Chapter 13 Proteinsynthesis PDF

Republic of the Philippines Leyte Normal University College of Arts and Sciences Science Unit Tacloban City Module 13: Proteinsynthesis: The Flow of Information from DNA to RNA to Protein Key Terms: Central dogma, Genes, Gene...

Republic of the Philippines Leyte Normal University College of Arts and Sciences Science Unit Tacloban City Module 13: Proteinsynthesis: The Flow of Information from DNA to RNA to Protein Key Terms: Central dogma, Genes, Genetic code, Transcription, Translation, Protein, Initiation, Elongation, Termination, Amino acid, RNA Polymerase, Promoter, mRNA, tRNA, mutation, Codon, Anticodon, Ribosomes, Promoters, mRNA splicing, intron, exon. From: https://meme.com/labome/status/1106822101154050048 Voice-Over PowerPoint Presentation The Central Dogma Prokaryotic Transcription Eukaryotic Transcription Learning Outcomes: 1. Explain the “central dogma” of DNA-protein synthesis. 2. Describe the genetic code and how the nucleotide sequence prescribes the amino acid and the protein sequence.  3. Describe and compare protein synthesis in prokaryotes and eukaryotes. 4. List the different steps in prokaryotic and eukaryotic transcription. 5. Discuss the role of promoters in prokaryotic transcription. 6. Discuss the role of RNA polymerase in eukaryotic transcription. 7. Describe how and when transcription is terminated. 8. Know the general functions of the three major types of RNA (mRNA, rRNA, tRNA). 9. Describe the DNA sequence motifs and proteins required to initiate transcription. 10. Predict the RNA transcribed from a DNA sequence identified as either the template strand or the coding strand. 11. Use the genetic code to predict the protein amino acid sequence translated from an mRNA sequence. 12. Describe the process of and key components required for translation. 13. Predict the likely effects of mutations in DNA on protein amino acid sequence, structure and function. 14. Describe the genetic code and how the nucleotide sequence prescribes the amino acid and the protein sequence Getting Started! Before starting your journey with this module, test yourself first with your current knowledge about chromosomes, mitosis and meiosis. Answer the following questions by encircling the correct letter that corresponds to the option that best answers the question. 1. Control of gene expression in eukaryotic cells occurs at which level(s)? A. only the transcriptional level B. epigenetic and transcriptional levels C. epigenetic, transcriptional, and translational levels D. epigenetic, transcriptional, post-transcriptional, translational, and post- translational levels 2. The binding of ________ is required for transcription to start. A. a protein C. RNA polymerase B. DNA polymerase D. a transcription factor 3. What will result from the binding of a transcription factor to an enhancer region? A. decreased transcription of an adjacent gene  B. increased transcription of a distant gene C. alteration of the translation of an adjacent gene D. initiation of the recruitment of RNA polymerase 4. Post-translational modifications of proteins can affect which of the following? A. protein function C. chromatin modification B. transcriptional regulation D. all of the above 5. Prokaryotic cells lack a nucleus. Therefore, the genes in prokaryotic cells are: A. all expressed, all of the time B. transcribed and translated almost simultaneously C. transcriptionally controlled because translation begins before transcription ends D. b and c are both true 6. Post-translational control refers to: A. regulation of gene expression after transcription B. regulation of gene expression after translation C. control of epigenetic activation D. period between transcription and translation 7. How does the regulation of gene expression support continued evolution of more complex organisms? A. Cells can become specialized within a multicellular organism. B. Organisms can conserve energy and resources. C. Cells grow larger to accommodate protein production. D. Both A and B. 8. Which of the following are involved in posttranscriptional control? A. control of RNA splicing C. control of RNA stability B. control of RNA shuttling D. all of the above 9. Binding of an RNA binding protein will ________ the stability of the RNA molecule. A. increase C. neither increase nor decrease B. decrease D. either increase or decrease 10. Cancer causing genes are called ________. A. transformation genes C. oncogenes B. tumor suppressor genes D. mutated genes Introduction Regardless of what kind of organism, (bacteria, archaea, or eukaryotes), the primary role of DNA is to store heritable information that encodes the instruction set required for creating the organism in question. While we have gotten much better at quickly reading the chemical composition like the sequence of nucleotides in a genome and some of the chemical modifications that are made to it, we still don't know how to reliably decode all of the information within and all of the mechanisms by which it is read and ultimately expressed. Since the rediscovery of Mendel’s work in 1900, the definition of the gene has progressed from an abstract unit of heredity to a tangible molecular entity capable of replication, expression, and mutation. Now, genes are composed of DNA and are linearly arranged on chromosomes. Further, genes specify the sequences of amino acids, which are the building blocks of proteins. In return, proteins are responsible for orchestrating nearly every function of the cell. Together, genes and the proteins they encode are absolutely essential to life. There are, however, some core principles and mechanisms associated with the reading and expression of the genetic code whose basic steps are understood and that need to be part of the conceptual toolkit for all biologists. Two of these processes are transcription and translation, which are the coping of parts of the genetic code written in DNA into molecules of the related polymer RNA and the reading and encoding of the RNA code into proteins, respectively. The basic flow of genetic information in biological systems is often depicted in a scheme known as "the central dogma". This states that information encoded in DNA flows into RNA via transcription and ultimately to proteins via translation. Processes like reverse transcription (the creation of DNA from and RNA template) and replication also represent mechanisms for propagating information in different forms. This scheme, however, doesn't say anything per se about how information is encoded or about the mechanisms by which regulatory signals move between the various layers of molecule types depicted in the model. Therefore, while the central dogma is a nearly required part of the lexicon of any biologist, perhaps left over from old tradition, students should also be aware that mechanisms of information flow are more complex as we'll learn about some as we go, and that "the central dogma" only represents some core pathways.  I. The Central Dogma: DNA Encodes RNA; RNA Encodes Protein As our understanding of biological molecules increased in the 20th century, researchers discovered that all living organisms share a genetic code. In 1956, Francis Crick proposed that DNA is an informational storage molecule capable of replicating itself. Thus, he proposed that the information that was transmitted had to be “read” by a manufacturing body within the cell which puts amino acids together in a specific sequence ultimately synthesizing a protein. This became known as the central dogma of molecular biology. Genes generally are the information for making proteins. The central dogma describes the flow of genetic information in cells from DNA to mRNA to protein (Figure 1). It also states that genes specify the sequence of mRNAs, which in turn specify the sequence of amino acids making up all proteins. With the help of specific proteins and RNAs, the decoding of one molecule to another is made possible. 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 collinear, such that nucleotides 1 to 3 correspond to amino acid 1, nucleotides 4 to 6 correspond to amino acid 2, and so on. In the central dogma, DNA serves as a template for the direct synthesis of a messenger RNA (mRNA) molecule, in a process known as transcription. Secondly, mRNA is “read” at a ribosome by transfer RNAs (tRNAs), which work together to assemble a specific chain of amino acids, which collectively assemble to generate a protein, in a process known as translation. Proteins are the cell’s internal machinery. Similar to parts of a car, each protein has a specific three-dimensional shape that determines its function. Any change in the shape potentially changes the function of the protein. LINKS TO LEARNING: View this link http://thebiologyprimer.com/the-central- dogma-lecture-video for additional information about the central dogma of biology.  Figure 1. Gene expression: The flow of genetic information from DNA via RNA to protein. In transcription, the enzyme RNA polymerase copies DNA to produce an RNA transcript. In translation, the cellular machinery uses instructions in mRNA to synthesize a polypeptide, following the rules of the genetic code II. The Genetic Code A code is a system of symbols that equates information in one language with information in another. In the Genetic Code, a triplet codon represents an amino acid. The language of nucleic acids is written in four nucleotides— A, G, C, and T in the DNA dialect; A, G, C, and U in the RNA dialect, while the language of proteins is written in amino acids. To understand how the sequence of bases in DNA or RNA encodes the order of amino acids in a polypeptide chain, it is a must to know how many distinct amino acids there are. Watson and Crick produced the now accepted list of the 20 amino acids that are genetically encoded by DNA or RNA sequence. They created the list by analyzing the amino-acid sequence of a variety of naturally occurring polypeptides. The protein sequences consist of 20 commonly occurring amino acids; therefore, it can be said that the protein alphabet consists of 20 “letters” (Figure 2). Different amino acids have different chemistries (such as acidic versus basic, or polar and nonpolar) and different structural constraints. Variation in amino acid sequence is responsible for the enormous variation in protein structure and function. Figure 2. Structures of the 20 amino acids found in proteins are shown. Each amino acid is composed of an amino group, a carboxyl group, and a side chain (blue). The side chain may be nonpolar, polar, or charged, 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.  The four nucleotides encode 20 amino acids through specific groupings of A, G, C, and T or A, G, C, and U. Each codon, designated by the bases defining its three nucleotides, specifies one amino acid. For example, GAA is a codon for glutamic acid (Glu), and GUU is a codon for valine (Val). Because the code comes into play only during the translation part of gene expression, that is, during the decoding of messenger RNA to polypeptide, geneticists usually present the code in the RNA dialect of A, G, C, and U, as depicted in Fig.3. When speaking of genes, they can substitute T for U to show the same code in the DNA dialect. Each amino acid is defined by a three- nucleotide sequence called the triplet codon. Given the different numbers of “letters” in the mRNA and protein “alphabets,” scientists theorized that single amino acids must be represented by combinations of nucleotides. 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 by Francis Crick and Sydney Brenner who 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, the normal proteins were not produced. When three nucleotides were inserted, the protein was synthesized and functional. This demonstrated that the amino acids must be specified by groups of three nucleotides. 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. 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. Further, in codons that instruct 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, performs a special function, 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. Following the start codon, the mRNA is read in groups of three until a stop codon is encountered. Figure 3. The genetic code: 61 codons represent the 20 amino acids, while 3 codons signify stop. To read the code, find the first letter in the left column, the second letter along the top, and the third letter in the right column; this reading corresponds to the 5′-to-3′ direction along the mRNA. The arrangement of the coding table reveals the structure of the code. There are sixteen "blocks" of codons, each specified by the first and second nucleotides of the codons within the block, e.g., the "AC*" block that corresponds to the amino acid threonine (Thr). Some blocks are divided into a pyrimidine half, in which the codon ends with U or C, and a purine half, in which the codon ends with A or G. Some amino acids get a whole block of four codons, like alanine (Ala), threonine (Thr) and proline (Pro). Some get the pyrimidine half of their block, like histidine (His) and asparagine (Asn). Others get the purine half of their block, like glutamate (Glu) and lysine (Lys). Note that some amino acids get a block and a half-block for a total of six codons. The specification of a single amino acid by multiple similar codons is called "degeneracy." 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. Moreover, amino acids with chemically similar side chains are encoded by similar codons. For example, aspartate (Asp) and glutamate (Glu), which occupy the GA* block, are both negatively charged. This nuance of the genetic code ensures that a single-nucleotide substitution mutation might specify the same amino acid but have no effect or specify a similar amino acid, preventing the protein from being rendered completely nonfunctional.  The genetic code is nearly universal. With a few minor 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. Having 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. A scientist sequencing mRNA identifies the following strand: CUAUGUGUCGUAACAGCCGAUGACCCG What is the sequence of the amino acid chain this mRNA makes when it is translated? LINKS TO LEARNING: Transcribe a gene and translate it to protein using complementary pairing and the genetic code at this site http://openstax.org/l/create_protein. A. Four general themes for gene expression (i) Pairing of complementary bases is the key to the transfer of information from DNA to RNA and from RNA to protein (ii) Polarities of DNA, RNA, and polypeptides help guide the mechanisms of gene expression (iii) Gene expression requires input of energy and participation of specific proteins and macromolecular assemblies (iv) Mutations that change genetic information or obstruct the flow of its expression can have dramatic effects on phenotype B. The Genetic Code: A Summary The following list summarizes the code’s main features: (i) The code consists of triplet codons, each of which specifies an amino acid. As written in Fig. 8.3 on p. 240, the code shows the 5′-to-3′ sequence of  the three nucleotides in each mRNA codon; that is, the first nucleotide depicted is at the 5′ end of the codon. (ii) The codons are nonoverlapping. In the mRNA sequence 5′ GAAGUUGAA 3′, for example, the first three nucleotides (GAA) form one codon; nucleotides 4 through 6 (GUU) form the second; and nucleotides 7 through 9 (GAA), the third. Each nucleotide is part of only one codon. (iii) The code includes three stop, or nonsense, codons: UAA, UAG, and UGA. These codons do not encode an amino acid and thus terminate translation. (iv) The code is degenerate, which means that in many cases more than one codon specifies the same amino acid. Despite its “degeneracy,” the code is unambiguous, because each codon specifies only one amino acid. (v) In reading the transcript of a gene, the cellular machinery scans a single strand of mRNA from a fixed starting point that establishes a reading frame. As we see later, the nucleotide triplet AUG, which specifies the amino acid methionine, serves in certain contexts as the initiation codon, marking the precise spot in the nucleotide sequence of an mRNA where the code for a particular polypeptide begins. (vi) Moving from the 5′ to the 3′ end of an mRNA, each successive codon is sequentially interpreted into an amino acid, starting at the N terminus and moving toward the C terminus of the resulting polypeptide. (vii) Mutations may modify the message encoded in a sequence of nucleotides in three ways. Frameshift mutations are nucleotide insertions or deletions that alter the genetic instructions for polypeptide construction by changing the reading frame. Missense mutations change a codon for one amino acid to a codon for a different amino acid. Nonsense mutations change a codon for an amino acid to a stop codon. C. Using Genetics to Verify the Genetic Code The experiments that cracked the genetic code by assigning codons to amino acids were all in vitro studies using cell-free extracts and synthetic mRNAs. A logical question thus arose: Do living cells construct polypeptides according to the same rules? (i) The study of Yanofsky, was used to verify the code. He found two trp auxotrophic mutations in the E. coli tryptophan synthetase gene that produced two different amino acids (arginine, or Arg, and glutamic acid, or Glu) at the same position—amino acid 211—in the polypeptide chain (Fig. 4a). According to the code, both of these mutations could have resulted from single-base changes in the GGA codon that normally inserts glycine (Gly) at position 211. (ii) Yanofsky obtained better evidence yet that cells use the genetic code in vivo by analyzing proflavin-induced frameshift mutations of the tryptophan synthetase gene (Fig. 4b). He first treated populations of E. coli with proflavin to produce trp mutants. Subsequent treatment of these mutants with more proflavin generated some trp+ revertants among the progeny. The most likely explanation for the revertants was that their tryptophan synthetase gene carried both a single-base-pair deletion and a single-base-pair insertion (+,- ). (iii) Yanofsky’s results helped confirm not only amino-acid codon assignments but other parameters of the code as well. His interpretations make sense only if codons do not overlap and are read from a fixed starting point with no pauses or commas separating the adjacent triplets. a b Figure 4a and b. (a) Altered amino acids in trp– mutations and trp+ revertants; (b) Amino acid alterations that accompany intragenic suppression. Experimental verification of the genetic code. (a) Single-base substitutions can explain the amino-acid substitutions of trp mutations and trp+ revertants. (b) The genetic code predicts the amino-acid alterations (yellow) that would arise from single-base-pair deletions and suppressing insertions. III. Prokaryotic Transcription The prokaryotes, represented Bacteria and Archaea, are mostly single-celled organisms that lack membrane bound nuclei and other organelles. A bacterial chromosome is a 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 region. In addition, prokaryotes often have abundant plasmids, which are shorter, circular DNA molecules that may only contain one or a few genes. These plasmids can be transferred independently from the bacterial chromosome during cell division and often carry traits such as those involved with 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 non-template strand, or the coding strand. The only nucleotide difference is that in mRNA, all of the T nucleotides are replaced with U nucleotides (Figure 5). In an RNA double helix, A can bind U via two hydrogen bonds, just as in A–T pairing in a DNA double helix. Figure 5. Messenger RNA is a copy of protein-coding information in the coding strand of DNA, with the substitution of U in the RNA for T in the coding sequence. However, new RNA nucleotides base pair with the nucleotides of the template strand. RNA is synthesized in its 5'-3' direction, using the enzyme RNA polymerase. As the template is read, the DNA unwinds ahead of the polymerase and then rewinds behind it 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 denoted with a “-” and are designated upstream nucleotides. Conversely, nucleotides following the initiation site are denoted with “+” numbering and are called downstream nucleotides. Steps in Prokaryotic Transcription (i) The Initiation of Transcription. RNA polymerase binds to double-stranded DNA at the beginning of the gene to be copied. RNA polymerase recognizes and binds to promoters, specialized DNA sequences near the beginning of a gene where transcription will start. Although the promoters of different genes vary substantially in size and sequence, all promoters contain two characteristic short sequences of 6–10 nucleotide pairs that help bind RNA polymerase. These short sequences are nearly identical in different promoters. In bacteria, the complete RNA polymerase (the holoenzyme) consists of a core enzyme, plus a α subunit involved only in initiation. The α subunit reduces RNA polymerase’s general affinity for DNA but simultaneously increases RNA polymerase’s affinity for the promoter. As a result, the RNA polymerase holoenzyme can hone in on a promoter and bind tightly to it, forming a so-called closed promoter complex. Prokaryotic Promoters A promoter is a DNA sequence onto which the transcription machinery, including RNA polymerase, 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 evolutionarily conserved in many species. 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 (Fig.6). The -10 sequence, called the -10 region, has the consensus sequence TATAAT. The -35 sequence has the consensus sequence TTGACA. These consensus sequences are 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 6. The σ subunit of prokaryotic RNA polymerase recognizes consensus sequences found in the promoter region upstream of the transcription start site. The σ subunit dissociates from the polymerase after transcription has been initiated. (ii) Elongation: Constructing an RNA copy of the gene. When the α subunit separates from the RNA polymerase, the enzyme loses its enhanced affinity for the promoter sequence and regains its strong generalized affinity for any DNA. These changes enable the core enzyme to leave the promoter yet remain bound to the gene. The region of DNA unwound by RNA polymerase is called the transcription bubble. Within the bubble, the nascent RNA chain remains base paired with the DNA template, forming a DNA-RNA hybrid. Once an RNA polymerase has moved off the promoter, other RNA polymerase molecules can move in to initiate transcription. If the promoter is very strong, that is, if it can rapidly attract RNA polymerase, the gene can undergo transcription by many RNA polymerases simultaneously. (See figure below) (iii) Termination: The End of Transcription. As shown in the illustration below, RNA sequences that signal the end of transcription are known as terminators. There are two types of terminators: intrinsic terminators, which cause the RNA polymerase core enzyme to terminate transcription on its own, and extrinsic terminators, which require proteins other than RNA polymerase—particularly a polypeptide known as rho protein—to bring about termination. All terminators, whether intrinsic or extrinsic, are specific sequences in the mRNA; they are transcribed from specific DNA regions and they signal the termination of transcription. Terminators often form hairpin loops in which nucleotides within the mRNA pair with nearby complementary nucleotides. Upon termination, RNA polymerase and a completed RNA chain are both released from the DNA. 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. In contrast, the presence of a nucleus in eukaryotic cells precludes simultaneous transcription and translation. Figure 7. 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. A fragment of bacterial DNA reads: 3’ –TACCTATAATCTCAATTGATAGAAGCACTCTAC– 5’ Assuming that this fragment is the template strand, what is the sequence of mRNA that would be transcribed? (Hint: Be sure to identify the initiation site). The product of transcription is a single-stranded primary transcript. The RNA produced by the action of RNA polymerase on a gene is a single strand of nucleotides known as a primary transcript. The bases in the primary transcript are complementary to the bases between the initiation and termination sites in the template strand of the gene. The nucleotides carrying these bases include groupings for a start codon, codons specifying all the amino acids in the polypeptide to be built, and a stop codon. (Fig. 8). Figure. 8. The Product of Transcription Is a Single-Stranded Primary Transcript  LINKS TO LEARNING: Visit this BioStudio animation (http://openstax.org/l/transcription2 ) to see the process of prokaryotic transcription. View this Molecular Movies animation (http://openstax.org/l/transcription ) to see the first part of transcription and the base sequence repetition of the TATA box. IV. Eukaryotic Transcription Prokaryotes and eukaryotes perform fundamentally the same process of transcription, with some key differences. The most important difference between prokaryote and eukaryote transcription is due to the presence of membrane-bound nucleus and organelles in eukaryotes. 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, which means that they specify a single protein. (i) 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 to help recruit the appropriate polymerase. (ii) The Three Eukaryotic RNA Polymerases. The features of eukaryotic mRNA synthesis are markedly more complex than 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. a) RNA polymerase I is located in the nucleolus, and a specialized nuclear substructure in which ribosomal RNA (rRNA) is transcribed, processed, and assembled into ribosomes. RNA polymerase I synthesizes all of the rRNAs from the tandemly duplicated set of 18S, 5.8S, and 28S ribosomal genes. b) RNA polymerase II is located in the nucleus and synthesizes all protein-coding nuclear pre-mRNAs. Eukaryotic pre-mRNAs undergo extensive processing after transcription but before translation. RNA  polymerase II is responsible for transcribing the overwhelming majority of eukaryotic genes. c) 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. (iii) RNA Polymerase II Promoters and Transcription Factors. Eukaryotic promoters are much larger and more intricate than prokaryotic promoters. However, both have a sequence similar to the -10 sequence of prokaryotes. But in eukaryotes, this sequence is called the TATA box, and has the consensus sequence TATAAA on the coding strand. It is located at -25 to -35 bases relative to the initiation (+1) site. This sequence is not identical to the E. coli -10 box, but it conserves the A–T rich element. (iv) 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. Although the enzymatic process of elongation is essentially the same in eukaryotes and prokaryotes, the DNA template is considerably 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 to 2,000 nucleotides beyond the end of the gene being transcribed. This pre-mRNA tail is subsequently  removed by cleavage during mRNA processing. Further, RNA polymerases I and 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. (v) In Eukaryotes, RNA processing after transcription produces a mature mRNA. Some RNA processing in eukaryotes modifies only the 5′ or 3′ ends of the primary transcript, leaving untouched the information content of the rest of the mRNA. Other processing deletes blocks of information from the middle of the primary transcript so that the content of the mature mRNA is related, but not identical, to the complete set of DNA nucleotide pairs in the original gene. Adding a Methylated Cap at the 5′ End and a Poly-A Tail at the 3′ End The nucleotide at the 5′ end of a eukaryotic mRNA is a G in reverse orientation from the rest of the molecule; it is connected through a triphosphate linkage to the first nucleotide in the primary transcript. This backward G is not transcribed from the DNA. Instead, a special capping enzyme adds it to the primary transcript after polymerization of the transcript’s first few nucleotides. Enzymes called methyl transferases adds methyl (–CH3) groups to the backward G and to one or more of the succeeding nucleotides in the RNA, forming a so-called methylated cap (Fig. 9). The methylated cap is critical for efficient translation of the mRNA into protein, even though it does not help specify an amino acid. Figure 9. How RNA processing adds a tail to the 3′ end of eukaryotic mRNAs. A ribonuclease recognizes AAUAAA in a particular context of the primary transcript and cleaves the transcript 11–30 nucleotides downstream to create a new 3′ end. The enzyme poly-A polymerase then adds 100–200 As onto this new 3′ end  The eukaryotic pre-mRNA undergoes extensive processing before it is ready to be translated. Eukaryotic protein-coding sequences are not continuous, as they are in prokaryotes. The coding sequences called exons are interrupted by noncoding introns, which must be removed to make a translatable mRNA. The additional steps involved in eukaryotic mRNA maturation also 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 (Fig. 10). 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 the introns. In rare cases, the mRNA transcript can be “edited” after it is transcribed. Figure 10. Eukaryotic mRNA contains introns that must be spliced out. A 5' cap and 3' poly-A tail is also added. 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 (intron 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. All of a pre-mRNA’s introns must be completely and precisely removed before the translation of 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 11). 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 complex proteins and RNA molecules called spliceosomes. Figure 11. 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 small nuclear RNAs (snRNAs). Spliceosomes recognize sequences at the 5' and 3' end of the intron  Using the figure above, 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. How do cells make a mature mRNA from a gene whose coding sequences are interrupted by introns? Figure 12 illustrates a more detailed process of RNA splicing. Three types of short sequences within the primary transcript—splice donors, splice acceptors, and branch sites—help ensure the specificity of splicing. These sites make it possible to sever the connections between an intron and the exons that precede and follow it, and then to join the formerly separated exons. The mechanism of splicing involves two sequential cuts in the primary transcript. The first cut is at the splice-donor site, at the 5′ end of the intron. The second cut is at the splice-acceptor site, at the 3′ end of the intron; this cut removes the intron. The discarded intron is degraded, and the precise splicing of adjacent exons completes the process of intron removal. Figure 12. How RNA processing splices out introns and joins adjacent exons. (a) Three short sequences within the primary transcript determine the specificity of splicing. (1) The splice-donor site occurs where the 3′ end of an exon abuts the 5′ end of an intron. In most splice-donor sites, a GU dinucleotide (arrows) that begins the intron is flanked on either side by a few purines (Pu; that is, A or G). (2) The splice-acceptor site is at the 3’ end of the intron where it joins with the next exon. The final nucleotides of the intron are always AG (arrows) preceded by 12–14 pyrimidines (Py; that is, C or U). (3) The branch site, which is located within the intron about 30 nucleotides upstream of the splice acceptor, must include an A (arrow) and is usually rich in pyrimidines. (b) Two sequential cuts, the first at the splice-donor site and the second at the splice-acceptor site, remove the intron (which is subsequently degraded), allowing precise splicing of adjacent exons.  LINKS TO LEARNING: See how introns are removed during RNA splicing at this website (http://openstax.org/l/RNA_splicing ). V. (Proteinsynthesis)Translation: Base Pairing Between mRNA and tRNAs Directs Assembly of a Polypeptide on the Ribosome Virtually all cells—both prokaryotic and eukaryotic—use the same basic genetic code to translate the sequence of nucleotides in a messenger RNA to the sequence of amino acids in the corresponding polypeptide. This process of translation takes place on ribosomes that coordinate the movements of transfer RNAs carrying specific amino acids with the genetic instructions of an mRNA. tRNA, ribosomes and aminoacyl tRNA synthetases are the machineries of proteinsynthesis. (i) Transfer RNAs Mediate the Translation of mRNA Codons to Amino Acids. There is no obvious chemical similarity or affinity between the nucleotide triplets of mRNA codons and the amino acids they specify. Rather, transfer RNAs (tRNAs) serve as adaptor molecules that mediate the transfer of information from nucleic acid to protein. tRNAs Carry an Anticodon at One End and an Amino Acid at the Other tRNAs are short, single-stranded RNA molecules 74–95 nucleotides in length. Several of the nucleotides in tRNAs contain modified bases produced by chemical alterations of the principal A, G, C, and U nucleotides (Fig. 13 ). Each tRNA carries one particular amino acid, and cells must have at least one tRNA for each of the 20 amino acids specified by the genetic code. The name of a tRNA reflects the amino acid it carries. Transfer RNAs serve as adaptor molecules. Each tRNA carries a specific amino acid and recognizes one or more of the mRNA codons that define the order of amino acids in a protein. Aminoacyl-tRNAs bind to the ribosome 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. 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. Figure 13. tRNAs mediate the transfer of information from nucleic acid to protein. (a) Many tRNAs contain modified bases produced by chemical alterations of A, G, C, and U. (b) The primary structures of tRNA molecules fold to form characteristic secondary and tertiary structures. The anticodon and the amino-acid attachment site are at opposite ends of the “L” formed by a tRNA. (ii) Ribosomes Are the Sites of Polypeptide Synthesis. A ribosome is a complex macromolecule composed of structural and catalytic rRNAs, and many distinct polypeptides. In E. coli, ribosomes consist of 3 different ribosomal RNAs (rRNAs) and 52 different ribosomal proteins (Fig. 14). These components associate to form two different ribosomal subunits called the 30S subunit and the 50S subunit (with S designating a coefficient of sedimentation related to the size and shape of the subunit; the 30S subunit is smaller than the 50S subunit). Ribosomes facilitate polypeptide synthesis in various ways. First, they recognize mRNA features that signal the start of translation. Second, they help ensure accurate interpretation of the genetic code by stabilizing the interactions between tRNAs and mRNAs; without a ribosome, codon-anticodon recognition, mediated by only three base pairs (one of which may wobble), would  be extremely weak. Third, they supply the enzymatic activity that links the amino acids in a growing polypeptide chain. Fourth, by moving 5′ to 3′ along an mRNA molecule, they expose the mRNA codons in sequence, ensuring the linear addition of amino acids. Finally, ribosomes help end polypeptide synthesis by dissociating from both the mRNA directing polypeptide construction and the polypeptide product itself. Figure 14. The ribosome: Site of polypeptide synthesis. (a) A ribosome has two subunits, each composed of rRNA and various proteins. (b) The small subunit initially binds to mRNA. The large subunit contributes the enzyme peptidyl transferase, which catalyzes the formation of peptide bonds in the growing polypeptide chain. The two subunits together form the A and P tRNA binding sites (iii) Aminoacyl tRNA Synthetases. The process of pre-tRNA synthesis by RNA polymerase III only creates the RNA portion of the adaptor molecule. It is the corresponding amino acid that 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 one of a group of enzymes called aminoacyl tRNA synthetases. VI. The Mechanism of Translation (Proteinsynthesis) Translation consists of an initiation phase that sets the stage for polypeptide synthesis; elongation, during which amino acids are added to a growing polypeptide; and a termination phase that brings polypeptide synthesis to a halt and enables the ribosome to release a completed chain of amino acids. The details of the three stages are illustrated and discussed below: Initiation: Setting the stage for polypeptide synthesis. The first three nucleotides of an mRNA do not serve as the first codon to be translated into an amino acid. Instead, a special signal indicates where along the mRNA translation should begin. In prokaryotes, this signal is called the ribosome binding site, and it has two important elements. The first is a short sequence of six nucleotides—usually 5′... AGGAGG... 3′—named the Shine-Dalgarno box after its discoverer. The second element in an mRNA’s ribosome binding site is the triplet 5′ AUG 3′, which serves as the initiation codon. A special initiator tRNA, whose 5′ CAU 3′ anticodon is complementary to AUG, recognizes an AUG preceded by the Shine-Dalgarno box of a ribosome binding site. The initiator tRNA carries N-formylmethionine (fMet), a modified methionine whose amino end is blocked by a formyl group. The specialized fMet tRNA functions only at an initiation site. During initiation, the 3′ end of the rRNA in the 30S ribosomal subunit binds to the mRNA’s Shine-Dalgarno box, the fMet tRNA binds to the mRNA’s initiation codon, and a large 50S ribosomal subunit associates with the small subunit to round out the ribosome. At the end of initiation, the fMet tRNA sits in the P site of the completed ribosome. Proteins known as initiation factors play a transient role in the initiation process. In eukaryotes, the small ribosomal subunit binds first to the methylated cap at the 5′ end of the mature mRNA. It then migrates to the initiation site—usually the first AUG it encounters as it scans the mRNA in the 5′-to-3′ direction. The initiator tRNA in eukaryotes carries unmodified methionine (Met) instead of fMet. Elongation: The addition of amino acids to a growing polypeptide. With the mRNA bound to the complete two subunit ribosome and with the initiating tRNA in the P site, the elongation of the polypeptide begins. A group of proteins known as elongation factors usher the appropriate tRNA into the A site of the ribosome. The anticodon of this charged tRNA must recognize the next codon in the mRNA. The ribosome holds the initiating tRNA at its P site and the second tRNA at its A site so that peptidyl transferase can catalyze formation of a peptide bond between the amino acids carried by the two tRNAs. Once formation of the first peptide bond causes the initiating tRNA in the P site to release its amino acid, the ribosome moves, exposing the next mRNA codon. Like the first steps of elongation, the ribosome’s movement requires the help of elongation factors and an input of energy. As the ribosome moves, the initiating tRNA in the P site, which no longer carries an amino acid, dissociates from the ribosome, and the other tRNA carrying the dipeptide shifts from the A site to the P site. The movement of ribosomes along the mRNA has important implications. Once a ribosome has moved far enough away from the mRNA’s ribosome binding site, that site becomes accessible to other ribosomes. In fact, several ribosomes can work on the same mRNA at one time. A complex of several ribosomes translating from the same mRNA is called a polyribosome. This complex allows the simultaneous synthesis of many copies of a polypeptide from a single mRNA. Termination: The ribosome releases the completed polypeptide No normal tRNAs carry anticodons complementary to the three nonsense (stop) codons UAG, UAA, and UGA. Thus, when movement of the ribosome brings a nonsense codon into the ribosome’s A site, no tRNAs can bind to that codon through complementary base pairing. Instead, proteins called release factors recognize the termination codons and bring polypeptide synthesis to a halt. During termination, three events must occur: (a) the tRNA specifying the C-terminal amino acid releases the completed polypeptide, (b) the same tRNA as well as the mRNA separate from the ribosome, and (c) the ribosome dissociates into its large and small subunits. 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? A. Protein Folding, Modification, and Targeting. During and after translation, individual amino acids may be chemically modified, signal sequences appended, and the new protein “folded” into a distinct three-dimensional structure as a result of intramolecular interactions. A signal sequence is a short sequence at the amino end of a protein that directs it to a specific cellular compartment. These sequences can be thought of as the protein’s “train ticket” to its ultimate destination, and are recognized by signal-recognition proteins that act as conductors. For instance, a specific signal sequence terminus will direct a protein to the mitochondria 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. B. There Are Significant Differences in Gene Expression Between Prokaryotes and Eukaryotes. The table below summarizes the differences of gene expression between prokaryotes and eukaryotes. Table 1. Differences Between Prokaryotes and Eukaryotes in the Details of Gene Expression C. How Mutations Affect Gene Expression As mentioned, the information in DNA is the starting point of gene expression. The cell transcribes that information into mRNA and then translates the mRNA information into protein. Mutations that alter the nucleotide pairs of DNA may modify any of the steps or products of gene expression (i) Silent Mutations Do Not Alter the Amino Acid Specified. One consequence of the code’s degeneracy is that some mutations, known as silent mutations, direct the inclusion of the same amino acid as the unmutated DNA and therefore have no effect on the amino-acid composition of the encoded polypeptide or on  phenotype. The majority of silent mutations change the third nucleotide of a codon, the position at which most codons for the same amino acid differ. For example, a change from GCA to GCC in a codon would still yield alanine in the protein product. (ii) Missense Mutations Replace One Amino Acid with Another. Missense mutations that cause substitution in the polypeptide of an amino acid with chemical properties similar to the one it replaces may have little or no effect on protein function. (iii) Nonsense Mutations Change an Amino-Acid-Specifying Codon to a Stop Codon. Nonsense mutations result in the production of proteins smaller than those encoded by wild-type alleles of the same gene. The shorter, truncated proteins lack all amino acids between the amino acid encoded by the mutant codon and the C terminus of the normal polypeptide. (iv) Frameshift Mutations Result from the Insertion or Deletion of Nucleotides Within the Coding Sequence. This happens if the number of extra or missing nucleotides is not divisible by 3, the insertion or deletion will skew the reading frame downstream of the mutation. As a result, frameshift mutations cause unrelated amino acids to appear in place of amino acids critical to protein function, destroying or diminishing polypeptide function. (v) Mutations Altering Genes Encoding Proteins or RNAs Involved in Gene Expression Are Usually Lethal. This occurs because such mutations adversely affect the synthesis of all proteins in a cell. In Drosophila, for example, null mutations in many of the genes encoding the various ribosomal proteins are lethal when homozygous. This same mutation in a heterozygote causes a dominant Minute phenotype in which the slow growth of cells delays the fly’s development. (vi) Mutations in tRNA Genes Can Suppress Mutations in Protein- Coding Genes. If more than one gene encoded a molecule with the same role in gene expression, a mutation in one of these genes would not be lethal and might even be useful. VII Summary of Gene Expression A. Gene expression is the process by which cells convert the DNA sequence of a gene to the RNA sequence of a transcript, and then decode the RNA sequence to the amino-acid sequence of a polypeptide. B. The nearly universal genetic code consists of 64 codons, each one composed of three nucleotides. Of these codons, 61 specify amino acids, while 3—UAA, UAG, and UGA—are nonsense or stop codons that do not specify an amino acid. C. Gene expression based on the genetic code produces the collinearity of a gene’s nucleotide sequence and a protein’s sequence of amino acids. D. Transcription is the first stage of gene expression. During transcription, RNA polymerase synthesizes a single-stranded primary transcript from a DNA template. (i) RNA polymerase initiates transcription by binding to the promoter sequence of the DNA and unwinding the double helix to expose bases for pairing; (ii) RNA polymerase extends the RNA in the 5′- to-3′ direction by catalyzing formation of phosphodiester bonds between successively aligned nucleotides; and (iii) Terminator sequences in the RNA cause RNA polymerase to dissociate from the DNA. d. In prokaryotes, the primary transcript is the mRNA that guides polypeptide synthesis. E. In eukaryotes, RNA processing after transcription produces a mature mRNA that travels from the nucleus to the cytoplasm to direct polypeptide synthesis. (i) RNA processing adds a methylated cap to the 5′ end and a poly-A tail to the 3′ end of the eukaryotic mRNA; (ii) The spliceosome removes introns from the primary transcript and precisely splices together the remaining exons. Alternative splicing makes it possible to produce different mRNAs from the same primary transcript. F. Translation is the stage of gene expression when the cell synthesizes protein according to instructions in the mRNA. (i) tRNAs carry amino acids to the translation machinery. Aminoacyl-tRNA synthetases connect amino acids to their corresponding tRNAs. Each tRNA molecule has an anticodon complementary to the mRNA codon specifying the amino acid it carries. Because of wobble, some tRNA anticodons recognize more than one mRNA codon; (ii) Translation occurs on complex molecular machines called ribosomes. Ribosomes have two binding sites for tRNAs—P and A—and an enzyme known as peptidyl transferase that catalyzes formation of a peptide bond between amino acids carried by the tRNAs at these two sites; (iii) Initiation: To start translation, part of the ribosome binds to a ribosome binding site on the mRNA, which includes the AUG initiation codon. Special initiating tRNAs with codons complementary to AUG carry the amino acid fMet in prokaryotes or Met in eukaryotes to the ribosomal P site. This amino acid will become the N terminus of the growing polypeptide; (iv) Elongation: When the carboxyl group of the amino acid connected to a tRNA at the ribosome’s P site  becomes attached through a peptide bond to the amino acid carried by the tRNA at the A site, the ribosome travels three nucleotides toward the 3′ end of the mRNA, and (V) Termination: When the ribosome encounters inframe nonsense (stop) codons, it ends translation by releasing the mRNA and disconnecting the complete polypeptide from the tRNA. G. Posttranslational processing may alter a polypeptide by adding or removing chemical constituents to or from particular amino acids, or by cleaving the polypeptide into smaller molecules. H. Mutations affect gene expression in several ways: (i) Mutations in a gene may modify the message encoded in a sequence of nucleotides. Silent mutations usually change the third letter of a codon and have no effect on polypeptide production. Missense mutations change the codon for one amino acid to the codon for another amino acid, and thereby direct incorporation of a different amino acid. Nonsense mutations change a codon for an amino acid to a stop codon, causing synthesis of a truncated polypeptide. Frameshift mutations change the reading frame of a gene, altering the identity of amino acids downstream of the mutation; (ii) Mutations outside of coding sequences that alter signals required for transcription, mRNA splicing, or translation can also disrupt gene expression; and (iii) Mutations in genes encoding molecules of the gene expression machinery are often lethal. Among the exceptions to this rule are mutations in tRNA genes that suppress mutations in polypeptide-encoding genes. LINKS TO LEARNING: Click through the steps of this PBS interactive (http://openstax.org/l/prokary_protein ) to see protein synthesis in action. _________________________________________________________________________________________  Activity 1: PART A. Read the following: Protein synthesis is the process used by the body to make proteins. The first step of protein synthesis is called Transcription. It occurs in the nucleus. During transcription, mRNA transcribes (copies) DNA. DNA is “unzipped” and the mRNA strand copies a strand of DNA. Once it does this, mRNA leaves the nucleus and goes into the cytoplasm. mRNA will then attach itself to a ribosome. The strand of mRNA is then read in order to make protein. They are read 3 bases at a time. These bases are called codons. tRNA is the fetching puppy. It brings the amino acids to the ribosome to help make the protein. The 3 bases on tRNA are called anti-codons. Remember, amino acids are the building blocks for protein. On the mRNA strand, there are start and stop codons. Your body knows where to start and stop making certain proteins. Just like when we read a sentence, we know when to start reading by the capitalized word and when to stop by the period. Ribosome mRNA DNA mRNA tRNA PART B. Answer the following questions: 1. What is the first step of protein synthesis? _________________________________ 2. What is the second step of protein synthesis? _________________________________ 3. Where does the first step of protein synthesis occur? _________________________________ 4. Where does the second step of protein synthesis occur? ____________________________ 5. Nitrogen bases are read ________ bases at a time. 6. The bases on the mRNA strand are called ______________________. 7. The bases on tRNA are called ______________________. 8. What is the start codon? ______________ 9. What are the stop codons? (Use your chart) _______________________________ 10. A bunch of amino acids attached together is called a _____________________________. PART C. Use your codon chart to determine the amino acid sequence. Remember to read through the strand and ONLY start on AUG and STOP when it tells you to stop. Follow example below: Example: DNA  AGA CGG TAC CTC CGG TGG GTG CTT GTC TGT ATC CTT CTC AGT ATC mRNA  UCU GCC AUG GAG GCC ACC CAC GAA CAG ACA UAG GAA GAG UCA UAG protein  start - glu – ala –thre – hist – asp –glu – threo - stop acid acid 1. DNA  CCT CTT TAC ACA CGG AGG GTA CGC TAT TCT ATG ATT ACA CGG TTG CGA TCC ATA ATC mRNA  protein  2. DNA  AGA ACA TAA TAC CTC TTA ACA CTC TAA AGA CCA GCA CTC CGA TGA ACT GGA GCA mRNA  protein  3. DNA  TAC CTT GGG GAA TAT ACA CGC TGG CTT CGA TGA ATC CGT ACG GTA CTC GCC ATC mRNA  protein  4. DNA  TAA ACT CGG TAC CTA GCT TAG ATC TAA TTA CCC ATC mRNA  protein  5. DNA  CTA TTA CGA TAC TAG AGC GAA TAG AAA CTT ATC ATC mRNA  protein  6. DNA  TAC CTT AGT TAT CCA TTG ACT CGA ATT GTG CGC TTG CTG ATC mRNA  protein  7. DNA  ACC CGA TAC CTC TCT TAT AGC ATT ACA AAC CTC CGA GCG mRNA  protein  8. DNA  TAC AGA CGG CAA CTC TGG GTG CTT TGT TCT CTT CTC AGT ATC mRNA  protein  Part D. Replication, Transcription & Translation Thinking Questions 1. Draw a DNA nucleotide & an RNA nucleotide. Label each of the 3 major parts. 2. What are the three major differences between DNA & RNA?  a) b) c) 3. What is the point of DNA replication? ____________________________ 4. When & where does replication occur? _____________________________ 5. What are three nucleotides together called on mRNA? _________ 6. The mRNA codons can be used in a chart to find: ____________________ 7. What molecule contains an anti-codon? _____________________________ 8. Why is this (answer to #13) molecule so important? 9. Translation takes place in the ______________ on a ________________. 10. Transcription and translation together is the process of _______________________ 11. What is any change in the DNA sequence called? ______________________________ 12. What are some examples of mutations in gene expression? Activity 2. Complete the diagram by supplying the missing parts/process. Write at least 2 paragraphs to describe the process.  Review Quiz A. Circle the correct choice within the parenthesis for 1 -18. 1. (DNA/RNA) can leave the nucleus. 2. mRNA is made during (transcription/translation). 3. mRNA is made in the (cytoplasm/nucleus). 4. DNA is located in the (nucleus/cytoplasm) 5. (Translation/Transcription) converts DNA into mRNA. 6. (mRNA/rRNA) is used to carry the genetic code from DNA to the ribosomes. 7. (tRNA/rRNA) makes up the ribosome. Look in the book for this. 8. (DNA/RNA) uses uracil instead of thymine. 9. (RNA/amino) acids make up a protein. 11. Transcription takes place in the (nucleus/cytoplasm). 12. tRNA is used in (translation/transcription). 13. tRNA uses (anticodons/codons) to match to the mRNA. 14. Proteins are made at the (nucleus/ribosome). 15. (tRNA/mRNA) attaches the amino acids into a chain. 16. tRNA is found in the (nucleus/cytoplasm). 17. (Translation/Transcription) converts mRNA into a protein. 18. Translation takes place in the (cytoplasm/nucleus). B. For each of the terms in the left column, choose the best matching phrase in the right column a. codon _______1. removing base sequences corresponding to introns from the primary transcript b. collinearity _______2. UAA, UGA, or UAG c. reading frame _______3. the strand of DNA that has the same base sequence as the primary transcript d. frameshift mutation ________4. a transfer RNA molecule to which the appropriate amino acid has been attached e. degeneracy of the ________5. a group of three mRNA bases signifying genetic code one amino acid f. nonsense codon ________6. most amino acids are not specified by a single codon g. initiation codon ________7. using the information in the nucleotide sequence of a strand of DNA to specify the nucleotide sequence of a strand of RNA h. template strand ________8. the grouping of mRNA bases in threes to be read as codons i. RNA-like strand ________9. AUG in a particular context j. intron ________10. the linear sequence of amino acids in the polypeptide corresponds to the linear sequence of nucleotide pairs in the gene k. RNA splicing ________11. produces different mature mRNAs from the same primary transcript l. transcription ________12. addition or deletion of a number of base pairs other than three into the coding sequence m. translation ________13. a sequence of base pairs within a gene that is not represented by any bases in the mature mRNA n. alternative splicing ________14. the strand of DNA having the base sequence complementary to that of the primary transcript o. charged tRNA ________15. using the information encoded in the nucleotide sequence of an mRNA molecule to specify the amino-acid sequence of a polypeptide molecule p. reverse transcription ________16. copying RNA into DNA C. Locate as accurately as possible those of the listed items which are shown on the following figure. Some items are not shown. You may draw and label them in the figure (a) 5′ end of DNA template strand; (b) 3′ end of mRNA; (c) ribosome; (d) promoter; (e) codon; (f) an amino acid; (g) DNA polymerase; (h) 5′ UTR; (i) centromere; (j) intron; (k) anticodon; (l) N terminus; (m) 5′ end of charged tRNA; (n) RNA polymerase; (o) 3′ end of uncharged tRNA; (p) a nucleotide; (q) mRNA cap; (r) peptide bond; (s) P site; (t) aminoacyl-tRNA synthetase; (u) hydrogen bond; (v) exon; (w) 5′ AUG 3′; (x) potential “wobble” interaction D. Read and explore the article entitled “HIV and Reverse Transcription: An Unusual DNA Polymerase Gives the AIDs Virus an Evolutionary Edge”. Make an insight paper about the article. Use the following as your guide in writing your insights. (a) summarize the main points; (b) explain the importance; (c) identify the different perspectives; (d) personal thoughts about the article. References Source Material: All images and some texts found in this module are a derivative of "Biology 2e" and “Genetics: Genes to Genome” by OpenStax CNX used under CC BY 4.0. 2020. Brown, T. (2012). Introduction to Genetics.A molecular approach. N.Y.:Garland Science, Taylor and Francis Grp. Campbell, N.A. & J.B.Reece. (2008). Biology 8th ed., Pearson Benjamin Cummings, California. Cummings, S. (2000). Current Perspecctives in Genetics. USA: Brooks/Cole Hartl, D.H. & Jones, E. (2002). Essential Genetics. A genomics perspective. (3rd ed.). MA: Jones and Barlett Publishers, Inc. Hartwell, L.H., Hood, L. Goldberg, M. Reynolds, Silver, L.M. & Jones, E. (2011). Genetics from genes to genomes. 4th ed. N.Y: McGraw Hill Karp, G.C. (2013). Cell and Molecular Biology, (7th ed.). U.S.:John Wiley and Sons, Inc. Klug, W.S., Cumming, M.R., Spencer, C. & Palladino, M.A. (2010). Essentials of genetics. (7th ed.) CA:Pearson Benjamin Cummings Lewis, R. (2003). Human genetics.Concepts and applications. 5th ed. N.Y: McGraw Hill Reece, J.B., et al. (2014). Biology, (10th ed.). U.S.:Pearson Publisher. Internet Sources: https://youtu.be/kiqhngnIZ5w (overview of the central dogma of molecular biology) https://youtu.be/gG7uCskUOrA (overview of gene expression in eukaryotes)