Chapter 21 Nucleic Acids and Protein Synthesis 21

Chapter 21 Nucleic Acids and Protein Synthesis 21. 1 Components of Nucleic Acids Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 1

Nucleic Acids Nucleic acids are § Molecules that store information for cellular growth and reproduction. § Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). § Large molecules consisting of long chains of monomers called nucleotides. 2

Nucleic Acids The nucleic acids DNA and RNA consist of monomers called nucleotides that consist of a § Pentose sugar. § Nitrogen-containing base. § Phosphate. nucleotide Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 3

Nitrogen Bases The nitrogen bases in DNA and RNA are § Pyrimidines C, T, and U § Purines A and G. 4 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Nitrogen-Containing Bases in DNA and RNA DNA contains the nitrogen bases § Cytosine (C) § Guanine (G) same in both DNA and RNA § Adenine (A) § Thymine (T) different in DNA than in RNA contains the nitrogen bases § Cytosine (C) § Guanine (G) same in both DNA and RNA § Adenine (A) § Uracil (U) different in DNA than in RNA 5

Pentose Sugars The pentose (five-carbon) sugar § In RNA is ribose. § In DNA is deoxyribose with no O atom on carbon 2′. § Has carbon atoms numbered with primes to distinguish them from the atoms in nitrogen bases. 6 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Nucleosides A nucleoside § Has a nitrogen base linked by a glycosidic bond to C 1′ of a sugar (ribose or deoxyribose). § Is named by changing the nitrogen base ending to -osine for purines and -idine for pyrimidines. HO Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 7

Nucleotides A nucleotide § Is a nucleoside that forms a phosphate ester with the C 5′ –OH group of a sugar (ribose or deoxyribose). § Is named using the name of the nucleoside followed by 5′-monophosphate. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 8

Formation of a Nucleotide A nucleotide forms when the −OH on C 5′ of a sugar bonds to phosphoric acid. 5’ 5’ 9

Nucleosides and Nucleotides with Purines Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 10

Nucleosides and Nucleotides with Pyrimidines Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 11

Names of Nucleosides and Nucleotides TABLE 21. 1 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 12

AMP, ADP, and ATP § Adding phosphate groups to AMP forms the diphosphate ADP and the triphosphate ATP. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 13

Learning Check Give the name and abbreviation for the following, and list its nitrogen base and sugar. 14

Solution Guanosine 5′-monophosphate; GMP nitrogen base: guanine sugar: ribose 15

Chapter 21 Nucleic Acids and Protein Synthesis 21. 2 Primary Structure of Nucleic Acids 21. 3 DNA Double Helix Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 16

Primary Structure of Nucleic Acids In the primary structure of nucleic acids, § Nucleotides are joined by phosphodiester bonds. § The 3’-OH group of the sugar in one nucleotide forms an ester bond to the phosphate group on the 5’-carbon of the sugar of the next nucleotide. 17

Primary Structure of Nucleic Acids Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 18

Structure of Nucleic Acids A nucleic acid polymer § Has a free 5’-phosphate group at one end a free 3’-OH group at the other end. § Is read from the free 5’-end using the letters of the bases. § This section is read as: 5’—A—C—G—T— 3’. 19 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Example of RNA The primary structure of RNA § Is a single strand of nucleotides. § Consists of the bases A, C, G, and U linked by 3’-5’ ester bonds between ribose and phosphate. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 20

Example of DNA In the primary structure of DNA, A, C, G, and T are linked by 3’-5’ ester bonds between deoxyribose and phosphate. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 21

Double Helix of DNA The DNA structure is a double helix that § Consists of two strands of nucleotides that form a double helix structure like a spiral stair case. § Has hydrogen bonds between the bases A–T and G–C. § Has bases along one strand that complement the bases along the other. 22 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Complementary Base Pairs DNA contains complementary base pairs in which § Adenine is always linked by two hydrogen bonds to thymine (A−T). § Guanine is always linked by three hydrogen bonds to cytosine (G−C). Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 23

DNA Double Helix Structure Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 24

Learning Check Write the complementary base sequence for the matching strand in the following DNA section: 5’—A—G—T—C—C —A—A—T—C— 3’ 25

Solution Write the complementary base sequence for the matching strand in the following DNA section: 5’—A—G—T—C—C—A—A—T—C— 3’ 3’—T—C—A—G—G—T—T—A—G— 5’ 26

Chapter 21 Nucleic Acids and Protein Synthesis 21. 4 DNA Replication Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 27

DNA Replication DNA replication involves § Unwinding the DNA. § Pairing the bases in each strand with new bases to form new complementary strands. § Producing two new DNA strands that exactly duplicate the original DNA. 28 Copyright © 2007 by Pearson Education, Inc Publishing as Benjamin Cummings

Hydrolysis Energy § Energy from the hydrolysis of each nucleoside triphosphate adding to the complementary strand is used to form the phosphodiester bond. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 29

Direction of Replication During DNA replication, § An enzyme helicase unwinds the parent DNA at several sections. § At each open DNA section called a replication fork, DNA polymerase catalyzes the formation of 5’-3’ester bonds of the leading strand. § The lagging strand growing in the 3’-5’ direction is synthesized in short sections called Okazaki fragments. § The Okazaki fragments are joined by DNA ligase to give a single 3’-5’ DNA strand. 30

Direction of Replication Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 31

Learning Check Match the following: 1) helicase 2) DNA polymerase 3) replication fork 4) Okazaki fragments A. Short segments formed by the lagging strand. B. The starting point for synthesis in unwound DNA sections. C. The enzyme that unwinds the DNA double helix. D. The enzyme that catalyzes the formation of phosphodiester bonds of complementary bases. 32

Solution Match the following: 1) helicase 2) DNA polymerase 3) replication fork 4) Okazaki fragments A. 4 Short segments formed by the lagging strand. B. 3 The starting point for synthesis in unwound DNA sections. C. 1 The enzyme that unwinds the DNA double helix. D. 2 The enzyme that catalyzes the formation of phosphodiester bonds of complementary bases. 33

Chapter 21 Nucleic Acids and Protein Synthesis 21. 5 RNA and Transcription Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 34

RNA § Transmits information from DNA to make proteins. § Has several types Messenger RNA (m. RNA) carries genetic information from DNA to the ribosomes. Transfer RNA (t. RNA) brings amino acids to the ribosome to make the protein. Ribosomal RNA (r. RNA) makes up 2/3 of ribosomes where protein synthesis takes place. 35

Types of RNA TABLE 21. 3 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 36

t. RNA Each t. RNA § Has a triplet called an anticodon that complements a codon on m. RNA. § Bonds to a specific amino acid at the acceptor stem. anticodon Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 37

Protein Synthesis Protein synthesis involves • Transcription m. RNA is formed from a gene on a DNA strand. • Translation t. RNA molecules bring amino acids to m. RNA to build a protein. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 38

Transcription: Synthesis of m. RNA In transcription § A section of DNA containing the gene unwinds. § One strand of DNA is copied starting at the initiation point, which has the sequence TATAAA. § A m. RNA is synthesized using complementary base pairing with uracil(U) replacing thymine(T). § The newly formed m. RNA moves out of the nucleus to ribosomes in the cytoplasm. 39

RNA Polymerase During transcription, § RNA polymerase moves along the DNA template in the 3’-5’direction to synthesize the corresponding m. RNA. § The m. RNA is released at the termination point. Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 40

Protein Synthesis: Transcription transcription 41 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

m. RNA Processing: Exons and Introns § The DNA of eukaryotes contains exons that code for proteins along with introns that do not. § The initial m. RNA called a pre-RNA includes the noncoding introns. § While in the nucleus, the introns are removed from the pre-RNA. § The exons that remain are joined to form the m. RNA that leaves the nucleus with the information for the synthesis of protein. 42

Removing Introns from Prem. RNA 43 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Regulation of Transcription is regulated by § A specific m. RNA synthesized when the cell requires a particular protein. § Feedback control, in which the end products speed up or slow the synthesis of m. RNA. § Enzyme induction, in which high levels of a reactant induce the transcription process to provide the necessary enzymes for that reactant. 44

Lactose Operon and Repressor § The lactose operon consists of a control site and the genes that produce m. RNA for lactose enzymes. § When there is no lactose in the cell, a regulatory gene produces a repressor protein that prevents the synthesis of lactose enzymes. § The repressor turns off m. RNA synthesis. 45

Lactose Operon Turned Off 46 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Lactose Operon and Inducer § When lactose is present in the cell, some lactose combines with the repressor, which removes the repressor from the control site. § Without the repressor, RNA polymerase catalyzes the synthesis of the enzymes by the genes in the operon. § The level of lactose in the cell induces the synthesis of the enzymes required for its metabolism. 47

Lactose Operon Turned On RNA Polymerase Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 48

Learning Check What is the sequence of bases in m. RNA produced from a section of the template strand of DNA that has the sequence of bases: 3’–C–T–A–A–G–G– 5’? 1. 5’–G–A–T–T–C–C– 3’ 2. 5’–G–A–U–U–C–C– 3’ 3. 5’–C–T–A–A–G–G– 3’ 49

Solution What is the sequence of bases in m. RNA produced from a section of the template strand of DNA that has the sequence of bases: 3’–C–T–A–A–G–G– 5’? 2. 5’–G–A–U–U–C–C– 5’ 50

Chapter 21 Nucleic Acids and Protein Synthesis 21. 6 The Genetic Code Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 51

Genetic Code The genetic code § Is a sequence of amino acids in a m. RNA that determine the amino acid order for the protein. § Consists of sets of three bases (triplet) along the m. RNA called codons. § Has a different codon for all 20 amino acids needed to build a protein. § Contains certain codons that signal the “start” and “end” of a polypeptide chain. 52

The Genetic Code: m. RNA Codons TABLE 21. 4 53 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Codons and Amino Acids Suppose we want to determine the amino acids coded for in the following section of a m. RNA. 5’—CCU —AGC—GGA—CUU— 3’ According to the genetic code, the amino acids for these codons are CCU = Proline AGC = Serine GGA = Glycine CUU = Leucine The m. RNA section codes for the amino acid sequence of Pro—Ser—Gly—Leu 54

Learning Check Write the order of amino acids coded for by a section of m. RNA with the base sequence 5’—GCC—GUA—GAC— 3’. Some possible codons to use are the following: GGC = Glycine GAC = Aspartic acid CUC = Leucine GUA = Valine GCC = Alanine CGC = Arginine 55

Solution GGC = Glycine CUC = Leucine GCC = Alanine GAC = Aspartic acid GUA =Valine CGC = Arginine 5’—GCC—GUA—GAC— 3’ Ala——Val——Asp 56

Chapter 21 Nucleic Acids and Protein Synthesis 21. 7 Protein Synthesis: Translation Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 57

t. RNA Activation The activation of t. RNA § Occurs when a synthetase uses energy of ATP hydrolysis to attach an amino acid to a specific t. RNA. § Prepares each t. RNA to use a triplet called an anticodon to complement a codon on m. RNA. Anticodon 58 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Initiation of Protein Synthesis For the initiation of protein synthesis § A m. RNA attaches to a ribosome. § The start codon (AUG) binds to a t. RNA with methionine. § The second codon attaches to a t. RNA with the next amino acid. § A peptide bond forms between the adjacent amino acids at the first and second codons. 59

Translocation During translocation § The first t. RNA detaches from the ribosome. § The ribosome shifts to the adjacent codon on the m. RNA. § A new t. RNA/amino acid attaches to the open binding site. § A peptide bond forms and that t. RNA detaches. § The ribosome shifts down the m. RNA to read the next codon. 60

Protein Synthesis Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings translation 61

Termination of Protein Synthesis In the terminiation of protein synthesis § After a polypeptide with all the amino acids for a protein is synthesized. § The ribosome reaches a “stop” codon: UGA, UAA, or UAG. § There is no t. RNA with an anticodon for the “stop” codons. § Protein synthesis ends. § The polypeptide detaches from the ribosome. 62

Learning Check Match the following: 1) Activation 3) Translocation A. B. C. D. 2) Initiation 4) Termination Ribosomes move along m. RNA adding amino acids to a growing peptide chain. A completed peptide chain is released. A t. RNA attaches to its specific amino acid. A t. RNA binds to the AUG codon of the m. RNA on the ribosome. 63

Solution Match the following: 1) Activation 3) Translocation 2) Initiation 4) Termination A. 3 Ribosomes move along m. RNA adding amino acids to a growing peptide chain. B. 4 A completed peptide chain is released. C. 1 A t. RNA attaches to its specific amino acid. D. 2 A t. RNA binds to the AUG codon of the m. RNA on the ribosome. 64

Summary of Protein Synthesis To summarize protein synthesis: § A m. RNA attaches to a ribosome. § t. RNA molecules bonded to specific amino acids attach to the codons on m. RNA. § Peptide bonds form between an amino acid and the peptide chain. § The ribosome shifts to each codon on the m. RNA until it reaches the STOP codon. § The polypeptide chain detaches to function as an active protein. 65

Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 66

Learning Check The following section of DNA is used to build a m. RNA for a protein. 3’—GAA—CCC—TTT— 5’ A. What is the corresponding m. RNA sequence? B. What are the anticodons on the t. RNAs? C. What is the amino acid order in the peptide? 67

Solution 3’—GAA—CCC—TTT— 5’ DNA A. What is the corresponding m. RNA sequence? 5’—CUU—GGG—AAA— 3’ m. RNA B. What are the anticodons for the t. RNAs? m. RNA codons CUU GGG AAA t. RNA anticodons GAA CCC UUU C. What is the amino acid order in the peptide? m. RNA 5’—CUU—GGG—AAA— 3’ Leu — Gly — Lys 68

Chapter 21 Nucleic Acids and Protein Synthesis 21. 8 Genetic Mutations 21. 9 Recombinant DNA Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 69

Mutations A mutation § Alters the nucleotide sequence in DNA. § Results from mutagens such as radiation and chemicals. § Produces one or more incorrect codons in the corresponding m. RNA. § Produces a protein that incorporates one or more incorrect amino acids. § Causes genetic diseases that produce defective proteins and enzymes. 70

Normal DNA Sequence § The normal DNA sequence produces a m. RNA that provides instructions for the correct series of amino acids in a protein. Correct order 71 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Mutation: Substitution In a substitution mutation, § A different base substitutes for the proper base in DNA. § There is a change in a codon in the m. RNA. § The wrong amino acid may be placed in the polypeptide. Incorrect order 72 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings Wrong amino acid

Mutation: Frame Shift In a frame shift mutation, § An extra base adds to or is deleted from the normal DNA sequence. § All the codons in m. RNA and amino acids are incorrect from the base change. 73 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings Incorrect amino acids

Genetic Diseases TABLE 21. 6 74 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Learning Check Identify each type of mutation as a substitution (S) frame shift (F) 1. Cytidine (C) enters the DNA sequence. 2. One adenosine is removed from the DNA sequence. 3. A base sequence of TGA in DNA changes to TAA. 75

Solution Identify each type of mutation as a substitution (S) frame shift (F) 1. F Cytosine (C) enters the DNA sequence. 2. F One adenosine is removed from the DNA sequence. 3. S A base sequence of TGA in DNA changes to TAA. 76

Recombinant DNA In recombinant DNA, § A DNA fragment from one organism is combined with DNA in another. § Restriction enzymes are used to cleave a gene from a foreign DNA and open DNA plasmids in E. coli. § DNA fragments are mixed with the plasmids in E. coli and the ends are joined by ligase. § The new gene in the altered DNA produces protein. 77

Recombinant DNA 78 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Products of Recombinant DNA TABLE 21. 7 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 79

DNA Fingerprinting In DNA fingerprinting (Southern transfer): § Restriction enzymes cut a DNA sample into smaller fragments (RFLPs). § The fragments are sorted by size. § A radioactive isotope that adheres to certain base sequences in the fragments produces a pattern on Xray film, which is the “fingerprint”. § The “fingerprint” is unique to each individual DNA. 80

Polymerase Chain Reaction A polymerase chain reaction (PCR): § Produces multiple copies of a DNA in a short time. § Separates the sample DNA strands by heating. § Mixes the separated strands with enzymes and nucleotides to form complementary strands. § Is repeated many times to produce a large sample of the DNA. 81

Polymerase Chain Reaction 82 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Chapter 21 Nucleic Acids and Protein Synthesis 21. 10 Viruses Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings 83

Viruses § Are small particles of DNA or RNA that require a host cell to replicate. § Cause a viral infection when the DNA or RNA enters a host cell. § Are synthesized in the host cell from the viral RNA produced by viral DNA. 84

Viruses 85 Copyright © 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Reverse Transcription In reverse transcription § A retrovirus, which contains viral RNA, but no viral DNA, enters a cell. § The viral RNA uses reverse transcriptase to produce a viral DNA strand. § The viral DNA strand forms a complementary DNA strand. § The new DNA uses the nucleotides and enzymes in the host cell to synthesize new virus particles. 86

Reverse Transcription Copyright © 2007 by Pearson Education, Inc Publishing as Benjamin Cummings 87

HIV Virus and AIDS The HIV-1 virus § Is a retrovirus that infects T 4 lymphocyte cells. § Decreases the T 4 level making the immune system unable to destroy harmful organisms. § Causes pneumonia and skin cancer associated with AIDS. HIV virus Copyright © 2007 by Pearson Education, Inc Publishing as Benjamin Cummings 88

AIDS Treatment § One type of AIDS treatment prevents reverse transcription of the viral DNA. § When altered nucleosides such as AZT and dd. I are incorporated into viral DNA, the virus is unable to replicate. 89

AIDS Treatment Azidothymine (AZT) Dideoxyinosine (dd. I) 90

AIDS Treatment § Another type of AIDS treatment involves protease inhibitors such as saquinavir, indinavir, and ritonavir. § Protease inhibitors modify the active site of the protease enzyme, which prevents the synthesis of viral proteins. Inhibited by AZT, dd. I reverse transcriptase Viral RNA Viral DNA Inhibited by protease inhibitors protease Viral proteins 91

Learning Check Match the following: 1) Virus 3) Protease inhibitor 2) Retrovirus 4) Reverse transcription A. A virus containing RNA. B. Small particles requiring host cells to replicate. C. A substance that prevents the synthesis of viral proteins. D. Using viral RNA to synthesize viral DNA. 92

Solution Match the following: 1) Virus 3) Protease inhibitor 2) Retrovirus 4) Reverse transcription A. 2 A virus containing RNA. B. 1 Small particles requiring host cells to replicate. C. 3 A substance that prevents the synthesis of viral proteins. D. 4 Using viral RNA to synthesize viral DNA. 93