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Chapter 7 notes (Exam II).docx

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Department
Biology
Course
BIO 190
Professor
Elaine Sia
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Chapter 7 Discribe Griffith;s experimetns showing the discovery of transformation. Griffiths used two strains that are distinguishable by the appearance of theoir colonies when grown in laboratory cultures—normal virulent type deadly to most laboraty animals and a nonvirulent type. The cells of this strain are enclosed in a polyssachride capsule—smooth, but the nonvirulent type is not enclosed giving colonies a rough appearance. Griffith killed some S cells by boiling them. He then injected killed cells in the cells into mice. The mice lived showi8ng carcasses of virulent type do not cause death. However, mice injected with a mixture of heat-killed virulent cells and live non-virulent cells did die. Live S cells could be recovered from the dead mice. However, these cells were S cells and virulent on subsequent injection, which must have occurred via transformation. How was the transforming agent identified? Avery et al. chemically destroyed all the major categories of chemicals in an extract of dead cells one at a time and find out if the extract had lost the ability to transform. They found out that the extract had only lo0st its ability ot transform when the donor mixture was treated with the enzyme DNase to break up the DNA. It is now known that fragm,emnts of the tansforming DNa that conger virulence enter the bacterial chromosome and replace their counterpoarts that confer nonvirulence. How did Hershey and Chase further prove that it is DNA that is the genetic material? They were stydying T2 phage and gave DNA and protein distinct labels (they labeled DNa with radioisotpope of Phosphorous 32P and proteins as sulfur 35S. They iunfected 2 E coli culture with many virus particles per cell: one e coli culture received 32P the other 35S. After allowing sufficient time for infection to take place, the sheared the empty phage carcasses off the bacterial cells by agaitation ina kitchen blender. They then separated the bacterial cells from the ghosts in a centrifuge and then measured the radioactivity in the two fractions. 32P labeled phjages used to infect E coli had most of the radioactivity inside indicationg that the phage DNA entered te cells. but 35S had most of radioactive material ending up in the phage ghosts indicating that the phage protein never entered the cells. Hence DNA is the hereditary material. The phage priteuns are mere structural packafinms that is discarded after delivering the viral DNA to the bacterial cell. 7.2 What are the key properties of hereditary material? 1. The structural features of DNA must allow faithful replication. 2. The genetic material must have informed content. 3. Genetic material must be able to change on rare occasions to allow for mutations to provide raw material for evolutionary selection. What are the building blocks of DNA? DNA contains three types of chemical components: 1)phosphate 2)sugar—deoxyribose 3) 4 nitrogenous bases Nucleoside: sugar + base= adenosine, guanosine//// Nucleotide= sugar + base + phosophate. Describe te structure of DNA The 3d structure is composed of two side by side chains of nucleotides twisted into the sjape of a double helix. In the double stranded DNA molecuile the 2 backnomes are antiparallel. The base pairs which are flat planar structures stack on top of one another at the center of the double helix. The loss of ‘OH 2’ OH stabilizes DNA. DNA is hydrophobic so excludes watrer molecules from the spaces between the base pairs. The double helix result in a double helix with 2 distince sizes of grooves running in a spiral: the major groove and the minor groove. What is meant by semiconservative replication? The unwinding of the two strands will expose single bases of each sstrand. Each exposed base will pair only with its complementary base. Thuse each of the single strands will act as a template to reform a double helix identical with the original. What kind of replication does DNA undergo and how did Meselson and Stahl demonstrate this? In order tofind this out they gre E. Coli cells in a mediyum containing the heavy isotope 15N rather than ligh 14N. After many cell division the DNa of the cells were labeled with 15N heavy isotope. The cells were then placed int o 14N medium and after one or two cell divsions samples were taken and the DNA was isolated from each sample. They were then able to distinguish DNA of different densities via cesium chloride gradient centrifugation. After centrifugation the cesium and chloride ions tend to be pushed by centrifugal force to the bottom of the tube. Highest ion concentration and density at bottom. After the first generation intermediate density DNA was ibserved. As this could be both disperisive or nd semiconservative the 2 generation was performed. Both intermediate and low density DNA were observed indicating that the DNA replication was semiconservative. How did John Cairns show the existence of the replication formk? Cairns tested this prediction by allowing replccating DNA in bacterial cells to incorporate tritiated thymidine 3H thymidine. After varying intervals and varying number of replication cycles in a “hot” medium, Cairns carefully lysed the bacteria and allowed the cell contents to settle onto grinds designed for electron microscopy. Cairns then covered the grid with photgraphic emulsion and exposed it in the dark for2 months. After one replication cucle in 3H thymidine a ring of dots appeared in the autoradiograph—Cairn interpreated this ring as a newlyy formedd radioactive strand in a curculalr daughter DNA molecue. In the second replication cycle, the replcation cycle was observed such that autoradiographic patterns revbealed the progressive movement of the replciation forks. What are the substrates and the function of DNA polymerase? The enzyme adds deoxyribonucleotides to the 3’ end of a growing nucleotide chain. The substrates for DNA polymerase are the triphosphate forms of the deoxyribonucleotide dATP, dGTP, dCTP, dTTP. This invloves the remobval of the two of the three phospophates in the form of pyrophasphate. The energy produced by cleaving this high-energy bond and the subsequent hydrolysis of pyrophasphate to two inorganic phosphate molecules helps drive the endo thermic process if building DNA polymer. What are the activities of DNA pol I? Pol 1 has the following activit8ies: 1) A polymerase activity, which catalyzes chain growth in the 5’-3’ direction 2) A 3’ to 5’ exonuclease activity, which removes mismatched bases and 3) A 5’ to 3’ exonuclease activity which degrades single strands of DNA or RNA. Too slow (~20 nucleotides per second) and too abundant (~400 nucleotides ) for DNA synthesis. It is distributivr. 7.4 Give an overview of DNA replication Synthesis of both the leading and lagging strands are intiated by a primer, or short chain of nucleotides, that binds with the template strand to form segments of duplex nucleic acid. RNA primase is distributive because it adds only a dew ribonucleotides before dissociating from the template. DNA polymerase III elongate
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