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Peter J. Russell CHAPTER 17 Regulation of Gene Expression in Eukaryotes edited by Yue-Wen Wang Ph. D. Dept. of Agronomy,台大農藝系 NTU 遺傳學 601 20000 Chapter 17 slide 1 Levels of Control of Gene Expression in Eukaryotes 1. Prokaryotes respond quickly to their environments mainly by transcriptional (regulatory proteins bind DNA) control. Translational control also occurs, mediated by stability of the mRNAs. 2. Eukaryotes have more complex means to regulate gene expression, because they have compartments (e.g., nucleus) within cells, and often multicellular structures that require differentiation of cells. 3. Levels at which expression of protein-coding genes is regulated in eukaryotes: a. Transcription. b. mRNA processing and transport. c. Translation. d. Degradation of mRNA. e. Protein processing. f. Protein degradation. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 2 Fig. 17.1 Levels at which gene expression can be controlled in eukaryotes Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 3 Transcriptional Control 1. Eukaryotic protein-coding gene expression is controlled by: a. Promoters situated just upstream of the transcription start site. i. Some promoter elements (e.g., TATA) are required to specify the start of transcription, through binding of transcription factor proteins (Figure 17.2). ii.Promoters may be either positively or negatively regulated. iii.Regulatory promoter elements are specialized, involving binding by regulatory proteins specific for control of one or a few genes. iv. A particular gene may have 1 to many regulatory promoter elements, and 1 to many regulatory proteins involved in controlling its function. v. Binding of regulatory proteins to promoters is highly specific to ensure that only the correct genes are activated. b. Enhancers are located some distance away, either upstream or downstream. i. Enhancers determine whether maximal transcription of the gene occurs. ii. Regulatory proteins bind specific enhancer elements. Which ones bind is determined by the DNA sequence recognized by each protein. iii. Specific transcription factors may form a loop in the DNA when they bind, by making contact with: (1)The enhancer element. (2)Transcription factors bound to the promoter. iv. Protein interactions determine whether transcription is activated or repressed. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 4 Fig. 17.2 Schematics of (a) positive regulation of gene transcription and (b) negative regulation of gene transcription Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 5 2. Promoters and enhancers bind specific regulatory proteins. a. Some regulatory proteins occur in most or all cell types, but others are very specific. b. Each promoter and enhancer has a particular set of proteins that can bind it, and the combination of proteins bound will determine its expression. c. If both positive and negative regulatory proteins are bound, interactions between them will control the rate of expression. d. When regulatory proteins bind an enhancer and have strong negative effect, the enhancer is a silencer element. e. Enhancers and promoters appear to bind many of the same proteins, implying interactions of regulatory proteins. i. Relatively few proteins are combined in a variety of ways, regulating the transcription of different arrays of genes. ii. A large number of cell types can be specified by combinatorial gene regulation. 3. Transcription factors and regulatory proteins bind specific DNA sequences and exert their effects. Structural motifs are defined for some DNA binding proteins. Examples of the DNA binding domains include (Figure 17.3): a. Helix-turn-helix. b. Zinc finger. c. Leucine zipper. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 6 台大農藝系 遺傳學 601 20000 Chapter 17 slide 7 Chromosome Changes and Gene Expression 1. The hypothesis that transcriptionally active genes have a looser chromosome structure than inactive genes was examined in DNase I digestion experiments. a. For example, erythroblasts synthesize globin, but not ovalbumin. i. Chromatin was extracted from erythroblasts and from control cells that don’t make globin. ii. Samples were treated with different concentrations of DNase I. iii. DNA was then extracted, digested with BamHI and Southern blotted with globin DNA as the probe. iv. The result was gradual disappearance of the globin DNA during DNase I treatment (indicating that the DNA was loosened enough to allow DNase I to make contact with the DNA). v. Chromatin from control cells not expressing the globin gene showed no decrease in globin DNA over exposure to DNase I. vi. The ovalbumin gene in erythroblast chromatin was also not degraded, corresponding with its lack of expression in these cells. b. Nearly all transcriptionally active genes have increased DNase I sensitivity. The DNA in these regions may still be organized into nucleosomes, but is less highly coiled than inactive regions. c. Regions hypersensitive to DNase I have also been identified. Most are upstream from transcription start sites, and include promoters and other sites of regulatory protein binding. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 8 台大農藝系 遺傳學 601 20000 Chapter 17 slide 9 2. Histones are highly conserved in both sequence and organization, and are distributed consistently on chromosomal DNA in all cell types. a. Uniform distribution of the histones on DNA means that they cannot account for the specific control of gene expression. Histones act as a general repressor of transcription, because they interfere with protein binding to DNA. b. However, histones may be acetylated and phosphorylated, altering their ability to bind DNA. This can free specific DNA regions to interact with proteins involved in gene expression. c. Experiments investigating histone influence on promoter DNA and gene expression show: i. If DNA is simultaneously mixed with both histones and promoter-binding proteins, it binds more readily to the histones, forming nucleosomes at the TATA box and preventing transcription. ii. If DNA is first mixed with promoter-binding proteins, adding histones does not produce nucleosomes, and transcription occurs. iii. If DNA is simultaneously mixed with histones, promoter-binding proteins and enhancer-binding proteins: (1) Enhancer-binding proteins bind the enhancer sequences. (2) Promoter-binding proteins bind the promoter sequences. (3) Histones are unable to bind, and so transcription occurs. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 10 d. Histones, therefore, are effective repressors, but other proteins can overcome that repression. e. A model for gene activation is that: i. Histones form nucleosomes on TATA boxes, blocking transcription. ii. Promoter-binding proteins cannot disrupt the nucleosomes. iii. Enhancer-binding proteins bind to enhancers, displacing any histones, and then cause the histones at the TATA box to free the DNA. f. Research has correlated histone acetylation with increased transcription. i. Histone are acetylated on lysines in regions on the outside of the nucleosome. ii. Acetylation destabilizes higher-order chromatin structure. iii. DNA becomes more accessible to transcription factors, overcoming histone repression of transcription. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 11 DNA Methylation and Transcriptional Control 1. A small proportion of DNA is chemically modified by enzymes. This may play a role in replication, repair or transcription, but information is lacking in most cases. 2. One modified base that has been correlated with gene activity is 5methylcytosine (5mC), which is generated shortly after DNA replication by DNA methylase (Figure 17.5). a. Higher eukaryotes like mammals have about 3% of their cytosines modified to 5mC, while lower eukaryotes have virtually 0%. b. 5mC is nonrandomly distributed, with most found in the symmetrical sequence CG. This allows patterns of methylation to be studied by using restriction enzymes that contain the CG sequence in their recognition sites. For example: i. HpaII cuts at 5’-CCGG-3’, but only if the cytosines are unmethylated. ii. MspI also cuts at 5’-CCGG-3’, regardless of methylation. iii. Differences in the array of DNA fragments produced by these enzymes on Southern blotting allow methylation patterns to be inferred (Figure 17.6). 台大農藝系 遺傳學 601 20000 Chapter 17 slide 12 Fig. 17.5 Production of 5-methylcytosine from cytosine in DNA by the action of the enzyme DNA methylase Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 13 Fig. 17.6a Effect of 5-methylcytosine on cleavage of DNA with HpaII and MspI Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 14 Fig. 17.6b Effect of 5-methylcytosine on cleavage of DNA with HpaII and MspI Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 15 c. Many systems have been examined with the HpaII/MspI approach, looking for relationships between gene methylation and transcriptional activity i. For at least 30 genes, DNA methylation and transcription are correlated, with lower levels of methylated DNA in transcriptionally active genes. ii. However, not all methylated C nucleotides in a gene region become demethylated when the gene is expressed. iii. No cause and effect information can be derived from these experiments. 3. For at least 30 genes, DNA methylation and transcription are correlated, with lower levels of methylated DNA in transcriptionally active genes. However, no cause and effect information can be derived from these experiments. 4. Other recent observations also indicate a role for methylation in gene expression: a. A methylase is essential for development in mice. b. Methylation is involved in fragile X syndrome, where expansion of a triplet repeat and abnormal methylation in the FMR-1 gene silence its expression. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 16 Transcriptional Control of Gene Expression in Eukaryotes 1. Short-term regulation of gene expression can be transcriptional and/or posttranslational. Transcriptional mechanisms are considered here. 2. An example is the galactose-utilization genes of yeast: a. Three genes encode enzymes for converting galactose to glucose 6-phosphate (Figure 17.7): i. GAL1 encodes galactokinase. ii. GAL7 encodes galactose transferase. iii. GAL10 encodes galactose epimerase. b. Glucose 6-phosphate enters the glycolytic pathway. Enzymes for the glycolytic pathway are constitutively produced. c. The GAL genes are not transcribed in the absence of galactose. d. When galactose is present and glucose is low or absent, the GAL genes are rapidly and coordinately induced. Glucose therefore exerts catabolite repression. e. The GAL genes are near each other, but do not constitute an operon. f. Another nearby gene, GAL4, encodes a protein similar to the phage λ repressor in its DNA binding and transcription activation domains. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 17 Fig. 17.7 The galactose metabolizing pathway of yeast Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 18 g. The DNA binding domain of Gal4p is a zinc finger that binds the promoter element called an upstream activator sequence (UASG). i. The UASG is located between the GAL1 and GAL10 genes. ii.UASG has four copies of a 17 bp sequence with 2-fold symmetry. iii. Gal4p binds each 17 bp sequence as a dimer (like λ repressor and Cro). h. Transcription occurs in both directions from UASG, with GAL10 to one side, and GAL1 to the other. i. A model for GAL gene transcription is shown in Figure 17.8. i. When galactose is absent, a Gal4p dimer binds UASG, along with another protein, Gal80p. No transcription can occur. ii. If galactose is added, its metabolite binds Gal80p, causing Gal4p to be phosphorylated at certain amino acids. iii. Phosphorylation changes the Gal4p-Gal80p complex conformation to the activating form needed for transcriptional activation. iv. Thus, Gal4p acts as a transcriptional activator, and galactose as an effector. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 19 Fig. 17.8 Model for the activation of the GAL genes of yeast Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 20 Animation: Regulation of Gene Expression by Steroid Hormones 3. Steroid hormone regulation in animals is another example of short-term control of gene expression. a. Cells of higher eukaryotes perform specialized functions, and are shielded from rapid changes in their environments. One way that a constant environment is maintained in a multicellular organism is by hormone signals. i. Levels of each hormone are maintained by complex feedback loops. ii. Hormones are effector molecules produced by one cell and causing a physiological response in another cell. b. Hormones may deliver their signals in different ways: i. Some (e.g., steroid hormones) bind cytoplasmic receptors (e.g., steroid hormone receptor, SHR) and then the complex binds directly to DNA, regulating gene expression. ii. Others (e.g., polypeptide hormones like insulin and vasopressin) work at the cell surface by activating a transmembrane enzyme such as adenylate cyclase. The cAMP then acts as a second messenger, transducing the signal to activate cellular events. c. Hormones act only on target cells that have receptors capable of binding the hormone. Receptors for polypeptide hormones are generally on the cell surface, while steroid hormone receptors are inside the cell. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 21 d. Steroid hormones are well studied. All have a common four-ring structure, and physiological effects derive from the differences in side groups (Figure 17.10). e. Steroid hormones show tissue-specific effects. Examples (Table 17.1): i. Estrogen induces prolactin in rat pituitary vitellogenin in frog liver, and conalbumin, lysozyme, ovalbumin and ovomucoid in the hen oviduct. ii. Glucocorticoids induce synthesis of growth hormone in rat pituitary and phosphoenolpyruvate carboxykinase in rat kidney. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 22 Fig. 17.9 Mechanisms of action of polypeptide hormones and steroid hormones Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 23 Fig. 17.10 Structures of some mammalian steroid hormones Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 24 f. Hormone receptors control the specificity of the response, because only cells with receptors can detect and respond to the hormone. i. Steroid hormones affect transcription and stability of mRNAs, and possibly processing of mRNA precursors. ii. SHRs have high affinity for their respective hormones, and all work in the same way. (1) If the hormone is absent, its SHR is found associated with chaperone proteins, including Hsp90. The SI-JR is inactive. (2) When hormone enters the cell, it binds its specific SHR, displacing Hsp90 and forming a glucocorticoid-SHR complex (Figure 17.11). (3) When steroid hormone binds SHR, the complex is found in the nucleus, where it binds specific DNA regulatory sequences with its zinc finger domains, activating or mactivating transcription within minutes. g. Genes responsive to a specific steroid hormone have a common DNA sequence, the steroid hormone response element (HRE) for steroid-receptor complex binding. i. The H is replaced with a letter to indicate the specific steroid hormone involved. ii. For example, GRE is glucocorticoid response element, and ERE is estrogen response element. iii. HREs are located in enhancer regions, often in multiple copies that show twofold symmetry. h. The mechanism of transcription regulation is not known, but presumably involves interactions between bound hormone-receptor complexes and transcription factors. i. In different types of cells, the same steroid hormone may activate different sets of genes, even with the same SHRs. Many regulatory proteins are involved, yielding different patterns of gene expression. j. In sum, steroid honnones act as effectors, and SHRs as regulatory molecules. Together, they bind DNA and regulate gene transcription. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 25 Fig. 17.11 Model for the action of steroid hormone glucocorticoid in mammalian cells Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 26 4. Plants also exhibit control by hormones. a. There are five main types of plant hormones (Figure 17.12): i. Ethylene. ii. Abscisic acid. iii. Auxins. iv. Cytokinins. v. Gibberellins. b. Each hormone is responsible for many activities. Gibberellins are an example: i. Gibberellins stimulate transcription, resulting in cell form and differentiation. ii. Gibberellins can make dwarf mutant plants grow tall, or normal plants grow taller. iii. They are also present in the aleurone layer of seeds, where they stimulate transcription of the gene for α-amylase, which breaks down endosperm, releasing nutrients for the embryo. iv. No gibberellin receptor has been found, and details of its activity are still unknown. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 27 Fig. 17.12 Chemical structures of the five plant hormones Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 28 Fig. 17.13 Effect of gibberellins on the germination of barley seeds Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 29 RNA Processing Control Animation: RNA Processing Control 1. RNA processing control regulates mRNA production from precursor-RNAs. a. Alternative processing options exist, including: i. Alternative polyadenylation. ii. Differential splicing (alternative splicing). b. Alternative polyadenylation and splicing occur independently of each other, and their activities may be tissue specific. c. Specific products depend on regulatory signals. Alternative polyadenylation and splicing produce proteins encoded by the same gene but structurally and functionally different (protein isoforms). 台大農藝系 遺傳學 601 20000 Chapter 17 slide 30 d. An example of alternative polyadenylation and splicing involves the human calcitonin gene (CALC), which has five exons and four introns (Figure 17.14). i. Alternative polyadenylation sites exist next to exon 4 (pA1, used in thyroid cells) and exon 5 (pA2, used in neurons). ii. Alternative splicing also occurs: (1) In the thyroid, pre-mRNA is spliced, bringing together exons 1, 2, 3 and 4. (2) In neurons, pre-mRNA is spliced to bring together introns 1, 2, 3 and 5. Exon 4 is excised and discarded. iii. The mRNAs are translated to produce prehormones, from which hormones are generated by protease cleavage. The products produced are: (1) Calcitonin in the thyroid, a circulating calcium-ion homeostatic hormone. (2) Calcitonin-gene related peptide (CGRP) in the hypothalamus, which has neuromodulatory and trophic (growth-promoting) activities. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 31 Fig. 17.14 Alternative polyadenylation and alternative splicing resulting in tissuespecific products of the human calcitonin gene, CALC Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 32 e. Alternative splicing is also important in sex-determination in Drosophila. i. Drosophila sex is determined by the X-chromosome : autosome (X:A) ratio. ii. Mutations disrupting sex determination have led to a regulation cascade model. (1) First, the X:A ratio is read and transmitted to the sex determination genes. (2) The choice between male and female developmental pathways starts with the master regulatory gene Sxl (sex lethal). (3) Both males and females transcribe Sxl into a pre-mRNA with eight exons. (4) In females, all introns are removed, along with exon 3, and functional Sxl protein is produced. (5) In males, all introns are removed. Exon 3, which contains stop codons, remains in the mature mRNA, and no functional Sxl protein is made. (6) In females, the Sxl protein causes splicing of the tra (transformer) gene, producing a protein that works with the tra-2 protein to cause splicing of the dsx RNA. (7) In females, dsx protein interacts with the ix (intersex) protein to turn off genes for male differentiation, resulting in female structures. (8) In males, no Sxl protein is produced, and alternative splicing of tra pre-mRNA gives no functional protein. Alternative splicing of the dsx pre-mRNA gives a malespecific protein that represses female differentiation genes, resulting in a male. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 33 Fig. 17.15 Regulatory cascade for sex determination in Drosophila Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 34 Transport Control 1. In eukaryotes, transport control regulates movement of transcripts from nucleus to cytoplasm. a. Studies show that about ½ of primary (hnRNA) transcripts never leave the nucleus and are degraded there. b. Mature mRNAs appear to exit through nuclear pores, but the mechanism and signals are not understood. c. The spliceosome retention model proposes that snRNPs complete with nuclear export by staying bound to spliced introns and preventing their export, while releasing the spliced exons and allowing them to interact with nuclear pores. d. A correct 5’ cap also appears to be required for export to the cytoplasm. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 35 mRNA Translation Control 1. Ribosomal translational control, selecting mRNAs for translation, also has an impact on gene expression. a. Unfertilized eggs are an example, in which mRNAs show significant increases in translation after fertilization, without new mRNA synthesis. b. Stored mRNAs are associated with proteins that both protect them and inhibit their translation. c. Poly(A) tails promote translation initiation, and stored mRNAs generally have shorter tails. i. In some mRNAs of mouse and frog oocytes, a normal-length poly(A) tail is added, and then trimmed enzymatically. ii. Particular mRNAs are marked for deadenylation by a region in the 3’ untranslated region, called the adenylate/uridylate (AU)-rich element (ARE), with the consensus sequence UUUUUAU. iii. Activation of the stored mRNA occurs when a cytoplasmic polyadenylation enzyme recognizes the ARE and adds about 150 A residues, making a fulllength poly(A) tail. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 36 mRNA Degradation Control 1. Breakdown (turnover) of mRNAs occurs in the cytoplasm, with mRNAs showing a wide range of stability (from minutes to months). Regulatory signals may modify the stability of an mRNA (Table 17.2). 2. mRNA degradation is believed to be a major control point in eukaryotic gene expression. Sequences or structures that affect the 1/2-life of mRNAs include: a. AU-rich elements (ARE, discussed above). b. Various secondary structures. c. Deadenylation-dependent mRNA decay involves removal A nucleotides from poly(A) tails, until they are too short to bind PAB (poly(A) binding protein). i. In yeast, PAB-dependent poly(A) nuclease (product of the PANJ gene) may catalyze deadenylation. ii. When the tail is almost removed, decapping removes the 5’cap. The yeast decapping enzyme is at least partially encoded by the DCPJ gene. iii. After decapping in yeast an enzyme (from,the XRNJ gene) aggressively degrades the mRNA from the 5’end by 5’-3’exonuclease activity. iv. Degradation still occurs in dcpl mutants, indicating that other mRNA degradation pathways exist. d. Deadenylation-independent mRNA decay includes two types of pathways: i. Direct decapping, exposing the 5’end to 5’-3’exonucleases. ii. Internal cleavage of the mRNA, and then degradation of the fragments. 3. Mammalian mRNA degradation mechanisms are less clear than those of yeast. Both deadenylation-dependent and deadenylation-independent pathways are found Chapter in 台大農藝系 遺傳學 601 20000 17 slide 37 mammals. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 38 Protein Degradation Control 1. Protein regulation occurs in many ways. Examples: a. A constitutively produced mRNA may be translated continuously, and so the protein degradation rate determines its level. b. A short-lived mRNA may make a very stable protein, so that it persists for long periods in the cell. 2. Protein stability varies from very stable (e.g., lens proteins in the eyes of higher vertebrates) to short-lived (e.g., steroid receptors and heat shock proteins). 3. Proteolysis (protein degradation) in eukaryotes requires ubiquitin, a protein cofactor. a. Ubiquitin bound to a protein identifies it for degradation by proteolytic enzymes. b. Ubiquitin is released intact, and able to tag other proteins for degradation. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 39 4. Protein stability is directly related to the amino acid at the N-terminus of the protein (the N-end rule). In yeast, stability of the same protein was measured with different N-terminal amino acids: a. The amino acids Arg, Lys, Phe, Leu and Trp all specified a 1/2-life of ≦ 3 minutes. b. The amino acids Cys, Ala, Ser, Thr, Gly, Val, Pro, and Met all specified 1/2-lives of ≧ 20 hours. c. Similar results are seen in experiments with E. coli. 5. The N-terminal amino acid directs the rate of ubiquitin binding which, in turn, determines the 1⁄2-life of the protein. 6. To summarize, prokaryotes control gene expression mainly at the transcriptional level, while eukaryotes regulate at transcriptional, posttranscriptional and post-translational levels. Eukaryotic systems control: a. b. c. d. e. f. Transcription. Precursor-RNA processing. Transport from the nucleus. Degradation of mature RNAs. Translation of mRNAs. Degradation of protein. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 40 Gene Regulation in Development and Differentiation 1. From one genotype, the cells of a complex eukaryote produce specialized tissues and organs. Cells with the same genotype develop a variety of phenotypes. 2. Long-term gene regulation is described by two terms: a. Development results from interaction of the genome with the cytoplasm and the cell’s external environment. It involves a programmed sequence of typically irreversible cellular events that characterize the life cycle of the organism. b. Differentiation involves formation of cell types, tissues and organs through specific gene regulation. Differentiated cells have characteristic structures and functions. 3. Events of development and differentiation are studied at several levels: a. Events at the level of morphology are well described. b. Biochemical events are understood to some extent. c. Activation-repression patterns are still in the early stages of study. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 41 Genomic Activity in Multicellular Eukaryotes 1. Eukaryotic genomes have a large amount of DNA that does not encode proteins. a. In the sea urchin, a maximum of about 6% of the unique sequences are active at any one time, with mature tissues dropping below 0.8% gene expression. b. The majority of the genome is not transcribed, and the function of the nontranscribed DNA (if any) is not known. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 42 Constancy of DNA in the Genome During Development 1. An early question in the study of development was whether the genome of an adult cell is the same as that of a zygote. Key studies were done, one in plants, the other in animals. a. Steward (1950s) dissociated carrot phloem tissue cells, and grew mature plants from them using tissue culture techniques. This showed that mature cells had all the DNA necessary for plant development, supporting the idea of a constant genome. b. Cloning of a sheep by nuclear transfer into an enucleated egg was reported in 1997 by Wilmut and colleagues, followed by reports of cloning of mice, cattle and monkeys. This means that a differentiated adult somatic nucleus is totipotent, carrying all genetic information needed for development (Box 17.1). 台大農藝系 遺傳學 601 20000 Chapter 17 slide 43 Fig. 17.16 Cloning of a mature carrot plant from a cell of a mature carrot Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 44 Box Fig. 17.1 Representation of Wilmot’s sheep cloning experiment, which showed the total potency of the nucleus of a differentiated, adult cell Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 45 Differential Gene Activity Among Tissues and During Development 1. Differences in morphology occur in cells with the same genome. The simplest explanation is that different cell types are produced by expression of different genes. Some examples: 2. Human adult hemoglobin Hb-A is a tetramer with 2 α and 2 β polypeptides, each coded by a separate gene (α globin and β globin). a. The two globin genes appear to be the result of a gene duplication. b. Both genes contain two introns that are transcribed but removed by splicing during mRNA processing. (Figure 17.17). 台大農藝系 遺傳學 601 20000 Chapter 17 slide 46 Fig. 17.17 Molecular organization of the human -globin and -globin genes Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 47 c. Other globins occur in humans at different times during development (Figure 17.18). i. Hemoglobin for human embryos is made in the yolk sac. (1) Embryonic hemoglobin consists of 2 ζ (zeta) subunits and ε (epsilon) subunits. (2) Amino acid comparisons show that ζ is like α polypeptide, and ε is like β. ii. After about 3 months of embryonic development, hemoglobin synthesis shifts to the liver and spleen, which make fetal hemoglobin (Hb-F). (1) Hb-F contains 2 α polypeptides and 2 γ polypeptides (either 2 γA or 2 γG). (2) γA and γG differ by 1 amino acid (out of 146). Each is encoded by a separate gene. iii. Just before birth, synthesis of the two types of γ chain stops, and hemoglobin synthesis begins in the bone marrow. (1) Polypeptides α and β are made, along with some β-like δ subunits. (2) Newborn through adult, most human hemoglobin is Hb-A (the α2β2 tetramer), while about 1 in 40台大農藝系 molecules is Hb-A2 (α2-δ2). 遺傳學 601 20000 Chapter 17 slide 48 Fig. 17.18 Comparison of synthesis of different globin chains at given stages of development Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 49 d. Globin gene expression shows a complex regulatory pattern during development. i. In the genome, the α-like genes (2 α genes and 1 ζ gene) are on chromosome 16, while the β-like genes (ε, γA, γG, δ and β) are on chromosome 11 (Figure 17.19). ii.Defective pseudogenes occur, 3 in the α genes and 1 in the β genes. iii. Gene arrangement in the chromosomes parallels the order of gene expression during development. iv. The relationship between gene arrangement and timing of expression is not yet well understood. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 50 Fig. 17.19 Linkage maps of human globin gene clusters Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 51 3. Polytene chromosome puffs occur during dipteran (2-winged fly) development. a. In certain tissues (e.g., larval salivary gland) repeated cycles of chromosome replication without nuclear division (endoreduplication) result in chromatid bundles (polytene chromosomes). b. Polytene chromosomes show distinct banding when stained (chromomeres), and are easily seen with a light microscope (Figure 17.21). c. Chromomeres puff (show localized unraveling) at characteristic times during development in a specific and repeatable pattern. Puffing is associated with very high levels of gene expression, as shown by incorporation of radioactive uridine. d. Puffing is often under hormonal control. A key hormone in fly larvae is ecdysone, and a model for ecdysone action has been proposed. i. Ecdysone binds a receptor protein, forming a complex that regulates both early and late genes. (1) Early genes are activated. (2) Late genes are repressed. ii. Early gene expression includes gene product(s) that accumulate during development. At a certain threshold, it displaces ecdysone-receptor complex from: (1) Early genes (repressing them). (2) Late genes (activating them). iii. Additional evidence supports this model: (1) Early gene products include DNA-binding proteins. (2) A cloned ecdysone receptor gene encodes a steroidlike receptor protein. e. Polytene puffs relate directly to development in dipteran insects, and at least some of the puffs are hormonally regulated. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 52 Fig. 17.20 Diagram of the complete set of Drosophila polytene chromosomes in a single salivary gland Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 53 台大農藝系 遺傳學 601 20000 Chapter 17 slide 54 Immunogenetics and Chromosome Rearrangements During Development 1. Antibodies, which mediate the humoral immune response, are produced by rearrangements in the genome of the cells that produce them, including loss of some DNA sequences. 2. B lymphocytes produce antibodies (immunoglobulins). Antibodies are inserted into the plasma membrane of the B cells, and also released into blood and lymph. a. Antibodies bind specifically to the antigens that stimulated their production. b. Immunity against a particular antigen results from clonal selection, where cells producing a specific antibody are stimulated to proliferate. c. In development, each B cell becomes committed to producing only one type of antibody, and therefore responding to only one type of antigen. d. A B cell responds to a specific antigen when it binds the antibodies on the cell’s plasma membrane, stimulating the B cell to divide and produce a clonal population of B cells. e. All antibodies made by a single B cell are identical, with the same protein chains and antigenbinding abilities. f. There are millions of B cell types, producing millions of types of antibodies. 3. IgG (immunoglobulin G) is a typical antibody type (Figure 17.22). IgG has: a. Two identical short polypeptide chains called light (L) chains. b. Two identical longer polypeptide chains called heavy (H) chains. c. Disulfide bonds (-S-S-) to hold the chains together and produce characteristic folding. d. An overall Y-shape to its structure, with each arm containing an antigen-binding site. e. The hinge region of the Y shape flexes, allowing easier binding. Each arm may bind a different molecule of the same antigen, allowing cross-linking.台大農藝系 遺傳學 601 20000 Chapter 17 slide 55 Fig. 17.22a IgG antibody molecule Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 56 4. The L chains of antibodies will be either 2 k (kappa) chains or 2 l (lambda) chains. Mammals have five major classes of antibodies, based on identity of their H chains: a. IgA, with α (alpha) H chains. b. IgD, with δ (delta) H chains. c. IgE, with ε (epsilon) H chains. d. IgG, with γ (gamma) H chains. These antibodies are the most abundant class in blood. e. 5. IgM, with μ (mu) H chains. These antibodies are used in the early stages of an immune response. Each antibody polypeptide chain is organized into domains of about 110 amino acids each. a. Each L chain (κ or λ) has 2 domains. b. The γ H chain of IgG has 4 domains. c. The μ H chain of IgM has 5 domains. d. The N-terminal ends of the H and L chains constitute the variable (V) regions (VL and VH), which are involved in binding to antigen. e. The rest of the amino acid sequence is constant in the H (CH) and L (CL) chains. i. For IgG, there are 3 CH domains, CH1, CH2, and CH3. ii. For IgM, there are 4 CH domains, CH1, CH2, CH3, and CH4. 6. Diversity of mammalian antibodies is far greater than the total number of genes in mammalian genomes. Diversity of antibodies is generated by somatic recombination of different gene segments, rather than entirely different genes for each antibody. a. Light chain gene recombination is well studied in mice. i. Gene segments are used to make a k chain, and 1 of each of these types is needed to make the k polypeptide. (1) L-Vk sequences have about 350 different options, each about 400 bp long. Vk encodes most of the light chain variable domain, and L encodes the leader sequence used for secretion of the IgG Chapter 17 slide 57 molecule. (The signal is cleaved at transport.)台大農藝系 遺傳學 601 20000 2) The Ck segment specifies the constant domain of the k chain. There is only 1 gene segment for Ck. (3) Jk segments join the Vk and Ck segments to produce a functional light chain. There are 4 Jk options, each about 30 bp long. ii. Pre-B cells have the L-Vk, Jk and Ck segments in that order, but widely separated on the chromosome. During B cell development, a particular L-Vk segment associates with one of the Jk segments, and with the Ck segment (Figure 17.23). iii. The rearranged DNA is transcribed to produce an RNA that is processed to form a mature mRNA. iv. The DNA rearrangement mechanism produces a great diversity of combinations, especially since the joining process may delete a few base pairs, resulting in still more possibilities. v. Thus, k light chain diversity arises from a combination of 3 mechanisms: (1) Variability in sequences of the multiple Vk gene segments. (2) Variability in the sequences of the 4 Jk gene segments. (3) Variability in the number of base pairs deleted at the Vk-Jk joints. vi. A similar mechanism assembles l light chains, although fewer options exist because of the limited number of gene segments: (1) There are 2 L-Vl segments. 遺傳學 601 20000 Chapter 17 slide 58 (2)There are 4 Cl segments, each台大農藝系 with its own Jl gene segment. Fig. 17.23 Production of the light chain gene in mouse by recombination of V, J, and C gene segments during development Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 59 b. Heavy chains are also encoded by the VH, JH and CH segments. Additional diversity comes from another segment, D, located between VH and JH.(Figure 17.24) i.Germ line mouse DNA has: (1) A tandem array of about 500 L-VH segments followed by a spacer. (2) Then 12 D segments followed by a spacer. (3) Then 4 JH segments and a spacer. (4) Then the constant region gene segments, arranged in the order m, d, g (4 related sequences), e and a. ii.Further diversity results from imprecise joining of the segments, as in light chain production. c. In combination, variation in the two possible chains can result in millions of different possible antibodies. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 60 Fig. 17.24 Production of heavy chain genes in mouse by recombination of V, D, J, and C gene segments during development Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 61 Genetic Regulation of Development in Drosophila 1. Eukaryotic organisms that have been important models for study include: a. Drosophila melanogaster, the fruit fly, was prominent in classical genetics, and has also become a model system for studying the genetics of development. b. Caenorhabditis elegans, a nematode, has a small genome, the lineage of each body cell is known, and the transparent worm is easily observed. c. Brachydanio rerio, the zebrafish, is useful for observing developmental steps, and techniques exist for identifying all genes involved in embryogenesis. d. Arabidopsis thaliana, a small plant, has been used for crosses that produce large numbers of progeny, and its development is well characterized. 2. The Drosophila system remains one of the most important, with large numbers of developmental mutants isolated and characterized in detail. Many genes in Drosophila have counterparts in higher organisms, 台大農藝系 遺傳學 601 20000 Chapter 17 slide 62 including humans. Drosophila Developmental Stages 1. A fertilized egg becomes an adult fly in a programmed series of events that are under strict genetic control. From egg to adult takes 10–12 days at 25°C (Figure 17.25). a. About 24 hours post-fertilization, the egg hatches into a larva. b. After three larval molts, a pupa forms. c. The pupa metamorphoses into an adult fly. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 63 Fig. 17.25 Development of an adult Drosophila from a fertilized egg Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 64 Embryonic Development 1. Development of a fertilized egg begins the process (Figure 17.26). a. Before fertilization, molecular gradients exist within the egg, with polar cytoplasm at the posterior end. b. After fertilization, the 2 nuclei fuse, forming a 2N zygote nucleus. c. The next 9 mitotic divisions occur without cytokinesis, making a multinucleate syncytium. d. After 7 divisions, some nuclei migrate to the polar cytoplasm, becoming germ-line precursors. e. Other nuclei migrate to the egg surface forming the syncytial blastoderm. f. After 4 more divisions, individual nuclei are separated by membranes, forming the somatic cells of the cellular blastoderm. 2. The next development depends on two processes (Figure 17.27): a. Gradients of molecules, both anterior-posterior and dorsal-ventral. b. Formation of parasegments and then embryonic segments, which give rise to adult body segments. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 65 Fig. 17.26 Embryonic development in Drosophila Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 66 3. Drosophila development genes are defined by mutations that are either lethal or cause abnormal structures during development. There are 3 major classes: a. Maternal effect genes, expressed by the mother during oogenesis. Key examples are: i. The bicoid gene, which regulates formation of anterior structures. (1) Mutants have posterior structures at each end. (2) The wild-type bicoid gene encodes a morphogen. (3) The gene is transcribed during oogenesis, the mRNA is at the anterior pole of the oocyte, and after fertilization the mRNA is polyadenylated and expressed. (4) The BICOID protein forms a gradient with highest concentration in the egg’s anterior and none at the posterior (an anterior-to-posterior gradient). ii. The nanos gene is involved in abdomen formation. (1) It produces mRNAs that collect in the posterior of the egg. (2) After fertilization, translation produces NANOS protein that forms a posterior-to-anterior gradient in the egg, acting as a morphogen for abdomen formation. iii. The torso gene controls anterior and posterior structures of the embryo. (1) Both transcription and translation occur during oogenesis. (2) The TORSO protein is found throughout the egg, but is only active at the two termini. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 67 Fig. 17.27 Drosophila development results from gradients in the egg that define parasegments in the cellular blastoderm and segments in the embryo and adult Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 68 b. Segmentation genes divide the embryo into regions. About 25 have been identified, and subclassified into three groups based on the effect of mutations in each gene. i. Gap gene mutations result in deletions of several adjacent segments (e.g., Krüppel, hunchback). ii. Pair rule gene mutations cause deletions of the same part of the pattern in every other segment (e.g., even-skipped, fushi tarazu). iii. Segment polarity gene mutants have portions of segments replaced by mirror images of adjacent half segments (e.g., gooseberry, engrailed). iv.Segmentation genes have a chain of action. Gap genes organize the embryo into broad regions, and maternal effect genes control gap genes. An important examples: (1) The hunchback gap gene is stimulated by the BICOID, but inhibited by NANOS.. (2) The result is an anterior-to-posterior gradient of HUNCHBACK protein.. (3) Gap genes (e.g., hunchback) encode transcription factors that control the pair-rule genes which divide the embryo into regions. (4) Pair-rule gene products regulate expression of segment polarity genes, defining the segments of larvae and adults. c. Homeotic (structure-determining) genes specify the body part that will develop at each segment during metamorphosis. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 69 Fig. 17.28 Functions of segmentation genes as defined by mutations Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 70 Imaginal Discs 1. Cellular blastoderm cells specify two types of cells: a. Those producing larval tissues. b. Those developing into adult tissues and organs. i. These include undifferentiated cells that form imaginal discs in the larvae. ii. Each imaginal disc differentiates into a specific adult structure (Figure 17.29). 台大農藝系 遺傳學 601 20000 Chapter 17 slide 71 Fig. 17.29 Locations of imaginal discs in a mature Drosophila larva and the adult structures derived from each disc Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 72 Homeotic Genes 1. Homeotic genes give each segment its developmental identity. Homeotic mutants will have abnormal segments. Some examples: a. The bithorax gene complex (BX-C) determines the posterior segments of the fly (Figure 17.30). i. The complex contains 3 complementation groups, Ubx (Ultrabithorax), abd-A (abdominal-A) and Abd-B (Abdominal-B), each encoding one protein. ii. Mutations in these genes are often lethal, but nonlethal mutations are useful for understanding normal development. b. The Antennapedia complex (ANT-C) determines the head and first 2 thoracic segments. i. At least four genes are involved: (1) Deformed (Dfd). (2) fushi tarazu (ftz). (3) Sex combs reduced (Scr). (4) Antennapedia (Amp). Examples of the effect of this gene: (a) One group of nonlethal mutations results in leg parts instead of an antenna growing out of cells near the eye (Figure 17.31). (b) Another Antp mutation causes the distal part of the antenna (the arista) to be a leg instead. ii. Most ANT-C mutations are lethal. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 73 台大農藝系 遺傳學 601 20000 Chapter 17 slide 74 Fig. 17.31 Effects of some mutations in the Antennapedia complex Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 75 2. Genes of the ANT-C and BX-C complexes have been cloned. Both are very large, containing both coding and regulatory regions (Figure 17.32). 3. Coding sequences for the proteins show a conserved 180 bp sequence, the homeobox. The corresponding 60 amino-acid region in each protein is the homeodomain. a. Homeodomains occur in over 20 Drosophila genes. Most regulate development. b. All bomoedomain proteins bind DNA, at an 8-bp consensus sequence upstream of genes they control, using helix-turn-helix motifs. c. Homeodomain proteins exert transcriptional regulation on these genes. 4. The complete set of homeotic genes and complexes in Drosophila (the Hox genes), consists of lab, pb, Dfd, Scr, Antp, Ubx, abdA and AbdB. 5. These complexes are arranged in the same order on the chromosomes as they are expressed along the antero-posterior body axis (the cohnearity rule). 6. Homeotic genes occur in all major animal phyla except sponges and coelenterates, and are highly conserved. a. In mammals there are 4 clusters of homeotic genes, HoxA-D, that specify the body plan. b. Homeotic genes also occur in plants, including Arabidopsis, where their role in flower development is well-studied. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 76 Fig. 17.32 Organization of the bithorax complex (BX-C) Peter J. Russell, iGenetics: Copyright © Pearson Education, Inc., publishing as Benjamin Cummings. 台大農藝系 遺傳學 601 20000 Chapter 17 slide 77