Genetics Final Exam Review
Genetics Final Exam Review Bio 2450
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Genetics Final Exam Review Genetics the study of heredity how characteristics or traits are passes from parent to offspring History of Genetics Aristotle 384 322 BC new structures arose spontaneously species were xed the product of special creation amp had very speci c purposes Ladder of Life scala naturae In 17005 Linnaeus thought each species was the product of special creation By 18005 the notion of biological change over time evolution was gaining acceptance 0 Jean Baptiste Lamarck Inheritance of Acquired Characters morphological changes that occurred during the life of an organism were passed on to offspring o Blending Inheritance tendency toward the mean of the parents 1858 Charles Darwin amp Alfred Russell Wallace published joint paper about Natural Selection but Blending inheritance was a problem 0 1859 Darwin publishes On the Origin of Species 1866 Gregor Mendel published a paper on the particulate nature of hereditary factors paper was ignored for 34 years 1900 Hugo de Vries Carl Correns Erich von Tshermak independently rediscovered Mendel published Proceedings of German Botanical Society Considered the Birth of Genetics 1906 William Bateson termed genetics 1909 Wilhelm Johanson termed gene Mendel s Experiments took a quantitative approach to study the inheritance in peas 1 P P PP N 7 characters of pea plants Seed coat color ower color Grey amp purple White amp white Seed color Yellow or green Seed shape Smooth or wrinkled Pod color Green or yellow Pod shape In ated or pinched Stem height tall or short Flower position Axial or terminal Isolat d truebreeding strains Looked at each trait individually initially Conducted careful breeding experiments Kept careful records Had large sample sizes Followed traits for several generations P P FP NE Mendel s Results Generation Phenotypes P Purple owers x White owers true breeding strains F1 All purple owers F1 X F1 F2 Some purple 705 Some white 224 315 to 1 ratio of purple to white 1 One phenotype disappears in F1 recessive trait 2 Reappears in F2 3 31 ratio dominantrecessive All results were 31 ratio for all other characteristics of pea plants Mendel s Hypothesis 1 Each trait is determined by a pair of discreet units genes 2 There are often 2 or more alternative forms of genes alleles 3 Truebreeding individuals have 2 of the same alleles homozygotes Hybrids have 2 different alleles heterozygotes 4 Dominant alleles can mask the expression of recessive alleles Genotypes the alleles carried by an individual Mendel s 1St Law Principle of Segregation the two members of an allele segregate from each other during gamete formation amp are randomly distributed to the offspring Offspring receive 1 allele from each parent Monohybrid cross two traits EX Gg X Gg Mendel s Model is a probabilistic model Multiplicative rule If the events A amp B are independent the probability that occur together PA and B PA PB Additive rule If the events A amp B are independent the probability of at least one of them occurring PA or B PA PB PAPB Binomial probability pqn P probability of event P Q probability of event Q N observations trials Pascal s Triangle EX anded binomial 1 2 3 2 4 6 4 5pq10pq 10pq 5 4 5 n x Generalized formula for binomial probability Why 4y X events P Y events Q X y n N nn1n2321 Note 0 1 As sample size increase the probability of random deviation from the expected decreases Results are less affected by chance as sample size increases X2 Test Statistic quotChi Squarequot 0 Used to compare data with the predicted values according to a hypothesis o If X2 gt critical value reject the hypothesis 0 If X2 lt critical value cannot reject the hypothesis We conclude that the deviation of observed values from predicted is due to chance 0 Critical value 005 obs exp 2 o Null hypothesis H0 there is not difference between observed amp X222 2 expected values 0 Degrees of freedom classes of data 1 Pvalues tell us the probability of seeing a deviation between observed amp expected values that is as great or greater that is due to chance Dihybrid cross 4 traits Pth X Pth 9331 ratio you only see this ratio if alleles at the 2 genes are assorting randomly with respect to each other Mendel s 2nOI Law Principle of Independent Assortment the alleles of different genes assort independently of each other all allele combinations are equally likely Implications of Mendel s Laws 1 Falsi ed the idea of Blending Inheritance 2 Genotype l Phenotype sometimes modi ed by the environment Pedigree Analysis D Male D Affected male 0 Female 0 Affected female Dominant Traits Guideline 1 Every affected individual must have at least 1 affected parent 2 Trait usually appears in every generation 3 Affected heterozygotes transmits mutant allele to 12 offspring on average 4 Most crosses are Aa X aa Recessive Traits Guideline 1 Most affected individuals have 2 normal parents 2 Trait often skips a generation 3 31 ratio expected in monohybrid crosses 4 When both parents are affected all offspring are affected 1902 Sutton amp Boveri proposed the quotChromosome Theory of Inheritancequot 1 Chromosomes are carriers of genes 2 Each species has a characteristic number of chromosomes which occur in pairs 3 Diploid 2N two sets of chromosomes 4 Homologous chromosomes members of a pair of chromosomes that carry the same genes Haploid 1N one set of chromosomes Autosomes vs Sex chromosomes Locus Loci the physical position of a gene on a chromosome Uquot Mitosis occurs in somatic cells 2N parents D 2N 2x daughter cells 1 lnterphase chromosomes replicate 2 Prophase chromosomes condense nucleus disappears spindle forms 3 Metaphase chromosomes align at the equatorial plate 4 Anaphase sister chromatids separate cytokinesis begins 5 Telophase nuclear membranes reappear chromatin expands cytoplasm divides Summary of Mitosis 1 Maintains the diploid condition 2 Produces 2 identical cells 3 The mechanism of cell proliferation Meiosis gamete formation 2 divisions Meiosis 1 2N parents D 1N 2x unique daughter cells 1 lnterphase chromosomes replicate 2 Prophase l chromosomes condense amp synapse crossing over occurs crossing over reciprocal exchange of segments between homologous chromosomes 3 Metaphase l chromosome pairs align at the equatorial plate 4 Anaphase l homologous chromosome pairs separate 5 Telophase l chromosomes each with two sister chromatids complete migration Meiosis ll 4 gametes 1N 1 Prophase ll 2 Metaphase ll 3 Anaphase ll 4 Telophase II Summary of Meiosis 1 A reduction division 2N l 1N 2 Each maternally amp paternally derived chromosome has an equal chance of aligning on one sid or the other during Metaphasel a Independent assortment amp segregation b Many possible combinations of chromosomes 2N 3 Crossing over creates even more variation 1910 Morgan s experiments with Drosophia Red eyes wild type amp White eyes mutant Males XY hemizygous amp Females XX Calvin Bridges noticed that about 12000 F1 ies wee whiteeyed females XLinked traits reciprocal crosses are not the same different ratios for the different sexes TriploX dies 0 X0 male sterile YO male dies Nondisjunction sister chromatids fail to separate amp move to opposite poles during meiosis Anueploidy abnormal chromosome condition Sex Determination basis mechanism 1 Genotypic sex determination the sex chromosomes play an active role a Ychromosome mechanism humans i Y is active ii Evidence comes from X chromosome iii Nondisjunction iv Turner syndrome XO female 45 chromosomes v Klinefelter syndrome XXY male 47 chromosomes amp XXXY male 48 chromosomes vi Molecular investigations revealed a sexdetermining region on the Y chromosome b X chromosome autosome balance system i Ratio of X chromosomes sets of autosomes ii XX female XY male XO male c Birds and Lepidoptera i ZW female heterogametic ii 22 male homogametic 2 Environmental sex determination a Marine worm Bonelia free swimming larvae are undifferentiated i Settle down alone D male ii Settle down near or on male D female b Turtles i Tgt32O male ii Tlt28O female iii In between T mixture c Some sh are male when small female when large lntersex in Humans the variety of expressions of sexual characteristics falling between standard male amp female Enlarged diminished or ambiguous external genitalia Absence of genitalia Combination of genitalia Frequency 1100 births not XX or XY 11666 births Xlinked recessive trait 1 Female must be homozygous to be affected 2 Affected males transmit to all daughters 3 Many more males than females are affected 4 All sons of affected mother are affected 5 11 ratio in sons of carriers 12 daughters will be carriers Xlinked dominant trait 1 Heterozygous females are affected more frequent in females 2 All daughters 7 none of the sons of affected father are affected Y linked trait 1 Every son of affected male is affected 2 No females are affected Extensions of Mendel Sex linkage Multiple alleles Dominance relations 0 Lethal alleles Gene interactions epistasis Pleiotropy 0 Environmental factors Dominance relationships 0 Complete dominance phenotype of heterozygotes phenotype of homozygous dominant Complete recessivity recessive allele expressed only in homozygote Incomplete or partial dominance heterozygote phenotype is intermediate Codominance heterozygotes exhibit the phenotypes of both homozygotes 0 Human blood groups ABO system lA amp lB are completely dominant to i lA amp lB are dominant with respect to each other Lethal alleles occur at essential genes they can be dominant or recessive lead to clear deviations from Mendelian ratios 1905 Discovered in mice coat color gene AY yellow allele dominant amp A agouti wild type Monohybrid cross AYA X NA 0 AYAY dies early in development 0 AY has dominant effect on coat color but is recessive lethal Dominant lethals are usually hard to study however sometimes death occurs after reproduction Huntington s Disease caused by a dominant lethal but effects nervous system degeneration are not seen in people until they re in their 30 s Epistasis a form of gene interaction in which one gene masks or otherwise affects the phenotypic expression of another Recessive epistasis presence of homozygous recessive at one gene masks the effect of the other gene Duplicate recessive epistasis homozygosity for the recessive allele at either gene or both genes gives white pigmented owers Pleiotropy a single gene affects multiple phenotypic traits Phenylketonuria PKU recessive allele causes a disorder in amino acid metabolism phenylalanine 1 Can have serious consequences early in life 2 Toxic substances phenylketones build up 3 Phenylketones cause brain damage mental impairment 4 Also interferes with melanin synthesis lightening of hair Penetrance frequency with which a genotype is manifested in the phenotypes of a population 100 penetrance in ABC group system Everybody with ii genotype have Otype blood Also observed in Mendel s 7 characters Incomplete penetrance individual with a particular genotype does not show the corresponding phenotypes Polydactyly shows incomplete penetrance p dominant mutation causes the presence of extra ngers amp toes Expressivity degree to which a penetrant gene is phenotypically expressed Osteogenesis imperfect autosomal dominant mutation Symptoms blue in whites of eyes fragile bones deafness 100 penetrance however there is variation in the severity Complete penetrance identical known genotypes yield 100 expected phenotype Incomplete penetrance identical known genotypes yield lt100 expected phenotype 3 Constant expressivity identical known genotype with no expressivity effect yield 100 expected phenotype 4 Variable expressivity identical known genotypes with an expressivity effect yield a range of phenotypes 5 Incomplete penetrance with variable expressivity identical know genotypes produce a broad range of phenotypes due to varying degrees of gene activation and expression N Environmental effects lnternal Environment Factors such as age amp sex may in uence the expression of traits Age all genes do no continually function some are deactivated or activated at speci c times or ages 0 Huntington s Disease not activated until 30 s 0 Pattern baldness at 20 30 years old Sex sexin uenced or sexlimited traits may be in uenced by hormones amp other factors 0 Pattern baldness External Environment Temperature Himalyan rabbit coat color Chemicals PKU 0 Low phenylalanine diet limits effects of disease Plasticity variable phenotypes depending on environment DNA Discovery Genetic material must have 3 principal characteristics 1 Must contain information 2 Must replicate accurately 3 Must be variable History 1800s an acid called quotnucleinquot was found in cells 1940s Beadle amp Tatum proposed the quotOne GeneOne Enzymequot hypothesis but DNA was thought to be a framework for proteins which carried genetic information 1928 Frederick Grif th demonstrated that noninfectious bacterial strains could be transformed to infectious strains 1944 Avery MacLeod amp McCarthy showed that the transformation occurred through exchange of DNA between bacterial strains not proteins 1944 DNA was localized in chromosomes 1949 Mirsky amp Ris showed that the amount of DNA in diploid cells was constant amp 2X amount in sperm 1952 Hershey amp Chase experimented with bacteriophages o Hershey amp Chase experiment Preparation of radioactively labeled T2 bacteriophage 1 T2 phage infected E coli amp grow in 32Pcontaining medium Progeny phages with 32plabeled DNA 2 Infected E coli amp grow in 35scontaing medium Progeny phages with 35sIabeIed protein Experiment that showed DNA to be the genetic material of T2 1 32p DNA injected into E coli blend brie y radioactivity recovered in host amp passed on to phage progeny phage progeny radioactive 2 355 protein injected into E coli blend brie y radioactivity recovered in phage ghosts amp not passed on to the progeny phage progeny no radioactive o Bacteriophage life cycle Inside a host bacterium E coli there is a bacterial chromosome Phage chromosome injected bacterial chromosome broken down Phage chromosome replicated using bacterial materials Expression of phage genes phage structures produced Phages assembled Phage particles released by ysis The structure of DNA amp RNA P P FP N DNA is a polymer Nucleotide sugar base phosphate group 1953 James Watson amp Francis Crick Available evidence 1 Chemical structure 2 Erwin Chargaff 1951 a A l T b G I C 3 Rosalind Franklin amp Maurice Wilkins used xray diffraction to detect helix pattern DNA chains are antiparallel Prokaryotic chromosomes E coli 4000000 nucleotides 1000 x length of the cell Topoisomerase enzymes that DNA from one form to another Eukaryotic chromosomes Humans 6x109 pairs 6 feet Chromatin DNA molecule with proteins and some RNA Histone proteins form nuceosomes Euchromatin most of the genome is genetically active Heterochromatin more condensed Constitutive form always inactive Faculatative form becomes condensed amp inactive Gene linkage amp linkage mapping Violation of the 2nOI law Independent assortment Occurs when genes are located on the same chromosome If genes are on the same chromosome assume no crossing over Physical linkage two genes are linked if they occur on the same chromosome amp do no assort independenUy Parental phenotypes Referring to the heterozygous parent 0 Most frequent phenotypes observed 0 Tell us how the alleles were arranged Recombinant phenotypes Less frequent among offspring Products of crossing over Genetic mapping 0 1913 Alfred Sturtevant suggested that the proportion of recombinants from crosses could be used as a measure of the distance between two genes 1 The further apart two genes are the greater the probability of a crossover between them 2 Quantitative measure 1 recombinants 1 centiMorgan cM 1 map unit mu recombinants recombinantstotal Sturtevant went further to discover that genetic map distance between a series of linked loci are additive Observation recombinants cannot exceed 50 1 Means genes are on different chromosomes independent assortment 2 Or genes are so far apart Pcrossover1 Genetic mapping 1 Large distance give inaccurate map 2 Double crossovers lead to an under estimation of distance 3 Having intervening genes allows detection of double crossovers 4 3point testcross can give more accurate map a Determine gene order b Calculate distance between genes 2 recombinants Eregion 1 2 double crossovers i39 total x 100 5 Double crossovers are rare for map distances lt 10mu Linkage maps provide the foundation for understanding how genomes are organized Linkage maps information on phenotypes used to identify regions of the genome that contain genes that in uence traits Forms the basis for understanding trait evolution of disease causes of diseases what happens in hybrid zones epistasis etc Population Genetics Population group of organism belonging to one species that live in a place amp share a common setofgenes Gene pool shared set of genes all of the alleles in a population Basic measures of population genetic variation total Allele frequency 2 copies of an allele Ea population 2 allele copies 6 a population 2 total Genotype freq uency 2 individuals with a genotypeE a population 6 individuals 6 a population 2 Evolution change in allele frequencies over time The HardyWeinburg Model P frequency of A amp q frequency of a HW conditions assumptions IF 1 Population is in nitely large 2 Random mating 3 No mutation 4 No migration gene ow a Emigration b Immigration 5 No natural selection THEN 1 Allele frequencies will not change 2 Genotype frequencies will be in these proportions a FAA p2 b FAa 2pq c Faa q2 HW equilibrium Causes of evolution Genetic drift random changes in allele frequencies Nonrandom mating Mutation Migration gene ow a Emigration b Immigration 5 Natural selection PWN How many degrees of freedom Df classes of data 1 1 for every parameter we estimate from the data Mutation spontaneous genetic changes ultimate source of genetic variation errors during replication A l a Deleterious mutations disadvantageous frequency will decrease in population due to natural selection Advantageous mutations bene cial frequency will increase in population due to natural selection 0 Neutral mutations neither deleterious nor advantageous neutral with respect to natural selection u mutation rate Al3 P1 P0 Al3 39lJPO The magnitude of the Ap depends on the mutation rate u amp the allele frequency p v reverse mutation rate AP VCl Equilibrium AP P1 P0 0 Apvqup rateofAaaA Ap Owhen vq up v u A and 2 D P5121 Solve for IUV v Mutation by itself is a relatively weak evolutionary force Genetic drift random changes in allele frequencies sampling error The HW assumption of in nite N is clearly not realistic but populations are often large enough that change does not play a large role Expected values are commonly observed Genetic drift is a phenomenon most often observes in a small population size N is the critical parameter Ne effective population size equivalent of adults contributing gametes to next generation lf females males Ne N If females 7 males Ne 4 X fo Nm N Nm A population is said to be xed when there is no genetic variation p 1 or p O 0 Random uctuations due to chance 0 Effects are more pronounced in small population sizes Fixation p l O or p l 1 0 Population is xed when there is no variation Populations can diverge just by drift meaning two populations can differ in allele frequencies simply due to drift Probability of xation of an allele depends on its frequency when p 05 amp q 05 equal probability When p 7 q the rarer allele is more likely lost Genetic drift always occurs in small populations Bottlenecks situations when a population is signi cantly reduced by catastrophe Founders Events dispersal amp colonization by a small group Migration changes in allele frequencies due to movements of individuals among populations gene ow Consequences 1 Spread of new alleles similar to mutation in that gene ow is a source of genetic variation within populations 2 Changes in allele frequencies when the allele frequencies of the two populations are different Migration homogenizes allele frequencies among populations Natural selection Necessary amp suf cient conditions 1 More offspring are produced than can survive amp reproduce 2 Organisms vary phenotypically 3 This variation is associated with differential survival amp reproduction 4 This variation is heritable Fitness measure of the average offspring produced by a genotypes probability of survival X reproductive output Antagonistic pleiotropy gene may yield high performance in one component of tness but low performance in another ZygotesD Adults D Ga metes D Zygotes Natural selection model of constant selection W measure of tness scaled to the most t genotype A1A1 A1A2 A2A2 breeding adults 25 50 25 offspring produced 200 200 50 by all the adults of the genotype in the next generation Average offspring 20025 8 20050 4 5025 2 genotype Absolute Fitness Relative tness W11 88 1 W12 48 05 W22 28 025 Selection coef cient 1 1 0 1 05 05 1 025 075 s 1w Alleles that comprise the genotypes with the highest tness survival X reproduction will increase in frequency due to natural selection 2 2 2 Mean relative tness W P Wu quu 4 W22 General formulas A1A1 A1A2 A2A2 lnitial genotype P2 2pq q2 frequencies Relative tness W11 W12 W22 Frequency after p2w11 2pqw12 q2w22 selection Relative genotype P pzwll w H 2qu12 w Q q2W22 w frequency after selection Allele frequencies after selection p P 12 H q 1 p PqPW11W12QW12W22 W Change in p in 1 generation Ap p p Sickle cell anemia Caused by mutation in the B hemoglobin gene Sickling prevents RBCs from carrying oxygen l anemia Partial dominance Hb A allele wild type amp Hb S sickle cell allele 0 Hb A Hb A nomal RBCs o Hb A Hb S mild form of sickle cell anemia amp resistance to malaria o Hb S Hb S sever form Fitness is dependent on environmental context 0 Despite deleterious nature of the mutation the Hb S persists in areas with malaria Combining Forces Mutation Selection Balance Consider a deleterious allele 1 Selection directional is acting against a deleterious recessive aee so that q decreases 2 But mutations A D a increases q Modek 1 For selection Asq Asp qu2 1sq2 selection against recessive complete dominance When q z 0 1qu D 1 Asq z qu2 2 For mutation ignore reverse mutations App 39Hp Auq App z lip Modek Asq z qu2 selection decreases q qu z Aup z up mutation increase q Equilibrium Aq Auq Asq At equilibrium Aq O Auq Asq ppqu2 ACl0lJl3SICgtCI2 Up qu2 A u I I I I I sq2 u 61 DeleterIous aee maintained In population by balance between mutation amp selection Quantitative Genetics Trait Mode of Inheritance Environmental Effects QualitativeDiscreteMendelian Simple single locus Little to none Blood type enzyme de ciencies Mendel s peas Nucleotide polymorphisms QuantitativeContinuous Complex polygenic Moderate to great Height milk yield IQ Normal Notice that the normal distribution arises from 2 sources 1 Many genes l many genotypes l many phenotypes 2 Then add environmental effects noise a The relationship between genotype amp phenotype is obscured Simpli ed Model of Polygenic Inheritance Simplifying Assumption 2 or more oci with 2 alleles at each ocus 1 allele contributes to the phenotype 1 allele is noncontributing All oci have equal effect a aees have equal amp additive effects No dominance no epistasis No environmental variation exists Mendel s laws apply P P PWN The model phenotype for a trait is determined by the number of contributing alleles Phenotype base value contributing aees contributionallele Where base value phenotype observed in a genotype with no contributing aees Estimating the number of genes Estimates can be made from the F2 distribution Phenotypic variance genetic variance environmental variance VP VG VE Studying quantitative traits in populations Sample subset of a population that is used to estimate population parameters Considerations 1 Sample must be large enough for accurate estimation 2 Sample must be random Statistics to describe the sample 2X n Mean x Xi fc z c vanance 26 varxs 22 Standard deviationx s S2 A few more tools 1 Correlation coef cient sometimes traits are correlated correlated traits means 2 traits vary together a Positive correlation positive slope b Negative correlation negative slope i Be careful correlation 2 causation ii To calculate the correlation coef cient X Zampampi XYu Y 1 Covariance n1 COVXYZZ COVXY 2 Correlation coef cient quot S S X Y 3 Range 1 to 1 higher R stronger correlation Heritability proportion of a population s phenotypic variance that is attributable to genetic factors VP VG VE Heritability can be measured by careful experiments Important for 1 Plant amp animal breeding 2 Studying evolution 3 Understanding treating diseases ANOVA analysis of variance Norms of reaction 1 All variance attributable to genetic factors 2 All variance attributable to environmental factors 3 Variance attributable to both genetic amp environmental factors 4 Covariance between genotype amp environmental factors EX Genotypes are not randomly distributed with respect to environment a VP VG VE COVGxE 5 Genotype by environment interactions Further complications interactions among genes 1 Additive genetic variance VA 2 Dominance relationships VD 3 Epistasis VI a VGVAVDV b VP VA VD V VE COVGxE VGxE V 2 G Broad sense heritability hB V proportion of phenotypic variance that consists of genetic P variance range 0 to 1 1 Ignores VA VD V 2 Ignores VGXg More often we want to measure the proportion of Vp that results only from VA This allows us to predict phenotype of offspring from the phenotypes of parents Narrow sense heritability VA h2 NVP proportion of Vp that results from additive genetic variance Ex That responds to selection Some cautionary notes 1 2 3 4 Heritability does not indicate the extent to which a quottrait is geneticquot Heritability is a population measure cannot be applied to individuals Heritability depends on the genetic makeup of the populations amp on the speci c environment in which h2 was measured Family resemblances familiality 2 heritability Calculating heritability 1 3 4 Parent offspring regression i Assume no common environmental effects between parents amp offspring ii The slope of the regression line provides information about the magnitude of heritability For regression of mean offspring phenotype amp mean parental phenotype a hi 2 b Slope of the regress ion line For regression of mean offspring phenotype amp one parental phenotype 2 b hNZZb Response to selection amount of phenotypic change that occurs due to natural selection L thixs a R response to selection b S selection differential c Depends on hN2 amp selection Traits continues to respond to selection as long as there is genetic variation or until detrimental effects accrue VG O or antagonistic peiotropy further increase in trait vaue decrease tness Molecular genetics Three models of DNA replication 1 2 3 Semiconservative Conservative Dispersive The Meselsohn Stahl Experiment demonstrated that replication proceeds in a semiconservative fashion 1 The grew E coli bacteria in a medium with ammonium chloride containing the heavy isotope of nitrogen 15N for several generations all DNA there contained 15N Then they switched the bacteria onto a medium containing the normal 14N isotope for two generations At each generation some bacteria were ysed DNA isolated amp centrifuged on a cesium chloride gradient This technique allowed the scientists to separate DNA according to density They found one band of DNA at the start corresponding to DNA made with 15N In the next generation they also found one band of DNA but t was less dense than in the previous generation This excluded the conservative model which predicts 2 kinds of DNA parental amp new 6 In the next generation they found two bands of DNA one light containing 14N amp one more dense a mixture of 14N amp 15N This nding excluded the dispersive model amp con rmed that replication occurs semiconservatively The replication reaction i u mti e DNA tfi h h t ily l rm dNMP dNTPs g e IHMPHH PPi TT a tT T Requires 1 All 4 dNTPs 2 Mg ions Mg 3 Template DNA 4 DNA polymerase enzyme Replication DNA polymerase also has 3 to 5 exonuclease activity proofreading error rate 10399 per generation Replication in prokaryotes Supercoiling relaxed by topoisomerase Initiator proteins bind to origin of replication i 9bp sequence repeats binding site ii 13bp sequence repeats ATrich easily denatured DNA helicase binds amp untwists DNA DNA primase binds to DNA helicase amp builds DNA primer 510 nucleotides DNA polymerase III binds synthesis of new strand 5 D 3 Singlestrand DNA binding proteins stabilize replication bubbe Nl P P FPquot Prokaryotic DNA polymerase I 3 5 amp 5 3 exonuclease activity primarily repairs DNA amp removes primers involved in DNA repair I 3 5 exonuclease synthesizes DNA during replication IV involved in DNA repairs V involved in DNA repair Joining Okazaki fragments Okazaki fragments are built on the lagging strand 1 DNA polymerase III leaves 3 end of new Okazaki fragment is next to 5 end of previous Okazaki fragment 2 DNA polymerase I binds amp simultaneously removes RNA primer on previous Okazaki fragment amp synthesizes DNA to replace it 3 When RNA primer is removed completely DNA polymerase I leaves A single stranded nick remains between the two fragments 4 DNA igase seals the nick with phosphodiester bonds amp then eaves Summary of replication in prokaryotes Step Enzyme or Protein 1 Relaxation of supercoiled DNA Topoisomerase 2 Denaturation amp untwisting Initiator proteins DNA helicase 3 Single stranded DNA is stabilized Single stranded DNA binding proteins SSBs 4 RNA primers synthesized 5 10 DNA primase nucleotides Helicase primase moving along primosome 5 Primers lengthened by addition of DNA polymerase III nucleotides complimentary to template a leading strand synthesis in direciotn of movement of replication fork 5 3 continuous replication b lagging strand synthesized by Okazaki fragment formation discontinuous 6 Primer section RNA removed amp replaced DNA polymerase I with DNA 7 Gap between Okazaki fragements closed DNA igase okazaki fragments joined Notes Replication rate 500 nucleotides second 10bp per turn means helix ahead of replication fork has to rotate at 50 revolutions second 3000 rpm DNA gyrase a topoisomerase introduces negative supercoils in ahead of replication fork to compensate relaxes the tension Cell cycle in eukaryotes Cdk cycin dependent kinase Replication in eukaryotes 15 DNA polymerases Semiconservative Semidiscontinuous Bidirectional Problem humans have gt3x109 bp haploid Mammal1n DNA Poiymerases t1 ti E E Haitians I39l Liil lfl ij Il 3 olyme tati n admit 4f 5 to 3 E o ucilease limiting 3 in El mofmading wasmnsible for DNA l mutation in mueleua DNA polymerase or acts with primase to start replication DNA polymerase s synthesizes DNA on leading strand DNA polymerase 6 synthesizes DNA on lagging strand Problems at the teomeres Telomeres ends of chromosomes with tandem repeats Telomerase extends teomeres Genetic diseases E m an nonfunctional emitng iifumul 39li til quot phenylpyrmic Heidi WMIEII E f phenylalanine hy f yl 2 f rr Philangglflnlmsi my i5 39F lF Film ihmiam r HE i ii i i K T Malia ruin a I 139 1 Alkaptonu ria acid It a rquot 1 Enzyme Helmiemcies I 39I see Table 41 iaHQD Beadle amp Tatum s 1942 Experiments with Neurospora crassa iii39um us Maple iiihurt life cycle cycle reman types 39 1 Ilia 1 11 manna HHW i r m wgP zer i r r Han eid 4 mamm 3 quotmquot Fl39lEllil l39 n 1 r if Elms aim maturation 4 39 1 Iainism Eli d a Euni isa 1 division H i l g1 kg 7 J v Ef i l r it Strontium Haggis quotml ii mr39irig mm MEI II hunting lymv magnum in I39ii ji lli g m Elli all gypsum mulling lung gn39m ther melaii 1 i k l i m nmlm integminla m liqrm a A r E Eumi a Hamlin ail Prototrophic can grow on a minima medium salts carbon source amp the vitamin biotin fungus synthesizes amino acids nucleic acids etc Auxotrophs nutritional mutants requires nutritional supplement to grow ElmHIE r l 1 Eifl i Sgtum ml i i a diililt r and Eil39ll lz iiil39 Willi lynx Hay wilt an by um buscuiluremiluha li lliiil lljp ll39r ili illirIIgri V i1 MIME 3 fl Hiquot g EWle Fe HHE l E l l I Hu r ia lflubir l n H il39 i IIi39 l iu l Iriaailiiuliii39liti yillh l h will Al ill summerr ll Elli j l l m n Hi im Elamdie Mau l slime H is lg w 5 rimm as li mitaria ii39iaiar Ill If i w mud ll r V Iquot ll llquot ll Wilma III39I il39nlr ill39l l mild ni Lirna Humilityi mu rllimrmil minim1 E El I 5 395 iii Fi ll i in 3911 i 39Ei39 m1 quotwanmi new lawnml is i j i39 ntrf lieuIgln a ri a Williinn 1i lrg l Him I n39lF Q Fiji 39139 H I iW f 39i i li IEl 1 up Fi inn g xi Hm 1 my 9 P Wim Iii i i 5 Hli m 11 39i39hrmrua ElMm rit humans E T WEI hi Elma El lawmakers Induced mutations with xrays Crossed strains Dissect ascospores Grow on complete medium Grow on minimal medium Grow on minimal medium supplement P P FWN Metabolic pathway Premmnr Empound E myrtmd Product Enzfm iii Enume lEIIIgiI E E Auxotrophs de cient in enzyme B should be able to grow on Minimal medium end product or Minimal medium compound B Will not grow on minimal medium compound A Beadle amp Tatum hypothesis 1 gene 1 enzyme grrmm Jm wilt Tallulah iii month Empai m I39l39 Bill iiii lii Jigmmmpl39iia illnl39i ll Hrhp r ler i39if Hymn lluiiinlli39ul ri 9 Ei Iai u rail imillililii la quot39 i lillhii H ii i l lli lJiE EEffrl i E 39I39 i 39illli liillf IIIdi 39llilll i39ia ilfll39kiquot llnigrri lialirl39llinvi 3H in I39m I 339 a 39U l Mil il l r 1 393 LT 5 7quot quot quotquotn E E 39 r g Inf q TVS 339quot Eii r n En mw I H in Huhquot vquot Tami M E i gt quota a a a 1 ii r 5 i M Hemmerm Il39fi r r 4 BE hr HT 5 l1 iquotla fl rf a Ha wf FHI W ef a 4 lid139a5r gr ama l39aija f ua ni Hm llil HE sin LN a Beadle amp Tatum worked out the details of many pathways Proposed a speci c gene encodes each enzyme D 1 gene 1 enzyme hypothesis Modi cation a speci c gene encodes each polypeptide D 1 gene 1 polypeptide hypothesis Sickle Cell anemia caused by changes in hemoglobin protein 0 4 heme groups 4polypeptidestetramer o 2 or polypeptides or globin gene 0 2 B polypeptides B globin gene 2 aees BA wildtype S mutant BABA make HbA 2 normal or polypeptides amp 2 normal B polypeptides BSBS make HbS 2 noma or polypeptides amp 2 defective B polypeptides Heterozygotes make both HbA amp HbS The rst seven Nterminal amino acids in normal amp sicked hemoglobin B polypeptides Narmal 1 2 l polypeptide wa t HallJVquot lhalal HIE LEui Sickleoell i y 7 lpciiypeptide Him15 HIGHquot 1llai His l Luau Fro l i Causes B polypeptide to fold in a different way amp sicking happens Kinds of RNA Messenger RNA mRNA Translated to produce a polypeptide Transfer RNA tRNA Brings amino acids to ribosome Ribosomal RNA rRNA Components of ribosome Small nuclear RNA snRNA Involved in processing mRNA MicroRNAs miRNAs Regulatory molecules Short interfering RNAs siRNAs Regulatory molecules Structural genes genes that code for a polypeptide protein coding enzymes amp structural proteins Transcripts mRNA RNA genes genes that code for tRNA rRNA or snRNA transcripts are the nal product RNA polymerases Prokaryotes RNA polymerase denatures DNA amp synthesizes RNA Eukaryotes RNA polymerase Location Products Nucleolus 285 185 585 rRNAs ll Nucleus mRNA some snRNAs lll Nucleus tRNA 5S rRNA some snRNAs Other proteins denature DNA Transcription in prokaryotes Initiation of transcription 1 2 factor binds to RNA polymerase holoenzyme 2 Holoenzyme binds to promoter regions a 35 box 5 39ITGACA 3 b 10 box Pribnow box 5 TATAAT 3 RNA polymerase denatures 25bp around 10 box Transcription begins at the 1 site The 0 factor falls off after synthesis of 8 9 nucleotides 25bp window moves 3O 50 nucleotides second P P FPquot Some nuance 10 box can vary in sequence different binding by holoenzyme which means rates of transcription vary from gene to gene Different 0 factors 070 regular one 032 produced under heat shock binds differently O54 when N is limiting 023 when infected by T4 phage Termination of transcription 2 ways 1 p dependent termination a p protein with 2 domains 1 binds RNA 1 binds to ATP 2 b binds to recognition site in termination region on mRNA c When RNA polymerase reaches the termination it pauses ATP is hydrolyzed by p amp RNA is unwound from DNA template RNA polymerase amp the RNA are released P independent termination a Sequence upstream from termination has 2fold symmetry b RNA transcript forms stemloop structure followed by series of Us in RNA c Stemloop plus U s causes termination of transcription may destabilize RNA DNA complex Key events in termination of transcription 1 2 3 RNA synthesis stops mRNA chain is released from DNA RNA polymerase is released from DNA Transcription in euka ryotes Regulatory elements DNA sequence involved in regulation of transcription Positive elements activate transcription Negative elements repress transcription 1 Regulatory elements bind a Transcription factors speci c proteins required for initiation b Regulatory factors proteins involved with stimulation or repression c Both are usually within a few hundred b usuay upstream 2 Promoter regulatory element adjacent to transcription start site 3 kinds Consensus Sequence Approximate location TATA element box 3 TATAAAA 3 3O CAAT element 5 CAAT 3 75 GC element 5 GGGCGG 3 9O 3 Also enhancer elements required for maximal transcription often gt1000 bp upstream of the promoter region Silencer elements repress gene transcription interacting with a repressor transcription factor mRNA products of transcription ll mlrary39ntes mature 539 llrrmwin lu39 rii r weirI m L 3 untranslated lend2r untranslated trailer sequent emumin 539 LITE l3 STRE 539 39 Em mm H Em 2 ll 3 not mlinear lEulmryntea l 111 HIRE ll Amine tei 51min swirl hailing region Iiiriding region Hon Ftding region Transcription amp translation are coupled in prokaryotes Posttranscriptional modi cation of premRNA in eukaryotes 5 capping catalyzed by capping element essential for ribosomal biding to 5 end of mRNA 0 3 poly A tail determines longevity amp stability of mRNA Cleavage amp polyadenylation speci city factor CPSF Cleavage stimulation factor CstF Poly A polymerase PAP Poly A binding proteins PAB Small nuclear ribonucleoprotein particles snRNPs snRNA proteins Eukaryotic rRNA transcription repeat unit 0 These rDNA repeat units occur in tandem repeats 100 1000 copies gene redundancy A nucleolus forms around these clusters Transcription amp ribosome assembly occur in the nucleolus NTS nontranscribed spacer contains promoter amp terminator of previous repeat unit ETS external transcribed spacer ITS internal transcribed spacer Processing of prerRNA is done by special ribonucleases All of the ribosomal proteins are transcribed amp translated 60 amp 40 subunits form migrate out of nucleolus into cytoplasm to form the ribosomes SS rRNA genes are transcribed by RNA polymerase Ill Promoter is located within the gene internal control region Initiation site is 50bp upstream from lCR Transfer RNA tRNAs bring amino acids to ribosomes 75 90 nucleotides tRNA genes have lCRs also gene redundancy Some processing is required 0 5 CCA 3 added after transcription 0 lntrons excised RNA ligase splices 0 Chemical modi cations Protein structure 0 1 amino acid sequence 2 folding amp twisting of polypeptide 3 3D structure into which 2 structure is folded 4 2 associated polypeptides The genetic code Crick et al s Experiments with T4 bacteriophages r wildtype rll mutants can t grow on certain E coli K12 1 Created rll mutants by treatment with chemical pro avin a Pro avin is a mutagen that causes additions or deletions of DNA bp 2 If rll result from addition or deletion treating rll with pro avin could cause reverse mutation reversion back to r a Some revertants resulted from exact correction of original mutation b Some mutants were more interesting i Wildtype r ii Frameshift mutation by deletion rll iii Reversion of deletion mutation by addition r 3 Genetic code is in the form of triplets a Also found that 3 additions recovered r but 4 or 5 additions did not Deciphering the code Nirenberg amp Khorana experiment 1 Built cellfree protein synthesizing system in a tube a From E coli i Ribosomes ii tRNAs with amino acids attached iii amino acids radioactively labeled 2 Introduce synthetic mRNA The genetic code 1 Codons read continuously from mRNA 2 Code is universal 3 Code is degenerate There is more than 1 codon for each AA except MET amp TRP 4 Start amp stop codons 5 Wobble when a base replaces another base but does not change amino acid Wobble in the genetic code Nucleotide at 5 end of Nucleotide at 3 end of an codon codon G Can pair with U orC C Can pair with G A Can pair with U U Can pair with AorG l inosine Can pair with A U or C The point 1 tRNA can pair with gt1 codon Translation overview 1 mRNA translated 5 to 3 2 Polypeptides constructed N terminal to C terminal a mRNA 5 AUG UUU ACA UCU 3 b Polypeptide 5 Met Phe Thr Ser3 3 Correct amino acid sequence is achieved by a Binding of each amino acid to its own speci c tRNA b Speci c match between tRNA anticodon and mRNA codon c Note that the speci city is in the anticodon not the amino acid Translation 1 Charging the tRNAs a Amino acid amp ATP bind to enzyme b Enzyme catalyzes coupling of amino acid to AMP to form aminoacylAMP Two phosphates are lost in the reaction c Uncharged tRNA binds to the enzyme aaAMPenzyme d Enzyme transfer amino acid from aminoacylAMP to tRNA to form aminoacyl tRNA aatRNA The aatRNA amp AMP are released from the enzyme e aatRNA amp AMP released aatRNA enzyme f Enzyme returns to its original state AminoacyltRNA synthetase Amino acid attached by carboxyl group to ribose of last nucleotide of tRNA chain Last 3 nucleotides of all tRNAs are CCA 3 2 Initiation Prokaryotes a Initiator fMet formylmethionine b Sequence at 3 end of 16S rRNA i 3 AUUCCUCCAUAG 5 c Example of sequence upstream of the AUG codon in an mRNA pairing with the 3 end of 16S rRNA i 539 UGUACUGUUGUAUGGAACAACGC 339 1 AGGAG ShineDalgarno sequence 3 Initiation Eukaryotes a Similar to prokaryotes but i The initiator Met is not fMet though there is a special tRNA ii There is no ShineDalgarno sequence b Instead the 5 cap in eukaryotic mRNAs serves essentially the same function c Eukaryotic initiation factors include several proteins plus cap binding protein d The initiator AUG codon is located in a short Kozak sequence i 539 UUAUGCCCC 339 1 GCCACCAUG Kozak sequence e MettRNAMet amp smaller ribosomal subunit 4OS bind to mRNA at the initiator codon 4 Elongation Elongation factors Tu binds to tRNA amp Ts recycles EFTu a Once 7OS initiation complex is formed fMettRNA fMet is bound to AUG codon in the P site of the ribosome b In a complex with elongation factor Tu EfTu amp GTP the next aminoacyltRNA molecule SertRNASer binds to the exposed codon UCC in the A site of the ribosome c Peptide bond forms between the two adjacent amino acids catalyzed by peptidyl transferase The linked amino acids are attached to the tRNA in the A site forming a peptidyltRNA i Peptidyl transferase 23S rRNA ribozyme a catalytic RNA d Transocation occurs as the ribosome moves one codon to the right requiring EFG amp GTP amp peptidyltRNA moves from the A site to the P site Uncharged tRNA moves from the P site to the E site e When translocation is complete amp the peptidyltRNA is in the P site uncharged tRNA is released from the E site amp the ribosome is ready for another elongation cycle f The elongation cycle repeats until stop codon is encountered 5 Termination a Stop codon is encountered b Release factor RF1 binds to stop codon c Polypeptide chain is released d RF3GDP binds causing RF1 release GTP replaces the GDP amp GTP hydrolysis release RF3 e Ribosome recycling factor RRF binds to A site f EFGGTP binds to ribosome Hydrolysis of GTP amp GDP causes translocation of the ribosome putting RRF in the P site amp the tRNA in the E site g RRF releases the uncharged tRNA EFG then release RRF amp the two ribosomal subunits dissociate from the mRNA i Release factors termination factors 1 Prokaryotes have 3 a RF1 UAA UAG b RF2 UAA UGA c RF3 release RF1 or RF2 2 Eukaryotes have 1 a Eukaryotic release factor eRF ii Ribosome recycling factor RRF 1 Stimulates disassembly of ribosome 2 Prokaryotes Mutation Types of Mutations 1 Chromosomal mutations a Nondisjunction single chromosome trisomy whole genome b Deletions of parts of chromosomes c Rearrangements translocations etc 2 Gene mutations alterations of DNA sequences spontaneous or induced Somatic mutations not passed on Germ line mutations carried by gametes a Point mutations affect a single base pair i Base pair substitutions one bp replaced with another 1 Transition mutation one purine pyrimidine replaced with another purine pyrimidine pair 2 Transversion mutation one purine pyrimidine pair replaced with a pyrimidine purine pair b Missense mutations mutations that cause changes in mRNA codon so that a different amino acid is inserted c Neutral mutation missense mutation but new amino acid does not change function of the protein i Ex Change from Arginine to Lysine involves a change from one basic amino acid to another basic amino acid that may be neutral d Silent mutation mutation that do not cause amino acid change e Nonsence mutation mutation to stop codon in mRNA leads to chain termination f Frameshift mutation addition or deletion of 1 or more bp that shifts the mRNA reading frame g Forward mutation genotype changes from wildtype to mutant h Reverse mutation genotype changes from mutant to wildtype reversions or back mutation i Suppressor mutation second site mutations compensates for effects of initial mutation i lntragenic suppressors occur in same gene but at different site 1 Different nucleotide in same codon or in a different codon ii lntergenic suppressor occurs in different gene 1 Ex Initial nonsense mutation is compensated by a mutation in a tRNA gene such that the anticodon recognizes former stop codon as an amino acid coding codon 2 Partial or complete recovery depends on the amino acid Mutation rate probability of a speci c mutation as a function of time mutations nucleotide generation mutations gene generation Roughly 10394 10397 per nucleotide Mutation frequency occurrences expressed as proportion of cells or individuals per 10000 individuals per 10000 gametes Causes of mutations Spontaneous mutations occur naturally 0 DNA replication error Mutations due to wobble mismatched bps during replication tautomeric shifts Tautomer alternative state Addition or deletion due to quotlooping outquot Mutations via spontaneous chemical changes Depurination chemical reaction of purine in which BNglycosidic bond is hydrolytically cleaved releasing a nucleic base A or G Deamination removal of an amine group from a molecule lnduced mutations caused by treatment with chemical or physical mutagens 0 Radiation Xrays ionizing radiation that can break chemical bonds Ultraviolet light nonionizing radiation that cause abnormal bonds Pyrimidine dimers impede replication also CCACT dimers but much less common 0 Base analogs 5bromouracil undergoes tautomeric shifts 0 Base modifying agents Nitrous acid deaminates bases Hydroxylamine adds OH 0 MMS is an alkylating agent adds CH3 0 lntercalating agents Pro avin Acridine Ethidium bromide Ames test designed to assess mutagenic properties in chemicals Uses histidine auxotrophic mutants of bacterium Salmonella typhimurium 1 Spread bacteria onto plates with minimal medium along with homogenized rat liver to stimulate effects of liver enzymes a Control plate b Experimental plate with the chemical being tested c Experimental plate with the chemical being tested plus rat liver enzymes 2 Incubate at 37 C for two days amp look for revertant colonies wildtype a Colonies on control plate represent spontaneous mutations b Colonies on experimental plate represent spontaneous revertants plus extra revertants resulting from mutagenic effects of the chemical DNA repair mechanisms 1 3 to 5 exonuclease activity DNA polymerase proofreading 2 Photoreactivation of UV induced pyrimidine dimers light repairs a Dimers revert to original form by exposure to light b Enzyme photolyase is activated by photons amp catalyzes reaction splitting dimers 3 Speci c enzymes to repair alkylation a O6 methylguanine methyltransferase i Can remove methyl group 4 Nucleotide excision repair dark repair a 4 proteins i uvrA ii uvrB iii uvrC iv uer Have endonuclease activity plus DNA polymerase I amp DNA ligase 5 Methyldirected mismatch repair in E coli dammethylase mutS mutL mutH a MutS bound to mismatch forms a complex with MutL amp MutH to bring the unmethylated GATC close to the mismatch b MutH nicks the unmethylated DNA strand amp a section of the new DNA strand is excised including the mismatch c DNA polymerase III amp ligase repairs the gap producing correct base pair 6 Translesion DNA synthesis amp SOS repair system an induced response to DNA damage lesions may block replication D lethal if not repaired a 17 genes involved all have SOS box 20bp regulatory sequence b Regulated by LexA amp RecA proteins c Uninduced state no DNA damage i LexA functions as a repressor by binding to SOS box d Induced state i RecA is activated by DNA damage causes LexA to cleave itself ii After DNA repair RecA is inactivated LexA again binds to SOS box e SOS response is itself mutagenic sometimes i One of the SOS gene products is a DNA polymerase for translesion DNA synthesis 1 Continues replication past lesion by incorporating nucleotides not speci ed by template 2 These nucleotides may not match wildtype l mutations but better than the lethal effects of stopping replication Gene Regulation controlling the expression of genes Constitutice genes housekeeping genes are always active 0 Ex Genes coding enzymes for protein synthesis glucometabolism etc Regulated genes controlled in response to needs of cell 0 Ex Genes expressed at different stages of development 0 Inducible genes turned on induced by some environmental factor 0 Repressible genes turned off repressed by some environmental factor 0 Gene product inducible genes are expressed only in the absence of a repressor andor presence of an effectorinducer molecule Gene regulation in prokaryotes The lac operon E coli can grow on medium with some salts amp a carbon source o Glucose is the preferred C course 7 metabolism occurs by constitutive gene products 0 If other sugars are provided instead of glucose a number of enzymes are synthesized induction of genes 0 If only lactose is provided 3 genes are turned on Bgalactosidase Catalyzes break down of lactose Catalyzes isomerization of lactose to allolactose the inducer Lactose permease Active transport of lactose Transacetylase Activates Bgalactosidase Coordinate induction all 3 turned on o 1961 Jacob amp Monod gured out how induction occurs 0 Induced mutations amp mapped 3 genes lacZ gene encodes Bgalactosidase lacY gene encodes lactose permease lacA gene encodes transacetylase Missense mutations only affect the function of each gene separately 0 Nonsense mutations acted differently Nonsense mutation in lacZ results in a loss of function in all 3 genes Nonsense mutation in lacY resulted in a loss of function in permease amp transacetylase but not Bgalactosidase Nonsense mutation in lacA affected only transacetylase Implication 3 genes transcribed as 1 mRNA polygenic mRNA polycistronic mRNA 0 Induced other mutations in which all 3 genes were expressed Constitutively lacO mutations map to small region adjacent to lacZ operator region lacI mutations map to a genesized region a short distance away 0 Jacob amp Monod used partial diploids to investigate F factor is a plasmid circular piece of double stranded DNA that is self replicating F factor is necessary for genetic exchange between bacteria one waY lacOC constitutive mutation affect only genes downstream on the same strand cisdominance Note lacZ indicates a missense mutation Absence of allolactose Bgalactosidase synthesized Presence of allolactose both Bgalactosidase amp permease Operator region does not produce a diffusible product lacl39 mutants also synthesize enzymes constitutively but have normal promoters amp operators lacI can overcome the presence of lacl39 transdominant since genes are on different chromosomes lacI gene must produce a diffusible product Promoter mutations PlaC39 stop synthesis of all 3 genes even when inducer is present Promoter is recognition sequence for RNA polymerase amp is cisdominant Jacob amp Monod s operon model 0 Operon cluster of genes regulated together by operator protein interactions plus the operator itself amp the promoter Wildtype E coli growing in the absence of lactose o Lac repressor proteins attach to operator amp prevent transcription of mRNA RNA polymerase cannot bind to the promoter when repressor is bound to the operator 0 o lacl encodes repressor protein Lac repressor protein is a homotetramer 4 polypeptides all the same Wildtype E coli growing in the presence of lactose If there is no repressor on operator RNA polymerase can transcribe the structural genes 0 Repressor is inactivated by inducer allolactose Partial diploid growing in the absence of lactose o Operator lacOC not bound by repressor lacl39 mutant growing in the presence or absence of lactose o lacl39 mutants produce inactive repressor proteins 0 Partial diploid mutant growing in the absence of lactose o lacl is transdominant Partial diploid mutant growing in the presence of lactose Other lacl mutants are known laclS superrepressor Positive control of the lac operon amp the quotglucose effectquot 0 This system ensures that the lac operon is expressed only when lactose is the only carbon source 0 Lac operon is shut off if lactose and glucose are present 0 Lactose only CAP protein is activated by cyclic AMP cAMP CAP then binds to CAP site bending DNA which facilitates holoenzyme binding 0 Lactose amp glucose Amount of cAMP is drastically reduced during glucose catabolism This inactivates CAP protein amp transcription of lac operon is lowered signi cantly 0 quotGlucose Effectquot catabolite repression Glucose catabolism results in inactivation of adenylate cyclase amp no cAMP is produced Also phosphodiesterase further reduces the concentration of cAMP Promoter region of the lacl gene Promoter amp operator regions of the lac operon Gene regulation in prokaryotes The trp operon in E coli 0 The trp operon is a repressible operon o Repression occurs in 2 ways 0 trpR gene encodes an aporepressor protein inactive until activated by tryptophan 70fold reduction in transcription When trp is present trpR is activated amp binds to operator reducing transcription 70X 0 Attenuation control occurs at low tryptophan concentrations Severe trp starvation l max transcription Less severe trp starvation D less transcription trpL plays a key role in attenuation control 8 10 fold reduction in transcription 0 Organization of trpL region 0 Attenuation control works because transcription amp translation are tightly coupled o Pairing of regions 1 amp 2 causes RNA polymerase to pause allowing ribosome to load onto mRNA tight coupling 0 During trp starvation TrptRNAtrp are in short supply ribosome stalls on trp codons regions 2 amp 3 form stem loop antitermination transcription continues 0 When trp is more abundant ribosome stalls on stop codon regions 3 amp 4 form stem loop termination o This is similar to rhoindependent termination of transcription o 2 other generalized regulatory mechanism 0 mRNA degradation mRNAs degrade rapidly they have a short halflife which allows for rapid response to changing trp levels 0 Feedback inhibition trp binds to the rst enzyme in the pathway causing an allosteric shift amp blocks enzyme function this reverses when trp concentration is low Gene regulation in prokaryotes other operons have attenuation control Predicted amino acid sequences of leader polypeptides from other operons Gene regulation in prokaryotes Phage A is a temperate phage Unlike virulent phages temperate phages have 2 possible cycles 0 The lytic cycle is similar to a virulent phage lifecycle Phage DNA infection A chromosome replication amp breakdown of host DNA Replicated A chromosome Assembly of progeny A phages Release of progeny A phages o The lysogeny cycle involves incorporation of the phage chromosome into the bacterial chromosome amp the lytic cycle is repressed Phage DNA infection lysogenic response Integration of A chromosome into host chromosome becomes prophage Multiple division Induction of lytic cycle excision of A chromosome Enters lytic cycle Gene regulation if prokaryotes Phage A chromosome Early right operon for replication during lytic cycle Late operon genes for phage components made during lytic cycle Early left operon genes for lysogenic cycle Big picture the A repressor protein amp the Cro protein are keys 0 Tehse proteins compete o If A repressor wins D lysogeny o If Cro protein wins l lytic pathway Upon initial infection neither cro protein not A repressor is present 1 Phage growth begins when RNA polymerase binds early promoters PL amp PR making mRNA for N amp cro genes a From PR transcription of cro control of repressor amp other b From PL transcription of N antiterminator 2 N protein acting at 3 sites extends RNA synthesis to other genes 0 amp P proteins permit DNA synthesis to begin cll protein is made amp acts as shown below a Antitermination allows cll protein to be produced 3 Two competing pathways occur a Toward lysogenic development dominance of repressor i cll protein stimulates synthesis of mRNA from PRE for repressor cl gene amp from P for intergrase int gene ii Suf cient repressor bind OR amp OL blocking RNA synthesis repressor bound to OR stimulates synthesis of more repressor lntegrase promotes integration of phage DNA Lysogeny is established For lysogeny cll promotes transcription from PRE repressor establishment l cl gene amp from P D integrase gene cl encodes A repressor cro NOT transcribed A repressor binds OL amp OR blocking transcription of cro amp N and stimulates transcription from PRM repressor maintenance Result Lysogeny b Toward lytic development dominance of Cro i Cro protein occupies operators OR amp OL prevents synthesis of repressor mRNA but allows enough rightward transcription for Q protein to accumulate ii Q protein stimulates late gene transcription these genes encode structural proteins that package DNA into new phage particles For lytic pathways from lysogeny Ultraviolet light DNA damage activates RecA 9SOS response which causes A repressor to cleave it self Cro protein blocks PRM amp stimulating transcription from PL amp PR Early right amp late operon transcribed Result Lytic pathway In healthy cells lots fo nutrients proteases break down cll protein l lytic pathway In unhealthy cells protease concentrations are low cll stable D lysogenic pathway Gene regulation in eukaryotes multiple levels of control Operons as we saw in the prokaryotes do not exist in eukaryotes though some genes are clustered Regulation occurs at many levels Gene regulation in eukaryotes transcription control Transcription factors play a key role at the initiation of transcription general transcription factors GTFs But with transcription factors alone only a low level of transcription is achieved Enhancers or silencer elements 0 Activators bind DNA amp a coactivator complex of multiple proteins which binds RNA polymerase l stimulate transcription 100x 0 Also silencer elements repress transcription interacting with a repressor transcription factors Transcription factors amp other DNAbinding proteins have specialized structures DNA binding motifs Helixturnhelix Zinc nger Leucine zipper O Combinatorial gene regulation different combinations of enhancers amp activators for different genes 0 Ex Galactose utilizing genes in yeast 0 Genes are transcribed unless galactose is present 0 When galactose is present amp glucose is absent there is coordinated induction Plus GAL4 amp GAL80 amp GAL3 0 Absence of galactose transcription 0 Presence of galactose Gal80p binds to Gal4p activation domain blocking it from activating Gal3p converts galactose to the inducer which binds to Gal80p causing it to move on Gal4p The now exposed Gal4p activation domain activates transcription Gal4p positive regulator Gal80p negative regulator Hormonal control used to transmit information to target cells 0 Hormones are effector molecules produced by one cell amp causing a physiological response in another cell 0 Peptide hormones bind to cell surface receptor which is activated causing signal transduction Cascade of protein kinases which phosphorylate other proteins lnsulin glucagon o Steroid hormones move across cell membrane amp bind to steroid hormone receptors SHRs This complex regulates transcription directly Steroid hormones also affect stability of mRNAs amp maybe are also involved with processing premRNAs Steroid hormone receptors are often found in association with chaperones Chaperones are displaced when the hormone arrives Chromosomal changes remodeling 0 Regions containing genes that are being transcribed are less tightly coiled Euchromatin less highly coiled looser Heterochromatin tightly coiled inactive o Histone proteins can be effective repressors nucleosomes form on TATA elements Acetylation amp phosphorylation of histones alters their conformation quotloosens their gripquot Nucleosome remodeling complexes use ATP to alter nucleosome position facilitating transcription 0 DNA methylation especially in promoters is negatively correlated with transcription gene silencing But 0 Correlation 2 causation but there is accumulating evidence 0 Not all methyl groups are removed for transcription 0 Some organisms do not have much methylation ex Yeast DNA methylation is also implicated in cases of genomic imprinting inherited patterns of methylation in uence gene regulation in offspring Gene regulation in eukaryotes RNA processing control 0 Two kinds 0 Alternative polyadenylation 0 Alternative splicing Gene regulation in eukaryotes mRNA transport control 0 mRNA transport control involves controlling movement out of the nucleus 0 only about 12 of all mRNAs move out of nucleus 0 Smell nuclear RNAs snRNAs play a role Gene regulation in eukaryotes translation control mRNAs can be translated at different rates 0 Ex mRNAs can be stored mRNAs associated with proteins to prevent degradation Stored mRNAs have shorter poly A tails Poly A tails play a role in initiation of translation ARE AUrich element acts as a signal for deadenylation later ARE acts as a recognition sequence for enzymes to add 150 A nucleotides Gene regulation in eukaryotes mRNA degradation control Deadenylation dependent decay o Poly A tails deadenylated over time 0 When short 5 15 As 5 cap is removed decapping 0 Then rapidly degraded by 5 to 3 exonulcease Regulatory molecules interact with ARE to control tail length Deadenylation independent decay o mRNAs decapped without deadenylation 0 Then rapidly degraded by 5 to 3 exonuclease Gene regulation in eukaryotes protein degradation control Ubiquitin a protein that binds to other proteins amp identi es them from degradation Ubiquitin tagged ubiquitinated proteins are transported to the proteasome for degradation by proteases Nend rule speci c Nterminal amino acids dictate the halflike o If Nterminal is Arg Lys Phe Leu or Trp l 3 minutes o If Nterminal is Cyt Ala Ser Thr Gly Val Pro or Met 20 hours RNAi RNA interference or RNA silencing Posttranscriptional regulation 0 Regulatory RNAs microRNAs miRNAs encoded by genes 21 33 nucleotides long DROSHA dsRNAspeci c endonuclease DICER dsRNAspeci c endonuclease SLICER RNA endonuclease MicroRNA induced silencing complex Translation repressed gene silencing Short interfering RNAs siRNAs not encoded by genes involved in RNA directed immune system 22 nucleotides long dsRNA ex From a virus DICER dsRNAspeci c endonuclease SLICERlike RNA endonuclease siRNA induced silencing complex 0 Target RNA cleaved ex Viral mRNA target RNAdirected immune system Both are untranslated RNAs that interfere with translation of mRNAs Gene regulation in development amp differentiation 0 Development regulated growth Differentiation formation of different types of cells tissues amp organs Note all cells have a complete genome totipotent Gene duplication events may allow one copy to evolve while the other copy maintains the original function Hemoglobins in Humans Polypeptides Stage a like 3 like Source Early embryo C zeta s Epsilon Yolk sac 3 month embryo Hb or y Gamma Liver amp spleen F After birth HbA or 3 Bone marrow Plus 6 Produced at low levels Globin genes arranged on the chromosome in the order of the timing of transcription 0 Pseudogenes similar to known functional genes but defective Pseudogenes are viewed as evidence of gene duplication Drosophia development 0 Sex determination o Genotypic sex determine The sex chromosome play an active role 0 X chromosome autosome balance system Drosophia C eegans Z of X Chr omosomes 39 Rat39o 2 sets of autosomes XX FemaleXY Male X0 Male How is the ratio X Autosomes detected Xlinked sisterless numerator genes 3 genes Autosomal deadpan denominator gene 0 Females excess numerator homodimers transcription factors that activate le Males insuf cient numerator homodimers le is not activated Transcriptional regulation of sex lethal Alternative promoters amp alternative splicing o In females numeratornumerator homodimers activate transcription from early promoter PE 0 Absence of early le protein leads to default splicing exons 3 has a stop codon Gene regulation in eukaryotes RNA processing control 0 Alternative splicing of le sex lethal gene plays a role in sex determination amp differentiation in Drosophia Drosophia Development 1 Fertilized egg with two parental nuclei 2 Parental nuclei fuse amp produce a diploid zygote nucleus 3 The nucleus divided for eight divisions in a common cytoplasm to give a multinucleate syncytium 4 The nuclei migrate to the periphery of the egg to produce a syncytial blastoderm Mitotic divisions continue 5 After the 13th division membranes form around the nuclei producing the somatic cells of the cellular blastoderm a 4 hours after fertilization b 6000 cell blastoderm Molecular gradients provide spatial location information o 3 major classes of genes regulate growth amp differentiation 1 Maternal effects genes transcribed during oogenesis in mother The protein products are called morphogens meaning that they are diffusible proteins that participate in controlling development a Bicoid regulates formation of anterior structures i Transcribed in mother mRNA stored in anterior of egg ii Translated after egg is laid iii BICOID protein forms a gradient greatest concentration in anterior iv Mutations produce embryos with posterior structures at each end v Bicoid protein acts as a transcription factor activating amp repressing genes along ant post axis vi Bicoid also represses translation of speci c genes b Nanos regulates formation of posterior structures i NANOS protein is localized in posterior end directs abdomen formation ii Translational repressor iii Mutations result in noabdomen phenotype c Torso regulates formation of terminal structures both ends i Torso is transcribed amp translated during oogenesis ii TORSO protein is activated at termini iii Mutations result in noterminals phenotype 2 Segmentation genes control subdivision of embryo 25 genes a Gap genes ex Kruppel hunchback mutations result in deletion of regions of several adjacent segments b Pair rule genes ex Evenskipped fushi tarazu mutations result in deletion of the same part of every other segment c Polarity genes ex Gooseberry engrailed mutations have portions of segments replaced by mirror images of adjacent half segments 3 Homeotic genes structure determining specify the identity of segments also called Hox genes a Homeotic mutants cause a segment to develop into a different body part than normal b All have 180bp homeobox which encodes 60 amino acid homeodomain helixturnhelix motif c The homeodomain recognizes an 8bp recognition sequence d There are 2 clusters of homeotic genes in Drosophia Organization of the bithorax complex 0 The bithorax complex consists of 3 genes UbX abd A amp abd B There determine the identities of segments T3A8 Mutations of the bithorax complex 0 Flies belong to the order Diptera meaning 2 wings Normal y with 1 pair of wings on T2 amp 1 pair of halters on T3 Mutant y with winds on T2 amp T3 Fly is homozygous for 3 mutant alleles Organization of the antennapedia complex 0 The antennapedia complex consists of 5 genes that determine the identities of the head segments amp T1 amp T2 Mutations of the antennapedia complex 0 Legs for antenna 0 Most Hox gene mutations are lethal Homeotic genes are conserved among animals this is especially true of the homeobox regions Vertebrates have 4 clusters of Hox genes also found in plants 0 In all known cases the genes are ordered on the chromosomes in the order they are expressed along the body axis colinearity rule Cancer biology Differentiation is highly regulated amp genetically programmed Deviations from the program can lead tot eh development of tissue masses called tumors or neoplasms quotnew growthsquot When tumors arise the cell has essentially lost the ability to control growth 0 Transformation the process by which a cell loses the ability to control growth Benign tumors tumors in which the cells stay together amp are not usually life threatening Malignant tumors tumors that invade amp disrupt adjacent tissues Metastasis cells from malignant tumor break off amp move around the body amp can form new tumors Oncogenesis process of tumor initiation genetic changes leading to uncontrolled growth o Spontaneous mutations 0 Exposure to mutagens or radiation 0 Cancerinducing viruses tumor viruses 0 How do tissues develop 0 Cell proliferation 0 Expression of tissuespeci c genes differentiation Stem cells undifferentiated ln malignant neoplasms cells divide without undergoing terminal differentiation cell cycle is out of control Terminally differentiated cells liver cells skin cells etc 0 Growth factors stimulate cell division 0 Growth inhibiting factors inhibit cell division 0 All cancers are genetic disorders 4 basic types of genes are involved 0 Protooncogenes 0 Tumor suppressor genes 0 Mutator genes 0 MicroRNA genes Familial Hereditary cancers inherited predisposition Sporadic Nonhereditary cancers arise from spontaneous or induced mutations most frequent type of cancer Retinoblastoma childhood eye cancer 0 2 forms Sporadic 60 of cases spontaneous in patients without family history Unilateral tumors 1 eye only Hereditary inherited predisposition Bilateral tumors 0 Usually develop earlier than sporadic retinoblastomas Pedigree analysis indicates 1 gene 0 1971 Alfred Knutson proposed the quot2 hit modelquot 2 mutational events are required P2 mutationsltltP1 mutation Sporadic form occurs alter in development Notes Retinoblastoma is one of very few cancers involving only one gene 0 RB is a tumor suppressor gene Oncogenes amp Retroviruses o Oncogenes mutated protooncogenes amp produce protein products that stimulate growth More active or active at inappropriate times 0 Tumor viruses induce tumors RNA viruses Retroviruses ex HIV1 o Retrovirus life cycle RSV genome RNA RSV proviral DNA Circular proviral DNA Integration of viral DNA begins Viral ends become joined to the ends of the cell s DNA by recombination Integration of viral DNA completed 0 Reverse Transcriptase reverse transcription produces double stranded DNA with long terminal repeats LTRs o LTRs have regulatory sequences 3 genes Gag encodes viral protein Pol encodes reverse transcriptase amp an enzyme needed for integration Env encodes glycoprotein for viral envelope 0 Genes are transcribed producing mRNAs for translation Progeny RNA genomes Transducing retroviruses or oncogenic retroviruses carry oncogenes ex Rous sarcoma virus Feline leukemia virus Mouse mammary tumor virus DNA viruses 0 Have DNA genomes Do not carry oncogenes normally viral genome is not integrated into host genome Transform cells by the action of genes that are essential components of their genome DNA viruses usually do not transform cells 0 Normally they produce a protein that activates replication in hot cell progeny are produced amp cell ysis occurs Rarely viral DNA is integrated into host amp replication is stimulated by the replicationactivating protein Viral oncogenes vonc have sequence homology to cellular oncogenes conc Transducing retroviruses have picked up a copy of a host cell s protooncogenes Transducing retroviruses form as a result of a genetic rearrangement usually a deletion event About 100 protooncogenes have been identi ed 0 Growth factors 0 Transcription factors 0 Protein kinases o Cytoplasmic regulators Protooncogene products stimulate cell proliferation encode a growthstimulating factor Mutations of protooncogenes are dominant o Mutations of protooncogenes produce oncogenes amp result in over expression or untimely synthesis of the protein 0 The result is uncontrolled cell proliferation Cancer induction can occur in 3 ways 0 Expression of viral oncogene vonc Mutation of protooncogene to produce a cellular oncogene conc point mutation deletions or gene ampli cations increased copy numberO lnsertional mutagenesis viral DNA integration next to a proto oncogene disrupts regulatory mechanisms Tumor suppressor genes encode a growthinhibiting factor 0 Suppress uncontrolled cell proliferation o Mutations inactivate these genes amp the inhibitory action if lost 0 Inactivation of both alleles is required for tumor formation Ex Retinoblastoma gene Mutations of a tumor suppressor gene cause loss of function amp inability to restrain cell growth recessive mutations An oncogene results from a mutation that causes an increase in function of a growth stimulating factor dominant mutations TP53 a tumor suppressor gene encodes p53 protein a transcription factor 0 Arrests cell cycle after DNA damage 0 Plays a role in apoptosis 0 2 mutations of TP53 result in no p53 being produced D no p21 D no arrest of cell cycle Mutator genes are normally involved with DNA replication amp DNA repair Mutations cause an increase in mutation rates for all genes including protooncogenes amp tumor suppressor genes MicroRNA genes are normally involved in regulating gene expression 0 Mutations cause an increase or decrease in miRNA gene expression many miRNAs are involved with regulating cell proliferation amp differentiation Ex miR155 is overexpressed in breast lung amp thyroid cancer 0 Overexpression could silence tumor suppressor genes Such miRNA genes are considered to be oncogenes o Underexpression could fail to silence protooncogenes Such miRNA genes are considered to be tumor suppressor genes The multistep nature of cancer 0 Unlike retinoblastoma most cancers require several mutations in several genes 0 Increased occurrence of cancer with age suggests that multiple mutations are needed for cancer development Carcinogens mutagens Chemical mutagens 0 Direct acting o Procarcinogens converted by metabolism to direct acting carcinogens Radiation direct action
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