Instant Download with all chapters and Answers
Sample Chapters
*you will get Solution manual in PDF in best viewable format after buy*
Molecular Microbiology
Summary
Chapter 4 focuses on the molecular biology of microorganisms and takes the student on a journey from DNA structure and replication to protein folding and secretion. Students should recall that cells are both biochemical catalysts and genetic entities. The chapter provides an excellent overview of the structure, storage, and replication of genetic material; the expression of genes through the process of transcription; and the translation of messenger RNA transcripts into protein products. Our understanding of these processes has been greatly enhanced through intensive study of the model organism Escherichia coli. Indeed, we know more about the structure and function of this bacterium than any other living organism, including humans.
4.1 | Macromolecules and Genes
Chapter 4 is organized from DNA structure and replication events to RNA synthesis and processing, and it concludes with protein synthesis, folding, and secretion. This first section describes the structure of nucleotides, the building blocks of nucleic acids (Figures 4.1 and 4.2), and the flow of genetic information from DNA to RNA to proteins (Figure 4.3). Using the figures, show that nucleotides consist of a nitrogenous base, a pentose sugar, and a phosphate group. Distinguish between pyrimidine bases and purine bases, pointing out the nucleic acid molecules (DNA or RNA) in which each of the five bases occurs. Explain that in nucleic
acids, individual nucleotides are covalently linked with a phosphate group between the
3¢-carbon of one sugar and the 5¢-carbon of the next sugar, forming a phosphodiester bond.
Introduce the terms gene (the functional unit of genetic information) and genetic elements (large nucleic acid molecules containing genes). Note that information flows from DNA to RNA to protein. These cellular components are referred to as informational macromolecules. By contrast, lipids and polysaccharides are macromolecules that are not considered informational since they have no coding functions.
Discuss the three stages involved in genetic information flow depicted in Figure 4.3:
(1) replication, (2) transcription, and (3) translation. All cells employ this central dogma (DNA ® RNA ® protein) when carrying out essential life functions. However, there are
differences in how these steps occur between eukaryotes and prokaryotes, and additional
differences exist between the prokaryotic domains Bacteria and Archaea. Point out that in contrast to eukaryotes, prokaryotes often transcribe multiple genes into a single primary mRNA transcript. Moreover, because prokaryotes do not have a compartmentalized nucleus, transcription and translation are coupled processes that occur simultaneously. This is not the case for eukaryotes, in which transcription is confined to DNA in the nucleus, and translation is catalyzed by ribosomes in the cytoplasm. Later, students will learn that some viruses do not adhere to the central dogma of molecular biology in that they have genomes composed of RNA rather than DNA (Chapter 9).
4.2 | The Double Helix
Use Figure 4.4 to review DNA structure, focusing on the sugar-phosphate backbone
containing phosphodiester bonds and the complementary base-pairing of nitrogenous bases between the two strands of the double helix. Point out that the two strands are held together by hydrogen bonding between the bases¾two hydrogen bonds between adenine and thymine (or uracil in RNA) and three hydrogen bonds between guanine and cytosine. Students should understand the 5¢ ® 3¢ orientation of nucleic acids and why the chain is synthesized in this direction. Note also that the DNA strands of the double helix must be antiparallel, which has important implications for how DNA is replicated (Sections 4.4–4.6). The major groove (and to some extent, the minor groove) shown in Figure 4.5 is an important contact point for DNA-binding proteins that play a role in controlling gene expression (Chapter 7).
The size of double-stranded DNA molecules is expressed as the number of nucleotide base pairs per molecule. A molecule of DNA containing 1500 base pairs is expressed as 1500 bp, which is equal to 1.5 kilobase pairs (1.5 kbp). Explain that the length of a single base pair in the helix is 0.34 nm; thus one can calculate the length of a DNA molecule when given the
total number of base pairs. You can reinforce this concept by having students calculate the length of the haploid human genome that is 3 billion base pairs long, and convert from metric units to inches or feet, as shown below.
(3 ´ 109 bp) ´ 0.34 nm/bp = 1.02 ´ 109 nm long = 102 cm (1 cm = 0.0328 ft.) = 3.35 feet (40.2 inches)
Section 4.2 also covers DNA supercoiling in the context of DNA structure and packaging. A good illustration of DNA supercoiling for your students is the ability of a bacterial cell to pack over 1 mm of DNA into less than 2 mm of cell length. The problem is even more complex in multicellular organisms and requires a much more elaborate packaging scheme employing histone proteins and nucleosome structures (see Figure 2.61). DNA can be supercoiled in either the positive (overwound) or negative (underwound) conformation. Enzymes called topoisomerases catalyze the formation or relaxation of supercoils in DNA by introducing either single-stranded breaks (class I topoisomerases) or double-stranded breaks (class II topoisomerases) in the DNA molecule (Figures 4.6 and 4.7). The breaks are resealed following the introduction of a supercoil. DNA gyrase, a class II topoisomerase, is found in Bacteria and Archaea and introduces negative supercoiling in DNA (Figure 4.7). Quinolones, fluoroquinolones, and novobiocin are antibiotics that inhibit DNA gyrase. Topoisomerases relax supercoils by making transient breaks that allow one strand to untangle. The relaxation of the DNA molecule is necessary due to the changes in the topology of DNA as it is being replicated. As the doubled-stranded helix is unwinding in one direction, the unwound region ahead of the replication fork experiences torsional stress that must be relieved for replication to continue. Supercoiling also plays a role in gene expression in both prokaryotes and eukaryotes.
4.3 | Genetic Elements: Chromosomes and Plasmids
Section 4.3 introduces students to the concept of chromosomal and extrachromosomal
genetic elements. In prokaryotes, the latter group includes virus genomes, plasmids, and transposable elements (Table 4.1). Transposons and insertion sequences are covered in more detail in Chapter 10.
The genomes of most prokaryotes consist of a single, circular chromosome ranging in size from 0.5 Mbp to 10 Mbp in most cases. The genome of E. coli is of approximately average size for a bacterium and depending on the strain, ranges from 4.5 Mbp to 5.5 Mbp. Using Figure 4.8, explain to students how genetic maps are used to illustrate the major properties of chromosomes. The size of the genome, the location of genes or operons, and the direction of transcription of those genes are all features that can be shown with genetic maps. Point out that the genes for some biochemical pathways are clustered into operons (e.g., the lac, gal, trp and his clusters on the right side of the chromosome figure), whereas others are not (e.g., the malT and malS genes are not included in the malKBM operon). Compared to eukaryotic genomes, which usually contain a large percentage of noncoding DNA, the genomes of prokaryotes are quite streamlined in that most of their DNA encodes a product necessary for cell function. About 88% of the genome of E. coli strain K-12 consists of protein-encoding genes, while another 1% encodes ribosomal and transfer RNA molecules. Students may find it interesting that sequenced genomes often contain up to 40% “hypothetical” protein coding sequences for which a function cannot be definitively assigned. Furthermore, it is estimated that a significant portion of genetic content in many genomes was acquired through horizontal gene transfer, a process by which genetic elements are mobilized and transferred from one cell to another (see Chapter 10).
Plasmids are also introduced in this section and are defined as small (typically less than 5% the size of the chromosome), usually circular segments of extrachromosomal double-stranded DNA having their own origin of replication (Figure 4.9). Some bacteria possess multiple plasmids. For example, Borrelia burgdorferi, the causative agent of Lyme disease, has 17 plasmids, some of which are even linear. Plasmid-encoded genes are not essential to cell survival but often confer useful properties on cells, including resistance to antibiotics
(via R plasmids; Figure 4.10) and the ability to produce toxins or other virulence factors
(Table 4.2). Mention to your students that the increased use of antibiotics has selected for pathogens for which few or no treatment options remain. This problem is discussed in more detail in Chapter 27. Plasmids replicate using the same enzymatic machinery as that which the cell uses for chromosomal replication, and like chromosomes, most circular plasmids
replicate bidirectionally (see Section 4.6).
Students may be interested to learn that many bacteria produce plasmid-encoded proteins called bacteriocins that kill related species, or even different strains of the same species, while at the same time producing proteins that make themselves immune to these toxins. Plasmids have also been engineered and artificially modified to use in research. The construction of artificial plasmids containing selected genes has played a key role in the exploitation of microorganisms for biotechnological and industrial purposes. The topic of genetic engineering will be covered in Chapter 11.
Use Table 4.2 to describe some of the phenotypes conferred by plasmids on their bacterial hosts, stressing that a given plasmid can be lost from the population if the genes it possesses do not provide a selective advantage to the host. The reasons for this stem from the fact that a bacterium will not expend the energy it takes to maintain an extrachromosomal element and transcribe and translate its genes unless the products of those genes are in some way beneficial. Consequently, a daughter cell produced without the plasmid may grow and reproduce faster, eventually curing the population of the plasmid.
4.4–4.6 | Transmission of Genetic Information: DNA Replication
Use Figure 4.11 to summarize the products of DNA replication, stressing the semiconservative process that occurs in both prokaryotes and eukaryotes. The precursor of each new
nucleotide to be added to the chain is a deoxyribonucleoside triphosphate. Hydrolysis of two of the phosphates is required by DNA polymerase to drive phosphodiester bond formation at the 3¢–OH of the preceding nucleotide (Figure 4.12). E. coli contains at least five distinct DNA polymerases (DNA Pol I–V). DNA Pol III functions as the main enzyme in DNA replication. DNA Pol I also plays a crucial role in replication, while the other three polymerases function primarily in DNA repair. Note that the initiation of DNA replication requires a primer. For circular molecules of DNA, the primer is a segment of RNA synthesized by the enzyme primase (Figure 4.13). Students must understand that a primer is necessary because DNA polymerase can begin synthesis only on DNA molecules having a free 3¢–OH. RNA primers inserted during replication are later removed and replaced with DNA nucleotides,
as described below.
Next, discuss the events that occur during the initiation phase at the replication fork and the enzymes involved (Figure 4.14). The origin of replication is an approximately 250-bp sequence that is recognized by the DnaA protein, which binds and opens the helix (Table 4.3). Two helicase (DnaB) enzymes bind with the help of the DnaC protein, and each attaches to one of the two strands in opposite directions (Figure 4.14 shows one of the two helicase complexes). Finally, single-stranded binding proteins bind to the two strands, stabilizing them and preventing them from re-annealing (Figure 4.15).
Continue your discussion of replication of circular prokaryotic chromosomes by clearly delineating the events involved in leading strand versus lagging strand synthesis (Figures 4.15 and 4.16). Note that the logistical problems inherent in synthesizing the lagging strand away from the replication fork necessitate the generation of Okazaki fragments and a “backstitching” type of mechanism by primase and DNA Pol I. RNA primers are removed by DNA Pol I and replaced with deoxyribonucleotides. The enzyme DNA ligase is required to seal the backbone (add a final phosphodiester bond) between Okazaki fragments on the lagging strand (Figure 4.16). Use Figure 4.17 to show students that there are two replication forks for a circular chromosome, resulting in the theta structure. The problem of excess torsion created by supercoils ahead of the replication forks is resolved by the topoisomerase DNA gyrase, which regulates the degree of supercoiling during replication (Figure 4.18). The proteins involved in replication do not act independently, but they aggregate into a large complex called the replisome (Figure 4.18). Table 4.3 describes the key enzymes involved in DNA replication in
Bacteria.
Students need to understand that although DNA polymerases III and I are remarkably efficient and precise in maintaining the fidelity of the DNA sequence during replication, they have an error (mutation) rate of ~10–8 and 10–11 errors per base pair inserted. Consequently, a backup system detects and corrects these errors. Most (but not all) of the errors are corrected by the proofreading capabilities of DNA Pol III and Pol I. Each of these enzymes possesses a 3¢ ® 5¢ exonuclease activity that is capable of excising a mismatched base and inserting the correct (complementary) base in its place (Figure 4.19). Other mechanisms for reducing
errors during DNA replication are discussed in Chapter 10.
4.7 | Transcription
Begin your discussion of transcription by reviewing the structural differences between RNA and DNA. Three major types of RNA are involved in protein synthesis: messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). Each of these RNAs is encoded
in the chromosome, so remind students that while most genes code for proteins, some code for structural RNA molecules (e.g., tRNA and rRNA molecules) that are not translated into proteins. Note that although the cell also produces other RNAs that mostly function in regulation (Chapter 7), this chapter focuses only on those involved in transcription and translation. Transcription and translation are often presented as a single topic, and students often get lost in the myriad details surrounding these processes. The approach here allows the instructor
to focus student attention on specific events in transcription before proceeding to translation. The processes discussed are those that occur in Bacteria, using E. coli as the model
organism.
Figure 4.20 presents an overview of transcription, focusing on the function of RNA polymerase in this process. The process of transcription can be approached in three stages: initiation, elongation, and termination. Note here that there are significant differences between RNA polymerases from Bacteria, Archaea, and Eukarya (Figure 4.21; archaeal and eukaryotic RNA polymerases are also discussed in Section 4.9). The Bacterial RNA polymerase holoenzyme consists of five different subunits, with the sigma factor subunit conferring promoter recognition specificity. Following initiation of transcription, the sigma subunit dissociates from RNA polymerase holoenzyme, and the remaining RNA polymerase core enzyme carries out elongation. E. coli makes seven different sigma factors that recognize different promoters. Discuss the interaction of sigma factor with specific recognition sequence contact points within the promoter (–10 and –35 regions; Figure 4.22). Sigma factors can bind tightly to specific promoter sequences called strong promoters or less specifically to weak promoters, thus affecting gene expression via the efficiency and rate of transcription. Inform students that strong promoters are used in genetic engineering to support high expression levels of desired genes (Chapter 11). Table 4.4 lists the common and alternative sigma factors used by E. coli, along with the function of the genes or operons upon which they act.
Following elongation of the RNA transcript, termination of transcription ensues. In Bacteria, this process is controlled by specific DNA sequences downstream of the 3¢ end of an open reading frame. These sequences, called inverted repeats, often consist of two GC-rich sequences with a nonrepeating sequence between them, allowing for the formation of a
stem-loop structure (Figure 4.23). This type of intrinsic termination occurs independently of catalysis by the termination-inducing enzyme Rho and is thus called Rho-independent termination. In contrast to this mechanism, Rho-dependent termination requires the specific protein Rho, which binds the newly synthesized RNA molecule and causes both it and RNA polymerase to be released from the DNA. Students should understand the difference between
intrinsic terminators and Rho-dependent termination.
4.8 | The Unit of Transcription
Emphasize for students the difference between a gene and a transcriptional unit—students often get these concepts confused. Figure 4.24 provides a helpful representation of an rRNA transcriptional unit that can be used to describe all of the elements a single unit contains. Note that an rRNA operon contains all three rRNA genes with spacers (short segments of noncoding DNA) between them. This cotranscription ensures that 16S, 23S, and 5S rRNA molecules are transcribed in equal proportions. Often, one or more tRNA genes are also included within an rRNA operon. Discuss the concept of operons and polycistronic mRNA (Figure 4.25) and the advantages of cotranscribing multiple related genes into a single RNA transcript.
4.9 | Transcription in Archaea and Eukarya
Transcription in Archaea shares features with the mechanisms of transcription in both the Bacteria and Eukarya. Recall that the RNA polymerase of Archaea is markedly similar to RNA Pol II of Saccharomyces cerevisiae, a eukaryote (Figure 4.21). As in eukaryotes but
unlike Bacteria, the RNA polymerase of Archaea is not sensitive to the antibiotic rifampicin. However, regulation of transcription in Archaea is similar to that in Bacteria
(Chapter 7).
Figure 4.26 shows features of an archaeal promoter region, including the 6–8-bp TATA box that is recognized by the transcription factor TATA-binding protein (TBP). Discuss the role(s) of each transcription factor. Little information is available regarding transcription
termination signals in Archaea, although some species appear to use inverted repeats, as
observed in Bacteria (Section 4.7). Eukaryotic transcription termination often requires termination factor proteins, but these are not related to the Rho protein of Bacteria.
Introns are intervening, noncoding DNA sequences within genes that are excised before producing the mature RNA transcript. As is the case in Bacteria, introns are rare in Archaea, although several tRNA and rRNA genes do possess them. Intron excision in Archaea is catalyzed by a specific endoribonuclease (Figure 4.27). Most eukaryotic transcripts contain introns, however, and they must be processed before they are functionally mature. Figure 4.28 illustrates the structure and mechanism of action of the spliceosome. This macromolecular complex contains small ribonucleoproteins and other proteins that facilitate both the removal of introns from primary transcripts and the splicing of exons to produce the mature mRNA. Discuss each of the steps shown in Figure 4.28a–e.
Next, describe two other unique RNA processing steps: (1) RNA capping and (2) the poly(A) tail (Figure 4.29). Note that both processes take place before intron excision. The 5′ guanosine cap is needed during formation of the mRNA-ribosome translation initiation complex, and the poly(A) tail stabilizes the transcript and signals its identity as mRNA to the
ribosome. This is in contrast to the function of poly(a) tails on mRNA molecules of chloroplasts and certain bacteria, which are usually interpreted as a signal for degradation of
the transcript.
4.10 | Polypeptides, Amino Acids, and the Peptide Bond
Proteins are important components in the chemistry of life, serving both structural and catalytic roles. Consequently, a general understanding of protein structure will allow students to better comprehend the tremendous variety of functions that these macromolecules perform in the cell as they proceed through the course. Figure 4.30 gives an excellent representation of amino acid structure and respective side chain (R group) diversity.
Discuss the dehydration synthesis that leads to peptide bond formation (Figure 4.31), and mention here that this reaction is catalyzed within a region of the ribosomal large subunit
during translation of an mRNA molecule into protein. The resulting string of amino acids specified by a gene is known as the primary structure of the protein, and the sizes of known proteins range from just 15 to 10,000 amino acids. Stress that proteins can vary in the number, types, and sequence of amino acids, which explains the enormous variety of protein structures possible. For these reasons, many early scientists believed that protein, not DNA, was the hereditary material in living systems.
4.11–4.12 | Translation and the Genetic Code and Transfer RNA
Protein synthesis, or translation, is a continuous process in the life of the cell. As in transcription, the various phases of translation can be considered as three discrete steps: initiation, elongation, and termination. Before discussing these details, students need to understand how genetic information in DNA dictates the amino acid sequence of a protein according to the genetic code (Table 4.5). The genetic code uses codons to specify amino acids. Each codon consists of a triplet of three nucleotide bases. Explain to students that a code of three-letter codons made up of four separate bases yields 64 possible codons (43) for encoding the twenty amino acids found universally in living organisms. Thus there is more than one codon for most amino acids. In addition, there is one start codon¾AUG, which codes for methionine (or N-formylmethionine in Bacteria)¾and three stop codons: UAA, UAG, and UGA
(Table 4.5). These signal the beginning or end of the encoded polypeptide, respectively.
Now is a good time to introduce the role of tRNA, which possesses a complementary anticodon that recognizes a specific codon (or codons) on mRNA and delivers the encoded amino acid (see Figure 4.34). There is usually more than one tRNA for a given amino acid because more than one codon can specify a single amino acid. However, some tRNAs can base pair with more than one codon due to the wobble phenomenon, in which complementary base pairing at the third nucleotide position of the codon is not required (Figure 4.32). For example, the anticodon for one lysyl tRNA (tRNA carrying lysine) can base pair with either AAA or AAG due to the wobble position effect (see Table 4.5).
Every gene needs to have a start signal and a stop signal that is recognized by the translational machinery and that defines the correct open reading frame. Use Figure 4.33 to show that while there are three possible reading frames on each strand in the 5¢ ® 3¢ direction, there is only one correct frame that specifies the desired polypeptide. It is important to point out that, in prokaryotes, DNA is not isolated from the cytoplasm by a nuclear membrane, and therefore transcription and translation are coupled processes (i.e., they occur concurrently). As soon as the ribosome-binding site (also called the Shine-Dalgarno sequence) near the
5¢ end of the growing mRNA transcript becomes accessible, ribosomes can bind and begin translation. Discuss the concept of codon bias and why this is such an important consideration in genetic engineering (see Chapter 11).
Students may be surprised to discover that over 100 amino acids that are not part of the universal genetic code are found in proteins. Most of these amino acids are generated by modification after being inserted into the protein chain. However, two genetically encoded amino acids that are not universally found in living organisms are selenocysteine and
pyrrolysine, both of which are encoded by stop codons. In such cases where a stop codon is intended to encode one of these rare amino acids, a recognition sequence just downstream of the codon is employed to indicate its role as a sense rather than nonsense (stop) codon.
Students should also be aware of the importance of secondary structure to the activity of transfer RNA molecules (Figure 4.34). Figure 4.35 nicely illustrates the mode of activity and charging mechanism of tRNAs by aminoacyl tRNA synthetases (Figure 4.35).
4.13 | Protein Synthesis
Begin your discussion of protein synthesis with the structure of the ribosome (Figure 4.36). Note the size of the small (30S) and large (50S) ribosomal subunits of prokaryotes, giving a total ribosome size of 70S. This is smaller than the 80S ribosomes of eukaryotes. Point out that ribosomes are ribonucleoprotein complexes (i.e., the E. coli ribosome contains three rRNA molecules and 52 distinct proteins).
Figure 4.36 shows the three stages of translation: initiation, elongation, and termination. The initiation stage begins with the attachment of a free 30S ribosomal subunit to the ribosome-binding site on the mRNA transcript. The initiator N-formylmethionine tRNA and several initiation proteins (IF1, IF2, and IF3) then attach, followed by the addition of the large (50S) subunit to form the active 70S ribosome. Describe the events occurring during elongation, including translocation of the ribosome one codon at a time along the mRNA transcript (Figure 4.36) and the eventual termination of translation at the stop codon. The termination event occurs when the ribosome reaches, and stalls, at a stop codon. This is followed by the dissociation of the polypeptide chain by release factor proteins.
Summarize the multiple roles of the ribosome during all stages of protein synthesis.
Explain that multiple ribosomes can translate a single mRNA transcript simultaneously, forming a polysome (Figure 4.37). Also discuss how truncated or defective mRNA molecules that may lack a stop codon can result in stalled ribosomes. This situation is remedied using a small RNA molecule called tmRNA, which is charged with alanine and contains its own stop codon (Figure 4.38). Binding of a release factor protein to the tmRNA stop codon frees the ribosome for use elsewhere.
You may want to conclude your discussion of translation by mentioning the fact that many antibiotics interfere with protein synthesis by binding to the large or small subunit of the
ribosome. Because prokaryotic ribosomes differ in structure from eukaryotic ribosomes, these antibiotics are medically useful because of their specificity. Examples include streptomycin, chloramphenicol, and tetracycline, and the mode of action of these and other antibiotics is discussed in Chapter 5.
4.14 | Protein Folding and Secretion
Discuss the important chemical interactions that confer secondary structure to a protein,
including the formation of α-helices and β-sheets (Figure 4.39). These structures often fold into functional regions, or domains, that are of utmost importance in the proper function of the final protein molecule. For example, transmembrane proteins, such as some sensor kinases involved in signal transduction, have both an extracellular and cytoplasmic face,
as well as a domain traversing the hydrophobic interior of the plasma membrane.
Examples of these and other protein domains and their function are discussed throughout
the text.
Higher order structure in proteins includes tertiary and quaternary structures. Tertiary structure is determined by both the primary and secondary structure and involves hydrogen bonding, covalent disulfide bonds, hydrophobic interactions, and other atomic interactions (Figure 4.40). Quaternary structure describes a protein composed of two or more polypeptide chains (either identical or nonidentical subunits) held together by either noncovalent interactions (hydrogen bonding, van der Waals forces, or hydrophobic interactions) or covalent linkages (intersubunit disulfide bridges).
Denaturation of a protein often results in a permanent loss of biological activity. This fact should demonstrate to students that the final three-dimensional conformation of a protein as determined by secondary, tertiary, and/or quaternary interactions is not the result of its
primary structure but rather is directed by it via the above interactions between amino acids within the polypeptide chain.
Although most proteins fold spontaneously into their active conformation, some require the aid of molecular chaperones (chaperonins). E. coli contains four essential chaperonins that use the energy released by ATP hydrolysis to properly fold proteins: DnaJ, DnaK,
GroEL, and GroES (Figure 4.41). The DnaKJ complex functions to slow spontaneous folding of newly synthesized proteins. When left unchecked, this process can result in an increased probability of folding errors. GroEL and GroES ensure the proper folding of only about 2% of proteins synthesized by E. coli. Nevertheless, at least one dozen of this small percentage is essential for cell survival, and therefore these molecular chaperones are indispensable.
Expression of molecular chaperones increases when cells experience heat stress because these proteins also play a key role in refolding partially heat-denatured proteins.
Protein transfer (secretion) across membranes should be mentioned because the process requires protein modification, usually in the form of a hydrophobic signal sequence that is cleaved after passage through the membrane. E. coli employs the Sec system to transport many proteins across the membrane. Using the membrane translocase SecYEG, proteins
containing a signal sequence are either bound by the SecA protein and secreted into the
external environment or bound by the signal recognition particle (SRP) and inserted into the membrane (Figure 4.42). Using the Sec system, proteins are transported in an unfolded state and fold into their final conformation only after being transported through the membrane. A second transport system in E. coli, the Tat system, allows transport of proteins already folded into their final conformation.
In regards to protein export, gram-negative bacteria have an additional logistical challenge because they have not one, but two membranes (the cytoplasmic and outer membranes) in which to deposit or through which to secrete proteins. Six different secretion systems (types I–VI) have been identified for this purpose. These complexes form protein channels that shuttle the transported protein through the membrane (or membranes) in either a single step (types I, III, IV, and VI) or by a two-step mechanism in which the Sec or Tat system is also employed (types II and V). Students should find it interesting that some pathogenic bacteria use single-step secretion systems called injectisomes (due to their structure and mechanism of action) to directly inject toxins into host cells (Figure 4.43).
Answers to Review Questions
- The central dogma holds that genetic information is contained in DNA. The flow of
information is from DNA ® RNA ® protein. - Before our chemical understanding of a gene, a gene was defined by its ability to breed a phenotypic trait and sometimes to mutate randomly. Today we define a gene as a
sequence of DNA specifying a product, such as a protein, a tRNA, or an rRNA. - Ease of separation upon heating is a function of the number of hydrogen bonds between paired bases in a DNA molecule. AT base pairs form two hydrogen bonds, while GC pairs form three hydrogen bonds. The additional hydrogen bond in a GC pair confers
increased thermostability to the DNA molecule. Therefore, the two strands of a GC-rich molecule of DNA resist heat denaturing (“melting”) to a greater degree than the two strands of a DNA molecule containing a higher percentage of AT pairs. - Linearized DNA is many times longer than the cell it came from because DNA in cells
is tightly packaged in supercoiled domains and, in the case of some Archaea and all
eukaryotes, also wrapped around protein complexes called histones. - The E. coli chromosome is ~4.6 Mbp and contains 4288 potential protein coding
sequences. Additional genetic elements that may be found in an E. coli cell include
plasmids and viral genomes. - R plasmids are extrachromosomal genetic elements in microorganisms that confer resistance to antibiotics and other growth inhibitors. The ability of these large plasmids to carry multiple antibiotic resistance genes and to spread rapidly by cell-to-cell contact makes them a serious medical problem.
- In reference to DNA, semiconservative refers to the fact that following DNA replication, each of the new DNA molecules contains one parental (old) strand and one progeny (new) strand. In DNA, the two strands are complementary to one another in that, based on the hydrogen bonding patterns dictated by the chemistry of the nitrogenous bases,
adenine always pairs with thymine, and guanine always pairs with cytosine. The two strands are also antiparallel, with one strand having a 5¢ to 3¢ orientation and its complement running in a 3¢ to 5¢ direction. - Replication of circular bacterial chromosomes is bidirectional at the origin. When replication is underway, newly formed DNA forces adjacent strands apart, leading to a
structure resembling the Greek letter theta. Eventually, as the round of replication is completed, two newly synthesized, semiconservative, circular chromosomes are
produced. - The rarity of errors in DNA replication is because a cell has two chances to incorporate the correct base at a given site. Complementary base pairing rules offer the initial opportunity. Secondly, because DNA Pol III has 3¢ ® 5¢ exonuclease activity, any mismatched nucleotide detected during proofreading may be replaced with the correct one.
- Because tRNA is synthesized by transcription, its genes must have promoters. However, tRNA is not translated, so a start codon as such would not exist. This does not preclude the existence of an AUG sequence on the tRNA.
- The start and stop signals for transcription are promoters and transcription terminators (Rho-dependent or Rho-independent). For translation, the start begins at a start codon
(in association with a ribosome-binding site) and ends with a stop codon. - An operon is a cluster of genes that are under the control of a single promoter and are therefore cotranscribed as a single, polycistronic mRNA molecule. This arrangement
allows related genes to be expressed in a coordinated fashion. - Processing of eukaryotic mRNA is necessary because the primary transcript produced by RNA polymerase contains noncoding sequences called introns that must be removed. Prokaryotic transcripts rarely contain introns.
- Amino acids are molecules that contain an —NH2 group (the amino portion) and a
—COOH group (the carboxylic acid portion); thus the term amino acid. Refer to
Figure 4.30 for the general structure of an amino acid. The R group gives an amino acid its identity. This identity may be as simple as an H atom (as in glycine) or as complicated as the double-ringed R group in tryptophan. The positions of the various R groups influence protein structure through bonding between side chains. Nonpolar side chains tend to
congregate at the protein’s core because of hydrophobic interactions. Hydrogen bonds and ionic bonds, though weak, also influence the structure of the protein. The R group of
cysteine contains a sulfur atom, which can combine with another cysteine R group, either on the same or another polypeptide chain, to form a disulfide bridge. Such covalent bonds are important in defining and maintaining protein shape. - The “wobble” effect refers to the requirement for some tRNAs to form stable base pairs only with the first two positions of the codon, which allows some tRNAs to recognize more than one codon.
- Aminoacyl-tRNA synthetases must be able to recognize an amino acid and the specific tRNA corresponding to that amino acid. The enzyme must first activate the amino acid with ATP:
Amino acid + ATP ® aminoacyl–AMP + P–P
and then transfer the activated amino acid to the free 3¢-OH of the terminal adenosine of the tRNA:
Aminoacyl–AMP + tRNA ® aminoacyl–tRNA + AMP
Transfer RNA recognition occurs through the anticodon loop, the D-loop, and the acceptor stem (the 5¢ end of the molecule).
- Peptidyl transferase activity appears to be catalyzed by the 23S rRNA component of
the ribosome. - Primary structure of a protein refers to the linear sequence of amino acids in a polypeptide. Secondary structure includes folding patterns resulting from hydrogen bonding
(i.e., the formation of a-helices and b-sheets). Tertiary structure involves additional chemical interactions, including ionic bonds, hydrophobic interactions, and disulfide bridges. Quaternary structure refers to proteins that have more than one polypeptide chain and describes the association of the different polypeptide subunits. - Proteins that aid other proteins in assuming their functional final conformation are called molecular chaperones, of which the major examples include the DnaKJ complex and the GroEL/GroES complex. Enzymes that degrade misfolded proteins are called proteases.
- Proteins designed to function outside of the cell contain signal sequences that can be
recognized by SecA protein or the signal recognition particle (SRP), both of which bind the signal sequence and direct the protein to the SecYEG translocase.
Answers to Application Questions
- If the Neisseria gonorrhoeae genome is 2220 kb in length, its total length would be 0.075 cm, according to the following calculation:
2,220,000 bp ´ 0.34 nm/bp = 754,800 nm = 754.8 mm = 0.7548 mm = 0.075 cm
If 85% of the genome is made up of open reading frames, then 1887 kbp would be open reading frames. If an average protein is 300 amino acids long, then 900-bp open reading frames would encode proteins. By using these estimates, one would predict that the N. gonorrhoeae genome could encode about 2097 proteins. The remaining 15% of the DNA might encode molecules such as stable RNA, promoters, transcription terminators, an origin of replication, and so on. A small percentage of this would also be noncoding DNA.
- Both DNA and RNA polymerase catalyze the formation of phosphodiester bonds, using deoxyribonucleotides and ribonucleotides, respectively. Both enzymes synthesize their polynucleotide chain in the 5¢ ® 3¢ direction. However, unlike DNA polymerase, RNA
polymerase can initiate synthesis de novo (not requiring a primer). Because both polymerases require a DNA template, they are more accurately referred to as a “DNA-dependent DNA polymerase” and “DNA-dependent RNA polymerase,” respectively. - Initiation of transcription one base upstream of the normal starting point by RNA
polymerase would result in the synthesis of an mRNA transcript one base longer. This would not affect protein synthesis because the 50S ribosome can bind only to the
mRNA-30S-tRNA complex when the start codon is recognized. If translation began one base downstream, it would result in a +1 reading frameshift at the beginning of the
process, and there would be no possibility of specifying a functional gene product. - The genetic code has evolved to minimize the impact of mutations by being a degenerate code. That is, in most cases, more than one codon can specify the same amino acid. This allows for a degree of DNA sequence mutation, especially in positions corresponding to the third position of the mRNA codon, without affecting the activity of the protein
product, a phenomenon referred to as silent mutation.