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Genetics, Exam 3 Study Guide

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by: Sarah Moran

Genetics, Exam 3 Study Guide BIO 235

Marketplace > Missouri State University > Biology > BIO 235 > Genetics Exam 3 Study Guide
Sarah Moran
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Exam 3 material
Lazlo Kovacs
Study Guide
Genetics, BIO 235
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This 14 page Study Guide was uploaded by Sarah Moran on Saturday January 23, 2016. The Study Guide belongs to BIO 235 at Missouri State University taught by Lazlo Kovacs in Summer 2015. Since its upload, it has received 117 views. For similar materials see Genetics in Biology at Missouri State University.


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Date Created: 01/23/16
EXAM 3 REVIEW Section 8.1 Variation in chromosome number pp 222 – 223  Aneuploidy: an organism gains or loses one or more chromosome but not a complete set o Varation in the number of particular chromosomes within a set o Monosomy: loss of a single chromosome o Trisomy: gain of one chromosome  Euploidy: complete haploid sets of chromosomes are present o Polyploidy: more than two sets of chromosomes are present  Variation in the number of complete sets of chromosomes  Not the normal diploid set but every set has the same amount  Ex: 3 sets of chromosomes = triploid  Klinefelter syndrome (47, XXY) Turners Syndrome (45,X)  Occurs more often in plants than animals/humans  Does not result in a gene dosage problem  everything would have equal amounts  Ex: 150% (triploid), 200% (tetraploid)  Nondisjunction: paired homologs fail to disjoin duing segregation – Meiosis o Fertilizing with a normal haploid gamete produces a zygote with ether 3 chromosomes or 1 chromosome Section 8.2 Monosomy and trisomy. pp 223-227  MONOSOMY – o Loss of one chromosome (2n-1) o Monosomy for any of the autosomes is usually not tolerated in humans/other animals  If one gene is represented by a lethal allele, the unpaired chromosome will result in the death of the organism  This is because monosomy unmasks recessive lethal alleles that would normally be covered by heterozygous alleles  Haploinsuficiency: a single copy of a recessive gene may be insufficient to provide adequate function for sustaining the organism  death  TRISOMY – o Gain of a chromosome (2n+1) o Typically cause severe effects and is usually lethal during development o Results in an unbalanced distribution  Ex: 150% alpha polypeptides and only 100% beta polypeptides (hemoglobin) o Down Syndrome – Trisomy 21  Only human autosomal trisomy in which a significant number of individuals survive longer than a year past birth  Primary down syndrome does not run in families  Occurs though nondisjunction of chromosome 21  Chance of having a child with down syndrome increase with maternal age  Familial down syndrome: involves a translocation of chromosome 21 – chromosome is carrying an extra piece attached to another chromosome Section 8.3 Polyploidy. pp 227-230  More than two multiples of the haploid chromosome set are found o Triploid (3n), tetraploid (4n) o Infrequent in many animals except amphibians/fish, common in plants o Odd numbers of chromosome sets are not maintained from generation to generation  polyploidy organisms w/ uneven # of homologs don’t produce genetically unbalanced gametes o Endopolyploidy: only certain cells in an otherwise diploid organism are polyploidy  The sets of chromosomes replicate repeatedly without nuclear division  EX: liver o Polyploidy originates in two ways:  Autopolyploidy: the addition of one or more extra sets of chromosomes identical to the parent spcies  EX: AAA (triploid) AAAA (tetraploid)  A failure of all chromosomes to segregate during division can produce a diploid gamete – if this is fertilized with a haploid gamete = zygote w/3 sets will be produced  Complete nondisjunction which is caused by the failure of mitosis @ anaphase  The sister chromatids separate but will not be pulled in separate daughter nuclei – instead they will all end up in the same nucleus  Tetra more common than tri b/c even # of sets = better reproduction  Allopolyploidy: hybridizing two closely related species  EX: AA is fertilized with sperm BB = AB, A = a1, a2, a3… B = b1, b2, b3…  Two species form a diploid hybrid which will function as a haploid bc chromosomes will fail to find a homologue or segregate properly  Complete nondisjunction in the hybrid will double the chromosome number and turn the plant into an amphidiploid which will become a functional diploid and can form balanced gametes  Distinction bw auto- vs. allo- is based on genetic origin of the extra chromosome  Polyploids having an odd number of chromosomes are usually sterile  Why: bc in meiosis, chromosomes want to pair, but there is an odd chromosome out, so it is still pulled by spindle fibers which will result in uneven number of copies in each chromosome GENE DOSAGE IMBALANCE Section 8.4 Variation in composition and arrangement of chromosomes. pp 230 -231  Add, delete, or rearrange substantial portions of one or more chromosomes o Deletions: loss of a chromosomal segment  Most deletions are lethal and not born  Cri-du-chat: deletion in chromosome 5 – talks like cat o Duplications: repetition of a chromosomal segment compared to the normal parent chromosome  Mechanism through which gene families arise from a single ancestral gene  Evolution o Rearrangements: a chromosome segment is inverted, exchanged with a segment or a nonhomologous chromosome, or transferred to another chromosome o Exchanges and transfers are called translocations: locations of genes are altered within the genome  Typically due to one or more breaks along the axis of the chromosome followed by either the loss or rearrangement of genetic material Section 8.7 Inversions rearrange the linear gene sequence. pp 235 – 236  Inversion: genetic rearrangements in which the order of the genes in a chromosome segment is reversed  Do not alter the genetic content – just rearranges the gene sequence  Requires breaking at two points along the length of the chromosome and reinsertion of the inverted segment o Paracentric inversion: in the centromere is not part of the rearranged chromosome segment  can result in zygotes that are not viable/fertile o Pericentric inversion: the centromere is part of the inverted segment o Inversion heterozygotes: organisms with one inverted chromosome and one noninverted homolog  pairing between such chromosomes is accomplished only if they form an inversion loop  you’ll end up with duplications and deletions in the chromosome but only if the recombination occurs in the region of the inversion  if crossing over does not occur within the inverted segment of the inversion loop, the homologs will segregate which results in two normal and two inverted chromatids that area distributed into gametes.  If crossing over does occur within the inversion loop, abnormal chromatids are produced o Dicentric chromatid: two centromeres  pulled in two directions during anaphase o Acentric chromatid: no centromere  move to one pole or the other  Which inversion causes dicentric centromeres….paracentric inversion  Crossing over within inversion: o You’ll end up with a dicentric chromosome and one that is acentric o Dicentric bridge between the two chromosomes…. Will eventually break Section 8.8 Translocation alters the location of chromosomal segments. pp 236 – 238  Translocation: movement of a chromosomal segment to a new location in the genome o A segment of one chromosome becomes attached to a different chromosome o Familial down syndrome is a product of translocation o Genetic information is not lost or gained – rearrangement of genetic material  Simple translocation o One way transfer  one piece of a chromosome is attached to another chromosome  Reciprocal translocation o Exchange of segments between two non-homologous chromosomes (2 way transfer) o Two different types of chromosomes exchange pieces  producing two abnormal chromosomes w/ translocations o When genetically unbalanced gametes participate in fertilization in animals, the offspring do not usually survive o Semisterility: fewer than 50% of the progeny of parents heterozygous for a reciprocal translocation survive  Impact on reproductive fitness plays role in evolution o Robertsonian Translocation:  Occurs in acrocentric chromosomes (centromere very close to end)  A small fragment results in a very short little chromosome which is often lost  form large submetacentric or metacentric chromosomes  Familial down syndrome: 14/21  Balanced translocation carrier- phenotypically normal but only has 45 chromosomes Section 8.9 Fragile sites in human chromosomes. pp 238 – 240  Fragile sites: a heritable gap, or non-staining region, of a chromosome that be inducted to generate chromosome breaks o Fragile-X Syndrome  Most common form of mental retardation  Trinucleotide repeats: a sequence of three nucleotides is repeated many times, expanding the size of the gene.  “CGG” is repeated in an untranslated area next to the coding sequence of the gene  6-54 repeats = normal  55-230 repeats = carriers but are unaffected  increased # in their children (mostly maternal)  More than 230 repeats = expression of the syndrome o Huntington’s Disease  “AGC” repetition Section 9.1 Genetic material must exhibit four characteristics . pp 251 – 252.  4 critical characteristics to serve as genetic material: o Replication o Storage of information – requires the molecules to act as a warehouse of genetic information that may or may not be expressed by the cell in which is resides  Most cells contain a complete copy of the organisms genome, at any point, they express only a part of it o Expression of information:  Information flow/ Central Dogma of Molecular genetics  Transcribe DNA  mRNA, rRNA, tRNA  ribosome  mRNA is translated into proteins via mediated by tRNA & rRNA  Translation: the chemical info in mRNA encodes for polypeptides which then folds into proteins o Variation by mutation  If a mutation occurs, the alteration is reflected in transcription and translation, affects the specific protein  Generations vary due to mutations Section 9.2: Until 1944, observations favored protein as the genetic material. PP 252  Fredrich Miescher – isolated cell nuclei and derived an acidic substance (DNA) that he called nuclein  Phoebus A. Levene: DNA contained approx.. equal amounts of 3 similar molecules called nucleotides o Incorrect assumption based in tetranucleotide hypothesis  Erwin Chargaff – showed that Levens proposal was wrong when he demonstrated that most organisms do not contain precisely equal proportions of the 4 nucleotides Section 9.3 Evidence that DNA is the genetic material: Experiments with bacteria and phage. pp 252 -258.  Oswald Avery, Colin MacLeod, Maclyn McCarty o The first direct experimental proof that DNA is the biomolecule responsible for heredity o Marked the beginning of the era of molecular genetics o Critical question: what molecule serves as the transforming principle? DNA o Added protease = protein destroyed (transformation), RNase = RNA destroyed (transformation), DNase = DNA destroyed (no transformation)  Frederick Griffith o Difference in virulence depends on the presence of polysaccharide capsule – virulent have them, Avirulent do not  Avirulent bacteria are engulfed and destroyed by phagocytic cells in the host animals circulatory system - produces rough colonies (r)  Virulent bacteria are not engulfed – multiply - Virulent produces smooth colonies (s)  Sterotypes: different strains of the bacteria – identified by immunological techniques – roman numerals  Inject mouse with heat-killed virulent form of bacteria– did not kill mouse  Injected mouse with heat-killed virulent form of bacteria with avirulent bacteria – mouse died  Concluded that the heat- killed IIIS bacteria converted live avirulent IIR cells into virulent IIIs cells = TRANSFORMATIONS & TRANSFORMATION PRINCIPLE  Hershey-Chase Experiment o E-coli o Bacteriophage T2 (phage) o Lytic cycle o T2 phage consists of 50% protein, 50% DNA o Infection is initiated by adsorption of the phage by its tail fibers to the bacteria cell o The production of new viruses occurs within the bacteria cell o Used radioactive labels 32P for DNA and 35S for protein *ingenious bc P is only found in DNA o PROCESS:  Grew e.coli in medium that contained phosphates – labeled phage 32P  Grew e.coli in medium that contained sulfur – labeled phage 35S  Introduced these phages into pure e.coli  Centrifuged them so that all the cells would sink to bottom of cuvette  E.coli sank to bottom and phage stayed towards the top  35S was always detected in the supernatant (within empty phage coats)  Bacteria cells were eventually lysed as new phages were produced  Proved that DNA (containing radioactive P) was always in cells – demonstrated that it was the infecting element = directs phage reproduction Section 9.6: Knowledge of Nucleic Acid Chemistry Is Essential to the Understanding of DNA Structure. PP 259-262 Nucleotides:  DNA is a nucleic acid – nucleotides are building blocks of all nucleic acids  Consist of three essential components – nitrogenous base, pentose sugar (5- carbon sugar) and a phosphate group  Purines – Adenine and Guanine  Pyrimidines – Cytosine, Thymine, and Uracil  DNA and RNA contain A, C, G, but DNA only contains T and RNA only contains U  If molecule is composed of a purine/pyrimidine base and a ribose/deoxyribose sugar = NUCLEOSIDE  Is a phosphate group is added to the nucleoside = NUCLEOTIDE  Purine = sugar bonded to N-9, Pyrimidine = sugar bonded to N-1 Nucleoside diphosphates and Triphosphates  The addition of one or two phosphate groups results in nucleoside diphosphates and triphosphates  Adenosine Triphosphate (ATP) and Guanosine Triphosphate (GTP) o Large amount of energy involved in adding or removing the terminal phosphate group Polynucleotides  Phosphodiester bond: linkage between two mononucleotides consist of a phosphate group linked to two sugars o Phosphoric acid has been joined to two alcohol by an ester linkage on both sides  Each structure has a C-5’ end and C-3’ end  The nitrogenous base is attached at the top, in the C-1’ position  Phosphate is attached to the C-3’ position of one sugar and the C-5’position of the neighboring sugar Section 9.7: The Structure of DNA Holds the Key to Understanding Its Function. PP 262-268 Watson and Crick  Proposed that the structure of DNA is in the form of a double helix  Two long polynucleotide chains are coiled around a central axis, forming a right-handed double helix  The two chains are antiparallel  Bases of both chains are flat structures lying perpendicular to the axis, they are “stacked” on one another .34 nm-3.4A apart on the inside of the double helix  The nitrogenous bases of opposite chains are paired as the result of the formation of hydrogen bonds, in DNA only A-T, C-G occur  Each complete turn of the helix is 3.4 nm long – 34A (length of 10 base pairs)  A larger major groove alternating with a smaller minor groove winds along the length of the molecule  The double helix has a diameter of 2.0 nm – 20A Chargaff  Quantitative methods were then used to determine the amounts of the four bases from each source  A-T ratio = 1, G-C Ratio = 1, ratios never change, percentages do  Crucial for Watson-Crick model  complementary base pairs (purines vs pyrimidines) X-Ray Diffraction Analysis  When fibers of DNA are subjected to X-rays, the X rays scatter in a pattern that depends on the molecules atomic structure  Can be captured as spots on photographic film  Rosalind Franklin – suggested that the structure of DNA was some sort of helix Section 10.1: DNA is reproduced by semi-conservative replication. PP 286-288  Semi-conservative replication: Each replicated DNA molecule would consist of one “old” and one “new” strand  Conservative replication: complementary polynucleotide chains are synthesized – the two newly created strands come together and the parental strand come together o The original helix is “conserved”  Dispersive replication: the parental strands are dispersed into two new double helices following replication – each strand consists of both old and new DNA o Involves cleavage of the parental strands during replication o Least likely to occur  Autoradiography: pinpoint the location of a radioisotope in a cell  DNA synthesis begins at a site termed the origin of replication o Each bacterial chromosome has only one origin of replication  Replication fork: at each point along the chromosome whre replication is occurring, the strands of the helix are unwound, which creates a replication fork. o Will initially appear at the point of origin of synthesis and then move along the DNA duplex as replication proceeds o If biodirectional – 2 forks will be present going in opposite directions away from origin o Will eventually meet at the opposite side of the bacterial chromosome  Replicon: length of DNA that is replicated following one initiation event at a single origin Section 10.2: DNA synthesis in bacteria involve five DNA polymerases. PP 290- 293.  KORNBERG!!  DNA polymerase I directs the synthesis of DNA  Highly coordinated process has two stages: o Initiation – proteins open up the double helix and prepare it for complementary base pairing o Elongation – proteins connects the correct sequence of nucleotides on newly formed DNA stands o DNA SYNTHESIS IS FROM 5’ – 3’ (5’-3’ of new strand) o DNA polymerase ADDS nucleotides to the 3’-OH of new strand  DNA polymerase II and III o Cannot initiate DNA synthesis on a template, but can elongate an existing DNA strand called a primer o DNA polymerase III extends the primer and makes a new daughter DNA strand based on the template. o DNA polymerase III catalyzes DNA synthesis  DNA synthesis must proceed in opposite directions, because the template strands run in opposite orientation (they are antiparallel).  DNA synthesis is continuous in one direction (leading strand), but non- continuous in the other (lagging strand). Section 10.3: Many complex issues must be resolved during DNA replication. PP 293-296.  The helix must undergo localized unwinding and the resulting “open” configuration must be stabilized so that synthesis may proceed along both strands o DnaA is responsible for the initial step in unwinding the helix o ATP hydrolysis provides energy to these proteins (known as Helicases) to break the hydrogen bonds and denature the double helix  As unwinding and DNA synthesis proceeds, increased coiling creates tension further down the helix, which must be reduced o Coiling tension is created ahead of the replication fork which produces supercoiling but is relaxed by DNA topoisomerase (gyrase)  A primer must be synthesized so that polymerization can start under the direction of DNA polymerase III o RNA serves as the primer that initiates synthesis o Synthesis of the RNA is directed by a form of RNA polymerase called primase, which does not require a free 3’ end to initiate synthesis  Once the RNA primers have been synthesized, DNA polymerase III begins to synthesize the DNA of both strands of the parent molecule – bc the two strands are anti-parallel, continuous synthesis in the direction that the replication fork moves is possible along only one of the two stands and on the other, synthesis must be discontinuous o Only one strand can serve as the template for continuous DNA synthesis – the newly synthesized DNA is called the leading strand o Many points of initiation are necessary on the opposite DNA template resulting in discontinuous DNA synthesis of the lagging strand  The RNA primers must be removed prior to completion of replication – the gaps that are created must be filled with DNA complementary to the template at each location o RNA primers are replaced by the Dnase I enzyme. o The sugar-phosphate backbones of the strands of the “finished” Okazaki fragments are covalently attached by DNA ligase.  The newly synthesized DNA strand that fills each temporary gap must be joined to the adjacent strand of DNA o DNA ligase joins the fragments  DNA polymerase is not perfect and sometimes inserts incorrect nucleotides to the strand – a proofreading mechanism that also corrects errors is critical to DNA synthesis o DNA polymerase all possess 3’-5’ exonuclease activity – potential for them to detect and excise a mismatched nucleotide in the 3’ – 5’ direction Section 10.6: Eukaryotic DNA replication . PP 297-298.  Similar o Double-stranded DNA is unwound at replication origins o Replication forks are formed o Bidirectional DNA synthesis creates leading and lagging strands from single-stranded DNA templates under the direction of DNA polymerase  Eukaryotic o More complex o Initiation at multiple replication origins  Greater amounts of DNA o Multiple DNA polymerases  To accommodate the increased number of replicons, there are more DNA polymerase molecules  More types of DNA polymerase o Replication through chromatin  Eukaryotic DNA is complexed with DNA- binding proteins, chromatin Section 10.7: Ends of linear chromosomes are problematic during replication. PP 300-302.  Telomeres preserve the integrity of linear chromosomes  Telomeres are protective caps on eukaryotic chromosomes. o Prevent fusion with other chromosomes o Maintains the integrity of the chromosome  Repetitive DNA sequences of TTAGGG o No protein coding genes in telomeres o TTAGGG repeats are highly conserved in distantly related organisms o Bind two types of proteins:  Protective proteins  The telomerase enzyme: a ribonucleoprotein (composed of RNA and protein)  Binding of telomerase to TTAGGG and addition of RNA extends the ends.  Telomerase is key in determining the life-span of cells o In immortalized cells, such as the cells of the germ line, cancer cells, and yeast cells, telomerase activity is present:  The telomeres are fully reconstructed with each DNA replication o In somatic cells, telomerase activity is virtually absent:  The telomeres are shortened with every cell division – somatic cells have a finite life span  Somatic cells are programmed to divide 30 to 50 times, then die o In yeast, deletion of the telomerase gene leads to shortening of chromosomes at each cell division  Telomerase is key in determining how many times a cell can divide o In cancer cells: telomerase activity is high o In somatic cells from which cancer cells derived: telomerase activity is very low or absent o If you can block telomerase activity, you can stop cancer Section 11.2 Supercoiling facilitates compaction of bacterial chromosomes. pp 317 -318  A DNA configuration in which the helix is coiled upon itself  Catalyzed by the enzyme topoisomerase. o Topoismerase can also “relax” the DNA molecule by nicking the single strand temporarily thereby relieving the molecule from the strain of supercoiling.  Can exist in stable forms only when the ends of the DNA are not free – as in covalently closed circular DNA molecule  Causes tighter packing and this an increase in sedimentation velocity  Positive supercoiling = over coiled/ increase tension and strain  Negative supercoiling = under coiled/ decreasing tension and strain Section 11.4 DNA is organized into chromatin in Eukaryotes. pp 320-324  Histone proteins abound in the chromatin of all eukaryotic cells. o Histones – small proteins with basic, positively charged amino acids lysine and arginine o Bind to and neutralize negatively charged DNA o Make up half of all chromatin protein by weight o Five types: H1, H2A, H2B, H3, and H4 o Core histones make up nucleosome: H2A, H2B, H3, and H4 o High level of similarity of histones among diverse organisms  DNA negatively charged and histones are positively charged = attraction together  Non-histone proteins are a heterogeneous group. o Half of proteins in chromatin are non-histone. o Large variety of non-histone protein molecules of each kind in diploid genomes o Large variety of functions • Scaffold – backbone of chromosome • DNA replications – e.g., DNA polymerases • Chromosome segregation – e.g., motor proteins of kinetochores • Transcriptional regulation – largest group regulates transcription during gene expression o Occur in different amounts in different tissues because of variety of function  The nucleosome: the fundamental unit of chromosomal packaging arises from DNA’s association with histones o Spacing of nucleosomes affects gene expression. • Regions between nucleosomes available for interactions with proteins involved in expression, regulation, and further compaction • Determines how and whether certain proteins interact with specific sequences o Packaging into nucleosomes condenses DNA sevenfold. • 2 meters of DNA shortens to less than 0.25 meters. Section 11.5 Chromosome banding. pp 324 – 326  The compaction level of interphase chromosomes is not completely uniform  Euchromatin o Less condensed regions of chromosomes o Transcriptionally active o The 30 nm fiber forms radial loop domains  Heterochromatin o Tightly compacted regions of chromosomes o Transcriptionally inactive (in general) o Radial loop domains compacted even further o There are two types of heterochromatin:  Constitutive heterochromatin  Regions that are always heterochromatic  Permanently inactive with regard to transcription  Usually contain highly repetitive sequences  Facultative heterochromatin  Regions that can interconvert between euchromatin and heterochromatin  Example: Barr body formation during development in female  Euchromatin is converted to heterochromatin and vice-versa  ATP-dependent chromatin remodeling refers to dynamic changes in chromatin structure  These changes range from a few nucleosomes to large scale changes  Carried out by diverse multiprotein machines that reposition and restructure nucleosomes  There are two types of signals that direct chromatin remodeling o DNA methylation - silences gene expression in two different ways  The presence of methylated CpG in the vicinity of the promoter interferes with transcription.  The presence of methylated CpG attracts methyl-CpG-binding protein to the DNA which in turn recruits ATP-dependent remodeling complexes o Histone protein modifications  influences the tightness of chromatin packaging  These modifications affect the level of transcription  May influence interactions between core histones and DNA  Occur in patterns that are recognized by other proteins o Called the histone code: o The pattern of modifications specify alterations to be made to chromatin structure o These proteins bind based on the code and affect transcription  Repetitive Sequence: o Sequence complexity: Refers to the number of times a particule base sequence appear in the genome o 3 main types:  Unique or non-repetitive  Moderately repetitive  Moderately repetitive sequences o Found once in a few times in the genome o Includes genes as well as intergenic areas o Make up roughly 41% of the genome  Found a few hundred to several thousand times  Tandemly repeated elements o Minisatellites (VNTRs): 15 to 100 bp units repeating for 100 to 20,000 bp stretches of DNA o Microsatellites (VNTRs): di-,tri-, tetra-, and penta- nucleotides repeated a few to tens of times o Multiple copy genes (rRNA genes)  Interspersed retrotransposons o LINEs: Long Interspersed elements, ~21% of the human genome o Example: L1 retrotransposon family o SINEs: Short Interspersed elements, ~13% of the human genome o Example: Alu family in humans  Approximately 300 bp long  Represents 10% of human genome  Found every 5000-6000 bp  Highly repetitive  Highly repetitive sequences – Satellite DNA  Found tens of thousands to millions of times  Each copy is relatively short (a few nucleotides to several hundred in length)  Satellite elements are clustered together in tandem arrays  The for constitutive heterochromatic DNA in the centromere and telomeres  Alphoid element (170 bp) forms up to 1 Mbp-long sequences in the centromere  Telomeric DNA sequence: 5’-TTAGGGG-3’ repeated up to 100 times 


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