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What is a gene? • A sequence of DNA nucleotides that specifies the primary structure of a polypeptide chain (tells the cell how to make it) • Genes-made of nucleotides • Proteins-made of amino acids • How does a nucleotide code (in the nucleus) specify an amino acid sequence (in the cytoplasm)? The Central Dogma • DNA is transcribed into RNAcharacteristics of RNA • RNA is translated into protein • Advantages • Exceptions LE 17-4 Gene 2 DNA molecule Gene 1 Gene 3 DNA strand (template) 5 3 TRANSCRIPTION mRNA 5 3 Codon TRANSLATION Protein Amino acid The Genetic code-characteristics • Triplet (3 nucleotides=codon=info for a specific amino acid);64 different codons (3 are stop codons) • Universal • Redundant (61 codons-20 amino acids)variability in third nucleotide of codon. Advantages of a redundant code? • Non-overlapping • Exceptions (ciliates; mito/chloroplasts) Third mRNA base (3 end) LE 17-5 Second mRNA base Figure 17-06 Gene Expression • If a gene is transcribed and the m-rna is translated (the gene is expressed); a protein is made. This often changes the phenotype of the cell that produces the protein. • Differential gene expression is involved in embryonic development and cell specialization. • Totipotency-each cell has the genetic information for an entire organism. • Differential gene expression results in cell specialization (differentiation) • Hormones often play a role in gene expression Transcription • • • • • • • The first step in gene expression Takes place in the nucleus Requirements A. RNA nucleotides B. DNA template (gene) C. Enzymes (RNA polymerase) Only one of the two DNA strands is copied (template strand) LE 17-7a-2 Promoter Transcription unit 5 3 Start point RNA polymerase DNA 3 5 Initiation 5 3 RNA Template strand Unwound tran- of DNA DNA script 3 5 LE 17-7a-3 Promoter Transcription unit 5 3 Start point RNA polymerase 3 5 DNA Initiation 5 3 3 5 RNA Template strand Unwound tran- of DNA DNA script Elongation Rewound DNA 5 3 3 5 RNA transcript 3 5 LE 17-7a-4 Promoter Transcription unit 5 3 Start point RNA polymerase 3 5 DNA Initiation 5 3 3 5 RNA Template strand Unwound tran- of DNA DNA script Elongation Rewound DNA 5 3 3 5 3 5 RNA transcript Termination 5 3 3 5 5 Completed RNA transcript 3 LE 17-7b Elongation Non-template strand of DNA RNA nucleotides RNA polymerase 3 3 end 5 Direction of transcription (“downstream”) 5 Newly made RNA Template strand of DNA LE 17-8 Promoter Eukaryotic promoters 5 3 3 5 TATA box Start point Template DNA strand Several transcription factors Transcription factors 5 3 3 5 Additional transcription factors RNA polymerase II 5 3 Transcription factors 3 5 5 RNA transcript Transcription initiation complex Transcription-some important details • Rate-30-60 nucleotides/second • RNA polymerase (Many forms in eucaryotes, 3 basic types in bacteria: type I transcribes r-rna, type II-mrna, types III-trna) • Promotors-(approximately 100 nucleotides)strong and weak promotors • Eukaryotes-transcription factors needed to help RNA polymerase to bind to TATA box (region of promotor 25 nucleotides upstream from initiation site). RNA products of transcription • • • • • • m-rna t-rna r-rna sn-RNA (small nuclear) mi-Rna (micro) Si-rna (small interfering) Not all genes code for proteins (m-rna)-Rrna and t-rna are obvious examples • Actually, recent discoveries indicate that a large part of the eukaryotic genome is non-coding RNA-Introns • Small rna (micro rna and small interfering rna)play a crucial role in the regulation of gene expression involving both transcription and translation. Rna interference (Rnai) • We’ll talk about regulation of gene expression in Chapter 15. Ribosomal RNA and ribosomes • R-rna; one of two important components of ribosomes (other is protein-some of the proteins are enzymes). 60% r-rna; 40% protein. • Ribosomes consist of 2 subunits • Ribosomes needed to translate proteins • “workbench of protein synthesis” • Position t-rna (which is attached to a specific amino acid) on the codon of a m-rna • Result is the synthesis of a protein (whose amino acid sequence is specified by the m-rna; which is transcribed from a gene) LE 17-16b P site (Peptidyl-tRNA binding site) A site (AminoacyltRNA binding site) E site (Exit site) E P A mRNA binding site Schematic model showing binding sites Large subunit Small subunit LE 17-16a tRNA molecules Growing polypeptide Exit tunnel Large subunit E P A Small subunit 5 3 mRNA Computer model of functioning ribosome Ribosomal –rna processing T-rna • Single polynucleotide chain folded into a complex 3-D shape (inter-chain H bonding). 75-80 nucleotides in length • Binds a specific amino acid (involvement of amino-acyl-trna-synthetase • Attaches to a specific m-rna codon via its anticodon • How many different t-rna’s are there? 61? Actually only 45 (wobble) LE 17-14a 3 Amino acid attachment site 5 Hydrogen bonds Anticodon Two-dimensional structure Amino acid attachment site 5 3 Hydrogen bonds 3 Anticodon Three-dimensional structure 5 Anticodon Symbol used in this book “Charging” t-rna with its specific amino acid • “charging” enzyme-amino acyl t-rna synthetase (20 different enzymes) • Requires ATP LE 17-15 Amino acid Aminoacyl-tRNA synthetase (enzyme) Pyrophosphate Phosphates tRNA AMP Aminoacyl tRNA (an “activated amino acid”) Messenger Rna (m-rna) • Contains the information for the primary sequence of a polypeptide chain • Consists of codons • Binds to ribosomes • T-rna binds to m-rna (codon/anticodon) LE 17-13 Amino acids Polypeptide tRNA with amino acid attached Ribosome tRNA Anticodon Codons 5 mRNA 3 Translation • • • • • • • Codons (m-rna) read by ribosomes/t-rna Polypeptide chain produced 3 steps in translationA. initiation B. elongation C. termination Translation is a process that consumes a tremendous amount of energy (ATP and GTP) LE 17-16c Amino end Growing polypeptide Next amino acid to be added to polypeptide chain E tRNA mRNA 5 3 Codons Schematic model with mRNA and tRNA Translation-Initiation • Initiation codon is AUG • T-rna that bonds to AUG has an anticodon UAC-this carries the amino acid methionine • Requires a GTP molecule • Requires proteins called initiation factors. LE 17-17 Large ribosomal subunit P site Initiator tRNA GTP GDP E A mRNA 5 3 5 3 Start codon mRNA binding site Small ribosomal subunit Translation initiation complex Translation-Elongation • The elongation cycle takes about 60 milliseconds • During elongation, one m-rna codon is read and then the ribosomes moves down the message to the next codon. • Binding of incoming t-rna to the A site of the ribosome requires a GTP • Translocation-requires a GTP LE 17-18 Amino end of polypeptide E 3 mRNA Ribosome ready for next aminoacyl tRNA P A site site 5 2 GTP 2 GDP E E P A P GDP GTP E P A A Translation-Termination • When the ribosome reaches a termination codon, it causes the m-rna/ribosome complex to separate • No t-rna binds to the termination codon. • Release factors • Newly made polypeptide chain is released (folds into its characteristic 3-D shape) LE 17-19 Release factor Free polypeptide 5 3 3 3 5 5 Stop codon (UAG, UAA, or UGA) When a ribosome reaches a stop codon on mRNA, the A site of the ribosome accepts a protein called a release factor instead of tRNA. The release factor hydrolyzes the bond between the tRNA in the P site and the last amino acid of the polypeptide chain. The polypeptide is thus freed from the ribosome. The two ribosomal subunits and the other components of the assembly dissociate. Summary of energy demands for protein synthesis • A. B. C. D. A rough estimate is that for every amino acid incorporated into a polypeptide chain, 3 ATP/GTP are consumed Charging the amino acid (1 ATP) Binding of incoming t-rna into the A site (1 GTP) Translocation (1 GTP) So a small protein (120 amino acids in length) would cost the cell 360 ATP/GTP to make (the equivalent of 12 glucose molecules going through aerobic cell respiration) Polyribosomes • A single ribosome can translate an average-sized polypeptide in about 1 minute • Several ribosomes can translate the same message one after the other. • Increases the efficiency of protein production LE 17-20a Growing polypeptides Completed polypeptide Incoming ribosomal subunits Start of mRNA (5 end) End of mRNA (3 end) An mRNA molecule is generally translated simultaneously by several ribosomes in clusters called polyribosomes. LE 17-20b Ribosomes mRNA 0.1 m m This micrograph shows a large polyribosome in a prokaryotic cell (TEM). M-rna modifications • Eukaryotic M-rna is modified extensively after transcription (while its still in the nucleus) • These modifications include A.Polyadenylation-added to 3’ end of m-rna B. 5’ cap C. Intron removal M-RNA modifications • • • • Poly A tail A. added to the 3’ end of the m-rna B.30-200 Adenine nucleotides C. roles-regulation of transport of m-rna out of the nucleus; regulation of degradation of m-rna in the cytoplasm; helps m-rna attach to small ribosomal subunit M-RNA modifications (continued) • 5’ cap • A. Modified guanine nucleotide stuck onto 5’ end of m-rna • B. Roles- positioning of m-rna on small ribosomal subunit in initiation; protects m-rna from degradation LE 17-9 Protein-coding segment Polyadenylation signal 5 5 Cap 5 UTR Start codon Stop codon 3 UTR Poly-A tail Introns • Many eukaryotic genes have nucleotide sequences that don’t code for amino acids (Introns) • Introns separate coding sequences (exons). Split genes • Introns must be removed from the m-rna before it is translated (introns have nucleotide sequences that indicate splicing sites) • Splicesomes are molecular machines that remove introns from m-rna LE 17-11-1 RNA transcript (pre-mRNA) 5 Exon 1 Intron Exon 2 Protein Other proteins snRNA snRNPs Spliceosome LE 17-11-2 Spliceosome 5 Spliceosome components Cut-out intron mRNA 5 Exon 1 Exon 2 Significance of introns • Why would chromosomes carry around extra DNA that isn’t used in the final mrna? A. Expensive to maintain (energy). B. Splicing out introns is a risky business (what if it’s done incorrectly) C. With these disadvantages, there must be an advantage or natural selection would not favor this arrangement Benefits of Introns • Evolution of protein diversity • One gene can be alternatively spliced in a number of different ways to form several different types of m-rna (alternative splicing) • Human antibody genes-about 500 genes can code for billions of different antibody molecules because of alternative splicing. Figure 15.12 Exons DNA 2 1 3 5 4 Troponin T gene Primary RNA transcript 2 1 3 4 5 RNA splicing mRNA 1 2 3 5 or 1 2 4 5 Summary of Transcription and Translation Mutation • An alteration in the nucleotide sequence of a DNA molecule (chromosome) • Chromosomal mutations (duplications; deletions; inversions) • Point mutations-alterations of one or a few nucleotides in a gene Point mutations • • • • • • Spontaneous mutations Induced mutations Consequences of mutationsA. no effect-”silent mutations” B. harmful mutations-(may be lethal) C. beneficial mutations (rare) Spontaneous mutations • Base pairing errors; why aren’t they corrected by DNA repair enzymes? • Effects: • A. no effect-silent mutation (redundancy of genetic code; alteration of a non-critical amino acid) • B. Positive effect-rare • C. negative effect-missense mutations; nonsense mutations LE 17-24a Wild-type mRNA 5 Protein Amino end 3 Stop Carboxyl end LE 17-24b Base-pair substitution No effect on amino acid sequence U instead of C Stop Missense A instead of G Stop Nonsense U instead of A Stop Sickle cell anemia • Results of a spontaneous missense mutation • Result-altered hemoglobin molecule • Effect-Depends on the environmental conditions and number of copies of the defective gene you inherited. LE 17-23 Wild-type hemoglobin DNA 3 Mutant hemoglobin DNA 5 3 mRNA 5 mRNA 3 Normal hemoglobin 5 5 3 Sickle-cell hemoglobin Induced mutations • Caused by environmental damage • Radiation (UV)- T-T dimers; excision repair enzymes; xerdoerma pigmentosa • Chemicals-Common result-base pair addition or deletion • Result of addition or deletion (frame shift mutation)-missense or nonsense • Worst scenario-addition/deletion of 1 or 2 nucleotides at the beginning of a gene LE 17-25 Wild type mRNA 5 Protein 3 Stop Carboxyl end Amino end Base-pair insertion or deletion Frameshift causing immediate nonsense Extra U Stop Frameshift causing extensive missense Missing Insertion or deletion of 3 nucleotides: no frameshift but extra or missing amino acid Missing Stop Mutations and Cancer • Many mutations make cells cancerous • 90% of known carcinogens are mutagens • Ames test-screens potential chemicals for being carcinogens by seeing if they are mutagens • Bacteria are the test subjects in the Ames test. Employee Resource Manual • Plasmids • Restriction Endonucleases • Agarose Gel Electrophoresis Plasmids • Small extrachromosomal pieces of DNA found in some bacterial species • May carry additional genes (such as antibiotic resistance) • Can be genetically modified and used as vectors for genetic engineering PUC 18-Plasmid Restriction Endonucleases • Produced by some bacteria as a defense against virus infection • Cleave DNA at specific bases sequences (different recognition site for each different enzyme) • Can be used to join DNA from 2 different sources (plasmid DNA and genomic DNA) ECOR1 Agarose Gel Electrophoresis • Separates DNA based upon size differences • DNA is pulled through a gel by an electric current • (-) charged DNA is pulled to the positive pole of the apparatus. • Smaller pieces of DNA migrate through the gel faster than larger pieces of DNA Agarose Gel Electrophoresis Procell Procell in Action What is your first job assignment? • Clone the H gene (use a bacteria to make copies of the gene for us) What kind of bacteria do we use to clone the H gene? • E.coli (lacZ(-), amp sensitve) Where is the H gene located? • Lambda virus How do you get the cloned gene into the bacteria so the bacteria can copy it? • Transform E.coli (lacZ(-), amp sensitve) with PUC 18-lambda plasmid (heat shock and osmotic shock) Lambda virus genes have been inserted into the plasmids here How do you get lambda genes into PUC 18 plasmid? • Incubate PUC-18 and lambda with EcoRI, ligate products How many different plasmids do you get when you mix PUC 18 and lambda, both of which have been ECOR1 and then ligated? 7 Would all 7 of the plasmids be recombinant (have lambda DNA)? No! How do you tell if bacteria have been transformed successfully with PUC-18 plasmid? • They will grow on amp agar. How can you distinguish whether plasmids that transformed bacteria were recombinant (lambda and PUC-18) or nonrecombinant (pUC-18 only)? Plate the transformed cells on Xgal-amp agar E.Coli transformed With nonrecombinant Plasmids (PUC-18) E.Coli transformed With recombinant Plasmids (PUC-18/lambda)