C6 - EXPRESSION OF BIOLOGICAL INFORMATION_edited latest selepas bengkel.pdf

CHAPTER 6 EXPRESSION OF BIOLOGICAL INFORMATION (Hours: 2L + 8T) CHAPTER 6: EXPRESSION OF BIOLOGICAL INFORMATION 6.1 DNA and genetic information 6.2 DNA Replication 6.3 Protein synthesis: transcription and translation 6.4 Gene regulation and expression – lac oper...

CHAPTER 6 EXPRESSION OF BIOLOGICAL INFORMATION (Hours: 2L + 8T) CHAPTER 6: EXPRESSION OF BIOLOGICAL INFORMATION 6.1 DNA and genetic information 6.2 DNA Replication 6.3 Protein synthesis: transcription and translation 6.4 Gene regulation and expression – lac operon UPS 1 PSPM 1 7 MCQ 14 marks  LEARNING OUTCOMES 6.1 DNA and genetic information a) State the concept of Central Dogma (C1) Learning Outcomes : 6.1 (a) : State the concept of Central Dogma Central Dogma Definition: The flow of genetic information from DNA to RNA to protein in living organisms. Learning Outcomes : 6.1 (a) : State the concept of Central Dogma Central Dogma Learning Outcomes : 6.1 (a) : State the concept of Central Dogma Central Dogma DNA information can be copied into mRNA during transcription. mRNA as a template carries coded information from DNA (nucleus) to cytoplasm & ribosomes use it for protein synthesis. This process is called translation.  LEARNING OUTCOMES 6.2 DNA Replication a) Explain semiconservative replication of DNA. (C3) b) Explain the enzymes and proteins involved in DNA replication. (C3) c) Explain the mechanism of DNA replication and the enzymes involved. (C3) Learning Outcome: 6.2 (a) : Explain semiconservative replication of DNA DNA Replication Definition: The process by which a DNA molecule is copied/synthesize. DNA Replication Models Conservative Semiconservative Dispersive Learning Outcome: 6.2 (a) : Explain semiconservative replication of DNA Semiconservative Model Semiconservative Model The two strands of the parental *Note: molecules separate, and each Blue (parental strand) functions as a template for synthesis Red (new strand) of a new, complementary strand. Learning Outcome: 6.2 (a) : Explain semiconservative replication of DNA Learning Outcomes : 6.2 (b) : Explain the enzymes and proteins involved in DNA Replication Enzymes and Protein in DNA Replication Unwinds parental double helix at replication DNA forks, separating the two parental strands helicase and making them available as template strands Binds to the unpaired DNA strand keeping them from re-pairing. The untwisting of the double helix causes tighter twisting ad strain ahead of replication fork Help relieve this strain by breaking, swiveling and rejoining DNA strands Synthesizes an RNA primer at 5’ end of leading strand and at 5’ end of each Okazaki fragment of lagging strand Learning Outcomes : 6.2 (b) : Explain the enzymes and proteins involved in DNA Replication Enzymes and Protein in DNA Replication DNA polymerase Using parental DNA as a template, synthesizes new DNA strand by adding DNA nucleotides to an III RNA primer or a pre-existing DNA strand DNA polymerase Remove RNA nucleotides of primer from I 5’ end and replaces with DNA nucleotides added to 3’ end of adjacent fragment Joining the sugar phosphate backbone of all Okazaki fragments into continuous DNA strand Learning Outcomes : 6.2 (b) : Explain the enzymes and proteins involved in DNA Replication Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved DNA Replication In eukaryotes, it occurs in nucleus. Replication of DNA begins at origins of replication, short DNA segment that have a specific sequence of nucleotides. Eukaryotic chromosome may have hundreds or thousand replication origins. Replication of DNA proceeds in both directions until the entire molecule is copied. At each end of a replication bubble is a replication fork, a Y-shaped region where the parental strands of DNA are being unwound. Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Figure 8 8 Origins of replication in a eukaryotic cell  Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Learning Outcomes : 4.1 (c) Explain the Mendel’s experiments on monohybrid cross Mechanism of DNA Replication The process involves are : Step 1: Unwinding the double helix strand of DNA Step 2: Synthesis of RNA primer Step 3: Assembling complementary strands Step 4: Formation of leading & lagging strands Step 5: Removing the RNA primer Step 6: Joining the Okazaki fragments Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Mechanism of DNA Replication Step 1: Unwinding the double helix strand of DNA ⮚ DNA helicase unwinds parental double helix at replication forks, by breaking hydrogen bond ⮚ Separating the two parental strands and making them available as template strands Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Step 1: Unwinding the double helix strand of DNA ⮚ Single-strand binding protein bind to the unpaired DNA strands, prevent them from re-pairing. Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Step 1: Unwinding the double helix strand of DNA ⮚ When DNA is unwound, it becomes more twisted / supercoiling ⮚ Topoisomerase helps relieve strain / tension ahead of the replication fork by breaking, swiveling and rejoining DNA strands ⮚ by cutting / nicking certain place on one of the DNA strand Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Mechanism of DNA Replication Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Mechanism of DNA Replication Step 2: Synthesis of RNA primer ⮚ Primase synthesizes an RNA primer at 5’ end of leading strand and at 5’ end of each Okazaki fragment of lagging strand ⮚ by adding RNA nucleotides that is complementary to the DNA template from 5’ to 3’ Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Step 2: Synthesis of RNA primer Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Mechanism of DNA Replication Step 3: Assembling complementary strands ⮚ DNA Polymerase III - catalyze the synthesis of new DNA strand by adding DNA nucleotides to the 3′ end of RNA primer ⮚ Nucleotides added is complementary to the DNA template Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Step 3: Assembling complementary strands Mechanism of DNA Replication ⮚ DNA replication occurs from 5’ to 3’ since new DNA nucleotides are added only to the 3’ end of the growing strand. Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA Replication and the enzymes involved Mechanism of DNA Replication Step 4: Formation of leading & lagging strands ⮚ Because of their antiparallel structure, DNA polymerase III can add DNA nucleotides only to the free 3′ end of RNA primer or growing DNA strand, never to the 5′ end. ⮚ Thus, a new DNA strand can elongate only in the 5′ to 3′ direction. Upon replication, two strands will be produced; leading and lagging strand. Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA Replication and the enzymes involved Step 4: Formation ofReplication leading & lagging strands Mechanism of DNA Leading strand ⮚ DNA polymerase III can synthesize a complementary strand continuously by elongating the new DNA in the 5′ to 3′ direction towards replication fork. ⮚ Only 1 RNA primer is required for DNA Polymerase III to synthesize the entire leading strand. Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA Replication and the enzymes involved Lagging strand ⮚ Synthesized discontinuously, away from the replication fork, as a series of segments called Okazaki fragments. ⮚ More than 1 RNA primer is needed. Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA Replication and the enzymes involved Mechanism of DNA Replication Step 5: Removing the RNA primer DNA Polymerase I remove RNA nucleotides of primer from 5’ end and replaces with DNA nucleotides added to 3’ end of adjacent fragment Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA Replication and the enzymes involved Mechanism Mechanism ofof DNADNA Replication Replication Step 6: Joining the Okazaki fragments ⮚ DNA Ligase join the Okazaki fragments into a continuous DNA strand by forming phosphodiester bond ⮚ Two identical copies of DNA are produced. Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA Replication and the enzymes involved CONCLUSION Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA Replication and the enzymes involved CONCLUSION  LEARNING OUTCOMES 6.3: Protein Synthesis: Transcription And Translation a) Explain briefly transcription and translation. (C3) b) Introduce codon and its relationship with sequence of amino acid using genetic code table. (C2) c) Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’. (C3) d) Explain translation and the stages involved in translation: (C3) i. Initiation ii. Elongation (codon recognition, peptide bond formation and translocation); and iii. Termination Learning Outcomes : 6.3 (a) : Explain briefly transcription and translation Transcription And Translation Transcription is the synthesis (production) of RNA using DNA template. In eukaryotes, it occurs in nucleus. Translation is the synthesis of a polypeptide using the information encoded in the mRNA. In eukaryotes, it occurs in cytoplasm. Learning Outcomes : 6.3 (b) : Introduce codon and its relationship with sequence of amino acid using genetic code table Codon and Genetic Code Table Information is transferred in the form of genetic code. A sequence of three bases (triplet code) in mRNA is called a codon and they are written in the 5′ to 3′ direction Learning Outcomes : 6.3 (b) : Introduce codon and its relationship with sequence of amino acid using genetic code table Learning Outcomes : 6.3 (b) : Introduce codon and its relationship with sequence of amino acid using genetic code table Characteristics of codon in Genetic code ⮚ 3 nucleotides in mRNA is called a codon A sequence of 3 bases is the most possible since it can give 64 (43) combinations of bases since there are 20 different amino acids ⮚ Each codon has only one meaning that encode for one specific amino acid ⮚ Amino acids may be coded by more than one codon. An amino acid can be specified by more than 1 triplet codon. Learning Outcomes : 6.3 (b) : Introduce codon and its relationship with sequence of amino acid using genetic code table ⮚ The genetic codes is nearly universal, shared by organisms from the simplest bacteria to the most complex plants and animals. ⮚ There is one start codon (AUG) & three stop codons (UAA, UAG, UGA). ⮚ Sequence of codon is non-overlapping. Read continuously from a fixed starting point until it reaches the termination point Learning Outcomes : 6.3 (b) : Introduce codon and its relationship with sequence of amino acid using genetic code table Figure 26 8 Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ Transcription Definition: Transcription is the synthesis (production) of RNA using DNA template. Transcription is a process by which genetic information contained in DNA is transcribed into mRNA using RNA polymerase II. There are THREE stages of transcription Step 3:1:Termination Stage Initiation Step 3:2:Termination Stage Elongation Stage 3: Termination Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ Transcription Stage 1: Initiation RNA polymerase II binds to the promoter on DNA The promoter of a gene typically extends “upstream” from the start point. **RNA polymerase II enzyme used in eukaryotes Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ Transcription - Initiation Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ Transcription - Initiation Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ Transcription - Initiation Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ Transcription Stage 2: Elongation RNA polymerase II unwinds the double helix as it moves along the DNA It breaks the hydrogen bond and separate the two DNA strands RNA synthesis begins at the start point on the template strand Only one DNA strand act as template. RNA Polymerase II add a free RNA nucleotide which are complementary to the bases on DNA template strand It continues to add free RNA nucleotides from 5’ to the 3’ end of the growing mRNA molecule Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ Transcription - Elongation As transcription proceeds, the newly synthesized mRNA molecule behind the RNA polymerase II detaches from DNA template, and the DNA double helix re-forms. Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ Transcription Stage 3: Termination Elongation proceeds until RNA polymerase II reaches a sequence on the DNA called polyadenylation signal sequence (AAUAAA) in the pre-mRNA RNA polymerase stop adding nucleotides to the mRNA strand A certain protein cut the RNA transcript free from the RNA polymerase, releasing the pre-mRNA from the DNA template Pre-mRNA undergoes splicing process Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ Transcription - Termination Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ RNA Processing Enzymes modify the two ends of a eukaryotic pre-mRNA molecule. During this RNA processing, both ends of the primary transcript are altered. These modifications produce a mature mRNA molecule ready for translation. Involves 2 steps: a. Alteration of mRNA ends b. Split Genes and RNA Splicing Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ RNA Processing a. Alteration of mRNA ends The 5′ end receives a 5′ cap, a modified form of a guanine (G) nucleotide added onto the 5′ end after transcription At the 3′ end, an enzyme then adds 50–250 more adenine (A) nucleotides, forming a poly-A tail. 39 Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ RNA Processing a. Alteration of mRNA ends Function 5’ end cap and poly-A tail: 1. to modify the two ends of a pre-mRNA molecules 2. to promote the export of mRNA from nucleus 3. Help to protect mRNA from degradation 39 Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ RNA Processing b. Split Genes and RNA Splicing Introns - noncoding segments of nucleic acid that lie between coding regions. Exons - coding segments of nucleic acid that lie between noncoding regions, that are eventually expressed by being translated into amino acid sequences. Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ RNA Processing b. Split Genes and RNA Splicing RNA splicing remove introns and join exons to create an mRNA molecule with a continuous coding sequence Conducted by a large complex made of proteins and small RNAs called spliceosome. Learning Outcomes : 6.3 (c) : Explain transcription and the stages involved (initiation, elongation and termination) in the formation of mRNA strand 5’ to 3’ RNA Processing b. Split Genes and RNA Splicing Final mRNA transcript consist of coding segments (exons) with 5’ cap and poly-A tail 41 Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation Translation Definition: Translation is a process synthesis of a polypeptide using the genetic information encoded in an mRNA molecule. The genetic information is a series of codons along an mRNA molecule The translator is called a transfer RNA (tRNA). Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation Translation The function of tRNA is to transfer an amino acid from the cytoplasm to a growing polypeptide in a ribosome. The 3’ end of tRNA acts as the attachment site for an amino acid. Anticodon is the nucleotide triplet in tRNA that base-pairs to a specific mRNA codon in the middle loop. Anticodons are conventionally written 3’ → 5’ to align properly codons written 5’→ 3’. Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation Translation – Activation of amino acid Activation of amino acid is a process in which a specific amino acid is attached to a tRNA molecule to form aminoacyl-tRNA catalysed by aminoacyl-tRNA synthetase Amino acids are linked to their respective tRNA molecules by the formation of ester bond at the 3’end of tRNA ATP is used as energy source Each type of aminoacyl-tRNA synthetase has unique active site that fits only a specific combination of amino acid and tRNA. To ensure accurate translation, tRNA that binds to the codon in mRNA must carry the amino acid encoded by that codon to the ribosome Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation Translation – Activation of tRNA Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation Translation The stages involved in translation: Stage 1: Initiation Stage 2: Elongation Stage 3: Termination Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: 1) Initiation Translation Stage 1: Initiation Small ribosomal subunit binds with mRNA. Initiator tRNA which carries amino acid methionine (with anticodon (3’UAC5’) attaches to the start codon (5’AUG3’) at mRNA by forming hydrogen bonds. Large ribosomal subunit attaches with these complex to form translation initiation complex. Energy is provided by the hydrolysis of GTP Initiator tRNA is now located in the P site of the ribosome, A site is empty Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: 1) Initiation Translation Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: Elongation (codon recognition, peptide bond formation and translocation) Translation Stage 2: Elongation Consists of a series of three step cycles as each amino acid is added one by one to the C-terminus of the growing chain a) Codon recognition b) Peptide bond formation c) Translocation Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: Elongation (codon recognition, peptide bond formation and translocation) Translation Stage 2: Elongation – codon recognition During codon recognition, aminoacyl-tRNA with anticodon which is complementary to the mRNA codon enters to the empty A site Energy is supplied by hydrolysis of GTP in this step Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: Elongation (codon recognition, peptide bond formation and translocation) Translation Stage 2: Elongation – peptide bond formation Peptide bond is formed between amino acid in the P site & the new amino acid in A site Catalyzes by peptidyl transferase, a ribozymes from large ribosomal subunit This step removes the growing polypeptide from the tRNA in the P site and binds it to the new amino acid on the tRNA in the A site by a peptide bond Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: Elongation (codon recognition, peptide bond formation and translocation) Translation Stage 2: Elongation - translocation During translocation, the ribosome moves one codon ahead (from 5’ to 3’) tRNA in the P site is translocated to the E site & then leaves the ribosome. The tRNA (with the attached polypeptide) in the A site is translocated to the P site Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: Elongation (codon recognition, peptide bond formation and translocation) Translation Stage 2: Elongation - translocation The next codon is now available at the A site. The three steps of elongation continue codon by codon to add amino acids until the polypeptide chain is completed. Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: Termination Translation Stage 3: Termination Termination occurs when the A site of ribosome reaches one of the three stop codons (UAG/UAA/UGA) on mRNA. The stop codon (UAG, UAA, UGA) do not code for amino acids; thus no tRNA will bind to the stop codon A release factor binds directly to the stop codon in the A site It causes the addition of water molecule, hydrolyzes the bond between the polypeptide and the tRNA in the P site. Newly synthesized protein is released Translation complex dissociate with the hydrolysis of 54 GTP molecules. Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: Termination Translation 55 Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation: Translation Polyribosomes A single mRNA is used to make many copies of a polypeptide. Multiple ribosomes may attached on the same mRNA at a time It forms a structure called polyribosomes / polysomes They enable a cell to make many copies of a polypeptide rapidly 55  LEARNING OUTCOMES 6.4 Gene regulation and expression – lac operon a) Explain the concept of operon and gene regulation (C3) b) State the components of operon (C1) c) Explain the components of lac operon and their function in E. coli (C3) d) Explain the mechanism of the operon in the absence and presence of lactose (C3) 72 Learning Outcomes : 6.4 a) Explain the concept of operon and gene regulation Operon A series of structural genes expressed as a group controlled by the single operator and promoter Consists of promoter, operator and structural genes Structural genes function as a single transcription unit transcribed into single mRNA Only in prokaryotes (e.g. bacteria) Learning Outcomes : 6.4 a) Explain the concept of operon and gene regulation Gene Regulation Genes of related functions are grouped into one transcription unit These genes are regulated such that they are all turned on or off together These genes are coordinately controlled The on-off switch is a segment of DNA called an operator Learning Outcomes : 6.4 a) Explain the concept of operon and gene regulation Gene Regulation Operator can be switched off by repressor protein Repressor binds to operator and blocks the attachment of RNA polymerase to the promoter, preventing transcription of the genes The repressor is encoded by regulatory gene (outside the operon, has its own promoter) Learning Outcomes : 6.4 b) State the components of operon Component of Operon Operon includes the following: 1) Promoter 2) Operator 3) Structural genes; lacZ, lacY, lacA Learning Outcomes : 6.4 c) Explain the components of lac operon and their function in E.coli lac operon The operon model was proposed by Francois Jacob and Jacques Monod (1961) in E. coli Some of the bacterial genes will be expressed only when it is necessary Lactose (lac) operon: ⮚ Regulates lactose metabolism in E. coli (contains genes that encode for enzymes used in the hydrolysis and metabolism of lactose) Learning Outcomes : 6.4 c) Explain the components of lac operon and their function in E.coli Components of lac operon lac operon consists of: 1. Structural genes ⮚lacZ ⮚lacY 2. Promoter ⮚lacA 3. Operator Learning Outcomes : 6.4 c) Explain the components of lac operon and their function in E.coli Components of lac operon 1. STRUCTURAL GENES lacZ : codes for /encodes β-galactosidase lacY : codes for /encodes permease lacA : codes for /encodes transacetylase Learning Outcomes : 6.4 c) Explain the components of lac operon and their function in E.coli Function of enzyme Enzyme Function β-galactosidase Catalyse the hydrolysis of lactose into glucose and galactose Permease Facilitate the transport of lactose into E. coli / Increase permeability of membrane to lactose Transacetylase Enzyme that detoxifies other molecules entering the cell via permease Learning Outcomes : 6.4 c) Explain the components of lac operon and their function in E.coli Components of lac operon 2. PROMOTER Binding site for RNA polymerase 3. OPERATOR Binding site for repressor protein Act as switch; which activate/ inactivate the operon Learning Outcomes : 6.4 c) Explain the components of lac operon and their function in E.coli REGULATORY GENE / lacI Encodes for repressor protein Located outside the operon, has its own promoter gene Learning Outcomes : 6.4 d) Explain the mechanism of the operon in the absence and presence of lactose Mechanism of lac operon LACTOSE ABSENT Repressor protein is synthesized by regulatory gene and in active form Repressor protein binds to operator Repressor protein blocks part of promoter RNA polymerase cannot bind to the promoter The lac operon is switched off No transcription of structural genes occur β-galactosidase, permease and transacetylase are NOT produced Learning Outcomes : 6.4 d) Explain the mechanism of the operon in the absence and presence of lactose Mechanism of lac operon LACTOSE ABSENT Learning Outcomes : 6.4 d) Explain the mechanism of the operon in the absence and presence of lactose Mechanism of lac operon LACTOSE PRESENT Some / small amount of lactose is converted to allolactose (inducer) Allolactose binds to repressor protein Repressor protein change its conformation to become inactive form Repressor protein cannot binds to the operator Learning Outcomes : 6.4 d) Explain the mechanism of the operon in the absence and presence of lactose Mechanism of lac operon LACTOSE PRESENT Promoter is unblocked RNA polymerase binds to promoter lac operon switched on Transcription of structural genes (lacZ, lacY and lacA) occur β-galactosidase, permease and transacetylase are produced Lactose hydrolysed into glucose and galactose Learning Outcomes : 6.4 d) Explain the mechanism of the operon in the absence and presence of lactose Mechanism of lac operon LACTOSE PRESENT Learning Outcomes : 6.4 d) Explain the mechanism of the operon in the absence and presence of lactose One mRNA is translated into three polypeptides  REFERENCES Campbell N.A & Reece, J.B., Biology, 12th ed. (2021), Pearson Education, Inc. Solomon E.P & Berg, L.R, Biology, 11th ed. (2019) Thomson Learning, Inc. Mason K.A & Losos J.B., Biology 12th ed. (2019), McGraw-Hill Education. Solomon, Martin, Martin, Berg (2018). Biology (11th ed). Cengage Reece, J., Urry, L., Cain, Wasserman, S. Minorsky, P., & Jackson, R.(2018). Biology (11th ed.), Pearson Benjamin Cummings  REFERENCES (FIGURE) Figure 1 – https://en.wikipedia.org/wiki/Three_prime_untranslated_region Figure 2 – https://byjus.com/biology/central-dogma-inheritance-mechanism/ Figure 3 – Copyright © 2010 Pearson Education, Inc Figure 4 & 5 – Campbell 12th ed page 372 Figure 6 – Adapted from Campbell 12th ed page 371 Figure 7 – Campbell 12th ed page 376 Figure 8 & 9 – Campbell 12th ed page 373 Figure 10 – https://www.slideshare.net/slideshow/dna-replication-17194030/17194030#24 Figure 11 – Solomon 11th ed page 264 Figure 12 – https://www.pinterest.com/pin/205406432982963372/ Figure 13 – Campbell 12th ed page 374 Figure 14 – https://www.pinterest.com/pin/205406432982963372/ Figure 15 – Solomon 11th ed page 267 Figure 16 & 17 – Campbell 12th ed page 375 Figure 18 – https://slideplayer.com/slide/14575879/ Figure 19 & 20 – Adapted from Campbell 12th ed page 375 Figure 21 – https://slcc.pressbooks.pub/humanbiology/chapter/14-dna-structure-protein-synthesis-and- gene-regulation-2/ Figure 22 – https://www.slideshare.net/slideshow/the-cell-cycle-229798422/229798422#13 Figure 23 – Raven 12th ed page 292 (e-book) Figure 24 – Campbell 12th ed page 391 Figure 25 – Campbell 12th ed page 390 Figure 26 – Mader 12th ed page 217 (e-book) Figure 27 – Campbell 12th ed page 393 Figure 28 – Campbell 12th ed page 394 Figure 29 – Solomon 11th ed page 280  REFERENCES (FIGURE) Figure 30 & 32 – Campbell 12th ed page 395 Figure 31 – Campbell 12th ed page 396 Figure 33 – Solomon 11th ed page 283 Figure 34 – Adapted from Campbell 12th ed page 397 Figure 35 – Campbell 12th ed page 398 Figure 36 – Campbell 12th ed page 399 Figure 37 – https://www.slideshare.net/slideshow/protein-syntesis/11499503#6 Figure 38 & 39 – Campbell 12th ed page 400 Figure 40 – Campbell 12th ed page 402 Figure 41 & 42 – Campbell 12th ed page 403 Figure 43 – Campbell 12th ed page 405 Figure 44 – Adapted from https://microbiochem.weebly.com/603uiiilacoperon.html Figure 45 – https://www.slideshare.net/slideshow/microphysio-5/18424488 Figure 46 – Adapted https://microbeonline.com/lac-operon-mechanism/ Figure 47 – http://academygenbioii.pbworks.com/f/Chapter13.pdf Figure 48 & 51 – Campbell 12th ed page 418 Figure 49 & 50 – Adapted from https://bio1151.nicerweb.com/Locked/media/ch18/lac_induced.html Figure 52 – https://bio3400.nicerweb.com/Locked/media/ch16/lac_mRNA.html