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This 9 page Study Guide was uploaded by Elisa Chaparro on Saturday November 7, 2015. The Study Guide belongs to BS 161 at Michigan State University taught by j. urbance in Summer 2015. Since its upload, it has received 37 views. For similar materials see Cell and molecular biology in Biology at Michigan State University.
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Date Created: 11/07/15
Chapter 14 Learn smart: - Helicases: enzymes that use ATP to unwind the DNA template o The single strands of DNA produces are unstable, because the process exposes the hydrophobic bases to water. o Cells solve this by using the protein (SSB) single-strand-binding protein, to coat exposed single strands. - DNA polymerases can unwind DNA as they synthesize new DNA - The unwinding of the two strands introduces torsional strain. - Supercoiling: this state describes how the double helix itself coils in space. - Topoisomerases: enzymes that can alter the topological state of DNA, they act to relieve the torsional strain and prevent supercoiling. - DNA gyrase is the topoisomerase involved in DNA replication - Nucleotide: phosphate group, sugar and a nitrogenous base. - Franklin used a technique called x-ray diffraction to suggest that DNA has a helical structure. Enzymes involved in DNA replication / function - Helicase/ causes DNA strand separation at the origin of replication - Topoisomerase/ relieves coiling in DNA strands ahead of the replication fork - Primase/ makes a 10-12 bp complementary primer to the DNA - Polymerase/ attaches a nucleotide to the 3’ end of the DNA strand Griffith performed experiments related to studying transformation in bacteria. The DNA phosphodiester backbone is composed of phosphate groups and sugars a-t,c-g prokaryotic DNA replication 1. Initiation at OriC 2. Replication proceeds bidirectionally 3. Replication ends at terminus The Watson-crick DNA molecule - Two complementary phosphodiester strands that each form a helix with a common axis. - Strands are antiparallel, w/ bases extending into interior of the helix - Complementary strands are formed - Although H-bonds b/w each individual base-pair are low-energy bonds, the sum of bonds of the polymer has enough energy that the entire molecule is stable. Chargraff’s experiments showed that the amount of adenine in a sample was always the same as the amount of thymine. Dna replication in eukaryotic cells is complicated by linear chromosomes The experiments used to distinguish between the 3 potential DNA replication mechanisms were performed by Meselson and Stahl. Strands/type of synthesis - Leading strand/ synthesized continuously - Lagging strand/ synthesized in small fragments that are later connected Repair mechanism/ how they work: - Specific/ target a single kind of lesion in DNA and repair that damage - Nonspecific/ repair multiple kinds of lesions in DNA The major significance of the Hershey Chase experiments is that DNA, not protein, constituted the genetic information that viruses inject into bacteria. DNA replication requires 3 things: - Something to copy (parental DNA) - Something to do the copying (enzymes) - Building blocks to assemble into the copy ( nucleoside triphosphates) - Nucleotides, polymerase, template Mutagen: agents that damade DNA can lead to mutations, and any agent that increases the number of mutation above background levels. Excision repair 1. Recognition of damage 2. Removal of damages region 3. Resynthesis using the information on the undamaged strand as a template. 4 nitrogenous bases in DNA 1. adenine 2. guanine 3. cytosine 4. thymine DNA replication that leads to the production of double helices with one parental strand and one newly synthesized strand is consistent with semiconservative replication Guanine forms 3 h-bonds with cytosine A thymine dimer is a site where 2 adjacent thymine bases become covalently cross- linked to one another. They all add new bases to the 3’ end of existing strands. That is , they synthesize in a 5’ to 3’ direction by extending a strand base paired to the template - once synthesis has begun, DNA can only be synthesized in the 5’ to 3’ direction. A chromosome is composed of DNA and proteins DNA polymerase 1. acts on the lagging strand to remove primers and replace them with DNA 2. not involved in replication but used for DNA repair 3. the main replication enzyme The stability of DNA molecules is due to the sum of the hydrogen bonds between base pairs Different models for DNA replication included: - semiconservative replication - dispersive replication - conservation replication Hershey chase: - identify the molecule that the phafe injects into the bacterial cells label both DNA and proteins to distinguish both from each other o nucleotides contain phosphorus, but proteins do not o some proteins contain sulfur but DNA does not 32 o radioact35e P isotope used to label DNA specifically o isotope S used to label proteins specifically o these isotopes are easily distinguished based on the energy of the particles they emit when they decay - 2 experistnts were performed 32 o 1 viruses were grown on a medium containing P ,others were grown in35S incorporated into t2 coat proteins o Both labeled viruses were then allowed to infect separate bacterial cultures o After infection, bacterial cell suspension was violently agitated in a blender to forcefully remove the infecting viral particles from the surfaces of the bacteria.--> this step ensured that only the part of the virus that had been injected into the bacterial cells- that is, the genetic material – would be detected when the cells were harvested. o Each bacterial suspension was then centrifuged to produce a pellet of cells for analysis. o In the 32P a large amount of radioactive phosphorous was found in the cell pellet but in the 35 S experiment, very little was Miescher’s nuclein was later termed nucleic acid because it was slightly acidic and found in the nucleus The nucleotides in the backbone of the DNA are held together by covalent bonds known as phosphodiester bonds. Griffith defined transformation as the transfer of virulence from one cell to another. In an aging cell, the activity of telomerase would be low Match each element of the nucleotide to the sugar carbon atom it is attached to 1’- carbon atom/ nitrogenous base 3’- OH group 5’- phosphate group The function of telomeres is to protect the ends of chromosomes from nucleases and are necessary to maintain the integrity of linear chromosomes. 2 h-bonds are formed between adenine and thymine in the double DNA helix Steps of lagging strand synthesis 1. Synthesize primers using primase 2. Synthesize DNA 3. Replace RNA primers with DNA 4. Seal nicks in the DNA Bacteriophages- viruses: a core of DNA surrounded by a coat of protein - When these viruses infect a bacterial cell, they first bind to the cell’s surface, and then inject their genetic information into the cell. There it is expressed by the bacterial cell’s gene expression machinery, leading to production of thousands of new viruses. The buildup of viruses eventually causes the cell to lyse, releasing progeny phage DNA repair systems: facilitates the reversal damage to our hereditary material before a permanent mutation can occur. An okazaki fragment is a short fragment of DNA created on the lagging strand of DNA. An origin of replication is a site within a chromosome where DNA replication begins. Genes are the chromosomal entities are where an individual’s traits specified Franklin: obtained x-ray diffraction patterns that confirmed helical structure of DNA Wilkins: prepared uniformly oriented DNA fibers to obtain the first diffraction information on DNA Chargraff: determined the amount of adenine in DNA always matches thymine and the amount of cytosine always matches guanine. Why do eukaryotic cells have multiple origins of replication to ensure timely replication of multiple, relatively large chromosomes Chapter 11 Gametes contain only one copy of each chromosome Fertilization joins two gametes - Gametes (eggs and sperm) each contain 2 chromosomes- haploid reproductive cell - Somatic cells: any of the cells of multicellular organisms except those that are destines to form gametes ( germ- line cells)- contain four chromosomes - Zygote: the diploid( 2n) cell resulting from the fusion of male and female gametes (fertilization)- contains two copies of each chromosome - Fertilization: the fusion of 2 haploid gamete nuclei to form a diploid zygote nucleus - Syngamy: same as fertilization Sexual Life cycles Alternate between haploid and diploid - Zygote undergoes mitotic divisions and gives ride to all cells of the adult body - Not all organisms reproduce sexually, some by mitosis, never forming gametes. - Only germ-line cells are capable of meiosis. Germ-line cells undergo meiosis, but how can the body maintain a constant supply of these cells? - Stem cells divide by mitosis to produce one cell that can undergo meiosis and another stem cell. So, stem cells provide the body with a constant supply of germ-line cells. Meiosis features two divisions with one round of DNA replication Meiosis Differs from Mitosis in two principal respects - 1. Pairing of homologous chromosomes: o Meiosis in a diploid organism consists of two rounds of division, called meiosis 1 and meiosis 2 o Meiosis 1: referred to as a” reduction division” because homologous chromosomes separate, and the daughter cells have only the haploid number of chromosomes. During early prophase1 of meiosis, homologous chromosomes find each other and become closely associated- synapsis: the point-by-point alignment (pairing0 of homologous chromosomes that occurs before the first meiotic division, crossing over takes place during synapsis. o Meiosis2: two haploid cells from meiosis 1 undergo a mitosis like division without DNA replication to produce four haploid daughter cells. o The synaptonemal complex: a protein lattice that forms b/w two homologous chromosomes in prophase 1 of meiosis, holding the replicated chromosomes in precise register with each other so that base-pairs can form between nonsister chromatids for crossing over that is usually exact within a gene sequence. o Crossing over: the exchange of corresponding chromatid segments between homologous chromosomes; responsible for genetic recombination between homologous chromosomes.- alleles that were formerly on separate homologues are not found on the same homologue. o Chiasmata: An x- shaped figure that can be seen in the light microscope during meiosis; evidence of crossing-over, where two chromatids have exchanged parts; chiasmata move to the ends of the chromosome arms as the homologues separate- maintained until anaphase 1 o The physical connection of homologues due to crossing over, as well as the continues joining of the sister chromatids y cohesion proteins, lock homologues together. - 2. Reduction division o Meiosis involves 2 successive divisions with no replication of genetic material between the two - In anaphase 1 homologues separate reducing the number of chromosome to the haploid number. This is followed without replication by a second division where sister chromatid separate to yield 4 haploid cells. - If sister chromatids separated at first division, would meiosis still work? o No, keeping sister chromatids together at the 1 division is the key to this division being reductive. Reducing the chromosome number from diploid to haploid. Homologues segregate at first division to accomplish this reduction. The process of meiosis involves intimate interactions between homologues - http://textflow.mheducation.com/figures/0077645960/mas32290_1104L_lg.jp g - http://textflow.mheducation.com/figures/0077645960/mas32290_1104L_lg.jp g - Homologous pairs then align closely side-by-side, apparently guided by heterochromatin sequences, in the process called synapsis. Crossing over - Synaptonemal complex - Recombination nodules- are thought to contain enzymatic machinery necessary to break and rejoin chromatids of homologous chromosomes. - DNA segments are exchanged between nonsister chromatids. - Reciprocal crossovers are controlled such that each chromosome arm has one or a few crossovers per meiosis, no matter what the size of the chromosome. - Humans have typically two or three Metaphase 1: - Kinetochore microtubules attach to HOMOLOGUES Anaphase 1; - Results from the differential loss of sister chromatid cohesion along the arms - Microtubules shortenbreak the chiasmatapull centromeres toward the poles= homologues are pulled apart - Release of chromatid cohesion along the chromosome arms, but not at the centromeres. - The difference is that the destruction is inhibited at the centromeres by a mechanism … - Because of the random orientation of homologous chromosomes on the metaphase plate, a pole may receive either the maternal or paternal homologue from each chromosome pair. - As a result meiosis 1 results in independent assortment of maternal and paternal chromosomes into the gametes. - Independent assortment: random assortment of alleles for each of the genes. o For genes on different chromosomes this results from the random orientations of different homologous pairs during metaphase 1 of meiosis. o For genes on the same chromosome, this occurs when the two loci are far enough apart for roughly equal numbers of odd- and even- numbered multiple crossover events. Telophase 1 - Cytokinesis, may or may not occur after telophase 1. - The second meiotic division, meiosis 2, occurs after an interval of variable length. - Achiasmate segregation: The lining up and subsequent separation of homologues during meiosis 1 without the formation of chiasmata between homologues; founing in Drosophelia males ( fruit flies) and some other species.- no crossing over (without chiasmata) Comparison of Meiosis and Mitosis Page 231 1. Homologous pairing and crossing over joins maternal and paternal homologues during meiosis I. 2. Sister chromatids remain connected at the centromere and segregate together during anaphase of meiosis I. 3. Kinetochores of sister chromatids are attached to the same pole in meiosis I, and to opposite poles in mitosis. 4. DNA replication is suppressed between the two meiotic divisions. Hw problems A cell biologist examines a diploid cell from a barley plant during prometaphase of mitosis and determines that 28 chromatids are present. The role of meiosis in this plant is to reduce the number of chromosomes per cell from 14 to 7 What would result of improper disjunction at anaphase 1? At anaphase 2? o An improper disjunction at anaphase 1 would result in 4 aneuploid gametes: 2 with an extra chromosome and 2 that are missing a chromosome. o Nondisjunction at anaphase2 would result in 2 normal gametes, and 2 aneuploid gametes: 1 with an extra and 1 with none. If a somatic cell from a frog contains 20 picograms of DNA during G of interphase, how 2 many picograms of DNA would be present in each gamete produced by this species? 5 If a germ-line cell from a frog contains 10 picograms of DNA during G1of interphase, then how many picograms of DNA would be present in each cell during prophase II of meiosis? 10 If a somatic cell from a mouse contains 40 picograms of DNA during G o2 interphase, then how many picograms of DNA would be present in each cell during anaphase I of meiosis? 40 How many molecules of DNA are present in the chromosomes of a single human cell (2n=46) during prophase I of meiosis? 92 A cell biologist determines that diploid cells from a particular species of Drosophila contain 16 chromatids during metaphase of mitosis. Based on this finding, how many molecules of DNA should be present in the chromosomes of a single cell from this species during anaphase II of meiosis? 1N-> 1 N = meiosis 2 in humans How can you compare the complement of DNA in an interphase (G1) diploid cell and a cell that has just complete meiosis 1? - They have the same amount of DNA, but the meiotic cell has half as many chromosomes. What is the necessity of two phases of meiosis? - In order to both reduce the chromosome number and separate sister chromatids
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