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Transcript
Chapter 1: The Genetic Approach to Biology Questions for Chapter 1 from text: 2, 4, 8, 11, 17 Gene - basic elements of the system of inherited information; the fundamental physical and functional unit of heredity, which carries information from one generation to the next; a segment of DNA composed of a transcribed region and a regulatory sequence that makes transcription possible Allele - one of the different forms of a gene that can exist at a single locus Phenotype - the form taken by some character or group of characters in a specific individual; the detectable outward manifestations of a specific genotype Genotype - the specific allelic composition of a cell, either of the entire cell or, more commonly, of a certain gene or a set of genes Mutation - the process that produces a gene or a chromosome set differing from that of the wild type; the gene or chromosome set that results from such a process Organisms Give rise to like organisms (reproduce) Respond to inputs: - In unicellular organisms, responses to different inputs possible Unicellular responds to environment by changes within the cell Eg) high temp = denatured proteins Eg) when low food = optimizes ability to absorb and retain nutrients - In multicellular organisms, cells derived from a single cell develop complex and diverse structures Multicellular organisms' cells are derived from a single cell develop complex and diverse structures Evolve over time Inherited units = genes Idea of inheritance by Gregor Mendel DNA is the molecule which makes a gene Idea of genes arose in early 1900s What key properties must hereditary material possess? 1. Ability to replicate faithfully (inheritance) 2. Provide an extraordinary diversity of information that can be 3. translated (changed) into structure and function of cells 4. Ability to change over time (evolution) Mutation combined with natural selection allows change over time Humans and chimps differ in 1% of DNA The genomes of humans and chimpanzees differ by only a small percentage of nucleotides 1. DNA Replication Each DNA strand is the template for the production of a new strand - result is two DNA molecules identical to the original DNA molecule is a double-stranded helix Nucleotides are joined by weak hydrogen bonds that can be separated by DNA polymerase or helicase Nucleotides (ATGC) are joined to strand at sulfur and phosphorus Strong covalent bonds occur along each strand (sulfur and phosporus) 2. Diversity of Genes There are four kinds of nucleotide (ATGC) within a molecule Diversity is possible because each gene is made up of thousands of nucleotides 3. Translation of information DNA > RNA > protein Transcription of DNA from a gene makes RNA Eg) Ribosomal RNA does work within cell as RNA Eg) Messanger RNA helps make proteins 4. Mutation Mutation combined with natural selection allows change over time - Errors during replication (enzymes recognize simple errors of Purine-Pyramidine pairing) - Mutation due to exposure to mutagens (chemical, x-ray, UV) Hereditary material (DNA) possesses 3 key properties: 1. DNA replication allows faithful replication that enables inheritance of information both between and within generations 2. Variations in DNA sequence provide an extraordinary diversity of information that can be translated via RNA and protein into cellular structure and function 3. Mutatation of DNA allows change over time Study of genes involves studying the effects of mutant forms of a gene (alleles) on the organism (Physiology, Biochem, development, disease, behaviour, ecology, evolution, etc.) Studying genetic diversity leads geneticists to understand biological processes genetic diversity (different genotypes) results in observable differences in organisms' morphology, physiology, development, ability to withstand environmental stress (phenotype) Albino -During the process genotype > transcription > translation > cellular phenotype result something goes wrong. -A mutant allele (genotype) produces a malfunctioning enzyme that leads to albinism (phenotype) (Figure 1-7) -A malfunctioning enzyme leads to albinism genetic diversity: alleles are different forms of a gene. An allelle is a form/ variant of a gene genotype: Normal DNA sequence = wild type allele Altered DNA sequence = mutant allele Phenotype: Normal morphology = wild type phenotype Abnormal morphology = mutant phenotype Model organisms are commonly used in genetic analysis Need enzymes to cut paste and arrange vectors of DNA Need ways of manipulating and testing organisms Tend to use same species of organism such as lab rats, small plants These model organisms share the following characteristics: Short generation time Small size Easily maintained Large numbers of progeny Small genome Representative (central to taxonomic group) Model Organisms in Genetics: Virus: bacteriophage lambda and its bacterial host Escherichia coli Fungus: Neurospora crassa Budding Yeast: Saccharomyces cerevisiae Plants: Zea maize; Arabidopsis thaliana Insect: Drosophila melanogaster Nematodes: Caenorhabditis elegans (completely transparent under microscope makes it easy to follow cell development) Vertebrates: Mus musculus (mouse), Xenopus laevis (frog) Clicker If a DNA molecule is 100 bases long, how many different combinations are possible? 4^100 33 amino acids are made from 100 bases Example: Consistency Let A = yellow and a= green; also, let the gene symbol be gene A We could also use: P, Y, X, Z Alleles: yel, YEL, yel-1, yel+, yela It doesn't matter what symbol you use, just be consistent Test Cross: When an individual of unknown genotype is crossed with a homozygous recessive individual to determine unknown's genotype through phenotypic result. The test cross progeny are in equal frequency, consistent with Mendels' hypothesis: Suggest parents contain 2 alleles, gametes contain Reproduction (meiosis specifically) accounts for this hypothesis because of alternating mitotic and meiotic divisions during eukaryotic life cycles Diploid > meiosis > haploid > fusion > diploid > mitosis Chapter 2: Single-Gene Inheritance The nuclear genome: all chromosomes of nuclear DNA, genes, and intergenic regions (not mitochondrial, chloroplast or plasmal DNA/RNA for nuclear genome) "-some" means nucleic acid plus proteins centromeres important in movement/ segregation of chromosomes Single-celled organism have smaller genomes, but also pack genes closer together, smaller intergenic regions Looking at a piece of DNA being transcribed: Introns taken out and remaining pieces spliced together DNA Start Exon Stop intron exon intron pre-mRNA Chromosomal DNA wrapped around histones 8 histones come together in a protein group called "octamer" These 8 polypeptides are for different types of histones: 2x(H2A, H213, H3, H4) Nucleosome: octamer wound around two times Chromatin: DNA and associated proteins Chromosomal DNA is wrapped around histone octamer (nucleosome), nucleosomes associate with and coil around H1 histone The nucleosome is 10 nm in diameter Histone octamers coil until 30 nm in diameter to made a solenoid Attachment to scaffold results in supercoiling Scaffold is a set of proteins eg) topoisomerase Telomere = chromosome end Centromere = constricted area of chromosome Nucleolar organizer - tandem repeats of genes encoding rRNA Nucleolus - RNA and protein associated; located around nucleolar organizer Heterochromatin - dense chromatin Euchromatin - less dense chromatin Diploid genome - 2 sets of chromosomes Eg) if 2n=6 there are 3 homologous pairs Gregor Mendel - founder/ father of genetics Firm understanding in physics and mathematics Garden pea: model organism Diploid fast generation time readily available, with multiple varieties easily grown Seedless watermelon is polyploid and infertile Wheat is a hexaploid result of three wild species coming together Three sets of chromosomes doubled in order to multiply Traits can be selected for such as gluten protein, stronger shells around seeds Bigger plants and bigger yield In peas, both cross=pollination and self-polination (selfing) are possible Mendel studied seven phenotypic pairs that were controlled by single genes: Round or wrinkled seed Yellow or green seeds Purple or white flowers Inflated or pinched ripe pods Green or yellow pods Axial or terminal flowers Long or short stem Mendel generated pure breeding lines If after 7 generations of selfing all populations were pure P Pure Yellow YY Pure Green yy F1 all yellow offspring: dominant phenotype Yy F2 Selfing of F1 3 dominant yellow : 1 recessive green YY or Yy yy Mendel's explanation - each parent contains 2 copies of each factor (each parent is diploid with 2 alleles of each gene) Parental lines have 2 identical alleles (homozygous) Each parent has 2 copies hereditary units 2 alleles of the gene Each parent contributes one unit to the offspring (diploid>haploid reproduction) Meiosis produces one haploid gamete from each parent YY - homozygous dominant Yy - heterozygous yy - homozygous recessive The F1 of Yy has 2 different alleles (heterozygous: monohybrid) Monohybrid: one gene involved for character Dominant allele - phenotype is expressed in a heterozygote Recessive allele - phenotype is not expressed in a heterozygote Male Y Yellow Parent y RESULT: Yellow Female Parent Y y YY Yy Yy 3: 1 Ratio of yellow to green yy 3x[½ Y ½ Y = ¼ Y] Yellow ¾ dominant 1x[ ½ y ½ y = ¼ y ] green ¼ recessive Frequency of 4 combinations is product of frequencies of each gamete frequency (product rule): two independently occurring events Independent assortment = separation CHAPTER 2 Definitions: genome - the complement of genetic material in the chromosomes Gene - the fundamental unit of heredity (in molecular terms, a transcribed segment of DNA and the region which enables transcription) Intergenic region - region of DNA between genes Intron - (intervening sequence) a segment of the gene that is initially transcribed, but is not in the final mRNA product Exon - a segment of the gene that is translated into the protein nucleosome - 8 histone polypeptides (octamer) wrapped in DNA Telomere - end of chromosome centromere - part of chromosome to which the spindle fibres attach via the kinetochore heterochromatin - densely staining, highly condensed regions of the chromosome that are usually not transcribed Euchromatin - a less condensed region, thought to contain genes that can be transcribed haploid - a cell or individual having one chromosome set diploid - a cell or individual having two chromosome sets homozygous (= true-breeding): an individual having identical alleles of a gene heterozygous: an individual having different alleles of a gene monohybrid: an individual heterozygous at one gene first filial (F1) generation - the first generation resulting from a controlled cross between two known parents (P) second filial (F1) generation - the second generation resulting from a controlled cross between two known parents (P) test cross - a cross between an individual of unknown genotype and an individual homozygous recessive for a particular gene(s) product rule: the probability that two independent events occur together is equal to the product of their individual probabilities sum rule: the probability that either of two mutually exclusive events occurs is equal to the sum of their individual probabilities genetic symbol: a symbol designating a gene which usually relates to the characteristic controlled by that gene (e.g. Y = yellow gene) allelic symbol: a symbol designating an allele which is a derivative of the gene symbol (e.g. Y = yellow allele, y = green allele) sex chromosome - a chromosome whose presence is associated with a particular sex autosome - a chromosome that is not a sex chromosome sex linkage - the location of a gene on a sex chromosome hemizygous - a gene present in only one copy in a diploid organism Text from Lectures: Single-Gene Inheritance The nuclear genome (Figure 2-2) DNA is condensed to form chromosomes Representative chromosomal landscapes in different organisms; note differences in gene density, presence of introns (Figure 2-7) 2 regions of human chromosome 21 showing banding patterns, gees with known transcription products (arrows) Figure 2-8 A diploid genome (Indian mutjac, 2n=6) visualized by dye specific to each chromosome (Figure 2-3) Interphase cell, chromosomes are in particular domains of the nucleus Cell beginning division phase, chromosomes are compact, clearly showing 2 members of each homologous pair Chromosomal DNA is wrapped around histones Chromatin = DNA + associated proteins (Figure 2-4a) Space-filling model shows DNA wrapped around histone octomer to form nucleosomes Histone octamer (green and blue) = 2x H2A, H2B, H3, H4 Yellow = histone H1 Chromosomal DNA is wrapped around histone octamer (nucleosome), nucleosomes associate with and coil around H1 histones (Figure 2-4b) 6 nuclesomes + 6 H1 histones = 1 coil note compaction Attachment to scaffold results in supercoiling (Figure 2-5) Some landmarks of tomato chromosome 2 (Figure 2-6) Telomere = chromosome end centromere = constricted area of chromosome Nucleolar organizer - tandem repeats of genes encoding rRNA Heterochromatin - dense chromatin euchromatin - less dense chromatin The founder of genetics, Gregor Mendel Figure 1-1 Mendel choose garden pea as a "model organism" Diploid Fast generation time Readily available, with multiple varieties Easily grown In peas, both cross-pollination and self-pollination (selfing) are possible (Figure 2-10) Mendel studied seven phenotypic pairs (Figure 2-9) with clearly contrasting and easily distinguished phenotypes Grew plants for several successive generations to be certain lines were true-breeding For each characteristic, Mendel crossed pairs of pure-breeding parents Figure 2-12 part 1 For all crosses the first filial generation (F1), showed only one phenotype; defined as the dominant phenotype, the other phenotype defined as recessive (Figure 2-12 part 2) For all crosses, the second filial generation (F2), showed only two phenotypes, and the ratio was always 3:1 (the larger class was the same phenotype as that in the F1; the dominant phenotype) (Figure 212 part 3) Mendel's explanation - each parent contains 2 copies of each factor; each parent is diploid with 2 alleles of each gene) Parental lines have two identical alleles, they are homozygous (Figure 2-12 part 5) Mendel's explanation - each parent contributes 1 copy of each factor to offspring with equal frequency (Mendel's law of equal segregation); gametes are haploid, fuse to form diploid offspring The F1 has 2 different alleles (heterozygous; monohybrid) Dominant allele - phenotype is expressed in a heterozygote Recessive allele - phenotype is not expressed in a heterozygote Figure 2-12 part 6 Mendel's explanation - F1 contributes 1 allele to F2 offspring - alleles fuse randomly with other (haploid gametes fuse to generate diploid offspring) Figure 2-12 part 7 Mendel's explanation - random fusion of gametes results in 4 combinations: Figure 2-12 part 11 Frequency of 4 combinations is product of frequencies of each gamete frequency (product rule): i.e. ½ x ½ = ¼ Yields a genotypic frequency of 1YY : 2Yy : 1 yy Yields a phenotypic frequency of 3 Y_ : 1 yy Figure 2-12 part 12 Next, Mendel performed a test cross - crossing one individual to an individual that is homozygous recessive Figure 2-12 part 4 If the F1 plant is heterozygous, it should generate two types of gametes with equal frequency I.e. F1 gametes are Y = ½, y= ½ Tester parent (y) generates only one type of gamete, y Therefore, phenotype of test cross progeny should reflect frequency of gametes from the F1 i.e. 1 Yy : 1 yy (Figure 2-12 part 13) Test cross progeny are in equal frequency, consistent with Mendel's hypothesis (Figure 2-12 part 15) Mendel's crosses are experiments that allow determination of number of genes, relationships amongst alleles, and genotypes of individuals (Figure 2-11) (note, the progeny of the left cross in Figure 2-11 is not the F2, rather it is the test cross progeny) Clicker Question #1: in cross 6, what is the dominant phenotype? a) axial b) terminal Clicker question #2: Two different alleles of the gene for plant height exist in peas. True breeding individuals may be either tall or short. Two tall individuals are crossed to one another, and the progeny are ¾ tall and ¼ short. The best explanation for this result is: a) both tall individuals are homozygous, the tall allele is dominant b) both tall individuals are homozygous, the short allele is dominant. c) both tall individuals are heterozygous, the short allele is dominant. d) both tall individuals are heterozygous, the tall allele is dominant. E) none of the above accurately explains the results Clicker Question #3 If a test cross is done to a heterozygous individual with yellow seeds, what phenotype(s) will the progeny be, and in what ratio? A) All yellow B) All green C) 3 yellow: 1 green D) 1 yellow : 1 green E) None of the above Stages of Mitosis Interphase Early mitotic prophase Late mitotic prophase Mitotic metaphase Mitotic anaphase Mitotic telophase Meiosis- two sequential nuclear divisions: MI - segregates homologous chromosomes MII - segregates chromatids Generates haploid daughter cells Stages of Meiosis Prophase I Chromosomes pair to form bivalent (2 dyads or double stranded = tetrad) Leptotene, zygotene, pachytene, diplotene Crossing over occurs Homologous chromosomes pair (fundamental difference between mitosis and meiosis, as well as between meiosis I and meiosis II) Homologous chromosomes: Same genes, possibly different alleles Metaphase I Paired homologous chromosomes (bivalent) align at metaphase plate Anaphase I Segregation Telophase I Reformation of nuclear membrane, cell division may occur Cells are haploid from this point onwards Prophase II Chromosomes recontract Metaphase II Chromosomes (dyads) line up Anaphase II Centromeres split and sister chromatids separate Telophase II Nuclei reform in 4 haploid cells Comparision of meiosis and mitosis Mitosis - separation of chromatids Meiosis I and II I: separation of homologous chromosomes II: separation of sister chromatids If we consider the behaviour of genes on chromosomes, alleles segregate during Meiosis I (separation of homologous chromosomes) Mendels' ratios predict the segregation Demonstration of equal segregation is more straightforward within a haploid life cycle such as in yeast Two mating types fuse to form meiocyte You cannot assess (score) the phenotype of the haploid generation Assigned problems: 9, 17, 20, 21, 22, 26, 28, 32, 36 Chapter 2b - Mitosis and Meiosis Definitions: Mitosis - cell division that produces 2 daughter cells with identical nuclei to the parent cell Meiosis - a process consisting of 2 consecutive cell divisions that produces daughter cells (gametes or sexual spores) with half the genetic material of the parent Meiocyte - a diploid cell about to undergo meiosis Chromatid - one of 2 daughter DNA molecules produced by chromosome replication Dyad - a chromosome with 2 chromatids; a double-stranded chromosomes Homologous pair - chromosomes that pair with each other at meiosis and contain the same genes (may have different alleles) Bivalent - two homologous chromosomes paired in Meiosis I Synaptonemal complex - complex that allows chromosome pairing during Prophase I Kinetochore - proteins that attach the centromere to the kinetochore microtubules Reduction division - division that results in cells with one member of each homologous pair (i.e. Meiosis I) Crossing over - the exchange between homologous chromosomes of parts of a chromatid through breakage and rejoining Mendel's hypotheses: suggest parents contain 2 alleles, gametes 1 allele gametes fuse to regenerate diploid in progeny what processes explain these ideas? what mechanism accounts for separation of alleles into gametes? Alternating mitotic and meiotic divisions during eukaryotic life cycles account for Mendel's diploid to haploid transitions (Figure 2-14) Stages of the asexual cell cycle (figure 2-13) During the S phase, DNA molecules replicate to form identical chromatids (Figure 2-18) Replicated chromosomes are segregated into daughter cells in an organized fashion during Mitosis - generates cells identical to mother cell Meiosis - generates haploid cells from diploid Stages of Mitosis (Box 2-1) Interphase = G1, S, G2 (growth + DNA replication) Figure 2-15 Prophase = chromosome condensation, break down of nuclear membrane (Figure 2-15) Metaphase = alignment of chromsomes along metaphase plate (Figure 2-15) Alignment requires action of kinetochore microtubules, attached to centromeres, (Figure 2-23) Anaphase = separation of chromosomes (Figure 2-15) Telophase = cytokinesis, reformation of nuclear membrane (Figure 2-15) Mitosis generates daughter cells of identical ploidy as original cell (Figure 2-15) Meiosis - two sequential nuclear divisions: MI - segregates homologous chromosomes MII - segregates chromatids Generates haploid daughter cells (Figure 2-15) Stages of Meiosis (Box 2-2) Prophase (Figure 2-15 part 7): Chromosomes pair to form bivalent (2 dyads = tetrad) Crossing over occurs Synaptonemal complexes at meiosis (Figure 2-16a) Figure 2-16b MetaphaseI - paired homologous chromosomes (bivalents) align at metaphase plate (Figure 2-15) Anaphase I - segregation of homologous chromosomes (dyads) due to kinetochore shortening (Figure 215) Telophase I - reformation of nuclear membrane, cell division may occur - cells are haploid from this point onward (Figure 2-15) Prophase II - chromosomes recontract (Figure 2-15) Metaphase II - chromosomes (dyads) align at the metaphase plate (Figure 2-15) Anaphase II - centromeres split and sister chromatids separate (Figure 2-15) Telophase II - nuclei reform in 4 haploid cells (Figure 2-15) Comparison of meiosis and mitosis Meiosis II - separation of sister chromatids • -iosisI-atseparoatichrnoomathoids ogouschromosomes tosis (dyads Me o f mol Mi separ ion f If we consider the behaviour of genes on chromosomes, alleles segregate during Meiosis I (separation of homologous chromosomes) Mendel's ratios predict the segregation of chromosomes during meiosis (Figure 2-12 part 15) Demonstration of equal segregation is more straightforward within a haploid life cycle such as in the yeast S. cerevisiae (Figure 2-17) Two mating types (MAT and MATa) fuse to form meiocyte MAT = r+ (white colonies) MATa = r (red colonies) Figure 2-17 part 2 Meiocyte undergoes meiosis to produce four products (2r, 2r+) - these form colonies directly Segregation can be followed based on phenotype or by DNA sequence differences (Figure 2-20) Assigned problems: 9, 17, 20, 21, 22, 26, 28, 32, 36 Ch 2 Questions 29, 31, 38, 40, 46, 48, 52 Meiocyte: undergoes meiosis to produce four products; these form colonies directly Segregation can be followed based on phenotype or by DNA sequence differences Probe - a piece of DNA that has a complementary sequence to the DNA of study Must be long enough Must be labelled in some way to visualize it (could be labelled with radioactivity to come up on Xray; flourescent probe; enzyme to change colour) genomic DNA - cut with an enzyme (restriction enzymes); cuts at specific sequence Genotipic Electrolysis Restriction enzyme cuts at a site along DNA strand gel in an electric field put DNA in a hole in the gel; DNA runs down to result in a smear of DNA with large DNA fragments on top and smaller fragments falling down to bottom of gel This smear is added to a membrane, where a probe can be used Forward genetics: Discovering genes controlling flower development Sex-linked single-gene inheritance patterns Genes that show different phenotypic ratios in the sexes Autosome - non sex chromosome 2 copies in all individuals Sex Chromosomes - sexes have different copies eg) XX = female; XY = male Homogametic sex (XX) female in humans and drosophilla Heterogametic sex (XY) male in humans and drosophilla Pseudoautosomal region - sequence similarity allows chromosome pairing Differential region - accounts for sex linkage Drosophilla Wild type red eyes = w+ Wild type white eyes = w Reciprocal Crosses: Yellow x green and green x yellow will always have same phenotypic results because coloration is an autosome In Mendels' monohybrid crosses.. Reciprocal Crosses (red(f)x white(m) and white(f) x red(m)) give different results for genders Xw+Xw+ by XwY red (f) x white (m) Result: all red eyes Pedigree analysis: applying Mendelian principles to human inheritance patterns Characteristics of autosomal dominant pedigree: Trait apprears in every generation Affected parents have unaffected children No correlation between sex and a particular phenotype Chapter 2 - Sex linkage and Pedigrees Definitions autosome - a chromosome that is not a sex chromosome sex linkage - the location of a gene on a sex chromosome hemizygous - a gene present in only one copy in a diploid organism Chapter 2 questions (second set): 29, 31, 38, 40, 46, 48, 52 • Sex-linked single-gene inheritance patterns genes that show different phenotypic ratios in the sexes Model Organism: Drosophila, as in humans chromosomal based sex determination, XX (homogametic sex) = female, XY (heterogametic sex) = male Human sex chromosomes - X and Y chromosomes Differential region - accounts for sex linkage (Figure 2-25) Pseudoautosomal region - sequence similarity allows chromosome pairing Red-eyed and white-eyed Drosophila (Figure 2-26), isolated by Thomas Morgan, 1910 Wild type = w+ (red eyes) Mutant = w (white eyes) In Mendel's monohybrid crosses, reciprocal crosses yielded identical results; in the case of red and white eyed flies, reciprocal crosses yielded different results Reciprocal crosses (red x white and white x red) give different F1 results (Figure 2-27) Reciprocal crosses (red x white and white x red) give different F2 results (Figure 2-27) Pedigree analysis - applying Mendelian principles to human inheritance patterns (Figure 2-28) Is the trait shown in generation IV autosomal or sex-linked? Dominant or recessive? (Figure 2-29) • • No tendency of phenotype with a particular sex (note small sample size): suggests autosomal Unaffected individuals give rise to affected individuals: recessive If affected individuals in generation IV are homozygous recessive, what genotype must their parents have been? - parents must be heterozygous in order to produce affected offspring (Figure 229) Phenotype of siblings = dominant phenotype Genotype of siblings in generation IV? •Must have A; may be AA or Aa (indicated as A/-); 2/3 Aa, 1/3 AA What about genotypes of grandparents (generation II)? •Both are unaffected, therefore both have dominant allele •At least one must carry the recessive allele •If condition is rare, unlikely that both carry recessive allele Siblings in generations II and III •Unaffected, therefore must be A •Other allele is unknown (either A or a); designate as A/In generation III, individuals have 50% chance of being AA, 50% chance of being Aa Genotypes of Great-grandparents (generation I)? •Unaffected, therefore must have A allele •Produce heterozygous progeny, therefore at least one must have allele a, unlikely that both have allele a Pseudoachondroplasia phenotype - inherited as an autosomal dominant (Figure 2-30) • • • Characteristics of autosomal dominant pedigree: Trait appears in every generation Affected parents have unaffected children No correlation between sex and a particular phenotype If trait is dominant, all unaffected individuals must be homozygous recessive Affected individuals must have dominant allele (figure 2-31) Examples of human dominant traits: Huntington disease: dominant autosomal, late onset means that affected individuals have often had children before disease manifests (Figure 2-32) Extra digits (Polydactyly) also a dominant autosomal trait (figure 2-33) Pedigree of family showing Polydactyly - phenotype of both hands (top L,R) and feet (bottom L,R) indicated (Figure 2-33b) • • • X-linked recessive inheritance (Figure 2-36-Figure 2-38) Phenotypeis more commoninmales thanfemales Affected fathershave no affected children Traitpassed from grandfather tograndson Inheritance in an X-linked dominant disorder (Figure 2-39) Fathers pass trait to all daughters but no sons Mothers pass trait to 1/2 sons and daughters Y-linkage: Phenotype passed from father to all sons but no daughters Maleness - primarily determined by SRY gene on the differential region of the Y chromosome Hairy ears: a phenotype proposed to be Y linked (Figure 2-40) Clicker Question #1: Which of the following events occurs in Meiosis but not Mitosis? a) Chromosome condensation b) Separation of chromatids c) Pairing of homologous chromosomes d) Alignment of chromosomes along metaphase plate e) None of the above Clicker question #1 In one cross, a female Drosophila with normal bristles is crossed to a male with forked bristles. The F1 has all normal bristles. In a second cross, a female with forked bristles is crossed to a male with normal bristles. The F1 is 50% wild type and 50% forked bristles. The best description of the forked allele is: a) It is dominant and autosomal b) It is recessive and autosomal c) It is recessive and sex-linked d) It is dominant and sex-linked CHAPTER 3: Independent Assortment of Genes First problem set: 1, 2, 6, 8, 9, 17, 19, 46 Second problem set: 13, 28, 37, 39, 41 dihybrid: an individual heterozygous at two genes product rule: the probability that two independent events occur together is equal to the product of their individual probabilities sum rule: the probability that either of two mutually exclusive events occurs is equal to the sum of their individual probabilities The green revolution in agriculture -Rice, wheat and corn that are shorter, disease resistant -Norman Borlaug - greatest man of the 20th century, awarded Nobel Peace Prize, thought to have saved lives of a billion people Mendel's law of independent assortment = Mendel's second law Mendel studied seven phenotypic pairs (Figure 2-9) what happens if pairs of traits are considered together? Mendel crossed round, green seeded individuals (R/r.y/y) by wrinkled yellow individuals (r/r.Y/Y) - "." indicates relative position of genes is unknown (Figure 3-4) F1 individuals show dominant traits for both characters - they are heterozygous at both genes (R/r.Y/y) = dihybrid Deriving 9:3:3:1 using a Punnett's square If genes segregate independently of one another, what gametes do we expect and in what frequencies? (Figure 3-4) ¼R.Y ¼ R.y ¼ r.Y ¼ r.y Punnett's square predicts that random fusion of four gamete types will give rise to a 9:3:3:1 ratio Mendel's second law: During gamete formation, the segregation of alleles of one gene is independent of the segregation of the alleles of another gene The product rule predicts a 9:3:3:1 ratio from two independent monohybrid (3:1) ratios Using a branch diagram to derive 9:3:3:1 Test-crossing a dihybrid Y/y.R/r x y/y.r/r If a dihybrid is testcrossed, expect the frequency of phenotypes in the test cross progeny to reflect the gamete frequency of the dihybrid Calculating genotypic ratios: Genotype ratior for F2 of a dihybrid will be (1:2:1 x 1:2:1) = 1:2:1:2:4:2:1:2:1 Calculating individual probabilities: If only one fusion event results in a particular genotype Use product rule: the probability of independent events is the product of their individual probabilities e.g. p(yyrr) = 1/4 x 1/4 = 1/16 If more than one fusion event results in a particular genotype -Apply both product rue and sum rule -Sum rule: The probability of either of two mutually exclusive events occurring is the sum of their individual probabilities e.g. P (YyRr) = [p(YR) x p(yr)] + [p(yr) x p(YR)] +[p(Yr) x p(yR)] + [p(yR) x p(Yr)] = 1/16 + 1/16 + 1/16 + 1/16 = 1/4 How to calculate the frequency of progeny from a cross where multiple genes are considered? E.g. in the cross: AaBbCcDdEe x AabbCCddEe what is the probability of an individual with genotype A_bbC_D_ee) = 3/4 x 1/2 x 1 x 1/2 x 1/4 = 3/64 The following pedigree concerns a rare disease: Is the disease causing allele: a)autosomal, dominant b)Autosomal recessive c)Sex-linked recessive d)X-linked dominant e)None of the above What is the probability that A and B will have an affected child?: a)1/4 b)1/12 c)1/16 d)1/8 e)None of the above Two dihybrid individuals (A/a B/b) are crossed. If A and B are independently segregating, what is the probability of progeny that is homozygous recessive at both A and B? A) 9/16 B) 1/4 C) 3/16 D) 1/16 E) none of the above Chapter 3 - Independent Assortment of Genes Mendel's second law predicts independent segregation of chromosomes during anaphase I of meiosis Interphase: chromosomes are unpaired Prophase I: chromosomes and centromeres have replicated, but centromeres have not split Metaphase I: chromosomes align along plate Anaphase I: centromeres attach to spindle and are pulled to poles of cell Telophase I: two cells form AnaphaseII: New spindles form and centromeres finally divide End of meiosis Four cells produced frm each meiosis Recombinants are meiotic output different from meiotic input Parentals - meiotic output identical to meiotic input Two ways to form a dihybrid 1 AABB x aabb AB ab P Meiotic input F1 or 2 AAbb x aaBB Ab aB AaBb Meiotic output ¼ AB ¼ab ¼Ab ¼aB 1 parental " recombinant " 2 ¼ AB recombinant ¼ ab " ¼ Ab parental ¼ aB " See textbook Fig3-12 Recombinant gamete frequency inferred from phenotypic frequency in test cross progeny In a haploid life cycle (eg. Neurospora) products of single meiocyte demonstrate principle of independent segregation. Fig 3-9. Eight asci are produced because after meiosis, each gamete undergoes mitosis. Segregation of homologous chromosomes and sister chromatids are reflected in the arrangement of spores If two strains are crossed, two octad types result in equal frequency Organelle Genomes The genomes within chloroplasts and mitochonrion Mitochondrial DNA and chloroplast DNA are relatively small Contain important genes; ribosomal RNA tRNA and genes controlling cellular energy Mitochondrial mutation Poky = slow growth phenotype in Neurospora - if poky females crossed with wide type males, all haploid progeny show poky phenotype, whereas nuclear gene (ad+) shows a 2:2 segregation In reciprocal cross (wild type femals crossed to poky males), all haploid progeny show wild type phenotype of offspring depends on maternal genotype (maternal inheritance) In reciprocal cross (wild type females crossed to poky males), all haploid progeny show wild type phenotype, nuclear gene (ad+) still shows 2:2 segregation - phenotype of offspring depends on maternal genotype (maternal inheritance) Variegated leaves caused by a mutation in cpDNA (Figure 3-21) Variegated leaves - mixture of pigments Due to segragation of organelles within mitotically dividing organism Some branches all white Some branches all green Main shoot and some branches both green and white = variegated (eg) this can be mutant and wild chloroplasts Cytoplasmic inheritance - phenotype of offspring depends on genotype of maternal cytoplasm (white, green, or variegated) Crosses using flowers from a variegated plant (Figure 3-22) - phenotype of offspring depends on cytoplasmic genotype of mother (cytoplasmic inheritance) Sites of mtDNA mutations in certain human diseases e.g. MERRF (mycoclonic epilepsy and ragged red fibre) single base change in mtDNA (Figure 3-24) Typical maternal inheritance in a human pedigree (Figure 3-25) Chi-Square (2) Test Use a statistical test to determine if the observed results are significantly different from the expected e.g. in a test cross AaBb x aabb 1) develop null hypothesis The genes, A and B, are independently assorting (i.e. they are not linked) and the test cross progeny do not deviate from the expected 1:1:1:1 ratio Calculate 2 value X2 = sum of (observed-expected)2/expected = (o - e)2/e degrees of freedom (df) = number of classes (n) -1 =3 compare the calculated chi-square value to the critical value from the chi-square table (Table 3-1) -if -if 2 is < critical value, P is > than 0.05, accept null hypothesis 2 is > critical value, P is < than 0.05, reject null hypothesis 100 > 7.815, reject hypothesis Clicker question 1: The following cross is done: Aa;bb;cc;DD;Ee;Ff;Gg x aa;Bb;CC;Dd Ee;ff;Gg In the progeny, what is the probability of an individual having the phenotype A_ bb C_ D_ ee ff G_? A) 0 B) 3/64 C) 3/128 D) 1/128 E) none of the above CHAPTER 4 - Mapping Eukaryote Chromosomes by Recombination Question set 1: 1, 2, 3, 6, 7, 8, 10, 12, 13, 16 Question set 2: 21, 22, 24, 25, 36, 37, 42, 50, 51, 53, 56 recombinant product: a product of meiosis that differs from the haploid genotypes that gave rise to the diploid meiocyte parental product: a product of meiosis that is the same as the haploid genotypes that gave rise to the diploid meiocyte dihybrid: an individual heterozygous at two genes coupling configuration - arrangement of linked alleles in a dihybrid such that the two dominant alleles are on one chromosome, the two recessive alleles are on the other chromosome (AB/ab) cis dihybrid - a dihybrid in which the alleles are coupled (AB/ab) repulsion configuration - arrangement of linked alleles in a dihybrid such that the a dominant and a recessive allele are together on each chromosome (Ab/aB) trans dihybrid - a dihybrid in which the alleles are in repulsion (Ab/aB) map unit (centimorgan, cM): the distance between two genes where one product of meiosis out of 100 is recombinant gene locus: the position of a gene within the genome linked - two genes that are on the same chromosome linkage group - a set of genes that are on the same chromosome William Bateson and Reginald Punnett Sweet Pea P - gene for flower colour L - gene for seed shape Expected independent segregation of P and L meaning: 1:1:1:1 ratio of dihybrid test cross progeny 9:3:3:1 ratio of dihybrid F2 progeny Instead, following selfing of a dihybrid, ratio was quite different from 9:3:3:1 and Bateson and Punnett observed: P and L are coupled Dominant alleles stay together and recessive alleles stay together Larger classes are those where P is together with L, or p together with l = configuration of alleles in parents Smaller classes are those where P is together with l, or p with L = different configuration from parents Thomas Morgan - Drosophila Eye colour - pr = purple, Wing shape - vg = vestigial, pr+ = red, where pr + > pr vg + = normal, where vg + >vg Like Bateson and Punnett, Morgan observed configuration of alleles in parental lines is more common in F2 progeny Hypothesis - pr and vg are on the same chromosome Experiment - test cross First cross: Parental cross: pr+/pr+.vg+/vg+ x pr/pr.vg/vg F1 progeny: pr+/pr.vg+/vg If genes assort independently, what phenotypic ratio do you expect in the test-cross progeny? Expected result is 1:1:1:1 Results: Configuration of alleles in parental lines is more common in test cross progeny Dominants are together, recessives together = coupling = cis (adjacent) conformation Annotation: pr+ vg+ /pr vg Second cross: Parental cross: pr+/pr+.vg/vg x pr/pr.vg+/vg+ F1 progeny: pr+/pr.vg+/vg Results: Configuration of alleles in parental lines is more common in test cross progeny Alleles are in repulsion = trans (opposite) configuration = trans dihybrid Annotation: pr+ vg /pr vg+ Conclusion: alleles that are together in the parents are more likely to stay together through subsequent generations Independent of dominance/recessive Conclusion: genes are linked (on the same chromosome) Linked alleles tend to be inherited together (figure 4-2) Morgan suggested that crossing over produces allelic combinations that are different from the meiotic input (recombinants) (Figure 4-3) Morgan suggested that chromosomes break and re-join during pairing, and that chiasmata are visual evidence of that event Chiasmata in grasshopper testes (Figure 4-4) 4.1 CROSSING OVER OCCURS DURING PROPHASE Morgan proposed that crossing over is between chromatids, not chromosomes (Figure 4-5) A second-division segregation pattern in a fungal octad demonstrates that crossing over must be between chromatids (Figure 4-21) Multiple crossovers can include more than two chromatids to generate multiple types of linear tetrads (figure 4-6) SUMMARY: Crossovers generate recombinant meiotic products (Figure 4-7) 4.2 MAPPING BY RECOMBINANT FREQUENCY Morgan's results: If test crosses were done involving different pairs of genes, saw different ratios of parental: recombinant Morgan's hypothesis: frequency of recombinant products may be related to distance between genes Alfred Sturtevant: If Morgan's hypothesis is true, should be able to use recombinant frequency to establish spatial relationship between genes on chromosomes - i.e. a chromosome map Alfred Sturtevant - data from Morgan's lab, generated formula map distance = percent recombinants Predicted that if calculated distance reflect linear position on chromosome, map distances should be additive (Figure 4-9) Sturtevant's map of the Drosophila X-chromosome Longer regions have more crossovers and thus higher recombinant frequencies (Figure 4-10) For linked genes, recombinant frequencies are less than 50 percent (Figure 4-8) (because it is more difficult for genes to separately cross over when they are positioned one on top of the other on a chromatid) If two genes are more than 50 cM apart on the same chromosome, cross-overs will occur so frequently that the genes will assort independently, therefore recombination frequency never exceeds 50% For these genes, we can only detect that they are on the same chromosome by showing that each is linked to a common gene Knowing recombinant frequency, we can calculate map distance: Map distance (mu or cM) = % recombinant products Knowing the map distance between genes, we can predict the frequency of recombinant products in a cross Calculated map distance is correlated to physical distance on DNA molecule - knowing the position of the gene in the chromosome allows us to find the sequence of gene and hence better understand its structure and function Clicker question #1 In Arabidopsis, seeds with the dominant TT8 allele are brown, whereas those with recessive tt8 allele are yellow; seedlings with dominant HY4 alelle have short hypocotyls, while those with recessive hy4 have long hypocotyls. A test cross is performed on dihybrid individual and the following progeny are obtained: TT8 HY4 24 TT8 hy4 178 tt8 HY4172 tt8 hy4 26 The best description of the dihybrid is: a) TT8/tt8.HY4/hy4 b) TT8/tt8;HY4/hy4 c) TT8 HY4/tt8 hy4 d) TT8 hy4/ tt8 HY4 e) None of the above Clicker question #2 A maize plant is dihybrid for two genes BR and DW. This dihybrid individual (BR/br DW/dw) is testcrossed, and progeny of the following phenotypes are seen: BR DW 240 br dw 234 br DW 64 BR dw 72 The best description of the genes in the dihybrid is: a) they are coupled b) they are sex-linked c) they are independently assorting d) none of the above Clicker question #3 In Arabidopsis, seeds with the dominant TTG allele are brown, whereas those with recessive ttg allele are yellow; seedlings with dominant HY4 alelle have short hypocotyls, while those with recessive hy4 have long hypocotyls. A test cross is performed on dihybrid individual and the following progeny are obtained: TTG HY4 24 TTG hy4 178 ttg HY4172 ttg hy4 26 The best description of the dihybrid is: a) TTG/ttg.HY4/hy4 b) TTG/ttg;HY4/hy4 c) TTG HY4/ttg hy4 d) TTG hy4/ ttg HY4 e) None of the above Clicker question #4 In Arabidopsis, seeds with the dominant TTG allele are brown, whereas those with recessive ttg allele are yellow; seedlings with dominant HY4 alelle have short hypocotyls, while those with recessive hy4 have long hypocotyls. A test cross is performed on dihybrid individual and the following progeny are obtained: TTG HY4 178 TTG hy4 24 ttg HY426 ttg hy4 172 The truebreeding parents of the dihybrid must have had genotypes a) TTG TTG hy4 hy4 x ttg ttg HY4 HY4 b) TTG TTG HY4 HY4 x ttg ttg hy4 hy4 c) TTG ttg HY4 hy4 x TTG ttg HY4 hy4 d) TTG HY4 x ttg hy4 e) None of the above Chapter 4 - Mapping Eukaryote Chromosomes by Recombination Questions: 1, 2, 3, 6, 6, 7, 10, 11, 12, 13, 16 Sweet Peas P - gene for flower colour L - gene for seed shape Expected segregation of P and L 1:1:1:1 ratio of dihybrid test cross 9:3:3:1 dihybrid progeny Cross: Purple Long PPLL F1: PpLl F2: 9 P_L_ 3 3 1 X x red round ppll (purple, long) Expect (215) Observe (284) increased decreased decreased increased William Bateson and Reginald Punnett observed: Larger classes are those where P is together with L or when p with l = configuration of alleles in parents Smaller classes are those where P is together with l, or when p with L = different configuration from parents Coupling: dominants together and the recessive together Thomas Morgan: Drosophila Nobel Prize 1933 Eye colour - pr = purple pr+ = red, pr+ > pr Wing shape - vg = vestigial vg+ = normal, vg+ > vg Like Bateson and Punnett, Morgan observed configuration of alleles in parental lines is more common in F2 progeny Hypothesis: pr and vg are on the same chromosome Morgan: If test crosses were done involving different pairs of genes, saw different ratios of parental : recombinant The frequency of recombinant products may be related to distance between genes Independently assorting results in 1:1 : 1:1 if they are liked then the ratio changes P increase : P decrease Alfred Sturtevant said that if this is true, should be able to use recombinant frequency to establish spatial relationship between genes on chromosomes - ie. a chromosome map Predicted that if calculated distance reflect linear position on chromosome, map distances should be additive % recombinant products = map unit (called m.u. or centiMorgans or cM) Transdihibrid AaBb x aabb 95 Ab 95 aB 6 ab 4 AB Parental phenotype " Recombinant Recombinant Map Distance = observed (1) + observed (2) x 100% total products Longer regions have more crossovers and thus higher recombinant frequencies see txt Fig 4-10 independently assorting 50% recombination frequency for linked genes, recombinant frequencies are always less than 50% if two genes are more than 50cM apart on the same chromosome, recombinations will occur so frequently that the genes will assort independently, therefore recombination frequency never exceeds 50 % Knowing recombinant frequency, we can calculate map distance: map distance (mu or cM) = %recombinant products Knowing the map distance between genes, we can predict the frequency of recombinant products in a cross Transdihibrid for A_C_ is test crossed. What is the frequency of "ac" phenotypes in the progeny? Ac / aC A c a C we would expect that if they are 20 mu apart; if the recombinant products (Ac/aC) are 20% then we want half of those for "ac" so the percent is 10% for "ac" Transdihibrid is selfed What is the frequency of "ac" phenotypes on F2? The allele "a" is recessive. Parentals: need specifically from mother and father Female "ac" x Male "ac" Aacc = (Ac)(ac) + (ac)(Ac) = 0.1 x 0.1 = 0.01 Therefore the frequency is 1% Calculated map distance is correlated to physical distance on DNA molecule - knowing the position of the gene in the chromosome allows us to find. CHAPTER 4 - Mapping Eukaryote Chromosomes by Recombination Question set i: 1, 2, 3, 6, 7, 8, 10, 12, 13, 16 Question set ii: 21, 22, 24, 25, 36, 37, 42, 50, 51, 53, 56 trihybrid: an individual heterozygous at three genes interference: a measurement of the degree to which a cross-over in one interval inhibits a cross-over in a second interval THREE POINT TEST-CROSS Figuring out the distance between 3 genes Mapping the distance between three genes on the same chromosome LFY - flowers lfy - leaves CER3+ - epidermal waxYI - green inflorescence cer3 - no wax yi - yellow inflorescence Following test crossing to a trihybrid individual, the following phenotypes were scored in the progeny: 450 lfy CER3 YI 462 LFY cer3 yi 1 lfy cer3 yi 1 LFY CER3 26 YI 32 lfy CER3 yi 17 LFY cer3 YI 11 LFY CER3 yi lfy cer3 YI From the data, we know that the true-breeding parents that were crossed to generate the F1: Parental cross: lfy CER3 YI x LFY cer3 yi F1: lfy CER3 YI / LFY cer3 yi METHOD 1 - Mapping 3 genes by calculating 3 pairwise distances: 1. Consider LFY- CER3 distance (ignore YI) distance between LFY and CER3 = % recombinants between LFY and CER3 Distance LFY and CER3 = ( lfycer3yi + LFYCER3YI + LFY CER3yi + lfycer3YI)/1000 recombinant frequency = (1 + 1 + 11 + 17)/1000 = 0.03 =3% map distance = 3.0 cM 2. Consider CER3 - YI distance (ignore LFY) Distance between CER3 and YI = % recombinants between CER3 and YI recombinant frequency = (26 + 32 + 11 + 17)/1000 = 0.086 = 8.6% map distance = 8.6 cM 3. Distance between LFY and YI (ignore CER3) distance between LFY and YI = % recombinants between LFY and YI recombinant frequency = (1 + 1+ 26 + 32)/1000 = 0.06 =6% map distance = 6 cM [----------------------- (8.6) ------------------------] CER3 LFY YI [-------- 3 --------][--------------- 6 ---------------] Notice that 3 + 6 is not 9 in this case, therefore we made a calculation error. We missed a recombinance when we calculated the distance between CER3 and YI. Therefore we counted something as parental instead of recombinance. Distance CER3 and YI = 26 + 32 + 17 + 11 + 2(1+1) x 100% 1000 = 9 cM We can avoid this problem by only calculating the two smaller distances. Next time we can just figure out that LFY is in the middle, and just calculate the distances from CER3 to LFY and then the LFY to YI. We recognize that LFY is in the middle because its alleles switch places relative to the outer two genes in the DCO class. Notice how the CER3 and YI stay same in the biggest two classes which are the parentals. Parental Parental CER3 lfy cer3 LFY YI yi Change: Change: CER3 LFY cer3 lf y YI yi Most likely arrangement based on distances: cer3 and yi are farthest apart Most accurate measurement of cer3 - yi distance is the sum of the other two pairs 3 cM + 6 cM = 9 cM So, why did we calculate a map distance of 8.6 cM? When we calculated distance between CER3 and YI, we considered lfy cer3 yi and LFY CER3 YI parentals In fact, these are double crossover (DCO) classes, resulting from one cross-over between YI and LFY and one cross-over between LFY and CER3 By comparing the parental (Largest) classes to the DCO (smallest) classes, we can recognize the gene in the middle, because its alleles switch places relative to the outer two genes in the DCO class Rules for a three point test cross: 1) there should be 8 phenotypic classes 2) progeny classes are grouped in pairs = reciprocal cross over events 3) largest classes are the parentals 4) smallest classes are the double cross-over classes (DCO) Recombinant frequency between outer genes = sum of frequency between two pairs = (sum of SCO classes) +2(sum of DCO classes) total = (26+32+11+18) + 2(1+1) 1000 = 0.09 Map distance = 9 cM Calculating interference: A measurement of the independence of two cross-over events (does one cross-over interfere with the possibility of another cross-over) If crossovers are independent of each other, frequency of two cross-overs should be the product of their individual probabilities Interference = 1 - coefficient of coincidence = 1 - observed DCO expected DCO = 1 - observed Double Cross Overs p(SCOI) x p(SCOII) x total progeny Interference = 1 - observed DCO p(SCOI) x p(SCOII) x total progeny Example lfy cer3 yi: Observed DCO = 2 Expected DCO = (0.06)(0.03)1000 =1.8 Interference =1 - 2/1.8 = - 0.11 Negative interference - more double-crossovers than expected (cross-over in one region promotes crossovers in another region) Positive interference (usually)- fewer double-crossovers than expected (cross-over in one region inhibits cross-overs in another region) Clicker question #1 In Arabidopsis, seeds with the dominant TTG allele are brown, whereas those with recessive ttg allele are yellow; seedlings with dominant HY4 alelle have short hypocotyls, while those with recessive hy4 have long hypocotyls. A test cross is performed on dihybrid individual and the following progeny are obtained: TTG HY4 24 TTG hy4 178 ttg HY4172 ttg hy4 26 The best description of the position of TTG and HY4 is: a) they are on different chromosomes b) They are more than 50 cM apart on the same chromosome c) They are 12.5 cM apart on the same chromosome d) None of the above BECAUSE: If a gene segregates independently; a test cross to the dihybrid will give a 1:1:1:1 ratio. Then, the alleles will be either more than 50cM apart on the same chromosome or they are on different chromosomes. Map distance = %recombinance TTG HY4 24 TTG hy4 178 ttg HY4172 ttg hy4 26 map distance between TTG and HY4 alleles = ( 24+26 ) x 100% 400 = 12.5 map units apart Clicker question #2 In tomato, genes R and WF are 15 cM apart on Chromosome 2. Gene R controls fruit colour (dominant allele R results in red fruit, recessive allele r in yellow fruit) while gene WF controls flower colour (dominant allele Wf results in yellow flowers, recessive allele wf in white flowers) An plant that is cis-dihybrid for R and WF is test-crossed. In the test cross progeny, what will the frequency of yellow fruit and white flowered plants be? a)85% b)42.5% c)15% d)7.5% e)None of the above ANSWER: We are looking for a parental class and since it is a cis-dihybrid with double recessive Map distance = % recombinance (b) 42.5% Three Point Test Cross Examples CLICKER 3 An individual trihybrid at three genes P, Y, and T is test-crossed for 500 progeny. PYt 144 pyT 141 PYT 7 pyt 8 Pyt 24 pYT 26 pYt 70 pyT 80 We can see that P and Y pair frequently, so T is in the middle on the chromosome. YT distance = [ ( 24+ 26 + 7 + 8 ) / 500] x 100% = 13 % = 13 cM apart TOTAL:PY distance = [ ( 24 + 26 + 70 + 80 + 2(7+8) ) .. 46 Clicker question 4 Within 1000 progeny bTS btS BTS BtS bts bTs BTs Bts 14 48 355 83 350 87 50 13 The 'B' allele is in the middle: TBS or SBT Find the distance between B and S where map distance is the per cent recombinance B-S distance = (14 + 48 + 50 + 13) x 100% = 12.5 cM 1000 Clicker Question 5 Test Cross Progeny ABC 450 abc 440 Abc 4 aBC 6 ABc 25 abC 30 aBc 25 AbC 20 The A allele is in the middle such that BAC or bac are parentals A-B distance = (4 + 6+ 25 + 20) x 100% 1000 = 5.5 cM A-C distance = (4 + 6 + 25 + 30) x 100% 1000 = 6.5 cM B-C distance = 5.5 cM + 6.5 cM [---------------12 cM-------------] = 12 cM Map B [ 5.5 A ][ interference = 1 - observed DCO expect DCO 6.5 C ] = 1 - ( 4 + 6) (0.055)(0.065) x 100 = -1.8 interference A trihybrid individual with Bad / bAD is test crossed. What is the probability of bad in the progeny? B A D 20 30 In order for this to happen, a cross over must occur between A and B. What is the probability of this happening? B b a A d D 20 cM = % recombinance of single combination 30 cM = % recombinance or cross over of A and D No recombinance is 70% or 0.7 between A and D p(bad) = ½ p(SingleCrossOverI) (NCO II) = ½ (0.2) (0.7) = 0.07 =7% * we make ½ apart of this probability because when B/b cross it causes two results (bad as well as BAD) we only want one What happens when Bad/bAD is selfed? What is the probability of bad in progeny? p(bad) = [ ½ p(SCOI)(NCOII) ]2 = [ ½ (0.2) (0.7) ]2 = 0.49 % Chapter 4 questions (set 1): 1, 2, 3, 6, 7, 8, 10, 12, 13, 16 Questions (set 2): 21, 22, 24, 25, 36, 37, 42, 50, 51, 53, 56 4.4 Centromere mapping with Linear Tetrads Chromosomes within spores of octad reflect position of chromatids during meiosis During two divisions: Division 1 - separation of homologues Division 2 - separation of sister chromatids 1st division (MI) segregation: meiosis in a heterozygote meiocyte; different alleles separate after MI - a 4:4 ratio (order +number) Indicates that 2 chromatids in each homologue were identical indicates that no crossover had occurred between gene and centromere 2nd division (MII) segregation: meiosis in a heterozygote meiocyte; different alleles separate after MII (2:2:2:2 ratio) Indicates that 2 chromatids linked by the centromere were not identical a cross-over had occurred between the gene and the centromere 4: 4 MI segregation pattern No recombinance 2:2:2:2 MII segregation pattern ½ MII spores are recombinants Distance = % recombinant products -MII segregation pattern indicates that a cross-over occurred within a meiocyte -of the spores produced by that meiocyte, ½ are recombinant Map Distance = % recombinant products = ½ MII segregation pattern frequency = ½ (MII octads / total )x 100 In linear octads, frequency of MII segregation (2:2:2:2) is a phenotypic indication of recombination between a gene and the centromere. We can use this phenotypic frequency to calculate the distance between the gene and the centromere The further the gene is from the centromere, the more frequently that MII segregation will occur. Example: Half the spores are recombinant and half parental in the MII asci Experiment: a diploid Sordaria, of genotype Aa undergoes meiosis Results: 1000 linear octads are scored as MI (4:4) or MII (2:2:2:2) patterns Distance A to centromere = ½(MII tetrads) x 100 1000 = ½ (13+9+12+11) x 100 1000 = 2.25 m.u. Map distance = % recombinants = [½ MII asci]/total x 100% *only situation where the one half ½ fraction is used Four different spindle attachments produce four second-division segregation patterns (Figure 4-22) 4.3 Mapping with Molecular markers Molecular markers: DNA sequence variation between individuals o May have no phenotypic consequence o Much more common than traditional phenotypic based markers e.g. in human genome, 1 SNP/1000 bases o Molecular heterozygotes can be identified o Can identify one recombinant, whereas if the phenotypic marker is a recessive allele, two recombinants must combine to be seen o Test crosses are not as critical Examples of Molecular markers: Single nucleotide polymorphisms (SNPs) o Silent in genes or intergenic regions o Causing a mutant phenotype Simple sequence length polymorphisms (SSLPs) o Mini- and micro-satellite markers Phenotypic and molecular markers mapped on human chromosome 1 (Figure 4-20) those highlighted in green are associated with a phenotype purple = SSLP blue = other polymorphism green = gene, other = intergenic region Detecting SNPs By sequencing - not always possible since this requires prior knowledge of sequence and is expensive By RFLP (restriction fragment length polymorphism) analysis By CAPS (cleaved amplified polymorphic sequence) analysis Restriction fragments: fragments generated by restriction ezymes/ endonucleases Nucleases: enzymes that cut nucleic acids / DNA and RNA Ectonuclease Endonuclease RFLPs Restriction enzymes - cleave DNA at a particular sequence (e.g. EcoRI cuts at 5'-GAATTC-3' 3'CTTAAG-5' If mutation changes sequence (AT to GC): 5'-GAGTTC-3' 3'-CTTCAG-5' Restriction enzyme is no longer able to cut IF a SNP changes a restriction enzyme recognition site, one sequence variant can be cut, the other cannot = RFLP RFLP is detected by a Southern blot Performing a Southern blot: 1. separate genomic DNA cut with restriction enzyme via electrophoresis (Figure 20-12) 2. DNA from gel is transferred to a membrane (Southern blotting, Figure 20-12) 3. the membrane (filter) is exposed to a labeled probe complementary to the region of interest (i.e. containing the RFLP) (Southern hybridization) 4. Bound, radioactively labeled probe will expose an x-ray film, and identify DNA fragments to which probe has bound SNP - single nucleotide polymorphism Detected by RFLP and CAPS (cleaved amplified polymorphic site) Amplify the DNA using PCR to isolate DNA before gel electrophoresis "amplified" - can isolate the DNA using polymerase chain reaction using primers Primers - provide beginnning and end points for DNA replication Results will be the same: - Allele 1 will be amplified - Allele 2 will be amplified with two band in the gel Less dangerous: no radioactivity An RFLP linked to a mouse disease gene (Figure 4-15a) 2 polymorphisms DNA morph1 = no RE cut site, one fragment recognized by probe DNA morph2 = RE site, two fragments recognized by probe Following a test cross, follow segregation of RFLP in individuals with disease (Dd) and those without (dd) (Figure 4-15b) morph2 (2 bands) is always associated with disease allele (D) except in the 8th individual In test cross progeny #8, a cross-over occurred between D and marker (Figure 4-15c) of 16 chromatids analyzed, one was recombinant, distance = 6.25 m.u. Using haplotypes (a set of linked markers) to deduce gene position (Figure 4-16) -closely linked markers will show linkage disequilibrium linkage disequilibrium: . -flanking markers = markers 4 and 5 do not dissociate from the mutation determined because mutant allele/gene of interest will not recombine with 4 and 5 although it does recombine with all other genes; therefore we deduce that our gene of interest lies between 4 and 5 • CAPS -primers flanking a polymorphic restriction enzyme site are used to amplify region by polymerase chain reaction (PCR) -amplified region is cut by restriction enzymes -compared to RFLP: avoids use of Southern blot, probes, requires less DNA SSLPs: Single sequence like polymorphism 1. Minisatellites Genomes contain large amount of repetitive DNA interspersed through genome Short sequences of repetitive DNA (15-100 nucleotides) = minisatellites subject to deletions and expansions Variable numbers of repeats = different alleles If DNA polymerase slips, deletion or expansion can occur where different alleles repeat Detection: cut with a restriction enzyme OUTSIDE the minisatellite (repeated area), use probe that recognizes repeat - see different sized bands on Southern blot Minisatellite regions can be used to generate a DNA fingerprint (Figure 4-18) 2. Microsatellites Very short sequences of repetitive DNA (2nt) = microsatellites subject to deletions and expansions Variable numbers of repeats = different alleles Detection: amplify by PCR using primers that bind to region flanking microsatellite - different size products reveal different alleles (Figure 4-19) Gel Electrophoresis Performing Readings: Omit: 4.5, 4.6, 4.7, 4.8 section 4.5 = chi-square to test for linkage, accounts for possible difference in gamete viability if you are asked to do this, simply use 1:1:1:1 ratio as the expected (null hypothesis = genes are independent of one another) Clicker question: •Three genes (TSB) are arranged on the chromosome with S in the middle, T 5 cM to one side and B 10 cM to the other side. An individual of genotype tSB/Tsb is test-crossed. What will the frequency of the tsb phenotype be in the progeny? •A) .05 •B) .1 •C) .025 •D) .0225 •E) none of the above Crossover between T and S t T S s B b 5cM 10cM Chance of a crossover to form tsb = ½ P(S (O TS) ) x P(Nco(SB)) = ½ (0.05) (0.9) = 0.225 2-3pm Wednesday office: D8 Tony Russel Chapter 5: The Genetics of Bacteria and their Viruses Bacteria and Viruses do have the ability to exchange DNA and undergo recombination. In both types, genes transfers occur: Bacteriophage inject DNA into bacteria While phage is replicating it acquires some host DNA in recombination Why is it important? The fruits of DNA technology mad possible by bacterial genetics -modern gene cloning techniques at 3 billion base pairs the first complete human genome was sequenced by cloning smaller sections of the genome Dividing bacterial cells -binary fission first make an identical copy of genome by DNA synthesis phase from single cell, production of two daughter cells -rapid exponential growth Mutation spontaneous mutation results in two different daughter cells recombination is a separate process that is NOT dependent on cell division Separation and visualization: bacterial cells with different genotypes and phenotypes It is easier to infer the genotype in bacteria because they are haploid bacterial cells are grown on plates each colony is a clone of cells with the same genotype Process: Suspension of bacterial cells is spread on a petri plate with agar gel and incubated from 1 to 2 days. Visible colonies are each a clone of the corresponding single cell from the original suspension. By looking at the phenotypes present on the plate, we infer the genotypes in the original liquid culture. Properties of Bacteria Most are prototrophic: self sufficient, ability to synthesize all essential molecules it requires for growth, can be grown on minimal media (carbon source and salts) Auxotrophic mutants require additional nutrient supplements (mutations affect the synthesis of a molecule such as amino acids or vitamins) The loss of an ability is denoted with a minus ( — ) sign Resistance and sensitivity can be denoted by (r) and (s) Eg) Symbol Character/ Phenotype arg requires arginine added as a supplement to minimal medium lac cannot utilize lactose as a carbon source strr resistant to the antibiotic streptomycin (mutant) sensitive to the antibiotic steptomycin (wild type) str s Bacterial Selection Selecting for auxotrophic: We can take an auxotrophic mutant into a suspension containing adenine and separate those which can survive on their own and grow Properties: -single - celled Prokaryotic (no nucleus or membrane-bound organelles) Have only 1 circular chromosomes Haploid organisms Do not undergo meiosis like eukaryotes (where recombination occurs in eukaryotes) Still undergo recombination Lederberg and Tatum (1946) Discovered gene transfer in E. coli Mixing bacterial genotypes produces rar recombinants A B Strain A is an auxotroph for 2 alleles Strain B is wild for what A is mutant Met- bio- thr+ leu+ thi+ met+ bio+ thr- leu- thiA+ BMixture Wild Type culture No spontaneous reversion Instead, need to mix the two in order for colonies to be produced Two strains, both auxotrophic Strain A = bio- metStrain B = thr- leu- thiPrototrophs can be detected at a low frequency ~1 in 10 million cells Is this recombination or do they cross feed from metabolic pathways? Answer: recombination Proof: U-tube experiment U-tube experiment Bernard Davis separated the two cultures with a filter to keep Strain A on the left and Strain B on the right with no ability for cell interaction or gene transfer, however molecules can move back and forth = no recombinants are produced without cell contact Bacteria conjugate by using pili Bacteria exchange DNA Only one we will look at that requires cell contact for gene transfer Plasmids - extrachromosomal elements - under certain conditions are non-essential Uni-directional: middle cell donates while other cells accept: Conjugation: Plasmid transfer during conjugation <-- Plasmids Genome Conjugation Partial genome tranfer during conjugation Question: is the transfer of genetic information bi-directional? Experiment 1. Strain A (Strs met- leu- bio+ thi+) X Strain B (StrR met+ leu+ bio- thi-) Allow conjugation Streptomycin to kill strain A Result: Prototrophs obtained 2. Stain A (StrR met- leu- bio+ thi+) X Strain B (StrR met+ leu+ bio- thi-) Conjugation Streptomycin to kill strain B Result: No proptotrophs obtained Conclusion: uni directional Only the strain B cells were able to produce prototrophs, therefore it is unidirectional from A to B . Conjugation Extra-Chromosomal DNA Plasmid - one member of a conjugating pair carries the fertility factor (F) on the F plasmid - F plasmid = small circular, extrachromosomal DNA - F+ strains: donor strain bacteria contains the F-factor - F- strains: acceptor that does not contain the F-factor Properties of F-factor (plasmid) - enables the production of pili (proteinaceous attachment tube between cells) - able to replicate during cell division and during conjugation - prevents conjugation between F+ cells - only get conjugation between F+ to F- cells F plasmids transfer during conjugation Donor F+ cell contains an F plasmid which can produce a pili Donor cell pulls acceptor cell with pili Cut in one of the two strands of DNA of F plasmid One DNA strand is pulled into recipient cell - transfer of single-stranded DNA - Replication in both donor and recipient - Rollin circle replication in donor The accepting cell will replicate the DNA as it accepts the new DNA Result: Donor and Recipient both have doublestranded F plasmid ICLICKER During Bacterial conjugation, which one of the following would not be expected to occur? A) an F- cell converted to F+ B) An F+ cell would generate pili to attach to F- cells C) DNA would be replicated in the F+ cell D) DNA would be replicated in the F- cell E) An F+ cell would be converted to FCavalli-Sforza Observations 1. Occaisionally F+ strains become strains with a high frequency of recombination: Hfr strains -1000x more frequent transfer of genes than regular F+ strains 2. Hfr x F-only rarely is F- converted to HFR or F+ Hfr can convert auxotrophs to prototrophs: In Hfr strains, F plasmid has been inserted into the bacterial chromosome IS = Insertion sequence The IS is similar to the bacterial chromosome in DNA composition The IS enables the F plasmid to be added to the bacterial chromosome The process, then, is reversible Integration of the F plasmid creates an Hfr strain F+ with an F plasmid in the cell Hfr with an integrated F plasmid Crossovers integrate parts of the transferred donor fragment - Cleavage in the Hfr DNA sequence transfers the single-stranded DNA copy - Exconjugant is formed with two homologous regions where we can get recombinantion; transferred fragment is converted into the double helix through DNA synthesis; however not all exconjugants will undergo recombination - Recombinant cell is formed where double crossover inserts donor DNA Exogenote: DNA coming from the donor Endogenote: DNA originally in the bacterial chromosome Summary: 1. Mechanism of DNA transfer is the same as F+ x Fa. replication in donor and host b. single-strand of NDA 2. A portion of Hfr genome gets transferred a. High recombination frequency 3. Recombination a. Exogenote b. Endogenote .. Two types of DNA transfer can take place during conjugation Chromosome transfer vs. Plasmid Transfer Start: F+ cell with a+ bacterial chromosome (1) Plasmid: conjugation and transfer of the F factor Recipient cell will have no chance to replace the a+ allele/ genotype (2) If there is enough Chromosome transfer then with enough transfer, can integrate the a+ allele but not the F plasmid Interrupted Mating Experiment Wolman and Jacob combined: Hfr strain: O thr+ leu+ azir tonr lac+ gal+ strS F- strain: thr- leu- aziS tonS lac- gal- strr Plate: media with streptomycin (kills Hfr) and no theonine or leucine provided (selects for recombinants that underwent conjugation) Then, scored for the underlined genotypes Tracking time of marker entry generates a chromosome map Frequency of Hfr genetic character among strr exconjugates vs. time Gives us a sense of distance between the alleles on the chromosomal DNA Origin is a specifice site where DNA cleavage and DNA replication begin Alleles that were closest to the origin are more likely to be transferred Time of entry is dependent on position relative to F plasmid insertion The first allele appearing in the new cell shows its close distance to origin The time taken to transfer is related to the distance between the gene and origin Therefore, we obtain a crude chromosome map of genes Questions: Is the relative position of genes, the position of the origin, and the direction of transger the same in different Hfr strains? Experiment: Consider 5 different Hfr strains derived. Example Isolated 5 different Hfr strains We infer the site of integration to be different for each strain Relative order between alleles in the strains are different Relative position of the alleles, however, are the same (eg) b is always associated with a or c on either side Results: 1) The first gene to enter may differ 2) The relative position of each marker (gene) is constant 3) The overall directionality of transfer can differ Conclusion: The difference between the Hfr strains is the relative position and orientation of the origin within a circular chromosome. This experiment proved that E.coli has circular DNA molecule ORIGIN (O) -- A B C D E -The only way to explain the transfer is if A and E are joined in a circle What controls these insertion patterns? Why are there different insertion sites? There are multiple insertion sequences (IS) in the E. coli genome. A single crossover inserts F at a specific locus Orientation of IS determines the order of gene transfer Homologous regions where pairing can take place on the E.coli chromosome: Location for IS The F integration site determines the order of gene transfer in Hfrs What is the difference between an Hfr strain and an F+ Possibility of recombination from conjugation Difference in crossing an Hfr with F- opposed to crossing an F+ with an F-? Some transferred DNA may be lost following Hfr x FF+ to F- cross cannot lose DNA because the F plasmid is circular, which is retained in the recipient Hfr cross outcome is with linear DNA that may not be maintained in a bacterial cell There are multiple insertion sequences (IS) in the E. coli genome -Orientation of IS determines the order of gene transfer Example: Figure of E. coli chromosome Fplasmid > Hfr > IS What do we expect if the orientation of the Fplasmid is flipped? The orientation of the alleles changes in the chromosome Then, the resulting linear DNA sequence will be the same sequence FINE-SCALE MAPPING BASED ON RECOMBINATION FREQUENCY Terminology Merozygote - partially diploid bacteria Endogenote - recipient genome Exogenote - transferred DNA To map chromosome 1. Ensure that recipients (F-) were myozygotes for all genes being considered 2. Select for presence of last gene being transferred Once DNA is transgerred, frequency of recombination depends on distance between genes Transferred DNA: gal+ arg+ met+ Host chromosome: galargmetRecipient cell is diploid for three alleles Integration depends on both transfer and recombination If gal+ is the last to be transferred (start at met+), select only those exconjugastes that are gal+ Time of entry show the direction of transfer and distance between alleles Example Hft srain . Therefore if we select cells that are leu+ we know that met+ and arg+ must also have been transferred although not always integrated Generation of various recombinants by crossing over in different regions Higher frequency of a genotype occuring: Because higher the crossover frequency, larger the distance between alleles Faulty outlooping produces F- plasmid a) insertion b) Integrated F factor on Hfr chromosome c) Excision d) F`lac with lac+ F plamid now containse chromosomal DNA -> lac+ region e) Meterozygote F`lac+/ lacFor a certain period of time, can exist with diploid for two lac alleles Dominant or recessive? -> relationships can be observed We expect to be better at transferring lac+ allele/converting genotype When F` plasmids acquire drug resistance genes usually called R plasmids Multiple drug resistances are common Cross-species conjugation Medical challenge - rapid transmittance Starting points for creating may gene cloning factors R plasmids have importance in cloning genes, and practical real world research applications A plasmid with segments from many former bacterial hosts Lactococcus lactis used in dairy products Plasmid contains genes that are acquired from many other bacterial species Many chromosomal alleles which are scattered throughout the plasmid shows many different integrations Could be a cause for concern if its antibiotic factors get transferred to bacterial DNA for pathogens Transformation: Acquiring DNA fragments from the environment through the bacterial wall Uptake system Convert DNA as it is imported into cell Takes up single stranded DNA to transform the genotype Artificially treated cells with calcium chloride or electroporation treatment enables cells to take up more DNA for recombination/transformation Transformation and linkage -only small DNA fragments are taken up naturally Therfore, only genes close together on chromosomal DNA will be co-transformed We can use this to tell us about distance on the chromosome Determine order of the three alleles on the DNA If the two alleles get transferred together frequenty, they are closer together on the chromosome Method: select for one marker and determine the frequency of the other two markers Results: Experiment Selected Marker Unselected Marker 1 met+ 20% gal+, 2% leu+ 2 leu+ 8% gal+, 2% met tells us that met and gal are close together; leu is closer to gal than to met met gal leu Bacteria exchange by several processes Transformation Conjugation Transduction Bacteriophages = viruses that attack bacteria Nucleic acid + protein coat -Phage T4 parts can spontaneously assemble individually -Free phage contains chromosomal DNA in its "hat" of protein coat -Phage infecting a host injects its genomic DNA into cell Bacteria exchange DNA by several processes Conjugation Transformation Transduction Transduction The lytic cycle -phage T4 infects bacterial cell, injects information (genomic DNA), degrades host chromosome. Assemble new phages, lysis of host cell -amplification of phages Plaques - are clear areas where all bacterial cells have been lysed by phage Eg) E. coli to low phage concentration to score phage genotypes on a bacterial cell We can map phage genes on phage "chromosome" because: 1. Several phages with different genotypes can infect the same bacterial cell simultaneously 2. Different phage genomes undergo recombination (ie) can h+r+ or h-r- phage be produced? Phage 1 Phage 2 h- r+ h+ rh-rh+r+ yes Doubling infecting a host with phages of different genotypes to produce recombinants P R h-r+ and h+rh-r- and h+r+ h- = infects two E.coli strains (dark plaque) h+ = infect one strain (cloudy plaque) r+ = normal cell lysis rate r- fast cell lysis rate Recombination Frequency (RF) = recombinants / total phages RF = h+r+ h-rAll genotypes Limitation: cannot detect recombination between phages with the same genotypes Advantage: We can screen large numbers of plaques and detect rare recombinants as those due to crossovers swapping portions of a single gene Transduction: phage-mediated transfer of bacterial chromosomal genes Two mechanisms (Generalized Transduction) 1. Faulty head transduction -bacterial genomic DNA instead of phage DNA 2. Specialized Transduction -prerequesite is integration of Generalized transduction by random incorporation of bacterial DNA into phage heads -no lysis -allows recombination A phage particle erroneously packages a host chromosomal DNA fragment -Co-transduction of bacterial genes indicates close linkage -Mapping is similar to transformation analysis -Size limitation of DNA package into phage particles Example: Three genes closely linked: leu+ arg+ lac+ What order are they in? Infect strain 1: leu+ arg+ lac+ with lytic phage Isolate phage progeny on a infect strain 2 leu- arg- lacResults: 1 2 3 Gene Marker arg+ leu+ lac+ 60% lac+ and 0% leu+ 8% lac+ and 0% arg+ 59% arg+ and 8% leu+ arg and leu are never packaged together because they are too far apart --l--------l-----------------l--arg lac leu Specialized Transduction Phage transfer bacterial genes through lysogeny-lytic cycle Lambda Phage ( ) Can integrate its genome into the E.coli chromosome and remain dormant = lysogeny - this depends on bacterial cell resources or Can actively replicate and cause cell lysis = lytic cycle Production of lysogen -targetted to a fixed site in the bacterial genome Single site of integration (*key difference) -phage factors are required for recombination -transfer of prophage during conjugation can trigger lysis Hfr () x F- lysis [zygotic induction] *Only genes transferred before entry will recombine and can be detected -transfer of prophage during conjugation will not trigger lysis of lysogenic FHfr () x F- () no lysis Cytoplasmic factors repress incoming phage from inducing lysis -faulty outlooping produces phage containing bacterial DNA production or initial lysate dgal cn infect but not integrate or recombine: needs help d = defective particle Lambda phage Can cause E. coli cell lysis Integrates at a single site in the E.coli bacterial genome Produces protein factors Circular genomic structure Can only be lytic or lysogenic at a time Production of initial lysate (i) Normal outlooping we expect to see a mixture of particles (ii) Rare abnormal outlooping dgal can infect but not integrate or recombine needs help Transduction by initial lysate Integration - helper phage provides necessary factors Recombination - homologous bacterial sequences allow double crossovers Transduction by initial lysate Example: gal-/gal+ (i) lysogenic transductions dgal is defective helper is wild type (has components to integrate) consequence: can integrate the gal+ allele but does not have to replace the gal- so a stable diplozygote is produced (ii) transductants produced by remcombination the consequence: replace gal- with gal+ haploid result DOMINANCE Complete dominance 3:1 phenotypic ratio 3 dominant : 1 recessive Incomplete Dominance 1:2:1 phenotypic ratio Heterozygote is different from parents Is a combination of characteristics together eg) Flower colour: Red+White parents = pink progeny Codominance 1:2:1 phenotypic ratio Heterozygote is different from parents Shows characteristics of both parents eg) calico cats eg) A-B blood group Blood Groups -categorized by how blood reacts with specific antibodies IA reacts with A antibody IB reacts with B antibody i - null allele (no antigen - recessive allele) called O type blood group - universal donor -heterozygote: IA IB - universal acceptor - both kinds of antigens IA IB blood group is an example of codominance Method of scoring the phenotype may result in assignment of different dominant/recessive relationships for the same allele Eg) Sickle Cell Anemea is caused by an Hb mutation (allele HbS) in hemoglobin A HbA HbA - wild type or normal; blood cells round; no anemia HbS HbS - blood cells sickle shaped; anemic Heterozygote: HbA HbS - blood cells normal except they are sickle shaped No anemia at sea level, therefore the wild type is dominant At higher elevation with low O2 the mutant allele is dominant= anemia Considering behaviour of allele products (A and S), heterozygous show characteristics of both homozygous individuals, HbA is codominant to HbS Lethal Alleles If a situation occurs where one of the homozygous classes will not survive, the phenotypic ratio changes from a 1:2:1 to a 2:1 because lethal alleles cause death of the individual Eg) yellow mice Eg) Tailess cats show heterozygousy for a lethal allele MM - wild type MLM - tailess Because the lethal allele is crucial for spinal cord development MLML - lethal Kitten dies due to spinal column not developing properly Therefore there will be a 2:1 phenotypic ratio in Manx cats for tailess individuals Archibald Garrod, discoverer of inborn errors of metabolism - connected a genetic defect to a defect in chemical reactions in the body Genes can control these chemical reactions George Beadle and Edward Tatum: one-gene-one-enzyme (polypeptide) model Now we know that multiple genes can contribute to one enzyme. Beadle and Tatum Isolated arginine auxotrophs in Neurospora Auxotroph - strain able to grow only on supplemented media Organisms unable to synthesize own compounds Prototroph (wild type) - able to grow on minimal media Minimal media is water, essential elements, and carbon-sugar source Able to synthesize biological compounds from the minimal media -Suggested that all biosythetic pathways consist of a series of steps -Genes control each step -Beatle and Tatum used this concept to look at intermediaries from each step in the production of argenine -They saw that each mutants could survive with some or all of the supplemented compounds (ornithine and citrulline can "rescue" the mutant) Supplements: ornithine, citrulline, and arginine if there is a mutation in arg 1 then it can be helped only by all three ------- ornithine---------------- citrulline ---------- arginine arg1 arg2 arg3 CHAPTER 6 - Gene Interaction questions: 4-8, 10-14, 21, 27, 28, 43, 49, 50, 57, 64 Definitions: pleiotrophic mutation: a mutation that affects several different phenotypic characteristics haplosufficiency: a single dose of the wild type allele can confer the wild type phenotype in a diploid cell/organism haploinsufficiency: a single dose of the wild type allele cannot confer the wild type phenotype in a diploid cell/organism dominant negative: a mutant allele that in one dose, as in a heterozygous individual, can eliminate the wild type function resulting in the mutant phentoype codominance: alleles which, when combined in the heterozygote show aspects of both homozygotes incomplete dominance: alleles which, when combined in the heterozygote show a phenotype intermediate between the homozygotes prototroph: a strain of organism that can grow on minimal media (media with water, basic elements and a Carbon-source) auxotroph; a strain of organism that cannot grow on minimal media but requires nutritional supplementation complementation: the production of a wild type phenotype when two mutations are combined in a diploid heterokaryon: cell with nuclei of two different types in a common cytoplasm epistasis: the phenotype of a mutant allele at one gene overrides the phenotype of a mutant allele at another gene, such that the double mutant has the same phenotype as the first mutant suppression: a mutation in one gene can cancel the effect of a mutation in a second gene, resulting in a wild type phenotype penetrance: the proportion of individuals with a specific genotype who show that genotype phenotypically expressivity: the degree to which a particular genotype is expressed in the phenotype Chapter 6 - Gene Interaction Situations where more than one gene is involved in a pathway Mendel's findings: one gene, two alleles each gene controls a single different character one allele is completely dominant to the other ratios: 3:1 phenotype = 1:2:1 genotype (monyhybrid selfed) 9:3:3:1 (dihybrid selfed) 1:1(monyhybrid test-crossed) 1:1:1:1(dihybrid test-crossed) IF genes in dihybrid are linked, frequencies shift towards parentals Complexity arises: -a gene may have multiple alleles -a gene may be pleiotropic (where a single gene controls more than one character) -several genes may control a single characteristic 6.1 Interactions between alleles of a single gene: variations on dominance So far, we have considered only cases where one allele is fully (completely) dominant over another i.e. phenotype of Aa = AA monohybrid selfed: 1:2:1 genotypic ratio becomes 3:1 phenotypic ratio Explanations for complete dominance Case I: wild type allele is fully dominant to the mutant allele One dose of the wild type allele is enough for wild type phenotype = haplosufficient (Figure 6-2) wild type allele is fully dominant to the mutant allele Example: Haplosufficient AA versus Aa will give the same phenotype Case II: mutant allele is fully dominant to the wild type allele explanation i: One dose of wild type allele is not enough for wild type phenotype=haploinsufficient explanation ii: The mutant allele (dominant negative) makes the wild type allele non-functional e.g. wild type allele makes 30 units of a protein, and 40 are required for wild type phenotype. A null mutation (0 units) will be dominant BB = No resulting functional gene product from B B+ = B makes none; resulting 30 units ++ = 60 functional b+=++ b makes 15 units plus 30 units from + makes 45 e.g. gene product forms a homodimer, and one defective polypeptide interferes with its function brittle bone disease - collagen is a trimer B+ = BB B makes + nonfunctional Two models for dominance of a mutation (figure 6-3) Incomplete dominance (Figure 6-4) In four-o'clocks true-breeding red (c+c+) x true-breeding white (cc) give 1 red (c+c+) : 2 pink (c+c) : 1 white (cc) Heterozygous individual has a phenotype intermediate between phenotype of two homozygous individuals Codominant allele: the heterozygous individual has a phenotype showing characteristics of both homozygous individuals example: blood groups Blood groups are categorized by how blood reacts to specific antibodies each antibody reacts with a antigen on the cell surface one class of antigen (complex sugar molecules) is controlled by gene I 3 alleles (IA, IB, i) that produce different antigens recognized by different antibodies IA reacts with A antibody IB reacts with B antibody i - null allele (no antigen) reacts with no antibody Homozygous genotypes and blood group designations: IA IA - A blood group; IB IB - B blood group; i i - O blood group Heterozygous combinations IA i - A blood group; IB i - B blood group; IA IB - AB blood group IA and IB are dominant to i IA and IB are codominant: the heterozygote has a phenotype showing characteristics of both homozygous individuals Method of scoring the phenotype may result in assignment of different dominant/recessive relationships e.g sickle cell anemia (Figure 6.5) Sickle cell anemea caused by a Hb mutation (allele HbS) in hemoglobin A HbA HbA -wild type, blood cells round, NO anemia HbS HbS - blood cells sickle shaped, ANEMIC HbA HbS - blood cells normal and some intermediate shaped cells, except sickle-shaped under low O2, NO anemia i) considering anemia HbA > HbS ii) Considering cell shape, HbA shows incomplete dominance to HbS (intermediate phenotype of some cells) under normal oxygen, but HbS > HbA under low oxygen iii) Considering behaviour (migration in an electrophoretic gel) of allele products (A and S), heterozygous show characteristics of both homozygous individuals, HbA is codominant to HbS (Figure 6-6) Lethal alleles - causes death of the individual e.g yellow coat in mice is a recessive lethal allele yellow x yellow results in 1 black : 2 yellow (Figure 6-8) Tailless, a recessive lethal allele in cats (Figure 6-9) ML/M = tailless ML/ML - lethal Archibald Garrod, discoverer of inborn errors of metabolism - connected a genetic defect to a defect in chemical reactions in the body (Figure 6-10) George Beadle and Edward Tatum (Nobel Prize, 1958): one-gene-one-enzyme (polypeptide) model Beadle and Tatum isolated arginine auxotrophs in Neurospora ((Figure 6-11) auxotroph - strain able to grow only on supplemented media prototroph (wild type) -able to grow on minimal media 1. mutagenized spores transferred to complete medium 2. asexual spores transferred to minimal media 3. auxotrophs cannot grow, tested for growth on various media Isolated auxotrophs for a variety of compounds all segregated as a single gene mutation (1:1 following cross to wild type) 3 arginine auxotrophs mapped to 3 chromosomes - 3 different genes Beadle and Tatum hypothesized that biosynthetic pathways = series of steps Arginine has 2 analogs citrulline and ornithine Do these represent different products of steps in the pathway? Do the mutations represent defects to the different steps? (Figure 6-12) Hypothesis: If the auxotrophs represent defects to different steps in the pathway, they should be rescued by a unique set of compounds (Table 6-1) compounds tested = arginine, ornithine and citrulline Based on data: 1. Mutations in different genes (auxotrophs) represent unique defects within the biosynthetic pathway 2. Steps within biosynthetic pathway are controlled by enzymes 3. Each mutation is defective in a different step (enzyme) 4. Each wild type gene controls a step (makes an enzyme) Biosynetic pathway consists of a series of steps, each step controlled by a unique enzyme that is made by a unique gene One-gene-one-enzyme hypothesis modified to one gene produces one polypeptide since some enzymes are multiple polypeptides 6.3 Inferring Gene Interactions The Complementation test Harebell plant, wild type flowers are blue, 3 plants are found, all with white flowers - mutations in 3 genes, 3 alleles of one gene? Consider the situation where 2 lines ($ and £) are alleles of one gene (w1), the third line (¥) is an allele of a second gene (w2) Figure 6-15 If we make pairwise crosses amongst the three lines ($ x £, $ x ¥, £ x ¥) progeny of crosses between alleles of the same gene will show no complementation, progeny of crosses between alleles of different genes will show complementation (Figure 6-15) In the first cross, the F1 progeny have no wild type copy of w1 gene In the second cross, the F1 progeny have one wild type copy of both the w1 and w2 genes (Figure 6-15) RULES FOR A COMPLEMENTATION TEST 1. can only be done with recessive mutations 2. If the mutations are in different genes, the two mutations will complement one another (progeny will be wild type) 3. If the mutations are alleles of the same gene, the two mutations will not complement one another (progeny will be mutant) Heterokaryon = variation on a complementation test -cells have two nuclei in a common cytoplasm -products of two nuclear genomes act together in the common cytoplasm (Figure 6-16) Using the complementation test to find a novel gene: A plant mutant for FORKED1 (FKD1) has an open vein pattern We mapped FKD1: 2 recombinations out of 3000 plants to a CAPS marker on chromsome 3 (0.067 cM), meaning that FKD1 must be within a region of about 40 genes One line, with a mutation in gene At3g63300, has a similar phenotype to fkd1-1 Crossing to this insertion line produces mutant seedlings (i.e. failed to complement), therefore these are alleles of the same gene and fkd1-1 must be an allele of At3g63300 Transformation of fkd1 mutant with the cloned FKD1 gene results in a wildtype phenotype = molecular complementation Clicker question 1: The gene wg controls formation of wings in Drosophila. For a cell to become part of the wing, it requires 50 units of the wg gene product; if a cell has less than 50 units, it will not become part of the wing. The wild type wg+ allele makes 30 units of the wg gene product; a mutant allele (wg) makes 0 units of the wg gene product. Which of the following statements are true: A) the gene is haplosufficient B) the mutant allele is dominant C) the wild type allele is dominant D) all of the above E) none of the above Clicker question 2: Nasturtium flowers are normally red, but orange, white and yellow varieties exist. You make crosses amongst different plants with the following results: cross progeny RxR all red WxW all white YxY all yellow OxO 1 red : 2 orange : 1 yellow RxY all orange OxY 1 orange : 1 yellow OxW 1 red : 1 yellow Which of the following is consistent with the above data? A) R and Y result from incompletely dominant alleles B) orange individuals are heterozygous C) the allele causing red is dominant to the allele causing white D) all of the above are consistent E) none of the above are consistent Mutants Harabells are usually blue. Two white mutants are crossed to give all white progeny. a) the mutants carry alleles of the same gene Chapter 3 - 9:3:3:1 independently assorting genes Chapter Four - shift towards parental gametes linkage not equal to 9331 Chapter Six - 9:3:3:1 genes interlinked ; where in same pathway the phenotype is indistinguishable meaning 0:9:9:3:4 ratio Analysing Double Mutants: variations on the 9:3:3:1 ratio 1. no gene interaction is indicated by 9:3:3:1 ratio eg) corn snakes with skin colour due to two unlinked genes o and b. o+ = presence of orange pigment o = absense of pigment b+ = presence of black pigment b = absense of black pigment o+/- ; +/o+/- ; b/b o/o ; b+/o/o ; b/b = camouflaged = orange = black = albino The 9:3:3:1 ratio indicates that the genes act in independent or separate pathways (although they are independently assorting) Colourless precursor plus b+ = black Colourless precursor plus o+ = orange Both o+ and b+ present = both expressed in camouflaged Neither o+ or b+ present = neither expressed = albino 2. The 9:7 phenotypic ratio 9:7 ratio indicates one of several explanations: - complementary gene action - same pathway (no distinguishable intermediary inter-individual) - one gene regulates the other 3. Genes in the same pathway may show an epistatic interaction blue-eyed Mary - wild type = blue two mutants: white (w/w) and magenta (m/m) Cross: w/w m + /m + x w+/w+ m/m White magenta F1: w+/w ; m+/m Blue F2: 9 w+_ m+_ (blue) 3 (magenta) 3 (white) 1 w/w m/m (white) Two possible pathways: w+ m+ white ------------ > magenta ---------- > blue m+ w+ magenta -------- > white --------------- > blue The first pathway is correct because the upstream gene covers up the downstream gene. Once w is defective, the second gene still does not matter and the individual is still white. We see in the progeny a 9 blue : 3 magenta : 4 white Epistasis = w is epistatic to m; w covers up m phenotype; w is upstream of the m in the gene pathway; progeny result in a 9:3:4 ratio where 3 is recessive 12:3:1 is dominant epistasis Labrador Retrievers Golden, Chocolate, and Black dogs Yellow gene (Y) required for deposition of pigment in fur - epistatic to the alleles B (black) and b (brow) Golden Labs are homozygous for e/e but you cannot tell if they have black or brown alleles because the pigment is not deposited in the fur Black B/B E/_ Chocolate b/b E/_ Golden _ /_ e/e Coat colour also controlled by B gene Siamese Cat - temperature sensitive allele of B B gene pigment is functional is cooler regions (face ears extremities tails paws) B is not expressed along body where the cat is white/beige Genes controlling spore colour in Neurospora Al+ = orange Al = albino Ylo+ = orange Ylo = yellow Create a diploid meiocyte by crossing the haploid parents Spore progeny created from the meiocyte are 1:2:1 ¼ al+ylo+ orange ¼ al+ ylo yellow ¼ al ylo+ albino ¼ al ylo albino Suggests that al is upstream of ylo 4. In foxglove flowers D= dark red, d = pink; w allows pigment throughout petal, W pigment is only in the throat of the flower; W is epistatic to alleles of D Dominant epistasis results in a 12:3:1 ratio W is controlling D 12 White - W_ ___ 3 1 Epistatic allele is upstream in the pathway so that is does not matter what the downstream allele is. 5. Suppression A mutation in one gene reverses the affect of a mutation in another gene, resulting in a wild type phenotype in the double mutant 10 3 3 10 6 13 3 Eg) purple eye colour in Drosophila and suppressor su Parents: p/d ; su+/su+ Purple x pd+/pd+ ; su/su Red F1 red F2 9 red: 3 red : 3 purple : 1 red Explanation: - one bypasses the other - suppressor modifies a binding partner Modified 9:3:3:1 ratios provide important information about the mechanisms of gene action within pathways PENETRANCE The percentage of individuals with a given genotype who exhibit the phenotype associated with that genotype is penetrance. Penetrance is expression within a population whereas expressivity is in the individual. Example B _ = normal vision bb = blind But of the 100 people of genotype bb, only 20 are blind, therefore allele b has the penetrance of 20% this means they have a tendency to being blind EXPRESSIVITY - the extent tow hich a given phenotype is expressed in an individual Example A _ = normal pigmentation aa = albino variable penetrance variable expressivity variable penetrance and expressivity Chapter 7: DNA Structure and Replication Introduce molecular biology/genetics Determination of genotypes at molecular level: Gene expression Gene replication Distinguishing classical Mendelian Genetics from modern Molec. Genetics What is genetic material? Main cellular components: Polysaccharides (sugars), lipids, proteins, DNA, RNA Pre-1944 it was believed that protein was more likely than DNA to be the genetic material because: - 20 standard amino acids known, only four bases in DNA, so proteins could house much information - DNA sequence thought to be repetitive (if same four bases repeated in same order so then little info content) Frederick Griffith 1928 Transforming Streptococcus R cells into S cells -His experiment showed that a subcomponent of a cell could house genetic information. Isolated two different strains of bacterial -S (wild = smooth; bacteria coating; when infect mice will kill mouse) -R (mutant = rough; bacteria is incapable of producing a polysacharide coat; unable to infect mice) a) b) c) d) S strain live cells mouse dies R strain mouse lives S strain heat-killed cells mouse lives S strain heat-killed and R strain mouse dies Conclusion: A component from dead S cells is taken up and transforms R cells with virulence properties The transformed R cells divide and propagate as S cells - hereditable material has been acquired R cells have acquired the genes for virulence. Avery MacLeod and McCarthy 1944 Which specific component of heat-killed S cells is required for transformation? Extract components from the S strain If enzymes were used to destroy polysaccharids, lipids, RNA, and proteins, the transforming ability was not affected and the bacteria still was able to kill the mice. In the case of destroying the DNA component of the S strain, the mouse lived; therefore the DNA contained the genetic material that was transformed Hershey and Chase 1952 Differential labeling of phage components: -proteins 35S-labeled -DNA 32P-labeled Is it the DNA or the protein coat that is responsible for transformation of genetic information? -proteins 35S-labeled allow radioactive phage to the E. coli cell then separate protein coats from E. coli Most of radioactivity recovered in phage ghosts (away from E. coli cell) -DNA 32P-labeled DNA is the radioactive component When phage particles attach to the cell, inject DNA Most of radioactivity recovered in E. coli cell Conclusion: The majority of radioactivity injected by phage into E.coli was 32P-labeled DNA. Phage DNA drives the production of new phage particles 1) What are the chemical and physical properties of DNA? 2) What is the 3-d structure of DNA? Requirements to store genetic information: 1. Stability -must maintain structure during the lifetime of an organism -most organisms use DNA; some viruses use RNA 2. Replication -accuracy and efficiency 3. Informational complexity -diversity of life forms -able to specify why each life form is different 4. Mutational capacity -balance between replication accurancy and ability to change -crucial to evolution Structure of the four DNA nucleotides -nitrogenous base difference between nucleotides -deoxyribose sugar -phosphate only one phosphate group individually two additional phosphates are dAMPs which are builing blocks of structure Purine nucleotides A- adenine nitrogenous base G- guanine nitrogenous base Pyrimidine nucleotides C- cytosine nitrogenous base Tthymine nitrogenous base Originally, biologists thought that DNA was too simple to house genetic information Chargaff's DNA base composition results from different organisms Accurately determining molar composition Different bases can be present in different relative ratios Shows that base content is dissimilar in different organisms Therefore, different species can have different genomes Rules: 1. Total base ratios A's + G's = C' + T's 2.A's=T's G's=C's This tells us that the purine ratio equals the pyrimidine bases Consistent patterns What is the 3-D structure of DNA? James Watson and Francis Crick, 1953 The era of molecular biology Watson and Crick deduced DNA structure from the information of earlier times - Chargaff's Rules - X-ray diffraction of DNA fibres - Molecular Modelling Rosalind Franklin's X-ray Diffraciton Data - repetitive structure: uniform geometry - Helical structure - Double helical: two strands Linus Pauling - working out protein structure - alpha helices produce a similar x-ray diffraction pattern to DNA therefore, DNA must be helical as well; but with two strands The structure of DNA -Double helix two strands in winding helix -Bases in the interior nitrogenous bases play critical role in holding strands together -Sugar-phosphate backbone on exterior exposed to aqueous solvent The Chemical Structure of DNA Strand polarity - 5` to 3` end -antiparrallel arrangement means the other strand has opposite orientation Hydrogen bonds between bases - G-C (3) - A-T (2) - stability maintains overall parameters/distance interactions between nitrogenous bases: hydrogen bonds hold together Phosphodiester bond links nucleotides - gives the strands their orientation - 5` and 3` ends have no more nucleotides attached -opposite orientation of two strands: antiparrellel arrangement is the only way to enable H-bond interaction Nucleic Acid: Formed by joining nucleotide units nucleotide units are joined by the phosphate groups Chapter 7 Practice Problems: 1, 2, 11, 16, 19, 24, 28 Representation of the DNA double helix -stacked and planar base-pairs -Space-filling representation: each atom as a sphere: shows 3-D orientation major groove: large space where no atoms as present as helix winds minot groove: also wrapping around length of the helix Grooves cause nitrogenous bases to be exposed chemicals and proteins can interact with DNA nitrogenous bases in grooves -Uniform geometry and helical width explain Chargaff's data The only way to accommodate the distance between phosphate backbone/ two DNA strands is if a purine is paired with a pyramidine This is the helical width (20A) The distance between adjacent bases is 3.4A The distance per revolution of the helic is 34 A Therefore, there are 10 base pairs per revolution of DNA double helix Question: Do the base-pairing rules observed within the double helical structure also control the process of DNA replication? Or is replication like a photocopying process? Alternative models for DNA replication Semiconservative replication Two blue strands One parental + one new strand (1 blue 1 yellow) one parental + one new strand (1 blue 1 yellow) Conservative replication Two parental strands (two blue) Two parental strands Two new daughter strands (two yellow) Dispersive replication Each strand is chimeric of Two parental strands ~50%old and ~50% new Patchwork of blue/yellow Meselson-Stahl Experiment Strategy: Distinguish newly synthesized strands from parental strands using different isotope labeling (different weight) Separate DNAs with different relative densities -labelled parental DNA by growing E.coli in 15N medium for many generations (heavy DNA) -transferred cells to 14N (which makes light DNA in replication) -extracted DNA after 1 and 2 generations (two rounds of DNA replication) -centrifuge in CsCl gradient -separate DNA of different densities -heavy nitrogen DNA will move more quickly to the bottom of tube Predictions of replication models Conservative Replication Would see first generation to have half heavy and half light DNA Second generation will see more light DNA to one heavy DNA Blue heavy DNA Yellow light DNA Parental Blue Blue Yellow Yellow light DNA Yellow light DNA Semiconservative Replication Predictions Would see first generation at middle of tube Second generation will have half in middle, half light DNA rising to top half Blue/Yellow Parental half Blue/Yellow Yellow light DNA Heavy Blue half Blue/Yellow Yellow light DNA half Blue/Yellow Dispersive Replication Predictions Would expect that after successive generations, the DNA becomes lighter and lighter when cultured in light nitrogen. ¼ Blue / ¾ Yellow Parental half Blue/Yellow ¼ Blue / ¾ Yellow Heavy Blue half Blue/Yellow ¼ Blue / ¾ Yellow ¼ Blue / ¾ Yellow RESULTS The cesium chloride centrifuging showed the semiconservative replication model Semiconservative DNA Replication Base-pairing to a template strand determines what is next added to a synthesized strand Both parental strands act as templates Strands at a replication fork Deoxyribo sugar and phosphate Reaction catalyzed by DNA polymerase Uses DNTP Hydroxyl group attacks phosphate Phosphodiester bond joins units *important consequence for replication Property #1 DNA synthesis always occurs 5` to 3` end But: As a replication fork proceeds in one direction along the DNA template Both strands are replicated simultaneously DNA replication at the growing fork Directional 5` to 3` synthesis of both strands Fork movement causes leading and lagging strands Leading strand: strand in which DNA synthesis proceeds in same direction as replication fork moves Lagging strand - DNA synthesis proceeds in opposite direction to replication fork movement Property #2 DNA polymerase requires a primer (short nucleic acid) Primase enzyme synthesizes an RNA primer 8-12 nucleotides long DNA polymerase adds first dNMP onto 3` end Synthesizing the lagging strand 1. primase synthesizes short RNA oligonucleotides (primer) copied from DNA 2. DNA polymerase iii elongates RNA primers with new DNA Okasaki fragment: has two components: RNA primer and .. ICLICKER example DNA sequence of two strands of one DNA molecule: 5` AGGGCTAAG 3` 3` TCCCGATTC 5` DNA pol I works as a 5` to 3` exonuclease DNA Polymerase First DNA polymerase activity characterized was E.coli DNA pol I - by Arthur Kornberg - not processive: only short DNAs synthesized; could not synthesize a long leading strand Later DNA polymerase III was discovered - processive: can synthesize long strands of DNA - replicates most of the chromosome Sythesizing the lagging strand 1. Primase synthesizes short RNA oligonucleotides (primer) copied from DNA 2. DNA pol III elongates RNA primers with new DNA 3. DNA pol I removes TNA at 5` end of neighbouring fragment and fills gap 4. DNA ligase connects adjacent fragments a. Ligation: taking two DNA molecule fragments and fusing to make one DNA polymerase I has a 5' to 3' exonuclease activity. This activity removes the RNA portion from the 5' end of an Okazaki fragment while the polymerase activity replaces the degraded RNA sequence with new DNA sequence - notice that it uses the adjacent 3' end of the neighboring Okazaki fragment as primer (DNA is not added to the Okazaki fragment whose RNA portion is being degraded) -DNA synthesis is continuous on the leading strand 1 RNA primer and synthesis to end of molecule -DNA synthesis discontinuous on the lagging strand Many Okazaki fragments and primers stitching together of many fragments. DNA ligase joins Okazaki fragments Creates new phosphodiester bonds This enzymes is very particular for the substrates it uses 5` phosphate and 3` hydroxyl required for joining to occur E. coli chromosome replication visualized using [3H] thymidine Chromosom after one round of replication Autoradiograph diagram Interpreation of Semi Conservative Single strand double strand of one radioactive DNA strand of radioactive DNA and one strand not radioactive Chromosome during second round of replication Autoradiograph diagram Interpreation of Semi Conservative Two strands replicated simultaneously Two radioactive strands This tells us both strands of DNA used in replication Therefore, a fork occurs because DNA strands separate . Step 1 - label synthesized DNA with low levels of radioactivity (early pulse) Step 2 - next label with higher radioactivity amount (late pulse) E. coli circular genome has a single replication start point - 1 origin of replication - 2 replication forks Bidirectional replication Bidirectional replication in eukaryotes 1. Pulse with high [3H] thymidine concentrations - Hot 2. Pulse with lower [3H] thymidine concentrations - Warm Replication of chromosomes in Drosophila cells. Only a very small region of a chromosome is illustrated here. Such studies have revealed clusters of active replicons Replicon - unit length of DNA that is replicated starting from a single origin of replication. Chromosomes in eukaryotic organisms such as humans may have thousands of origins of replication. Replication of linear eukaryotic chromosomes -Replication forks from adjacent replicons will eventually meet and the synthesized DNA strands will be ligated. -Replication time for the entire chromosome is greatly reduced by starting at multiple origins. .. SUMMARY E. coli (and other prokaryotes) use a single origin of replication - bidirectional replication - new replication initiation may occur before the previous round of replication is complete (faster cell growth) Eukaryotes use many origins - bidirectional replication - one initiation event at each origin; new initiation may occur before the previous. Opening the helix at the origin in prokaryotes Steps: 1. Proteins specifically recognize sequences (box elements) unique to the origin 2. DNA is melted (strand separation) 3. Proteins keep the strands separated Opening the helix at the origin Origin = recognition sequences (box elements) + AT richregion •DnaA - protein that recognizes box elements - makes sequence-specific interactions •DnaB - protein (helicase) that recognizes DnaA - further helix opening during DNA synthesis •Single-stranded DNA binding proteins - keep the strands apart 1. DnaAbindsboxes 2. Co-operativefurtherbinding of DnaA coats the origin and leads to melting in the AT-rich region 3. DnaB helicase recognizes DnaA - displacement of DnaA 4. Binding of replication machinery - replication forks are now established Opening the helix at the origin in eukaryotes Yeast model • Origin recognition complex (ORC) - multi-subunit - binds the lone DNA box element • Cdc6 and Cdt1 - bind ORC and regulate initiation • Helicase - activity is inhibited by Cdc6 and Cdt1 - strands will not be unwound and separated and DNA polymerase will not bind without functional helicase activity DNA replication in eukaryotes What is the importance of the regulatory Cdc6 and Cdt1 proteins? 1. DNA replication must only occur during S phase 2. Chromosomes must only be replicated once 2n 4n or n 2n (in haploid fungi) Additional Chapter 7 Problems: 3, 5, 14, 18, 21 Opening the helix at the origin in prokaryotes Steps: 1. Proteins specifically recognize sequences (box elements) unique to the origin. 2. DNA is melted (strand separation) 3. Proteins keep the strand separated Helicase protein at each replication fork DNA Replication in Eukaryotes 1. DNA replication must only occur during S phase 2. Chromosomes must only be replicated once 2n 4n ONCE! Additional Chapter 7 Problems: 8, 13, 15, 22, 26 ICLICKER Which one of the following statements about lagging strand synthesis is incorrect? d) DNA pol III is used to replace the RNA primer sequence with DNA sequence Correct statements of lagging strands a) an RNA primer is used and eventually removed from the 5` end of the Okazaki fragments b) Laggging strand synthesis requires the production of many Okazaki fragments c) Lagging strand synthesis occurs in a 5` to 3` direction d) DNA pol I is used to replace the RNA primer sequence with DNA sequence e) DNA pol III synthesizes MOST DNA REPLICATION Regulatory proteins: Cdc6 and cdt1 There are specific stages in the cell cycle; DNA can only be replicated in S phase and only once so that only two copies result Gap Phase synthesizes regulatory proteins Pre-replication complex forms: -Origin Recognition Complex (ORC) -Cdc6 -cdt1 -helicase (not active for replication in G1, G2 or M, activates in S phase) G1 Phase -Cdc6 and Cdt1 inhibit the helicase Transition between G1 and S phases -removal of protein components: -Cdc6 degraded, Cdt1 inhibited Helicase activated S phase -Helicase creates a replication fork -DNA pol complex can begin recruiting -only initiate replication once, since regulatory proteins have been removed, helicase can only be activated at the one previous time The Replisome -All the components required for DNA replication Functions: 1) Progressively unwind the helix and keep strands separated 2) Catalyze DNA synthesis (phosphodiester bond formation) on both leading and lagging strands -Replisome has both kinds of catalytic activity -two DNA pol III complexes for each lagging and leading strand 3) Join the Okazaki fragments -Replisome has activities that can remove RNA primer and join the frags Polymeraze components Two associated DNA pol III complexes Beta clamp for processivity DNA looping (portion of DNA is pulled around to be read by DNA pol III so that DNA pols can move in the same direction, but can still synthesize in a 5` to 3` direction.) Primase for lagging strand synthesis Replisome at work Unwinding components Helicase - strand separation Single-stranded binding proteins (ensure each individual strand stays apart after helicase separates them) Topoisomerase Progressive fork movement and DNA unwinding creates helical strain Parental Duplex is unwound -remove helical twist In circlular DNA, such as in E. coli: -DNA molecule does not like this, so what happens to compensate is an over-winding of upper DNA regions = Supercoils -as replication fork moves around circle, the supercoils would create enough tension to stop helicase working, therefore topoisomerases are required to unwind the supercoils as they form Topoisomerases relieve the strain from supercoils DNA gyrase is a type of topoisomerase 1) DNA gyrase cuts DNA strands 2) DNA can rotate to remove the twists of the coil 3) DNA gyrase rejoins the DNA strands Replisome at work Finally, must deal with Okazaki fragments to finish off lagging strand synthesis DNA pol I and DNA ligase are components of the Replisome as well TEST QUESTION ON ABOVE MATERIAL **** what would be the role of this enzyme activity and what is it a component of*** In eukaryotic organisms the chromosomal DNA exist as chromatin Complex of DNA + histones = nucleosome Nucleosomes must be displaced and then replaced behind the replication fork on the daughter molecules There are not enough nucleosomes from the parent molecule for both daughters Assembling nucleosomes during DNA replication Chromatin assembly factor -binds to newly synthesized histones -assembles nucleosome structure The replication problem at chromosome ends Gap cannot be filled because there is no primer Observation: Eukaryotic chromosomes have repetitive sequences at their ends called telomeres Telomerase: A protein-RNA complex adds the repetitive sequences to the 3` ends of linear DNA molecules 1. Provides a template (RNA) for DNA synthesis 2. Enzyme activity that uses RNA as template for DNA synthesis (reverse transcriptase) Lengthening of the 3` Overhang A region of the telomerase RNA acts as a template -enzyme binds to ends of linear chromosomes -base pair interaction between RNA/DNA -this leaves an overhang region telomerase enzyme can add on bases there is a repeated sequence at both ends (eg. AAC ------- AAC) Reverse transcriptase: 5` to 3` synthesis Repetitive template sequence allows re-positioning Replication of complementary strand A primer is synthesized Polymerase fills in the gap The primer is removed and ligase seals the gap Only a small repeated sequence is lost at the end of the chromosome during this replication. Chromosome spread -Telomeres can be recognized by a fluorescently labeled DNA probe complementary to the repeated sequence -eg picture: see four telomeres DNA replication has occurred Two telomeres per DNA molecule -Intact nuclei Telomerase: Links to human disease and cell aging Germ cells: lots of telomerase activity Must maintain chromosomal activity to pass on genetic information to sex cells and gametes to offspring Somatic cells: little telomerase activity -cells enter into a senescence phase -loss of telomere integrity Cancer cells: many have over-active telomerase -immortalization Deficiencies in telomerase -Werner Syndrome -Dyskeratosis congenita =Premature aging Chapter 8 RNA: Transcription and Processing Questions: How is genetic information encoded in the nucleus (DNA) transmitted to the cytoplasm where proteins are synthesized? What molecule is synthesized in the nucleus and then moves into cytoplasm that could carry such genetic information? Pulse-Chase Experiment Pulse label a eukaryotic cell with radioactive uracil (incorporate radioactivity into a newly synthesized molecule in a cell) (use uracil because it is in RNA and will give newly synthesized only) Then chase with excess non-radioactive uracil (we can follow the molecule as it moves through cell) The molecule will breakdown Radioactivity disappears This experiment tells us where the molecule is synthesized, where it goes, and how long it survives as that molecule in the cell Pulse: Shows us where RNA is made We see radioactivity in the nuclues where the molecule is synthesized Chase: Eukaryotic RNA moves from nucleus to cytoplasm The molecule (RNA is "chased" or followed by seeing radioactivity) temporarily accumulates in the cytoplasm Let us see the half-life of RNA in the cell - because can track where radioactive uracil is going without continuing to regenerate new mRNA without more r.U. Results We see some RNA in cytoplasm and some (few) left in nucleus snRNA in nucleus tRNA, mRNA, and rRNA in cytoplasm RNA = ribonucleic acid RNA is a precursor to protein -messanger RNA Carries the "genetic message" from genes in the nucleus to ribosomes in the cytoplasm (eukaryotes) Properties of RNA Structural features in RNA are similar and different to DNA RNA DNA Sugar: Ribose Deoxyribose Difference that RNA has a 2` hydroxyl group; DNA lacks that O at 2` C Nucleotides: Purines: Pyrimidines: Adenosine Guanine Cytosine Uracil Adenosine Guanine Cytosine Thymine Summary of Structural Differences - Contains ribose sugar o More reactive and less stable due to hydroxyl group - Contains uracil (U) base instead of thymine base o U pairs with A - Can be single-stranded or double-stranded o Structural complexity o Internal base pairing within one strand Classes of RNA 1. Coding RNA Informational RNA - provide informational template for protein synthesis a. Messenger RNAs (mRNA) 2. Non-coding RNAs Functional RNAs - is not translated into a protein Do not code for protein sequence Functional/Non-Coding RNAs - Transfer RNAs (tRNAs) - transport aa to the ribosome - Ribosomal RNAs (rRNAs) - component of ribosome - Small nuclear RNAs (snRNA) - RNA processing Splisosome used to remove introns from RNA species - Micro RNA (miRNA) - inhibit gene expression Ability to bind to prevent protein synthesis Functional RNAs can vary in size/length and therefore overall complexity and roles Transcription = production of RNA from DNA template Transcription of DNA translation protein synthesis Eukaryotic Cell -mRNA moves from nucleus to cytoplasm -functional RNAs may reside in any cell compartment RNA synthesis Same as DNA synthesis: -always synthesize 5` to 3` -complementary basepairing same -template read 3` to 5` -NTPs serve as the substrates -catalytic mechanism of breaking phosphodiester bond similar Ability to base pair to different nucleotides dictates the sequence of RNA - Uracil to Adenosine Comparison: Transcription of RNA from DNA template Similarity -5` to 3` synthesis and basepairing rules specify the addition of the next nucleotide Differences -RNA polymerase does not require a primer -Only one strand is synthesized (continuous leading strand) and therefore one template strand only -for each gene, one DNA strand acts as a template -other strand = non-template strand Different strands serve as templates for different genes Sequences of DNA and transcribed RNA Non-template strand = Coding strand RNA is complementary to the template strand, identical to the non-template strand (except U for T) Coding strands predicts amino acid -based on the direction of the RNA transcript arising from the gene we can equate the directionality of the RNA to the directionality of the DNA gene 5` end of the RNA defines 5` end of the gene RNA transcription "flows" from upstream to downstream 5` to 3` direction a DNA region is "upstream" of the 5` end a DNA region is "downstream" of the 3` end Rule: For most genes there will be a conserved transcription start point and a conserved end point Sequence signals must define the start and end points Steps in transcription: 1. Initiation - rate-limiting step, dictates amount of RNA that can be produced - Start making the RNA chain (slow step) i. Bind the template ii. First phosphodiester bonds are formed 2. Elongation - Continuous synthesis of the RNA chain (faster) - RNA polymerase movement along the template 3. Termination - Stop synthesis i. Phosphodiester bond formation ceases ii. RNA transcript released Bacteria have one RNA synthesis complex: the polymerase -multi-subunit complex -good model is the RNA pol complex in E. coli Core enzyme Contains four different subunits (alpha beta beta` and gamma) Elongation and termination Has catalytic activity that can create phosphodiester bonds Holoenzyme Same composition of core enzyme plus the sigma factor Initiation Recognizes sequence elements to do the initial binding Which of the following statements about bacterial RNA polymerase is correct? ANS: It exists in one form called the holoenzyme which contains a subunit called the sigma factor How does RNA polymerase know where to start transcription? Approach: Compare many genes and many start sites. What is in common? Conserved DNA sequences? E.coli promoters Promoter - set if DNA sequences that are required to initiate transcription - recognized by RNA polymerase consensus sequence most common nucleotide among collection, but a gene may have some variation box elements: a conserved secquence serve as binding sites for RNA pol complex RNA pol will start transcription at the +1 position Summary E.coli promoter genes contain two minimal elements (core elements) - the -10 and -35 box - are located upstream of transcriptional start site - conserved sequences (consensus) - conserved distances from start site, which is +1 - strong promoters (high level RNA synthesis) are closer matches to the consensus Transcription in Prokaryotes Transcription Initiation a) RNA pol binding to promoter -the -35 and -10 region - recognized by sigma factor b) Initiation -of phosphodiester bond formation c) transcription bubble forms d) RNA synthesis sigma factor can then dissociate Transcription Elongation Transcription buble moves along with the RNA pol complex Short RNA-DNA hybrid is maintained (10 base-pairs) throughout elongation Anything that disrupts this will trigger premature termination As RNA pol moves along, it needs to let DNA rewind after Unwind with helicase then rewind the DNA double helix Many RNAs can be simultaneously transcribed from a single gene -initiation before elongation complete -looks like christmas tree light structure -longest RNA transcripts towards end of gene -we see RNA strands emanating off the gene - multiple RNA pol on same gene -example: rRNA transcription in frog oocytes - very active transcription Question: What defines a transcription stop point? Answer: Sequence and structure in the RNA just synthesized Two elements required in RNA transcript: 1) Hairpin structure (stem-loop) 2) Poly U stretch There is an intrinsic terminator on the RNA transcript Intrinsic: RNA pol complex on its own without any other RNA factors recognizes this RNA feature The RNA transcript has ability to base pair with itself, antiparrallel base pair interactions, so are forming a helical structure in the hairpin stem-loop At the 3` end of the RNA there needs to be a consequetive stretch of U's This must be positioned at the base of the stem-loop structure Termination: intrinsic mechanism -RNA-DNA hybrid is weak The base pairing between U's of RNA and the A's of DNA is weak -Hairpin helps destabilize the hybrid -RNA then dissociates from DNA template Transcription in Eukaryotes -More complex than in prokaryotes -larger genes and genomes -more complex gene regulation -polymerases are larger -accessory factors -3 RNA pol (eukaryotes) vs. 1 in prokaryotes -RNA pol I - ribosomal RNAs -RNA pol II - mRNAs (and snRNAs) -RNA pol III - tRNAs and 5SRNA (small RNA component of the ribosome) Initiation in Eukaryotes RNA pol complex does not recognize promoters on its own -unlike the bacterial RNA polymerase holoenzyme General transcription factors -several large complexes that recognize promoter elements and/or modify RNA pol activity -Somewhat analogous to the sigma factor -Stimulate RNA pol II transcription -Transcription Factors II (TF IIs) TFIID, TFIIE, TFIIF etc Transcription Initiation in Eukaryotes TFIID - a complex containing several proteins including: TATA box binding protein The TATA box - contains the sequence TATA - correctly positions the RNA pol complex - more transcription factors bind to form the pre-initiation complex Pre-initiation complex -DNA is melted -RNA pol has bound -no phosphodiester bond formation yet (not activated yet) Initiation to Elongation -RNA polymerase structural change -RNA pol II begins elongation 1. Phosphorylation of RNA polymerase (by TFIIH) 2. Continuous RNA synthesis begins 3. Transcription factors dissociate from the polymerase RNA processing - in the eukaryotic nucleus All eukaryotic RNAs undergo processing events to convert a precursor (primary) transcript into its mature form Prokaryotic and eukaryotic transcription and translation Prokaryote: coupled transcription-translation Ribosomes can start synthesizing the proteins on one end of mRNA while transcription is still occuring on that mRNA Eukaryote: transcription in nucleus The transcript is converted into a mature form of mRNA Transportation of mRNA from nucleus to cytoplasm Translation can then occur because ribosome recognizes mRNA Transcription Elongation Termination: Addition of a polymeric (poly) A tail Endoneclease recongizes a sequence element and then cuts 3` of it Eukaryotic genes encoding proteins contain intervening sequences Intervening sequences = introns -don't usually code for amino acids in proteins -removed after transcription Protein-coding portions of genes = Exons -present in final mature mRNA Splicing -Introns are transcribed along with exons in the primary transcript -Introns are removed and the exons are spliced together Question: How are the correct sites for splicing chosen? Conserved sequences specify intron splice sites GU-AG"rule" GU at 5` end of pre-mRNA and AG at 3` end The mechanism of Splicing (write this out a few times/memorize/final) -moving of phosphodiester bonds Reactions in exon splicing First transesterification Branch point "A" is involved in a 3` to 5` phosphodiester bond PLUS a 2`-5` phosphodiester bond To perform this reaction need a free hydroxyl group; the only free one in this case is the 2` hydroxyl from branch point "A" There is a transfer of a phosphodiester bond This can only be done in RNA Because of the 2` hydroxyl, RNA is more reactive; DNA lacks the hydroxyl Splicing must be an accurate process Splicing at same intron position Complex pattterns of eukaryotic mRNA splicing occur Highly dependent on spliceosome During splicing, the mRNA associated with a large RNA-protein complex is called a spliceosome Eukaryotic mRNA introns are called spliceosomal introns Roles of the Spliceosome 1) Recognize the conserved intron elements 2) Help fold the intron into the correct 3d shape - RNA catalysis 3) Regulate the splicing process - Alternative splicing - Controls which extrons are getting spliced together Spliceosome composition 1) Spliceosomal proteins and small nuclear RNAs (snRNAs) 2) snRNAs are called U RNAs (uracil rich) - U1, U2, U3.. 3) snRNAs recognize the intron elements - RNA-RNA base pairing Spliceosome assembly and function Splice site and branch point recognition SNPs U1 and U2 bind to the 5` splice site and internal A The U4-U5-U6 complex joins Intron folding Makes catalytically active Splicing step 1 First splicing reaction: one intron end attaches to A Splicing step 2 Second splicing reaction: other intron end cleaved' exons join Summary Spliceosome recognizes the conserved intron elements Splice site definition Different snRNPs assemble at different stages Dynamic complex Only a subset of snRNPs are present during the trans-esterification reactions (U2, U5 and U6) mRNA in eukaryotes: -Sequences present in the primary transcript may not be present in the mature mRNA. -Sequences present on the 5' and 3' ends of the mature mRNA were not present in the DNA template. -Removal of introns generates a lariat intron product. -Exons from a single gene may be spliced (joined) together in different combinations. Cotranscriptional processing of RNA in eukaryotes The phosphorylation form of RNA polymerase II interacts with the complexes involved in capping, splicing, and polyadenylation -efficiency of mRNA production a) Capping - Interacts with phosphorylation - Activates catalytic activity b) Splicing - Phosphorylation status changes c) cleavage and polyadenylaiton - phosphorylation status changes again (functionality) Self-splicing introns -spliceosomal introns require the spliceosome for catalysis -other intron classes can splice themselves no protein factors required -first identified in a rRNA gene in Tetrahymena (Cech, 1989) also found in mitochondrial, chloroplast and bacterial genes Self-splicing reaction (autocatalytic) Group I intron Free GTP is used as the first nucleophile -attacks the phosphodiester bond at the 5` splice site using the 3` hydroxyl -now have a new 5` end which has an added G from GTP -5` exon joined to 3` end Products are ligated exons and a linear intron The RNA World Walter Gilbert - coinventor of DNA sequencing; looked at introns - RNA can store genetic information (viruses, mRNA) and RNA can undergo enzymaticlike reactions - Catalytic RNAs are called ribozymes (RNA-based enzymes) Therefore, RNA can act both like DNA genetic material and as protein (storing of information and enzymatic activity) However, both DNA and proteins separately are better at their roles DNA is more stable than RNA Proteins can have more enzymatic variety than RNA RNA could have been the ancestral genetic material Small RNAs were the breakthrough of the year (non-coding RNAs) Small RNAs control gene expression at all levels: DNA - structure of chromation RNA - both informational an functional RNAs affected Protein synthesis Similar suppression phenotypes can be induced by injecting cells with double-stranded RNA molecules Eg) pigment inhibition in petunia flowers Different strands serves as templates for different genes but only one strand of RNA is produced. Any gene in the chromosome is only producing one RNA strand. Trans-genetics -unusual insertion sites for introduced genes may generate double-stranded RNA mRNA produced from the transgene, and gene adjacent to the transgene also has a promoter so another RNA strand is produced RNA Interference (RNAi) Gene silencing mediated through small RNA molecules Double-stranded RNA acts as the trigger mechanism Precursor - double stranded RNA RNA cleavage occurs RNA interference A way to silence genes using small RNA molecules We looked at the mechanism where you can silence a gene at the molecular level Example: petunia flowers, suppression of purple pigment artificially Pathway starts with double stranded RNA molecules RNAi - interfering RNA Trans-genetics Unusual insertion sites for introduced genes may generate double-stranded RNA mRNA ---Transgene------Gene--- --Promoter-- siRNAs - small interfering RNAs Mechanism of action of RNAi Dicer - an endoribonuclease A double-strand specific The enzyme "dicer" cuts the double stranded RNA into smaller sections One of the siRNAs will be incorporated into risc complex (RISC) RISC - RNA induced silencing complex: -RISC binds to the single-stranded siRNA and then binds to the mRNA -mRNA cleavage and degraded Therefore inactivates mRNA Gene is silenced RISC - RNA induced silencing complex Binds a single-stranded siRNA and binds the mRNA mRNA cleavage need a strand of DNA that is complementatary to the mRNA In order for this to happen, one strand of the double-stranded siRNA must be complementary to the target mRNA Specific base-pairing between an siRNA and mRNA target mRNA cleaved and inactivated The RNAi technique is particular useful because: 1) Gene replacement or knockout is not required to eliminate gene expression - RNAi is fast to employ in a lab setting and can be used to inactivate gene expression at different development stages 2) Specific - target only one mRNA for destruction - possible treatment for viral infections or cancer cells? Additional Chapter 8 Problems: # 6 , 15 Chapter 9 Proteins and their Synthesis -Protein synthesis occurs through the process called translation -Translation occurs on RNP particles called ribosomes Year 2000 - first high resolution 3D structures solved for ribosomes: How does the ribosome work? How is the ribosome inhibited - antibiotics? The binding of a drug to the ribosome prevents translations Eg) -Large ribosomal subunit Erythromycin (drug) binds to the middle of the ribosome The binding site is the exit point of the amino acid chain - channel where the polypeptide leaves the ribosome Translation components -Ribosome - RNA complex - catalyze peptide bond formation -Transfer RNAs (tRNAs) - bring amino acids to the ribosome - decode information on the mRNA -Messenger RNA . Protein Structure -amino acids with different properties -protein sequence and 3D shape influences function -molecules that can be bound -cellular location -protein-protein interactions -Protein 3D structure is stabilized by: -ionic interactions -H bonding -hydrophobic and Van der Waals forces Amino Acid Structure There are 20 different standard amino acids in proteins Side chain R group is different for each one Contains a central carbon, an amino group, and a carboxylic acid group. Different side chain (R group) properties: -charged -Protein = chain of amino acids linked by peptide bonds = polypeptide chain Peptide bond: carbox joined to amino Synthesized from the amino end towards the end that will be the carboxyl end Atoms in the peptide bond are co-planer Oxygen and nitrogen atoms of the peptide bond can engage in H-bonds with other amino acid positions The peptide bond is stable at physiological pH Levels of protein structure 1. Primary structure = linear sequence (order) of amino acids within the polypeptide 2. Secondary structure - arises from interactions between "nearby" amino acids (e.g. H-bonds form an -helix). The amino acids aren't necessarily directly adjacent but are relatively close in the linear sequence. 3. Tertiary structure - the overall 3-d shape of the entire protein. Does the protein adopt a globular structure or is it elongated? Where are the precise positions of all the atoms of the protein located in 3-d space? How are all the secondary structural elements positioned relative to each other?The illustrated hemoglobin complex contains: 2 subunits and 2 subunits (4 polypeptides total). Therefore, this structure is a heterotetramer. 4. Quaternary structure - interactions between multiple polypeptides E.g. dimer = two polypeptides, tetramer - 4 polypeptides Homo - identical (protein products from same gene) Hetero - different (from at least 2 genes - different polypeptides) The Genetic Code How is the sequence in DNA converted into amino acid sequence in proteins? - How is mRNA read as a template during translation? 1) How many nucleotides specify one amino acid in a protein? 2) What is the directionality of reading an mRNA? The Genetic Code The length of mRNA sequence that specifies one amino acid = Codon Codon length 1 nt = 4 possible combinations (4 amino acids) 2 nt = 4 X 4 = 16 amino acids could be specified 3 nt = 4 X 4 X 4 = 64 possible amino acids could be specified Therefore, codons of 3 nt length contain enough information to specify the 20 different standard amino acids found in protein structure Overlapping versus non-overlapping genetic codes Overlapping - each nucleotide is part of more than one codon Non-overlapping - each nucleotide is only part of one codon Mutants and Suppressors demonstrate a non-overlapping triplet code (Crick and Brenner, 1961) Observation #1: A single substitution mutation (1 nt change) only ever changes 1 amino acid within a protein. Therefore, code is non-overlapping Observation #2: Only certain combinations of addition and deletion mutants suppress mutant phenotypes (re-establish protein functionality) Additions in multiples of 3 Deletions in multiples of 3 or equal numbers of additions and deletions For example portion of mRNA sequence wild - type 5' --------------- GAU CGA CUU GGA AAA UUU GCA CCA ------------------3' 1 addition Asp- Arg- Leu- Gly - Lys - Phe - Ala - Pro mutant 5' --------------- GAU UCG ACU UGG AAA AUU UGC ACC A -----------------3' Asp- Ser- Thr- Trp - Lys - Ile - Cys - Thr Changed portion of mRNA sequence SYNTHETIC POLYRIBONUCLEOTIDES DIRECT SYNTHESIS OF POLYPEPTIDES IN A CELL-FREE SYSTEM (Nirenberg & Matthaei, 1961) mRNA template - UUUUUUUUUUUUUUUIn vitro peptide synthesized polyphenylalanine: (Phe)n polylysine: (Lys)n Therefore, UUU is the codon for phenylalanine (Phe) AAA is the codon for lysine (Lys) Likewise: poly(G): Gly (only) poly(Gly) GGG = Gly poly(C): Pro (only) poly(Pro) CCC = Pro -AAAAAAAAAAAAAAAAACHEMICALLY-SYNTHESIZED POLYRIBONUCLEOTIDES OF DEFINED SEQUENCE DIRECT THE SYNTHESIS OF POLYPEPTIDES OF DEFINED SEQUENCE (Khorana) e.g., poly(UG) ... UGU|GUG|UGU|GUG|UGU|GUG|UGU|GUG ... ... Cys-Val-Cys-Val-Cys-Val-Cys-Val ... The same alternating sequence of two amino acids (Cys and Val) results regardless of where translation begins. USE OF SYNTHETIC POLYRIBONUCLEOTIDES WITH REPEATING SEQUENCES TO DECIPHER THE CODE This example shows how polypeptides derived from the (AAG)n polymer were used to confirm the triplet code and help to identify the codons. The (AAG)n polymer can specify three different polypeptides, depending on which reading frame is used. Chapter 9 Practice Problems: 1, 2, 3, 5, 7, 9, 16, 17, 27, 28 The Genetic Code (written as RNA) Standard Genetic Code - the vast majority of organisms use the same codes for amino acids Chain termination (`stop`) codons are shown in orange UAA UAG UGA The usual intial (`start`) codon is shown in green AUG For Example 5` CAC AGU 3` His-Ser NH3+ Hist - Ser - COO- Degeneracy and the Genetic Code The genetic code is: Unambiguous: one codon, one amino acid Eg) methionine encoded by AUG Degenerate: one amino acid, more than one codon Transfer RNA (tRNA) is the adaptor molecule Base-pairing mRNA codon: tRNA anti-codon At least 20 different tRNAs Amino acids are temporarily covalently attached to tRNAs - "charging" The accuracy of attaching the correct amino acid (cognate) to the correct tRNA is critical for ensuring translation accuracy Accuracy - same amino acid to same tRNA species - called the cognate species The structure of tRNA -many base pairing interactions -small, but highly structured -amino acids are always added onto the very 3` end of the tRNA -the other end interacts with the mRNA codon sequence called the tRNA "anticodon" Specificity -Binds only one amino acid (cognate) -Binds only cognate tRNA(s) -ATP is required during the charging reaction Aminoacyl-tRNA synthetase is specific for an amino acid Eg) alanine Binding site for alanine Binding site for tRNAAla Enzyme joins carboxylic acid carbon of alanine onto the 3` end of tRNA Requires ATP The structure of tRNA Four stems of base-pairing interactions in a 2-D cloverleaf representation Inverted L shape in the 3-D representation -3` end is amino acid attachment site -DHU loop -TyC loop -Anticodon loop Two superimposed yeast tRNAs -notice how different species of tRNAs have the same path of phosphodiester backbone; this nearsuperimposable structure must be highly stable -tRNA structure is different in anticodon region, and A.A. binding site Genetic code is degenerate Amino acid serine is specified by 6 different codons. How does the cellular tRNA set relate to the degeneracy? Eg) Different tRNAs that can service codons for Serine tRNA Anticodon Codon tRNASer1 AGG+ wobble UCC .. How does the cellular tRNA set relate to the degeneracy? 1) More than one tRNA species can be charged with the same amino acid Isoacceptors Most organisms contain at least 3 different tRNAs that are charged with serine: tRNASer1 2) A single tRNA species can recognize more Wobble allows one tRNA to recognize multiple codons RNA can engage in non-canonical base pairs such as G-Upairs This tRNA can recognize either 5`UCC3` or 5`UCU3` codons Wobble position Non-colonical? pairings; G can base pair to a U This is restricted to only the anticodon's 5` end, so the mRNAs 3` codon end This explains the genetic table; Wobble pairing explains patterns why multiple codons can encode for the same amino acid Wobble poisiton is 5` in the anticodon 3` in the codon on the mRNA (third position) Codon-Anticodon Pairing Allowed by Wobble Rules: 5` end of anticodon tRNA 3` end of codon mRNA G C or U C G only A U only U A or G I U, C, or A I= Inosine, a modified base Modified nucleotide forms from converting an A into an I, present in tRNA Similar structure to guanine Ribosome structure -small subunit (SSU) and large unite (LSU) -conservation of structure - particularly in the ribosomal RNAs (rRNA) Prokaryotic vs. Eukaryotic Ribosomes -prokaryotic are smaller -fewer ribosomal proteins and shorter rRNAs Prokaryotic Ribosomes -the 70 S ribosome -two subunits, small (30 S) and large (50 S) -small 16S rRNA and 21 proteins only found in the small subunit -large 23S rRNA plus 5S rRNA plus 31 proteins Eukaryotic Ribosome -the 80S ribosome is larger -small ribosomal subunit is 40S -large subunit is 60S -produces more proteins; has conterparts to the prokaryotic rRNAs -Large produces three rRNAs Ribosomal RNAs (rRNAs) are highly structured -engage in lots of base pairing -intramolecular base-pairing interactions -secondary structure -RNA helices -conserved tertiary structure -3D structural conservation (prokaryotes and eukaryotes) -binding sites for ribosomal proteins, mRNA, .. Secondary structure of prokaryotic 16S (small subunit) rRNA -rRNA folds up by intramolecular base pairing Short range interaction -stem-loop element Long range interaction -base-pairing between distant sites Ribosome subunit roles -Small subunit -bind the mRNAs -decoding site -Large subunit -catalyze peptide bond formation peptidyl transferase activity Key sites of interaction in the ribosome Computer model 50S large subunit -polypeptide exit tunnel -peptidyl transferase center 30S small subunit -mRNA binding site 3 tRNA binding sites: E site, P site, and the A site As the polypeptide grows, (from amino terminus to carboxy terminus) it becomes too large for the ribosome and must be released Key sites of interaction in the ribosome tRNA movement A site P site E site Opposite direction of motion of the ribosome along the mRNA strand Schematic model When we form a peptide bond, always have two tRNAs binded tRNA binding sites A site - Amino acyl site Next incoming charged tRNA binds here -next amino acid to be added Incoming tRNA that matches the mRNA enters ribosome at A site P site Peptidyl site tRNA temporarily carrying the elongated peptide chain resides here E site - Exite site tRNA temporarily resides here just prior to leaving the ribosome Deacylated tRNA released from the E site Initiation of translation 1. Binding of mRNA first to the small subunit (SSU) 2. Recognition of a start codon a. Usually AUG b. Utilizes a specialized initiator tRNA i. Prokaryotes - tRNAfMet ii. Eukaryotes - tRNAiMet 3. Requires protein translation factors called initiation factors How is the correct AUG codon chosen? In prokaryotes, sequences upstream of an AUG start codon define the start site by binding to the small subunit of the ribosome. We use the plank sequence (upstream) which makes interactions to serve as a binding site. Shine Dalgarno Sequence Upstream of an AUG codon Preference with a purine nucleotide Small subunit of mRNA positioned upstream of AUG codon Base-pairing between Shine Dalgarno sequence in mRNA and anti-Shine-Dalgarno sequence at 3` end of small subunit RNA Translation initiation in prokaryotes 30S ribosomal subunit + initiation factors 1. Keep subunits separated IF2-fMet-tRNAf + mRNA Gives fMet and IF3 2. Recruit the initiatior tRNA and position it at a start codon 3. Binding of the large subunit a. Initiation factor release b. tRNAfMet is positioned in the P site Once the initiation process occurs, it is essential that the initiator tRNA is in P site Translation initiation in Eurakyotes Difference: no Shine-Daglarno sequence Instead, take advantage of 5` cap structure for recognition during initiation 1. Recognition of the mRNA cap structure 2. Scanning to find an appropriate start codon a. AUG and flanking nucleotides Not necessarily going to be close to the 5` end Ribosome moves 5` to 3` b. Energy dependent Requires hydrolysis of ATP 3. Binding of the large subunit a. Initiation factors released b. tRNAfMet is positioned Translation Elongation -Covers all steps where we make peptide bonds -Progressive creation of peptide bonds until a stop codon Cycle: 1. aminoacylated-tRNA binds to the A site of the ribosome 2. formation of peptide bonds through peptidyl transferase reaction 3. Translocation a. Ribosome re-positions itself b. . Steps in translation elongation in prokaryotes -tRNA is binded to the P site -Translocation factor EF-Tu brings aminoacylated tRNA to the ribosome -Peptidyltransferase reaction moves peptide (or first amino acid) to the A site tRNA -Elongation factor EF-G displaces A site tRNA GTP dependent mRNA moves through the ribosome Translocation Termination 1. Stop codon not recognized by a tRNA anticodon a. Recognized by a release factor Prokaryotes: RF1, RF2 (specifically to recognize stop), and RF3 2. polypeptide is released from tRNA and ribosome 3. ribosome subunits dissociate from each other and mRNA template Termination of translocation -Release factor recognizes a stop codon assisted by other RF's and protein factors hydrolysis breaks the bond joining the polypeptide to tRNA Post-translational processing events -Covalent additions of chemical groups or peptides phosphates, sugars, ubiquitin covalent bonds are formed -Protein sorting and targeting signal sequences -Protein folding (also co-translational) chaperonin complexes associate with unfolded proteins to allow folding in some cases the chaperonin complex can occur at the ribosome and therefore be either co-translational or post-translational Phosphorylation -protein kinases use ATP as a substrate and add it onto a specific amino acid position to convert it from inactive to active enzyme Phosphorylation Regulates - enzymatic activity - protein-protein interactions or the association of protein complexes - protein-nucleic acid interactions phosphates are charged groups that can interact with acids - cellular localization Ubiquitin may target a protein for degeneration Ubiquitin is a highly-conserved small protein added onto lysine side chains Ubiquitin chain is degraded Ubiquitin is attached to a different protein Ubiquitin protein is formed addition of multiple ubiquitins (ubiquitin chain) is a signal for protein digestion called a 26S protease protease degrades proteins results in oligopeptides and ubiquitin chain Signal sequences (protein sorting) - short peptide sequences within the protein with distinctive chemical properties - used to direct proteins to the nucleus, mitochondria, chloroplasts, plasma membrane, for excretion - signal sequence may be removed (cleaved) or retained Signal sequences target proteins for secretion Signal recognition particle - an RNP - directs the ribosome to the translocation complex in the endoplasmic reticulum membrane
