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Genetics Wk 7 Notes

by: Anna Ballard

Genetics Wk 7 Notes Bisc 336

Anna Ballard
GPA 3.33

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About this Document

These notes cover the last little bit that will be on the test plus a little bit of review
Ryan Garrick
Class Notes
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This 5 page Class Notes was uploaded by Anna Ballard on Thursday October 13, 2016. The Class Notes belongs to Bisc 336 at University of Mississippi taught by Ryan Garrick in Summer 2016. Since its upload, it has received 7 views. For similar materials see Genetics in Biology at University of Mississippi.


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Date Created: 10/13/16
Bisc 336 Test 2 Review Test on Friday, October 14 DNA Replication in Eukaryotes Several similarities to prokaryotes • 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) - Use histone proteins – prevent interactions between templates and enzymes o Temporarily unwind DNA from histones • Chromosomes linear (centromere and telomeres) - Mitochondrial DNA is not linear Multiple Replication Origins • 100s of origins in yeast… 1000s in mammals • Replication “bubbles” each make 2 replication forks Initiation of Replication Must keep track of part of chromosome that needs to be replicated vs those that have already been replicated • Eukaryotic origins control timing of replication • Origin Recognition Complex (ORC) – proteins that tag the origin locations, early in G1 phase • Pre-replication Complex (pre-RC) – proteins that assemble at origins… presence enables replication • When replication occurs (via DNA polymerases) pre-RC proteins disperse and re-assemble at next G1 Multiple DNA Polymerases • 3 different types of polymerase important for DNA synthesis, essential that they work together - Alpha = make RNA; delta and epsilon = extension • 4 DNA polymerase exclusive to mitochondria, others involved in post- synthesis repair - Details not important except for number of enzymes involved Eukaryotic DNA Binding Proteins • Chromatin: DNA + binding proteins (e.g., histones) • Must remove/modify binding proteins before DNA polymerases can proceed with new strand synthesis • Following synthesis, binding proteins re-associate Problems at Telomeres • Open ends of linear eukaryotic chromosomes resemble “breaks”, which oculd fuse or degrade • Telomeres are unique DNA sequences that distinguish the ends of chromosomes from breaks • Telomeric sequences = short tandem repeats: - 5’ – … (TTAGGG) ..n – 3’ G-rich strand (triple G) o longer than C-rich strand because it has a long overhang that forms a loop - 3’ – … (AATCCC) …n– 5’ C-rich strand (triple C) • G-rich strand has 3’ overhang (i.e., is ssDNA) - 5’ – … (TTAGGG) 2100.. – 3’ G-rich strand (triple G) - 3’ – … (AATCCC) 2000 – 5’ C-rich strand (triple C) • 3’ overhang folds on itself and forms C-G bonds (G- quartet), making an inert closed loop Replication at the Telomere • Following removal of RNA primer, lagging strand left with a 3’ gap • Ultimately shortens the dsDNA helix • Will be exacerbated each generation of DNA replication FIG. 10-15 Telomerase • Telomerase – can add extra repeats onto the G-rich strand’s 3’ end (despite no DNA template) • This enzyme contains a short piece of RNA that is complementary to the G- rich strand’s DNA repeat • Reverse transcription helps DNA polymerase fill the 3’ gap • Telomerase enzyme = large blob • Telomerase RNA = green strand • Repeating DNA on the G-rich strand = pink FIG 10-16 • telomerase binds to lagging strand template • also use reverse transcription REVIEW CH. 17 –> Recombinant DNA During Cloning, the same sticky end restriction enzyme is used to cut DNA from two different sources so that __________ overhangs are _______. A) DNA ligase; complementary B) Single-stranded DNA; transformed C) Single-stranded DNA; complementary D) DNA ligase; transformed • When using RE to cut DNA for cloning action… - DNA from different sources - Same RE recognition site occurs in both eukaryotic and vector sequence - RE creates “sticky ends” o Palindrome (read the same front and backwards) o GAATTC and CTTAAG  If cut between AA and TT, it would make a blunt end because the ends wouldn’t be complementary anymore In the Polymerase Chain Reaction, ____ adds nucleotides to the _____ during the ______ step A) Primer, DNA polymerase; extension B) DNA polymerase, primer, denaturation C) DNA polymerase; primer; annealing D) DNA polymerase; primer extension - Annealing: primers form bonds with single-stranded DNA (~50°C) - Extension: Taq adds to 3’ of primer (~70°C) o Forward and reverse primer - 1 cycle doubles number of ‘target DNA” templates… If following a cloning experiment you observe a mixture of blue and white colonies on a media plate, you would conclude that _______ of the bacterial host cells _______. A) Only some; took up a recombinant plasmid B) all; took up a non-recombinant plasmid C) only some; had the ampicillin resistance gene D) none; took up a recombinant plasmid • in case of cloning, lots going on - FIG 17-5 - Cloning experiments can fail when... o plasmids do not become recombinant o bacterial competent cells don’t take up the plasmids through their receptor sites; thus fail to be transformed - polylinker region – where RE cuts host cell open • Mixture of blue and white, some took up normal plasmid and others took up the recombinant plasmid • only blue colonies means that there are competent cells that took up only the recombinant plasmid and there are competent cells that took up no plasmid at all and so they died Ch. 19 –> GE and Ethics Ex. From Class Goal of GE How it was done Humulin Make lots of pure Synthesize a & B, then insulin mix Vaccine Protect against HPV Synthesize pathogen surface proteins Her. Resist. Protect crop from Over-express plant (cp) Roundup gene in chloroplast genome (overexpress chloroplast gene) Insect resist. Protect crop form Express bacterial (Bt) herbiv. gene Nutrient enhancement Prevent malnutrition Express gene for (golden rice) vitamin A precursor Dairy cows Increase milk … production Ch. 07 – Euk Mapping If a doubly heterozygous individual (Aa,Bb) undergoes meiosis and produces only Ab and aB gametes, you would conclude that the two genes are on __________. A) Different chromosomes B) The same chromosome, and far apart C) The same chromosome, and close together D) The Y chromosome CH. 08 Characteristic Conjugation Transformation Transduction No. of cells 2 (donor, 1 (competent) 1 (virus not a involved recipient) cell) Time span: Within 2 gen. Across 2 gen Within 1 gen (binary fission) shared Nicking e/z cuts Nicking e/z X at start; unwindsinvolved in chromosome creating hetero- duplex structure Shared X dsDNA uptake dsDNA uptake Unique Pilus Receptor site Defective phage unique … … …


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