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12. Regulation of Genome Activity. Learning outcomes. When you have read Chapter 12 , you should be able to: Distinguish between differentiation and development, and outline how regulation of genome expression underlies these two processes
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Learning outcomes When you have read Chapter 12, you should be able to: Distinguish between differentiation and development, and outline how regulation of genome expression underlies these two processes Describe, with examples, the various ways in which extracellular signaling compounds can bring about transient changes in genome activity, making clear distinction between those signaling compounds that enter the cell and those that bind to a cell surface receptor Describe, with examples, the various ways in which permanent and semipermanent changes in genome activity can be brought about, making clear distinction between those processes that involve rearrangement of the genome, those that involve changes in chromatin structure, and those that involve feedback loops Discuss how studies of sporulation in Bacillus subtilis, vulva development in Caenorhabditiselegans, and embryogenesis in Drosophila melanogaster have contributed to our understanding of genome regulation during development, and explain why lower organisms can act as models for development in higher eukaryotes such as humans
Figure 12.1. Two ways in which genome activity is regulated. The genes on the left are subject to transient regulation and are switched on and off in response to changes in the extracellular environment. The genes on the right have undergone a permanent or semipermanent change in their expression pattern, resulting in the same three genes being expressed continuously.
12.1. Transient Changes in Genome Activity 12.2. Permanent and Semipermanent Changes in Genome Activity 12.3. Regulation of Genome Activity During Development
12.2. Permanent and Semipermanent Changes in Genome Activity
Figure 12.2. Two ways in which an extracellular signaling compound can influence events occurring within a cell.
Figure 12.3. Three ways in which an extracellular signaling compound could influence genome activity
Figure 12.4. Copper-regulated gene expression in Saccharomyces cerevisiae. Yeast requires low amounts of copper because a few of its enzymes (e.g. cytochrome c oxidase and tyrosinase) are copper-containing metalloproteins, but too much copper is toxic for the cell. When copper levels are low, the Mac1p protein factor is activated by copper binding and switches on expression of genes for copper uptake. When the copper levels are too high, a second factor, Ace1p, is activated, switching on expression of a different set of genes, these coding for proteins involved in copper detoxification.
Figure 12.5. Gene activation by a steroid hormone. Estradiol is one of the estrogen steroid hormones. After entering the cell, estradiol attaches to its receptor protein and the complex enters the nucleus where it binds to the 15-bp estrogen response element (abbreviation: N, any nucleotide), which is located upstream of those genes activated by estradiol and other estrogens. Other steroid hormone receptors recognize other response elements. For example, glucocorticoid hormones target the sequence 5′-AGAACANNNTGTTCT-3′. Note that this sequence, and that of the estrogen response element, is an inverted palindrome. The response element for vitamin D3, which is a steroid derivative that activates transcription via a nuclear receptor (see the text), has the sequence 5′-AGGTCANNNAGGTCA-3′, which is a direct repeat rather than an inverted palindrome.
Figure 12.6. All steroid hormone receptor proteins have similar structures. Three receptor proteins are compared. Each one is shown as an unfolded polypeptide with the two conserved functional domains aligned. The DNA-binding domain is very similar in all steroid receptors, displaying 50–90% amino acid sequence identity. The hormone-binding domain is less well conserved, with 20–60% sequence identity. The activation domain (Section 9.3.2) lies between the N terminus and the DNA-binding domain, but this region displays little sequence similarity in different receptors.
Figure 12.7. Catabolite repression. (A) A typical diauxic growth curve, as seen when Escherichia coli is grown in a medium containing a mixture of glucose and lactose. During the first few hours the bacteria divide exponentially, using the glucose as the carbon and energy source. When the glucose is used up there is a brief lag period while the lac genes are switched on before the bacteria return to exponential growth, now using up the lactose. (B) Glucose overrides the lactose repressor. If lactose is present then the repressor detaches from the operator and the lactose operon should be transcribed, but it remains silent if glucose is also present. Refer to Figure 9.24Bfor details of how the lactose repressor controls expression of the lactose operon. (C) Glucose exerts its effect on the lactose operon and other target genes by controlling the activity of adenylate cyclase and hence regulating the amount of cAMP in the cell. The catabolite activator protein (CAP) can attach to its DNA-binding site only in the presence of cAMP. If glucose is present, the cAMP level is low, so CAP does not bind to the DNA and does not activate the RNA polymerase. Once the glucose has been used up, the cAMP level rises, allowing CAP to bind to the DNA and activate transcription of the lactose operon and its other target genes.
Figure 12.8. The role of a cell surface receptor in signal transduction. Binding of the extracellular signaling compound to the outer surface of the receptor protein causes a conformational change that results in activation of an intracellular protein, for example by phosphorylation. The events occurring ‘downstream' of this initial protein activation are diverse, as described in the text. ‘P' indicates a phosphate group, PO32-.
Figure 12.9. Signal transduction involving STATs. (A) If the receptor is a member of the tyrosine kinase family then it can activate the STAT directly. (B) If the receptor is a tyrosine-kinase-associated type then it acts via a Janus kinase (JAK), which autophosphorylates when the extracellular signal binds and then activates the STAT. Note that activation of the JAK usually involves dimerization, the extracellular signal inducing two subunits to associate, resulting in the version of the JAK with phosphorylation activity. Dimerization is also central to activation of a STAT, phosphorylation causing two STATs, not necessarily of the same type, to form a dimer. This dimer is able to act as a transcription activator. ‘P' indicates a phosphate group, PO32-
Figure 12.10. Signal transduction by the MAP kinase pathway. See the text for details. ‘MK' is the MAP kinase and ‘P' indicates a phosphate group, PO32-. Elk-1, c-Myc and SRF (serum response factor) are examples of transcription factors activated at the end of the pathway.
Figure 12.11. The Ras signal transduction system. See the text for details. Abbreviations: GAP, GTPase activating protein; GNRP, guanine nucleotide releasing protein. ‘P' indicates a phosphate group, PO32-
Figure 12.12. Induction of the calcium second messenger system. See the text for details. Abbreviations: DAG, 1,2-diacylglycerol; Ins(1,4,5)P3, inositol-1,4,5-trisphosphate; PtdIns(4, 5)P2, phosphatidylinositol-4,5-bisphosphate