kaplan-biochemistry-2006.pdf

TM* Biochemistry & Me...

TM* Biochemistry & Medical Genetics *USMLETM is a joint program of the Federation of State Medical Boards of the United States, Inc. and the National Board of Medical Examiners.  USMLE Step 1 Biochemistry Lecture Notes 2006-2007-- Edition KAPLA~. I meesea 'USMLE is a joint program of the Federation of State Medical Boards of the United States, Inc. and the National Board of Medical Examiners. ©2006 Kaplan, Inc. All rights reserved. No part of this book may be reproduced in any form, by photostat, microfilm, xerography or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of Kaplan, Inc.  BIOCHEMISTRY MEDICAL GENETICS Author Author Barbara Hansen, Ph.D. Lynne B. Jorde, Ph.D. Biochemistry/Medical Genetics Professor and Associate Chairman Kaplan Medical Department of Human Genetics Chicago, IL University of Utah Health Sciences Center Salt Lake City, UT Coauthor Sam Turco, Ph.D. Coauthor Professor, Department of Biochemistry Barbara Hansen, Ph.D. University of Kentucky College of Medicine Biochemistry/Medical Genetics Lexington, KY Kaplan Medical Chicago, IL Contributors Director of Medical Curriculum Roger Lane, Ph.D. Sonia Reichert, M.D. Associate Professor University of South Alabama College of Medicine Medical Illustrator Mobile, AL Christine Schaar Vernon Reichenbacher, Ph.D. Associate Professor Editorial Director. Department of Biochemistry and Molecular Biology Kathlyn McGreevy Marshall University School of Medicine Huntington, wv Production Manager Mary Ruebush, Ph.D. Michael Wolff Microbiology/Immunology Adjunct Professor of Microbiology Production Editor Montana State University William Ng Bozeman, MT. Cover Design Joanna Myllo Cover Art Christine Schaar Rich LaRocco  Contents Preface.......................................................... vii Section I: Molecular Biology and Biochemistry Chapter 1: Nucleic Acid Structure and Organization 3 Chapter 2: DNA Replication and Repair............................... 17 Chapter 3: Transcription and RNA Processing 31 Chapter 4: The Genetic Code, Mutations, and Translation 47 Chapter 5: Genetic Regulation...................................... 69 Chapter 6: Recombinant DNA 81 Chapter 7: Techniques of Genetic Analysis............................ 99 Chapter 8: Amino Acids, Proteins, and Enzymes 115 Chapter 9: Hormones 129 Chapter 10: Vitamins 141 Chapter 11: Overview of Energy Metabolism 153 Chapter 12: Glycolysis and Pyruvate Dehydrogenase 161 Chapter 13: Citric Acid Cycle and Oxidative Phosphorylation 179 Chapter 14: Glycogen, Gluconeogenesis, and the Hexose Monophosphate Shunt 193 Chapter 15: Lipid Synthesis and Storage 211 KAPLA~. I v meulCa 'I Chapter 16: Lipid Mobilization and Catabolism 231. Chapter 17: Amino Acid Metabolism 249 Chapter 18: Purine and Pyrimidine Metabolism 275 Section II. Medical Genetics Chapter 1: Single-Gene Disorders 289 Chapter 2: Population Genetics..................................... 317 Chapter 3: Cytogenetics 331 Chapter 4: Genetics of Common Diseases 353 Chapter 5: Gene Mapping 365 Chapter 6: Genetic Diagnosis 381 Index : 397 VI UPLANd' me lea I  Preface These seven volumes of Lecture Notes represent a yearlong effort on the part of the Kaplan Medical faculty to update our curriculum to reflect the most-likely-to-be-tested material on the current USMLE Step 1 exam. Please note that these are Lecture Notes, not review books. The Notes were designed to be accompanied by faculty lectures-live, on video, or on the web. Reading these Notes without accessing the accompanying lectures is not an effective way to review for the USMLE. To maximize the effectiveness of these Notes, annotate them as you listen to lectures. To facilitate this process, we've created wide, blank margins. While these margins are occasionally punctuated by faculty high-yield "margin notes;' they are, for the most part, left blank for your notations. Many students find that previewing the Notes prior to the lecture is a very effective way to prepare for class. This allows you to anticipate the areas where you'll need to pay particular attention. It also affords you the opportunity to map out how the information is going to be presented and what sort of study aids (charts, diagrams, etc.) you might want to add. This strategy works regardless of whether you're attending a live lecture or watching one on video or the web. Finally, we want to hear what you think. What do you like about the notes? What do you think could be improved? Please share your feedback by [email protected]. Thank you for joining Kaplan Medical, and best of luck on your Step 1 exam! Kaplan Medical KAP LA meulCa '!._ I vii  SECTION I Molecular Biology and Biochemistry  Nucleic Acid Structure and Organization OVERVIEW: THE CENTRAL DOGMA OF MOLECULAR BIOLOGY An organism must be able to store and preserve its genetic information, pass that information along to future generations, and express that information as it carries out all the processes of life. The major steps involved in handling genetic information are illustrated by the central dogma of molecular biology (Figure I-1-1). Genetic information is stored in the base sequence of DNA molecules. Ultimately, during the process of gene expression, this information is used to synthesize all the proteins made by an organism. Classically, a gene is a unit of the DNA that encodes a particular protein or RNA molecule. Although this definition is now complicated by our increased appreciation of the ways in which genes may be expressed, it is still useful as a general, working definition. Replication 0 1,. Transcription 1~lmt~II ~ 11tD11_1i_ra_n_s_la_ti_on_~ HN I CH ,[ ~N N O~N 3 5' CH20H 5' CH20H i:. Adenosine Deoxythymidine Figure 1-1-4. Examples of Nucleosides Nucleotides are formed when one or more phosphate groups is attached to the 5' carbon of a nucleoside (Figure 1-1-5). Nucleoside di- and triphosphates are high-energy compounds because of the hydrolytic energy associated with the acid anhydride bonds (Figure 1-1-6). NH2 High-Energy N.)-yNI): o o Bonds \ ~.~.: N. _ II 5' _ II 5' a ( a \ a. 0- P- 0-CH2 0- P- 0-CH2 6-~-o-M-o-M-O-C~2 a I I b- b" b- 0- 4' o aH H ATP Figure 1-1-6. High-Energy Bonds in a Nucleoside Triphosphate ~ Uridine Monophosphate Deoxyguanosine (UMP) Monophosphate (dGMP) Figure 1-1-5. Examples of Nucleotides The nomenclature for the commonly found bases, nucleosides, and nucleotides is shown in Table 1-1-2. Note that the "deoxy" part of the names deoxythymidine, dTMP, etc., is sometimes understood, and not expressly stated, because thymine is almost always found attached to deoxyribose. KAPLAN'd_ I 6 me lea  Nucleic Acid Structure and Organization Table 1-1-2. Nomenclature of Important Bases, Nucleosides, and Nucleotides ,...-._---_._--.--..--»_..__._.__.--_.. Base Nucleoside Nucleotides In a Nutshell Adenine Adenosine AMP (dAMP) ADP (dADP) ATP (dATP) I Nucleic Acids I ! (Deoxyadenosine) GTP (dGTP) Nucleotides linked by 3', 5' Guanine Guanosine GMP (dGMP) GDP (dGDP) I Cytosine (Deoxyguanosine) Cytidine CMP (dCMP) CDP (dCDP) CTP (dCTP) phosphodiester bonds Have distinct 3' and 5' ends, thus polarity I (Deoxycytidine) ! Sequence is always specified I Uracil Uridine UMP (dUMP) UDP (dUDP) UTP (dUTP) as 5'~3' ! (Deoxyuridine) (dTTP) Thymine (Deoxythymidine) (dTMP) (dTDP) ~._N __ ""'''''''_Y~~'~.,.. 'v"""'w ___ ·__ ".,..V"'y.......,,.,,.,NN"''''''''''''''', Names of nucleosides and nucleotides attached to deoxyribose are shown in parentheses. NUCLEIC ACIDS Nucleic acids are polymers of nucleotides joined by 3',5' -phospho diester bonds; that is, a phos- phate group links the 3' carbon of a sugar to the 5' carbon of the next sugar in the chain. Each strand has a distinct 5' end and 3' end, and thus has polarity. A phosphate group is often found at the 5' end, and a hydroxyl group is often found at the 3' end. The base sequence of a nucleic acid strand is written by convention, in the 5'~3' direction (left to right). According to this convention, the sequence of the strand on the left in Figure 1-1-7 must be written 5'- TCAG-3' or TCAG: If written backward, the ends must be labeled: 3'-GACT-5' The positions of phosphates may be shown: pTpCpApG In DNA, a "d" (deoxy) may be included: dTdCdAdG In eukaryotes, DNA is generally double-stranded (dsDNA) and RNA is generally single-strand- ed (ssRNA). Exceptions occur in certain viruses, some of which have ssDNA genomes and some of which have dsRNA genomes. KAPLA~. I meulCa 7  USMLE Step 1: Biochemistry 5'- Phosphate 3'- Hydroxyl o- r H o-p=o HC I N ? 5' 5'CH2 3 M-H----- O------H-NH~ N~ A)--{ rJ OH 3' ~N~O ~N "'~ o 5'CH2 -O-p I = 0 H -o-p=o t I o ~ I ~N-H-------o~~ N 0 I 5,dH ~N--"--H-N G N~ 20 \~O H_N~N 3' 3' I 0 H ~ o 5~~ I. H" CH3 b -O-P=O ~~N \ N-H------ 0tli I -O-P=O b N f A'\ ------H- N T~ I 5~ N=d o~~ I H 5'CH2 -O-P-O 5'CH2 0 1 '- r HG N N-1~-H------~' o------H-~C\\ / -o-~-o I I- N- l-H------O N~O H 3' o 3' OH ? 5'CH2 I 5' -O-P=O I 0- 3'- Hydroxyl 5'- Phosphate Figure 1-1-7. Hydrogen-Bonded Base Pairs in DNA KAPlAlf - I 8 me dlea  Nucleic Acid Structure and Organization Note DNA Structure Figure 1-1-8 shows an example 'of a double-stranded DNA molecule. Some of the features of Using Chargaff's Rules double-stranded DNA include: In dsDNA (or dsRNA) (ds = The two strands are antiparallel (opposite in direction). double-stranded) The two strands are complementary. A always pairs with T (two hydrogen bonds), and G always pairs with C (three hydrogen bonds). Thus, the base sequence on one strand 0/0 A = % T (% U) defines the base sequence on the other strand. %G=%C Because of the specific base pairing, the amount of A equals the amount of T, and the amount of G equals the amount of C. Thus, total purines equals total pyrimidines. % purines = % pyrimidines These properties are known as Chargaff's rules. With minor modification (substitution of U for T) these rules also apply to dsRNA. A sample of DNA has 10% G; what is the % 17 Most DNA occurs in nature as a right-handed double-helical molecule known as Watson-Crick DNA or B-DNA (Fig 1-1-8). The hydrophilic sugar-phosphate backbone of each strand is on 10% G + 10% C = 20% the outside of the double helix. The hydrogen-bonded base pairs are stacked in the center of therefore, % A + % T the molecule. There are about 10 base pairs per complete turn of the helix. A rare left-handed must total 80% double-helical form of DNA that occurs in G-C-rich sequences is known as Z-DNA. The bio- logic function of Z-DNA is unknown, but may be related to gene regulation. 40% A and 40% T Ans: 40% T Daunorubicin and doxorubicin are antitumor drugs that are used in the treatment of leukemias. They exert Major Groove ssDNA-binding protein ssDNA-binding protein I Primase (an RNA polymerase) Primase (an RNA polymerase) DNA pol III (leading/lagging) DNA pol (5 (leading) DNA pol a. (lagging) DNA poll DNA ligase DNA ligase DNA gyrase (Topo II) DNA topoisomerase II Telomerase DNA REPAIR Gl phase of eukaryotic cell cycle: UV radiation: thymine (pyrimidine) dimers; excinuclease Deaminations (C becomes U); uracil glycosylase Loss of purine or pyrimidine; AP endonuclease G2 phase of eukaryotic cell cycle: I. L... __ Mismatch repair: hMSH2, hMLHl (HPNCC) ,..__.,......._,_. ,''_....__..... _.._,. __..~._.,,.. ",.__ ,."'_,, ,"' 1 Review Questions Select the ONE best answer. 1. It is now believed that a substantial proportion of the single nucleotide substitutions causing human genetic disease are due to misincorporation of bases during DNA replica- tion. Which proofreading activity is critical in determining the accuracy of nuclear DNA replication and thus the base substitution mutation rate in human chromosomes? A. 3' to 5' polymerase activity of DNA polymerase (5 B. 3' to 5' exonuclease activity of DNA polymerase y ~I I D. C. Primase activity of DNA polymerase a. 5' to 3' polymerase activity of DNA polymerase III E. 3' to 5' exonuclease activity of DNA polymerase (5 28 meCtical  DNA Replication and Repair 2. The proliferation of cytotoxic T-cells is markedly impaired upon infection with a newly discovered human immunodeficiency virus, designated HIV- V. The defect has been traced to the expression of a viral-encoded enzyme that inactivates a host-cell nuclear protein required for DNA replication. Which protein is a potential substrate for the viral enzyme? A. TATA-box binding protein (TBP) B. Cap binding protein (CBP) C. Catabolite activator protein (CAP) D. Acyl-carrier protein (ACP) E. Single-strand binding protein (SBP) 3. The deficiency of an excision endonuclease may produce an exquisite sensitivity to ultraviolet radiation in Xeroderma pigmentosum. Which of the following functions would be absent in a patient deficient in this endonuclease? A. Removal of introns B. Removal of pyrimidine dimers C. Protection against DNA viruses D. Repair of mismatched bases during DNA replication E. Repair of mismatched bases during transcription 4. The anti-Pseudomonas action of norfloxacin is related to its ability to inhibit chromosome duplication in rapidly dividing cells. Which of the following enzymes participates in bac- terial DNA replication and is directly inhibited by this antibiotic? A. DNA polymerase I B. DNA polymerase II C. Topoisomerase I D. Topoisomerase II E. DNA ligase Answers 1. Answer: E. The 3' to 5' exonuclease activity of DNA pol b represents the proofreading activity of an enzyme required for the replication of human chromosomal DNA. DNA pol 'Y(mitochondrial) and DNA pol III (prokaryotic) do not participate in this process, short RNA primers are replaced with DNA during replication, and new DNA strands are always synthesized in the 5' to 3' direction. 2. Answer: E. TBP and CBP participate in eukaryotic gene transcription and mRNA trans- lation, respectively. CAP regulates the expression of prokaryotic lactose operons. ACP is involved in fatty acid synthesis. 3. Answer: B. Nucleotide excision repair of thymine (pyrimidine) dimers is deficient in XP patients. 4. Answer: D. Norfloxacin inhibits DNA gyrase (topoisomerase II). KAPlA~. I meulCa 29 I II I I I I' I, I, II 1 I ' I ' r ,  Transcription and RNA Processing OVERVIEW OF TRANSCRIPTION The first stage in the expression of genetic information is transcription of the information in the base sequence of a double-stranded DNA molecule to form the base sequence of a single- stranded molecule of RNA. For any particular gene, only one strand of the DNA molecule, called the template strand, is copied by RNA polymerase as it synthesizes RNA in the 5' to 3' direction. Because RNA polymerase moves in the 3' to 5' direction along the template strand of DNA, the RNA product is antiparallel and complementary to the template. RNA polymerase recognizes start signals (promoters) and stop signals (terminators) for each of the thousands of transcription units in the genome of an organism. Figure 1-3-1 illustrates the arrangement and direction of transcription for several genes on a DNA molecule. CD CD "0 "0 E E UO!ldp::>SUBJl. e 0.. e 0.. oil -.. - 3' 5' -.. - Spacer Spacer Spacer Spacer DNA DNA DNA DNA.-_.- 5 3'. -- -0 ~ ~ a Transcription Transcription 3 sa. m """' Figure 1-3-1.Transcription of Several Genes on a Chromosome TYPES OF RNA RNA molecules playa variety of roles in the cell. The major types of RNA are: Ribosomal RNA (rRNA), which is the most abundant type of RNA in the cell. It is used as a structural component of the ribosome. Ribosomal RNA associates with ribo- somal proteins to form the complete, functional ribosome. Transfer RNA (tRNA), which is the second most abundant type of RNA. Its function is to carry amino acids to the ribosome, where they will be linked together during pro- tein synthesis. Messenger RNA (mRNA), which carries the information specifying the amino acid sequence of a protein to the ribosome. Messenger RNA is the only type of RNA that is translated. The mRNA population in a cell is very heterogeneous in size and base sequence, as the cell has essentially a different mRNA molecule for each of the thou- sands of different proteins made by that cell. KAPLAlf - I me illea 31  I I USMLE Step 1: Biochemistry ,I Heterogeneous nuclear RNA (hnRNA or pre-mRNA), which is found only in the Ii nucleus of eukaryotic cells. It represents precursors of mRNA, formed during its post- transcriptional processing. Small nuclear RNA (snRNA), which is also only found in the nucleus of eukaryotes. One of its major functions is to participate in splicing (removal of introns) mRNA. Ribozymes, which are RNA molecules with enzymatic activity. They are found in both prokaryotes and eukaryotes. TRANSCRIPTION: IMPORTANT CONCEPTS AND TERMINOLOGY RNA is synthesized by a DNA-dependent RNA polymerase (uses DNA as a template for the synthesis of RNA). Important terminology used when discussing transcription is illustrated in Figure 1-3-2. RNA polymerase locates genes in DNA by searching for promoter regions. The pro- moter is the binding site for RNA polymerase. Binding establishes where transcription begins, which strand of DNA is used as the template, and in which direction transcrip- tion proceeds. No primer is required. RNA polymerase moves along the template strand in the 3' to 5' direction as it synthe- sizes the RNA product in the 5' to 3' direction using NTPs (ATP, GTP, CTP, UTP) as substrates. RNA polymerase does not proofread its work. The RNA product is comple- mentary and antiparallel to the template strand. The coding (antitemplate) strand is.not used during transcription. It is identical in sequence to the RNA molecule, except that RNA contains uracil instead of the thy- mine found in DNA. By convention, the base sequence of a gene is given from the coding strand (5' ~3'). In the vicinity of a gene, a numbering system is used to identify the location of impor- tant bases. The first base transcribed as RNA is defined as the + 1 base of that gene region. To the left (5', or upstream) of this starting point for transcription, bases are -1, -2, -3, etc. To the right (3', or downstream) of this point, bases are +2, +3, etc. Transcription ends when RNA polymerase reaches a termination signal. Upstream Downstream (TranscriPtion Unit~ Start Terminator Site -10 j+ 10 Coding (Antitemplate) StrJnd ~:~~- DNA : ---- - - - - - - - - - - - - - - - - - - _-_ Template Strand ----~_I~: RNA Polymerase Transcribes DNA Template Strand , RNA 5" ~3' RNA Transcript Is Synthesized 5' -- 3' I Figure 1-3-2. Transcription of DNA I Ii KAPLAH'd- I 32 me lea  Transcription and RNA Processing Flow of Genetic Information From DNA to Protein :1 For the case of a gene coding for a protein, the relationship among the sequences found in dou- II ble-stranded DNA, single-stranded mRNA, and protein is illustrated in Figure 1-3-3. Messenger I RNA is synthesized in the 5' to 3' direction. It is complementary and antiparallel to the template \1 I. strand of DNA. The ribosome translates the mRNA in the 5' to 3' direction, as it synthesizes the protein from the amino to the carboxyl terminus. Coding Strand DNA , ·I~~~~""'~w R~~U~~'JJ 5······ ~ ·3 , DNA Coding Strand Is Identical to the mRNA (except T for U) DNA Template Strand Is Complementary TranJriPtiOn3'..... T ACe C eGA GTe GeT G ·5' and Anti arallel to the mRNA. Template Strand mRNA ' 5......, iI!l'-:i~.~..... 3' _ Direction of ~ Transcription Tranl'ation ~ ~ ~ ~ ~ Codons Protein NH2....~ ®~ @) ~... COOH _ Directio~ of ~ Translation Figure 1-3-3. Flow of Genetic Information From DNA to Protein Sample Questions 1. During RNA synthesis, the DNA template sequence TAGC would be transcribed to pro- duce which of the following sequences? A. ATCG B. GCTA C. CGTA D. AUCG E. GCUA The answer is E. RNA is antiparallel and complementary to the template strand. Also remem- ber that, by convention, all base sequences are written in the 5' to 3' direction regardless of the direction in which the sequence may actually be used in the cell. Approach: Cross out any option with a T (RNA has U). Look at the 5' end of DNA (T in this case). What is the complement of this base? (A) Examine the options given. A correct option will have the complement (A in this example) at the 3' end. Repeat the procedure for the 3' end of the DNA. This will usually leave only one or two options. - I KAPUlf medlea 33 1'1 USMLE Step 1: Biochemistry 2. Transcription of the following sequence of the tryptophan operon occurs in the direction indicated by the arrow. What would be the base sequence of the mRNA produced? 3'... CGCCGCTGCGCG... 5' Transcription ~ Which product? 5'... GCGGCGACGCGC... 3' A. 5' GCGGCGACGCGC... 3' B. 5' GCGCGUCGCCGC... 3' C. 5' GCGCGTGCGGCG 3' D. 5' GCGGCGUCGCGC 3' E. 5' CGCGCTCGCCGC 3' The answer is A. Because all nucleic acids are synthesized in the 5' to 3' direction, mRNA and the coding strand of DNA must each be oriented 5' to 3', i.e., in the direction of transcription. This means that the bottom strand in this example is the coding strand. The top strand is the template strand. Approach: Cross out any option with a T. Identify the coding strand of DNA from the direction of transcription. Find the option with a sequence identical to the coding strand (remember to substi- tute U for T, if necessary). Alternatively, if you prefer to find the complement of the template strand, you will get the same answer. RNA POLYMERASES There is a single prokaryotic RNA polymerase that synthesizes all types of RNA in the cell. The core polymerase responsible for making the RNA molecule has the subunit structure a BW. 2 A protein factor called sigma (c) is required for the initiation of transcription at a promoter. Sigma factor is released immediately after initiation of transcription. Termination of transcrip- tion sometimes requires a protein called rho (p) factor. This enzyme is inhibited by rifampin. Actinomycin D binds to the DNA, preventing transcription. There are three eukaryotic RNA polymerases, distinguished by the particular types of RNA they produce: RNA polymerase I is located in the nucleolus and synthesizes 28S, 18S, and 5.8S rRNAs. RNA polymerase II is located in the nucleoplasm and synthesizes hnRNNmRNA and somesnRNA. RNA polymerase III is located in the nucleoplasm and synthesizes tRNA, some snRNA, and 5S rRNA. Transcription factors (such as TFIID for RNA polymerase II) help to initiate transcription. The requirements for termination of transcription in eukaryotes are not well understood. All transcription can be inhibited by actinomycin D. In addition, RNA polymerase II is inhibited by n-arnanitin (a toxin from certain mushrooms). 34 iiieClical r Transcription and RNA Processing PRODUCTION OF PROKARYOTIC MESSENGER RNA The structure and expression of a typical prokaryotic gene coding for a protein are illustrated in Figure 1-3-4. The following events occur during the expression of this gene: 1. With the help of sigma factor, RNA polymerase recognizes and binds to the promoter region. The bacterial promoter contains two "consensus" sequences, called the Pribnow box (or TATAbox) and the -35 sequence. The promoter identifies the start site for tran- scription and orients the enzyme on the template strand. The RNA polymerase separates the two strands of DNA as it reads the base sequence of the template strand. 2. Transcription begins at the + 1 base pair. Sigma factor is released as soon as transcription is initiated. 3. The core polymerase continues moving along the template strand in the 3' to 5' direction, synthesizing the mRNA in the 5' to 3' direction. 4. RNA polymerase eventually reaches a transcription termination signal, at which point it will stop transcription and release the completed mRNA molecule. There are two kinds of transcription terminators commonly found in prokaryotic genes: Rho-independent termination occurs when the newly formed RNA folds back on itself to form a GC-rich hairpin loop closely followed by 6-8 U residues. These two struc- tural features of the newly synthesized RNA promote dissociation of the RNA from the DNA template. This is the type of terminator shown in Figure 1-3-4. Rho-dependent termination requires participation of rho factor. This protein binds to the newly formed RNA and moves toward the RNA polymerase that has paused at a termination site. Rho then displaces RNA polymerase from the 3' end of the RNA. 5. Transcription and translation can occur simultaneously in bacteria. Because there is no processing of prokaryotic mRNA (no introns), ribosomes can begin translating the message even before transcription is complete. Ribosomes bind to a sequence called 'the Shine-Dalgarno sequence in the 5' untranslated region (UTR) of the message. Protein synthesis begins at an AUG codon at the beginning of the coding region and continues until the ribosome reaches a stop codon at the end of the coding region. 6. The ribosome translates the message in the 5' to 3' direction, synthesizing the protein from amino terminus to carboxyl terminus. n1eClical 35  USMLE Step~,1.: JJiochemistry Transcription ATG TGA GC rich I I I I I I DNA 5' Untranslated -----3' Untranslated Region (UTR) Region (UTR) +1 Transcription Terminates Transcription Shine-Dalgarno GC rich Sequence -. AVUG UGA _ ,~ Coding Region V mRNA 5' en I Luuuuuu3' ~ 5' UTR 3' UTR.'Translation H2N - Protein - COOH Figure 1-3-4.A Prokaryotic Transcription Unit The mRNA produced by the gene shown in Figure 1-3-4 is a monocistronic message. That is, it is transcribed from a single gene and codes for only a single protein, The word cistron is another name for a gene, Some bacterial operons (for example, the lactose operon, Chapter 5) produce polycistronic messages, In these cases, related genes grouped together in the DNA are transcribed as one unit. The mRNA in this case contains information from several genes and codes for several different proteins (Figure 1-3-5). I I I I I I I" 'I KAPlACi- leaI I I I 36 me  Transcription and RNA Processing Promoter 5' UTR U U 5' UTR 1:ene 1 UI.. 2:ene 2 I.... 2:ene 3 ;G 3' UTR SL / Shine- / Shine- / Shine- Dalgarno Dalgarno Dalgarno ~N-Protein-COOH H~-Protein-COOH H~-protein-COOH 2 3 Figure 1-3-5. Polycistronic Gene Region Codes for Several Different Proteins PRODUCTION OF EUKARVOTIC MESSENGER RNA In eukaryotes, most genes are composed of coding segments (exons) interrupted by noncoding segments (introns). Both exons and introns are transcribed in the nucleus. Introns are removed during processing of the RNA molecule in the nucleus. In eukaryotes, all mRNA is monocis- tronic. The mature mRNA is translated in the cytoplasm. The structure and transcription of a typical eukaryotic gene coding for a protein is illustrated in Figure 1-3-6. Transcription of this gene occurs as follows: 1. With the help of proteins called transcription factors, RNA polymerase II recognizes and binds to the promoter region. The basal promoter region of eukaryotic genes usually has two consensus sequences called the TATA box (also called Rogness box) and the CAAT box. 2. RNA polymerase II separates the strands of the DNA over a short region to initiate tran- scription and read the DNA sequence. The template strand is read in the 3' to 5' direction as the RNA product (the primary transcript) is synthesized in the 5' to 3' direction. Both exons and introns are transcribed. 3. RNA polymerase II ends transcription when it reaches a termination signal. These signals are not well understood in eukaryotes. KAPLA~. I meulCa 37 ';'i !/' I I USMLE Step 1: Biochemistry Transcription Promoter Poly-A Addition Site ATG Signal AATAAA 5' DNA 3' 3' Untranslated Region (UTR) 3' Splice Transcription Site Terminates Transcription Poly-A Addition Primary Exon 1 Intron Signal AAUAAA Transcript V 3' RNA i ~ 5' Splice. Site 3'UTR 3' Splice Site Figure 1-3-6. A EUkaryotic Transcription Unit Processing of Eukaryotic Messenger RNA The primary transcript must undergo extensive posttranscriptional processing inside the nucleus to form the mature mRNA molecule (Figure 1-3-7). These processing steps include the following: 1. A 7-methylguanosine cap is added to the 5' end while the RNA molecule is still being synthesized. The cap structure serves as a ribosome-binding site and also helps to protect the mRNA chain from degradation. 2. A poly-A tail is attached to the 3' end. In this process, an endonuclease cuts the molecule 1,_, on the 3' side of the sequence AAUAAA (poly-A addition signal), then poly-A polymerase I adds the poly-A tail (about 200 As) to the new 3' end. The poly-A tail protects the message against rapid degradation and aids in its transport to the cytoplasm. A few mRNAs (for example, histone mRNAs) have no poly-A tails. 38 m~Clical  Transcription and RNA Processing Poly-A Addition AUG UAG Signal AAUAAA Primary Transcript RNA 5' -IIifiI i Intron i '-- V --..,.-- --' 3' Y 5' Splice 3' Splice 3'UTR 5'UTR Site Site I Capping and Poly-A Addition (Nucleus) + Poly-A Addition Signal AAUAAA AAAAAAAA3' Poly-A Tail i 3' Splice ~ 3'UTR Site hnRNA Excised Intran (Lariat) Degraded in Nucleus AUG UAG © e AAAAAAAA3' mRNA 5' \ Poly-A Tail Gppp Cap Y 5'UTR I Transport to Cytoplasm ~ 3'UTR and Translation H2N-Protein-COOH Figure 1-3-7. Processing Eukaryotic mRNA ineCtical 39  'I ! USMLE Step 1: Biochemistry I r 'I Note 3. Introns are removed from hnRNA by splicing, accomplished by spliceosomes (also known as an snRNP, or snurp), which are complexes of snRNA and protein. The hnRNA mol- Mutations in splice sites can ecule is cut at splice sites at the 5' (donor) and 3' (acceptor) ends of the intron. The intron lead to abnormal proteins. is excised in the form of a lariat structure and degraded. Neighboring exons are joined For example, mutations together to assemble the coding region of the mature mRNA. that interfere with proper splicing of f3-globin mRNA are 4. All of the intermediates in this processing pathway are collectively known as hnRNA. responsible for some cases of 5. The mature mRNA molecule is transported to the cytoplasm, where it is translated to f3-thalassemia. form a protein. ALTERNATIVE SPLICING OF EUKARYOTIC PRIMARY TRANSCRIPTS For some genes, the primary transcript is spliced differently to produce two or more variants of a protein from the same gene. This process is known as alternative splicing and is illustrated in Figure I-3-8. Variants of the muscle proteins tropomyosin and troponin T are produced in this way. The synthesis of membrane-bound immunoglobulins by unstimulated B lymphocytes, as opposed to secreted immunoglobulins by antigen-stimulated B lymphocytes, also involves alternative splicing. The primary transcripts from a large percentage of genes undergo alternative splicing. This may occur within the same cell, or the primary transcript of a gene may be alternatively spliced in differ- II. I ent tissues, giving rise to tissue-specific protein products. By alternative splicing, an organism can make many more different proteins than it has genes to encode. A current estimate of the number of human proteins is about 100,000, whereas. the current estimate of human genes is about only 30,000. These figures should not be memorized because they may change upon more research. Alternative splicing can be detected by Northern blot, a technique discussed in Chapter 7. Primary RNA Transcript Exon 1 Intron 2 Intron 3 Exon 3 Exon 4 5' --1 Exon 1 Exon4 r- 3' 5' --1 Exon 1 I Exon 3 I Exon 4 ~ 3' mRNA#1 mRNA#2 Figure 1-3-8. Alternative Splicing of Eukaryotic hnRNA (pre-mRNA) to Produce Different Proteins PRODUCTION OF OTHER CLASSES OF RNA Genes encoding other classes of RNA are also expressed. The RNA products are not translated to produce proteins, but rather serve different roles in the process of translation. KAPLAH"d- I 40 me lea  Transcription and RNA Processing RIBOSOMAL RNA (rRNA) IS USED TO CONSTRUCT RIBOSOMES Figure 1-3-9 shows the components of prokaryotic and eukaryotic ribosomes. Eukaryotic ribosomal RNA is transcribed in the nucleolus by RNA polymerase 1 as a single piece of 455 RNA, which is subsequently cleaved to yield 285 rRNA, 185 rRNA, and 5.85 rRNA. RNA polymerase III transcribes the 55 rRNA unit from a separate gene. The ribosomal subunits assem- ble in the nucleolus as the rRNA pieces combine with ribosomal proteins. Eukaryotic ribosomal subunits are 605 and 405. They join during protein synthesis to form the whole 805 ribosome. Figure 1·3·9. The Composition of Prokaryotic and Eukaryotic Ribosomes The large and small prokaryotic ribosomal subunits are 505 and 305, respectively. The complete prokaryotic ribosome is a 705 particle. (Note: The 5 values are determined by behavior of the particles in an ultracentrifuge. They are a function of both size and shape, and therefore the numbers are not additive.) TRANSFER RNA (tRNA) CARRIES ACTIVATED AMINO ACIDS FOR TRANSLATION There are many different specific tRNAs. Each tRNA carries only one type of activated amino acid for making proteins during translation. The genes encoding these tRNAs in eukaryotic cells are transcribed by RNA polymerase III. The tRNAs enter the cytoplasm where they combine with their appropriate amino acids (see Chapter 4, Amino Acid Activation). Although all tRNAs have the same general shape shown in Figure 1-3-10, small structural features distinguish among them. KAPLA~. I meulCa 41 , ~~' -~t l 1  USMLE Step 1: Biochemistry 3'end A OH f- Activated amino acid is attached C to 3'OH. C :! 5'end I A p C A G - C C - G G - C G C G - C U C - G G - C G - C G - C f- Anticodon sequence (CAU) pairs with codon in mRNA. Figure 1-3-10. Transfer RNA (tRNA) I' " 42 iiie&ical  Transcription and RNA Processing Table 1-3-1. Summary of Important Points About Transcription and RNA Processing r---- ---- Prokaryotic -- Eukaryotic I I Always monocistronic Gene regions May be polycistronic Genes are continuous coding regions Genes have exons and introns Very little spacer (noncoding) DNA Large spacer (noncoding) DNA between genes I RNA polymerase between genes Core enzyme: a2~W RNA polymerase I: rRNA RNA polymerase II: mRNA; snRNA \ RNA polymerase III: tRNA, 5S RNA I I r Initiation of Promoter (-10) TATAATand Promoter (-25) TATA and (-70) CAAT I transcription (-35) sequence Sigma initiation subunit required to Transcription factors (TFIID) bind promoter I \ recognize promoter I I mRNA synthesis Template read 3' to 5'; mRNA synthesized 5' to 3'; Gene sequence specified from coding strand 5' I to 3'; Transcription begins at + 1 base Stem and loop + UUUUU Not well characterized Termination of transcription Stem and loop + rho factor I I I Relationship of RNA is antiparallel and complementary to DNA template strand; RNA is identical (except U substitutes for T) to DNA coding strand RNA transcript \ I to DNA Posttranscrip- None In nucleus: I tional processing ofhnRNA 5' cap (7-MeG) 3' tail (poly-A sequence) '-I Removal of introns from hnRNA (pre-mRNA) Alternative splicing yields variants of protein product II I I Ribosomes 70S (30S and 50S) 80S (40S and 60S) rRNA and protein rRNA and protein \ tRNA Cloverleaf secondary structure Acceptor arm (CCA) carries amino acid Anticodon arm; anticodon complementary and antiparallel to codon in mRNA KAPLAlf - I med lea 43  USMLEStep I: Biochemistry Review Questions Select the ONE best answer. 1. The base sequence of codons 57-58 in the cytochrome ~5 reductase gene is CAGCGC. The mRNA produced upon transcription of this gene will contain the sequence: A. GCGCTG B. CUGCGC C. GCGCUG D. CAGCGC E. GUCGCG 2. A gene encodes a protein with 150 amino acids. There is one intron of 1,000 bps, a 5' -untranslated region of 100 bp, and a 3' -untranslated region of 200 bp. In the final processed mRNA, how many bases lie between the start AUG codon and the final termi- nation codon? A. 1,750 B. 750 C. 650 D. 450 E. 150 Items 3-5: Identify the nuclear location. :1 E o ~,I I: ! A c B 3. Transcription of Class 1 genes by RNA polymerase 1 4. Euchromatin 5. Polyadenylation of pre-mRNA by poly-A polymerase KAPLA~. I 44 meulCa  III '·'!, I: Transcription and RNA Processing I:t 1,[ ;~ If Answers r;! I 1. Answer: D. Since the sequence in the stem represents the coding strand, the mRNA I sequence must be identical (except U for T). No T in the DNA means no U in themRNA. 2. Answer: D. Only the coding region remains to be calculated 3 x 150 =:: 450. 3. Answer: B. rRNA genes are transcribed by this enzyme in the nucleolus. 4. Answer: A. Less condensed chromatin, lighter areas in the nucleus. Darker areas are het- erochromatin. 5. Answer: A. Polyadenylation of pre-mRNA occurs in the nucleoplasm. Generallyassoci- ated with active gene expression in euchromatin. KAPLA~. I meulCa 45  I. ~. r I , I l. I  The Genetic Code, Mutations, and Translation OVERVIEW OF TRANSLATION The second stage in gene expression is translating the nucleotide sequence of a messenger RNA molecule into the amino acid sequence of a protein. The genetic code is defined as the relation- ship between the sequence of nucleotides in DNA (or its RNA transcripts) and the sequence of amino acids in a protein. Each amino acid is specified by one or more nucleotide triplets (codons) in the DNA. During translation, mRNA acts as a working copy of the gene in which the codons for each amino acid in the protein have been transcribed from DNA to mRNA. tRNAs serve as adapter molecules that couple the codons in mRNA with the amino acids they each specify, thus aligning them in the appropriate sequence before peptide bond formation. Translation takes place on ribosomes, complexes of protein and rRNA that serve as the molecu- lar machines coordinating the interactions between mRNA, tRNA, the enzymes, and the protein factors required for protein synthesis. Many proteins undergo posttranslational modifications as they prepare to assume their ultimate roles in the cell. THE GENETIC CODE Most genetic code tables designate the codons for amino acids as mRNA sequences (Figure 1-4-1). Important features of the genetic code include: Each codon consists of three bases (triplet). There are 64 codons. They are all written in the 5' to 3' direction. 61 codons code for amino acids. The other three (UAA, UGA, UAG) are stop codons (or nonsense codons) that terminate translation. There is one start codon (initiation codon), AUG, coding for methionine. Protein syn- thesis begins with methionine (Met) in eukaryotes, and formylmethionine (fmet) in prokaryotes. The code is unambiguous. Each codon specifies no more than one amino acid. The code is degenerate. More than one codon can specify a single amino acid. All amino acids, except Met and tryptophan (Trp), have more than one codon. For those amino acids having more than one codon, the first two bases in the codon are usually the same. The base in the third position often varies. The code is universal (the same in all organisms). Some minor exceptions to this occur in mitochondria. The code is commaless (contiguous). There are no spacers or "commas" between codons on an mRNA. Neighboring codons on a message are nonoverlapping. mettlcal 47  USMLE Step 1: Biochemistry Second Position First Pos ition Third Position (5' End) U C A G (3' E nd) U UUU} UUC Phe ueu} UCC Ser UAU} Tyr UAC UGU} Cys UGC C U UUA} UCA UAA} UGA Stop A UUG Leu UCG UAG Stop UGG Tr"p G C euu} CUC Leu eeu} CCC CCA Pro CAUl His CAC eGu} CGC Arg U C CUA CGA A CAA} CUG CCG CAG Gin CGG G I, A AUU} lie AUC AeU} ACC Thr AAU} Asn AAC AGU} Ser AGC U C AUA ACA AAA} AGA} AUG Met AAG Lys AGG Arg A ACG G G GUU} Geu} GUC Val GCC Ala GAU} Asp GAC GGU} GGC Gly U C GUA GCA GAA} Glu GGA GUG GCG GAG A GGG G Figure 1-4-1. The Genetic Code MUTATIONS A mutation is any permanent, heritable change in the DNA base sequence of an organism. This altered DNA sequence can be reflected by changes in the base sequence of mRNA, and, sometimes, by changes in the amino acid sequence of a protein. Mutations can cause genetic diseases. They can also cause changes in enzyme activity, nutritional requirements, antibiotic susceptibility, morphology, antigenicity, and many other properties of cells. A very common type of mutation is a single base alteration or point mutation. A transition is a point mutation that replaces a purine-pyrimidine base pair with a dif- ferent purine-pyrimidine base pair. For example, an A-T base pair becomes a G-C base pair. A transversion is a point mutation that replaces a purine-pyrimidine base pair with a pyrimidine-purine base pair. For example, an A-T base pair becomes a T-A or a C-G base pair. Mutations are often classified according to the effect they have on the structure of the gene's protein product. This change in protein structure can be predicted using the genetic code table in conjunction with the base sequence of DNA or mRNA. A variety of such mutations is listed in Table 1-4-1. Point mutations and frameshifts are illustrated in more detail in Figure 1-4-2. KAPLAN.' I 48 medlea ii@!i----- The Genetic Code, Mutations, and Translation Table 1-4-1. Effects of Some Common Types of Mutations on Protein Structure Type of Mutation Effect on Protein I I i! Silent: new codon specifies same amino acid None I Ii II Missense: new codon specifies Possible decrease in function; variable effects different amino acid I l Nonsense: new codon is stop codon Shorter than normal; usually nonfunctional i I Frameshift: deletion or addition of a base Usually nonfunctional; often shorter than normal Loss of function; shorter than normal or entirely missing !! Large segment deletion (unequal I! crossover in meiosis) Variable effects ranging from addition or deletion of a few I ! Splice donor or acceptor amino acids to deletion of an entire exon I ! Triplet repeat expansion Expansions in coding regions cause protein product to be longer than normal and unstable. Disease often shows anticipation in pedigree. Figure 1-4-2. Some Common Types of Mutations in DNA KAPLAlf - I medlea 49 USMLE Step·): Biochemistry large Segment Deletions Large segments of DNA can be deleted from a chromosome during an unequal crossover in meiosis. Crossover or recombination between homologous chromosomes is a normal part of meiosis I that generates genetic diversity in reproductive cells (egg and sperm), a largely beneficial result. In a normal crossover event, the homologous maternal and paternal chromo- somes exchange equivalent segments, and although the resultant chromosomes are mosaics of maternal and paternal alleles, no genetic information has been lost from either one. On rare occasions, a crossover can be unequal and one of the two homologs loses some of its genetic information. Both normal and unequal crossing over are shown in Figure 1-4-3. a-Thalassemia is a well-known example of a genetic disease in which unequal crossover has deleted one or more a-globin genes from chromosome 16. Cri-du-chat (mental retardation, microcephaly, wide-set eyes, and a characteristic kittenlike cry) results from a terminal deletion of the short arm of chromosome 5 (see Section II, Chapter 3; Deletions). Maternal Paternal Normal Crossover Deletion from Paternal Chromosome ~ Figure 1-4-3. Large Segment Deletion During Crossing-Over in Meiosis KAPLA~. I 50