Download CHAPTER 17 Regulation of Gene Expression in Eukaryotes

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.
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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.
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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.
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Chapter 17 slide 6
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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.
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Chapter 17 slide 8
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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.
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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).
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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.
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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.
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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.
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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.
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Chapter 17 slide 22
Fig. 17.9 Mechanisms of action of polypeptide hormones and steroid hormones
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Chapter 17 slide 23
Fig. 17.10 Structures of some mammalian steroid hormones
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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.
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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.
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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).
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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.
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Chapter 17 slide 31
Fig. 17.14 Alternative polyadenylation and alternative splicing resulting in tissuespecific products of the human calcitonin gene, CALC
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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.
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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.
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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.
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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.
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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.
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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.
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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).
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Chapter 17 slide 46
Fig. 17.17 Molecular organization of the human -globin and -globin genes
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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).
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Chapter 17 slide 48
Fig. 17.18 Comparison of synthesis of different globin chains at given stages of
development
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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.
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Chapter 17 slide 50
Fig. 17.19 Linkage maps of human globin gene clusters
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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.
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Chapter 17 slide 52
Fig. 17.20 Diagram of the complete set of Drosophila polytene chromosomes in a
single salivary gland
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Chapter 17 slide 53
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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
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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.
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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
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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.
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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
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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,
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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.
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Fig. 17.25 Development of an adult Drosophila from a fertilized egg
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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.
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Chapter 17 slide 65
Fig. 17.26 Embryonic development in Drosophila
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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.
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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
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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.
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Chapter 17 slide 69
Fig. 17.28 Functions of segmentation genes as defined by mutations
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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).
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Chapter 17 slide 71
Fig. 17.29 Locations of imaginal discs in a mature Drosophila larva and the adult
structures derived from each disc
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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.
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Chapter 17 slide 73
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Chapter 17 slide 74
Fig. 17.31 Effects of some mutations in the Antennapedia complex
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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.
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Chapter 17 slide 76
Fig. 17.32 Organization of the bithorax complex (BX-C)
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Chapter 17 slide 77