Gas Mileage.A Nissan Sentra gets approximately 32 mpg on the highway and 18 mpg in the city. Suppose 265 miles were driven on a full tank (12 gallons) of gasoline. Approximately how many miles were driven in the city and how many on the highway?
W. Lu 1 BIO 302 Bieberic and Eisenman Lecture 17: DNA Repair; Chromosomal Abnormalities; Transposons I. Chemically induced mutations a. Oxidative reaction: i. superoxide radicals: O ga2ns an extra electron 1. becomes radical (like to oxidize substances) ii. Example: iii. iv. Guanine 8- dihydroguanine ( is dihydrodeoxyguanine only when attached to deoxyribose sugar) 1. Can mispair with adenine b. DNA Intercalating agents: substances that squeezes in between DNA i. Planar molecules (flat) fit in between DNA base pairs 1. Distort DNA duplex/ form bulky abductsCause distortion a. Leads to DNA nicking (not efficiently repaired) i. Results: add or loss of DNA (mutations) ii. Issue: when polymerase reached intercalating agents it stalls 1. Result: Non-replicated DNAmutations iii. Examples: 1. Proflavin, 2. acridine orange 3. ethidium bromide 4. benzo(a)-pyrene (mutagenic) W. Lu 2 BIO 302 Bieberic and Eisenman iv. II. Repair Systems Correct Some DNA Damage a. The integrity of DNA is under continuous assault form spontaneous change and from mutagens b. Organisms preserve the fidelity of DNA using multiple repair systems c. These either directly repair DNA damage or allow the organism to circumvent the problems caused by unrepaired damage III. Direct Repair of DNA Damage a. The most direct way to repair DNA lesions is to identify and the reverse the DNA damage i. In other words: Direct Repair : “unmodify the modified base” b. One direct mechanism i. proofreading activity of DNA polymerase 1. 3’-5’ exonuclease activity to proofread (first line of defense) c. Additional direct repair systems i. Repair of Damage by Alkylating Agents 1. enzymes remove the added chemical groups restoring the nucleotide to its normal form 6 a. Example: O -methylg6anine is converted back to guanine by the enzyme : O -methylguanine methyltransferase i. Remove CH3 group IV. Repair of UV induced Photoproducts a. UV rays can cause adjacent thymine to covalently bind together (pyrimidine dimers) i. Use energy from UV rayscarbon groups covalently bind together W. Lu 3 BIO 302 Bieberic and Eisenman ii. Pyrimidine dimers can be directly repaired by photoreactive repair, 1. Only in a. Bacteria b. single-celled eukaryotes c. plants and some animals (not humans) 2. The enzyme photolyase uses energy from visible light to break the bonds between pyrimidine dimers 3. Example: a. Photolyase is encoded by the E.coli phr (photoreactive repair) gene 4. The thymines are adjacent to one another. (introstrand DNA modification) UV rays covalently bind thymine Photolyase use visible light to break the covalent bond 5. V. UV Repair (bacteria) a. UV- induced DNA damage can be addressed by UV repair i. Can repair DNA without light b. A short segment of DNA surrounding the photoproduct (damaged DNA) is excised i. Instead of breaking the covalent bond between the thymines, a section of DNA is cut out on one strand c. New DNA is synthesized to replace the removed nucleotides; i. in E.coli, the enzymes responsible are encoded by the genes W. Lu 4 BIO 302 Bieberic and Eisenman 1. uvr-A (2 molecules of uvr-A) bind to region where DNA is damaged 2. uvr-Bbinds with uvr-A 3. uvr-C (UV- Cut)- cuts the thymine dimer out from the DNA 4. uvr-D helicase: breaks hydrogen bonds so damages DNA strand can leave and DNA polymerase can enter and fill the missing nucleotides VI. Process of UV Repair a. two molecules of UVR-A protein and one UVR-B bind the DNA strand opposite the site of the photoproduct (dimer), i. 2 uvr-A and 1 uvr-B bind on the opposite strand of the thymine dimer b. uvr-A proteins dissociate i. uvr-A leaves c. A molecule of UVR-C joins UVR-B to form a complex i. Breaks a pair of phosphodiester bond on damaged DNA strand on both side of the photo product (dimer) 1. 5’ cut and a 3’ cut ii. About 12 nucleotides are removed d. UVR-D (a helicase) and DNA polymerase I bind and the missing nucleotide are filled in and ligase seals the nick 1) 2 UVR-A + 1 UVR- B on 2) 2 UVR-A leaves opposite side of the dimer 3) UVR- C joins UVR- B and binds on the opposite side of the dimer 4) UVR-C cuts the 12 nucleotide portion of DNA out 5) UVR- D binds to by making a 3’ cut release damaged strand of DNA and a 5’ cut DNA pol fills in the gap (red strand = template) 6) Ligase seals the e. nick W. Lu 5 BIO 302 Bieberic and Eisenman VII. Nucleotide excision and Replacement (both bacteria and eukaryotic) a. if a modified nucleotide cannot be directly repaired, it can be excised and replace with the correct nucleotide i. cut out the entire nucleotide 1. cut out modified base 2. cut sugar at AP site 3. fill it back in with DNA polymerase b. this exploits the nature of the double stranded DNA ; the undamaged strand serves as a template to guide incorporation of the correct complementary nucleotide c. Nucleotide base excision repair is a multistep process i. DNA glycosylases recognize and remove modified bases leaving an AP site ii. AP nuclease then removes the remainder of the nucleotide iii. DNA polymerase and DNA ligase replace the gap with the appropriate nucleotide iv. v. W. Lu 6 BIO 302 Bieberic and Eisenman vi. DNA Damage DNA Repair mechanism Modified nucleotide Proofreading activity of DNA polymerase (exonuclease) Remove added chemical groups to modified bases Pyrimidine dimers (bacteria) Photoactive repair ( in visible light) (bacteria)break covalent bond b/w thymine dimers (not for humans) UV repair (without light) (bacteria) short section of DNA is cut out Modified nucleotide that cannot be corrected by Nucleotide excision and replacement(both direct repair (both bacteria and eukaryotes) bacteria and eukaryotes)cut out the modified base vii. VIII. DNA Damage Signaling Systems (eukaryotes) a. Biochemical mechanisms to recognize the presence of DNA lesions and initiate a repair response consist of a tightly regulated genetic process b. In humans and other animals, a multiprotein complex acts as a genomic sentry to identify mutations i. Guardian of the Genome: p53 can help identify mutations ii. Tumor suppressor genes: genes that are important in blocking and preventing the progression of cancer W. Lu 7 BIO 302 Bieberic and Eisenman 1. No p53 = no one watching the “hen house” aka mutations will go out of control c. Mutated in 90%-95% in late stage cancer d. This process is in action throughout the cell cycle e. Key molecules: i. ATM is a protein kinase that phosphorylated p53 (and other proteins) ii. ATM activates the p53 repair pathway iii. P53 pathway can make 2 choices 1. Kill the cell a. To prevent mutations from dividing 2. Delay cell division and repair the mutation IX. Role of the p53 Repair Pathway a. The p53 repair pathway controls cell responses to mutation by deciding to either i. Pause the cell cycle at the G1-to-S transition to allow time to repair 1. P53 levels are low in healthy cells but increases as ATM levels increase in response to DNA damage a. Healthy cells: low p53 b. Non healthy cells : high (active) ATM high phosphorylated p53 i. Phosphorylated p53: 1. Higher half life 2. accumulates 2. P53 initiates G 1rrest a. induces synthesis of p21, i. P21: inhibits formation of cyclin-CDK complexes ( require for cell to progress through cell cycle) b. allows time to repair damaged DNA 3. Completed repair reduced p53 levels and allows the cell cycle to proceed a. Repair DNA mutation kinase less active less phosphorylated p53 ( does not accumulated anymore because of less phosphorylated p53 lower half-life) ii. OR! iii. Direct the cell to undergo programmed cell death 1. P53 repair pathway induces apoptosis (programmed cell death) 2. P53 also activated the transcription of the BAX gene, which encodes a slowly acting inhibitor of BCL2 W. Lu 8 BIO 302 Bieberic and Eisenman 3. BCL2 represses apoptosis a. BCL2: prevents your cells from dying (apoptosis) b. healthy cells : BCL2 protein maintains this repression (prevents cell from dying) c. damages cells: if the p53 induced pause of the cell cycle persists too long, the apoptotic pathway is induced when BCL2 is inhibited by binding to BAX i. BCL2 binds with BAX BAX accumulatesinhibits BCL2 cell death 4. BCL2 is LIFE!!! Literally… 5. In summary: a. Super High p53 transcribe BAX gene BCL2 inhibited RIP cell ATM directly phosphorylate p53 there is DNA damage… P21 binding site has a higher p53 affinity because it requires a lower concentration of p53 to bind Choice 1: and cause a reaction. Choice 2: Turn on Higher level P21 is like “gum on a shoe” only takes one piece p21repair of p53 gum for it to stick cell life Transcribe BAX is like a “sticky note”] BAX gene does not stick as inhibits BCL2 cell easily death 6. X. Nondisjunction leads to changes in chromosome number a. Nondisjunction: the process of failed sister chromatid segregation i. Failure of homologous chromosomes or sister chromatids to separate as they normally do during cell division b. It can result in abnormalities in chromosome number c. In somatic cells, it can result in one daughter cell with an W. Lu 9 BIO 302 Bieberic and Eisenman i. extra chromosome (2n+1) ii. missing one chromosome (2n-1) d. The relatively poor survival of these cells normally limits their number in animals XI. Euploidy and Aneuploidy a. The number of chromosomes in a nucleus and the relative size and shape of each chromosome are species-specific characteristics b. Euploid number of chromosomes: the number of complete set of chromosomes an individual has c. Aneuploid: if cells contain a number of chromosomes that is not euploid XII. Non disjunction in Germ-Line cells a. The result of nondisjunction in germ-line cells is aneuploidy gametes that can produce aneuploidy zygotes b. Nondisjunction in meiosis I failure of homologs to separate gametes produced are either (n+1) or (n-1) Meiosis I homologs i. Fusion of these gametes with normal (n) gametes produces trisomic did not (2n+1) or monosomic (2n-1) offspring separate c. Meiosis II d. Nondisjunction in Meiosis II sister i. Failure of sister chromatids to separate normally chromatids ii. Among the 4 gametes produced by meiosis, only two will be affected in did not this case separate 1. Two gametenormal 2. One gamete n+1 3. One gameten-1 W. Lu 10 BIO 302 Bieberic and Eisenman iii. XIII. Gene Balance a. Changes in gene dosage lead to an imbalance of gene products from the affected chromosome relative to the unaffected chromosomes i. Cells care how many copy of gene there are 1. If they have extra genes, they do not know what to do and can cause problems b. Most animals are highly sensitive to changes in gene dosage (number of copy of genes a cell has) c. In contrast, plants tolerate gene dosage changes more readily XIV. Aneuploidy in Humans a. Humans are highly sensitive to gene dosage and aneuploids usually do not survive b. Only trisomies of chromosomes 13,18 and 21 are seen in newborn infants i. no autosomal monosomies are observed fatal c. Multiple forms of sex- chromosome trisomies, and one type of sex-chromosome monosomy occur i. Does not affect mental capacity, only secondary sex characteristics or fertility W. Lu 11 BIO 302 Bieberic and Eisenman d. Monosomy e. Trisomies and monosomies other than those found in newborn infants are known to occur in humans i. Studies that monitor human pregnancies indicate that about half of all conceptions spontaneously abort during the first trimester 1. Half of these pregnancies carry abnormalities of chromosome number or structure f. Trisomy 21: Down Syndrome i. Research has identified a link between the risk of trisomy 21 and maternal age 1. Older age= higher risk of trisomy 21 ii. A small number of gene on chromosome 21 are responsible for the cognitive abilities and heart abnormalities that are principal symptoms XV. Critical region on Chromosome 21 a. A portion of the chromosome called DSCR (Down syndrome Critical Region) i. correlated Down Syndrome symptoms ii. 3 chromosomes, 3 copies of DSCR Down Syndrome symptoms iii. Worse if inherit the whole chromosome b. A candidate gene called DYRK, known to produce dosage sensitive learning defects in mice and flies, makes a major contribution to Down syndrome XVI. Chromosome Breakage Causes Mutation by Loss, Gain, and Rearrangement of Chromosomes a. Mutations that result in loss or gain of chromosome segments can produce abnormalities due to gene dosage balances b. Changes may be large enough to detect microscopically c. Others may be too small for microscopy but may be detectable with molecular methods (i.e FISH) i. Example: W. Lu 12 BIO 302 Bieberic and Eisenman 1. Interstitial deletions like DSCR XVII. Partial Chromosome Deletion a. chromosome break point: the location in which both DNA strands are severed when a chromosome breaks b. The broken chromosome ends can adhere to other broken ends or the termini of intact chromosomes i. Can stick to end of another chromosome c. Or, since part of the broken chromosome is acentric (lacks a centromere), i. it may be loss during cell division 1. because it is not attached to the mitotic spindle when the cell divides ii. Acentric: a broken chromosome without a centromere d. What can break chromosomes: i. Ionization radiation ii. Gamma rays iii. Cosmic radiation e. Breakage can result in Partial Chromosome Deletion f. Partial Chromosome Deletion: loss of a portion of a chromosome i. Small deletions, or microdeletions are too small to be seen microscopically, and remove one or a few genes g. Larger Chromosome Deletions i. Larger chromosome deletions can be detected microscopically and affect more genes ii. Detachment of one chromosome arm leads to a terminal deletion; 1. Terminal deletion: the broken fragment lacks a centromere a. lost during cell division iii. Partial deletion heterozygotes: organisms with one normal and one terminally deleted chromosome 1. Usually no phenotypic consequences in somatic cells because if that part of the chromosome gets attached to another chromosome the gene dosage is still the same ( same number of genes) W. Lu 13 BIO 302 Bieberic and Eisenman 2. 3. XVIII. Interstitial Deletions a. Interstitial deletions: the loss of an internal portion of a chromosome that results from two chromosome breaks i. Observed in many organisms, including humans ii. Occurs frequently in cancers 1. Because p53 (tumor suppressor genes) can be lost in interstitial deletion ( two p53 genes from parents) a. One p53 = mutation b. Second p53= undergoes interstitial deletion c. Mutations accumulate d. The loss of p53 gene just gives the cell an advantage i. Because the loss of the p53 gene allows mutated cells the ability to continue to reproduce despite having a lot of mutations. 1. Normally BAX would have activated apoptosis but it is not in this case resulting in a mini evolution within the organism where mutated cells can survive iii. W. Lu 14 BIO 302 Bieberic and Eisenman If this was a cancer case: Common deletion region (CDR): where the p53 gene is If this was a Down syndrome case: CDR: where DSCR is XIX. Unequal Crossover a. Unequal crossovers occurs occasionally between two homologs b. This can result in partial duplication on one homolog and partial deletion on the other i. Leads to changes in the copy number of genes that can lead to a mutant phenotype ii. iii. XX. Detecting Duplication and Deletion a. Large deletions or duplications can be detected by microscopy that reveals altered chromosome banding patterns b. Microdeletions and micro-duplication W. Lu 15 BIO 302 Bieberic and Eisenman i. Too small to use banding technique to detect ii. Use FISH (fluorescent in situ hybridization) a molecular technique 1. Used to detect the presence or absence of a DNA sequence a. Use a gene specific probe that only lights up a specific gene XXI. Inversion and Translocation a. Caused by chromosome breakage b. No critical genes mutated + dose sensitive genes in balance = may be no phenotypic consequences c. Inversion: reattachment of chromosomes in the wrong orientation to the same chromosome i. Two types of inversion (depending on the position of the centromere) 1. Paracentric inversion: the centromere is outside of the inverted region (no centromere) 2. Pericentric inversion: the centromere is within the inverted region (have a centromere) ii. Inversion heterozygotes: heterozygotes with one normal and one inverted homolog iii. W. Lu 16 BIO 302 Bieberic and Eisenman iv. v. Inversion loop (inversion during meiosis) form from the alignment of a normal chromosome with its inverted homologs at the synapsis vi. Crossing over that occurs outside the inverted region takes place as normal vii. Crossing over within the inverted region results in duplications and deletions in the recombinant chromosomes 1. Big trouble! viii. Crossing over within a paracentric inversion 1. Crossing over that occurs within a paracentric inversion results a. a dicentric chromosome (2 centromeres) + acentric fragment (no centromeres) 2. The dicentric chromosome is pulled toward both spindle poles of the cell and eventually breaks at a random point; a. both product of the break are missing genetic material b. The acentric fragment is lost W. Lu 17 BIO 302 Bieberic and Eisenman Normal: ABCDEFGHI Inverted: ADCBEFGHI To synapse during meiosis, the genes have to line up forming the inversion loop Crossing over leads to the formation of the dicentric chromosomes Crossing over leads to the formation of the dicentric bridge. Acentric fragment is lost d. XXII. Translocation: reattachment to a non-homologous chromosome a. occurs when broken ends of the non-homologous chromosomes are reattached b. common in cancers of blood cells (‘liquid’ tumors), very infrequent in most solid tumors c. Three types of Translocations i. Unbalanced translocations 1. One piece of one chromosome is translocated to a non- homologous chromosome 2. No reciprocal event W. Lu 18 BIO 302 Bieberic and Eisenman 3. ii. Reciprocal balanced translocations 1. pieces of two non-homologous chromosomes switch places 2. iii. Robertsonian translocation 1. Aka chromosome fusions 2. Fusion of two non-homologous chromosomes a. The entire chromosome actually fuses together 3. 4. W. Lu 19 BIO 302 Bieberic and Eisenman d. Transposable genetic elements move throughout the genome i. Transposable genetic elements (both in eukaryotes and bacteria): DNA sequences that can be moved within the genome by an enzyme driven process, transposition ii. Transposition: 1. the enzyme process that move transposable genetic elements within the genome iii. Transposition moves transposable genetic elements iv. Exist in many forms, lengths and copy numbers v. Insertional inactivation: 1. a transposable element that causes a wild-type allele and disrupts its function e. General characteristics of transposable elements: i. Flanking direct repeats 1. Repeats in the genome 2. Not part of transposon 3. A consequence on how the transposable element moves in the genome a. Involve cleaving the genomic DNA with an enzyme and leaves an overhang b. And when the DNA polymerase fills in the gap, it leads to the flanking direct repeats ii. Terminal inverted repeats 1. I AM PART OF THE TRANSPOSABLE ELEMENT. a. The DNA sequence that can move around 2. W. Lu 20 BIO 302 Bieberic and Eisenman When the transposable element moves it involves cleaving the DNA with an enzyme. The flanking direct repeats form when DNA polymerase fills in the gap. f. Conservative (non-replicative) Transposition i. excises a transposable element from one position and inserts it into a new location ii. similar to cut and paste (CTRL C, CTRL V) iii. moves transposable element around the genome 1. NO increase in number of transposable elements 2. Just changed the location g. Transposition Modifies Bacterial Genomes i. Transposons carry transposase gene 1. Transposase gene encodes the enzyme required for movement ii. IS (Insertion sequences) 1. Simple transposon elements a. Only have the genes and sequences needed for autonomous transposition 2. About 1000bp of DNA 3. Contain a transposase gene that is bracketed by a short inverted repeated (IR) sequence 4. Different IS = different IR sequences 5. Inverted repeats are important for transposase- mediated transposition 6. W. Lu 21 BIO 302 Bieberic and Eisenman The purple section is 1057- 36 = 1021bp Terminal inverted repeat= same number of base pairs 7. 8. Insertion sequence transposition process a. Cleavage at ends of IS elements b. Transposase recognizes and cleaves a sequence at the new integration site c. IS element is ligated into the new site d. DNA repair fills in the gaps e. A pair of Direct repeats is produced upon integration f. W. Lu 22 BIO 302 Bieberic and Eisenman g. h. Transposition Modifies Eukaryotic Genomes i. Eukaryotic transposable elements are divided into two groups ii. DNA transposons are transposed through conservative or replicative transposition iii. Retrotransposons (unique to eukaryotes) are transcribed 1. Then reverse transcriptase produces a double-stranded copy of the element a. The one strand of RNA becomes a double stranded DNA 2. Then inserted into the genome iv. Retroviruses 1. Infect eukaryotic cells 2. Genome= single stranded RNA 3. On infection, the RNA is transcribed into double-stranded by reverse transcriptase a. Allow DNA to integrate into the host’s genome 4. the viral genes gag and env encoded by the integrated virus are needed to produce new retroviral particles; pol encodes reverse transcriptase W. Lu 23 BIO 302 Bieberic and Eisenman a. retrovirus have genes gag, pol (codes for polymerase- reverse transcriptase) and env (envelope that forms the outside of the virus) i. Retrotransposons are related to Retroviruses i. Both carry pol and contain the same gag ii. Retrotransposons do not have env 1. So they can be reverse transcribed and inserted into the host DNA 2. BUT 3. Unable to produce viral particles a. Retrotransposons cannot reproduce while retroviruses can iii. The gene(s) carried on retrotransposons are flanked by long terminal repeats (LTRs) 1. Not inverted Retrotransposon Retrovirus Carry pol and gag Carry pol, gag and env (maybe) Can reproduce No env Cant reproduce Flanked by LTRs (not inverted) iv. W. Lu 24 BIO 302 Bieberic and Eisenman v. j. Discovery of transposition i. Barbara McClintock discovered transposition using crosses involving three linked genes C, Sh, Wx in maize ii. C= purple kernels, c 1 colorless, Sh= plump, sh= shrunken; Wx= shiny kernels, wx= waxy (linked- close together on the chromosome) iii. In experiments with trihybrid C Sh WxlC 1 1. ( first observation) Observed there were some kernels that were mostly purple but had certain sections that were colorless, the colorless sectors were also shrunken and waxy iv. Results: 1. Observed that the nuclei of cells from the colorless sectors there was terminal deletion of one chromosome 9 homolog 2. The chromosome 9 homologs in the purple sectors were both intact 3. Observed that the break points of chromosome 9 all occurred in the same position of the affect chromosomes v. Interpretations of Results: 1. A genetic element (dissociation (Ds) element) was located at the site of the chromosome breakage 2. Ds needs activator element (Ac) to cause chromosome breakage W. Lu 25 BIO 302 Bieberic and Eisenman 3. 4. k. Unstable Mutants i. (second observation): colorless kernels with varied patterns of purple spotting ii. Conclusion: 1. Unstable mutant alleles were caused by insertion of Ds into the C locus to produce a colorless kernel 2. Random transposition of Ds out of the gene caused reversion to wild type of occasional purple spots W. Lu 26 BIO 302 Bieberic and Eisenman 3. XXIII. How is control of a gene expression achieved at the transcriptional level a. Cis- acting regulatory elements: i. A piece of DNA Yellow/ circle: promotor 1. Example: a. Promotors Act in Cis on the Pink/rectangle: gene 1 b. Operators same DNA molecule Blue rectangle: DNA molecule The promotor is in cis with gene 1 b. Trans-acting factor i. Can move around the cell freely ii. DNA- Binding protein 1. Can bind to promotor or operators 2. Does not have to be on the same DNA molecule can be on different DNA molecules 3. Domain: 60-90 amino acids, responsible for binding to DNA, forming hydrogen bonds with DNA a. Only part of the protein 4. Motif: a simple structure that fits into the major groove of the DNA W. Lu 27 BIO 302 Bieberic and Eisenman i. Part of DNA binding protein that will recognize the DNA sequence b. Within the binding domain c. Distinctive types of DNA- binding proteins based on the motif 5. DNA- binding proteins can be grouped based on the presence of conserved DNA- binding motifs a. Help regulate gene expression 6. XXIV. Transcriptional Control of Gene Expression requires DNA- Protein Interaction a. constitutive (ie, continuous) transcription: i. always transcribed ii. for bacterial genes needed to perform routine tasks b. Regulated transcription: i. Not always transcribed ii. For genes that are needed for responses to changing environmental condition c. Regulation of transcription includes control of both initiation and amount of transcription d. Negative control of transcription i. Binding of a repressor protein to a regulatory DNA sequence and preventing transcription of a gene or gene cluster 1. Repressor protein bind to DNA Prevent transcription e. Positive control of transcription i. Binding of an activator protein to regulatory DNA and initiating gene transcription W. Lu 28 BIO 302 Bieberic and Eisenman