Unit 4 - Gene Expression and Module 4A – Control of Gene Expression Genetic Technology Every cell contains thousands of genes 4A. Control of Gene Expression which code for proteins. However, every gene is not actively 4B. Biotechnology producing proteins at all times. 4C. Genomics In this module, we will examine some 4D. Genes and Development of the factors that help regulate when a gene is active, and how strongly it is expressed. 1 2 Objective # 1 Objective 1 Prokaryotes: Explain the importance of gene ¾ are unicellular or colonial regulation in both prokaryotes ¾ evolved to quickly exploit transient and eukaryotes. resources ¾ main role of gene regulation is to allow cells to adjust to changing conditions ¾ in the same cell, different genes are active at different times 3 4 Objective 1 Objective # 2 Eukaryotes: ¾ mostly complex multicellular organisms List and describe the different ¾ evolved the ability to maintain a stable levels of gene regulation, and internal environment (homeostasis) identify the level where genes ¾ main role of gene regulation is to allow are most commonly regulated. specialization and division of labor among cells ¾ at the same time, different genes are active in different cells 5 6 1 Objective 2 Objective 2 We can classify levels of gene To be expressed, a gene must be regulation into 2 main categories: transcribed into m-RNA, the m-RNA must be translated into a protein, and ¾ Transcriptional controls - factors that the protein must become active. regulate transcription ¾ Posttranscriptional controls – factors Gene regulation can theoretically occur that regulate any step in gene at any step in this process. expression after transcription is complete 7 8 Levels of 4. Destruction of 5.Translation. Many gene control the transcript. Many enzymes proteins take part in in degrade mRNA, and translation, and gene expression can regulating the activity Eukaryotes be regulated by PO Amino modulating the 4 or availability of any of acid DNA degree to which the transcript is them can alter the rate protected. of gene expression. Nuclear 3′ membrane PO4 3′ RNA Completed polymerase 5′ Cap polypeptide tRNA 1. Initiation of 5′ transcription. Small Most control of gene 3′ ribosomal 5′ expression is achieved 5′ Primary subunit by regulating the RNA transcript Nuclear frequency of pore Ribosome moves transcription initiation. 5′ toward 3′ end 3′ Large Exons Cap ribosomal 3′ Poly-A mRNA subunit tail Poly-A Cytoplasm Introns tail Ribosome mRNA 3′ 3. Passage through the 2. RNA splicing. Gene nuclear membrane. 6. Post-translational modification. Phosphorylation or expression can be controlled Gene expression can be other chemical modifications can alter the activity of by altering the rate of splicing regulated by controlling access in eukaryotes. to or efficiency of transport a protein after it is produced. channels. 9 10 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Objective 2 Objective # 3 It is most efficient to regulate genes at the first step of gene expression – Explain what regulatory namely transcription. proteins do and describe how Both prokaryotes and eukaryotes rely they identify specific sequences primarily on transcriptional controls to regulate gene expression. on the DNA double helix. 11 12 2 Objective 3 Objective 3 In order to regulate transcription, Biologists used to think the DNA special proteins called regulatory double helix had to unwind and proteins must bind to specific regions separate before proteins could “read” of the DNA called regulatory the sequence of base pairs. sequences. However, careful study has revealed 2 How does each regulatory protein helical grooves winding around the recognize the appropriate regulatory DNA molecule. The deeper one is sequence on the DNA? called the major groove and the shallower one is the minor groove. 13 14 Objective 3 The Helix-Turn-Helix Motif Regulatory proteins can read DNA α Helix Turn sequences by inserting bent regions (Recognition helix) of the protein chain, called “DNA- binding motifs”, into the major α Helix groove. Turn α Helix 3.4 nm Several common DNA-binding 90° motifs are shared by many different regulatory proteins. One example is the helix-turn-helix motif: Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 15 16 Objective 3 Once the DNA-binding motif is inserted into the major groove, it examines the chemical groups that project into the groove. Since each of the 4 possible base pairs has a different set of chemical groups that extend into the groove, the regulatory protein can determine the base pair sequence by examining these groups: 17 18 3 Objective # 4 Objective 4 Gene regulation relies on both negative Distinguish between negative and positive control systems: control systems and positive ¾ In negative control systems, the regulatory protein is a repressor which control systems. binds to DNA and blocks transcription. ¾ In positive control systems, the regulatory protein is an activator which binds to DNA and promotes transcription. 19 20 Objective 4 Objective 4 We will look at 3 regulatory systems. 2) In regulation by induction: 1) In regulation by repression: ¾ The repressor alone can bind to ¾ The repressor alone cannot bind to DNA and block transcription. DNA and block transcription. ¾ When an inducer binds to the ¾ However, when a co-repressor binds repressor, the repressor is to the repressor, the repressor is inactivated and can no longer bind activated. Once activated, the repressor can bind to DNA and to DNA and block transcription. block transcription. 21 22 Objective 4 3) In regulation by activation: ¾ The activator alone cannot bind to DNA and stimulate transcription. ¾ When an inducer joins with the activator, the activator can bind to DNA and stimulate transcription. 23 24 4 Objective # 5 Objective 5 Prokaryotes can quickly turn production Describe the process of gene of specific proteins on or off in regulation in prokaryotic cells, response to environmental conditions. and provide examples of both This rapid response is facilitated by 3 negative and positive important mechanisms: transcriptional control in a ¾ simultaneous transcription and translation prokaryotic cell such as E. coli. ¾ short-lived m-RNAs ¾ 25 operons 26 Objective 5 Objective 5 Operons: ¾ A single promoter serves all 5 genes. ¾ In prokaryotes, functionally related Recall that the promoter is the region genes are often located next to each where RNA polymerase binds to other and are transcribed as a unit. DNA and begins transcription ¾ For example, in E. coli 5 different ¾ The genes are transcribed as a unit, enzymes are needed to synthesize the producing one long mRNA molecule amino acid tryptophan. The genes (polycistronic mRNA) which contains that code for these enzymes are the code to make all 5 enzymes. located together. 27 28 Objective 5 Objective 5 ¾ There is also a single regulatory switch, called the operator. Transcription of the 5 coding genes in the tryptophan operon is blocked ¾ The operator is positioned within the when a transcriptional repressor binds promoter, or between the promoter and to the operator. the protein coding genes. It controls access of RNA polymerase to the genes. The repressor binds to the operator only when it is activated by the ¾ All together, the operator, promoter, presence of tryptophan: and the genes they control is called an operon. 29 30 5 Tryptophan repressor The tryptophan Objective 5 repressor is activated when it joins to tryptophan Because transcription of the coding genes in the tryptophan operon can be Tryptophan blocked by a repressor, it is an example of negative transcriptional DNA control: 3.4 nm 31 32 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Objective 5 Other regulatory proteins act as transcriptional activators by enhancing the ability of RNA polymerase to join with the promoter. For example, CAP (catabolite activator protein) stimulates transcription of genes that allow E. coli to use other food sources when glucose is not present: 33 34 Objective 5 cAMP binds to CAP, altering its shape so ¾ Low levels of glucose lead to high it can bind to DNA: levels of cAMP. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. ¾ The cAMP binds to CAP, altering its DNA shape so it can bind to DNA near any of several promoters. This bends the DNA, making it easier for RNA polymerase to join with the promoter and initiate transcription: CAP cAMP 35 36 6 Objective 5 lac Operon of E. coli The lactose operon is present in Gene for repressor protein Gene for E. coli. It controls the production of permease CAP-binding Operator Promoter site 3 enzymes needed to digest lactose for I gene Promoter for Gene for Gene for (a disaccharide made of glucose and lac operon β-galactosidase transacetylase P I CAP Plac O I Z galactose). Y A The lac operon consists of a coding Regulatory region Coding region region and a regulatory region: lac control system Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 37 38 Objective 5 Objective 5 Since glucose is the preferred food for E. coli, it makes sense to produce the 3 How lactose affects the lac operon: enzymes needed to digest lactose only ¾ When lactose is absent, the lac when lactose is present AND glucose repressor binds to the operator and is absent. blocks transcription of the genes Therefore, the lac operon is affected needed to digest lactose: by the levels of both lactose and glucose in the cell. 39 40 When lactose is absent, the lac repressor Objective 5 binds to the operator and blocks transcription: ¾ When lactose is present, a metabolite lac repressor polypeptide mRNA lac Repressor of lactose called allolactose binds to (No lactose present) CAP–binding site the repressor and inactivates it.