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Unlike prokaryotic chromosomes, eukaryotic chromosomes are linear. As you’ve learned, the enzyme DNA pol can add nucleotides only in the 5' to 3' direction. In the leading strand, synthesis continues until the end of the chromosome is reached. 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 for a primer to be made for the DNA fragment to be copied at the end of the chromosome. These ends thus remain unpaired, and over time these ends may get progressively shorter as cells continue to divide. The ends of the linear chromosomes are known as telomeres, which have repetitive sequences that code for no particular gene. In a way, these telomeres protect the genes from getting deleted as cells continue to divide. In humans, a six base pair sequence, TTAGGG, is repeated 100 to 1000 times. The discovery of the enzyme telomerase (Figure 14.16) helped in the understanding of how chromosome ends are maintained. The telomerase enzyme 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. Thus, the ends of the chromosomes are replicated. 568 Chapter 14 | DNA Structure and Function Figure 14.15 The ends of linear chromosomes are maintained by the action of the telomerase enzyme. Telomerase is typically active in germ cells and adult stem cells. It is not active in adult somatic cells. For her discovery of telomerase and its action, Elizabeth Blackburn (Figure 14.16) received the Nobel Prize for Medicine and Physiology in 2009. Figure 14.16 Elizabeth Blackburn, 2009 Nobel Laureate, is the scientist who discovered how telomerase works. (credit: US Embassy Sweden) Telomerase and Aging Cells that undergo cell division continue to have their telomeres shortened because most somatic cells do not make telomerase. This essentially means that telomere shortening is associated with aging. With the advent of modern medicine, preventative health care, and healthier lifestyles, the human life span has increased, and there is an increasing demand for people to look younger and |
have a better quality of life as they grow older. In 2010, scientists found that telomerase can reverse some age-related conditions in mice. This may have potential in regenerative medicine. Telomerase-deficient mice were used in these studies; these mice have tissue atrophy, stem cell depletion, organ system failure, and impaired tissue injury responses. Telomerase reactivation in these mice caused [2] This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 569 extension of telomeres, reduced DNA damage, reversed neurodegeneration, and improved the function of the testes, spleen, and intestines. Thus, telomere reactivation may have potential for treating age-related diseases in humans. Cancer is characterized by uncontrolled cell division of abnormal cells. The cells accumulate mutations, proliferate uncontrollably, and can migrate to different parts of the body through a process called metastasis. Scientists have observed that cancerous cells have considerably shortened telomeres and that telomerase is active in these cells. Interestingly, only after the telomeres were shortened in the cancer cells did the telomerase become active. If the action of telomerase in these cells can be inhibited by drugs during cancer therapy, then the cancerous cells could potentially be stopped from further division. Difference between Prokaryotic and Eukaryotic Replication Property Prokaryotes Eukaryotes Origin of replication Single Multiple Rate of replication 1000 nucleotides/s 50 to 100 nucleotides/s DNA polymerase types 5 Telomerase Not present RNA primer removal DNA pol I 14 Present RNase H Strand elongation DNA pol III Pol δ, pol ε Sliding clamp Sliding clamp PCNA Table 14.2 14.6 | DNA Repair In this section, you will explore the following questions: • What are different types of mutations in DNA and the significance of mutations? • What are examples of mechanisms that repair mutations in DNA? Connection for AP® Courses DNA polymerase is an efficient enzyme but it can make mistakes while adding nucleotides during replication. It edits the DNA by proofreading every newly added base. An incorrect base is removed and replaced by the correct base. If a base remains mismatched, special repair enzymes can often recognize the wrongly incorporated base, excise it from the DNA, and replace it with the correct base. Most mistakes are corrected, but if |
they are not they may result in a mutation, which is defined as a permanent change in a DNA sequence. A mutation can be passed to daughter cells through DNA replication and cell division. There are several types of DNA mutations, including substitution, deletion, insertion, and translocation. Mutations in repair genes may lead to serious consequences, such as cancer. Mutations can be induced by environmental factors, such as UV radiation, or they can occur spontaneously. (We will explore the effects of mutation in more detail in a later chapter. Remember that mutations are not always detrimental. They can be beneficial, too. Changes in DNA increase genetic variation—the foundation of evolution by natural selection.) Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Living systems store, retrieve, transmit, and respond to information essential to life processes. Enduring Understanding 3.C The processing of genetic information is imperfect and is a source of genetic variation. 2. Jaskelioff et al., “Telomerase reactivation reverses tissue degeneration in aged telomerase-deficient mice,” Nature 469 (2011): 102-7. 570 Essential Knowledge Science Practice Science Practice Chapter 14 | DNA Structure and Function 3.C.1 Changes in genotype can result in changes in phenotype. 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. Learning Objective 3.24 The student is able to predict how a change in genotype, when expressed as a phenotype, provides a variation that can be subject to natural selection. Essential Knowledge 3.C.1 Changes in genotype can result in changes in phenotype. Science Practice 1.1 The student can create representations and models of natural or man-made phenomena and systems in the domain. Learning Objective 3.25 The student can create a visual representation to illustrate how changes in a DNA nucleotide sequence can result in a change in the polypeptide produced. Essential Knowledge Science Practice Learning Objective 3.C.2 |
Biological systems have multiple processes that increase genetic variation. 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices. 3.28 The student is able to construct an explanation of the multiple processes that increase variation within a population. DNA replication is a highly accurate process, but mistakes can occasionally occur, such as a DNA polymerase inserting a wrong base. Uncorrected mistakes may sometimes lead to serious consequences, such as cancer. Repair mechanisms correct the mistakes. In rare cases, mistakes are not corrected, leading to mutations; in other cases, repair enzymes are themselves mutated or defective. Most of the mistakes during DNA replication are promptly corrected by DNA polymerase by proofreading the base that has been just added (Figure 14.17). In proofreading, the DNA pol reads the newly added base before adding the next one, so a correction can be made. The polymerase checks whether the newly added base has paired correct<|endoftext|>ly with the base in the template strand. If it is the right base, the next nucleotide is added. If an incorrect base has been added, the enzyme makes a cut at the phosphodiester bond and releases the wrong nucleotide. This is performed by the exonuclease action of DNA pol III. Once the incorrect nucleotide has been removed, a new one will be added again. Figure 14.17 Proofreading by DNA polymerase corrects errors during replication. Some errors are not corrected during replication, but are instead corrected after replication is completed; this type of repair is known as mismatch repair (Figure 14.18). The enzymes recognize the incorrectly added nucleotide and excise it; this is then replaced by the correct base. If this remains uncorrected, it may lead to more permanent damage. How do mismatch repair enzymes recognize which of the two bases is the incorrect one? In E. coli, after replication, the nitrogenous base This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 571 adenine acquires a methyl group; the parental DNA strand will have methyl groups, whereas the newly synthesized strand lacks them. Thus, DNA polymerase is able to remove the wrongly incorporated bases from the newly synthesized, nonmethylated strand. In eukaryotes, the mechanism is not very well understood, but it is believed to involve recognition of unsealed nicks in the new strand, as well as a short |
-term continuing association of some of the replication proteins with the new daughter strand after replication has completed. Figure 14.18 In mismatch repair, the incorrectly added base is detected after replication. The mismatch repair proteins detect this base and remove it from the newly synthesized strand by nuclease action. The gap is now filled with the correctly paired base. In another type of repair mechanism, nucleotide excision repair, enzymes replace incorrect bases by making a cut on both the 3' and 5' ends of the incorrect base (Figure 14.19). The segment of DNA is removed and replaced with the correctly paired nucleotides by the action of DNA pol. Once the bases are filled in, the remaining gap is sealed with a phosphodiester linkage catalyzed by DNA ligase. This repair mechanism is often employed when UV exposure causes the formation of pyrimidine dimers. Figure 14.19 Nucleotide excision repairs thymine dimers. When exposed to UV, thymines lying adjacent to each other can form thymine dimers. In normal cells, they are excised and replaced. A well-studied example of mistakes not being corrected is seen in people suffering from xeroderma pigmentosa (Figure 14.20). Affected individuals have skin that is highly sensitive to UV rays from the sun. When individuals are exposed to UV, pyrimidine dimers, especially those of thymine, are formed; people with xeroderma pigmentosa are not able to repair the damage. These are not repaired because of a defect in the nucleotide excision repair enzymes, whereas in normal individuals, the thymine dimers are excised and the defect is corrected. The thymine dimers distort the structure of the DNA double helix, and this may cause problems during DNA replication. 572 Chapter 14 | DNA Structure and Function Figure 14.20 Xeroderma pigmentosa is a condition in which thymine dimerization from exposure to UV is not repaired. Exposure to sunlight results in skin lesions. (credit: James Halpern et al.) Errors during DNA replication are not the only reason why mutations arise in DNA. Mutations, variations in the nucleotide sequence of a genome, can also occur because of damage to DNA. Such mutations may be of two types: induced or spontaneous. Induced mutations are those that result from an exposure to chemicals, UV rays, x-rays, or some other environmental agent. Spontaneous mutations occur without any exposure to any environmental agent; they are a result |
of natural reactions taking place within the body. Mutations may have a wide range of effects. Some mutations are not expressed; these are known as silent mutations. Point mutations are those mutations that affect a single base pair. The most common nucleotide mutations are substitutions, in which one base is replaced by another. These can be of two types, either transitions or transversions. Transition substitution refers to a purine or pyrimidine being replaced by a base of the same kind; for example, a purine such as adenine may be replaced by the purine guanine. Transversion substitution refers to a purine being replaced by a pyrimidine, or vice versa; for example, cytosine, a pyrimidine, is replaced by adenine, a purine. Mutations can also be the result of the addition of a base, known as an insertion, or the removal of a base, also known as deletion. Sometimes a piece of DNA from one chromosome may get translocated to another chromosome or to another region of the same chromosome; this is also known as translocation. These mutation types are shown in Figure 14.22. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 573 Sometimes a nucleotide is overlooked by the DNA repair system for no known reason. This malignant melanoma is the result of DNA not undergoing repair after too much UV exposure. Figure 14.21 Which statement about the above malignant melanoma is most likely true? a. b. c. d. it was the result of a spontaneous mutation it was caused by thymine dimer formation it was caused by a transition substitution it was caused by a transversion substitution 574 Chapter 14 | DNA Structure and Function Figure 14.22 Mutations can lead to changes in the protein sequence encoded by the DNA. A frameshift mutation that results in the insertion of three nucleotides is often less deleterious than a mutation that results in the insertion of one nucleotide. Why? a. Addition of three nucleotides does not shift the reading frame. b. Addition of three nucleotides shifts the reading frame. c. Addition of three nucleotides incorporates two amino acids. d. Addition of three nucleotides removes two amino acids. Mutations in repair genes have been known to cause cancer. Many mutated repair genes have been implicated in certain forms of pancreat |
ic cancer, colon cancer, and colorectal cancer. Mutations can affect either somatic cells or germ cells. If many mutations accumulate in a somatic cell, they may lead to problems such as the uncontrolled cell division observed in cancer. If a mutation takes place in germ cells, the mutation will be passed on to the next generation, as in the case of hemophilia and xeroderma pigmentosa. Think About It Infertility can sometimes be explained by chromosome translocations. Explain how chromosome translocations can cause infertility. Are there times when a chromosome translocation might not result in infertility? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 575 KEY TERMS electrophoresis technique used to separate DNA fragments according to size helicase during replication, this enzyme helps to open up the DNA helix by breaking the hydrogen bonds induced mutation mutation that results from exposure to chemicals or environmental agents lagging strand during replication, the strand that is replicated in short fragments and away from the replication fork leading strand strand that is synthesized continuously in the 5'-3' direction which is synthesized in the direction of the replication fork ligase enzyme that catalyzes the formation of a phosphodiester linkage between the 3' OH and 5' phosphate ends of the DNA mismatch repair type of repair mechanism in which mismatched bases are removed after replication mutation variation in the nucleotide sequence of a genome nucleotide excision repair type of DNA repair mechanism in which the wrong base, along with a few nucleotides upstream or downstream, are removed Okazaki fragment DNA fragment that is synthesized in short stretches on the lagging strand point mutation mutation that affects a single base primase enzyme that synthesizes the RNA primer; the primer is needed for DNA pol to start synthesis of a new DNA strand primer short stretch of nucleotides that is required to initiate replication; in the case of replication, the primer has RNA nucleotides proofreading function of DNA pol in which it reads the newly added base before adding the next one replication fork Y-shaped structure formed during initiation of replication silent mutation mutation that is not expressed single-strand binding protein during replication, protein that binds to the single-stranded DNA; this helps in keeping the two strands of DNA apart so that they may serve as templates sliding clamp ring-shaped protein that holds the DNA pol on the DNA strand spontaneous mutation mutation that takes place in the cells as a |
result of chemical reactions taking place naturally without exposure to any external agent telomerase enzyme that contains a catalytic part and an inbuilt RNA template; it functions to maintain telomeres at chromosome ends telomere DNA at the end of linear chromosomes topoisomerase enzyme that causes underwinding or overwinding of DNA when DNA replication is taking place transformation process in which external DNA is taken up by a cell transition substitution when a purine is replaced with a purine or a pyrimidine is replaced with another pyrimidine transversion substitution when a purine is replaced by a pyrimidine or a pyrimidine is replaced by a purine 576 Chapter 14 | DNA Structure and Function CHAPTER SUMMARY 14.1 Historical Basis of Modern Understanding DNA was first isolated from white blood cells by Friedrich Miescher, who called it nuclein because it was isolated from nuclei. Frederick Griffith's experiments with strains of Streptococcus pneumoniae provided the first hint that DNA may be the transforming principle. Avery, MacLeod, and McCarty proved that DNA is required for the transformation of bacteria. Later experiments by Hershey and Chase using bacteriophage T2 proved that DNA is the genetic material. Chargaff found that the ratio of A = T and C = G, and that the percentage content of A, T, G, and C is different for different species. 14.2 DNA Structure and Sequencing The currently accepted model of the double-helix structure of DNA was proposed by Watson and Crick. Some of the salient features are that the two strands that make up the double helix are complementary and anti-parallel in nature. Deoxyribose sugars and phosphates form the backbone of the structure, and the nitrogenous bases are stacked inside. The diameter of the double helix, 2 nm, is uniform throughout. A purine always pairs with a pyrimidine; A pairs with T, and G pairs with C. One turn of the helix has ten base pairs. During cell division, each daughter cell receives a copy of the DNA by a process known as DNA replication. Prokaryotes are much simpler than eukaryotes in many of their features. Most prokaryotes contain a single, circular chromosome. In general, eukaryotic chromosomes contain a linear DNA molecule packaged into nucleosomes, and have two distinct regions that can be distinguished by staining, reflecting different states of packaging and compaction. 14.3 Basics of DNA Replication |
The model for DNA replication suggests that the two strands of the double helix separate during replication, and each strand serves as a template from which the new complementary strand is copied. In conservative replication, the parental DNA is conserved, and the daughter DNA is newly synthesized. The semi-conservative method suggests that each of the two parental DNA strands acts as template for new DNA to be synthesized; after replication, each double-stranded DNA includes one parental or “old” strand and one “new” strand. The dispersive mode suggested that the two copies of the DNA would have segments of parental DNA and newly synthesized DNA. 14.4 DNA Replication in Prokaryotes Replication in prokaryotes starts from a sequence found on the chromosome called the origin of replication—the point at which the DNA opens up. Helicase opens up the DNA double helix, resulting in the formation of the replication fork. Single-strand binding proteins bind to the single-stranded DNA near the replication fork to keep the fork open. Primase synthesizes an RNA primer to initiate synthesis by DNA polymerase, which can add nucleotides only in the 5' to 3' direction. One strand is synthesized continuously in the direction of the replication fork; this is called the leading strand. The other strand is synthesized in a direction away from the replication fork, in short stretches of DNA known as Okazaki fragments. This strand is known as the lagging strand. Once replication is completed, the RNA primers are replaced by DNA nucleotides and the DNA is sealed with DNA ligase, which creates phosphodiester bonds between the 3'-OH of one end and the 5' phosphate of the other strand. 14.5 DNA Replication in Eukaryotes Replication in eukaryotes starts at multiple origins of replication. The mechanism is quite similar to prokaryotes. A primer is required to initiate synthesis, which is then extended by DNA polymerase as it adds nucleotides one by one to the growing chain. The leading strand is synthesized continuously, whereas the lagging strand is synthesized in short stretches called Okazaki fragments. The RNA primers are replaced with DNA nucleotides; the DNA remains one continuous strand by linking the DNA fragments with DNA ligase. The ends of the chromosomes pose a problem as polymerase is unable to extend them without a primer. Telomerase, an enzyme with an inbuilt RNA template, extends the ends by copying the |
RNA template and extending one end of the chromosome. DNA polymerase can then extend the DNA using the primer. In this way, the ends of the chromosomes are protected. 14.6 DNA Repair DNA polymerase can make mistakes while adding nucleotides. It edits the DNA by proofreading every newly added base. Incorrect bases are removed and replaced by the correct base, and then a new base is added. Most mistakes are corrected during replication, although when this does not happen, the mismatch repair mechanism is employed. Mismatch repair enzymes recognize the wrongly incorporated base and excise it from the DNA, replacing it with the correct base. In yet This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 577 another type of repair, nucleotide excision repair, the incorrect base is removed along with a few bases on the 5' and 3' end, and these are replaced by copying the template with the help of DNA polymerase. The ends of the newly synthesized fragment are attached to the rest of the DNA using DNA ligase, which creates a phosphodiester bond. Most mistakes are corrected, and if they are not, they may result in a mutation defined as a permanent change in the DNA sequence. Mutations can be of many types, such as substitution, deletion, insertion, and translocation. Mutations can be induced or may occur spontaneously. REVIEW QUESTIONS 1. Who was the first person to isolate the material that came to be known as nucleic acids? a. Frederick Griffith b. Friedrich Miescher c. James Watson d. Oswald Avery 2. What is bacterial transformation? sequence: a. 3’-AATGCTAC-5’ b. 3’-CATCGTAA-5’ c. 3’-TTACGATG-5’ d. 3’-GTAGCATT-5’ 7. The DNA double helix does not have which of the following? a. The transformation of a bacterium occurs during a. antiparallel configuration replication. b. It is the transformation of a bacterium into a pathogenic form. c. Transformation of bacteria involves changes in its chromosome. d. Transformation is a process in which external DNA is taken up by a cell, thereby changing morphology and physiology. 3. What type of nucleic acid material is analyzed the most frequently in forensics cases? |
a. cytoplasmic rRNA b. mitochondrial DNA c. nuclear chromosomal DNA d. nuclear mRNA 4. The experiments by Hershey and Chase helped confirm that DNA was the hereditary material on the basis of the finding of what? a. Radioactive phages were found in the pellet. b. Radioactive cells were found in the supernatant. c. Radioactive sulfur was found inside the cell. d. Radioactive phosphorus was found in the cell. 5. If DNA of a particular species was analyzed and it was found that it contains 27% A, what would be the percentage of C? a. b. c. d. 23% 27% 30% 54% b. complementary base pairing c. major and minor grooves d. uracil 8. What is a purine? a. a double ring structure with a six-membered ring fused to a five-membered ring b. a single six-membered ring c. a six-membered ring d. three phosphates covalently bonded by phosphodiester bonds 9. What is the name of the method developed by Fred Sanger to sequence DNA? a. Dideoxy Chain Termination method b. Double Helix Determination c. Polymerase Chain Reaction d. Polymer Gel Electrophoresis 10. What happens when a dideoxynucleotide is added to a developing DNA strand? a. The chain extends to the end of the DNA strand. b. The DNA stand is duplicated. c. The chain is not extended any further. d. The last codon is repeated. 11. In eukaryotes, what is DNA wrapped around? a. histones b. polymerase c. d. single-stranded binding proteins sliding clamp 6. If the sequence of the 5’ to 3’ strand is AATGCTAC, then the complementary sequence has the following 12. Which enzyme is only found in prokaryotic organisms? 578 Chapter 14 | DNA Structure and Function DNA strand during replication? a. DNA gyrase b. helicase c. d. ligase telomerase 20. Which enzyme is most directly responsible for the main process of producing a new DNA strand? a. DNA pol I b. DNA pol II c. DNA pol III d. DNA pol I, DNA pol II, and DNA pol III 21. Which portion of a chromosome contains Okazaki fragments? a. helicase b. c. lagging strand |
leading strand d. primer 22. Which of the following does the enzyme primase synthesize? a. DNA primer b. Okazaki fragments c. phosphodiester linkage d. RNA primer 23. The ends of the linear chromosomes are maintained by what? a. DNA polymerase b. helicase c. primase d. telomerase 24. What is the difference in the rate of replication of nucleotides between prokaryotes and eukaryotes? a. Eukaryotes are 50 times slower. b. Eukaryotes are 20 times faster. c. Prokaryotes are 100 times slower. d. Prokaryotes are 10 times faster. 25. What are Autonomously Replicating Sequences (ARS)? a. DNA gyrase b. helicase c. d. ligase telomerase 13. Uracil is found where? a. chromosomal DNA b. helicase c. mitochondrial DNA d. mRNA 14. What prevents the further development of a DNA strand in Sanger sequencing? a. b. c. d. the addition of DNA reductase the addition of dideoxynucleotides the elimination of DNA polymerase the addition of uracil 15. Which of the following is not one of the proteins involved during the formation of the replication fork? a. helicase b. ligase c. origin of replication d. single-strand binding proteins 16. In which direction does DNA replication take place? a. 5’ to 3’ b. 3’ to 5’ c. 5’ d. 3’ 17. Meselson and Stahl’s experiments proved that DNA replicates by which mode? a. conservative b. converse c. dispersive d. semi-conservative 18. Which set of results was found in the Meselson and Stahl’s experiments? a. The original chromosome was kept intact and a duplicate was made. b. The original chromosome was split and half went to each duplicate. c. The original chromosome was mixed with new material and each duplicate strand contained both old and new. d. The original chromosome was used as a template for two new chromosomes and discarded. 19. Which enzyme initiates the splitting of the double This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 579 a. areas of prokaryotic chromosomes that initiate a |
. frameshift copying b. portions of prokaryotic chromosomes that can be transferred from one organism to another c. areas of eukaryotic chromosomes that are equivalent to the origin of replication in E. coli d. portions of eukaryotic chromosomes that replicate independent of the parent chromosome 26. What type of body cell does not exhibit telomerase activity? a. adult stem cells b. embryonic cells c. germ cells d. liver cells 27. During proofreading, which of the following enzymes reads the DNA? a. DNA polymerase b. helicase c. topoisomerase d. primase 28. If a prokaryotic cell is replicating nucleotides at a rate of 100 per second, how fast would a eukaryotic cell be replicating nucleotides? a. b. c. d. 1000 per second 100 per second 10 per second 1 per second 29. Which type of point mutation would have no effect on gene expression? CRITICAL THINKING QUESTIONS 34. Explain Griffith’s transformation experiments. What did he conclude from them? b. missense c. nonsense d. silent 30. Which type of point mutation would result in the substitution of a stop codon for an amino acid? a. frameshift b. missense c. nonsense d. silent 31. You have developed skin cancer and you are pregnant. You are worried that your child will be born with the cancer you have while carrying the baby. Should you be worried? a. Yes, the cancer can spread to the baby. b. No, the mutations causing the cancer are in somatic cells, not reproductive germ cells. c. Yes, the mutations can be passed on to the child through the placenta. d. No, UV light only affects adult, somatic cells. 32. What is the initial mechanism for repairing nucleotide errors in DNA? a. DNA polymerase proofreading b. mismatch repair c. nucleotide excision repair d. thymine dimers 33. Nucleotide excision repair is often employed when UV exposure causes the formation of what? a. phosphodiester bonds b. purine conjugates c. pyrimidine dimers d. tetrad disassembly 580 Chapter 14 | DNA Structure and Function a. The amount of adenine, thymine, guanine, and cytosine varies from species to species and are not found in equal quantities. They do not vary between individuals of the same species and can |
be used to identify different species. b. The amount of adenine, thymine, guanine, and cytosine varies from species to species and is found in equal quantities. They do not vary between individuals of the same species and can be used to identify different species. c. The amount of adenine and thymine is equal to guanine and cytosine and is found in equal quantities. They do not vary between individuals of the same species and can be used to identify different species. d. The amount of adenine, thymine, guanine, and cytosine varies from species to species and they are not found in equal quantities. They vary between individuals of the same species and can be used to identify different species. 37. In the Avery, Macleod, and McCarty experiments, what conclusion would the scientists have drawn if the use of proteases prevented the transformation of R strain bacteria? 38. Describe the structure and complementary base pairing of DNA. a. Two strains of S. pneumoniae were used for the experiment. Griffith injected a mouse with heatinactivated S strain (pathogenic) and R strain (non-pathogenic). The mouse died and S strain was recovered from the dead mouse. He concluded that external DNA is taken up by a cell that changed morphology and physiology. b. Two strains of Vibrio cholerae were used for the experiment. Griffith injected a mouse with heatinactivated S strain (pathogenic) and R strain (non-pathogenic). The mouse died and S strain was recovered from the dead mouse. He concluded that external DNA is taken up by a cell that changed morphology and physiology. c. Two strains of S. pneumoniae were used for the experiment. Griffith injected a mouse with heatinactivated S strain (pathogenic) and R strain (non-pathogenic). The mouse died and R strain was recovered from the dead mouse. He concluded that external DNA is taken up by a cell that changed morphology and physiology. d. Two strains of S. pneumoniae were used for the experiment. Griffith injected a mouse with heatinactivated S strain (pathogenic) and R strain (non-pathogenic). The mouse died and S strain was recovered from the dead mouse. He concluded that mutation occurred in the DNA of the cell that changed morphology and physiology. 35. Explain why radioactive sulfur and phosphorous were used to label bacteriophages in the Hershey and Chase experiments. a. Protein was labeled with radioactive |
sulfur and DNA was labeled with radioactive phosphorous. Phosphorous is found in DNA, so it will be tagged by radioactive phosphorous. b. Protein was labeled with radioactive phosphorous and DNA was labeled with radioactive sulfur. Phosphorous is found in DNA, so it will be tagged by radioactive phosphorous. c. Protein was labeled with radioactive sulfur and DNA was labeled with radioactive phosphorous. Phosphorous is found in DNA, so DNA will be tagged by radioactive sulfur. d. Protein was labeled with radioactive sulfur and DNA was labeled with radioactive phosphorous. Phosphorous is found in DNA, so DNA will be tagged by radioactive sulfur. 36. How can Chargaff’s rules be used to identify different species? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 581 a. DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with thymine and cytosine pairs with guanine. The two strands are anti-parallel in nature; that is, the 3’ end of one strand faces the 5’ end of other strand. Sugar, phosphate and nitrogenous bases contribute to the DNA structure. b. DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with cytosine and thymine pairs with guanine. The two strands are anti-parallel in nature; that is, the 3’ end of one strand faces the 5’ end of other strand. Sugar, phosphate and nitrogenous bases contribute to the DNA structure. c. DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with thymine and cytosine pairs with guanine. The two strands are parallel in nature; that is, the 3’ end of one strand faces the 3’ end of other strand. Sugar, phosphate and nitrogenous bases contribute to the DNA structure. d. DNA is made up of two strands that are twisted around each other to form a helix. Adenine pairs up with thymine and cytosine pairs with guanine. The two strands are anti-parallel in nature; that is, the 3’ end of one strand faces the 5’ end of other strand. Only sugar contributes to the DNA structure. a. Frederick |
Sanger’s sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner. b. Frederick Sanger’s sequencing is a chain elongation method that is used to generate DNA fragments that elongate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner. c. Frederick Sanger’s sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is joined together by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a laser scanner. d. Frederick Sanger’s sequencing is a chain termination method that is used to generate DNA fragments that terminate at different points using dye-labeled dideoxynucleotides. DNA is separated by electrophoresis on the basis of size. The DNA sequence can be read out on an electropherogram generated by a magnetic scanner. 40. Compare and contrast the similarities and differences between eukaryotic and prokaryotic DNA. 39. Provide a brief summary of the Sanger sequencing method. 582 Chapter 14 | DNA Structure and Function a. Eukaryotes have a single, circular chromosome, a. DNA polymerase reads the template strand in while prokaryotes have multiple, linear chromosomes. Prokaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that create heterochromatin and euchromatin, which is not present in prokaryotes. the 3’ to 5’ direction and adds nucleotides only in the 5’ to 3’ direction. The leading strand is synthesized in the direction of the replication fork. Replication on the lagging strand occurs in the direction away from the replication fork in short stretches of DNA called Okazaki fragments. b. Prokaryotes have a single, circular chromosome, b. DNA polymerase reads the template strand in while eukaryotes have multiple, linear |
chromosomes. Prokaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that could form heterochromatin, which is not present in prokaryotes. the 5’ to 3’ direction and adds nucleotides only in the 5’ to 3’ direction. The leading strand is synthesized in the direction of the replication fork. Replication on the lagging strand occurs in the direction away from the replication fork in short stretches of DNA called Okazaki fragments. c. Prokaryotes have a single, circular chromosome, c. DNA polymerase reads the template strand in while eukaryotes have multiple, linear chromosomes. Eukaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Prokaryotes chromosomes are wrapped around histone proteins that could form heterochromatin, which is not present in eukaryotes. the 3’ to 5’ direction and adds nucleotides only in the 5’ to 3’ direction. The leading strand is synthesized in the direction away from the replication fork. Replication on the lagging strand occurs in the direction of the replication fork in short stretches of DNA called Okazaki fragments. d. Prokaryotes have a single, circular chromosome, d. DNA polymerase reads the template strand in while eukaryotes have multiple, linear chromosomes. Prokaryotes pack their chromosomes by super coiling, managed by DNA gyrase. Eukaryote chromosomes are wrapped around histone proteins that could form heterochromatin, which is present in prokaryotes. the 5’ to 3’ direction and adds nucleotides only in the 3’ to 5’ direction. The leading strand is synthesized in the direction of the replication fork. Replication on the lagging strand occurs in the direction away from the replication fork in long stretches of DNA called Okazaki fragments. 41. DNA replication is bidirectional and discontinuous; explain your understanding of those concepts. 42. Discuss how the scientific community learned that DNA replication takes place in a semiconservative fashion. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 583 a. Replication of the lagging strand occurs in the direction away from the replication fork in short stretches of DNA, since |
access to the DNA is always from the 5’ end. This results in pieces of DNA being replicated in a discontinuous fashion. b. Replication of the leading strand occurs in the direction away from the replication fork in short stretches of DNA, since access to the DNA is always from the 5’ end. This results in pieces of DNA being replicated in a discontinuous fashion. c. Replication of the lagging strand occurs in the direction of the replication fork in short stretches of DNA, since access to the DNA is always from the 5’ end. This results in pieces of DNA being replicated in a discontinuous fashion. d. Replication of the lagging strand occurs in the direction away from the replication fork in short stretches of DNA, since access to the DNA is always from the 3’ end. This results in pieces of DNA being replicated in a discontinuous fashion. 44. Explain the events taking place at the replication fork. If the gene for helicase is mutated, what part of replication will be affected? a. Meselson and Stahl experimented with E. coli. DNA grown in 15 N was heavier than DNA grown in 14 N. When DNA in 15 N was switched to 14 N media, DNA sedimented halfway between the 15 N and 14 N levels after one round of cell division, indicating fifty percent presence of 14 N. This supports the semi-conservative replication model. b. Meselson and Stahl experimented with S. pneumonia. DNA grown in 15 N was heavier than DNA grown in 14 N. When DNA in 15 N was switched to 14 N media, DNA sedimented halfway between the 15 N and 14 N levels after one round of cell division, indicating fifty percent presence of 14 N. This supports the semi-conservative replication model. c. Meselson and Stahl experimented with E. coli. DNA grown in 14 N was heavier than DNA grown in 15 N. When DNA in 15 N was switched to 14 N media, DNA sedimented halfway between the 15 N and 14 N levels after one round of cell division, indicating fifty percent presence of 14 N. This supports the semi-conservative replication model. d. Meselson and Stahl experimented with S. pneumonia. DNA grown in 15 N was heavier than DNA grown in 14 N. When DNA in 15 N was switched to 14 N media, DNA sedimented halfway between the 15 N and 14 N levels after one round of cell division, indicating complete presence of 14 N. This supports the semiconservative replication model. 43. |
Explain why half of DNA is replicated in a discontinuous fashion. 584 Chapter 14 | DNA Structure and Function a. Helicase separates the DNA strands at the origin and DNA ligase in DNA replication. a. DNA polymerase I removes the RNA primers from the developing copy of DNA. DNA ligase seals the ends of the new segment, especially the Okazaki fragments. b. DNA polymerase I adds the RNA primers to the already developing copy of DNA. DNA ligase separates the ends of the new segment, especially the Okazaki fragments. c. DNA polymerase I seals the ends of the new segment, especially the Okazaki fragments. DNA ligase removes the RNA primers from the developing copy of DNA. d. DNA polymerase I removes the enzyme primase from the developing copy of DNA. DNA ligase seals the ends of the old segment, especially the Okazaki fragments. 47. If the rate of replication in a particular prokaryote is 900 nucleotides per second, how long would it take to make two copies of a 1.2 million base pair genome? a. b. c. d. 22.2 minutes 44.4 minutes 45.4 minutes 54.4 minutes 48. How do the linear chromosomes in eukaryotes ensure that their ends are replicated completely? a. The ends of the linear chromosomes are maintained by the activity of the telomerase enzyme. b. The ends of the linear chromosomes are maintained by the formation of a replication fork. c. The ends of the linear chromosomes are maintained by the continuous joining of Okazaki fragments. d. The ends of the linear chromosomes are maintained by the action of the polymerase enzyme. 49. Compare and contrast prokaryotic and eukaryotic DNA replication. of replication. Topoisomerase breaks and reforms DNA’s phosphate backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will not be separated at the beginning of replication. b. Helicase joins the DNA strands together at the origin of replication. Topoisomerase breaks and reforms DNA’s phosphate backbone after the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes RNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated |
, the DNA strands will not be joined together at the beginning of replication. c. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s sugar backbone ahead of the replication fork, thereby increasing the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by DNA polymerase to form a daughter strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication. d. Helicase separates the DNA strands at the origin of replication. Topoisomerase breaks and reforms DNA’s sugar backbone ahead of the replication fork, thereby relieving the pressure. Single-stranded binding proteins prevent reforming of DNA. Primase synthesizes DNA primer which is used by RNA polymerase to form a parent strand. If helicase is mutated, the DNA strands will be separated at the beginning of replication. 45. What are Okazaki fragments and how they are formed? a. Okazaki fragments are short stretches of DNA on the lagging strand, which is synthesized in the direction away from the replication fork. b. Okazaki fragments are long stretches of DNA on the lagging strand, which is synthesized in the direction of the replication fork. c. Okazaki fragments are long stretches of DNA on the leading strand, which is synthesized in the direction away from the replication fork. d. Okazaki fragments are short stretches of DNA on the leading strand, which is synthesized in the direction of the replication fork. 46. Compare and contrast the roles of DNA polymerase I This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 585 a. A prokaryotic organism’s rate of replication is ten times faster than that of eukaryotes. Prokaryotes have a single origin of replication and use five types of polymerases, while eukaryotes have multiple sites of origin and use fourteen polymerases. Telomerase is absent in prokaryotes. DNA pol I is the primer remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes. b. A prokaryotic organism’s rate of replication is ten times slower than that of eukary |
otes. Prokaryotes have a single origin of replication and use five types of polymerases, while eukaryotes have multiple sites of origin and use fourteen polymerases. Telomerase is absent in eukaryotes. DNA pol I is the primer remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes. c. A prokaryotic organism’s rate of replication is ten times faster than that of eukaryotes. Prokaryotes have five origins of replication and use a single type of polymerase, while eukaryotes have a single site of origin and use fourteen polymerases. Telomerase is absent in prokaryotes. DNA pol I is the primer remover in prokaryotes, while in eukaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes. d. A prokaryotic organism’s rate of replication is ten times slower than that of eukaryotes. Prokaryotes have a single origin of replication and use five types of polymerases, while eukaryotes have multiple sites of origin and use fourteen polymerases. Telomerase is absent in prokaryotes. DNA pol I is the primer remover in eukaryotes, while in prokaryotes it is RNase H. DNA pol III performs strand elongation in prokaryotes and pol δ and pol ε do the same in eukaryotes. 50. What would be the consequence of a mutation in a mismatch repair enzyme? How will this affect the function of a gene? a. Mismatch repair corrects the errors after the replication is completed by excising the incorrectly added nucleotide and adding the correct base. Any mutation in a mismatch repair enzyme would lead to more permanent damage. b. Mismatch repair corrects the errors during the replication by excising the incorrectly added nucleotide and adding the correct base. Any mutation in the mismatch repair enzyme would lead to more permanent damage. c. Mismatch repair corrects the errors after the replication is completed by excising the added nucleotides and adding more bases. Any mutation in the mismatch repair enzyme would lead to more permanent damage. d. Mism |
atch repair corrects the errors after the replication is completed by excising the incorrectly added nucleotide and adding the correct base. Any mutation in the mismatch repair enzyme would lead to more temporary damage. 51. A mutation has occurred in the DNA and in the mRNA for a gene. Discuss which would have a more significant effect on gene expression. Why? a. Both will result in the production of defective proteins. The DNA mutation, if not corrected, is permanent, while the mRNA mutation will only affect proteins made from that mRNA strand. Production of defective protein ceases when the mRNA strand deteriorates. b. Both will result in the production of defective proteins. The DNA mutation, if not corrected, is permanent, while the mRNA mutation will not affect proteins made from that mRNA strand. Production of defective protein continues when the mRNA strand deteriorates. c. Only DNA will result in the production of defective proteins. The DNA mutation, if not corrected, is permanent. Production of defective protein ceases when the DNA strand deteriorates. d. Only mRNA will result in the production of defective proteins. The mRNA mutation will only affect proteins made from that mRNA strand. Production of defective protein ceases when the mRNA strand deteriorates. 52. Discuss the effects of point mutations on a DNA strand. 586 Chapter 14 | DNA Structure and Function a. Mutations can cause a single change in an amino acid. A nonsense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in non-functional proteins. b. Mutations can cause a single change in amino acid. A missense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in non-functional proteins. c. Mutations can cause a single change in amino acid. A nonsense mutation can stop the replication or reading of that strand. Substitution mutations can cause a frame shift. This can result in non-functional proteins. d. Mutations can cause a single change in amino acid. A nonsense mutation can stop the replication or reading of that strand. Insertion or deletion mutations can cause a frame shift. This can result in functional proteins. TEST PREP FOR AP® COURSES 54. What would Chase and Hershey have concluded if the supernatant contained radioactive labeled-phosphorus? a. DNA was the primary source of heritable information. b. RNA was the primary source of heritable information. c. Protein was the primary source of |
heritable information. d. Phages were the primary source of heritable information. 55. Which piece of evidence supports that the material Miescher discovered was DNA? a. The precipitate contained sulfur. b. The precipitate contained oxygen. c. The precipitate contained phosphorus. d. The precipitate contained protein. 53. Discuss the significance of mutations in tRNA and rRNA. a. Mutations in tRNA and rRNA would lead to the production of defective proteins or no protein production. b. Mutations in tRNA and rRNA would lead to changes in the semi-conservative mode of replication of DNA. c. Mutations in tRNA and rRNA would lead to production of a DNA strand with a mutated single strand and normal other strand. d. Mutations in tRNA and rRNA would lead to skin cancer in patients of xeroderma pigmentosa. a. Comparison of DNA from a known source or individual with analysis of the sequence of an unknown sample of DNA allows scientists to find out if both of them are similar or not. b. DNA from the unknown sample is sequenced and analyzed. The result of the analysis is then matched with any random population. The matching individual then helps in forensics. c. Comparison of DNA from a known source or individual with analysis of the sequence of bases in strands of an unknown sample of RNA allows scientists to find out if both of them are similar or not. d. Comparison of DNA from a known source or individual with analysis of the sugars and phosphates in strands of an unknown sample of DNA allows scientists to find out if both of them are similar or not. 56. Explain how forensic scientists are able to use DNA analysis to identify individuals. 57. Discuss the contributions of Francis Crick, James Watson, and Rosalind Franklin to the discovery of the structure of DNA. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 587 a. Rosalind Franklin used X-ray diffraction methods to demonstrate the helical nature of DNA, while Watson and Crick formulated the double stranded structural model of DNA. b. Rosalind Franklin, Watson and Crick first employed the technique of X-ray diffraction to understand the storage of DNA. Since it did not work out, Watson and Crick then ran experiments to ascertain the DNA structure. c. Rosalind Franklin, Watson and Crick used X-ray |
diffraction methods to demonstrate the helical nature of DNA, while Rosalind Franklin formulated the double stranded structural model of DNA. d. Watson and Crick used X-ray diffraction methods to demonstrate the helical nature of DNA, while Rosalind Franklin formulated the double stranded structural model of DNA. 58. What do RNA and DNA have in common? a. Both contain four different nucleotides. b. Both are usually double-stranded molecules. c. Both contain adenine and uracil. d. Both contain ribose. 59. Which of the following would be a good application of plasmid transformation? a. b. c. d. to make copies of DNA to isolate a change in a single nucleotide to separate DNA fragments to sequence DNA 60. Explain how the components of DNA fit together. a. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with thymine and guanine pairs with cytosine. Adenine and thymine form two hydrogen bonds and cytosine and guanine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other. b. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form two hydrogen bonds and guanine and thymine form three hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run antiparallel to each other. c. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run |
antiparallel to each other. d. DNA is composed of nucleotides, consisting of a 5 carbon sugar, a phosphate, and a nitrogenous base. DNA is a double helical structure in which complementary base pairing occurs. Adenine pairs with cytosine and guanine pairs with thymine. Adenine and cytosine form three hydrogen bonds and guanine and thymine form two hydrogen bonds. The two individual strands of DNA are held together by covalent bonds between the phosphate of one nucleotide and sugar of the next. The two strands run parallel to each other. 61. Describe the Sanger DNA sequencing method used for the human genome sequencing project. 588 Chapter 14 | DNA Structure and Function a. A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is added to the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPS’s. b. A DNA sample is denatured by heating and then put into four tubes. A primer, RNA polymerase and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is added to the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different ddNTPS’s. c. A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase and all four nucleotides are added. Limited quantities of one of the four dideoxynucleotides (ddNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent ddNTP is removed from the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths |
specific to the four different ddNTPS’s. d. A DNA sample is denatured by heating and then put into four tubes. A primer, DNA polymerase and all four nucleotides are added. Limited quantities of one of the four deoxynucleotides (dNTPs) are added to each tube respectively. Each one of them carries a specific fluorescent label. Chain elongation continues until a fluorescent dNTP is added the growing chain, after which chain termination occurs. Gel electrophoresis is performed and the length of each base is detected by laser scanners with wavelengths specific to the four different dNTPS’s. 62. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 What process is illustrated in the figure? a. transcription b. mutation c. excision d. translation 63. Describe how the model of DNA replication illustrates the function of topoisomerase. a. Topoisomerase relieves the pressure that results from supercoiling by breaking and reforming DNA’s phosphate backbone ahead of the replication fork. b. Topoisomerase increases the pressure to increase supercoiling by breaking and reforming DNA’s phosphate backbone ahead of the replication fork. c. Topoisomerase relieves the pressure that results from supercoiling by breaking and reforming DNA’s nucleotide base pairs ahead of the replication fork. d. Topoisomerase relieves the pressure that results from separation of DNA strands by breaking and reforming DNA’s phosphate backbone ahead of the replication fork. 64. Flamingos have genotypes for white feathers yet often appear with pink feathers within the same population. What is most likely affecting the phenotype of some flamingos, causing their feathers to turn pink in an isolated population? a. weather variations b. dietary changes c. DNA mutations d. translation failure 65. What can be the result of DNA failing to undergo repair after too much UV exposure? a. second degree burns b. a malignant melanoma c. a breakdown of deep layers of the skin d. a sun burn 66. Identify the type of change that can occur in the DNA of a chromosome that is termed a chromosomal mutation. Chapter 14 | DNA Structure and Function 589 a. b. substitution translocation c. missense d. transversion 67. Explain why patients with Xeroderma Pigmentosa are more prone to cancer than the rest of |
the population. a. Xeroderma Pigmentosa patients cannot employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, thymine dimers are formed and they are not able to repair this defect. These dimers distort the structure of DNA and cause them to have a high risk of contracting skin cancer. b. Xeroderma Pigmentosa patients can employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, the thymine dimers are formed and they are able to repair this defect. These dimers do not distort the structure of DNA and they have moderate risk of contracting skin cancer. c. Xeroderma Pigmentosa patients cannot employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, the adjacent adenine forms dimers and they are not able to repair this defect. These dimers distort the structure of DNA and they have high risk of contracting skin cancer. d. Xeroderma Pigmentosa patients cannot employ the nucleotide excision repair mechanism. When these patients are exposed to UV light, the adjacent thymine cannot form thymine dimers and they are not able to repair this defect. The non-formation of dimers distorts the structure of DNA and they have high risk of contracting skin cancer. 68. You are looking at two fragments of DNA. Both have the sequence CATTCTG on one strand and GTAAGAC on the other. One of the fragments is exposed to UV light, the other is not. What will happen to the fragments and how might these mutations be repaired? a. The fragment exposed to UV light contains thymine dimers. Thymines lying adjacent to each other can form thymine dimers when exposed to UV light. They can be repaired by nucleotide excision. b. The fragment exposed to UV light contains adenine dimers. Adenines lying adjacent to each other can form dimers when exposed to UV light. They can be repaired by nucleotide excision. c. The fragment exposed to UV light contains thymine dimers. Thymines lying parallel to each other can form thymine dimers when exposed to UV light. They can be repaired by nucleotide excision. d. The fragment exposed to UV light contains thymine dimers. Thymines lying adjacent to each other can form thymine dimers when exposed to UV light. They can be synthesized by nucleotide excision. 69. Discuss how mutations can increase variation within |
a population. a. Substitution mutations may cause a different amino acid to be placed at a specific location, causing small changes in the protein. Frameshift mutations usually cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring. b. Substitution mutations may cause multiple amino acid changes, increasing chances that a new protein will form, leading to radically different characteristics in the offspring. Frameshift mutations may cause a different amino acid to be placed at a specific location, causing small changes in a protein. c. Substitution mutations may cause a different amino acid to be placed at a specific location, resulting in major changes to the protein and leading to radically different characteristics in the offspring. Frameshift mutations cause multiple amino acid differences in a protein, leading to small changes in the protein. d. Substitution mutations result in a different amino acid being placed at a specific position in a protein, causing small changes. Silent mutations could result in new characteristics possessed by an offspring when a stop codon is substituted for an amino acid. SCIENCE PRACTICE CHALLENGE QUESTIONS 70. The proof that DNA, not protein, is the carrier of genetic information involved a number of historical experiments, including transformation or horizontal gene transfer (HGT), which is the uptake and expression of extracellular DNA. A. As described in Figure 14.3, transformation or HGT 590 Chapter 14 | DNA Structure and Function density, or heat shock. Such conditions are often difficult to model in the laboratory, where competence can be induced by high concentrations of divalent cations, Ca+2 or Mg+2, or electrical shock. In either setting, extracellular DNA can be transported into the cell, and (to a good approximation) uptake is proportional to the concentration of extracellular DNA. D. Identify a factor that might affect transformation or HGT. Then, design a plan to evaluate the dependence of transformational efficiency (defined as the number of transformations per gram of extracellular DNA) of plasmids that transfer antibiotic resistance to a particular strain of Escherichia coli that is not resistant on that factor. 71. Prior to the work of Hershey and Chase, scientists thought that inheritance involved “nucleoproteins.” The amount of information to be transmitted between generations did not seem consistent with the chemical simplicity of the few nucleotides found in polymers of deoxyribonucleic acids in comparison to the diversity of |
protein polymers. Briefly explain: • • • the relationship between the structure of polymeric DNA and the information stored the relationship between the interactions between base pairs on complementary strands of the double helix and Chargaff’s observation on the relative abundance of nucleotides in DNA the meaning of the statement from the Nature publication on the structure of DNA by Watson and Crick: “It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.” 72. In 1977, Fred Sanger developed a method to determine the order of nucleotides in a strand of DNA. Sanger won a Nobel Prize for his work, and his method of sequencing based on dideoxy chain termination (Figure 14.8) has been foundational to the rapid development of more modern, rapid, and cheap methods of sequencing. The challenge of the $1,000 in one-day sequencing of the human genome was achieved in 2016 by next-generation sequencing (NGS), a “catch-all” term describing several sequencing methods. was first reported by Griffith in 1928 in an experiment in which the following occurred: 1. heat-treated, pathogenic bacteria recovered their pathogenicity when incubated with nonpathogenic bacteria 2. plasmids were transferred to nonpathogenic bacteria from pathogenic bacteria through conjugation 3. nonpathogenic bacteria acquired pathogenicity when incubated in a broth containing heattreated, pathogenic bacteria 4. polysaccharide cell capsules from pathogenic bacteria were transferred to nonpathogenic bacteria B. Griffith’s experiment, however, left undetermined the identity of the cellular component that encoded genetic information. The identity of DNA as the carrier of genetic information was resolved through the experiments by Martha Chase and Alfred Hershey because they observed the following: 1. injections with a serum containing chemically isolated polysaccharides and nonpathogenic bacteria were not lethal 2. pathogenic bacterial DNA that was radioactively labeled using a phosphorus isotope was not present in mice that died 3. bacteriophages from a bacterial culture grown in a nutrient-containing medium and radioactively labeled using a sulfur isotope transferred the label to bacteria incubated in an unlabeled nutrient-containing medium 4. bacteriophages from a bacterial culture grown in a nutrient-containing medium and radioactively labeled using a sulfur isotope did not transfer the label to bacteria incubated in an unlabeled nutrient-containing medium C. Transformation and transduction |
increase variation within populations of bacteria and archaebacteria by the following: a. b. c. transferring DNA among different species transferring free DNA across the cell membrane without energy expenditure transferring DNA between different strains of the same species of bacteria d. phagocytosis of bacteriophages The evolution of antibiotic resistance via HGT poses a challenge to medical technology. On the other hand, transformation is often assayed by incorporating an antibiotic-resistance gene in the plasmid to be transferred into the host organism. In natural environments, bacterial and archaebacterial cells become competent (able to transport DNA through the cytoplasmic membrane) in response to stress such as UV radiation, high population This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 591 insurers from denying coverage or increasing costs of premiums based on genetic information. It also prohibits employers from making use of these data for hiring, firing, or promotion. The act passed in the House with a vote of 420 to 3, although it was lobbied against by organizations representing business (human resources, health insurance, and manufacturers), including the U.S. Chamber of Commerce. The act does not cover life, long-term care, or disability insurance. Pose three questions that are relevant to the use of whole-genome data. 73. Our understanding of the mechanisms of DNA replication is important to research on cancer and aging. Additionally, the molecular basis of Mendelian genetics was established. A. The mechanism of DNA replication was investigated by Meselson and Stahl. The diagram below from their 1958 paper summarizes their findings. Describe how this representation illustrates the manner in which DNA is copied for transmission between generations. Figure 14.23 A. Using the diagrams shown above for reference, explain the effect of the addition of dideoxynucleotides on chain growth of the DNA strand that is copied during sequencing in terms of the structures of dideoxyribose and deoxyribose. B. Suppose that a single strand to be sequenced is 5’CGAGTACG3’. In the presence of each of the four deoxynucleotides and the dideoxynucleotide ddCTP, describe the strands that would be formed from this template. Include in your description an annotation indicating the 3’ and 5’ ends of the fragments resulting from the procedure. C. Next-generation |
sequencing makes termination technology very rapid and relatively inexpensive. All babies born in the U.S. are currently screened by statemandated tests for several genetic conditions. The number of conditions tested ranges from 29 (GA and KS) to 59 (IL and MS). It is proposed that whole-genome sequencing should be mandatory for all newborns. The Genetic Information Nondiscrimination Act (2008) prevents health Figure 14.24 B. During the synthesis of new strands of DNA from the 592 Chapter 14 | DNA Structure and Function parent strands, DNA polymerase can only add nucleotides at the terminal 3’ of a growing strand. Using the diagram below, describe the similarities and differences between the DNA replication of both strands. Figure 14.25 C. Shown at the left end of the upper parent strand is the six-base repeat sequence TTAGGG. In humans, this is the repeated, telomeric sequence that is attached to the telomere. The RNA primer in humans spans 10 base pairs, unlike in the drawing where it spans only three. In somatic cells, an enzyme called telomerase no longer functions. Explain the function of telomerase in the development of stem cells and cancer cells, and the inhibition of telomerase in programmed cell death or apoptosis. 74. The mitochondria of eukaryote cells contain their own circular DNA (mtDNA), consistent with their origin according to the theory of endosymbiosis. The mitochondrial genome is highly conserved in Eukarya. In humans, the 50 to 100 mitochondria in each of the cells in most tissues have 5 to 10 copies of the genome. Each has 37 genes that primarily encode proteins of the electron transport chain. Point mutations in which a single nucleotide is incorrectly placed is not repaired because the error-checking provided by DNA polymerase is not present in the mitochondria. The mutation rate for mtDNA is approximately 100 times higher than the mutation rate for nuclear DNA. The simultaneous existence of multiple alleles in each cell is likely, a condition called heteroplasmy. In mammals, sperm mitochondria are destroyed prior to fertilization. A. Explain how point mutations in mtDNA can result in a loss of function in critical cellular components such as cytochrome c yet not be lethal to the cell. B. Oocyte mitochondria are randomly segregated during meiosis, resulting in variation in the frequency of mtDNA mutations in offspring relative to the parent. Explain how a loss of function does not accumulate, |
lowering the metabolic performance from generation to generation. As described in the Evolution Connection in this chapter of the text, a fossil fingertip found in a Siberian cave revealed an evolutionary link between Neanderthals and Denisovans. Fossils from 28 individuals were located in the “pit of bones,” Sima de los Huesos, in Spain, thousands of miles from the Siberian cave. In 2013, mtDNA from a femur of one of these individuals was compared with mtDNA of Denisovans, Neanderthals, and modern humans. It was found that the Sima fossil shared many more alleles with Denisovans than with either Neanderthals or modern humans. In 2016, the same group of scientists who sequenced the mtDNA from the femur of one of the Sima fossils partially sequenced the DNA from that fossil, showing a clear connection to Neanderthals. C. Analyze these data to draw alternative conclusions regarding the relatedness of the three fossils and support each with evidence. D. Design a plan to differentiate or resolve these alternative conclusions. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 593 15 | GENES AND PROTEINS Figure 15.1 Genes, which are carried on (a) chromosomes, are linearly organized instructions for making the RNA and protein molecules that are necessary for all of processes of life. The (b) interleukin-2 protein and (c) alpha-2uglobulin protein are just two examples of the array of different molecular structures that are encoded by genes. (credit “chromosome: National Human Genome Research Institute; credit “interleukin-2”: Ramin Herati/Created from PDB 1M47 and rendered with Pymol; credit “alpha-2u-globulin”: Darren Logan/rendered with AISMIG) Chapter Outline 15.1: The Genetic Code 15.2: Prokaryotic Transcription 15.3: Eukaryotic Transcription 15.4: RNA Processing in Eukaryotes 15.5: Ribosomes and Protein Synthesis Introduction The definition of gene has progressed from being an abstract unit of heredity in Mendel’s time to our current concept of a tangible molecular entity capable of replication, expression, and mutation (Figure 15.1). |
Currently, we can perform tests for many genetic diseases, but these tests create ethical and legal issues. For example, would you want to be tested for a debilitating genetic disease if there was the possibility insurance companies could use that information to deny you coverage? Fortunately, the Genetic Information Nondiscrimination Act of 2008 protects American citizens from discrimination from both insurance companies and employers based on genetic information. More information about policy, legal, and ethical issues in genetic research can be found here (http://openstaxcollege.org/l/32genomegov). 15.1 | The Genetic Code In this section, you will explore the following questions: • What is the “Central Dogma” of protein synthesis? • What is the genetic code, and how does nucleotide sequence prescribe the amino acid and polypeptide sequence? 594 Chapter 15 | Genes and Proteins Connection for AP® Courses Since the rediscovery of Mendel’s work in the 1900s, scientists have learned much about how the genetic blueprints stored in DNA are capable of replication, expression, and mutation. Just as the 26 letters of the English alphabet can be arranged into what seems to be a limitless number of words, with new ones added to the dictionary every year, the four nucleotides of DNA—A, T, C, and G—can generate sequences of DNA called genes that specify tens of thousands of polymers of amino acids. In turn, these sequences can be transcribed into mRNA and translated into proteins which orchestrate nearly every function of the cell. The genetic code refers to the DNA alphabet (A, T, C, G), the RNA alphabet (A, U, C, G), and the polypeptide alphabet (20 amino acids). But how do genes located on a chromosome ultimately produce a polypeptide that can result in a physical phenotype such as hair or eye color—or a disease like cystic fibrosis or hemophilia? The Central Dogma describes the normal flow of genetic information from DNA to mRNA to protein: DNA in genes specify sequences of mRNA which, in turn, specify amino acid sequences in proteins. The process requires two steps, transcription and translation. During transcription, genes are used to make messenger RNA (mRNA). In turn, the mRNA is used to direct the synthesis of proteins during the process of translation. Translation also requires two other types of RNA: transfer RNA (tRNA) and ribosomal RNA (rRNA). The genetic code is |
a triplet code, with each RNA codon consisting of three consecutive nucleotides that specify one amino acid or the release of the newly formed polypeptide chain; for example, the mRNA codon CAU specifies the amino acid histidine. The code is degenerate; that is, some amino acids are specified by more than one codon, like synonyms you study in your English class (different word, same meaning). For example, CCU, CCC, CCA, and CCG are all codons for proline. It is important to remember the same genetic code is universal to almost all organisms on Earth. Small variations in codon assignment exist in mitochondria and some microorganisms. Deviations from the simple scheme of the central dogma are discovered as researchers explore gene expression with new technology. For example the human immunodeficiency virus (HIV) is a retrovirus which stores its genetic information in single stranded RNA molecules. Upon infection of a host cell, RNA is used as a template by the virally encoded enzyme, reverse transcriptase, to synthesize DNA. The viral DNA is later transcribed into mRNA and translated into proteins. Some RNA viruses such as the influenza virus never go through a DNA step. The RNA genome is replicated by an RNA dependent RNA polymerase which is virally encoded. The content presented in this section supports the Learning Objectives outlined in Big Idea 1 and Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives merge Essential Knowledge content with one or more of the seven Science Practices. These Learning Objectives provide a transparent foundation for the AP® Biology course, along with inquiry-based laboratory experiences, instructional activities, and AP® Exam questions. Big Idea 1 The process of evolution drives the diversity and unity of life. Enduring Understanding 1.B Organisms are linked by lines of descent from common ancestry. Essential Knowledge 1.B.1 Organisms share many conserved core processes and features that evolved and are widely distributed among organisms today. Science Practice Science Practice Learning Objective Big Idea 3 Enduring Understanding 3.A 3.1 The student can pose scientific questions. 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. 1.15 The student is able to describe specific examples of conserved core biological processes and features shared by all domains or within one domain of life, and how these shared, conserved core processes and |
features support the concept of common ancestry for all organisms. Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 595 Essential Knowledge Science Practice Learning Objective 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. 6.5 The student can evaluate alternative scientific explanations. 3.1 The student is able to construct scientific explanations that use the structure and functions of DNA and RNA to support the claim that DNA and, in some cases, that RNA are the primary sources of heritable information. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 3.4][APLO 3.25] The cellular process of transcription generates messenger RNA (mRNA), a mobile molecular copy of one or more genes with an alphabet of A, C, G, and uracil (U). Translation of the mRNA template converts nucleotide-based genetic information into a protein product. Protein sequences consist of 20 commonly occurring amino acids; therefore, it can be said that the protein alphabet consists of 20 letters (Figure 15.2). Each amino acid is defined by a three-nucleotide sequence called the triplet codon. Different amino acids have different chemistries (such as acidic versus basic, or polar and nonpolar) and different structural constraints. Variation in amino acid sequence gives rise to enormous variation in protein structure and function. Figure 15.2 Structures of the 20 amino acids found in proteins are shown. Each amino acid is composed of an amino + ), a carboxyl group (COO-), and a side chain (blue). The side chain may be nonpolar, polar, or charged, group ( NH3 as well as large or small. It is the variety of amino acid side chains that gives rise to the incredible variation of protein structure and function. 596 Chapter 15 | Genes and Proteins The Central Dogma: DNA Encodes RNA; RNA Encodes Protein The flow of genetic information in cells from DNA to mRNA to protein is described by the Central Dogma (Figure 15.3), which states that genes specify the sequence of mRNAs, which in turn specify the sequence of proteins. The |
decoding of one molecule to another is performed by specific proteins and RNAs. Because the information stored in DNA is so central to cellular function, it makes intuitive sense that the cell would make mRNA copies of this information for protein synthesis, while keeping the DNA itself intact and protected. The copying of DNA to RNA is relatively straightforward, with one nucleotide being added to the mRNA strand for every nucleotide read in the DNA strand. The translation to protein is a bit more complex because three mRNA nucleotides correspond to one amino acid in the polypeptide sequence. However, the translation to protein is still systematic and colinear, such that nucleotides 1 to 3 correspond to amino acid 1, nucleotides 4 to 6 correspond to amino acid 2, and so on. Figure 15.3 Instructions on DNA are transcribed onto messenger RNA. Ribosomes are able to read the genetic information inscribed on a strand of messenger RNA and use this information to string amino acids together into a protein. The Genetic Code Is Degenerate and Universal Given the different numbers of “letters” in the mRNA and protein “alphabets,” scientists theorized that combinations of nucleotides corresponded to single amino acids. Nucleotide doublets would not be sufficient to specify every amino acid because there are only 16 possible two-nucleotide combinations (42). In contrast, there are 64 possible nucleotide triplets (43), which is far more than the number of amino acids. Scientists theorized that amino acids were encoded by nucleotide triplets and that the genetic code was degenerate. In other words, a given amino acid could be encoded by more than one nucleotide triplet. This was later confirmed experimentally; Francis Crick and Sydney Brenner used the chemical mutagen proflavin to insert one, two, or three nucleotides into the gene of a virus. When one or two nucleotides were inserted, protein synthesis was completely abolished. When three nucleotides were inserted, the protein was synthesized and functional. This This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 597 demonstrated that three nucleotides specify each amino acid. These nucleotide triplets are called codons. The insertion of one or two nucleotides completely changed the triplet reading frame, thereby altering the message for every subsequent amino acid (Figure 15.4). Though insertion |
of three nucleotides caused an extra amino acid to be inserted during translation, the integrity of the rest of the protein was maintained. Figure 15.4 The deletion of two nucleotides shifts the reading frame of an mRNA and changes the entire protein message, creating a nonfunctional protein or terminating protein synthesis altogether. Scientists painstakingly solved the genetic code by translating synthetic mRNAs in vitro and sequencing the proteins they specified (Figure 15.5). Figure 15.5 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. (credit: modification of work by NIH) In addition to instructing the addition of a specific amino acid to a polypeptide chain, three of the 64 codons terminate protein synthesis and release the polypeptide from the translation machinery. These triplets are called nonsense codons, or stop codons. Another codon, AUG, also has a special function. In addition to specifying the amino acid methionine, it also serves as the start codon to initiate translation. The reading frame for translation is set by the AUG start codon near the 5' end of the mRNA. The genetic code is universal. With a few exceptions, virtually all species use the same genetic code for protein synthesis. Conservation of codons means that a purified mRNA encoding the globin protein in horses could be transferred to a tulip cell, and the tulip would synthesize horse globin. That there is only one genetic code is powerful evidence that all of life on Earth shares a common origin, especially considering that there are about 1084 possible combinations of 20 amino acids and 64 triplet codons. 598 Chapter 15 | Genes and Proteins Transcribe a gene and translate it (http://openstaxcollege.org/l/create_protein). to protein using complementary pairing and the genetic code at this site Some hereditary and age-related diseases are caused by translation errors. Explain why an error in translation may cause disease. a. If there is an error in translation, the correct lipids will not be made for signaling, storage of energy or to perform vital functions. This can cause hereditary and age-related diseases. b. Translation is the process in which a particular segment of DNA is copied into RNA (mRNA) by the enzyme RNA polymerase. Error in such copying can lead to various hereditary and age-related diseases. c. Translation is the process used by ribosomes |
to synthesize proteins from amino acids. If there is an error in this process, the correct proteins will not be made to build important body tissue or perform vital functions thus leading to hereditary and age-related diseases. d. Translation is the process Golgi bodies use to synthesize proteins from amino acids. If there is an error in this process, the correct proteins will not be made to build important body tissue or perform vital functions. Think About It • A strand of DNA has the nucleotide sequence 3'……GCT GTC AAA TTC GAT……5'. What is the sequence of mRNA that is complementary to this DNA sequence? Using the chart of codons in the text, determine the sequence of amino acids which can be generated from this strand of DNA. • How does degeneracy of the genetic code make cells less vulnerable to mutations? What is an advantage of degeneracy with respect to the negative impact of random mutations on natural selection and evolution? Degeneracy is believed to be a cellular mechanism to reduce the negative impact of random mutations. Codons that specify the same amino acid typically only differ by one nucleotide. In addition, amino acids with chemically similar side chains are encoded by similar codons. This nuance of the genetic code ensures that a single-nucleotide substitution mutation might either specify the same amino acid but have no effect or specify a similar amino acid, preventing the protein from being rendered completely nonfunctional. Which Has More DNA: A Kiwi or a Strawberry? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 599 Figure 15.6 Do you think that a kiwi or a strawberry has more DNA per fruit? (credit “kiwi”: "Kelbv"/Flickr; credit: “strawberry”: Alisdair McDiarmid) Question: Would a kiwifruit and strawberry that are approximately the same size (Figure 15.6) also have approximately the same amount of DNA? Background: Genes are carried on chromosomes and are made of DNA. All mammals are diploid, meaning they have two copies of each chromosome. However, not all plants are diploid. The common strawberry is octoploid (8n) and the cultivated kiwi is hexaploid (6n). Research the total number of chromosomes in the cells of each of these fruits and think about how |
this might correspond to the amount of DNA in these fruits’ cell nuclei. Read about the technique of DNA isolation to understand how each step in the isolation protocol helps liberate and precipitate DNA. Hypothesis: Hypothesize whether you would be able to detect a difference in DNA quantity from similarly sized strawberries and kiwis. Which fruit do you think would yield more DNA? Test your hypothesis: Isolate the DNA from a strawberry and a kiwi that are similarly sized. Perform the experiment in at least triplicate for each fruit. 1. Prepare a bottle of DNA extraction buffer from 900 mL water, 50 mL dish detergent, and two teaspoons of table salt. Mix by inversion (cap it and turn it upside down a few times). 2. Grind a strawberry and a kiwifruit by hand in a plastic bag, or using a mortar and pestle, or with a metal bowl and the end of a blunt instrument. Grind for at least two minutes per fruit. 3. Add 10 mL of the DNA extraction buffer to each fruit, and mix well for at least one minute. 4. Remove cellular debris by filtering each fruit mixture through cheesecloth or porous cloth and into a funnel placed in a test tube or an appropriate container. 5. Pour ice-cold ethanol or isopropanol (rubbing alcohol) into the test tube. You should observe white, precipitated DNA. 6. Gather the DNA from each fruit by winding it around separate glass rods. Record your observations: Because you are not quantitatively measuring DNA volume, you can record for each trial whether the two fruits produced the same or different amounts of DNA as observed by eye. If one or the other fruit produced noticeably more DNA, record this as well. Determine whether your observations are consistent with several pieces of each fruit. Analyze your data: Did you notice an obvious difference in the amount of DNA produced by each fruit? Were your results reproducible? Draw a conclusion: Given what you know about the number of chromosomes in each fruit, can you conclude that chromosome number necessarily correlates to DNA amount? Can you identify any drawbacks to this procedure? If you had access to a laboratory, how could you standardize your comparison and make it more quantitative? 600 Chapter 15 | Genes and Proteins 15.2 | Prokaryotic Transcription In this section, you will explore the following questions: • What are the steps, in order, in prokaryotic transcription? • How and when is transcription terminated |
? Connection for AP® Courses During transcription, the enzyme RNA polymerase moves along the DNA template, reading nucleotides in a 3′ to 5′ direction, with U pairing with A and C with G, and assembling the mRNA transcript in a 5′ to 3′ direction. In prokaryotes, mRNA synthesis is initiated at a promoter sequence on the DNA template. Transcription continues until RNA polymerase reaches a stop or terminator sequence at the end of the gene. Termination frees the mRNA and often occurs by the formation of an mRNA hairpin. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® Exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.A Essential Knowledge Science Practice Learning Objective Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. 6.5 The student can evaluate alternative scientific explanations. 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA and RNA to support the claim that DNA and, in some cases, that RNA are the primary sources of heritable information. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 2.23][APLO 3.28][APLO 4.8][APLO 4.24] The prokaryotes, which include bacteria and archaea, are mostly single-celled organisms that, by definition, lack membranebound nuclei and other organelles. A bacterial chromosome is a covalently closed circle that, unlike eukaryotic chromosomes, is not organized around histone proteins. The central region of the cell in which prokaryotic DNA resides is called the nucleoid. In addition, prokaryotes often have abundant plasmids, which are shorter circular DNA molecules that may only contain one or a few genes. Plasmids can be transferred independently of the bacterial chromosome during cell division and often carry traits such as antibiotic resistance. Transcription in prokaryotes |
(and in eukaryotes) requires the DNA double helix to partially unwind in the region of mRNA synthesis. The region of unwinding is called a transcription bubble. Transcription always proceeds from the same DNA strand for each gene, which is called the template strand. The mRNA product is complementary to the template strand and is almost identical to the other DNA strand, called the nontemplate strand. The only difference is that in mRNA, all of the T nucleotides are replaced with U nucleotides. In an RNA double helix, A can bind U via two hydrogen bonds, just as in A–T pairing in a DNA double helix. The nucleotide pair in the DNA double helix that corresponds to the site from which the first 5' mRNA nucleotide is transcribed is called the +1 site, or the initiation site. Nucleotides preceding the initiation site are given negative numbers and are designated upstream. Conversely, nucleotides following the initiation site are denoted with “+” numbering and are This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 601 called downstream nucleotides. Initiation of Transcription in Prokaryotes Prokaryotes do not have membrane-enclosed nuclei. Therefore, the processes of transcription, translation, and mRNA degradation can all occur simultaneously. The intracellular level of a bacterial protein can quickly be amplified by multiple transcription and translation events occurring concurrently on the same DNA template. Prokaryotic transcription often covers more than one gene and produces polycistronic mRNAs that specify more than one protein. Our discussion here will exemplify transcription by describing this process in Escherichia coli, a well-studied bacterial species. Although some differences exist between transcription in E. coli and transcription in archaea, an understanding of E. coli transcription can be applied to virtually all bacterial species. Prokaryotic RNA Polymerase Prokaryotes use the same RNA polymerase to transcribe all of their genes. In E. coli, the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted α, α, β, and β' comprise the polymerase core enzyme. These subunits assemble every time a gene is transcribed, and they disassemble once transcription is complete. Each subunit has a unique |
role; the two α-subunits are necessary to assemble the polymerase on the DNA; the β-subunit binds to the ribonucleoside triphosphate that will become part of the nascent “recently born” mRNA molecule; and the β' binds the DNA template strand. The fifth subunit, σ, is involved only in transcription initiation. It confers transcriptional specificity such that the polymerase begins to synthesize mRNA from an appropriate initiation site. Without σ, the core enzyme would transcribe from random sites and would produce mRNA molecules that specified protein gibberish. The polymerase comprised of all five subunits is called the holoenzyme. Prokaryotic Promoters A promoter is a DNA sequence onto which the transcription machinery binds and initiates transcription. In most cases, promoters exist upstream of the genes they regulate. The specific sequence of a promoter is very important because it determines whether the corresponding gene is transcribed all the time, some of the time, or infrequently. Although promoters vary among prokaryotic genomes, a few elements are conserved. At the -10 and -35 regions upstream of the initiation site, there are two promoter consensus sequences, or regions that are similar across all promoters and across various bacterial species (Figure 15.7). The -10 consensus sequence, called the -10 region, is TATAAT. The -35 sequence, TTGACA, is recognized and bound by σ. Once this interaction is made, the subunits of the core enzyme bind to the site. The A–T-rich -10 region facilitates unwinding of the DNA template, and several phosphodiester bonds are made. The transcription initiation phase ends with the production of abortive transcripts, which are polymers of approximately 10 nucleotides that are made and released. Figure 15.7 The σ subunit of prokaryotic RNA polymerase recognizes consensus sequences found in the promoter region upstream of the transcription start sight. The σ subunit dissociates from the polymerase after transcription has been initiated. 602 Chapter 15 | Genes and Proteins View this MolecularMovies animation (http://openstaxcollege.org/l/transcription) to see the first part of transcription and the base sequence repetition of the TATA box. Mutations can occur in any part of the DNA. What can happen if there is a mutation in the promoter sequence? a. All processes will carry on as usual. Nothing will be |
affected. b. RNA polymerase will not be able to attach. c. RNA polymerase will bind upstream of the promoter sequence. d. RNA polymerase will bind downstream of the promoter sequence. Elongation and Termination in Prokaryotes The transcription elongation phase begins with the release of the σ subunit from the polymerase. The dissociation of σ allows the core enzyme to proceed along the DNA template, synthesizing mRNA in the 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 15.8). The base pairing between DNA and RNA is not stable enough to maintain the stability of the mRNA synthesis components. Instead, the RNA polymerase acts as a stable linker between the DNA template and the nascent RNA strands to ensure that elongation is not interrupted prematurely. Figure 15.8 During elongation, the prokaryotic 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. Prokaryotic Termination Signals Once a gene is transcribed, the prokaryotic polymerase needs to be instructed to dissociate from the DNA template and liberate the newly made mRNA. Depending on the gene being transcribed, there are two kinds of termination signals. One is protein-based and the other is RNA-based. Rho-dependent termination is controlled by the rho protein, which tracks along behind the polymerase on the growing mRNA chain. Near the end of the gene, the polymerase encounters a run of G nucleotides on the DNA template and it stalls. As a result, the rho protein collides with the polymerase. The interaction with rho releases the mRNA from the transcription bubble. Rho-independent termination is controlled by specific sequences in the DNA template strand. As the polymerase nears the end of the gene being transcribed, it encounters a region rich in C–G nucleotides. The mRNA folds back on itself, and the complementary C–G nucleotides bind together. The result is a stable hairpin that causes the polymerase to stall as soon as it begins to transcribe a region rich in A–T nucleotides. The complementary U–A region of the mRNA transcript forms only a weak interaction with the template DNA. This, coupled with the stalled |
polymerase, induces enough instability for the core enzyme to break away and liberate the new mRNA transcript. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 603 Upon termination, the process of transcription is complete. By the time termination occurs, the prokaryotic transcript would already have been used to begin synthesis of numerous copies of the encoded protein because these processes can occur concurrently. The unification of transcription, translation, and even mRNA degradation is possible because all of these processes occur in the same 5' to 3' direction, and because there is no membranous compartmentalization in the prokaryotic cell (Figure 15.9). In contrast, the presence of a nucleus in eukaryotic cells precludes simultaneous transcription and translation. Figure 15.9 Multiple 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. this BioStudio animation (http://openstaxcollege.org/l/transcription2) to see the process of prokaryotic Visit transcription. Why is the stop codon necessary for translation? a. The stop codon is the first step in a series of steps to end translation. b. The stop codon is necessary to initiate translation. c. The stop codon ends translation which allows the polypeptide strand to be released. d. The stop codon ends translation in order to initiate transcription. Activity Working in small groups, use a model of DNA to demonstrate synthesis transcription of mRNA to other groups in your class. In your demonstration, be sure to distinguish the differences between DNA and RNA, the template and non-template strands of the DNA, the directionality of transcription, and the significance of promoters. Think About It If mRNA is complementary to the DNA template strand, and the DNA template strand is complementary to the DNA non-template strand, are the base sequences of mRNA and the DNA non-template strand ever identical? Justify your answer. 604 Chapter 15 | Genes and Proteins 15.3 | Eukaryotic Transcription In this section, you will explore the following questions: • What are the steps in eukaryotic transcription? • What are the structural and functional similarities and differences among the three RNA polymerases? Connection for AP® Courses As expected |
, transcription in eukaryotes is more complex than transcription in prokaryotes. First, transcription in eukaryotes involves one of three types of RNA polymerase, depending on the gene being transcribed. Second, the initiation of transcription involves the binding of several transcription factors to complex promoters which are usually located upstream of the gene being copied. Transcription factors can either activate or inhibit gene expression. Termination of transcription involves the RNA polymerases. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® Exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.A Essential Knowledge Science Practice Learning Objective Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. 6.5 The student can evaluate alternative scientific explanations. 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA and RNA to support the claim that DNA and, in some cases, that RNA are the primary sources of heritable information. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 3.3][APLO 3.22][APLO 2.36][APLO 1.14][APLO 2.22][APLO 4.5] Prokaryotes and eukaryotes perform fundamentally the same process of transcription, with a few key differences. The most important difference between prokaryotes and eukaryotes is the latter’s membrane-bound nucleus and organelles. With the genes bound in a nucleus, the eukaryotic cell must be able to transport its mRNA to the cytoplasm and must protect its mRNA from degrading before it is translated. Eukaryotes also employ three different polymerases that each transcribe a different subset of genes. Eukaryotic mRNAs are usually monogenic, meaning that they specify a single protein. Initiation of Transcription in Eukaryotes Unlike the prokaryotic polymerase that can bind to a DNA |
template on its own, eukaryotes require several other proteins, called transcription factors, to first bind to the promoter region and then help recruit the appropriate polymerase. The Three Eukaryotic RNA Polymerases The features of eukaryotic mRNA synthesis are markedly more complex those of prokaryotes. Instead of a single polymerase comprising five subunits, the eukaryotes have three polymerases that are each made up of 10 subunits or more. Each eukaryotic polymerase also requires a distinct set of transcription factors to bring it to the DNA template. RNA polymerase I is located in the nucleolus, a specialized nuclear substructure in which ribosomal RNA (rRNA) is This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 605 transcribed, processed, and assembled into ribosomes (Table 15.1). The rRNA molecules are considered structural RNAs because they have a cellular role but are not translated into protein. The rRNAs are components of the ribosome and are essential to the process of translation. RNA polymerase I synthesizes all of the rRNAs except for the 5S rRNA molecule. The “S” designation applies to “Svedberg” units, a nonadditive value that characterizes the speed at which a particle sediments during centrifugation. Locations, Products, and Sensitivities of the Three Eukaryotic RNA Polymerases RNA Polymerase Cellular Compartment Product of Transcription α-Amanitin Sensitivity I II III Nucleolus Nucleus Nucleus All rRNAs except 5S rRNA Insensitive All protein-coding nuclear premRNAs Extremely sensitive 5S rRNA, tRNAs, and small nuclear RNAs Moderately sensitive Table 15.1 RNA polymerase II is located in the nucleus and synthesizes all protein-coding nuclear pre-mRNAs. Eukaryotic premRNAs undergo extensive processing after transcription but before translation. For clarity, this module’s discussion of transcription and translation in eukaryotes will use the term “mRNAs” to describe only the mature, processed molecules that are ready to be translated. RNA polymerase II is responsible for transcribing the overwhelming majority of eukaryotic genes. RNA polymerase III is also located in the |
nucleus. This polymerase transcribes a variety of structural RNAs that includes the 5S pre-rRNA, transfer pre-RNAs (pre-tRNAs), and small nuclear pre- RNAs. The tRNAs have a critical role in translation; they serve as the adaptor molecules between the mRNA template and the growing polypeptide chain. Small nuclear RNAs have a variety of functions, including “splicing” pre-mRNAs and regulating transcription factors. A scientist characterizing a new gene can determine which polymerase transcribes it by testing whether the gene is expressed in the presence of a particular mushroom poison, α-amanitin (Table 15.1). Interestingly, α-amanitin produced by Amanita phalloides, the Death Cap mushroom, affects the three polymerases very differently. RNA polymerase I is completely insensitive to α-amanitin, meaning that the polymerase can transcribe DNA in vitro in the presence of this poison. In contrast, RNA polymerase II is extremely sensitive to α-amanitin, and RNA polymerase III is moderately sensitive. Knowing the transcribing polymerase can clue a researcher into the general function of the gene being studied. Because RNA polymerase II transcribes the vast majority of genes, we will focus on this polymerase in our subsequent discussions about eukaryotic transcription factors and promoters. Structure of an RNA Polymerase II Promoter Eukaryotic promoters are much larger and more complex than prokaryotic promoters, but both have a TATA box. For example, in the mouse thymidine kinase gene, the TATA box is located at approximately -30 relative to the initiation (+1) site (Figure 15.10). For this gene, the exact TATA box sequence is TATAAAA, as read in the 5' to 3' direction on the nontemplate strand. This sequence is not identical to the E. coli TATA box, but it conserves the A–T rich element. The thermostability of A–T bonds is low and this helps the DNA template to locally unwind in preparation for transcription. 606 Chapter 15 | Genes and Proteins Figure 15.10 A generalized promoter of a gene transcribed by RNA polymerase II is shown. Transcription factors recognize the promoter. RNA polymerase II then binds and forms the transcription initiation complex. Figure 15.11 Eukaryotic mRNA contains introns that must be spliced out. A 5 |
' cap and 3' poly-A tail are also added. A scientist splices a eukaryotic promoter in front of a bacterial gene and inserts the gene in a bacterial chromosome. Would you expect the bacterium to transcribe the gene? a. b. Initially the bacterium will not transcribe it, but will transcribe after 5’ capping. It will not transcribe it because there is no poly A tail in prokaryotes. c. No, prokaryotes use different promoters than eukaryotes. d. Yes, the bacterium would transcribe the eukaryotic gene. The mouse genome includes one gene and two pseudogenes for cytoplasmic thymidine kinase. Pseudogenes are genes that have lost their protein-coding ability or are no longer expressed by the cell. These pseudogenes are copied from mRNA and This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 607 incorporated into the chromosome. For example, the mouse thymidine kinase promoter also has a conserved CAAT box (GGCCAATCT) at approximately -80. This sequence is essential and is involved in binding transcription factors. Further upstream of the TATA box, eukaryotic promoters may also contain one or more GC-rich boxes (GGCG) or octamer boxes (ATTTGCAT). These elements bind cellular factors that increase the efficiency of transcription initiation and are often identified in more “active” genes that are constantly being expressed by the cell. Transcription Factors for RNA Polymerase II The complexity of eukaryotic transcription does not end with the polymerases and promoters. An army of basal transcription factors, enhancers, and silencers also help to regulate the frequency with which pre-mRNA is synthesized from a gene. Enhancers and silencers affect the efficiency of transcription but are not necessary for transcription to proceed. Basal transcription factors are crucial in the formation of a preinitiation complex on the DNA template that subsequently recruits RNA polymerase II for transcription initiation. The names of the basal transcription factors begin with “TFII” (this is the transcription factor for RNA polymerase II) and are specified with the letters A–J. The transcription factors systematically fall into place on the DNA template, with each one further stabilizing the preinitiation complex and contributing to |
the recruitment of RNA polymerase II. The processes of bringing RNA polymerases I and III to the DNA template involve slightly less complex collections of transcription factors, but the general theme is the same. Eukaryotic transcription is a tightly regulated process that requires a variety of proteins to interact with each other and with the DNA strand. Although the process of transcription in eukaryotes involves a greater metabolic investment than in prokaryotes, it ensures that the cell transcribes precisely the pre-mRNAs that it needs for protein synthesis. 608 Chapter 15 | Genes and Proteins During human embryonic development, a transcription factor encoded by the SRY gene starts a chain of events, causing the embryo to develop male sex characteristics. This gene is on the Y chromosome in humans and many other mammals. A deletion or mutation of the SRY gene can cause the human embryo to not develop into a male even though the individual has an XY genotype, a condition called Swyer syndrome. Figure 15.12 The SYR gene of the Y chromosome produces proteins that lead to the expression of primary sex characteristics, as shown. The protein product of the SRY gene is a DNA binding protein. Together with a protein called SF1, the SRY protein acts as a transcription factor that “turns on” certain genes. Which of the following statements best describes how a change in these two proteins would affect male sexual development? a. A mutation that abolished activity of SF1 would increase the effect of a SRY mutation, making the person more feminine. b. A mutation that abolished activity of SF1 would cancel out a mutation in SRY, so if both mutations occur together male sex characteristics would develop normally. c. A mutation in the SRY protein that abolished activity would result in abnormal development of male sex characteristics but a mutation of SF1 would not. d. Both a mutation in the SRY protein and a mutation in SF1 that abolished activity would result in a lack of development of male sex characteristics. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 609 The Evolution of Promoters The evolution of genes may be a familiar concept. Mutations can occur in genes during DNA replication, and the result may or may not be beneficial to the cell. By altering an enzyme, structural protein, or some other factor, the process of mutation can transform functions or physical features. However, euk |
aryotic promoters and other gene regulatory sequences may evolve as well. For instance, consider a gene that, over many generations, becomes more valuable to the cell. Maybe the gene encodes a structural protein that the cell needs to synthesize in abundance for a certain function. If this is the case, it would be beneficial to the cell for that gene’s promoter to recruit transcription factors more efficiently and increase gene expression. Scientists examining the evolution of promoter sequences have reported varying results. In part, this is because it is difficult to infer exactly where a eukaryotic promoter begins and ends. Some promoters occur within genes; others are located very far upstream, or even downstream, of the genes they are regulating. However, when researchers limited their examination to human core promoter sequences that were defined experimentally as sequences that bind the preinitiation complex, they found that promoters evolve even faster than protein-coding genes. It is still unclear how promoter evolution might correspond to the evolution of humans or other higher organisms. However, the evolution of a promoter to effectively make more or less of a given gene product is an intriguing alternative to the evolution of the genes themselves. [1] According to this passage, which of the following has been shown to evolve faster than protein-coding genes? a. core promoters that bind the preinitiation complex b. core promoters that occur within genes c. promoters that occur far upstream of the gene d. promoters that occur downstream of a gene Promoter Structures for RNA Polymerases I and III In eukaryotes, the conserved promoter elements differ for genes transcribed by RNA polymerases I, II, and III. RNA polymerase I transcribes genes that have two GC-rich promoter sequences in the -45 to +20 region. These sequences alone are sufficient for transcription initiation to occur, but promoters with additional sequences in the region from -180 to -105 upstream of the initiation site will further enhance initiation. Genes that are transcribed by RNA polymerase III have upstream promoters or promoters that occur within the genes themselves. Eukaryotic Elongation and Termination Following the formation of the preinitiation complex, the polymerase is released from the other transcription factors, and elongation is allowed to proceed as it does in prokaryotes with the polymerase synthesizing pre-mRNA in the 5' to 3' direction. As discussed previously, RNA polymerase II transcribes the major share of eukaryotic genes, so this section will focus on how this |
polymerase accomplishes elongation and termination. Although the enzymatic process of elongation is essentially the same in eukaryotes and prokaryotes, the DNA template is more complex. When eukaryotic cells are not dividing, their genes exist as a diffuse mass of DNA and proteins called chromatin. The DNA is tightly packaged around charged histone proteins at repeated intervals. These DNA–histone complexes, collectively called nucleosomes, are regularly spaced and include 146 nucleotides of DNA wound around eight histones like thread around a spool. For polynucleotide synthesis to occur, the transcription machinery needs to move histones out of the way every time it encounters a nucleosome. This is accomplished by a special protein complex called FACT, which stands for “facilitates chromatin transcription.” This complex pulls histones away from the DNA template as the polymerase moves along it. Once the pre-mRNA is synthesized, the FACT complex replaces the histones to recreate the nucleosomes. The termination of transcription is different for the different polymerases. Unlike in prokaryotes, elongation by RNA polymerase II in eukaryotes takes place 1,000–2,000 nucleotides beyond the end of the gene being transcribed. This premRNA tail is subsequently removed by cleavage during mRNA processing. On the other hand, RNA polymerases I and 1. H Liang et al., “Fast evolution of core promoters in primate genomes,” Molecular Biology and Evolution 25 (2008): 1239–44. 610 Chapter 15 | Genes and Proteins III require termination signals. Genes transcribed by RNA polymerase I contain a specific 18-nucleotide sequence that is recognized by a termination protein. The process of termination in RNA polymerase III involves an mRNA hairpin similar to rho-independent termination of transcription in prokaryotes. 15.4 | RNA Processing in Eukaryotes In this section, you will explore the following questions: • What are the steps in eukaryotic transcription? • What are the structural and functional similarities and differences among the three RNA polymerases? Connection for AP® Courses Scientists discovered a strand of mRNA translated into a sequence of amino acids (polypeptide) shorter than the mRNA molecule transcribed from DNA. Before the information in eukaryotic mRNA is translated into protein, it is modified or edited in several ways. A 5′ methylguanosine |
(or GTP) cap and a 3′ poly-A tail are added to protect mature mRNA from degradation and allow its export from the nucleus. Pre-mRNAs also undergo splicing, in which introns are removed and exons are reconnected. Exons can be reconnected in different sequences, a phenomenon referred to as alternative gene splicing, which allows a single eukaryotic gene to code for different proteins. (We will explore the significance of alternative gene splicing in more detail in other chapters.) Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® Exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.A Essential Knowledge Science Practice Learning Objective Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. 6.5 The student can evaluate alternative scientific explanations. 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA and RNA to support the claim that DNA and, in some cases, that RNA are the primary sources of heritable information. After transcription, eukaryotic pre-mRNAs must undergo several processing steps before they can be translated. Eukaryotic (and prokaryotic) tRNAs and rRNAs also undergo processing before they can function as components in the protein synthesis machinery. mRNA Processing The eukaryotic pre-mRNA undergoes extensive processing before it is ready to be translated. The additional steps involved in eukaryotic mRNA maturation create a molecule with a much longer half-life than a prokaryotic mRNA. Eukaryotic mRNAs last for several hours, whereas the typical E. coli mRNA lasts no more than five seconds. Pre-mRNAs are first coated in RNA-stabilizing proteins; these protect the pre-mRNA from degradation while it is processed and exported out of the nucleus. The three most important steps of pre-mRNA processing are the addition of stabilizing and signaling factors at the 5' and 3' ends of the molecule, and the removal of |
intervening sequences that do not specify the appropriate amino acids. In rare cases, the mRNA transcript can be “edited” after it is transcribed. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 611 RNA Editing in Trypanosomes The trypanosomes are a group of protozoa that include the pathogen Trypanosoma brucei, which causes sleeping sickness in humans (Figure 15.13). Trypanosomes, and virtually all other eukaryotes, have organelles called mitochondria that supply the cell with chemical energy. Mitochondria are organelles that express their own DNA and are believed to be the remnants of a symbiotic relationship between a eukaryote and an engulfed prokaryote. The mitochondrial DNA of trypanosomes exhibit an interesting exception to The Central Dogma: their pre-mRNAs do not have the correct information to specify a functional protein. Usually, this is because the mRNA is missing several U nucleotides. The cell performs an additional RNA processing step called RNA editing to remedy this. Figure 15.13 Trypanosoma brucei is the causative agent of sleeping sickness in humans. The mRNAs of this pathogen must be modified by the addition of nucleotides before protein synthesis can occur. (credit: modification of work by Torsten Ochsenreiter) Other genes in the mitochondrial genome encode 40- to 80-nucleotide guide RNAs. One or more of these molecules interacts by complementary base pairing with some of the nucleotides in the pre-mRNA transcript. However, the guide RNA has more A nucleotides than the pre-mRNA has U nucleotides to bind with. In these regions, the guide RNA loops out. The 3' ends of guide RNAs have a long poly-U tail, and these U bases are inserted in regions of the pre-mRNA transcript at which the guide RNAs are looped. This process is entirely mediated by RNA molecules. That is, guide RNAs—rather than proteins—serve as the catalysts in RNA editing. RNA editing is not just a phenomenon of trypanosomes. In the mitochondria of some plants, almost all pre-mRNAs are edited. RNA editing has also been identified in mammals such as rats, rabbits, and even humans. What could be |
the evolutionary reason for this additional step in pre-mRNA processing? One possibility is that the mitochondria, being remnants of ancient prokaryotes, have an equally ancient RNAbased method for regulating gene expression. In support of this hypothesis, edits made to pre-mRNAs differ depending on cellular conditions. Although speculative, the process of RNA editing may be a holdover from a primordial time when RNA molecules, instead of proteins, were responsible for catalyzing reactions. In eukaryotes, pre-mRNAs are processed to form mature mRNAs. How does the mRNA editing that occurs in Trypanosoma brucei differ from mRNA processing that occurs in all eukaryotes? a. mRNA editing changes the coding sequence of the mRNA, but mRNA processing does not. b. mRNA editing splices out noncoding RNA, but mRNA processing does not. c. mRNA editing adds a cap of 5’-methylguanosine to the mRNA, but mRNA processing does not. d. mRNA editing adds a 3’ poly-A tail, but mRNA processing does not. 612 5' Capping Chapter 15 | Genes and Proteins While the pre-mRNA is still being synthesized, a 7-methylguanosine cap is added to the 5' end of the growing transcript by a phosphate linkage. This moiety (functional group) protects the nascent mRNA from degradation. In addition, factors involved in protein synthesis recognize the cap to help initiate translation by ribosomes. 3' Poly-A Tail Once elongation is complete, the pre-mRNA is cleaved by an endonuclease between an AAUAAA consensus sequence and a GU-rich sequence, leaving the AAUAAA sequence on the pre-mRNA. An enzyme called poly-A polymerase then adds a string of approximately 200 A residues, called the poly-A tail. This modification further protects the pre-mRNA from degradation and signals the export of the cellular factors that the transcript needs to the cytoplasm. Pre-mRNA Splicing Eukaryotic genes are composed of exons, which correspond to protein-coding sequences (ex-on signifies that they are expressed), and intervening sequences called introns (int-ron denotes their intervening role), which may be involved in gene regulation but are removed from the pre-mRNA during processing. Intron sequences in mRNA do not encode functional proteins. The discovery of introns came as a surprise to researchers in |
the 1970s who expected that pre-mRNAs would specify protein sequences without further processing, as they had observed in prokaryotes. The genes of higher eukaryotes very often contain one or more introns. These regions may correspond to regulatory sequences; however, the biological significance of having many introns or having very long introns in a gene is unclear. It is possible that introns slow down gene expression because it takes longer to transcribe pre-mRNAs with lots of introns. Alternatively, introns may be nonfunctional sequence remnants left over from the fusion of ancient genes throughout evolution. This is supported by the fact that separate exons often encode separate protein subunits or domains. For the most part, the sequences of introns can be mutated without ultimately affecting the protein product. All of a pre-mRNA’s introns must be completely and precisely removed before protein synthesis. If the process errs by even a single nucleotide, the reading frame of the rejoined exons would shift, and the resulting protein would be dysfunctional. The process of removing introns and reconnecting exons is called splicing (Figure 15.14). Introns are removed and degraded while the pre-mRNA is still in the nucleus. Splicing occurs by a sequence-specific mechanism that ensures introns will be removed and exons rejoined with the accuracy and precision of a single nucleotide. The splicing of pre-mRNAs is conducted by complexes of proteins and RNA molecules called spliceosomes. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 613 Figure 15.14 Pre-mRNA splicing involves the precise removal of introns from the primary RNA transcript. The splicing process is catalyzed by protein complexes called spliceosomes that are composed of proteins and RNA molecules called snRNAs. Spliceosomes recognize sequences at the 5' and 3' end of the intron. Errors in splicing are implicated in cancers and other human diseases. What kinds of mutations might lead to splicing errors? Think of different possible outcomes if splicing errors occur. a. Mutations in the spliceosome recognition sequence at each end of an intron, or in the proteins and RNAs that make up the spliceosome, may occur. Mutations may also add new spliceosome recognition sites |
. b. Mutations in the spliceosome recognition sequence at each end of an exon, or in the proteins and RNAs that make up the spliceosome, may occur. Mutations may also add new spliceosome recognition sites. c. Mutations in the spliceosome recognition sequence at each end of an intron, or in the proteins and RNAs that make up the spliceosome, may occur. Mutations may also delete existing spliceosome recognition sites. d. Mutations at the each end of intron and exon, or in the proteins and RNAs that make up the spliceosome, may occur. Mutations may also add new spliceosome recognition sites and delete existing sites. Note that more than 70 individual introns can be present, and each has to undergo the process of splicing—in addition to 5' capping and the addition of a poly-A tail—just to generate a single, translatable mRNA molecule. 614 Chapter 15 | Genes and Proteins See how introns are removed during RNA splicing at this website (http://openstaxcollege.org/l/RNA_splicing). Explain why helper proteins are necessary for the formation of the final protein during RNA splicing in higher organisms. a. Helper proteins attach themselves to the ends of introns so that they can be spliced out during RNA splicing and coded areas are spliced together to form mRNA which then codes for the final protein. b. Helper proteins attach themselves to the ends of exons so that they can be spliced out during RNA splicing and coded areas are spliced together to form mRNA which encodes the final protein. c. Helper proteins attach themselves to mRNA in order to remove the non-coded areas and thus form the pre- mRNA which codes for the final protein. d. Helper proteins help the pre-mRNA to recruit various other components which splice out the non-coded regions and form mRNA which codes for the final protein. Processing of tRNAs and rRNAs The tRNAs and rRNAs are structural molecules that have roles in protein synthesis; however, these RNAs are not themselves translated. Pre-rRNAs are transcribed, processed, and assembled into ribosomes in the nucleolus. Pre-tRNAs are transcribed and processed in the nucleus and then released into the cytoplasm where they are linked to free amino |
acids for protein synthesis. Most of the tRNAs and rRNAs in eukaryotes and prokaryotes are first transcribed as a long precursor molecule that spans multiple rRNAs or tRNAs. Enzymes then cleave the precursors into subunits corresponding to each structural RNA. Some of the bases of pre-rRNAs are methylated; that is, a –CH3 moiety (methyl functional group) is added for stability. PretRNA molecules also undergo methylation. As with pre-mRNAs, subunit excision occurs in eukaryotic pre-RNAs destined to become tRNAs or rRNAs. Mature rRNAs make up approximately 50 percent of each ribosome. Some of a ribosome’s RNA molecules are purely structural, whereas others have catalytic or binding activities. Mature tRNAs take on a three-dimensional structure through intramolecular hydrogen bonding to position the amino acid binding site at one end and the anticodon at the other end (Figure 15.15). The anticodon is a three-nucleotide sequence in a tRNA that interacts with an mRNA codon through complementary base pairing. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 615 Figure 15.15 This is a space-filling model of a tRNA molecule that adds the amino acid phenylalanine to a growing polypeptide chain. The anticodon AAG binds the Codon UUC on the mRNA. The amino acid phenylalanine is attached to the other end of the tRNA. 15.5 | Ribosomes and Protein Synthesis In this section, you will explore the following questions: • What are the different sequential steps in protein synthesis? • What is the role of ribosomes in protein synthesis? Connection for AP® Courses After the information in the gene has been transcribed to mRNA, it is ready to be translated to polypeptide. The players in translation include the mRNA template, ribosomes, tRNA molecules, amino acids, and various enzymes. Ribosomes consist of small and large subunits of protein and rRNA which bind with mRNA; many ribosomes can move along the same mRNA at a time. Translation begins at the initiating AUG on mRNA, specifying methionine, the first amino acid in any poly |
peptide. Each amino acid is carried to the ribosome by attaching to a specific molecule of tRNA. A tRNA molecule often is depicted as a cloverleaf, with an anticodon on one end, and the amino acid attachment site at the other. Amino-acid charging enzymes ensure that the correct amino acid is attached to the correct tRNA. The anticodons on tRNA are complementary to the codons on mRNA; for example, the anticodon AAA on tRNA corresponds to TTT on mRNA. Sequential amino acids are linked by peptide bonds. The mRNA is translated, elongating the polypeptide, until a STOP or nonsense codon is reached. When this happens, a release factor dissociates the components and frees the new polypeptide. Folding of the protein occurs during and after translation. Once a polypeptide is synthesized, its role as a protein is established, such as determining a physical phenotype of an organism. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® Exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. 616 Chapter 15 | Genes and Proteins Big Idea 3 Enduring Understanding 3.A Essential Knowledge Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. Science Practice 1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain. Learning Objective 3.4 The student is able to describe representations and models illustrating how genetic information is translated into polypeptides. Essential Knowledge 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. Science Practice 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. Learning Objective 3.6 The student can predict how a change in a specific DNA or RNA sequence can result in changes in gene expression. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 1. |
16][APLO 4.22][APLO 3.6] 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 component of living organisms (with the exception of water), and proteins perform virtually every function of a cell. The process of translation, or protein synthesis, involves the decoding of an mRNA message into a polypeptide product. Amino acids are covalently strung together by interlinking peptide bonds in lengths ranging from approximately 50 amino acid residues to more than 1,000. Each individual amino acid has an amino group (NH2) and a carboxyl (COOH) group. Polypeptides are formed when the amino group of one amino acid forms an amide (i.e., peptide) bond with the carboxyl group of another amino acid (Figure 15.16). This reaction is catalyzed by ribosomes and generates one water molecule. Figure 15.16 A peptide bond links the carboxyl end of one amino acid with the amino end of another, expelling one water molecule. For simplicity in this image, only the functional groups involved in the peptide bond are shown. The R and R' designations refer to the rest of each amino acid structure. 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 may vary across species; for instance, ribosomes may consist of different numbers of rRNAs and polypeptides depen<|endoftext|>ding 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. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 617 Click through the steps of this PBS interactive (http://openstaxcollege.org/l/prokary_protein) to see protein synthesis in action. A lack of protein in the diet can cause hair loss. Explain why this occurs. a. Due to lack of protein in the diet, our body will not be able to form other proteins; thus, it will conserve the protein it has for critical use, leading to |
hair loss. b. Lack of protein in the diet can weaken the immune system, thus leading to hair loss. c. Due to lack of protein in the diet, energy will be lost, thus leading to hair loss. d. Lack of protein in the diet will lead to breakage of disulfide bonds between proteins, thus leading to hair loss. Ribosomes Even before an mRNA is translated, a cell must invest energy to build each of its ribosomes. In E. coli, there are between 10,000 and 70,000 ribosomes present in each cell at any given time. A ribosome is a complex macromolecule composed of structural and catalytic rRNAs, and many distinct polypeptides. In eukaryotes, the nucleolus is completely specialized for the synthesis and assembly of rRNAs. Ribosomes exist in the cytoplasm in prokaryotes and in the cytoplasm and rough endoplasmic reticulum in eukaryotes. Mitochondria and chloroplasts also have their own ribosomes in the matrix and stroma, which look more similar to prokaryotic ribosomes (and have similar drug sensitivities) than the ribosomes just outside their outer membranes in the cytoplasm. 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, and the large subunit is 50S, for a total of 70S (recall that Svedberg units are not additive). Mammalian ribosomes have a small 40S subunit and a large 60S subunit, for a total of 80S. The small subunit is responsible for binding the mRNA template, whereas the large subunit sequentially binds tRNAs. 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 mRNA/poly-ribosome structure is called a polysome. tRNAs The tRNAs are structural RNA molecules that were transcribed from genes by RNA polymerase III. Depending on the species, 40 to 60 types of tRNAs exist in the cytoplasm. Serving as adaptors, specific tRNAs bind to sequences |
on the mRNA template and add 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. Of the 64 possible mRNA codons—or triplet combinations of A, U, G, and C—three specify the termination of protein synthesis and 61 specify the addition of amino acids to the polypeptide chain. Of these 61, one codon (AUG) also encodes the initiation of translation. Each tRNA anticodon can base pair with one of the mRNA codons and add an amino acid or terminate translation, according to the genetic code. For instance, if the sequence CUA occurred on an mRNA template in the proper reading frame, it would bind a tRNA expressing the complementary sequence, GAU, which would be linked to the amino acid leucine. As the adaptor molecules of translation, it is surprising that tRNAs can fit so much specificity into such a small package. Consider that tRNAs need to interact with three factors: 1) they must be recognized by the correct aminoacyl synthetase (see below); 2) they must be recognized by ribosomes; and 3) they must bind to the correct sequence in mRNA. Aminoacyl tRNA Synthetases The process of pre-tRNA synthesis by RNA polymerase III only creates the RNA portion of the adaptor molecule. The corresponding amino acid must be added later, once the tRNA is processed and exported to the cytoplasm. Through the process of tRNA “charging,” each tRNA molecule is linked to its correct amino acid by a group of enzymes called 618 Chapter 15 | Genes and Proteins aminoacyl tRNA synthetases. At least one type of aminoacyl tRNA synthetase exists for each of the 20 amino acids; the exact number of aminoacyl tRNA synthetases varies by species. These enzymes first bind and hydrolyze ATP to catalyze a high-energy bond between an amino acid and adenosine monophosphate (AMP); a pyrophosphate molecule is expelled in this reaction. The activated amino acid is then transferred to the tRNA, and AMP is released. The Mechanism of Protein Synthesis As with mRNA synthesis, protein synthesis can be divided into three phases: initiation, elongation, and termination. The process of translation is similar in prokaryotes and |
eukaryotes. Here we’ll explore how translation occurs in E. coli, a representative prokaryote, and specify any differences between prokaryotic and eukaryotic translation. Initiation of Translation 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 (IFs; IF-1, IF-2, and IF-3), and a special initiator tRNA, called Met. The initiator tRNA interacts with the start codon AUG (or rarely, GUG), links to a formylated methionine tRNAf called fMet, and can also bind IF-2. Formylated methionine is inserted by fMet − tRNAf polypeptide chain synthesized by E. coli, but it is usually clipped off after translation is complete. When an in-frame AUG is encountered during translation elongation, a non-formylated methionine is inserted by a regular Met-tRNAMet. Met at the beginning of every In E. coli mRNA, a sequence upstream of the first AUG codon, called the Shine-Dalgarno sequence (AGGAGG), interacts with the rRNA molecules that compose the ribosome. This interaction anchors the 30S ribosomal subunit at the correct location on the mRNA template. Guanosine triphosphate (GTP), which is a purine nucleotide triphosphate, acts as an energy source during translation—both at the start of elongation and during the ribosome’s translocation. In eukaryotes, a similar initiation complex forms, comprising mRNA, the 40S small ribosomal subunit, IFs, and nucleoside triphosphates (GTP and ATP). The charged initiator tRNA, called Met-tRNAi, does not bind fMet in eukaryotes, but is distinct from other Met-tRNAs in that it can bind IFs. Instead of depositing at the Shine-Dalgarno sequence, the eukaryotic initiation complex recognizes the 7-methylguanosine cap at the 5' end of the mRNA. A cap-binding protein (CBP) and several other IFs assist the movement of the ribosome to the 5' cap. Once at the cap, the initiation complex tracks along the mRNA in |
the 5' to 3' direction, searching for the AUG start codon. Many eukaryotic mRNAs are translated from the first AUG, but this is not always the case. According to Kozak’s rules, the nucleotides around the AUG indicate whether it is the correct start codon. Kozak’s rules state that the following consensus sequence must appear around the AUG of vertebrate genes: 5'-gccRccAUGG-3'. The R (for purine) indicates a site that can be either A or G, but cannot be C or U. Essentially, the closer the sequence is to this consensus, the higher the efficiency of translation. Once the appropriate AUG is identified, the other proteins and CBP dissociate, and the 60S subunit binds to the complex of Met-tRNAi, mRNA, and the 40S subunit. This step completes the initiation of translation in eukaryotes. Translation, Elongation, and Termination In prokaryotes and eukaryotes, the basics of elongation are the same, so we will review elongation from the perspective of E. coli. The 50S ribosomal subunit of E. coli consists of three compartments: 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 exception to this assembly line of Met is capable of entering the P site directly without first entering the A site. Similarly, tRNAs: in E. coli, fMet − tRNAf the eukaryotic Met-tRNAi, with help from other proteins of the initiation complex, binds directly to the P site. In both cases, this creates an initiation complex with a free A site ready to accept the tRNA corresponding to the first codon after the AUG. During translation elongation, the mRNA template provides specificity. As the ribosome moves along the mRNA, each mRNA codon comes into register, and specific binding with the corresponding charged tRNA anticodon is ensured. If mRNA were not present in the elongation complex, the ribosome would bind t |
RNAs nonspecifically. Elongation proceeds with charged tRNAs entering the A site and then shifting to the P site followed by the E site with each single-codon “step” of the ribosome. Ribosomal steps are induced by conformational changes that advance the ribosome by three bases in the 3' direction. The energy for each step of the ribosome is donated by an elongation factor that hydrolyzes GTP. 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 enzyme that is integrated into the 50S ribosomal subunit. The energy for each peptide bond formation is This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 619 derived from GTP hydrolysis, which is catalyzed by a separate elongation factor. 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 (Figure 15.17). 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. 620 Chapter 15 | Genes and Proteins Figure 15.17 Translation begins when an initiator tRNA anticodon recognizes a codon on mRNA. The large ribosomal subunit joins the small subunit, and a second tRNA is recruited. As the mRNA moves relative to the ribosome, the polypeptide chain is formed. Entry of a release factor into the A site terminates translation and the components dissociate. Many antibiotics inhibit bacterial protein synthesis. For example, tetracycline blocks the A site on the bacterial ribosome, and chloramphenicol blocks peptidyl transfer. What specific effect would you expect each of these antibiotics to have on protein synthesis? Tetracycline would directly affect: 1. tRNA binding to the ribosome 2. Ribosome assembly 3. Growth of |
the protein chain Chloramphenicol would directly affect: 1. tRNA binding to the ribosome 2. Ribosome assembly 3. Growth of the protein chain a. Tetracycline would directly affect tRNA binding to the ribosome. Chloramphenicol would affect the growth of the protein chain. b. Tetracycline would directly affect ribosome assembly. Chloramphenicol would affect the growth of the protein chain. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 621 c. Tetracycline would directly affect the growth of the protein chain. Chloramphenicol would affect the tRNA binding to the ribosome. d. Tetracycline would directly affect mRNA binding to the ribosome. Chloramphenicol would affect the ribosome assembly. Termination of translation occurs when a nonsense codon (UAA, UAG, or UGA) is encountered. Upon aligning with the A site, these nonsense codons are recognized by release factors in prokaryotes and eukaryotes that instruct peptidyl transferase to add a water molecule to the carboxyl end of the P-site amino acid. This reaction forces the P-site amino acid to detach from its tRNA, and the newly made protein is released. The small and large ribosomal subunits dissociate from the mRNA and from each other; they are recruited almost immediately into another translation initiation complex. After many ribosomes have completed translation, the mRNA is degraded so the nucleotides can be reused in another transcription reaction. Protein Folding, Modification, and Targeting During and after translation, individual amino acids may be chemically modified, signal sequences may be appended, and the new protein “folds” into a distinct three-dimensional structure as a result of intramolecular interactions. A signal sequence is a short tail of amino acids that directs a protein to a specific cellular compartment. These sequences at the amino end or the carboxyl end of the protein can be thought of as the protein’s “train ticket” to its ultimate destination. Other cellular factors recognize each signal sequence and help transport the protein from the cytoplasm to its correct compartment. For instance, a specific sequence at the amino terminus will direct a protein to the mitochond |
ria or chloroplasts (in plants). Once the protein reaches its cellular destination, the signal sequence is usually clipped off. Many proteins fold spontaneously, but some proteins require helper molecules, called chaperones, to prevent them from aggregating during the complicated process of folding. Even if a protein is properly specified by its corresponding mRNA, it could take on a completely dysfunctional shape if abnormal temperature or pH conditions prevent it from folding correctly. Activity • Working in a small group, create a simple board game to model the key steps of transcription and translation and have classmates spend ten minutes playing the game. • Provided with incomplete or incorrect diagrams illustrating transcription and translation in prokaryotes, have students refine or revise the diagrams and share the edited versions with classmates for critical review. Think About It • Many antibiotics inhibit protein synthesis. For example, tetracycline blocks the A site on the ribosome. What is the likely effect of tetracycline on protein synthesis? • Using a chart of codons, transcribe and translate the following DNA sequence (non-template strand): 5′ATGGCCGGTTATTAAGCA-3′. How can a single nucleotide change affect the protein produced from this sequence and its function? 622 Chapter 15 | Genes and Proteins KEY TERMS 7-methylguanosine cap modification added to the 5' end of pre-mRNAs to protect mRNA from degradation and assist translation aminoacyl tRNA synthetase corresponding amino acid enzyme that “charges” tRNA molecules by catalyzing a bond between the tRNA and a anticodon three-nucleotide sequence in a tRNA molecule that corresponds to an mRNA codon CAAT box (GGCCAATCT) essential eukaryotic promoter sequence involved in binding transcription factors Central Dogma states that genes specify the sequence of mRNAs, which in turn specify the sequence of proteins codon three consecutive nucleotides in mRNA that specify the insertion of an amino acid or the release of a polypeptide chain during translation colinear in terms of RNA and protein, three “units” of RNA (nucleotides) specify one “unit” of protein (amino acid) in a consecutive fashion consensus DNA sequence that is used by many species to perform the same or similar functions core enzyme prokaryotic RNA polymerase consisting of α, α, β, and β' but missing σ; this complex performs elongation degeneracy (of |
C, G), and the polypeptide alphabet (20 amino acids). The Central Dogma describes the flow of genetic information in the cell from genes to mRNA to proteins. Genes are used to make mRNA by the process of transcription; mRNA is used to synthesize proteins by the process of translation. The genetic code is degenerate because 64 triplet codons in mRNA specify only 20 amino acids and three nonsense codons. Almost every species on the planet uses the same genetic code. 15.2 Prokaryotic Transcription In prokaryotes, mRNA synthesis is initiated at a promoter sequence on the DNA template comprising two consensus sequences that recruit RNA polymerase. The prokaryotic polymerase consists of a core enzyme of four protein subunits 624 Chapter 15 | Genes and Proteins and a σ protein that assists only with initiation. Elongation synthesizes mRNA in the 5' to 3' direction at a rate of 40 nucleotides per second. Termination liberates the mRNA and occurs either by rho protein interaction or by the formation of an mRNA hairpin. 15.3 Eukaryotic Transcription Transcription in eukaryotes involves one of three types of polymerases, depending on the gene being transcribed. RNA polymerase II transcribes all of the protein-coding genes, whereas RNA polymerase I transcribes rRNA genes, and RNA polymerase III transcribes rRNA, tRNA, and small nuclear RNA genes. The initiation of transcription in eukaryotes involves the binding of several transcription factors to complex promoter sequences that are usually located upstream of the gene being copied. The mRNA is synthesized in the 5' to 3' direction, and the FACT complex moves and reassembles nucleosomes as the polymerase passes by. Whereas RNA polymerases I and III terminate transcription by protein- or RNA hairpin-dependent methods, RNA polymerase II transcribes for 1,000 or more nucleotides beyond the gene template and cleaves the excess during pre-mRNA processing. 15.4 RNA Processing in Eukaryotes Eukaryotic pre-mRNAs are modified with a 5' methylguanosine cap and a poly-A tail. These structures protect the mature mRNA from degradation and help export it from the nucleus. Pre-mRNAs also undergo splicing, in which introns are removed and exons are reconnected with single-nucleotide accuracy. Only finished mRNAs that have undergone 5' |
capping, 3' polyadenylation, and intron splicing are exported from the nucleus to the cytoplasm. Pre-rRNAs and pretRNAs may be processed by intramolecular cleavage, splicing, methylation, and chemical conversion of nucleotides. Rarely, RNA editing is also performed to insert missing bases after an mRNA has been synthesized. 15.5 Ribosomes and Protein Synthesis The players in translation include the mRNA template, ribosomes, tRNAs, and various enzymatic factors. The small ribosomal subunit forms on the mRNA template either at the Shine-Dalgarno sequence (prokaryotes) or the 5' cap (eukaryotes). Translation begins at the initiating AUG on the mRNA, specifying methionine. The formation of peptide bonds occurs between sequential amino acids specified by the mRNA template according to the genetic code. Charged tRNAs enter the ribosomal A site, and their amino acid bonds with the amino acid at the P site. The entire mRNA is translated in three-nucleotide “steps” of the ribosome. When a nonsense codon is encountered, a release factor binds and dissociates the components and frees the new protein. Folding of the protein occurs during and after translation. REVIEW QUESTIONS 1. What is the flow of information for the synthesis of proteins according to the central dogma? a. DNA to mRNA to protein b. DNA to mRNA to tRNA to protein c. DNA to protein to mRNA to protein d. mRNA to DNA to mRNA to protein 2. The DNA of virus A is inserted into the protein coat of virus B. The combination virus is used to infect E. coli. The virus particles produced by the infection are analyzed for DNA and protein contents. What results would you expect? a. DNA and protein from B b. DNA and protein from A c. DNA from A and protein from B d. DNA from B and protein from A 3. The AUC and AUA codons in mRNA both specify isoleucine. What feature of the genetic code explains this? a. Complementarity b. Degeneracy c. Nonsense codons d. Universality 4. How many nucleotides are in 12 mRNA codons? a. b. c. d. 12 24 36 48 5. Which of the following molecules does not contain genetic information? a. DNA b. mRNA c. |
Protein d. RNA 6. Which molecule in the central dogma can be compared to a disposable photocopy of a book kept on reserve in the This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 625 library? 5S rRNA? a. DNA b. mRNA c. Protein d. tRNA a. polymerase I b. polymerase II c. polymerase III d. ribonuclease I 7. Which subunit of the E. coli polymerase confers specificity to transcription? 14. What transcripts will be most affected by low levels of α -amanitin? a. α b. c. β β ’ d. σ 8. Why are the −10 and −35 regions of prokaryotic promoters called consensus sequences? a. 18S and 28S rRNAs b. 5S rRNAs and tRNAs c. other small nuclear RNAs d. pre-mRNAs 15. Which of the following features distinguishes eukaryotic transcription from bacterial transcription? a. Eukaryotic transcription does not start at a a. They are identical in all bacterial species. consensus sequence. b. They are similar in all bacterial species. b. Eukaryotic transcription does not require an c. They exist in all organisms. d. They have the same function in all organisms. 9. The sequence that signals the end of transcription is called the: a. promoter b. stop codon c. TATA box d. terminator 10. If the ρ protein is missing, will a prokaryotic gene be terminated? a. It depends on the gene. b. No, the rho protein is essential. c. Transcription termination is not required. d. Yes, the rho protein is not involved in transcription. 11. Which feature of promoters can be found in both prokaryotes and eukaryotes? a. GC box b. octamer box c. TATA box d. -10 and -35 sequences initiation complex. c. Eukaryotic transcription and translation do not take place at the same time. d. Eukaryotic transcription does not require a termination sequence. 16. A poly-A sequence is added at the: a. 5’ end of a transcript in the nucleus b. 3’-end of a transcript in the nucleus c. 5’ end of |
a transcript in the cytoplasm d. 3’-end of a transcript in the cytoplasm 17. Which pre-mRNA processing step is important for initiating translation? a. poly-A tail b. RNA editing c. splicing d. 7-methylguanosine cap 18. Where are the RNA components of ribosomes synthesized? a. cytoplasm b. endoplasmic reticulum c. nucleus d. nucleolus 12. At what stage in the transcription of a eukaryotic gene would TFII factors be active? 19. What processing step enhances the stability of pretRNAs and pre-rRNAs? a. elongation b. initiation c. processing d. termination 13. Which polymerase is responsible for the synthesis of a. cleavage b. methylation c. nucleotide modification d. splicing 626 Chapter 15 | Genes and Proteins 20. What are introns? a. DNA sequences to which polymerases bind b. c. the processed mRNA translated DNA sequences in a gene d. untranslated DNA sequences in a gene 21. What is often the first amino acid added to a polypeptide chain? a. adenine b. leucine c. methionine d. thymine 22. In any given species, there are at least how many types of aminoacyl tRNA synthetases? a. b. c. d. 20 40 100 200 23. In prokaryotic cells, ribosomes are found in/on the: a. cytoplasm b. mitochondrion c. nucleus d. endoplasmic reticulum CRITICAL THINKING QUESTIONS 27. If mRNA is complementary to the DNA template strand and the DNA template stand is complementary to the DNA non-template strand, why are base sequences of mRNA and the DNA non-template strand not identical? Could they ever be? a. No, they cannot be identical because the T nucleotide in DNA is replaced with U nucleotide in RNA and AUG is the start codon. b. No, they cannot be identical because the T nucleotide in RNA is replaced with U nucleotide in DNA. c. They can be identical if methylation of the U nucleotide in RNA occurs and gives T nucleotide. d. They can be identical if de-methylation of the U nucleotide in RNA occurs and gives T nucleotide. 28. Imagine if there were |
200 commonly occurring amino acids instead of 20. Given what you know about the genetic code, what would be the shortest possible codon length? Explain. 24. The peptide bond synthesis in prokaryotic translation is catalyzed by: a. a ribosomal protein b. a cytoplasmic protein c. mRNA itself d. ribosomal RNA 25. What would happen if the 5’ methyl guanosine was not added to an mRNA? a. The transcript would degrade when the mRNA moves out of the nucleus to the cytoplasm. b. The mRNA molecule would stabilize and start the process of translation within the nucleus of the cell. c. The mRNA molecule would move out of the nucleus and create more copies of the mRNA molecule. d. The mRNA molecule would not be able to add the poly-A tail on its strand at the 5’ end. 26. Which of the following is associated with the docking of mRNA on a ribosome in eukaryotic cells? a. Kozak’s sequence b. poly-A sequence c. Shine-Dalgarno sequence d. TATA box a. Four b. Five c. Two d. Three 29. What part of central dogma is not always followed in viruses? a. The flow of information in HIV is from RNA to DNA, then back to RNA to proteins. Influenza viruses never go through DNA. b. The flow of information is from protein to RNA in HIV virus, while the influenza virus converts DNA to RNA. c. The flow of information is similar, but nucleic acids are synthesized as a result of translation in HIV and influenza viruses. d. The flow of information is from RNA to protein. This protein is used to synthesize the DNA of the viruses in HIV and influenza. 30. Suppose a gene has the sequence ATGCGTTATCGGGAGTAG. A point mutation changes the gene to read ATGCGTTATGGGGAGTAG. How This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 627 would the polypeptide product of this gene change? 31. Explain the initiation of transcription in prokaryotes. Include all proteins involved. a. b. c. d. In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of |
these subunits, denoted α, α, β, and β ’, comprise the polymerase core enzyme. The fifth subunit, σ, is involved only in transcription initiation. The polymerase comprised of all five subunits is called the holoenzyme. In prokaryotes the polymerase is composed of four polypeptide subunits, two of which are identical. These subunits, denoted α, α, β, and β ’, comprise the polymerase core enzyme. There is a fifth subunit that is involved in translation initiation. The polymerase comprised of all four subunits is called the holoenzyme. In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted α, α, β, and β ’, comprise the polymerase holoenzyme. The fifth subunit, σ, is involved only in transcription initiation. The polymerase comprised of all five subunits is called the core enzyme. In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted α, α α, β, and β ’, comprise the polymerase core enzyme. The fifth subunit, σ, is involved only in termination. The polymerase comprised of all five subunits is called the holoenzyme. 32. In your own words, describe the difference between ρ -dependent and ρ -independent termination of transcription in prokaryotes. a. Rho-dependent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in C-G nucleotides the mRNA folds into a hairpin loop that causes the polymerase to stall. b. Rho-independent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-dependent termination, when the polymerase encounters a region rich in C-G nucleotides, the mRNA folds into a hairpin loop that causes polymerase to stall. c. Rho-dependent termination is controlled by rho protein and the polymerase begins near the end of the gene at a run of G nucleotides on the DNA |
template. In rho-independent termination, when the polymerase encounters a region rich in C-G nucleotides, the mRNA creates a hairpin loop that causes polymerase to stall. d. Rho-dependent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in A-T nucleotides, the mRNA creates a hairpin loop that causes polymerase to stall. 33. What is the main structure that differentiates between ρ -dependent and ρ -independent termination in prokaryotes? a. Rho-independent termination involves the formation of a hairpin. b. Rho-dependent termination involves the formation of a hairpin. c. Rho-dependent termination stalls when the polymerase begins to transcribe a region rich in A-T nucleotides. d. Rho-independent termination stalls when the polymerase begins to transcribe a region rich in G nucleotides. 34. Which step in the transcription of eukaryotic RNA differs the most from its prokaryotic counterpart? 628 Chapter 15 | Genes and Proteins a. The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved. b. The initiation step in prokaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved. c. The elongation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved. d. The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are not involved. 35. Would you be able to determine which RNA polymerase you isolated from a eukaryotic cell without analyzing its products? a. No, because they have the same α -amanitin sensitivity in all products. b. No, quantitative analysis of products is done to determine the type of polymerase. c. Yes, they can be determined as they differ in α -amanitin sensitivity. d. Yes, they can be determined by the number of molecules that bind to DNA. 36. Can you |
predict how alternative splicing may lead to an economy of genes? Do you need a different gene for every protein that the cell can produce? a. Alternative splicing can lead to the synthesis of several polypeptides from a single gene. b. Alternative splicing can lead to the synthesis of several forms of mRNA from a single gene. c. Alternative splicing can lead to the synthesis of several forms of codons from a set of genes. d. Alternative splicing can lead to the synthesis of several forms of ribosomes from a set of genes. 37. What is the major challenge in the production of RNA in eukaryotes compared to prokaryotes? added to an mRNA? a. The transcript would degrade when the mRNA moves out of the nucleus to the cytoplasm. b. The mRNA molecule would stabilize and start the process of translation within the nucleus of the cell. c. The mRNA molecule would move out of the nucleus and create more copies of the mRNA molecule. d. The mRNA molecule would not be able to add the poly-A tail on its strand at the 5’ end. 39. Transcribe and translate the following DNA sequence (nontemplate strand): 5’-ATGGCCGGTTATTAAGCA-3’ a. The mRNA would be 5’- AUGGCCGGUUAUUAAGCA-3’ and the protein will be MAGY. b. The mRNA would be 3’- AUGGCCGGUUAUUAAGCA-5’ and the protein will be MAGY. c. The mRNA would be 5’- ATGGCCGGTTATTAAGCA-3’ and the protein will be MAGY. d. The mRNA would be 5’- AUGGCCGGUUAUUAAGCA-3’ and the protein will be MACY. 40. The RNA world hypothesis proposes that the first complex molecule was RNA and it preceded protein formation. Which major function of the ribosomal RNA supports the hypothesis? a. b. c. d. rRNA has catalytic properties in the large subunit and it assembles proteins. rRNA is a protein molecule that helps in the synthesis of other proteins. rRNA is essential for the transcription process. rRNA plays a major role in post-translational processes. 41. A tRNA is chemically modified so that the amino acid bound is different than the one specified |
by its anticodon. Which codon in the mRNA would the tRNA recognize: the one specified by its anticodon or the one that matches the modified amino acid it carries? a. exporting the mRNA across the nuclear a. The anticodon will match the codon in mRNA. b. c. d. membrane importing the mRNA across the nuclear membrane b. The anticodon will match with the modified amino acid it carries. c. The anticodon will lose the specificity for the the mRNA staying inside the nuclear membrane tRNA molecule. the mRNA translating into proteins within seconds d. The enzyme amino acyl tRNA synthetase would lose control over the amino acid. 38. What would happen if the 5’ methyl guanosine was not This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 629 TEST PREP FOR AP® COURSES 42. What characteristic of the genetic code points to a common ancestry for all organisms? a. The code is degenerate b. The code contains 64 codons. c. The genetic code is almost universal. d. The code contains stop codons 43. What process transfers heritable material to the next generation? a. b. c. d. replication splicing transcription translation 44. When comparing transcription of heritable information in prokaryotes and eukaryotes, which events are the same? a. Transcription by polymerase, recognition of a consensus sequence in the promoter, and termination by a hairpin loop are conserved. b. Translation by polymerase, recognition of a consensus sequence in the promoter, and termination by a hairpin loop are conserved. c. Transcription by polymerase, recognition of a highly variable sequence in the promoter, and termination by a hairpin loop are conserved. d. Transcription by polymerase, recognition of a consensus sequence in the promoter, and elongation by a hairpin loop are conserved. 45. Which of the following cell structures does not contain heritable information? a. chloroplast b. cytoplasmic membrane c. mitochondria d. nucleus 46. How does the enzyme reverse transcriptase violate the central dogma of molecular biology in HIV? a. The enzyme reverse transcriptase reverse transcribes the RNA in the genome of HIV to DNA. b. The enzyme reverse transcriptase translates the RNA of the HIV into protein and then back to DNA. c. |
The enzyme reverse transcriptase transcribes the DNA straight into the protein molecules. d. The enzyme reverse transcriptase transcribes DNA to RNA, then again to DNA. There is no protein synthesis. 47. Radioactive deoxythymidine triphosphate is supplied to the protist Euglena. After an interval of time, the cells are homogenized and different fractions are analyzed for radioactivity content in large nucleic acid molecules. Which fraction will not be labeled? a. nucleus b. mitochondrion c. chloroplast d. plasma membrane 48. You sequence a gene of interest and isolate the matching mRNA. You find that the mRNA is considerably shorter than the DNA sequence. Why is that? a. There was an experimental mistake. The mRNA should have the same length as the gene. b. The mRNA should be longer than the DNA sequence because the promoter is also transcribed. c. The processed mRNA is shorter because introns were removed. d. The mRNA is shorter because the signal sequence to cross the nuclear membrane was removed. 49. A mutation in the promoter region of the gene for the beta-globin can cause beta-thalassemia, a hereditary condition which causes anemia. Why would mutations in the promoter region lead to low levels of hemoglobin? a. The globin chains produced are too long to form functional hemoglobin. b. The globin chains are too short to form functional hemoglobin. c. Fewer globin chains are synthesized because less mRNA is transcribed. d. Globin chains do not fold properly and are non- functional. 50. You are given three mRNA sequences: 630 Chapter 15 | Genes and Proteins 1. 5’-UCG-GCA- AAU-UUA -GUU-3’ 2. 5’-UCU-GCA- AAU-UUA -GUU-3’ 3. 5’-UCU-GCA- AAU-UAA -GUU-3’ Using the table, write the peptide encoded by each of the mRNA sequences. a. 1. Serine-alanine-asparagine-leucine-valine 2. Serine-alanine-asparagine-leucine-valine 3. Serine-alanine-asparagine(-stop) b. 1. Serine-phenylalanine-asparagine-leucine- valine 2. Serine |
-alanine-asparagine-leucine-valine 3. Serine-alanine-asparagine (-stop) c. 1. Serine-alanine-asparagine-leucine-valine 2. Serine-alanine-asparagine (-stop) 3. Serine-alanine-asparagine-leucine-valine d. 1. Serine-alanine-asparagine-leucine-valine 2. Serine-arginine-asparagine-leucine-valine 3. Serine-alanine-asparagine(-stop) 51. You are given three mRNA sequences: 1. 5’-UCG-GCA- AAU-UUA -GUU-3’ 2. 5’-UCU-GCA- AAU-UUA -GUU-3’ 3. 5’-UCU-GCA- AAU-UAA -GUU-3’ Using the peptide encoded by each of the above, compare the three peptides obtained. How are peptides 2 and 3 different from 1? What would be the consequence for the cell in each case? a. There is a silent mutation in peptide 2 and peptide 3 has a stop codon due to mutation. b. There is a silent mutation in peptide 3 and peptide 2 has a stop codon due to mutation. c. There is a different amino acid in peptide 2 and peptide 3 has a stop codon due to mutation. d. There isn’t a mutation in peptide 2 and peptide 3 has a stop codon due to mutation. SCIENCE PRACTICE CHALLENGE QUESTIONS 52. Gamow (1954) proposed that the structure of DNA deduced by Watson and Crick (1953) could be interpreted as a way of forming roughly 20 "words" of the common amino acids from the four "letters" A, T, C, and G that represent DNA nucleotides. Crick and coworkers (1961) used a method developed by Benzer to induce mutations in the DNA of a virus by the insertion of a single nucleotide. The mutant could not infect the bacterium Escherichia coli and neither could viruses with a second insertion of a second DNA nucleotide. However, a third nucleotide insertion restored the ability of the virus to infect the bacterium. In 1961 |
, Nirenberg and Matthaei conducted a series of experiments to better understand the flow of genetic information from gene to protein. They discovered that in solutions containing the contents of ruptured E. coli bacterial cells from which DNA had been removed, polymers containing only one repeating amino acid, phenylalanine, would be synthesized if synthetic mRNA composed of only the single nucleotide, uracil (U), was added to the solution in which phenylalanine was also present. In solutions containing mRNA with only adenine (A) or cytosine (C) and the amino acids lysine or proline, polymers containing only these amino acids would be synthesized. The researchers found that when ribosomes were removed by filtration, these polymers did not form. Nirenberg and Leder (1964) extended this work to include other nucleotides. A. Summarize the conclusions regarding the encoding and decoding of heritable information supported by these studies. Explain how these studies provided evidence to support the Triplet Code. Khorana (1960) developed a technique for synthesizing RNA composed of predictable distributions of repeated pairs or triplets of nucleotides. He found, for example, that RNA synthesized when A and U were present in relative concentrations of 4:1, respectively, will produce RNA sequences with these distributions determined by their relative probabilities: AAU:AAA, AUA:AAA, and UAA:AAA; 0.82 × 0.2/0.83 = 1/4 [calculated as follows: i) 4/5 of the bases are A, so the likelihood of selecting A is 0.8; ii) the selection is repeated to determine the second letter of the three-letter codon; iii) the likelihood of selecting a U is 1 in 5; iv) the probability of selecting the This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 15 | Genes and Proteins 631 set AUU is the product; v) similarly, the probability of AAA is (4/5)3; and vi) the ratio of these probabilities is their relative likelihood]: AUU:AAA, UUA:AAA, and UAU:AAA; 0.8 × 0.22/0.83 = 1/16; and UUU:AAA; 0.23/0.83 = 1/64. B. Based on Khor |
ana’s findings, calculate the relative distributions of the following ratios of concentrations of RNA triplet sequences from mixtures in which the relative concentrations of guanine and cytosine, G:C, are 5:1. Ratio Relative Probabilities GGC:GGG GCG:GGG CGG:GGG GCC:GGG CGC:GGG CCG:GGG CCC:GGG Table 15.2 C. Based on the work of Nirenberg, Matthaei, Leder, and Khorana, the following table was constructed (taken from Khorana's Nobel Prize address): Figure 15.18 A solution containing the amino acids shown in the table above and equal concentrations of the two nucleotides C and G is prepared. Predict the proteins that can be synthesized from this mixture in terms of each possible codon and their relative concentrations in terms of their amino acid repeat sequences. D. Describe the effects of the codons UAA, UAG, and UGA on protein synthesis. 53. The yeast life cycle is usually dominated by haploid cells, each with a single set of unpaired chromosomes. The cell propagates asexually, and the genetic material is replicated through mitosis. Cell division occurs every 2–4 hours, leading to 60–100 generations in a single day. Yeast also reproduce sexually, particularly under adverse environmental conditions. When two haploid cells—with DNA containing complementary mating-type alleles—conjugate, a diploid zygote results. The diploid zygote can then complete the sexual segment of the life cycle through meiosis. After meiosis, four haploid spores are produced, which can germinate. Researchers can grow yeast easily on nutrient-containing plates. Because both asexual and sexual reproduction is rapid, yeast has become an important organism for the experimental investigation of mutagenesis and evolution among eukaryotes. Environmental factors, such as chemicals or radiation, induce mutations. High-energy UV-c radiation of less than 1 minute in duration will result in many mutated yeast cells. UV-c can be used to mutate a strain of yeast in which the synthesis of adenine is blocked. This mutation is observable because the ade-2 mutant has a red color when cultured on nutrientcontaining plates. Exposure to uv-c also can result in additional mutations. In particular, one mutant, ade-7, changes the color of the ade-2 mutant |
to white. A. You have a uv-c lamp, culture plates, and growth chambers at 23 °C and 37 °C. You also have available known haploid strains that are (ade-2,+,+), where + denotes the wild type. Design a plan to determine the rate of uv-c-induced mutations in nutrient-containing plates inoculated with yeast. Earth's ozone layer removes high-energy ultraviolet radiation, uv-c, from the solar radiation received at the surface. Lower-energy ultraviolet radiation, uv-b, strikes Earth’s surface. Damage to DNA induced by ultraviolet radiation occurs with the formation of bonds between an adjacent pair of pyrimidine nucleotides, thymine and cytosine, on the same strand of DNA. A repair enzyme, photolyase, which is activated by visible light, is present in plants and most animals, but not in humans. In characterizing the relationship between environmental mutagens and cell damage, a useful assumption is often made and referred to as the linear hypothesis. This assumption states that the extent of damage is proportional to the amount of radiation received. Mutation rates for a strain (preac) that does not produce photolyase and a wild-type (+) strain were studied. Cultures of the two strains of yeast were diluted, and nutrient-containing plates were inoculated in triplicate at 23 °C. The plates were exposed to bright sunlight for varying time intervals. After exposure, the plates were incubated in the dark at 23 °C. After incubation between 1 and 8 hours, data shown in the table below were collected by counting the density of living cells relative to the control, and averaging these among replicates. B. Using the data table below, graph the average survival fraction, relative to the wild-type control. Predict the number of mutations in a sample of 1,000 cells of the preac type that are exposed to bright sunlight for 15 632 seconds. Chapter 15 | Genes and Proteins Figure 15.19 This is a 5 column table, showing Incubation time, in hours in the left most column, ranging from 1 to 8. A 10 second exposure has the following values for an incubation time of 1 to 8: 0.83, 1.00, 0.92, 0.75, 0.99, 0.81, 0.80, 1.05 and 0.89 with a standard deviation of 0.11. A 20 second |
exposure has the following values for an incubation time of 1 to 8: 0.58, 0.43, 0.38, 0.35, 0.49, 0.42, 0.32, 0.59, 0.45, with a standard deviation of 0.10. A 30 second exposure has the following values for an incubation time of 1 to 8: 0.33, 0.17, 0.12, 0.08, 0.11, 0.12, 0.09, 0.11, 0.14, with a standard deviation of 0.08. A 40 second exposure has the following values for an incubation time of 1 to 8: 0.17, 0.09, 0.03, 0.01, 0.01, 0.01, 0.01, 0.01, 0.04, with a standard deviation of.0.06. A 50 second exposure has the following values for an incubation time of 1 to 8: 0.08, 0.04, 0.01, 0.00, 0.00, 0.00, 0.00, 0.00, 0.02 with a standard deviation of 0.03 Yeast can also be used to study sexual reproduction, a somewhat puzzling phenomenon. Cloning of cells through mitosis is molecularly much less complex than meiosis, consumes less energy, and is less risky. Two alternative explanations for the evolution of sexual reproduction are popular. In one model, through assortment of genes, meiosis leads to an increase in the frequency of beneficial mutations. In the second model, detrimental mutations are purged from a population through sex. Studies using yeast (Gray and Goddard, Evol. Biol., 2012 and McDonald et al., Nature 2012) have provided a mechanism to study these models. As shown below, the fitness (defined as the log of the ratio of the number of cells in successive generations) of yeast is graphed as a function of number of mitotic reproductions in yeast grown in low-stress and high-stress environments, and with and without alternating induction of sexual reproduction. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Figure 15.20 C. Based on these data, evaluate the merits of the alternative theories of the adaptive advantage provided by sexual reproduction. 54. A. Describe the storage and retrieval of genetic information with the following model. Use the list to fill in the |
blanks with the letter corresponding to the correct term. a. amino acid b. tRNA c. DNA d. transcription e. mRNA f. translation g. protein h. RNA polymerase i. rRNA Within the cytoplasm, __ is synthesized from __ bound to __ in a sequence that corresponds to information provided by __. This process is called __. Within the nucleus, information originating in __ is encoded as a sequence of bases in __, which is synthesized by the enzyme __ that is embedded in the __. This process is called __. B. During development, cell differentiation occurs, and the expression of genes is permanently switched off. Using the model summarized above, explain where information flow is most effectively blocked. C. A chemical message is received by the cell regulating the timing of events controlled by gene expression. Using the model summarized above, explain where information flow is most effectively managed. Chapter 15 | Genes and Proteins 633 55. Structure and function in biology result from both the presence of genetic information and the expression of that information. Some genes are continually expressed, whereas the expression of most genes is regulated, commonly at the level of transcription. At the initiation of transcription, the TATA-binding protein (TBP) provides access to the DNA strand to be transcribed. The 5’TATAAA3’ sequence called the TATA box is found in prokaryotes, archaebacteria, and eukaryotes. Even among eukarya, when the TATA box is not present among eukaryotes, the initiation of transcription involves TBP. Scientists attribute this common characteristic to the relative thermostability of the A-T interaction. Hydrogen bonds hold the two strands of the DNA double helix together. This type of bond has the smallest interaction energy of all intermolecular forces; as temperature increases, these bonds are broken. A. Explain the advantage, in terms of the energy required, which is provided by an AT-rich region in the sequence where transcription is initiated. B. The fact that the TATA box or the associated TBP are common to all domains provides evidence of common ancestry among all life. Pose a scientific question that would need to be addressed by a valid alternative explanation of this fact. C. A whole-genome survey of prokaryotes (Zheng and Wu, BMC Bioinformatics, 2010) showed that the relative amounts of guanine and cytosine in DNA poorly predicted the temperature |
range conditions that are suitable for an organism. Refine the question posed in part B, taking this result into account. 56. Only a fraction of DNA encodes proteins. The noncoding portion of a gene is referred to as the intron. The intron fraction depends upon the gene. Introns are rare in prokaryotic and mitochondrial DNA; in human nuclear DNA, this fraction is about 95%. The intron is transcribed into mRNA, but this noncoding mRNA is edited out before translation of the coding portion, or exon, of a gene. The edited exon segments are then spliced together by a spliceosome, a very large and complex collection of RNAs and proteins. Although introns do not encode proteins, they have functions. In particular, they amplify expression of the exon, although the mechanism is unknown. When introns are very long, which is common among mammalian genes with roles in development, they can significantly extend the time required to complete transcription. Analysis of genes common to different plant and animal species shows many shared intronic positions and base sequences, although in some organisms, such as yeast, many introns have been deleted. Because introns do not encode proteins, mutations can remain silent and accumulate. A. As described above, introns are ancestral remnants that are replicated because they do not disadvantage the organism. Consider the claim that introns are “junk DNA.” Evaluate the claim with supporting evidence. B. Introns may be retained during transcription. Explain how the retention of a transcribed intron between two transcribed exons within a gene could do the following: • block expression of one polypeptide sequence • increase expression of a polypeptide • alter the polypeptide expressed 634 Chapter 15 | Genes and Proteins This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 16 | Gene Regulation 635 16 | GENE REGULATION Figure 16.1 The genetic content of each somatic cell in an organism is the same, but not all genes are expressed in every cell. The control of which genes are expressed dictates whether a cell is (a) an eye cell or (b) a liver cell. It is the differential gene expression patterns that arise in different cells that give rise to (c) a complete organism. Chapter Outline 16.1: Regulation of Gene Expression 16.2: Prokaryotic Gene |
Regulation 16.3: Eukaryotic Epigenetic Gene Regulation 16.4: Eukaryotic Transcriptional Gene Regulation 16.5: Eukaryotic Post-transcriptional Gene Regulation 16.6: Eukaryotic Translational and Post-translational Gene Regulation 16.7: Cancer and Gene Regulation Introduction Most people know that regular exercise is important to maintain good health. It promotes cardiovascular health and helps to prevent obesity. Scientists have now discovered that long-term endurance training also changes how genes are expressed in muscle tissue. In a recent study, 23 healthy people each exercised one leg for 45 minutes four days a week while resting the other leg. After three months, muscles from participants’ legs were biopsied, and scientists analyzed the activity level of over 20,000 genes in the tissue samples. They found that for each participant the exercised leg had reduced inflammation and improved metabolism compared with the non-exercised leg. These differences were accompanied by changes in genes associated with metabolism and inflammation. However, the actual nucleotide sequences of the genes weren’t changed. Instead, some genes were methylated, which simply means methyl groups were attached to certain nucleotides along the sequence. This, essentially, turned the genes “off” or otherwise changed how they were expressed. DNA methylation is an example of epigenetics, which is a process that alters genes without affecting the nucleotide sequence of the genes. The full research article can be found here (http://openstaxcollege.org/l/32endurance). 636 Chapter 16 | Gene Regulation 16.1 | Regulation of Gene Expression In this section, you will explore the following question: • How does prokaryotic gene regulation differ from eukaryotic gene regulation? Connection for AP® Courses Structure and function in biology result from the presence of genetic information and the correct expression of this information. In the chapter on DNA structure and function, we explored how genes are translated into proteins, which in turn determine the nature of the cell. But how does a cell know when to “turn on” its DNA? With few exceptions, each cell in your body contains identical genetic information. If each cell has the same exact DNA make up, how is it that a liver cell differs from a nerve or muscle cell? As we will discover, although each cell shares the same genome and DNA sequence, each cell does not express exactly the same genes. Many factors determine when and how genes are expressed |
in a given cell. Even the type of chromosome a gene is located on, like whether it is a sex chromosome or not, can determine its expression pattern, as can mutations or changes in DNA sequence and other external factors. In prokaryotes, gene expression is regulated primarily at the level of transcription, when DNA is copied into RNA. However, eukaryotes have evolved regulatory mechanisms in gene expression at multiple levels. In all cases, regulation of gene expression determines the type and amount of protein produced in the cell. Errors in regulatory processes can result in many human diseases and conditions, including cancer. Gene expression regulation occurs at different points in prokaryotes and eukaryotes. Prokaryotic organisms express their entire genome in every cell, but not necessarily all at the same time. In general, a gene is expressed only when its specific protein product is needed. Remember that each cell in an organism carries the same DNA as every other cell. Yet cells of eukaryotic organisms each express a unique subset of DNA depending on cell type. To express a protein, DNA is first transcribed into RNA, which is then translated into proteins. In prokaryotic cells, transcription and translation occur almost simultaneously. In eukaryotic cells, transcription occurs in the nucleus, separate from the translation that occurs in the cytoplasm along ribosomes attached to endoplasmic reticulum. As stated above, gene expression in prokaryotes is regulated at the level of transcription, whereas in eukaryotes, gene expression is regulated at multiple levels, including the epigenetic (DNA), transcriptional, pre- and post-transcriptional, and translational levels. The science of epigenetics studies heritable changes in the genome that do not affect the underlying DNA gene sequences. The content presented in this section supports the learning objectives outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The AP® learning objectives merge essential knowledge content with one or more of the seven science practices. These objectives provide a transparent foundation for the AP® Biology course, along with inquiry-based laboratory experiences, instructional activities, and AP® exam questions. Big Idea 3 Enduring Understanding 3.B Essential Knowledge Science Practice Learning Objective Living systems store, retrieve, transmit and respond to information essential to life processes. Expression of genetic information involves cellular and molecular mechanisms. 3.B.1 Gene regulation results in differential gene expression, leading to cell specialization 7.1 The student can connect phenomena and models across |
spatial and temporal scales. 3.18 The student is able to describe the connection between the regulation of gene expression and observed differences between different kinds of organisms. For a cell to function properly, necessary proteins must be synthesized at the proper time. All cells control or regulate the synthesis of proteins from information encoded in their DNA. The process of turning on a gene to produce RNA and protein is called gene expression. Whether in a simple unicellular organism or a complex multi-cellular organism, each cell controls This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 16 | Gene Regulation 637 when and how its genes are expressed. For this to occur, there must be a mechanism to control when a gene is expressed to make RNA and protein, how much of the protein is made, and when it is time to stop making that protein because it is no longer needed. Although genetic differences between species and between individuals within a species are often responsible for phenotypic differences, another mechanism that can create phenotypic differences is differences in gene expression. For example, although every cell in an organism contains the same genes, the bone cells in the organism appears different from the fat cells due to differences in which genes are expressed by which cell. Similarly, although mice and humans share approximately 97.5% of their genes, they are very different organisms because different genes are turned on at different times during development and in different cells. Even organisms that share 100% identity in their genomes (a.k.a clones) can eventually appear different if they express their genes differently in response to different environmental conditions, for example. Even among humans, identical twins can possess different birthmarks, wrinkles, or other features that arise during development sometimes due to differential gene expression. The regulation of gene expression conserves energy and space. It would require a significant amount of energy for an organism to express every gene at all times, so it is more energy efficient to turn on the genes only when they are required. In addition, only expressing a subset of genes in each cell saves space because DNA must be unwound from its tightly coiled structure to transcribe and translate the DNA. Cells would have to be enormous if every protein were expressed in every cell all the time. The control of gene expression is extremely complex. Malfunctions in this process are detrimental to the cell and can lead to the development of many diseases. Prokaryotic versus Eukaryotic Gene Expression To understand how gene |
expression is regulated, we must first understand how a gene codes for a functional protein in a cell. The process occurs in both prokaryotic and eukaryotic cells, just in slightly different manners. Prokaryotic organisms are single-celled organisms that lack a cell nucleus, and their DNA therefore floats freely in the cell cytoplasm. To synthesize a protein, the processes of transcription and translation occur almost simultaneously. When the resulting protein is no longer needed, transcription stops. As a result, the primary method to control what type of protein and how much of each protein is expressed in a prokaryotic cell is the regulation of DNA transcription. All of the subsequent steps occur automatically. When more protein is required, more transcription occurs. Therefore, in prokaryotic cells, the control of gene expression is mostly at the transcriptional level. Eukaryotic cells, in contrast, have intracellular organelles that add to their complexity. In eukaryotic cells, the DNA is contained inside the cell’s nucleus and there it is transcribed into RNA. The newly synthesized RNA is then transported out of the nucleus into the cytoplasm, where ribosomes translate the RNA into protein. The processes of transcription and translation are physically separated by the nuclear membrane; transcription occurs only within the nucleus, and translation occurs only outside the nucleus in the cytoplasm. The regulation of gene expression can occur at all stages of the process (Figure 16.2). Regulation may occur when the DNA is uncoiled and loosened from nucleosomes to bind transcription factors ( epigenetic level), when the RNA is transcribed (transcriptional level), when the RNA is processed and exported to the cytoplasm after it is transcribed ( post-transcriptional level), when the RNA is translated into protein (translational level), or after the protein has been made ( post-translational level). 638 Chapter 16 | Gene Regulation Figure 16.2 Prokaryotic transcription and translation occur simultaneously in the cytoplasm, and regulation occurs at the transcriptional level. Eukaryotic gene expression is regulated during transcription and RNA processing, which take place in the nucleus, and during protein translation, which takes place in the cytoplasm. Further regulation may occur through post-translational modifications of proteins. The differences in the regulation of gene expression between prokaryotes and eukaryotes are summarized in Table 16 |
.1. The regulation of gene expression is discussed in detail in subsequent modules. Differences in the Regulation of Gene Expression of Prokaryotic and Eukaryotic Organisms Prokaryotic organisms Eukaryotic organisms Lack nucleus Contain nucleus DNA is found in the cytoplasm DNA is confined to the nuclear compartment RNA transcription and protein formation occur almost simultaneously Gene expression is regulated primarily at the transcriptional level Table 16.1 RNA transcription occurs prior to protein formation, and it takes place in the nucleus. Translation of RNA to protein occurs in the cytoplasm. Gene expression is regulated at many levels (epigenetic, transcriptional, nuclear shuttling, post-transcriptional, translational, and post-translational) This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 16 | Gene Regulation 639 Prokaryotic cells can only regulate gene expression by controlling the amount of transcription. As eukaryotic cells evolved, the complexity of the control of gene expression increased. For example, with the evolution of eukaryotic cells came compartmentalization of important cellular components and cellular processes. A nuclear region that contains the DNA was formed. Transcription and translation were physically separated into two different cellular compartments. It therefore became possible to control gene expression by regulating transcription in the nucleus, and also by controlling the RNA levels and protein translation present outside the nucleus. Some cellular processes arose from the need of the organism to defend itself. Cellular processes such as gene silencing developed to protect the cell from viral or parasitic infections. If the cell could quickly shut off gene expression for a short period of time, it would be able to survive an infection when other organisms could not. Therefore, the organism evolved a new process that helped it survive, and it was able to pass this new development to offspring. Cytochrome c oxidase is a highly conserved protein found in bacteria and in the mitochondria of eukaryotes. Based on your knowledge of evolutionary relationships, which of the following statements would you expect to be true of the cytochrome c oxidase protein sequence? a. The bacterial protein will be more similar to the human protein than the yeast protein. b. The yeast protein will be more similar to the human protein than the bacterial protein. c. The bacterial protein will be more similar to the yeast protein than the human protein. d. The bacterial and yeast proteins will share a similar sequence, but |
the human protein will be unrelated. Think About It How does controlling gene expression alter the overall protein level in the cell? 16.2 | Prokaryotic Gene Regulation In this section, you will explore the following question: • What are operons and what are the roles of activators, inducers, and repressors in regulating operons and gene expression? Connection for AP® Courses The regulation of gene expression in prokaryotic cells occurs at the transcriptional level. Simply stated, if a cell does not transcribe the DNA’s message into mRNA, translation (protein synthesis), does not occur. Bacterial genes are often organized into common pathways or processes called operons for more coordinated regulation of expression. For example, in E. coli, genes responsible for lactose metabolism are located together on the bacterial chromosome. (The operon model includes several components, so when studying how the operon works, it is helpful to refer to a diagram of the model. See Figure 16.3 and Figure 16.4.) The operon includes a regulatory gene that codes for a repressor protein that binds to the operator, which prevents RNA polymerase from transcribing the gene(s) of interest. An example of this is seen in the structural genes for lactose metabolism. However, if the repressor is inactivated, RNA polymerase binds to the promoter, and transcription of the structural genes occurs. There are three ways to control the transcription of an operon: inducible control, repressible control, and activator control. 640 Chapter 16 | Gene Regulation The lac operon is an example of inducible control because the presence of lactose “turns on” transcription of the genes for its own metabolism. The trp operon is an example of repressible control because it uses proteins bound to the operator sequence to physically prevent the binding of RNA polymerase. If tryptophan is not needed by the cell, the genes necessary to produce it are turned off. Activator control (typified by the action of Catabolite Activator Protein) increases the binding ability of RNA polymerase to the promoter. Certain genes are continually expressed via this regulatory mechanism. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The learning objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam |
questions. A learning objective merges required content with one or more of the seven science practices. Big Idea 3 Enduring Understanding 3.B Living systems store, retrieve, transmit and respond to information essential to life processes. Expression of genetic information involves cellular and molecular mechanisms. Essential Knowledge 3.B.1 Gene regulation results in differential gene expression, leading to cell specialization Science Practice 1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively Learning Objective 3.21 The student can use representations to describe how gene regulation influences cell products and function. Essential Knowledge 3.B.2 A variety of intercellular and intracellular signal transmissions mediate gene expression. Science Practice 1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. Learning Objective 3.23 The student can use representations to describe mechanisms of the regulation of gene expression. The DNA of prokaryotes is organized into a circular chromosome supercoiled in the nucleoid region of the cell cytoplasm. Proteins that are needed for a specific function, or that are involved in the same biochemical pathway, are encoded together in blocks called operons. For example, all of the genes needed to use lactose as an energy source are coded next to each other in the lactose (or lac) operon. In prokaryotic cells, there are three types of regulatory molecules that can affect the expression of operons: repressors, activators, and inducers. Repressors are proteins that suppress transcription of a gene in response to an external stimulus, whereas activators are proteins that increase the transcription of a gene in response to an external stimulus. Finally, inducers are small molecules that either activate or repress transcription depending on the needs of the cell and the availability of substrate. The trp Operon: A Repressor Operon Bacteria such as E. coli need amino acids to survive. Tryptophan is one such amino acid that E. coli can ingest from the environment. E. coli can also synthesize tryptophan using enzymes that are encoded by five genes. These five genes are next to each other in what is called the tryptophan (trp) operon (Figure 16.3). If tryptophan is present in the environment, then E. coli does not need to synthesize it and the switch controlling the activation of the genes in the trp operon is switched off. However, when |
tryptophan availability is low, the switch controlling the operon is turned on, transcription is initiated, the genes are expressed, and tryptophan is synthesized. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 16 | Gene Regulation 641 Figure 16.3 The five genes that are needed to synthesize tryptophan in E. coli are located next to each other in the trp operon. When tryptophan is plentiful, two tryptophan molecules bind the repressor protein at the operator sequence. This physically blocks the RNA polymerase from transcribing the tryptophan genes. When tryptophan is absent, the repressor protein does not bind to the operator and the genes are transcribed. A DNA sequence that codes for proteins is referred to as the coding region. The five coding regions for the tryptophan biosynthesis enzymes are arranged sequentially on the chromosome in the operon. Just before the coding region is the transcriptional start site. This is the region of DNA to which RNA polymerase binds to initiate transcription. The promoter sequence is upstream of the transcriptional start site; each operon has a sequence within or near the promoter to which proteins (activators or repressors) can bind and regulate transcription. A DNA sequence called the operator sequence is encoded between the promoter region and the first trp coding gene. This operator contains the DNA code to which the repressor protein can bind. When tryptophan is present in the cell, two tryptophan molecules bind to the trp repressor, which changes shape to bind to the trp operator. Binding of the tryptophan–repressor complex at the operator physically prevents the RNA polymerase from binding, and transcribing the downstream genes. When tryptophan is not present in the cell, the repressor by itself does not bind to the operator; therefore, the operon is active and tryptophan is synthesized. Because the repressor protein actively binds to the operator to keep the genes turned off, the trp operon is negatively regulated and the proteins that bind to the operator to silence trp expression are negative regulators. Watch this video (http://openstaxcollege.org/l/trp_operon) to learn more about the trp operon. What would happen if bacteria did not have trp R? a. The cell would not |
be able to break down tryptophan. b. The cell will gradually produce more tryptophan over time. c. The cell would not be able to make tryptophan. d. The cell would make tryptophan when it was not needed. 642 Chapter 16 | Gene Regulation Catabolite Activator Protein (CAP): An Activator Regulator Just as the trp operon is negatively regulated by tryptophan molecules, there are proteins that bind to the operator sequences that act as a positive regulator to turn genes on and activate them. For example, when glucose is scarce, E. coli bacteria can turn to other sugar sources for fuel. To do this, new genes to process these alternate genes must be transcribed. When glucose levels drop, cyclic AMP (cAMP) begins to accumulate in the cell. The cAMP molecule is a signaling molecule that is involved in glucose and energy metabolism in E. coli. When glucose levels decline in the cell, accumulating cAMP binds to the positive regulator catabolite activator protein (CAP), a protein that binds to the promoters of operons that control the processing of alternative sugars. When cAMP binds to CAP, the complex binds to the promoter region of the genes that are needed to use the alternate sugar sources (Figure 16.4). In these operons, a CAP binding site is located upstream of the RNA polymerase binding site in the promoter. This increases the binding ability of RNA polymerase to the promoter region and the transcription of the genes. Figure 16.4 When glucose levels fall, E. coli may use other sugars for fuel but must transcribe new genes to do so. As glucose supplies become limited, cAMP levels increase. This cAMP binds to the CAP protein, a positive regulator that binds to an operator region upstream of the genes required to use other sugar sources. The lac Operon: An Inducer Operon The third type of gene regulation in prokaryotic cells occurs through inducible operons, which have proteins that bind to activate or repress transcription depending on the local environment and the needs of the cell. The lac operon is a typical inducible operon. As mentioned previously, E. coli is able to use other sugars as energy sources when glucose concentrations are low. To do so, the cAMP–CAP protein complex serves as a positive regulator to induce transcription. One such sugar source is lactose. The lac operon encodes the genes necessary to acquire and process the |
lactose from the local environment. CAP binds to the operator sequence upstream of the promoter that initiates transcription of the lac operon. However, for the lac operon to be activated, two conditions must be met. First, the level of glucose must be very low or non-existent. Second, lactose must be present. Only when glucose is absent and lactose is present will the lac operon be transcribed (Figure 16.5). This makes sense for the cell, because it would be energetically wasteful to create the proteins to process lactose if glucose was plentiful or lactose was not available. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 16 | Gene Regulation 643 Figure 16.5 Transcription of the lac operon is carefully regulated so that its expression only occurs when glucose is limited and lactose is present to serve as an alternative fuel source. In E. coli, the trp operon is on by default, while the lac operon is off. Why do you think that this is the case? a. The trp operon is inducible and is positively regulated. Therefore it is ON by default whereas the lac operon is repressible and is OFF by default. b. The trp operon synthesizes tryptophan which is essential for the cell and therefore remains ON, whereas the lac operon synthesizes enzymes for the breakdown of a sugar that is not always available and remains OFF by default. c. The trp operon is constitutive and remains ON by default, whereas the lac operon is repressible and therefore is OFF by default. d. The lac operon undergoes transcriptional attenuation and therefore is OFF by default, whereas the trp operon is not regulated by any such mechanism and is ON by default. If glucose is absent, then CAP can bind to the operator sequence to activate transcription. If lactose is absent, then the repressor binds to the operator to prevent transcription. If either of these requirements is met, then transcription remains off. 644 Chapter 16 | Gene Regulation Only when both conditions are satisfied is the lac operon transcribed (Table 16.2). Signals that Induce or Repress Transcription of the lac Operon Glucose CAP binds Lactose Repressor binds Transcription + + - - Table 16.2 - - + + - + - + + - + - No Some No |
Yes Watch an animated tutorial (http://openstaxcollege.org/l/lac_operon) about the workings of lac operon here. The E. coli bacteria can have several mutations that affect the lac operon system. One mutation inhibits the ability of RNA polymerase to bind to the lac operon. How would this affect the cell? a. The cell would make more lactose. b. There would be no lactose outside of the cell. c. The cell would not be able to process tryptophan. d. The cell would not be able to process lactose. Activity Modeling the Operon. Use construction paper or more elaborate materials, such as Styrofoam noodles, electrical tape, and Velcro tabs, to create a model of the lac and trp operons that include a regulator, promoter, operator, and structural genes. Then use the model to show how the presence of substrate, e.g., allolactose or tryptophan, can change the activity of the operons. As an extension of the activity, use the model to make predictions about the effects of mutations in any of the regions on gene expression. Think About It In E. coli, the trp operon is on by default, while the lac operon is off by default. Why do you think this is the case? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 16 | Gene Regulation 645 16.3 | Eukaryotic Epigenetic Gene Regulation In this section, you will explore the following question: • What is the science of epigenetics and how is this process regulated? Connection for AP® Courses One reason that eukaryotic gene expression is more complex than prokaryotic gene expression is because the processes of transcription and translation are physically separated within the eukaryotic cell. Eukaryotic cells also package their genomes in a more sophisticated way compared with prokaryotic cells. Consequently, eukaryotic cells can regulate gene expression at multiple levels, beginning with control of access to DNA. Because genomic DNA is folded around histone proteins to create nucleosome complexes, nucleosomes physically regulate the access of proteins, such as transcription factors and enzymes, to the underlying DNA. Methylation of DNA and histones causes nucleosomes to pack tightly together, preventing transcription factors from binding to the DNA. Methylated nucleosomes |
contain DNA that is not expressed. On the other hand, histone acetylation results in loose packing of nucleosomes, allowing transcription factors to bind to DNA. Acetylated nucleosomes contain DNA that may be expressed. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.B Essential Knowledge Science Practice Learning Objective Living systems store, retrieve, transmit and respond to information essential to life processes. Expression of genetic information involves cellular and molecular mechanisms. 3.B.1 Gene regulation results in differential gene expression, leading to cell specialization. 7.1 The student can connect phenomena and models across spatial and temporal scales 3.19 The student is able to describe the connection between the regulation of gene expression and observed differences between individuals in a population Epigenetic Control: Regulating Access to Genes within the Chromosome As stated earlier, one reason why eukaryotic gene expression is more complex than prokaryotic gene expression is because the processes of transcription and translation are physically separated. Unlike prokaryotic cells, eukaryotic cells can regulate gene expression at many different levels. Eukaryotic gene expression begins with control of access to the DNA. This form of regulation, called epigenetic regulation, occurs even before transcription is initiated. The human genome encodes over 20,000 genes; each of the 23 pairs of human chromosomes encodes thousands of genes. The DNA in the nucleus is precisely wound, folded, and compacted into chromosomes so that it will fit into the nucleus. It is also organized so that specific segments can be accessed as needed by a specific cell type. The first level of organization, or packing, is the winding of DNA strands around histone proteins. Histones package and order DNA into structural units called nucleosome complexes, which can control the access of proteins to the DNA regions (Figure 16.6a). Under the electron microscope, this winding of DNA around histone proteins to form nucleosomes looks like small beads on a string (Figure 16.6b). These beads (histone proteins) can move along the string (DNA) and change the structure of the molecule. 646 |
Chapter 16 | Gene Regulation Figure 16.6 DNA is folded around histone proteins to create (a) nucleosome complexes. These nucleosomes control the access of proteins to the underlying DNA. When viewed through an electron microscope (b), the nucleosomes look like beads on a string. (credit “micrograph”: modification of work by Chris Woodcock) If DNA encoding a specific gene is to be transcribed into RNA, the nucleosomes surrounding that region of DNA can slide down the DNA to open that specific chromosomal region and allow for the transcriptional machinery (RNA polymerase) to initiate transcription (Figure 16.7). Nucleosomes can move to open the chromosome structure to expose a segment of DNA, but do so in a very controlled manner. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 16 | Gene Regulation 647 Figure 16.7 Nucleosomes can slide along DNA. When nucleosomes are spaced closely together (top), transcription factors cannot bind and gene expression is turned off. When the nucleosomes are spaced far apart (bottom), the DNA is exposed. Transcription factors can bind, allowing gene expression to occur. Modifications to the histones and DNA affect nucleosome spacing. In females, one of the two X chromosomes is inactivated during embryonic development because of epigenetic changes to the chromatin. What impact do you think these changes would have on nucleosome packing? a. Methylation of DNA and hypo-acetylation of histones causes the nucleosomes to pack loosely together, inactivating one of the X chromosomes. b. Methylation of histones and hyper-acetylation of DNA causes the nucleosomes to pack tightly together, inactivating one of the X chromosome due to transcriptional repression. c. Methylation of DNA and hypo-acetylation of histones causes the nucleosomes to pack tightly together, inactivating one of the X chromosomes. d. Acetylation of DNA and hyper-methylation of histones causes the nucleosomes to pack tightly together, inactivating one of the X chromosomes. How the histone proteins move is dependent on signals found on both the histone proteins and on the DNA. These signals are tags added to histone proteins and DNA that tell the histones if a chromosomal region should be open or closed (Figure 16. |
8 depicts modifications to histone proteins and DNA). These tags are not permanent, but may be added or removed as needed. They are chemical modifications (phosphate, methyl, or acetyl groups) that are attached to specific amino acids in the protein or to the nucleotides of the DNA. The tags do not alter the DNA base sequence, but they do alter how tightly wound the DNA is around the histone proteins. DNA is a negatively charged molecule; therefore, changes in the charge of the histone will change how tightly wound the DNA molecule will be. When unmodified, the histone proteins have a large positive charge; by adding chemical modifications like acetyl groups, the charge becomes less positive. The DNA molecule itself can also be modified. This occurs within very specific regions called CpG islands. These are stretches with a high frequency of cytosine and guanine dinucleotide DNA pairs (CG) found in the promoter regions of genes. When this configuration exists, the cytosine member of the pair can be methylated (a methyl group is added). This modification changes how the DNA interacts with proteins, including the histone proteins that control access to the region. 648 Chapter 16 | Gene Regulation Highly methylated (hypermethylated) DNA regions with deacetylated histones are tightly coiled and transcriptionally inactive. Figure 16.8 Histone proteins and DNA nucleotides can be modified chemically. Modifications affect nucleosome spacing and gene expression. (credit: modification of work by NIH) This type of gene regulation is called epigenetic regulation. Epigenetic means “around genetics.” The changes that occur to the histone proteins and DNA do not alter the nucleotide sequence and are not permanent. Instead, these changes are temporary (although they often persist through multiple rounds of cell division) and alter the chromosomal structure (open or closed) as needed. A gene can be turned on or off depending upon the location and modifications to the histone proteins and DNA. If a gene is to be transcribed, the histone proteins and DNA are modified surrounding the chromosomal region encoding that gene. This opens the chromosomal region to allow access for RNA polymerase and other proteins, called transcription factors, to bind to the promoter region, located just upstream of the gene, and initiate transcription. If a gene is to remain turned off, or silenced, the histone proteins and DNA have different modifications that signal a closed chromosomal configuration. In this closed |
configuration, the RNA polymerase and transcription factors do not have access to the DNA and transcription cannot occur (Figure 16.7). Think About It In females, one of the two X chromosomes is inactivated during embryonic development because of epigenetic changes to the chromatin. What impact do you think these changes will have on nucleosome packaging and, consequently, gene expression? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 16 | Gene Regulation 649 View this video (http://openstaxcollege.org/l/epigenetic_reg) that describes how epigenetic regulation controls gene expression. Explain how the study of epigenetics can lead to improved treatment of cancer. a. Epigenetics would allow new body parts to be synthesized that could replace those damaged by cancer. b. Epigenetics could change the genetic code of all cells in the body to prevent them from becoming cancerous. c. New therapies could be made that changes the genetic code of harmful cancer genes. d. New therapies could be made that do not require altering the cancer cell’s DNA. 16.4 | Eukaryotic Transcriptional Gene Regulation In this section, you will explore the following question: • What is the role of transcription factors, enhancers, and repressors in gene regulation? Connection for AP® Courses To start transcription, general transcription factors must first bind to a specific area on the DNA called the TATA box and then recruit RNA polymerase to that location. In addition, other areas on the DNA called enhancer regions help augment transcription. Transcription factors can bind to enhancer regions to increase or prevent transcription. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices Big Idea 3 Enduring Understanding 3.B Essential Knowledge Science Practice Learning Objective Essential Knowledge Living systems store, retrieve, transmit and respond to information essential to life processes. Expression of genetic information involves cellular and molecular mechanisms. 3.B.1 Gene regulation results in differential gene expression, leading to cell specialization 7.1 The student can connect phenomena and models across spatial and temporal scales. 3.18 The |
student is able to describe the connection between the regulation of gene expression and observed differences between different kinds of organisms 3.B.1 Gene regulation results in differential gene expression, leading to cell specialization 650 Science Practice Learning Objective Essential Knowledge Science Practice Learning Objective Chapter 16 | Gene Regulation 7.1 The student can connect phenomena and models across spatial and temporal scales 3.19 The student is able to describe the connection between the regulation of gene expression and observed differences between individuals in a population 3.B.1 Gene regulation results in differential gene expression, leading to cell specialization. 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices 3.20 The student is able to explain how the regulation of gene expression is essential for the processes and structures that support efficient cell function. Essential Knowledge 3.B.1 1 Gene regulation results in differential gene expression, leading to cell specialization. Science Practice Learning Objective 1.4 The student can use representations and models to analyze situations or solve problems qualitatively and quantitatively. 3.21 The student can use representations to describe how gene regulation influences cell products and function. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 3.18] Like prokaryotic cells, the transcription of genes in eukaryotes requires the actions of an RNA polymerase to bind to a sequence upstream of a gene to initiate transcription. However, unlike prokaryotic cells, the eukaryotic RNA polymerase requires other proteins, or transcription factors, to facilitate transcription initiation. Transcription factors are proteins that bind to the promoter sequence and other regulatory sequences to control the transcription of the target gene. RNA polymerase by itself cannot initiate transcription in eukaryotic cells. Transcription factors must bind to the promoter region first and recruit RNA polymerase to the site for transcription to be established. The activity of transcription factors can regulate differential gene expression in cells, resulting in the development of different cell products and functions. For example, scientists have found that primary sexual characteristics is regulated by several genes Figure 16.9. In the fruit fly Drosophila, the slx gene determines sex. This gene is expressed when the organism has two copies of the X chromosome. The gene product for slx binds to the mRNA of the tra gene and regulates its splicing. In the presence of slx, tra is spliced into its female form and influences the expression of dsx and fru to result |
in female sexual characteristics. In the absence of slx, tra is spliced into its male form and male sexual characteristics result. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 16 | Gene Regulation 651 Figure 16.9 In Drosophila melanogaster, the sex is determined by a series of splicing events in sex determination genes on a cell-by-cell basis without any involvement of sex hormones (which circulate throughout the entire body). The primary sex-determination gene is Sex lethal (Sxl), which is transcribed only when the X/A ratio (the X chromosome-to-autosome ratio) equals or exceeds 1. As females have two X chromosomes and males have one, Sxl is transcribed only in females (see the figure, left part). Sxl is a splicing repressor and binds to its target, the primary RNA of the transformer (tra) gene, which undergoes differential splicing depending on the presence (female) or absence (male) of Sxl, yielding a protein-coding tra mRNA only in females. The Tra protein binds to the primary RNAs of doublesex (dsx) and fruitless (fru). In females, Tra promotes splicing to occur near its binding site, whereas in males it uses an alternative, default splice site. The dsx primary RNA thus produces female-specific mRNA and malespecific mRNA, both of which encode functional Dsx proteins, DsxF and DsxM, respectively. The presence (female) and absence (male) of Tra similarly results in female-type fru mRNA and male-type fru mRNA, but here, only the maletype fru mRNA encodes a functional protein. View the process of transcription—the making of RNA from a DNA template—at this site (http://openstaxcollege.org/ l/transcript_RNA). Describe the major events that occur during eukaryotic transcription. a. DNA unwinds, transcription factors bind, the termination complex forms, and DNA polymerase adds nucleotides to the mRNA. b. DNA unwinds, transcription factors bind, and RNA polymerase adds nucleotides to the mRNA. c. The transcription complex forms, transcription factors add nucleotides to the forming mRNA, and the mRNA disconnects from the DNA. d. Elongation occurs, followed by |
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