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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 Concept Definition: The flow of genetic information from DNA to RNA to protein in living organisms. 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 : 6.1 (a) : State the concept of Central Dogma Central Dogma 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. 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 Outcomes : 6.2 (b) : Explain the enzymes and proteins involved in DNA Replication Enzymes and Protein in DNA Replication Learning Outcomes : 6.2 (b) : Explain the enzymes and proteins involved in DNA Replication Enzymes and Protein in DNA Replication 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 Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Mechanism of DNA Replication ⮚ Replication of DNA begins at origins of replication ⮚ Helicase unwinds DNA double helix at replication forks, by breaking hydrogen bond ⮚ Separating the two parental strands and making them available as template strands ⮚ Forming replication bubble Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved ⮚ Single-strand binding protein bind to the unpaired DNA strands, prevent them from repairing. Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved ⮚ When DNA is unwound, it becomes more twisted / supercoiling ⮚ Topoisomerase help relieve strain of overwinding ahead of replication fork 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 ⮚ Primase synthesize RNA primer by adding RNA nucleotides that are complementary to the DNA template Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved Mechanism of DNA Replication ⮚ DNA Polymerase III - synthesis new DNA strand by adding DNA nucleotides to the 3′ end of RNA primer or growing DNA strand ⮚ Nucleotides are added complementary to the DNA template Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA replication and the enzymes involved 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 DNA strand. Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA Replication and the enzymes involved Mechanism of DNA Replication ⮚ 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 Mechanism of DNA Replication Leading strand ⮚ New DNA strand that synthesized continuously, towards replication fork from 5’ to 3’ direction. ⮚ 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 ⮚ New DNA strand that synthesized discontinuously, away from the replication fork from 5’ to 3’ direction. ⮚ Forming a series of short 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 DNA Polymerase I remove RNA primer and replaces with DNA nucleotides Learning Outcomes : 6.2 (c) : Explain the mechanism of DNA Replication and the enzymes involved Mechanism Mechanism ofof DNADNA Replication Replication ⮚ DNA Ligase join the Okazaki fragments into a continuous lagging strand by forming phosphodiester bond ⮚ Two identical copies of DNA are produced, consisting of one parental strand and one newly strand 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 encode 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 (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/ RNA polymerase II. There are THREE stages of transcription Step 3:1:Termination Stage Initiation Enzyme RNA polymerase II is used Step 3:2:Termination Stage Elongation in the transcription of eukaryotes 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 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 typically extends upstream from the start point. RNA polymerase (II) unwinds the double helix of DNA by breaking the hydrogen bond and separating the two DNA strands Only one DNA strand acts as a template (3’ to 5’) RNA synthesis begins at the start point on the template strand 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 As RNA polymerase (II) moves along the DNA, it adds free RNA nucleotide complementary to the bases on the DNA template strand at the 3’ end of the growing RNA transcript/ pre-mRNA. The synthesis of pre-mRNA occurs from 5’ to 3’ direction. 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 pre-mRNA molecule behind the RNA polymerase (II) peels away / detaches from the 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 RNA polymerase (II) transcribe polyadenylation signal sequence on DNA. Pre-mRNA transcribed contains polyadenylation signal (AAUAAA) 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 pre-mRNA 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 End of pre-mRNA is modified by adding : 5′ end receives a 5′ cap 3′ end 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’ cap and poly-A tail: 1. To promote the export of mRNA from nucleus 2. Help to protect mRNA from degradation 3. Help ribosome to attach to 5’ end of the mRNA in cytoplasm 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 segments. Exons - coding segments of nucleic acid that lie between noncoding segments, 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 exons (coding segments) 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 Component of translation ❖ tRNA / aminoacyl tRNA - to transfer an amino acid from the cytoplasm to a growing polypeptide in a ribosome ❖ mRNA – template for translation ❖ Ribosome – site for translation 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 Activation of amino acid Amino acid activation is a process in which a specific amino acid is attached to a tRNA molecule to form aminoacyl-tRNA catalyzed by aminoacyl-tRNA synthetase Amino acids are linked to their respective tRNA molecules by the formation of covalent bond (ester bond) at the 3’end of tRNA ATP is used as energy source Learning Outcomes : 6.3 (d) : Explain translation and the stages involved in translation Ribosome 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 with initiator tRNA binds at 5’ cap of mRNA And then move along the mRNA until reaches start codon. Initiator tRNA which carries amino acid methionine (with anticodon (3’-UAC-5’) reach to the start codon(5’-AUG-3’) at mRNA by forming hydrogen bonds. Large ribosomal subunit attaches with these complex to form translation initiation complex. 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 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 a ribozymes which is peptidyl transferase or rRNA molecule from large subunit of ribosomal. 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/ moved 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 the ribosome reaches one of the three stop codons (5’-UAG-3’/ 5’-UAA-3’/ 5’-UGA- 3’) on mRNA. The stop codon does 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 molecules, hydrolyzes the bond between the polypeptide and the tRNA in the P site. Newly synthesized protein is released Translation complex dissociates with the hydrolysis of GTP molecules. 54 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) 66 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. E. coli) 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 hydrolyze/ breakdown 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 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 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 the structural genes (lacZ, lacY, lacA) No translation 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 the promoter lac operon switched on Transcription of structural genes (lacZ, lacY and lacA) occur Translation occurs β-galactosidase, permease and transacetylase are produced Lactose hydrolyzed 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. 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