Nucleic Acids Module (Group Reporting) PDF

1 22.1 Nucleosides and Nucleotides 761 W hether you are tall or short, fair-skinned or dark-complexioned, blue-eyed or brown-eyed,...

1 22.1 Nucleosides and Nucleotides 761 W hether you are tall or short, fair-skinned or dark-complexioned, blue-eyed or brown-eyed, your unique characteristics are determined by the nucleic acid polymers that reside in the chromosomes of your cells. The nucleic acid DNA stores the genetic information of a particular organism, while the nucleic acid RNA translates this genetic information into the synthesis of proteins needed by cells for proper function and development. Even minor alterations in the nucleic acid sequence can have significant effects on an organism, sometimes resulting in devas- tating diseases like sickle cell anemia and cystic fibrosis. In Chapter 22, we study nucleic acids and learn how the genetic information stored in DNA is translated into protein synthesis. 22.1 Nucleosides and Nucleotides Nucleic acids are unbranched polymers composed of repeating monomers called nucleo- tides. There are two types of nucleic acids. DNA, deoxyribonucleic acid, stores the genetic information of an organism and transmits that information from one generation to another. RNA, ribonucleic acid, translates the genetic information contained in DNA into proteins needed for all cellular functions. The nucleotide monomers that compose DNA and RNA consist of three components—a mono- saccharide, a nitrogen-containing base, and a phosphate group. A nucleotide NH2 N O N −O P O CH2 O N N nitrogen-containing base phosphate O− OH monosaccharide DNA molecules contain several million nucleotides while RNA molecules are much smaller, containing perhaps a few thousand nucleotides. DNA is contained in the chromosomes of the nucleus, each chromosome having a different type of DNA. The number of chromosomes differs from species to species. Humans have 46 chromosomes (23 pairs). An individual chromosome is composed of many genes. A gene is a portion of the DNA molecule responsible for the synthesis of a single protein. We begin our study of nucleic acids with a look at the structure and formation of the nucleotide monomers. 22.1A Nucleosides—Joining a Monosaccharide and a Base The nucleotides of both DNA and RNA contain a five-membered ring monosaccharide, often called simply the sugar component. In RNA, the monosaccharide is the aldopentose D-ribose. In DNA the monosaccharide is D-2-deoxyribose, an aldopentose that lacks a hydroxyl group at C2. 5 HOCH2 O OH HOCH2 O OH 4 1 3 2 OH OH OH no OH at C2 The prefix deoxy means without oxygen. D-ribose D-2-deoxyribose (present in RNA) (present in DNA) 2 762 Chapter 22 Nucleic Acids and Protein Synthesis Only five common nitrogen-containing bases are present in nucleic acids. Three bases with one ring (cytosine, uracil, and thymine) are derived from the parent compound pyrimidine. Two bases with two rings (adenine and guanine) are derived from the parent compound purine. Each base is designated by a one-letter abbreviation as shown. NH2 O O 4 CH3 5 N3 N NH NH 6 2 N N O N O N O 1 H H H pyrimidine cytosine uracil thymine (parent compound) C U T NH2 O 7 6 N 5 N N N1 N NH 8 2 9N 4 N N N N N NH2 H 3 H H purine adenine guanine (parent compound) A G Uracil (U) occurs only in RNA, while thymine (T) occurs only in DNA. As a result: DNA contains the bases A, G, C, and T. RNA contains the bases A, G, C, and U. A nucleoside is formed by joining the anomeric carbon of the monosaccharide with a nitro- gen atom of the base. A nucleoside is an N-glycoside, because it is a nitrogen derivative of the glycosides discussed in Chapter 20. Primes (') are used to number the carbons of the monosac- charide in a nucleoside, to distinguish them from the atoms in the rings of the base. NH2 N NH2 5' HOCH2 OH HO CH2 O N O O N 1 + 4' 1' 1 N-glycoside N O 3' 2' join H OH OH OH OH D-ribose cytosine cytidine C a ribonucleoside NH2 N NH2 N 5' HOCH2 OH N HO CH2 O 9N O N N + 4' 1' N-glycoside 9N N 3' 2' join H OH OH D-2-deoxyribose adenine deoxyadenosine A a deoxyribonucleoside With pyrimidine bases, the nitrogen atom at the 1 position bonds with the 1' carbon of the sugar. With purine bases, the nitrogen atom at the 9 position bonds with the 1' carbon of the sugar. For example, joining cytosine with ribose forms the ribonucleoside cytidine. Joining adenine with 2-deoxyribose forms the deoxyribonucleoside deoxyadenosine. 3 22.1 Nucleosides and Nucleotides 763 Nucleosides are named as derivatives of the bases from which they are formed. To name a nucleoside derived from a pyrimidine base, use the suffix -idine (cytosine cytidine). To name a nucleoside derived from a purine base, use the suffix -osine (adenine adenosine). For deoxyribonucleosides, add the prefix deoxy-, as in deoxyadenosine. SAMPLE PROBLEM 22.1 Identify the base and monosaccharide used to form the following nucleoside, and then name it. O CH3 NH HO CH2 O N O OH Analysis The sugar portion of a nucleoside contains the five-membered ring. If there is an OH group at C2', the sugar is ribose, and if there is no OH group at C2', the sugar is deoxyribose. The base is joined to the five-membered ring as an N-glycoside. A pyrimidine base has one ring and is derived from either cytosine, uracil, or thymine. A purine base has two rings and is derived from either adenine or guanine. Nucleosides derived from pyrimidines end in the suffix -idine. Nucleosides derived from purines end in the suffix -osine. Solution O The sugar contains no OH at C2', so it is derived CH3 from deoxyribose. The base is thymine. To name NH the deoxyribonucleoside, change the suffix of the base to -idine and add the prefix deoxy; thus, from thymine HO CH2 N O thymine deoxythymidine. O 2' OH no OH group deoxyribose PROBLEM 22.1 Identify the base and monosaccharide used to form the following nucleosides, and then assign names. O O NH N NH a. HO CH2 O N O b. HO CH2 O N N NH2 OH OH OH PROBLEM 22.2 Draw the structure of guanosine. Number the carbon atoms in the monosaccharide. Classify the compound as a ribonucleoside or a deoxyribonucleoside. 4 764 Chapter 22 Nucleic Acids and Protein Synthesis 22.1B Nucleotides—Joining a Nucleoside with a Phosphate Nucleotides are formed by adding a phosphate group to the 5'-OH of a nucleoside. Nucleo- tides are named by adding the term 5'-monophosphate to the name of the nucleoside from which they are derived. Ribonucleotides are derived from ribose, while deoxyribonucleotides are derived from 2-deoxyribose. NH2 NH2 Add phosphate here. N N O O 5' 5' HO CH2 N O −O CH2 N O O P O O CMP −O O− P − O O− phosphate OH OH OH OH cytidine cytidine 5'-monophosphate a ribonucleoside a ribonucleotide NH2 NH2 Add phosphate here. N N N O N O 5' 5' HO CH2 N −O CH2 N O P O O −O N N dAMP P O− − O O− phosphate OH OH deoxyadenosine deoxyadenosine 5'-monophosphate a deoxyribonucleoside a deoxyribonucleotide Because of the lengthy names of nucleotides, three- or four-letter abbreviations are commonly used instead. Thus, cytidine 5'-monophosphate is CMP and deoxyadenosine 5'-monophosphate is dAMP. Figure 22.1 summarizes the information about nucleic acids and their components learned thus far. Table 22.1 summarizes the names and abbreviations used for the bases, nucleosides, and nucleotides needed in nucleic acid chemistry. Figure 22.1 Type of Compound Components Summary of the Components Nucleoside A monosaccharide + a base of Nucleosides, Nucleotides, A ribonucleoside contains the monosaccharide ribose. and Nucleic Acids A deoxyribonucleoside contains the monosaccharide 2-deoxyribose. Nucleotide A nucleoside + phosphate = a monosaccharide + a base + phosphate A ribonucleotide contains the monosaccharide ribose. A deoxyribonucleotide contains the monosaccharide 2-deoxyribose. DNA A polymer of deoxyribonucleotides The monosaccharide is 2-deoxyribose. The bases are A, G, C, and T. RNA A polymer of ribonucleotides The monosaccharide is ribose. The bases are A, G, C, and U. 5 22.1 Nucleosides and Nucleotides 765 Table 22.1 Names of Bases, Nucleosides, and Nucleotides in Nucleic Acids Base Abbreviation Nucleoside Nucleotide Abbreviation DNA Adenine A Deoxyadenosine Deoxyadenosine 5'-monophosphate dAMP Guanine G Deoxyguanosine Deoxyguanosine 5'-monophosphate dGMP Cytosine C Deoxycytidine Deoxycytidine 5'-monophosphate dCMP Thymine T Deoxythymidine Deoxythymidine 5'-monophosphate dTMP RNA Adenine A Adenosine Adenosine 5'-monophosphate AMP Guanine G Guanosine Guanosine 5'-monophosphate GMP Cytosine C Cytidine Cytidine 5'-monophosphate CMP Uracil U Uridine Uridine 5'-monophosphate UMP PROBLEM 22.3 Identify the base and monosaccharide in each nucleotide. Name the nucleotide and give its three- or four-letter abbreviation. O O CH3 NH NH O O a. −O CH2 N O b. −O CH2 N O P O O P O O O− O − OH OH OH PROBLEM 22.4 Which nucleic acid (DNA or RNA) contains each of the following components? a. the sugar ribose c. the base T e. the nucleotide GMP b. the sugar deoxyribose d. the base U f. the nucleotide dCMP PROBLEM 22.5 Label each statement about the compound deoxycytidine as true or false. a. Deoxycytidine is a nucleotide. b. Deoxycytidine is a nucleoside. c. Deoxycytidine contains a phosphate at its 5'-OH group. d. Deoxycytidine contains a pyrimidine base. Di- and triphosphates can also be prepared from nucleosides by adding two and three phos- phate groups, respectively, to the 5'-OH. For example, adenosine can be converted to adenosine 5'-diphosphate and adenosine 5'-triphosphate, abbreviated as ADP and ATP, respectively. We will learn about the central role of these phosphates, especially ATP, in energy production in Chapter 23. NH2 NH2 N N O O N O O O N 5' −O CH2 N −O CH2 N P O P O O N P O P O P O O N − − − − − O O O O O OH OH OH OH adenosine 5'-diphosphate adenosine 5'-triphosphate ADP ATP 6 766 Chapter 22 Nucleic Acids and Protein Synthesis SAMPLE PROBLEM 22.2 Draw the structure of the nucleotide GMP. Analysis Translate the abbreviation to a name; GMP is guanosine 5'-monophosphate. First, draw the sugar. Since there is no deoxy prefix in the name, GMP is a ribonucleotide and the sugar is ribose. Then draw the base, in this case guanine, bonded to C1' of the sugar ring. Finally, add the phosphate. GMP has one phosphate group bonded to the 5'-OH of the nucleoside. Solution O Add phosphate here. Add base here. N O NH HO CH2 O OH 5' −O CH2 N P O O N NH2 O − 1' OH OH OH OH ribose guanosine 5'-monophosphate GMP PROBLEM 22.6 Draw the structure of each nucleotide: (a) UMP; (b) dTMP; (c) AMP. PROBLEM 22.7 Give the name that corresponds to each abbreviation: (a) GTP; (b) dCDP; (c) dTTP; (d) UDP. 22.2 Nucleic Acids O Nucleic acids—both DNA and RNA—are polymers of nucleotides, formed by joining the R O P O R 3'-OH group of one nucleotide with the 5'-phosphate of a second nucleotide in a phosphodiester linkage (Section 19.6). O− phosphodiester For example, joining the 3'-OH group of dCMP (deoxycytidine 5'-monophosphate) and the 5'-phosphate of dAMP (deoxyadenosine 5'-monophosphate) forms a dinucleotide that contains a 5'-phosphate on one end (called the 5' end) and a 3'-OH group on the other end (called the 3' end). NH2 N A dinucleotide O 5' −O NH2 P O CH2 O N O 5'-phosphate O− N 3' O 5' OH dCMP −O P O CH2 N O O NH2 + NH2 O− 3' N N O N O N 5' 5' phosphodiester CH2 N −O CH2 N O P O O N P O O N linkage − − O O 3' 3' dAMP OH OH 3'-OH As additional nucleotides are added, the nucleic acid grows, each time forming a new phospho- diester linkage that holds the nucleotides together. Figure 22.2 illustrates the structure of a poly- nucleotide formed from four different nucleotides. Several features are noteworthy. 7 22.2 Nucleic Acids 767 A polynucleotide contains a backbone consisting of alternating sugar and phosphate groups. All polynucleotides contain the same sugar–phosphate backbone. The identity and order of the bases distinguish one polynucleotide from another. A polynucleotide has one free phosphate group at the 5' end. A polynucleotide has a free OH group at the 3' end. The primary structure of a polynucleotide is the sequence of nucleotides that it contains. This sequence, which is determined by the identity of the bases, is unique to a nucleic acid. In DNA, the sequence of bases carries the genetic information of the organism. Polynucleotides are named by the sequence of the bases they contain, beginning at the 5' end and using the one-letter abbreviation for the bases. Thus, the polynucleotide in Figure 22.2 contains the bases cytosine, adenine, thymine, and guanine, in order from the 5' end; thus, it is named CATG. Figure 22.2 Primary Structure of a Polynucleotide 5'-phosphate sugar–phosphate backbone O 5' cytosine −O 5'-phosphate P P O CH2 O O O− C 3' O P 5' adenine O P O CH2 O O A O− = 3' P O O 5' T thymine O P O CH2 O O− P 3' O G O 5' guanine O P O CH2 The phosphodiester linkage joins O 3'-OH HO the 3' C of one nucleotide to the O− 5' C of another nucleotide. 3' OH 3'-OH In a polynucleotide, phosphodiester bonds join the 3'-carbon of one nucleotide to the 5'-carbon of another. The name of a polynucleotide is read from the 5' end to the 3' end, using the one-letter abbreviations for the bases it contains. Drawn is the structure of the polynucleotide CATG. SAMPLE PROBLEM 22.3 (a) Draw the structure of a dinucleotide formed by joining the 3'-OH group of AMP to the 5'-phosphate in GMP. (b) Label the 5' and 3' ends. (c) Name the dinucleotide. Analysis Draw the structure of each nucleotide, including the sugar, the phosphate bonded to C5', and the base at C1'. In this case the sugar is ribose since the names of the mononucleotides do not contain the prefix deoxy. Bond the 3'-OH group to the 5'-phosphate to form the phosphodiester bond. The name of the dinucleotide begins with the nucleotide that contains the free phosphate at the 5' end. 8 768 Chapter 22 Nucleic Acids and Protein Synthesis Solution a. and b. NH2 NH2 N N 5'-phosphate O N adenine O N 5' 5' A −O CH2 N −O P O O N P O CH2 O N N O − O O− 3' 3' N AMP NH O OH OH OH 5' Form a phosphodiester O O P O CH2 O N + N NH2 using the 3'-OH and 5'-phosphate. N O− O NH 3' guanine 5' OH OH G −O P O CH2 O N N NH2 O− 3'-OH 3' GMP OH OH c. Since polynucleotides are named beginning at the 5' end, this dinucleotide is named AG. PROBLEM 22.8 Draw the structure of a dinucleotide formed by joining the 3'-OH group of dTMP to the 5'-phosphate in dGMP. PROBLEM 22.9 Draw the structure of each polynucleotide: (a) CU; (b) TAG. PROBLEM 22.10 Label each statement about the polynucleotide ATGGCG as true or false. a. The polynucleotide has six nucleotides. b. The polynucleotide contains six phosphodiester linkages. c. The nucleotide at the 5' end contains the base guanine. d. The nucleotide at the 3' end contains the base guanine. e. The polynucleotide could be part of a DNA molecule. f. The polynucleotide could be part of an RNA molecule. 22.3 The DNA Double Helix Our current understanding of the structure of DNA is based on the model proposed initially by James Watson and Francis Crick in 1953 (Figure 22.3). DNA consists of two polynucleotide strands that wind into a right-handed double helix. The sugar–phosphate backbone lies on the outside of the helix and the bases lie on the inside, perpendicular to the axis of the helix. The two strands of DNA run in opposite directions; that is, one strand runs from the 5' end to the 3' end, while the other runs from the 3' end to the 5' end. The double helix is stabilized by hydrogen bonding between the bases of the two DNA strands as shown in Figure 22.4. A purine base on one strand always hydrogen bonds with a pyrimidine base on the other strand. Two bases hydrogen bond together in a predictable manner, forming complementary base pairs. Adenine pairs with thymine using two hydrogen bonds, forming an A–T base pair. Cytosine pairs with guanine using three hydrogen bonds, forming a C–G base pair. 9 22.3 The DNA Double Helix 769 Figure 22.3 The Three-Dimensional Structure of DNA—A Double Helix a. b. sugar–phosphate bases backbone G C T A bases DNA consists of a double helix of polynucleotide chains. In view (a), the three-dimensional molecular model shows the sugar–phosphate backbone with the red (O), black (C), and white (H) atoms visible on the outside of the helix. In view (b), the bases on the interior of the helix are labeled. Figure 22.4 Hydrogen Bonding in the DNA Double Helix 5' 3' H O H N N N N G–C base pair N N H N O N H G C G C H T A H N N CH3 O H N T–A base pair N N N H A N O T 3' 5' hydrogen bonding between base pairs Hydrogen bonding of base pairs (A–T and C–G) holds the two strands of DNA together. 10 770 Chapter 22 Nucleic Acids and Protein Synthesis Because of this consistent pairing of bases, knowing the sequence of one strand of DNA allows us to write the sequence of the other strand, as shown in Sample Problem 22.4. SAMPLE PROBLEM 22.4 Write the sequence of the complementary strand of the following portion of a DNA molecule: 5'–TAGGCTA–3'. Analysis The complementary strand runs in the opposite direction, from the 3' to the 5' end. Use base pairing to determine the corresponding sequence on the complementary strand: A pairs with T and C pairs with G. Solution Original strand: 5'–T A G G C T A–3' Complementary strand: 3'–A T C C G A T–5' PROBLEM 22.11 Write the complementary strand for each of the following strands of DNA. a. 5'–AAACGTCC–3' c. 5'–ATTGCACCCGC–3' b. 5'–TATACGCC–3' d. 5'–CACTTGATCGG–3' The enormously large DNA molecules that compose the human genome—the total DNA content of an individual—pack tightly into the nucleus of the cell. The double-stranded DNA helices wind around a core of protein molecules called histones to form a chain of nucleosomes, as shown in Figure 22.5. The chain of nucleosomes winds into a supercoiled fiber called chromatin, which composes each of the 23 pairs of chromosomes in humans. In Section 22.2 we learned that the genetic information of an organism is stored in the sequence of bases of its DNA molecules. How is this information transferred from one genera- tion to another? How, too, is the information stored in DNA molecules used to direct the synthe- sis of proteins? Identical twins have the same genetic To answer these questions we must understand three key processes. makeup, so that characteristics determined by DNA—such as hair color, Replication is the process by which DNA makes a copy of itself when a cell divides. eye color, or complexion—are also Transcription is the ordered synthesis of RNA from DNA. In this process, the genetic identical. information stored in DNA is passed onto RNA. Translation is the synthesis of proteins from RNA. In this process, the genetic message contained in RNA determines the specific amino acid sequence of a protein. parent transcription translation DNA RNA protein replication daughter DNA Each chromosome contains many genes, those portions of the DNA molecules that result in the synthesis of specific proteins. We say that the genetic message of the DNA molecule is expressed in the protein. Only a small fraction (1–2%) of the DNA in a chromosome contains genetic mes- sages or genes that result in protein synthesis. 11 22.4 Replication 771 Figure 22.5 The Structure of a Chromosome chromosome nucleus chromatin fiber cell nucleosome histones DNA (double helix) base pairs 22.4 Replication How is the genetic information in the DNA of a parent cell passed onto new daughter cells during replication? The structure of the double helix and the presence of complementary base pairs are central to the replication process. During replication, the strands of DNA separate and each serves as a template for a new strand. Thus, the original DNA molecule forms two DNA molecules, each of which contains one strand from the parent DNA and one new strand. This process is called semiconservative 12 772 Chapter 22 Nucleic Acids and Protein Synthesis replication. The sequence of both strands of the daughter DNA molecules exactly matches the sequence in the parent DNA. parent DNA daughter DNA daughter DNA 5' 5' 5' 3' 3' 3' A T A T A T T A T A T A T A T A T A replication C G C G + C G G C G C G C A T A T A T 3' 3' 3' 1 2 5' 5' 1 2 5' 1 2 strand new strand new strand Although the semiconservative nature of replication has been known since the elegant experi- ments of Matthew Meselson and Franklin Stahl were reported in 1958, the details of replica- tion have only slowly been determined over the last 50 years. The first step in replication is the unwinding of the DNA helix to expose the bases on each strand. Unwinding occurs at many places simultaneously along the helix, creating “bubbles” where replication can occur. The point at which unwinding occurs is called the replication fork. Unwinding breaks the hydrogen bonds that hold the two strands of the double helix together. Once bases have been exposed on the unwound strands of DNA, the enzyme DNA polymerase cata- lyzes the replication process using the four nucleoside triphosphates (derived from the bases A, T, G, and C) that are available in the nucleus. Three features are key and each is illustrated in Figure 22.6. The identity of the bases on the template strand determines the order of the bases on the new strand: A must pair with T, and G must pair with C. A new phosphodiester bond is formed between the 5'-phosphate of the nucleoside triphosphate and the 3'-OH group of the new DNA strand. Replication occurs in only one direction on the template strand, from the 3' end to the 5' end. Since replication proceeds in only one direction—that is, from the 3' end to the 5' end of the template—the two new strands of DNA must be synthesized by somewhat different techniques. One strand, called the leading strand, grows continuously. Since its sequence is complementary to the template, its nucleotide sequence grows in the 5' to 3' direction. The other strand, called the lagging strand, is synthesized in small fragments, which are then joined together by a DNA ligase enzyme. The end result is two new strands of DNA, one in each of the daughter DNA mol- ecules, both with complementary base pairs joining the two DNA strands together. SAMPLE PROBLEM 22.5 What is the sequence of a newly synthesized DNA segment if the template strand has the sequence 3'–TGCACC–5'? Analysis The newly synthesized strand runs in the opposite direction, from the 5' end to the 3' end in this example. Use base pairing to determine the corresponding sequence on the new strand: A pairs with T and C pairs with G. Solution Template strand: 3'–T G C A C C–5' New strand: 5'–A C G T G G–3' 13 22.4 Replication 773 Figure 22.6 DNA Replication thymine adenine cytosine guanine replication fork template strand template strand The lagging strand replicates in segments. DNA polymerase nucleoside triphosphate 3' Replication proceeds 5' in the 3'-to-5' direction of the template for The leading strand 5' both the leading and grows continuously. the lagging strands. Replication proceeds along both strands of unwound DNA. Replication always occurs in the same direction, from the 3' to the 5' end of the template strand. The leading strand grows continuously, while the lagging strand must be synthesized in fragments that are joined together by a DNA ligase enzyme. PROBLEM 22.12 What is the sequence of a newly synthesized DNA segment if the template strand has each of the following sequences? a. 3'–AGAGTCTC–5' c. 3'–ATCCTGTAC–5' b. 5'–ATTGCTC–3' d. 5'–GGCCATACTC–3' 14 774 Chapter 22 Nucleic Acids and Protein Synthesis 22.5 RNA While RNA is also composed of nucleotides, there are important differences between DNA and RNA. In RNA, The sugar is ribose. U (uracil) replaces T (thymine) as one of the bases. RNA is single stranded. RNA molecules are much smaller than DNA molecules. Although RNA contains a single strand, the chain can fold back on itself, forming loops, and intermolecular hydrogen bonding between paired bases on a single strand can form helical regions. When base pairing occurs within an RNA molecule (or between RNA and DNA), C and G form base pairs, and A and U form base pairs. There are three different types of RNA molecules. Ribosomal RNA (rRNA) Messenger RNA (mRNA) Transfer RNA (tRNA) Ribosomal RNA, the most abundant type of RNA, is found in the ribosomes in the cytoplasm of the cell. Each ribosome is composed of one large subunit and one small subunit that contain both RNA and protein. rRNA provides the site where polypeptides are assembled during protein synthesis. Messenger RNA is the carrier of information from DNA (in the cell nucleus) to the ribosomes (in the cytoplasm). Each gene of a DNA molecule corresponds to a specific mRNA molecule. The sequence of nucleotides in the mRNA molecule determines the amino acid sequence in a particu- lar protein. mRNA is synthesized from DNA on an as-needed basis, and then rapidly degraded after a particular protein is synthesized. Transfer RNA, the smallest type of RNA, interprets the genetic information in mRNA and brings specific amino acids to the site of protein synthesis in the ribosome. Each amino acid is recognized by one or more tRNA molecules, which contain 70–90 nucleotides. tRNAs have two important sites. The 3' end, called the acceptor stem, always contains the nucleo- tides ACC and has a free OH group that binds a specific amino acid. Each tRNA also con- tains a sequence of three nucleotides called an anticodon, which is complementary to three bases in an mRNA molecule, and identifies what amino acid must be added to a growing polypeptide chain. tRNA molecules are often drawn in the cloverleaf fashion shown in Figure 22.7a. The acceptor stem and anticodon region are labeled. Folding creates regions of the tRNA in which nearby complementary bases hydrogen bond to each other. A model that more accurately depicts the three-dimensional structure of a tRNA molecule is shown in Figure 22.7b. Table 22.2 summarizes the characteristics of the three types of RNAs. Table 22.2 Three Types of RNA Molecules Type of RNA Abbreviation Function Ribosomal RNA rRNA The site of protein synthesis, found in the ribosomes Messenger RNA mRNA Carries the information from DNA to the ribosomes Transfer RNA tRNA Brings specific amino acids to the ribosomes for protein synthesis 15 22.6 Transcription 775 a. tRNA–Cloverleaf representation b. tRNA–Three-dimensional representation Figure 22.7 amino acid A Transfer RNA C 3' C acceptor stem 5' hydrogen bonding between complementary base pairs anticodon Folding of the tRNA molecule creates regions in In the three-dimensional model of a which complementary base pairs hydrogen bond tRNA, the binding site for the amino acid to each other. Each tRNA binds a specific amino is shown in yellow and the anticodon is acid to its 3' end and contains an anticodon that shown in red. identifies that amino acid for protein synthesis. 22.6 Transcription The conversion of the information in DNA to the synthesis of proteins begins with transcription— that is, the synthesis of messenger RNA from DNA. RNA synthesis begins in the same manner as DNA replication: the double helix of DNA unwinds (Figure 22.8). Since RNA is single stranded, however, only one strand of DNA is needed for RNA synthesis. Figure 22.8 Transcription partially unwound DNA double helix gene that codes for a specific protein RNA polymerase non-template strand of DNA 5' 3' end + 3' RNA nucleotides 5' template strand of DNA re-formed RNA newly made RNA DNA helix direction of transcription In transcription, the DNA helix unwinds and RNA polymerase catalyzes the formation of mRNA along the DNA template strand. Transcription proceeds from the 3' end to the 5' end of the template, forming an mRNA molecule with base pairs complementary to the DNA template strand. 16 776 Chapter 22 Nucleic Acids and Protein Synthesis The template strand is the strand of DNA used for RNA synthesis. The informational strand (the non-template strand) is the strand of DNA not used for RNA synthesis. Each mRNA molecule corresponds to a small segment of a DNA molecule. Transcription begins at a particular sequence of bases on the DNA template using an RNA polymerase enzyme, and proceeds from the 3' end to the 5' end of the template strand. Complementary base pairing deter- mines what RNA nucleotides are added to the growing RNA chain: C pairs with G, T pairs with A, and A pairs with U. Thus, the RNA chain grows from the 5' to 3' direction. Transcription is completed when a particular sequence of bases on the DNA template is reached. The new mRNA molecule is released and the double helix of the DNA molecule re-forms. In bacteria, the new mRNA molecule is ready for protein synthesis immediately after it is pre- pared. In humans, the mRNA molecule first formed is modified before it is ready for protein synthesis. Portions of the mRNA molecule are removed and pieces of mRNA are spliced together by mechanisms that will not be discussed here. Transcription forms a messenger RNA molecule with a sequence that is complementary to the DNA template from which it is prepared. Since the informational strand of DNA is complementary to the template strand, the mRNA molecule is an exact copy of the informational strand, except that the base U replaces T on the RNA strand. How the base sequence of an RNA molecule compares with the informational and template strands of a DNA molecule is shown in Sample Problem 22.6. SAMPLE PROBLEM 22.6 If a portion of the template strand of a DNA molecule has the sequence 3'–CTAGGATAC–5', what is the sequence of the mRNA molecule produced from this template? What is the sequence of the informational strand of this segment of the DNA molecule? Analysis mRNA has a base sequence that is complementary to the template from which it is prepared. mRNA has a base sequence that is identical to the informational strand of DNA, except that it contains the base U instead of T. Solution complementary Template strand of DNA: 3'–C T A G G A T A C–5' mRNA sequence: 5'–G A U C C U A U G–3' complementary Informational strand of DNA: 5'–G A T C C T A T G–3' PROBLEM 22.13 For each DNA segment: What is the sequence of the mRNA molecule synthesized from each DNA template? What is the sequence of the informational strand of the DNA molecule? a. 3'–TGCCTAACG–5' c. 3'–TTAACGCGA–5' b. 3'–GACTCC–5' d. 3'–CAGTGACCGTAC–5' PROBLEM 22.14 What is the sequence of the DNA template strand from which each of the following mRNA strands was synthesized? a. 5'–UGGGGCAUU–3' c. 5'–CCGACGAUG–3' b. 5'–GUACCU–3' d. 5'–GUAGUCACG–3' 17 22.7 The Genetic Code 777 22.7 The Genetic Code Once the genetic information of DNA has been transcribed in a messenger RNA molecule, RNA can direct the synthesis of an individual protein. How can RNA, which is composed of only four different nucleotides, direct the synthesis of polypeptides that are formed from 20 different amino acids? The answer lies in the genetic code. A sequence of three nucleotides (a triplet) codes for a specific amino acid. Each triplet is called a codon. For example, the codon UAC in an mRNA molecule codes for the amino acid serine, and the codon UGC codes for the amino acid cysteine. The same genetic code occurs in almost all organ- isms, from bacteria to whales to humans. UAC serine UGC cysteine mRNA amino acid mRNA amino acid Given four different nucleotides (A, C, G, and U), there are 64 different ways to combine them into groups of three, so there are 64 different codons. Sixty-one codons code for specific amino acids, so many amino acids correspond to more than one codon, as shown in Table 22.3. For example, the codons GGU, GGC, GGA, and GGG all code for the amino acid glycine. Three codons—UAA, UAG, and UGA—do not correspond to any amino acids; they are called stop codons because they signal the termination of protein synthesis. Table 22.3 The Genetic Code—Triplets in Messenger RNA First Base Third Base Second Base (5' end) (3' end) U C A G UUU Phe UCU Ser UAU Tyr UGU Cys U UUC Phe UCC Ser UAC Tyr UGC Cys C U UUA Leu UCA Ser UAA Stop UGA Stop A UUG Leu UCG Ser UAG Stop UGG Trp G CUU Leu CCU Pro CAU His CGU Arg U CUC Leu CCC Pro CAC His CGC Arg C C CUA Leu CCA Pro CAA Gln CGA Arg A CUG Leu CCG Pro CAG Gln CGG Arg G AUU Ile ACU Thr AAU Asn AGU Ser U AUC Ile ACC Thr AAC Asn AGC Ser C A AUA Ile ACA Thr AAA Lys AGA Arg A AUG Met ACG Thr AAG Lys AGG Arg G GUU Val GCU Ala GAU Asp GGU Gly U GUC Val GCC Ala GAC Asp GGC Gly C G GUA Val GCA Ala GAA Glu GGA Gly A GUG Val GCG Ala GAG Glu GGG Gly G 18 778 Chapter 22 Nucleic Acids and Protein Synthesis PROBLEM 22.15 What amino acid is coded for by each codon? a. GCC b. AAU c. CUA d. AGC e. CAA f. AAA PROBLEM 22.16 What codons code for each amino acid? a. glycine b. isoleucine c. lysine d. glutamic acid Codons are written from the 5' to 3' end of mRNA. The 5' end of the mRNA molecule codes for the N-terminal amino acid in a protein, and the 3' end of the mRNA molecule codes for the C-terminal amino acid. Sample Problem 22.7 illustrates the conversion of a sequence of bases in mRNA to a sequence of amino acids in a peptide. SAMPLE PROBLEM 22.7 Derive the amino acid sequence that is coded for by the following mRNA sequence. 5' CAU AAA ACG GUG UUA AUA 3' Analysis Use Table 22.3 to identify the codons that correspond to each amino acid. Codons are written from the 5' to 3' end of an mRNA molecule and correspond to a peptide written from the N-terminal to C-terminal end. Solution 5' CAU AAA ACG GUG UUA AUA 3' His Lys Thr Val Leu Ile N-terminal C-terminal amino acid amino acid PROBLEM 22.17 Derive the amino acid sequence that is coded for by each mRNA sequence. a. 5' CAA GAG GUA UCC UAC AGA 3' b. 5' GUC AUC UGG AGG GGC AUU 3' c. 5' CUA UGC AGU AGG ACA CCC 3' PROBLEM 22.18 Write a possible mRNA sequence that codes for each of the following peptides. a. Met–Arg–His–Phe c. Gln–Asn–Gly–Ile–Val b. Gly–Ala–Glu–Gln d. Thr–His–Asp–Cys–Trp PROBLEM 22.19 Considering the given sequence of nucleotides in an mRNA molecule, (a) what is the sequence of the DNA template strand from which the RNA was synthesized? (b) What peptide is synthesized by this mRNA sequence? 5' GAG CCC GUA UAC GCC ACG 3' 22.8 Translation and Protein Synthesis The translation of the information in messenger RNA to protein synthesis occurs in the ribo- somes. Each type of RNA plays a role in protein synthesis. mRNA contains the sequence of codons that determines the order of amino acids in the protein. Individual tRNAs bring specific amino acids to add to the peptide chain. rRNA contains binding sites that provide the platform on which protein synthesis occurs. 19 22.8 Translation and Protein Synthesis 779 Each individual tRNA contains an anticodon of three nucleotides that is complementary to the codon in mRNA and identifies individual amino acids (Section 22.5). For example, a codon of UCA in mRNA corresponds to an anticodon of AGU in a tRNA molecule, which identifies serine as the amino acid. Other examples are shown in Table 22.4. mRNA tRNA amino acid UCA AGU serine codon anticodon Table 22.4 Relating Codons, Anticodons, and Amino Acids mRNA Codon tRNA Anticodon Amino Acid ACA UGU threonine GCG CGC alanine AGA UCU arginine UCC AGG serine PROBLEM 22.20 For each codon: Write the anticodon. What amino acid does each codon represent? a. CGG b. GGG c. UCC d. AUA e. CCU f. GCC There are three stages in translation: initiation, elongation, and termination. Figure 22.9 depicts the main features of translation. Initiation Translation begins when an mRNA molecule binds to the smaller subunit of the ribosome and a tRNA molecule carries the first amino acid of the peptide chain to the binding site. Translation always begins at the codon AUG, which codes for the amino acid methionine. The arriving tRNA contains an anticodon with the complementary base sequence UAC. The large and small subunits of the ribosome combine to form a tight complex where protein synthesis takes place. Elongation The next tRNA molecule containing an anticodon for the second codon binds to mRNA, deliver- ing its amino acid, and a peptide bond forms between the two amino acids. The first tRNA mol- ecule, which has delivered its amino acid and is no longer needed, dissociates from the complex. The ribosome shifts to the next codon along the mRNA strand and the process continues when a new tRNA molecule binds to the mRNA. Protein synthesis always occurs on two adjacent sites of the ribosome. Termination Translation continues until a stop codon is reached. There is no tRNA that contains an anti- codon complementary to any of the three stop codons (UAA, UAG, and UGA), so protein synthesis ends and the protein is released from the ribosome. Often the first amino acid in the chain, methionine, is not needed in the final protein and so it is removed after protein synthesis is complete. Figure 22.10 shows a representative segment of DNA, and the mRNA, tRNA, and amino acid sequences that correspond to it. 20 780 Chapter 22 Nucleic Acids and Protein Synthesis Figure 22.9 Translation—The Synthesis of Proteins from RNA Initiation small codon A UG UC A ribosomal subunit U A C A GU mRNA A UG UC A UAC anticodon Elongation Met Ser tRNA A 2nd tRNA binds. Met The peptide bond forms. large ribosomal subunit amino acid A UG UC A GAG U A C AGU mRNA Ser A UG UC A GA G Met UAC new peptide A UG UC A GAG bond Termination U AGU tRNA A C One tRNA is released. Ser A new tRNA binds. Met Glu Ser The process continues. Met completed protein C CU Glu Initiation consists of the binding of the ribosomal subunits to mRNA and the arrival of the first tRNA carrying its amino acid. Hydrogen bonding occurs between the complementary bases of the codon of mRNA and the anticodon of the tRNA. The protein is synthesized during elongation. One by one a tRNA with its designated amino acid binds to a site on the ribosome adjacent to the first tRNA. A peptide bond forms and a tRNA is released. The ribosome shifts to the next codon and the process continues. Termination occurs when a stop codon is reached. The synthesis is complete and the protein is released from the complex. Figure 22.10 Comparing the Sequence of DNA, mRNA, tRNA, and a Polypeptide DNA informational strand 5' end ATG TTG GGA GCC GGA TCA 3' end DNA template strand 3' end TAC AAC CCT CGG CCT AGT 5' end mRNA 5' end AUG UUG GGA GCC GGA UCA 3' end tRNA anticodons UAC AAC CCU CGG CCU AGU Polypeptide Met Leu Gly Ala Gly Ser 21 22.9 Mutations and Genetic Diseases 781 SAMPLE PROBLEM 22.8 What sequence of amino acids would be formed from the following mRNA sequence: 5' CAA AAG ACG UAC CGA 3'? List the anticodons contained in each of the needed tRNA molecules. Analysis Use Table 22.3 to determine the amino acid that is coded for by each codon. The anticodons contain complementary bases to the codons: A pairs with U, and C pairs with G. Solution Amino acid sequence: Gln Lys Thr Tyr Arg mRNA codons: 5' CAA AAG ACG UAC CGA 3' tRNA anticodons: GUU UUC UGC AUG GCU PROBLEM 22.21 What sequence of amino acids would be formed from each mRNA sequence? List the anticodons contained in each of the needed tRNA molecules. a. 5' CCA CCG GCA AAC GAA GCA 3' b. 5' GCA CCA CUA AGA GAC 3' SAMPLE PROBLEM 22.9 What polypeptide would be synthesized from the following template strand of DNA: 3' CGG TGT CTT TTA 5'? Analysis To determine what polypeptide is synthesized from a DNA template, two steps are needed. First use the DNA sequence to determine the transcribed mRNA sequence: C pairs with G, T pairs with A, and A (on DNA) pairs with U (on mRNA). Then use the codons in Table 22.3 to determine what amino acids are coded for by a given codon in mRNA. Solution DNA template strand: 3' CGG TGT CTT TTA 5' mRNA: 5' GCC ACA GAA AAU 3' Polypeptide: Ala Thr Glu Asn PROBLEM 22.22 What polypeptide would be synthesized from each of the following template strands of DNA? a. 3' TCT CAT CGT AAT GAT TCG 5' b. 3' GCT CCT AAA TAA CAC TTA 5' 22.9 Mutations and Genetic Diseases Although replication provides a highly reliable mechanism for making an exact copy of DNA, occasionally an error occurs, thus producing a DNA molecule with a slightly different nucleotide sequence. A mutation is a change in the nucleotide sequence in a molecule of DNA. If the mutation occurs in a nonreproductive cell, the mutation is passed on to daughter cells within the organism, but is not transmitted to the next generation. If the mutation occurs in an egg or sperm cell, it is passed on to the next generation of an organism. Some mutations are random events, while others are caused by mutagens, chemical substances that alter the structure of DNA. Exposure to high-energy radiation such as X-rays or ultraviolet light can also produce mutations. Mutations can be classified according to the change that results in a DNA molecule. 22 782 Chapter 22 Nucleic Acids and Protein Synthesis A point mutation is the substitution of one nucleotide for another. Original DNA: GAGT TC replacement of G by C Point mutation: GACT TC A deletion mutation occurs when one or more nucleotides is lost from a DNA molecule. Original DNA: GAGT TC loss of G Deletion mutation: GAT TC An insertion mutation occurs when one or more nucleotides is added to a DNA molecule. Original DNA: GAGT TC addition of C Insertion mutation: GAGC T T C A mutation can have a negligible, minimal, or catastrophic effect on an organism. To understand the effect of a mutation, we must determine the mRNA sequence that is transcribed from the DNA sequence as well as the resulting amino acid for which it codes. A point mutation in the three-base sequence CTT in a gene that codes for a particular protein illustrates some possible outcomes of a mutation. The sequence CTT in DNA is transcribed to the codon GAA in mRNA, and using Table 22.3, this triplet codes for the amino acid glutamic acid. If a point mutation replaces CTT by CTC in DNA, CTC is transcribed to the codon GAG in mRNA. Since GAG codes for the same amino acid—glutamic acid—this mutation does not affect the protein synthesized by this segment of DNA. Such a mutation is said to be silent. Original: CT T GAA Glu DNA mRNA same amino acid The mutation has no effect. Point mutation: CTC GAG Glu Alternatively, suppose a point mutation replaces CTT by CAT in DNA. CAT is transcribed to the codon GUA in mRNA, and GUA codes for the amino