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by: Marcella Lebsack


Marcella Lebsack
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This 96 page Class Notes was uploaded by Marcella Lebsack on Thursday October 22, 2015. The Class Notes belongs to BIOL 3416 at Texas Tech University taught by Rodgers in Fall. Since its upload, it has received 57 views. For similar materials see /class/226500/biol-3416-texas-tech-university in Biology at Texas Tech University.


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Date Created: 10/22/15
Key Concepts Chapter 2 Monohybrid Crosses and Mendel s Principle of Segregation Equot W 5 U 9 gt 00 0 O Terminology used in breeding experiments a Parental generation is the P generation b Progeny of P generation is the rst lial generation designated F1 c When F1 interbreed the second lial generation F2 is produced d Subsequent interbreeding produces F3 F4 and F5 generations A monohybrid cross involves truebreeding strains that differ in a single trait To determine whether both parents contribute equally to the phenotype of a particular trait in offspring a set of reciprocal crosses is performed By convention the female parent is given rst In Mendelian genetics offspring of a monohybrid cross will exactly resemble only one of the parents This is the principle of uniformity in F1 Figure 25 Traits that disappear in the F1 reappear in the F2 The F2 has a ratio of about one individual with the reappearing phenotype to three individuals with the phenotype of the F1 Mendel reasoned that information to create the trait was present in the F1 in the form of factors now called genes Figure 26 Each gene exists in alternative forms now called alleles that control a speci c trait Truebreeding strains contain identical genes The F1 contain one of each but since the trait is just like one of the parents rather than a mix one dominant allele has masked expression of the other recessive one By convention letters may be used to designate alleles with the dominant a capital letter 8 and the recessive in lowercase s Individuals with identical alleles eg genotypes SS and ss are said to be homozygous for that gene because all their gametes will have the same allele for this trait Individuals with different alleles eg Ss are heterozygous because 12 of their gametes will contain one allele and 12 the other Figure 27 Diagrams of a smoothwrinkled cross appear in Figure 28 The Punnett square is a diagram showing all possible combinations of the gametes produced by each parent Note that it accounts for the 31 ratio in the F2 generation When Mendel had conducted experiments for the seven different traits in garden peas Table 21 he made these conclusions a Results of reciprocal crosses are always the same b The F1 resembled only one of the parents c The trait missing in the F1 reappeared in about 14 of the F2 individuals The Principle of Segregation 1 The rst Mendelian law the principle of segregation states Recessive characters which are masked in the F1 from a cross between two true breeding strains reappear in a speci c proportion in the F2 This is because alleles segregate during anaphase I of meiosis and progeny are then produced by random combination of the gametes 2 A summary of terms and concepts appears in Box 21 Representing Crosses with a Branch Diagram 1 The branch diagram is an alternative approach to predicting the outcome of crosses Figure 29 2 Results from a branch diagram will be identical to those obtained with a Punnett square Continuing the Principle of Segregation The Use of Testcrosses Mendel observed that plants with the recessive phenotype are all true breeding When plants with the dominant phenotype are selfed 13 are truebreeding and 23 produce progeny with both phenotypes Figure 210 2 A more common method to determine whether an individual with the dominant phenotype is homozygous or heterozygous is to perform a testcross by crossing the individual with one that is homozygous recessive Figure 211 Dihybrid Crosses and Mendel s Principle of Independent Assortment The Principle of Independent Assortment 1 After Mendel analyzed crosses involving two pairs of traits dihybrid crosses he formulated his second law the principle of independent assortment which says that the factors for different traits assort independently of one another This allows for new combinations of the traits in the offspring Figure 212 2 A dihybrid cross will produce four possible phenotypic classes in a 9331ratio Branch Diagram of Dihybrid Crosses 1 Dihybrid crosses can be represented with either a Punnett square or a branch diagram With practice the student should be able to calculate results of crosses simply by applying Mendelian principles Figure 213 2 Individuals of unknown genotype that show the dominant phenotype can be represented by a dominant allele and a dash eg S to indicate that the second allele is unknown If alleles assort independently all possible phenotypes will be represented in the F2 in a ratio of9 3 3 l Ifthe F1 are testcrossed all types of offspring in a ratio ofl l l 1 will be produced In the F2 of a dihybrid cross there will be four phenotypic classes and nine genotypic classes Table 22 Trihybrid Crosses l Crosses involving three independently assorting character pairs are called trihybrid There are 64 possible combinations of the eight different gamete types contributed by each parent creating 27 different genotypes There will be eight different phenotypes in a predicted ratio of 27 9993331Figure2l4 Some useful generalizations about Mendelian genetics with the following assumptions 1 The parents are two different truebreeding strains for the genes under study 2 The F1 selffertilize or interbreed Under these assumptions The F1 will be heterozygous for each gene involved in the cross 32 When the F1 interbreed the F2 will contain 34 dominant phenotype and 14 recessive phenotype individuals with genotype frequencies of 14 for AA12 forAa and 14 for aa There are two phenotypic classes in the F2 of a monohybrid cross four in a dihybrid cross and eight in a trihybrid cross The general rule is that there are 211 phenotypic classes in the F2 where n is the number of independently assorting heterozygous gene pairs Table 23 57 o 9 There are three genotypic classes in the F2 of a monohybrid cross while there are nine in a dihybrid cross and 27 in a trihybrid cross The general rule is that there are 311 phenotypic classes in the F2 where n is the number of independently assorting heterozygous gene pairs The phenotypic rule 211 is also used to predict the number of genotypic classes produced in the testcross of a multiply heterozygous F1 because the number of genotypic classes will match the number of phenotypic ones 9 Statistical Analysis of Genetic Data The ChiSquare Test One of the advantages of the Mendelian approach is that it is quantitative and may be tested mathematically A hypothesis is presented as a null hypothesis proposing that the observed data are the same as the predicted results A chisquare f test checks for goodness of t between the eXpected and observed results to determine whether differences are likely owed to chance alone See an example using the chisquare test to analyze a hypothesis Tables 24 and 25 4 Chisquare analysis cannot tell us that a hypothesis is correct or incorrect only whether the observed result is a good t with predictions of the hypothesis Figure 215 a If differences between the results and the prediction are unlikely to be due to chance alone the hypothesis will be rejected and another one tried b Typically if the probability of obtaining the observed F values is greater than 5 Pgt005 the hypothesis is not rejected Pedigree Analysis 1 The study of the phenotypic records of a family over several generations is pedigree analysis The individual upon whom the study focuses is the propositus male or proposita female 2 The symbols of pedigree analysis are summarized in Figure 216 and Figure 217 shows a sample pedigree Note that generations are numbered with Roman numerals while individuals are numbered with Arabic numerals Examples of Human Genetic Traits 1 Recessive traits are well documented in humans and are usually the result of a mutation causing loss or modi cation of a gene product Albinism Figure 218 is an example 2 Deleterious recessive alleles persist in the population because heterozygous individuals carry the allele without developing the phenotype and so are not at a selective disadvantage 3 Characteristics of recessive inheritance of a relatively rare trait a Parents of most affected individuals have normal phenotypes but are heterozygous If the allele is rare the trait will skip generations 57 Mating of heterozygotes will produce 34 normal progeny and 14 with the recessive phenotype 0 If both parents have the recessive trait all their progeny will usually also have the trait 4 Dominant traits are also well documented in humans A mutation may produce a dominant phenotype by causing a function to be gained because of an altered gene product capable of a new activity Examples a Woolly hair Figure 219 b Achondroplasia c Brachydactyly d Marfan syndrome 5 Dominant alleles produce a distinct phenotype when in a heterozygote whose other allele is wild type Due to the rarity of dominant mutant alleles causing recognizable traits homozygous dominant individuals are very unusual Most pairings are between a heterozygote and a homozygous recessive wildtype individual Characteristics of recessive inheritance of a relatively rare trait 9 Affected individuals have at least one affected parent 6 The trait is present in every generation Offspring of an affected heterozygote will be 12 affected and 12 wild type 0 Key Concepts Chapter 3 Chromosomes and Cellular Reproduction Cytology and genetics together were used to determine the association of genes and chromosomes Eukaryotic chromosomes are transmitted during cell division by mitosis and during reproduction by meiosis Eukaryotic Chromosomes l Eukaryotes have multiple linear chromosomes in a number characteristic of the species Most have two versions of each chromosome and so are diploid 2N Diploid cells are produced by haploid N gametes that fuse to form a zygote The zygote then undergoes development forming a new individual 9 57 Examples of diploid organisms are humans 23 pairs and Drosophila melanogaster 4 pairs The yeast Saccharomyces cerevisiae is haploid 16 chromosomes Chromosome pairs in diploid organisms are homologous chromosomes One member of each pair homolog is inherited from each parent Chromosomes that have different genes and do not pair are nonhomologous chromosomes Figure 31 Animals and some plants have male and female cells with distinct chromosome sets due to seX chromosomes One seX has a matched pair eg human females with XX and the other has an unmatched pair human male with XY Autosomes are chromosomes other than seX chromosomes Chromosomes differ in size and morphology Each has a constriction called a centromere that is used in segregation during mitosis and meiosis The centromere location is useful for identifying chromosomes Figure a M etacentric means the centromere is approximately in the center of the chromosome producing two equal arms b Submetacentric means one arm is somewhat longer than the other c Acrocentric chromosomes have one long arm and a short stalk and often a bulb satellite as the other arm d Telocentric chromosomes have only one arm because the centromere is at the end 5 A karyotype shows the complete set of chromosomes in a cell Metaphase chromosomes are used because they are easiest to see under the microscope after staining The karyotype is speciesspeci c a The karyotype for a normal human male has 22 pairs of autosomes and 1 each ofX and Y Figure 33 b Human chromosomes are numbered from largest l to smallest although 21 is actually smaller than 22 c Human chromosomes with similar morphologies are grouped A through G d Staining produces bands on the chromosomes allowing easier identi cation G banding is an example i Chromosomes are partially digested with proteolytic enzymes or treated with mild heat and then stained with Giemsa stain The dark bands produced are G bands ii In humans metaphase chromosomes show about 300 G bands while about 2000 can be distinguished in prophase iii Drawings ideograms show the G banding pattern of human chromosomes iV Standard nomenclature is used to reference speci c regions of the chromosomes 1 The two arms are separated by the centromere with the smaller one designated p and the larger q 2 Regions and subregions are numbered from the centromere outward l is closest 3 An example is the BRCA1 breast cancer susceptibility gene at l7q2l long arm of chromosome 17 in region 21 4 If a gene spans subregions both are given For example the human cystic brosis gene is at 7q3l27q3l3 spanning both subregions 2 and 3 on the long arm of chromosome 7 Mitosis 1 Both unicellular and multicellular eukaryotes show a cell cycle with growth mitosis and cell diVision Figure 34 a The cycle of somatic cells consists of i Mitosis ii Interphase composed of l Gap 1 G1 when the cell prepares for chromosome replication b 0 N V Synthesis S when DNA replicates and new chromosomes are formed W V Gap 2 G2 when the cell prepares for mitosis and cell division Relative time in each phase varies among cell types with duration of G1 generally the deciding factor Some cells eXit G1 and enter a nondividing state called G0 Interphase chromosomes are elongated and hard to see with light microscopy Sister chromatids are held together by replicated but 1 39 The 39 39 39 become visible in prophase and metaphase of mitosis When the centromeres separate they become daughter chromosomes 2 Mitosis is a continuous process but geneticists divide it into 4 cytologically distinguishable stages Figures 35 and 36 a 57 o d Prophase is 39 39 39by 39 39 toaform visible by light microscopy The mitotic spindle composed of microtubules made of tubulins begins to form In animal cells the centrioles replicate and become the focus for the aster radial array of microtubules During prophase asters move from near each other and the nuclear envelope to the poles of the cell spanned by the mitotic spindle iii The nucleoli in the nucleus cease to be discrete areas iv The nuclear envelope breaks down Kinetochores form on the centromeres and attach the chromosomes to spindle microtubules lt Metaphase begins when the nuclear envelope has completely disappeared 39 The kinetochore microtubules orient the chromosomes with their centromeres in a plane between the spindle poles the metaphase plate A protein scaffold causes the chromosomes to reach a highly condensed state Figure 37 Anaphase begins when the centromeres of the sister chromatids separate The chromatids separate disjunction and daughter chromosomes move toward opposite poles by kinetochore microtubules Shape of the chromosomes moving toward the poles is de ned by their centromere locations iii Cytokinesis usually begins near the end of anaphase Telophase is the period when migration of daughter chromosomes is completed i Chromosomes begin to uncoil and form interphase chromosomes Nuclear envelope forms around each chromosome group ii Spindle microtubules disappear iv Nucleoli reform lt Nuclear division is complete Cytokinesis is division of the cytoplasm compartmentalizing the new nuclei into separate daughter cells Figure 38 In animal cells cytokinesis begins with a constriction in the center of the cell which develops until two new cells are produced 3 57 Most plant cells form a cell plate membrane and wall between the two nuclei resulting in two progeny cells Gene segregation in mitosis is highly ordered so that each new cell receives a complete set of chromosomes pairs in a diploid cell and one of each type in a haploid cell Meiosis l Meiosis is two successive divisions of a diploid nucleus after only one DNA replication cycle The result is haploid gametes animals or meiospores plants The two rounds of division in meiosis are meiosis I and meiosis II each with a series of stages Figure 39 Cytokinesis usually accompanies meiosis producing four haploid cells from a single diploid cell Meiosis I is when the chromosome information is reduced from diploid to haploid It has four stages a Prophase I is very similar to prophase of mitosis except that homologous chromosomes pair and undergo crossingover Leptonema is when chromosomes begin to coil committing the cell to the meiotic process Homologous chromosomes pair during the leptomene stage ii In zygonema chromosomes are at maXimum condensation and synapsis a tight association between homologous chromosomes occurs Telomeres mediate formation of the 4chromatid synaptonemal tetrad Pachynema follows when the synaptonemal compleX is disassembled and chromosomes start to elongate Crossover occurs during pachynema iv Crossingover is reciprocal exchange of chromosome segments between homologous chromosomes If the homologs are not identical new gene combinations recombinant chromosomes can result but usually no genetic material is added or lost V Diplonema is the period when chromosomes begin to move apart V and chiasmata singular is chiasma formed by crossingover become visible Figure 310 1 Human oocytes arrest in diplonema in the 7th month of fetal development and remain there until an oocyte is activated to prepare for ovulation 2 Preparation for ovulation takes the oocyte through meiosis I 3 Fertilization causes meiosis II to occur allowing fusion with the sperm nucleus to form a zygote Diakinesis involves breakdown of the nucleoli and nuclear envelope and assembly of the spindle This is the phase where chromosomes are most easily counted SeX chromosomes are not homologous but in some mammals behave as if they were due to a pseudoautosomal region PAR shared between X and Y The PAR region enables crossing over b Metaphase I starts with the nuclear envelope completely broken down bivalents pairs of homologs aligned at the equatorial plane the spindle completely formed and microtubules attached to kinetochores It is distinguishable from metaphase of mitosis because homologous chromosomes form bivalents c Anaphase Iis when bivalents separate with chromosomes of each homologous pair disjoining Resulting dyads migrate toward opposite poles where new nuclei will form This migration assumes that Centromeres derived from each parent will migrate randomly toward each pole Each pole will receive a haploid complement of replicated centromeres with associated chromosomes Sister chromatids will remain attached to each other the major difference from mitosis d Telophase I has dyads completing migration to the poles and usually formation of a nuclear envelope around each haploid grouping Cytokinesis follows in most species forming two haploid cells Meiosis II is very similar to mitotic division a Prophase II involves chromosome condensation b Metaphase 11 includes spindle formation with centromeres lining up on the equator c Anaphase II involves splitting of the centromeres with chromosomes pulled to opposite poles d Telophase 11 takes place as a nuclear envelope forms around each set of chromosomes e Cytokinesis usually takes place and chromosomes become elongated and invisible with light microscopy 4 After both rounds of meiotic division four haploid cells gametes in animals are usually produced Each has one chromosome from each homologous pair but these are not exact copies due to crossingover Figure 311 compares mitosis and meiosis V39 Meiosis has three signi cant results a Haploid cells are produced because two rounds of division follow only one round of chromosome replication Fusion of haploid cells restores the diploid number maintaining a constant chromosome number through generations in sexually reproducing organisms 57 Alignment of paternally and maternally derived chromosomes is random in metaphase I resulting in random combinations of chromosomes in each nucleus generated Figure 312 ii 0 The number of possible chromosome arrangements at the meiosis I metaphase plate is 2quotD1 n is the number of chromosome pairs The number of possible chromosome combinations in nuclei produced by meiosis is 211 Due to differences between paternally and maternally derived chromosomes many possibilities exist Nuclei produced by meiosis will be genetically distinct from parental cells and from one another Crossingover between maternal and paternal chromatid pairs during meiosis I provides still more variation making the number of possible progeny nuclei extremely large 6 Meiosis in animals and plants is somewhat distinct a In diploid animals the only haploid cells are gametes produced by meiosis and used in sexual reproduction Gametes are produced by specialized cells Figure 313 i In males spermatogenesis produces spermatozoa within the testes l Primordial germ cells primary spermatogonia undergo mitosis to produce secondary spermatogonia 2 Secondary spermatogonia transform into primary spermatocytes meiocytes which undergo meiosis I giving rise to two secondary spermatocytes 3 Each secondary spermatocyte undergoes meiosis II producing haploid spermatids that differentiate into spermatozoa In females oogenesis produces eggs oocytes in the ovary l Primordial germ cells primary oogonia undergo mitosis to produce secondary oogonia 2 Secondary oogonia transform into primary oocytes which grow until the end of oogenesis 3 Primary oocytes undergo meiosis I and unequal cytokinesis producing a large secondary oocyte and a small cell called the rst polar body 4 The secondary oocyte produces two haploid cells in meiosis II One is a very small cell the second polar body and the other rapidly matures into an ovum U V The rst polar body may or may not divide during meiosis I Polar bodies have no function in most species and degenerate so that a round of meiosis produces only one viable gamete the ovum Human oocytes form in the fetus completing meiosis only after fertilization Sex Chromosomes 1 Behavior of seX chromosomes offers support for the chromosomal theory In many animals seX chromosome composition relates to seX while autosomes are constant 2 Independent work of McClung Stevens and Wilson indicated that chromosomes are different in male and female insects Stevens named the eXtra chromosome found in females X lt7 9 In grasshoppers all eggs have an X and half of the sperm produced have an X and the other half do not After fertilization an unpaired X produces a male while paired X chromosomes produce a female 3 Other insects have a partner for the X chromosome Stevens named it Y In mealworms for example XX individuals are female and XY are male 4 In both humans and fruit ies Drosophila melanogaster females have two X chromosomes while males have X and Y Figure 316 9 Males produce two kinds of gametes with respect to seX chromosomes X or Y and are called the heterogametic sex 57 Females produce gametes with only one kind of seX chromosome X and are called the homogametic seX 0 In some species the situation is reversed with heterogametic females and homogametic males 5 Random fusion of gametes Figure 317 produces an F1 that is 12 female XX and 12 male XY Evidencefor the YC 7 l 7 39 ofSex D t Mt 1 Understanding of the Y chromosome mechanism of seX determination came from the study of individuals with unusual chromosome complements In humans these aneuploidies include a XO individuals who are sterile females eXhibiting Turner syndrome Figure 324 Most XO fetuses die before birth Surviving Turner syndrome individuals become noticeable at puberty when secondary sexual characteristics fail to develop Other traits include i Belowaverage height ii Weblike necks 6 0 F F iii Poorly developed breasts iv Immature internal sexual organs V Reduced ability to interpret spatial relationships XXY individuals who are male and have Klinefelter syndrome Figure 325 Other traits include i Aboveaverage height ii Breast development in about 50 of XXY individuals iii Subnormal intelligence in some cases XYY individuals are male and tend to be taller than average Fertility is sometimes affected XXX individuals are usually normal women although they may be slightly less fertile and a few have belowaverage intelligence Higher numbers of X andor Y chromosomes are sometimes found including XXXY XXXXY and XXYY The effects are similar to Klinefelter syndrome F of seX 39 39 39 39 in humans are summarized in Table 32 1 Dosage Compensation Mechanism for X Linked Genes in Mammals 1 Gene dosage varies between the sexes in mammals because females have two copies of X while males have one Early in development gene eXpression from the X chromosome must be equalized to avoid death Different dosage compensation systems have evolved in different organisms 2 In mammals female somatic cell nuclei contain a Barr body highly condensed chromatin while male nuclei do not Figure 326 The Lyon hypothesis eXplains the phenomenon a 6 0 F F A Barr body is a condensed and mostly inactivated X chromosome Lyonization of one chromosome leaves one transcriptionally active X equalizing gene dose between the sexes An X is randomly chosen in each cell for inactivation early in development in humans day 16 postfertilization Descendants of that cell will have the same X inactivated making female mammals genetic mosaics Examples are Calico cats in which differing descendant cells produce patches of different color on the animal Figure 327 ii Women heterozygous for an Xlinked allele responsible for sweat glands who have a mosaic of normal skin and patches lacking sweat glands anhidrotic ectodermal displasia Lyonization allows eXtra seX chromosomes to be tolerated well No such mechanism eXists for autosomes and so an eXtra autosome is usually lethal The number of Barr bodies is the number of X chromosomes minus one f X inactivation involves three steps i Chromosome counting determining number of Xs in the cell ii Selection of an X for inactivation iii Inactivation itself Counting the chromosomes involves the Xinactivation center XI C in humansXic in mice Experiments in transgenic mice show that P Inactivation requires the presence of at least two X 1390 sequences one on each X chromosome Autosomes with an X 1390 inserted are randomly inactivated showing that X 1390 is suf cient for chromosome counting and initiation of lyonization P Selection of an X for inactivation is made by the Xcontrolling element Xce in the Xic region There are different alleles of Xce and each allele has a different probability that the X chromosome carrying it will be inactivated The gene X ist is required for X inactivation Uniquely it is expressed from the inactive X The Xist gene transcript is l7kb Although it has no ORFs it receives splicing and a polyA tail During X inactivation this RNA coats the chromosome to be inactivated and silences most of its genes Inactivation itself is not well understood but it is known that it initiates at the Xic and moves in both directions ultimately resulting in heterochromatin ii Analysis of SexLinked Traits in Humans l Xlinked traits like autosomal ones can be analyzed using pedigrees Human pedigree analysis however is complicated by several factors Data collection often relies on family recollections 57 3 If the trait is rare and the family small there may not be enough affected individuals to establish a mechanism of inheritance o Expression of the trait may vary resulting in affected individuals being classi ed as normal 9 More than one mutation may result in the same phenotype and comparison of different pedigrees may show different inheritance for the same trait X Linked Recessive Inheritance 1 Human traits involving recessive alleles on the X chromosome are Xlinked recessive traits A famous example is hemophilia A among Queen Victoria s descendants Figure 328 2 Xlinked recessive traits occur much more frequently among males who 3 4 are hemizygous A female would express a recessive Xlinked trait only if she were homozygous recessive at that locus Some characteristics of Xlinked recessive inheritance a Affected fathers transmit the recessive allele to all daughters who are therefore carriers and to none of their sons 57 Fathertoson transmission of Xlinked alleles generally does not occur 0 Many more males than females exhibit the trait 9 All sons of affected homozygous recessive mothers are expected to show the trait With a carrier mother about 12 of her sons will show the trait and 12 will be free of the allele 9 quot7 A carrier female crossed with a normal male will have 12 carrier and 12 normal daughters Other Xlinked recessive traits are Duchenne muscular dystrophy and two forms of color blindness X Linked Dominant Inheritance l 2 3 Only a few Xlinked dominants are known Examples include 9 Hereditary enamel hypoplasia faulty and discolored tooth enamel ST Webbing to the tips of the toes 0 Constitutional thrombopathy severe bleeding due to lack of blood platelets Patterns of inheritance are the same as Xlinked recessives except that heterozygous females show the trait although often in a milder form Y Linked Inheritance Ylinked holandric traits except for maleness itself resulting from SRY on the Y chromosome have not been con rmed but many genes on the Y chromosome have been identi ed The hairy ears trait may be Y linked but it is a complex phenotype and might also be the result of autosomal genes andor effects of testosterone Chapter 4 Key Concepts Determining the Number of Genes for Mutations with the Same Phenotype Relationship between phenotype and gene can be studied through mutants identi ed by phenotype distinct from wild type Complementation test cis trans test determines whether independently isolated mutations for the same phenotype are in the same or different genes by crossing two mutants Figure 41 3 a If mutations are in different genes phenotype will be wild type complementation b If mutations are in the same gene phenotype will be mutant no complementation Drosophila provides an example Wildtype body color is greyyellow If two truebreeding mutant blackbodied strains are crossed all F1 are wild type Figure 42 m Genes are e ebony and 17 black Black parents are homozygous mutant but in different genes ee blbl and 66 1717 57 F1 are heterozygous at both loci ele blb and therefore wild type showing complementation Multiple Alleles Not all genes have only two forms alleles many have multiple alleles Figure 43 No matter how many alleles for the gene exist in the multiple allelic series however a diploid individual will have only two alleles one on each homologous chromosome ABO Blood Groups 1 ABO blood groups are important in blood transfusions and result from a series of three alleles IA IE and i that combine to produce four phenotypes A B AB and 0 Figures 44 and 45 Both A and B are dominant to 139 while A and B are codominant to each other The resulting phenotypes are Table 41 a People with genotype ii are blood type 0 b People with genotype IAIA or IAi are blood type A c People with genotype IBIB or IBi are blood type B d People with genotype IAIB are blood type AB ABO inheritance follows Mendelian principles For example a type 0 individual s genotype is ii Possible genotypes of the parents could be ii and ii both blood type 0 IAi and ii one type A the other type 0 IAi and IAi both type A IBi and ii one type B the other type 0 IBi and IBi both type B IAi and IBi one type A the other type B Woo0579 Bloodtyping may be used in cases of disputed parentage Blood typing does not prove the identity of a parent It can however eliminate individuals who are not biological parents ofa particular child Example a A child with blood type AB IA1B could not have a parent with type 0 ii b Bloodtype data are not considered adequate legal proof for parenthood in most states and DNA ngerprinting is generally used Biochemical Genetics of the Human ABO Blood Group E 3 V39 Cellular antigens are important in blood transfusions since recipient antibodies may respond to antigens on donor cells Karl Landsteiner discovered human ABO blood groups in the early 1900s and received the 1930 Nobel Prize in Physiology or Medicine for this work Properties of the human ABO blood group There are three alleles at the ABO locus IA IE and i From these three alleles four phenotypes are produced 9 Type A individuals have the A antigen on their RBCs and antiB antibodies in their blood Their genotype is IAIA or IAi Type B individuals have the B antigen on their RBCs and antiA antibodies in their blood Their genotype is IBIB or IBz39 iii Type AB individuals have both the A and the B antigen on their RBCs and neither antiA nor antiB antibodies in their blood Their genotype is IAIB iv Type 0 individuals have neither the A nor the B antigen on their RBCs and both antiA and antiB antibodies in their blood Their genotype is ii Antigeniantibody relationships and their impact in blood transfusions are summarized in Figure 44 Summary of the relationship between the ABO alleles and RBC antigens The ABO locus produces RBC antigens by encoding glycosyltransferases which add sugars to eXisting polysaccharides on membrane glycolipid molecules These polysaccharides act as the antigen in the ABO system Figure 45 9 6 In most people the glycolipid is the H antigen Activity of the A gene product aNacetylgalactosamyl transferase converts the H antigen to the A antigen Activity of the B gene product aDgalactosyltransferase converts the H antigen to the B antigen Both enzymes are present in an IAIB individual and some H antigens will be modi ed to the A antigen While others are modi ed to the B antigen ii iv Neither enzyme is present in an ii individual and so the H antigen remains unmodi ed Production of the H antigen is controlled by a different genetic locus from the ABO enzymes Rarely an individual lacks the dominant allele H needed for H antigen production This hh genotype results in the Bombay blood type which is similar to type 0 except that Bombay blood type individuals produce antiO antibodies that are not seen in true type 0 individuals Modi cations of Dominance Relationships Complete dominance and complete recessiveness are two extremes in the range of dominance possible between pairs of alleles Many allelic pairs are less extreme in their expression showing incomplete dominance or codominance 9 In incomplete dominance a heterozygote s phenotype will be intermediate between the two possible homozygous phenotypes 6 In codominance the heterozygote shows the phenotypes of both homozygotes 0 At the molecular level these relationships between pairs of alleles depend upon patterns of gene expression Incomplete Dominance l Incomplete dominance is an allelic relationship where dominance is only partial In a heterozygote the recessive allele is not expressed The one dominant allele is unable to produce the full phenotype seen in a homozygous dominant individual The result is a new intermediate phenotype An example is plumage color in chickens Figure 47 a Crossing a truebreeding black chicken CECE with a truebreeding white one CWCW produces an Andalusian blue F1 CBCW b When the F1 interbreed the F2 include black CECE Andalusian blue CBCW and white CWCW birds in a ratio ofl 2 l c At the molecular level two copies of CB produce black while one copy is su icient to produce only the grey Andalusian blue phenotype Palomino horses goldenyellow body with nearly white mane and tail are another example When palominos are interbred the progeny are a 14 cremello cream colored with genotype C c C c b 12 palomino with genotype CCquot c 14 light chestnut with genotype CC Incomplete dominance often occurs in plants An example is ower color in snapdragons involving two alleles CR and CW Red owered plants CRCR crossed with white owered ones CWCW produce all pink progeny CRCW Codominance 1 In codominance the heterozygote s phenotype includes the phenotypes of both homozygotes Examples include a The ABO blood series in which a heterozygous IAIB individual will express both antigens resulting in blood type AB 57 The human MN blood group involves red blood cell antigens that are less important in transfusions There are three types i Type M with genotype LMLM ii Type MN with genotype LMLN iii Type N with genotype LNLN Gene Interactions and Modi ed Mendelian Ratios Phenotypes result from complex interactions of molecules under genetic control Genetic analysis can often detect the patterns of these reactions For example 32 In the dihybrid cross Aa BbAa Bb nine genotypes will result 6 If each allelic pair controls a distinct trait and exhibits complete dominance a 9 3 3 l phenotypic ratio results 0 Deviation from this ratio indicates that interaction of two or more genes is involved in producing the phenotype Two types of interactions occur 32 Different genes control the same general trait collectively producing a phenotype One gene masks the expression of others epistasis and alters the phenotype Examples here are dihybrid but in the real world larger numbers of genes are often involved in forming traits 6 The molecular explanations offered here are currently hypothetical models and await rigorous analysis using the tools of molecular biology Gene Interactions That Produce New Phenotypes Nonallelic genes that affect the same characteristic may interact to give novel phenotypes and often modi ed phenotypic ratios Examples include 32 Comb shape in chickens in uenced by two gene loci to produce four different comb types Each will breed true if parents are homozygous Figure 48 1 In a cross between a homozygous rosecombed RR pp bird and a singlecombed rr pp bird 1 The F1 Rr 1717 will all have rose combs 2 The F2 will be 3 rose RDpp 1 single rrpp Similarly pea comb rr PP is dominant over single rr 1717 with F1 rr Pp all showing pea combs and a 3 1 ratio of pea to single in the F2 iii Crossing truebreeding rose RR 1717 and pea rr PP results in Figure 49 1 An F1 with all walnut combs Rr Pp 2 An F2 showing a ratio of9 walnut RDPD 3 rose RDpp 3 pea rr PD 1 single rr pp iv These interactions t the expected ratios for a Mendelian dihybrid cross The molecular basis for each phenotype is unknown but it appears that the dominant alleles R and P each produce a factor that modi es comb shape from single to a more compleX form 6 Fruit shape in summer squash shows a 9 6 1 ratio Two genes are involved each completely dominant Interaction between the two loci produces a new phenotype Figure 410 i Long fruit aa bb are always truebreeding ii Sphereshaped fruit AD 1717 or aa BD are not always true breeding and sometimes produce long aa 1717 or diskshaped ADBD fruit iii A cross between truebreeding spherical strains AA bbaa BB produces a diskshaped F1 The F2 will be 916 diskshaped ADBD 616 spherical ADbb or aa BD and 116 long aa 1212 This modi cation of the Mendelian ratio indicates that two loci are involved iv The precise molecular basis of these phenotypes is unknown Epistasis 1 In epistasis one gene masks the eXpression of another but no new phenotype is produced a A gene that masks another is epistatic b A gene that gets masked is hypostatic Several possibilities for interaction eXist all producing modi cations in the 9 3 3 1 dihybrid ratio a Epistasis may be caused by recessive alleles so that aa masks the effect of B recessive epistasis b Epistasis may be caused by a dominant allele so thatA masks the effect of B c Epistasis may occur in both directions between genes requiring bothA and B to produce a particular phenotype duplicate recessive epistasis In recessive epistasis AD 1717 and aa 17 have the same phenotype producing an F2 ratio of 9 3 4 Examples include a Coat color determination in rodents Figures 411 and 412 i Wild mice have individual hairs with an agouti pattern bands of 57 black or brown and yellow pigment Agouti hairs are produced by a dominant allele A Mice with genotype aa do not produce yellow bands and have solidcolored hairs ii The B allele produces black pigment while 17 mice produce brown pigment TheA allele is epistatic over B and b in that it will insert bands of yellow color between either black or brown iii The C allele is responsible for development of any color at all and so it is epistatic over both the agouti A and the pigment B gene loci A mouse with genotype cc will be albino regardless of its genotype at the A and B loci iv In the cross Aa CcAa Cc the offspring will be 1 916 agouti AD CD 2 316 solid aa CD 3 416 albino 316AD cc115aa cc Coat color determination in labrador retriever dogs Figure 413 i Gene BD makes black pigment while 17 makes brown ii Another gene ED allows eXpression of the B gene while ee does not iii Genotypes and their corresponding phenotypes 1 BDED is black 2 1717 ED is brown chocolate 3 DD ee produces yellow with nose and lips either dark BD ee or pale bb ee Essential Genes and Lethal Alleles 1 Some genes are required for life essential genes and mutations in them lethal alleles may result in death Dominant lethal alleles result in death of both homozygotes and heterozygotes while recessive lethal alleles cause death only when homozygous An example is the yellow body color gene in mice Cuenot 1905 a 57 o d Yellow crossed with nonyellow results in a ratio of 1 yellow 1 nonyellow This suggests yellow is heterozygous Yellow mice never breed true another indication of heterozygosity When yellow is bred with yellow the result is about 2 yellow 1 nonyellow instead of the predicted 3 1 Castle and Little 1910 proposed that yellow homozygotes die in utero and are therefore missing from the progeny The yellow allele has a dominant effect on coat color but also acts as a recessive lethal allele Yellow is an allele of the agouti locus designatedAY Figure 418 shows the yellow 3 yellow cross i The cross is AYAlAYAl and death of the homozygous yellow animals AYAY results in a 2 1 ratio ii When two heterozygotes are crossed and produce a 2 1 ratio of progeny a recessive lethal allele is suspected 9 Molecular cloning of the agouti locus assists in analysis of these phenotypes Wildtype agouti mice eXpress the agouti gene only during hair development in the days after birth and when plucked hair is being regenerated Gene expression is seen in no other tissues and at no other time Heterozygous mice AVAl eXpress the AY allele at high levels in all tissues during all developmental stages Tissuespeci c regulation appears to be lost in the AY allele The AY allele transcript RNA is 50 longer than that of the wild type allele Al This is because ii39 l The AY allele results from deletion of an upstream sequence removing the normal promoter of the agouti gene 2 The gene is transcribed from the promoter of an upstream gene called Raly The beginning of the sequence encoding Raly is fused with the agouti gene producing a longer transcript iv Embryonic lethality of AYAY mice probably results from lack of Raly gene activity rather than from the defective agouti gene 5 Essential genes probably occur in all diploid organisms including snapdragons and Drosophila 4 Human examples of recessive lethal alleles include 32 TaySachs disease resulting from an inactive gene for the enzyme hexosaminidase Homozygous individuals develop neurological symptoms before 1 year of age and usually die within the rst 374 years of life 57 Hemophilia results from an Xlinked recessive allele and is lethal if untreated 5 A dominant lethal gene causes Huntington disease characterized by progressing central nervous system degeneration The phenotype is not eXpressed until individuals are in their 30s Dominant lethals are rare since death before reproduction would eliminate the gene from the pool Gene Expression and the Environment 1 Development of a multicellular organism from a zygote is a series of generally irreversible phenotypic changes resulting from interaction of the genome and the environment Four major processes are involved 9 Replication of genetic material Growth 57 o Differentiation of cells into types 9 Arrangement of cell types into de ned tissues and organs 2 Internal and external environments interact with the genes by controlling their expression and interacting with their products Penetrance and EXpressivity l Penetrance describes how completely the presence of an allele corresponds with the presence of a trait It depends on both the genotype eg epistatic genes and the environment of the individual Figure 419 a If all those carrying a dominant mutant allele develop the mutant phenotype the allele is completely 100 penetrant 57 If some individuals with the allele do not show the phenotype penetrance is incomplete If 80 of individuals with the gene show the trait the gene has 80 penetrance 0 Human examples include Brachydactyly involves abnormalities of the ngers and shows 50 7 80 penetrance ii Many cancer genes are thought to have low penetrance making them harder to identify and characterize 2 Expressivity describes variation in expression of a gene or genotype in individuals a Two individuals with the same mutation may develop different phenotypes due to variable expressivity of that allele b Like penetrance expressivity depends on both genotype and environment and may be constant or variable c A human example is osteogenesis imperfecta inherited as an autosomal dominant with nearly 100 penetrance i Three traits are associated with the allele 1 Blueness of the sclerae whites of eyes 2 Very fragile bones 3 Deafness Osteogenesis imperfecta shows variable expressivity because an individual with the allele may have one two or all three of its symptoms in any combination Bone fragility is also highly variable 3 Some genes have both incomplete penetrance and variable expressivity An example is neuro bromatosis Figure 420 a The allele is an autosomal dominant that shows 50780 penetrance and variable expressivity b Individuals with the allele show a wide range of phenotypes i The mildest form of the disease is a few pigmented areas on the skin caf au lait spots ii More severe cases may include 1 Neuro broma tumors of various sizes 2 High blood pressure 3 Speech impediments 4 Headaches 5 Large head 6 Short stature 7 Tumors of eye brain or spinal cord 8 Curvature of the spine 4 Incomplete penetrance and variable expressivity complicate medical genetics and genetic counseling Effects of the Environment 1 Age of onset is an effect of the individual s internal environment Different genes are expressed at different times during the life cycle and programmed activation and inactivation of genes in uences many traits Human examples include a Pattern baldness appearing in males aged 20730 years b Duchenne muscular dystrophy appearing in children aged 27 5 years 2 Sex of the individual affects the expression of some autosomal genes a Sexlimited traits appear in one sex but not the other Examples include Milk production in dairy cattle where both sexes have milk genes but only females express them ii Horn formation in some sheep species where only males express the genes used to produce horns iii Facial hair distribution in humans b Sexin uenced traits appear in both sexes but the sexes show either a difference in frequency of occurrence or an altered relationship between genotype and phenotype Human examples include i Pattern baldness controlled by an autosomal gene that is dominant in males and recessive in females Figure 421 l The genotype 17 produces pattern baldness in both men and women 2 The genotype bllf gives a nonbald phenotype in both sexes 3 The genotype be will lead to the bald phenotype in men and the nonbald phenotype in women 4 This gene shows variable expressivity in a Age of onset b Site of baldness crown or forehead c Degree ofhair loss 5 The pattern baldness gene interacts with the individual s hormonal environment and with other genes involved in hair production ii Cleft lip and palate 2 1 ratio of males to females iii Clubfoot 2 1 iv Gout 8 l v Rheumatoid arthritis 1 3 vi Osteoporosis l 3 vii Systemic lupus erythematosus l 9 Temperature may alter the activity of enzymes so that they function normally at one temperature but are nonfunctional at another An example is fur color in Himalayan rabbits Figure 422 a These white rabbits develop darker fur on the cooler parts of their bodies ears nose and paws b Since all body cells have the same genotype this fur pattern might result from environmental in uences This was tested by raising Himalayan rabbits under different temperature conditions Rabbits reared above 30 C were entirely white Rabbits raised at 25 C had the typical Himalayan phenotype Rabbits raised at 25 C with part of the body experimentally cooled to below 25 C had a dark spot on the experimentally cooled part ii39 Nature versus Nurture Phenotypes seen for many traits are in uenced by both genes and environment Some human examples a Human height has both genetic and environmental components i Genetically children tend to have about the same stature as their parents and several genetic forms of dwar sm are known achondroplasia is an example ii Environmentally diet and health care are probably responsible for the increase in human height of about 1 inch per generation over the last century b Alcoholism is an example of a behavioral trait in uenced by both genes and environment i A genetic in uence is shown in studies of adopted children Those with alcoholic biological fathers are signi cantly more likely to become alcoholics than those with nonalcoholic biological fathers Environment plays a key role also since alcoholism can develop only if alcohol is available iii Genes make individuals more or less susceptible to alcohol abuse perhaps by affecting metabolism of alcohol or development of personality traits involved in drinking but the genes alone do not produce the phenotype o Human intelligence is an example of a very complex relationship between genes and environment Genetic disorders are known to produce mental retardation Examples are PKU and Down syndrome Genes also in uence IQ among nonretarded people with adopted children scoring closer to their biological parents than to their adoptive parents Environmental in uences are seen in studies of identical twins who frequently differ in IQ scores ii39 Interactions between many genes and all aspects of the environment are involved in forming human intelligence Genes can t be changed but the environment can be altered to affect this very complex phenotype Key Concepts Chapter 6 1 Genes on nonhomologous chromosomes assort independently but genes on the same chromosome may instead be inherited together linked and belong to a linkage group Equot Classical genetics analyzes the frequency of allele recombination in progeny of genetic crosses a New associations of parental alleles are recombinants produced by genetic recombination 6 Testcrosses determine which genes are linked and a linkage map genetic map is constructed for each chromosome 0 Genetic maps are useful in recombinant DNA research and experiments dealing with genes and their anking sequences 3 Current highresolution maps include both gene markers from testcrosses and DNA markers composed of genomic regions that differ detectably between individuals Early Studies of Genetic Linkage Morgan s Experiments with Drosophila Morgan proposed that i During meiosis alleles of some genes assort together because they are near each other on the same chromosome ii Recombination occurs when genes are exchanged between the X chromosomes of the F1 females 9 A series of experiments supported Morgan s hypothesis In each case parental phenotypes were the most frequent while recombinant phenotypes occurred less frequently 9 Some relevant terminology i A chiasma plural chiasmata is the site on the homologous chromosomes where crossover occurs Crossingover is the reciprocal exchange of homologous chromatid segments involving the breaking and rejoining of DNA Crossingover is also the event leading to genetic recombination between linked genes in both prokaryotes and eukaryotes ii39 quot7 Crossingover occurs at the fourchromatid stage of prophase I in meiosis Each crossover event involves two of the four chromatids All chromatids may be involved in crossingover as chiasmata form along the aligned chromosomes Corn Experiments 3 Creighton and McClintock worked with corn Zea mays plants in which the two chromosomes under study differed cytologically The study used a corn strain heterozygous for two genes on chromosome 9 Figure 62 a One gene determines seed color C for colored seeds 0 for colorless b The other gene is involved in starch synthesis The wildtype allele Wx produces amylose and the combination of amylose and amylopectin forms normal starch in a corn seed The waxy mutant wx lacks amylose and has waxy starch containing only amylopectin In this corn strain the appearance of each chromosome 9 homolog correlated with its genotype a One chromosome 9 had the genotype 0 Wx and a normal appearance b Its homolog had the genotype C wx and cytological markers at each end of the chromosome The end near the C locus had a darkly staining knob and the other end nearer the wx locus had a translocated piece of chromosome 8 When testcrossed recombinant phenotypes were evident and could be correlated with cytological features a Whenever the genes had recombined the cytological features had also recombined b In the parental nonrecombinant type progeny no exchange of cytological markers was evident 5 This was direct evidence of physical exchange between homologs resulting in genetic recombination Constructing Genetic Maps 1 Genetic recombination experiments can be used in genetic or linkage mapping Detecting Linkage through Testcrosses l Linked genes are used for mapping They are found by looking for deviation from the frequencies expected from independent assortment 2 A testcross one parent is homozygous recessive works well for analyzing linkage a If the alleles are not linked and the second parent is heterozygous all four possible combinations of traits will be present in equal numbers in the progeny b A signi cant deviation in this ratio more parental and fewer recombinant types indicates linkage 3 Chisquare analysis is used to analyze testcross data and determine whether a deviation is signi cant A null hypothesis the genes are not linked is used because it is not possible to predict phenotype frequencies produced by linked genes The Concept Ufa Genetic Map 1 In an individual heterozygous at two loci there are two arrangements of alleles a The cis coupling arrangement has both wildtype alleles on one homologous chromosome and both mutants on the other eg w ml and w m 57 The trans repulsion arrangement has one mutant and one wild type on each homolog eg wlm and w ml 0 A crossover between homologs in the cis arrangement results in a homologous pair with the trans arrangement A crossover between homologs in the trans arrangement results in cis homologs Gene Mapping with Two Point Testcrosses With autosomal recessive alleles when a double heterozygote is testcrossed four phenotypic classes are expected If the genes are linked the two parental phenotypes will be about equally frequent and more abundant than the two recombinant phenotypes Figure 64 2 Mapping of genes with other mechanisms of inheritance is also done with twopoint testcrosses 9 For Xlinked recessives a female double heterozygote al bla b is crossed with a male hemizygous for the recessive alleles a bY 57 For either Xlinked case it is possible to cross the females with males of any type As long as only male progeny are analyzed the father s X will be irrelevant 0 Phenotypes obtained in any of these crosses will depend on whether the alleles are arranged in coupling cis or repulsion trans F Recombination frequency is used directly as an estimate of map units i The measure is more accurate when the alleles are close together ii Scoring large numbers of progeny increases the accuracy 9 Mapping in all types of organisms shows genes arranged with a l l correspondence between linkage groups and chromosomes Generating a Genetic Map l V39 A genetic map is generated from estimating the crossover rate in a particular segment of the chromosome It may not exactly match the physical map because crossover is not equally probable at all sites on the chromosome Recombination frequency is also used to predict progeny in genetic crosses For example a 20 crossover rate between two pairs of alleles in a heterozygote albla b will give 10 gametes of each recombinant type all and a 17 A recombination frequency of 50 means that genes are unlinked There are two ways in which genes may be unlinked a They may be on separate chromosomes b They may be far apart on the same chromosome If the genes are on the same chromosome multiple crossovers can occur The further apart two loci are the more likely they are to have crossover events take place between them The chromatid pairing is not always the same in crossover so that 2 3 or 4 chromatids may participate in multiple crossover Figure 65 To determine whether the genes are on the same chromosome or on different ones other genes in the linkage group may be mapped in relation to a and b and used to deduce their locations Gene Mapping Using Three Point Testcrosses Typically geneticists design experiments to gather data on several traits in one testcross An example of a threepoint testcross would be 4 p rprjprjprjFigure66 In the progeny each gene has two possible phenotypes For three genes there are 238 expected phenotypic classes in the progeny Establishing the Order of Genes 1 The order of genes on the chromosome can be deduced from results of the cross Of the eight expected progeny phenotypes a Two classes are parental Q7 r j p r j andp r j p r j and will be the most abundant 57 Of the siX remaining phenotypic classes two will be present at the lowest frequency resulting from apparent double crossover 17 ffp r j andp r fp r j This establishes the gene order asp j r Figure 67 Calculating the Recombination Frequencies for Genes Cross data is organized to re ect the gene order and in this example the region between genes 17 andj is called region I and that betweenj and r is region 11 Figure 68 Recombination frequencies are now calculated for two genes at a time It includes single crossovers in the region under study and double crossovers since they occur in both regions Recombination frequencies are used to position genes on the genetic map each 1 recombination frequency1 map unit for the chromosomal region Figure 69 Recombination frequencies are not identical to crossover frequencies and typically underestimate the true map distance KEY CONCEPTS CHAPTER 8 Variations in Chromosome Structure Mutations involving changes in chromosome structure occur in four common types a Deletions ST Duplications 0 Inversions changing orientation of a DNA segment 9 Translocations moving a DNA segment All chromosome structure mutations begin with a break in the DNA leaving ends that are not protected by telomeres but are sticky and may adhere to other broken ends Polytene chromosomes bundles of chromatids produced by DNA synthesis without mitosis or meiosis are useful for studying chromosome structure mutations Figure 81 a Polytene chromosomes are easily detectable microscopically b Homologs are tightly paired joined at the centromeres by a proteinaceous chromocenter c Detailed banding patterns are characterized for the four polytene chromosomes with each band averaging 30kb of DNA enough to encode several genes Deletion 1 In a deletion part of a chromosome is missing a Deletions start with chromosomal breaks induced by Heat or radiation especially ionizing ii Viruses iii Chemicals iv Transposable elements V Errors in recombination b Deletions do not revert because the DNA is missing 2 The effect of a deletion depends on what was deleted a A deletion in one allele ofa homozygous wildtype organism may give a normal phenotype while the same deletion in the wildtype allele of a heterozygote would produce a mutant phenotype ST Deletion of the centromere results in an acentric chromosome that is lost usually with serious or lethal consequences No known living human has an entire autosome deleted from the genome o Large deletions can be detected by unpaired loops seen in karyotype analysis Figure 82 Human disorders caused by large chromosomal deletions are generally seen in heterozygotes since homozygotes usually die a The number of gene copies is important b Syndromes result from the loss of several to many genes 3 Examples of human disorders caused by large chromosomal deletions a Criduchat cry of the cat syndrome OMIM 123450 resulting from deletion of part of the short arm of chromosome 5 Figure 84 The deletion results in severe mental retardation and physical abnormalities 57 PraderWilli syndrome OMIM 176270 occurring in heterozygotes with part of the long arm of one chromosome 15 homolog deleted The deletion results in feeding dif culties poor male sexual development behavioral problems and mental retardation Duplication l Duplications result from doubling of chromosomal segments and occur in a range of sizes and locations Figure 85 a Tandem duplications are adjacent to each other b Reverse tandem duplications result in genes arranged in the opposite order of the original c Tandem duplication at the end of a chromosome is a terminal tandem duplication Figure 86 d Heterozygous duplications result in unpaired loops and may be detected cytologically 2 An example is the Drosophila eye shape allele Bar that reduces the number of eye facets giving the eye a slitlike rather than oval appearance Figure 87 a The Bar allele resembles an incompletely dominant mutation i Females heterozygous for Bar have a kidneyshaped eye that is larger and more faceted than that in a female homozygous for Bar ii Males hemizygous for Bar have slitlike eyes like those of a BarBar female 6 Cytological examination of polytene chromosomes showed that the Bar allele results from duplication of a small segment 16A of the X chromosome Inversion Inversion results when a chromosome segment excises and reintegrates oriented 180 from the original orientation There are two types Figure a Pericentric inversions include the centromere b Paracentric inversions do not include the centromere 2 Inversions generally do not result in lost DNA but phenotypes can arise if the breakpoints are in genes or regulatory regions 3 Linked genes are often inverted together The meiotic consequence depends on whether the inversion occurs in a homozygote or a heterozygote a A homozygote will have normal meiosis b The effect in a heterozygote depends on whether crossingover occurs i If there is no crossingover no meiotic problems occur ii If crossingover occurs in the inversion unequal crossover may produce serious genetic consequences 4 Different recombinant chromosomes are produced by crossover in a heterozygote depending on centromere involvement a Paracentric inversions no centromere result in visible inversion loops between homologous chromosomes Figure 89 b Pericentric inversions that undergo a single crossover will result in i Two viable gametes one with genes in normal order the other with the inversion ii Two inviable gametes each with some genes deleted and others duplicated Translocation 1 A change in location of a chromosome segment is a translocation No DNA is lost or gained Simple translocations are of two types Figure 811 9 ST Intrachromosomal with a change of position within the same chromosome Interchromosomal with transfer of the segment to a nonhomologous chromosome i If a segment is transferred from one chromosome to another it is nonreciprocal ii If segments are exchanged it is reciprocal 2 Gamete formation is affected by translocations a 57 0 F In homozygotes with the same translocation on both chromosomes altered gene linkage is seen Gametes produced with chromosomal translocations often have unbalanced duplications andor deletions and are inviable or produce disorders such as familial Down syndrome Strains that are homozygous for a reciprocal translocation form normal gametes Strains that are heterozygous for a reciprocal translocation must pair a set of normal chromosomes N with a set of translocated ones T Chromosomal Mutations and Human Tumors 1 Most human malignant tumors have chromosomal mutations a 6 s o The most common are translocations There is much variation in chromosome abnormalities however and they include simple rearrangements to compleX changes in chromosome structure and number Many tumor types show a variety of mutations Some however are associated with speci c chromosomal abnormalities 2 Examples of speci c abnormalities associated with tumors a Chronic myelogenous leukemia CML OMIM 151410 involves a reciprocal translocation of chromosomes 9 and 22 Figure 813 i Myeloblasts stem cells of white blood cells replicate uncontrollably ii 90 of CML patients have the Philadelphia chromosome Phl reciprocal translocation iii The reciprocal translocation causes transition from a differentiated cell to a tumor cell by translocating a protooncogene from chromosome 9 to chromosome 22 and probably converting it to the ABL oncogene iv The hybrid gene arrangement causes eXpression of a leukemia producing gene product 57 Burkitt lymphoma BL involves a reciprocal translocation of chromosomes 8 and 14 i Induced by a virus this disease is common in Africa ii B cells are affected and secrete antibodies as they proliferate iii The reciprocal translocation positions the MY C protooncogene neXt to an active immunoglobulin gene resulting in over eXpression of MY C and development of the lymphoma Fragile Sites and Fragile X Syndrome l Chromosomes in cultured human cells develop narrowings or unstained areas gaps called fragile sites over 40 human fragile sites are known A wellknown example is fragile X syndrome in which a region at position Xq273 is prone to breakage and mental retardation may result Figure 814 a Fragile X syndrome has an incidence in the United States of about ll250 in males and l2500 in females heterozygotes Figure 815 6 Inheritance follows Mendelian patterns but only 80 of males with a fragile X chromosome are mentally retarded The 20 with fragile X chromosome but a normal phenotype are called normal transmitting males i A normal transmitting male can pass the chromosome to his daughters ii Sons of those daughters frequently show mental retardation o About 33 of carrier heterozygous females show mild mental retardation i Sons of carrier females have a 50 chance of inheriting the fragile X ii Daughters of carrier females have a 50 chance of being carriers 9 Molecular analysis shows a repeated 3bp sequence CGG in the FMR I fragile X mental retardationl gene at the fragile X site 1 Normal individuals have 6 to 54 CGG repeats with an average of ii Normal transmitting carrier males their daughters and some other carrier females have 55 to 200 copies but do not show symptoms 2 iii Individuals with fragile X syndrome have 200 to 1300 copies indicating that tandem ampli cation of this sequence is tolerated until a threshold number of copies is reached iv Ampli cation of CGG repeats occurs only in females perhaps during a slipped mispairing process during DNA replication V The FMR I product is an RNAbinding protein The triplet repeat eXpansion in FMR I affects eXpression rather than protein coding resulting in loss of gene activity Variations in Chromosome Number 1 An organism or cell is euploid when it has one complete set of chromosomes or exact multiples of complete sets Eukaryotes that are normally haploid or diploid are euploid as are organisms with variable numbers of chromosome sets Aneuploidy results from variations in the number of individual chromosomes not sets so that the chromosome number is not an exact multiple of the haploid set of chromosomes Changes in One or a Few Chromosomes Aneuploidy can occur due to nondisjunction during meiosis 9 Nondisjunction during meiosis I will produce four gametes two with a chromosome duplicated and two that are missing that chromosome Fusion of a normal gamete with one containing a chromosomal duplication will produce a zygote with three copies of that chromosome and two of all others Fusion of a normal gamete with one missing a chromosome will result in a zygote with only one copy of that chromosome and two of all others 6 Nondisjunction during meiosis 11 produces two normal gametes and two that are abnormal one with two sibling chromosomes and one with that chromosome missing Fusion of abnormal gametes with normal ones will produce the genotypes discussed above ii Normal gametes are also produced and when fertilized will produce normal zygotes 0 More compleX gametic chromosome composition can result when i One chromosome is involved ii Nondisjunction occurs in both meiotic divisions iii Nondisjunction occurs in mitosis result is somatic cells with unusual chromosome complements Autosomal aneuploidy is not well tolerated in animals and in mammals is detected mainly after spontaneous abortion Aneuploidy is much better tolerated in plants Human examples of aneuploidy in autosomes and seX chromosomes are summarized in Table 81 a 6 0 9 Sex chromosome aneuploidy is found more often than is autosome aneuploidy because 39 39 39 for 39 dosage 1 Autosomal monosomies are rarely found in humans presumably because they are lost early in pregnancy Autosomal trisomies account for about half of fetal deaths and only a few are seen in live births Most trisomy8 13 and 18 result in early death with only trisomy21 Down syndrome surviving to adulthood T1isomy21 occurs in an estimated 3510 106 conceptions and 1430106 births Figure 818 i Down syndrome OMIM 190685 individuals are characterized by 1 Low IQ 2 Epicanthal folds over eyes 3 Short and broad hands 4 Belowaverage height Table 82 shows the correlation of maternal age and probability of t1isomy21 1 A female fetus before birth produces primary oocytes in her ovaries that stop their development at prophase I of meiosis 2 After puberty secondary oocytes develop from the primary ones entering the second meiotic division but again arresting this time at metaphase II If fertilization occurs the second meiotic division is completed The probability of nondisjunction increases with the length of time the primary oocyte is in the ovary 5 Amniocentesis or chorionic villus sampling can deter mine whether the fetus has a normal complement of chromosomes Additional risks for Down syndrome include 2 Increased paternal age Smoking in mothers who have an error in meiosis 11 especially if they use oral contraceptives oral contraceptives alone do not increase risk iv Robertsonian translocation centric fusion produces three copies of the long arm of chromosome 21 resulting in familial Down syndrome Figure 819 1 In this nonreciprocal translocation two nonhomologous acrocentric centromeres near end chromosomes break at centromeres a Both long arms become attached to the same centromere creating a chromosome with the long arm of chromosome 21 and the long arm of chromosome 14 or 15 b V The short arms also fuse forming a reciprocal product that is usually lost within a few cell divisions c The heterozygous carrier of this chromosome is phenotypically normal since the two copies of each major chromosome arm supply two copies of all essential genes d Trisomy13 Patau syndrome occurs in 2104 live births and most die within the rst three months Characteristics include Figure 821 Cleft lip and palate ii Small eyes iii Polydactyly eXtra ngers and toes iv Mental and developmental retardation v Cardiac and other abnormalities quot7 Trisomy18 Edwards syndrome occurs in 25 104 live births and 90 die within 6 months About 80 of Edwards syndrome infants are female Characteristics include Figure 822 Small size with multiple congenital malformations throughout the body Clenched sts iii Elongated skull iv Lowset ears v Mental and developmental retardation Chapter 10 Key Concepts The Search for the Genetic Material 1 Some substance must be responsible for passage of traits from parents to offspring For a substance to do this it must be a Stable enough to store information for long periods b Able to replicate accurately c Capable of change to allow evolution In the early 1900s chromosomes were shown to be the carriers of hereditary information In eukaryotes they are composed of both DNA and protein and most scientists initially believed that protein must be the genetic material Grif th s Transformation Experiment Frederick G1iffith s 1928 with i I I bacteria in mice showed that something passed from dead bacteria into nearby living ones allowing them to change their cell surface Figures 101 and 102 He called this agent the transforming principle but did not know what it was or how it worked Avery s Transformation Experiments 1 5 In 1944 Avery MacLeod and McCarty published results of a study that identi ed the transforming principle from S pneumoniae Their approach was to break open dead cells chemically separate the components eg protein nucleic acids and determine which was capable of transforming live S pneumoniae cells Figure 103 Only the nucleic acid fraction was capable of transforming the bacteria Critics noted that the nucleic acid fraction was contaminated with proteins The researchers treated this fraction with either RNase or protease and still found transforming activity but when it was treated with DNase no transformation occurred indicating that the transforming principle was The Hershey Chase Bacteriophage Experiments 1 In 1953 more evidence for DNA as the genetic material resulted from Alfred Hershey and Martha Chase s work on E coli infected with bacteriophage T2 In one part of the experiment T2 proteins were labeled with 35S and in the other part T2 DNA was labeled with 32R Then each group of labeled viruses was mixed separately with the E coli host After a short time phage attachment was disrupted with a kitchen blender and the location of the label determined Figure 104 The 35Slabeled protein was found outside the infected cells while the 32Plabeled DNA was inside the E coli indicating that DNA carried the information needed for viral infection This provided additional support for the idea that genetic inheritance occurs via DNA The Composition and Structure of DNA and RNA 1 2 DNA and RNA are polymers composed of monomers called nucleotides Each nucleotide has three parts a A pentose 5carbon sugar b A nitrogenous base V39 c A phosphate group The pentose sugar in RNA is ribose and in DNA it is deoxyribose The only difference is at the 22 position where RNA has a hydroxyl OH group while DNA has only a hydrogen Figure 106 There are two classes of nitrogenous bases Figure 107 a Purines doublering ninemembered structures include adenine A and guanine G b Pyrimidines onering siXmembered structures include cytosine C thymine T in DNA and uracil U in RNA The structure of nucleotides has these features a The base is always attached by a covalent bond between the 12 carbon of the pentose sugar and a nitrogen in the base speci cally the 9 nitrogen in purines and the 1 nitrogen in pyrimidines ST The sugarbase combination is a nucleoside When a phosphate is added always to the 52 carbon of the pentose sugar it becomes a nucleoside phosphate or simply nucleotide Nucleotide naming conventions are given in Table 101 0 Polynucleotides of both DNA and RNA are formed by stable covalent bonds phosphodiester linkages between the phosphate group on the 52 carbon of one nucleotide and the 32 hydroxyl on another nucleotide Figure 108 This creates the backbone of a nucleic acid molecule The asymmetry of phosphodiester bonds creates 3 27 52 polarity within the nucleic acid chain The DNA Double Helix 1 James Watson and Francis Crick published the famous doubleheliX structure in 1953 Figure 109 When they began their work it was known that DNA is composed of nucleotides but how the nucleotides are assembled into nucleic acid was unknown Two additional sources of data assisted Watson and Crick with their model Erwin Chargaff s ratios obtained for DNA derived from a variety of sources showed that the amount of purine always equals the amount of pyrimidine and further that the amount of G equals C and the amount ofA equals T Table 102 Rosalind Franklin s Xray diffraction images of DNA showed a helical structure with regularities at 034nm and 34nm along the aXis of the molecule Figure 1010 m 57 Watson and Crick s threedimensional model Figure 1011 has these main features a It is two polynucleotide chains wound around each other in a right handed heliX b The two chains are antiparallel E 0 Q The sugarphosphate backbones are on the outside of the heliX and the bases are on the inside stacked perpendicularly to the long aXis like the steps of a spiral staircase The bases of the two strands are held together by hydrogen bonds between complementary bases two for A T pairs and three for GO pairs Individual Hbonds are relatively weak and so the strands can be separated by heating for example Complementary base pairing means that the sequence of one strand dictates the sequence of the other strand Figure 1012 The base pairs are 034nm apart and one full turn of the DNA heliX takes 34nm so there are 10bp in a complete turn The diameter of a dsDNA heliX is 2nm Because of the way the bases Hbond with each other the opposite sugarphosphate backbones are not equally spaced resulting in a major and minor groove This feature of DNA structure is important for protein binding 9 quot7 The 1962 Nobel Prize in Physiology or Medicine was awarded to Francis Crick James Watson and Maurice Wilkins the head of the lab in which Franklin worked Franklin had died and so was not eligible posthumously Different DNA Structures 1 Xray diffraction studies show that DNA can eXist in different forms Figure 1013 a ADNA is the dehydrated form and so it is not usually found in cells It is a righthanded heliX with 109bptum with the bases inclined 13 from the heliX aXis ADNA has a deep and narrow major groove and a wide and shallow minor groove BDNA is the hydrated form of DNA the kind normally found in cells It is also a righthanded heliX with only 100bptum and the bases inclined only 2 from the heliX aXis BDNA has a wide major groove and a narrow minor groove and its major and minor grooves are of about the same depth 57 ZDNA is a lefthanded heliX with a zigzag sugarphosphate backbone that gives it its name It has 120bptum with the bases inclined 88 from the heliX aXis ZDNA has a deep minor groove and a very shallow major groove Its eXistence in living cells has not been proven 0 DNA in the Cell 1 All known cellular DNA is in the B form ADNA would not be eXpected because it is dehydrated and cells are aqueous Some organisms show evidence of ZDNA but its physiological role if any is unknown RNA Structure RNA structure is very similar to that of DNA a It is a polymer of ribonucleotides the sugar is ribose rather than deoxyribose 6 Three of its bases are the same A G and C While it contains U rather than T 2 RNA is singlestranded but internal base pairing can produce secondary structure in the molecule 3 Some viruses use RNA for their genomes In some it is dsRNA While in others it is ssRNA Doublestranded RNA is structurally very similar to dsDNA The Organization of DNA in Chromosomes 1 Cellular DNA is organized into chromosomes A genome is the chromosome or set of chromosomes that contains all the DNA of an organism 2 In prokaryotes the genome is usually a single circular chromosome In eukaryotes the genome is one complete haploid set of nuclear 39 39 39 ial and 39 39 l 39 DNA are not included Prokaryotic Chromosomes l The typical prokaryotic genome is one circular dsDNA chromosome but some prokaryotes are more exotic with a main chromosome and one or more smaller ones When a minor chromosome is dispensable to the life of the cell it is called aplasmid Some examples a Borrelia burgdorferi Lyme disease in humans has a 09lMb linear chromosome plus an additional 053Mb of DNA in 17 different linear and circular molecules Eukaryotic Chromosomes l The genome of most prokaryotes consists of one chromosome While most eukaryotes have a diploid number of chromosomes 2 A genome is the information in one complete haploid chromosome set The total amount of DNA in the haploid genome of a species is its C value Table 104 The structural complexity and the C value of an organism are not related creating the C value paradox 3 The form of eukaryotic chromosomes changes through the cell cycle 9 In G1 each chromosome is a single structure 57 In S chromosomes duplicate into sister chromatids but remain joined at centromeres through G2 c At M phase sister chromatids separate into daughter chromosomes 4 In G1 eukaryotic chromosomes are linear dsDNA and contain about twice as much protein as DNA by weight The DNAprotein complex is called chromatin and it is highly conserved in all eukaryotes Chromatin Structure 1 Both histones and nonhistones are involved in physical structure of the chromosome 2 Histones are abundant small proteins with a net charge The ve main types are H1 H2A H2B H3 and H4 By weight chromosomes have equal amounts of DNA and histones 3 Histones are highly conserved between species H1 less than the others 4 Histones organize DNA condensing it and preparing it for further condensation by nonhistone proteins This compaction is necessary to t large amounts of DNA 2m65ft in humans into the nucleus of a cell 5 Nonhistone is a general name for other proteins associated with DNA This is a big group with some structural proteins and some that bind only transiently Nonhistone proteins vary widely even in different cells from the same organism Most have a net D charge and bind by attaching to histones 6 Chromatin formation involves histones and condenses the DNA so it will t into the cell making a 10nm ber Figure 1020 Chromatin formation has two components a Two molecules each of histones H2A H2B H3 and H4 associate to form a nucleosome core and DNA wraps around it 134 times for a 7 fold condensation factor Nucleosome cores are about 11 nm in diameter 57 H1 further condenses the DNA by connecting nucleosomes to create chromatin with a diameter of 30nm for an additional 6fold condensation The solenoid model proposes that the nucleosomes form a spiral with 6 nucleosomes per turn Figures 1021 and 1022 7 Beyond the 30nm lament stage electron microscopy shows 30790 loops of DNA attached to a protein scaffold Figure 1023 Each loop is 1807 300 nucleosomes of the 30nm ber SARs scaffoldassociated regions bind nonhistone proteins to form loops that radiate out in spiral fashion Figure 1024 8 Fully condensed chromosome is 10000fold shorter and 400fold thicker than DNA alone Euchromatin and Heterochromatirt 1 The cell cycle affects DNA packing with DNA condensing for mitosis and meiosis and decondensing during interphase being most dispersed at S phase Chromosomes are most condensed at metaphase when the looped domains are further coiled and the chromatin has a diameter of about 700nm Nonhistone proteins form the scaffold for this additional condensation 2 Staining of chromatin reveals two forms a Euchromatin condenses and decondenses with the cell cycle It is actively transcribed and lacks repetitive sequences Euchromatin accounts for most of the genome in active cells u u b I 39 39 remains 4 4 the cell cycle It replicates later than euchromatin and is transcriptionally inactive There are two types based on activity 39 Constituitive heterochromatin occurs at the same sites in both homologous chromosomes of a pair and consists mostly of repetitive DNA eg centromeres Facultative heterochromatin varies between cell types or developmental stages or even between homologous chromosomes It contains condensed and thus inactive euchromatin eg Barr bodies Centromeric and Telomeric DNA Centromeres and telomeres are eukaryotic chromosomal regions with special functions 2 Centromeres are the site of the kinetochore where spindle bers attach during mitosis and meiosis They are required for accurate segregation of chromatids 3 Yeast Saccharomyces cerevisiae centromeres are wellstudied Called CEN regions their sequence and organization are similar but not identical between the chromosomes Other eukaryotes have different centromere sequences so while function is conserved it is not due to a single type of DNA sequence Figure 1025 4 Proteins interact with the centromere and the spindle microtubule to form the kinetochore structure Figure 1026 V39 Telomeres are needed for chromosomal replication and stability Generally composed of heterochromatin they interact with both the nuclear envelope and each other All telomeres in a species have the same sequence a Simple telomeric sequences are short speciesspeci c and tandemly repeated Examples Tetrahymena is 52 TTGGGG 32 and human is 52 TTAGGG 32 A new model suggests that the singlestranded end of the chromosome folds back to form a tloop and then invades the doublestranded region to form a Dloop Figure 1027 57 Telomereassociated sequences are internal to the simple telomeric sequences These compleX repetitive sequences may eXtend many kb into the chromosome 0 Drosophila is unusual in having telomeres composed of transposons rather than the short repeats seen in most eukaryotes Unique Sequence and Repetitive Sequence DNA 1 Sequences vary widely in how often they occur within a genome The categories are a Uniquesequence DNA present in one or a few copies b Moderately repetitive DNA present in a few to 105 copies c Highly repetitive DNA present in about 1057107 copies 2 Prokaryotes have mostly uniquesequence DNA with repeats only of sequences like rRNAs and tRNAs Eukaryotes have a mix of unique and repetitive sequences 3 Uniquesequence DNA includes most of the genes that encode proteins as well as other chromosomal regions Human DNA contains about 65 unique sequences 4 Repetitivesequence DNA includes the moderately and highly repeated sequences They may be dispersed throughout the genome or clustered in tandem repeats V39 Dispersed repetitive sequences occur in families that have a characteristic sequence O en the same few sequences are highly repeated and comprise most of the dispersed repeats in the genome Little is known of their function or indeed whether they actually serve a function There are two types of interspersion patterns found in all eukaryotic organisms 9 LINEs long interspersed repeated sequences with sequences of 5 kb or more The common example in mammals is LINEl with sequences up to 7kb in length that can act as transposons ST SINEs short interspersed repeated sequences with sequences of 1007 500bp An example is the Alu repeats found in some primates including humans where these repeats of 2007300bp make up 9 of the genome SINEs are also transposons but are dependent on LINES for transposase genes 6 Tandemly repetitive sequences are common in eukaryotic genomes ranging from very short sequences 17 lObp to genes and even longer sequences This group includes centromere and telomere sequences and rRNA and tRNA genes KEY CONCEPTS CHAPTER 11 Semiconservative DNA Replication 1 Watson and Crick DNA model implies a mechanism for replication a Unwind the DNA molecule b Separate the two strands c Make a complementary copy for each strand 2 Three possible models were proposed for DNA replication a Conservative model proposed both strands of one copy would be entirely old DNA while the other copy would have both strands of new DNA Dispersive model was that dsDNA might fragment replicate dsDNA and then reassemble creating a mosaic of old and new dsDNA regions in each new chromosome 6 c Semiconservative model is that DNA strands separate and a complementary strand is synthesized for each so that sibling chromatids have one old and one new strand This model was the winner in the Meselson and Stahl experiment Figure 111 The Meselson Stahl Experiment 1 Meselson and Stahl 1958 grew E coli in a heavy not radioactive isotope of nitrogen 15N in the form oflsNH4 Cl Because it is heavier DNA containing 15N is more dense than DNA with normal 14N and so can be separated by CsCl density gradient centrifugation Once the E coli were labeled with heavy 15N the researchers shifted the cells to medium containing normal 14N and took samples at time points DNA was eXtracted from each sample and analyzed in CsCl density gradients Figure 112 After one replication cycle in normal 14N medium all DNA had density intermediate between heavy and normal After two replication cycles there were two bands in the density gradient one at the intermediate position and one at the position for DNA containing entirely 14N Results compared with the three proposed models a Does not t conservative model because after one generation there is a single intermediate band rather than one with entirely 15N DNA and another with entirely 14N DNA 57 The dispersive model predicted that a single band of DNA of intermediate density would be present in each generation gradually becoming less dense as increasing amounts of 14N were incorporated with each round of replication Instead Meselson and Stahl observed two bands of DNA with the intermediate form decreasing over time c The semiconservative model ts the data very well Roles of DNA Polymerases All DNA polymerases link dNTPs into DNA chains Figure 114 Main features of the reaction a An incoming nucleotide is attached by its 52phosphate group to the 32OH of the growing DNA chain Energy comes from the release of two phosphates from the dNTP The DNA chain acts as a primer for the reaction 6 The incoming nucleotide is selected by its ability to hydrogen bond with the complementary base in the template strand The process is fast and accurate c DNA polymerases synthesize only from 52 to 32 The enzyme Kornberg isolated was believed to be the only DNA polymerase in E 001139 However mutations in this gene pol11 were not V39 lethal indicating that other DNA polymerases must eXist in E coli Temperaturesensitive mutants are used to study essential genes At 37 C pol19x1 strains are normal and the protein has normal activity in vitro At 42 C however the protein lacks 5232 exonuclease activity and bacterial cells with this mutation are dead Additional DNA polymerases have been isolated including DNA polymerase II 1970 DNA polymerase III 1971 DNA polymerase IV and DNA polymerase V The properties of known E coli DNA polymerases are m DNA polymerase Iis a single peptide encoded by MM and used for DNA replication Replicates DNA in the 52D 32 direction Has 52D 32 exonuclease activity to remove nucleotides from 52 end of DNA or from an RNA primer 57 DNA polymerase II is a single peptide encoded by 17013 Used for DNA repair 0 DNA polymerase III has three polypeptide subunits in the catalytic core of the enzyme lt encoded by the dnaE gene 2 dnaQ and k holE Holoenzyme has an additional siX different polypeptides Replicates DNA in the 52 D 32 direction DNA polymerase IV is encoded by the dinB gene and is used in DNA repair F 9 DNA polymerase V is encoded by umuDC and is used in DNA repair E coli DNA polymerases used for DNA replication DNA polymerase I and DNA polymerase 111 have 32 D 52 exonuclease proofreading activity Molecular Model of DNA Replication 1 Table 111 shows key genes and DNA sequences involved in replication Initiation of Replication Replication starts when DNA at the origin of replication denatures to eXpose the bases creating a replication fork Replication is usually bidirectional from the origin E 001139 has one origin oriC which has a A minimal sequence of about 245 bp required for initiation b Three copies ofa 13bp ATrich sequence c Four copies ofa 9bp sequence Events in E coli initiating DNA synthesis derived from in vitro studies Figure 115 a Initiator proteins attach E coli s initiator protein is DnaA from the dnaA gene DNA helicase from dnaB binds initiator proteins on the DNA and denatures the ATrich region using ATP as an energy source 57 o DNA primase from dnaG binds helicase to form a primosome which synthesizes a short 5 710nt RNA primer Semidiscontinuous DNA Replication 1 5 V39 9 gt1 When DNA denatures at the ori replication forks are formed DNA replication is usually bidirectional but will consider events at just one replication fork Figure 116 9 helicase preventing reannealing ST Primase synthesizes a primer on each template strand DNA polymerase III adds nucleotides to the 32 end of the primer synthesizing a new strand complementary to the template and displacing the SSBs DNA is made in opposite directions on the two template strands 0 New strand made 52 D 32 in the same direction as movement of the replication fork is leading strand while new strand made in the opposite direction is lagging strand Leading strand needs only one primer while lagging needs a series of primers 9 Helicase denaturing DNA causes tighter winding in other parts of the circular chromosome Gyrase relieves this tension Leading strand is synthesized continuously while lagging strand is synthesized discontinuously in the form of Okazaki fragments DNA replication is therefore semidiscontinuous Each fragment requires a primer to begin and is eXtended by DNA polymerase III Okazaki data show that these fragments are gradually joined together to make a fulllength dsDNA chromosome DNA polymerase I uses the 32 OH of the adjacent DNA fragment as a primer and simultaneously removes the RNA primer while resynthesizing the primer region in the form of DNA The nick remaining between the two fragments is sealed with DNA ligase Figure 117 Key proteins are associated to form a replisome Template DNA probably bends to allow synthesis of both leading and lagging strands at the replication fork Figure 118 In bidirectional replication the two replisomes are mirror images moving away from each another Figure 119 Replication of Circular DNA and the Supercoiling Problem 1 Some circular chromosomes eg E coll are circular throughout Singlestrand DNAbinding proteins SSBs bind the ssDNA formed by replication creating a thetalike k shape As the strands separate on one side of the circle positive supercoils form elsewhere in the molecule Replication fork moves about 500ntsecond so at 10bpturn replication fork rotates at 3000rpm 2 Topoisomerases relieve the supercoils allowing the DNA strands to continue separating as the replication forks advance Rolling Circle Replication 1 Another model for replication is the rolling circle Figure 1110 which is used by several bacteriophages including X174 after a complement is made for the genomic ssDNA and 1 after circularization by base pairing between the sticky ssDNA cos ends 2 Rolling circle replication begins with a nick singlestranded break at the origin of replication The 52 end is displaced from the strand and the 32 end acts as a primer for DNA polymerase III which synthesizes a continuous strand using the intact DNA molecule as a template 3 The 52 end continues to be displaced as the circle rolls and is protected by SSBs until discontinuous DNA synthesis makes it a dsDNA molecule again 4 A DNA molecule many genomes in length can be made by rolling circle replication During viral assembly it is cut into individual viral chromosomes and packaged into phage head 5 Bacteriophage L regardless of whether entering the lytic or lysogenic pathway 39 39 39 its 39 39 quot 39 after infection Figure 11 11 a In a lysogenic infection the circular DNA integrates into a speci c site in the E coli chromosome by a crossover event 6 In a lytic infection rolling circle replication produces a long concatamer of one DNA and then a viral endonuclease product of the ter gene recognizes the cos sites and makes the staggered cuts that generate the 12nt sticky ends The linear chromosomes are then used to assemble new virus particles DNA Replication in Eukaryotes 1 DNA replication is very similar in both prokaryotes and eukaryotes except that eukaryotes have more than one chromosome Replicons 1 Eukaryotic chromosomes generally contain much more DNA than those of prokaryotes and their replication forks move much more slowly If they were like typical prokaryotes with only one origin of l quot quot per 39 DNA 1 quot would take many days 2 Instead eukaryotic chromosomes contain multiple origins at which DNA denatures and replication then proceeds bidirectionally until an adjacent replication fork is encountered The DNA replicated from a single origin is called a replicon or replication unit Figure 1112 In eukaryotes replicon size is smaller than it is in prokaryotes replication is slower and each chromosome contains many replicons Number and size of replicons vary with cell type Not all origins within a genome initiate DNA synthesis simultaneously Cellspeci c patterns of origin activation are observed so that chromosomal regions are replicated in a predictable order in each cell cycle Figure 1113 Roles of DNA Polymerases Equot V39 All DNA polymerases link dNTPs into DNA chains Figure 114 Main features of the reaction a An incoming nucleotide is attached by its 52phosphate group to the 32OH of the growing DNA chain Energy comes from the release of two phosphates from the dNTP The DNA chain acts as a primer for the reaction 6 The incoming nucleotide is selected by its ability to hydrogen bond with the complementary base in the template strand The process is fast and accurate c DNA polymerases synthesize only from 52 to 32 The enzyme Kornberg isolated was believed to be the only DNA polymerase in E 001139 However mutations in this gene pol11 were not lethal indicating that other DNA polymerases must eXist in E coli Temperaturesensitive mutants are used to study essential genes At 37 C pol19x1 strains are normal and the protein has normal activity in vitro At 42 C however the protein lacks 5232 exonuclease activity and bacterial cells with this mutation are dead Additional DNA polymerases have been isolated including DNA polymerase II 1970 DNA polymerase III 1971 DNA polymerase IV and DNA polymerase V The properties of known E coli DNA polymerases are a DNA polymerase I is a single peptide encoded by 17011 and used for DNA replication Replicates DNA in the 52 D 32 direction Has 52 D 32 exonuclease activity to remove nucleotides from 52 end of DNA or from an RNA primer 57 DNA polymerase II is a single peptide encoded by 17013 Used for DNA repair 0 DNA polymerase III has three polypeptide subunits in the catalytic core of the enzyme lt encoded by the dnaE gene 2 dnaQ and holE Holoenzyme has an additional siX different polypeptides Replicates DNA in the 52 D 32 direction d DNA polymerase IV is encoded by the dinB gene and is used in DNA repair e DNA polymerase V is encoded by umuDC and is used in DNA repair 6 E coli DNA polymerases used for DNA replication DNA polymerase I and DNA polymerase 111 have 32 D 52 exonuclease proofreading activity Molecular Model of DNA Replication 1 Table 111 shows key genes and DNA sequences involved in replication Initiation of Replication 1 Replication starts when DNA at the origin of replication denatures to eXpose the bases creating a replication fork Replication is usually bidirectional from the origin E 001139 has one origin oriC which has a A minimal sequence of about 245 bp required for initiation b Three copies ofa 13bp ATrich sequence c Four copies ofa 9bp sequence Events in E coli initiating DNA synthesis derived from in vitro studies Figure 115 m Initiator proteins attach E coli s initiator protein is DnaA from the dnaA gene 6 DNA helicase from dnaB binds initiator proteins on the DNA and denatures the ATrich region using ATP as an energy source 0 DNA primase from dnaG binds helicase to form a primosome which synthesizes a short 5 710nt RNA primer Semidiscontinuous DNA Replication When DNA denatures at the ori replication forks are formed DNA replication is usually bidirectional but will consider events at just one replication fork Figure 116 Singlestrand DNAbinding proteins SSBs bind the ssDNA formed by helicase preventing reannealing 9 ST Primase synthesizes a primer on each template strand 0 DNA polymerase III adds nucleotides to the 32 end of the primer synthesizing a new strand complementary to the template and displacing the SSBs DNA is made in opposite directions on the two template strands F New strand made 52 D 32 in the same direction as movement of the replication fork is leading strand while new strand made in the opposite direction is lagging strand Leading strand needs only one primer while lagging needs a series of primers Helicase denaturing DNA causes tighter winding in other parts of the circular chromosome Gyrase relieves this tension 3 Leading strand is synthesized continuously while lagging strand is synthesized discontinuously in the form of Okazaki fragments DNA replication is therefore semidiscontinuous 4 Each fragment requires a primer to begin and is extended by DNA polymerase III V39 Okazaki data show that these fragments are gradually joined together to make a fulllength dsDNA chromosome DNA polymerase I uses the 32 OH of the adjacent DNA fragment as a primer and simultaneously removes the RNA primer while resynthesizing the primer region in the form of DNA The nick remaining between the two fragments is sealed with DNA ligase Figure 117 6 Key proteins are associated to form a replisome Template DNA probably bends to allow synthesis of both leading and lagging strands at the replication fork Figure 118 7 In bidirectional replication the two replisomes are mirror images moving away from each another Figure 119 DNA Replication in Eukaryotes 1 DNA replication is very similar in both prokaryotes and eukaryotes except that eukaryotes have more than one chromosome Replicons 1 Eukaryotic chromosomes generally contain much more DNA than those of prokaryotes and their replication forks move much more slowly If they were like typical prokaryotes with only one origin of l quot quot per 39 DNA 1 quot quot would take many days 2 Instead eukaryotic chromosomes contain multiple origins at which DNA denatures and replication then proceeds bidirectionally until an adjacent replication fork is encountered The DNA replicated from a single origin is called a replicon or replication unit Figure 1112 3 In eukaryotes replicon size is smaller than it is in prokaryotes replication is slower and each chromosome contains many replicons Number and size of replicons vary with cell type 4 Not all origins within a genome initiate DNA synthesis simultaneously Cellspeci c patterns of origin activation are observed so that chromosomal regions are replicated in a predictable order in each cell cycle Figure 1113 Initiation of Replication 1 Eukaryotic origins are generally not well characterized those of the yeast Saccharomyces cerevisiae are among the best understood 2 Chromosomal DNA fragments about 100bp that are able to replicate autonomously when introduced into yeast as eXtracellular circular U gt1 DNA are known as ARSs autonomously replicating sequences ARSs are yeast replicators The three sequence elements typically found in ARSs are A B1 and B2 Initiator protein in yeasts is the multiprotein origin recognition complex ORC which binds to A and B1 Other replication proteins join including one that unwinds DNA at B2 The yeast origin of replication is between regions B1 and B2 DNA and histones must be doubled in each cell cycle G1 prepares the cell for DNA replication chromosome duplication occurs during S phase G2 prepares for cell division and segregation of progeny chromosomes occurs during M phase allowing the cell to divide Cell cycle control is complex and only outlined here Yeasts in which chromosomal replication is well studied serve as a eukaryotic model organism Initiation of replication has two separate steps controlled by cyclin dependent kinases Cdks that are present throughout the cell cycle except during G1 a In the absence of Cdsk during G1 replicator selection occurs ORC and other proteins assemble on each replicator to from prereplicative complexes preRC 57 When cell enters S phase Cdks are present and activate preRCs to initiate replication 0 Cdk activity inhibits another round of preRC formation until the cell again enters G1 when Cdks are absent Eukaryotic Replication Enzymes l 2 Enzymes of eukaryotic DNA replication aren t as well characterized as their prokaryotic counterparts The replication process is similar in both groups iDNA denatures replication is semiconservative and semidiscontinuous and primers are required Fifteen DNA polymerases are known in mammalian cells a Three DNA polymerases are used to replicate nuclear DNA Pol lt alpha extends the lOnt RNA primer by about 30nt Pol TM delta and Pol Z epsilon extend the RNADNA primers one on the leading strand and the other on the lagging it is not clear which synthesizes which Other DNA pols replicate mitochondrial or chloroplast DNA or are used in DNA repair 57 Replicating the Ends of Chromosomes 1 When the ends of chromosomes are replicated and the primers are removed from the 52 ends there is no adjacent DNA strand to serve as a primer and so a singlestranded region is left at the 52 end of the new strand If the gap is not addressed chromosomes would become shorter 2 3 4 5 6 with each round of replication Figure 1114 Most eukaryotic chromosomes have short speciesspeci c sequences tandemly repeated at their telomeres Blackburn and Greider have shown that chromosome lengths are maintained by telomerase which adds telomere repeats without using the cell s regular replication machinery In the ciliate Tetrahymena the telomere repeat sequence is 52TTGGGG 32 Figure 1115 a Telomerase an enzyme containing both protein and RNA binds to the terminal telomere repeat when it is single stranded synthesizing a 3nt sequence TTG 6 The 32 end of the telomerase RNA contains the sequence AAC which binds the TTG positioning telomerase to complete its synthesis of the T TGGGG telomere repeat 0 Additional rounds of telomerase activity lengthen the chromosome by adding telomere repeats After telomerase adds telomere sequences chromosomal replication proceeds in the usual way Any shortening of the chromosome ends is compensated by the addition of the telomere repeats If the sequence of the telomerase RNA is mutated telomeres will correspond to the mutant sequence rather than the organism s normal telomere sequence Using an RNA template to make DNA telomerase functions as a reverse transcriptase called TERT telomerase reverse transcriptase Telomere length may vary but organisms and cell types have characteristic telomere lengths Mutants affecting telomere length have been identi ed and data indicate that telomere length is genetically controlled Shortening of telomeres eventually leads to cell death and this may be a factor in the regulation of normal cell death Assembling Newly Replicated DNA into Nucleosomes 1 2 3 4 When eukaryotic DNA is replicated it complexes with histones This requires synthesis of histone proteins and assembly of new nucleosomes Transcription of histone genes is initiated near the end of G1 phase and translation of histone proteins occurs throughout S phase Assembly of newly replicated DNA into nucleosomes is shown in Figure 1116 m Parental histone cores separate into an H3H4 tetramer and two H2A HZB dimers 6 H3H4 tetramer preeXisting or newly made binds to replicated dsDNA and begins nucleosome assembly 0 H2AH2B dimers preexisting or newly made are added in an assembly process that requires histone chaperone proteins to direct it Selfassembly of nucleosomes has been observed only in vitro CHAPTER 12 KEY CONCEPTS Gene Control of Enzyme Structure 1 Genes encode proteins including enzymes 2 Genes work in sets to accomplish biochemical pathways 3 Genes often work in cooperation with other genes 4 These discoveries are the foundation of modern molecular genetics Genetically Based Enzyme De ciencies in Humans 1 Single gene mutations are responsible for many human genetic diseases Some mutations create a simple phenotype while others are pleiotropic Table 122 Phenylketonuria l Phenylketonuria PKU is commonly caused by a mutation on chromosome 12 in the phenylalanine hydrolase gene preventing the conversion of phenylalanine into tyrosine Figure 121 2 Phenylalanine is an essential amino acid but in excess it is harmful and so it is normally converted to tyrosine Excess phenylalanine affects the CNS causing mental retardation slow growth and early death 3 PKU s effect is pleiotropic Some symptoms result from excess phenylalanine Others result from inability to make tyrosine these include fair skin and blue eyes even with browneye genes and low adrenaline levels 4 Diet is used to manage PKU by providing just enough phenylalanine for protein synthesis but not enough that it accumulates To be effective the special diet must commence in the rst two months a er birth continue at least throughout childhood and be resumed before pregnancy in PKU women to avoid phenylalanine levels that would affect the fetus 5 All US newborns are screened for PKU using the Guthrie test 32 A drop of blood on lter paper is placed on solid media containing 2thienylalanine and the bacterium Bacillus subtilis 57 Normally 2thienylalanine inhibits growth of Bacillus subtilis o Phenylalanine allows Bacillus subtilis to grow in the presence of 2 thienylalanine so bacterial growth indicates high phenylalanine levels in the blood and the possibility that the infant has PKU 6 NutraSweet is aspartame which breaks down to aspartic acid and phenylalanine with serious consequences for a phenylketonuric Albinism 1 Classic albinism results from an autosomal recessive mutation in the gene for tyrosinase Tyrosinase is used to convert tyrosine to DOPA in the melanin pathway Without melanin individuals have white skin and hair 2 and red eyes due to lack of pigmentation in the iris Two other forms of albinism are known resulting from defects in other genes in the melanin pathway A cross between parents with different forms of albinism can produce normal children Lesch Nyhan Syndrome 1 2 3 4 LeschNyhan syndrome Figure 124 results from a recessive mutation on the X chromosome in the gene for hypoxanthineguanine phosphoribosyl transferase HGPRT The fatal disease is found in males 57 33 Heterozygous carrier females may show symptoms when lyonization of the normal X chromosome leaves the X chromosome with the defective HGPRT gene in control of cells HGPRT is an enzyme essential to purine utilization In LeschNyhan syndrome this pathway is highly impaired Purines accumulate and are converted to uric acid Symptoms of LeschNyhan syndrome a Infants develop normally for several months Orange uric acid crystals in diapers of males are the only clue of disease 57 At 37 8 months motor development delays lead to weak muscles Muscle tone is altered producing uncontrollable movements and involuntary spasms 0 9 At 273 years children show bizarre activity such as compulsive self mutilation that is dif cult to control and painful as well as aggression toward others 9 LeschNyhan individuals score severely retarded on intelligence tests possibly due to poor communications skills quot7 Most LeschNyhan individuals die before their 20s typically from infection kidney failure or uremia In the case of LeschNyhan syndrome a defect in a single enzyme HGPRT has very pleiotropic effects giving rise to uremia kidney failure mental de ciency and so far ineXplicably selfmutilation Tay Sachs Disease 1 2 3 TaySachs Figure 125 is one ofa group of diseases called lysosomal storage diseases Generally caused by recessive mutations these diseases result from mutations in genes encoding lysosomal enzymes TaySachs disease infantile amaurotic idiocy results from a recessive mutation in the gene hex1 which encodes the enzyme N acetylhexosaminidase A The HeXA enzyme cleaves a terminal N acetylgalactosamine group from a brain ganglioside Figure 126 Infants homozygous recessive for this gene will have nonfunctional HeXA enzyme Unprocessed ganglioside accumulates in brain cells and causes various clinical symptoms 9 Infants have enhanced reaction to sharp sounds ST A cherrycolored spot surrounded by a white halo may be visible on the retina 0 Rapid neurological degeneration begins about age 1 as brain loses control of normal functions due to accumulation of unprocessed ganglioside Q Progress is rapid with blindness hearing loss and serious feeding problems leading to immobility by age 2 9 Death often occurs at 374 years of age often from respiratory infection 4 The disease is incurable Carriers and affected individuals can be detected by genetic testing Gene Control of Protein Structure 1 Genes also make proteins that are not enzymes Structural proteins such as hemoglobin are often abundant making them easier to isolate and purify Sickle Cell Anemia Outline of the genetics and gene products involved in sicklecell anemia and trait a Wildtype chain allele is A which is codominant with S b Hemoglobin of A A individuals has normal subunits while hemoglobin of those with the genotype S S has subunits that sickle at low Oz tension 0 Hemoglobin of A S individuals is 12 normal and 12 sickling form The two chains of an individual hemoglobin molecule will be of the same type rather than mixed These heterozygotes may experience sicklecell symptoms after a sharp drop in the oxygen content of their environment Other Hemoglobin Mutants Screening of hemoglobin for altered electrophoretic mobility has identi ed over 200 hemoglobin mutants showing a variety of amino acid substitutions in both the lt and the chains Each appears to derive from a single amino acid change Figure 1211 2 Most effects are not as severe as those seen in sicklecell anemia Cystic Fibrosis Cystic brosis CF affects the pancreas lungs and digestive system and sometimes the vas deferens in males The disease is characterized by abnormally viscous secreted mucus and lung complications are managed by percussion and antibiotics to treat infections Life expectancy with current treatments is about 40 years Figure 1212 The affected gene is on the long arm of chromosome 7 and encodes a protein called cystic brosis transmembrane conductance regulator CFTR Comparing DNA sequences of the cloned gene from normal and CF individuals shows that the CF mutation commonly is the deletion of a speci c 3bp region removing one amino acid from the protein product The structure of the protein has been deduced from its sequence Figure 1213 CFTR has homology with a large family of active transport membrane proteins Functional analysis shows that CFTR normally forms a chloride channel in the cell membrane The mutated gene results in an abnormal CFTR protein preventing chlorine ion transport and resulting in CF symptoms Genetic Counseling Genetic testing can detect many inherited enzyme and protein defects yielding information about whether an individual has a disease or is a carrier Chromosomal abnormalities can also be detected Genetic counseling is advice based on genetic analysis focusing either on the probability that an individual has a genetic defect or the probability that prospective parents will produce a child with a genetic defect Genetic counselors have the task of eXplaining diseases probabilities and options to affected individuals or parents Some aspects of human heredity are well understood others not yet as well Effective genetic counseling requires uptotheminute knowledge of genetic research and the ability to offer clients unbiased and nonprescriptive information from two main sources 32 Pedigree analysis is an important tool of genetic counseling considering phenotypes found in both families over several generations This is particularly useful for identifying suspected carriers 57 Fetal analysis includes assays for enzyme activity or protein level or detection of changes in the DNA itself For most defective alleles there is currently no way to change the resulting phenotype and so genetic counseling focuses primarily on informing clients of risks and probabilities Carrier Detection 1 A carrier is heterozygous for a recessive gene mutation In a cross between two carrier parents 14 of the offspring are eXpected to develop the disease and 12 to also be carriers The carrier s phenotype is normal but if levels of the affected protein are determined they may be well below those of a normal individual Detection of carriers may be performed by direct DNA testing Fetal Analysis Genetic counseling is also concerned with whether a fetus is normal A sample of fetal cells is needed for the analysis There are currently two methods of obtaining the necessary sample a Amniocentesis Figure 1214 is removal of a sample of amniotic uid using a syringe needle inserted through the uterine wall The procedure is seldom done before the 12th week of pregnancy due to small amounts of amniotic uid and risk to the fetus ST Chorionic villus sampling Figure 1215 can be done in the 8th712th weeks of pregnancy by removal of chorionic villus tissue either through the abdomen as in amniocentesis or via the vagina Once fetal cells are obtained they are usually cultured in the laboratory although chorionic villus sampling may provide enough tissue to assay directly They are examined for protein or enzyme alterations or de ciencies DNA changes and chromosomal abnormalities Amniocentesis is costly and cannot be performed until the second trimester removing early abortion as an option in cases of severe genetic defects Chorionic villus sampling can be done earlier but carries a higher risk of fetal death and inaccurate diagnosis due to the presence of maternal cells CHAPTER 13 KEY CONCEPTS Gene Expression An Overview 1 The ow of information from DNA D RNA D protein was called the Central Dogma by Francis Crick in 1956 Synthesis of an RNA molecule using a DNA template is called transcription Only one of the DNA strands is transcribed The enzyme used is RNA polymerase There are four major types of RNA molecules a Messenger RNA mRNA encodes the amino acid sequence of a polypeptide 57 Transfer RNA tRNA brings amino acids to ribosomes during translation 0 Ribosomal RNA rRNA combines with proteins to form a ribosome the catalyst for translation 9 Small nuclear RNA snRNA combines with proteins to form complexes used in eukaryotic RNA processing The Transcription Process RNA Synthesis 1 Transcription or gene expression is regulated by gene regulatory elements 4 associated with each gene DNA unwinds in the region next to the gene due to RNA polymerase in prokaryotes and other proteins in eukaryotes In both RNA polymerase catalyzes transcription Figure 131 RNA is transcribed 52 to32 The template DNA strand is read 32to52 Its complementary DNA the nontemplate strand has the same polarity as the RNA RNA polymerization is similar to DNA synthesis Figure 132 except a The precursors are NTPs not dNTPs b No primer is needed to initiate synthesis c Uracil is inserted instead of thymine Initiation of Transcription at Promoters E 3 V39 9 Transcription is divided into three steps for both prokaryotes and eukaryotes They are initiation elongation and termination The process of elongation is highly conserved between prokaryotes and eukaryotes but initiation and termination are somewhat different This section is about initiation of transcription in prokaryotes E coli is the model organism Figure 133 A prokaryotic gene is a DNA sequence in the chromosome The gene has three regions each with a function in transcription a A promoter sequence that attracts RNA polymerase to begin transcription at a site speci ed by the promoter 6 The transcribed sequence called the RNAcoding sequence The sequence of this DNA corresponds with the RNA sequence of the transcript c A terminator region that speci es where transcription will stop Promoters in E coli generally involve two DNA sequences centered at D35bp and D10bp upstream from the 1 start site of transcription The common E coli promoter that is used for most transcription has these consensus sequences a For the D35 region the consensus is 52 TTGACA32 b For the D10 region previously known as a Pribnow box the consensus is 52TATAAT32 Transcription initiation requires the RNA polymerase holoenzyme to bind to the promoter DNA sequence Holoenzyme consists of a Core enzyme ofRNAl 39 four I 39 I I 39 39 two lt one and one 2 b Sigma factor f Sigma factor binds the core enzyme and confers ability to recognize promoters 00 0 O RNA polymerase holoenzyme binds promoter in two steps Figure 134 First it loosely binds to the D35 sequence of dsDNA closed promoter complex 9 57 Second it binds tightly to the D10 sequence untwisting about 17bp of DNA at the site and in position to begin transcription open promoter compleX Promoters often deviate from consensus The associated genes will show different levels of transcription corresponding with sigma s ability to recognize their sequences E 001139 has several sigma factors with important roles in gene regulation Each sigma can bind a molecule of core RNA polymerase and guide its choice of genes to transcribe Most E coli genes have a 70 promoter and 70 is usually the most abundant sigma factor in the cell Other sigma factors may be produced in response to changing conditions Examples ofE coli sigma factors a 70 recognizes the sequence TTGACA at D35 and TATAAT at D10 b 32 recognizes the sequence CCCCC at D39 and TATAAATA at D15 Sigma32 arises in response to heat shock and other forms of stress c 54 recognizes the sequence GTGGC at D26 and TTGCA at D14 Sigma54 arises in response to heat shock and other forms of stress d 23 recognizes the sequence TATAATA at position D15 Sigma23 is present in cells infected with phage T4 12 E 001139 has additional sigma factors Other bacterial species also have multiple sigma factors Elongation and Termination of an RNA Chain 3 Once initiation is completed RNA synthesis begins After 879 NTPs have been joined in the growing RNA chain sigma factor is released and reused for other initiations Core enzyme completes the transcript Figure 134 Core enzyme untwists DNA heliX locally allowing a small region to denature Newly synthesized RNA forms an RNADNA hybrid but most of the transcript is displaced as the DNA heliX reforms The chain grows at 307 50ntsecond RNA polymerase has two types of proofreading 9 Similar to DNA polymerase editing newly inserted nucleotide is removed by reversing synthesis reaction 57 Enzyme moves back one or more nucleotides cleaves RNA then resumes synthesis in forward direction Terminator sequences are used to end transcription In prokaryotes there are two types a Rhoindependent gtindependent or type Iterminators have twofold 57 symmetry that would allow a hairpin loop to form Figure 135 The palindrome is followed by 47 SU residues in the transcript and when these sequences are transcribed they cause termination Rhodependent gtdependent or type II terminators lack the polyU region and many also lack the palindrome The protein gt is required for termination It has two domains one binding RNA and the other binding ATP ATP hydrolysis provides energy for gt to move along the transcript and destabilize the RNADNA hybrid at the termination reglon Transcription in Eukaryotes Prokaryotes contain only one RNA polymerase which transcribes all RNA for the cell Eukaryotes have three different polymerases each transcribing a different class of RNA Processing of transcripts is also more complex in eukaryotes Eukaryotic RNA Polymerases l Eukaryotes contain three different RNA polymerases 9 ST 0 RNA polymerase I located in the nucleolus transcribes the three major rRNAs 28S 18S and 58S RNA polymerase 11 located in the nucleoplasm transcribes mRNAs and some snRNAs RNA polymerase 111 located in the nucleoplasm transcribes tRNAs 5S rRNA and the remaining snRNAs Transcription of Protein Coding Genes by RNA Polymerase II 1 Equot When proteincoding genes are rst transcribed by RNA pol II the product is a precursormRNA premRNA The premRNA will be modi ed to produce a mature mRNA Results of promoter analysis reveal two types of elements a 57 Core promoter elements are located near the transcription start site and specify where transcription begins Examples include i The initiator element Inr a pyramidinerich sequence that spans the transcription start site The TATA box also known as a TATA element or Goldberg Hogness box at D30 its full sequence is TATAAAA This element aids in local DNA denaturation and sets the start point for transcription Promoter proximal elements are required for high levels of transcription They are further upstream from the start site at positions between D50 and D200 These elements generally function in either orientation Examples include i The CAAT box located at about D75 ii The GC box consensus sequence GGGC GG located at about D90 3 Various combinations of core and proximal elements are found near different genes Promoter proximal elements are key to gene expression 9 Activators proteins important in transcription regulation are recognized by promoter proximal elements b Housekeeping used in all cell types for basic cellular functions genes have common promoter proximal elements and are recognized by activator proteins found in all cells c Genes expressed only in some cell types or at particular times have promoter proximal elements recognized by activator proteins found only in speci c cell types or times 5 Enhancers are another cisacting element They are required for maximal transcription of a gene 9 Enhancers are usually upstream of the transcription initiation site but may also be downstream They may modulate from a distance of thousands of base pairs away from the initiation site 57 Enhancers contain short sequence elements some similar to promoter sequences 0 Activators bind these sequences and other protein complexes form bringing the enhancer complex close to the promoter and increasing transcription U Transcription initiation requires assembly of RNA polymerase II and binding of general transcription factors GTFs on the core promoter a GTFs are needed for initiation by all three RNA polymerases 6 GTFs are numbered to match their corresponding RNA polymerase and lettered in the order of discovery eg TFIID was the fourth GTF discovered that works with RNA polymerase II Eukaryotic mRNAs l Eukaryotic mRNAs have three main parts Figure 138 a The 52 leader sequence or 52 untranslated region 52 UTR varies in length b The coding sequence which speci es the amino acid sequence of the protein that will be produced during translation It varies in length according to the size of the protein that it encodes c The trailer sequence or 32 untranslated region 32 UTR also varies in length and contains information in uencing the stability of the mRNA 2 Eukaryotes and prokaryotes produce mRNAs somewhat differently Figure 139 a Prokaryotes use the RNA transcript as mRNA without modi cation Transcription and translation are coupled in the cytoplasm Messages may be polycistronic b Eukaryotes modify preRNA into mRNA by RNA processing The processed mRNA migrates from nucleus to cytoplasm before translation Messages are always monocistronic Production of Mature mRNA in Eukaryotes 52 and 32 Modi cations 1 The newly made 52 end of the mRNA is modi ed by 52 capping A capping enzyme adds a guanine usually 7methyl guanosine m7G to the 52 end using a 52to52 linkage Sugars of the two adjacent nt are also methylated The cap is used for ribosome binding to the mRNA during translation initiation Figure 1310 2 The 32 end of mRNA is marked by a polyA tail Figure 1311 a Transcription of mRNA continues through the polyA consensus sequence AAUAAA 6 Proteins bind and cleave RNA These include i CPSF cleavage and polyadenylation speci city factor ii CstF cleavage stimulation factor iii Two cleavage factor proteins CFI and CFII 0 After cleavage the enzyme polyA polymerase PAP adds A nucleotides to the 32 end of the RNA using ATP as a substrate PAP is bound to CPSF during this process 9 PABII polyA binding protein 11 binds the polyA tail as it is produced Processing of Pre mRNA t0 Mature mRNA 1 Events in eukaryotic mRNA production are summarized in Figure 1312 They include a Transcription of the gene by RNA polymerase II b Addition ofthe 52 cap c Addition of the polyA tail 9 Splicing to remove introns 2 Introns typically begin with 52GU and end with AG32 but mRNA splicing signals involve more than just these two small sequences a are small nuclear 1quot 39 l 39 particles snRNPs associated with premRNAs b Spliceosome principal snRNAs are U1 U2 U4 U5 and U6 i Each snRNA is associated with several proteins ii U4 and U6 are part of the same snRNP Others are in their own snRNPs iii Each snRNP type is abundant 8 105 copies per nucleus The steps of splicing are outlined in Figure 1313 9 U1 snRNP binds the 52 splice junction of the intron as a result of base pairing of the U1 snRNA to the intron RNA U2 snRNP binds the branchpoint sequence upstream of the 32 splice junction U4U6 and U5 snRNPs interact then bind the U1 and U2 snRNPs creating a loop in the intron 57 o 9 U4 snRNP dissociates from the complex forming the active spliceosome F The spliceosome cleaves the intron at the 52 splice junction freeing it from exon 1 The free 52 end of the intron bonds to a speci c nucleotide usually A in the branchpoint sequence to form an RNA lariat quot7 The spliceosome cleaves the intron at the 32 junction liberating the intron lariat Exons 1 and 2 are ligated and the snRNPs are released Self Splicing Introns V39 The rRNA genes of most species do not contain introns Some species of the protozoan Tetrahymena have a 413bp intron in their 28S rRNA sequence Tom Cech 1982 showed that splicing of this intron called a group I intron is protein independent The intron selfsplices by folding into a secondary structure that catalyzes its own excision Group I introns are rare but examples occur in a Nuclear rRNA genes b Some mitochondrial mRNA genes c Some tRNA and mRNA genes in bacteriophages The steps in selfsplicing of a group I intron in Tetrahymena are shown in Figure 1314 a The prerRNA is cleaved at the 52 splice junction and guanosine is added to the 52 end of the intron 6 The intron is cleaved at the 32 splice junction 0 The two exons are joined together 9 The excised intron forms a lariat structure which is cleaved to produce a circular RNA and a short linear piece of RNA Selfsplicing is not an enzyme activity because the RNA is not regenerated in its original form at the end of the reaction However the discovery of ribozymes catalytic RNAs has signi cantly altered our view of the biochemistry involved in the origin of life Transcription of Other Genes 1 Genes that do not encode proteins are also transcribed including genes for rRNA tRNA and snRNA Ribosomal RNA and Ribosomes 1 Ribosomes are the catalyst for protein synthesis facilitating binding of charged tRNAs to the mRNA so that peptide bonds can form A cell contains thousands of ribosomes Ribosome Structure 1 Ribosomes in both prokaryotes and eukaryotes consist of two subunits of unequal size large and small each with at least one rRNA and many ribosomal proteins 2 E coli is the model for a prokaryotic ribosome It is 70S with 50S and 30S subunits Figure 1316 a The 50S subunit contains the 23S rRNA 2904nt and 5S rRNA 120nt plus 34 different proteins b The 30S subunit contains the 16S rRNA 1542nt plus 20 different proteins 3 Eukaryotic ribosomes are larger and more compleX than prokaryotic ones and vary in size and composition among organisms Mammalian ribosomes are an example they are 80S with 60S and 40S subunits Figure 1317 a The 60S subunit contains the 28S rRNA 4700nt the 58S rRNA 156nt and the 5S rRNA 120nt plus about 50 proteins b The 40S subunit contains the 18S rRNA 1900nt plus about 35 proteins Transcription of rRNA Genes 1 DNA regions that encode rRNA are called ribosomal DNA rDNA or rRNA transcription units 2 E coli is a typical prokaryote with seven rRNA coding regions designated rm scattered in its chromosome Figure 1318 a Each rm contains the rRNA genes 16S23S5S in that order with tRNA sequences in the spacers b A single prerRNA transcript is produced from rm and cleaved by RNases to release the rRNAs Cleavage occurs in a compleX of rRNA and ribosomal proteins resulting in functional ribosomal subunits 3 Eukaryotes generally have many copies of the rRNA genes Figure 1319 a The three rRNA genes with homology to prokaryotic rRNA genes are 18S58S28S in that order In the chromosome these genes are tandemly repeated 10071000 times to form rDNA repeat units The SS rRNA gene copies are located elsewhere in the genome b A nucleolus forms around each rDNA repeat unit and then they fuse to make one nucleolus Ribosomal subunits are produced in this structure by addition of the 5S rRNA and ribosomal proteins 0 RNA polymerase I transcribes the rDNA repeat units producing a pre rRNA molecule containing the 18S 58S and 28S rRNAs separated by spacer sequences EXternal transcribed sequences ETS located upstream of 18S and downstream of 28S ii Internal transcribed spacers ITS located on each side of 58S iii Nontranscribed spacer NTS sequence is between each rDNA repeat unit and includes the promoter 4 Speci c cleavage steps free the rRNAs from their transcript as part of pre rRNA processing that takes place in the compleX of prerRNA 5S rRNA and ribosomal proteins The result is formation of 40S and 60S ribosomal subunits which are then transported to the cytoplasm Transcription of Genes by RNA Polymerase III 1 Genes transcribed by RNA polymerase 111 include a The eukaryotic 5S rRNA 120nt found in the 60S ribosomal subunit This rRNA has no counterpart in the prokaryotic ribosome ST The tRNAs 75 790nt which occur in repeated copies in the eukaryotic genome Each tRNA has a different sequence ii All tRNAs have CCA added posttranscriptionally at their 32 ends iii Extensive chemical modi cations are performed on all tRNAs after transcription iv All tRNAs can be shown in a cloverleaf structure with complementary base pairing between regions to form four stems and loops Figure 1320 Loop 11 contains the anticodon used to recognize mRNA codons during translation Folded tRNAs resemble an upsidedown L c Some snRNAs the others are transcribed by RNA polymerase II 2 Promoter sequences for SS rRNA and tRNA genes are typically within the sequences that will be transcribed hence internal control regions IRC Promoters for the snRNA genes transcribed by RNA pol III are typically upstream of the genes 3 Transcription of 5S rDNA produces a mature 5S rRNA and no sequences need to be removed 4 Some tRNA genes contain introns About 10 of the 400 yeast tRNA genes have introns If present introns usually are found just 32 to the anticodon and they are removed by a speci c endonuclease with splicing completed by RNA ligase Chemical Structure of Proteins V39 9 Proteins are built from amino acids held together by peptide bonds The amino acids confer shape and properties to the protein Two or more polypeptide chains may associate to form a protein complex Each cell type has characteristic proteins that are associated with its function All amino acids except proline have a common structure Figure 141 a The carbon is bonded to An amino group NH2 which is usually charged at cellular pH NH A carboxyl group COOH which is also usually charged at cellular pH COOD A hydrogen atom H ii39 iv An R group which is different for each amino acid and confers distinctive properties The R groups in an amino acid chain give polypeptides their structural and functional properties There are 20 amino acids used in biological proteins They are divided into subgroups according to the properties of their R groups acidic basic neutral and polar or neutral and nonpolar Figure 142 Polypeptides are chains of amino acids joined by covalent peptide bonds A peptide bond forms between the carboxyl group of one amino acid and the amino group of another Figure 143 Polypeptides are unbranched and have a free amino group at one end the N terminus and a carboxyl group at the other the C terminus The N terminal end de nes the beginning of the polypeptide Molecular Structure of Proteins 1 2 Proteins have up to four levels of organization Figure 144 a Primary structure is the amino acid sequence of the polypeptide This is determined by the nucleotide sequence of the corresponding gene 6 Secondary structure is folding and twisting of regions within a polypeptide resulting from electrostatic attractions andor hydrogen bonding Common examples are helix and pleated sheet 0 Tertiary structure is the threedimensional shape of a single polypeptide chain often called its conformation Tertiary structure arises from interactions between R groups on the amino acids of the polypeptide and thus relates to primary structure 9 Quaternary structure occurs in multisubunit proteins as a result of the 39 39 o I I 39 39 chains 1 is an example with two 141aminoacid polypeptides and two 146aminoacid polypeptides each associated with a heme group u More than amino acid sequence alone determines the folding of a polypeptide into a functional protein Cell biology experiments show that proteins in the molecular chaperone family assist other proteins in folding Characteristics of the Genetic Code 1 Characteristics of the genetic code a 6 0 F F quot7 P It is a triplet code Each threenucleotide codon in the mRNA speci es one amino in the polypeptide It is comma free The mRNA is read continuously three bases at a time without skipping any bases It is nonoverlapping Each nucleotide is part of only one codon and is read only once during translation It is almost universal In nearly all organisms studied most codons have the same amino acid meaning Examples of minor code differences include the protozoan Tetrahymena and mitochondria of some organisms It is degenerate Of 20 amino acids 18 are encoded by more than one codon Met AUG and Trp UGG are the exceptions all other amino acids correspond to a set of two or more codons Codon sets often show a pattern in their sequences variation at the third position is most common The code has start and stop signals AUG is the usual start signal for protein synthesis and de nes the open reading frame Stop signals are codons with no corresponding tRNA the nonsense or chainterminating codons There are generally three stop codons UAG amber UAA ochre and UGA opal Wobble occurs in the anticodon The third base in the codon is able to basepair less speci cally because it is less constrained three dimensionally It wobbles allowing a tRNA with base modi cation of its anticodon eg the purine inosine to recognize up to three different codons Figure 148 and Table 141 Translation The Process of Protein Synthesis 1 Ribosomes translate the genetic message of mRNA into proteins 2 The mRNA is translated 52D 32 producing a corresponding N terminal D Cterminal polypeptide 3 Amino acids bound to tRNAs are inserted in the proper sequence due to a b Speci c binding of each amino acid to its tRNA Speci c basepairing between the mRNA codon and tRNA anticodon The mRNA Codon Recognizes the tRNA Anticodon 1 tRNACys normally carries the amino acid cysteine Ehrenstein Weisblum and Benzer attached cysteine to tRNACys making CystRNACys and then chemically altered it to alanine making AlatRNACys When used for in vitro synthesis of hemoglobin the tRNA inserted alanine at sites where cysteine was expected They concluded that the speci city of codon recognition lies in the tRNA molecule and not in the amino acid it carries Adding an Amino Acid t0 tRNA V39 AminoacyltRNA synthetase attaches amino acids to their speci c tRNA molecules The charging process aminoacylation produces a charged tRNA aminoacyltRNA using energy from ATP hydrolysis There are 20 different aminoacyltRNA synthetase enzymes one for each amino acid and each recognizes the structure of the speci c tRNAs it charges The amino acid and ATP bind to the speci c aminoacyltRNA synthetase enzyme ATP loses two phosphates and the resulting AMP is bound to the amino acid forming aminoacylAMP Figure 149 The tRNA binds to the enzyme and the amino acid is transferred onto it displacing the AMP The aminoacyltRNA is released from the enzyme The amino acid is now covalently attached by its carboxyl group to the 32 end of the tRNA Every tRNA has a 32 adenine and the amino acid is attached to the 3 27 OH or 227 OH ofthis nucleotide Figure 1410 Initiation of Translation 5 Protein synthesis is similar in prokaryotes and eukaryotes Some signi cant differences do occur and are noted below In both it is divided into three stages a Initiation b Elongation c Termination Initiation of translation requires a An mRNA b Aribosome c A speci c initiator tRNA d Initiation factors e Mg magnesium ions Prokaryotic translation begins with binding of the 30S ribosomal subunit to mRNA near the AUG codon Figure 1411 The 30S comes to the mRNA bound to a All three initiation factors IF1 IF2 and IF3 b GTP c Mg V39 9 Ribosome binding to mRNA requires more than the AUG a RNase protection experiments have shown that the ribosome binds at a ribosomebinding site where it is oriented to the correct reading frame for protein synthesis 6 The AUG is clearly identi ed in these studies 0 An additional sequence 8 712 nucleotides upstream from the AUG is commonly involved Discovered by Shine and Delgamo these purine rich sequences eg AGGAG are complementary to the 32 end of the 16S rRNA Figure 1412 Complementarity between the ShineDelgamo sequence and the 32 end of 16S rRNA appears to be important in ribosome binding to the mRNA NeXt the initiator tRNA binds the AUG to which the 30S subunit is bound AUG universally encodes methionine Newly made proteins begin with Met which is often subsequently removed 9 m Initiator methionine in prokaryotes is formylmethionine VIet It is carried by a speci c tRNA with the anticodon 52 CAU 32 6 The tRNA rst binds a methionine and then transforrnylase attaches a forrnyl group to the methionine making IettRNAfMET a charged initiator tRNA 0 Methionines at sites other than the beginning of a polypeptide are inserted by tRNAMet a different tRNA which is charged by the same aminoacyltRNA synthetase as tRNA Iet When MettRNA Iet binds the 30SmRNA compleX IF3 is released and the 50S ribosomal subunit binds the compleX GTP is hydrolysed and IFl and IF2 are released The result is a 70S initiation compleX consisting of a mRNA b 70S ribosome 30S and 50S subunits with a vacant A site c fMettRNA in the ribosome s P site The main differences in eukaryotic translation are a Initiator methionine is not modi ed As in prokaryotes it is attached to a special tRNA 57 Ribosome binding involves the 52 cap rather than a ShineDelgamo sequence i Eukaryotic initiator factor eIF4F is a multimer of proteins including the capbinding protein CBP It binds the 52 mRNA cap ii Then the 40S subunit complexed with initiator MettRNA several eIFs and GTP binds the cap compleX along with other eIFs iii The initiator compleX scans the mRNA for a Kozak sequence that includes the AUG start codon This is usually the rst AUG in the transcript iv When the start codon is located 40S binds and then 60S binds displacing the eIFs and creating the 80S initiation complex with initiator MettRNA in the ribosome s P site 0 The eukaryotic mRNA s 32 polyA tail also interacts with the 52 cap PolyA binding protein PABP binds the polyA and also binds a protein in eIF4F on the cap circularizing the mRNA and stimulating translation Elongation 0f the Polypeptide Chain 1 Elongation of the amino acid chain has three steps Figure 1413 a Binding of aminoacyltRNA to the ribosome b Formation of a peptide bond c Translocation of the ribosome to the neXt codon Binding of Aminoacyl tRNA 1 Protein synthesis begins with fMettRNA in the P site of the ribosome The neXt charged tRNA approaches the ribosome bound to EFTuGTP When the charged tRNA hydrogen bonds with the codon in the ribosome s A site hydrolysis of GTP releases EFTuGDP 2 EFTu is recycled with assistance from EFTs which removes the GDP and replaces it with GTP preparing EFTuGTP to escort another aminoacyl tRNA to the ribosome Peptide Bond Formation l The two aminoacyltRNAs are positioned by the ribosome for peptide bond formation which occurs in two steps Figure 1414 a In the P site the bond between the amino acid and its tRNA is cleaved b Peptidyl transferase forms a peptide bond between the nowfree amino acid in the P site and the amino acid attached to the tRNA in the A site Experiments indicate that the 23S rRNA is most likely the catalyst for peptide bond formation 0 The tRNA in the A site now has the growing polypeptide chain attached to it Translocation l The ribosome now advances one codon along the mRNA EFG is used in translocation in prokaryotes EFGGTP binds the ribosome GTP is hydrolyzed and the ribosome moves one codon while the uncharged tRNA leaves the P site Eukaryotes use a similar process with a factor called eEF2 2 Release of the uncharged tRNA involves the 50S ribosomal E for Exit site Binding of a charged tRNA in the A site is blocked until the spent tRNA is released from the E site 3 During translocation the peptidyltRNA remains attached to its codon but is transferred from the ribosomal A site to the P site by an unknown mechanism 4 The vacant A site now contains a new codon and an aminoacyltRNA with the correct anticodon can enter and bind The process repeats until a stop codon is reached 5 Elongation and translocation are similar in eukaryotes except for differences in number and type of elongation factors and the exact sequence of events 6 In both prokaryotes and eukaryotes simultaneous translation occurs New ribosomes may initiate as soon as the previous ribosome has moved away from the initiation site creating a polyribosome polysome an average mRN A might have 8710 ribosomes Figure 1415 Termination of Translation 1 Termination is signaled by a stop codon UAA UAG UGA which has no corresponding tRNA Figure 1416 2 Release factors RF assist the ribosome in recognizing the stop codon and terminating translation a In E 001139 i RFl recognizes UAA and UAG ii RF2 recognizes UAA and USA iii RF3 stimulates termination b In eukaryotes eRF is the only termination factor recognizing all three stop codons and stimulating termination 3 Termination events triggered by release factors are a Peptidyl transferase releases the polypeptide from the tRNA in the ribosomal P site b The tRNA is released from the ribosome c The two ribosomal subunits and RF dissociate from the mRNA d The initiator amino acid VIet or Met is usually cleaved from the polypeptide CHAPTER 15 KEY CONCEPTS MUTATIONS amp DNA REPAIR Mutations De ned 1 A mutation is a change in a DNA base pair or a chromosome a Somatic mutations affect only the individual in which they arise b Germline mutations alter gametes affecting the neXt generation 2 Mutations are quanti ed in two different ways a Mutation rate is the probability of a particular kind of mutation as a function of time eg number per gene per generation 6 Mutation frequency is the number of times a particular mutation occurs in proportion to the number of cells or individuals in a population eg number per 100000 organisms Types of Point Mutations Equot 5 There are two general categories of point mutations basepair substitutions and basepair deletions or insertions A basepair substitution replaces one base pair with another There are two types Figure 153 a Transitions convert a purinepyrimidine pair to the other purine pyrimidine pair eg AT to GO or TA to CG b T convert a pmiu r pair eg AT to TA or AT to CG 39 pair to a pyrimidinepurine Basepair substitutions in ORFs are also de ned by their effect on the protein sequence Effects vary from none to severe a Nonsense mutations change a codon in the ORF to a stop nonsense codon resulting in premature termination of translation and a truncated often nonfunctional protein Figure 154 b Missense mutations have a basepair change resulting in a different mRNA codon and therefore a different amino acid in the protein c Phenotypic effects may or may not occur depending on the speci c amino acid change Neutral mutations change a codon in the ORF but the resulting amino acid substitution produces no detectable change in the function of the protein eg AAA to AGA substitutes arginine for lysine The amino acids have similar properties so the protein s function may not be altered Silent mutations occur when the mutant codon encodes the same amino acid as the wildtype gene so that no change occurs in the protein produced eg AAA and AAG both encode lysine so this transition would be silent Deletions and insertions can change the reading frame of the mRNA downstream of the mutation resulting in a frameshift mutation a When the reading frame is shifted incorrect amino acids are usually incorporated b Frameshifts may bring stop codons into the reading frame creating a shortened protein c Frameshifts may also result in readthrough of stop codons resulting in a longer protein d Frameshift mutations result from insertions or deletions when the number of affected base pairs is not divisible by three Reverse Mutations and Suppressor Mutations 1 Point mutations are divided into two classes based on their effect on phenotype a Forward mutations change the genotype from wild type to mutant b Reverse mutations reversions or back mutations change the genotype from mutant to wild type or partially wild type 1 A reversion to the wildtype amino acid in the affected protein is a true reversion A reversion to some other amino acid that fully or partly restores protein function is a partial reversion Suppressor mutations occur at sites different from the original mutation and mask or compensate for the initial mutation without actually reversing it Suppressor mutations have different mechanisms depending on the site at which they occur a Intragenic suppressors occur within the same gene as the original mutation but at a different site Two different types occur 57 1 A different nucleotide is altered in the same codon as the original mutation A nucleotide in a different codon is altered eg an insertion frameshift is suppressed by a nearby deletion event Intergenic suppressors occur in a different gene the suppressor gene from the original mutation Many work by changing mRNA translation Each suppressor gene works on only one type of nonsense missense or frameshift mutation A given suppressor gene suppresses all mutations for which it is speci c Suppressor genes often encode tRNAs that recognize stop codons and insert an amino acid preventing premature termination of translation 1 Full or partial function of the polypeptide may be restored 2 This suppression process is inef cient but does allow functional polypeptides to be produced Nonsense suppressors fall into three classes one for each stop codon UAG UAA and UGA Figure 155 Typical tRNA suppressor mutations are in redundant tRNA genes so the wildtype tRNA activity is not lost Vi Nonsense suppression occurs by competition between release factors and suppressor tRNAs 1 UAG and UGA suppressor tRNAs do well in competition with release factors 2 UAA suppressor tRNAs are only 1 to 5 ef cient Vii Suppression by a tRNA occurs at all of its speci c stop codons eg UGA or UAG not just the mutant one This may produce readthrough proteins but the inefficiency of nonsense suppressor tRNAs means they rarely affect phenotype Spontaneous and Induced Mutations 1 Most mutations are spontaneous rather than induced by a mutagen Spontaneous Mutations 1 All types of point mutations can occur spontaneously during S G1 and G2 phases of the cell cycle or by the movement of transposons The spontaneous mutation rate in eukaryotes is between 10D4 and 10D6 per gene per generation and in bacteria and phages between 10D5 and 10D7 per gene per generation Many spontaneous errors are corrected by the cellular repair systems and so do not become xed in DNA DNA Replication Errors 1 DNA replication errors can be either point mutations or small insertions or deletions Basepair substitution mutations can result from wobble pairing Bases are normally in the keto form but sometimes can undergo tautomeric shift to form the abnormal enol form The enol form can hydrogen bond with an incorrect partner due to different spatial positioning of the atoms involved in Hibonding Figure 156 An example is a GCtoAT transition Figure 157 m During DNA replication the keto form of G could wobble pair with a normal T producing a GT pair 57 In the neXt round of replication G is likely to be back in its keto form and so both G and A are likely to pair normally producing one progeny DNA with a GC pair and another with an AT pair 0 GT pairs are targets for correction by proofreading during replication and by other repair systems Only mismatches uncorrected before the neXt round of replication lead to mutations Additions and deletions can occur spontaneously during replication Figure 15 8 a DNA loops out from the template strand generally in a run of the same ST 0 9 base DNA polymerase skips the loopedout bases creating a deletion mutation If DNA polymerase adds untemplated bases new DNA looping occurs resulting in additional mutation Insertions and deletions in structural genes generate frameshift mutations especially if they are not multiples of three 4 Spontaneous chemical changes include depurination and deamination of particular bases creating lesions in the DNA a 57 Depurination removes the purine A or G from DNA by breaking the bond with its deoxyribose in the backbone i Depurination is common ii If not repaired before the neXt round of replication DNA polymerase may stall or dissociate Deamination removes an amino group from a base eg cytosine to uracil Figure 159 i Uracil is an abnormal base in DNA and it will usually be repaired ii If uracil is not replaced it will pair with an A during replication resulting in a CG toTA transition iii Both prokaryotic and eukaryotic DNA have small amounts of 5 methylcytosine 5mC in place of the normal C 2 Deamination of 5mC produces T T is a normal nucleotide in DNA so it is not detected by repair mechanisms Deamination of 5mC results in CG toTA transitions 4 W V Locations of 5mC in the chromosome are often detected as mutational hot spots Induced Mutations 1 Exposure to physical mutagens plays a role in genetic research where they are used to increase mutation frequencies to provide mutant organisms for study Radiation eg Xrays and UV induces mutations a Xrays are an example of ionizing radiation which penetrates tissue and collides with molecules knocking electrons out of orbits and creating ions Ions can break covalent bonds including those in the DNA sugar phosphate backbone Ionizing radiation is the leading cause of human gross chromosomal mutations iii Ionizing radiation kills cells at high doses and lower doses produce point mutations iv Ionizing radiation has a cumulative effect A particular dose of radiation results in the same number of mutations whether it is received over a short or a long period of time 6 Ultraviolet UV causes photochemical changes in the DNA i UV is not energetic enough to induce ionization ii UV has lowerenergy wavelengths than do Xrays and so has limited penetrating power iii However UV in the 2547 260nm range is strongly absorbed by purines and pyrimidines forming abnormal chemical bonds 1 A common effect is dimer formation between adjacent pyrimidines commonly thymines designated TAT Figure 1510 2 CC CAT and TC dimers also occur but at lower frequency Any pyrimidine dimer can cause problems during DNA replication 3 Most pyrimidine dimers are repaired because they produce a bulge in the DNA heliX If enough are unrepaired cell death may result Chemical Mutagens 1 Chemical mutagens may be naturally occurring or synthetic They form different groups based on their mechanism of action a Base analogs depend upon replication which incorporates a base with alternate states tautomers that allow it to basepair in alternate ways depending on its state i Analogs are similar to normal nitrogen bases and so are incorporated into DNA readily ii Once in the DNA a shift in the analog s form will cause incorrect basepairing during replication leading to mutation iii 5bromouracil SBU is an example SBU has a bromine residue instead of the methyl group of thymine Figure 1511 1 Normally SBU resembles thymine pairs with adenine and is incorporated into DNA during replication 2 In its rare state SBU pairs only with guanine resulting in a TAtoCG transition mutation 3 If SBU is incorporated in its rare form the switch to its normal state results in a CGtoTA transition Thus SBU induced mutations may be reverted by another eXposure to iv Only those base analogs that cause basepair changes are mutagens eg AZT is a stable analog that does not shift b Basemodifying agents can induce mutations at any stage of the cell cycle They work by modifying the chemical structure and properties of the bases Three types are Figure 1512 i Deaminating agents remove amino groups An example is nitrous acid HNOZ which deaminates G C and A 1 HNO2 deaminates guanine to produce xanthine which has the same base pairing as G No mutation results 2 HNOZ deaminates cytosine to produce uracil which produces a CGtoTA transition 3 HNOZ deaminates adenine to produce hypoxanthine which pairs with cytosine causing an ATtoGC transition 4 Mutations induced by HNO2 can revert with a second treatment Hydroxylating agents include hydroxylamine NHZOH 1 NH2 OH speci cally modi es C with a hydroxyl group OH so that it pairs only with A instead of with G 2 NHZ OH produces only CGto TA transitions and so revertants do not occur with a second treatment 3 NHZ OH mutants however can be reverted by agents that do cause TAtoCG transitions eg SBU and HNOZ ii39 Alkylating agents are a diverse group that add alkyl groups to bases Usually alkylation occurs at the 6oxygen of G producing O alkylguanine 1 An example is methylmethane sulfonate MNIS which methylates G to produce O alkyl G O alkylG pairs with T rather than C causing GCtoAT transitions N V 0 Intercalating agents insert themselves between adjacent bases in dsDNA Figure 1513 They are generally thin platelike hydrophobic molecules At replication a template that contains an intercalated agent will cause insertion of a random eXtra base ii The basepair addition is complete after another round of replication during which the intercalating agent is lost iii If an intercalating agent inserts into new DNA in place of a normal base the neXt round of replication will result in a deletion mutation iv Point deletions and insertions in ORFs result in frameshift mutations These mutations show reversion with a second treatment Repair of DNA Damage 1 Both prokaryotes and eukaryotes have enzymebased DNA repair systems that prevent mutations and even death from DNA damage 2 Repair systems are grouped by their repair mechanisms Some directly reverse damage while others excise the damaged area and then repair the gap Direct Reversal of DNA Damage 1 DNA polymerase proofreading corrects most of the incorrect nucleotide insertions that occur during DNA synthesis which stalls until the wrong nucleotide is replaced with a correct one a The role of 32 to52 exonuclease activity is illustrated by mutator mutations in E coli which confer a much higher mutation rate on the cells that carry them 6 The mutD gene encoding the e subunit of DNA polymerase III is an example Cells mutant in mutD are defective in proofreading 2 UVinduced pyrimidine dimers are repaired using photoreactivation light repair 32 Near UV light 3207370nm activates photolyase product of the phr gene to split the dimer 6 Photolyases are found in prokaryotes and simple eukaryotes but not in humans 3 Damage by alkylation usually methyl or ethyl groups can be removed by speci c DNA repair enzymes 32 For example 06methylguanine methyltransferase from the ada gene recognizes 06methylguanine in DNA and removes the methyl group restoring the base to its original form 6 A similar system repairs alkylated thymine 4 Mutations of repair enzyme genes increase the organism s rate of spontaneous mutations Base Excision Repair 1 Base excision repair uses a repair glycosylase enzyme to recognize and remove damaged bases 32 Bond between base and deoxyribose is cleaved by the glycosylase 6 Other enzymes cleave sugarphosphate from backbone leaving a gap in the DNA 0 Repair DNA polymerase and DNA ligase use the opposite strand as template to ll the gap Repair Involving Excision ofNucleotides 1 A repair system that does not require light was discovered in 1964 It is called dark repair the excision repair system or more recently the nucleotide excision repair NER system a In E coli NER corrects pyrimidine dimers and other damageinduced distortions of the DNA heliX Methyldirected mismatch repair recognizes mismatched base pairs excises the incorrect bases and then carries out repair synthesis 9 In E 001139 initial stages involve products of the mutS mutL and mutH genes Figure 1517 57 Eukaryotes also have mismatch repair but it is not clear how old and new DNA strands are identi ed Translesion DNA synthesis is used to allow a cell to survive when speci c basepairing cannot occur Survival is usually at the cost of incurring new mutations a Bacteria use a system called SOS In E coli SOS is controlled by two genes lexA and recA Mutants in either of these genes have their SOS response permanently turned on ST The SOS system is a mutagenic bypass synthesis system DNA polymerase for translesion synthesis is made in SOS response It replicates over and past the lesion Nucleotides are incorporated at the lesion that may not match wild type leading to mutations ii This is better than the alternative death due to incompletely replicated DNA KEY CONCEPTS CHAPTER 16 DNA Cloning The goal of molecular cloning is large amounts of pure DNA that can be further manipulated and studied Summary of the procedure 9 Isolate DNA from the organism 57 Use restriction enzymes to cut the DNA and ligate fragments into a cloning vector 0 Transform recombinant DNA into a host which will replicate the DNA molecular cloning and pass copies to all progeny Restriction Enzymes 1 Restriction endonucleases restriction enzymes each recognize a speci c DNA sequence restriction site and break a phosphodiester linkage V39 9 between a 32 carbon and phosphate within that sequence Restriction enzymes are used to create DNA fragments for cloning and to analyze positions of restriction sites in cloned or genomic DNA Restriction enzymes are a bacterial defense against viral DNA Restriction sites in the bacterial chromosome are methylated and thus protected A restriction enzyme has a threeletter name derived from the genus and species of the organism from which it was isolated it is underlined or italicized Roman numerals and sometimes letters designating a particular bacterial strain may follow Many restriction sites are palindromes of 4 6 or 8 base pairs but others are not completely symmetrical Figure 161 and Table 161 All copies of a chromosome will contain the same restriction sites and will be cut into identical fragments Based on probability a speci c short DNA sequence occurs more frequently than a longer one a In a 50 G C organism with random distribution of bases the probability of a speci c base at a given position is 14 b Therefore the frequency of a particular restriction site is 14 n where n is the number of base pairs in the recognition sequence One major class of restriction enzymes recognizes and cuts DNA at speci c sequences Two types of DNA ends can be generated Figure 162 a Some restriction enzymes produce blunt ends where both DNA strands are cut between the same base pairs b Others create sticky staggered ends Sticky ends are useful in cloning because complementary sequences hydrogen bond anneal and are held together so that DNA ligase can covalently link them Figure 163 Cloning Vectors and DNA Cloning Plasmid Cloning Vectors Several types of cloning vectors have been constructed each with different molecular properties and cloning capacity Plasmid cloning vectors are derived from natural plasmids circles of dsDNA that include origin sequences on needed for replication in bacterial cells An E coli plasmid vector for example must contain these features 9 An ori sequence for replication 6 A selectable marker such as antibiotic resistance 0 Unique restriction sites so that a particular restriction enzyme cuts only once in the plasmid A fragment of insert DNA cut with the same enzyme is commonly inserted into the unique restriction site 3 An example ofa typical E coli cloning vector is pUCl9 2686bp The pUCl9 plasmid features Figure 164 9 High copy number in E coli with nearly 100 copies per cell provides a good yield of cloned DNA 57 Its selectable marker is ampR 0 It has a cluster of unique restriction sites called the polylinker multiple cloning site 9 The polylinker is part of the lacZ galactosidase gene The pUCl9 plasmid will complement a lacZDE 001139 allowing it to become lacZ When DNA is cloned into the polylinker lacZ is disrupted preventing complementation from occurring 9 Xgal a chromogenic analog of lactose turns blue when galactosidase is present and remains white in its absence so bluewhite screening can indicate which colonies contain recombinant plasmids 4 DNA can be inserted into a cloning vector by restriction digestion and then ligation Figure 165 a Cut pUCl9 the vector plasmid with a restriction enzyme that has a unique site in the polylinker Cut the DNA to be cloned insert DNA with the same enzyme Mix insert DNA with pUCl9 DNA and allow random joining of fragments to occur 57 o F Resulting plasmids are transformed into E coli either through chemical treatment of the cells or by electroporation The cells are grown on media plates containing ampicillin and Xgal FD Ampicillinresistant colonies result from pUCl9 sequences Blue colonies contain only the vector with its ends rejoined while white colonies often contain pUCl9 with its lacZ gene inactivated by insert quot7 If the 52phosphates of vector DNA are removed by alkaline phosphatase DNA ligase will not rejoin their ends and fewer blue colonies will result Expression Vectors 1 Expression vectors have sequences to allow transcription and translation of the cloned genes Pharmaceuticals produced by biotechnology are an example 2 Plasmid cloning vectors are modi ed to include a Promoter and transcription terminator if needed suitable to the host organism 6 Any modi cations needed to cross prokaryoticeukaryotic boundary eg ShineDelgamo sequence is added for translation of eukaryotic sequences in E coll Arti cial Chromosomes l Cloning vectors that can accommodate very large pieces of DNA produce molecules resembling small chromosomes Two examples are YACs and ACs Yeast Arti cial Chromosomes YACs l YAC vectors function as arti cial chromosomes in yeast Their features include Figure 166 57 3 0 F 6 Linear structure with a yeast telomere TEL at each end Ayeast centromere sequence CEN A marker gene on each arm that is selectable in yeast eg TRPI and URA3 A yeast origin of replication known as an autonomously replicating sequence ABS Unique restriction sites for inserting foreign DNA 2 Several hundred kb of insert DNA can be cloned in a YAC However frequent RNA rearrangements in the host make YACs unsuitable for genome sequencing 3 YAC clones are made by a 6 a o 39D Propagating the DNA in E coli as a circular plasmid with telomeres endtoend Cut with restriction enzymes in multiple cloning site and another between the two T ELs generating two arms Ligating long insert DNA fragment with the two arms Transforming into yeast Selecting for markers eg TRPI and URA3 to ensure that both arms are present Bacterial Artificial Chromosomes BACs l BACs are used for cloning fragments up to about 200kb in E coli BAC vectors contain a b C The ori of an E coli plasmid called the F factor A multiple cloning site A selectable marker 2 BACs can be handled like regular bacterial plasmids but the F factor ori keeps copy number at one BAC molecule per cell 3 BACs do not undergo rearrangements in the host 4 BACs are commonly used to study gene regulation in vertebrates Polymerase Chain Reaction PCR PCR starts with a mixture of DNA molecules and produces many copies of one speci c DNA sequence amplimers Mullis developed the technique in the 1980s 2 PCR begins with DNA containing the sequence to be ampli ed and a pair of synthetic primers that ank the sequence Steps in the process include Figure 1621 a 57 o F 39D qozn Heat denature DNA 94795 C Cool 37765 C and anneal primers to complementary sequences with their 32 ends facing each other anking the target DNA Extend the primers 70775 C with heatresistant DNA polymerase from the thermophilic bacterium T hermus aquaticus Repeat the cycle of denaturation and primer binding Extend again with T aq DNA polymerase Products the length of the target sequence begin to be produced Repeat the process of denaturing and annealing new primers Repeat the primer extension with Taq polymerase doubling the amount of target DNA with each round 3 In 20 PCR cycles millionfold ampli cation of the target sequence occurs The temperature changes are automated in a thermal cycler Advantages and Limitations of PCR 1 PCR is more sensitive and faster than cloning but there are limitations 57 3 0 Speci c primers require that sequence information be known Taq polymerase does not proofread meaning that mismatches go uncorrected Alternative polymerases such as Vent polymerase do proofread decreasing errors The sensitivity can result in ampli cation of contaminating sequences a special hazard in forensic applications 2 Applications of PCR include 9957 9 quot7 P Amplifying DNA for cloning Amplifying DNA from genomic DNA for sequencing Without cloning Mapping DNA segments Disease diagnosis Subcloning segments of cloned DNA eg the yeast ARGI gene i Individual genes may be ampli ed from a cloned multigene DNA fragment ii Complementation is used to determine functions of each gene Forensics the analysis of legal evidence in samples including hair blood or semen The study of molecular evolution CHAPTER 19 REGULATION OF GENE EXPRESSION PROKARYOTES The lac Operon of E coli 1 Growth and division genes of bacteria are regulated genes Their expression is controlled by the needs of the cell as it responds to its environment with the goal of increasing in mass and dividing 2 Genes that generally are continuously expressed are constitutive genes housekeeping genes Examples include protein synthesis and glucose metabolism 3 All genes are regulated at some level so that as resources dwindle the cell can respond with a different molecular strategy 4 Prokaryotic genes are often organized into operons that are cotranscribed A regulatory protein binds an operator sequence in the DNA adjacent to the gene array and controls 1 39 39 39 mRNA 39 of the I I V39 Gene regulation in bacteria and phage is similar in many ways to the emerging information about gene regulation in eukaryotes including humans Much remains to be discovered even in E 001139 one of the most closely studied organisms on earth 35 percent of the genomic ORFs have no attributed function The lac Operon of E coli 1 An inducible operon responds to an inducer substance eg lactose An inducer is a small molecule that joins with a regulatory protein to control transcription of the operon 2 The regulatory event typically occurs at a speci c DNA sequence controlling site near the proteincoding sequence Figure 191 3 Control of lactose metabolism in E coli is an example of an inducible operon Lactose as a Carbon Source for E coli 1 E coli expresses genes for glucose metabolism constitutively but the genes for metabolizing other sugars are regulated in a sugar speci c sort of way Presence of the sugar stimulates synthesis of the proteins needed 2 Lactose is a disaccharide glucosegalactose If lactose is E coli s sole carbon source three genes are expressed 32 galactosidase has two functions Breaking lactose into glucose and galactose Galactose is converted to glucose and glucose is metabolized by constitutively produced enzymes Converting lactose to allolactose an isomerization Allolactose is involved in regulation of the lac operon Figure 192 b Lactose permease M protein is required for transport of lactose across the cytoplasmic membrane c Transacetylase is poorly understood 3 The lac operon shows coordinate induction a In glucose medium E coli normally has very low levels of the lac gene products 6 When lactose is the sole carbon source levels of the three enzymes increase coordinately simultaneously about 1000fold i Allolactose is the inducer molecule ii The mRNA for the enzymes has a short halflife When lactose is gone lac transcription stops and enzyme levels drop rapidly Experimental Evidence for the Regulation of lac Genes l The experiments of Jacob and Monod produced an understanding of arrangement and control of the lac genes Mutations in the Protein coding Genes l Mutagens produced mutations in the lac structural genes that were used to map their locations 32 galactosidase is lacZ ST Permease is lacY 0 Transacetylase is lacA F The genes are tightly linked in the order lacZ lacY lacA 2 The type of mutation made a difference in eXpression of the downstream genes a Missense mutations affect only the product of the gene with the mutation ST Nonsense mutations show polarity polar mutations and affect translation of the downstream genes as well 3 The interpretation of gene polarity is that ribosomes translate the rst gene in the polycistronic polygenic RNA and nish in proper position to initiate and translate the neXt gene Premature translation termination prevents this by reducing translation of the downstream genes Figure 193 Jacob and Monod s Operon Model for the Regulation of lac Genes 1 Jacob and Monod s model of regulation with more recent information follows a 6 0 An operon is a cluster of genes that are regulated together The order of the lac genes is shown in Figure 194 and Figure 195 shows the operon when lactose is absent The ac gene has its own constitutive weak promoter and terminator and repressor protein is always present in low concentration The repressor functions as a tetramer Figure 196 Repressor protein binds the operator 151001 and prevents RNA polymerase initiation to transcribe the operon genes negative control ii Binding of the repressor to the operator is not absolute and so an occasional transcript is made resulting in low levels of the structural proteins galactosidase in wildtype E coli growing with lactose as the sole carbon source converts lactose into allolactose Figures 192 and 197 Repressor bound with allolactose bound changes shape allosteric shift and dissociates from the lac operator Free repressor allolactose complexes are unable to bind the operator Allolactose induces eXpression of the lac operon by removing the repressor and allowing transcription to occur 2 The lacOC mutations result in constitutive gene eXpression They are cis dominant to lacol because repressor cannot bind to the lacOC operator sequence Figure 198 5 The lacD mutations change the repressor protein s conformation and prevent it from binding the operator resulting in constitutive eXpression of the operon Figure 199 9 57 In a partial diploid lacF 1ac0 zacz Igorlad 1ac0 zacz lacYD the wild type repressor lacF is dominant over lacID mutants Defective lacD repressor can t bind either operator but normal repressor from lacF binds both operators and regulates transcription resulting in functional galactosidase and permease Additional lac mutants have been identi ed a Superrepressor lads mutants produce no lac enzymes 1 The mutant repressor cannot bind allolactose so lactose does not induce the operon The lads allele is transdominant in partial diploids lacFlacls Figure 1910 The superrepressor protein binds both operators and transcription cannot occur ii39 Normal repressor cannot compete because superrepressor cannot be induced to fall off iv Low levels of transcription will occur superrepressor is not covalently bound to the DNA operator sequence but lacISE coli cannot use lactose as a carbon source These mutants indicate that repressor has three different recognition interactions a Binding to the operator region b Binding with the inducer allolactose c Binding of repressor polypeptide subunits to form an active tetramer Positive Control of the lac Operon 1 Repressor exerts negative control by preventing transcription Positive control of this operon also occurs when lactose is E coli s sole carbon source with no glucose present Figure 1911 a Catabolite activator protein CAP binds cyclic AMP cAMP Figure 1912 b CAPcAMP complex is a positive regulator of the lac operon It binds the CAPsite a DNA sequence upstream of the operon s promoter c Binding of CAPcAMP complex recruits RNA polymerase to the promoster leading to transcription When both glucose and lactose are in the medium E coli preferentially uses glucose due to catabolite repression a Glucose metabolism greatly reduces cAMP levels in the cell b The CAPcAMP level drops and is insuf cient to maintain high transcription of the lac genes c Even when allolactose has removed the repressor protein from the operator lac gene transcription is at very low levels Without CAP cAMP complex bound to the CAPsite d Experimental evidence supports this model Adding cAMP to cells restored transcription of the lac operon even when glucose was present The model is that catabolite repression targets adenylate cyclase the enzyme that makes cAMP Figure 1912 9 In E 001139 the phosphorylated form of IIIGlc enzyme activates adenylate cyclase ST Glucose transport into the cell triggers events including Glc dephosphorylation of III With IIIGlc protein dephosphorylated adenylate cyclase is inactivated and no new cAMP is produced 0 Catabolic genes for other sugars are also regulated by catabolite repression In all cases a CAP site in their promoters is bound by a CAP cAMP complex increasing RNA polymerase binding Molecular Details of lac Operon Regulation V39 The ac DNA sequence shows the expected transcription and translation signals except that the start codon is GUG not AUG The single basepair mutation of 151le is also characterized Figure 1913 The lac operon promoter region begins at D84 immediately next to the lac gene stop codon and ends at D8 just upstream from the transcription start site Features of the promoter region include a The consensus sequence for CAPcAMP binding is in two regions D54 to D58 and D65 to D69 6 The RNA polymerase binding site including a Pribnow box spans DNA from D47 to D8 with consensus sequence matches at D10 and D35 The operator overlaps the promoter with repressor protein protecting DNA from D3 to 21 With repressor bound to the operator RNA polymerase cannot transcribe Figure 1914 The operon transcript begins at 1 which is within the operator region bound by repressor Part of the operator is transcribed 9 The galactosidase gene has a leader region before the start codon 57 Start codon for galactosidase AUG is at 39 to 41 0 Several lacOC mutations have been characterized All are single base pair substitutions The lac operon was the rst molecular model for gene regulation Operons are common in bacteria and phages but very rare in eukaryotes Key Concepts CHAPTER 20 EUKARYOTIC GENE REGULATION Levels of Control of Gene Expression in Eukaryotes Prokaryotes respond quickly to their environments mainly by transcriptional regulatory proteins bind DNA control Translational control also occurs mediated by stability of the mRNAs Eukaryotes have more complex means to regulate gene expression because they have compartments eg nucleus within cells and often multicellular structures that require differentiation of cells Levels at which expression of proteincoding genes is regulated in eukaryotes Figure 202 Transcription g 32 mRNA processing and transport Translation Degradation of mRNA 9 o e Protein processing f Protein degradation Control of Transcription Initiation In eukaryotes most control of protein gene expression is at the level of transcription initiation controlled by promoter immediately upstream and enhancers distal from the gene a Expression from the promoter alone is at basal level b For maximal transcription activator proteins bind to i Promoter proximal elements ii Enhancer elements 2 Binding of activators a Recruits proteins that make the chromatin accessible to the transcription machinery b Increases binding of the transcription machinery to the promoter 3 Variations occur in different genes Chromatin Remodeling In eukaryotes binding of histones to form chromatin generally represses gene expression making speci c repressor proteins unnecessary 2 Evidence for the role of chromatin structure includes a Increased sensitivity to DNaseI of transcriptionally active genes b Hypersensitive DNaseI digestion sites upstream of transcription start sites corresponding to promoter regions c In vitro experiments showing directly that histones can repress gene expression i If DNA is simultaneously mixed with both histones and promoter binding proteins it binds more readily to the histones forming nucleosomes at the TATA box and preventing transcription ii If DNA is rst mixed with promoterbinding proteins adding histones does not produce nucleosomes and transcription occurs iii If DNA is simultaneously mixed with histones binding proteins 1 Enhancerbinding proteins bind the enhancer sequences 2 Promoterbinding proteins bind the promoter sequences 3 Histones are unable to bind and so transcription occurs d Histones therefore are effective repressors but other proteins can overcome that repression Activating Genes by Remodeling Chromatin Activation of eukaryotic genes requires alteration off the chromatin structure near the core promoter a process called chromatin remodeling Two classes of protein complexes cause chromatin remodeling Figure 203 a Acetylating and deacetylating enzymes act on core histones Histone acetyl transferases HATs are part of multiprotein complexes recruited to chromatin when activators bind DNA i HATs acetylate lysines in the aminoterminus of core histones ii The negative charges of acetyl groups decrease the positive charges of the histones reducing their af nity for DNA iii Acetylation of histones changes 30nm chromatin to lOnm ber making promoter more accessible for transcription iv The effect is reversible When histone deacetylases HDACs remove acetyl groups 30nm chromatin reforms Activation of Transcription by Activator and Coactivators 1 Three classes of proteins are involved in transcription activation a General transcription factors GTFs discussed earlier are required for basal transcription but do not change the rate of transcription initiation b Activators transactivators are involved in chromatin remodeling to activate transcription i There are two key domains DNAbinding and transcription activation with a exible region between Homodimers are often used ii Structural motifs for DNA binding regions include Figure 205 1 Helixtumhelix 2 Zinc nger 3 Leucine zipper iii Activation domains are variable They stimulate transcription initiation up to lOOfold c Coactivators are multiprotein complexes that bind to activators and transcription factors creating loops in DNA i Their presence recruits RNA polymerase II to initiate transcription ii Several types of coactivators exist in cells and their large numbers of proteins make their study dif cult iii An example of a coactivator is the mediator complex consisting of 20 or more proteins that bind to activators and to the carboxy terminal domain of RNA polymerase II Blocking Transcription with Repressors l Repressors counteract activators for some genes blocking transcription a Two domains occur in repressors a DNAbinding region and a repressing domain ST Repressors work in a variety of ways Examples Repressor binds near activator s binding site and repressor domain interacts with activation domain of the activator preventing activation Repressor binding site overlaps activator binding site preventing activator binding Chromatin remodeling can also block transcription if repressor binds its site and recruits HDAC histone deacetylase to cause chromatin compaction Combinatorial Gene Regulation 1 Eukaryotic proteincoding gene eXpression is controlled by a Promoters situated just upstream of the transcription start site Some promoter elements eg TATA are required to specify the start of transcription through binding of transcription factor proteins Regulatory promoter elements are specialized involving binding by regulatory proteins speci c for control of one or a few genes A particular gene may have 1 to many regulatory promoter elements and one to many regulatory proteins involved in controlling its function Binding of regulatory proteins to promoters is highly speci c to ensure that only the correct genes are activated b Enhancers located some distance away either upstream or downstream Enhancers determine whether maXimal transcription of the gene occurs Regulatory proteins bind speci c enhancer elements Which ones bind is determined by the DNA sequence recognized by each protein Protein interactions determine whether transcription is activated or repressed 2 Promoters and enhancers bind speci c regulatory proteins a Some regulatory proteins occur in most or all cell types but others are very speci c b Each promoter and enhancer has a particular set of proteins that can bind it and the combination of proteins bound will determine its eXpression c If both positive and negative regulatory proteins are bound interactions F 9 between them will control the rate of expression When regulatory proteins bind an enhancer and have strong negative effect the enhancer is a silencer element Enhancers and promoters appear to bind many of the same proteins implying interactions of regulatory proteins Relatively few proteins are combined in a variety of ways regulating the transcription of different arrays of genes A large number of cell types can be speci ed by combinatorial gene regulation PostTranscriptional Control RNA Processing Control 1 RNA processing control regulates mRNA production from precursor RNAs a 57 0 9 Alternative processing options exist including i Alternative polyadenylation ii Differential splicing alternative splicing Alternative polyadenylation and splicing occur independently of each other and their activities may be tissue speci c Speci c products depend on regulatory signals Alternative polyadenylation and splicing produce proteins encoded by the same gene but structurally and functionally different protein isoforms An example of alternative polyadenylation and splicing involves the human calcitonin gene CALC which has ve exons and four introns Figure 2016 Alternative polyadenylation sites exist next to exon 4 pAl used in thyroid cells and exon 5 pA2 used in neurons Alternative splicing also occurs 1 In the thyroid premRNA is spliced bringing together exons 1 2 3 and 4 2 In neurons premRNA is spliced to bring together introns l 2 3 and 5 Exon 4 is excised and discarded iii The mRNAs are translated to produce prehormones from which hormones are generated by protease cleavage The products produced are l Calcitonin in the thyroid a circulating calciumion homeostatic hormone 2 Calcitoningene related peptide CGRP in the hypothalamus which has neuromodulatory and trophic growthpromoting activities mRNA Transport Control 1 In eukaryotes transport control regulates movement of transcripts from nucleus to cytoplasm a Studies show that about 12 of primary hnRNA transcripts never leave the nucleus and are degraded there b Mature mRNAs exit through nuclear pores A 52 cap is required for exit which involves proteins that bind mRNA and interact with proteins of the nuclear pore complex to move the mRNA into the cytoplasm c The spliceosome retention model proposes that snRNPs stay bound to spliced introns and prevent their export while releasing the spliced exons and allowing them to interact with nuclear pores mRNA Translation Control 1 Ribosomal translational control selecting mRNAs for translation also has an impact on gene expression a Unfertilized eggs are an example in which mRNAs show signi cant increases in translation after fertilization without new mRNA synthesis b Stored mRNAs are associated with proteins that both protect them and inhibit their translation c PolyA tails promote translation initiation and stored mRNAs generally have shorter tails i In some mRNAs of mouse and frog oocytes a normallength polyA tail is added and then trimmed enzymatically ii Particular mRNAs are marked for deadenylation by a region in the 32 untranslated region called the adenylateuridylate AUrich element ARE with the consensus sequence UUUUUAU iii Activation of the stored mRNA occurs when a cytoplasmic polyadenylation enzyme recognizes the ARE and adds about 150 A residues making a fulllength polyA tail mRNA Degradation Control Breakdown turnover of mRNAs occurs in the cytoplasm with mRNAs showing a wide range of stability from minutes to months Regulatory signals may modify the stability of an mRNA 2 mRNA degradation is believed to be a major control point in eukaryotic gene expression Sequences or structures that affect the halflife of mRNAs include a AUrich elements ARE discussed above b Various secondary structures c Deadenylationdependent mRNA decay involves removal A nucleotides from polyA tails until they are too short to bind PAB polyA binding protein In yeast PABdependent polyA nuclease product of the PAN gene may catalyze deadenylation When the tail is almost removed decapping removes the 52 cap The decapping enzyme in yeast is at least partially encoded by the DCPI gene iii After decapping an enzyme in yeast from the XRNI gene aggressively degrades the mRNA from the 52 end by 52to32 exonuclease activity iv Degradation still occurs in dcpl mutants indicating that other mRNA degradation pathways exist 9 Deadenylationindependent mRNA decay includes two types of pathways i Direct decapping eXposing the 52 end to 52to32 exonucleases ii Internal cleavage of the mRNA and then degradation of the fragments 3 Mammalian mRNA degradation mechanisms are less clear than are those of yeast Both deadenylationdependent and deadenylationindependent pathways are found in mammals Protein Degradation Control 1 Protein regulation occurs in many ways Examples 32 A constitutively produced mRNA may be translated continuously and so the protein degradation rate determines its level 6 A shortlived mRNA may make a very stable protein so that it persists for long periods in the cell 2 Protein stability varies from very stable eg lens proteins in the eyes of higher vertebrates to shortlived eg steroid receptors and heat shock proteins 3 Proteolysis protein degradation in eukaryotes requires ubiquitin a protein cofactor m Ubiquitin bound to a protein identi es it for degradation by proteolytic enzymes 6 Ubiquitin is released intact and able to tag other proteins for degradation 4 Protein stability is directly related to the amino acid at the N terminus of the protein the N end rule In yeast stability of the same protein was measured with different N terminal amino acids a The amino acids Arg Lys Phe Leu and Trp all speci ed a halflife of 53 minutes b The amino acids Cys Ala Ser Thr Gly Val Pro and Met all speci ed 5 halflives of 820 hours c Similar results are seen in experiments with E coli The N terminal amino acid directs the rate of ubiquitin binding which in turn determines the halflife of the protein To summarize prokaryotes control gene expression mainly at the transcriptional level While eukaryotes regulate at transcriptional posttranscriptional and posttranslational levels Eukaryotic systems control Transcription 57 33 PrecursorRNA processing Transport from the nucleus 9 0 Degradation of mature RNAs Translation of mRNAs Hz 0 Degradation of protein RNA Interference A Mechanism for Silencing Gene Expression 1 2 5 Small dsRNA fragments can silence the expression of a matching gene This is RNA interference RNAi recently discovered in C elegans 9 Injecting dsRNA into adult worms results in speci c loss of the corresponding mRNA in the worm and its progeny ST RNAi also occurs in many other organisms where it protects against viral infection and regulates developmental processes RNAi is highly speci c and sensitive with only a few molecules of dsRNA needed making it an excellent research tool The mechanism of RNAi silencing of a gene includes Figure 2017 9 Speci c matching of the mRNA exon and dsRNA dsRNA matching the promoter or introns does not cause silencing ST The enzyme Dicer cleaves dsRNA into short interfering RNA siRNA fragments with 32 overhangs 0 The siRNADicer complex recruits other proteins which have the siRNA transferred to them to form the RNAinduced silencing complex RISC The RISC is activated in an ATPdependent manner leading to unwinding of the siRNA F F The singlestranded siRNA is then paired with the complementary mRNA by the activated RISC quot7 An endoribonuclease as yet unidenti ed cleaves the mRNA silencing the gene Activated RISC may also function by P Allowing the singlestranded siRNA to remain bound to the mRNA preventing translation Migration of the complex into the nucleus where siRNA directs binding to the complementary DNA recruiting a chromatin remodeling compleX and silencing transcription The activated RISC may also cause ampli cation of RNAi When the siRNA binds mRNA it may prime RNA synthesis generating a dsRNA molecule that becomes a new substrate for Dicer repeating the cycle and amplifying interference


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