Lecture 15: Translation and Transcription

Similar documents
Transcription and Translation of DNA

DNA Replication & Protein Synthesis. This isn t a baaaaaaaddd chapter!!!

From DNA to Protein. Proteins. Chapter 13. Prokaryotes and Eukaryotes. The Path From Genes to Proteins. All proteins consist of polypeptide chains

Protein Synthesis How Genes Become Constituent Molecules

Molecular Genetics. RNA, Transcription, & Protein Synthesis

Genetic information (DNA) determines structure of proteins DNA RNA proteins cell structure enzymes control cell chemistry ( metabolism )

Structure and Function of DNA

a. Ribosomal RNA rrna a type ofrna that combines with proteins to form Ribosomes on which polypeptide chains of proteins are assembled

Chapter 17: From Gene to Protein

Lecture Series 7. From DNA to Protein. Genotype to Phenotype. Reading Assignments. A. Genes and the Synthesis of Polypeptides

Specific problems. The genetic code. The genetic code. Adaptor molecules match amino acids to mrna codons

Coding sequence the sequence of nucleotide bases on the DNA that are transcribed into RNA which are in turn translated into protein

RNA & Protein Synthesis

Translation. Translation: Assembly of polypeptides on a ribosome

PRACTICE TEST QUESTIONS

Translation Study Guide

Name Class Date. Figure Which nucleotide in Figure 13 1 indicates the nucleic acid above is RNA? a. uracil c. cytosine b. guanine d.

13.2 Ribosomes & Protein Synthesis

BCH401G Lecture 39 Andres

2. The number of different kinds of nucleotides present in any DNA molecule is A) four B) six C) two D) three

DNA, RNA, Protein synthesis, and Mutations. Chapters

Provincial Exam Questions. 9. Give one role of each of the following nucleic acids in the production of an enzyme.

Lecture 4. Polypeptide Synthesis Overview

Name Date Period. 2. When a molecule of double-stranded DNA undergoes replication, it results in

Transcription: RNA Synthesis, Processing & Modification

Basic Principles of Transcription and Translation

Ms. Campbell Protein Synthesis Practice Questions Regents L.E.

The sequence of bases on the mrna is a code that determines the sequence of amino acids in the polypeptide being synthesized:

The Steps. 1. Transcription. 2. Transferal. 3. Translation

To be able to describe polypeptide synthesis including transcription and splicing

From DNA to Protein

RNA and Protein Synthesis

Lecture Transfer of proper aminoacyl-trna from cytoplasm to A-site of ribosome.

CHAPTER 30: PROTEIN SYNTHESIS

Page 1. Name:

Genetics Module B, Anchor 3

Protein Synthesis. Page 41 Page 44 Page 47 Page 42 Page 45 Page 48 Page 43 Page 46 Page 49. Page 41. DNA RNA Protein. Vocabulary

Lecture 1 MODULE 3 GENE EXPRESSION AND REGULATION OF GENE EXPRESSION. Professor Bharat Patel Office: Science 2, b.patel@griffith.edu.

Sample Questions for Exam 3

Thymine = orange Adenine = dark green Guanine = purple Cytosine = yellow Uracil = brown

Nucleotides and Nucleic Acids

Announcements. Chapter 15. Proteins: Function. Proteins: Function. Proteins: Structure. Peptide Bonds. Lab Next Week. Help Session: Monday 6pm LSS 277

TRANSCRIPTION TRANSLATION - GENETIC CODE AND OUTLINE OF PROTEIN SYNTHESIS

GENE REGULATION. Teacher Packet

Lecture 26: Overview of deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) structure

Cellular Respiration Worksheet What are the 3 phases of the cellular respiration process? Glycolysis, Krebs Cycle, Electron Transport Chain.

Problem Set 3 KEY

Lecture 6. Regulation of Protein Synthesis at the Translational Level

Module 3 Questions. 7. Chemotaxis is an example of signal transduction. Explain, with the use of diagrams.

Basic Concepts of DNA, Proteins, Genes and Genomes

AP BIOLOGY 2009 SCORING GUIDELINES

A disaccharide is formed when a dehydration reaction joins two monosaccharides. This covalent bond is called a glycosidic linkage.

Academic Nucleic Acids and Protein Synthesis Test

The Nucleus: DNA, Chromatin And Chromosomes

Complex multicellular organisms are produced by cells that switch genes on and off during development.

Lecture 8. Protein Trafficking/Targeting. Protein targeting is necessary for proteins that are destined to work outside the cytoplasm.

1 Mutation and Genetic Change

Multiple Choice Write the letter that best answers the question or completes the statement on the line provided.

Lab # 12: DNA and RNA

Chem 465 Biochemistry II

DNA (genetic information in genes) RNA (copies of genes) proteins (functional molecules) directionality along the backbone 5 (phosphate) to 3 (OH)

Proteins and Nucleic Acids

Problem Set 1 KEY

Answer: 2. Uracil. Answer: 2. hydrogen bonds. Adenine, Cytosine and Guanine are found in both RNA and DNA.

Central Dogma. Lecture 10. Discussing DNA replication. DNA Replication. DNA mutation and repair. Transcription

CCR Biology - Chapter 8 Practice Test - Summer 2012

Control of Gene Expression

4. Which carbohydrate would you find as part of a molecule of RNA? a. Galactose b. Deoxyribose c. Ribose d. Glucose

CHAPTER 40 The Mechanism of Protein Synthesis

13.4 Gene Regulation and Expression

The Molecules of Cells

Chapter 5. The Structure and Function of Macromolecule s

NO CALCULATORS OR CELL PHONES ALLOWED

Lecture Overview. Hydrogen Bonds. Special Properties of Water Molecules. Universal Solvent. ph Scale Illustrated. special properties of water

ISTEP+: Biology I End-of-Course Assessment Released Items and Scoring Notes

Replication Study Guide

DNA. Discovery of the DNA double helix

Chapter 18 Regulation of Gene Expression

Chapter 6 DNA Replication

Chapter 5: The Structure and Function of Large Biological Molecules

DNA is found in all organisms from the smallest bacteria to humans. DNA has the same composition and structure in all organisms!

Disaccharides consist of two monosaccharide monomers covalently linked by a glycosidic bond. They function in sugar transport.

Modeling DNA Replication and Protein Synthesis

Biological molecules:

How To Understand The Chemistry Of Organic Molecules

Chemical Basis of Life Module A Anchor 2

2. True or False? The sequence of nucleotides in the human genome is 90.9% identical from one person to the next. False (it s 99.

T C T G G C C G A C C T;

MUTATION, DNA REPAIR AND CANCER

BioBoot Camp Genetics

Gene Models & Bed format: What they represent.

Given these characteristics of life, which of the following objects is considered a living organism? W. X. Y. Z.

Protein Synthesis CHAPTER OUTLINE

CHAPTER 6: RECOMBINANT DNA TECHNOLOGY YEAR III PHARM.D DR. V. CHITRA

Transcription in prokaryotes. Elongation and termination

Preliminary MFM Quiz

12.1 The Role of DNA in Heredity

Proteins. Proteins. Amino Acids. Most diverse and most important molecule in. Functions: Functions (cont d)

Chapter 3 Molecules of Cells

Transcription:

Lecture 15: Translation and Transcription Making proteins requires that genetic information is transcribed from DNA into RNA and then translated into protein. The overall process can be viewed as two separate processes transcription and translation. I. Background A. RNA 1. RNA is a nucleic acid a. Polymer nucleotides 2. RNA structure differs from DNA a. Five-carbon sugar in RNA nucleotides is ribose (not deoxyribose) b. Pyrimidine in RNA is uracil instead of thymine 3. The linear sequence of DNA ultimately determines the primary structure of proteins a. Unique combination of nucleotides code for the 20 amino acids that make up proteins B. Terminology 1. Transcription synthesis of RNA from a DNA template a. Unique nucleotide sequence of a gene is transcribed from DNA into a complementary nucleotide sequence in messenger RNA (mrna) b. mrna carries the protein-building instructions to the cellular machinery that synthesizes protein 2. Translation polypeptide synthesis under the direction of mrna a. Linear sequence of mrna is translated into a linear sequence of amino acids in a polypeptide b. Translation occurs on a complex particle known as a ribosome. i. Ribosomes are comprised of ribosomal RNA (rrna) and protein. ii. Ribosomes facilitate the orderly linking of amino acids into polypeptide chains C. Differences in protein synthesis between prokaryotes and eukaryotes 1. Prokaryotes do not have nuclei a. DNA is not segregated from ribosomes and protein machinery. b. Translation and transcription occur in rapid succession 2. Eukaryotes have nuclear envelopes that segregate DNA segregated from ribosomes and protein machinery. a. Transcription takes place in the nucleus b. Translation takes place in the cytoplasm c. mrna serves as an intermediary d. mrna is modified (i.e., RNA processing) before it leaves the nucleus

D. Nucleotide triplets specify amino acids 1. There is not a one-to-one correspondence between nucleotides and amino acids a. There are only 4 nucleotides and 20 amino acids b. A two-to-one correspondence would only specify 16 amino acids (4 2 ) c. A three-to-one correspondence would specify 64 amino acids (4 3 ) (i) Nucleotide triplets are known as codons (ii) Basic unit of the genetic code (iii) Specify either individual amino acids or termination

2. Genes are not directly translated into amino acids a. DNA is transcribed into mrna first b. For each gene, only one of the two DNA strands (i.e., template strand) is transcribed c. The complementary non-template strand is the parental strand for making a new template when DNA replicates d. The same DNA strand can serve as the template for some genes and the non-template strand for others

3. mrna is complementary to the DNA strand from which it is transcribed DNA A C C A A A C C G G C A mrna U G G U U U G G C C G U Protein Tryptophan Phenylalanine Glycine Arganine 4. During translation, the linear sequence of codons along mrna is translated into the linear sequence of amino acids in a polypeptide a. There is redundancy in the code but no ambiguity i. Redundancy exists because two or more codons differing only in their third base (e.g., UUU and UUC each code for phenylalanine) ii. Codons only code for one amino acid, therefore there is no ambiguity II. Transcription A. DNA-directed synthesis of RNA 1. RNA polymerase catalyzes this synthesis a. Two functions of RNA polymerase: i. Separates DNA strands ii. Links RNA nucleotides from the 5 3 direction as they base-pair along the DNA template

2. Types of RNA polymersae a. Prokaryotes have a single type of polymerase that synthesizes all types of RNA (e.g., trna, rrna, mrna) b. Eukaryotes have three types: i. RNA polymerase II catalyzes mrna synthesis 3. Transcription unit a. A specific sequence of DNA that is transcribed into a single mrna molecule b. Includes sequences with specific functions i. Region where transcription begins (initiation) ii. Region where transcription ends (termination) iii. Intervening nucleotides (i.e., gene) c. In eukaryotes, the transcription unit contains a single gene i. Resulting mrna codes for only one peptide d. In prokaryotes, the transcription unit can contain several genes i. Resulting mrna may code for different proteins ii. Functionally related B. Stages of transcription 1. Polymerase binding and initiation 2. Elongation 3. Termination C. Polymerase binding and initiation of transcription 1. Promotor site where RNA polymerase binds and where transcription begins (i.e., initiation site) a. Approximately 100 nucleotides in eukaryotes b. Includes nucleotide sequences that are recognized by specific DNA-binding proteins (transcription factors) that help initiate transcription 2. Transcription factors a. In eukaryotes, RNA polymerase cannot recognize the promotor without transcription factors b. Transcription factors bind to TATA box i. TATA box is a region rich in thiamine and adenine located approximately 25 nucleotides upstream from the initiation site ii. RNA polymerase recognizes the complex formed by the binding of the iii. transcription factor to the TATA box Other transcription factors associate with the RNA polymerase/tata sequence complex 3. Activated RNA polymerase binds to the downstream promotor, separates the strands at the initiation site and transcription begins D. Elongation of the RNA strand 1. Functions of RNA polymerase

a. Untwists and opens DNA to reveal approximately 10 nucleotide bases the exposed region is the template for mrna b. Links incoming RNA nucleotides to the 3 end of the elongating strand i. 30-60 nucleotides/second ii. Newly formed mrna peels away from the template spontaneously iii. DNA-DNA double helix re-forms; non-template strand pairs with the template 2. Several RNA pol II can simultaneously transcribe the gene a. Allows an increased production of particular proteins b. Length of each strand reflects how far the enzyme has traveled from the initiation site

E. Post-transcriptional modification 1. In eukaryotes, mrna is modified before it leaves the nucleus a. Includes covalent alteration of both the 3 and 5 ends b. Removal of intervening sequences 2. Terms a. Primary transcript i. Initial RNA transcribed from DNA b. Pre-mRNA i. Primary transcript that will be processed to functional mrna 3. Alteration of pre-mrna ends a. 5 cap modified G (guanine tri-phosphate) that is added after transcription i. Protects from hydrolytic enzymatic degradation ii Helps small ribosomal subunit recognize attachment site on mrna 5 end b. Leading sequence non-coding (untranslated) sequence of mrna i. From 5 end to start codon c. Poly (A) tail i. Sequence of 30-200 adenines added to the 3 end d. Trailer sequence i. Non-coding sequence of mrna from the stop codon to poly (A) tail

4. Split genes and RNA splicing in eukaryotes, genes that code for proteins may not be continuous sequences a. Introns (intervening sequences) non-coding sequence in DNA that intervenes between coding sequences i. Transcribed but not translated ii. Excised before leaving nucleus iii. Not all genes have introns b. Exons coding sequence of a gene that is transcribed and expressed c. RNA splicing RNA processing that removes introns and joins exons i. Produces mature RNA that will pass into the cytoplasm ii. Not limited to mrna, also t and r d. Small nuclear ribonucleoproteins (snrnps snurps) complex of proteins and small nuclear RNA found in the nucleus; some participate in RNA splicing i. Small nuclear RNA is about 300 nucleotides ii. Protein III. Synthesis of protein A. Translation RNA directed synthesis of a polypeptide 1. mrna contains a genetic message consisting of sequential codons (mrna codon codes for one amino acid) 2. trna is the interpreter between the two forms of information base sequence in mrna and amino acid sequence in the polypeptide

a. trna aligns the appropriate amino acid to form a new polypeptide i. Collects appropriate amino acid from cytoplasmic pool ii. Recognizes codon in mrna iii. Each trna recognizes only one amino acid and its unique mrna codon b. Physical properties of trna i. Ability to bind amino acid and read codon depends on its structure ii. Site that attaches to mrna codon iii. Site that attaches to amino acid c. Anticodon nucleotide triplet in trna that base pairs with complementary triplet (codon) in mrna trna Anticodon mrna Codon Amino Acid AAA UUU Phenylalanene d. There are 45 different trna s but 64 codon (4 3 ) i. Some trna s can recognize 2-3 mrna codon specifying the same amino acid ii. Wobble base pairing rules are relaxed between the 3 rd base of the mrna and the corresponding anticodon iii. 5 end (3 rd base) can hydrogen bond with more than one kind of base in the 3 rd position (3 ) of the codon iv. Some trnas contain a modified base called inosine (I) which is in the antocodon s wobble position v. Inosine can base pair with U, C and A (not G) trna Anticodon mrna Codon Amino Acid CCI GGU, GGA, GGC Glycine B Aminoacyl-tRNA synthetase

1. Amino acids must first be correctly linked with trna before the trna anticodon pairs with its complementary mrna codon 2. Aminoacyl-tRNA synthetase catalyzes the attachment of the amino acid to its trna a. Each amino acid has a unique aminoacyl-trna synthetase b. Reaction requires energy (i.e., endergonic reaction) provided by the hydrolysis of ATP c. Reaction forming aminoacyl-trna complex requires two steps: i. Activation of amino acid with AMP enzyme binds AMP to amino acid with net release of pyrophoshate (i.e., ATP to AMP + pyrophosphate) ii. trna is covalently bonded to amino acid, displacing AMP d. Aminoacyl-tRNA complex releases from the enzyme and transfers its amino acid to growing polypeptide on the ribosome

C. Ribosomes 1. Ribosomes coordinate the pairing of trna anticodons to mrna codons. 2. Structure a. Two subunits small and large i. Subunits are separated when not involved in protein synthesis b. Comprised of 60% ribosomal RNA (rrna) and 40% protein c. Constructed in the nucleolus d. Enters cytoplasm through pores in the nucleus e. Assembled into functional ribosomes only when attached to mrna f. Four binding sites:

i. mrna ii. P site holds the trna carrying the polypeptide iii. A site holds the trna carrying the next amino acid to be added iv. E site discharged trna exits from this site g. Ribosomes do not catalyze the synthesis of bonds between the amino acid and the growing polypeptide; this is done by other enzymes (see below) IV. Building a peptide A. Translation occurs in three steps: 1. Initiation 2. Elongation 3. Termination B. Initiation 1. Initiation brings together the mrna and the first amino acid (always methionine) and the two ribosomal subunits a. Small ribosomal subunit is bonded to mrna and initiator trna b. In eukaryotes, 5 cap of the mrna aids in binding the leader sequence to the small ribosomal subunit c. Small ribosomal subunit facilitates binding of initiator trna to the start code

i. AUG is the start code ii. AUG is the codon for methionine (i.e., that s why the first amino acid is always methionine) d. Initiation Complex small ribosomal subunit, initiator trna and mrna

Requires: i. Initiator factors that are bound to small ribosomal subunit ii. One GTP molecule that when hydrolyzed drives the attachment of the large ribosomal subunit e. Functional translational complex large ribosomal subunit binds to initiation complex (i.e., to the small ribosome)

i. Initiation factors attached to the small subunit are released, allowing the large subunit to bind with small subunit ii. Initiator trna fits into the P site iii. Vacant A site is ready for the next aminoacyl-trna C. Elongation 1. Three step cycle to add amino acids one by one to the initial amino acid and subsequent polypeptide

a. Codon recognition b. Peptide bond is formed c. Translocation 2. Codon recognition a. mrna codon in A site of the ribosome forms hydrogen bonds with the anticodon of the entering trna carrying the next amino acid 3. Peptide bond is formed a. Peptide bond is formed between the amino acid in the P site and neighboring amino acid in the A site b. Formation of peptide bond is catalyzed by peptidyl transferase c. Polypeptide separates from its trna and is transferred to the new amino acid carried by the trna in the A site

4. Translocation trna in the A site carrying the growing peptide is translocated to the P site while the trna in the P site is simultaneously translocated to the E site (where it exits the ribosome complex) a. mrna and trna move as a unit maintaining codon and anticodon bonding and bringing the next codon into the A site b. mrna is moved through the ribosome in the 5 to 3 direction c. Requires the hydrolysis of GTP for energy d. Elongation takes about 1/10 sec per amino acid D. Termination 1. Elongation continues until a termination codon is reached 2. Termination codon (stop codon) base triplet (codon) on mrna that signals translation to stop a. UAA, UAG, UGA b. Stop codons do not code for amino acids 3. When reached, release factor binds to the codon and initiates the following events: a. Bond between the polypeptide and trna in the P site is hydrolyzed b. Polypeptide and trna are released from the ribosome c. Translocation complex dissociates E. Polyribosomes 1. Cluster of ribosomes simultaneously translating an mrna 2. Once the ribosome complex passes the initiation codon, a second ribosome can attach to the leader sequence a. Multiple ribosomes can translate an mrna at once V. Post-translational modification 1. Biological activity of a protein is conferred by its three-dimensional conformation. 2. Genes determine protein primary structure (i.e., linear sequence of amino acids) 3. Primary structure determines how a polypeptide chain will spontaneously coil and fold to form a three-dimensional shape (i.e., secondary and tertiary conformation)

4. Some proteins undergo post-translational modification to become functional a. Types of modification i. Chemical modification sugars, lipids, phosphate groups and other molecules attached to specific amino acids in the polypeptide ii. Chain length modification one or more amino acids may be cleaved from the end of a peptide iii. iv. Single peptide may be divided into two or more pieces (e.g., proinsulin to insulin) Two or more peptides may be joined as subunits of a protein and determine quaternary structure VI. Point Mutations A. Terms: 1. Mutation change in the genetic material of a cell 2. Point mutation mutation limited to one or a few base pairs in a single gene B. Types of point mutations: 1. Substitution a. Base-pair substitution replacement of one base-pair with another; nucleotide and its partner in the complementary DNA are replaced with another pair of nucleotides according to base-pairing rules i. Silent substitution because of redundancy in the genetic code, a base pair change may simply transform one codon into another that codes for the same amino acid

ii. iii. Conservative substitution change of base pair results in a change in amino acid but the new amino acid has similar properties or the substitution occurs in a nonessential area of the protein Base pair changes may change a single amino acid in a crucial area, altering protein activity either enhancing (rare) protein activity or impairing protein activity b. Base-pair substitutions are usually missense mutations or nonsense mutations i. Missense mutation base-pair substitution that alters an amino acid codon (sense codon) to a new codon that codes for a different amino acid; these are translated; most common type of base-pair substitution ii. Nonsense mutation base-pair substitution that changes an amino acid codon to a termination codon, or vice versa; usually results in a non-functioning protein 2. Insertions or deletions usually have negative effects

a. Base-pair insertion insert one or more base-pairs into a gene b. Base-pair deletion removes one or more base-pairs out of a gene c. If insertion or deletion is not a multiple of 3 nucleotides, the base-pair insertion or deletion results in a frame shift mutation d. Frame shift mutation insertion or deletion of base pairs that results in a change in reading frame i. Change causes nucleotide to be improperly grouped ii. Results in missense that will ultimately result in nonsense iii. Translated protein is usually not functional