Microbial Genetics

Genetics is the study of inheritance and inheritable traits as expressed in an organism’s genetic material. Geneticists study many aspects of inheritance, including the physical structure and function of genetic material, mutations, and the transfer of genetic material among organisms. In this chapter, we will examine these topics as they apply to microorganisms, the study of which has formed much of the basis of our understanding of human, animal, and plant genetics. The Structure and Replication of Genomes LE A R N I N G O U TC O M E 7.1 Compare and contrast the genomes of prokaryotes and eukaryotes. The genome (jē’nōm) of a cell or virus is its entire genetic complement, including both its genes—specific sequences of nucleotides that code for RNA or polypeptide molecules—and nucleotide sequences that connect genes to one another. The genomes of cells and DNA viruses are composed solely of molecules of deoxyribonucleic acid (DNA), whereas RNA viruses use ribonucleic acid instead. (We will examine the genomes of viruses in more detail in Chapter 13.) The remainder of this chapter focuses on bacterial genomes—their structure, replication, function, mutation, and repair and how they compare and contrast with eukaryotic genomes and with the genomes of archaea. We begin by examining the structure of nucleic acids. The Structure of Nucleic Acids LE A R N I N G O U TC O M E 7.2 Describe the structure of DNA, and discuss how it ­ facilitates the ability of DNA to act as genetic material. Nucleic acids are polymers of basic building blocks called nucleotides. Each nucleotide is made up of phosphate attached to a nucleoside. A nucleoside is in turn made up of a pentose sugar (ribose in RNA and deoxyribose in DNA) attached to one of five nitrogenous bases: guanine (G), cytosine (C), thymine (T), adenine (A), or uracil (U) (FIGURE 7.1). The bases of nucleotides hydrogen-bond to one another in specific ways called complementary base pairs (bp): In DNA, the complementary bases thymine and adenine bond to one another with two hydrogen bonds (FIGURE 7.1a), whereas in RNA, uracil, not thymine, forms two hydrogen bonds with adenine (FIGURE 7.1b). In both DNA and RNA, the complementary bases guanine and cytosine bond to one another with three hydrogen bonds (FIGURE 7.1c). The critical function of DNA is to carry information in sequences of nucleotide bases in DNA and RNA molecules. Just as computer code is a series of zeros and ones with rules to transmit information, so DNA and RNA use specific base sequences to transmit information. They are the “language of life.” M07_BAUM7206_05_SE_C07.indd 193 The Structure and Replication of Genomes       Microbial Genetics Deoxyribonucleotides are linked through their sugars and phosphates to form the two backbones of the familiar, helical, double-stranded DNA (dsDNA) molecule (FIGURE 7.1d). The carbon atoms of deoxyribose are numbered 1’ (pronounced “one prime”) through 5’ (“five prime”). One end of a DNA strand is called the 5’ end because it terminates in a phosphate group attached to a 5’ carbon. The opposite end of that DNA strand terminates with a hydroxyl group bound to a 3’ carbon of deoxyribose, so that end is called the 3’ end. The two strands are constructed similarly but are oriented in opposite directions to each other; one strand runs in a 5’ to 3’ direction, while the other runs 3’ to 5’. Scientists say the two strands are antiparallel. The base pairs extend into the middle of the molecule in a way reminiscent of the steps of a spiral staircase. The hydrophobic bases are tucked inside the molecule—away from water—while the hydrophilic sugars and phosphates are outside. This provides an explanation for the double-helical structure of cellular DNA. The lengths of DNA molecules are not usually given in metric units; instead, the length of a DNA molecule is expressed in base pairs. For example, the genome of an insect-dwelling bacterium tentatively named Nasuia deltocephalinicola (nă-soo’ē-ă del-to-sef’ă-lĭn-ĭ-kō-lă) is 112,091 bp long, making it the smallest known cellular genome discovered to date. Most known bacteria have genomes of 1.5 to 6 million base pairs. The structure of DNA helps explain its ability to act as genetic material. First, the linear sequence of nucleotides carries the instructions for the synthesis of polypeptides and RNA molecules—in much the way a sequence of letters carries information used to form words and sentences. Second, the complementary structure of the two strands allows a cell to make exact copies to pass to its progeny. We will examine the genetic code and DNA replication shortly. The amount of DNA in a genome can be extraordinary, as some examples will illustrate. A cell of the bacterium Escherichia coli (esh-ĕ-rik’ē-ă kō’lī) is approximately 2 mm long, but its genome consists primarily of a 4.6 * 106 bp DNA molecule that is about 1600 mm long—800 times longer than the cell. The human genome has about 6 billion bp in 46 nuclear DNA molecules and numerous copies of a unique mitochondrial DNA molecule, and the entire genome would be about 3 meters (3 million mm) long if all DNA molecules from a single cell were laid end to end. Most of a human cellular genome is packed into a nucleus that is typically only 5 mm in diameter. This is like packing 45 miles of thread into a golf ball and still being able to access any particular section of the thread. To understand how cells package such prodigious amounts of DNA into such small spaces, we must first understand that bacteria, archaea, and eukaryotes package DNA in different ways. We begin by examining the structure of prokaryotic genomes. The Structure of Prokaryotic Genomes LE A R N I N G O UTC O M E S 7.3 Describe the structure, shape, and number of chromo- somes in typical prokaryotic cells. 7.4 Describe the structure and function of plasmids. 11/23/16 3:01 AM r t h       C ap e 7   Microbial Genetics Hydrogen bond H H N H H N N N H O CH3 N H N O H H H N H G N Uracil (U) nucleoside Guanine (G) nucleoside (b) A–U base pair (RNA) N H O H Cytosine (C) nucleoside H (c) G–C base pair (DNA and RNA) 5¿ end 3¿ end H O C O T H H N O Adenine (A) nucleoside O P O– G N N N A T H O N O 3¿ end N N H N (a) A–T base pair (DNA) 5¿ end H H N N H Thymine (T) nucleoside Adenine (A) nucleoside A N N N Sugar H 5¿ CH2 H H O 4¿ 1¿ 3¿ N H O N N N H H N H O– N N 2¿ N H O H O CH2 O H Guanine H O P CH2 O Cytosine O N N H O N H N H N O– H 3¿ end (d) Double-stranded DNA CH3 2¿ Adenine 3¿ end 3¿ H N N 5¿ end O H O O P O Thymine 1¿ H O– 4¿ O CH2 5¿ O P O O 5¿ end Thymine nucleoside Thymine nucleotide ▲ FIGURE 7.1 The structure of nucleic acids. Nucleic acids are polymers of nucleotides A A A A A A consisting of nucleosides (a pentose sugar and a nitrogenous base) bound to a phosphate. (a) Base pairing between the complementary bases adenine ( ) and thymine (T) formed by two hydrogen bonds, found in DN only. (b) Base pairing between adenine and uracil (U), found in RN only. Notice the structural similarities between thymine and uracil. (c) Base pairing between the complementary bases guanine (G) and cytosine (C) formed by three hydrogen bonds, found in both DN and RN . (d) Double-stranded DN , which consists of antiparallel strands of nucleotides held to one another by the hydrogen bonding between complementary bases. Note that scientists number the carbon atoms in a nucleotide’s sugar component as 1 prime through 5 prime. What structures do DNA nucleotides and RNA nucleotides have in common? Figure 7.1 DNA and RNA nucleotides are each composed of a pentose sugar, a phosphate, and a nitrogenous base. Prokaryotic Chromosomes Bacterial cells package the main portion of their DNA, along with associated molecules of protein and RNA, as distinct chromosomes.1 Each cell has a single copy of the genome and is 1From Greek chroma, meaning “color” (because they typically stain dark in eukaryotes, where they were first discovered), and soma, meaning “body.” M07_BAUM7206_05_SE_C07.indd 194 t herefore called a haploid cell. A t ypical chromosome (FIGURE 7.2a) consists of a circular molecule of DNA local­ The DNA of prokaryotic genomes is found in two structures: chromosomes and plasmids. ized in a region of the cytoplasm called the nucleoid. With few exceptions, no membrane surrounds a nucleoid, though the chromosome is packed in such a way that a distinct boundary is visible between a nucleoid and the rest of the cytoplasm. Chromosomal DNA is folded into loops that are 50,000 to 100,000 bp long (FIGURE 7.2b) held in place by molecules of protein and RNA. Archaeal DNA is also haploid, located in a nucleoid, and typically circular; however, in an archaeal chromosome, DNA is wrapped around globular proteins called histones, which makes it different from a bacterial chromosome. 11/23/16 3:01 AM       The Structure and Replication of Genomes for normal metabolism, for growth, or for cellular reproduction but can confer advantages to the cells that carry them. Researchers have identified many types of plasmids (sometimes called factors), including the following: Nucleoid TEM (a) 0.25 μm Bacterium Chromosome Plasmid SEM (b) 1 μm A E ▲ FIGURE 7.2 Bacterial genome. (a) Bacterial chromosomes are packaged in a region of the cytosol called the nucleoid, which is not surrounded by a membrane. (b) The packing of a circular bacterial chromosome into loops, as seen after the cell was gently broken open to release the chromosome. xtrachromosomal DN in the form of plasmids is also visible. • Fertility (F) plasmids carry instructions for conjugation, a process by which some bacterial cells transfer DNA to other bacterial cells. We will consider conjugation in more detail near the end of this chapter. • Resistance (R) plasmids carry genes for resistance to one or more antimicrobial drugs or heavy metals. By processes we will discuss shortly, certain cells can transfer resistance plasmids to other cells, which then acquire resistance to the same antimicrobial chemicals. One example of the effects of an R plasmid involves strains of E. coli that have acquired resistance to the antimicrobials ampicillin, tetracycline, and kanamycin from a strain of bacteria in the genus Pseudomonas (soo-dō-mō’nas). • Bacteriocin (bak-tēr’ē-ō-sin) plasmids carry genes for proteinaceous toxins called bacteriocins, which kill bacterial cells of the same or similar species that lack the plasmid. In this way, a bacterium containing this plasmid can kill its competitors. • Virulence plasmids carry instructions for structures, enzymes, or toxins that enable a bacterium to become pathogenic. For example, E. coli, a normal resident of the human gastrointestinal tract, can cause diarrhea when it carries plasmids that code for certain toxins. Now that we have examined the structure of prokaryotic genomes, we turn to the structure of eukaryotic genomes. The Structure of Eukaryotic Genomes LE A R N I N G O UTC O M E 7.5 Compare and contrast prokaryotic and eukaryotic chromosomes. For many years, scientists thought that each prokaryote had only a single circular chromosome, but we now know that there are many exceptions. For example, Epulopiscium (ep’yoolō-pis’sē-ŭm), a giant bacterium, has as many as hundreds or thousands of identical chromosomes. Other bacterial species have linear chromosomes. For example, Agrobacterium tumefa­ ciens (ag’rō-bak-tēr’ē-um tū’me-fāsh-enz), a bacterium used to transfer genes into plants, has two chromosomes, one circular and one linear. Plasmids In addition to chromosomes, many prokaryotic cells contain one or more plasmids, which are small molecules of DNA that replicate independently of the chromosome. Plasmids are usually circular and 1% to 20% of the size of a prokaryotic chromosome (see Figure 7.2b), ranging in size from a few thousand base pairs to a few million base pairs. Each plasmid carries information required for its own replication and often for one or more cellular traits. Typically, genes carried on plasmids are not essential M07_BAUM7206_05_SE_C07.indd 195 Eukaryotic genomes consist of both nuclear and extranuclear chromosomes. Some eukaryotic cells contain plasmids. Nuclear Chromosomes Eukaryotic cells have more than one nuclear chromosome in their genomes, though some eukaryotic cells, such as mammalian red blood cells, lose their chromosomes as they mature. Eukaryotic cells are often diploid; that is, they have two copies of each chromosome. Eukaryotic nuclear chromosomes differ from typical prokaryotic chromosomes in that nuclear chromosomes are all linear (recall that many prokaryotic chromosomes are circular) and are sequestered within a nucleus. A nucleus is an organelle surrounded by two membranes, which together are called the nuclear envelope. Given that a typical eukaryotic cell must package substantially more DNA than its prokaryotic counterpart, it is not surprising that its nuclear chromosomes are more elaborate than those of prokaryotes. 11/23/16 3:01 AM r t h       C ap e 7   Microbial Genetics 10 nm Nucleosome Active (loosely packed) Histones Linker DNA Inactive (tightly packed) DNA 10 nm 30 nm (a) Nucleosomes (b) Chromatin fiber 700 nm (c) Euchromatin and heterochromatin 1400 nm (d) Highly condensed, duplicated chromosome of dividing nucleus A A ▲ FIGURE 7.3 Eukaryotic nuclear chromosomal packaging. (a) Histones stabilize and package DN to form nucleosomes connected by linker DN . (b) Nucleosomes clump to form chromatin fibers. (c) Chromatin fibers fold and are organized into active euchromatin and inactive heterochromatin. (d) During nuclear division (mitosis), duplicated chromatin fully condenses into a mitotic chromosome that is visible by light microscopy. If the nucleosomes were actually the size shown in the artist’s illustration (a), the chromosome in (d) would be 20 m (about 65 feet) long. Most eukaryotic chromosomes have globular eukaryotic histones,2 which are similar to archaeal histones. DNA, which has an overall negative electrical charge, wraps around the positively charged histone proteins to form 10-nm-diameter beads called nucleosomes (FIGURE 7.3a). Nucleosomes clump with other proteins to form chromatin fibers that are about 30 nm in diameter (FIGURE 7.3b). Except during mitosis (nuclear division), chromatin fibers are dispersed throughout the nucleus and are too thin to be resolved without the extremely high magnification of electron microscopes. In regions of the chromosome where genes are active, chromatin fibers are loosely packed to form euchromatin (yū-krō’mă-tin); inactive DNA is more tightly packed and is called heterochromatin (het’er-ōkrō’mă-tin) (FIGURE 7.3c). 2Dinoflagellates, a group of single-celled aquatic microorganisms, are the only eukaryotes without histones. M07_BAUM7206_05_SE_C07.indd 196 Prior to mitosis, a cell replicates its chromosomes and then condenses them into pairs of chromosomes visible by light microscopy (FIGURE 7.3d). One molecule of each pair is destined for each daughter nucleus. (Chapter 12 discusses mitosis in more detail.) The net result is that each DNA molecule is packaged as a mitotic chromosome that is 50,000 times shorter than its extended length. Extranuclear Chromosomes in Eukaryotes Not all of the DNA of a eukaryotic genome is contained in its nuclear chromosomes: most eukaryotic cells also have mitochondria; and plant, algal, and some protozoan cells have chloroplasts. Both mitochondria and chloroplasts contain chromosomes. Chromosomes of mitochondria are circular, and though researchers had thought the chromosomes of chloroplasts were also circular, recent research has shown that chloroplast chromosomes are linear. The extranuclear chromosomes of both mitochondria and chloroplasts resemble the chromosomes of prokaryotes. 11/23/16 3:01 AM       The Structure and Replication of Genomes TABLE 7.1 Characteristics of Microbial Genomes Bacteria Archaea Eukarya Number of Chromosomes Single (haploid) copies of one or more One (haploid) With one exception, two or more, typically diploid Plasmids Present? In some cells; frequently more than one per cell In some cells In some fungi, algae, and protozoa Type of Nucleic Acid Circular or linear dsDNA Circular dsDNA Linear dsDNA in nucleus and chloroplasts; circular dsDNA in mitochondria and plasmids Location of DNA In nucleoid of cytoplasm and in plasmids In nucleoid of cytoplasm and in plasmids In nucleus and in mitochondria, chloroplasts, and plasmids in cytosol Histones Present? No, though chromosome is associated with a small amount of nonhistone protein Yes Yes, in nuclear chromosomes; not in extranuclear chromosomes Original DNA First replication Second replication Original strand New strands ­ Genes located on extranuclear chromosomes code for about 5% of the RNA and polypeptides required for each organelle’s replication and function; nuclear DNA codes for the remaining 95% of the organelle’s RNA molecules and polypeptides. Some proteins have a quaternary structure formed from the association of individual polypeptides (see Chapter 2, Figure 2.24). Interestingly, mitochondrial or chloroplast chromosomes do not alone code for functional proteins. Rather, mitochondria-coded and chloroplast-coded polypeptides become functional only when associated with polypeptides coded by nuclear chromosomes. In addition to the extranuclear DNA in their mitochondria and chloroplasts, some fungi, algae, and protozoa carry plasmids. For instance, most strains of brewer’s yeast (Saccharomyces cerevisiae, sak-ă-rō-mī’sēz se-ri-vis’ē-ī) contain about 70 copies of a plasmid known as a 2@mm circle. Each 2@mm circle is about 6300 bp long and has four protein-encoding genes that are involved solely in replicating the plasmid and confer no other traits to the cell. In summary, the genome of a prokaryotic cell consists of haploid chromosomal DNA and all extrachromosomal DNA in the form of plasmids that are present. In contrast, a eukaryotic genome consists of nuclear chromosomal DNA in one or more linear chromosomes, plus all the extranuclear DNA in mitochondria, chloroplasts, and any plasmids that are present. The genomes of prokaryotes and eukaryotes are compared and contrasted in TABLE 7.1. DNA Replication LE A R N I N G O U TC O M E S 7.6 Describe the replication of DNA as a semiconservative process. 7.7 Compare and contrast the synthesis of leading and lag- ging strands in DNA replication. 7.8 Contrast bacterial DNA replication with that of eukaryotes. DNA replication is an anabolic polymerization process that allows a cell to make copies of its genome. Though M07_BAUM7206_05_SE_C07.indd 197 ­ ▲ FIGURE 7.4 Semiconservative model of DNA replication. ­ Each of the two strands of the original molecule (shown dark) serves as a template for the synthesis of a new, complementary strand (shown lighter). ­ bacterial, archaeal, and eukaryotic cells package DNA differently, all three types employ similar mechanisms for DNA replication. ▶ ANIMATIONS: DNA Replication: Overview The key to DNA replication is the complementary structure of the two strands: Adenine and guanine in one strand bond with thymine and cytosine, respectively, in the other. DNA replication is a simple concept—a cell separates the two original strands and uses each as a template for the synthesis of a new complementary strand. Biologists say that DNA replication is semiconservative because each daughter DNA molecule is composed of one original strand and one new strand (FIGURE 7.4). All polymerization processes require monomers (building blocks) and energy. The monomers of DNA are four triphosphate deoxyribonucleotides composed of a base, deoxyribose sugar, and three phosphate groups. DNA nucleotides differ from one another only in the kind of base present. Guanosine triphosphate deoxyribonucleotide (dGTP) has a guanine base (FIGURE 7.5a). Cytidine triphosphate deoxyribonucleotide (dCTP) has a cytosine base. Thymine is the base in thymidine triphosphate deoxyribonucleotide (dTTP), and adenine is the base in adenosine triphosphate deoxyribonucleotide (dATP). The structure of dATP is similar to that of the energy storage molecule ATP, except that ATP uses ribose sugar; it is a ribonucleotide rather than a deoxyribonucleotide (see Figure 2.25b). 06/03/2018 08:40 r ◀ FIGURE 7.5 The dual role of triphosphate deoxyribonucleotides Guanosine triphosphate deoxyribonucleotide (dGTP) H N N P O P H H2N N O 5¿ CH2 N 1¿ H H H 2¿ H Guanine base O O 4¿ P O– O P O A O– OH O Figure 7.6 This molecule (dGTP) contains deoxyribose; rGTP contains ribose (see Figure 2.25b). O– O A as building blocks and energy sources in DNA synthesis. (a) Guanosine triphosphate deoxyribonucleotide (dGT ), like all the triphosphate monomers of DN , is a nucleotide to which two additional phosphate groups are attached. (b) The energy required for DN polymerization (the addition of nucleotide building blocks to a DN strand) is carried by each triphosphate nucleotide in the high-energy bonds between phosphate groups. What is the difference between dGTP and guanosine triphosphate ribonucleotide (rGTP)? Guanine nucleotide (dGMP) A t h       C ap e 7   Microbial Genetics High-energy bond H 3¿ OH Deoxyribose Guanosine (nucleoside) (a) Pyrophosphate P P P 3¿ OH P + 5¿ A C P 3¿ OH P P P G Triphosphate nucleotide P P P 3¿ OH 5¿ 5¿ T Existing DNA strand P Pyrophosphate (diphosphate) removed; released energy is used for synthesis A C T G Longer DNA strand (b) The three phosphate groups of a nucleotide are linked together by high-energy bonds. During DNA synthesis, energy is released when two of the phosphate groups are removed from a nucleotide as a pyrophosphate (also called diphosphate) (FIGURE 7.5b). Thus, a nucleotide is both a building block and an energy source. The following sections focus on bacterial DNA replication and then consider small differences in the process in eukaryotes. Archaeal processes are not as well characterized and are not examined here. Initial Processes in Bacterial DNA Replication DNA replication begins at a specific sequence of nucleotides called an origin. An enzyme called DNA helicase locally “unzips” the DNA molecule by breaking the hydrogen bonds between complementary nucleotide bases, which exposes the nucleotide bases in a replication fork (FIGURE 7.6a). Other protein molecules stabilize the separated single strands so that they do not rejoin while replication proceeds. ▶ ANIMATIONS: DNA Replication: Forming the Replication Fork After helicase untwists and separates the strands, a molecule of an enzyme called DNA polymerase (po-lim’er-ās) binds to each strand. Scientists have identified several kinds of bacterial DNA polymerase. These vary in their specific functions, but all of them share one important feature—they catalyze synthesis of DNA by the addition of new nucleotides only to a hydroxyl group at the 3’ end of a nucleic acid. Beneficial Microbes: Life M07_BAUM7206_05_SE_C07.indd 198 in a Hot Tub on p. 200 describes a type of DNA polymerase that has become a mainstay of genomic investigations. All DNA polymerases replicate DNA by adding nucleotides in only one direction—5′ to 3′—like a jeweler stringing pearls to make a necklace, adding them one at a time, always moving from one end of the string to the other. DNA polymerase III is the usual enzyme of DNA replication in bacteria. Because the two original strands are antiparallel, cells synthesize new strands in two different ways. One new strand, called the leading strand, is synthesized continuously—5’ to 3’—as a single long chain of nucleotides. The other new strand, called the lagging strand, is also synthesized 5’ to 3’, but in short segments that are later joined. We will consider synthesis of the leading strand before examining replication of the lagging strand, but you should understand that the two processes occur simultaneously. ▶ ANIMATIONS: DNA Replication: Replication Proteins Synthesis of the Leading Strand A cell synthesizes a leading strand toward the replication fork in the following series of five steps, the first three of which are shown in FIGURE 7.6b: 1 An enzyme called primase synthesizes a short RNA molecule that is complementary to the DNA strand being copied. This RNA primer provides a 3’ hydroxyl group, which is required by DNA polymerase III. 11/23/16 3:01 AM       The Structure and Replication of Genomes Chromosomal proteins (histones in eukaryotes and archaea) removed DNA polymerase III 3¿ Replication fork 5¿ DNA helicase Stabilizing proteins (a) Initial processes DNA polymerase III 4 DNA polymerase proofreads as it extends (not shown) 3 Primase 3 C 1 A T 3¿ Replication fork 2 5¿ Leading strand P+P Triphosphate nucleotide RNA primer A G A A T C T T A P P Pyrophosphate (b) Synthesis of leading strand Replication fork 8b Triphosphate nucleotide RNA primer 5 7 Okazaki fragment Lagging strand 3¿ 5¿ 6 Primase DNA polymerase III 8a 9 DNA ligase DNA polymerase I (c) Synthesis of lagging strand ▲ FIGURE 7.6 DNA replication. (a) Initial processes. The cell removes proteins (called histones in A A A A A A eukaryotes and archaea) from the DN molecule. Helicase unzips the double helix—breaking hydrogen bonds between complementary base pairs—to form a replication fork. (b) Continuous synthesis of the leading strand. DN synthesis always moves in the 5′ to 3′ direction, so the leading strand is synthesized toward the replication fork. The numbers refer to the steps in the process, which are described in the text. The proofreading function of DN polymerase III occurs at step 4, though proofreading is not detailed in the artwork. Replacement of the leading strand’s RN primer with DN is shown in step 8b. (c) Discontinuous synthesis of the lagging strand, which proceeds moving away from the replication fork. ctual Okazaki fragments are about 1000 nucleotides long. Why is DNA replication termed “semiconservative”? A go to the MasteringMicrobiology Study rea and watch the MicroFlix. A Figure 7.6 “Semiconservative” refers to the fact that each of the daughter molecules retains one parental strand and has one new strand; in other words, each is half new and half old. M07_BAUM7206_05_SE_C07.indd 199 ▶ To see a 3-D animation on DN replication, 11/23/16 3:01 AM r t h       C ap e 7   Microbial Genetics E A E E B N FICI L MICROB S Life in a Hot Tub 2 3 4 Triphosphate deoxyribonucleotides form hydrogen bonds with their complements in the parental strand. Adenine nucleotides bind to thymine nucleotides, and guanine nucleotides bind to cytosine nucleotides. Using the energy in the high-energy bonds of the triphosphate deoxyribonucleotides, DNA polymerase III covalently joins them one at a time to the leading strand. DNA polymerase III can add about 500 to 1000 nucleotides per second to a new strand. DNA polymerase III also performs a proofreading function. About one out of every 100,000 nucleotides is mismatched with its template; for instance, a guanine might become incorrectly paired with a thymine. DNA polymerase III recognizes most of these errors and removes the incorrect nucleotides before proceeding with synthesis. This role, known as the proofreading exonuclease function, acts like the backspace key on a keyboard, removing the most recent error. Because of this proofreading exonuclease function and other repair strategies beyond the scope of this discussion, only about one error remains for every 10 billion (1010) bp replicated, so cells make an almost exact copy of their genome. As replication of the leading strand continues, another DNA polymerase—DNA polymerase I—replaces the RNA primer with DNA nucleotides (as shown in FIGURE 7.6c, step 8a and 8b). Note that researchers named DNA polymerase enzymes in the order of their discovery, not the order of their actions. Synthesis of the Lagging Strand Because DNA polymerase III adds nucleotides only to the 3’ end of the new strand, the enzyme moves away from the replication fork as it synthesizes a lagging strand. As a result, the lagging strand is synthesized discontinuously and always lags M07_BAUM7206_05_SE_C07.indd 200 P A A P E A A P The temperature required to perform CR is about 94°C. This temperature, which is almost that of boiling water, is the temperature required to break the hydrogen bonds of DN and unzip the double helix, but this temperature also permanently denatures most DN polymerase enzymes. nter Thermus aquaticus, a bacterium that thrives in hot springs such as those of Yellowstone National ark. Since this bacterium loves hot water, it is not surprising that its enzymes are heat stable, and its DN polymerase—called Taq polymerase or Taq—was the first polymerase used for CR replication of DN . Though Science magazine declared Taq “Molecule of the Year” in 1989, scientists now have many other heat-stable polymerases from bacterial and archaeal hyperthermophiles available for CR. P A A P A P P A A A A Scientists replicate DN of cells for a variety of tasks: studying gene action and regulation, elucidating relationships among various kinds of cells, detecting hereditary diseases, and determining “genetic fingerprints” for such purposes as paternity tests, detection of pathogens, diagnosis of infectious diseases, and identification of DN found at crime scenes. ll such studies use millions or billions of identical copies of DN produced using a process called polymerase chain reaction (PCR). CR enzymatically replicates DN without using living cells. The concept of CR is relatively simple. s its inventor wrote in Scientific American, “Beginning with a single molecule of the genetic material DN , the CR can generate 100 billion similar molecules in an afternoon. The reaction is easy to execute. It requires no more than a test tube, a few simple reagents, and a source of heat.” In the latter, however, lies a problem. behind the process occurring in the leading strand. The steps in the synthesis of a lagging strand are as follows (FIGURE 7.6c): 5 Primase synthesizes RNA primers, but in contrast to its action on the leading strand, primase synthesizes multiple primers—one every 1000 to 2000 DNA bases of the template strand. 6 DNA polymerase III joins neighboring nucleotides and proofreads. In contrast to synthesis of the leading strand, however, the lagging strand is synthesized in discontinuous segments called Okazaki fragments, named for the Japanese scientist Reiji Okazaki (1930–1975), who first identified them. Each Okazaki fragment uses one of the new RNA primers, so each fragment consists of 1000 to 2000 nucleotides. 7 Nucleotides pair up with their complements in the template—adenine with thymine, and cytosine with guanine. 8 DNA polymerase I replaces the RNA primers of Okazaki fragments as well as the RNA primer in the leading strand with DNA and proofreads the short DNA segment it has just synthesized. DNA ligase seals the nicks between adjacent Okazaki fragments to form a continuous DNA strand. 9 In summary, synthesis of the leading strand proceeds continuously toward the replication fork from a single RNA primer at the origin, following helicase and the replication fork down the DNA. The lagging strand is synthesized away from the replication fork discontinuously as a series of Okazaki fragments, each of which begins with its own RNA primer. All the primers are eventually replaced with DNA nucleotides, and ligase joins the Okazaki fragments. As noted earlier, DNA replication is semiconservative; each daughter molecule is composed of one parental strand and 11/23/16 3:01 AM       The Structure and Replication of Genomes Origin Parental strand Replication forks Daughter strand Replication proceeds in both directions Termination of replication Termination point ▲ FIGURE 7.7 The bidirectionality of DNA replication in prokaryotes. Replication begins at an origin and proceeds in both directions to a termination sequence. The origin in E. coli is 80 bp in size. Bacterial chromosomes (shown here) have a single origin, but eukaryotic chromosomes have thousands of origins. one daughter strand. The replication process produces doublestranded daughter molecules with a nucleotide sequence virtually identical to that in the original double helix, ensuring that the integrity of an organism’s genome is maintained each time it is copied. ▶ ANIMATIONS: DNA Replication: Synthesis Other Characteristics of Bacterial DNA Replication DNA replication is bidirectional; that is, DNA synthesis proceeds in both directions from the origin. In bacteria, the process of replication typically proceeds from a single origin, so it involves two sets of enzymes and two replication forks each having a leading strand and a lagging strand (FIGURE 7.7). One replication fork moves clockwise and the other moves counterclockwise until replication ends about 180° from the origin at termination sequences in the DNA molecule. The unzipping and unwinding action of helicase introduces supercoils into the DNA molecule ahead of the replication forks. Excessive supercoiling creates tension on the DNA molecule—like your grandmother’s overwound phone cord— and would stop DNA replication. The enzymes gyrase and topoisomerase remove such supercoils by cutting the DNA, rotating the cut ends in the direction opposite the supercoiling, and then rejoining the cut ends. DNA replication is further complicated by methylation of the daughter strands, in which an enzyme adds a methyl group to one or two bases that are part of specific nucleotide sequences. Bacteria typically methylate adenine bases and only rarely a cytosine base. Methylation plays a role in a variety of cellular processes, including the following: • Control of genetic expression. In some cases, genes that are methylated are “turned off” and are not transcribed, M07_BAUM7206_05_SE_C07.indd 201 whereas in other cases methylated genes are “turned on” and are transcribed. • Initiation of DNA replication. In many bacteria, methylated nucleotide sequences play a role in initiating DNA replication. • Protection against viral infection. Methylation at specific sites in a nucleotide sequence enables cells to distinguish their DNA from viral DNA, which lacks methylation. The cells can then selectively degrade viral DNA. • Repair of DNA. The role of methylation in some DNA repair mechanisms is discussed later in the chapter (p. 220). Replication of Eukaryotic DNA Eukaryotes replicate DNA in much the same way as do bacteria; helicases and topoisomerases unwind DNA, protein molecules stabilize single-stranded DNA, and molecules of DNA polymerase synthesize leading and lagging strands simultaneously. However, eukaryotic replication differs from prokaryotic replication in some significant ways: • Eukaryotic cells use different DNA polymerases to replicate DNA. There are four basic types: One initiates replication, including synthesis of a primer—the function performed by primase in bacteria. A second type of eukaryotic DNA polymerase elongates the leading strand, and a third appears to be responsible for replicating the lagging strand. A fourth type replicates mitochondrial DNA. • The large size of eukaryotic chromosomes necessitates thousands of origins per molecule, each generating two replication forks; otherwise, the replication of eukaryotic genomes would take days instead of hours. 11/23/16 3:01 AM r t h       C ap e 7   Microbial Genetics • Eukaryotic Okazaki fragments are shorter than those of bacteria—100 to 400 nucleotides long. • Plant and animal cells methylate cytosine bases exclusively. The Transfer of Genetic Information LE A R N I N G O UTC O M E 7.10 State the central dogma of genetics, and explain the roles of DNA and RNA in polypeptide synthesis. We have examined the physical structure of cellular genes— the specific sequences of DNA nucleotides—and the way cells replicate their genes. Now we will consider how genes function and how cells control genetic expression. Gene Function The first topic we must consider if we are to understand gene function is the relationship between an organism’s genotype and its phenotype. The Relationship Between Genotype and Phenotype LE A R N I N G O U TC O M E ∆IAITHMAπITEXPHπOMAIEΠΩ≥E∂EIN KAMNONTΩNKATA∆YNAMINKAIKPIπINEMHN EΠI∆H∂HπEI∆EKAIA∆IKIH Explain how the genotype of an organism determines its phenotype. Greek genos, meaning “race,” and typos, meaning “type.” 4From Greek phainein, meaning “to show.” 3From M07_BAUM7206_05_SE_C07.indd 202 3¿ 5¿ T A C G C C A T G A A T A T G C G G T A C T T A U A 5¿ DNA (genotype) 3¿ Transcription 5¿ A U G C G G U A C U 3¿ mRNA Translation by ribosomes NH2 Methionine Arginine Tyrosine Leucine … Polypeptide Phenotype ▲ FIGURE 7.8 The central dogma of genetics. A cell transcribes RN from a DN gene and then synthesizes polypeptides by translating the code carried by the RN molecules. olypeptides determine phenotype by acting as structural, enzymatic, and regulatory proteins. P A A The genotype3 (jēn’ō-tīp) of an organism is the actual set of genes in its genome. A genotype differs from a genome in that a genome also includes nucleotides that are not part of genes, such as the nucleotide sequences that link one gene to another. At the molecular level, the genotype consists of all the series of DNA nucleotides that carry instructions for an organism’s life. Phenotype4 (fē’nō-tīp) refers to the physical features and functional traits of an organism, including characteristics such as structures, morphology, and metabolism. For example, the shape of a cell, the presence and location of flagella, the enzymes and cytochromes of electron transport chains, and membrane receptors that trigger chemotaxis are all phenotypic traits. Genotype determines phenotype by specifying what kinds of RNA and which structural, enzymatic, and regulatory protein molecules are produced. Though genes do not code directly for such molecules as phospholipids or for behaviors such as chemotaxis, ultimately phenotypic traits result from the actions of RNA and protein molecules that are themselves coded by DNA. Not all genes are active at all times; that is, the information of a genotype is not always expressed as a phenotype. For example, E. coli activates genes for lactose catabolism only when it detects lactose in its environment. A 7.9 Cells must continually synthesize proteins required for growth, reproduction, metabolism, and regulation. This synthesis requires that they accurately transfer the genetic information contained in DNA nucleotide sequences to the amino acid sequences of proteins. Cells do not convert the information coded in DNA directly into proteins but first make an RNA copy of the gene. In this copying process, called transcription, 5 the information in DNA is copied as RNA nucleotide sequences; RNA molecules in ribosomes then synthesize polypeptides in a process called translation.6 These processes make up the central dogma of genetics: DNA is transcribed to RNA, which is translated to form polypeptides (FIGURE 7.8). An analogy serves to illustrate the central dogma. Suppose you were trying to understand the following message (which is a portion of the oath of Hippocrates, written in the Greek alphabet): ­ DNA replication requires a large amount of energy, yet none of a cell’s ATP energy supply is used. Why isn’t it? ­ TELL ME WHY Latin trans, meaning “across,” and scribere, meaning “to write”—that is, to transfer in writing. 6From Latin translatus, meaning “transferred.” 5From 11/23/16 3:01 AM       Gene Function If the Greek alphabet is foreign to you, you might have the Greek characters transcribed into the familiar English alphabet as a first step in understanding the message: Diaiteimasi te chreisomai ep ophelein kamnonton kata dunamin kai krisin emein epi deileisei de kai adikiei eirzein Then you could begin the process of having the Greek words, now expressed in English letters, translated into English words: I will prescribe treatment to the best of my ability and judgment to help the sick and never for a harmful or illicit purpose To a ribosome, DNA is like a foreign language written in a foreign alphabet. Thus, a cell must use processes analogous to those just described: It must first transcribe the “foreign alphabet” of DNA nucleotides (genes) into the more “familiar alphabet” of RNA nucleotides; then it must translate the message formed by these “letters” into the “words” (amino acids) that make up the “message” (a polypeptide). In this way, a genotype can be expressed as a phenotype. There are a few exceptions to the central dogma. For example, some RNA viruses transcribe DNA from an RNA template—a process that is the reverse of cellular transcription. In the following sections, we will examine the processes of transcription and translation. ▶ ANIMATIONS: Transcription: Overview; Translation: Overview The Events in Transcription LE A R N I N G O U TC O M E S 7.11 Describe three steps in RNA transcription, mentioning the following: DNA, RNA polymerase, promoter, initiation, 5’ to 3’ direction, terminator, and Rho. 7.12 Contrast bacterial transcription with that of eukaryotes. Cells transcribe six main types of RNA from DNA: • RNA primer molecules for DNA polymerase to use during DNA replication • Messenger RNA (mRNA) molecules, which carry genetic information from chromosomes to ribosomes • Ribosomal RNA (rRNA) molecules, which combine with ribosomal polypeptides to form ribosomes—the organelles that synthesize polypeptides • Transfer RNA (tRNA) molecules, which deliver the correct amino acids to ribosomes based on the sequence of nucleotides in mRNA • Regulatory RNA molecules, which interact with DNA to control gene expression • Ribozymes, which are RNA molecules that function as metabolic enzymes in a cell We have already considered the role of RNA primer in DNA replication and will more closely examine the functions M07_BAUM7206_05_SE_C07.indd 203 of the other types of RNA shortly. Next we examine transcription in bacteria and contrast it with eukaryotic transcription; archaeal processes are not as well known. Transcription occurs in the nucleoid region of the cytoplasm in bacteria. The three steps of RNA transcription are 1 initiation of transcription, 2 elongation of the RNA transcript, and 3 termination of transcription. FIGURE 7.9 depicts the events in transcription. ▶ ANIMATIONS: Transcription:The Process Initiation of Transcription RNA polymerases—the enzymes that synthesize RNA—bind to specific DNA nucleotide sequences called promoters, each of which is located near the beginning of a gene and serves to initiate transcription ( 1a in FIGURE 7.9a). In bacteria, a polypeptide subunit of RNA polymerase called a sigma factor is necessary for recognition of a promoter. Once it adheres to a promoter sequence, an RNA polymerase unzips and unwinds the DNA molecule in the promoter region and then travels along the DNA, unzipping the double helix to form a “bubble” as it moves 1b . Sigma factor often drops off the RNA polymerase as transcription proceeds, though it remains attached in some bacteria. Cells may use different sigma factors and different promoter sequences to provide some control over transcription. RNA polymerases using different sigma factors do not adhere equally strongly to all promoters; there is about a millionfold difference between the strongest attraction and the weakest one. The greater the attraction between a particular sigma factor and a promoter, the more likely that a particular gene will be transcribed; thus, variations in sigma factors and promoters affect the amounts and kinds of polypeptides produced. Elongation of the RNA Transcript RNA transcription does not actually begin in the promoter region but, rather, at a spot 10 nucleotides away. There, triphosphate ribonucleotides (ATP, UTP, GTP, and CTP) align opposite their complements in the open DNA “bubble.” RNA polymerase links together two adjacent ribonucleotide molecules using energy from the phosphate bonds of the first ribonucleotide ( 2 in FIGURE 7.9b). The enzyme then moves down the DNA strand, elongating RNA by repeating the process. Only one of the separated DNA strands is transcribed. In prokaryotes, many molecules of RNA polymerase may concurrently transcribe the same gene (FIGURE 7.10). In this way, a cell simultaneously produces numerous identical copies of RNA from a single gene—much as many identical prints can be made from a single photographic negative. Like DNA polymerase, RNA polymerase links nucleotides only to the 3’ end of the growing molecule; that is, transcription is in the 5’ to 3’ direction. However, RNA polymerase differs from DNA polymerase in the following ways: • RNA polymerase unwinds and opens DNA by itself; helicase is not required. • RNA polymerase does not need a primer. • RNA polymerase copies only one of the DNA strands. 11/23/16 3:01 AM 1a RNA polymerase attaches nonspecifically to DNA and travels down its length until it recognizes a promoter sequence. Sigma factor enhances promoter recognition in bacteria. 1b Upon recognition of the promoter, RNA polymerase unzips the DNA molecule beginning at the promoter. Terminator RNA polymerase 5¿ 3¿ 3¿ 5¿ Promoter DNA Sigma factor Attachment of RNA polymerase “Bubble” 5¿ 3¿ 3¿ 5¿ Template DNA strand Unzipping of DNA, movement of RNA polymerase (a) Initiation of transcription “Bubble” 2 Triphosphate ribonucleotides align with their DNA 5¿ complements, and RNA 3¿ polymerase links them together, synthesizing RNA. Growing RNA molecule No primer is needed. The (transcript) triphosphate ribonucleotides also provide the energy required for RNA synthesis. 3¿ 3¿ 5¿ 5¿ 3¿ 5¿ C G A U T A G C 3¿ G G C U A C G C A 5¿ Template DNA strand (b) Elongation of the RNA transcript Terminator 3¿ 5¿ 5¿ 3¿ 5¿ 3¿ RNA transcript released 3a Self-termination: transcription of GC-rich terminator region produces a hairpin loop, which creates tension, loosening the grip of the polymerase on the DNA. 3b Rho-dependant termination: Rho pushes between polymerase and DNA. This causes release of polymerase, RNA transcript, and Rho. RNA polymerase Rho termination protein GGCC GCCCG GC-rich hairpin loop CGGGC Rho protein moves along RNA UU U UUU C G G TAT C C G G G C C A A A UU AAA U Terminator Terminator AA 3¿ Template strand (c) Termination of transcription: release of RNA polymerase ▲ FIGURE 7.9 The events in the transcription of RNA in prokaryotes. (a) Initiation of transcription. (b) Elongation of the RNA transcript. (c) Termination of transcription: release of RNA polymerase by one of two methods. What is the difference between a promoter sequence and an origin? GGCCCGGTATCCGGGCCAAAAAAA Figure 7.9 A promoter is a DNA sequence that initiates transcription; an origin is a point where DNA replication begins. M07_BAUM7206_05_SE_C07.indd 204 11/23/16 3:01 AM       Gene Function RNA polymerases Promoter 5¿ 3¿ 5¿ Terminator 5¿ 3¿ 3¿ 3¿ 3¿ 3¿ 3¿ 3¿ 5¿ 5¿ Template DNA strand Sigma factor RNA 5¿ 5¿ A ▲ FIGURE 7.10 Concurrent RNA transcription. Once an RN polymerase 5¿ A A molecule has cleared the promoter, another molecule can recognize the promoter and initiate transcription. In this manner, multiple copies of RN are transcribed simultaneously. The art shows six molecules of RN being transcribed at the same time along a gene. Termination of Transcription Transcription terminates when RNA polymerase and the transcribed RNA are released from DNA (FIGURE 7.9c). RNA polymerase is tightly associated with the DNA molecule and cannot be removed easily; therefore, the termination of transcription is complicated. Scientists have elucidated two types of termination processes in bacteria—those that are self-terminating and those that depend on the action of an additional protein called Rho. These processes of transcription termination should not be confused with termination of translation, which we examine in a later section. Self-Termination Self-termination occurs when RNA polymerase transcribes a terminator sequence of DNA composed of two regions: one that is symmetrical in guanine and cytosine bases, followed by a region rich in adenine bases (see 3a in Figure 7.9c). RNA polymerase slows down during transcription of the GC-rich portion of the terminator because the three hydrogen bonds between each guanine and cytosine base pair make unwinding the DNA helix more difficult. This pause in transcription, which lasts about 60 seconds, provides enough time for the RNA molecule to form hydrogen bonds between its own symmetrical sequences, forming a hairpin loop structure that puts tension on the union of RNA polymerase and the DNA. M07_BAUM7206_05_SE_C07.indd 205 When RNA polymerase transcribes the adenine-rich portion of the terminator, the relatively few hydrogen bonds between the adenine bases of DNA and the uracil bases of RNA cannot withstand the tension, and the RNA transcript breaks away from the DNA, releasing RNA polymerase. Rho-Dependent Termination The second type of termination depends on the termination protein called Rho. Rho binds to a specific RNA sequence near the end of an RNA transcript. Rho moves toward RNA polymerase at the 3’ end of the growing RNA molecule, pushing between RNA polymerase and the DNA strand and forcing them apart; this releases RNA polymerase, the RNA transcript, and Rho (see 3b in Figure 7.9c). Transcriptional Differences in Eukaryotes Eukaryotic transcription differs from bacterial transcription in several ways. First, a eukaryotic cell transcribes RNA inside its nucleus, primarily in the region of the nucleus called the nucleolus, as well as inside any mitochondria and chloroplasts that are present. In contrast, transcription in prokaryotes occurs in the cytosol. Another difference between eukaryotes and bacteria is that eukaryotes have three types of nuclear RNA polymerase—one for transcribing mRNA, one for transcribing the major rRNA gene, and one for transcribing tRNA and smaller rRNA molecules. Mitochondria use a fourth type of RNA polymerase. Further, several separate protein transcription factors at a time assist in binding eukaryotic RNA polymerase to promoter sequences, in contrast to a single sigma factor in bacteria. After initiating transcription, eukaryotic RNA polymerases shed most of the transcription factors and recruit another set of polypeptides called elongation factors. Finally, eukaryotic cells must process mRNA before beginning polypeptide translation (FIGURE 7.11). In general, the functions of RNA processing are to aid in export of RNA from ­ • RNA polymerase is slower than DNA polymerase III, proceeding at a rate of about 50 nucleotides per second. • RNA polymerase incorporates ribonucleotides instead of deoxyribonucleotides. • Uracil nucleotides are incorporated instead of thymine nucleotides. • The proofreading function of RNA polymerase is less efficient, leaving a base-pair error about every 10,000 nucleotides. 11/23/16 3:01 AM r t h       C ap e 7   Microbial Genetics Exons (polypeptide coding regions) 5¿ Template DNA strand 3¿ Introns (noncoding regions) Transcription Exon 1 Exon 2 Exon 3 Exon 4 Pre-mRNA AAAAAAAAAAAAAA 5¿ cap Intron 1 Intron 2 Intron 1 Poly-A tail Processing Spliceosomes 5¿ Intron 3 AAAAAAAAAAAAAA Exon 1 Exon 2 Exon 3 3¿ mRNA splicing Exon 4 5¿ AAAAAAAAAAAAAA 3¿ mRNA (codes for one polypeptide) Nuclear envelope Nucleoplasm Nuclear pore A major advantage of splicing is the ability to form multiple proteins from one gene, called alternative splicing. For example, Figure 7.11 shows the joining of exons 1, 2, 3, and 4. However, a different protein could be made by splicing together 1, 2, and 4 (skipping exon 3). This is how human cells generate over 150,000 proteins from only 25,000 genes. Now that we have discussed how cells use DNA as the genetic material, maintain the integrity of their genomes through semiconservative replication, and transcribe RNA from DNA genes, we turn to the process of translation and the role of each type of RNA. Translation LE A R N I N G O UTC O M E S 7.13 Describe the genetic code in general, and identify the relationship between codons and amino acids. 7.14 Describe the synthesis of polypeptides, identifying the Cytosol roles of three types of RNA. mRNA AAAA AA AA A A AA AA A A E A ▲ FIGURE 7.11 Processing eukaryotic mRNA. Within the nucleus, transcription produces pre-mRN , which contains coding exons and noncoding introns. nzymes cap the 5′ end with a modified guanine nucleotide and add hundreds of adenine nucleotides to the 3′ end, a process known as polyadenylation. Ribozymes further process pre-mRN by removing introns and splicing together exons to form a molecule that codes for a single polypeptide. ukaryotic mRN then moves from the nucleus to the cytoplasm. E refers to intervening sequences (i.e., they lie between coding regions), whereas the “ex” in exon refers to the fact that these regions are expressed. Five small RNA molecules associate with about 300 polypeptides to form a spliceosome that acts as a ribozyme (ribosomal enzyme) to splice pre-mRNA into mRNA—it removes introns and splices the exons to produce a functional mRNA molecule that exits the nucleus. the nucleus, to stabilize mRNA in the cytoplasm, and to aid in translation. RNA processing involves three events: 1. Capping. The cell adds a modified guanine nucleotide to the 5’ end of the mRNA when the RNA molecule is about 30 nucleotides long. 2. Polyadenylation. When RNA polymerase reaches the end of a gene, termination proteins cleave the RNA molecule and add 100 to 250 adenine nucleotides, depending on the organism, to the 3’ end. Polyadenylation occurs without a DNA template. 3. Splicing. Newly capped and polyadenylated mRNA molecules are called pre­messenger RNA (pre­mRNA) because they contain introns, which are noncoding sequences that may be thousands of nucleotides long. (Few prokaryotic mRNA molecules contain introns.) A cell removes introns to make functional mRNA containing only coding regions called exons, each of which is about 150 nucleotides long. The “in” in intron M07_BAUM7206_05_SE_C07.indd 206 7.15 Contrast translation in bacteria from that in eukaryotes. Translation is the process whereby ribosomes use the genetic information of nucleotide sequences to synthesize polypeptides composed of specific amino acid sequences. Some proteins are single polypeptides, whereas other proteins are composed of several polypeptides bound together in a quaternary structure (see Figure 2.24). Ribosomes can be thought of as “polypeptide factories,” so consider the following analogy between translation and a hypothetical automobile factory. Trucks deliver auto parts to the factory at the correct times and in the correct order to manufacture one of a large variety of automobile models, depending on instructions from corporate headquarters delivered by special courier. Similarly, molecules of tRNA (the trucks) deliver preformed amino acids (the parts) to a ribosome (the factory), which can manufacture an infinite variety of polypeptides (the car models) by assembling amino acids in the correct order according to the instructions from DNA (corporate headquarters) delivered via mRNA (the special courier). How do ribosomes interpret the nucleotide sequence of mRNA to determine the correct order in which to assemble amino acids? To answer this question, we will consider the genetic code, examine in more detail the RNA molecules that participate in translation, and describe the specific steps of translation. The Genetic Code When geneticists in the early 20th century began to consider that DNA might be the genetic molecule, they were confronted with a problem: How can four kinds of DNA 11/23/16 3:01 AM       Gene Function nucleotide bases—adenine, thymine, guanine, and cytosine (abbreviated A, T, G, and C)—specify the 20 common amino acids found in proteins? If each nucleotide coded for a single amino acid, only four amino acids could be specified. Eventually scientists showed that genes are composed of sequences of three nucleotides where each triplet specifies a particular amino acid. For example, the DNA nucleotide sequence AAA specifies the amino acid phenylalanine, and GTA codes for histidine. There are 64 possible arrangements of the four nucleotides in triplets (43)—more than enough to code all 20 amino acids. These examples are DNA triplets, but ribosomes do not directly access genetic information on a DNA molecule. Instead, molecules of mRNA carry the code to the ribosomes; therefore, scientists define the genetic code (FIGURE 7.12) as triplets of mRNA nucleotides called codons (kō’donz), which code for specific amino acids. UUU is a codon for phenylalanine, and CAU is a codon for histidine. In most cases, 61 codons specify amino acids, and three codons—UAA, UAG, and UGA—instruct ribosomes to stop translating; however, under some conditions, UGA codes for a 21st amino acid, selenocysteine. Codon AUG also has a dual function, acting both as a start signal and as the codon for the amino acid methionine. In bacteria, mitochondria, and chloroplasts, AUG as a start codon codes for N-formylmethionine (fMet), a modified amino acid: O H H H C N C COOH H H H N C CH2 CH2 CH2 CH2 S S CH3 N-formylmethionine (fMet) COOH CH3 Methionine (Met) As you examine the genetic code, notice that it is redundant; that is, more than one codon is associated with every amino acid except methionine and tryptophan. With most redundant codons, the first two nucleotides determine the amino acid, and the third nucleotide is inconsequential. For example, the codons GUU, GUC, GUA, and GUG all specify the amino acid valine. Second nucleotide base UUU UUC U UUA First nucleotide base (5’ position) UUG UCU Phenylalanine (Phe) UCC CUC CUA AUU AUC A UAG STOP* UGG CCU CAU Histidine (His) CGU Glutamine (Gln) CGA Asparagine (Asn) AGU Lysine (Lys) AGA Aspartic acid (Asp) GGU Glutamic acid (Glu) GGA CAC Proline (Pro) CAA CCG CAG ACU AAU ACC AAC Threonine (Thr) ACA AUG START & Methionine (Met) ACG AAG GUU GCU GAU GUA GCC Valine (Val) GCA GUG UGC Cysteine (Cys) UCG AUA GUC G Isoleucine (Ile) UGU STOP CCA CUG UAC Serine (Ser) Tyrosine (Tyr) UAA CCC Leucine (Leu) UAU G Selenocysteine UGA STOP & (SeCys) UCA Leucine (Leu) CUU C A C AAA GAC Alanine (Ala) GAA GCG GAG CGC Tryptophan (Trp) AGG GGC GGG C A G U Arginine (Arg) C A G CGG AGC U Serine (Ser) Arginine (Arg) U C A G Third nucleotide base (3’ position) U U Glycine (Gly) C A G *Also codes for a 22nd amino acid, pyrrolysine, in some prokaryotes. M07_BAUM7206_05_SE_C07.indd 207 A A AA A A ▲ FIGURE 7.12 The genetic code. The table shows the set of mRN codons and the amino acids for which they code. UG not only is the start codon but also specifies methionine (Met) in eukaryotes and N-formylmethionine (fMet) in prokaryotes, mitochondria, and chloroplasts. Two codons (U and U G) are stop codons that do not typically specify amino acids. UG functions as a stop codon and also specifies selenocysteine. 11/23/16 3:01 AM r 5¿ 5¿ Template DNA strand Transcription UAA Ribosomebinding site (RBS) Stop codon 3¿ mRNA Untranslated mRNA Translation Polypeptide 1 A ▲ FIGURE 7.13 Prokaryotic mRNA molecules typically lack introns and exons. lmost all prokaryotic genes, and therefore E A ­ prokaryotic mRN molecules, lack intervening sequences. ach codes for a single polypeptide. Interestingly, the genetic code is nearly universal; that is, with few exceptions, ribosomes in archaeal, bacterial, plant, fungal, protozoan, and animal cells use the same genetic code. ▶ ANIMATIONS: Translation: Genetic Code Participants in Translation As we have discussed, transcription produces messenger RNA, transfer RNA, and ribosomal RNA—each of which is involved in translation. We now discuss each kind of RNA in turn. Transfer RNA A tRNA molecule is a sequence of about 75 ribonucleotides that curves back on itself to form three main hairpin loops held in place by hydrogen bonding between complementary nucleotides (FIGURE 7.14a). Although transfer RNA molecules can be modeled simplistically by a cloverleaf structure, their three-dimensional shape is more complex (FIGURE 7.14b). For simplicity, tRNA will be represented in subsequent figures by an icon shaped like the curved cylinder in Figure 7.14b. Amino acids have no direct affinity for or recognition of mRNA codons; a molecule of tRNA carries the correct amino acid to a ribosome during polypeptide synthesis. To this end, tRNA has an anticodon (an-tē-kō’don) triplet in its bottom loop that is complemetary to a codon and an acceptor stem for the corresponding amino acid at its 3’ end. Specific enzymes in the cytoplasm, called aminoacyl tRNA synthetases, each “charge” a specific tRNA molecule; that is, they attach the appropriate amino acid to the acceptor stem. Anticodons are complementary to mRNA codons, and each acceptor stem is designed to carry one particular amino acid, which varies with the tRNA. In other words, each transfer RNA carries a specific amino acid and recognizes mRNA codons only for that amino acid. A tRNA molecule is designated by a superscript abbreviation of its amino acid. For example, tRNAPhe carries phenylalanine, and tRNASer carries serine. The fact that 61 codons specify the 20 amino acids commonly used by cells does not mean that there must be 61 different types of tRNA, because many tRNA molecules recognize more than one codon. E. coli, for example, has only about 40 different tRNAs. The variability in codon recognition by tRNA Messenger RNA Messenger RNA molecules carry genetic information from chromosomes to ribosomes as codons (triplet sequences of RNA nucleotides) that encode the order of amino acid sequences in a polypeptide. In prokaryotes, a basic mRNA molecule contains codons that are recognized by ribosomes: an AUG start codon, sequential codons for other amino acids in the polypeptide, and one of the three stop codons. A single molecule of prokaryotic mRNA often contains start codons and instructions for more than one polypeptide arranged in series (FIGURE 7.13). Because both transcription and the subsequent events of translation occur in the cytosol of prokaryotes, prokaryotic ribosomes can begin translation before transcription is finished. Eukaryotic mRNA differs from prokaryotic mRNA in several ways: M07_BAUM7206_05_SE_C07.indd 208 Acceptor stem 3’ Amino acid is attached here 5’ Hydrogen bonds 5’ 3’ tRNA icon Hairpin loops A C C Anticodon Anticodon (a) (b) A A A ▲ FIGURE 7.14 Transfer RNA. (a) two-dimensional “cloverleaf” representation of tRN showing three hairpin loops held in place by intramolecular hydrogen bonding. (b) three-dimensional drawing of the same tRN . specific amino acid attaches to the 3′ acceptor stem; the anticodon is a nucleotide triplet that is complementary to the mRN codon for that amino acid. Which amino acid would be attached to the acceptor stem of this tRNA? A • As we have seen, eukaryotic cells extensively process premRNA to make mRNA (see Figure 7.11). • A processed molecule of eukaryotic mRNA contains instructions for only one polypeptide. • Eukaryotic mRNA is not translated until it is fully transcribed and processed and has left the nucleus. In other words, transcription and translation of a molecule of eukaryotic mRNA cannot occur simultaneously because transcription occurs in the nucleus, and translation by eukaryotic ribosomes occurs in the cytoplasm. OH A Start codon AUG Terminator ­ Gene 1 A Promoter ­ t h 3¿ Figure 7.14 UGG is the codon for tryptophan, which is complementary to the anticodon ACC.       C ap e 7   Microbial Genetics 11/23/16 3:01 AM       Gene Function Ribosomes and Ribosomal RNA Prokaryotic ribosomes, which are also called 70S ribosomes based on their sedimentation rate in an ultracentrifuge, are extremely complex associations of ribosomal RNAs and polypeptides. Each ribosome is composed of two subunits: 50S and 30S (FIGURE 7.15a). The 50S subunit is in turn composed of two rRNA molecules (23S and 5S) and about 34 different polypeptides, whereas the 30S subunit consists of one molecule of 16S rRNA and about 21 ribosomal polypeptides. The ribosomes of mitochondria and chloroplasts are composed of similar subunits and polypeptides. In contrast, both the cytosol and the rough endoplasmic reticulum (RER) of eukaryotic cells have 80S ribosomes composed of 60S and 40S subunits (FIGURE 7.15b). These subunits contain larger molecules of rRNA and more polypeptides than the corresponding prokaryotic subunits, though Large subunit mRNA Nucleotide bases 5¿ Small subunit 3¿ E P A site site site tRNAbinding sites Small subunit mRNA Prokaryotic ribosome (angled view) attached to mRNA Prokaryotic ribosome (schematic view) showing tRNA-binding sites A E A E P A A ▲ FIGURE 7.16 Assembled ribosome and its tRN -binding sites. The site accepts the tRN carrying the next amino acid to be added to the growing polypeptide, whereas the site holds the tRN carrying the polypeptide. mpty tRN molecules exit from the site. researchers do not agree on the exact number of polypeptides. The term eukaryotic ribosome is understood to mean only the 80S ribosomes of the cytosol and RER. Because the ribosomes of mitochondria and chloroplasts are 70S, they are “prokaryotic” ribosomes even though they are in eukaryotic cells. The structural differences between prokaryotic and eukaryotic ribosomes play a crucial role in the efficacy and safety of antimicrobial drugs. Because erythromycin, for example, binds only to the 23S rRNA found in prokaryotic ribosomes, it has no effect on eukaryotic 80S ribosomes and thus little deleterious effect on a patient. (Chapter 10 discusses antimicrobial drugs in more detail.) The smaller subunit of a ribosome is shaped to accommodate three codons at one time—that is, nine nucleotide bases of a molecule of mRNA. Each ribosome also has three tRNA-binding sites that are named for their function (FIGURE 7.16): • The A site accommodates a tRNA delivering an amino acid. • The P site holds a tRNA and the growing polypeptide. • Discharged tRNAs exit from the E site. 5S rRNA + 34 polypeptides 50S subunit 23S rRNA Large subunit A is due to “wobble” of the anticodon’s third nucleotide. Wobble, which is a change of angle from the normal axis of the molecule, allows the third nucleotide to hydrogen bond to a nucleotide other than its usual complement. For example, a guanine nucleotide in the third position normally bonds to cytosine, but it can wobble and also pair with uracil; therefore, whether a codon has cytosine or uracil in the third position makes no difference because the same tRNA recognizes either nucleotide in the third position. For example, the codons UUU and UUC specify the amino acid phenylalanine because the anticodon AAG recognizes both of them. Similarly, UCU and UCC code for serine, and UAU and UAC code for tyrosine. This redundancy in the genetic code helps protect cells against the effects of errors in replication and transcription. Prokaryotic 70S ribosome 16S rRNA + 21 polypeptides We will now examine translation, the process whereby ribosomes actually synthesize polypeptides using amino acids delivered by tRNAs in the sequence dictated by the order of codons in mRNA. 30S subunit (a) Events in Translation Molecular biologists divide translation into three stages: initi­ ation, elongation, and termination. All three stages require additional protein factors that assist the ribosomes. Initiation and elongation also require energy provided by molecules of the ribonucleotide GTP, which are free in the cytosol (i.e., they are not part of an RNA molecule). Here, we consider bacterial translation. ▶ ANIMATIONS: Translation:The Process 5S rRNA + 5.8S rRNA + + 49 polypeptides 60S subunit 28S rRNA Eukaryotic 80S ribosome 18S rRNA + >33 polypeptides (b) 40S subunit A A ▲ FIGURE 7.15 Ribosomal structures. (a) The 70S prokaryotic ribosome, which is composed of polypeptides and three rRN molecules arranged in 50S and 30S subunits. (b) The 80S eukaryotic ribosome, which is composed of molecules of rRN and polypeptides, arranged in 60S and 40S subunits. M07_BAUM7206_05_SE_C07.indd 209 Initiation During initiation, the two ribosomal subunits, mRNA, several protein factors, and tRNAfMet form an initiation complex. Initiation in prokaryotes may occur while the cell is still transcribing mRNA from DNA. The events of initiation in a bacterium are as follows (FIGURE 7.17): 1 The smaller ribosomal subunit attaches to mRNA at a ribosome-binding site (also known as a Shine­Dalgarno 11/23/16 3:01 AM 1 r t h       C ap e 7   Microbial Genetics fMet 3 2 Initiator tRNA fMet Large ribosomal subunit fMet fMet U mRNA 5¿ A C GTP tRNA GDP + P Anticodon Start codon A U G U U U A C G P E U A C A A U G U U U A C G 3¿ P U A C A U G U U U A C G A P Small ribosomal subunit A Initiation complex o e utor id A P V T A A A A ▲ FIGURE 7.17 The initiation of translation in prokaryotes. 1 The smaller ribosomal subunit attaches to mRN at a ribosome-binding site near a start codon ( UG). 2 The anticodon of tRNafMet aligns with the start codon on the mRN ; energy from GT is used to bind the tRN in place. 3 The larger ribosomal subunit attaches to form an initiation complex—a complete ribosome attached to mRN . 2 3 sequence after its discoverers) so as to position a start codon (AUG) at the ribosomal subunit’s P site. tRNAfMet (whose anticodon, UAC, is complementary to the start codon, AUG) attaches at the ribosome’s P site. The larger ribosomal subunit then attaches to form a complete initiation complex. Elongation Elongation of a polypeptide is a cyclical process that involves the sequential addition of amino acids to a polypeptide chain growing at the P site. FIGURE 7.18 illustrates several cycles of the process. The steps of each cycle occur as follows: 4 5 6 7 The tRNA whose anticodon is complementary to the next codon—in this example, AAA complementary to the codon UUU—delivers its amino acid, in this case, phenylalanine (Phe), to the A site. Proteins called elongation factors escort the tRNA along with a molecule of GTP (not shown). Energy from GTP is used to stabilize each tRNA as it binds at the A site. An enzymatic RNA molecule—a ribozyme—in the larger ribosomal subunit forms a peptide bond between the terminal amino acid of the growing polypeptide chain (in this case, N-formylmethionine) and the newly introduced amino acid. The polypeptide is now attached to the tRNA occupying the A site. Using energy supplied by more GTP, the ribosome moves one codon down the mRNA. This transfers each tRNA to the adjacent binding site; that is, the first tRNA moves from the P site to the E site, and the second tRNA (with the attached polypeptide) moves to the vacated P site. The ribosome releases the “empty” tRNA from the E site. In the cytosol, the appropriate enzyme recharges the empty tRNA with another molecule of the type of amino acid carried by that tRNA. M07_BAUM7206_05_SE_C07.indd 210 8 The cycle repeats, each time adding another amino acid, at a rate of about 15 amino acids per second (in this example, threonine, then alanine, and then glutamine). As elongation proceeds, ribosomal movement exposes the start codon, allowing another ribosome to attach behind the first one. In this way, one ribosome after another attaches at the start codon and begins to translate identical polypeptide molecules from the same message. Such a group of ribosomes, called a polyribosome, resembles beads on a string (FIGURE 7.19). Termination Termination does not involve tRNA; instead, proteins called release factors halt elongation. It appears that release factors somehow recognize stop codons and modify the larger ribosomal subunit in such a way as to activate another of its ribozymes that severs the polypeptide from the final tRNA (resident at the P site). The ribosome then dissociates into its subunits. Termination of translation should not be confused with termination of transcription, covered in a previous section. The polypeptides released from a ribosome at termination may function alone as proteins, or they may function with other polypeptides in quaternary protein structures. Translational Differences in Eukaryotes Eukaryotic translation is similar to that of bacteria, with some notable differences, including the following: • Initiation of translation in eukaryotes occurs when the small ribosomal subunit binds to the 5’ guanine cap rather than a specific nucleotide sequence. • The first amino acid in eukaryotic polypeptides is methionine rather than formylmethionine. • Ribosomes attached to membranes of endoplasmic reticulum (ER), forming rough ER (RER), can synthesize polypeptides into the cavity of the RER. 11/23/16 3:01 AM       Gene Function Peptide bond formed by ribozyme 4 fMet E E A U A C A A A A U G U U U A C G G C A C A A G C U G G G 5¿ P A 3¿ o P 3¿ Phe utor id e fMet U A C A A A A U G U U U A C G G U A 5¿ V T 5 Phe ▶ FIGURE 7.18 The elongation stage of translation. Transfer A 6 fMet Phe P A A A RN s sequentially deliver amino acids as directed by the codons of the mRN . Ribosomal RN in the large ribosomal subunit catalyzes a peptide bond between the amino acid at the site and the growing polypeptide at the site. The steps in the process are described in the text. How would eukaryotic translation differ? Movement of ribosome one codon toward 3¿ end Figure 7.18 The initial amino acid in eukaryotic polypeptide is methionine. Archaeal translation is more similar to that of eukaryotes than to that of bacteria; however, archaea lack ER. The processes we have examined thus far—how a cell replicates DNA, transcribes RNA, and translates RNA into polypeptides—are summarized in TABLE 7.2 on p. 212. Next, we examine the way cells control the process of transcription. U A C A A A A U G U U U A C G G C A C A A G C U G G G 5¿ 7 Regulation of Genetic Expression E fMet U A C LE A R N I N G O U TC O M E S 7.16 Explain the operon model of transcriptional control in P A Thr Phe E U G C A A A A U G U U U A C G G C A C A A G C U G G G 5¿ 3¿ P prokaryotes. A 3¿ 7.17 Contrast the regulation of an inducible operon with that of a repressible operon, and give an example of each. 7.18 Describe the use of microRNA, small interfering RNA, 8 and riboswitches in genetic control. Many of a bacterium’s genes are expressed at all times; that is, they are constantly transcribed and translated and play a persistent role in the phenotype. Such genes code for RNAs and polypeptides that are needed continuously by the cell—for example, integral proteins of the cytoplasmic membrane, structural proteins of ribosomes, and enzymes of glycolysis. Other genes are regulated so that the polypeptides they encode are synthesized only in response to a change in the environment; after all, protein synthesis requires a large amount of energy, which can be conserved if a cell forgoes production of unneeded polypeptides. For example, Pseudomonas aeruginosa (ā-roo-ji-nō’să), a potential pathogen of cystic fibrosis patients, can synthesize harmful proteins. Early in an infection, Pseudomonas does not produce the proteins: The body would respond defensively and eliminate the bacterium. Instead, the pathogen uses quorum sensing—a process whereby cells secrete quorum-sensing molecules into their environment and other cells detect these signals so as to measure their density. The result is that Pseudomonas cells synthesize harmful proteins only after there are numerous bacterial cells, overwhelming M07_BAUM7206_05_SE_C07.indd 211 fMet Phe E Thr A A A U G C A U G U U U A C G G C A C A A G C U G G G 5¿ P A 3¿ Two more cycles fMet Phe Thr Growing polypeptide Gln C G U 5¿ Ala Ala C G A E G U U A U G U U U A C G G C A C A A G C U G G G P A 3¿ 11/23/16 3:01 AM mRNA r t h       C ap e 7   Microbial Genetics Ribosomes Polypeptides mRNA Ribosomes Polypeptides Direction of transcription (a) TEM (b) 50 nm A A A ▲ FIGURE 7.19 Polyribosome in prokaryotes—one mRNA and many ribosomes and polypeptides. (a) s each ribosome moves down the mRN , the start codon ( UG) becomes available to another A A ribosome. In this manner, numerous identical polypeptides are translated simultaneously from a single mRN molecule. (b) polyribosome in a prokaryotic cell. (such as channel proteins), and an adjacent regulatory element called an operator (FIGURE 7.20), which controls movement of RNA polymerase. Operons can be polycistronic; that is, they may code for more than one polypeptide. Inducible operons are usually inactive and must be induced, or “turned on,” by inducers. Such operons often code for proteins that transport and catabolize nutrients that are only sometimes available to a cell. Repressible operons operate in reverse fashion—they are typically continually active until repressors deactivate them. Repressible operons often produce proteins that are involved in anabolic (synthesis) pathways. To clarify these concepts, we next examine an inducible operon and a repressible operon found in E. coli. ­ ­ the body’s defenses. Emerging Disease Case Study: Vibrio vulnificus Infection describes another case of quorum sensing. Cells regulate polypeptide synthesis in many ways. They may initiate or stop transcription of mRNA or may stop translation directly. Much of our knowledge of regulation has come from the study of microorganisms such as E. coli. We will examine two types of regulation of transcription in this bacterium—induction and repression. But before we examine these processes, we must first consider operons—special arrangements of prokaryotic genes that play roles in gene regulation. The Nature of Prokaryotic Operons As originally described, a prokaryotic operon consists of a promoter, a series of genes that code for enzymes and structures TABLE 7.2 Comparison of Genetic Processe Enzyme DN polymerases RN polymerases Ribosomes Template Both parental strands of DN One strand of DN mRN Start Site Origin of replication Fidelity Mechanism olymerase proofreading, followed by mismatch repair systems None Specificity of enzymes that charge tRN s Termination Termination sequences Terminator U rokaryotes: cytosol; can be simultaneous with replication rokaryotes: cytosol; can be simultaneous with transcription Ribonucleotides Direction of Polymerization 5′ to 3′ 5′ to 3′ A A A A AA P E E ukaryotes: cytosol or on R R olypeptides; some function alone as proteins, and others work together as a single protein T for charging tRN s A Deoxyribonucleotides , U G, or UG stop codons P Energy Source for Process N terminus (end with amino group) to C terminus (end with carboxyl group; see FIGUR 7.23) E A RN , single stranded A Two daughter DN strands, each paired with one original strand (semiconservative replication) UG start codon P A P E ukaryotes: Nucleolus in nucleus Product A rokaryotes: cytosol ukaryotes: nucleus P A P P E Location romoter A Translation A Transcription A Replication M07_BAUM7206_05_SE_C07.indd 212 11/23/16 3:01 AM E A E S C S STUDY EA M RGING DIS E E       Gene Function Vibrio vulnificus Infection A Greg enjoyed Florida’s beaches; swimming in the warm water was his favorite pastime. Of course, the salt water did sting his leg where he had cut himself on some coral, but it didn’t sting enough to stop Greg from enjoying the beach. He spent the afternoon jogging in the pure sand, throwing a disc, watching people, drinking a few beers, and of course spending more time in the water. That evening, he felt chilled, a condition he associated with having gotten too much sun during the day, but by midnight he thought he must have caught a rare summertime flu. He was definitely feverish, extremely weak, and tired. His leg felt strangely tight, as though the underlying muscles were trying to burst through his skin. The next morning, he felt better, except for his leg. It was swollen, dark red, tremendously painful, and covered with fluid-filled blisters. The ugly sight motivated him to head straight for the hospital, a decision that likely saved his life. Greg was the victim of an emerging pathogen, Vibrio vulnificus—a slightly curved, Gram-negative bacterium with DN similar to that of V. cholerae (cholera bacterium). V. vulnificus lives in salty, warm water around the globe. Unlike the cholera bacterium, V. vulnificus is able to infect a person by penetrating directly into a deep wound, a cut, or even a tiny scratch. In a person, the multiplying bacterium secretes quorum-sensing molecules. When the cells sense that there is a certain population size (a quorum), they “turn on” some of their genes, allowing them to thrive in the body and cause disease. Greg’s doctor cut away the dead tissue and prescribed doxycycline and cephalosporin for two weeks. Greg survived and kept his leg. Half of the victims of V. vulnificus are not so lucky; they lose a limb or die. Who knew that a beach could be so dangerous? (For more about quorum sensing, see pp. 167–169.) 1. Why is it necessary for Vibrio vulnificus to turn on different genes when the microbe invades a human? 2. What does the term “turn on” mean in relation to transcription and translation? 3. Why do you think the related microbe V. cholerae is unable to infect through the skin? The lactose (lac) operon of E. coli is an inducible operon and the first operon whose structure and action were elucidated. The lac operon includes a promoter, an operator, and three genes that encode proteins involved in the transport and catabolism of the disaccharide sugar, lactose. E. coli can catabolize lactose to release glucose and galactose, which can then be further catabolized as energy sources. Two events induce, or activate, the lac operon: (1) positive regulation by a protein called CAP and (2) deactivation of a repressor molecule. Let’s examine these two events. ▶ ANIMATIONS: Operons: Overview lactose when glucose is available. When glucose levels are low, a small molecule called cyclic adenosine monophosphate (cAMP) accumulates in the cell. Buildup of cAMP acts as a glucose starvation signal. Cyclic AMP binds to a regulatory protein called catabolite activator protein (CAP). CAP then attaches to CAP-binding sites on DNA near operons controlling sugar catabolism, including the lac operon. With CAP attached to the nearby CAP-binding site, RNA polymerase is more attracted to the lac operon’s promoter (FIGURE 7.21). Thus, CAP positively enhances lac transcription in the absence of glucose. Positive Regulation by CAP E. coli uses glucose more efficiently than other sugar molecules, so the bacterium avoids using Repression and Induction The lac operon is controlled by a regulatory gene located outside the operon (FIGURE 7.22a) . The Lactose Operon, an Inducible Operon Operon Promoter and regulatory gene Promoter Operator Genes 1 3¿ 2 3 4 5¿ DNA strand A A ▲ FIGURE 7.20 An operon. n operon consists of genes, their promoter, and an operator. The genes code for enzymes and structures such as channel and carrier proteins. separate regulatory gene that is not part of the operon codes for a protein that controls the operon. M07_BAUM7206_05_SE_C07.indd 213 11/23/16 3:01 AM r Ribosomes then translate the newly synthesized mRNA to produce enzymes that catabolize lactose (not shown). Once the lactose supply has been depleted, there is no more allolactose inducer, so the repressor once again becomes active, binds to the operator, and suppresses the lac operon. In this manner, E. coli cells synthesize enzymes for the catabolism of lactose only when two conditions are met: glucose is absent (CAP is bound), and lactose is present. Inducible operons are often involved in controlling catabolic pathways whose polypeptides are not needed unless a particular nutrient is available. A different situation occurs with anabolic pathways, such as those that synthesize amino acids. ▶ ANIMATIONS: Operons: Induction RNA polymerase cAMP bound to CAP Transcription can proceed CAPbinding site Lactose catabolism genes Promoter P A AP A A AP AP P A ▲ FIGURE 7.21 CAP-cAMP enhances lac transcription. Cyclic adenosine monophosphate (c M ), accumulating in E. coli when glucose is absent, binds to catabolic activator protein (C ). C -c M binds to a C -binding site of DN , allowing RN polymerase to effectively bind to the lac promoter and begin transcription. The Tryptophan Operon, a Repressible Operon Usually, the regulatory gene is constantly transcribed and translated to produce a repressor 1 protein that attaches to DNA at the lac operator. The presence of this repressor prevents RNA polymerase from binding to the promoter. Thus, the lac operon is usually inactive even if CAP is present 2 . When lactose is available, E. coli converts it to allolactose. Allolactose induces the lac operon by inactivating the repressor so that it no longer attaches to DNA (FIGURE 7.22b 3 ). With the operator unblocked, RNA polymerase binds to the promoter and transcribes the three lac genes—the operon has been induced 4 . ◀ FIGURE 7.22 The lac operon, an example of an inducible operon. The lac operon cAMP bound to CAP Promoter and regulatory gene E. coli can synthesize all the amino acids it needs for polypeptide synthesis; however, it can save energy by using amino acids available in its environment. In such cases, E. coli represses the genes for a given amino acid’s synthetic pathway. The tryptophan (trp) operon is a repressible polycistronic operon consisting of a promoter, an operator, and five genes that code for the enzymes involved in the synthesis of tryptophan. Just as with the lac operon, a regulatory gene codes for a repressor molecule that is constantly synthesized. In contrast to an inducible operon’s repressor, this repressor is normally Promoter Operator (blocked) 1 3¿ 2 3 5¿ DNA strand Lactose catabolism genes Continual transcription 2 RNA polymerase cannot bind mRNA Continual translation 1 Repressor repressor, a protein encoded by a regulatory gene, is constantly synthesized. (a) When lactose is absent from the cell’s environment, the repressor binds to the operator, blocking the movement of RN polymerase and halting transcription. (b) When lactose is present in the cell’s environment, its derivative, allolactose, acts as an inducer by inactivating the repressor so that the repressor cannot bind to the operator, allowing transcription to proceed. A t h       C ap e 7   Microbial Genetics (a) lac operon repressed RNA polymerase 1 2 Transcription proceeds 4 3¿ Repressor mRNA Repressor cannot bind to operator 3 5¿ DNA strand mRNA for lactose catabolism 5¿ Repressor 3 Inactivated repressor Inducer (allolactose from lactose) (b) lac operon induced M07_BAUM7206_05_SE_C07.indd 214 11/23/16 3:01 AM       Gene Function trp operon with five genes Regulatory gene Promoter Operator 3¿ 1 2 3 4 5 5¿ DNA strand Transcription 3¿ mRNA 5¿ 5¿ Inactive repressor mRNA coding multiple polypeptides 3¿ Enzymes of tryptophan biosynthetic pathway (a) trp operon active Trp Tryptophan Movement of RNA polymerase ceases 1 3¿ Inactive repressor Trp 2 3 4 5 5¿ Operator blocked Trp Trp inactive. Thus, the trp operon is usually active: the cell transcribes mRNA, translates the enzymes for tryptophan synthesis, and synthesizes tryptophan (FIGURE 7.23a). However, when tryptophan is available, it binds to the repressor, thus activating it. The activated repressor then binds to the operator, halting the movement of RNA polymerase and transcription (FIGURE 7.23b). In other words, tryptophan stops its own synthesis. TABLE 7.3 summarizes how the basic characteristics of inducible and repressible operons relate to the regulation of transcription. ▶ ANIMATIONS: Operons: Repression RNA Molecules Can Control Translation Besides using inducible and repressible operons, cells can also use molecules of RNA to regulate translation of polypeptides. Regulatory RNA molecules include microRNA, small interfering RNA, and riboswitches. Eukaryotic cells transcribe single-stranded RNA molecules about 22 nucleotides long called microRNAs (miRNAs), which are not translated by ribosomes. Instead, miRNA joins with regulatory proteins to form a miRNA­induced silencing complex (miRISC). miRISC binds to messenger RNA that is complementary to the microRNA within the miRISC. Once bound, miRISC M07_BAUM7206_05_SE_C07.indd 215 TABLE 7.3 Basic Roles of Operons in Regulating Transcription Type of Regulation Type of Metabolic Pathway Regulated Regulating Condition Inducible Operons Catabolic pathways resence of substrate of pathway nabolic pathways resence of product of pathway Repressible Operons P (b) trp operon repressed P Activated repressor operon. (a) When tryptophan is absent from the cell’s environment, the repressor is inactive; so, the structural genes are transcribed and translated, and the five enzymes needed in the synthesis of tryptophan are produced. (b) When tryptophan is present in the cell’s environment, it acts as a corepressor, activating the repressor and inhibiting its own synthesis. A Tryptophan (corepressor) ▲ FIGURE 7.23 The trp operon, an example of a repressible Trp ­ Trp performs one of two functions: In some cases, miRISC cuts the messenger RNA molecule, rendering it useless. In other cases, miRISC remains bound to messenger RNA, effectively hiding the mRNA molecule from ribosomes. In both cases, no polypeptide is formed from the messenger RNA molecule; thus, miRNAs (associated with RISC) regulate gene expression by blocking translation. Eukaryotic cells use miRISC to regulate a number of processes, including embryogenesis, cell division, apoptosis (programmed cell death), blood cell formation, and development of cancer. Another method of regulation involving RNA uses small interfering RNA (siRNA). siRNAs are about the same length 11/23/16 3:01 AM       Chapter 7   Microbial Genetics as miRNAs but differ from miRNAs in that siRNAs are double stranded. Further, strands of siRNAs may be complementary to mRNA, tRNA, or DNA. siRNAs unwind and join RISC proteins to form siRISC, which appears to always bind to and cut the target nucleic acid. This result is called gene silencing. Scientists can create siRNA molecules so as to artificially regulate gene expression in laboratory studies. A riboswitch is another RNA molecule that helps regulate translation. Riboswitches change shape in response to environmental conditions such as changes in temperature or shifts in the concentration of specific nutrients, including vitamins, nucleotide bases, or amino acids. Some messenger RNA molecules themselves act as riboswitches. When conditions warrant, riboswitch mRNA folds to either favor or block translation, depending on the need by the cell for the polypeptide it encodes. For example, mRNA for the virulence regulator of the plague bacterium Yersinia pestis (yer-sin’ē-ă pes’tis) folds in such a way as to prevent translation when the temperature is below human body temperature (37°C). When the bacterium enters a human, the mRNA refolds into a shape that allows translation, and virulence regulator is synthesized; plague ensues. TELL ME WHY In bacteria, polypeptide translation can begin even before mRNA transcription is complete. Why can’t this happen in eukaryotes? Point mutations include substitutions and frameshift mutations, which result from either insertions or deletions. The following analogy illustrates some types of point mutations. Suppose that the DNA code was represented by the letters THECATATEELK. Grouping the letters into triplets (like codons) yields THE CAT ATE ELK. The substitution of a single letter at one point could either change the meaning of the sentence, as in THE RAT ATE ELK, or result in a meaningless phrase, such as THE CAT RTE ELK. Insertions or deletions cause frameshift mutations because nucleotide triplets following the mutation are displaced, creating new sequences of codons that result in vastly altered polypeptide sequences. Such point mutations produce more serious changes, such as TRH ECA TAT EEL K (insertion) or TEC ATA TEE LK (deletion). Frameshift mutations can affect a cell much more seriously than mere substitutions because a frameshift affects all codons subsequent to the mutation. Mutations can also involve inversion (THE ACT ATE KLE), duplication (THE CAT CAT ATE ELK ELK), or transposition (THE ELK ATE CAT). Such mutations and even larger deletions and insertions are gross mutations. ▶ ANIMATIONS: Mutations: Types Effects of Point Mutations LE A R N I N G O UTC O M E 7.21 List three effects of point mutations. Mutations of Genes LE A R N I N G O U TC O M E 7.19 Define mutation. The life of a cell is dependent on both the integrity and accurate control of the cell’s genes; however, the nucleotide sequences of genes are not always accurately maintained. A mutation is a heritable change in the nucleotide base sequence of a genome. Mutations of genes are almost always deleterious, though a few make no difference to an organism. Even more rarely, a mutation leads to a novel property that might improve the ability of the organism and its descendants to survive and reproduce. This is evolution in action. Mutations in unicellular organisms are passed on to the organism’s progeny, but mutations in multicellular organisms typically are passed to offspring only if a mutation occurs in gametes (sex cells) or gamete-producing cells. Types of Mutations LE A R N I N G O U TC O M E 7.20 Define point mutation, and describe three types. Mutations range from large changes in an organism’s genome, such as the loss or gain of an entire chromosome, to point mutations, in which just a single nucleotide base pair is affected. M07_BAUM7206_05_SE_C07.indd 216 Some base-pair substitutions produce silent mutations because redundancy in the genetic code prevents the substitution from altering the amino acid sequence (compare FIGURE 7.24a and b). For example, when the DNA triplet AAA is changed to AAG, the mRNA codon will be changed from UUU to UUC; however, because both codons specify phenylalanine, there is no change in the phenotype—the mutation is silent because it affects the genotype only. Of greater concern are substitutions that change a codon for one amino acid into a codon for a different amino acid. A change that specifies a different amino acid is called a missense mutation (FIGURE 7.24c); what gets transcribed and translated makes sense, but not the right sense. The effect of missense mutations depends on where in the protein the changed amino acid occurs. When the different amino acid is in a critical region of a protein, the protein becomes nonfunctional; however, when the different amino acid is in a less important region, the mutation may have no adverse effect. A third type of mutation occurs when a base-pair substitution changes an amino acid codon into a stop codon. This is called a nonsense mutation (FIGURE 7.24d). Nearly all nonsense mutations result in nonfunctional proteins. Frameshift mutations (i.e., insertions or deletions) typically result in drastic missense and nonsense mutations (FIGURE 7.24e and f) except when the insertion or deletion is very close to the end of a gene. TABLE 7.4 on p. 218 summarizes the types and effects of point mutations. 06/03/2018 09:03       Mutations of Genes Normal DNA Template DNA strand A A A A T A C G T G C A Normal mRNA Normal polypeptide U U U U A U G C A C G U mRNA Phe Tyr Ala ◀ FIGURE 7.24 The effects of the various types of point mutations. (a) Normal gene. Base-pair substitutions can result in (b) silent mutations, (c) missense mutations, or (d) nonsense mutations. (e) Frameshift insertions and (f) frameshift deletions usually result in severe missense or nonsense mutations because all codons downstream from the mutation are altered. Arg Mutagens (a) Normal LE A R N I N G O UTC O M E S Mutated template DNA strand A A G A T A C G T G C A U U C U A U G C A C G U Mutated mRNA Phe Tyr Ala Arg (b) Silent mutation No change in amino acid sequence of polypeptide Mutated template DNA strand A A A A T A C C T G C A U U U U A U G G A C G U Mutated mRNA Phe Tyr Gly Arg Slightly different amino acid sequence (c) Missense mutation Mutated template DNA strand A A A A T T C G T G C A U U U U A A G C A C G U Mutated mRNA STOP Phe CODON (d) Nonsense mutation Polypeptide synthesis ceases Frameshift mutations: Insertion A A A T A T A C G T G C A Mutated template DNA strand U U U A U A U G C A C G U Mutated mRNA Phe Ile Cys Thr (e) Frameshift insertion Major difference in amino acid sequence T A A A A A C G T G C A Mutated template DNA strand U U U U U G C A C G U Mutated mRNA Phe (f) Frameshift deletion M07_BAUM7206_05_SE_C07.indd 217 Leu His Val Major difference in amino acid sequence 7.22 Discuss how different types of radiation cause mutations in a genome. 7.23 Describe three kinds of chemical mutagens and their effects. Mutations occur naturally during the life of an organism. Such spontaneous mutations result from errors in replication and repair as well as from recombination in which relatively long stretches of DNA move among chromosomes, plasmids, and viruses, introducing frameshift mutations. For example, we have seen that mismatched base pairing during DNA replication results in one error in every 10 billion (1010) base pairs. Since an average gene has 103 base pairs, about one of every 107 (10 million) genes contains an error. Further, though cells have repair mechanisms to reduce the effect of mutations, the repair process itself can introduce additional errors. Physical or chemical agents called mutagens (myū’tă-jenz), which include radiation and several types of DNA-altering chemicals, induce mutations. ▶ ANIMATIONS: Mutagens Radiation In the 1920s, Hermann Muller (1890–1967) discovered that X rays increased phenotypic variability in fruit flies by causing mutations. Gamma rays also damage DNA. X rays and gamma rays are ionizing radiation; that is, they energize electrons in atoms, causing some of the electrons to escape from their atoms (see Chapter 9). These free electrons strike other atoms, producing ions that can react with the structure of DNA, thereby creating mutations. More seriously, electrons and ions can break the covalent bonds between the sugars and phosphates of a DNA backbone, causing physical breaks in chromosomes and complete loss of cellular control. Nonionizing radiation in the form of ultraviolet (UV) light is also mutagenic because it causes adjacent pyrimidine bases to covalently bond to one another, forming pyrimidine dimers (FIGURE 7.25). The presence of dimers prevents hydrogen bonding with nucleotides in the complementary strand, distorts the sugar-phosphate backbone, and prevents proper replication and transcription. Cells have several methods of repairing dimers (discussed shortly). Chemical Mutagens Here we consider three of the many basic types of mutagenic chemicals. Nucleotide Analogs Compounds that are structurally similar to normal nucleotides are called nucleotide analogs 11/23/16 3:01 AM r t h       TABLE 7.4 C ap e 7   Microbial Genetics The Types of Point Mutations and Their Effects Description Effects Substitution Mismatching of nucleotides or replacement of one base pair by another Silent mutation if change results in redundant codon, as amino acid sequence in polypeptide is not changed. Missense mutation if change results in codon for a different amino acid; effect depends on location of different amino acid in polypeptide. Nonsense mutation if codon for an amino acid is changed to a stop codon. Frameshift (insertion) ddition of one or a few nucleotide pairs creates new sequence of codons Missense and nonsense mutations Frameshift (deletion) Removal of one or a few nucleotide pairs creates new sequence of codons Missense and nonsense mutations A Type of Point Mutation Ultraviolet light Normal nucleotide base CH3 Thymine dimer G C T G T=T G Br O N H Analog N H H N H N O G T A C G A C A A C C A T O O Thymine 5¿-bromouracil (bU) (a) ▲ FIGURE 7.25 A pyrimidine (in this case, thymine) dimer. Ultraviolet light causes adjacent pyrimidine bases to covalently bond to each other, preventing hydrogen bonding with bases in the complementary strand. The resulting distortion of the sugar-phosphate backbone prevents proper replication and transcription. Normal DNA T A Incorporation of 5¿-bromouracil in place of thymine (FIGURE 7.26a) . When nucleotide analogs are available to replicating cells, they may be incorporated into DNA in place of normal nucleotides, where their structural differences either inhibit DNA replication or result in mismatched base pairing. For example, when thymine is replaced by the analog 5’-bromouracil, a wrong complement can form during subsequent replications—5’-bromouracil can pair with guanine rather than with adenine, resulting in a point mutation (FIGURE 7.26b). (Figure 10.7 illustrates other nucleotide analogs.) Nucleotide (or nucleoside) analogs make potent antiviral and anticancer drugs. Because viruses and cancer cells typically replicate faster than normal cells, they more quickly incorporate nucleotide or nucleoside analogs into their DNA and RNA. The presence of such analogs is deleterious; therefore, nucleotide analogs kill viruses and cancer cells more quickly than they do normal cells. (Chapter 10 considers the use of nucleotide analogs as antimicrobial agents.) M07_BAUM7206_05_SE_C07.indd 218 A First replication cycle: bU pairs with guanine T bU A G Second replication cycle: guanine pairs with cytosine T T C bU A A G G Normal daughter DNA molecules Base-pair substitution (b) ▲ FIGURE 7.26 The structure and effects of a nucleotide analog. (a) The structures of thymine and one of its nucleotide analogs, A A A A 5′-bromouracil (bU). (b) When 5′-bromouracil is incorporated into DN , the result is a point mutation: a single replication cycle results in one normal DN molecule and one mutated molecule in which bU pairs with guanine. fter a second replication cycle, a complete base-pair substitution has occurred in one of the four DN molecules. CG has been substituted for T . A Nucleotide-Altering Chemicals Some chemical mutagens alter the structure of nucleotides. For example, Aspergillus (as-per-jil’ŭs) molds growing on grains and nuts produce nucleotide-altering chemicals called aflatoxins. When the liver catabolizes aflatoxins, a by-product can convert guanine nucleotides into thymine nucleotides so that a GC base pair is converted to a TA base pair, resulting in missense mutations and possibly cancer. The Food and Drug Administration prohibits excessive amounts of aflatoxins in human and animal food. Another example of a bU 11/23/16 3:01 AM       Mutations of Genes nucleotide-altering chemical is nitrous acid (HNO2), which converts adenine into a guanine analog by removing the amine group from adenine. When a cell replicates DNA containing this guanine analog, an AT base pair is changed to a GC base pair in one daughter molecule—a base-pair-substitution mutation. Frameshift Mutagens Still other mutagenic chemical agents insert or delete nucleotide base pairs, resulting in frameshift mutations. Examples of frameshift mutagens are benzopyrene, which is found in smoke; ethidium bromide, which is used to stain DNA; and acridine, one of a class of dyes commonly used as mutagens in genetic research. These chemicals are exactly the right size to slip between adjoining nucleotides in DNA, producing a bulge in the molecule (FIGURE 7.27). When DNA polymerase copies the misshapen strands, one or more base pairs may be inserted or deleted in the daughter strands. Frequency of Mutation seen, about 1 of every 10 million (107) genes contains an error. Mutagens typically increase the mutation rate by a factor of 10 to 1000 times; that is, mutagens induce an error in one of every 106 to 104 genes. Many mutations are deleterious because they code for nonfunctional proteins or stop transcription entirely. Cells without functional proteins cannot metabolize; therefore, deleterious mutations are removed from the population when the cells die. Rarely, however, a cell acquires a beneficial mutation that allows it to survive, reproduce, and pass the mutation to its descendants. Evolution involves a change in gene frequency in a population. For example, the tuberculosis bacterium has acquired a mutation that confers resistance to the antimicrobial drug rifampin. In patients taking rifampin, mutated bacterial cells survive and reproduce, increasing the frequency of the mutated gene in the population because cells without the mutation die. As long as rifampin is present, cells with such a mutation have an advantage over cells without the mutation, and the population evolves resistance to rifampin. LE A R N I N G O U TC O M E 7.24 Discuss the relative frequency of deleterious and useful mutations. DNA Repair LE A R N I N G O UTC O M E Mutations are rare events. If they occurred often, organisms could not live or effectively reproduce themselves. As we have G T A C C A A T T A C G C C A T G G T T A A T G C C Normal DNA Acridine 7.25 Describe, base-excision repair, light repair of pyrimidine dimers, nucleotide-excision repair, mismatch repair, and error-prone repair, such as the SOS response. We have seen that mutations rarely convey an advantage; most mutations are deleterious. To respond to the dangers mutations pose, cells have numerous methods for repairing damaged DNA. Here we consider three categories of repair: direct repair, single-strand repair, and repair of double-strand damage. The latter is highly error prone. ▶ ANIMATIONS: Mutations: Repair Direct Repair C A A T T A C A T G G T T A A T ­ G T A C Direct repair systems are highly accurate because they correct damage to nucleotides in cases where an error exists in one strand of DNA, while the correct base can be ascertained from the complementary strand. An example of direct repair is base-excision repair in which one of several enzymes recognizes a specific incorrect base and removes and replaces only that base. For example, an enzyme replaces hypoxanthine (H), which is an altered adenine base, with a normal adenine base (FIGURE 7.28a). Another form of direct repair is the repair of pyrimidine dimers by so-called light repair. Pyrimidine dimers form when adjoining pyrimidines (thymine or cytosine nucleotides) absorb ultraviolet light (UV) and covalently bond to one another, which prevents them from hydrogen bonding to their complementary purines. Many cells contain photolyase, an enzyme that is activated by visible light to break pyrimidine dimers, reversing the mutation and restoring the original DNA sequence (FIGURE 7.28b). Replication Deletion of CC Insertion of TAT C A T G T A T G T T A A T G T A A A T T A C G C or G T A A A T T A C G C G T A C A T A C A A T T A C A T T T A A T G C C C A T G T A T G T T A A T Daughter DNA A A ▲ FIGURE 7.27 The action of a frameshift mutagen. When DN polymerase III passes the bulge caused by the insertion of acridine between the two DN strands, it incorrectly synthesizes a daughter strand that contains either a deletion or an insertion mutation. M07_BAUM7206_05_SE_C07.indd 219 Single-Strand Repair When damage cannot be repaired directly, then a cell cuts away a section of a single strand containing the damage and 11/23/16 3:01 AM r t h       C ap e 7   Microbial Genetics Incorrect base (here hypoxanthine instead of adenine) ◀ FIGURE 7.28 DNA repair mechanisms. Base excision repair enzymes remove incorrect nucleotide G G C T T A H C G T G G C T T A C G T G G C T T A A C G T C C G A A T T G C A C C G A A T T G C A C C G A A T T G C A A Visible light Thymine dimer T T A G A C T T A C T G A A T C T G A A T G A C A A A (a) Base-excision repair (a type of direct repair) (a) Base-excision repair (type of direct repair), in which enzymes remove incorrect bases. Here, hypoxanthine (H) is corrected with adenine. (b) In another type of direct repair, called light repair, a light-activated enzyme, photolyase, breaks pyrimidine-to-pyrimidine bonds. (c) Nucleotide-excision repair is a type of single-strand repair in which a repair enzyme removes the entire damaged section from one strand of DN , and DN polymerase I and DN ligase repair the breach. (d) nother type of single-strand repair is mismatch repair, which involves excision of a segment of a single strand. The repair enzyme distinguishes the original strand from the new strand because the original strand is methylated. Photolyase, a light-activated repair enzyme (b) Light repair (a type of direct repair) Cut G A C T T A C Repair enzyme C T G A A T G G A C T T A C DNA polymerase and ligase repair the gap C T G A A T G G A C T T A C C T G A A T G (c) Nucleotide-excision repair (a type of single-strand repair) Mismatch repair enzyme removes incorrect segment New strand Original strand (methylated) Mutated DNA (incorrect nucleotide pair) T A C T DNA polymerase III correctly repairs the gap C T T A G C G T G C G T C C T A G C G T G G A T C G C A G G A T C G C A G G A T C G C A (d) Mismatch repair (a type of single-strand repair) uses the complementary strand as a template to repair the gap. An example is nucleotide-excision repair (sometimes called dark repair because it can correct a pyrimidine dimer without using light). In this repair system, an enzyme removes a short section of the strand (about a dozen nucleotides) containing the dimer and then DNA polymerase I and DNA ligase repair the breach using the complementary, undamaged strand as a template (FIGURE 7.28c). Another type of single-strand repair is mismatch repair, which is a repair mechanism that corrects errors produced during replication. Mismatch repair enzymes scan newly synthesized DNA looking for mismatched bases, which they remove and replace (FIGURE 7.28d). How does the mismatch repair system determine which of the two strands is in error? M07_BAUM7206_05_SE_C07.indd 220 If it chose randomly, 50% of the time it would choose the wrong strand and introduce mutations. Mismatch repair enzymes, however, do not choose randomly. They distinguish a new DNA strand from an old strand because old strands are methylated. Recognition of an error as far as 1000 bp away from a methylated portion of DNA triggers mismatch repair enzymes. Once a new DNA strand is methylated, mismatch repair enzymes cannot correct any errors that remain, so mismatch repair must immediately follow DNA replication. Error-Prone Repair Sometimes damage to DNA is so extensive, perhaps even corrupting both strands of DNA, that regular repair mechanisms cannot cope with the damage. The repair system of last resort 11/23/16 3:01 AM       Mutations of Genes is error­prone repair (not shown in Figure 7.28). Geneticists best understand error-prone repair in E. coli, where it is called an SOS response. This form of repair is a last-ditch effort by the cell to salvage enough DNA following severe damage to allow the cell to remain alive. SOS repair involves a variety of processes, such as the production of novel DNA polymerases (called IV and V) capable of copying less-than-perfect DNA. These polymerases replicate DNA with little regard to the base sequence of the strand being repaired. Of course, this introduces many new and potentially fatal mutations, but presumably SOS repair allows a few offspring of these bacteria to survive. Penicillinresistant cell Identifying Mutants, Mutagens, and Carcinogens Medium with penicillin (only penicillin-resistant cell grows into colony) LE A R N I N G O U TC O M E S Penicillinsensitive cells Medium without penicillin (both types of cells form colonies) (a) 7.26 Contrast positive and negative selection techniques for isolating mutants. Mutagen induces mutations 7.27 Describe the Ames test, and discuss its use in discover- ing carcinogens. If a cell does not successfully repair a mutation, it and its descendants are called mutants. In contrast, cells normally found in nature (in the wild) are called wild-type cells. Scientists distinguish mutants from wild-type cells by observing or testing for altered phenotypes. Because mutations are rare and nonfatal mutations are even rarer, mutants can easily be “lost in the crowd.” Therefore, researchers have developed methods to recognize mutants amidst their wild-type neighbors. Penicillin-resistant mutants indistinguishable from nonmutants Positive Selection Medium with penicillin Medium without penicillin (b) ▲ FIGURE 7.29 Positive selection of mutants. Only mutants that are resistant to penicillin can survive on the plate containing this antibiotic. (a) Normally, a population includes very few mutants. (b) Introduction of a mutagen increases the number of mutants and thus the number of colonies that grow in the presence of penicillin. What is the rate of mutation induced by the mutagen in (b)? Figure 7.29 5 - 1 * 100% = 400% 1 Positive selection involves selecting a mutant by eliminating wild-type phenotypes. Assume, for example, that researchers want to isolate penicillin-resistant bacterial mutants from a liquid culture. To do so, they spread the liquid medium, which contains mostly penicillin-sensitive cells but also the few penicillin-resistant mutants, onto a medium that includes penicillin. Only the penicillin-resistant mutants multiply on this medium and produce colonies (FIGURE 7.29a). When a mutagenic agent is added to a liquid culture, it increases the number of mutants (FIGURE 7.29b). While the researchers are isolating mutants, they can also determine the rate of mutation by comparing the number of mutant colonies formed after use of the mutagen with the number formed before treatment. The rate of mutation can be calculated as follows: Negative (Indirect) Selection (awk’sō-trōf).7 For example, a mutant bacterium that has lost the ability to synthesize tryptophan is auxotrophic for this amino acid—it must acquire tryptophan from the environment. Obviously, if a researcher attempts to grow tryptophan auxotrophs on media lacking tryptophan, it will not be possible; the bacteria will die because tryptophan is a necessary amino acid for proper synthesis of proteins. Therefore, to isolate such auxotrophs, scientists can use a technique called negative (indirect) An organism that requires different nutrients than its wildt y p e phe not y p e r e q u i r e s i s k now n a s a n au xot ro ph 7From number of colonies seen with use of mutagen - number of colonies seen without use of mutagen number of colonies seen without the use of mutagen M07_BAUM7206_05_SE_C07.indd 221 : 100% Greek auxein, meaning “to increase,” and trophe, meaning “nutrition.” 11/23/16 3:01 AM r t h ­ 1 2 3 4 5 6 The researcher inoculates a sample of a bacterial suspension containing potential mutants onto a plate containing complete media (including tryptophan). The sample is diluted such that the plate receives only about 100 cells. Both auxotrophs and wild-type cells reproduce and form colonies on the plate, though the colonies are indistinguishable. The researcher simultaneously picks up cells from all the colonies on the plate with a sterile velvet pad by pressing the pad onto the plate. The researcher inoculates two new plates—one containing tryptophan, the other lacking tryptophan—by pressing the pad onto each of them. This technique is called replica plating. After the plates have incubated for several hours, the researcher compares the two replica plates. Tryptophan auxotrophs growing on medium containing tryptophan are revealed by the absence of a corresponding colony on the plate lacking tryptophan. The researcher takes cells of the auxotroph colony from the replica plate and inoculates them into a complete medium. The auxotroph is now isolated. 1 Inoculate bacteria onto complete medium containing tryptophan. Mutagen Bacterial suspension X Incubation 2 Bacterial colonies grow. A few may be tryptophan auxotrophs. Most are wild type. X 3 Stamp sterile velvet onto plate, picking up cells from each colony. Sterile velvet surface X X selection. The process by which a researcher uses negative selection to culture a tryptophan auxotroph is as follows (FIGURE 7.30). X       C ap e 7   Microbial Genetics Bacteria 4 Stamp replica plates with velvet. The Ames Test for Identifying Mutagens Numerous chemicals in food, the workplace, and the environment in general have been suspected of being carcinogenic (kar’si-nōjen’ik) mutagens; that is, of causing mutations that result in cancer. Because animal tests to prove that they are indeed carcinogenic are expensive and time consuming, researchers have used a fast and inexpensive method of screening for mutagens called an Ames test, which is named for its inventor, Bruce Ames (1928–). An Ames test uses mutant Salmonella (sal’mŏ-nel’ă) bacteria possessing a point mutation that prevents the synthesis of the amino acid histidine; in other words, they are histidine auxotrophs, indicated by the abbreviation his- . To perform the test, an investigator mixes his- mutants with the substance suspected to be a mutagen and with liver extract (FIGURE 7.31). The presence of liver extract simulates the conditions in the body under which liver enzymes can turn harmless chemicals into mutagens. The researcher then spreads the mixture on a solid medium lacking histidine. If the suspected substance does in fact cause mutations, some of the mutations will likely reverse the effect of the original point mutation, producing revertant cells (designated his+ ), which are cells that have regained the wildtype phenotype; in this case, they have regained the ability to synthesize histidine and thus can survive on a medium lacking histidine. Thus, the presence of colonies during an Ames test reveals that the suspected substance is mutagenic in Salmonella. Given that DNA in all cells is very similar, the ability of a chemical to cause mutations in Salmonella indicates that it is likely to cause mutations in humans as well. Some mutations cause cancers, so the mutagenic chemical may also be carcinogenic. To prove that the substance can in fact cause cancer, M07_BAUM7206_05_SE_C07.indd 222 X X Complete medium containing tryptophan Medium lacking tryptophan Incubation 5 Identify auxotroph as colony growing on complete medium but not on lacking medium. X X All colonies grow. Tryptophan auxotroph cannot grow. 6 Inoculate auxotroph colony into complete medium. ▲ FIGURE 7.30 The use of negative (indirect) selection to isolate a tryptophan auxotroph. The plates are marked (in this case, with an X) so that their orientation can be maintained throughout the procedure. Researchers may have to inoculate hundreds of such plates to identify a single mutant. 11/23/16 3:01 AM       Genetic Recombination and Transfer Experimental tube Liver extract nucleotide sequences called homologous sequences. Scientists have discovered a number of molecular mechanisms for genetic recombination, one of which is illustrated simply in FIGURE 7.32. In this type of recombination, enzymes nick one strand of DNA at the homologous sequence, and another enzyme inserts the nicked strand into the second DNA molecule. Ligase then reconnects the strands in new combinations, and the molecules resolve themselves into novel molecules. DNA molecules that contain such new arrangements of nucleotide sequences are called recombinant DNA. An example of genetic recombination is crossing over during gamete formation, part of sexual reproduction in eukaryotes. Cells that contain recombinant DNA are called recombinants. (Chapter 12 discusses crossing over in more detail.) Control tube Suspected mutagen Liver extract Culture of his– Salmonella Medium lacking histidine Homologous sequences Incubation 3¿ DNA A 5¿ 3¿ DNA B 5¿ Colony of revertant (his+) Salmonella No growth Enzyme nicks one strand of DNA at homologous sequence. ▲ FIGURE 7.31 The Ames test. mixture containing his-Salmonella mutants, rat liver extract, and the suspected mutagen is inoculated onto a plate lacking histidine. Colonies will form only if a mutagen reverses the his-mutation, producing revertant his+organisms with the ability to synthesize histidine. control tube that lacks the suspected mutagen demonstrates that reversion did not occur in the absence of the mutagen. What is the purpose of liver extract in an Ames test? A A A B Recombination enzyme inserts the cut strand into second molecule, which is nicked in the process. Figure 7.31 Liver extract simulates conditions in the body by providing enzymes that may degrade harmless substances into mutagens. scientists must test it in laboratory animals, a process that usually is warranted only if a chemical is mutagenic in Salmonella. Ames testing reduces the cost and time required to assay chemicals for carcinogenicity in animals. Ligase anneals nicked ends in new combinations. TELL ME WHY Changes in RNA sequences resulting from poor transcription are not as deleterious to an organism as changes to DNA resulting from mutations. Why is this the case? Molecules resolve into recombinants. Genetic Recombination and Transfer Genetic recombination refers to the exchange of nucleotide sequences between two DNA molecules and often involves segments that are composed of identical or nearly identical M07_BAUM7206_05_SE_C07.indd 223 A A A ▲ FIGURE 7.32 Simplified depiction of one type of recombination between two DN molecules. fter an enzyme nicks one strand (here, strand ), a recombination enzyme rearranges the strands, and ligase seals the gaps to form recombinant molecules. What is the normal function of ligase (during DNA replication)? Figure 7.32 Ligase functions to anneal Okazaki fragments during replication of the 7.28 Define genetic recombination. Recombinant B lagging strand. LE A R N I N G O U TC O M E Recombinant A 11/23/16 3:01 AM r t h       C ap e 7   Microbial Genetics Horizontal Gene Transfer Among Prokaryotes L E A R N I N G O U TCO M E S 7.29 Contrast vertical gene transfer with horizontal gene transfer. 7.30 Explain the roles of an F factor, F + cells, and Hfr cells in bacterial conjugation. 7.31 Describe the structures and actions of simple and com- plex transposons. 7.32 Compare and contrast crossing over, transformation, transduction, and conjugation. ­ ­ As we have discussed, both prokaryotes and eukaryotes replicate their genomes and supply copies to their descendants. This is known as vertical gene transfer—the passing of genes to the next generation. Many prokaryotes can acquire genes from other microbes of the same generation—a process termed horizontal (lateral) gene transfer. In horizontal gene transfer, a donor cell contributes part of its genome to a recipient cell, which may be of a different species from the donor. Typically, a recipient cell inserts part of a donor’s DNA into its own chromosome, becoming a recombinant cell. Cellular enzymes then usually degrade remaining unincorporated donor DNA. Horizontal gene transfer is a rare event, typically occurring in less than 1% of a population of prokaryotes. ▶ ANIMATIONS: Horizontal Gene Transfer: Overview Here, we consider the three types of horizontal gene transfer: transformation, transduction, and bacterial conjugation. Transformation In transformation, a recipient cell takes up DNA from the environment, such as DNA that might be released by dead organisms. Frederick Griffith (1879–1941) discovered this process in 1928 while studying pneumonia caused by Streptococ­ cus pneumoniae (strep-tō-kok’ŭs nū-mō’nē-ī). Griffith worked with two strains of Streptococcus. Cells of the first strain have protective capsules that enable them to escape a body’s defensive white blood cells; thus, these encapsulated cells cause deadly septicemia when injected into mice (FIGURE 7.33a). They are called strain S because they form smooth colonies on an agar surface. The application of heat kills such encapsulated cells and renders them harmless when injected into mice (FIGURE 7.33b). In contrast, cells of the second strain (called strain R because they form rough colonies) are mutants that cannot make capsules. The unencapsulated cells of strain R are much less virulent, only infrequently causing disease (FIGURE 7.33c) because a mouse’s defensive white blood cells quickly devour them. Griffith discovered that when he injected both heat-killed strain S and living strain R into a mouse, the mouse died even though neither of the injected strains was harmful when administered alone (FIGURE 7.33d). Further, and most significantly, Griffith isolated numerous living bacterial cells from the dead mouse, cells that had capsules. He realized that harmless, unencapsulated strain R bacteria had been transformed into deadly, encapsulated strain S bacteria. Subsequent M07_BAUM7206_05_SE_C07.indd 224 investigations showed that transformation, by which cells take up DNA f rom t he env iron ment, a lso occ urs in vitro 8 (FIGURE 7.33e). The fact that the living encapsulated cells retrieved at the end of this experiment outnumbered the dead encapsulated cells injected at the beginning indicated that strain R cells were not merely appropriating capsules released from dead strain S cells. Instead, strain R cells had acquired the capability of producing their own capsules by assimilating the capsule-coding genes of strain S cells. In 1944, Oswald Avery (1877–1955), Colin MacLeod (1909–1972), and Maclyn McCarty (1911–2005) extracted various chemicals from S cells and determined that the transforming agent was DNA. This discovery was one of the conclusive pieces of evidence that DNA is the genetic material of cells. Cells that have the ability to take up DNA from their environment are said to be competent. Competence results from expression of a transporter protein that crosses the cytoplasmic membrane and brings DNA into the cell. Scientists have observed natural competence in a few types of bacteria, including pathogens belonging to the genera Streptococcus, Haemophilus (hē-mof’i-lŭs), Neisseria (nī-se’rē-ă), Bacillus (ba-sil’ŭs), Staphylococcus (staf’i-lō-kok’ŭs), and Pseudomonas. Scientists can also generate competence artificially in Escherichia and other bacteria by manipulating the temperature and salt content of the medium. Because competent cells take up DNA from any donor genome, competence and transformation are important tools in recombinant DNA technology, commonly called genetic engineering. ▶ ANIMATIONS: Transformation Transduction A second method of horizontal gene transfer, called transduction, involves the transfer of DNA from one cell to another via a replicating virus. Transduction can occur either between prokaryotic cells or between eukaryotic cells; it is limited only by the availability of a virus that can infect both donor and recipient cells. Here, we will consider transduction in bacteria. A virus that infects bacteria is called a bacteriophage or simply a phage (fāj).9 The process by which a phage participates in transduction is depicted in FIGURE 7.34. To replicate, a bacteriophage attaches to a bacterial host cell and injects its genome into the cell 1 . Phage enzymes (enzymes coded by the phage’s genome even though translated by bacterial ribosomes) degrade the cell’s DNA 2 . The phage genome now controls the cell’s functions and directs the cell to synthesize new phage DNA and phage proteins. Normally, phage proteins assemble around phage DNA to form new daughter phages, but some phages mistakenly incorporate remaining fragments of bacterial DNA. This forms transducing phages 3 . Eventually, the host cell lyses, releasing daughter and transducing phages. Transduction occurs when a transducing phage injects donor DNA into a new host cell (the recipient) 4 . The recipient host cell incorporates the donated DNA into its chromosome by recombination 5 . 8Latin, 9From meaning “within glassware.” Greek phagein, meaning “to eat.” 11/23/16 3:01 AM       Genetic Recombination and Transfer Observations of Streptococcus pneumoniae Live cells (a) Griffith‘s experiment: Living strain R Injection Capsule Heat-treated dead cells of strain S Mouse dies + In vitro transformation XX XX Heat-treated dead cells of strain S DNA broken into pieces Heat-treated dead cells of strain S DNA fragment from strain S Injection Living strain R Some cells take up DNA from the environment and incorporate it into their chromosomes Injection XX Mouse dies (b) Mouse lives Strain R live cells (no capsule) Culture of Streptococcus from dead mouse Transformed cells acquire ability to synthesize capsules Injection Living cells with capsule (strain S) (c) Mouse lives (d) (e) ▲ FIGURE 7.33 Transformation of Streptococcus pneumoniae. Griffith’s observations revealed that (a) encapsulated strain A S killed mice, (b) heating renders strain S harmless to mice, and (c) unencapsulated strain R did not harm mice. (d) In Griffith’s experiment, a mouse injected concurrently with killed strain S and live strain R (each harmless) died and was found to contain numerous living, encapsulated bacteria. (e) demonstration that transformation of R cells to S cells also occurs in vitro. In generalized transduction, the transducing phage carries a random DNA segment from a donor host cell’s chromosome or plasmids to a recipient host cell. Generalized transduction is not limited to a particular DNA sequence. In specialized transduction, only certain host sequences are transferred. In nature, specialized transduction is important in transferring genes encoding certain bacterial toxins—including those responsible for diphtheria, scarlet fever, and the bloody, life-threatening diarrhea caused by E. coli O157:H7—into cells that would otherwise be harmless. (Chapter 8 discusses the use of specialized transduction to intentionally insert genes into cells.) ▶ ANIMATIONS: Transduction: Generalized Transduction, Specialized Transduction Bacterial Conjugation A third method of genetic transfer in bacteria is conjugation.10 Unlike the typical donor cells in transformation and transduction, a donor cell in conjugation remains alive. Further, conjugation requires physical contact between donor and recipient cells. Scientists discovered conjugation between cells of E. coli, and it is best understood in this species. Thus, the remainder of our discussion will focus on conjugation in this bacterium. ▶ ANIMATIONS: Conjugation: Overview 10From Latin conjugatus, meaning “yoked together.” M07_BAUM7206_05_SE_C07.indd 225 Conjugation is mediated by pili (pī’ lī; singular: pilus, pī’lus), also called conjugation or sex pili,11 which are thin, proteinaceous tubes extending from the surface of a cell. The gene coding for conjugation pili is located on a plasmid called an F (fertility) plasmid. (Recall that a plasmid is a small, circular, extrachromosomal molecule of DNA; see Figure 7.2b.) Cells that contain an F plasmid are called F + cells (“ef plus”), and they serve as donors during conjugation. Recipient cells are F - ; that is, they lack an F plasmid and therefore have no pili. Despite the historical use of the terms sex and fertility, conjugation is completely unrelated to sexual reproduction, which involves mixing DNA from two organisms. Conjugation is a unidirectional method of horizontal gene transfer from a donor to a recipient. ▶ ANIMATIONS: Conjugation: F Factor The process of bacterial conjugation is illustrated in FIGURE 7.35. First, a sex pilus connects a donor cell (F + ) to a recipient cell (F - ) 1 . The pilus may draw the cells together, though scientists have filmed DNA transfer between cells 12 mm apart 2 . An enzyme, coded by the plasmid, nicks one strand of the plasmid DNA, and this single strand transfers into the recipient, beginning with a section called the origin of 11Bacterial conjugation may resemble intercourse, but because no gametes are involved, it is not true sex. 11/23/16 3:01 AM r t h       C ap e 7   Microbial Genetics Cell 1 Bacteriophage Host bacterial cell (donor cell) Bacterial chromosome 1 Phage injects its DNA. 2 Phage enzymes degrade host DNA. Phage DNA Phage with donor cell's DNA (transducing phage) 3 Cell synthesizes new phages that incorporate phage DNA and, mistakenly, some host DNA. Transducing phage Cell 2 Recipient host cell 4 Transducing phage injects donor DNA. Conjugation: Hfr Conjugation Because an F plasmid can integrate into a chromosome at only a few locations, the order in which donor genes are transferred into the recipient cell is consistent. Scientists produced the first gene maps of bacterial chromosomes by noting the time, in minutes, required for a particular gene to transfer from donor cell to recipient cell. ▶ ANIMATIONS: Conjugation: Chromosome Mapping Conjugation occurs in several species of bacteria, which can be quite promiscuous; that is, conjugation can occur among bacteria of widely varying kinds and even between a bacterium and a yeast cell or between a bacterium and a plant cell. For example, the crown gall bacterium, Agrobacterium, transfers some of its genes into the chromosome of its host plant. The natural transfer of genes by conjugation among diverse organisms heightens some scientists’ concerns about the spread of resistance (R) plasmids among pathogens. These scientists note that antibiotic resistance developed by one pathogen can spread to several other pathogens. This is of particular concern in health care settings. TABLE 7.5 summarizes the mechanisms of horizontal gene transfer in bacteria. Transposons and Transposition Transduced cell 5 Donor DNA is incorporated into recipient‘s chromosome by recombination. Inserted DNA A ▲ FIGURE 7.34 Transduction. fter a virus called a bacteriophage A A (phage) attaches to a host bacterial cell, it injects its genome into the cell and directs the cell to synthesize new phages. During assembly of new phages, some host DN may be incorporated, forming transducing phages, which subsequently carry donor DN to a recipient host cell. transfer 3 . Both cells synthesize complementary strands of F plasmid DNA to form complete double-stranded DNA circles. The F - recipient is now an F + cell 4 . In some bacterial cells, the F plasmid does not remain independent in the cytosol but instead integrates into the cellular chromosome. Such cells, which are called Hfr (high M07_BAUM7206_05_SE_C07.indd 226 frequency of recombination) cells, can conjugate with F - cells (FIGURE 7.36). After the F plasmid has integrated 1 and the Hfr and F - cells join via a sex pilus 2 , DNA transfer begins at the F plasmid’s origin of transfer and begins to carry a copy of the donor’s chromosome into the recipient 3 . In most cases, movement of the cells breaks the intercellular connection before the entire donor chromosome is transferred 4 . Since the recipient receives only a portion of the F plasmid, it remains an F - cell; however, it acquires some chromosomal genes from the donor. Recombination can integrate the donor’s DNA into the recipient’s chromosome 5 . The recipient is now a recombinant cell because it contains some donor genes. ▶ ANIMATIONS: LE A R N I N G O UTC O M E S 7.33 Describe a transposon and a complex transposon. 7.34 Describe transposition and its effects. Transposons12 are segments of DNA, 700 to 40,000 bp in length, that transpose (move) themselves from one location in a DNA molecule to another location in the same or a different molecule. American geneticist Barbara McClintock (1902–1992) discovered these “jumping genes” through a painstaking analysis of the colors of the kernels of corn. She discovered that the genes for kernel color were turned on and off by the insertion of transposons. Subsequent research has shown that transposons are found in many, if not all, prokaryotes and eukaryotes and in many viruses. 12From transposable elements. 11/23/16 3:02 AM       Genetic Recombination and Transfer TABLE 7.5 Natural Mechanisms of Horizontal Genetic Transfer in Bacteria TEM (a) Requirements Transformation Free DN in the environment and a competent recipient Transduction Bacteriophage Conjugation Cell-to-cell contact and F plasmid, which is either in cytosol or incorporated into chromosome of donor (Hfr) cell A Mechanism 1 µm Pilus Cell 1 Cell 2 F plasmid Origin of transfer Chromosome 1 Donor cell attaches to a recipient cell with its pilus. F+ cell F– cell 2 Pilus draws cells together. Origin of transfer 3 One strand of F plasmid DNA transfers to the recipient. Pilus 4 The recipient synthesizes a complementary strand to become an F+ cell with a pilus; the donor synthesizes a complementary strand, restoring its complete plasmid. F+ cell Now F+ cell (b) A A A ▲ FIGURE 7.35 Bacterial conjugation. conjugation pilus connecting two cells mediates the transfer of DN between the cells. (a) Two E. coli cells are connected by a pilus. (b) rtist’s rendition of the process. The result of the action of a transposon is termed transposition; in effect, it is a kind of frameshift insertion (see Figure 7.24e). This “illegitimate” recombination does not need a region of homology, unlike other recombination events. Transposition occurs between plasmids and chromosomes and within and M07_BAUM7206_05_SE_C07.indd 227 among chromosomes (FIGURE 7.37 ) . Further, plasmids can carry transposons from one cell to another. Fortunately, although transposons are common, transposition and the frameshift mutations it causes are relatively rare occurrences. ▶ ANIMATIONS: Transposons: Overview 11/23/16 3:02 AM r t h       C ap e 7   Microbial Genetics Cell 1 A ◀ FIGURE 7.36 Conjugation involving an Hfr cell. n Hfr cell is formed when an F + cell integrates its F plasmid into its chromosome. Hfr cells donate a partial copy of their DN and a portion of the F plasmid to a recipient, which is rendered a recombinant cell but remains F - . A Donor chromosome Pilus F+ cell Hfr cell 1 F plasmid integrates into chromosome by recombination. Cell 2 Pilus F+ cell (Hfr) F plasmid 2 Cells join via a pilus. F– recipient Donor DNA Origin of transfer 3 Portion of F plasmid partially moves into recipient cell trailing a strand of donor’s DNA. Incomplete F plasmid; cell remains F– 4 Conjugation ends with pieces of F plasmid and donor DNA in recipient cell; cells synthesize complementary DNA strands. 5 Portions of the plasmid and donor DNA combine with recipient DNA to make a recombinant cell, which remains F–, because it does not contain a complete F plasmid. Recombinant cell (still F–) Transposons vary in their nucleotide sequences, but all of them contain palindromic sequences at each end. A palindrome13 is a word, phrase, or sentence that has the same sequence of 13From Greek palin, meaning “again,” and dramein, meaning “to run.” M07_BAUM7206_05_SE_C07.indd 228 letters when read backward or forward—for example, “Madam, I’m Adam.” In genetics, a palindrome is a region of DNA in which the sequence of nucleotides is identical to an inverted sequence in the complementary strand. For example, GAATTC is the palindrome of CTTAAG. Such a palindromic sequence is also known as an inverted repeat (IR). 11/23/16 3:02 AM       Genetic Recombination and Transfer Transposon Plasmid with transposon DNA (a) Jumping transposons. Transposons move from one place to another on a DNA molecule. (b) Replicating transposons. Transposons may replicate while moving, resulting in more transposons in the cell. (c) Transposons can move onto plasmids. Transposons moving onto plasmids can be transferred to another cell. P ▲ FIGURE 7.37 Transposition. Transposons move from place to place within, among, and between chromosomes and plasmids. lasmids can carry transposons to and from cells. Transposase gene Inverted repeat (IR) A (a) A Inverted repeat (IR) A T C A G T A A G T T A G T C A T T C A A A C T T A C T G A T T G A A T G A C T A ◀ FIGURE 7.38 Transposons. (a) simple transposon, or insertion sequence. Shown here is insertion sequence 1 (IS1), which consists of a gene for the enzyme transposase bounded by identical (though inverted) repeats of nucleotides. (b) Transposase recognizes its target site (elsewhere in the same DN molecule, as shown here, or in a different DN molecule); then it moves the transposon (or, as in the case of IS1, a copy of the transposon) to its new site. The target site is duplicated in the process. (c) complex transposon, which contains genes not related to transposition. Shown here is Tn5, which consists of a gene for kanamycin resistance between two IS1 transposons. A Transposon: Insertion sequence IS1 DNA molecule Target site IS1 Target IS1 Transposase Target site Copy of IS1 Copy of target site Original IS1 Transposons: Insertion Sequences, Complex Transposons (Chapter 8 discusses how scientists have adapted the natural processes of transformation, transduction, conjugation, and transposition to manipulate the genes of organisms.) (b) Complex transposon IR (c) M07_BAUM7206_05_SE_C07.indd 229 IR IS1 Kanamycin- IR resistance gene The simplest transposons, called insertion sequences (IS), consist of no more than two inverted repeats and a gene that encodes the enzyme transposase (FIGURE 7.38a). Transposase recognizes its own inverted repeat in a target site, cuts the DNA at that site, and inserts the transposon (or a copy of it) into the DNA molecule at that site (FIGURE 7.38b). Such transposition also produces a duplicate copy of the target site. Complex transposons contain one or more genes not connected with transposition, such as genes for antibiotic resistance (FIGURE 7.38c). R plasmids often contain transposons. R factors are of great clinical concern because they spread antibiotic resistance among pathogens. ▶ ANIMATIONS: IR IS1 TELL ME WHY Why is the genetic ancestry of microbes much more difficult to ascertain than the ancestry of animals? 11/23/16 3:02 AM r Deadly Horizontal Gene Transfer A P A acquire genes that convey resistance to antimicrobial drugs. s a result of such horizontal gene transfer, E. faecium has become an MDR bacterium—that is, a strain resistant to many different kinds of antimicrobials. Drug-resistant enterococci are a leading cause of healthcare-associated infections (nosocomial infections). hysicians treated Sarah with a series of antimicrobial drugs, including metronidazole, penicillin, erythromycin, ciprofloxacin, and vancomycin, but to no avail. fter 90 days of hospitalization, she died. 1. Define healthcare-associated infection. 2. What is the likely source of Sarah’s infection? 3. List three ways by which E. faecium might have acquired genes for drug resistance. 4. How can hospital personnel prevent the spread of resistant E. faecium throughout the hospital? E A Sarah might have been 85 years old, but her mind was still sharp; it was her body that was failing. fter she broke her hip by falling in the bathroom, she had succumbed to a series of complications that ultimately required admittance to a large universityassociated hospital in New York City. Treatment necessitated a urinary catheter, a LM feeding tube, and intensive 10 µm antibiotic therapy. While hospitalized, she acquired a life-threatening urinary tract infection with multi-drug-resistant (MDR) Enterococcus faecium. nterococci are normal members of the microbiota of the colon, yet these Gram-positive bacteria have a propensity to ­ CLINICAL CASE STUDY t h       C ap e 7   Microbial Genetics 1. The prevalence of CRE infections in health care settings has increased significantly in recent years, in part because these organisms are able to share plasmid-encoded resistance genes. Which method of horizontal gene transfer is most likely responsible for transfer of the plasmid between bacteria? Explain your choice. 2. Ben most likely acquired the infection while he was in the hospital recovering from his stroke. What are some possible sources of infection in the hospital that might have led to Ben’s infection? E E E ­ ­ The doctor and nurses are able to stabilize Ben, and he is transferred to the hospital. Upon arrival, Ben has a fever of 103.1°F, and his white blood cell (WBC) count is elevated. Blood and urine cultures are collected and sent for lab analysis. The elevated WBC count suggests an infection, so the doctor starts Ben on meropenem, an ultra-broad-spectrum antibiotic. The next day, Ben does not appear to be responding to the meropenem treatment. The lab results indicate multi-drug-resistant Klebsiella pneumoniae present in Ben’s blood sample. This diagnosis is cause for concern because this organism is likely a carbapenemresistant member of the nterobacteriaceae, a family of bacteria that have high levels of resistance to multiple antibiotics. The doctor puts Ben on a course of high-dose tigecycline, an antibiotic that has shown effectiveness against carbapenem-resistant nterobacteriaceae (CR ). The next few days bring little improvement to Ben’s condition. His condition continues to decline, and he ultimately suffers multiple organ failure and dies. ­ Another Stroke? M07_BAUM7206_05_SE_C07.indd 230 11/23/16 3:02 AM       Chapter Summary CHAPTER SUMMARY 5. DNA replication is semiconservative; that is, each newly synthesized strand of DNA remains associated with one of the parental strands. After helicase unwinds and unzips the original molecule, synthesis of each of the two daughter strands—called the leading strand and the lagging strand—occurs from 5’ to 3’. Synthesis is mediated by enzymes that prime, join, and proofread the pairing of new nucleotides. ▶ ANIMATIONS: DNA Replication: Overview, Forming the Replication Fork, Replication Proteins, Synthesis 6. After DNA replication, methylation occurs. Methylation plays several roles, including the control of gene expression, the initiation of DNA replication, recognition of a cell’s own DNA, and repair. Gene Function (pp. 202–216) 1. The genotype of an organism is the actual set of genes in its genome, whereas its phenotype refers to the physical and functional traits expressed by those genes. 2. RNA has several forms. These include RNA primer; messenger RNA (mRNA), which carries genetic information from DNA to a ribosome; transfer RNA (tRNA), which carries amino acids to the ribosome; ribosomal RNA (rRNA), which, together with M07_BAUM7206_05_SE_C07.indd 231 6. In translation, the sequence of genetic information carried by mRNA is used by ribosomes to construct polypeptides with specific amino acid sequences. Initiation of translation involves three steps. ▶ ANIMATIONS: Translation: Overview, Genetic Code, The Process V T e V T e o 4. In addition to DNA, eukaryotic chromosomes contain eukaryotic histones and are arranged as nucleosomes (beads of DNA) that clump to form chromatin fibers. Eukaryotic cells may also contain extranuclear DNA in mitochondria, chloroplasts, and plasmids. 5. Eukaryotic mRNA is synthesized as pre-messenger RNA. Before translation can occur, a spliceosome removes noncoding introns from pre-mRNA and splices together the exons, which are the coding sections. id 3. Bacterial and archaeal genomes consist of one (or, rarely, two or more) chromosomes, which are typically circular molecules of DNA associated with protein and RNA molecules, localized in a region of the cytosol called the nucleoid. Archaeal DNA organizes around globular proteins called histones. Prokaryotic cells may also contain one or more extrachromosomal DNA molecules called plasmids, which contain genes that regulate nonessential life functions, such as bacterial conjugation and resistance to antibiotics. 4. The transfer of genetic information begins with transcription of the genetic code from DNA to RNA, in which RNA polymerase links RNA nucleotides that are complementary to genetic sequences in DNA. Transcription begins at a region of DNA called a promoter (recognized by RNA polymerase) and ends with a sequence called a terminator. In bacteria, Rho protein may assist in termination, or termination may depend solely on the nucleotide sequence of the transcribed RNA. ▶ ANIMATIONS: Transcription: Overview, The Process utor 2. The two strands of DNA are held together by hydrogen bonds between complementary base pairs (bp) of nucleotides. Adenine bonds with thymine, and guanine bonds with cytosine. 3. The central dogma of genetics states that genetic information is transferred from DNA to RNA to polypeptides, which function alone or in conjunction as proteins. o 1. Genetics is the study of inheritance and inheritable traits. Genes are composed of specific sequences of nucleotides that code for polypeptides or RNA molecules. A genome is the sum of all the genes and linking nucleotide sequences in a cell or virus. Prokaryotic and eukaryotic cells use DNA as their genetic material; some viruses use DNA, and other viruses use RNA. id (pp. 193–202) utor The Structure and Replication of Genomes polypeptides, makes up the structure of ribosomes; regulatory RNA, which interacts with DNA to control gene expression; and ribozymes, which are RNA molecules that function as metabolic enzymes. 7. The genetic code consists of triplets of mRNA nucleotides, called codons. During translation, these bind with complementary anticodons on transfer RNAs (tRNAs), which are molecules that carry specific amino acids. A ribozyme (rRNA), which is part of a ribosome, catalyzes the bonding of one amino acid to another to form a polypeptide. A sequence of nucleotides thus codes for a sequence of amino acids. 8. A ribosome contains three tRNA binding sites: an A site (associated with incoming amino acids), a P site (associated with elongation of the polypeptide), and an E site from which tRNA exits the ribosome. 9. Prokaryotes measure their cellular density by a process called quorum sensing in which cells secrete quorum-sensing molecules that neighboring cells detect. 10. An operon consists of a series of prokaryotic genes, a promoter, and, as originally described, an operator sequence, all controlled by one regulatory gene. The operon model explains gene regulation in prokaryotes. ▶ ANIMATIONS: Operons: Overview 11. Inducible operons are normally “turned off” and are activated when the repressor no longer binds to the operator site, whereas 11/23/16 3:02 AM r t h       C ap e 7   Microbial Genetics repressible operons are normally “on” and are deactivated when the repressor binds to the operator site. ▶ ANIMATIONS: Operons: Induction, Repression 12. Cells and some viruses use short microRNAs (miRNAs) in conjunction with proteins to form RNA silencing complexes (RISC) that attach to mRNA sequences to inhibit polypeptide translation. 13. Genetic expression can also be controlled with riboswitches or with small interfering RNA (siRNA) associated with protein in RISC. Mutations of Genes (pp. 216–223) 1. A mutation is a change in the nucleotide base sequence of a genome. Point mutations involve a change in a single nucleotide base pair and include substitutions and two types of frameshift mutations: insertions and deletions. 2. Gross mutations include major changes to the DNA sequence, such as inversions, duplications, transpositions, and large deletions or insertions. 3. Mutations can be categorized by their effects as silent mutations, missense mutations, or nonsense mutations. Most mutations are spontaneous. ▶ ANIMATIONS: Mutations: Types 4. Physical or chemical agents called mutagens can increase the normal rate of mutation. Physical mutagens include ionizing radiation, such as X rays and gamma rays, and nonionizing ultraviolet light. Ultraviolet light causes adjacent pyrimidine bases to bond to one another to form pyrimidine dimers. Mutagenic chemicals include nucleotide analogs, chemicals that are structurally similar to nucleotides and can result in mismatched base pairing, and chemicals that insert or delete nucleotide base pairs, producing frameshift mutations. ▶ ANIMATIONS: Mutagens 5. Cells repair damaged DNA via direct repair, including baseexcision repair and light repair of pyrimidine dimers; single-strand repair, such as nucleotide-excision repair (dark repair) and mismatch repair; and error-prone repair adopted when damage is extensive such as the SOS response of E. coli. ▶ ANIMATIONS: Mutations: Repair 6. Researchers have developed methods to distinguish mutants, which carry mutations, from normal wild-type cells. These methods include positive selection, negative (indirect) selection, and the Ames test, which is used to identify mutagens, which may be potential carcinogens. Genetic Recombination and Transfer (pp. 223–229) 1. Organisms acquire new genes through genetic recombination, which is the exchange of segments of DNA. Crossing over occurs during gamete formation, part of sexual reproduction in eukaryotes. 2. Vertical gene transfer is the transmission of genes from parents to offspring. In horizontal (lateral) gene transfer, DNA from a donor cell is transmitted to a recipient cell. A recombinant cell results from genetic recombination between donated and recipient DNA. Transformation, transduction, and bacterial conjugation are types of horizontal gene transfer. ▶ ANIMATIONS: Horizontal Gene Transfer: Overview 3. In transformation, a competent recipient prokaryote takes up DNA from its environment. Competence is found naturally or can be created artificially in some cells. ▶ ANIMATIONS: Transformation 4. In transduction, a virus such as a bacteriophage, or phage, carries DNA from a donor cell to a recipient cell. Donor DNA is accidentally incorporated in such transducing phages. ▶ ANIMATIONS: Transduction: Generalized Transduction, Specialized Transduction 5. In conjugation, an F + bacterium—that is, one containing an F (fertility) plasmid—forms a pilus (conjugation pilus) that attaches to an F - recipient bacterium. Plasmid genes are transferred to the recipient, which becomes F + as a result. ▶ ANIMATIONS: Conjugation: Overview, F Factor, Hfr Conjugation, Chromosome Mapping 6. Hfr (high frequency of recombination) cells result when an F plasmid integrates into a prokaryotic chromosome. Hfr cells form conjugation pili and transfer cellular genes more frequently than regular F + cells do. 7. Transposons are DNA segments that code for the enzyme transposase and have palindromic sequences known as inverted repeats (IR) at each end. Transposons move among locations in chromosomes in eukaryotes and prokaryotes—a process called transposition. The simplest transposons, known as insertion sequences (IS), consist only of inverted repeats and transposase. Complex transposons contain other genes as well. ▶ ANIMATIONS: Transposons: Overview, Insertion Sequences, Complex Transposons E E E QU STIONS FOR R VI W Answers to the Questions for Review (except Short Answer Questions) begin on p. A-1. Multiple Choice 1. Which of the following is most likely the number of base pairs in a bacterial chromosome? a. 4,000,000 b. 4000 c. 400 d. 40 M07_BAUM7206_05_SE_C07.indd 232 2. Which of the following is a true statement concerning prokaryotic chromosomes? a. They typically have two or three origins of replication. b. They contain single-stranded DNA. c. They are located in the cytosol. d. They are associated in linear pairs. 11/23/16 3:02 AM       Questions for Review 3. A plasmid is __________________. a. a molecule of RNA found in bacterial cells b. distinguished from a chromosome by being circular c. a structure in bacterial cells formed from plasma membrane d. extrachromosomal DNA 4. Which of the following forms ionic bonds with eukaryotic DNA and stabilizes it? a. chromatin c. histone b. bacteriocin d. nucleoid 5. Nucleotides used in the replication of DNA __________________. a. carry energy b. are found in four forms, each with a deoxyribose sugar, a phosphate, and a base c. are present in cells as triphosphate nucleotides d. All of the above are correct. 6. Which of the following molecules functions as a “proofreader” for a newly replicated strand of DNA? a. DNA polymerase III c. helicase b. primase d. ligase 7. The addition of —CH3 to a cytosine nucleotide after DNA replication is called __________________. a. methylation c. transcription b. restriction d. transversion 8. In translation, the site through which tRNA molecules leave a ribosome is called the __________________. a. A site c. P site b. X site d. E site 9. The Ames test __________________. a. uses auxotrophs and liver extract to reveal mutagens b. is time intensive and costly c. involves the isolation of a mutant by eliminating wild-type phenotypes with specific media d. proves that suspected chemicals are carcinogenic 10. Which of the following methods of DNA repair involves enzymes that recognize and correct nucleotide errors in unmethylated strands of DNA? a. light repair of T dimers c. mismatch repair b. dark repair of P dimers d. SOS response 11. Which of the following is not a mechanism of genetic transfer between cells? a. transduction c. transcription b. transformation d. conjugation 12. Cells that have the ability to take up DNA from their environment are said to be __________________. a. Hfr cells c. genomic b. transposing d. competent 13. Which of the following statements is true? a. Conjugation requires a sex pilus extending from the surface of a cell. b. Conjugation involves a C factor. c. Conjugation is an artificial genetic engineering technique. d. Conjugation involves DNA that has been released into the environment. 14. Which of the following are called “jumping genes”? a. Hfr cells c. palindromic sequences b. transducing phages d. transposons M07_BAUM7206_05_SE_C07.indd 233 15. Although two cells are totally unrelated, one cell receives DNA from the other cell and incorporates this new DNA into its chromosome. This process is __________________. a. crossing over of DNA c. horizontal gene transfer from the two cells d. transposition b. vertical gene transfer 16. Transcription produces __________________. a. DNA molecules c. polypeptides b. RNA molecules d. palindromes 17. A nucleotide is made of __________________. a. a five-carbon sugar c. a nitrogenous base b. phosphate d. all of the above 18. In DNA, adenine forms __________________ hydrogen bonds with __________________. a. three / uracil c. two / thymine b. two / uracil d. three / thymine 19. A sequence of nucleotides formed during replication of the lagging DNA strand is a(n) __________________. a. palindrome c. coding strand b. Okazaki fragment d. operon 20. Which of the following is not part of an operon? a. operator c. origin b. promoter d. gene 21. Repressible operons are important in regulating prokaryotic __________________. a. DNA replication c. rRNA processing b. RNA transcription d. sugar catabolism 22. Which of the following is part of each molecule of mRNA? a. palindrome c. anticodon b. codon d. base pair 23. Ligase plays a major role in __________________. c. polypeptide synthesis by a. replication of lagging ribosomes strands d. RNA transcription b. mRNA processing in eukaryotes 24. Before mutations can affect a population permanently, they must be __________________. a. lasting c. beneficial b. inheritable d. all of the above 25. The trp operon is repressible. This means it is usually __________________ and is directly controlled by a(n) __________________. a. active / inducer c. inactive / inducer b. active / repressor d. inactive /repressor Fill in the Blanks 1. The three steps in RNA transcription are __________________, __________________, and __________________. 2. A triplet of mRNA nucleotides that specifies a particular amino acid is called a __________________. 3. Three effects of point mutations are __________________, __________________, and __________________. 4. Insertions and deletions in the genetic code are also called __________________ mutations. 11/23/16 3:02 AM r t h       C ap e 7   Microbial Genetics 5. An operon consists of __________________, __________________, and __________________ and is associated with a regulatory gene. 6. In general, __________________ operons are inactive until the substrate of their genes’ polypeptides is present. 7. A daughter DNA molecule is composed of one original strand and one new strand because DNA replication is __________________. VISUALIZE IT! 1. On the figure below, label DNA polymerase I, DNA polymerase III, helicase, lagging strand, leading strand, ligase, nucleotide (triphosphate), Okazaki fragment, primase, replication fork, RNA primer, and stabilizing proteins. 8. A gene for antibiotic resistance can move horizontally among bacterial cells by __________________, __________________, and __________________. c b d a 9. __________________ are nucleotide sequences containing palindromes and genes for proteins that cut DNA strands. 10. __________________ __________________ is a recombination event that occurs during gamete formation in eukaryotes. e 11. __________________ RNA carries amino acids. f 12. __________________ RNA and __________________ RNA are antisense; that is, they are complementary to another nucleic acid molecule. g h i j k l Short Answer 1. How does the genotype of a bacterium determine its phenotype? Use the terms gene, mRNA, ribosome, and polypeptide in your answer. 2. This bacteriophage DNA molecule has been warmed. Label the portions that likely have a higher ratio of GC base pairs and the portions that have a higher ratio of AT base pairs. 2. List several ways in which eukaryotic messenger RNA differs from prokaryotic mRNA. 18. 3. Compare and contrast introns and exons. 4. Polypeptide synthesis requires large amounts of energy. How do cells regulate synthesis to conserve energy? Describe one specific example. 5. Describe the operon model of gene regulation. 6. Compare and contrast the structure and components of DNA and RNA in prokaryotes. 7. Besides the fact that it synthesizes RNA, how does RNA polymerase differ in function from DNA polymerase? SEM 8. Describe the formation and function of mRNA, rRNA, and tRNA in prokaryotes and eukaryotes. 20 nm 9. Describe how DNA is packaged in both prokaryotes and eukaryotes. 3. The drugs ddC and AZT are used to treat AIDS. 10. Explain the central dogma of genetics. 11. Compare and contrast the processes of transformation, transduction, and conjugation. NH2 12. Fill in the following table: O Process Purpose Beginning Point Replication Ending Point Origin or end of molecule P Transcription Translation HO Synthesis of polypeptides M07_BAUM7206_05_SE_C07.indd 234 romoter CH3 H O H H N N H O O HO H H H H H ddC (2¿,3¿-dideoxycytidine) CH3 H CH3 N N H O H H N3 H H AZT (3¿-azido-2¿,3¿-dideoxythymidine) Based on their chemical structures, what is their mode of action? 11/23/16 3:02 AM       Questions for Review A CRITIC L THINKING 1. If molecules of mRNA have the following nucleotide base sequences, what will be the sequence of amino acids in polypeptides synthesized by eukaryotic ribosomes? Remember, transcription starts with an AUG codon. a. AUGGGGAUACGCUACCCC b. CCGUACAUGCUAAUCCCU c. CCGAUGUAACCUCGAUCC d. AUGCGGUCAGCCCCGUGA 2. A scientist uses a molecule of DNA composed of nucleotides containing radioactive deoxyribose as a template for replication and transcription in a nonradioactive environment. What percentage of DNA strands will be radioactive after three DNA replication cycles? What percentage of RNA molecules will be radioactive? 3. Explain why an insertion of three nucleotides is less likely to result in a deleterious effect than an insertion of a single nucleotide. 4. How could scientists use siRNA to turn off a cancer-inducing gene? 5. The chromosome of Mycobacterium tuberculosis is 4,411,529 bp long. A scientist who isolates and counts the number of nucleotides in its DNA molecule discovers that there are 2,893,963 molecules of guanine. How many molecules of the other three nucleotides are in the original DNA? 6. Suppose that the E. coli gene for the lac operon repressor has a mutation that makes the repressor ineffective. What effect would this have on the bacterium’s ability to catabolize lactose? Would the mutant strain have an advantage over wild-type cells? Explain your answer. 7. A student claims that nucleotide analogs can be dangerously carcinogenic. Another student in the study group disagrees, insisting that nucleotide analogs are used to treat cancer. Explain why both students are correct. 8. Why is DNA polymerase so named? 9. Corynebacterium diphtheriae, the causative agent of diphtheria, secretes a toxin that enzymatically inactivates all molecules of elongation factor in a eukaryotic cell. What immediate and longterm effects does this have on cellular metabolism? M07_BAUM7206_05_SE_C07.indd 235 10. How can knowledge of nucleotide analogs be useful to a cancer researcher? 11. The endosymbiotic theory states that mitochondria and chloroplasts evolved from prokaryotes living within other prokaryotes (see p. 87). What aspects of the eukaryotic genome support this theory? What aspects do not support the theory? 12. Hydrogen bonds between complementary nucleotides are crucial to the structure of dsDNA because they hold the two strands together. Why couldn’t the two strands be effectively linked by covalent bonds? 13. On average, RNA polymerase makes one error for every 10,000 nucleotides it incorporates in RNA. By contrast, only one basepair error remains for every 10 billion bp during DNA replication. Explain why the accuracy of RNA transcription is not as critical as the accuracy of DNA replication. 14. We have seen that wobble makes the genetic code redundant in the third position for C and U. After reexamining the genetic code in Figure 7.12, state what other nucleotides in the third position appear to accommodate anticodon wobbling. 15. If a scientist synthesizes a DNA molecule with the nucleotide base sequence TACGGGGGAGGGGGAGGGGGA and then uses it for transcription and translation, what would be the amino acid sequence of the product? 16. What DNA nucleotide triplet codes for codon UGU? Identify a base-pair substitution that would produce a silent mutation at this codon. Identify a base-pair substitution that would result in a missense mutation at this codon. Identify a base-pair substitution that would produce a nonsense mutation at this codon. 17. Suppose you want to insert into your dog a gene that encodes a protein that protects dogs from heartworms. A dog’s cells are not competent, so they cannot take up the gene from the environment; but you have a plasmid, a competent bacterium, and a related (though incompetent) F + bacterium that lives as an intracellular parasite in dogs. Describe a possible scenario by which you could use natural processes to genetically alter your dog to be heartworm resistant. 11/23/16 3:02 AM M APP CONC EPT       Chapter 7   Microbial Genetics ING Using the following terms, fill in the following concept map that describes point mutations. Substituted Substitution mutation Nonsense mutation One or few nucleotide base pairs Silent mutation Incorrect amino acid substituted Inserted Insertion mutation Change in DNA sequence Deleted Effect on amino acid sequence Point mutations can be classified by affect 1 8 2 in which nucleotide base pairs may be 3 known as 6 4 5 9 known as known as in which there is Deletion mutation may result in M07_BAUM7206_05_SE_C07.indd 236 such as 7 No change in amino acid sequence 10 Missense mutations in which there is 11 in which there is Premature termination of polypeptide Frameshift 06/03/18 4:45 PM