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Unit #3 Schedule: • Previously: – Sanger Sequencing – Central Dogma Overview – Mutation – Transcription, RNA Processing, Translation • Last Class: – Central Dogma Sculpting • Today: – Regulation of Gene Expression + Trivia – StudyNotes 9 Due • Tutorial (Apr 5) • Review (Apr 9) • EXAM 3 (Apr 11) • Homework 6 Due Regulation of Gene Expression There are at least 300 different kinds of cells in the human body. Most of them have identical DNA. Regulation of Gene Expression We examine this at a very general level. Prokaryotes vs. Eukaryotes Definition: Operon • A region of DNA that codes for a series of functionally related genes and is transcribed from a single promoter into mRNA. Negative Control and Positive Control • Transcription can be regulated via negative control or positive control. • Negative control occurs when a regulatory protein binds to DNA and shuts down transcription. • Positive control occurs when a regulatory protein binds to DNA and triggers transcription. © 2011 Pearson Education, Inc. © 2011 Pearson Education, Inc. Negative Control Negative control occurs when a regulatory protein binds to DNA and shuts down transcription. Fig. 18-2 PROKARYOTIC REGULATION OF GENE EXPRESSION: Campbell Fig. 18-3a trp operon Promoter Promoter Genes of operon DNA trpR Regulatory gene mRNA Protein 5 trpE 3 Operator Start codon mRNA 5 RNA polymerase Inactive repressor (a) Tryptophan absent, repressor inactive, operon on E trpD trpC trpB trpA B A Stop codon D C Polypeptide subunits that make up enzymes for tryptophan synthesis Fig. 18-3b-2 DNA No RNA made mRNA Protein Active repressor Tryptophan (corepressor) (b) Tryptophan present, repressor active, operon off © 2011 Pearson Education, Inc. Fig. 18-4b lac operon DNA lacZ lacI 3 mRNA 5 lacA RNA polymerase mRNA 5 -Galactosidase Protein Allolactose (inducer) lacY Inactive repressor (b) Lactose present, repressor inactive, operon on Permease Transacetylase © 2011 Pearson Education, Inc. Negative Control Transcription of Trp Operon in the absence of Tryptophan. Tryptophan activates the repressor. Transcription of the Lac Operon in the presence of Lactose. Lactose deactivates the repressor. Positive Control Positive control occurs when a regulatory protein binds to DNA and triggers transcription. Fig. 18-5 CAP: catabolite activator protein What is cAMP? © 2011 Pearson Education, Inc. mc1r Gene Sequence 5’TGCCCACCCAGGGGCCTCAGAAGAGGCTTCTGGGTTCTCTCAACTCCACC TCCACAGCCACCCCTCACCTTGGACTGGCCACAAACCAGACAGGGCCTTGGT GCCTGCAGGTGTCTGTCCCGGATGGCCTCTTCCTCAGCCTGGGGCTGGTGAG TCTGGTGGAGAATGTGCTGGTCGTGATAGCCATCACCAAAAACCGCAACCTG CACTCGCCCATGTATTCCTTCATCTGCTGTCTGGCCCTGTCTGACCTGATGG TGAGTATAAGCTTGGTGCTGGAGACGGCTATCATCCTGCTGCTGGAGGCAGG GGCCCTGGTGACCCGGGCCGCTTTGGTGCAACAGCTGGACAATGTCATTGAC GTGCTCATCTGTGGCTCCATGGTGTCCAGTCTTTGCTTCCTTGGTGTCATTG CCATAGACCGCTACATCTCCATCTTCTATGCATTACGTTATCACAGCATTGT GACGCTGCCCCGGGCACGACGGGCCATCGTGGGCATCTGGGTGGCCAGCATC TTCTTCAGCACCCTCTTTATCACCTACTACAACCACACAGCCGTCCTAATCT GCCTTGTCACTTTCTTTCTAGCCATGCTGGCCCTCATGGCAATTCTGTATGT CCACATGCTCACCCGAGCATACCAGCATGCTCAGGGGATTGCCCAGCTCCAG AAGAGGCAGGGCTCCACCCGCCAAGGCTTCTGCCTTAAGGGTGCTGCCACCC TTACTATCATTCTGGGAATTTTCTTCCTGTGCTGGGGCCCCTTCTTCCTGCA TCTCACACTCATCGTCCTCTGCCCTCAGCACCCCACCTGCAGCTGCATCTTT AAGAACTTCAACCTCTACCTCGTTCTCATCATCTTCAGCTCCATCGTCGACC CCCTCATCTATGCTTTTCGGAGCCAGGAGCTCCGCATGACACTCAGGGAGGT GCTGCTGTGCTCCTGGTGA 3’ mc1r Gene Sequence 5’TGCCCACCCAGGGGCCTCAGAAGAGGCTTCTGGGTTCTCTCAACTCCACC TCCACAGCCACCCCTCACCTTGGACTGGCCACAAACCAGACAGGGCCTTGGT GCCTGCAGGTGTCTGTCCCGGATGGCCTCTTCCTCAGCCTGGGGCTGGTGAG TCTGGTGGAGAATGTGCTGGTCGTGATAGCCATCACCAAAAACTGCAACCTG CACTCGCCCATGTATTCCTTCATCTGCTGTCTGGCCCTGTCTGACCTGATGG TGAGTATAAGCTTGGTGCTGGAGACGGCTATCATCCTGCTGCTGGAGGCAGG GGCCCTGGTGACCCGGGCCGCTTTGGTGCAACAGCTGGACAATGTCATTGAC GTGCTCATCTGTGGCTCCATGGTGTCCAGTCTTTGCTTCCTTGGTGTCATTG CCATAGACCGCTACATCTCCATCTTCTATGCATTACGTTATCACAGCATTGT GACGCTGCCCCGGGCACGACGGGCCATCGTGGGCATCTGGGTGGCCAGCATC TTCTTCAGCACCCTCTTTATCACCTACTACAACCACACAGCCGTCCTAATCT GCCTTGTCACTTTCTTTCTAGCCATGCTGGCCCTCATGGCAATTCTGTATGT CCACATGCTCACCCGAGCATACCAGCATGCTCAGGGGATTGCCCAGCTCCAG AAGAGGCAGGGCTCCACCCGCCAAGGCTTCTGCCTTAAGGGTGCTGCCACCC TTACTATCATTCTGGGAATTTTCTTCCTGTGCTGGGGCCCCTTCTTCCTGCA TCTCACACTCATCGTCCTCTGCCCTCAGCACCCCACCTGCAGCTGCATCTTT AAGAACTTCAACCTCTACCTCGTTCTCATCATCTTCAGCTCCATCGTCGACC CCCTCATCTATGCTTTTCGGAGCCAGGAGCTCCGCATGACACTCAGGGAGGT GCTGCTGTGCTCCTGGTGA 3’ Consequence of Mutation • A single nucleotide mutation from a Cytosine to a Thymine leads to… • An amino acid change from an Arginine to a Cysteine Amino Acid Sequence Dark Fur: MPTQGPQKRLLGSLNSTSTATPHLGLATNQTGPWCLQVSIPDGLFLSLGLVSLVENVLVVIAI TKNRNLHSPMYSFICCLALSDLMVSISLVLETAIILLLEAGALVTRAALVQQLDNVIDVLICG SMVSSLCFLGVIAIDRYISIFYALRYHSIVTLPRARRAIXGIWVASIFFSTLFITYYNHTAVL ICLVTFFLAMLALMAXLYVHMLTRAYQHAQGIAQLQKRQGSTXQGFCLKGAXTLTIILGIFFL CWGPFFLHLTLIVLCPQHPTCSCIFKNFNLYLVLIIFSSIVDPLIYAFRSQELRMTLREVLLC SW Amino Acid Sequence Light Fur: MPTQGPQKRLLGSLNSTSTATPHLGLATNQTGPWCLQVSVPDGLFLSLGLVSLVENVLVVIAI TKNCNLHSPMYSFICCLALSDLMVSISLVLETAIILLLEAGALVTRAALVQQLDNVIDVLICG SMVSSLCFLGVIAIDRYISIFYALRYHSIVTLPRARRAIVGIWVASIFFSTLFITYYNHTAVL ICLVTFFLAMLALMAILYVHMLTRAYQHAQGIAQLQKRQGSTRQGFCLKGAATLTIILGIFFL CWGPFFLHLTLIVLCPQHPTCSCIFKNFNLYLVLIIFSSIVDPLIYAFRSQELRMTLREVLLC SW Changing 1 amino acid: • Arginine: – Strongest +charge – Very hydrophilic • Cysteine: – Not hydrophilic – Forms disulfide bonds Beach Mice - Missense substitution mutation of one nucleotide CT Changes one amino acid: Arginine Cysteine Changes the function of the MC1R protein How is eumelanin produced? • When the MC1R protein is stimulated, cAMP is produced • Lots of cAMP within a melanocyte cell will facilitate the expression of at least four genes: c(tyr), Tyrp1, Tyrp2, p How is eumelanin produced? • When cAMP is plentiful, c(tyr), Tyrp1, Tyrp2 and p are all expressed and their enzymes facilitate the biosynthetic pathway that leads to eumelanin production. How is eumelanin produced? • When cAMP is scarce, c(tyr), Tyrp1, Tyrp2 and p are not as readily expressed. • If only small amounts of cAMP are present, c(tyr) may still be expressed and its enzyme may facilitate the biosynthetic pathway leading pheomelanin production. How is eumelanin produced? • If c(tyr) is not adequately expressed it is possible that neither biosynthetic pigment production pathway may occur. This would result in no pigment production. The Melanocortin-1-Receptor Effectively stimulated by hormone Ineffectively stimulated by hormone MC1RR67 MC1RC67 Results in lots of cAMP production Results in little cAMP production c(tyr), p tyrp1, tyrp2 <<activated>> c(tyr) <<activated>> p, tyrp1, tyrp2 <not activated> LOTS of EUMELANIN produced EUMELANIN not produced Fig. 18-5 CAP: catabolite activator protein EUKARYOTIC REGULATION OF GENE EXPRESSION Activators and Enhancers of Transcription Campbell 8e, Fig. 18.8 Fig. 17-8 1 Promoter A eukaryotic promoter includes a TATA box Template 5 3 TATA box Start point 2 Transcription factors 3 5 Template DNA strand Several transcription factors must bind to the DNA before RNA polymerase II can do so. 5 3 3 5 3 Additional transcription factors bind to the DNA along with RNA polymerase II, forming the transcription initiation complex. RNA polymerase II Transcription factors 5 3 3 5 5 RNA transcript Transcription initiation complex Campbell 8e, Fig. 18.9 Controlling Gene Expression: Enhancers and Activators Provide a way to turn on specific genes in specific cells Different genes have different enhancers Different cells have different activators Campbell 8e, Fig. 18.10 Controlling Gene Expression: Enhancers and Activators Tissue- and cell-type specific gene expression Liver cells make albumin, but not crystallin Lens cells make crystallin, but not albumin Campbell 8e, Fig. 18.10 Coming Up: Friday: • Tutorial (3-5pm in C-3) Tuesday: • Interactive Review (WhiteBoards + Clickers) Thursday: • Midterm Exam 3 Review Part 1 Clicker Review With respect to nucleotide bonds: (A)A-T is stronger than C-G (B) C-G is stronger than A-T (C) A-T and C-G have approximately equal strength Which mode of information transfer usually does not occur? (A)DNA to DNA (B) DNA to RNA (C) DNA to protein (D)RNA to protein (E) All occur in a working cell In replication of DNA, the helix is opened and untwisted by (A)DNA polymerase (B) ligase (C) helicase (D) telomerase (E) topoisomerase _______________ joins DNA fragments to the lagging strand. (A) Telomere (B) DNA Polymerase I (C) Helicase (D) DNA Polymerase III (E) Ligase In a nucleic acid, the phosphate group, nitrogenous base and free hydroxyl group are attached to the _______________ carbons of ribose (respectively). (A) 1', 3', 5' (B) 5', 3', 1' (C) 3', 5', 1' (D) 5', 1', 3' (E) 3', 1', 5' DNA polymerase III is thought to add nucleotides (A) to the 5' end of the RNA primer (B) to the 3' end of the RNA primer (C) in the place of the primer RNA after it is removed (D) on single stranded templates without need for an RNA primer (E) in the 3' to 5' direction Considering the structure of double stranded DNA, what kinds of bonds hold one complementary strand to the other? (A)peptide (B) covalent (C) hydrogen (D) phosphodiester (E) ionic The nitrogenous base adenine is found in all members of which group? (A)proteins, ATP, and DNA (B) proteins, carbohydrates and ATP (C) glucose, ATP and DNA (D) ATP, RNA and DNA (E) proteins, glycerol and hormones Where and how are Okazaki fragments synthesized? (A)on the leading strand, in a 5’ 3’ direction (B) on the leading strand, in a 3’ 5’ direction (C) on the lagging strand, in a 5’ 3’ direction (D)on the lagging strand, in a 3’ 5’ direction Which of the following types of mutation, resulting in an error in the mRNA just after the AUG start of translation, is likely to have the most serious effect on the polypeptide product? (A) insertion of a codon (B) deletion of two codons (C) substitution of the third nucleotide in an ACC codon (D) deletion of a nucleotide (E) insertion of 9 nucleotides Regarding beach mice: The C T substitution at position 199 of the mc1r gene: A. arose by a mutation in the beach mouse populations in response to a need for protection from predation. B. leads to the failure of melanocytes to make an MC1R protein. C. arose by a mutation then increased in frequency because it was selectively advantageous in the beach mouse populations. D. had no effect on the beach mouse populations. E. produced an alternate allele that was detrimental to mice on the white sand beaches Regarding beach mice: What was the reason for the lighter coat colors of the mice on the white sand beaches? (A) Owls and other carnivores prey on beach mice that do not carry the mutant allele. (B) A substitution of cysteine for arginine at position 67 of the MC1R protein. (C) A substitution of thymine for cystosine at position 199 of the mc1r gene nucleotide sequence. (D) The failure of melanocytes to lay down melanin pigment in the cortex of hairs of the lighter colored beach mice. (E) The poorer binding affinity for α-MSH and the lower amount of cAMP produced by individuals with the mutated MC1R protein. The template strand of DNA at the beginning of a protein-coding region has the sequence: 5'–TACTGGGATAGCC*TACAT–3' The “*” indicates the position of a point mutation: a T originally present at this location has been deleted. This deletion will most likely result in _____. (A) mRNA codons preceding the mutation being misread (B) mRNA codons following the mutation being misread (C) no change in the polypeptide coded by this gene (D)the AUC triplet functioning as a chain terminator Biologists use the terms transcription and translation to describe the two steps in genetic information flow from DNA to protein. Which of the following is correct? (A) Transcription is the synthesis of protein from mRNA by ribosomes; translation is the synthesis of mRNA from DNA by RNA polymerase. (B) Transcription is the synthesis of mRNA from DNA by ribosomes; translation is the synthesis of protein from mRNA by RNA polymerase. (C) Transcription is the synthesis of protein from mRNA by RNA polymerase; translation is the synthesis of mRNA from DNA by ribosomes. (D) Transcription is the synthesis of mRNA from DNA by RNA polymerase; translation is the synthesis of protein from mRNA by ribosomes. Transcription in eukaryotes requires which of the following in addition to RNA polymerase (A) several transcription factors (B) the protein product of the promoter (C) start and stop codons (D) ribosomes and tRNA (E) a signal recognition particle Transcription occurs along a ____ template forming an mRNA in the ____ direction. (A) 5' to 3'; 5' to 3' (B) 5' to 3'; 3' to 5' (C) 3' to 5'; 5' to 3' (D) 3' to 5'; 3' to 5’ (E) All of the above could be correct depending on the orientation. Assume that RNA polymerase transcribes a gene containing the section of DNA shown below: 5'–GATGCGAATCGT–3' 3'–CTACGCTTAGCA–5' If the top strand were the template strand, the RNA corresponding to this section would be _____. (A) 5'–GATGCGAATCGT–3’ (B) 5'–GAUGCGAAUCGU–3’ (C) 5'–ACGATTCGCATC–3’ (D) 5'–ACGAUUCGCAUC–3' What happens when RNA polymerase reads a stop codon? (A)The RNA transcript is cleaved off. (B) RNA polymerase detaches from the DNA. (C) Both of the above. (D)Neither of the above. How is transcription terminated in eukaryotes? (A)A stop codon is read. (B) A hairpin loop forms. (C) A polyadenylation sequence is read. (D)A termination sequence is read. (E) None of the above. The chicken ovalbumin gene is 7,700 base pairs in length, yet the mature messenger RNA is only 1872 nucleotides. Which of the following correctly describes the composition of the mature messenger RNA: (A) It has, in part, the complementary sequence to the poly-A tail of the gene. (B) It lacks the complementary sequence to the exons of the gene. (C) It has, in part, the complementary sequence to the TATA box of the promoter. (D) It has, in part, the complementary sequence to the 5' cap of the gene. (E) It lacks the complementary sequence to the introns of the gene. A spliceosome: (A)Splices out introns from mRNA (B) Splices out exons from pre-mRNA (C) Splices out exons from mRNA (D)Splices out introns from pre-MRNA (E) None of the above Why is a 5’ poly-A tail added to an mRNA transcript in eukaryotes? (A)To help the mRNA bind to the ribosome. (B) To prevent the mRNA from degrading on its journey from the nucleus to the cytosol. (C) To promote the binding of a signal recognition particle. (D)Both (A) and (B) (E) A 5’ poly-A tail is not added to an mRNA transcript in eukaryotes. What modifications are made to an mRNA transcript in prokaryotes prior to translation? (A)A 5’ modified guanine cap is added. (B) A poly-A tail is added. (C) Introns are spliced out. (D)All of the above. (E) None of the above. Which site in a ribosome accepts a charged tRNA? (A)The aminoacyl tRNA binding site (B) The peptidyl tRNA binding site (C) The exit site (D)The initiation site (E) The start codon A particular triplet of bases in the template strand of DNA is 5'–TGA–3'. Which of the following is the anticodon component of the tRNA that binds the mRNA codon transcribed from this DNA? (Note: By convention, the 3' end of each anticodon is written on the left, and the 5' on the right.) (A) ACU (B) AGU (C) AGT (D) UGA What type of bond forms between amino acids in a growing polypeptide chain: (A)peptide (B) covalent (C) hydrogen (D) phosphodiester (E) ionic How is translation terminated? (A)A stop codon is read by the ribosome. (B) A hairpin loop forms in the polypeptide chain. (C) A signal recognition particle cleaves the chain off. (D)A release factor adds water to the growing polypeptide chain. (E) The small and large subunit of the ribosome break apart. Which of the following pieces of mRNA would be successfully translated into a polypeptide chain in a prokaryote? (mRNA shown in its entirety) (A) (B) (C) (D) (E) 5’-AUGAUCCCGUCCCGGGCACCUUAG-3’ 5’-UGAGCUGCGCCCAAUGCUUGGCAA-3’ 5’-CGACGACCCGGUUACGAAUCUAAC-3’ 5’-GGCUAAGAGUCUAGUAUCUGGAAG-3’ Any/all of the above Which component is not directly involved in translation? (A)mRNA (B) tRNA (C) ribosomes (D)GTP (E) All of the above are directly involved. What is the function of a signal recognition particle? (A)To help mRNA bind to the ribosome. (B) To help mRNA find a ribosome. (C) To help a ribosome bind to the RER. (D)To block transcription inhibitors. (E) None of the above. Muscle cells differ from nerve cells mainly because they (A)express different genes (B) contain different genes (C) use different genetic codes (D)have unique ribosomes (E) have different chromosomes