6.1, 6.2, 6.3 & 6.4 expression of Biological Information

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6.0 EXPRESSION OF BIOLOGICAL INFORMATION:

6.0 EXPRESSION OF BIOLOGICAL INFORMATION 6.1 DNA and Genetic Information. 6.2 DNA Replication. 6.3 Protein Synthesis: Transcription & Translation. 6.4 Gene Regulation and Expression- lac operon.

OBJECTIVES:

OBJECTIVES Explain DNA as the carrier of genetic information. Explain gene concept: one gene one polypeptide. Describe semi-conservative replication of DNA. Describe DNA replication during cell division.

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Griffith’s experiment (1931). Avery, MacLeod & McCarty’s experiment (1944). Hershey & Chase’s experiment (1952). 3 DNA AS THE CARRIER OF GENETIC INFORMATION

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4 Transformation of Bacteria In 1931, bacteriologist Frederick Griffith experimented with Streptococcus pneumoniae (pneumococcus) that causes pneumonia in mammals. 2. Griffith injected mice with two strains of pneumococcus: an encapsulated (S) strain and non-encapsulated (R) strain.

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5 The experiment done by Frederick Griffith Source: Campbell & Reece 6th edition

RESULT:

RESULT a) The S strain is virulent (the mice died); it has a mucous capsule and forms shiny colonies (smooth surface). b. The R strain is not virulent (the mice lived); it has no capsule and forms dull colonies (rough surface).

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c) In an effort to determine if the capsule alone was responsible for the virulence of the S strain, he injected the mice with heat-killed S strain bacteria; the mice lived. d) Finally, he injected mice with a mixture of heat-killed S strain and live R strain bacteria. The mice died and living S strain pneumococcus were recovered from their bodies.

conclusion from the experiment:

conclusion from the experiment Griffith concluded some substance necessary to synthesis of the capsule and, therefore, virulence must pass from dead S strain bacteria to living R strain bacteria so the R strain were transformed. This change in phenotype of the R strain bacteria must be due to a change in their genotype, which suggested that the transforming substance may have passed from S strain to R strain.

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Question: Was the “transforming agent” protein, RNA or DNA, lipids or carbohydrate?

Avery, MacLeod & McCarty (1944):

Avery, MacLeod & McCarty (1944)

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Dr Oswald T. Avery Maclyn McCarty Colin M. MacLeod 1944 O.T.Avery et. al- to identify the chemical substance causing the transformations.

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12 Source: www.visionlearning.com

RESULT:

RESULT The “transforming agent” in the liquid was DNA. To further demonstrate this, the scientists took liquid extracted from heat-killed S. pneumoniae (S strain) and subjected it to extensive preparation and purification, isolating only the pure DNA from the mixture. This pure DNA was also able to transform the R strain into the S strain and generate pathogenic S. pneumoniae . 13

CONCLUSION FROM THE EXPERIMENT:

Their experimental results demonstrated DNA is genetic material and DNA controls biosynthetic properties of a cell. CONCLUSION FROM THE EXPERIMENT

HERSHEY & CHASE (1956):

HERSHEY & CHASE (1956)

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16 Source: Campbell & Reece 7th edition

RESULT:

RESULT When only phage protein coats were labeled, most of the radioactivity was detected outside the cells. But when phage DNA was labeled, most of the radioactivity was detected inside the cells. 17

CONCLUSION FROM THE EXPERIMENT:

CONCLUSION FROM THE EXPERIMENT The phage's DNA entered the bacterial cell during infection, but the proteins did not. DNA must carry the genetic information responsible for producing new phages. Their results convinced the scientific world that DNA was the hereditary/genetic material. 18

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19 GENE CONCEPT: ONE GENE ONE POLYPEPTIDE

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20

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21 The Beadle and Tatum experiment that suggested the one gene one enzyme hypothesis Source: www.emc.maricopa.edu

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22 Conclusion: only the medium containing the amino acid arginine supports growth. this indicates that mutations effect some part of arginine biosynthesis.

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23 Biosynthesis of amino acids (the building blocks of proteins) is a complex process with many chemical reactions mediated by enzymes, which if mutated would shut down the pathway, resulting in no-growth.

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24 Beadle and Tatum began to search for mutants of a bread mold, Neurospora crassa . They discovered mutants that differed from the wild-type mold in their nutritional needs ( Fig. 5.1.3 ) Source: Campbell & Reece 6th edition

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25 Beadle & Tatum deduced that each mutant was unable to carry out one step in the pathway for synthesizing Arginine. Source: Campbell & Reece 6th edition

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26 Beadle and Tatum proposed the “ one gene one enzyme “ hypothesis. One gene codes for the production of one protein( one gene one protein hypothesis). "One gene one protein" has since been modified to "one gene one polypeptide" since many proteins (such as hemoglobin) are made of more than one polypeptide. They won The Nobel prize in 1958.

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27

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LECTURE 2 & 3 DNA REPLICATION

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INTRODUCTION DNA replication is the process of copying a DNA molecule. We will approach the study of the molecular mechanism of DNA replication from the point of view of the machinery that is required to accomplish it.

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Three possible ways in which DNA can replicate are illustrated. The two original strands of DNA are shown in yellow (light); newly synthesized DNA is blue (dark) Conservative replication Dispersive replication Semiconservative replication Possible Models for DNA Replication

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6 Source: www.accessexcellence.org

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Conservative replication would leave intact the original DNA molecule and generate a completely new molecule. Dispersive replication would produce two DNA molecules with sections of both old and new DNA interspersed along each strand. Semiconservative replication would produce molecules with both old and new DNA, but each molecule would be composed of one old strand and one new strand.

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DNA replication is semiconservative meaning that the replicated DNA consists of one old strand (derived from the old molecule) and one new strand.—Jackie H. Pink Team DNA replication is semiconservative meaning that the replicated DNA consists of one old strand (derived from the old molecule) and one new strand. Source: www.sparknotes.com

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Source: Campbell & Reece 6th edition

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Meselson and Stahl experiment : using the radioisotope labelling technique to show the semiconservative model of DNA replication.

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Source: Campbell & Reece 6th edition

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REPLICATION OF DNA: THE MECHANISM

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The Enzymes & Proteins of DNA Replication Topoisomerase/ DNA Gyrase. DNA Helicase DNA polymerase Primase DNA Ligase Single-stranded binding proteins

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Topoisomerase/ DNA Gyrase is the enzyme that relieves tension to the DNA molecule by nicking and cutting certain placed on the phosphate backbone. DNA Helicase is the next enzyme that is involved in "unwinding" the double strand DNA to produce two single strands of DNA. Primase synthesize the RNA primer. DNA polymerase III uses the RNA primer to start adding new nucleotides on the single strand of DNA. DNA polymerase I remove the RNA primer and fill the gap with DNA nucleotides. DNA Ligase joins the Okazaki fragments by forming phosphodiester bond to form continuous DNA strands. Single Stranded Binding Protein (SSBp) help to stabilize the unwound DNA strand.

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A few things to remember for synthesizing DNA 1. Both strands of parental DNA serve as templates for the synthesis of new DNA. DNA replication starts at a unique origin ( oriC ) and proceeds sequentially in opposite direction (bidirectional)  Prokaryotic cell usually have only one origin of replication on each circular DNA molecule. In eukaryotic cell, the process is speeded up by having multiple origin of replication. 3. Parental DNA is unwound and new DNA is synthesized at replication fork.

Origin of replication:

Replication begins at specific sites where two parental strands separate & form replication bubbles. The bubbles expand laterally, as DNA replication proceeds in both direction. Eventually, the replication bubbles fuse, & synthesis of daughter strands is complete. Origin of replication Source: Campbell & Reece 6th edition

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4. The overall direction of DNA synthesis must be 5’  3’ because the DNA polymerases only synthesize DNA in the 5’  3’ direction. 5. DNA synthesis is primed by RNA (RNA primer). 6. One strand of DNA is synthesized continously (leading strand). 7 Meanwhile the other strand of DNA is synthesized discontinously (lagging strand).

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STEPS OF DNA REPLICATION

UNWINDING THE DOUBLE HELIX STRAND OF DNA:

UNWINDING THE DOUBLE HELIX STRAND OF DNA DNA Helicase will unwind the double strand of DNA by breaking the hydrogen bonds between both strand. Each strand will act as a template. The unwound strands is stabilized by single- strand binding proteins – prevents the double helix from reforming until the strands are copied. In advance of the replication fork, topoisomerase/ DNA gyrase diminishes the tension that is created as the helix supercoils.

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Helicase Single-stranded binding proteins Source: Campbell & Reece 6th edition

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Source: http://openwetware.org/wiki/IGEM:IMPERIAL/2007 Source: www.biochem.arizona.edu

BUILDING A PRIMER:

RNA primase adds a short RNA primer. New DNA cannot be synthesized on the exposed templates until a primer is constructed, as DNA polymerases require 3’ primers to initiate replication. BUILDING A PRIMER Source: http://classes.midlandstech.edu

ASSEMBLING COMPLEMENTARY STRANDS:

ASSEMBLING COMPLEMENTARY STRANDS DNA Polymerase III add free DNA nucleotides to the RNA primer complementary to the bases of the template. Source: www.gatewaycoalition.org

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Fig. 14.14

The movement of DNA polymerase…:

Along one template strand, DNA polymerase III can synthesize a complementary strand continuously by elongating the new DNA in the 5’3’ direction (toward the replication fork). The DNA strand made by this mechanism is called leading strand. To elongate the other new strand in the 5’3’ direction, DNA polymerase III must work along the other template strand in the direction away from the replication fork. The DNA strand synthesized in this direction is called the lagging strand. The movement of DNA polymerase…

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Both strands require RNA primer for initiation of synthesis because DNA can be elongated only by addition to 3’ end of existing polynucleotide strand. Source: www.valuemd.com

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Lagging strand is synthesize as short Okazaki fragments. Okazaki fragment synthesis begins with the synthesis of RNA primer. Note that first Okazaki fragment synthesized is now at far left. Source: www.valuemd.com

REMOVING THE PRIMER:

REMOVING THE PRIMER The enzyme DNA polymerase I now removes the RNA primer and fills in the gap with DNA nucleotides, as well as any gaps between Okazaki fragments.

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Source: www.uic.edu

JOINING THE OKAZAKI FRAGMENTS:

JOINING THE OKAZAKI FRAGMENTS DNA ligase joins the Okazaki fragments with phosphodiester bonding forming a single new DNA strand. Two identical copies of DNA is produced.

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Source: Campbell & Reece 8th edition

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CONCLUSION DNA replication involves many different proteins that open and unwind the DNA double helix, stabilize the single strands, synthesize RNA primers, assemble new complementary strands on each exposed parental strand – one of them discontinuously – remove the RNA primer, and join new discontinuous segments on the lagging strand.

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LECTURE 4 & 5 PROTEIN SYNTHESIS

OBJECTIVES:

OBJECTIVES Overview the roles of transcription & translation in the flow of genetic information. Explain transcription. Describe the stages involved: i) initiation. ii) elongation. iii) termination. State the formation of mRNA strand from 5’ to 3’.

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Fig. 15.5

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CENTRAL DOGMA OF MOLECULAR BIOLOGY (EUKARYOTES) Source: Campbell & Reece 6th edition

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CENTRAL DOGMA OF MOLECULAR BIOLOGY (proKARYOTES) Source: Campbell & Reece 6th edition

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TRANSCRIPTION

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TRANSCRIPTION DEFINITION The process in which mRNA is synthesized from a DNA template.  Occurs inside the nucleus. The enzyme-catalyzed assembly of an RNA molecule complementary to a strand of DNA.

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STEP 1 (TRANSCRIPTION) INITIATION

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The presence of a promoter sequence determines which strand of the DNA helix is the template. Within the promoter is the starting point for the transcription of a gene. The promoter also includes a binding site for RNA polymerase several dozen nucleotides upstream of the start point. In prokaryotes, RNA polymerase can recognize and bind directly to the promoter region. Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

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Source: Campbell & Reece 5th edition

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In eukaryotes, proteins called transcription factors recognize the promoter region, a TATA box, and bind to the promoter. After they have bound to the promoter, RNA polymerase binds to transcription factors to create a transcription initiation complex. RNA polymerase then starts transcription. Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings Source: Campbell & Reece 5th edition

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STEP 2 (TRANSCRIPTION) ELONGATION

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As RNA polymerase moves along the DNA, it continues to untwists the double helix, exposing about 10 to 20 DNA bases at a time for pairing with RNA nucleotide. The enzyme adds nucleotides to the 3’ end of the growing RNA strand. Behind the point of RNA synthesis, the double helix re-forms and the RNA molecule peels away. Fig. 17.6b Source: Campbell & Reece 6th edition

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Fig. 15.8 15 25

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STEP 3 (TRANSCRIPTION) TERMINATION

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Transcription proceeds until after the RNA polymerase transcribes a terminator sequence in the DNA. In prokaryotes, RNA polymerase stops transcription right at the end of the terminator. Both the RNA and DNA is then released. Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

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In eukaryotes, the polymerase continues transcribing for hundreds of nucleotides past the terminator sequence, AAUAAA.  At a point about 10 to 35 nucleotides past this sequence, AAUAA, the pre-mRNA is cut from the enzyme.

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POST-TRANSCRIPTIONAL PROCESS

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The modification of mRNA within a eukaryotic nucleus is known as post-transcriptional processing (PTM).  During this RNA processing, both ends of the mRNA are usually altered.  Also, certain sections of the molecule (mRNA) are cut out & the remaining parts spliced together. It results in a fully functional mRNA ready for export to the cytoplasm and ribosome.

Alteration of mRNA ends:

Alteration of mRNA ends At the 5’ end of the pre-mRNA molecule, a modified form of guanine is added, the 5’ cap . This helps protect mRNA from degradation by hydrolytic enzymes. It also functions as an “attach here” signal for ribosomes. Source: Campbell & Reece 8th edition

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At the 3’ end, an enzyme adds 50 to 250 adenine nucleotides, the poly(A) tail . In addition to inhibiting hydrolysis and facilitating ribosome attachment, the poly(A) tail also seems to facilitate the export of mRNA from the nucleus. The mRNA molecule also includes nontranslated leader and trailer segments. Source: Campbell & Reece 6th edition

Split Genes & RNA Splicing:

The most remarkable stage of RNA processing occurs during the removal of a large portion of the RNA molecule during RNA splicing. Most eukaryotic genes and their RNA transcripts have long noncoding stretches of nucleotides. Noncoding segments, introns , lie between coding regions. The final mRNA transcript includes coding regions, exons , that are translated into amino acid sequences, plus the leader and trailer sequences. Split Genes & RNA Splicing

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Fig. 15.15a

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Fig. 15.15b

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Fig. 17.9 RNA splicing removes introns and joins exons to create an mRNA molecule with a continuous coding sequence. POST-TRANSCRIPTIONAL PROCESS Source: Campbell & Reece 6th edition

GENETIC CODE:

GENETIC CODE

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Information are transferred in the form of codes. This is ‘language’ of the genes, also known as the genetic code. There are 20 different amino acids that make up of proteins in the body. But there are only 4 different bases in the DNA and RNA. This would mean that one base cannot code for one amino acid since there are 20 amino acids available. If taking only 2 bases give 16 (42) combination which does not account for all the amino acid available. A sequence of 3 bases seem the most probable since it can give 64 (43) possible combination of bases, this could count for the twenty amino acids available in organism.

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A sequence of three bases (triplet code) in mRNA is called a codon = one codon = one amino acid but one amino acid may be coded by more than one codon. Codon not only specify for protein but also responsible as codes for starting or stopping polypeptide sequence.

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Starting codon The starting point of synthesis is determined by the first Codon AUG on the mRNA. 2. Stop codon / nonsense codon Three triplet bases of stop signal: UGA , UAG,UAA . The newly synthesized protein and the mRNA are released when the ribosome encounters a stop codon. Nonsense codon because do not code for any amino acid.

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Fig. 15.6b

Characteristic of genetic code:

Characteristic of genetic code Genetic code is almost universal. Same for all organism (bacteria, plant or animal. However, the genetic code is not quiet universal. eg: Mitochondrial genome (human, mice, cattle). UGA read as the tryptophan.

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In laboratory experiments, genes can be transcribed and translated after they are transplanted from one species to another. This tobacco plant is expressing a transpired firefly gene. Source: Campbell & Reece 6th edition

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Genetic code is degenerate. Amino acids may be coded by more than one codon. From table; more codons than amino acids. Eg : UCU ,UCC, UCA, UCG, AGU, AGC – code for serine.

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3. The genetic code is unambiguous  each triplet codon has only one meaning. 4. The code has start and stop signals :  there is one start codon and three stop codons.

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DEFINITION The process by which the transcribed information carried in the base sequence of mRNA ( codon ) is used to produce a sequence of amino acids in a polypeptide chain. Occurs inside the cytoplasm. TRANSLATION

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Types of RNA involve in translation process mRNA tRNA rRNA

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TRANSFER RNA ( tRNA)

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A tRNA molecule consists of a strand of about 80 nucleotides that folds back on itself to form a three-dimensional structure. It includes a loop containing the anticodon and an attachment site at the 3’ end for an amino acid. Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings Source: Campbell & Reece 6th edition

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Each amino acid is joined to the correct tRNA by aminoacyl-tRNA synthetase. The 20 different synthetases match the 20 different amino acids. Each has active sites for only a specific tRNA and amino acid combination. The synthetase catalyzes a covalent bond between them, forming aminoacyl-tRNA or activated amino acid. Fig. 17.14 Source: Campbell & Reece 6th edition

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RIBOSOMAL RNA ( rRNA)

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formed inside the nucleolus. large single-stranded RNA. consists of : - small subunit (40S) - large subunit (60S) 80% from the whole RNA produced –the most abundant RNA in cell. composed of proteins. Ribosomal RNA (rRNA) Source: www.emc.maricopa.edu

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Each ribosome has a binding site for mRNA and three binding sites for tRNA molecules. The P site holds the tRNA carrying the growing polypeptide chain. The A site carries the tRNA with the next amino acid. Discharged tRNAs leave the ribosome at the E site . Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings Source: Campbell & Reece 6th edition

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TRANSLATION

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STAGE 1 (TRANSLATION) INITIATION

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Initiation brings together mRNA, a tRNA with the first amino acid, and the two ribosomal subunits. First, a small ribosomal subunit binds with mRNA and a special initiator tRNA, which carries methionine ( Met - UAC ) and attaches to the start codon (AUG). -The union of mRNA, initiator tRNA and small ribosomal subunit is followed by the attachment of a large ribosomal subunit completing a translation initiation complex. Initiation factors are required to bring all the translation complex together. Initiator tRNA occupies in the P site of ribosome and vacant A site is ready for next tRNA. Fig. 17.17

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INITIATION PROCESS OF TRANSLATION Source: Campbell & Reece 6th edition

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STAGE 2 (TRANSLATION) ELONGATION

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Elongation consists of a series of three step cycles as each amino acid is added to the proceeding one. i) codon recognition ii) peptide bond formation iii) translocation Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings

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62 63 Source: Campbell & Reece 6th edition

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A tRNA molecule with an anticodon complementary to the exposed mRNA codon binds at A site. mRNA codon in A site forms hydrogen bonds with the anticodon of an incoming molecule of tRNA carrying its appropriate amino acid. - This step requires the hydrolysis of two GTP. i) Codon Recognition Source: www.emc.maricopa.edu

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During peptide bond formation , an rRNA molecule catalyzes the formation of a peptide bond between the polypeptide in the P site with the new amino acid in the A site. This step separates the tRNA at the P site from the growing polypeptide chain and transfers the chain, now one amino acid longer, to the tRNA at the A site. ii) Peptide Bond Formation Source: Campbell & Reece 6th edition

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During translocation , the ribosome moves the tRNA with the attached polypeptide from the A site to the P site ( the anticodon remains bonded to the mRNA codon). The next codon is now available at the A site. The tRNA that had been in the P site is moved to the E site and then leaves the ribosome. Translocation is fueled by the hydrolysis of GTP. Effectively, translocation ensures that the mRNA is “read” 5’ -> 3’ codon by codon. Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings (iii) Translocation

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Fig. 15.13

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The three steps of elongation continue codon by codon to add amino acids until the polypeptide chain is completed. Source: Campbell & Reece 6th edition

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STAGE 3 (TRANSLATION) TERMINATION

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Termination occurs when one of the three stop codons reaches the A site. A release factor binds to the stop codon and hydrolyzes the bond between the polypeptide and its tRNA in the P site. This frees the polypeptide and the translation complex disassembles. Copyright © 2002 Pearson Education, Inc., publishing as Benjamin Cummings Source: Campbell & Reece 6th edition

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POLYRIBOSOME

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Typically a single mRNA is used to make many copies of a polypeptide simultaneously. Multiple ribosomes, polyribosomes , may trail along the same mRNA. A ribosome requires less than a minute to translate an average-sized mRNA into a polypeptide. Source: Campbell & Reece 6th edition

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Fig. 15.10

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Increase the rate of same protein/polypeptide synthesis. More of same protein can be made simultaneously. More of same protein in a shorter time/rapidly.

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LACTOSE OPERON

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OBJECTIVES DESCRIBE the components of lac operon and its function in E . coli . DESCRIBE the mechanism of the operon in the absence and presence of lactose.

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INTRODUCTION What is operon? In 1961, French microbiologist Francois Jacob and Jacques Monod proposed operon model to explain the regulation of gene expression in prokaryotes; they received a Nobel prize for this. a) Operon model: several gene codes for an enzyme in same metabolic pathway and are located in sequence on chromosome; expression of structural genes controlled by same regulatory genes. www. https://bms.ncl.ac.uk/wiki

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Operon models Source: Campbell & Reece 6th edition

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b) An operon is a functioning unit of nucleotide sequences of DNA including an operator, a promoter, and one or more structural genes, which is controlled as a unit to produce messenger RNA (mRNA), in the process of transcription by an RNA polymerase. www. https://bms.ncl.ac.uk/wiki

THE COMPONENTS OF OPERON:

THE COMPONENTS OF OPERON i) A promoter is a sequence of DNA where RNA polymerase attaches when a gene is transcribed/ where RNA polymerase first binds. Source: http://en.wikipedia.org

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ii) An operator is a short sequence of DNA where repressor binds, preventing RNA polymerase from attaching to the promoter. A region between the promoter and the first gene which acts as an “on-off switch”. iii) Structural genes code for enzymes of a metabolic pathway; are transcribed as a unit. Source: http://en.wikipedia.org

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The operon may also contain regulatory genes which codes for a repressor protein that binds to the operator and inhibits transcription. www. https://bms.ncl.ac.uk/wiki

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The operon that regulates lactose metabolism in the bacterium Escherichia coli. In bacteria, the genes coding for the enzymes of a particular pathway are clustered together and transcribed into a single polycistronic mRNA molecule. The Lactose Operon ( in E.coli ) * polycistronic= carries the information for more than one type of protein.

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The Lactose Operon ( in E.coli ) The lactose operon consists of three structural genes (lac Z, lacY, and lacA ), lacP (promoter) and lacO (operator) . If E. coli is denied glucose and given lactose instead, it makes three enzymes to metabolize lactose.

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These three enzymes are encoded by three genes ( lacZ, lacY and lacA ). lacZ gene codes for β-galactosidase that hydrolyzes lactose to glucose and galactose. lacY gene codes for a permease that transports lactose into the cell. lacA gene codes for enzyme transacetylase , whose function in lactose metabolism is still unclear.

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Fig. 16.9

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The lac operon is said to be inducible operon because its transcription is usually off but can be stimulated (induced) when a specific small molecule (inducer) interacts with a repressor protein (product of regulatory gene, lacI ).

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LacI is a gene encodes for repressor and will be transcribed into mRNA and translated into repressor protein. Its binds to the operator site and no transcription occurs.

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Fig. 16.11a

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Source: www.phschool.com

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THE MECHANISM OF THE LAC OPERON a) In the absence of lactose In the absence of lactose, the repressor protein is able to bind to the operator ( lac O ). Prevent the binding of RNA polymerase to the promoter. No transcription of the genes lac Z, lac Y & lac A. No mRNA is produced. The enzymes β-galactosidase, permease & transacetylase are not produced. Source: Campbell & Reece 6th edition

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Source: www.phschool.com

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Fig. 18.21b When lactose is present in the cell, allolactose, an isomer of lactose are formed. b) In the presence of lactose Source: Campbell & Reece 6th edition

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Allolactose binds to an allosteric site on the repressor protein causing a conformational change. The repressor can’t longer bind to the operator. RNA polymerase can then bind to the promoter & transcribe the lac genes ( lac Z, lac Y & lac A). mRNA is produced & translation occurs. The enzymes β-galactosidase, permease & transacetylase are produced. Source: Campbell & Reece 6th edition

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Source: www.phschool.com

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CONCLUSION

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If an E. Coli cell growing in the absence of lactose , a repressor protein binds to the operator, preventing RNA polymerase from transcribing the lac operon's genes. The operon is OFF.

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Source: Campbell & Reece 6th edition

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When the inducer, lactose , is added, its isomer, allolactose, binds to the repressor and changes the repressor's shape so as to eliminate binding to the operator. As long as the operator remains free of the repressor, RNA polymerase that recognizes the promoter can transcribe the operon's structural genes into mRNA. The operon is ON.

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Source: Campbell & Reece 6th edition

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