BSCI222- Genetics BSCI222
Popular in Genetics
verified elite notetaker
Popular in Biological Sciences
This 28 page Class Notes was uploaded by Brittany Taylor on Thursday January 28, 2016. The Class Notes belongs to BSCI222 at University of Maryland taught by Dr. O'Brochta in Spring 2015. Since its upload, it has received 10 views. For similar materials see Genetics in Biological Sciences at University of Maryland.
Reviews for BSCI222- Genetics
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 01/28/16
1. Agriculture A) Based on understanding of heredity B) Genetically modified foods C) Green Revolution I. Increased dramatically the amount of food 2. Eugenics A) Using genetics to improve man B) Failed attempts such as with nazis and forced sterilization C) Can prove innocence/guilt in crime D) Find relationships through time E) Gene therapy I. Personalized medicine 3. Meiosis A) Critical for genome integrity between generations I. Transmission genetics need for transmission of genetic materials in somatic material between/within generations B) Stages I. Meiosis 1 (separates homologous chromosomes from same parent) (a) Prophase 1 (i) Replicates chromosomes, doubles DNA molecules (ii) Crossing over (iii) Longest, most complex phase Mitotic prophase= ~under 60 minutes Male meiotic prophase= ~days Female meiotic prophase= ~many years (b) Metaphase 1 (c) Anaphase 1 (i) Divides into two daughter cells (d) Telophase 1 II. Meiosis 2 (separates sister chromatids)/Mitosis (a) Prophase 2 (b) Metaphase 2 (c) Anaphase 2 (i) Divides into two more daughter cells (d) Telophase 2 C) Pieces I. Centromere= where sister chromatids bind together II. Chromatids= individual strands of DNA that bind together to form a chromosome III. Sisters= when on same chromosome IV.Homologs= chromosomes with same number V. Chiasmata= where physical exchange of DNA between homologous chromosomes occurs D) Processes I. Segregation of homologs (a) Will align on center plate during metaphase, and then are pulled to opposite ends in anaphase (b) Homologs are being separated into separate cells (c) Can follow genes and make outcome predictions based on where these chromosomes/genes go II. Independent assortment of nonhomologs (a) Different alignments during metaphase (b) Completely random and independent assortment of pairs of homologs III. Crossing over (recombination) between nonsister chromatids 4. Karyotype A) Shapes of chromosomes vary I. Metacentric= centromere in middle II. Submetacentric= centromere slightly above middle III. Acrocentric= centromere is at top IV.Telocentric= centromere is top 5. Genetic variation A) Independent assortment I. 2 sets of homologs, 2 of each homolog, 4 different combinations (gametes) x II. In general, 2 possible gametes (x= sets of homologs) 6. Genes A) Locus= location of gene in a genome B) Alleles forms of genes I. Not constant in their order; can differ II. Homozygous= 2 identical alleles at same locus III. Heterozygous= 2 different alleles at same locus (a) Only with respect to the locus/gene we are talking about (b) At one locus can only be two different alleles (one from each parent), but can be many in a population C) Segregation from meiosis I. Parent to gamete (a) Sister chromatids arise out of replication, so they have to have same allele (b) Meiosis one, simple segregation into gametes with on or the other allele (c) Can predict gametes based on parents' genes (d) Can also predict frequency II. Gamete to parent (a) Could also deduce what the genome would look like III. Must consider all possible combinations/gametes D) Crossing over I. More complex because of possibility of shuffling of chromatids 1. Genotype/phenotype A) Predicting outcomes I. Monohybrid crosses (one locus only) (a) Consider all combinations (b) Calculate expected probabilities of each combo (i) Use basic probability rules II. Dihybrid crosses (two loci, no cross over) B) If you know genotype/phenotype relationship, then you can predict ratios of phenotypes 2. Probability rules A) Multiplication rule (the “and” rule) B) Addition rule (the “or” rule) 3. Data often don’t march expectations A) Chance variations B) Wrong expectations 4. Chi squared A) Difference between observed and expected results due to random sampling error B) Graph tells you how probable your deviation from the expectations would be I. Cutoff point for rejection of the model at 5% (less than 5%= reject) C) In genetics, use results to run chi square problem I. Look on table to see where value falls II. If above 5%, then accept 5. Hypotheses are tested, NOT proven A) Testing hypotheses DOES NOT tell us which hypothesis is true B) Testing tells us the PROBABILITY of observing our data IF our hypothesis is true 6. Dominance A) Depends on gene and allele I. Relationship to other alleles; i.e. can be both dominant and recessive depending on other alleles B) Complete dominance I. One is dominant when that allele determines the phenotype no matter what the other allele says (a) Only need one copy of the “big” allele to make phenotype C) Incomplete dominance I. Mix of two alleles; blended (a) Ex: flower color II. May still have one dominant over the other D) Codominance I. Both alleles equally manifested (a) Ex: blood type antigens or horse coat color E) Dominance series I. Involves multiple alleles at one locus (a) Each more dominant than the other II. Can figure out which alleles in a species are dominant 7. Phenotype variations A) Same genotype may not represent same phenotype I. Penetrance (a) Physical presence in gene (b) How many alleles show phenotype? II. Expressivity (a) “Strength” of gene; “opacity” (b) How variable is the phenotype? 8. Genotype interactions A) With environment I. Can impact penetrance and expressivity II. Ex: temperature can impact rabbit fur color B) With other genes/loci I. Epistasis II. Two genes affecting same phenotype can compete/mask one another (a) Ex: baldness genotype masks hair color genotype III. Two loci 1. Human genetics A) Relationship between genotype and phenotype is most important thing B) Use organisms with short life cycle, lots of progeny, and easily controlled crosses C) Family studies through pedigree analysis I. Diagram of a family’s relevant genetic features through multiple generations II. Help detect patterns of inheritance (a) Can infer genotypes bases on these patterns 2. Pedigrees A) Proband person who initiates the discussion B) Autosomal dominant trait I. Never skips a generation II. Affected persons have at least one affected parent III. Appear with equal frequency in men and women IV.Approximately ½ of each generation afflicted, assuming afflicted people are heterozygous due to low frequency V. Xlinked (a) More females than males (i) Both can be affected (b) Every generation (c) Males pass on trait only to daughters, not to sons (d) Half of affected mother’s son will be affected C) Autosomal recessive trait I. Skip a generation II. Xlinked (a) More males than females (b) Male with one x will display recessive gene (only one copy) (i) Can’t pass from father to son (ii) Father passes trait to daughter, which becomes carrier, which will pass it on the males 50% of the time (c) Skips a generation D) Ylinked I. Father to son only II. Males only E) Advantages I. Counseling for reproduction and health II. Mapping genes on the genome 3. Recombination A) Map distances using frequency of recombination 4. Linkage A) Use LOD score (log [probability of linkage/probability of independence]) I. Want greater than or equal to 3 to claim linkage 5. Structure A) Four bases of DNA/RNA I. Purine (a) Adenine (b) Guanine II. Pyrimidine (a) Cytosine (b) Thymine (only in DNA) (c) Uracil (only in RNA) B) Simple ring structures I. Flat in spacefilling model C) Sugars I. Ribose (2,3OH) (a) In RNA II. Deoxyribose (3OH, 2H) (a) In DNA D) Arrangement I. Bases attached to carbon 1 of ribose ring II. Carbon 5 is phosphorylated III. Backbones of sugars and phosphates (a) 3 prime hydroxyl joined through phosphate bond to 5 prime carbon (b) Bases strung along like a ladder IV.DNA has two strands in opposite order that allows bases to pair (a) Antiparallel one strand runs 3 prime to 5 prime, the other runs 5 prime to 3 prime V. RNA has only one strand 6. Forms of DNA A) A form B) B form I. Most commonly found in cells C) Z form I. Turns the other way II. Some in cells, but not a lot of significance 7. Information (Central Dogma) A) DNA replication B) Transcription C) RNA D) Translation E) Protein 8. Genomes A) C Value Paradox I. Genome size doesn’t directly correlate to complexity B) Packaging Problem I. Twisting and turning can only do so much II. 3 feet of DNA in each cell III. Nucleosomes help solve it (histones H2A, H2B, H3, H4) (a) Neutralize charges on phosphates to overcome repulsive forces, allowing for tighter packaging (b) Have eight “tails”, ends of histones, protein structures (c) Change the way that they interact with the genetic material (i) Can change charges and gene expression (ii) Help in regulating genes (iii) Banding patterns C) Junk DNA I. Highly repeated II. Moderately repeated III. Tandem or interspersed IV.Transposable genes 1. Human genome A) 2% of our genes are protein coding B) Packaging problems I. Histones and their tails neutralize repulsive charges (a) Also help in regulation II. Still faced with the problem of reading the compacted genome C) 45% are transposable elements I. Repeated, nonsensical sequences (a) “Genetic parasites” II. 2 types (a) Retrotransposons (i) Copy and paste (i.e. now two copies of it) (b) Transposons (i) Cut and paste (i.e. the one copy moved) III. Enlargement of genome (a) Differences in amounts of genomes made of transposable elements contribute to c factor paradox IV.Transposon characteristics (a) Short inverted repeats (10100 bp) (i) Act as protein binding sites (b) Codes for transposase, an element specific enzyme (cuts and pastes) (c) Many different types, but appear evolutionarily related (i) ITR Inverted terminal repeats (ii) TSD Target site duplication (iii) Transposase “cut and paste” –ase (d) Consequences (i) Increase genome size (ii) Mutagenesis .3% of all human mutations are from de novo jumping of transposable elements 65 cases of heritable diseases (iii) Rearrangements (iv)Gene formation (v) Modulate gene expression V. Cut and paste (a) D) DNA Replication I. Fast: consider genome size and length of cells II. Accurate: consider the number of replications needed to make organisms and the consequence of errors III. Many models (a) Conservative complete strand of DNA is completely replicated for a completely new, identical one (b) Dispersive little bits of original DNA are replicated and mixed up with new DNA (c) Semiconservative one strand is copied and another strand is created IV.DNA Polymerase (a) Synthesis is not just during S phase (i) Repair, recombination, transposition, etc. V. Basic Chemistry (a) Strand growth by addition to 3’ hydroxyl (i) Must unwind DNA to serve as a template (b) Base pairing rules determine base addition (i) Polymerase can only add bases to an existing primer (c) Leading vs lagging strand (i) Antiparallel structure results in continuous and discontinuous synthesis Discontinues forms little continuous bits (ii) Bubbles formed (d) Melting DNA at origins of replication (i) A/T rich helicases (ii) Single strand binding proteins (iii) Helicasesseparate DNA duplex, take energy (iv)Gyrase relieves strain ahead of fork (e) Fidelity (i) High good for heredity (ii) Low good for evolution (iii) Proofreading by DNA polymerase (iv)One error every 10 or better (v) Processivity tendency to not “fall of” many polymerases are highly processive (vi)Fidelity tendency to not misincorporate bases 1. Synthesis A) Can only add onto 3’ side B) DNA needs to build on a primer I. Eventually gets taken out C) Antiparallel structure results in continuous and discontinuous synthesis D) Little “nick” between lagging strand parts I. Ligase glues them together 2. Replication A) End remains unreplicated following primer removal I. Will start degrading chromosomes throughout multiple replications II. Telomeres on the end, telomerase can polymerize DNA on the end 3. Transcription A) DNA RNAPROTEIN B) What gets transcribed? How do we read DNA without unwinding it? I. Surfaces of major and minor groove (a) Chemistry of edge groups II. On protein coding gene (a) Promotor tells RNA what gene to transcribe (i) Site is called +1 (ii) 10 and 35 nucleotides upstream are sequences that represent key codes in the sequence that proteins will recognize Little bit of variation in sequence is okay without changing the meaning (iii) Eukaryote promoters tend to be richer in information than prokaryote promoters Lots of different sequences provide different information (b) Transcription termination site tells where to stop transcribing (c) Protein coding region gets turned into an RNA molecule III. Both strands carry gene info (a) Only one strand will be used as a template for any given gene C) Chemistry I. Always add to 3’ OH end II. Use uracil instead of thymidine III. Prokaryotes (a) Core enzyme= RNA polymerase (b) Sigma= specifically for promotor recognition (c) Holoenzyme= core enzyme + sigma (d) Initiation (i) Promoter recognition, 10 ad 35 binding (“closed complex”) (ii) Unwind DNA around 10 to start site region (“open complex”) (iii) Transition to the elongation stage, loss of sigma factor (iv)TATA box 4. In mRNA A) Nucleotides carry a message I. 5’ end won’t be translated, but translation machinery construction and translation initiation II. 3’ end plays a role in translation initiation and message persistence 5. Modification A) After being made, the 5’ nucleotides mRNA is modified I. Most common is methylation, adding of a methylated guanine (a) Added 5’ to 5’ (b) Cap, helps in formation of ribosomes B) 3’ end is also modified I. Cleaved, and a long string of A’s is added to make a polyA tail (a) Protein recognition for proteins that move message into cytoplasm (b) Resist degradation (c) Plays a role in initiating translation C) Message is ordered into blocks of coding and noncoding sequences I. Exons must be separated from noncoding introns (a) Boundaries marked by consensus sequences (splice sites) (i) 3’ and 5’ sites differ (b) Size and number of introns varies extensively (c) Two step process (i) 5’ splice site is cleaved, then joined at 2’ hydroxyl of nucleotide that is at branch point (ii) 3’ splice site cleaved, joins exons together (iii) Spliceosome (iv)Large, multi protein (v) Must find splice sites Done by ribonucleoprotein > RNA in it is a sequences finder, complementary II. Alternate splicing (a) Not all exons are necessarily used at one time (i) Some are linked, but not others (ii) One gene can make multiple proteins with different combos of exons (b) Relative order is never changed (c) Choosing process is critical and highly regulated III. Transsplicing (a) Can use exons from different strands 6. Processing A) Ribosomal RNA are a structural component of ribosomes I. RNA polymerase 1 II. Prokaryotes (a) Three ribosomal RNAs are made by cleaving the primary transcript at precise points (b) Some nucleotides are methylated beforehand III. Eukaryotes (a) Ribosomes are constantly being replenished (i) Constantly transcribed (b) Primary transcript that is processed by methylation and cleavage B) tRNA I. Must be processed in order to create a mature DNA (a) Chemical modification of certain nucleotides that create unusual bases II. Each tRNA can have amino acid on 3’ end at amino acid attachment site C) Structures I. rRNA and tRNA don’t exist naturally as linear molecules (a) Regions of the molecules that are complementary can base pair to form secondary structures II. rRNA (a) Diverse as proteins (i) Many different processes Ex: splicing= snRNAs (b) Can have enzymatic activities (i) Ribozymes (ii) Cleavage activities, etc (c) RNA world 7. 1. Initiation A) In prokaryotes I. Nucleic acid signals (nucleotide sequences) and specific protein factors (initiation factors) (a) AUG codon (ShineDelgarno complex) (b) fMET allows addition of large subunit II. Source of energy B) Eukaryotic I. Process (a) Small ribosomal subunit and (many more) initiation factors associate with Met charged tRNA (b) Binds to 5’ 7methyl guanine cap (c) Begins to move and scan until it finds Met codon (i) Not the first one it encounters Also need signaling Kozak sequences for the “good” AUG (d) Large subunit is added (e) All factors dissociate and are recycled II. Separated in time and space III. mRNA longevity is highly variable IV.Ribosome structure is different 2. Genetic code A) Degenerate more than one codon per acid B) Only about 40 tRNAs I. Some anticodons can recognize more than one codon (a) First and second must match (b) Third position doesn’t have to match exactly (i) “Wobble position” (ii) Enables cells to use one tRNA to recognize more than one codon despite limit on number of available anticodons 3. Elongation A) Process I. Starts with Met in P site II. Adds next appropriately charge tRNA to A site (a) Determined by sequence of RNA and anticodon or T=RMA III. Peptide bond between incoming amino acid and first Met/existing chain (a) Transferred to incoming acid in A site IV.Ribosome shifted one codon (a) tRNA in A site moves to P site (b) tRNA in P site moves to E site and exits in the next step (c) A site is now empty B) Requires energy and elongation factors I. 1 step= eftu and efts (a) Shepherd tRNA with their amino acids into A site (b) GTP converted to GDP II. Ribosomal large subunit catalyzing peptide bond formation (a) Ribozyme (RNA with enzymatic activity) (b) Peptide chain attached to tRNA in P sight to amino acid on tRNA in A site (i) P site is now free III. EFG moves amino acid chain down one codon C) Ends at stop codon I. No tRNAs that recognize them II. Proteins (release factors) recognize them (a) RF1. RF2, and RF3 to A site III. Peptide is released from the last tRNA and ribosome dissociates 4. Regulation A) Each cell does not express every single gene in the body B) Can selectively turn genes on and off I. Some are on all the time= constitutive gene expression (a) Ribosomal proteins, elongation factors, basic processes, etc II. Expensive, inefficient, and chaotic sometimes C) Circuit, switch, feedback loop D) Can be regulated at many different levels 5. Lac operon has two switches A) Glucose switch I. Upstream of lactose switch II. CAP protein binds to region of Lac operon, turning on system (a) Normally it can’t bind to DNA (b) When glucose is high, cAMP is low, it is switched off; when glucose is low, cAMP is high, it is switched on B) Lactose switch I. Downstream of glucose switch II. Allolactose binds to repressor, keeping it form binding to operator and vice verse (a) When lactose is high, allolactose is high, it is switched on; when lactose is low, allolactose is low, it is switched off C) 2 switches in series I. Glucose is master regulator (a) CAP must bind to DNA to turn on system (positive system) (b) In order for that, cAMP must induce it (positive inducible system) II. Lactose is subordinate to it III. Some sort of sensors that flip the switch 6. Negative repressible switches as well A) Trp operon 1. Eukaryotic gene regulation A) Diversity I. Number of genes II. Size of genomes and gene density III. Organization of genome/chromatin IV.Organization of eukaryotic cells B) Differences between pro and euk I. Nucleys II. Large amounts of noncoding DNA III. Intonrs IV.Coupled vs uncouples transcription/translation V. Polycistronic vs monocystronic (a) Cistrons make regulation more efficient in prokaryotes VI.Lots of RNA modification VII. Multicellularity VIII. Chromatin C) Modes I. Transcriptional control (a) Primary method (b) Steps (i) Initiation (ii) Elongation (iii) Termination (c) Controls (i) Chromatin/access to promoters (ii) Rate of assembly of initiation complex Activator and enhancers > Facilitate assembly or initiation complex > Gene specific, bind sequences in promoters > Recruit components to promoter (iii) Rate of transition to elongation phase Mediator > Integrates signals that affect Pol II Pol II > Carboxyl terminal domain is phosphorylated Insulators > Prevent activators from interacting with wrong initiation complex II. Posttranscriptional control (a) Chromatin modification histone tails can be modified (i) Acetylation Change in charge or chromatin structure Make protein associate more strongly with DNA (ii) Remodelling rearranging nucleosomes Slide it or disassemble it to expose sequence Mechanism chromatin remodeling complexes > Alter and reshape chromatin structure III. Translational control 2. Insulators A) Prevent activators form interacting with the wrong initiation complex 3. Regulation A) Transcriptional control I. 3 steps (a) Initiation (b) Elongation (c) Termination II. Controls (a) Access to promotors controlling chromatin (b) Rate of assembly of initiation complex (c) Rate of transition to elongation phase B) Posttranscriptional controls I. Alternative splicing (a) Splicing activators and repressors (b) Different splice sites uses (c) Spliceosomes C) Translational control I. Phosphorylation (a) Inactivates translation 4. Epigenetics A) DNA Methylation I. Regulated II. Happens at CpGs III. Patterns can be inherited IV.Imprinting B) Chromatin Modification 1. Methylation occurs at CpGs A) Patterns can be copied following replication, so therefore can be inherited 2. Epigenetics A) Everything we do can affect future generation(s) 3. RNAbased gene regulation A) dsRNA silencing I. Mechanism (a) Dicer cuts dsRNA into 2125 bp fragments (associated with RNA induced silencing complex) (b) One strand retained as a guide RNA (c) This guides RISC to homologous mRNA, which it cuts (so no gene) II. Multiple ways (a) Alter chromatin (b) Block translation (c) Destroy mRNA III. Sources (a) Viruses (b) Transcribed microRNA (noncoding, transcribed, regulate gene expression through RNAi (c) Therapeutics 4. Genome sequencing A) Old Sanger (dideoxymethod) I. “Sequencing by synthesis” II. DideoxyNTP no 2’ or 3’ OH III. Chain termination technology (a) Separation technology based on charge and size (i) Capillary electrophoresis B) New Next gen sequencing I. Uses micro and nanotechnologies to reduce the size of sample components, reduce reagent costs, and enables massively parallel sequencing reactions (a) Simultaneous sequencing and analysis of millions of samples II. Multiple technologies (a) Illumina (b) 454 1. DNA sequencing A) Chain termination technology (Sanger method) I. Initiation of strand synthesis II. A deoxynucleotide III. Strand synthesis terminates when a ddNTP is added IV.Electrophoresis (separation based on size and charge) B) Next generation sequencing I. employs micro and nanotechnologies to reduce the size of the sample components, reducing reagent costs, and enabling massively parallel sequencing reactions II. Take templates and fix them onto a solid substrate III. Replicate template IV.Cross section through the flow cell and third cycle 2. Whole genomes A) Readlimits of technologies I. Sanger= about 1000 bases, Illumina= about 100 bases B) Clone by clone I. Ordered bits of genomes (subclones) II. Each sequenced; choose minimal number to cover entire genome III. Put everything back together (“topdown” approach) IV.Slower and harder C) Shotgun I. Unordered sequencing clones II. Sequences all of them III. Put all bits together with computer; overlapping sequences IV.Faster and easier D) Assembly problem I. Annotation (a) Ab initio methods predictions based on known “spelling and grammar” (b) Based on functional evidence (e.g. gene expression) (i) Collect mRNA, sequence, compare to genome (probably are genes) 3. Gene expression profiling A) Patterns of gene expression can be insightful B) RNAseq “reads” to genome I. DNA hybridization as a way to ask questions II. Complementary sequences will anneal to each other III. Can use this to see if a population of DNA/RNA contains any that are complementary to reference sample C) Microarray I. Orderly arrangement of genome DNA fragments II. Can see which genes are expressed (under certain conditions) 4. Genotyping individuals A) SNPs differences in one base pairs I. Microarrays can detect specific SNPs II. Can use them to find genes associated with phenotypes III. Can compare between groups or do genome wide association studies (GWAS) 5. Association A) Three genes only (some alleles associated with disease) B) Can see possible cause/correlation 6. Metagenomics A) Sequencing DNA from environmental samples I. See what genes are and what they do B) Community composition and metabolism 7. Chromosomal mutations A) Types I. Deletions/rearrangements (a) Duplications (i) Region gets duplicated (b) Deletions (i) Region gets deleted (ii) Ex: Cri du chat (c) Translocations (i) Break off of region and reattachment to other chromosomes (ii) Reciprocal vs nonreciprocal Reciprocal= preserves all the total information, but some gametes formed are unstable (iii) Like crossing over, but in the wrong place (d) Inversions (i) Region gets broken off, switched/inverted, and stuck back in II. Aneuploidy (single chromosome) (a) Change in individual chromosome copy number (b) Nondisjunction in meiosis 1 or 2 III. Polyploidy (wholegenome) (a) Change in total chromosome copy (more common in plants) B) Consequences I. Disrupt copy number of genes abnormal level of gene products II. Disrupt genes at breakpoints abolish gene expression and create new fusion proteins III. Rearrange gene regulatory elements abnormal patterns of gene expression IV.In meiosis (inversion and translocation) (a) Affect homologous pairings reduced recombination, leading to defective meiotic products 8. Gene mutation A) Places I. Somatic cells (a) Passed from cell to cell during mitosis II. Germ cells (a) Passed to offspring by sex B) Types I. DNA (a) Base substitution, Insertions, Deletions, Point, or Frameshift II. Protein (a) Missense (nonsynonymous) (i) Completely different amino acid (change the sequence) (b) Silent (synonymous) (i) Change one base pair, but not the amino acid (c) Nonsense (i) Codes for a stop codon (terminates it) III. Function (a) Silent, Forward, Reverse, Gainoffunction, Lossoffunction, Conditional, or Lethal IV.Evolution (a) Harmful, Adaptive, or Neutral 9. Causes A) DNA replication errors I. Misincorporation error wrong nucleotide, not corrected by proofreader B) Chemistry I. Depurination, deamination, base analogs, alkylation, hydroxylation, oxidative reactions, radiation (ionizing, xray, or UV) II. Many chemical modifications alter base paring properties (a) Ex: deamination CG to UA to TA (b) Can be fixed before replication (c) Happen every day 10. DNA Repair A) Mismatch repair during replication B) Base excision repair repairs damaged base I. Finds damaged bas, AP endonuclease cuts phosphodiester bond, other enzymes remove ribose, DNA Pol 1 in bacteria/DNA Pol beta in eukaryotes repair hole C) Nucleotide excision repair removes damage that distorts DNA D) Double strand DNA break repair I. Non homologous end joining: just stick ends back together II. Homologous recombination repair: cross over involved A) What is cancer? I. Disease causing cells to escape normal controls of cell division II. Genetic diseases III. Diseases caused by some viruses and environmental factors B) Failures in external, internal, or apoptotic signaling can lead to cancer I. Internal signals (a) Intracellular cyclin protein concentrations oscillate during cell cycles (i) Cyclins interact with cyclindependentkinases (CDKs) to phosphorylate target proteins that regulate gene expression (b) Ex: G1 to S transition (i) RB and E2F (transcription factor) interact. CDKs 4, 6, and 2 phosphorylate RB in the presence of cyclins D and E. RB then dissociates from E2F II. External signals (a) Ex: RAS pathway (i) Ras signal transduction pathway is involved in cell survival and death C) Evidence I. Pedigree analysis II. Chromosomal abnormalities D) Big ideas I. Clonal nature of tumors (a) Genotype of tumor ells the same II. Multihit somatic mutations (a) Driver mutations contribute to cancer (b) Passenger mutations no role in cancer E) Classes of genes I. Oncogenes (a) Mutation leads to activation of genes (b) Gene activation promotes growth advantage (c) “Go” genes (d) Dominant cells “go” continuously II. Tumor suppressor genes (a) Mutations inactivate genes (b) Inactivation promotes growth advantage (c) “Stop” genes (d) Recessive cell does not “stop” III. Stability (caretaker) genes (a) Repair genes (b) Recombination (c) Rates of mutations increases (d) Easier to get multiple hits needed for cancer 2. Chronic myelogenous leukemia A) Reciprocal translocation at chromosomes 9 and 22 3. BRCA1 and BRCA2 A) Lots of different signaling pathways (repair) B) Found using pedigree analysis and positional cloning I. Physical gene mapping onto region of chromosome II. Cloning and transcript identification C) Large genes I. Common mutations (even more common in Ashkenazi Jews) 4. Genetic testing A) Sequence specific areas B) In vitro fertilization 5. Cancer genomics A) Whole genome sequencing B) Copynumber variation (insertions, deletions, inversions, duplications, etc) C) Number of somatic mutations in representative human cancers I. Synonymousno change in amino acid sequence II. Nonsynonymous changes in amino acid sequence III. Number/distribution of driver mutations change cancer/tumor types 6. Epigenetics A) CPG islands I. Increased methylation B) Modifying histones/chromatin structure C) Driver mutations (recurrent somatic mutations found in cancer genomes) I. Some histone demethylase and histone methyltransferase genes contain these D) Therapeutics I. Restore the correct pattern to the gene at DNA methyltransferase 1. Quantitative genetics= genetics of traits controlled by multiple genes A) Continuous/quantitative traits (skin color) I. Controlled by multiple genes II. Complex relationship between genotype and phenotype B) Discrete phenotypes (fruit shapes) C) Threshold characteristic= appears like a discrete phenotype but is actually continuous 2. Questions I. How much of a phenotype is heritable (a) Predict risk of disease and outcomes of breeding (b) Heritability= how much of the observed variance in phenotype is due to genetics (i) Can have multiple sources (ii) Total variance= results from genes + results from environment + interactions between results from genes and environment (iii) Heritability= results from genes divided by total variance Control genes or environment so that the results are zero (iv)Heritability= 2(r monozygotic dizygotic (v) Accounts for the variance in that phenotype within a group Doesn’t tell us about individuals, other groups, or the degree to which genes play a role (vi)Doesn’t discredit environmental factors II. what/were are the genes responsible 1. Herita2ility (broad sense) A) H =V /G P I. Proportion of total variance that is due to genetics B) Studying in humans I. Twin studies (a) H =2(R /MZ) DZ (i) R= correlation coefficient C) Does not mean environment is not important 2. Studies for identifying genes responsible for quantitative trait A) Mapping I. Cross two lines with “extreme” phenotypes of interest II. Determine phenotype and genotype with respect to markers (a) For each marker, calculate probability observed data due to linkage and due to random assortment B) Association I. GWAS II. Use population (no controlled crosses) III. Look for associations between SNPs and phenotypes 3. Population genetics= study of the frequency and interaction of alleles and genes in populations A) Calculating gene frequencies I. Number of organisms showing genotypes/all organisms B) Calculating allele frequencies I. Number of copies of the allele/number of copies of all alleles at the locus II. p and q (a) p+q=1 2 2 2 (b) (p+q) = p +2pq+q =1 C) If you know one, you can calculate for the other D) HardyWeinberg Law I. No change in allele frequency between generations if population is ideal (a) Large population, randomly mating, no mutations or migration, no natural selection 1. Population genetics= study of frequency and interaction of alleles and genes in populations A) Can predict frequency of people in a population carrying a recessive allele responsible for a disease B) Evolution occurs when there are changes in frequencies of alleles within a population 2. Not HardyWeinberg A) Specific set of conditions B) Small populations are re subject to sampling errors C) Genetic drift= changes in allele frequencies due to sampling errors I. Variance in allele frequency II. Causes (a) Bottleneck (b) Founder Effect III. Reduces genetic variability within populations D) Assortative mating I. Positive (like individuals mate) increases homozygosity (a) Inbreeding increases it very quickly II. Negative (unlike individuals) increases heterozygosity E) Mutation I. Forward and reverse II. Rates are low, effects are small in short term F) Migration/gene flow I. Diversifies individual populations, homogenizes all populations G) Natural selection=when certain genotypes show differential reproduction I. Fitness= relative reproductive success of genotype (a) Relative fitness (W) ranges from 0 to 1 (b) Can be calculated if you know the reproductive success of genotypes (c) Impacts genotype frequencies; frequency in next generation determined by initial frequency and W 3. Drift= change in allele frequencies due to sampling errors 4. Genomes as historical records A) Homology= evolved from the same character in a common ancestor I. Orthologs= homologous genes in different species II. Paralogs= homologous genes within a species (from gene duplication) B) Genome is mutable and replicated I. Changes are maintained over time II. Shared traits reflect common ancestry III. Compare variation to infer relationships similarities and differences C) Relationships can be represented by a “tree” I. Phylogenetic tree= a model for a set of genetic relationships reflecting evolution D) Molecular clock I. If the rate of a mutation of a gene is constant and we know the rate then we can use the gene as a clock to determine when cladogenesis occurred within a lineage 5. Genome altering A) Gene duplication I. Translocation or unequal crossing over B) Exon shuffling 1. Exon shuffling A) New combinations of exons B) Transposons C) Unequal crossing over 2. Death of a gene A) Pseudogenization (becoming a pseudogene) B) Through mutation 3. Genome modification A) Control/alter phenotypes I. Bigger/better crops and animals II. Preventing/curing disease B) Considerations I. Deliver transgenes to cells/nuclei (a) Somatic vs germline (b) Methods (i) Transvection (ii) Injection (iii) Biolistics (iv)Bacteria/viruses II. Integrate transgene into genome (a) Integration system (b) Methods (i) Homologous recombination (random) (ii) Viruses/transposons III. Find transgenic cells/progeny (a) Genetic marker system
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'