Chapter Fundamental Molecular Genetic Mechanisms

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Chapter 5-2 - Fundamental Molecular Genetic Mechanisms 5.1 Structure of Nucleic Acids 5.2 Transcription of Protein-Coding Genes and Formation of Fun ctional mrna 5.

1 Chapter Fundamental Molecular Genetic Mechanisms 5.1 Structure of Nucleic Acids 5.2 Transcription of Protein-Coding Genes and Formation of Fun ctional mrna 5.3 The Decoding of mrna by trnas 5.4 Stepwise Synthesis of Proteins on Ribosomes 5.5 DNA Replication 5.6 DNA Repair and Recombination 5.7 Viruses: Parasites of the Cellular Genetic System

2 Question: How to read it for translation? ssdna ssdna RNA RNA Peptides

3 Fundamental Molecular Genetic Mechanisms 5.3 The Decoding of mrna by trnas DNA/RNA codon triplet code Coding system trna structure and function

4 Three roles of RNA in protein synthesis. Three RNAs involved in protein synthesis: mrna, trna, rrna mrna: nucleotide sequence encodes the order of amino acids a ribosome assembles into polypeptide chain trnas: Each amino acid type is covalently bound to a subset of trnas containing a specific threenucleotide anticodon sequence. Each anticodon base-pairs with its complementary mrna codon to position the encoded amino acid in the ribosome A site where it is covalently linked to the C-terminus of the growing peptide. rrna: Ribosomes are composed of numerous proteins and three (bacterial) or four (eukaryotic) ribosomal RNA (rrna) molecules (not shown). One of the rrnas catalyzes peptide bond formation between incoming aa-trna amino-group and the carboxy-terminus of the growing protein chain.

6 Multiple reading frames in an mrna sequence. How many frames are possible?

7 Multiple reading frames in an mrna sequence.

8 Known Deviations from the Universal Genetic Code Universal code established early in evolution. Minor differences were later evolutionary developments.

9 trna

10 How to get the specificity? 5-20 Structure of trnas. (a) All trna structures: fold into four base-paired stems and three loops have a CCA sequence at the 3ʹ end (acceptor stem), to which an amino acid is attached by an aminoacyl trna transferase have an anticodon triplet at the tip of the anticodon loop have post-transcriptionally modified A, C, G, and U residues (b) 3D model of the generalized L-shaped backbone of all trnas

11 Translating nucleic acid sequence into amino acid sequence. Enzyme! Step 1: An aminoacyl-trna synthetase couples a specific amino acid via a high-energy ester bond (making the amino acid activated) to either the 2ʹ or 3ʹ hydroxyl of the terminal adenosine in the trna that has the proper anticodon (cognate trna). The energy of the ester bond subsequently drives the formation of the peptide bonds linking adjacent amino acids in a growing polypeptide chain in the ribosome. Proofreading by the aminoacyl-trna synthetase maintains a very low error. The anticodon three nucleotides are located in a loop where they are accessible for codon-anticodon base pairing. Step 2: The anticodon three base sequence in the trna base-pairs with a complementary codon in the mrna specifying the attached amino acid.

12 Nonstandard base pairing at the wobble position. 1. How many trnas are existed for amino acid codon? Many cells have fewer than the 61 different trnas required to perfectly base pair each amino acid codon. Since most organisms have fewer than 45 species of trna, some trna species must pair with more than one codon. I (inosine) in the wobble position can pair with three different but synonymous codons (C, A, U in codon third base position) bearing the same amino acid. G or U in the wobble position each can pair with two codons. A in the wobble position is rare in nature.

13 Fundamental Molecular Genetic Mechanisms 5.4 Stepwise Synthesis of Proteins on Ribosomes Ribosome structure and function Translation mechanism protein synthesis

14 Structure of the bacterial ribosome. Ribosome: protein synthesis organelle/machine rrna in ribosome: third RNA required for protein synthesis (in addition to mrna, and trna) Ribosomes differ in bacteria, archaea, and eukaryotes, but share structural and functional similarities. Ribosome (T. thermophilus) structure: small (30S) subunit 16S rrnaand proteins large (50S) subunit 23S and 5S rrnas and proteins internal positions of trnas in the A, P, and E sites and elongating peptide attached to the trna in the P site. The S is an abbreviation for "Svedberg" the unit of molecular size named after the Nobel prize winner who developed methods of determining molecular size by centrifugation.

15 RNA gel electrophoresis Reason?

16 1.Aminoacyl-tRNA binding, or A-site 2.Peptidyl-tRNA binding, or P-site 3.tRNA exit site, or E-site

17 Ribosome Components Ribosomes share structural and functional similarities, but differ in number of rrnas and proteins in bacteria, archaea, and eukaryotes. S (svedberg) units are a measure of the sedimentation rate of macromolecules centrifuged under standard conditions.

18 Comparison of the common core structure at the center of ribosomes from all domains of life and bacterial, yeast, and human ribosomes. (a) RNA in the common core structure (light blue) and protein domains common to all ribosomes (pink).

19 Initiation of translation in eukaryotes. Eight initiation steps involving eukaryotic translation initiation factors (eifs) and GTP hydrolysis, which stabilizes some complexes:

20 Chain elongation in eukaryotes. During chain elongation each incoming trna moves through three sites - A, P, and E.

21 Termination of translation in eukaryotes. Translation is terminated by release factors when a stop codon (UAA, UGA, UAG) is in A site.

Circular structure of mrna increases translation efficiency. Circular mrna, polysomes, and rapid ribosome recycling increase the efficiency of translation.

23 Circular structure of mrna increases translation efficiency. Circular mrna, polysomes, and rapid ribosome recycling increase the efficiency of translation. Polysome: structure with multiple individual ribosomes simultaneously translating a eukaryotic mrna (a) force-field electron micrograph (b) model of protein synthesis on circular polysomes and recycling of ribosomal subunits

24 Fundamental Molecular Genetic Mechanisms 5.5 DNA Replication Semiconservative replication mechanism

25 Two Hypothesis!! Meselson-Stahl experiment. Experiment showed that DNA replicates by a semiconservative mechanism: E. coli cells were grown in a medium containing heavy/h nitrogen ( 15 N) ammonium salts until all cellular DNA was labeled.

26 Two Hypothesis!!

27 DNA polymerases require a primer to initiate replication. DNA synthesis always proceeds in the 5ʹ 3ʹ direction, because chain growth results from formation of a phosphoester bond between the 3ʹ oxygen of a growing strand and the α phosphate of a dntp. DNA polymerases require a short, preexisting RNA or DNA primer strand (with free 3 hydroxyl) that is base-paired to the template strand to begin second strand growth.

28 Leading-strand and lagging-strand DNA synthesis. Duplex DNA is unwound, and daughter strands are formed at the DNA replication fork. DNA polymerase adds nucleotides to a growing daughter strand in the 5ʹ 3ʹ direction. Leading strand: synthesized continuously from a single RNA primer (red) at its 5ʹ end. Lagging strand

Think what s necessary for replication!!! (1) A viral large T-antigen helicase hexamer unwinds the parental DNA strands, beginning at a replication origin.

29 Think what s necessary for replication!!! (1) A viral large T-antigen helicase hexamer unwinds the parental DNA strands, beginning at a replication origin. (2) DNA polymerase ε (Pol ε) extends the leading strand up to the replication fork. (3) The PCNA sliding clamp keeps Pol ε stably associated with the replication fork. (4) Multiple copies of the heterotrimeric protein RPA bind the lagging strand. (5) A primase RNA polymerase-dna polymerase α (Pol α) complex synthesizes primers for lagging-strand synthesis. (6) A PCNA Pol δ complex binds the 3ʹ end of each primer and extends the primer to synthesize most of an Okazaki fragment. (Proliferating cell nuclear antigen)

Bidirectional replication in SV40 DNA. Replication is initiated by binding two large T-antigen hexameric helicases to the single SV40 origin and formation of two oppositely oriented replication forks.

31 Bidirectional replication in SV40 DNA. Replication is initiated by binding two large T-antigen hexameric helicases to the single SV40 origin and formation of two oppositely oriented replication forks. EM: centers of various sized replication bubbles were a constant distance from each end of the EcoRI cut SV40 DNA, consistent with chain growth in two directions from a common origin located at the center of a bubble.

32 Bidirectional mechanism of DNA replication. Eukaryotic chromosomal DNA molecules contain multiple replication origins separated by tens to hundreds of kilobases.

33 Fundamental Molecular Genetic Mechanisms 5.6 DNA Repair and Recombination DNA sequence changes copying errors and the effects of var ious physical and chemical agents. Three excision repair systems base excision repair nonhomologous end joining homologous recombination Defects in DNA repair are associated with cancers

34 Proofreading by DNA polymerase.

36 Fundamental Molecular Genetic Mechanisms 5.7 Viruses: Parasites of the Cellular Genetic System Virus genome and structure Virus invasion and replication in host cells Virus life cycles lytic and nonlytic

39 Bacteriophage DNA genome

Step 1: When viral capsid proteins at the tip of the tail in T4 interacts with specific receptor proteins on the exterior of the host cell (adsorption), the viral genome is injected into the host

40 Step 1: When viral capsid proteins at the tip of the tail in T4 interacts with specific receptor proteins on the exterior of the host cell (adsorption), the viral genome is injected into the host cell (penetration). Step 2: Host-cell enzymes transcribe viral early genes into mrnas, which host-cell ribosomes translate into viral early proteins. Step 3: The early proteins replicate the viral DNA (replication) and induce expression of viral late proteins, including capsid and assembly proteins and enzymes that degrade the host-cell DNA, supplying nucleotides for synthesis of more viral DNA. Step 4: Progeny virions assemble in the cell (assembly). Step 5: Viral proteins lyse the cell (release), and newly liberated viruses initiate another cycle of infection in other host cells.

41 Lytic replication cycle of an enveloped animal virus. What s difference?

42 Release of progeny virions by budding

43 Retroviral life cycle. Step 1: Viral envelope glycoproteins interact with a specific host-cell membrane protein, causing the envelope to fuse directly with the plasma membrane, allowing entry of the nucleocapsid into the cytoplasm of the cell. Step 2: Viral reverse transcriptase copies the viral ssrna genome into a double-stranded DNA. Step 3: The viral dsdna is transported into the nucleus and integrated into one of many possible sites in the host-cell chromosomal DNA. Step 4: The integrated viral DNA (provirus) is transcribed by the hostcell RNA polymerase, generating viral genomic RNA molecules (bright red) and viral mrnas (dark red), which are translated by host-cell ribosomes into glycoproteins and nucleocapsid proteins. Step 5: Progeny virions assemble and are released by budding.

44 Virus diseases Retroviruses infect only certain cell types HTLV : leukemia HIV-1 : AIDS DNA viruses integrate into host-cell genome HPV expression of oncogenes cervical cancer

45 Discussion with friends You synthesized DNA sequences using DNA polymerase, but got several wrong sequences. Which step you will check first? Different from eukaryotic genome, prokaryote has circular genome. The circular genome has one problem in its replication process, especially at the end of replication. What s the problem and how does prokaryote solve this? If you want to use the viruses for exogenous gene expression in cell line, which processes should be removed for the safe and stable gene expression?

46 참고 (etc)

Name 10 Molecular Biology of the Gene Test Date Study Guide You must know: The structure of DNA. The major steps to replication.

Name 10 Molecular Biology of the Gene Test Date Study Guide You must know: The structure of DNA. The major steps to replication. The difference between replication, transcription, and translation. How