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Genetics Notes

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Genetics Notes BIO 340


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notes consist of weeks 3 and 4. will be on midterm 2
Jianmin Zhong
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This 9 page Bundle was uploaded by . on Wednesday February 24, 2016. The Bundle belongs to BIO 340 at Humboldt State University taught by Jianmin Zhong in Spring 2016. Since its upload, it has received 28 views. For similar materials see Genetics in Biology at Humboldt State University.

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Date Created: 02/24/16
When genes are on the same chromosome they don’t follow the Law of Independent  Assortment Linked genes/alleles tend to be inherited together  can form 2 gametes  linked genes means genes are located close together on the same chromosome Crossing Over:  produces new allelic combinations  crossover→ the breakage of 2 DNA molecules at the same position and their rejoining in  two reciprocal recombinant combinations  occurs in meiosis  segments of nonsister chromatids can be exchanged, creating recombinant chromatids  in prophase one of meiosis Evidence of crossing over: 1. Chiasmata→ the sites of crossing over  cross shape configuration structure that can be seen in prophase one 2. Counting phenotypes of progeny of testcross  unlinked genes→ different chromosomes  linked genes→ same chromosome I) Two genes on two different chromosomes (A/a;B/b)  Gametes have a 1:1:1:1 ratio → AB:Ab:aB:ab II) Two genes on same chromosomes (AB/ab)  Gametes o AB and ab  → parental types >50% o Ab and aB → recombination type <50% ­ parental must occupy majority of progeny… fewer recombination types because recombination is considered a rare event. III) Two genes on same chromosome (Ab/aB)  Gametes o Ab and aB → parental types >50% o AB and ab → recombination types <50% Recombination Frequency:  % recombinant progeny in a cross = (# recombinants/total progeny)X 100% Application of Recombination Frequency: #1 Gene Mapping= identify the location of a gene and the distance between genes. Methods: ­ testcross ­ # of parentals, # of recombinants ­ recombination frequency ­ determine distance **Examples** #2 Predicting outcomes of crosses with linked genes **2 Examples** #3 If genes are completely linked (no crossing over)  produce 2 progeny rather than 4  cannot produce recombination type  distribution = 50%/50% Biological pathway = a series of reactions in a cell that leads to a certain product Genes are haplosufficient **Arginine Example** Chi Square  when p­value is > or < 0.5 Summary Application of Recombination Frequency  gene mapping  predicting outcome of crosses with linked genes  constructing recombination map 1) Recombination map  r.f. is proportional to distance of genes  hypothetical map  unit = m.u. (map unit) 2) Physical map  based on results of DNA sequencing  a map of actual genomic DNA  unit = base pairs Relationships:  two maps roughly proportional to each other  distance calculated between 2 maps is not always similar > variance across chromosome in rate of recombination  hotspots  blocks Hotspots:   elevated rate of r.f.  trinucleotide repeats  cause DNA to break Blocks:  crossing over suppressed   centromere  condensed DNA Existence of double triple crossover 1. in linkage, null hypothesis says that genes are not linked 2. ���2 = ∑ (observed­expected)^2/expected 3. data interpretation: ���2 > 3.84 → p< 0.05 then the differences between observed and  expected is not due to chance → reject null hypothesis → two genes are linked! 4. ���2 < 3.84 → p>0.05 → difference is due to chance alone → accept the null hypothesis→  genes are independently assorted (not linked!) **degree of freedom always equals one for this class** Molecular Model of Crossing Over Step 1 = Double Strand Break  prophase I in meiosis  Enzyme endonuclease breaks open phosphodiester bond Step 2 = Resection  remove some nucleotides at the end of break  only occurs on one of two strands  produce two 3’­ single stranded tails Step 3 = Stand Invasion  single stranded nucleoproteins (= 3’­single strand tails and different proteins)  invade non­sister chromatids  single stranded tails form double helix with the non­sister chromatid  the replaced strand of non sister chromatid becomes single stranded Step 4 = Form double Holliday junction  invading 3’­single stranded tail join to free end on the right of the break  the displaced strand forms a D­loop that base pairs with the broken DNA  Two holliday junctions are formed Step 5 = Branch migration  heteroduplex is formed  expand the region Step 6 = Resolution of Holliday Junction  enzyme endonuclease cuts holliday junction  cut 2 ways  1. horizontal and horizontal vertical and vertical ­ no crossing over occurs 2) horizontal and vertical vertical and horizontal ­ crossing over occurs DNA: Structure Deoxyribonucleic acid 1. codes for protein 2. genetic element/inheritance 3 chemicals 1. deoxyribose 2. phosphate 3. four nitrogenous bases 1. Purines: adenine and guanine 2. pyrimidine: thymine, cytosine Nucleoside: sugar and DNA base Nucleotide: phosphate group Primary Structure: string of nucleotides joined together by phosphodiester bonds Secondary Structure: double helix  hydrogen bonds between A­T and G­C base pairs Ribbon Diagram 1. two DNA helix 2. backbone (phosphate and deoxyribose) 3. bases 4. hydrogen bond (formed between 2 DNA bases) 1. 2 A=T 2. 3 G C→ stronger interaction because 3 Hydrogen bond 5) Stacking interaction­ non­covalent interactions between aromatic rings of bases  strength: GC stacks > AT stacks DNA absorbs UV light  method: spec machine at 260 nm  why? exposed purine and prymidines Single stranded DNA vs. Double stranded DNA ssDNA: strong absorb because more exposed purine and prymidines dsDNA: less strongly absorb→ stronger stacking interactions  stacking interaction is stronger than hydrogen bonds Denaturation  heat destroys the hydrogen bonds and stacking interactions Salt concentration effects on DNA denaturation 1. DNA carry negative charge → repel each other 2. high concentration of salt → salt eliminate repulsive force between 2 negative charges→  not favoring denaturation 3. low concentration of salt→ don’t remove negative charge→ favoring denaturation Tertiary Structure of DNA:  dependent on either eukaryotes or prokaryotes o eukaryotes: chromatin­­>nucleosome­­>basic unit ­histone­ protein that carries a positive charge→ can remove repulsive force of DNA­­>DNA  compacted ­ prokaryotes: twisted loops ­ DNA­binding proteins→ bind to DNA and form twisted loop→ 3rd strand DNA Replication only occurs in the S­phase… Why?  Proteins that recognize origin of replication are... 1. only made in G1 2. bind to origin of replication→ replication occurs once 3. protein destroyed by proteolysis  Three Models of DNA Replication: 1. Semiconservative→ 2 double helix; each with original and new strand 2. Conservative→ one with original strand and one with new strands 3. Dispersive→ each with new and original strands  Conservative and Dispersive models are wrong: 1. because no research data to support 2. does not immediately suggest a DNA replication mechanism What are the three functions of DNA polymerase? 1. 5’ to 3’ DNA polymerase: makes daughter strand 2. 5’ to 3’ exonuclease activity  RNA primer is removed 3. 3’ to 5’ exonuclease activity  proofreading: removes mispaired bases  connect nucleotide together by phosphodiester bond **Tautomeric Shift** Amine (C­G and A­T) and Keto (T­A and G­C) provide correct base pairs Imino and Enol provide wrong base pair Requirements for DNA replication in Prokaryotes:  DNA Polymerase III  template  primer  deoxynucleoside triphosphates Requirements for DNA replication in Eukaryotes:  DNA Polymerase ��� (delta)  template  primer  deoxynucleoside triphosphates Mechanism of DNA replication in Bacteria: a. Prokaryotic Initiation i. DNA 1. OriC (replication origin) is composed of…  2. AT­rich DNA 3. DnaA boxes ii. Proteins 1. DnaA protein (=initiator protein) 1. binds to DnaA boxes 2. unwind dsDNA→ ssDNA at AT­rich b. single­stranded binding proteins: keeps the structure open c. helicase: binds to dsDNA and stimulates the separation of the two  strands d. gyrase­relaxes supercoils 1. problem without gyrase→ DNA replication halt at over­wound  regions 2. solution; DNA gyrase Function of DNA gyrase:  cuts DNA strand  rotates to remove coils  rejoins DNA strands (b) Elongation Synthesis of leading and lagging strands: 1. Function of DNA polymerase (5’ to 3’ DNA polymerase) 2. leading and lagging strand synthesis  template of DNA 3’ to 5’  template of DNA 5’ to 3’ → lagging strand: discontinuously (Okazaki fragments) Proteins involved in elongation: 1. leading strand synthesis 1. primase → synthesizes RNA primer (ssDNA) and provides 3’­OH 2. 3’OH allows DNA polymerase III to synthesize the leading strand b. lagging strand synthesis  1. primase synthesizes the RNA primer 2. DNA polymerase III synthesizes Okazaki fragments 3. DNA polymerase I → removes RNA primer and fills gap by replacing w/ DNA 4. DNA ligase → join different Okazaki fragments into continuous lagging strand DNA Replication in Eukaryotes 1. initiate at multiple places 2. linear chromosome → problems of replicating the ends of linear 3. high order structure → to remove and to reform chromatin structure a. Eukaryotic Initiation i. DNA 1. AT rich 2. consensus sequence ii. Proteins 1. ORC­ origin recognition complex → binds to origin and unwinds AT rich 2. Recruit Cdc6 and Cdt1 (regulators of replication) 3. ORC Cdc6 Cdt1 (recruit helicase)  Newly synthesized histone binds to CAF­1 which binds to PCNA; this complex of  histone, CAF­1 and PCNA allows histones to be deposited on replicating DNA and  wraps around  histone to form nucleosomes→ chromatin      (b) Eukaryotic Elongation Leading Strand:  primase: lays down RNA primer which provides 3’OH important because…   DNA polymerase ��� (delta): needs double stranded region to bind ^  Beta clamp: helps polymerase III stay bound longer Lagging Strand synthesis in Eukaryotes 1. primase 2. DNA polymerase ��� (delta)­ synthesizes okazaki fragments 3. RNase  H (ribonuclease H) → remove RNA primer 4. DNA polymerase ��� (alpha) → replace RNA primer with DNA 5. ligase­joins different okazaki fragments together Reform the High Order Structure 1. Newly synthesised Histone → binds to CAF­1 → complex binds to PCNA Complex: histone, CAF­1, PCNA → allows histones to be deposited on replicating DNA → wraps around Histone to form nucleosomes → chromatin Telomeres = end of chromosome ­ once RNA primer is removed there is a gap of genetic material (2) Solution of avoiding losing genetic materials at telomeres a. Telomerase has 2 components: i. polymerase ii. RNA → “AACCCCAAC” b. Telomere i. 3’ overhang ii. repeat: “TTGGGG” b. Mechanism of replicating telomeres i. RNA telomerase (AACCCCAAC) paired with TTGGGG of 3’ overhang of  telomeres ii. 1st elongation of 3’ overhang by polymerase activity of telomerase (add TTG) iii. telomerase hops over iv. 2nd elongation of 3’ overhang by polymerase of telomerase (add GGGTTG)  Complementary Strand Synthesis a. primase­makes RNA primer and provides the 3’OH b. DNA polymerase ��� (delta)­ synthesized daughter strand c. RNase H­ removes RNA primer d. DNA ligase­ join DNA fragments together Telomeric Cap Structure  made by telomere DNA and lots of proteins  purpose is to prevent degradation and recombination Chemical Constituents of DNA There are 4 major differences between DNA and RNA: 1. sugar: ribose instead of deoxyribose ­hydroxyl group 2. have U instead of T (uracil/thymine) 2. Most RNA are single stranded 2. only RNA can carry out biochemical reactions **Know the 3 major classes of RNA** 1. rRNA (ribosomal RNA) 1. function: structural and catalytic components of ribosomes 2. highly abundant b. mRNA (messenger RNA) b. RNA polymerase III 1. transcribes tRNAs, small rRNAs, some miRNAs, some snRNAs


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