1. 5. Mapping Genomes

2. Learning outcomes When you have read Chapter 5, you should be able to:
Explain why a map is an important aid to genome sequencing
Distinguish between the terms ‘genetic map' and ‘physical map'
Describe the different types of marker used to construct genetic maps, and state how each type of marker is scored
Summarize the principles of inheritance as discovered by Mendel, and show how subsequent genetic research led to the development of linkage analysis
Explain how linkage analysis is used to construct genetic maps, giving details of how the analysis is carried out in various types of organism, including humans and bacteria
State the limitations of genetic mapping
Evaluate the strengths and weaknesses of the various methods used to construct physical maps of genomes
Describe how restriction mapping is carried out
Describe how fluorescent in situ hybridization (FISH) is used to construct a physical map, including the modifications used to increase the sensitivity of this technique
Explain the basis of sequence tagged site (STS) mapping, and list the various DNA sequences that can be used as STSs
Describe how radiation hybrids and clone libraries are used in STS mapping

3. 5.1. Genetic and Physical Maps
5.2. Genetic Mapping
5.3. Physical Mapping

4. Figure 5. 1. The shotgun approach to sequence assembly
Figure 5.1. The shotgun approach to sequence assembly. The DNA molecule is broken into small fragments, each of which is sequenced. The master sequence is assembled by searching for overlaps between the sequences of individual fragments. In practice, an overlap of several tens of base pairs would be needed to establish that two sequences should be linked together.

5. Figure 5. 2. Problems with the shotgun approach
Figure 5.2. Problems with the shotgun approach. (A) The DNA molecule contains a tandemly repeated element made up of many copies of the sequence GATTA. When the sequences are examined, an overlap is identified between two fragments, but these are from either end of the tandem repeat. If the error is not recognized then the internal region of the tandem repeat will be omitted from the master sequence. (B) In the second example, the DNA molecule contains two copies of a genome-wide repeat. When the sequences are examined, two fragments appear to overlap, but one fragment contains the left-hand part of one repeat and the other fragment has the right-hand part of the second repeat. In this case, failure to recognize the error would lead to the segment of DNA between the two genome-wide repeats being left out of the master sequence. If the two genome-wide repeats were on different chromosomes, then the sequences of these chromosomes might mistakenly be linked together.

6. Figure 5. 3. Alternative approaches to genome sequencing
Figure 5.3. Alternative approaches to genome sequencing. A genome consisting of a linear DNA molecule of 2.5 Mb has been mapped, and the positions of eight markers (A-H) are known. On the left, the clone contig approach starts with a segment of DNA whose position on the genome map has been identified because it contains markers A and B. The segment is sequenced by the shotgun method and the master sequence placed at its known position on the map. On the right, the whole-genome shotgun approach involves random sequencing of the entire genome. This results in pieces of contiguous sequence, possibly hundreds of kb in length. If a contiguous sequence contains a marker, then it can be positioned on the genome map. Note that, with either method, the more markers there are on the genome map the better. For more details of these sequencing strategies, see Section 6.2.

7. 5.1. Genetic and Physical Maps
The convention is to divide genome mapping methods into two categories.
Genetic mapping is based on the use of genetic techniques to construct maps showing the positions of genes and other sequence features on a genome. Genetic techniques include cross-breeding experiments or, in the case of humans, the examination of family histories (pedigrees). Genetic mapping is described in Section 5.2.
Physical mapping uses molecular biology techniques to examine DNA molecules directly in order to construct maps showing the positions of sequence features, including genes. Physical mapping is described in Section 5.3.

8. 5.2. Genetic Mapping

9. Figure 5. 4. A restriction fragment length polymorphism (RFLP)
Figure 5.4. A restriction fragment length polymorphism (RFLP). The DNA molecule on the left has a polymorphic restriction site (marked with the asterisk) that is not present in the molecule on the right. The RFLP is revealed after treatment with the restriction enzyme because one of the molecules is cut into four fragments whereas the other is cut into three fragments.

10. Figure 5. 5. Two methods for scoring an RFLP
Figure 5.5. Two methods for scoring an RFLP. (A) RFLPs can be scored by Southern hybridization. The DNA is digested with the appropriate restriction enzyme and separated in an agarose gel. The smear of restriction fragments is transferred to a nylon membrane and probed with a piece of DNA that spans the polymorphic restriction site. If the site is absent then a single restriction fragment is detected (lane 2); if the site is present then two fragments are detected (lane 3). (B) The RFLP can also be typed by PCR, using primers that anneal either side of the polymorphic restriction site. After the PCR, the products are treated with the appropriate restriction enzyme and then analyzed by agarose gel electrophoresis. If the site is absent then one band is seen on the agarose gel; if the site is present then two bands are seen.

11. Figure 5. 6. SSLPs and how they are typed
Figure 5.6. SSLPs and how they are typed. (A) Two alleles of a microsatellite SSLP. In allele 1 the motif ‘GA' is repeated three times, and in allele 2 it is repeated five times. (B) How the SSLP could be typed by PCR. The region surrounding the SSLP is amplified and the products loaded into lane A of the agarose gel. Lane B contains DNA markers that show the sizes of the bands given after PCR of the two alleles. The band in lane A is the same size as the larger of the two DNA markers, showing that the DNA that was tested contained allele 2.

12. Figure 5.7. A single nucleotide polymorphism (SNP)

13. Figure 5. 8. Oligonucleotide hybridization is very specific
Figure 5.8. Oligonucleotide hybridization is very specific. Under highly stringent hybridization conditions, a stable hybrid occurs only if the oligonucleotide is able to form a completely base-paired structure with the target DNA. If there is a single mismatch then the hybrid does not form. To achieve this level of stringency, the incubation temperature must be just below the melting temperature or Tm of the oligonucleotide. At temperatures above the Tm, even the fully base-paired hybrid is unstable. At more than 5 °C below the Tm, mismatched hybrids might be stable. The Tm for the oligonucleotide shown in the figure would be about 58 °C. The Tm in °C is calculated from the formula Tm= (4 × number of G and C nucleotides) + (2 × number of A and T nucleotides). This formula gives a rough indication of the Tm for oligonucleotides of 15–30 nucleotides in length.

14. Figure 5. 9. One way of detecting an SNP by solution hybridization
Figure 5.9. One way of detecting an SNP by solution hybridization. The oligonucleotide probe has two end-labels. One of these is a fluorescent dye and the other is a quenching compound. The two ends of the oligonucleotide base-pair to one another, so the fluorescent signal is quenched. When the probe hybridizes to its target DNA, the ends of the molecule become separated, enabling the fluorescent dye to emit its signal. The two labels are called ‘molecular beacons'.

15. Figure 5. 10. Homozygosity and heterozygosity
Figure Homozygosity and heterozygosity. Mendel studied seven pairs of contrasting characteristics in his pea plants, one of which was violet and white flower color, as shown here. (A) Pure-breeding plants always give rise to flowers with the parental color. These plants are homozygotes, each possessing a pair of identical alleles, denoted here by VV for violet flowers and WW for white flowers. (B) When two pure-breeding plants are crossed, only one of the phenotypes is seen in the F1 generation. Mendel deduced that the genotype of the F1 plants was VW, so V was the dominant allele and W was the recessive allele.

16. Figure Mendel's Laws enable the outcome of genetic crosses to be predicted. (A) Mendel's First Law states that alleles segregate randomly. The example shows inheritance of alleles A and a in a cross involving two heterozygous parents. Each member of the F1 generation has an equal chance of inheriting A or a from either of the parents. (B) The Second Law says that pairs of alleles segregate independently. One parent is shown - a plant that is heterozygous for both genes A and B. Each member of the F1 generation has an equal chance of inheriting either allele of either gene from this parent. (C) Two crosses with their predicted outcomes. In a monohybrid cross, the alleles of a single gene are followed, in this case allele T for tall pea plants and allele t for short pea plants. T is dominant and t is recessive. The grid shows the predicted genotypes and phenotypes of the F1 generation. When Mendel carried out this cross he obtained 787 tall pea plants and 277 short plants, a ratio of 2.84 : 1. In the dihybrid cross, two genes are followed. The second gene determines the shape of the peas, the alleles being R (round, the dominant allele) and r (wrinkled, which is recessive). Again, the predicted genotypes and phenotypes are shown

17. Figure 5. 12. Genes on the same chromosome should display linkage
Figure Genes on the same chromosome should display linkage. Genes A and B are on the same chromosome and so should be inherited together. Mendel's Second Law should therefore not apply to the inheritance of A and B, but holds for the inheritance of A and C, or B and C. Mendel did not discover linkage because the seven genes that he studied were each on a different pea chromosome.

18. Figure 5. 13. Partial linkage
Figure Partial linkage. Partial linkage was discovered in the early 20th century. The cross shown here was carried out by Bateson, Saunders and Punnett in 1905 with sweet peas. The parental cross gives the typical dihybrid result (see Figure 5.11C ), with all the F1 plants displaying the same phenotype, indicating that the dominant alleles are purple flowers and long pollen grains. The F1 cross gives unexpected results as the progeny show neither a 9 : 3 : 3 : 1 ratio (expected for genes on different chromosomes) nor a 3 : 1 ratio (expected if the genes are completely linked). An unusual ratio is typical of partial linkage.

19. Figure Mitosis. During interphase (the period between nuclear divisions) the chromosomes are in their extended form (Section 2.2.1). At the start of mitosis the chromosomes condense and by late prophase have formed structures that are visible with the light microscope. Each chromosome has already undergone DNA replication but the two daughter chromosomes are held together by the centromere. During metaphase the nuclear membrane breaks down (in most eukaryotes) and the chromosomes line up in the center of the cell. Microtubules now draw the daughter chromosomes towards either end of the cell. In telophase, nuclear membranes re-form around each collection of daughter chromosomes. The result is that the parent nucleus has given rise to two identical daughter nuclei. For simplicity, just one pair of homologous chromosomes is shown; one member of the pair is red, the other is blue.

20. Figure Meiosis. The events involving one pair of homologous chromosomes are shown; one member of the pair is red, the other is blue. At the start of meiosis the chromosomes condense and each homologous pair lines up to form a bivalent. Within the bivalent, crossing-over might occur, involving breakage of chromosome arms and exchange of DNA. Meiosis then proceeds by a pair of mitotic nuclear divisions that result initially in two nuclei, each with two copies of each chromosome still attached at their centromeres, and finally in four nuclei, each with a single copy of each chromosome. These final products of meiosis, the gametes, are therefore haploid. The molecular basis of recombination is described in Section 14.3.

21. Figure 5. 16. The effect of a crossover on linked genes
Figure The effect of a crossover on linked genes. The drawing shows a pair of homologous chromosomes, one red and the other blue. A and B are linked genes with alleles A, a, B and b. On the left is a meiosis with no crossover between A and B: two of the resulting gametes have the genotype AB and the other two are ab. On the right, a crossover occurs between A and B: the four gametes display all of the possible genotypes: AB, aB, Ab and ab.

22. Figure 5. 17. Working out a genetic map from recombination frequencies
Figure Working out a genetic map from recombination frequencies. The example is taken from the original experiments carried out with fruit flies by Arthur Sturtevant. All four genes are on the X chromosome of the fruit fly. Recombination frequencies between the genes are shown, along with their deduced map positions

23. Figure 5. 18. Two examples of the test cross
Figure Two examples of the test cross. In Scenario 1, A and B are genetic markers with alleles A, a, B and b. The resulting progeny are scored by examining their phenotypes. Because the double homozygous parent (Parent 2) has both recessive alleles - a and b - it effectively makes no contribution to the phenotypes of the progeny. The phenotype of each individual in the F1 generation is therefore the same as the genotype of the gamete from Parent 1 that gave rise to that individual. In Scenario 2, A and B are DNA markers whose allele pairs are codominant. In this particular example, the double homozygous parent has the genotype Ab/Ab. The alleles present in each F1 individual are directly detected, for example by PCR. These allele combinations enable the genotype of the Parent 1 gamete that gave rise to each individual to be deduced.

24. Figure 5. 19. An example of human pedigree analysis
Figure An example of human pedigree analysis. (A) The pedigree shows inheritance of a genetic disease in a family of two living parents and six children, with information about the maternal grandparents available from family records. The disease allele (closed symbols) is dominant over the healthy allele (open symbols). The objective is to determine the degree of linkage between the disease gene and the microsatellite M by typing the alleles for this microsatellite (M1, M2, etc.) in living members of the family. (B) The pedigree can be interpreted in two different ways: Hypothesis 1 gives a low recombination frequency and indicates that the disease gene is tightly linked to microsatellite M; Hypothesis 2 suggests that the gene and microsatellite are much less closely linked. In (C), the issue is resolved by the reappearance of the maternal grandmother, whose microsatellite genotype is consistent only with Hypothesis 1. See the text for more details.

25. Figure 5. 20. Three ways of achieving DNA transfer between bacteria
Figure Three ways of achieving DNA transfer between bacteria. (A) Conjugation can result in transfer of chromosomal or plasmid DNA from the donor bacterium to the recipient. Conjugation involves physical contact between the two bacteria, with transfer thought to occur through a narrow tube called the pilus. (B) Transduction is the transfer of a small segment of the donor cell's DNA via a bacteriophage. (C) Transformation is similar to transduction but ‘naked' DNA is transferred. The events illustrated in (B) and (C) are often accompanied by death of the donor cell. In (B), death occurs when the bacteriophages emerge from the donor cell; in (C), release of DNA from the donor cell is usually a consequence of the cell's death through natural causes.

26. Figure 5. 21. The basis of gene mapping in bacteria
Figure The basis of gene mapping in bacteria. (A) Transfer of a functional gene for tryptophan biosynthesis from a wild-type bacterium (genotype described as trp+) to a recipient that lacks a functional copy of this gene (trp-). The recipient is called a tryptophan auxotroph (the word used to describe a mutant bacterium that can survive only if provided with a nutrient - in this case, tryptophan - not required by the wild type; see Section ). After transfer, two crossovers (shown as green crosses) are needed to integrate the transferred gene into the recipient cell's chromosome, converting the recipient from trp- to trp+. (B) During conjugation, DNA is transferred from donor to recipient in the same way that a string is pulled through a tube. The relative positions of markers on the DNA molecule can therefore be mapped by determining the times at which the markers appear in the recipient cell. In the example shown, markers A, B and C are transferred 8, 20 and 30 minutes after the beginning of conjugation, respectively. The entire Escherichia coli chromosome takes approximately 100 minutes to transfer. (C) To be co-transferred during transduction and transformation, two or more markers must be closely linked, because these processes usually result in less than 50 kb of DNA being passed from donor to recipient. Transduction and transformation mapping are used to determine the relative positions of markers that are too close together to be mapped precisely by conjugation analysis. For more details on bacterial gene mapping see Freifelder (1987).

27. Figure Comparison between the genetic and physical maps of Saccharomyces cerevisiae chromosome III. The comparison shows the discrepancies between the genetic and physical maps, the latter determined by DNA sequencing. Note that the order of the upper two markers (glk1 and cha1) is incorrect on the genetic map, and that there are also differences in the relative positioning of other pairs of markers. Reprinted with permission from Oliver SG et al., Nature, 357, 38–46. Copyright 1992 Macmillan Magazines Limited.

28. Box 5.2. Multipoint crosses

29. Technical Note 5.1. DNA microarrays and chips

30. 5.3. Physical Mapping

31. Figure 5.23. Not all restriction sites are polymorphic

32. Figure 5. 24. Restriction mapping
Figure Restriction mapping. The objective is to map the EcoRI (E) and BamHI (B) sites in a linear DNA molecule of 4.9 kb. The results of single and double restrictions are shown at the top. The sizes of the fragments given after double restriction enable two alternative maps to be constructed, as explained in the central panel, the unresolved issue being the position of one of the three BamHI sites. The two maps are tested by a partial BamHI restriction (bottom), which shows that Map II is the correct one.

33. Figure The sequence 5′-CG-3′ is rare in human DNA because of methylation of the C, followed by deamination to give T.

34. Figure Conventional and non-conventional agarose gel electrophoresis. (A) In standard agarose gel electrophoresis the electrodes are placed at either end of the gel and the DNA molecules migrate directly towards the positive electrode. Molecules longer than about 50 kb cannot be separated from one another in this way. (B) In OFAGE, the electrodes are placed at the corners of the gel, with the field pulsing between the A pair and the B pair. OFAGE enables molecules up to 2 Mb to be separated.

35. Figure 5. 27. Optical mapping. The image shows a 2
Figure Optical mapping. The image shows a 2.4-Mb segment of the Deinococcus radiodurans genome after treatment with the restriction endonuclease NheI. The positions of the cut sites are visible as gaps in the white strand of DNA. Reprinted with permission from Lin et al., Science, 285, 1558–1562. Copyright 1999 American Association for the Advancement of Science.

36. Figure 5. 28. Gel stretching and molecular combing
Figure Gel stretching and molecular combing. (A) To carry out gel stretching, molten agarose containing chromosomal DNA molecules is pipetted onto a microscope slide coated with a restriction enzyme. As the gel solidifies, the DNA molecules become stretched. It is not understood why this happens but it is thought that fluid movement on the glass surface during gelation might be responsible. Addition of magnesium chloride activates the restriction enzyme, which cuts the DNA molecules. As the molecules gradually coil up, the gaps representing the cut sites become visible. (B) In molecular combing, a cover slip is dipped into a solution of DNA. The DNA molecules attach to the cover slip by their ends, and the slip is withdrawn from the solution at a rate of 0.3 mm s-1, which produces a ‘comb' of parallel molecules.

37. Figure 5. 29. Fluorescent in situ hybridization
Figure Fluorescent in situ hybridization. A sample of dividing cells is dried onto a microscope slide and treated with formamide so that the chromosomes become denatured but do not lose their characteristic metaphase morphologies (see Section 2.2.1). The position at which the probe hybridizes to the chromosomal DNA is visualized by detecting the fluorescent signal emitted by the labeled DNA.

38. Figure A method for blocking repetitive DNA sequences in a hybridization probe. In this example the probe molecule contains two genome-wide repeat sequences (shown in green). If these sequences are not blocked then the probe will hybridize non-specifically to any copies of these genome-wide repeats in the target DNA. To block the repeat sequences, the probe is prehybridized with a DNA fraction enriched for repetitive DNA.

39. Figure 5. 31. A fragment collection suitable for STS mapping
Figure A fragment collection suitable for STS mapping. The fragments span the entire length of a chromosome, with each point on the chromosome present in an average of five fragments. The two blue markers are close together on the chromosome map and there is a high probability that they will be found on the same fragment. The two green markers are more distant from one another and so are less likely to be found on the same fragment.

40. Figure 5. 32. One method for preparing cDNA
Figure One method for preparing cDNA. Most eukaryotic mRNAs have a poly(A) tail at their 3′ end (Section ). This series of A nucleotides is used as the priming site for the first stage of cDNA synthesis, carried out by reverse transcriptase - a DNA polymerase that copies an RNA template (Section 4.1.1). The primer is a short synthetic DNA oligonucleotide, typically 20 nucleotides in length, made up entirely of Ts (an ‘oligo(dT)' primer). When the first strand synthesis has been completed, the preparation is treated with ribonuclease H, which specifically degrades the RNA component of an RNA-DNA hybrid. Under the conditions used, the enzyme does not degrade all of the RNA, instead leaving short segments that prime the second DNA strand synthesis reaction, this one catalyzed by DNA polymerase I.

41. Figure 5. 33. Radiation hybrids
Figure Radiation hybrids. (A) The result of irradiation of human cells: the chromosomes break into fragments, smaller fragments generated by higher X-ray doses. In (B), a radiation hybrid is produced by fusing an irradiated human cell with an untreated hamster cell. For clarity, only the nuclei are shown.

42. Figure 5. 34. Separating chromosomes by flow cytometry
Figure Separating chromosomes by flow cytometry. A mixture of fluorescently stained chromosomes is passed through a small aperture so that each drop that emerges contains just one chromosome. The fluorescence detector identifies the signal from drops containing the correct chromosome and applies an electric charge to these drops. When the drops reach the electric plates, the charged ones are deflected into a separate beaker. All other drops fall straight through the deflecting plates and are collected in the waste beaker.

43. Figure 5. 35. The value of clone libraries in genome projects
Figure The value of clone libraries in genome projects. The small clone library shown in this example contains sufficient information for an STS map to be constructed, and can also be used as the source of the DNA that will be sequenced