New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

AGRY 32000 Exam 1 Study Guide

by: Gayatri

AGRY 32000 Exam 1 Study Guide AGRY 32000

Marketplace > Purdue University > Agriculture and Forestry > AGRY 32000 > AGRY 32000 Exam 1 Study Guide
GPA 3.91

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Includes learning objectives from each lecture and explanations of specific concepts from the posted review document
Dr. Brenda Owens
Study Guide
50 ?




Popular in Genetics

Popular in Agriculture and Forestry

This 14 page Study Guide was uploaded by Gayatri on Friday February 26, 2016. The Study Guide belongs to AGRY 32000 at Purdue University taught by Dr. Brenda Owens in Spring 2016. Since its upload, it has received 54 views. For similar materials see Genetics in Agriculture and Forestry at Purdue University.

Similar to AGRY 32000 at Purdue

Popular in Agriculture and Forestry


Reviews for AGRY 32000 Exam 1 Study Guide


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: 02/26/16
AGRY 32000 Exam 1 Study Guide Introduction to genetics and genetic analyses • To know the basic definition of a gene and their functional properties o Gene: the fundamental physical and functional unit of heredity o A sequence of DNA composed of a transcribed region that codes for protein, a site where RNA polymerase complex binds, and regulatory sequences that control transcription • To understand the flow of information from DNA to protein o (replication) à DNA à (transcription) à RNA à (translation) à Protein • To understand the basis of a genetic approach to a biological problem o Use of model organisms: models that are simple, idealized systems that are easily manipulated. Characteristics: § Rapid development with short life cycles § Small adult size § Ready availability § Tractability o Applications in agriculture, medicine, society, and science o Disease identification: using genetic family trees § Tracking mutations – single nucleotide polymorphism (one base is changed) DNA Structure and Replication • Before DNA was established as the genetic material, the following was known… o Genes and mutations in genes affect phenotype o One gene one enzyme hypothesis o Genes reside on chromosomes o Chromosomes are made of DNA and protein • Understand the experiments that led to the conclusion that DNA is the genetic material o Griffith Experiments, 1928 § Used two strains; one was S (virulent) and one was R (a-virulent) § Live S à mouse dies § R à mouse lives § Dead S à mouse lives § Dead S + R à mouse dies (R transformed into S!) *Transformation: directed modification of a genome by the external application of DNA from a cell of a different genotype o Avery, MacLeod, and McCarty, 1944 § Transformation of S in Griffith’s experiment involved transmission of DNA encoding gene lacking in the R strain § Naked DNA enters cell, recombines into chromosomes of R cells, which then survive, replicate, and are now recovered with S phenotype o Hershey & Chase, 1952 § Virus protein and DNA were labeled with radioactivity to determine which one replicated virus inside host cells § Protein coat had Sulfur, packaged DNA had Phosphorus § Sulfur recovered in phage ghosts § Phosphorus recovered in bacteria (genetic material is DNA!) • Know the basic structural features of the nucleotides and DNA o DNA = deoxyribonucleic acid o Four types of DNA nucleotides § Purines: adenine, guanine (2 rings) § Pyrimidines: cytosine, thymine (1 ring) o Nucleotides are made of: § Phosphate: negative charge, cleavage of triphosphate bonds provides energy for DNA synthesis § Deoxyribose sugar: link between base and phosphate, missing an OH compared to RNA § Nitrogenous base: A, G, C, T o Primary structure consists of a polymer of nucleotides joined by 5’ and 3’ phosphodiester bonds o Chargaff’s Rule: (T+C) = (A+G), and A = T just as G = C o 3D structure is a double helix with a sugar phosphate backbone and rungs made of base pairs joined by hydrogen bonds (A-T, C-G), arranged antiparallel • Understand the basic process of DNA replication. o DNA is replicated through mitosis or meiosis o Semiconservative replication: each new double stranded DNA has one original and one new strand • Bacterial Replication o Circular, continuous chromosome o Enzymes involved § Helicase: disrupts H bonds between 2 DNA strands § Single stranded binding proteins: keep strands apart § Primase: initiates short RNA strand complementary to DNA § Ligase: seals gaps between DNA fragments after they are made by DNA polymerase III § Topoisomerase: cuts DNA and rotates it to relieve tension as it unwinds (e.g. gyrase) § DNA polymerase: links nucleotides in a specific order based on single stranded DNA template • Polymerase III: replication 5’ à 3’ • Polymerase I: replication and repair • Polymerase III holoenzyme: made of 2 polymerase III units, 2 Beta clamps, and accessory proteins o Initiation § Origin (DnaA boxes made of 13 bp and AT region) is recognized by DnaA protein and unwinds § Helicase (DnaB protein is loaded onto chromosome § Replication fork opens up and replisome is brought in § Replication occurs 5’ à 3, with one leading and one lagging strand o Elongation § Primase makes RNA primer § DNA polymerase III builds on to primers with DNA § DNA polymerase I removes RNA, puts in DNA § Ligase seals gaps between Okazaki fragments on lagging strand § DNA gyrase relieves tension while this occurs • Eukaryotic Replication o Multiple origins of replication o Initiation § Origin (11 bp sequence with AT region) is recognized by origin recognition complex (ORC) § Helicase and replisome are loaded on to chromosome by Cdc6 and Cdt1 § Helicase slides and unwinds double helix, DNA polymerase comes in o Elongation (same process as bacteria) o Telomeres § DNA polymerase cannot finish up last part of lagging strand, left with overhang on 3’ end § Telomerase uses RNA template to extend the DNA § TRF1, TRF2, and WRN proteins bind to the telomeric repeat sequence and form telomeric cap Transcription • DNA molecules encode many genes • RNA: ribonucleic acid o Ribose sugar has an extra O on one carbon and uses U instead of T o Single stranded, ‘ à 3’ polarity o Some RNAs are ribozymes o Types of RNA: § mRNA: messengers § rRNA: ribosomal § tRNA: transfer § snRNA: small nuclear (splicing) § miRNA/siRNA: expression regulation § lncRNA: non-coding • Transcription: synthesis of RNA from DNA o Base pairing directs the correct base to the 3’ end o Primer is not needed o Coding strand (sense) is the non-template strand (RNA like, just need to replace T’s with U’s), and the template one (antisense) is the one complementary to RNA o Transcription steps (same for eukaryotes and prokaryotes) § Initiation: RNA polymerase complex binds near transcription start site • Promoters (at -35 and -10 boxes) define the site of initiation and direction of transcription • Transcription begins at +1 site • Sigma subunit recognizes promoter region and then dissociates • Beta subunit has transcription function § Elongation • Unwinding, transcription, rewinding of DNA, bubble moves § Termination • Intrinsic mechanism causes hairpin loop to form a section of U’s, makes weaker bonds, dissociation occurs • Rho mediated termination: Rho binds to rut, C-rich region, RNA pol. Pauses at AC rich region, Rho causes RNA pol to dissociate o Prokaryotes vs. Eukaryotes § In prokaryotes, transcription takes place in the cytoplasm § In eukaryotes, the process is a lot more complicated • Existence of exons and introns • Translation takes place in the nucleus, mRNA is transported to cytoplasm, and translation takes place here • Lots of proteins are involved • The process o Primary transcript is known as pre-mRNA (unstable and has exons), which then becomes completed mRNA which is ready for export to the cytoplasm § It is processed through three steps: 1. Addition of 5’ cap to increase stability and promote translation 2. Splicing to remove introns (have GU at 5’ and AG at 3’ ends, plus a branch point A) and join exons (This is done by a spliceosome, which has 100+ proteins, including ribonucleoproteins with snRNA, two snRNAs (U1 and U2) 3. Additions of poly A tail when polyadenylation signal from conserved AAUAAA or AUUAAA is received o Transcription factors/proteins bind in two places: § TATA box: -25 to -30 (TATAAAAA) § CAAT box: -75 to -80 (GCAAT) o The Pre-initiation complex is made of six GTFs, TFFIIA, B, D (has the TATA binding protein), F, H, and I, and the RNA polymerase II core • RNA polymerases o RNA polymerase I transcribes rRNA genes (except 5S rRNA) o RNA polymerase II transcribes all protein coding genes and some snRNAs, and has a carboxy-terminal domain protein tail o RNA polymerase III transcribes small functional RNA genes, some snRNAs, and 5S rRNA • Activators, proteins which bind to enhancers (DNA sequence that is distant from transcription start site that indirectly promotes initiation) and basal transcription complex to promote initiation are needed • Examples of small functional RNAs o microRNA: part of longer RNA o siRNA: antisense/interfering, promoted by insertion of foreign DNA o piRNA o ALL have dsRNA (double stranded RNA): complementary to region of mRNA which inhibit translation • Alternative Splicing o The process in which the same gene codes for multiple proteins o Different mRNAs and therefore different proteins are made from the same primary transcript by splicing together different combos of exons o More than 70% of human genes are alternatively spliced, while the process is rare in plants o Proteins produced by alternative splicing are usually related since they contain subsets of the same exons, and are often used in different cell types/different stages of development Translation • To know the definition and the role of translation in the central dogma of molecular genetics o Translation is the process of converting RNA into amino acids, which come together to form proteins • Protein structure o Proteins are polymers made of monomers called amino acids, which have a carboxyl group, an amino group, an H, and a variable R group o Levels of protein structure § Primary: linear sequence of AAs § Secondary: specific shapes, called alpha helices and beta sheets which are made by hydrogen bonds between amino acids § Tertiary: folding of secondary structures into peptides (subunits), heme group in middle § Quaternary: interaction of peptides to make polypeptides o Globular proteins have a compact structure (e.g. enzymes, antibodies) o Fibrous proteins have a linear structure (e.g. components of skin, hair, tendons) o Domains: AAs/protein folds associated with particular functions • tRNA: “adapter”, responsible for translating the three nucleotide codon in the mRNA into the corresponding AA o Structural components: § 3’ OH amino acid attachment site § Anticodon loop, which carries the anticodon that is complementary to the codon for the AA carried by mRNA • Aminoacyl tRNA synthetase: an enzyme which attaches AAs to tRNA (each AA has its own) o Has two binding sites, one for its specific AA and one for tRNA • To know the mRNA code and its characteristics o the triplet codon: three nucleotides code for one amino acid o degeneracy of the code: multiple codons can code for one amino acid § Most amino acids can be brought to the ribosome by several different tRNAs § Certain tRNAs can bring specific AAs to any one of several codons, because they bind to alternative codons, not just the ones that are complementary to them à Wobble pairing • Wobble pairing refers to the third nucleotide of the codon being able to form hydrogen bonds with its complementary nucleotide or another one: • 5’ end of anticodon G can pair with C or U • 5’ end of anticodon C can pair with G ONLY • 5’ end of anticodon A can pair with U ONLY • 5’ end of anticodon U can pair with A or G • 5’ end of anticodon I can pair with C, U, or A o non-overlapping: there is no bias for adjacent amino acids, and one DNA sequence change affects one amino acid o start and stop codons: start codon is Methionine (AUG), stop codons are UAA, UGA, and UAG, and are noncoding o universal: all organisms have the same genetic code • To know the function and major structural features of the translation machinery o mRNA: read 5’ à 3’ by the ribosome o ribosome: made of two subunits, small and large; has three sites, Aminoacyl, Peptidyl, and Exit which tRNAs pass through o tRNA: bring their specific amino acid to the codons being read, amino acid is attached to its tRNA by aminoacyl tRNA synthetase • To understand the 3 basic steps of translation and how translation initiation differs between prokaryotes and eukaryotes o Initiation § In prokaryotes, 1) The small ribosomal subunit binds to the shine-delgarno sequence in the mRNA, 2) The unique tRNA for methionine initiation codon binds, and 3) the large subunit binds § In eukaryotes, 1) The 5’ cap is reached and the large subunit comes in, 2)the subunit scans for the start codon AUG, and 3) the ribosome is fully assembled when AUG is reached o Elongation § Aminoacyl tRNA bonds to A site § Peptide bond forms between the amino acids § Translocation of the ribosome occurs § New Aminoacyl tRNA binds to A site, old tRNA in E site leaves o Termination § Stop codons are reached, and have no tRNA § Release factors occupy the A site and disrupt peptidyl transferase activity (Rf1 fits UAA or UAG, Rf2 fits UAA or UGA) PCR and DNA Sequencing • Polymerase Chain Reaction, meaning and process o A simple, effective method of amplifying or producing many copies of a specific piece of DNA o Ingredients: § Template/target DNA § dNTP (dATP, dCTP, dGTP, dTTP) § Thermostable DNA polymerase: Taq polymerase from Thermus aquaticus, able to withstand high temperature encountered in protein denaturation § Oligonucleotide primers § Buffer o Steps § Denaturation: DNA is separated into single strands, allowing for primers to have access to it (92 – 95 degrees C) § Annealing: primer is base-paired with the template DNA to define specific regions of interest and provide 3’ OH group for DNA polymerase to extend from (55 – 65 degrees C) § Elongation: primer DNA is extended complementary to template (72 degrees C) § Cycle is repeated 25 – 30x • Sanger method for DNA sequencing o Uses 4 dideoxyribonucleotide triphosphates (ddNTPs) that correspond with each dNTP, and have: § A cap that blocks further synthesis when dNTP § A fluorescent dye o Steps § Denaturation of strands § Annealing (primer creation) § Adding of a mix of: 1) DNA polymerase, 2) dNTPS, 3) One of four ddNTPs § Synthesis halts when dNTP is detected for that specific ddNTP on the strand § Each ddNTP has a unique fluorescent tag, and these can be detected and read to come up with the molecular sequence of the strand • In the past, bases were read off the gel manually, but now, they are read by automatic capillary sequencing machines o Used in forensics, DNA sequencing, finger printings, etc. Bacterial Genetics • Heredity in bacteria o Bacteria can undergo sexual and asexual reproduction § Mutation occurs in asexual reproduction as much as it does in eukaryotes § In asexual reproduction, the genomic DNA is replicated and separated between daughter cells, but it is different than mitosis § In sexual reproduction, two DNA molecules from two different sources are brought together, but usually instead of two complete chromosomes combining together, one chromosome and one fragment of another are o Processes of DNA exchange § Transformation: uptake and integration into the chromosome of a piece of DNA from the environment • The genome of the recipient is permanently changed as a result of this process since the DNA of taken up is of a different genotype than the recipient • Double transformation = the uptake and carrying of two donor genes together, occurs if they are located close enough on the chromosome § Transduction: transfer and integration into the chromosome of DNA between bacterial cells through phages. Two types: • Generalized transduction: when a donor cell is lysed by P1, the bacterial chromosome is broken up into small pieces, and occasionally the phage particles will incorporate a piece of bacteria DNA into itself instead of its own DNA. This then infects another bacteria, and can either incorporate the DNA into the host bacterial cells chromosome (P22) or remain like a plasmid (P1) • Specialized transduction: the phage inserts into the bacterial chromosome at a specific position (can trigger lysis in non-immune host cell) **Both processes above do not require cell division, “horizontal transmission” § Conjugation with plasmid transfer: transfer of genetic material between cells as a result of physical contact, where a plasmid is transferred from one cell to another independently from the chromosome • In terms of F plasmids, F+ donates to F-, and F- becomes F+ (no recombination) § Conjugation with partial genome transfer: transfer of genetic material between cells as a result of physical contact, where a plasmid is inserted into the genome and then transferred from one cell to another • In terms of F plasmids, F is integrated into Hfr, and conjugation of F- and Hfr leads to recombination but further passing on does not occur (non-transferable), and the DNA is integrated into the chromosome in this case. The bacteria ends up with a portion of diploid DNA (partial diploid = merozygote), and when crossover occurs, a portion of it is lost **Conjugation in both cases occurs through bacterial cells extending projections (pilli) that attach to other cells and pull the two cells together • Auxotrophy vs. prototrophy o Prototrophic bacteria: wild type, can grow and divide on minimal medium (inorganic salts, a carbon source, and water) o Auxotrophic bacteria: mutants, cannot grow unless the medium contains specific cellular building blocks (adenine, biotin, threonine) § Mutants can also be resistant to antibiotics, have the ability to use a specific carbon source § These mutants are helpful in determining genetic markers to keep track of different bacterial phenotypes and genotypes • Discovery of Conjugation: the fertility factor o 1953: William Hayes discovered that conjugation occurs as an unequal process, and only “donor” cells could donate DNA (exogenote, incomplete genome), just as only “recipient” cells could receive DNA (endogenote, complete genome) § Different from eukaryotic crosses where parents contribute genes equally to a progeny o A variant of the donor strain also existed, that would not product recombinants when crossed with recipient strain, it had lost its ability to donate § But it could regain this ability by associating with other donor strains à the donor ability was transmittable between strains! § This was called the fertility factor (F), and was a type of plasmid: donators were F+, recipients were F- o Cavalli: another derivative of F was discovered, called Hfr § When crossed with F-, it produced lots more recombinants than a F+ x F- cross § Hfr x F- did not convert F- to F+ or Hfr (no infectious transfer of donor ability) o F’ plasmid: an F plasmid carrying bacterial genomic DNA, comes from the removal of the F factor from the chromosome by reversal of recombination § Can be used to establish partial diploids for studies of bacterial dominance and allele interaction • Mapping o Linear, time dependent gene transfer is observed, generates a chromosome map o DNA is transferred in a linear fashion, and begins at the origin (this is where the F plasmid is inserted) § The farther a gene is from O, the later it gets transferred § This is why F- exconjugants (cells that take part in conjugation) are rarely converted to Hfr or F+, because the F gene is one of the last to be inserted, and by then the transmission process is likely over o The orientation in which F is inserted determines the order of entry of donor alleles • Bacteriophage genetics o A bacteriophage (virus that infects bacteria) consists of a chromosome (either DNA or RNA) surrounded by a protein coat o Components: § Head § Neck and collar § Core § Sheath § End plate § Fibers o Infection process: § Phage attaches to bacterium and injects its genetic material into the through the cell wall into the cytoplasm § It takes over the machinery of the cell by turning off synthesis of bacterial components and turning on synthesis of phage components § New phages are made inside the bacteria, and are released when the cell wall breaks open through a process called lysis • Phage lysate = population of newly made phages o Phage cross = infecting a host cell with two varieties of parental phages, each with different alleles (mixed infection) § Progeny phages have recombinant as well as parental genotypes o 2 types of phage cycle: § Virulent phages: those that immediately lyse and kill host § Temperate phages: can remain in the host cell for a period without killing it • DNA integrates into the host chromosome to replicate with it or it replicates separately like a plasmid • Prophage = phage integrated into a bacterial chromosome • The bacteria in this case is known as a lysogen (is lysogenic) Recombinant DNA • Amplification: utilization of DNA machinery to replicate a DNA segment o In vivo: DNA Cloning § uses donor DNA (usually an entire genome) which is cut using restriction enzymes (called endonucleases) § A plasmid is cleaved with the same restriction enzyme, so they have complementary ends/overhangs § Fragments of donor are inserted into a plasmid or phage (called vectors) • They hybridize because the ends are complementary § Ligase seals gaps between plasmid and donor DNA § Recombinant plasmids are inserted into bacterial cells § Normal replication takes place o In vitro: a specific gene of interest is isolated and amplified by DNA polymerase (PCR) • Restriction enzymes: cut DNA at specific sequences, create sticky ends o EcoR1 o HindIII • cDNA: complementary DNA, a DNA version of an mRNA molecule, used because RNAs are less stable, cannot be manipulated by enzymes, and cannot be amplified purified o Made using reverse transcriptase, isolated from retroviruses • DNA palindrome: when two strands of DNA have the same nucleotide sequence in an antiparallel orientation, used as cut sites by restriction enzymes, creating sticky ends • Characteristics of a plasmid vector o Restriction site (polylinker): DNA fragments are inserted here o Selectable market: encodes protein which imparts antibiotic resistance o Origin of replication: ensures plasmid is replicated o Plasmids enter through transformation • Example of a plasmid vector: PUC18 o Allows for simple screening of recombinant plasmids o DNA insertion into this plasmid disrupts a gene lacZ that encodes for the beta-galactosidase enzyme, which is needed to cleave X-gal, a component of the plating media so that it produces a blue pigment § White colonies = contain DNA insert § Blue colonies = do not contain insert • Other types of vectors o Bacteriophage vectors: viral vectors for large fragments (10 – 15 kb), harbors DNA as an insert packaged inside the phage § E.g. phage lambda § Central part of genome is not needed for replication or packaging of phage lambda DNA molecules, so it can be cut out by restriction enzymes, and is replaced by inserts of donor DNA § Enter through infection o Fosmids: vectors that hold 35 – 40 kb inserts, hybrids of phage lambda and bacterial F plasmid DNA § Enter through transduction o BAC: bacterial artificial chromosome: most popular, derived from F plasmid, hold 100 – 200 kb inserts § Enter through infection • Genomic libraries: collections of bacterial clones that contain fragments of donor DNA • Distinguishing different alleles of a gene o This is done by using probes, which find and mark the desired clone containing the gene of interest. Two types: § Those that recognize a specific nucleic acid sequence § Those that recognize a specific protein o Southern Blot process § DNA fragments are run through electrophoresis, and migrate according to their size on the gel § The gel is placed in a buffer and covered with a membrane and towels § The fragments are denatured into single strands so they can stick to the membrane § The membrane is incubated with a radioactively labeled probe complementary to target sequence § Unbound probe is washed away § The membrane is exposed to X-ray film § Since the probe is hybridized only with the complementary restriction fragments, they can be visualized with the X-ray and compared to labeled markers to see their sizes and numbers • Sanger Sequencing o Uses 4 dNTPs (to pair with bases), and 4 ddNTPs (to block synthesis, have dye and cap in addition to base) o Steps: § Strands are denatured § Primer is made § Make mixture of dNTPs and one of four ddNTPS o Synthesis of strand stops when that dNTP that corresponds to the selected ddNTP is reached o Fragments made by this process are separated by a gel o Bases are read by automated capillary sequencing machine Prokaryotic Gene Regulation • Criteria for cell mechanisms: o Must recognize environmental conditions in which they should activate or repress transcription of relevant genes o Must be able to toggle on and off, like a switch, the transcription of each specific gene or group of genes • Regulation depends on protein – DNA interactions o RNA polymerase protein & promoter DNA: when RNA polymerase binds to the promoter, transcription can begin a few bases away from the promoter site § Every gene must have a promoter, or it cannot be transcribed o Repressor protein & operator DNA site o Activator protein & activator-binding DNA site o The proteins involved in these processes have two sites: § DNA binding domain § Allosteric site (acts as a sensor that sets DNA binding domain in functional or nonfunctional mode) • Interacts with allosteric effectors, which causes change in shape of the protein • Lac induction system o Presence of lactose is needed to synthesize the genes needed to process it (relief of repression) o Structural genes (segments that encode proteins) involved: § (Y) Permease: to transport lactose into the cell § (Z) Beta-galatosidase: to modify lactose into allolactose and cleave lactose à glucose and galactose § (A) transacetylase **All three genes are on one mRNA and are coordinately controlled (their transcription and therefore expression is controlled by one common regulator) o Regulatory components § (I) Gene for lac repressor: can block expression of Y, Z, and A • Its proximity is not important because it is a diffusible protein • Not technically part of the LacOperon • Lac repressor protein has DNA binding site and allosteric site that binds allolactose/analogs § (P) Lac promoter site: where RNA binds to initiate transcription of Y, Z, A § (O): Lac operator site: where lac repressor (I) binds (located between P and Z) o The P, O, Z, Y, A segments all together make an operon, which is a segment of DNA that encodes multigenic RNA as well as a common promoter and regulatory region o Genetic Evidence § Z+ is dominant to Z- in operator § Oc is always expressed § Operator is cis acting: mutations only affect the same chromosome § I+ is wild type and inducible, and trans acting § I- is constitutive § Is always represses § Is is dominant to I+ • To understand the concept of negative and positive transcriptional regulation (look over images in Ch. 11!) o In negative regulation a repressor protein binds to an operator to prevent a gene from being expressed. o In positive regulation a transcription factor is required to bind at the promoter in order to enable RNA polymerase to initiate transcription. Genetic Inference Eukaryotic Gene Expression • Characteristics of eukaryotic transcriptional that are different from bacteria: o In bacteria, all genes are transcribed in RNA by same RNA polymerase, which eukaryotes have three RNA polymerases (focus on RNA pol II) o RNA transcripts go through a lot of processing in eukaryotes o RNA pol II is larger and more complex than bacterial RNA • Criteria for gene regulation o Expression of most genes must be turned off at any one time while activating a subset of gene o Thousands of patterns of gene expression must be able to be generated • Regulatory proteins must have one of the following domains, that can: o Recognize DNA sequence o Interact with RNA pol or a protein associated with it o Interact with proteins bound to nearby regulatory sequences to act cooperatively to regulate transcription o Influence chromatin condensation (directly or indirectly) o Act as a sensor of physiological conditions in the cell • GAL system in yeast o Main genes: GAL1, GAL2, GAL7, and GAL10 à encode enzymes o Other genes: GAL3, GAL4 (main regulator) and GAL80 à encode proteins that regulate expression of enzymes o GAL4 regulates multiple genes through upstream activation sequences (UAS) § GAL1, GAL2, GAL7, and GAL10 all have two or more GAL4 binding sites § It has an activation domain required for regulatory activity and a DNA binding domain, and it binds to the gene as a dimer • It can provide the activation portion for the DNA binding domain on the LexA site § GAL80 binds to GAL4 and makes it inactive, but in the presence of GAL3, GAL80 undergoes a conformation change and so GAL4 can proceed • Chromosome structure o DNA double helix < nucleosomes < chromatin fiber of packed nucleosomes < octamer histone core with DNA wrapped around core • Altering chromatin structure o Chromatin remodeling: moving nucleosomes along the DNA § Exposes regulatory sequences o Histone modification in the nucleosome core § Acetylation: opens up chromatin, DNA is exposed to protein activity (deacetylation can turn off transcription) § Methylation creates binding sites for proteins that activate or repress gene expression o Replacing common histones with histone variants Mutation • Mutation: refers to a heritable change in DNA • Types of mutations o Point mutation: single base pair change (can alter mRNA splicing – if splice site is changed, can trigger creation “retention” or deletion of introns) § Substitution: can be a transition (pur-pur (A – G) or pyr-pyr (C –T) switch) or inversion (pur-pyr) § Insertion or deletion: an cause a frameshift mutation, change that results in altered reading frame o Spontaneous mutation § Bases can be removed (depurination/depyrimidination) § Deamination: cytosine à uracil § Superoxide radicals (O2, H2O2, OH) • Can cause OH to block replication, cell dies (thymidine glycol) • Mispairing with A occurs, because of C=O § Replication slippage, due to CAG, causes repeat expansion • Loop is made o Induced mutation § Caused by ionizing or non-ionizing radiation (change in bases, replication is hindered) § Caused by chemical mutagens • Intercalating regions – insertions/deletions • Deamination – substitutions • Hydroxylation – substitutions • Aflatoxin – deletions (blockage in replication and translation) • Alkylation – insertions/deletions/substitutions (EMS changes base structure, T pairs with G) • To know the different types of mutations that occur and how they affect amino acid sequence o Same sense/silent/synonymous: doesn’t affect gene o Missense/non-synonymous: affects amino acid sequence § Conservative: change to amino acid with similar chemistry § Non-conservative: change to amino acid with different chemistry o Nonsense: replacement of amino acid with stop codon • Mutation Repair Mechanisms o Nucleotide excision repair: needs dozens of proteins to carry out these four steps § Recognition of damaged base(s) § Assembly of multiprotein complex at the site § Cutting of the damaged strand several nucleotides upstream and downstream of damage site and removal of nucleotides between cuts § Use of undamaged strand as a template for DNA pol to build new DNA, and then ligation o Base excision repair: damaged bases are removed and repaired through these actions: § DNA glycosylase cleaves the base sugar bond § AP endonuclease makes the cut § dRpase removed stretch of DNA § DNA pol synthesizes new DNA § Ligase seals gap (In base excision repair, nonbulky damage to DNA is recognized by the DNA glycosylases) Genomics • Genome sequencing o Cut many genome copies into random fragments o Sequence each fragment o Overlap sequence reads to get contigs o Overlap contigs for complete sequence • Pyrosequencing reactions o Nucleotides are added to form the complementary strand of the single stranded template, to which a sequencing primer has been annealed. o The reactions are carried out in the presence of DNA polymerase, sulfurylase, and luciferase. o One molecule of pyrophosphate (PPi) is released for every nucleotide added to the strand, and converted to ATP by sulfurase, and this ATP is converted into light by luciferase • To know the basic of a few genome sequencing projects o Human Genome Sequencing project § 1990-2003 § Determined sequences of 3 billion chemical base pairs § 20,000 - 25,000 genes in our DNA o International Rice Genome Sequencing Project § 1997 – 2004 § 370 million base pairs § 35,000 genes o Soybean Genome Sequencing Project § 2006 – 2009 § 1.1 billion base pairs § 25,000 – 30,000 genes • To know the strategies for sequencing an entire genome of an organism o Sequencing by synthesis is used at Purdue § Fluorescent dye on nucleotides § All can be added together § Overcomes homopolymer issue § But, error rate is increased because read length is increased (background noise increased because dye removal, residues) • Sequence contigs = sequences of overlapping reads are assembled into units (contiguous, touching) o Paired end reads may be used to join two sequence contigs Transposable Elements • To know how transposable elements were discovered o Discovered in 1940s by Barbara McClintock while studying Indian corn o She found that one strain one maize (chr 9) broke quite often and at one particular locus o This was due to two factors: § Ds: dissociation, at site of break (nonautonomous – can’t move without Ac in the cell) § Ac: activator, needed to activate break at Ds locus (autonomous – can move from one position to another) o Phenotypes produced: § C gene (wild type) à pigmented § C gene with Ds but no Ac à colorless § C gene with Ds and Ac à (Ds is removed) à spotted § C gene with Ac à (Ac is removed) à spotted • To learn the general features of transposable elements o Autonomous: encode transposase needed for their own movement, can turn into non-autonomous by mutation o Non-autonomous: do not encode transposase but can borrow it from autonomous elements of their family • Mechanisms responsible for transposition o Replicative mechanism § Prokaryotes only § Occurs between 2 plasmids, which are cut by transposase and integrated to form a “cointegrate” § The two are resolved, separated o Cut and paste mechanism § Prokaryotes and eukaryotes § Transposon is cut out of genome and moves into the target site o Eukaryotes have retrotransposons § Long terminal repeats (LTRs) § Non-LTRs • Long interspersed (LINEs) – autonomous • Short interspersed (SINEs) – non-autonomous § Mechanisms • Retrotransposition: DNA à RNA à DNA (using reverse transcriptase) • Conservation transposition: Ac catalyzed excision and integration of Ds • Types of transposable elements o LINEs: auto --- 21% o SINEs: nonauto --- 13% o DNA transposons --- 3% (most are turned off) • Many insertions are found in intergenic or intronic regions o Transposable elements insert into exons and introns, but only those in introns stay, because they are less likely to cause deleterious mutations o Transposable elements in grasses are responsible for differences in genome size § Maize: 70% of the genome is transposons o Can be used to make new daughter cells that are NOT isogenic, vary from parents (cut and paste mechanism used during replication) • Distribution of transposons is not random o Small genome: tRNA, rRNA o Large genome: rested insertion o Very few are actually active, and can be activated by tissue culture, environmental stress


Buy Material

Are you sure you want to buy this material for

50 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


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'

Why people love StudySoup

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

Kyle Maynard Purdue

"When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the I made $280 on my first study guide!"

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."


"Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.