Microbiology: Structure and Function of Cellular Genomes PDF

10.4 Structure and Function of Cellular Genomes 387 CHECK YOUR UNDERSTANDING What is the role of noncoding DNA? Extrachromosomal DNA Although most DNA is contained within a cell’s chromosomes, many cells have additional molecules of DNA outside the chromosomes, called extrachrom...

10.4 Structure and Function of Cellular Genomes 387 CHECK YOUR UNDERSTANDING What is the role of noncoding DNA? Extrachromosomal DNA Although most DNA is contained within a cell’s chromosomes, many cells have additional molecules of DNA outside the chromosomes, called extrachromosomal DNA, that are also part of its genome. The genomes of eukaryotic cells would also include the chromosomes from any organelles such as mitochondria and/or chloroplasts that these cells maintain (Figure 10.26). The maintenance of circular chromosomes in these organelles is a vestige of their prokaryotic origins and supports the endosymbiotic theory (see Foundations of Modern Cell Theory). In some cases, genomes of certain DNA viruses can also be maintained independently in host cells during latent viral infection. In these cases, these viruses are another form of extrachromosomal DNA. For example, the human papillomavirus (HPV) may be maintained in infected cells in this way. FIGURE 10.26 The genome of a eukaryotic cell consists of the chromosome housed in the nucleus, and extrachromosomal DNA found in the mitochondria (all cells) and chloroplasts (plants and algae). The cells shown in (b) represent cells obtained from a pap smear. The cells on the left are normal squamous cells whereas the cells on the right are infected with human papillomavirus and show enlarged nuclei with increased staining (hyperchromasia). Besides chromosomes, some prokaryotes also have smaller loops of DNA called plasmids that may contain one or a few genes not essential for normal growth (Figure 3.12). Bacteria can exchange these plasmids with other bacteria in a process known as horizontal gene transfer (HGT). The exchange of genetic material on plasmids sometimes provides microbes with new genes beneficial for growth and survival under special conditions. In some cases, genes obtained from plasmids may have clinical implications, encoding virulence factors that give a microbe the ability to cause disease or make a microbe resistant to certain antibiotics. Plasmids are also used heavily in genetic engineering and biotechnology as a way to move genes from one cell to another. The role of plasmids in horizontal gene transfer and biotechnology will be discussed further in Mechanisms of Microbial Genetics and Modern Applications of Microbial Genetics. CHECK YOUR UNDERSTANDING How are plasmids involved in antibiotic resistance? CASE IN POINT Lethal Plasmids Maria, a 20-year-old anthropology student from Texas, recently became ill in the African nation of Botswana, where she was conducting research as part of a study-abroad program. Maria’s research was focused on 388 10 Biochemistry of the Genome traditional African methods of tanning hides for the production of leather. Over a period of three weeks, she visited a tannery daily for several hours to observe and participate in the tanning process. One day, after returning from the tannery, Maria developed a fever, chills, and a headache, along with chest pain, muscle aches, nausea, and other flu-like symptoms. Initially, she was not concerned, but when her fever spiked and she began to cough up blood, her African host family became alarmed and rushed her to the hospital, where her condition continued to worsen. After learning about her recent work at the tannery, the physician suspected that Maria had been exposed to anthrax. He ordered a chest X-ray, a blood sample, and a spinal tap, and immediately started her on a course of intravenous penicillin. Unfortunately, lab tests confirmed the physician’s presumptive diagnosis. Maria’s chest X- ray exhibited pleural effusion, the accumulation of fluid in the space between the pleural membranes, and a Gram stain of her blood revealed the presence of gram-positive, rod-shaped bacteria in short chains, consistent with Bacillus anthracis. Blood and bacteria were also shown to be present in her cerebrospinal fluid, indicating that the infection had progressed to meningitis. Despite supportive treatment and aggressive antibiotic therapy, Maria slipped into an unresponsive state and died three days later. Anthrax is a disease caused by the introduction of endospores from the gram-positive bacterium B. anthracis into the body. Once infected, patients typically develop meningitis, often with fatal results. In Maria’s case, she inhaled the endospores while handling the hides of animals that had been infected. The genome of B. anthracis illustrates how small structural differences can lead to major differences in virulence. In 2003, the genomes of B. anthracis and Bacillus cereus, a similar but less pathogenic bacterium of 21 the same genus, were sequenced and compared. Researchers discovered that the 16S rRNA gene sequences of these bacteria are more than 99% identical, meaning that they are actually members of the same species despite their traditional classification as separate species. Although their chromosomal sequences also revealed a great deal of similarity, several virulence factors of B. anthracis were found to be encoded on two large plasmids not found in B. cereus. The plasmid pX01 encodes a three-part toxin that suppresses the host immune system, whereas the plasmid pX02 encodes a capsular polysaccharide that further protects the bacterium from the host immune system (Figure 10.27). Since B. cereus lacks these plasmids, it does not produce these virulence factors, and although it is still pathogenic, it is typically associated with mild cases of diarrhea from which the body can quickly recover. Unfortunately for Maria, the presence of these toxin-encoding plasmids in B. anthracis gives it its lethal virulence. FIGURE 10.27 Genome sequencing of Bacillus anthracis and its close relative B. cereus reveals that the pathogenicity of B. anthracis is due to the maintenance of two plasmids, pX01 and pX02, which encode virulence factors. What do you think would happen to the pathogenicity of B. anthracis if it lost one or both of its plasmids? CLINICAL FOCUS Resolution Within 24 hours, the results of the diagnostic test analysis of Alex’s stool sample revealed that it was positive for 21 N. Ivanova et al. “Genome Sequence of Bacillus cereus and Comparative Analysis with Bacillus anthracis.” Nature 423 no. 6935 (2003):87–91. Access for free at openstax.org  10.4 Structure and Function of Cellular Genomes 389 heat-labile enterotoxin (LT), heat-stabile enterotoxin (ST), and colonization factor (CF), confirming the hospital physician’s suspicion of ETEC. During a follow-up with Alex’s family physician, this physician noted that Alex’s symptoms were not resolving quickly and he was experiencing discomfort that was preventing him from returning to classes. The family physician prescribed Alex a course of ciprofloxacin to resolve his symptoms. Fortunately, the ciprofloxacin resolved Alex’s symptoms within a few days. Alex likely got his infection from ingesting contaminated food or water. Emerging industrialized countries like Mexico are still developing sanitation practices that prevent the contamination of water with fecal material. Travelers in such countries should avoid the ingestion of undercooked foods, especially meats, seafood, vegetables, and unpasteurized dairy products. They should also avoid use of water that has not been treated; this includes drinking water, ice cubes, and even water used for brushing teeth. Using bottled water for these purposes is a good alternative. Good hygiene (handwashing) can also aid the prevention of an ETEC infection. Alex had not been careful about his food or water consumption, which led to his illness. Alex’s symptoms were very similar to those of cholera, caused by the gram-negative bacterium Vibrio cholerae, which also produces a toxin similar to ST and LT. At some point in the evolutionary history of ETEC, a nonpathogenic strain of E. coli similar to those typically found in the gut may have acquired the genes encoding the ST and LT toxins from V. cholerae. The fact that the genes encoding those toxins are encoded on extrachromosomal plasmids in ETEC supports the idea that these genes were acquired by E. coli and are likely maintained in bacterial populations through horizontal gene transfer. Go back to the previous Clinical Focus box. Viral Genomes Viral genomes exhibit significant diversity in structure. Some viruses have genomes that consist of DNA as their genetic material. This DNA may be single stranded, as exemplified by human parvoviruses, or double stranded, as seen in the herpesviruses and poxviruses. Additionally, although all cellular life uses DNA as its genetic material, some viral genomes are made of either single-stranded or double-stranded RNA molecules, as we have discussed. Viral genomes are typically smaller than most bacterial genomes, encoding only a few genes, because they rely on their hosts to carry out many of the functions required for their replication. The diversity of viral genome structures and their implications for viral replication life cycles are discussed in more detail in The Viral Life Cycle. CHECK YOUR UNDERSTANDING Why do viral genomes vary widely among viruses? MICRO CONNECTIONS Genome Size Matters There is great variation in size of genomes among different organisms. Most eukaryotes maintain multiple chromosomes; humans, for example have 23 pairs, giving them 46 chromosomes. Despite being large at 3 billion base pairs, the human genome is far from the largest genome. Plants often maintain very large genomes, up to 150 billion base pairs, and commonly are polyploid, having multiple copies of each chromosome. The size of bacterial genomes also varies considerably, although they tend to be smaller than eukaryotic genomes (Figure 10.28). Some bacterial genomes may be as small as only 112,000 base pairs. Often, the size of a bacterium’s genome directly relates to how much the bacterium depends on its host for survival. When a bacterium relies on the host cell to carry out certain functions, it loses the genes encoding the abilities to carry out those functions itself. These types of bacterial endosymbionts are reminiscent of the prokaryotic origins of mitochondria and chloroplasts. From a clinical perspective, obligate and facultative intracellular pathogens also tend to have small genomes (some 390 10 Biochemistry of the Genome around 1 million base pairs). Because host cells can supply most of their nutrients, they tend to have a reduced number of genes encoding metabolic functions, making their cultivation in the laboratory difficult if not impossible Due to their small sizes, the genomes of organisms like Mycoplasma genitalium (580,000 base pairs), Chlamydia trachomatis (1.0 million), Rickettsia prowazekii (1.1 million), and Treponema pallidum (1.1 million) were some of the earlier bacterial genomes sequenced. Respectively, these pathogens cause urethritis and pelvic inflammation, chlamydia, typhus, and syphilis. Whereas obligate intracellular pathogens have unusually small genomes, other bacteria with a great variety of metabolic and enzymatic capabilities have unusually large bacterial genomes. Pseudomonas aeruginosa, for example, is a bacterium commonly found in the environment and is able to grow on a wide range of substrates. Its genome contains 6.3 million base pairs, giving it a high metabolic ability and the ability to produce virulence factors that cause several types of opportunistic infections. Interestingly, there has been significant variability in genome size in viruses as well, ranging from 3,500 base pairs to 2.5 million base pairs, significantly exceeding the size of many bacterial genomes. The great variation observed in viral genome sizes further contributes to the great diversity of viral genome characteristics already discussed. FIGURE 10.28 There is great variability as well as overlap among the genome sizes of various groups of organisms and viruses. LINK TO LEARNING Visit the genome database (https://www.openstax.org/l/22NCBIgendata) of the National Center for Biotechnology Information (NCBI) to see the genomes that have been sequenced and their sizes. Access for free at openstax.org  10 Summary 391 Summary 10.1 Using Microbiology to Discover the each of which contains a pentose sugar, a Secrets of Life phosphate group, and a nitrogenous base. Deoxyribonucleotides within DNA contain DNA was discovered and characterized long deoxyribose as the pentose sugar. before its role in heredity was understood. DNA contains the pyrimidines cytosine and Microbiologists played significant roles in thymine, and the purines adenine and guanine. demonstrating that DNA is the hereditary Nucleotides are linked together by information found within cells. phosphodiester bonds between the 5ʹ phosphate In the 1850s and 1860s, Gregor Mendel group of one nucleotide and the 3ʹ hydroxyl experimented with true-breeding garden peas to group of another. A nucleic acid strand has a demonstrate the heritability of specific free phosphate group at the 5ʹ end and a free observable traits. hydroxyl group at the 3ʹ end. In 1869, Friedrich Miescher isolated and purified Chargaff discovered that the amount of adenine a compound rich in phosphorus from the nuclei is approximately equal to the amount of thymine of white blood cells; he named the compound in DNA, and that the amount of the guanine is nuclein. Miescher’s student Richard Altmann approximately equal to cytosine. These discovered its acidic nature, renaming it nucleic relationships were later determined to be due to acid. Albrecht Kossell characterized the complementary base pairing. nucleotide bases found within nucleic acids. Watson and Crick, building on the work of Although Walter Sutton and Theodor Boveri Chargaff, Franklin and Gosling, and Wilkins, proposed the Chromosomal Theory of proposed the double helix model and base Inheritance in 1902, it was not scientifically pairing for DNA structure. demonstrated until the 1915 publication of the DNA is composed of two complementary strands work of Thomas Hunt Morgan and his colleagues. oriented antiparallel to each other with the Using Acetabularia, a large algal cell, as his phosphodiester backbones on the exterior of model system, Joachim Hämmerling the molecule. The nitrogenous bases of each demonstrated in the 1930s and 1940s that the strand face each other and complementary nucleus was the location of hereditary bases hydrogen bond to each other, stabilizing information in these cells. the double helix. In the 1940s, George Beadle and Edward Tatum Heat or chemicals can break the hydrogen bonds used the mold Neurospora crassa to show that between complementary bases, denaturing DNA. each protein’s production was under the control Cooling or removing chemicals can lead to of a single gene, demonstrating the “one renaturation or reannealing of DNA by allowing gene–one enzyme” hypothesis. hydrogen bonds to reform between In 1928, Frederick Griffith showed that dead complementary bases. encapsulated bacteria could pass genetic DNA stores the instructions needed to build and information to live nonencapsulated bacteria and control the cell. This information is transmitted transform them into harmful strains. In 1944, from parent to offspring through vertical gene Oswald Avery, Colin McLeod, and Maclyn McCarty transfer. identified the compound as DNA. The nature of DNA as the molecule that stores 10.3 Structure and Function of RNA genetic information was unequivocally Ribonucleic acid (RNA) is typically single demonstrated in the experiment of Alfred stranded and contains ribose as its pentose Hershey and Martha Chase published in 1952. sugar and the pyrimidine uracil instead of Labeled DNA from bacterial viruses entered and thymine. An RNA strand can undergo significant infected bacterial cells, giving rise to more viral intramolecular base pairing to take on a three- particles. The labeled protein coats did not dimensional structure. participate in the transmission of genetic There are three main types of RNA, all involved in information. protein synthesis. 10.2 Structure and Function of DNA Messenger RNA (mRNA) serves as the intermediary between DNA and the synthesis of Nucleic acids are composed of nucleotides, protein products during translation. 392 10 Review Questions Ribosomal RNA (rRNA) is a type of stable RNA Prokaryotes are typically haploid, usually having that is a major constituent of ribosomes. It a single circular chromosome found in the ensures the proper alignment of the mRNA and nucleoid. Eukaryotes are diploid; DNA is the ribosomes during protein synthesis and organized into multiple linear chromosomes catalyzes the formation of the peptide bonds found in the nucleus. between two aligned amino acids during protein Supercoiling and DNA packaging using DNA synthesis. binding proteins allows lengthy molecules to fit Transfer RNA (tRNA) is a small type of stable inside a cell. Eukaryotes and archaea use histone RNA that carries an amino acid to the proteins, and bacteria use different proteins with corresponding site of protein synthesis in the similar function. ribosome. It is the base pairing between the Prokaryotic and eukaryotic genomes both tRNA and mRNA that allows for the correct contain noncoding DNA, the function of which is amino acid to be inserted in the polypeptide not well understood. Some noncoding DNA chain being synthesized. appears to participate in the formation of small Although RNA is not used for long-term genetic noncoding RNA molecules that influence gene information in cells, many viruses do use RNA as expression; some appears to play a role in their genetic material. maintaining chromosomal structure and in DNA packaging. 10.4 Structure and Function of Cellular Extrachromosomal DNA in eukaryotes includes Genomes the chromosomes found within organelles of The entire genetic content of a cell is its genome. prokaryotic origin (mitochondria and Genes code for proteins, or stable RNA chloroplasts) that evolved by endosymbiosis. molecules, each of which carries out a specific Some viruses may also maintain themselves function in the cell. extrachromosomally. Although the genotype that a cell possesses Extrachromosomal DNA in prokaryotes is remains constant, expression of genes is commonly maintained as plasmids that encode a dependent on environmental conditions. few nonessential genes that may be helpful A phenotype is the observable characteristics of under specific conditions. Plasmids can be a cell (or organism) at a given point in time and spread through a bacterial community by results from the complement of genes currently horizontal gene transfer. being used. Viral genomes show extensive variation and may The majority of genetic material is organized into be composed of either RNA or DNA, and may be chromosomes that contain the DNA that controls either double or single stranded. cellular activities. Review Questions Multiple Choice 1. Frederick Griffith infected mice with a identify the location of genetic material? combination of dead R and live S bacterial A. It lacks a nuclear membrane. strains. What was the outcome, and why did it B. It self-fertilizes. occur? C. It is a large, asymmetrical, single cell easy A. The mice will live. Transformation was not to see with the naked eye. required. D. It makes a protein capsid. B. The mice will die. Transformation of genetic material from R to S was required. 3. Which of the following best describes the results C. The mice will live. Transformation of from Hershey and Chase’s experiment using genetic material from S to R was required. bacterial viruses with 35S-labeled proteins or 32P-labeled DNA that are consistent with protein D. The mice will die. Transformation was not required. being the molecule responsible for hereditary? A. After infection with the 35S-labeled 2. Why was the alga Acetabularia a good model viruses and centrifugation, only the pellet organism for Joachim Hämmerling to use to would be radioactive. Access for free at openstax.org  10 Review Questions 393 B. After infection with the 35S-labeled D. Base pairing occurs at the interior of the viruses and centrifugation, both the pellet double helix. and the supernatant would be radioactive. C. After infection with the 32P-labeled 9. If a DNA strand contains the sequence 5ʹ- viruses and centrifugation, only the pellet ATTCCGGATCGA-3ʹ, which of the following is the would be radioactive. sequence of the complementary strand of DNA? D. After infection with the 32P-labeled A. 5ʹ-TAAGGCCTAGCT-3ʹ viruses and centrifugation, both the pellet B. 5ʹ-ATTCCGGATCGA-3ʹ and the supernatant would be radioactive. C. 3ʹ-TAACCGGTACGT-5ʹ D. 5ʹ-TCGATCCGGAAT-3ʹ 4. Which method did Morgan and colleagues use to show that hereditary information was carried on 10. During denaturation of DNA, which of the chromosomes? following happens? A. statistical predictions of the outcomes of A. Hydrogen bonds between complementary crosses using true-breeding parents bases break. B. correlations between microscopic B. Phosphodiester bonds break within the observations of chromosomal movement sugar-phosphate backbone. and the characteristics of offspring C. Hydrogen bonds within the sugar- C. transformation of nonpathogenic bacteria phosphate backbone break. to pathogenic bacteria D. Phosphodiester bonds between D. mutations resulting in distinct defects in complementary bases break. metabolic enzymatic pathways 11. Which of the following types of RNA codes for a 5. According to Beadle and Tatum’s “one gene–one protein? enzyme” hypothesis, which of the following a. dsRNA enzymes will eliminate the transformation of b. mRNA hereditary material from pathogenic bacteria to c. rRNA nonpathogenic bacteria? d. tRNA A. carbohydrate-degrading enzymes B. proteinases 12. A nucleic acid is purified from a mixture. The C. ribonucleases molecules are relatively small, contain uracil, D. deoxyribonucleases and most are covalently bound to an amino acid. Which of the following was purified? 6. Which of the following is not found within DNA? A. DNA A. thymine B. mRNA B. phosphodiester bonds C. rRNA C. complementary base pairing D. tRNA D. amino acids 13. Which of the following types of RNA is known for 7. If 30% of the bases within a DNA molecule are its catalytic abilities? adenine, what is the percentage of thymine? A. dsRNA A. 20% B. mRNA B. 25% C. rRNA C. 30% D. tRNA D. 35% 14. Ribosomes are composed of rRNA and what 8. Which of the following statements about base other component? pairing in DNA is incorrect? A. protein A. Purines always base pairs with B. carbohydrates pyrimidines. C. DNA B. Adenine binds to guanine. D. mRNA C. Base pairs are stabilized by hydrogen bonds. 15. Which of the following may use RNA as its 394 10 Review Questions genome? A. genotype A. a bacterium B. phenotype B. an archaeon C. change in DNA base composition C. a virus D. adaptation to the environment D. a eukaryote 19. Which of the following genes would not likely be 16. Which of the following correctly describes the encoded on a plasmid? structure of the typical eukaryotic genome? A. genes encoding toxins that damage host A. diploid tissue B. linear B. genes encoding antibacterial resistance C. singular C. gene encoding enzymes for glycolysis D. double stranded D. genes encoding enzymes for the degradation of an unusual substrate 17. Which of the following is typically found as part of the prokaryotic genome? 20. Histones are DNA binding proteins that are A. chloroplast DNA important for DNA packaging in which of the B. linear chromosomes following? C. plasmids A. double-stranded and single-stranded DNA D. mitochondrial DNA viruses B. archaea and bacteria 18. Serratia marcescens cells produce a red pigment C. bacteria and eukaryotes at room temperature. The red color of the D. eukaryotes and archaea colonies is an example of which of the following? True/False 21. The work of Rosalind Franklin and R.G. Gosling 24. Double-stranded RNA is commonly found inside was important in demonstrating the helical cells. nature of DNA. 25. Within an organism, phenotypes may change 22. The A-T base pair has more hydrogen bonding while genotypes remain constant. than the C-G base pair. 26. Noncoding DNA has no biological purpose. 23. Ribosomes are composed mostly of RNA. Matching 27. Match the correct molecule with its description: ___tRNA A. is a major component of ___rRNA ribosome ___mRNA B. is a copy of the information in a gene C. carries an amino acid to the ribosome Fill in the Blank 28. The element ____________ is unique to nucleic 30. The end of a nucleic acid strand with a free acids compared with other macromolecules. phosphate group is called the ________. 29. In the late 1800s and early 1900s, the 31. Plasmids are typically transferred among macromolecule thought to be responsible for members of a bacterial community by ________ heredity was ______________. gene transfer. Access for free at openstax.org  10 Review Questions 395 Short Answer 32. Why do bacteria and viruses make good model 40. How do complementary base pairs contribute to systems for various genetic studies? intramolecular base pairing within an RNA molecule? 33. Why was nucleic acid disregarded for so long as the molecule responsible for the transmission of 41. If an antisense RNA has the sequence hereditary information? 5ʹAUUCGAAUGC3ʹ, what is the sequence of the mRNA to which it will bind? Be sure to label the 34. Bacteriophages inject their genetic material into 5ʹ and 3ʹ ends of the molecule you draw. host cells, whereas animal viruses enter host cells completely. Why was it important to use a 42. Why does double-stranded RNA (dsRNA) bacteriophage in the Hershey–Chase stimulate RNA interference? experiment rather than an animal virus? 43. What are some differences in chromosomal 35. What is the role of phosphodiester bonds within structures between prokaryotes and the sugar-phosphate backbone of DNA? eukaryotes? 36. What is meant by the term “antiparallel?” 44. How do prokaryotes and eukaryotes manage to fit their lengthy DNA inside of cells? Why is this 37. Why is DNA with a high GC content more difficult necessary? to denature than that with a low GC content? 45. What are some functions of noncoding DNA? 38. What are the differences between DNA nucleotides and RNA nucleotides? 46. In the chromatin of eukaryotic cells, which regions of the chromosome would you expect to 39. How is the information stored within the base be more compact: the regions that contain sequence of DNA used to determine a cell’s genes being actively copied into RNA or those properties? that contain inactive genes? Critical Thinking 47. In the figure shown, if the nuclei were contained within the stalks of Acetabularia, what types of caps would you expect from the pictured grafts? 50. In considering the structure of the DNA double helix, how would you expect the structure to differ if there was base pairing between two purines? Between two pyrimidines? 51. Identify the location of mRNA, rRNA, and tRNA in the figure. 48. Why are Hershey and Chase credited with identifying DNA as the carrier of heredity even though DNA had been discovered many years before? 49. A certain DNA sample is found to have a makeup consisting of 22% thymine. Use Chargaff’s rules to fill in the percentages for the other three nitrogenous bases. 52. Why does it make sense that tRNA and rRNA molecules are more stable than mRNA molecules? 396 10 Review Questions 53. A new type of bacteriophage has been isolated bacteriophage is A (15%), C (20%), T (35%), and and you are in charge of characterizing its G (30%). What can you conclude about the genome. The base composition of the genome of the virus? Access for free at openstax.org CHAPTER 11 Mechanisms of Microbial Genetics FIGURE 11.1 Escherichia coli (left) may not appear to have much in common with an elephant (right), but the genetic blueprints for these vastly different organisms are both encoded in DNA. (credit left: modification of work by NIAID; credit right: modification of work by Tom Lubbock) CHAPTER OUTLINE 11.1 The Functions of Genetic Material 11.2 DNA Replication 11.3 RNA Transcription 11.4 Protein Synthesis (Translation) 11.5 Mutations 11.6 How Asexual Prokaryotes Achieve Genetic Diversity 11.7 Gene Regulation: Operon Theory INTRODUCTION In 1954, French scientist and future Nobel laureate Jacques Monod (1910–1976) famously said, “What is true in E. coli is true in the elephant,” suggesting that the biochemistry of life was maintained throughout evolution and is shared in all forms of known life. Since Monod’s famous statement, we have learned a great deal about the mechanisms of gene regulation, expression, and replication in living cells. All cells use DNA for information storage, share the same genetic code, and use similar mechanisms to replicate and express it. Although many aspects of genetics are universally shared, variations do exist among contemporary genetic systems. We now know that within the shared overall theme of the genetic mechanism, there are significant differences among the three domains of life: Eukarya, Archaea, and Bacteria. Additionally, viruses, cellular parasites but not themselves living cells, show dramatic variation in their genetic material and the replication and gene expression processes. Some of these differences have allowed us to engineer clinical tools such as antibiotics and antiviral drugs that specifically inhibit the reproduction of pathogens yet are harmless to their hosts. 11.1 The Functions of Genetic Material LEARNING OBJECTIVES By the end of this section, you will be able to: Explain the two functions of the genome Explain the meaning of the central dogma of molecular biology Differentiate between genotype and phenotype and explain how environmental factors influence phenotype 398 11 Mechanisms of Microbial Genetics CLINICAL FOCUS Part 1 Mark is 60-year-old software engineer who suffers from type II diabetes, which he monitors and keeps under control largely through diet and exercise. One spring morning, while doing some gardening, he scraped his lower leg while walking through blackberry brambles. He continued working all day in the yard and did not bother to clean the wound and treat it with antibiotic ointment until later that evening. For the next 2 days, his leg became increasingly red, swollen, and warm to the touch. It was sore not only on the surface, but deep in the muscle. After 24 hours, Mark developed a fever and stiffness in the affected leg. Feeling increasingly weak, he called a neighbor, who drove him to the emergency department. Did Mark wait too long to seek medical attention? At what point do his signs and symptoms warrant seeking medical attention? What types of infections or other conditions might be responsible for Mark’s symptoms? Jump to the next Clinical Focus box. DNA serves two essential functions that deal with cellular information. First, DNA is the genetic material responsible for inheritance and is passed from parent to offspring for all life on earth. To preserve the integrity of this genetic information, DNA must be replicated with great accuracy, with minimal errors that introduce changes to the DNA sequence. A genome contains the full complement of DNA within a cell and is organized into smaller, discrete units called genes that are arranged on chromosomes and plasmids. The second function of DNA is to direct and regulate the construction of the proteins necessary to a cell for growth and reproduction in a particular cellular environment. A gene is composed of DNA that is “read” or transcribed to produce an RNA molecule during the process of transcription. One major type of RNA molecule, called messenger RNA (mRNA), provides the information for the ribosome to catalyze protein synthesis in a process called translation. The processes of transcription and translation are collectively referred to as gene expression. Gene expression is the synthesis of a specific protein with a sequence of amino acids that is encoded in the gene. The flow of genetic information from DNA to RNA to protein is described by the central dogma (Figure 11.2). This central dogma of molecular biology further elucidates the mechanism behind Beadle and Tatum’s “one gene-one enzyme” hypothesis (see Using Microorganisms to Discover the Secrets of Life). Each of the processes of replication, transcription, and translation includes the stages of 1) initiation, 2) elongation (polymerization), and 3) termination. These stages will be described in more detail in this chapter. FIGURE 11.2 The central dogma states that DNA encodes messenger RNA, which, in turn, encodes protein. A cell’s genotype is the full collection of genes it contains, whereas its phenotype is the set of observable characteristics that result from those genes. The phenotype is the product of the array of proteins being produced by the cell at a given time, which is influenced by the cell’s genotype as well as interactions with the cell’s environment. Genes code for proteins that have functions in the cell. Production of a specific protein encoded by an individual gene often results in a distinct phenotype for the cell compared with the phenotype without that protein. For this reason, it is also common to refer to the genotype of an individual gene and its phenotype. Although a cell’s genotype remains constant, not all genes are used to direct the production of their proteins simultaneously. Cells carefully regulate expression of their genes, only using genes to make specific proteins when those proteins are needed (Figure 11.3). Access for free at openstax.org  11.1 The Functions of Genetic Material 399 FIGURE 11.3 Phenotype is determined by the specific genes within a genotype that are expressed under specific conditions. Although multiple cells may have the same genotype, they may exhibit a wide range of phenotypes resulting from differences in patterns of gene expression in response to different environmental conditions. CHECK YOUR UNDERSTANDING What are the two functions of DNA? Distinguish between the genotype and phenotype of a cell. How can cells have the same genotype but differ in their phenotype? EYE ON ETHICS Use and Abuse of Genome Data Why can some humans harbor opportunistic pathogens like Haemophilus influenzae, Staphylococcus aureus, or Streptococcus pyogenes, in their upper respiratory tracts but remain asymptomatic carriers, while other individuals become seriously ill when infected? There is evidence suggesting that differences in susceptibility to infection between patients may be a result, at least in part, of genetic differences between human hosts. For example, genetic differences in human leukocyte antigens (HLAs) and red blood cell antigens among hosts have been implicated in different immune responses and resulting disease progression from infection with H. influenzae. Because the genetic interplay between pathogen and host may contribute to disease outcomes, understanding differences in genetic makeup between individuals may be an important clinical tool. Ecological genomics is a relatively new field that seeks to understand how the genotypes of different organisms interact with each other in nature. The field answers questions about how gene expression of one organism affects gene expression of another. Medical applications of ecological genomics will focus on how pathogens interact with specific individuals, as opposed to humans in general. Such analyses would allow medical professionals to use knowledge of an individual’s genotype to apply more individualized plans for treatment and prevention of disease. With the advent of next-generation sequencing, it is relatively easy to obtain the entire genomic sequences of 1 pathogens; a bacterial genome can be sequenced in as little as a day. The speed and cost of sequencing the human genome has also been greatly reduced and, already, individuals can submit samples to receive extensive reports on their personal genetic traits, including ancestry and carrier status for various genetic diseases. As sequencing technologies progress further, such services will continue to become less expensive, more extensive, and quicker. However, as this day quickly approaches, there are many ethical concerns with which society must grapple. For example, should genome sequencing be a standard practice for everybody? Should it be required by law or by employers if it will lower health-care costs? If one refuses genome sequencing, do they forfeit their right to health insurance coverage? For what purposes should the data be used? Who should oversee proper use of these data? If genome sequencing reveals predisposition to a particular disease, do insurance companies have the right to increase rates? Will employers treat an employee differently? Knowing that environmental influences 1 D.J. Edwards, K.E. Holt. “Beginner’s Guide to Comparative Bacterial Genome Analysis Using Next-Generation Sequence Data.” Microbial Informatics and Experimentation 3 no. 1 (2013):2. 400 11 Mechanisms of Microbial Genetics also affect disease development, how should the data on the presence of a particular disease-causing allele in an individual be used ethically? The Genetic Information Nondiscrimination Act of 2008 (GINA) currently prohibits discriminatory practices based on genetic information by both health insurance companies and employers. However, GINA does not cover life, disability, or long-term care insurance policies. Clearly, all members of society must continue to engage in conversations about these issues so that such genomic data can be used to improve health care while simultaneously protecting an individual’s rights. 11.2 DNA Replication LEARNING OBJECTIVES By the end of this section, you will be able to: Explain the meaning of semiconservative DNA replication Explain why DNA replication is bidirectional and includes both a leading and lagging strand Explain why Okazaki fragments are formed Describe the process of DNA replication and the functions of the enzymes involved Identify the differences between DNA replication in bacteria and eukaryotes Explain the process of rolling circle replication The elucidation of the structure of the double helix by James Watson and Francis Crick in 1953 provided a hint as to how DNA is copied during the process of replication. Separating the strands of the double helix would provide two templates for the synthesis of new complementary strands, but exactly how new DNA molecules were constructed was still unclear. In one model, semiconservative replication, the two strands of the double helix separate during DNA replication, and each strand serves as a template from which the new complementary strand is copied; after replication, each double-stranded DNA includes one parental or “old” strand and one “new” strand. There were two competing models also suggested: conservative and dispersive, which are shown in Figure 11.4. FIGURE 11.4 There were three models suggested for DNA replication. In the conservative model, parental DNA strands (blue) remained associated in one DNA molecule while new daughter strands (red) remained associated in newly formed DNA molecules. In the semiconservative model, parental strands separated and directed the synthesis of a daughter strand, with each resulting DNA molecule being a hybrid of a parental strand and a daughter strand. In the dispersive model, all resulting DNA strands have regions of double- stranded parental DNA and regions of double-stranded daughter DNA. Matthew Meselson (1930–) and Franklin Stahl (1929–) devised an experiment in 1958 to test which of these models correctly represents DNA replication (Figure 11.5). They grew E. coli for several generations in a medium Access for free at openstax.org  11.2 DNA Replication 401 containing a “heavy” isotope of nitrogen (15N) that was incorporated into nitrogenous bases and, eventually, into the DNA. This labeled the parental DNA. The E. coli culture was then shifted into a medium containing 14N and allowed to grow for one generation. The cells were harvested and the DNA was isolated. The DNA was separated by ultracentrifugation, during which the DNA formed bands according to its density. DNA grown in 15N would be expected to form a band at a higher density position than that grown in 14N. Meselson and Stahl noted that after one generation of growth in 14N, the single band observed was intermediate in position in between DNA of cells grown exclusively in 15N or 14N. This suggested either a semiconservative or dispersive mode of replication. Some cells were allowed to grow for one more generation in 14N and spun again. The DNA harvested from cells grown for two generations in 14N formed two bands: one DNA band was at the intermediate position between 15N and 14N, and the other corresponded to the band of 14N DNA. These results could only be explained if DNA replicates in a semiconservative manner. If DNA replication was dispersive, a single purple band positioned closer to the red 1414 would have been observed, as more 14 was added in a dispersive manner to replace 15. Therefore, the other two models were ruled out. As a result of this experiment, we now know that during DNA replication, each of the two strands that make up the double helix serves as a template from which new strands are copied. The new strand will be complementary to the parental or “old” strand. The resulting DNA molecules have the same sequence and are divided equally into the two daughter cells. FIGURE 11.5 Meselson and Stahl experimented with E. coli grown first in heavy nitrogen (15N) then in 14N. DNA grown in 15N (blue band) was heavier than DNA grown in 14N (red band), and sedimented to a lower level on ultracentrifugation. After one round of replication, the DNA sedimented halfway between the 15N and 14N levels (purple band), ruling out the conservative model of replication. After a second round of replication, the dispersive model of replication was ruled out. These data supported the semiconservative replication model. CHECK YOUR UNDERSTANDING What would have been the conclusion of Meselson and Stahl’s experiment if, after the first generation, they had found two bands of DNA? 402 11 Mechanisms of Microbial Genetics DNA Replication in Bacteria DNA replication has been well studied in bacteria primarily because of the small size of the genome and the mutants that are available. E. coli has 4.6 million base pairs (Mbp) in a single circular chromosome and all of it is replicated in approximately 42 minutes, starting from a single origin of replication and proceeding around the circle bidirectionally (i.e., in both directions). This means that approximately 1000 nucleotides are added per second. The process is quite rapid and occurs with few errors. DNA replication uses a large number of proteins and enzymes (Table 11.1). One of the key players is the enzyme DNA polymerase, also known as DNA pol. In bacteria, three main types of DNA polymerases are known: DNA pol I, DNA pol II, and DNA pol III. It is now known that DNA pol III is the enzyme required for DNA synthesis; DNA pol I and DNA pol II are primarily required for repair. DNA pol III adds deoxyribonucleotides each complementary to a nucleotide on the template strand, one by one to the 3’-OH group of the growing DNA chain. The addition of these nucleotides requires energy. This energy is present in the bonds of three phosphate groups attached to each nucleotide (a triphosphate nucleotide), similar to how energy is stored in the phosphate bonds of adenosine triphosphate (ATP) (Figure 11.6). When the bond between the phosphates is broken and diphosphate is released, the energy released allows for the formation of a covalent phosphodiester bond by dehydration synthesis between the incoming nucleotide and the free 3’-OH group on the growing DNA strand. FIGURE 11.6 This structure shows the guanosine triphosphate deoxyribonucleotide that is incorporated into a growing DNA strand by cleaving the two end phosphate groups from the molecule and transferring the energy to the sugar phosphate bond. The other three nucleotides form analogous structures. Initiation The initiation of replication occurs at specific nucleotide sequence called the origin of replication, where various proteins bind to begin the replication process. E. coli has a single origin of replication (as do most prokaryotes), called oriC, on its one chromosome. The origin of replication is approximately 245 base pairs long and is rich in adenine-thymine (AT) sequences. Some of the proteins that bind to the origin of replication are important in making single-stranded regions of DNA accessible for replication. Chromosomal DNA is typically wrapped around histones (in eukaryotes and archaea) or histone-like proteins (in bacteria), and is supercoiled, or extensively wrapped and twisted on itself. This packaging makes the information in the DNA molecule inaccessible. However, enzymes called topoisomerases change the shape and supercoiling of the chromosome. For bacterial DNA replication to begin, the supercoiled chromosome is relaxed by topoisomerase II, also called DNA gyrase. An enzyme called helicase then separates the DNA strands by breaking the hydrogen bonds between the nitrogenous base pairs. Recall that AT sequences have fewer hydrogen bonds and, hence, have weaker interactions than guanine-cytosine (GC) sequences. These enzymes require ATP hydrolysis. As the DNA opens up, Y-shaped structures called replication forks are formed. Two replication forks are formed at the origin of replication, allowing for bidirectional replication and formation of a structure that looks like a bubble when viewed with a transmission electron microscope; as a result, this structure is called a replication bubble. The DNA near each replication fork is coated with single-stranded binding proteins to prevent the single- stranded DNA from rewinding into a double helix. Once single-stranded DNA is accessible at the origin of replication, DNA replication can begin. However, DNA pol III is able to add nucleotides only in the 5’ to 3’ direction (a new DNA strand can be only extended in this direction). This is because DNA polymerase requires a free 3’-OH group to which it can add nucleotides by forming a covalent phosphodiester bond between the 3’-OH end and the 5’ phosphate of the next nucleotide. This also means that it cannot add nucleotides if a free 3’-OH group is not available, which is the case for a single strand of DNA. The problem is solved with the help of an RNA sequence that provides the free 3’-OH end. Because this sequence allows the start of DNA synthesis, it is appropriately called the primer. The primer is five to 10 nucleotides long and Access for free at openstax.org  11.2 DNA Replication 403 complementary to the parental or template DNA. It is synthesized by RNA primase, which is an RNA polymerase. Unlike DNA polymerases, RNA polymerases do not need a free 3’-OH group to synthesize an RNA molecule. Now that the primer provides the free 3’-OH group, DNA polymerase III can now extend this RNA primer, adding DNA nucleotides one by one that are complementary to the template strand (Figure 11.4). Elongation During elongation in DNA replication, the addition of nucleotides occurs at its maximal rate of about 1000 nucleotides per second. DNA polymerase III can only extend in the 5’ to 3’ direction, which poses a problem at the replication fork. The DNA double helix is antiparallel; that is, one strand is oriented in the 5’ to 3’ direction and the other is oriented in the 3’ to 5’ direction (see Structure and Function of DNA). During replication, one strand, which is complementary to the 3’ to 5’ parental DNA strand, is synthesized continuously toward the replication fork because polymerase can add nucleotides in this direction. This continuously synthesized strand is known as the leading strand. The other strand, complementary to the 5’ to 3’ parental DNA, grows away from the replication fork, so the polymerase must move back toward the replication fork to begin adding bases to a new primer, again in the direction away from the replication fork. It does so until it bumps into the previously synthesized strand and then it moves back again (Figure 11.7). These steps produce small DNA sequence fragments known as Okazaki fragments, each separated by RNA primer. Okazaki fragments are named after the Japanese research team and married couple Reiji and Tsuneko Okazaki, who first discovered them in 1966. The strand with the Okazaki fragments is known as the lagging strand, and its synthesis is said to be discontinuous. The leading strand can be extended from one primer alone, whereas the lagging strand needs a new primer for each of the short Okazaki fragments. The overall direction of the lagging strand will be 3’ to 5’, and that of the leading strand 5’ to 3’. A protein called the sliding clamp holds the DNA polymerase in place as it continues to add nucleotides. The sliding clamp is a ring-shaped protein that binds to the DNA and holds the polymerase in place. Beyond its role in initiation, topoisomerase also prevents the overwinding of the DNA double helix ahead of the replication fork as the DNA is opening up; it does so by causing temporary nicks in the DNA helix and then resealing it. As synthesis proceeds, the RNA primers are replaced by DNA. The primers are removed by the exonuclease activity of DNA polymerase I, and the gaps are filled in. The nicks that remain between the newly synthesized DNA (that replaced the RNA primer) and the previously synthesized DNA are sealed by the enzyme DNA ligase that catalyzes the formation of covalent phosphodiester linkage between the 3’-OH end of one DNA fragment and the 5’ phosphate end of the other fragment, stabilizing the sugar-phosphate backbone of the DNA molecule. 404 11 Mechanisms of Microbial Genetics FIGURE 11.7 At the origin of replication, topoisomerase II relaxes the supercoiled chromosome. Two replication forks are formed by the opening of the double-stranded DNA at the origin, and helicase separates the DNA strands, which are coated by single-stranded binding proteins to keep the strands separated. DNA replication occurs in both directions. An RNA primer complementary to the parental strand is synthesized by RNA primase and is elongated by DNA polymerase III through the addition of nucleotides to the 3’-OH end. On the leading strand, DNA is synthesized continuously, whereas on the lagging strand, DNA is synthesized in short stretches called Okazaki fragments. RNA primers within the lagging strand are removed by the exonuclease activity of DNA polymerase I, and the Okazaki fragments are joined by DNA ligase. Termination Once the complete chromosome has been replicated, termination of DNA replication must occur. Although much is known about initiation of replication, less is known about the termination process. Following replication, the resulting complete circular genomes of prokaryotes are concatenated, meaning that the circular DNA chromosomes are interlocked and must be separated from each other. This is accomplished through the activity of bacterial topoisomerase IV, which introduces double-stranded breaks into DNA molecules, allowing them to separate from each other; the enzyme then reseals the circular chromosomes. The resolution of concatemers is an issue unique to prokaryotic DNA replication because of their circular chromosomes. Because both bacterial DNA gyrase and topoisomerase IV are distinct from their eukaryotic counterparts, these enzymes serve as targets for a class of antimicrobial drugs called quinolones. Access for free at openstax.org  11.2 DNA Replication 405 The Molecular Machinery Involved in Bacterial DNA Replication Enzyme or Function Factor DNA pol I Exonuclease activity removes RNA primer and replaces it with newly synthesized DNA DNA pol III Main enzyme that adds nucleotides in the 5’ to 3’ direction Helicase Opens the DNA helix by breaking hydrogen bonds between the nitrogenous bases Ligase Seals the gaps between the Okazaki fragments on the lagging strand to create one continuous DNA strand Primase Synthesizes RNA primers needed to start replication Single- Bind to single-stranded DNA to prevent hydrogen bonding between DNA strands, reforming stranded double-stranded DNA binding proteins Sliding clamp Helps hold DNA pol III in place when nucleotides are being added Topoisomerase Relaxes supercoiled chromosome to make DNA more accessible for the initiation of II (DNA replication; helps relieve the stress on DNA when unwinding, by causing breaks and then gyrase) resealing the DNA Topoisomerase Introduces single-stranded break into concatenated chromosomes to release them from each IV other, and then reseals the DNA TABLE 11.1 CHECK YOUR UNDERSTANDING Which enzyme breaks the hydrogen bonds holding the two strands of DNA together so that replication can occur? Is it the lagging strand or the leading strand that is synthesized in the direction toward the opening of the replication fork? Which enzyme is responsible for removing the RNA primers in newly replicated bacterial DNA? DNA Replication in Eukaryotes Eukaryotic genomes are much more complex and larger than prokaryotic genomes and are typically composed of multiple linear chromosomes (Table 11.2). The human genome, for example, has 3 billion base pairs per haploid set of chromosomes, and 6 billion base pairs are inserted during replication. There are multiple origins of replication on each eukaryotic chromosome (Figure 11.8); the human genome has 30,000 to 50,000 origins of replication. The rate of replication is approximately 100 nucleotides per second—10 times slower than prokaryotic replication. 406 11 Mechanisms of Microbial Genetics FIGURE 11.8 Eukaryotic chromosomes are typically linear, and each contains multiple origins of replication. The essential steps of replication in eukaryotes are the same as in prokaryotes. Before replication can start, the DNA has to be made available as a template. Eukaryotic DNA is highly supercoiled and packaged, which is facilitated by many proteins, including histones (see Structure and Function of Cellular Genomes). At the origin of replication, a prereplication complex composed of several proteins, including helicase, forms and recruits other enzymes involved in the initiation of replication, including topoisomerase to relax supercoiling, single-stranded binding protein, RNA primase, and DNA polymerase. Following initiation of replication, in a process similar to that found in prokaryotes, elongation is facilitated by eukaryotic DNA polymerases. The leading strand is continuously synthesized by the eukaryotic polymerase enzyme pol δ, while the lagging strand is synthesized by pol ε. A sliding clamp protein holds the DNA polymerase in place so that it does not fall off the DNA. The enzyme ribonuclease H (RNase H), instead of a DNA polymerase as in bacteria, removes the RNA primer, which is then replaced with DNA nucleotides. The gaps that remain are sealed by DNA ligase. Because eukaryotic chromosomes are linear, one might expect that their replication would be more straightforward. As in prokaryotes, the eukaryotic DNA polymerase can add nucleotides only in the 5’ to 3’ direction. In the leading strand, synthesis continues until it reaches either the end of the chromosome or another replication fork progressing in the opposite direction. On the lagging strand, DNA is synthesized in short stretches, each of which is initiated by a separate primer. When the replication fork reaches the end of the linear chromosome, there is no place to make a primer for the DNA fragment to be copied at the end of the chromosome. These ends thus remain unpaired and, over time, they may get progressively shorter as cells continue to divide. The ends of the linear chromosomes are known as telomeres and consist of noncoding repetitive sequences. The telomeres protect coding sequences from being lost as cells continue to divide. In humans, a six base-pair sequence, TTAGGG, is repeated 100 to 1000 times to form the telomere. The discovery of the enzyme telomerase (Figure 11.9) clarified our understanding of how chromosome ends are maintained. Telomerase contains a catalytic part and a built-in RNA template. It attaches to the end of the chromosome, and complementary bases to the RNA template are added on the 3’ end of the DNA strand. Once the 3’ end of the lagging strand template is sufficiently elongated, DNA polymerase can add the nucleotides complementary to the ends of the chromosomes. In this way, the ends of the chromosomes are replicated. In humans, telomerase is typically active in germ cells and adult stem cells; it is not active in adult somatic cells and may be associated with the aging of these cells. Eukaryotic microbes including fungi and protozoans also produce telomerase to maintain chromosomal integrity. For her discovery of telomerase and its action, Elizabeth Blackburn (1948–) received the Nobel Prize for Medicine or Physiology in 2009. Access for free at openstax.org  11.2 DNA Replication 407 FIGURE 11.9 In eukaryotes, the ends of the linear chromosomes are maintained by the action of the telomerase enzyme. Comparison of Bacterial and Eukaryotic Replication Property Bacteria Eukaryotes Genome structure Single circular chromosome Multiple linear chromosomes Number of origins per chromosome Single Multiple Rate of replication 1000 nucleotides per second 100 nucleotides per second Telomerase Not present Present RNA primer removal DNA pol I RNase H Strand elongation DNA pol III pol δ, pol ε TABLE 11.2 LINK TO LEARNING This animation (https://openstax.org/l/22DNAreplicani) compares the process of prokaryotic and eukaryotic DNA replication. 408 11 Mechanisms of Microbial Genetics CHECK YOUR UNDERSTANDING How does the origin of replication differ between eukaryotes and prokaryotes? What polymerase enzymes are responsible for DNA synthesis during eukaryotic replication? What is found at the ends of the chromosomes in eukaryotes and why? DNA Replication of Extrachromosomal Elements: Plasmids and Viruses To copy their nucleic acids, plasmids and viruses frequently use variations on the pattern of DNA replication described for prokaryote genomes. For more information on the wide range of viral replication strategies, see The Viral Life Cycle. Rolling Circle Replication Whereas many bacterial plasmids (see Unique Characteristics of Prokaryotic Cells) replicate by a process similar to that used to copy the bacterial chromosome, other plasmids, several bacteriophages, and some viruses of eukaryotes use rolling circle replication (Figure 11.10). The circular nature of plasmids and the circularization of some viral genomes on infection make this possible. Rolling circle replication begins with the enzymatic nicking of one strand of the double-stranded circular molecule at the double-stranded origin (dso) site. In bacteria, DNA polymerase III binds to the 3’-OH group of the nicked strand and begins to unidirectionally replicate the DNA using the un-nicked strand as a template, displacing the nicked strand as it does so. Completion of DNA replication at the site of the original nick results in full displacement of the nicked strand, which may then recircularize into a single- stranded DNA molecule. RNA primase then synthesizes a primer to initiate DNA replication at the single-stranded origin (sso) site of the single-stranded DNA (ssDNA) molecule, resulting in a double-stranded DNA (dsDNA) molecule identical to the other circular DNA molecule. FIGURE 11.10 The process of rolling circle replication results in the synthesis of a single new copy of the circular DNA molecule, as shown here. CHECK YOUR UNDERSTANDING Is there a lagging strand in rolling circle replication? Why or why not? 11.3 RNA Transcription LEARNING OBJECTIVES By the end of this section, you will be able to: Explain how RNA is synthesized using DNA as a template Distinguish between transcription in prokaryotes and eukaryotes Access for free at openstax.org  11.3 RNA Transcription 409 During the process of transcription, the information encoded within the DNA sequence of one or more genes is transcribed into a strand of RNA, also called an RNA transcript. The resulting single-stranded RNA molecule, composed of ribonucleotides containing the bases adenine (A), cytosine (C), guanine (G), and uracil (U), acts as a mobile molecular copy of the original DNA sequence. Transcription in prokaryotes and in eukaryotes requires the DNA double helix to partially unwind in the region of RNA synthesis. The unwound region is called a transcription bubble. Transcription of a particular gene always proceeds from one of the two DNA strands that acts as a template, the so-called antisense strand. The RNA product is complementary to the template strand of DNA and is almost identical to the nontemplate DNA strand, or the sense strand. The only difference is that in RNA, all of the T nucleotides are replaced with U nucleotides; during RNA synthesis, U is incorporated when there is an A in the complementary antisense strand. Transcription in Bacteria Bacteria use the same RNA polymerase to transcribe all of their genes. Like DNA polymerase, RNA polymerase adds nucleotides one by one to the 3’-OH group of the growing nucleotide chain. One critical difference in activity between DNA polymerase and RNA polymerase is the requirement for a 3’-OH onto which to add nucleotides: DNA polymerase requires such a 3’-OH group, thus necessitating a primer, whereas RNA polymerase does not. During transcription, a ribonucleotide complementary to the DNA template strand is added to the growing RNA strand and a covalent phosphodiester bond is formed by dehydration synthesis between the new nucleotide and the last one added. In E. coli, RNA polymerase comprises six polypeptide subunits, five of which compose the polymerase core enzyme responsible for adding RNA nucleotides to a growing strand. The sixth subunit is known as sigma (σ). The σ factor enables RNA polymerase to bind to a specific promoter, thus allowing for the transcription of various genes. There are various σ factors that allow for transcription of various genes. Initiation The initiation of transcription begins at a promoter, a DNA sequence onto which the transcription machinery binds and initiates transcription. The nucleotide pair in the DNA double helix that corresponds to the site from which the first 5’ RNA nucleotide is transcribed is the initiation site. Nucleotides preceding the initiation site are designated “upstream,” whereas nucleotides following the initiation site are called “downstream” nucleotides. In most cases, promoters are located just upstream of the genes they regulate. Although promoter sequences vary among bacterial genomes, a few elements are conserved. At the –10 and –35 positions within the DNA prior to the initiation site (designated +1), there are two promoter consensus sequences, or regions that are similar across all promoters and across various bacterial species. The –10 consensus sequence, called the TATA box, is TATAAT. The –35 sequence is recognized and bound by σ. Elongation The elongation in transcription phase begins when the σ subunit dissociates from the polymerase, allowing the core enzyme to synthesize RNA complementary to the DNA template in a 5’ to 3’ direction at a rate of approximately 40 nucleotides per second. As elongation proceeds, the DNA is continuously unwound ahead of the core enzyme and rewound behind it (Figure 11.11). FIGURE 11.11 During elongation, the bacterial RNA polymerase tracks along the DNA template, synthesizes mRNA in the 5’ to 3’ direction, and unwinds and rewinds the DNA as it is read. 410 11 Mechanisms of Microbial Genetics Termination Once a gene is transcribed, the bacterial polymerase must dissociate from the DNA template and liberate the newly made RNA. This is referred to as termination of transcription. The DNA template includes repeated nucleotide sequences that act as termination signals, causing RNA polymerase to stall and release from the DNA template, freeing the RNA transcript. CHECK YOUR UNDERSTANDING Where does σ factor of RNA polymerase bind DNA to start transcription? What occurs to initiate the polymerization activity of RNA polymerase? Where does the signal to end transcription come from? Transcription in Eukaryotes Prokaryotes and eukaryotes perform fundamentally the same process of transcription, with a few significant differences (see Table 11.3). Eukaryotes use three different polymerases, RNA polymerases I, II, and III, all structurally distinct from the bacterial RNA polymerase. Each transcribes a different subset of genes. Interestingly, archaea contain a single RNA polymerase that is more closely related to eukaryotic RNA polymerase II than to its bacterial counterpart. Eukaryotic mRNAs are also usually monocistronic, meaning that they each encode only a single polypeptide, whereas prokaryotic mRNAs of bacteria and archaea are commonly polycistronic, meaning that they encode multiple polypeptides. The most important difference between prokaryotes and eukaryotes is the latter’s membrane-bound nucleus, which influences the ease of use of RNA molecules for protein synthesis. With the genes bound in a nucleus, the eukaryotic cell must transport protein-encoding RNA molecules to the cytoplasm to be translated. Protein-encoding primary transcripts, the RNA molecules directly synthesized by RNA polymerase, must undergo several processing steps to protect these RNA molecules from degradation during the time they are transferred from the nucleus to the cytoplasm and translated into a protein. For example, eukaryotic mRNAs may last for several hours, whereas the typical prokaryotic mRNA lasts no more than 5 seconds. The primary transcript (also called pre-mRNA) is first coated with RNA-stabilizing proteins to protect it from degradation while it is processed and exported out of the nucleus. The first type of processing begins while the primary transcript is still being synthesized; a special 7-methylguanosine nucleotide, called the 5’ cap, is added to the 5’ end of the growing transcript. In addition to preventing degradation, factors involved in subsequent protein synthesis recognize the cap, which helps initiate translation by ribosomes. Once elongation is complete, another processing enzyme then adds a string of approximately 200 adenine nucleotides to the 3’ end, called the poly-A tail. This modification further protects the pre-mRNA from degradation and signals to cellular factors that the transcript needs to be exported to the cytoplasm. Eukaryotic genes that encode polypeptides are composed of coding sequences called exons (ex-on signifies that they are expressed) and intervening sequences called introns (int-ron denotes their intervening role). Transcribed RNA sequences corresponding to introns do not encode regions of the functional polypeptide and are removed from the pre-mRNA during processing. It is essential that all of the intron-encoded RNA sequences are completely and precisely removed from a pre-mRNA before protein synthesis so that the exon-encoded RNA sequences are properly joined together to code for a functional polypeptide. If the process errs by even a single nucleotide, the sequences of the rejoined exons would be shifted, and the resulting polypeptide would be nonfunctional. The process of removing intron-encoded RNA sequences and reconnecting those encoded by exons is called RNA splicing and is facilitated by the action of a spliceosome containing small nuclear ribonucleo proteins (snRNPs). Intron-encoded RNA sequences are removed from the pre-mRNA while it is still in the nucleus. Although they are not translated, introns appear to have various functions, including gene regulation and mRNA transport. On completion of these modifications, the mature transcript, the mRNA that encodes a polypeptide, is transported out of the nucleus, destined for the cytoplasm for translation. Introns can be spliced out differently, resulting in various exons being included or excluded from the final mRNA product. This process is known as alternative splicing. The advantage of alternative splicing is that different types of mRNA transcripts can be generated, all derived from the same DNA sequence. In recent years, it has been shown that some archaea also have the ability to splice their pre- Access for free at openstax.org  11.3 RNA Transcription 411 mRNA. Comparison of Transcription in Bacteria Versus Eukaryotes Property Bacteria Eukaryotes Number of polypeptides encoded per mRNA Monocistronic or polycistronic Exclusively monocistronic Strand elongation core + σ = holoenzyme RNA polymerases I, II, or III Addition of 5’ cap No Yes Addition of 3’ poly-A tail No Yes Splicing of pre-mRNA No Yes TABLE 11.3 LINK TO LEARNING Visualize how mRNA splicing (https://openstax.org/l/22mrnasplice) happens by watching the process in action in this video. See how introns are removed during RNA splicing (https://openstax.org/l/22rnasplice) here. CHECK YOUR UNDERSTANDING In eukaryotic cells, how is the RNA transcript from a gene for a protein modified after it is transcribed? Do exons or introns contain information for protein sequences? CLINICAL FOCUS Part 2 In the emergency department, a nurse told Mark that he had made a good decision to come to the hospital because his symptoms indicated an infection that had gotten out of control. Mark’s symptoms had progressed, with the area of skin affected and the amount of swelling increasing. Within the affected area, a rash had begun, blistering and small gas pockets underneath the outermost layer of skin had formed, and some of the skin was becoming gray. Based on the putrid smell of the pus draining from one of the blisters, the rapid progression of the infection, and the visual appearance of the affected skin, the physician immediately began treatment for necrotizing fasciitis. Mark’s physician ordered a culture of the fluid draining from the blister and also ordered blood work, including a white blood cell count. Mark was admitted to the intensive care unit and began intravenous administration of a broad-spectrum antibiotic to try to minimize further spread of the infection. Despite antibiotic therapy, Mark’s condition deteriorated quickly. Mark became confused and dizzy. Within a few hours of his hospital admission, his blood pressure dropped significantly and his breathing became shallower and more rapid. Additionally, blistering increased, with the blisters intensifying in color to purplish black, and the wound itself seemed to be progressing rapidly up Mark’s leg. What are possible causative agents of Mark’s necrotizing fasciitis? What are some possible explanations for why the antibiotic treatment does not seem to be working? Jump to the next Clinical Focus box. Go back to the previous Clinical Focus box. 412 11 Mechanisms of Microbial Genetics 11.4 Protein Synthesis (Translation) LEARNING OBJECTIVES By the end of this section, you will be able to: Describe the genetic code and explain why it is considered almost universal Explain the process of translation and the functions of the molecular machinery of translation Compare translation in eukaryotes and prokaryotes The synthesis of proteins consumes more of a cell’s energy than any other metabolic process. In turn, proteins account for more mass than any other macromolecule of living organisms. They perform virtually every function of a cell, serving as both functional (e.g., enzymes) and structural elements. The process of translation, or protein synthesis, the second part of gene expression, involves the decoding by a ribosome of an mRNA message into a polypeptide product. The Genetic Code Translation of the mRNA template converts nucleotide-based genetic information into the “language” of amino acids to create a protein product. A protein sequence consists of 20 commonly occurring amino acids. Each amino acid is defined within the mRNA by a triplet of nucleotides called a codon. The relationship between an mRNA codon and its corresponding amino acid is called the genetic code. The three-nucleotide code means that there is a total of 64 possible combinations (43, with four different nucleotides possible at each of the three different positions within the codon). This number is greater than the number of amino acids and a given amino acid is encoded by more than one codon (Figure 11.12). This redundancy in the genetic code is called degeneracy. Typically, whereas the first two positions in a codon are important for determining which amino acid will be incorporated into a growing polypeptide, the third position, called the wobble position, is less critical. In some cases, if the nucleotide in the third position is changed, the same amino acid is still incorporated. Whereas 61 of the 64 possible triplets code for amino acids, three of the 64 codons do not code for an amino acid; they terminate protein synthesis, releasing the polypeptide from the translation machinery. These are called stop codons or nonsense codons. Another codon, AUG, also has a special function. In addition to specifying the amino acid methionine, it also typically serves as the start codon to initiate translation. The reading frame, the way nucleotides in mRNA are grouped into codons, for translation is set by the AUG start codon near the 5’ end of the mRNA. Each set of three nucleotides following this start codon is a codon in the mRNA message. The genetic code is nearly universal. With a few exceptions, virtually all species use the same genetic code for protein synthesis, which is powerful evidence that all extant life on earth shares a common origin. However, unusual amino acids such as selenocysteine and pyrrolysine have been observed in archaea and bacteria. In the case of selenocysteine, the codon used is UGA (normally a stop codon). However, UGA can encode for selenocysteine using a stem-loop structure (known as the selenocysteine insertion sequence, or SECIS element), which is found at the 3’ untranslated region of the mRNA. Pyrrolysine uses a different stop codon, UAG. The incorporation of pyrrolysine requires the pylS gene and a unique transfer RNA (tRNA) with a CUA anticodon. Access for free at openstax.org  11.4 Protein Synthesis (Translation) 413 FIGURE 11.12 This figure shows the genetic code for translating each nucleotide triplet in mRNA into an amino acid or a termination signal in a nascent protein. The first letter of a codon is shown vertically on the left, the second letter of a codon is shown horizontally across the top, and the third letter of a codon is shown vertically on the right. (credit: modification of work by National Institutes of Health) CHECK YOUR UNDERSTANDING How many bases are in each codon? What amino acid is coded for by the codon AAU? What happens when a stop codon is reached? The Protein Synthesis Machinery In addition to the mRNA template, many molecules and macromolecules contribute to the process of translation. The composition of each component varies across taxa; for instance, ribosomes may consist of different numbers of ribosomal RNAs (rRNAs) and polypeptides depending on the organism. However, the general structures and functions of the protein synthesis machinery are comparable from bacteria to human cells. Translation requires the input of an mRNA template, ribosomes, tRNAs, and various enzymatic factors. Ribosomes A ribosome is a complex macromolecule composed of catalytic rRNAs (called ribozymes) and structural rRNAs, as well as many distinct polypeptides. Mature rRNAs make up approximately 50% of each ribosome. Prokaryotes have 70S ribosomes, whereas eukaryotes have 80S ribosomes in the cytoplasm and rough endoplasmic reticulum, and 70S ribosomes in mitochondria and chloroplasts. Ribosomes dissociate into large and small subunits when they are not synthesizing proteins and reassociate during the initiation of translation. In E. coli, the small subunit is described as 30S (which contains the 16S rRNA subunit), and the large subunit is 50S (which contains the 5S and 23S rRNA subunits), for a total of 70S (Svedberg units are not additive). Eukaryote ribosomes have a small 40S subunit (which contains the 18S rRNA subunit) and a large 60S subunit (which contains the 5S, 5.8S and 28S rRNA subunits), for a total of 80S. The small subunit is responsible for binding the mRNA template, whereas the large subunit binds tRNAs (discussed in the next subsection). Each mRNA molecule is simultaneously translated by many ribosomes, all synthesizing protein in the same direction: reading the mRNA from 5’ to 3’ and synthesizing the polypeptide from the N terminus to the C terminus. The complete structure containing an mRNA with multiple associated ribosomes is called a polyribosome (or polysome). In both bacteria and archaea, before transcriptional termination occurs, each protein-encoding transcript is already being used to begin synthesis of numerous copies of the encoded polypeptide(s) because the processes of transcription and translation can occur concurrently, forming polyribosomes (Figure 11.13). The reason why transcription and translation can occur simultaneously is because both of these processes occur in the same 5’ 414 11 Mechanisms of Microbial Genetics to 3’ direction, they both occur in the cytoplasm of the cell, and because the RNA transcript is not processed once it is transcribed. This allows a prokaryotic cell to respond to an environmental signal requiring new proteins very quickly. In contrast, in eukaryotic cells, simultaneous transcription and translation is not possible. Although polyribosomes also form in eukaryotes, they cannot do so until RNA synthesis is complete and the RNA molecule has been modified and transported out of the nucleus. FIGURE 11.13 In prokaryotes, multiple RNA polymerases can transcribe a single bacterial gene while numerous ribosomes concurrently translate the mRNA transcripts into polypeptides. In this way, a specific protein can rapidly reach a high concentration in the bacterial cell. Transfer RNAs Transfer RNAs (tRNAs) are structural RNA molecules and, depending on the species, many different types of tRNAs exist in the cytoplasm. Bacterial species typically have between 60 and 90 types. Serving as adaptors, each tRNA type binds to a specific codon on the mRNA template and adds the corresponding amino acid to the polypeptide chain. Therefore, tRNAs are the molecules that actually “translate” the language of RNA into the language of proteins. As the adaptor molecules of translation, it is surprising that tRNAs can fit so much specificity into such a small package. The tRNA molecule interacts with three factors: aminoacyl tRNA synthetases, ribosomes, and mRNA. Mature tRNAs take on a three-dimensional structure when complementary bases exposed in the single-stranded RNA molecule hydrogen bond with each other (Figure 11.14). This shape positions the amino-acid binding site, called the CCA amino acid binding end, which is a cytosine-cytosine-adenine sequence at the 3’ end of the tRNA, and the anticodon at the other end. The anticodon is a three-nucleotide sequence that bonds with an mRNA codon through complementary base pairing. An amino acid is added to the end of a tRNA molecule through the process of tRNA “charging,” during which each tRNA molecule is linked to its correct or cognate amino acid by a group of enzymes called aminoacyl tRNA synthetases. At least one type of aminoacyl tRNA synthetase exists for each of the 20 amino acids. During this process, the amino acid is first activated by the addition of adenosine monophosphate (AMP) and then transferred to the tRNA, making it a charged tRNA, and AMP is released. Access for free at openstax.org  11.4 Protein Synthesis (Translation) 415 FIGURE 11.14 (a) After folding caused by intramolecular base pairing, a tRNA molecule has one end that contains the anticodon, which interacts with the mRNA codon, and the CCA amino acid binding end. (b) A space-filling model is helpful for visualizing the three- dimensional shape of tRNA. (c) Simplified models are useful when drawing complex processes such as protein synthesis. CHECK YOUR UNDERSTANDING Describe the structure and composition of the prokaryotic ribosome. In what direction is the mRNA template read? Describe the structure and function of a tRNA. The Mechanism of Protein Synthesis Translation is similar in prokaryotes and eukaryotes. Here we will explore how translation occurs in E. coli, a representative prokaryote, and specify any differences between bacterial and eukaryotic translation. Initiation The initiation of protein synthesis begins with the formation of an initiation complex. In E. coli, this complex involves the small 30S ribosome, the mRNA template, three initiation factors that help the ribosome assemble correctly, guanosine triphosphate (GTP) that acts as an energy source, and a special initiator tRNA carrying N-formyl-methionine (fMet-tRNAfMet) (Figure 11.15). The initiator tRNA interacts with the start codon AUG of the mRNA and carries a formylated methionine (fMet). Because of its involvement in initiation, fMet is inserted at the beginning (N terminus) of every polypeptide chain synthesized by E. coli. In E. coli mRNA, a leader sequence upstream of the first AUG codon, called the Shine-Dalgarno sequence (also known as the ribosomal binding site AGGAGG), interacts through complementary base pairing with the rRNA molecules that compose the ribosome. This interaction anchors the 30S ribosomal subunit at the correct location on the mRNA template. At this point, the 50S ribosomal subunit then binds to the initiation complex, forming an intact ribosome. In eukaryotes, initiation complex formation is similar, with the following differences: The initiator tRNA is a different specialized tRNA carrying methionine, called Met-tRNAi Instead of binding to the mRNA at the Shine-Dalgarno sequence, the eukaryotic initiation complex recognizes the 5’ cap of the eukaryotic mRNA, then tracks along the mRNA in the 5’ to 3’ direction until the AUG start codon is recognized. At this point, the 60S subunit binds to the complex of Met-tRNAi, mRNA, and the 40S subunit. 416 11 Mechanisms of Microbial Genetics FIGURE 11.15 Translation in bacteria begins with the formation of the initiation complex, which includes the small ribosomal subunit, the mRNA, the initiator tRNA carrying N-formyl-methionine, and initiation factors. Then the 50S subunit binds, forming an intact ribosome. Elongation In prokaryotes and eukaryotes, the basics of elongation of translation are the same. In E. coli, the binding of the 50S ribosomal subunit to produce the intact ribosome forms three functionally important ribosomal sites: The A (aminoacyl) site binds incoming charged aminoacyl tRNAs. The P (peptidyl) site binds charged tRNAs carrying amino acids that have formed peptide bonds with the growing polypeptide chain but have not yet dissociated from their corresponding tRNA. The E (exit) site releases dissociated tRNAs so that they can be recharged with free amino acids. There is one notable exception to this assembly line of tRNAs: During initiation complex formation, bacterial fMet−tRNAfMet or eukaryotic Met-tRNAi enters the P site directly without first entering the A site, providing a free A site ready to accept the tRNA corresponding to the first codon after the AUG. Elongation proceeds with single-codon movements of the ribosome each called a translocation event. During each translocation event, the charged tRNAs enter at the A site, then shift to the P site, and then finally to the E site for removal. Ribosomal movements, or steps, are induced by conformational changes that advance the ribosome by three bases in the 3’ direction. Peptide bonds form between the amino group of the amino acid attached to the A- site tRNA and the carboxyl group of the amino acid attached to the P-site tRNA. The formation of each peptide bond is catalyzed by peptidyl transferase, an RNA-based ribozyme that is integrated into the 50S ribosomal subunit. The amino acid bound to the P-site tRNA is also linked to the growing polypeptide chain. As the ribosome steps across the mRNA, the former P-site tRNA enters the E site, detaches from the amino acid, and is expelled. Several of the steps during elongation, including binding of a charged aminoacyl tRNA to the A site and translocation, require energy derived from GTP hydrolysis, which is catalyzed by specific elongation factors. Amazingly, the E. coli translation apparatus takes only 0.05 seconds to add each amino acid, meaning that a 200 amino-acid protein can be translated in just 10 seconds. Termination