Lecture 23: Gene Expression Part 3 - PDF

 A cell does NOT express all of its genes, all of the time. Cells are very selective about 1. which genes they express, 2. how strongly they are expressed (i.e., level), and 3. Gene Chips: DNA Microarrays when they are expressed. allow us t...

 A cell does NOT express all of its genes, all of the time. Cells are very selective about 1. which genes they express, 2. how strongly they are expressed (i.e., level), and 3. Gene Chips: DNA Microarrays when they are expressed. allow us to look at gene expression patterns for thousands of genes in cells Why is this important? 2  Regulation of gene Regulation of gene expression in unicellular expression in multicellular organisms allows a cell to: organisms allows a cell to: A. Change cell activities in A. Change cell activities in response to signals from other cells and the cell’s response to its environment. environment. B. Develop into a specific cell type B. Use limited resources and during development. energy efficiently. C. To participate in a coordinated response in the adult body. 3  Change P Change Change protein frequency of mRNA life activity by translation span chemical initiation or rate (stability) modifications Change frequency of transcription initiation Protein Change protein life mRNA Ribosome span RNA RNA polymerase DNA Transcriptional control Translational control Post-translational control Regulation can be positively or negatively controlled by: 1. Transcriptional control: affects the level of mRNA made; slow but energetically efficient. 2. Translational control: affects the type/level of polypeptide made; intermediate speed and energy efficiency. 3. Post-translational control: affects the level of active protein; most rapid response but is energetically expensive. 4 (a) Negative control: Repressor regulatory protein shuts (b) Positive control: Activator regulatory protein triggers down transcription. transcription. TRANSCRIPTION No, or low levels of, transcription No negative No positive control... control... RNA polymerase Transcription unit RNA polymerase Repressor protein Activator protein No transcription TRANSCRIPTION With negative With positive control... control... Transcription unit Transcription can be positively or negatively controlled by regulatory proteins. Influence the frequency of transcription initiation à number of mRNA copies made. 5  Common regulation model in prokaryotes to regulate a number of genes that produce proteins involved in a common cell activity to coordinate a response. Co-transcription makes a polycistronic mRNA: Genes are adjacent and transcribed into one mRNA from a single promoter. A regulatory gene produces a regulatory protein influence transcription of the operon by positive or negative control based on cell conditions. Prevents transcription Cleaves Membrane of lacZ and lacY when lactose to transport lactose is absent glucose and protein, galactose imports lactose β-Galactosidase Galactoside Repressor permease Segment of E. coli lacI mRNA Polycistronic (multiple gene) mRNA genome lacl lacl Promoter Operator lacZ lacY lacA promoter of lac operon (Regulatory Sequence) Regulation gene lac operon The lac operon coordinates expression proteins involved in lactose metabolism when lactose is present. 6  (a) Lactose absent inside cell; repressor active: Repressor binds to DNA. Transcription is blocked. Repressor bound to operator blocks transcription Outside Promoter Inside cell cell Repressor is synthesized Galactoside permease lacI lacZ lacY Plasma Repressor (Normal gene) membrane RNA polymerase (b) Lactose present in cell; repressor inactive: Lactose (the inducer) binds to repressor. Repressor releases from DNA. Transcription occurs. β-Galactosidase Permease Repressor is mRNA synthesized lacI lacZ lacY Lactose (Normal gene) Repressor Repressor releases allosteric regulation A repressor regulatory protein inhibits transcription in the absence of a stimulus. An inducer (lactose) binds and inactivates the repressor to activate transcription. 7  The ara operon controls the expression of genes that produce proteins involved in arabinose metabolism: araB, araA, and araD. Under positive regulation by the AraC regulatory protein produced from the araC gene. AraC acts as a activator when bound to arabinose. The ara operon coordinates expression proteins involved in arabinose metabolism when arabinose is present. 8  Gene expression and its regulation is more complex in eukaryotes than in prokaryotes. Several steps in gene expression are unique to eukaryotes and can be used as targets to fine-tune gene expression. Which steps? Differential Gene Expression: the expression of a different set of genes by different tissue cells with the same genome in a multicellular organism. The need for each cell type to have a unique pattern of gene expression may explain why control of gene expression is so much more complex in multicellular eukaryotes than in bacteria. 9 Gene Expression Signal Regulation Targets in Eukaryotes Chromatin Transcriptional control 1. Chromatin Post-transcriptional / Translational control Modification Post-translational control DNA Gene available for transcription 2. Transcription Initiation Frequency RNA Exon Primary transcript Intron 3. RNA Processing Tail Cap mRNA in nucleus 4. Transport to NUCLEUS Cytoplasm CYTOPLASM 10  CYTOPLASM mRNA 5. Lifespan of 6. Translation Transcriptional control Initiation and Rate mRNA Post-transcriptional / Translational control Post-translational control Polypeptide 7. Protein Processing Functional Protein 8. Lifespan of Protein 9. Transport to Cellular Destination Protein is active Which level of control does alternative RNA splicing fall into? 11  Enhancer for albumin gene Promoter Albumin gene DNA in both liver and lens cells Enhancer for contains the albumin gene and crystallin gene Promoter Crystallin gene the crystallin gene. Why is albumin expressed in a liver cell but not in a lens cell and why is crystallin expressed LENS CELL NUCLEUS in a lens cell but not a liver LIVER CELL NUCLEUS Available Available cell? activators activators A. Chromatin structure of those genes differ in the two cells. How? B. Available activators Albumin gene (regulatory transcription Albumin gene not expressed factors) differ in the two expressed cells. Crystallin gene not expressed Crystallin Liver cell Lens cell gene expressed 12  Transcriptional control is the Increase in Transcription primary control strategy in Regulatory Chromatin-remodeling transcription complex (or HATs) factor gene expression. What is the 1. Chromatin remodeling. advantage over other strategies? Primarily due to control of: Exposed DNA (chromatin is relaxed) A. Chromatin remodeling link Exon Intron Exon Intron Exon Distal with gene expression? Distal Proximal Promoter Enhancer element Enhancer Element Transcription Unit Element B. Frequency of transcription 2. Exposure of promoter and control elements. initiation at the promoter cer 1. Regulatory En an h transcription an En h cer factors (pink) DNA loop Chromatin remodeling 2. Mediator DNA loop complexes 3. Assembly of proteins. (purple) factors bind to chromatin 3. Basal transcription complex (blue) Promoter tags to alter chromatin structure. 4. Attachment of RNA TRANSCRIPTION Transcription factors bind polymerase and initiation of transcription. to control elements and RNA polymerase II Transcription Initiation influence the frequency of transcription initiation. 13  Small-scale mutations within a gene can be divided into three classes based on their effect on DNA structure. 1. Nucleotide pair __________ 2. Nucleotide pair __________ 3. Nucleotide pair __________ Small-scale mutations that involve one nucleotide pair are referred to as __________ mutations. 14  A substitution is the change of a nucleotide pair to another one. 15  Base pair substitutions can be classified into three types based on their effects on protein structure and function: 1. Silent mutations 2. Missense mutations 3. Nonsense mutations 16  Wild type (Normal Sequence) DNA template strand 3! T A C T T C A A A C C G A T T 5! 5! A T G A A G T T T G G C T A A 3! mRNA5! A U G A A G U U U G G C U A A 3! Protein Met Lys Phe Gly Stop Amino end Carboxyl end (a) Nucleotide-pair substitution: silent A instead of G 3! T A C T T C A A A C C A A T T 5! 5! A T G A A G T T T G G T T A A 3! U instead of C 5! A U G A A G U U U G G U U A A 3! Met Lys Phe Gly Stop 1. Silent mutations: the same amino acid is added to a polypeptide because of the redundancy in the genetic code. Has no effect on protein function. 17  Wild type (Normal Sequence) DNA template strand 3! T A C T T C A A A C C G A T T 5! 5! A T G A A G T T T G G C T A A 3! mRNA5! A U G A A G U U U G G C U A A 3! Protein Met Lys Phe Gly Stop Amino end Carboxyl end (a) Nucleotide-pair substitution: missense T instead of C 3! T A C T T C A A A T C G A T T 5! 5! A T G A A G T T T A G C T A A 3! A instead of G 5! A U G A A G U U U A G C U A A 3! Met Lys Phe Ser Stop 2. Missense mutations: still code for an amino acid, but a different amino acid in a polypeptide. Can have varying effects of protein structure and function. 18  Wild type (Normal Sequence) DNA template strand 3! T A C T T C A A A C C G A T T 5! 5! A T G A A G T T T G G C T A A 3! mRNA5! A U G A A G U U U G G C U A A 3! Protein Met Lys Phe Gly Stop Amino end Carboxyl end (a) Nucleotide-pair substitution: nonsense A instead of T T instead of C 3! T A C A T C A A A C C G A T T 5! 5! A T G T A G T T T G G C T A A 3! U instead of A 5! A U G U A G U U U G G C U A A 3! Met Stop 3. Nonsense mutations: change an amino acid codon into a stop codon, leading to a truncated (shortened) polypeptide and, most likely, a nonfunctional protein. Effect depends on how many of the total amino acids are missing (determined by the position of the new stop codon). 19  An insertion is an addition of one (or more) nucleotide pair(s). A deletion is the loss of a one (or more) nucleotide pair(s). 20  These mutations have a disastrous effect on protein structure and function. Produce a frameshift and is referred to as a frameshift mutation. The addition or deletion shifts the codon reading frame. A frameshift changes all of the codons after the mutation and results in extensive missense. If the polypeptide does not contain any of the right amino acids, it is a non-functional protein. Extent of missense and effect on polypeptide depends on the position of the mutation in the gene. 21 Wild type (Normal Sequence) DNA template strand 3! T A C T T C A A A C C G A T T 5! 5! A T G A A G T T T G G C T A A 3! mRNA5! A U G A A G U U U G G C U A A 3! Protein Met Lys Phe Gly Stop Amino end Carboxyl end (b) Nucleotide-pair insertion or deletion: frameshift causing extensive missense A missing 3! T A C T T C A A C C G A T T 5! 5! A T G A A G T T G G C T A A 3! U missing 5! A U G A A G U U G G C U A A 3! Met Lys Leu Ala 1 nucleotide-pair deletion 22 What would be the consequence of adding or deleting 3 nucleotide pairs in series. 23  Virtually all organisms on the tree of life use the same genetic code. Differences within the code are minor modifications: e.g., some protists use UAA and UAG to code for glutamine instead of stop. Indicates that the genetic code is ancient and inherited from the ancestor of all current organisms. 24  The structure of the code does not seem to represent a random assemblage of bases. Seems to be a product of natural selection. The code is structured in a way that minimizes the effects of __________ mutations. How? substitution 25 The sequence below represents the coding (non-template) strand of the first exon of a gene that encodes a polypeptide. The shaded, italicized nucleotide represents the +1 site (transcription start site) and the underlined region is part of the promoter. Wild-type (normal) exon 1 sequence: 5’-TAGTTAATTATCGTGGCAATGGTAAATCAGACCGCTGGATCTTTTGTG-3’ a) What would be the sequence of the template strand of this gene region? N.B. Be sure to include the orientation of the sequence (i.e., 5’ and 3’ ends). b) What would be the mRNA sequence produced for this exon? N.B. Be sure to include the orientation of the sequence (i.e., 5’ and 3’ ends). c) Identify the 5’ untranslated region (5’UTR) in the mRNA sequence above. d) What would be the amino acid sequence of the peptide produced for this exon? N.B. You can use the 3-letter codes for the amino acids used in the genetic code table above. Be sure to include the orientation of the sequence (i.e., N-terminus and C-terminus). 26 The following sequences represent possible mutations of the wild-type sequence of exon 1 discussed above. Mutation 1: 5’-TAGTTAATTATCGTGGCAATGGTAAAATCAGACCGCTGGATCTTTTGTG-3’ Mutation 2: 5’-TAGTTAATTATCGTGGCAATGGTAAATCATACCGCTGGATCTTTTGTG-3’ Which of the mutations (1 or 2) would have the most harmful effect on the structure and function of the polypeptide? To answer this question, answer the following guiding questions for each mutation: i) identify the mutation in the sequence ii) list the name of the mutation with respect to the change at the DNA level iii) list the name used to describe the mutation based on its effect on the polypeptide iv) describe the possible effects of the mutation on the structure and function of the polypeptide. 27