14 Mutation Repair and Recombination Learning outcomes When

14. Mutation, Repair and Recombination

Learning outcomes When you have read Chapter 14, you should be able to 1. Distinguish between the terms ‘mutation' and ‘recombination', and define the various terms that are used to identify different types of mutation 2. Describe, with specific examples, how mutations are caused by spontaneous errors in replication and by chemical and physical mutagens 3. Recount, with specific examples, the effects of mutations on genomes and organisms 4. Discuss the biological significance of hypermutation and programmed mutations 5. Distinguish between the various types of DNA repair mechanism, and give detailed descriptions of the molecular events occurring during each type of repair 6. Outline the link between DNA repair and human disease 7. Draw diagrams, with detailed annotation, illustrating the processes of homologous recombination, gene conversion, site-specific recombination, conservative and replicative transposition, and retrotransposition, and discuss the biological significance of each of these mechanisms

14. 1. Mutations 14. 2. DNA Repair 14. 3. Recombination

14. 1. Mutations

Figure 14. 1. Mutation, repair and recombination. (A) A mutation is a small-scale change in the nucleotide sequence of a DNA molecule. A point mutation is shown but there are several other types of mutation, as described in the text. (B) DNA repair corrects mutations that arise as errors in replication and as a result of mutagenic activity. (C) Recombination events include exchange of segments of DNA molecules, as occurs during meiosis (see Figure 5. 15 ), and the movement of a segment from one position in a DNA molecule to another, for example by transposition (Section 14. 3. 3).

Figure 14. 2. Examples of mutations. (A) An error in replication leads to a mismatch in one of the daughter double helices, in this case a T-to-C change because one of the As in the template DNA was miscopied. When the mismatched molecule is itself replicated it gives one double helix with the correct sequence and one with a mutated sequence. (B) A mutagen has altered the structure of an A in the lower strand of the parent molecule, giving nucleotide X, which does not base -pair with the T in the other strand so, in effect, a mismatch has been created. When the parent molecule is replicated, X base-pairs with C, giving a mutated daughter molecule. When this daughter molecule is replicated, both granddaughters inherit the mutation.

Figure 14. 3. Mechanisms for ensuring the accuracy of DNA replication. (A) The DNA polymerase actively selects the correct nucleotide to insert at each position. (B) Those errors that occur can be corrected by ‘proofreading' if the polymerase has a 3′→ 5′ exonuclease activity. If the last nucleotide that was inserted is base-paired to the template then the polymerase activity predominates, but if the last nucleotide is not base-paired then the exonuclease activity is favored.

Figure 14. 4. The effects of tautomerism on base-pairing. In each of these three examples, the two tautomeric forms of the base have different pairing properties. Cytosine also has amino and imino tautomers but both pair with guanine

Figure 14. 5. Replication slippage. The diagram shows replication of a five-unit CA repeat microsatellite. Slippage has occurred during replication of the parent molecule, inserting an additional repeat unit into the newly synthesized polynucleotide of one of the daughter molecules. When this daughter molecule replicates it gives a granddaughter molecule whose microsatellite array is one unit longer than that of the original parent.

Figure 14. 6. 5 -Bromouracil and its mutagenic effect. See the text for details

Figure 14. 7. Hypoxanthine is a deaminated version of adenine. The nucleoside that contains hypoxanthine is called inosine (see Table 10. 5)

Figure 14. 8. The mutagenic effect of ethidium bromide. (A) Ethidium bromide is a flat platelike molecule that is able to slot in between the base pairs of the double helix. (B) Ethidium bromide molecules are shown intercalated into the helix: the molecules are viewed sideways on. Note that intercalation increases the distance between adjacent base pairs.

Figure 14. 9. Photoproducts induced by UV irradiation. A segment of a polynucleotide containing two adjacent thymine bases is shown. (A) A thymine dimer contains two UVinduced covalent bonds, one linking the carbons at position 6 and the other linking the carbons at position 5. (B) The (6 -4) lesion involves formation of a covalent bond between carbons 4 and 6 of the adjacent nucleotides.

Figure 14. 10. The mutagenic effect of heat. (A) Heat induces hydrolysis of β-N-glycosidic bonds, resulting in a baseless site in a polynucleotide. (B) Schematic representation of the effect of heat-induced hydrolysis on a double-stranded DNA molecule. The baseless site is unstable and degrades, leaving a gap in one strand.

Figure 14. 11. Effects of point mutations on the coding region of a gene. Four different effects of point mutations are shown, as described in the text. The readthrough mutation results in the gene being extended beyond the end of the sequence shown here, the leucine codon created by the mutation being followed by AAA = lys, TAT = tyr and ATA = ile. See Figure 3. 20 for the genetic code.

Figure 14. 12. Deletion mutations. In the top sequence three nucleotides comprising a single codon are deleted. This shortens the resulting protein product by one amino acid but does not affect the rest of its sequence. In the lower section, a single nucleotide is deleted. This results in a frameshift so that all the codons downstream of the deletion are changed, including the termination codon which is now read through. See Figure 3. 20 , for the genetic code. Note that if a three-nucleotide deletion removes parts of adjacent codons then the result is more complicated than shown here. Consider, for example, deletion of the trinucleotide GCA from the sequence … ATGGGCAAATAT … coding for metgly-lys-tyr. The new sequence is … ATGGAATAT … , coding for met-glu-tyr. Two amino acids have been replaced by a single, different one.

Figure 14. 13. Two possible effects of deletion mutations in the region upstream of a gene

Figure 14. A loss-of-function mutation is usually recessive because a functional version of the gene is present on the second chromosome copy

Figure 14. 15. A tryptophan auxotrophic mutant. Two Petri-dish cultures are shown. Both contain minimal medium, which provides just the basic nutritional requirements for bacterial growth (nitrogen, carbon and energy sources, plus some salts). The medium on the left is supplemented with tryptophan but the medium on the right is not. Unmutated bacteria, plus tryptophan auxotrophs, can grow on the plate on the left, the auxotrophs growing because the medium supplies the tryptophan that they cannot make themselves. Tryptophan auxotrophs cannot grow on the plate on the right, because this does not contain tryptophan. To identify a tryptophan auxotroph, colonies are first grown on the minimal medium + tryptophan plate and then transferred to the minimal medium plate by replica plating (see Figure 4. 18 ). After incubation, colonies appear on the minimal medium plate in the same relative positions as on the plate containing tryptophan, except for the tryptophan auxotrophs which do not grow. These colonies can therefore be identified and samples of the tryptophan auxotrophic bacteria recovered from the minimal medium + tryptophan plate.

Figure 14. 16. The effect of a constitutive mutation in the lactose operator. The operator sequence has been altered by a mutation and the lactose repressor can no longer bind to it. The result is that the lactose operon is transcribed all the time, even when lactose is absent from the medium. This is not the only way in which a constitutive lac mutant can arise. For example, the mutation could be in the gene coding for the lactose repressor, changing the tertiary structure of the repressor protein so that its DNA-binding motif is disrupted and it can no longer recognize the operator sequence, even when the latter is unmutated. See Figure 9. 24 , for more details about the lactose repressor and its regulatory effect on expression of the lactose operon.

Figure 14. 17. Hypermutation of the V-gene segment of an intact immunoglobulin gene. See Figure 12. 15 , for a description of the events leading to assembly of an immunoglobulin gene.

Technical Note 14. 1. Mutation detection

Research Briefing 14. 1. Programmed mutations?

14. 2. DNA Repair

Figure 14. 18. Four categories of DNA repair system. See the text for details

Figure 14. 19. Repair of a nick by DNA ligase

Figure 14. 20. Base excision repair. (A) Excision of a damaged nucleotide by a DNA glycosylase. (B) Schematic representation of the base excision repair pathway. Alternative versions of the pathway are described in the text.

Figure 14. 21. Short patch nucleotide excision repair in Escherichia coli. The damaged nucleotide is shown distorting the helix because this is thought to be one of the recognition signals for the Uvr. AB trimer that initiates the short patch process. See the text for details of the events occurring during the repair pathway.

Figure 14. 22. Outline of the events involved during nucleotide excision repair in eukaryotes. The endonucleases that remove the damaged region make cuts specifically at the junction between single-stranded and double-stranded regions of a DNA molecule. The DNA is therefore thought to melt either side of the damaged nucleotide, as shown in the diagram, possibly as a result of the helicase activity of TFIIH.

Figure 14. 23. Methylation of newly synthesized DNA in Escherichia coli does not occur immediately after replication, providing a window of opportunity for the mismatch repair proteins to recognize the daughter strands and correct replication errors.

Figure 14. 24. Long patch mismatch repair in Escherichia coli. See the text for details.

Figure 14. 25. Single- and double-strand-break repair. (A) A single-strand break does not disrupt the integrity of the double helix. The exposed single strand is coated with PARP 1 proteins, which protect this intact strand from breaking and prevent it from participating in unwanted recombination events. The break is then filled in by the enzymes involved in the excision repair pathways. (B) A double-stranded break is more serious because the double helix is cleaved into two segments. Various proteins bind to the broken ends, notably Ku (see the text), to protect the ends and initiate double-strand break repair.

Figure 14. 26. Non-homologous end-joining (NHEJ) in humans. (A) The repair process. Additional proteins not shown in the diagram are also involved in NHEJ. These include the protein kinases ATM and ATR (Section 13. 3. 2), whose main role may be to signal to the cell the fact that a double-strand break has occurred and the cell cycle should be arrested until the break is repaired. If the cell enters mitosis with a broken chromosome then part of that chromosome will be lost, because only one of the fragments will contain a centromere, and centromeres are essential for distribution of chromosomes to the daughter nuclei during anaphase (see Figure 5. 14 ). (B) Structure of the Ku-DNA complex. On the left is the view looking down onto the broken end of the DNA double helix, and on the right is the side view, with the broken end of the DNA molecule on the left. Ku is a heterodimer made up of the Ku 70 and Ku 80 subunits, the numbers indicating the molecular masses in k. Da. Ku 70 is colored and Ku 80 orange. The DNA molecule is shown in gray. Reprinted with permission from Walker et al. , Nature, 412, 607– 614. Copyright Macmilllan Magazines Limited.

Figure 14. 27. The SOS response of Escherichia coli. See the text for details.

14. 3. Recombination

Figure 14. 28. The Holliday model for homologous recombination

Figure 14. 29. Two schemes for initiation of homologous recombination. (A) Initiation as described by the original model for homologous recombination. (B) The Meselson. Radding modification, which proposes a more plausible series of events formation of the heteroduplex.

Figure 14. 30. The Rec. BCD pathway for homologous recombination in Escherichia coli. The events leading to formation of the heteroduplex are shown. In the bottom structure the Rec. A-coated DNA filament has formed a triplex structure, which is thought to be an intermediate in the process. See the text for details

Figure 14. 31. The role of the Ruv proteins in homologous recombination in Escherichia coli. Branch migration is induced by a structure comprising four copies of Ruv. A bound to the Holliday junction with an Ruv. B ring on either side. After Ruv. AB has detached, two Ruv. C proteins bind to the junction, the orientation of their attachment determining the direction of the cuts that resolve the structure. Reprinted with permission from Rafferty JB et al. (1996)Science, 274, 415– 421. Copyright 1996 American Association for the Advancement of Science.

Figure 14. 32. Gene conversion. One gamete contains allele A and the other contains allele a. These fuse to produce a zygote that gives rise to four haploid spores, all contained in a single ascus. Normally, two of the spores will have allele A and two will have allele a, but if gene conversion occurs the ratio will be changed, possibly to 3 A : 1 a as shown here.

Figure 14. 33. The double-strand break model for recombination in yeast. This model explains how gene conversion can occur. See the text for details.

Figure 14. 34. Integration of the bacteriophage λ genome into Escherichia coli chromosomal DNA. Both λ and E. coli DNA have a copy of the att site, each one comprising an identical central sequence called ‘O' and flanking sequences P and P′ (for the phage att site) or B and B′ (bacterial att site). Recombination between the O regions integrates the λ genome into the bacterial DNA.

Figure 14. 35. Integrated transposable elements are flanked by short direct repeat sequences. This particular transposon is flanked by the tetranucleotide repeat 5′-CTGG-3′. Other transposons have different direct repeat sequences

Figure 14. 36. Replicative and conservative transposition. DNA transposons use either the replicative or conservative pathway (some can use both). Retroelements transpose replicatively via an RNA intermediate.

Figure 14. 37. A model for the process resulting in replicative and conservative transposition. See the text for details.

Figure 14. 38. Transposition of a retroelement - Part 1. This diagram shows how an integrated retroelement is copied into a free doublestranded DNA version. The first step is synthesis of an RNA copy, which is then converted to double-stranded DNA by a series of events that involves two template switches, as described in the text

Figure 14. 39. Transposition of a retroelement - Part 2. Integration of the retroelement into the genome results in a four-nucleotide direct repeat either side of the inserted sequence. With retroviruses, this stage in the transposition pathway requires both integrase and the DNA-PKCS protein kinase and Ku proteins that are also involved in NHEJ (Section 14. 2. 4; Daniel et al. , 1999).