An enzyme attaches ubiquitin to the protein.
Ubiquitin and the complex is recognized by a proteasome.
and the complex is recognized by a proteasome.
An enzyme attaches ubiquitin to the protein.
Ubiquitin is released and recycled.
14.22 The Proteasome Breaks Down Proteins Proteins targeted for breakdown are bound to ubiquitin, which "leads" them to the proteasome, a complex composed of many polypeptides.
Ubiquitin is released and recycled.
a 76-amino acid protein called ubiquitin (so called because it is ubiquitous, or widespread) is covalently linked to a protein targeted for breakdown. The protein-ubiquitin complex then binds to a huge complex of several dozen polypeptide chains called a proteasome (Figure 14.22). The entryway to this "molecular chamber of doom" is a hollow cylinder. This part of the complex has ATPase activity, and it uses the released energy to cut off the ubiquitin for recycling and unfold its targeted protein "victim." The protein then passes by three different proteases (thus the name of the complex), which digest it into small peptides and amino acids.
The cellular concentrations of many proteins are determined not by differential transcription of their genes, but by their degradation in proteasomes. Cyclins, for example, are degraded at just the right time during the cell cycle (see Figure 9.4). Transcription regulators are broken down after they are used, lest the affected genes be always "on." Abnormal proteins are often targeted for destruction by a quality control mechanism. Human papillomavirus, which causes cervical cancer, targets the cell division inhibitory protein p53 for pro-teasomal degradation, so that unregulated cell division— cancer—results.
► Although eukaryotes have more DNA in their genomes than prokaryotes, there is no apparent relationship between genome size and organism complexity within eukaryotes.
► There are many differences between prokaryotic and eukaryotic genomes and their mechanisms of expression. Review Table 14.1
► Unlike prokaryotic DNA, eukaryotic DNA is contained within a nucleus, so that transcription and translation are physically separated. Review Figure 14.1. See Web/CD Activity 14.1
► The genome of the single-celled budding yeast contains genes for the same metabolic machinery found in prokaryotes, with the addition of genes for protein targeting in the cell. Review Table 14.2
► The genome of the multicellular roundworm Caenorhabditis elegans contains genes required for intercellular interactions. Review Table 14.3
► The genome of the fruit fly has fewer genes than that of the roundworm. Many of its genes are homologs of genes found in the roundworm and mammalian genomes.
► The puffer fish genome is the most compact vertebrate genome known.
► The compact genome of the simple plant Arabidopsis is often used in the study of plant genomes. Review Table 14.4
► The rice genome is similar to that of Arabidopsis, and its sequence holds a key to feeding the increasing human population. Review Table 14.5
Repetitive Sequences in the Eukaryotic Genome
► Highly repetitive DNA is present in up to millions of copies of short sequences. It is not transcribed.
► Some moderately repetitive DNA sequences, such as those that code for rRNAs, are transcribed. Review Figure 14.2
► Some moderately repetitive DNA sequences are transposons, which are able to move about the genome. Review Figure 14.3
► A typical eukaryotic protein-coding gene is flanked by promoter and terminator sequences and contains noncoding internal sequences, called introns. Review Figure 14.4
► Nucleic acid hybridization is an important technique for analyzing eukaryotic genes. Review Figure 14.5, 14.6
► Some eukaryotic genes exist as families of related genes, which have similar sequences and code for similar proteins. These related proteins may be made at different times and in different tissues. Some sequences in gene families are pseudogenes, which code for nonfunctional mRNAs or proteins. Review Figure 14.7
► Differential expression of different genes in the P-globin cluster of the globin family ensures important physiological changes during human development. Review Figure 14.8
► The transcribed pre-mRNA is altered by the addition of a G cap at the 5' end and a poly A tail at the 3' end. Review Figure 14.9
► The introns are removed from the mRNA precursor by the spliceosome, a complex of snRNPs and proteins. Review Figure 14.10. See Web/CD Tutorial 14.1
► Eukaryotic gene expression can be regulated at the transcrip-tional, posttranscriptional, translational, and posttranslational levels. Review Figure 14.11. See Web/CD Activity 14.2
► The major method of regulation of eukaryotic gene expression is selective transcription, which results from the binding of specific proteins to regulatory sequences on DNA.
► A series of transcription factors must bind to one another to form a transcription complex before RNA polymerase can bind. Whether RNA polymerase initiates transcription also depends on the binding of regulator proteins, activator proteins (which bind to enhancers and stimulate transcription), and repressor proteins (which bind to silencers and inhibit transcription). Review Figures 14.12, 14.13. See Web/CD Tutorial 14.2
► The simultaneous regulation of widely separated genes is possible through common sequences in their promoters, to which the same regulatory proteins bind. Review Figure 14.14
► The DNA-binding domains of most DNA-binding proteins have one of four structural motifs: helix-turn-helix, zinc finger, leucine zipper, or helix-loop-helix. Review Figure 14.15
► Chromatin remodeling allows the transcription complex to bind DNA and to move through the nucleosomes. Review Figure 14.16
► Heterochromatin is a condensed form of DNA that cannot be transcribed. It is found in the inactive X chromosome of female mammals. Review Figure 14.17
► Interference RNA (RNAi) is important in inhibiting transcription of the inactive X chromosome. Review Figure 14.18
► The movement of a gene to a new location on a chromosome may alter its ability to be transcribed, as in the change from one mating type to another in budding yeast.
► Some genes are selectively amplified in some cells. The extra copies of these genes result in increased transcription of their protein product. Review Figure 14.19
► Alternative splicing of pre-mRNA can be used to produce different proteins. The transcripts of over half the genes in the human genome are alternatively spliced, which increases the number of proteins that can be encoded by a single gene. Review Figure 14.20
► The stability of mRNA in the cytoplasm can be regulated.
► Mature mRNA can be edited by the addition of new nucleo-tides or by the alteration of existing nucleotides. Review Figure 14.21
Translational and Posttranslational Regulation
► Translational repressors can inhibit the translation of mRNA.
► Proteasomes degrade proteins targeted for breakdown by attachment of ubiquitin. Review Figure 14.22
1. Eukaryotic protein-coding genes differ from their prokary-otic counterparts in that only eukaryotic genes a. are double-stranded.
b. are present in only a single copy.
c. contain introns.
d. have a promoter.
e. transcribe mRNA.
2. Comparison of the genomes of yeast and bacteria shows that only yeast has many genes for a. energy metabolism.
b. cell wall synthesis.
c. intracellular protein targeting.
d. DNA binding proteins.
e. RNA polymerase.
3. The genomes of a fruit fly and nematode work are similar to that of yeast, except that the former have many genes for a. intercellular signaling.
b. synthesis of polysaccharides.
c. cell cycle regulation.
d. intracellular protein targeting.
e. transposable elements.
4. Which of the following does not occur after mRNA is transcribed?
a. binding of RNA polymerase II to the promoter b. capping of the 5' end c. addition of a poly A tail to the 3' end d. splicing out of the introns e. transport to the cytosol
5. Which statement about RNA splicing is not true?
a. It removes introns.
b. It is performed by small nuclear ribonucleoprotein particles (snRNPs).
c. It always removes the same introns.
d. It is usually directed by consensus sequences.
e. It shortens the RNA molecule.
6. Eukaryotic transposons a. always use RNA for replication.
b. are approximately 50 bp long.
c. are made up of either DNA or RNA.
d. do not contain genes coding for transposition.
e. make up about half of the human genome.
7. Which statement about selective gene transcription in eukaryotes is not true?
a. Different classes of RNA polymerase transcribe different parts of the genome.
b. Transcription requires transcription factors.
c. Genes are transcribed in groups called operons.
d. Both positive and negative regulation occur.
e. Many proteins bind at the promoter.
8. Heterochromatin a. contains more DNA than does euchromatin.
b. is transcriptionally inactive.
c. is responsible for all negative transcriptional control.
d. clumps the X chromosome in human males.
e. occurs only during mitosis.
9. Translational control a. is not observed in eukaryotes.
b. is a slower form of regulation than transcriptional control.
c. can be achieved by only one mechanism.
d. requires that mRNA be uncapped.
e. ensures that heme synthesis equals globin synthesis.
10. Control of gene expression in eukaryotes includes all of the following except a. alternative splicing of RNA transcripts.
b. binding of proteins to DNA.
c. transcription factors.
d. feedback inhibition of enzyme activity by allosteric control.
e. DNA methylation.
1. In rats, a gene 1,440 bp long codes for an enzyme made up of 192 amino acid units. Discuss this apparent discrepancy. How long would the initial and final mRNA transcripts be?
2. The genomes of rice, wheat, and corn are similar to each other and to the weed, Arabidopsis. Discuss how these plants might nevertheless have very different proteins.
3. The activity of the enzyme dihydrofolate reductase (DHFR) is high in some tumor cells. This activity makes the cells resistant to the anticancer drug methotrexate, which targets DHFR. Assuming that you had the complementary DNA for the gene that encodes DHFR, how would you show whether this increased activity was due to increased transcription of the single-copy DHFR gene or to amplification of the gene?
4. Describe the steps in the production of a mature, translatable mRNA from a eukaryotic gene that contains introns. Compare this to the situation in prokaryotes (see Chapter 13).
5. A protein-coding gene has three introns. How many different proteins can be made from alternate splicing of the pre-mRNA transcribed from this gene?
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