BCH3033 Biochemistry 1 Chapter 27 Exam 3 PDF

BCH3033: Biochemistry 1 Chapter 27 Exam 3 Friday 04/19/24 Ch. 8, 25-27 04.12.2024 Donella Beckwith, Ph.D. [email protected] 1 Important Concepts (1 of 3) Protein synthesis: consumes more cellular resources than any other process in most cells. – Overall, almost 300 different macromolecules cooperate...

BCH3033: Biochemistry 1 Chapter 27 Exam 3 Friday 04/19/24 Ch. 8, 25-27 04.12.2024 Donella Beckwith, Ph.D. [email protected] 1 Important Concepts (1 of 3) Protein synthesis: consumes more cellular resources than any other process in most cells. – Overall, almost 300 different macromolecules cooperate to synthesize polypeptides. – Protein synthesis can account for up to 90% of the chemical energy used by a cell for all biosynthetic reactions. – Why? The enzymatic synthesis of an amide bond should require little energetic input. However, the sequence of amino acids in a protein is a form of biological information. Synthesis of each amide (peptide) bond between two particular amino acids is ensured by the investment of more than four nucleoside triphosphates (NTPs). 2 Important Concepts (2 of 3) The genetic code is nearly universal and arose early in evolution. This is one of the many characteristics of living systems that ties all of them to a common ancestor. Even the rare exceptions to code universality reinforce this rule. Although each organism’s DNA is unique, all DNA is composed of the same nitrogen-based molecules It is simply the order in which these smaller molecules are arranged that differs among species. Amino acid codons are identical in nearly all species (exceptions: some bacteria, some single celled eukaryotes, and minor variations in mitochondria) The genetic code functions via linker molecules. The tRNAs are the crucial adaptor, matching amino acids with DNA codons. 3 Important Concepts (3 of 3) Proteins are synthesized by RNAs. The study of protein synthesis offers another important reward: a look at a world of RNA catalysts that may have existed in an RNA world before the dawn of life “as we know it.” Proteins are synthesized by a gigantic RNA enzyme. 4 Crick’s Adaptor Hypothesis adaptor hypothesis = postulation that a small nucleic acid could act as an adaptor, binding to both a specific amino acid and the mRNA sequence encoding that amino acid – verified with the discovery of tRNA 5 The Process of Translation translation = the overall process of mRNA-guided protein synthesis the tRNA adaptor “translates” the nucleotide sequence of an mRNA into the amino acid sequence of a polypeptide 6 Aminoacyl-tRNAs amino acids are “activated” for protein synthesis. aminoacyl-tRNAs = tRNA attached to an amino acid aminoacyl-tRNA synthetases = catalyze the formation of aminoacyl-tRNAs 7 Question 1 Crick’s adaptor hypothesis was verified with the discovery of: A. B. C. D. DNA. mRNA. rRNA. tRNA. 8 Response Crick’s adaptor hypothesis was verified with the discovery of: D. tRNA. Translation requires adaptor molecules, the tRNAs, that recognize codons and insert amino acids into their appropriate sequential positions in the polypeptide. 9 Three Nucleotides are Necessary to Encode Each Amino Acid the four code letters of DNA (A, T, G, and C) in groups of two yields only 42 = 16 different combinations – insufficient to encode 20 amino acids groups of three yield 43 = 64 different combinations START 10 The Genetic Code Was Cracked Using Artificial mRNA Templates codon = a triplet of nucleotides that codes for a specific amino acid in all living systems, translation occurs in such a way that codons are read in a successive, nonoverlapping fashion 11 The Triplet, Nonoverlapping Code Nonoverlapping Code inserting or deleting one base pair alters the sequence coded by the mRNA, not necessarily the protein inserting or deleting three nucleotides leaves the remaining triplets intact – provides evidence that a codon has only three nucleotides Change a codon to 1 that encodes a different amino acid, causes a small change in the protein (sickle cell anemia) Change codon to 1 that encodes the same amino acid, causes NO change in protein Change an amino acid coding codon to a STOP codon, cause an incomplete protein Most detrimental change, codon changes occur near the start of a sequence 12 Reading Frames in the Genetic Code reading frame = method of dividing nucleotides such that a new codon begins every three nucleotide residues – established by the first codon – no punctuation between codons – in principle, any given ssDNA or mRNA sequence has three possible reading frames 13 Setting and Maintaining the Reading Frame the reading frame is set when translation of an mRNA molecule begins the reading frame is maintained as triplets are read sequentially a “missense” protein with a garbled amino acid sequence forms if: – the initial reading frame is off by one or two bases – translation skips a nucleotide in the mRNA 14 Question 2 In principle, how many reading frames does any given singlestranded DNA or mRNA sequence possess? A. B. C. D. 1 2 3 4 15 Response In principle, how many reading frames does any given singlestranded DNA or mRNA sequence possess? C. 3 In principle, any given single-stranded DNA or mRNA sequence has three possible reading frames. Each reading frame gives a different sequence of codons, but only one is likely to encode a given protein. 16 Effect of a Termination Codon there are 3 codons that disrupt amino acid coding patterns when they occur in a synthetic RNA polymer these codons were identified as termination codons termination codons 17 “Dictionary” of Amino Acid Code Words in mRNAs 61 codons code for amino acids 3 codons (UAA, UGA, UAG) are (nonsense) – Have no amino acids 1 codon (AUG) is the start codon (as well as the Met codon in internal positions) 5’ AACGCGAUGGGGGCUAUUUCAUAGCUG 3’ 18 Question 3 How many codon combinations actually encode amino acids? A. B. C. D. 20 32 61 64 19 Response How many codon combinations actually encode amino acids? C. 61 Consolidation of the results from many experiments permitted assignment of 61 of the 64 possible codons. The other three were identified as termination codons. 20 The Genetic Code is Degenerate degenerate = an amino acid may be specified by more than one codon – degeneracy of the code is not uniform each codon specifies only one amino acid Table 27-3 Degeneracy of the Genetic Code Amino acid Number of codons Amino acid Number of codons Met 1 Tyr 2 Trp 1 Ile 3 Asn 2 Ala 4 Asp 2 Gly 4 Cys 2 Pro 4 Gln 2 Thr 4 Glu 2 Val 4 His 2 Arg 6 Lys 2 Leu 6 Phe 2 Ser 6 21 Question 4 Why is the genetic code said to be “degenerate”? A. The first base in a codon is the most important for coding, followed by the second base, and the third base is least important. B. Each amino acid can be specified by more than one codon. C. Fewer amino acids are coded by the four bases than was expected. D. The less complexity in an organism, the fewer codons are specified. E. The genetic code deteriorates with each replication (generation), unless proofreading functions correct for errors. 22 Response Why is the genetic code said to be “degenerate”? A. The first base in a codon is the most important for coding, followed by the second base, and the third base is least important. B. Each amino acid can be specified by more than one codon. C. Fewer amino acids are coded by the four bases than was expected. D. The less complexity in an organism, the fewer codons are specified. E. The genetic code deteriorates with each replication (generation), unless proofreading functions correct for errors. A striking feature of the genetic code is that an amino acid may be specified by more than one codon, so the code is described as degenerate. 23 Pairing Relationship of Codon and Anticodon anticodon = a three-base sequence on the tRNA that base pairs with mRNA codons base pairing occurs via hydrogen bonding the alignment of the two RNA segments is antiparallel 24 Wobble Allows Some tRNAs to Recognize More than One Codon when several different codons specify one amino acid, the difference usually lies at the third base position – for example, alanine is encoded by GCU, GCC, GCA, and GCG the third position in each codon is much less specific and is said to “wobble” – this allows certain tRNAs to recognize more than one codon 25 Wobble: Take-Away the first two bases of the codon form strong Watson-Crick base pairs with the anticodon – confers most of the coding specificity the first base of the anticodon (read in the 5′⟶3′ direction) determines the number of codons recognized by the tRNA when an amino acid is specified by several different codons, the codons that differ in either of the first two bases require different tRNAs 26 Question 5 Which statement does NOT describe a feature of the genetic code? A. It consists of nonoverlapping triplet codons, with three possible reading frames. B. Bacteria have a slightly different code than do eukaryotes. C. It is degenerate. D. Base-pairing of the 3′ base of the mRNA codon to the 5′ base of the tRNA anticodon is often not “normal” WatsonCrick base-pairing. 27 Response Which statement does NOT describe a feature of the genetic code? B. Bacteria have a slightly different code than do eukaryotes. The genetic code is nearly universal. With the intriguing exception of a few minor variations in mitochondria, some bacteria, and some single-celled eukaryotes, amino acid codons are identical in all species examined so far. 28 Question 6 What mediates the interaction between the nucleotide triplets present on tRNA molecules and triplet codons? A. B. C. D. ionic bonds hydrogen bonds covalent bonds peptide bonds 29 Response What mediates the interaction between the nucleotide triplets present on tRNA molecules and triplet codons? B. hydrogen bonds Transfer RNAs base-pair with mRNA codons at a threebase sequence on the tRNA called the anticodon. As in DNA, base pairing occurs via hydrogen bonding. 30 Question 7 Mutation of the codon GAU to which codon would MOST likely result in a protein with unaltered or minimally altered function? A. B. C. D. GAG GUG GCG GGG 31 Response Mutation of the codon GAU to which codon would MOST likely result in a protein with unaltered or minimally altered function? A. GAG In the third, or wobble, position of the codon, single base substitutions produce a change in the encoded amino acid only about 25% of the time. Most such changes are thus silent mutations, in which the nucleotide is different but the encoded amino acid remains the same. 32 BCH3033: Biochemistry 1 Chapter 27b Exam 3 Friday 04/19/24 Ch. 8, 25-27 04.17.2024 Donella Beckwith, Ph.D. [email protected] 33 The Genetic Code Is Mutation-Resistant missense mutations = mutations in which a single new base pair replaces another – most common type of mutation – in the third position of the codon, single base substitutions change the amino acid ~25% of the time silent mutations = mutations in which the nucleotide is different but the encoded amino acid stays the same 34 Transition Mutations transition mutation = a missense mutation in which a purine is replaced by a purine, or a pyrimidine by a pyrimidine – all three codon positions have some resistance to these mutations mutation in the first base of a codon usually produces a conservative substitution – for example: changing GUU (Val) to AUU (Ile) substitutes one hydrophobic amino acid for another 35 Question 8 Which mutation is an example of a transition mutation? A. B. C. D. G→A G→T A→C A→T 36 Response Which mutation is an example of a transition mutation? A. G→A A transition mutation is when a purine is replaced by another purine or a pyrimidine is replaced by another pyrimidine. 37 An Overview of the Five Stages of Protein Synthesis rRNA catalyzes rRNA decodes Do you know what small and large ribosomal units do? 38 Stage 1: Aminoacyl-tRNA Synthetases Attach the Correct Amino Acids to Their tRNAs stage 1: – occurs in the cytosol – activates the carboxyl group of each amino acid – establishes a link between each new amino acid and the information encoding it in the mRNA tRNAs are “charged” when attached to their amino acid (aminoacylated) 39 Stage 2: A Specific Amino Acid Initiates Protein Synthesis the AUG initiation codon specifies an amino-terminal methionine residue. all organisms have two tRNAs for methionine: – one for when (5′)AUG is the initiation codon – one when a Met residue in an internal position in a polypeptide 40 Bacterial Ribosomes Have Three Sites that Bind tRNAs the aminoacyl (A) site = site where incoming aminoacyltRNAs bind the peptidyl (P) site = site where amino acids are added to the growing chain the exit (E) site = binds uncharged tRNAs 41 Question 9 Why do a great many eukaryotic proteins start with a Met residue on the amino terminus? A. At the start of synthesis, only the Met tRNA can bind to the regulatory site on the small ribosomal subunit. B. Most proteins are soluble and the signal sequence always starts with a Met. C. The sulfur of Met forms a thioester linkage to the small ribosomal subunit during synthesis. D. The initiation codon also codes for Met. E. There is currently no accepted hypothesis for this observation. 42 Response Why do a great many eukaryotic proteins start with a Met residue on the amino terminus? D. The initiation codon also codes for Met. The initiation codon AUG is the most common signal for the beginning of a polypeptide in all cells, in addition to coding for Met residues in internal positions of polypeptides. 43 Stage 3: Peptide Bonds Are Formed in the Elongation Stage (many steps) elongation requires: – the initiation complex – aminoacyl-tRNAs – elongation factors – GTP (energy) 44 Elongation Step in Peptide Bond Formation the N-formylmethionyl group is transferred to the amino group of the aminoacyl-tRNA in the A site to form a dipeptidyl-tRNA – α-amino group of the amino acid in the A site acts as the nucleophile tRNA shift to a hybrid binding state – the 3′ and 5′ ends of tRNAfMet are in the E site – the 3′ and 5′ ends of the peptidyl-tRNA are in the P site 45 Elongation Step: Translocation in translocation, the ribosome moves one codon toward the 3′ end of the mRNA – shifts the anticodon of the dipeptidyl-tRNA from the A site to the P site (a-c) – shifts the deacylated tRNA from the P site to the E site (a-c) – leaves the A site open for a new aminoacyl-tRNA (c-e) ribosome translocation requires EF-G and energy provided GTP hydrolysis (pink) 46 The Elongation Cycle Repeats the attachment of a third amino acid residue occurs in the same way as addition of the second residue two GTPs are hydrolyzed for each amino acid residue added to the growing polypeptide 47 The Eukaryotic Elongation Cycle elongation cycle steps are similar to those in bacteria eukaryotic elongation factors have analogous functions to the bacterial elongation factors when a new aminoacyl-tRNA binds to the A site, an allosteric interaction leads to ejection of the uncharged tRNA from the E site 48 Question 10 Given the initiation and elongation phases of protein synthesis, how many nucleoside triphosphates (or their equivalents) are hydrolyzed to nucleoside diphosphates in order to make one peptide bond? A. B. C. D. E. one two more than 4 more than 8 There is no hydrolysis. All the reactions are phosphoryl transfers. 49 Response Given the initiation and elongation phases of protein synthesis, how many nucleoside triphosphates (or their equivalents) are hydrolyzed to nucleoside diphosphates in order to make one peptide bond? C. more than 4 Formation of each aminoacyl-tRNA uses two high-energy phosphate groups. An additional ATP is consumed each time an incorrectly activated amino acid is hydrolyzed by the deacylation activity of an aminoacyl-tRNA synthetase as part of its proofreading activity. A GTP is cleaved to GDP and Pi during the first elongation step, and another 50 during the translocation step. Stage 4: Termination of Polypeptide Synthesis Requires a Special Signal termination is signaled by a termination codon in the mRNA (UAA, UAG, UGA) occupying the A site termination factors (release factors) = the proteins RF1, RF2, and RF3 which function to: – hydrolyze the terminal peptidyl-tRNA bond – release the polypeptide and the last uncharged tRNA – cause dissociation of the 70S ribosome into its subunits eukaryotes have a single release factor, eRF 51 Energy Cost of Fidelity in Protein Synthesis at least four high-energy phosphate equivalents are required to generate each peptide bond – aminoacyl-tRNA formation uses two high-energy phosphate groups – a GTP is cleaved during the first elongation step – a GTP is cleaved during translocation – ATP is consumed each time an incorrectly activated amino acid is hydrolyzed during proofreading the energy investment is required to guarantee fidelity 52 Stage 5: Newly Synthesized Polypeptide Chains Undergo Folding and Processing during or after synthesis, a polypeptide progressively assumes its native conformation chaperones and chaperonins assist by restricting formation of unproductive aggregates and limiting the conformational space – correct chaperones prevent proteins having pathogenic consequences – ~750 different eukaryotic proteins ???? 53 Posttranslational Modifications many proteins require further processing by posttranslational modifications to attain their final active conformation 54 Amino-Terminal and CarboxylTerminal Modifications forming the final functional protein may include enzymatic removal of residues, including: – the formyl group – the amino-terminal Met residue – additional amino-terminal residues – carboxyl-terminal residues the amino group of the amino-terminal residue is N-acetylated in up to 50% of eukaryotic proteins carboxyl-terminal residues may also be modified 55 Question 11 Which statement about posttranslational modifications of newly synthesized proteins is false? A. The amino-terminal fMet or Met residue may be enzymatically removed. B. The amino-terminal Met residue in eukaryotes may be N-acetylated. C. Carboxyl-terminal residues are never modified or removed. D. Sequences that help direct the protein to its ultimate destination in the cell may be removed by specific peptidases. 56 Response Which statement about posttranslational modifications of newly synthesized proteins is false? C. Carboxyl-terminal residues are never modified or removed. In some cases, carboxyl-terminal residues may be removed enzymatically or modified in formation of the final functional protein. 57 Protein Synthesis Is Inhibited by Many Antibiotics and Toxins puromycin = an inhibitory antibiotic that binds to the A site and terminates polypeptide synthesis – made by the mold Streptomyces alboniger – structure similar to the 3′ end of an aminoacyl-tRNA 58 Tetracycline and Chloramphenicol tetracyclines = inhibit bacterial protein by blocking the A site – prevent aminoacyl-tRNA binding chloramphenicol = inhibits bacterial, mitochondrial, and chloroplast protein synthesis blocking peptidyl transfer – does not affect eukaryotic protein synthesis 59 Cycloheximide and Streptomycin cycloheximide = blocks the peptidyl transferase of 80S eukaryotic ribosomes – does not affect 70S bacterial, mitochondrial, and chloroplast ribosomes streptomycin = causes misreading of the genetic code in bacteria at low concentrations and inhibits initiation at higher concentrations 60 Do you want one? DNA and RNA Metabolism and Protein Synthesis Outline: Know bases, sugars, and phosphates involvement within nucleotides and the DNA structure, their roles, stability involvement DNA replication (all steps, components and what each does, order, location) Errors, Mismatches, mutations, proofreading, misreading: how they happen, how they are fixed, what other changes may occur, components involved, causes RNA synthesis and how it happens, polymerases, proteins involved and where Know how to use the codon chart (will be provided), steps of formation and order (mRNA and tRNA), codons, anticodons, order of protein synthesis and where steps occur and how they happen PCR Be able to make connection between the chapters since all topic are interrelated 61