Protein sequences will evolve and diverge more slowly because amino acid changes may change the structure and function of the protein. Since the amino acid sequence is functionally constrained, natural selection will eliminate many deleterious amino acid changes. This will reduce the rate of change in the amino acid sequence and lead to greater sequence conservation of the amino acid sequence compared to the nucleotide sequence.
354 Chapter Fourteen 28. To inactivate a gene by RNAi, what information do you need? Do you need the map position of the target gene? Answer: Gene position is not a necessary requirement for using RNAi. What is needed is the sequence of the gene that you intend to inactivate, as you need to start with double-stranded RNA that is made with sequences that are homologous to part of the gene. 29. What is the purpose of generating a phenocopy? Answer: Phenocopying is the mimicking of a mutant phenotype by inactivating the gene product rather than the gene itself. It can be used regardless of how well-developed the genetic technology has been elaborated for a particular organism and can provide meaningful results when gene knockout or sitedirected mutagenesis is not possible. Three methods of producing phenocopies are: the prevention of gene-specific translation by the introduction of anti-sense RNA (RNA complementary to the gene-specific mrna); the introduction of double-stranded RNA with sequences homologous to part of a specific mrna resulting in major reduction in the level of the mrna (called double-stranded RNA interference); and inhibition of a specific protein through high-affinity binding of compound(s) identified through chemical genetics. 30. What is the difference between forward and reverse genetics? Answer: Forward genetics identifies heritable differences by their phenotypes and map locations and precedes the molecular analysis of the gene products. Reverse genetics starts with an identified protein or RNA and works toward mutating the gene that encodes it (and in the process, discovered the phenotype when the gene is mutated). CHALLENGING PROBLEMS 31. You have the following sequence reads from a genomic clone of the Homo sapiens genome: Read 1: ATGCGATCTGTGAGCCGAGTCTTTA Read 2: AACAAAAATGTTGTTATTTTTATTTCAGATG Read 3: TTCAGATGCGATCTGTGAGCCGAG Read 4: TGTCTGCCATTCTTAAAAACAAAAATGT Read 5: TGTTATTTTTATTTCAGATGCGA Read 6: AACAAAAATGTTGTTATT a. Use these six sequence reads to create a sequence contig of this part of the H. sapiens genome.
Chapter Fourteen 355 b. Translate the sequence contig in all possible reading frames. c. Go to the BLAST page of the National Center for Biotechnology Information, or NCBI (http://www.ncbi.nlm.nih.gov/blast/, Appendix B) and see if you can identify the gene of which this sequence is a part by using each of the reading frames as a query for protein protein comparison (BLASTp). Answer: a. This is just a matter of aligning the sequences to determine their overlap. Read 1: ATGCGATCTGTGAGCCGAGTCTTTA Read 2: AACAAAAATGTTGTTATTTTTATTTCAGATG Read 3: TTCAGATGCGATCTGTGAGCCGAG Read 4: TGTCTGCCATTCTTAAAAACAAAAATGT Read 5: TGTTATTTTTATTTCAGATGCGA Read 6: AACAAAAATGTTGTTATT And this creates the contig 5 -TGTCTGCCATTCTTAAAAACAAAAATGTTGTTATTTTTATTTCAGATGCGATCTGTGAGCCGAGTCTTTA-3 Since you do not know if the DNA sequence represents the template strand for the mrna or is the complementary strand, there are two possible transcripts. 5 -UGUCUGCCAUUCUUAAAAACAAAAAUGUUGUUAUUUUUAUUUCAGAUGCGAUCUGUGAGCCGAGUCUUUA-3 and 3 -ACAGACGGUAAGAAUUUUUGUUUUUACAACAAUAAAAAUAAAGUCUACGCUAGACACUCGGCUCAGAAAU-5 b. Translation of the first possible transcript starting at the first letter reads CLPFLKTKMLLFLFQMRSVSRVF Translation starting at the second letter reads VCHS stop Translation starting at the third reads SAILKNKNVVIFISDAICRPSL For the second possible transcript, starting at the first letter reads stop For the second possible transcript, starting at the second letter reads KDSAHRSHLK stop
356 Chapter Fourteen For the second possible transcript, starting at the third letter reads KTRLTDRI stop c. Using the nucleotide sequence of the contig and performing BLASTn or the possible translation products and performing tblastn, you will discover that this sequence and the translation product above listed first match perfectly with a region of exon 19 of the human CFTR gene. 32. Some sizable regions of different chromosomes of the human genome are more than 99 percent nucleotide identical with one another. These regions were overlooked in the production of the draft genome sequence of the human genome because of their high level of similarity. Of the techniques discussed in this chapter, which would allow genome researchers to identify the existence of such duplicate regions? Answer: The correct assembly of large and nearly identical regions is problematic with either method of genomic sequencing. However, the whole genome shotgun method is less effective at finding these regions than the clonebased strategy. This method also has the added advantage of easy access to the suspect clone(s) for further analysis. (Comparative genomic hybridization has been used effectively and is superior to sequencing for this determination. See Chapter 17, Figure 17-36 in the companion text.) 33. Some exons in the human genome are quite small (less than 75 bp long). Identification of such microexons is difficult because these distances are too short to reliably use ORF identification or codon bias to determine if small genomic sequences are truly part of an mrna and a polypeptide. What techniques of gene finding can be used to try to assess if a given region of 75 bp constitutes an exon? Answer: Assessing whether a short sequence constitutes an exon is difficult. The best way to determine if a suspected micro-exon is actually used is to look for a cdna or an EST that includes it. Alternatively, identification of consensus donor and acceptor splice site sequences can be tried and also the use of comparative genomics, that is, the conservation of the predicted amino acid encoded by the micro-exon in the same or other genomes. 34. You are studying proteins having roles in translation in the mouse. By BLAST analysis of the predicted proteins of the mouse genome, you identify a set of mouse genes that encode proteins with sequences similar to those of known eukaryotic translation-initiation factors. You are interested in determining the phenotypes associated with loss-of-function mutations of these genes.
Chapter Fourteen 357 a. Would you use forward- or reverse-genetics approaches to identify these mutations? b. Briefly outline two different approaches that you might use to look for lossof-function phenotypes in one of these genes. Answer: a. Forward genetics identifies heritable differences by their phenotypes and map locations and precedes the molecular analysis of the gene products. Reverse genetics starts with an identified protein or RNA and works toward mutating the gene that encodes it (and in the process, discovers the phenotype when the gene is mutated). Because you have known proteins and want to determine the phenotypes associated with loss-of-function mutations in the genes that encoded them, a reverse genetic approach is the answer. b. The two general approaches would be either directed mutations in the gene of interest or the generation of phenocopies of the mutant phenotype by inactivating the gene product rather than the gene itself. 35. The entire genome of the yeast Saccharomyces cerevisiae has been sequenced. This sequencing has led to the identification of all the open reading frames (ORFs, gene-size sequences with appropriate translational initiation and termination signals) in the genome. Some of these ORFs are previously known genes with established functions; however, the remainder are unassigned reading frames (URFs). To deduce the possible functions of the URFs, they are being systematically, one at a time, converted into null alleles by in vitro knockout techniques. The results are as follows: 15 percent are lethal when knocked out. 25 percent show some mutant phenotype (altered morphology, altered nutrition, and so forth). 60 percent show no detectable mutant phenotype at all and resemble wild type. Explain the possible molecular-genetic basis of these three mutant categories, inventing examples where possible. Answer: Fifteen percent are essential gene functions (such as enzymes required for DNA replication or protein synthesis). Twenty-five percent are auxotrophs (enzymes required for the synthesis of amino acids or the metabolism of sugars, genes required for normal growth rate and normal colony morphology or other traits evident in the laboratory, etc.). Sixty percent are redundant or pathways not tested (genes for histones, tubulin, ribosomal RNAs, etc., are present in multiple copies; the yeast may require
358 Chapter Fourteen many genes under only unique or special situations or in other ways that are not necessary for life in the lab where the yeast will be exposed to competition from other microbes and a wide array of conditions not found in the laboratory). 36. Different strains of E. coli are responsible for enterohemorrhagic and urinary tract infections. Based on the differences between the benign K-12 strain and the enterohemorrhagic O157:H7 strain, would you predict that there are obvious genomic differences: a. Between K-12 and uropathogenic strains? b. Between O157:H7 and uropathogenic strains? c. What might explain the observed pair-by-pair differences in genome content? d. How might the function of strain-specific genes be tested? Answer: a. The two E. coli strains K-12 and O157:H7 have 3574 protein-coding genes in common and among these orthologous genes, share 98.4 percent nucleotide identity. Even so, the complete genomes and proteomes of these species differ enormously. The genome of O157:H7 contains 1387 genes not found in the K-12 genome, and there are 528 genes in the K-12 genome not found in the O157:H7 genome. Most of the O157:H7-specific sequences are horizontally transferred foreign DNAs and would be expected to contain candidate genes that encode virulence factors, cell-invasion proteins, adherence proteins, secretion proteins, and possible metabolic genes required for nutrient transport, antibiotic resistance, and other activities that may confer the ability to survive in different hosts. It would certainly be expected that comparisons between K-12 and uropathogenic strains of E. coli would show similar enormous differences in their genomes and proteomes.
Chapter Fourteen 359 b. In comparing O157:H7 to uropathogenic strains of E. coli, you would expect that they would share the same backbone genes as K-12 and also possibly share some of the candidate genes that encode virulence factors, cell-invasion proteins, adherence proteins, secretion proteins, and other activities that help confer the ability to survive in different hosts. However, you would also expect significant differences between the two genomes as the urinary tract and colon represent separate niches that likely require unique gene products. c. The various strains of E. coli have evolved through a complex process. The ancestral backbone genes that define E. coli have undergone slow accumulation of vertically acquired sequence changes. This can be seen in the 98.4 percent nucleotide identity in the orthologous genes shared by K-12 and O157:H7. However, the sequences not in common have been introduced via numerous, independent horizontal gene-transfer events at many discrete sites. These sequences are from the genomes of viruses and other bacterial species. Some differences between the strains may also be the result of deletion. d. Functions of genes may be determined by various mechanisms. Comparison of the strain-specific genes to all other known genes using a BLAST search might provide insight to function. Reverse genetic analysis to disrupt gene function and observation of the resulting phenotype may allow you to assess the role of the normal gene product. Finally, transforming E. coli K-12 with pathogenic-specific genes and analysis of the resulting transgenic strains may also allow gene function to be assessed.