Genetics Test 2 Study Guide
Genetics Test 2 Study Guide Bisc 336
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This 33 page Study Guide was uploaded by Anna Ballard on Monday October 10, 2016. The Study Guide belongs to Bisc 336 at University of Mississippi taught by Ryan Garrick in Summer 2016. Since its upload, it has received 61 views. For similar materials see Genetics in Biology at University of Mississippi.
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Date Created: 10/10/16
LECTURE 10 Explain the role of restriction enzymes (Res), vectors, and host cells in cloning of DNA fragments RECOMBINANT DNA (Fill in the blank) • Creation of ____________________ that don’t occur naturally - IS NOT a product of ______________________ in meiosis • Tool used to isolate “_______________” and make ____________________ of it - Ex: __________ from a fruit fly put into bacterial __________ (Matching vocab) A. Restriction enzymes B. Vectors/plasmids C. Competent bacterial cells D. Receptor Sites 1. _______________ take up DNA from the environment (lab)… ends sealed to make complete circular piece 2. _______________ transformed, then clonal growth, contain receptor sites 3. _______________ take plasmids up and bring them into cell and DNA replication can proceed 4. _______________ cut ‘target’ DNA at recognition sites RESTRICTION ENZYMES (REs) Q: What is the purpose of Restriction Enzymes? A: Cut DNA: “________” DNA for __________________________, and when found, cleave both strands - Cuts in __________ fashion • Create __________________ … may have ‘______________’ (i.e., ______________________) • Original source from bacteria as a defense mechanism against bacteriophages. VECTORS (= PLASMIDS) • A vehicle for transporting a __________________ into a __________ + a replicator of the inserted DNA • Most only handle fairly small fragments (________) Good vectors… • ___________________– region that would be cut by enzymes so that it opens up to insert DNA fragment of target gene in it - it is __________ and can be cut by many different enzymes - the region is a ________________ called ____________ and is involved in _________ ______________ and changes _______ of cell when it is functioning properly o only works when there is ____________ in genes • bacterial cell with this plasmid will break down food and turn _________ • _____________________ – bacteria takes this up and will be able to grow on a medium Vectors/Plasmids… • Replicate themselves + inserted DNA, once in _________ • Have several known RE cuts sites (= versatile) • DNA inserts at a location that disrupts a color gene • carry ___________________, only transformed hosts survive Sticky ends and DNA ligase FIG. 17-2 • Start with cleavage with EcoRI from eukaryotic gene and one from bacterial plasmid (vector) • Fragments with complementary Tails • Annealing allows recombinant DNA molecules to form by complementary base pairing • DNA ligase seals the gap A recombinant DNA molecule • if intact, blue colonies form • when broken white colonies • host cell needs this gene to survive ampicillin treated growth medium Q: What would happen if one of this vector failed to take up the plasmids at all? A: They would die before they’d even be able to be seen because they would not be resistant to ampicillin and would die from the growth medium Transforming the Host Cell • lab strain of ___________ often used as bacterial ‘________’ • ‘__________ ’ = take up DNA from environment. (including _____________________)… when ___________ (heat or electric) • Modified ________________(yeast) used as host cells to study expression of cloned eukaryotic genes Cloning – summary • Host-cell chromosome • the two __________ are __________ to form a recombinant molecule • DNA to be cloned is _________ with same _______________ • introduction into host cell • host cell makes copies Know what is happening in each of the major steps in the Polymerase Chain Reaction (PCR) POLYMERASE CHAIN REACTION (PCR) • Uses a DNA Pol. Enzyme (______) that has high copying accuracy and happily operates at ___________________________ • _______-step cycle makes identical copies of ‘__________ ‘ • Use synthetic primers to set boundaries of ‘target’ PCR: first cycle 1) __________________: heat to roughly 95°C –> becomes ssDNA - Separating two strands of DNA and each are available to act as a template for synthesis of new strands - _________: roughly 20-30bp long, synthesized commercially - Strands have __________: 3’ and 5’ ends 2) __________: primers form bonds with _____________ (roughly 50°C) 3) ____________: Taq adds to 3’ end of primer (roughly 70°C) • 1 cycle ____________ number of ‘target DNA’ templates PCR: second cycle • 1 cycle has 3 parts (_________, __________, ____________) • Each cycle doubles number of template from previous cycle leads to ____________________ … 25-30 cycles usually all that’s needed - Before first cycle: 1 strand - After first cycle: 2 strands - After second cycle: 4 strands Advantages (v. Cloning) • easy (no host cells) and fast (hours, not days) • uses ___________ rather than ________________________ to isolate the ‘target DNA’ - more versatile • Requires very little DNA as starting material - E.g., forensics (single blood drop, or hair) Limitations (v. cloning) • ________________: requires some baseline information about DNA sequencing of an Organism (or close relative) • prone to ____________: tiny amount of starting DNA needed, so require several negative controls LECTURE 11 Cloning: recap • cloning isn’t always perfect, sometimes the bacteria fails to take up the inserted target DNA - If something goes horribly wrong – all those bacterial cells will fail to grow because they lack the gene that makes them resistant to ampicillin Know what is happening in each of the major steps in Polymerase Chain Reaction (PCR) Which of the following steps are involved in PCR, and correct order (within a single cycle)? A) annealing –> denature –> extension B) Denature –> duplicate –> replicate C) Polymerase –> chain –> reaction D) Denature –> annealing –> extension **More for fun Learning** …Even PCR of tardigrade DNA! • “Water bears” (small and tough) survive at -273°C to 151 °C for 10 years without water • “Cheps” Sands (tall and spooky) molecular ecologist British Antarctic Survey Understand how PCR amplicons can be assayed for DNA mutations using and restriction enzymes More applications… • Species identification (i.e., illegal whale meat sales) • Chromosome walking (i.e., sequence unknown DNA regions, adjacent to known/well-characterized ones) - Essentially start at a particular place on a chromosome and extend outward; generating an entire chromosome sequence • Screen known mutations (e.g., allele-specific PCR) Following cloning or PCR • __________________: use a variety of Res to cut ‘target’ DNA, and electrophonically separate fragments • Establish ___________, ___________, and ____________ between cut sites • different alleles at a locus (____________) uncovered by ________________________ ________________ (RFLPs) Electrophoresis • Technique for ________________________ according to differences in ____________using a gel to act as a viscous matrix - electric charge pass through this gel - Send electric current from negative –> positive end; DNA migrates towards positive end (Anode) - Bands with longer fragments will take more time to move through the gel o Aka they will travel further or shorter distances - Bands with shorter fragments will take less time o Aka they will travel further or shorter distances Medical Diagnosis • Screening disease-associated alleles via restriction fragment length polymorphism (RFLP) analysis • _____________: single DNA _______________ that leads to a different amino acid, and ß-globin protein - Homozygous: detrimental; heterozygous: effects show later • leads to change in (and loss of) RE recognitions sites… provides diagnostic RFLP banding pattern Sickle cell diagnosis using RFLP • MSTII cuts normal ß-globin 3x –> ___________DNA fragments • MSTII cuts a single mutant ß-globin 2x –> ___________ fragment • Heterozygous vs. homozygous (diploid genotype) Pharmacogenomics • Medication side-effects: common and sometimes fatal • medication effectiveness: variable, often works for only about 60% of the population • Pharmacogenomics: drug selection based on an individuals’ _______________ (rather than trial and error) - Individuals with same disease metabolize 6 MP differently - Differences due to genetic make-up (TPMT genotype) - _______________ allows customized treatment Recall why RT-PCR generates ‘snapshots’ of tissue-specific gene expression Complimentary DNA (cDNA) • ________: protein-coding DNA only… not ____________ DNA (including regulatory regions, e.g., promotors) 1. Extract ________ from a particular tissue type, at a particular time (e.g., developmental stage) 2. Use _____________________ PCR (RT-PCR) to convert mRNA back to DNA • The resulting pool of DNA represents only those _____________ - Complimentary DNA: ________________ Reverse Transcription PCR • Mature mRNA = coding • Can use as ____________ for reverse transcription • __________ primer anneals to _____________ (double stranded molecule) • __________________ enzyme extends, makes complimentary DNA enzyme extends, makes complementary DNA • mRNA-DNA hybrid molecule created • __________ enzyme nicks and digests mRNA strand • Remaining mRNA used __________________ • ….dsDNA = expressed coding DNA (‘snap shot’) DNA Sequencing • Direct way to assay ____________ and _____________ of genes, and levels of variation within/among organisms • Determine the _________ and _________ of ___________ • Until recently, almost exclusively done via the “chain termination” (aka Sanger) sequencing method Chain Termination Sequencer • _________ = set of colored peaks on a chromatogram • 1 peak per nucleotide (100bp fragment = 100 peaks) - Colored peaks tell us which is A, T, C, or G in a nucleotide sequence • Usually around 800bp reads • but…. Fairly slow and expensive LECTURE 12 Recall the steps involved in sequencing a genome Understand the role of sequence alignment Describe approaches used to annotate a genome Give an example of comparative genomics “Next-generation” Sequencers • ________________________ – 150-bp reads, fast and cheap • ________________________ – 400-bp reads, fast and cheap ** relevant to align and assemble nucleotide sequences** Ch. 18 –> Genomics and Bioinformatics NEW-ISH FIELDS A) Genomics B) Proteomics C) Bioinformatics 1. ______________: filtering and analyses of very large genetic datasets; software development 2. ______________: identify proteins present in a cell at a given time/location; modifications and interactions 3. ______________: structure and function of genomes (all of the DNA carried in an organism), often comparative Genomics • The study of the complete genetic information of small creatures that live in the depths of the earth, who guard buried treasure Karyotype is highly variable Organism Diploid number (2n) African Wild Dog 78 Badger 32 Carp 104 Adders-Tongue 1262 (!) Cotton 52 Organismal “complexity” Not proportional to the number of chromosomes… …but what about ____________? (absolute number of base pairs) - Total amount of DNA within a haploid genome - Measured as the total number of base pairs (Kb – thousand, Mb – million, Gb – billion) Genome size also variable Organism Billion bp (1000 million) African Wild Dog 2.66 Snapping Shrimp 15.45 Carp 1.61 Poison Dart frog 8.70 Hugh-man 3.03 (human) How does number of chromosomes relate to absolute size of genome? - Size of African wild dog genome is quite larger than carps’ - Chromosomes are of different size (African Wild Dog’s are larger than carps) Sequencing a Genome • ____________ – complete set of all the DNA in a cell; size is variable (usually: virus < prokaryote < eukaryote) • ____________________ – most methods have short “read lengths” (< 800-bp of continuous nucleotides) • Whole genome “shotgun” sequencing - ________________________ - ________________________ - ________________________ Genomic Libraries • large collection of DNA fragments (i.e., broken up whole genome), inserted/stored in vectors • Does not involve PCR… use RE digests of genomic DNA to make the fragments (varying size) • Try to get at least one copy of every sequence in the fewest number of fragments (use BAC or YAC clones) Shotgun Sequencing Technique • ____________ the genome (using REs), in slightly different ways • e.g., use a few different REs (but ____________) • Gives ____________ overlapping fragments • ______________ overlapping fragments (computationally) • Eventually, ____________ whole chromosome sequence and repeat for all chromosomes ASSEMBLING “CONTIGS” • ____________: a continuous stretch of DNA sequence (overlap) - Within just one gene - Across adjacent genes HP Computing (align, assemble) • ____________________________________ – never done before, no baseline data •____________ – already have a reference genome or “scaffold” Diploid Genomes • An individual can have a _____________ genotype - Allele 1 v. Allele 2 - E.g., 1/1 (hom) vs. ½ (het) vs. 2/2 (hom) … so may need to align two variants of a given gene DNA Sequence Alignment • Scenario 1 – 2 reads do not start in same place - Align sequences to see a match-up - Mismatch between 2 nucleotides – point mutation/DNA sequence substitution - Q: are they otherwise identical? o A: NO Gene Pools • ____________ – gene pool can contain many alleles/allelic variants e.g., alleles 1, 2, 3, 4 … etc. … so may need to align >2 variants of a given gene - Look out for different kinds of polymorphisms DNA Sequence Alignment • Scenario 2 – many individuals; potentially many alleles - Align sequences to see a match-up - Q: but how many? o A: at least 5 Take multiple sequence alignment and count the number of different sequences that exist • Bioinformatics allows us to figure all of this quickly Annotating a Genome Interpretation and analysis of genome sequences relies on having names and locations assigned to genes Protein-coding genes have some hallmarks: - Have start codons (mRNA = “AUG”, DNA = “TAC”) o T pairs with A, A pairs with U, and C pairs with G - Open Reading Frame (no internal “stop” codons) - Have conserved upstream regulatory regions (e.g., TATA-box) o _________________ – often have conserved genetic DNA sequences Can also compare to annotated sequences in databases: - _________ searches to determine identity o Query data base with stretch of DNA or amino acid sequence and find the “best” match that exists in data base… can infer the likely chromosome or origin or function of the portion of DNA that is included in the search - Can query using DNA sequencing, or amino acid sequencing Public databases • National Center for Biotechnology Information (NCBI) - Contains DNA sequences, annotated DNA sequences from whole organisms and model organisms • Human chromosome maps, including known polymorphisms - Blue text – protein coding genes • Role of bioinformatics is not just to align and assemble existing short reads of DNA sequences to generate whole genomes, but also in comparison of whole genomes across individuals within same species or across different species • ____________: can query NCBI database with an “unknown” stretch of DNA sequencing… find closest matches BLAST SEARCHES • Query sequence –> stretch of sequence in which we generated and don’t know its chromosomal origin or its function; going to compare it to its closest match in the blast search - Best match highlighted in blue - FIG 18-3 • Mouse (subject) sequence already annotated, matching portion is an insulin receptor gene on Chromosome 8 • Rat (query) sequence was from an unknown genomic location, but identity and function can now be inferred - Can infer that the 280 base pair sequence from the rat probably originated from the same genomic location in the rat genome - Can also infer that the 280 rat sequence does not contain stop codons where they don’t belong if we expect this to be a protein coding sequence… o or if we were able to identify an AUG start codon or even an upstream regulatory region * stronger inference if rat sequence also has an ORF, etc. • Comparing genomes of different organisms, some close relatives, others distantly-related - considerable similarity among organisms you would not have thought • Understand structure, function and expression of (mutant) genes involved in human diseases - 60% of genes from 300+ human diseases also in ____________ - 60% of inherited dog diseases similar to those in humans o comparative genomics provides a way to identify useful model organisms that can help in experimental crosses that may inform us how to treat human genetic diseases • Important evolutionary insights from comparative genomics - direct comparison between genome sizes between different classes of organisms o prokaryotes, eukaryotes, viruses - ____________ – genome sizes small relatively; moderately strong correlation between genome size and number of protein coding genes - ____________ – strong linear relationship between genome size and number of protein coding genes in the genome - ____________ – diffuse relationship (but still correlated) but not that tight; some large genomes with relatively few protein coding genes (large portion of genome is comprised of non-coding DNA) LECTURE 13 ALL ABOUT IN CLASS MICE VS. BEACH ARTICLE From Article: • Adaptive change evolved rapidly (w/in 6,000 years) • Predation rate increases for white mice in dark backgrounds and dark mice in lighter backgrounds - Suggests that phenotype is adaptive and depends on the context of the environment in which the mice lives • Single DNA sequence change in protein coding portion – amino acid replacement - Different protein variant that functions differently - How they figured out allelic variant with white coat colored is derived – A SNP is 1 nucleotide along a stretch of protein coding region… this is the only variant of focus that exists in natural populations…derived or inherited? o Used phylogenic approach o Derived because most mice in lineage is dark, only every now and then a white phenotype will arise o Tells us what the gene does • Describes details to controlled cross and gives indication of what the results mean - Dominance/recessiveness - 1 gene or more than 1 - gene is pleiotropic and RR is dark and CC is light so RC is intermediate - copy of normal allele and copy of mutant allele o look at figure 4 - proportion of chromosomes that carry the allelic variants - mosaic chromosomes – different phenotypes o 75 dominant phenotype and 25 recessive – standard Mendelian assumption does not hold in this case (varying phenotypes) not always equal fitness - Table 1 – shows results of the controlled cross focusing on the F2 generation and their phenotypes; genotypes have been sequenced o PVE is main column of interest because it shows the percentage of variancts explained • Figure 2 – basically what we get is 2 cell culture lines identical in every way except for which allelic variants of MC2R gene they have - Alleles were expressed - Figure shows how the 2 different cell lines perform with respect to generating melanin - MC1R has an activator which shows dark pigment and an antagonist which produces a lighter color o Lots of activating protein – normal allele produces lots of melanin while mutant allele does not produce as much Shows that DNA sequence mutation that alters amino acid is encoded to the protein produced –> respond differently in cell • The light coloration on the Gulf Coast is due to the MC1R allele while the light coloration of the mice on Atlantic coast is not due to the MC1R allele. This suggests that there are different ways to obtain that phenotype depending on the environment of the population - White population on Atlantic coast evolved differently – nothing to do with MC1R allele • Does evolutionary change proceed gradually through many small mutational steps or can adaptation occur via a few large leaps? • Does adaptation generally proceed through dominant or recessive mutations? Any of the above • Do beneficial mutations tend ot affect protein function, or ists spatial or temporal expression? • Are same genes and mutations responsible for similar traits in different poulations or species LECTURE 14 ** will be exam questions about the Research Experiment we read and studied Which of the following were part of the controlled crosses that Koekstra et al. (2006 performed? A) Mainland pale X beach brown B) Mainland brown X F1 intermediate C) F1 intermediate X F1 intermediate D) Beach pale X peach pale Hoekstra et al. (2006) used controlled rosses to determine the association between A) Strength of selection and allele frequency B) MC1R genotype and fur color phenotype - Main focus C) Allele expression and protein function D) A and B but not C From their mouse study, what was Hoekstra et al.’s (2006) conclusion about adaptive evolution? A) It may occur more slowly than first assumed B) It must involve mutations outside of genes C) An advantageous phenotype can only arise once D) None of the above Describe applications of genetic engineering (GE) to pharmaceutical and vaccine production Ch. 19 –> Applications and Ethics of Genetic Engineering and Biotechnology Modified Organisms • ____________: alteration of a genome (adding/removing genes) using recombinant technology - E.g., insect resistance/pesticide in crops (Bt cotton) • ____________: use living organisms (and their biochemical reactions) to “manufacture” a product - E.g., synthesis of pharmaceuticals by GE species Insulin-Producing Bacteria • ________________________: therapeutic proteins produced by an organism (now usually has recombinant DNA) • ____________: hormone produced by the pancreas, regulates carbohydrate and fat metabolism st • …1 human gene product made by a GE organism - Can make large quantities of insulin in a _________ way • Insulin has 2 ____________ chains (A and B), bonded - Originally produced A and B ____________ and then fused them together • Human genes encoding A and B in ____________ • Inserted genes are next to the bacterial lacZ gene • Transcribed and translated, creates a ____________ - ____________: where polymerase is binded - lacZ is bacterial gene not relevant to producing insulin - insulin gene is important and of eukaryotic origin - white colonies are the ones that have taken up the plasmid - fusion protein – string of amino acids encoded by insulin gene A • Fusion proteins extracted • A/B chains then are separated from lacZ • Subunits spontaneously unite and bond… becomes active insulin molecule - first widespread application of genetic engineering for human health - no risk of passing on diseases …and applications to crop plants (e.g., herbicide and pest resistance, nutritional enhancement) Transgenic Animals • Bacteria useful, but don’t always work well - can’t have successful translation of mRNA transcript generated by electrolytic gene when happening with prokaryotic machinery o bacterial machinery for transcription and translation doesn’t always work well when dealing with mRNA transcripts from eukaryotic DNA • Not able process/modify ________________________ - May have reduced activity or be completely inactive • So can use ____________ as “bioreactors/bio-factories” Vaccines from GE organisms • Stimulate the immune system to produce antibodies • Normal vaccines are injections of weak but still living (attenuated) or a dead (inactivated) pathogen - Heat kill a virus – virus still has all its surface proteins • GE organisms now used to make subunit vaccines - Contain only surface proteins of the pathogen - Still stimulates immune response - This approach can ____________________________________ Subunit Vaccines • ____________: protects against sexually transmitted HPV, including strains that are the major cause of cervical cancer • Provides immune protection prior to infection, but not against existing infections • FDA recommend vaccination before adolescence, nut some strong political resistance GE and Agriculture • Different phenotypic variants occur naturally • ________________________: cross individuals that share a desirable phenotype - ____________ –> thousands of generations of selective breeding causes shift of phenotype • Farmers have been doing this for 1000’s of years • Recombinant DNA technology can fast-track what could have been done through selective breeding… • Can also create GE organisms could not have ever arisen via selective breeding, irrespective of duration • GE food plants and/or animals: insect resistance, herbicide resistance, nutritional enhancements GE Crops are Global • > 300 million acres planted with GE crops in 2008 • roughly 800 GE crop trials, mostly corn and cotton Herbicide-resistant crops • Traditional herbicides (e.g., sprays) often kill weeds and crop plants, whereas some GE crops unaffected • Roundup = effective herbicide with nice properties (use in low concentration, effective, rapid degradation) - Inhibits photosynthetic pathways so must be put directly on weeds not the crop • Roundup inhibits EPSP synthase production - Shuts down a gene crucial for photosynthesis • Create Roundup-resistant fusion gene (EPSP syn. From E. coli, viral promotor) • Make recombinant plasmid, transport into crop plant via infective soil microbe Plant Resistant Crops • Bacillus thuringiensis produces a protein that kills some insect herbivores when ingested (crystallizes) • Previously needed recurrent spraying, but now many GE crops have Bt genes (= continually protected) - Cotton growers - Cause of declines in Monarch butterfly (non-target)? o Pollen plants produced travel far and wide but the protein is still on the pollen, killing butterflies due to the GE cotton o Accidental____________ ____________ LECTURE 15 Learning Goals Describe applications of genetic engineering to agricultural crop plants and livestock Ch. 19 –> Applications and Ethics of Genetic Engineering and Biotechnology Nutritional Enhancements • Vitamin A deficient human diet in much Asia and Africa, preventable permanent blindness, >500M people/year - Malnutrition is a worldwide problem • Rice is staple food in these regions but has no vitamin A rd - 3 world countries have lots of vitamin A deficiencies • Golden rice engineered to have enhanced ß-carotene (A precursor) - ethical consideration: companies that put a lot of money in research and development want to recoup their expenditures o farmers need to buy this golden rice specifically from one company every time o paten on genes unethical? o Using humans as research subjects for effects of golden rice is not good Transgenic Cows • Mammary glands susceptible to infection by Staphylococcus aureus, block ducts/contaminate milk - Also leads to death of cow itself • introduce lysostaphin gene from S. simulana, expressed in milk - Natural enemy of staphylococcus aureus because it produces a protein detrimental to the growth and production of protein • This enzyme breaks down S. aureus cell wall • Acts as a natural antibiotic in milk, wards off infection - Helps in longevity of dairy cows Ch. 7 –> Linkage and Mapping in Eukaryotes Recall how crossing-over frequency relates to physical distance between a pair of genes Physical Linkage • Genes on the same chromosome (i.e., basic unit of inheritance) may not th show independent assortment (Mendel’s 4 Postulate) • ____________ (the outcome of crossing over in prophase I of meiosis I) can decouple linked genes • Probability of decoupling depends on ________________________ between a pair of genes on a chromosome Basic Expectations of No Linkage FIG 7-1a • Genes are on different chromosomes (____________) • Together, gametes have different combinations of alleles from the 2 genes • No crossing-over required What is being represented here? (FIG 7-1A) A) 2 individuals, 2n = 4 chromos each B) 1 individual, 2n = 8 chromos C) 1 individual double heterozygote D) 4 pairs of homologous chromos Basic Expectation: Strong Linkage FIG 7-1b • Two Genes on the same chromosome., extremely close • Very ____________ to be affected by crossing-over • Gametes have the same combos of alleles that were seen in the parent Basic Expectation: Weak Linkage FIG 7-1c • Two genes on the same chromosome, very far apart • Very ____________ to be affected by crossing over • Some gametes have combinations of alleles not seen in the parent - outside chromatids ____________ Linkage and Recombination (FIG a, b, & c on slide) ____________ - genes on different chromosomes - Gametes have all possible combos (AB, Ab, aB, ab) ____________ - On same chromosome, no recombination - Gametes without all combos (just AB, ab) ____________ - On same chromosome with recombination - Gametes have all possible combos Recombination and Mapping • As distance between 2 genes increases, frequency of crossing-over between them also increases* ** need consider multiple meiotic events; not just 1!** • can be measured in terms of the number of recombinant v. parental gametes that are formed • This information can be used to determine the distances between a pair of genes (mapping) LECTURE 16 Recall how crossing-over frequency relates to physical distance between a pair of genes Ch. 7 –> Linkage and Mapping in Eukaryotes Linkage and Mapping Recap • Genes on same chromosome segregate together, linkage alters phenotypic ratio • Gametes can be parental (________________________) vs. recombinant (________________________) • At least 50% of gametes produced by meiosis are parental (non-crossover) gametes • % of recombinant (____________) gametes depends on distance between genes • the ratio of parental vs. cross-over gametes reflects physical distance separating 2 genes - proportion increases the further apart they are and decreases the closer they are ** Need consider multiple meiotic events, not just one** Linkage and Recombination Following Meiosis, several outcomes possible: • ________________________ gametes: 2 genes are very close together, a cross-over point unlikely to fall between them • ________________________ gametes: 2 genes are fairly close together, some crossing over occurs, but rare •________________________: 2 genes are far apart, always decoupled by crossing-over Mapping • Chance of a crossing-over point ____________ forming between 2 genes that are close together is very low • This principle can be used to map the locations and order of genes on a chromosome - How often the 2 traits appear to “separate” from one another • Distances measured in “map units” (mu), where 1 mu = 1% recombination between a pair of genes • Alternate name for mu = centiMorgen (cM) • Chromosome map: estimated locations of genes on a chromosome (incl. distances between them) Run through a worked example (old exam Q) Worked Example A cross is made between a male AaBb mosquito and a female AABB mosquito, and the genotypes of all of their offspring were then determined… if these two genes occur on the same chromosome, which of the following gametes have an allele combination that can only be produced by crossing over? A) Ab B) ab C) aB and Ab D) aB and AB E) AB From which individuals would gametes that can only be produced by crossing over have originated from? (first one is male, second one is female) A) Female parent only B) Male or female parent C) Female offspring only D) Male parent only E) Male offspring only Know the difference between single- vs. multiple cross-overs and how it affects mapping accuracy Single Crossovers During prophase I: • Amount of crossing-over is limited - When A and B X a and b are close to each other, segments of two non- sister chromatids are exchanged, but the linkage between the A and B alleles and a and b alleles is unchanged • Only 1 pair of non-sister chromatids are involved • Locations of cross-over points are random - Could cross over at any point along the chromosomes Multiple Crossovers During ____________ I: • ____________ ____________ also possible - Can occur concurrently during same round of meiosis o Aka simultaneously at 2 locations at the same time • Detectable only when following inheritance of alleles at 3 genes • Again, only 1 pair of non-sister chromatids involved • In a double cross-over, 2 events are independent and simultaneous (i.e., apply the product law) - Takes probability of each event and puts it together –> ________________________ • Probability of a particular double cross-over is ____________ than each of its components (2 single cross-overs) • In mapping experiments, many offspring needed to determine the location and order of neighboring genes Mapping Accuracy • Ideally, crossover frequency is directly proportional to distance between genes (linear relationship) • Deviations can arise from double (or multiple) exchanges that restore the original state o Leads to ________________________ • These undetectable events are a source of error; lead to underestimation of distance between genes • This error is most pronounced for pairs of genes that are very far apart (more undetected dbl. crossovers) • Solution: avoid long distance mapping…focus on genes that are fairly close - When distances are small, the theoretical v. actual relationship are ________________________, but when distances are larger, the relationship can be ________________________ LECTURE 17 Learning Goals • Describe Lederberg and Tatum’s classic experiment • Contrast conjugation and transformation • Recall the steps involved in phage reproduction • Describe how conjugation, transformation and transduction can be used to map bacterial genes Ch. 8 –> Genetic Analysis and Mapping in Bacteria and Phage Bacteria and Phage Bacteria = ____________ • Single celled, 1 haploid chromosome (often circular), +/- plasmid(s) - May or may not have a plasmid Bacteriophages = Viruses • Infect bacteria, DNA, or RNA - Original source of Restriction Enzymes (used as a ________________________) Bacteria, Mutations, and Growth • DNA mutations are spontaneous, constantly arise (inducement by environmental agents not required) • Primary source of new genetic variation are expressed, not ‘masked’ as in diploid organisms) - No dominant allele overriding recessive • ________________________ – liquid broth or semi-solid agar, nutrient content can be experimentally manipulated - Important: minimal medium has just bare essentials - Enriched media – has lots of nutrients Synthesizing Compounds Different strains of same species of bacteria: • ____________ – just needs carbon and ions to make all necessary compounds (AAs, sugars, vitamins, etc.) - Can biosynthesize its necessary amino acids • ____________ – unable to synthesize one (or several) organic compounds required for growth, usually due to a mutation (designated his- if can’t make histidine) - Have sustained these mutations - Growth medium must be spiked so that they get the right nutrients they need • Only ____________ can grow on minimal media; ____________ require enriched (=complete) media - Enriched media spiked with whatever amino acids are needed Recombination via Conjugation All inform us about gene order • ____________ – process in which one bacterial cell passes genetic information to another - Happens within their lifetime - Recipient cell gains genetic variance • Followed by ________________________ – replacement of one or more genes (NOT reciprocal exchange) • Still leads to ‘mosaic-like’ novel chromosomes… altered genotypes, and altered phenotypes Lederberg and Tatum’s Experiment • Strains A & B are both auxotrophs (several null allele mutations each) - Those non functional in strain B do not match those that are in strain A • Grown together for several generations, some (not all) descendants = prototrophs • Gain-of-function mutations unlikely… Q: Why are gain-of-function mutations unlikely? - Mutations strike randomly, should also be seen in controls - Need a particular and complex combo of gains (product law) Davis’s Experiment • Q: release of DNA directly into the media? • A: No… bacterial cells must come in contact with each other - Not enough for the two strains to be in close proximation –> must have contact Conjugation, Donors, and Recipients • Unidirectional transfer of DNA between strains • ____________ = donors of genes, F- = recipients of genes (i.e., become recombinants) • Physical contact between F+ and F- strains involves the formation of an extension of the cell (F pilus) - Recombination and sharing of material to take place • F+ cells are able to donate genes because they carry an ____________ (a mobile DNA element, or plasmid) F-factor Transmission 1) Conjugation occurs between F+ and F- cell 2) One strand of F factor is nicked by endonuclease and moves across conjugation tube - Outside strand that was nicked is unwinding a single strand of the DNA and moving into recipient cell 3) The DNA complement is synthesized on both single strands - Occurs concurrently in donor and recipient cells Following conjugation: • A copy of the F-factor is always passed on • The F-factor is also retained in original F+ • Recipient (F-) cells become donors (F+) - FIG 8-4 The F-factor is a Plasmid • ____________: separate units of inheritance, autonomous • Circular dsDNA, carry one or few genes - Usually involved in resistance • Name indicates function - F-factor –> ____________ - R plasmid –> ____________ - Col plasmid –> ____________ protein • In lab, this produced super weird results Becoming a Hfr Strain… • F-factor enters and integrates, converts cell to Hfr cell - Hfr (high frequency recombinant cell) • The new Hfr cell and “normal” cell conjugate and restriction enzyme cuts F- factor • RE cut site creates origin - Origin – location where RE nicks DNA (binds to DNA then gets cut - Cut out internal portion and clean the rest • Transfer of ssDNA via conjugation tube, starting with genes near origin • DNA replication in Hfr cell (replace donated material) and in F- (copy received genes) - Conjugation did not happen lnon - F- cell • Conjugation interrupted prior to complete transfer • A & B genes got in, can recombine with homologous region of host chromosome • F-factor did not make it, recipient cell remains F-/ - Changes in allelic composition of recombinant - Tells us ordering of genes as well as physical distance between them o Distance –> amount of time it takes to get new allelic variant Hfr Bacteria and Mapping • Interrupting conjugation between Hfr and normal strains showed ordered transfer = a map of gene order - (first) …azi –> ton –> lac –> gal… (last) LECTURE 18 Learning Goals • Contrast Conjugation, transformation, and transduction • Describe how they are used to map bacterial genes • Recall key experiments that determined DNA (not protein) is the genetic material **Make own conjugation, transformation, and transduction compare/contrast chart to study** conjugation: http://highered.mheducation.com/sites/dl/free/0072835125/126997/animatio n6.html transformation: https://highered.mheducation.com/sites/9834092339/student_view0/chapter 28/bacterial_transformation.html transduction: https://www.youtube.com/watch?v=VX1Ze5edmkE Ch. 8 –> Genetic Analysis and Mapping in Bacteria and Phage Conjugation: Recap • Mutant (Hfr) strain, has F-factor plasmid integrated into main bacterial chromosome - once have strains, allow them to come in contact with F- cells and conjugation takes place - FIG 8-8 Conjugation and Mapping: Recap • Recombination in recipient, no change in donor • Interrupting conjugation between Hfr and normal strains showed ordered transfer = a map of gene order - (first)…azi –> ton –> lac –> gal…(last) Transformation • alternative approach to getting bacteria with ________________________ - haploid bacterial cell with main bacterial chromsome - complement bacterial cell has receptor on surface • Recombination mechanism; introduce ___________________ into a bacterial chromosome • Small extracellular dsDNA enters a competent cell via receptor site on the surface • One DNA strand digested, ____________________________________ • Foreign ssDNA aligns and pairs with its homolog in the bacterial host chromosome • Excision and replacement • _____________________ formation (i.e., complementary DNA strands with different origins) • Transforming DNA often carries novel mutations • After DNA replication: - 1 recombinant chromosome - 1 original chromosome • After cell division: - 1 transformed cell o TGC with ACG Recombinant cell - 1 untransformed cell o TAC with ATG Transformation and Mapping • Fragments of transforming DNA ~10-20 Kb long, so carry several neighboring genes • Two ____________________ and _____________________ transformation events are very rare (product law • Close genes move together, distant genes do not • Compare transformed v. untransformed cells to map linkage groups First…azi –> ton –> lac –> gal… (last Those close to each other co-transform a lot (ex: azi –> ton) Those further apart rarely do (ex: azi & lac) Those furthest apart NEVER do (ex: azi & gal) ** time is our measure for how far apart genes are in Conjugation ** - No heteroduplex - Single stranded DNA moves into recipient cell ** in transformation, frequency with which they co-transform is our measure distance** - Heteroduplex - Double stranded DNA moves through receptor site T4 Bacteriophage (___________________) • Virus that infect bacteria • Genetic material is DNA, ~150 genes total • Tail fibers have binding sites, recognize E. coli T4 Lytic Cycle • Attaches via __________________ (receptor-protein binding) - Recognizes its appropriate host… likes to attack E. coli bacteria • Phage DNA crosses cell membrane and replicates - Squats and injects DNA into cytoplasm of bacteria cell - Cascade of changes in bacterial cell o Bacteriophage chops up main bacterial chromosome o Uses cellular machinery to replicate its own DNA within the cell • Viral molecule synthesis and assembly …. Lysozyme breaks open host cell - Must reassemble the newly formed bacteriophages within cells - These new bacteriophages lyse and break out of host cell Transduction • Genetic recombination in bacteria can be mediated by the phage that infect them! • During the lytic cycle, _________________ package bacterial DNA in their head, then exit via lysis - __________________: head capsules packaged with fragmented bacteria chromosomes instead of DNA from virus • These phages find another host and infect with bacterial (not phage) DNA… host gains new genes - Small proportion of defective viruses are now able to find new host cells • Adsorption of phage, infect host with DNA • Replicate and assemble (some phage defective) • Bacterial genes injected and integrated Transduction and Mapping Genetic composition of defective phage: • (1) phage DNA + bacterial DNA (___________) • (2) only bacterial DNA (____________) Standard principles apply: • Close genes move together (co-transduction) • Distant genes only move via 2 independent events … transduction segments represent linkage groups Ch. 9 –> DNA Structure and Analysis Genetic Material ________________________ • Complete genome in cells • Copied during cell cycle (mitosis and meiosis) _______________________ • Only part of the genome “on” at any given time • On-off switches drive information flow Evidence for DNA (not protein) • Avery et al. experimented with 2 strains of Diplococcus (staphylococcus) bacteria: ____________ (non virulent) and ____________ (virulent) • Built on the observation that IIR became virulent when mixed with heat- killed IIIS (transformation) • From heat-killed IIIS cells, they removed one component at a time before mixing with IIR FIG 9-2 More Evidence • Hershey and Chase experimented with bacteria and phage (E. coli and T2) • Built on the knowledge that a component of a phage entered the host cell and directed viral reproduction • Radioisotopes differentially labeled E. coli DNA ( P) and Protein ( S)… Q: Why not stop after eliminating protein? A: Could have been something other than DNA that the experimenters did not think about - Experimental procedure also could have been messed up and allowed transformation in all of them • First allow T2 infection of the _________ • Then allow T2 progeny to infect ________ • Determine type __________ Indirect Evidence in Eukaryotes • Prediction 1: there should be more DNA in ________ cells • Prediction 2: absorption spectrum of DNA should match action spectrum of UV (mutation inducer) …. Predictions validated Direct Evidence in Eukaryotes • Recombinant techniques: ___________ synthesized by bacteria ____________ • Not just bacteria, also with transgenic animals (e.g., human genes inserted and expressed in mice) RNA (NOT DNA) in Some Viruses • Retro-viruses use reverse transcription to convert their RNA back into DNA once in the host cell • The newly-synthesized DNA is incorporated into the host cell’s genome, codes for production of viral RNA LECTURE 19 Learning Goals • Recall key experiments that determined DNA (not protein) is the genetic material • Describe the observations/data that led to Watson and Crick’s proposal of the DNA helix model • Know how mode of DNA replication was determined Ch. 9 –> Structure and Analysis Chargaff et al. determined DNA base composition of many different species. They repeatedly found that: A) Amount of A ≠ T B) Amount of C ≠ G C) Amount of (C+G) = (A+T) D) None of the above ** Needed to account for the fact that there were similar amounts of A&Ts as well as C&Gs More Evidence (DNA, Not Protein) • ____________ and ____________ experimented with bacteria and phage (E. coli and T2) • Built on the knowledge that a component of phage entered the host cell and directed viral reproduction 32 35 • Radioisotopes differentially labeled E. coli DNA ( P) and protein ( S)… • First allow T2 infection of the radio-labeled host • Then allow T2 progeny to infect unlabeled E. coli • Determine type of radioactivity Indirect Evidence in Eukaryotes • Prediction 1: there should be more DNA in diploid than haploid cells • Prediction 2: absorption spectrum of DNA should match action spectrum of UV (mutation inducer) …. Predictions validated Direct Evidence in Eukaryotes • Recombinant techniques: - eukaryotic gene produces synthesized by bacteria following DNA insertion • Not just bacteria, also with ______________________ (e.g., human genes inserted and expressed in mice) RNA (NOT DNA) in Some Viruses • Retro-viruses use reverse transcription to convert their RNA back into DNA once in the host cell • The newly-synthesized DNA is incorporated into the host cell’s genome, codes for production of viral RNA DNA Molecules • Nucleotides – the basic units of DNA molecules, composed of a nitrogenous base, sugar, and phosphate • Types of bases – purine (A, G); pyrimidine (C, T = U) DNA Structure • ___________________ determined DNA base composition in diverse organisms, showed repeated patterns: - Amount of A = T and C + G; amount of (C+G) ≠ (A+T) • Franklin’s x-ray diffraction (i.e., images of scatter patterns) suggested a helical structure of DNA … indicated overall arrangement (even though she didn’t recognize it at the time) The DNA Model • Watson and Crick formulated a model to account for earlier observations (published and some unpublished) • Proposed that base complementary & pairing provides a mechanism for DNA replication - ____________ ____________ bond holding A and T together - ____________ ____________ bond holding C and G together • Recognized that this could help us realize how DNA is replicated RNA Structure and Function • Usually, single-stranded, except when folded back on itself, or in some animal viruses - Loops back on itself and hydrogen bonds between complementary bases can occur causing it to look doubled (stems) with single parts (loops) o Stems and loops • U replaces T and sugar ribose replaces deoxyribose DNA codes for main 3 types of RNA: - ____________ –> a structural component of ribosomes - ____________ –> key to expression of gene products - ____________ –> carries AA’s to ribosome, assembling protein Ch. 10 –> DNA Replication and Recombination DNA Replication • Main function of DNA is storage and replication • Replication is critical for maintaining genetic continuity across generations (cells, and organsims) - when replication happens, either of the 2 strands both become a template - synthesis of a complementary strand happens with lots of accuracy via DNA polymerase • Requires extremely high copying fidelity, although some errors still do occur (the basis for evolution) • Many enzymes and proteins involved in DNA copying and synthesis of new strands - _______________________ Animation: www.yourgenome.org/video/dnareplication Mode of DNA Replication • Semiconservative: each strand of the helix is a template for synthesis of a complementary strand - New helix = 50/50 mix of new and old DNA Other Possible Replication Modes • Conservative: after replication, the old strands re-associate and newly synthesized strands pair-up - 2 new strands separate separate and associate with one another o original template goes back together • Dispersive: helix fragmented (cleaved); after replication, each strand consists of varying amounts of old and new DNA - each of the 2 DNA strands serve as a template for a new complementary strand for each - chopped up and arranged into mosaic like structure helix - along the length of a new single strand you have old + new mosaic like double helix Meselson-Stahl Experiment • 3 alternative modes make contrasting predictions about helix composition (proportion of old vs. new) - realized that the 3 alternatives make different predictions 15 • ____14______ = nitrogen-containing molecule; N (heavy) distinguishable from N (light) via centrifugation - incorporates nitrogen and these versions were available to them - after centrifuging, you can distinguish between the two • First grew E. coli on ____________, then transferred to ____________ and isolated DNA from cells each generation Predictions • After 1 generation, a DNA band of intermediate density - _____________ - 2 helixes that are both light and heavy 50/50 • After 1 generation, 2 DNA bands; heavy and light - ____________ o one helix that is heavy and one helix that is light • After 1 generation: a DNA band of intermediate density - ____________ o mosaic like structure o 50/50 heavy and light • After 2 generations, 2 DNA bands: _______________________________________ • After 2 generations, 2 DNA bands: ________________________ • After 2 generations, a DAN band of intermediate density Predicted • if semi-conservative mode is correct Observed • CsCI gradient centrifugation LECTURE 20 Learning Goals • Know how mode of DNA replication was determined • Compare and contrast DNA replication in bacteria vs. DNA replication in eukaryotes • Recall the challenges associated with DNA replication at telomeres, and how they are overcome Replicaiton Origin and Direction • _______________ – appears at origin, complementary strands unwind and separate into single-stranded DNA (ssDNA) • _______________ – the length of DNA that is replicated following one initiation event • E. coli have a single origin, replicon = whole chromosome and replication is bi-directional For DNA replication to take place - first separate the strands o each single strand serves as a template - complimentary strand attaches - “unzip” DNA DNA Synthesis in Bacteria DNA Polymerase • Adds complementary nucelotides (_______________) • maintains copying accuracy (_______________) • Needs dNTPs, DNA template and magnesium Unwinding the Helix • Origin (oriC) region = repetitive DNA sequence • DnaA (a helicase) initially opens helix by breaking H bonds and denaturing strands • DnaB and DnaC (helicases) then help destabilize strands • Single-stranded binding proteins keep helix open • DNA gyrase alleviate twists (supercoiling) that form ahead of the replication fork via “cut and re-paste” Elongation from a Primer • RNA (not DNA) is the origin of DNA pol. Extension • RNA primers are short (~10-12 bp) and complementary • Synthesis of RNA primers directed by primase Leading v. Lagging Strands For the lagging strand: • DNA polymerase I removes primers and replaces missing bases • DNA ligase “pastes” short fragments together For both strands: • DNA polymerase III synthesis of complements is concurrent 3’ ––> 5’ Exonuclease Activity • Polymerases can pause during synthesis, reverse direction, and excise wrongly incorporated nucleotides • _______________ increases overall copying accuracy - checking for polymerases error bc every now and then a mutation will occur DNA Replication in Eukaryotes Several similarities to _______________: • Helix unwinds at origin, replication fork forms • Leading and Lagging ssDNA (bi-directional synthesis) • Elongation mediated by DNA polymerase Also some important differences: • Much more DNA • DNA is packaged (nucleosome) • Chromosome is linear (centromere and telomeres) Multiple Replication Origin • 100s of origins in yeast … 1000’s in mammal • Replication “bubbles” each make 3 2 replication forks Initiation of Replicatoin • Eukaryotic origins control timing of replication • origin recognition complex (ORC): proteins that tag the origin locations, early in G1 wstage • pre-replication complex (pre-RC: proteins that assemble the ori…. • When replication occurs (via DNA polymereases), pre-RC proteins disperase and re-assemble at next G1
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