13 Recombinant DNA.Q

13 Recombinant DNA.Q - Central Dogma DNA RNA Protein...

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DNA RNA Protein Central Dogma
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RECOMBINANT DNA TECHNOLOGY Purification Amplification
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Principles of Cloning To achieve a purification and amplification of genomic sequences, restriction fragments inserted into Escherichia coli bacterial cells. This achieves purification because only one or two restriction fragments enters each bacterium, so the fragments are physically separated in different cells. This achieves amplification because the bacteria replicate the DNA fragment as they grow, so that from a single cell carrying a single genomic fragment, an unlimited number of clones can be grown. The process of cloning involves a series of steps. 1. The genomic restriction fragments must be joined, or ligated , to a bacterial vector in order to be replicated and segregated during bacterial growth. 2. The DNA must be transformed into bacterial host to create the genomic library . 3. The transformants must be identified and physically separated. 4. The specific gene of interest must be identified among the members of the library.
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A/C
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Restriction Enzymes
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Variety of Restriction Enzyme Sites
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Gel Electrophoresis Separating DNA Fragments by Size
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Agarose Gel Electrophoresis When complexed with DNA, the stain, ethidium bromide, fluoresces orange when excited by uv light. The stain can be added to the gel, and the ethidium bromide-DNA complex can be visualized by placing the gel on a uv light box. The DNA migrates as a function of size, with smaller fragments migrating through the matrix faster than larger fragments. The intensity of the staining of fragments is a function of size also, with larger fragments absorbing more stain per molecule than smaller fragments. Therefore, it a large DNA molecule is digested into fragments with a restriction enzyme, the larger fragments will stain more heavily than the smaller ones. _ + small large
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The genome of phage consists of approximately 48,000 bp. If it were digested with a restriction enzyme that recognizes a 6 bp site, how many fragments would be expected? a)8000 b)4000 c)1000 d)48 e)12
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The genome of consists of approximately 48,000 bp. If it were digested with a restriction enzyme that recognizes a 6 bp site, how many fragments would be expected? a)8000 b)4000 c)1000 d)48 e)12 The likelihood of any 6 bp sequence matching a 6 bp site is (1/4) 6 or 1/4096.
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The genome of consists of approximately 48,000 bp. If it were digested with a restriction enzyme that recognizes a 6 bp site, how many fragments would be expected? a)8000 b)4000 c)1000 d)48 e)12 The likelihood of any 6 bp sequence matching a 6 bp site is (1/4) 6 or 1/4096. The number of sites that would appear in a 48,000 bp sequence would be: (48,000)(1/4096) = 11.7 or 12 fragments.
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The genome of consists of approximately 48,000 bp. If it were digested with a restriction enzyme that recognizes a 6 bp site, how many fragments would be expected?
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  • Fall '08
  • ZITOMER
  • molecular biology, Genetics, Bacteria, DNA, RNA, Stop Stop, T L T C Y F R P T T R F S Stop Y W H L C T N T L Q L D Q S V C Q R Y H A S L D Q P Y V G H P F P R C G, R F G L L P S S V N P G Q R D R K D R Stop K H L D L A P G N F R F, E Q W N R K L Stop T K K G K Q E R K K F L G S R T I G R K N R Q R T N N G K T K K K K K N T E K N A E R C Q L K K K K V N T G K K I K K K K K K G G

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