3333outline</TI= TLE> <META http-equiv=3DContent-Type content=3D"text/html; = charset=3Diso-8859-1"> <META content=3DWord.Document name=3DProgId> <META content=3D"MSHTML 6.00.6000.16414" name=3DGENERATOR> <META content=3D"Microsoft Word 11" name=3DOriginator><LINK=20 href=3D"3333outlineWITHtrackchanges_files/filelist.xml" = rel=3DFile-List><LINK=20 href=3D"3333outlineWITHtrackchanges_files/editdata.mso" = rel=3DEdit-Time-Data><o:SmartTagType name=3D"Sn"=20 namespaceuri=3D"urn:schemas:contacts"></o:SmartTagType><o:SmartTagType=20 name=3D"country-region"=20 namespaceuri=3D"urn:schemas-microsoft-com:office:smarttags"></o:SmartTagT= ype><o:SmartTagType=20 name=3D"place"=20 namespaceuri=3D"urn:schemas-microsoft-com:office:smarttags"></o:SmartTagT= ype><o:SmartTagType=20 name=3D"City"=20 namespaceuri=3D"urn:schemas-microsoft-com:office:smarttags"></o:SmartTagT= ype><o:SmartTagType=20 name=3D"State"=20 namespaceuri=3D"urn:schemas-microsoft-com:office:smarttags"></o:SmartTagT= ype> <STYLE>@font-face { font-family: Tahoma; } @page Section1 {size: 8.5in 11.0in; margin: 1.0in 1.25in 1.0in 1.25in; = mso-header-margin: .5in; mso-footer-margin: .5in; mso-paper-source: 0; } P.MsoNormal { FONT-SIZE: 12pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: "Times New Roman"; = mso-style-parent: ""; mso-pagination: widow-orphan; = mso-fareast-font-family: "Times New Roman" } LI.MsoNormal { FONT-SIZE: 12pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: "Times New Roman"; = mso-style-parent: ""; mso-pagination: widow-orphan; = mso-fareast-font-family: "Times New Roman" } DIV.MsoNormal { FONT-SIZE: 12pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: "Times New Roman"; = mso-style-parent: ""; mso-pagination: widow-orphan; = mso-fareast-font-family: "Times New Roman" } A:link { COLOR: blue; TEXT-DECORATION: underline; text-underline: single } SPAN.MsoHyperlink { COLOR: blue; TEXT-DECORATION: underline; text-underline: single } A:visited { COLOR: purple; TEXT-DECORATION: underline; text-underline: single } SPAN.MsoHyperlinkFollowed { COLOR: purple; TEXT-DECORATION: underline; text-underline: single } P { FONT-SIZE: 12pt; MARGIN-LEFT: 0in; MARGIN-RIGHT: 0in; FONT-FAMILY: = "Times New Roman"; mso-pagination: widow-orphan; = mso-fareast-font-family: "Times New Roman"; mso-margin-top-alt: auto; = mso-margin-bottom-alt: auto } P.MsoAcetate { FONT-SIZE: 8pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: Tahoma; = mso-pagination: widow-orphan; mso-fareast-font-family: "Times New = Roman"; mso-style-noshow: yes } LI.MsoAcetate { FONT-SIZE: 8pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: Tahoma; = mso-pagination: widow-orphan; mso-fareast-font-family: "Times New = Roman"; mso-style-noshow: yes } DIV.MsoAcetate { FONT-SIZE: 8pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: Tahoma; = mso-pagination: widow-orphan; mso-fareast-font-family: "Times New = Roman"; mso-style-noshow: yes } SPAN.msoIns { COLOR: teal; TEXT-DECORATION: underline; text-underline: single; = mso-style-type: export-only; mso-style-name: "" } SPAN.msoDel { COLOR: red; TEXT-DECORATION: line-through; mso-style-type: export-only; = mso-style-name: "" } SPAN.msoChangeProp { mso-style-type: export-only; mso-style-name: "" } SPAN.SpellE { mso-style-name: ""; mso-spl-e: yes } SPAN.GramE { mso-style-name: ""; mso-gram-e: yes } DIV.Section1 { page: Section1 } OL { MARGIN-BOTTOM: 0in } UL { MARGIN-BOTTOM: 0in } </STYLE>  <META content=3DC:\MSOFFICE\OFFICE\html.dot name=3DTemplate></HEAD> <BODY lang=3DEN-US style=3D"tab-interval: .5in" vLink=3Dpurple = link=3Dblue> <DIV class=3DSection1> <SPAN=20 style=3D"FONT-SIZE: 13.5pt">This outline is designed to provide you with = a general=20 summary of the topics that will be discussed in class. It is not = designed to=20 substitute for your own in-class notes, or for the detailed discussions = of these=20 topics in the assigned readings. Interspersed within this outline are = references=20 to readings in the Hartwell text, where this subject material can be=20 found. = I<SPAN=20 style=3D"FONT-SIZE: 10pt">. Course Description A. Text B. Grading C. Course Schedule II. =20 What is a Gene? A. It must be = transmitted between=20 generations and each individual must have a physical copy of this = material. B. It must provide information to its = carriers with=20 respect to biological structure and function. III. =20 Historical Perspectives:PreMendelian = concepts=20 of inheritance A. Spontaneous = generation (till=20 1861!) B. Concepts emphasizing continuity between = generations 1. Early Greeks (~500 = B.C.) a. Pangenesis: somatic = tissue=20 contributes to heritable information b. Heritable contributions of males and = females=20 differ. 2. <SPAN=20 class=3DSpellE>Preformationism (1600's) 3. Studies of early embryogenesis and = the concept=20 of epigenesis (~1750-1880) 4. Charles Darwin's concepts of = inheritance (1859)=20 a. <SPAN=20 class=3DSpellE>gemmule theory b. the = observation of=20 continuous variation across the biological spectrum and an emphasis on = blending=20 of traits in offspring. C. <SPAN=20 class=3DSpellE>Gregor Mendel: a contemporary of Charles Darwin = (1822-1884)=20 1. Entered monastery and = was=20 educated at <st1:City w:st=3D"on"><st1:place=20 w:st=3D"on">Vienna</st1:place></st1:City>, Brunn. 2. Started experiments in peas in 1854 = 3. Published experiments in 1866 in a = journal=20 distributed to over 100 libraries in <st1:place = w:st=3D"on">Europe</st1:place>;=20 corresponded with major biologists of his day. 4. Assumed administrative responsibilities = as abbot=20 (1868); died 1884 in scientific obscurity. IV. = Principles of=20 Transmission Genetics: Mendel's experiments in Peas. A. His experimental = organism, the=20 common garden pea, very useful for genetic analysis. 1. Chose characteristics = that were=20 unambiguous. 2. Relatively quick reproductive cycle; = large=20 numbers of progeny. 3. Organism could be either = self-pollinated or <SPAN=20 class=3DSpellE>outcrossed (hybridized). 4. Took a quantitative approach to data=20 collection. B. The monohybrid = <SPAN=20 class=3DGramE>cross: principles of unit inheritance and = segregation. 1. If two plants with = differing=20 traits were crossed (P generation) the next generation (F1) = always=20 gave rise to plants displaying only one parental character. If=20 the F1 plants are now allowed to self-fertilize, the other = parental=20 character reappears in the next generation (F2), representing = 25% of=20 the offspring. 2. Generated a hypothesis consistent with = his=20 results. a. Traits could be = represented as=20 discrete, particulate entities or factors (represented by symbol). b. Factors appear to be paired, although = some=20 traits are not observed in hybrids even though the factor associated = with their=20 expression may be present in the hybrid (concept of dominant and = recessive=20 traits). c. Factors will randomly segregate = during=20 production of gametes; one factor is received from each parent; = restoration of=20 paired factors upon fertilization. C. The <SPAN=20 class=3DSpellE>dihybrid cross: <SPAN=20 class=3DSpellE>prinicple of independent assortment 1. Expanded his = observations,=20 watching the segregation of two traits simultaneously in the same cross. = 2. Results confirmed earlier observations. = a. Two traits observed = together=20 will partition independently of one another. b. observed = F2=20 ratios (9:3:3:1) were two independent 3:1 ratios. D. The testcross = (backcross):=20 Supported his basic hypothesis. 1. Backcross of = F1=20 monohybrid to single recessive parent produced expected 1:1 ratios. 2. Backcross of F1 <SPAN=20 class=3DSpellE>dihybrid to double recessive parent produced = expected=20 1:1:1:1 ratios. E. Basic Terminology = 1. Merkmal=20 (Mendel's German term) =3D factor =3D gene. 2. alleles: = variant genes=20 for any given trait; capital letter (A) represents dominant; small = letter (a)=20 represents recessive. 3. homozygote: = identical=20 pair at a given locus; AA or aa. (<SPAN=20 class=3DGramE>locus =3D location on a chromosome) 4. heterozygote: a <SPAN=20 class=3DSpellE>nonidentical pair at a given locus; <SPAN=20 class=3DSpellE>Aa. 5. genotype: = the genetic=20 composition or complement (for any given gene pair or locus) 6. phenotype: = the physical=20 appearance observed based on expression of the genotype. V. 19th Century Advances = in Cell=20 Biology: Indentifying the Physical Basis for = Mendel's=20 Laws. A. Nucleus and = discovery of cell=20 structure. (Brown, 1833) B. Fertilization and pronuclear fusion. = (<SPAN=20 class=3DSpellE>Hertwig et al. ~1875) C. Mitosis (Fleming et al. ~1880) D. Germplasm = theory=20 (Weismann, ~1880) E. Meiosis (Van Beneden,=20 Boveri et al. 1880s-early 1900s) F. Rediscovery of Mendel's Laws (<SPAN=20 class=3DSpellE>deVries, Correns, von = <SPAN=20 class=3DSpellE>Tschermak; 1900) VI. The Sutton-<SPAN=20 class=3DSpellE>Boveri Hypothesis (1903) A. Both chromosomes = and <SPAN=20 class=3DSpellE>Mendelian factors are found in pairs; of both = maternal and=20 paternal origin. B. Chromosomes of each pair separate = during=20 meiosis, as do Mendelian factors. C. On fertilization, the "<SPAN=20 class=3DSpellE>doubleness" or the chromosome complement is = restored, as are=20 Mendelian factors. VII. Segregation, = Independent=20 Assortment and Probability. A. Probability =3D = number of=20 anticipated events/total number of possibilities; e.g. the = probability of=20 pulling an ace out of a straight deck of cards: P(ace) =3D 4/52 or 1/13. = B. Product Rule: For independent = events,=20 the probability of several events occurring is the product of = their=20 individual probabilities; P(A and B) = =3D P(A) *=20 P(B). e.g. P (ace = of spades) =3D=20 P (ace) * P (spade) =3D 1/13 * 1/4 =3D 1/52 C. Sum Rule: For mutually exclusive = events,=20 the probability represents the sum of the individual = probabilities. P (A=20 or B) =3D P(A) + P(B). e.g. P (ace = or king) =3D=20 1/13 + 1/13 =3D 2/13 D. For <SPAN=20 class=3DSpellE>Mendelian patterns of inheritance, the events that = occur=20 during the process of meiosis and fertilization (segregation, = independent=20 assortment) can be expressed in probabilistic terms. 1. In a monohybrid = cross, the=20 fraction of progeny with a given phenotype follows the sum rule: <SPAN=20 class=3DGramE>P(AA or Aa) =3D 1/4 +=20 1/2 =3D 3/4; P(A_ or aa) =3D = 3/4 + 1/4. 2. Assuming independent assortment, a = <SPAN=20 class=3DSpellE>dihybrid cross would be expected to yield a = 9:3:3:1=20 phenotypic ratio- applying the product rule to two independent events. = P( A_ and=20 B_) =3D P(A_) * P(B_) =3D 9/16 A_B_ P (A_ and bb) =3D P(A_) *=20 P(bb) =3D 3/16 A_bb P (aa and = B_) =3D <SPAN=20 class=3DGramE>P(aa) * P(B_) =3D 3/16 = <SPAN=20 class=3DSpellE>aaB_ P (aa and = bb) =3D <SPAN=20 class=3DGramE>P(aa) * P(bb) =3D 1/16 = <SPAN=20 class=3DSpellE>aabb 3. In the cross <SPAN=20 class=3DSpellE>AaBbCc x <SPAN=20 class=3DSpellE>AaBbcc, what is the probability of = producing progeny=20 with the same phenotype as the first parent? (ans.=20 9/32) I. Documenting Patterns = of=20 Transmission for Single Gene Traits in Humans. A. Importance of Familiy Histories for Human Genetic Analysis B. Pedigree Symbols and Guidelines C. Pedigree Patterns for <SPAN=20 class=3DSpellE>Autosomal Dominant Inheritance. 1. Trait appears in = every generation=20 (exceptions; new mutation or incomplete dominance). 2. Trait is passed from affected person to = 1/2 of=20 progeny, regardless of sex. 3. Unaffected members will not pass on the = trait. D. Pedigree Patterns = for <SPAN=20 class=3DSpellE>Autosomal Recessive Inheritance. 1.Trait=20 appears to skip generations, often observed in sibs but not parents. = 2. On the average, the trait will show up = in 25% of=20 the sibs of an affected proband, <SPAN=20 class=3DSpellE>regarless of sex. 3. The parents may be consanguineous. E. Probability and = Risk=20 Assessment. <IMG id=3D_x0000_i1025 height=3D245=20 src=3D"http://www.ou.edu/cas/zoology/Courses/3333/Image3.gif" = width=3D192> 1. From pedigree, trait = appears=20 inherited as an autosomal recessive. 2. From phenotype of II-3, parents are = obligate=20 heterozygotes. 3. If parents are obligate <SPAN=20 class=3DSpellE>heterozygotes, III-2 and III-3 have a 2/3 = probability of=20 being heterozygotes. 4. If parents of IV-1 and IV-2 are <SPAN=20 class=3DSpellE>heterozygotes, IV-1 and IV-2 each stand a 1/2 = probability of=20 inheriting disease gene allele from their parents. 5. If they are <SPAN=20 class=3DSpellE>heterozygotes, they have a 1/4 probability of = having an=20 affected child. 6. The risk to IV-1 and IV-2 of having an = affected=20 child is therefore: 2/3*2/3*1/2*1/2*1/4 =3D = 1/36 II. Evidence for the = Chromosomal=20 Theory of Inheritance A. Sex Chromosome = Dimorphism: homo-=20 and heterogametic sexes. B. The white locus in <SPAN=20 class=3DSpellE>Drosphila: Thomas Hunt Morgan (1910). 1. Characteristic = segregation=20 pattern: white female x normal male shows criss-cross=20 inheritance. 2. Calvin Bridges: Correlation of = exceptional=20 progeny with nondisjunction during = meiosis. C. Sex Determination: = Nature of the=20 "Primary" Switch; mechanisms vary. (See Hartwell, = pp. <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:20>605-610</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:20>669, 672-677</INS> = for detailed=20 discussion of molecular mechanisms in Drosophila.) 1. Drosophila: = X<SPAN=20 class=3DGramE>:A ratio. <P=20 style=3D"MARGIN-LEFT: 2.5in; mso-list: none; mso-list-ins: MHE = 20070208T1120">2.=20 Mammals: TDF/SRY gene on Y chromosome. (See Hartwell, pp. <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:20>653</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:20>720</INS><SPAN=20 style=3D"BACKGROUND: windowtext; mso-highlight: windowtext"><SPAN=20 style=3D"BACKGROUND: yellow; mso-highlight: yellow">)<INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:20> Please note: part of this = discussion=20 was deleted from this chapter in the 3/e. =20 Please check</INS><INS=20 cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:20><o:p></o:p></INS> <o:p> </o:p> III. Patterns of = Sex-Linked=20 Expression in Humans. A. X-linked Recessive = Inheritance.=20 1. Incidence much = higher in males=20 than in females. 2. Affected male will give rise to = daughters who=20 are carriers. On average, 1/2 of a carrier female's sons will be = affected; half=20 of her daughters will also be carriers.. 3. Never father to son transmission. B. X-linked Dominant = Inheritance.=20 1. Affected males will = have no=20 affected sons and no normal daughters. 2. Transmission by females is not = distinguishable=20 from an autosomal dominant. 3. Since most females for <SPAN=20 class=3DSpellE>autosomal dominants are heterozygous and most = males will be=20 hemizygous, the trait is often more severe = in=20 males. C. <SPAN=20 class=3DSpellE>Holandric Inheritance. 1. Relatively rare in = human=20 populations. Y chromosome, relative to X, is gene poor (constitutive=20 heterochromatin). 2. TDF/SRY gene. D. Sex-limited and = Sex-Influenced=20 Traits. 1. Sex-limited; e.g. = precocious=20 puberty. 2. Sex-influenced; e.g. pattern = baldness. VI. Probability = Revisited. <SPAN=20 class=3DGramE>(N.B. See Chapter 5 of Thompson; this is not = formally=20 discussed in Hartwell) A. Use of the Binomial = Expansion.=20 1. In a three child = family, what=20 is the probability of two boys and 1 girl? P(boy) =3D = p=3D 1/2; P(girl)=20 =3D q=3D 1/2; n =3D number of events (births) =3D 3. All potential possiblities are represented by the binomial = expansion, with=20 the specific case in question emphasized: (p + q)3 =3D 1=20 or p3 <BLINK>+3p2q </BLINK>+ = 3pq2 +=20 q3 =3D 1; 3p2q =3D 3(1/2)(1/2)(1/2) =3D 3/8 = For any given term in the expansion = (e.g. in a=20 three child family, the odds of two boys and a girl) is also represented = by the=20 equation: <IMG id=3D_x0000_i1026 height=3D41=20 src=3D"http://www.ou.edu/cas/zoology/Courses/3333/Image4.gif" = width=3D58> where n =3D = number of=20 children =3D 3, P(boy) =3D p =3D 1/2; P(girl) =3D q =3D 1/2; s =3D = number of boys=3D 2; t =3D=20 number of girls =3D 1. Note that p + q =3D 1; s + t =3D n. n!/<SPAN=20 class=3DSpellE>s!t! =3D (3*2*1)/(2*1)(1) =3D 3; and=20 <SPAN=20 class=3DGramE>psqt =3D = (1/2) (1/2)=20 (1/2) =3D 1/8; therefore; = 3* 1/8 =3D 3/8 <P class=3DMsoNormal=20 style=3D"MARGIN-LEFT: 1.5in; TEXT-INDENT: -0.25in; mso-margin-top-alt: = auto; mso-margin-bottom-alt: auto; mso-list: l0 level3 lfo1; = mso-list-change: '%3:1:0:.' MHE 20070208T1119; tab-stops: list = 1.5in"><![if !supportLists]><SPAN=20 style=3D"mso-list: Ignore">1.<SPAN=20 style=3D"FONT: 7pt 'Times New Roman'"> =20 <![endif]>The binomial expansion is a special case of the = more=20 general multinomial distribution, where the model can be expanded to = include=20 more than two alternatives: <IMG id=3D_x0000_i1027 = height=3D41=20 src=3D"http://www.ou.edu/cas/zoology/Courses/3333/Image5.gif" = width=3D186> B. Testing Experimental = Data for=20 Conformity to an Experimental Hypothesis: the Chi Square Test. (See = Thompson,=20 ch. 5; discussed in Hartwell with regard to = linkage=20 analysis, pp. <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:21>117-120</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:21>127-131</INS>) 1. Using Mendel's data = set as an=20 example of observed vs. experimental data. 2. Concept of Degrees of Freedom; for = genetic=20 tests like these, the degree of freedom is one less than the number of = classes.=20 3. The chi square table and assessment = of the=20 probability the differences between the observed and expected values are = due to=20 chance. 4. The 0.05 criterion for level of=20 significance. I. Dominance = Relationships A. Complete Dominance = 1. Mendel's pea = experiments. 2. A. Garrod:=20 (1902-1909) a. Demonstrated that = human disorders=20 (e.g. alcaptonuria) were transmitted in = human=20 pedigrees according to Mendelian laws of = inheritance=20 (e.g. alcaptonuria shows a=20 autosomal recessive pattern of inheritance.) = b. Postulated that the reason for the = disease was a=20 metabolic defect that was genetically inherited- that genes are = ultimately=20 responsible for making enzymes and the genetic defect in <SPAN=20 class=3DSpellE>alcaptonuria was related to the absence of = functional enzyme=20 ("inborn error of metabolism"). Like Mendel, this postulate was ahead of = its=20 time, and not rigorously tested until the 1030s-40s. B. Incomplete Dominance = 1. Transmission = genetics could=20 demonstrate that some heterozygotes = displayed a=20 phenotype different from either parent, although pattern of segregation = followed=20 Mendel's laws. e.g. flower color in 4 o'clock = plants.=20 Pure breeding red and white individuals produce pink F1s. = Pink=20 F1 x Pink F1 produce 1 red: 2 pink: 1 white = offspring; red=20 and white breed true; pink phenotype due to heterozygote state. 2. If phenotype, in certain instances, = is=20 reflected by level of enzymatic activity, then some phenotypes may show = a=20 threshold effect (e.g. alcaptonuria) where = any enzyme=20 activity above a certain level confers a wild type phenotype, and some = may show=20 a saturation effect, where incremental amounts of the enzyme may lead a = gradient=20 of phenotypes (e.g. sugar deposition in some plant seeds). C. Multiple Alleles and = <SPAN=20 class=3DSpellE>codominance 1. Although any = individual in a=20 population may contain only two allelic variants (<SPAN=20 class=3DGramE>Aa), within the population there may be may = different=20 variants (A, a, A1, A2, A3 etc.) 2. Allelic variation can be detected = through a=20 number of means: a.=20 electrophoresis (protein level) b. nucleic acid = analysis=20 (DNA level) 3. <SPAN=20 class=3DGramE>examples: beta globin,=20 alpha1-anti-trypsin. 4. <SPAN=20 class=3DGramE>codominance: the heterozygote exhibits a = phenotype=20 based on the expression of both alleles. e.g. = ABO blood=20 group locus. 5. Molecular discussion of beta <SPAN=20 class=3DSpellE>globin variants and ABO blood group locus: = (Hartwell, <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:21>277-78; 308-310;=20 46-7; 56-7</DEL><INS cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:21>301-302; 336-339; 48-49; = 60-61</INS>) II. Deviation from = <SPAN=20 class=3DSpellE>Mendelian Ratios due to Single Gene, Polygenic, or = Environmental Interactions. A. = lethal=20 genes. e.g. yellow allele in mice = leads to 2:1=20 ratio if two yellow animals are crossed. B. genetic = heterogeneity.=20 e.g. albinism can be caused by a defect at = more than=20 one genetic locus. C. <SPAN=20 class=3DSpellE>phenocopy. e.g. = <SPAN=20 class=3DSpellE>kwashiorkhor- environmental factors mimic genetic = disorder=20 D. Variable Expressivity and <SPAN=20 class=3DSpellE>Penetrance. 1. Variable = Expression: single=20 gene effects can be variable in severity of expression and <SPAN=20 class=3DSpellE>pleiotropic (have several different phenotypic = effects).=20 e.g. anonychia = and=20 neurofibromatosis. 2. Penetrance: the=20 frequency with which a gene manefests itself = in the=20 appropriate genotype: some genes are incompletely <SPAN=20 class=3DSpellE>penetrant (do not express the appropriate = phenotype due to=20 gene or environmental modifiers). III. Interactions of = Genes with=20 other Genes (epistatic interactions). A. 4 distinct phenotypes = (comb shape=20 in chickens). B. complementary gene action (<SPAN=20 class=3DSpellE>anthocyanin pigment formation in pea flowers) C. recessive epistasis=20 (albinism in mice) D. recessive suppression of a recessive = phenotype in=20 Drosophila; purple eye color E. duplicate genes (fruit shape in <SPAN=20 class=3DSpellE>shepard's purse) I. Linkage and = Recombination A. Two genes close = together on the=20 same chromosome will not assort independently; they will be transmitted = to the=20 same gamete >50% of the time since they are attached to a common = <SPAN=20 class=3DSpellE>centromere. B. In heterozygotes,=20 linked genes may be arrayed in two possible configurations (phasing=20 relationships). 1. <SPAN=20 class=3DGramE>cis (or coupling): both dominant loci on = same=20 chromosome and both recessive loci on the same chromosome: (AB/<SPAN=20 class=3DSpellE>ab). 2. trans (or = repulsion):=20 A dominant gene at one locus associated on the same chromosome with a = recessive=20 gene at a second locus: (Ab/aB). C. Linkage can be = followed in a=20 testcross. II. Use of testcrosses = to=20 demonstrate linkage (e.g. T.H. Morgan). A. e.g. The=20 white and minature genes on = the X=20 chromosome of Drosophila. w=20 m/ + + x w m/Y or The white=20 and yellow genes on the X chromosome of Drosophila. = y +/+ w =20 x y=20 w/Y B. Parental types = recovered in=20 greater frequency than on basis of random assortment. 1. Results deviated = from expected=20 1:1:1:1 ratio; non-recombinant (parental) classes recovered in greater = frequency=20 than the recombinant (non-parental) classes. Hypothesized that=20 the non-parental gene combinations arose from physical exchange of = chromosomal=20 material during meiosis in the heterozygote. 2. The frequency of recombination is a = function of=20 how close two loci are linked. 3. The crossover frequency can be = directly related=20 to physical distances between genes on the chromosome (A.H. Sturtevant, = 1911).=20 a) the=20 farther apart two loci are located from one another, the more likely a = crossover=20 will occur between them. b) 1% recombination in gametes of = heterozygote=20 represents 1 map unit ( m.u.=20 or CentiMorgan, cM). c) for = white and=20 miniature, distance is large (~37.6 m.u.); for=20 yellow and white, distance between genes in much smaller (~1.4 <SPAN=20 class=3DSpellE>m.u.); d) the closer = the loci, the=20 greater the number of progeny that must be examined to detect a = crossover=20 event. 4. Conclusions from = Linkage=20 Experiments. a) Genes are linked = through all=20 generations, no random mixing. b) Genes are arrayed in a specific linear = order=20 along the length of a chromosome, which is a species characteristic. = c) Homologs = carry the same=20 array of gen loci, but not necessarily the = same=20 alleles. During meiosis, homologs could = exchange parts=20 by crossing over. Chiasmata represent = crossover points=20 between homologs (Janssens,=20 1909). III. Three point = testcrosses. A. = <SPAN=20 class=3DSpellE>eg. Corn: Mapping the genes for the <SPAN=20 class=3DSpellE>rececessive traits lazy growth, glossy leaf, and = sugary=20 endosperm. Progeny <SPAN=20 class=3DGramE>From a Three Point Testcross in Corn <o:p> </o:p> <DIV align=3Dcenter> <TABLE class=3DMsoNormalTable style=3D"BACKGROUND: #80ff80; = mso-cellspacing: 0in"=20 cellSpacing=3D0 cellPadding=3D0 border=3D0> <TBODY> <TR style=3D"mso-yfti-irow: 0; mso-yfti-firstrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"=20 colSpan=3D3> <o:p> </o:p></TD></TR> <TR style=3D"mso-yfti-irow: 1"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Phenotype of testcross = progeny</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Genotype of Gamete from Hybrid = Parent</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Number</TD></TR> <TR style=3D"mso-yfti-irow: 2"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> normal (<SPAN=20 class=3DSpellE>wildtype)</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Lz Gl=20 Su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 286</TD></TR> <TR style=3D"mso-yfti-irow: 3"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> lazy</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lz Gl=20 Su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 33</TD></TR> <TR style=3D"mso-yfti-irow: 4"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> glossy</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Lz gl=20 Su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 59</TD></TR> <TR style=3D"mso-yfti-irow: 5"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> sugary</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Lz Gl = <SPAN=20 class=3DSpellE>su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 4</TD></TR> <TR style=3D"mso-yfti-irow: 6"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lazy,glossy</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lz gl=20 Su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2</TD></TR> <TR style=3D"mso-yfti-irow: 7"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> lazy,=20 sugary</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lz Gl = <SPAN=20 class=3DSpellE>su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 44</TD></TR> <TR style=3D"mso-yfti-irow: 8"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>glossy,sugary</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Lz gl = <SPAN=20 class=3DSpellE>su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 40</TD></TR> <TR style=3D"mso-yfti-irow: 9; mso-yfti-lastrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> lazy, <SPAN=20 class=3DSpellE>glossy,sugary</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lz gl = <SPAN=20 class=3DSpellE>su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 272</TD></TR></TBODY></TABLE></DIV> B. Review of data set = shows=20 deviation from 1:1:1:1:1:1:1:1 ratio that would be expected for = independent=20 assortment; instead, two classes are in very high frequency, two classes = are in=20 very low frequency and 4 classes are recovered in intermediate = frequencies. C. The parental class represents the most = frequent=20 phenotypic classes amongst the progeny. This allows the determination of = the=20 orientation of the dominant and recessive alleles in the heterozygous = parent.=20 D. The double recombinant classes = represent the=20 least frequent phenotypic classes amongst the progeny. The 4 = intermediate=20 classes represent the single recombinant classes in each interval. E. Comparison of parental and double = recombinant=20 classes allows the determination of which gene is in the middle; from = the data=20 set given in class, the phasing relationship and order of the genes in = the=20 heterozygous parent was: <SPAN=20 class=3DSpellE>Lz = Su =20 Gl/lz su <SPAN=20 class=3DSpellE>gl F. Once gene array in = the parental=20 heterozygote is determined, identify the phenotypic classes in the = progeny that=20 represent the single and double crossovers which occur within each gene=20 interval: RF =3D (#SCO + #DCO)/total progeny for each gene interval. = <SPAN=20 class=3DSpellE>eg. for=20 the example given in class, The gene interval between <SPAN=20 class=3DSpellE>Lz and Su could be calculated as : = RF =3D ([33 +=20 40] + [4 + 2])/740 =3D 0.107 =3D 10.7 m.u. = E. Interference. 1. A crossover in one = area may=20 interfere with crossing over in an adjacent area of the chromosome. Male Progeny=20 from the mating of Drosophila melanogaster=20 females heterozygous for three X-linked Genes with <SPAN=20 class=3DSpellE>Wildtype Males <o:p> </o:p> <DIV align=3Dcenter> <TABLE class=3DMsoNormalTable style=3D"BACKGROUND: #ffff80; = mso-cellspacing: 0in"=20 cellSpacing=3D0 cellPadding=3D0 border=3D0> <TBODY> <TR style=3D"mso-yfti-irow: 0; mso-yfti-firstrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"=20 colSpan=3D3> <o:p> </o:p></TD></TR> <TR style=3D"mso-yfti-irow: 1"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Phenotype</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Genotype of=20 female gamete</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> number</TD></TR> <TR style=3D"mso-yfti-irow: 2"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>wildtype</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> + +=20 +</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 5</TD></TR> <TR style=3D"mso-yfti-irow: 3"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>crossveinless</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>cv + +</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2207</TD></TR> <TR style=3D"mso-yfti-irow: 4"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> echinus</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> + <SPAN=20 class=3DSpellE>ec +</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 217</TD></TR> <TR style=3D"mso-yfti-irow: 5"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> cut</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> + +=20 ct</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 265</TD></TR> <TR style=3D"mso-yfti-irow: 6"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>crossveinless, echinus</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>cv ec = +</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 273</TD></TR> <TR style=3D"mso-yfti-irow: 7"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>crossveinless,cut</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>cv + ct</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 223</TD></TR> <TR style=3D"mso-yfti-irow: 8"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>echinus,cut</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> + <SPAN=20 class=3DSpellE>ec ct</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2125</TD></TR> <TR style=3D"mso-yfti-irow: 9; mso-yfti-lastrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>crossveinless,echinus,cut</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>cv ec=20 ct</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 3</TD></TR></TBODY></TABLE></DIV> 2. This may be = observed in three=20 point testcrosses, where the number of expected double crossovers that = are=20 actually recovered is less than the number that are=20 expected. e.g. for the example taken from a=20 Drosophila cross in class (see above data set), the calculated = map=20 distances between the echinus, crossveinless=20 and cut loci were: <SPAN=20 class=3DGramE>ec-------10.3 <SPAN=20 class=3DSpellE>m.u.--------<SPAN=20 class=3DSpellE>cv------8.4 <SPAN=20 class=3DSpellE>m.u.------ct and out of = 5318 progeny,=20 8 double crossovers were actually observed. The expected number of = double=20 crossovers was therefore: (0.103)(0.084) =3D (0.0086<SPAN=20 class=3DGramE>)(5318) =3D 46 expected double crossovers. coefficient = of=20 coincidence =3D observed double crossovers/expected double crossovers = =3D 0.174 Interference =3D 1- coefficient of = coincidence =3D=20 0.826 IV. Ordered tetrads in = fungi. A. Some fungi produce = meiotic=20 products, haploid spores, in a saclike structure (<SPAN=20 class=3DSpellE>ascus). All meiotic products from the same = division cycle=20 can be recovered; in some species, each member of the tetrad giving rise = to=20 these spores are arrayed in the order of = their=20 segregation. Likewise, in some species, a mitotic division follows = meiosis=20 giving rise to two spore pairs for each tetrad. B. Other advantages to genetic analysis. = 1. Organism is haploid = for majority=20 of life cycle. 2, They produce = large=20 numbers of spores- rare events are easily detected. C. Life cycle of Ascomycetes: e.g. <SPAN=20 class=3DSpellE>Neurospora crassa,=20 bread mold. 1. Haploid spores = generate <SPAN=20 class=3DSpellE>hyphal structures. 2. Haploid nuclei of opposite mating types = can=20 undergo nuclear fusion and meiosis in fruiting bodies (<SPAN=20 class=3DSpellE>perithecia). 3. The zygote is the only diploid stage; = this=20 transient diploid state undergoes meiosis and then mitosis; 8 <SPAN=20 class=3DSpellE>ascospores are produced. 4. The meiotic products (tetrads) are = maintained in=20 a linear order. V. Mapping by Ordered = Tetrad=20 Analysis. All meiotic products are recovered in the same structure in an = ordered=20 array. This permits the mapping of genes relative to the <SPAN=20 class=3DSpellE>centromere. A. Calculating a Map = Distance=20 Between the Gene and <SPAN=20 class=3DSpellE>Centromere. 1. If no recombination = occurs, you=20 will expect to see a 1st division segregation pattern: a 4:4=20 array of spores relative to phenotype. 2. If a single crossover occurs between = the gene and=20 the centromere, you will expect to see a 2nd = division=20 segregation pattern: a 2:2:2:2 or a 2:4:2 array (fig 5-<DEL=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:22>23 </DEL><SPAN=20 class=3DmsoIns><INS cite=3Dmailto:MHE dateTime=3D2007-02-08T11:22>21 = </INS>in=20 Hartwell). 3. The percentage of 2nd division = segregation=20 patterns is related to the distances between the gene and <SPAN=20 class=3DSpellE>centromere. 1/2(asci with=20 2nd division segregation pattern) X 100 =3D map distance=20 total number of = <SPAN=20 class=3DSpellE>asci Remember that you a scoring <SPAN=20 class=3DSpellE>asci here, and that the "1/2" in the above = equation corrects=20 for the fact that only 1/2 of the chromosomes arising from that meiosis = will be=20 recombinant. B. Gene Linkages. = Suppose two loci=20 are present in the same cross. In the example given in class, the cross = AB x=20 ab yields the following data set: Hypothetical Data Set from the <SPAN=20 class=3DSpellE>Neuospora Cross AB x <SPAN=20 class=3DSpellE>ab <o:p> </o:p> <DIV align=3Dcenter> <TABLE class=3DMsoNormalTable style=3D"BACKGROUND: aqua; = mso-cellspacing: 0in"=20 cellSpacing=3D0 cellPadding=3D0 border=3D0> <TBODY> <TR style=3D"mso-yfti-irow: 0; mso-yfti-firstrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"=20 colSpan=3D5> <o:p> </o:p></TD></TR> <TR style=3D"mso-yfti-irow: 1"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <o:p> </o:p> <o:p> </o:p></TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Ascus = Type</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Ascus = Type</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Ascus = Type</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Ascus = Type</TD></TR> <TR style=3D"mso-yfti-irow: 2"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Spore=20 pair</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 1</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 3</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 4</TD></TR> <TR style=3D"mso-yfti-irow: 3"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> </TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> (PD)</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> (TT)</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> (PD)</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> (NPD)</TD></TR> <TR style=3D"mso-yfti-irow: 4"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 1</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>aB</TD></TR> <TR style=3D"mso-yfti-irow: 5"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>aB</TD></TR> <TR style=3D"mso-yfti-irow: 6"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 3</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>aB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Ab</TD></TR> <TR style=3D"mso-yfti-irow: 7"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 4</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Ab</TD></TR> <TR style=3D"mso-yfti-irow: 8; mso-yfti-lastrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Total number</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 108</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 60</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 30</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2</TD></TR></TBODY></TABLE></DIV> 1. 3 types of tetrads = are produced:=20 a) <SPAN=20 class=3DGramE>parental ditype (PD): = <st1:place=20 w:st=3D"on"><st1:City w:st=3D"on">AB</st1:City> <st1:State = w:st=3D"on"><SPAN=20 class=3DSpellE>AB</st1:State></st1:place> <SPAN=20 class=3DSpellE>ab ab b) <SPAN=20 class=3DGramE>nonparental ditype (NPD):=20 aB aB = <SPAN=20 class=3DSpellE>Ab Ab c) <SPAN=20 class=3DGramE>tetratype (TT): AB Ab=20 aB ab 2. If genes are = unlinked, then the=20 A and B loci might be expected to assort independently; = therefore=20 PD =3D NPD (approximately the same numbers). 3. If the two genes are linked, then the = only way to=20 generate a NPD would be through a 4 strand double crossover; and = PD>> NPD.=20 In the above data set, the A and B loci are linked. 4. Centromere = order can be=20 determined by examining frequency of classes showing 1st and second = degree=20 segregation patterns in relation to frequency of NPD classes. For = example, the=20 type 3 ascus pattern above is informative; = both the=20 A and B loci show a second degree segregation pattern, = indicating=20 a cross over between both genes and the centromere. If=20 genes were on opposite sides of the centromere, you=20 would need a double crossover to generate this class. Its frequency = relative to=20 the other classes indicates that it is a single crossover class, and = that=20 A is closer to the centromere than = B:=20 centromere--------A----------B. 5. Recombination between <SPAN=20 class=3DSpellE>centromere and B: 1/2 (60 + 30) X 100 =3D 22.5 map = units=20 200 6. Recombination between <SPAN=20 class=3DSpellE>centromere and A: 1/2 (30) X 100 =3D 7.5 map units=20 200 The deduced difference between the genes = is=20 therefore 15 map units. 7. The above calculation will lead to an=20 underestimate of distance, since the above technique is not accurately=20 estimating the number of double crossovers. The number of double = crossovers can=20 be estimated, however from the NPD class (presented in class: diagram = found in=20 ZAP room notes). From this class, we can estimate the total number of = single and=20 double crossovers occurring in the interval between A and = B: DCO =3D 4 NPD T =3D SCO + 1/2 DCO; therefore SCO =3D T - 2 NPD RF =3D 1/2 SCO + DCO RF =3D 1/2[TT-2NPD] + 4NPD=20 &nb= sp; =20 total asci for the above = data set:=20 RF =3D 1/2(60-4) + 8=20 &nb= sp; =20 200 =3D 36/200 =3D = 0.18 =3D 18 map=20 units between the A and B loci. (note our=20 previous estimate was 15 map units, an underestimate). VII. Mitotic = Recombination. A. Although much rarer = than meiotic=20 recombination during gametogenesis, = recombination=20 events can be detected in somatic tissue. A classic example is the = formation of=20 twin spots during Drosophila development (Stern, 1936). = B. Twin spots have proven useful in the = analysis of=20 the developmental effects of mutations which may otherwise cause = lethality if=20 inherited in homozygous form in the zygote. VIII. Human Gene = Linkage.=20 (Hartwell, pp. <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:22>392-33</DEL><INS = cite=3Dmailto:MHE dateTime=3D2007-02-08T11:22>416-417</INS>; = Thompson,=20 Chapters 10-12) A. Information from = Traditional=20 Linkage Data. It is obviously difficult to generate large numbers of = progeny and=20 perform selective matings in humans. = Nevertheless,=20 information on linkage can be obtained from pedigree data using = statistical=20 techniques. B. Lod Scores. = <SPAN=20 class=3DSpellE>Lod score analysis is a statistical technique that = permits=20 linkage relationships to be derived based on relative probabilities of = linkage=20 given the segregation patterns observed in pedigrees. The statistical = data=20 handling is beyond the scope of this course; for those who are = interested, I can=20 provide you with a more detailed description of the analysis method. = Briefly,=20 linkage data can be represented as a likelihood ratio, which represents = the=20 probability that two alleles are linked at some given recombination = frequency=20 (q = ), divided=20 by the probability that the alleles assort independently (RF =3D 0.5). = The data is=20 usually presented as the log of the likelihood ratios (<SPAN=20 class=3DSpellE>Lod Score): Zq =3D log=20 10 (Probability of family for <SPAN=20 style=3D"FONT-FAMILY: Symbol">q =3D 0.01, 0.02, etc.)=20 &nb= sp; =20 (Probability of family for q = =3D 0.50)=20 Lod scores = from individual=20 experiments can be calculated and are additive; it is possible to arrive = at=20 linkage probabilities without knowledge of the phasing relationships in = the=20 heterozygous parent. A likelihood ratio ratio of=20 1000:1 (Lod score =3D 3) is usually = considered "proof"=20 of linkage. IX. Assigning Genes to = Chromosomes=20 Using Somatic Cell Genetics. (see Thompson, = chapter 10)=20 A. Production of Hybrid = Cells.=20 Usually a rodent tumor cell line is fused with human fibroblast cells. = Cell=20 fusion is promoted by either viral infection or chemicals (polyethylene = glycol;=20 PEG). 1. HAT Selection = System. Used to=20 enrich for hybrid cells and eliminate parental cells and same-parent = fusions=20 from the culture. Growth medium contains a chemical (<SPAN=20 class=3DSpellE>aminopterin) to shut down the major pathway for = nucleotide=20 precursor synthesis- forcing the use of a salvage pathway to produce the = subunits needed for DNA. The "salvage" molecules hypoxanthine and <SPAN=20 class=3DSpellE>thymidine are also present in the growth medium = and can be=20 incorporated into DNA, provided the cells have two salvage pathway = enzymes-=20 HGPRT (hypoxanthine-guanine phosphoribosyltransferase)=20 and TK (thymidine kinase)=20 necessary to convert them into the appropriate DNA precursors. By using = parental=20 lines deficient (i.e. mutant) for one or the other of the two enzyme = activities,=20 (e.g. human <SPAN=20 style=3D"FONT-SIZE: 10pt">HGPRT-/TK+; = rodent /<SPAN=20 style=3D"FONT-SIZE: 10pt">HGPRT+/TK-; = hybrid <SPAN=20 style=3D"FONT-SIZE: 10pt">HGPRT+/TK+) one = can select=20 for the rare cell fusions that will produce hybrid cells = (complementation). 2. The hybrid cells will eventually = undergo=20 nuclear fusion to form heterokaryons. This = process=20 will lead to a spontaneous loss of human chromosomes from the hybrid = cells. The=20 chromosomes will be lost at random, giving rise to colonies which will = contain=20 different portions of the human chromosome complement. 3. The presence or absence of a human = gene product=20 can then we assayed in the hybrid cells. Only those hybrid cells = containing the=20 human gene (and the chromosome on which it resides) would be expected to = produce=20 the human product. Concordance and discordance between the presence or = absence=20 of the human gene product (enzyme or protein) and the human chromosomes = a hybrid=20 cell line carries establishes what chromosome the gene resides on (synteny). 4. Once synteny is=20 established, the cell lines carrying fragments of the <SPAN=20 class=3DSpellE>chromsome in question can also be assayed. 5. The above technique requires that we = be able to=20 identify individual human chromosomes. It also assumes that we can = identify a=20 human gene product in the hybrid cell. We can do both. Indeed, the = nucleic acid=20 sequence of the gene itself is sufficiently different to distinguish = mouse from=20 human genes, allowing DNA gene probes to be used for mapping = purposes. X. Aberrant <SPAN=20 class=3DSpellE>Asci Patterns, Gene Conversion and Recombination = at the=20 Molecular Level.(Hartwell, <DEL=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:22>178-186</DEL><SPAN=20 class=3DmsoIns><INS cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:22>191-200</INS>) A. Aberrant <SPAN=20 class=3DSpellE>asci patterns (e.g. 3:1:1:3; 6:2; 5:3) have led to = insights=20 into the molecular nature of recombination. The recovery of products of = a=20 meiotic division in a heterozygote at some ratio other than the expected = 1a:1A=20 resulting from reciprocal recombination is known as gene = conversion.=20 Because we can trap all the meiotic products of the same division cycle, = and=20 because the mitotic division following meiosis allows us to analyze the=20 individual DNA molecules that underwent recombination in each tetrad, we = have=20 been able to develop models of how recombination occurs. Many of these = models=20 have common features which can explain the origin of aberrant <SPAN=20 class=3DSpellE>asci during fungal meiosis. These models suggest a = process=20 of DNA nicking, strand invasion, heteroduplex=20 formation and DNA repair occurring at the recombination site. = Aspects of=20 these models have been verified by observing recombination in other = model=20 systems (e.g. bacteria). The Holliday Model is one such model = developed=20 to explain gene conversion. Features of the model and how it can = explain an=20 aberrant ascus pattern are covered in pp. = <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>178-186</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:23>191-200</INS> of=20 Hartwell. I. Overview of Mutation = Section A. Mutation: there is = no genetics=20 without variation B. Gene or "Point" Mutation: allelic = variations=20 occuring within a single gene locus, usually = resulting=20 in a change or loss of a small number of nucleotides. (Hartwell, = Chapter=20 7) C. Chromosomal Mutation: a change in a = segment,=20 chromosome or set of chromosomes, which may, or may not include a change = within=20 a single gene locus. 1. Aberrations due to = chromosome=20 structure (Hartwell, Chapter <DEL cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:23>13</DEL><INS = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>14</INS>) 2. Aberrations due to chromosome number = (Hartwell,=20 Chapter <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>13</DEL><INS = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>14</INS>) D. We will cover this = section in=20 the order given below, first discussing chromosome structure and changes = that=20 can occur at the chromosome level. We will return to a discussion of = mutation at=20 the allele level prior to a discussion of bacterial genetics, and = emphasize the=20 important role bacterial genetics had in laying the groundwork for = molecular=20 genetics- the study of information transfer at the molecular level. II. Techniques Used to = Visualize=20 Chromosome Structure: Karyotype analysis = (Hartwell,=20 pp. <DEL cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:23>81;=20 422-24</DEL><INS cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>84; 472-474</INS>) A. Chromosomes are a = complex of=20 primarily DNA and protein (chromatin). <SPAN=20 class=3DGramE>euchromatin: lightly staining, genetically = active;=20 heterochromatin, darkly staining, genetically inactive. <SPAN=20 class=3DGramE>(Hartwell, <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:24>422-24</DEL><INS = cite=3Dmailto:MHE dateTime=3D2007-02-08T11:24>472-474</INS>) for = definitions=20 and description). B. During interphase,=20 chromosome structure is not resolved due to <SPAN=20 class=3DSpellE>decondensed state of chromatin. C. DNA becomes highly compacted during = process of=20 mitosis; chromosome structure can be visualized. D. In humans, a cell population may need = to be=20 stimulated to start mitosis before analysis can be performed (e.g. white = blood=20 cells (lymphocytes) from peripheral blood; or fetal fibroblasts (<SPAN=20 class=3DSpellE>amniocytes) taken from amniotic fluid). 1. <SPAN=20 class=3DSpellE>Interphase cells are induced to enter mitosis by = stimulation=20 with a mitogen (e.g. <SPAN=20 class=3DSpellE>phytohemagluttinin). 2. Cell may be stopped in metaphase by = disrupting=20 spindle fiber formation (tubulin = polymerization) with=20 the chemical colchicine. 3. The cells must be prepared to = preserve=20 chromosome structure and specifically disrupted to spread the = chromosomes. III. Basic Chromosome = Morphology=20 A. Size B. Centromere = position=20 (metacentric, <SPAN=20 class=3DSpellE>submetacentric, acrocentric, <SPAN=20 class=3DSpellE>telocentric). C. Other distinguishing features such as = nucleolus=20 organizer regions (secondary constrictions, site of ribosomal RNA = genes. IV. Differential = Banding=20 Techniques. A. It is possible to = identify every=20 human chromosome on the basis of a distinctive banding pattern revealed = by=20 partially disrupting chromosome and staining with specific dyes. B. Several Methodologies: require = treatments with a=20 variety of agents that will partially disrupt chromatin structure = (acids, bases,=20 proteolytic enzymes, heating). C. Most common techniques: 1. G Banding: standard = for=20 comparisons (Paris Convention, 1971). 2. Q Banding: fluorescent staining: = similar to G=20 Band pattern. 3. R Banding: banding pattern appears = reversed=20 from G Band pattern. 4. C Banding: will stain only the = heterochromatin=20 surrounding the centromere. D. Normal Standard of = <SPAN=20 class=3DGramE>Resolution : ~ 300-500 bands at metaphase; ~1200 = bands at=20 prometaphase, when chromatin is less = condensed. (The=20 most recent estimates of the number of genes within human genome <SPAN=20 class=3DGramE>is on the order of 25,000 (revised since Hartwell = was printed=20 in 2004 and still not a certainty); so this level of cytological = resolution is=20 obviously limited in its ability to detect structural defects that may = exist in=20 chromosome.) E. Convention for designating sites on = chromosomes.=20 Sites are designated by first giving the chromosome number, followed by = the=20 chromosome arm (p=3D "petite" or short arm; q =3D long arm). Chromosome = regions are=20 identified on the basis of "landmark bands" which define <SPAN=20 class=3DSpellE>chromsome regions; these regions are numbered away = from the=20 centromere. The band found within a region = is given=20 last; the individual bands are also numbered away from the <SPAN=20 class=3DSpellE>centromere:<IMG id=3D_x0000_i1028 height=3D259=20 src=3D"http://www.ou.edu/cas/zoology/Courses/3333/Image7.gif" = width=3D142> e.g. In the = above example,=20 the arrow would be located at 8q22 in the human chromosome = complement. V. Morphology of Human = Sex=20 Chromosomes. A. Mammalian Y = Chromosomes.=20 (Hartwell, pp. <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:24>79-82; 103-4</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:24>82-86; = 110-112</INS>) 1. Apparently does not = contain=20 many genes, other than those involved in human sex determination, or = sperm=20 fertility factors. 2. The chromosome contains much = chromatin that=20 appears genetically inert (consitutive=20 heterochromatin). There is much population polymorphism in the size of = the long=20 arm of the Y. 3. Pairs with the X chromosome during = meiosis=20 along a small region that is homologous to the X (<SPAN=20 class=3DSpellE>pseudoautosomal region). 4. TDF gene serves as a major = developmental switch=20 for determining maleness. B. = Mammalian X=20 Chromosomes. (Hartwell, pp. <DEL = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:24>79-82; 431-32</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:24>82-86; 4</INS><INS cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:25>81-<SPAN=20 class=3DGramE>482;) </INS> 1. Females have two = copies of the=20 X; males have only one; Is there a dosage = compensation=20 mechanism that operates to control and normalize levels of X chromosome = gene=20 products? 2. Interphase cells of=20 females show a dark body (Barr body) usually associated with the = periphery of=20 nucleus; males have no Barr body; individuals with extra X chromosomes = show=20 extra Barr bodies, such that the number of Barr bodies observed are 1 = less that=20 the total number of X chromosomes. 3. <st1:place = w:st=3D"on">Lyon</st1:place>=20 Hypothesis: In a normal female (46; XX), one X chromosome becomes = functionally=20 inactive early in embryogenesis (facultative heterochromatin). The = inactivation=20 process is random, and the chromosome that is inactivated remains so in = all=20 progeny cells. Since inactivation is a random event, all females are=20 functionally mosaic; some cells contain one inactivated X, some cells = the other.=20 e.g. X linked coat color genes in mice; anhydrotic dysplasia = in=20 humans. C. Genomic Imprinting. = (Hartwell,=20 pp. <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>596-602</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:25>660-664</INS>) 1. In mammals, a small = number of=20 genes are passed on from the parent to the zygote in an inactive form; = embryos=20 therefore inherit only one active gene. 2. Some of = these genes=20 are inactivated during spermatogenesis; some during <SPAN=20 class=3DSpellE>oogenesis. 3. If one parent passes a wild-type = inactive gene=20 and the other parent passes a mutant allele, the resultant individual = will show=20 a mutant phenotype. e.g. <SPAN=20 class=3DSpellE>Praeder-Willi Syndrome (maternal gene inactive); = <SPAN=20 class=3DSpellE>Angelman Syndrome (paternal gene inactive). 4. Why would one gene be specifically = inactivated=20 by a parent? One theory suggests that imprinting is related to = competition=20 between embryo and mother for resources. According to this theory, genes = that=20 are passed in an active form by males would lead to increased growth = during=20 embryogenesis; genes that are passed in an active form by females would = tend to=20 retard embryonic growth. VI. Departures from = <SPAN=20 class=3DSpellE>Diploidy: Heritable Aberrations in Chromosome = Number;=20 Humans. (Hartwell, Chapter <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>13</DEL><INS = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>14</INS>) A. <SPAN=20 class=3DSpellE>Aneuploidy: a chromosome number that is not an = exact=20 multiple of the haploid chromosome number. <SPAN=20 class=3DSpellE>Aneuploidies usually arise from de novo errors = (<SPAN=20 class=3DSpellE>nondisjunction) during cell division. B. In humans, aneuploidy=20 involving sex chromosomes are usually the least severe of the possible = <SPAN=20 class=3DSpellE>aneuploidies: 1. Turner Syndrome = (45, X) 2. Klinefelter Syndrome=20 (47, XXY). 3. Multiple Y males (47, XYY). 4. Poly X females (47, XXX; 48, XXXX, = etc.) C. In humans, <SPAN=20 class=3DSpellE>aneuploidy involving autosomes is=20 usually fatal; all autosomal <SPAN=20 class=3DSpellE>monoploidy is fatal; most <SPAN=20 class=3DSpellE>trisomies are fatal with the exception of <SPAN=20 class=3DSpellE>Patau, Edwards and Down Syndrome. 1. Patau=20 Syndrome (47, +13): seen approximately 1/4000 births; severe mental = retardation=20 and physiological defects. Usually fatal in early=20 infancy. 2. Edwards Syndrome (47, +18): seen in=20 approximately 1/7000 births; severe mental retardation and other = physiological=20 defects. Usually fatal in early infancy. 3. Down = Syndrome (47,=20 +21): seen in approximately 1/800 births; physical signs in infants are = lowered=20 muscle tone, protruding tongue, characteristic folds at eye; spots on = iris,=20 characteristic fingerprints (dermatoglypics). Children=20 are at risk for increased cardiac defects, severe mental retardation; = older=20 individuals will develop early onset Alzheimer disease. 4. There is a pronounced increase in = risk for=20 chromosomal aneuplodies in older women. VII. Structural = Chromosomal=20 Aberrations (Hartwell, Chapter <DEL = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>13</DEL><INS = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>14</INS>) A. Factors Generating = Chromosomal=20 Damage (clastogenic agents) 1. <SPAN=20 class=3DGramE>ionizing radiation. 2. some viral = infections.=20 3. chemicals. B. Breakage Events. 1. Telomeres <SPAN=20 class=3DGramE>represents "proper" chromosomal ends; they are = specialized=20 structures that with a highly conserved DNA sequence which marks the = natural end=20 of a chromosome. 2. If a break is introduced, new "ends" = are not=20 properly terminated and will be subject to DNA repair mechanisms. The = joining of=20 inappropriate ends can lead to structural abnormalities. C. Reciprocal = Translocations: a=20 reciprocal exchange between non-homologous chromosomes. 1. If no genes are = damaged at the=20 site of breakage, then individuals heterozygous for the reciprocal = translocation=20 (i.e. carrying two translocation chromosomes and two normal chromosomes) = may=20 show no overt abnormal phenotype. 2. Pairing problems during meiosis, = however, will=20 lead to the formation of tetravalents; = abnormal=20 segregation from this structure (adjacent segregation) will lead to = unbalanced=20 gametes carrying duplications and deficiencies for certain genes. The = result is=20 semisterility in the translocation = heterozygote. 3. Robertsonian=20 Translocations: a specific type of reciprocal translocation, where = breaks are=20 introduced to opposite sides of the centromeres of two=20 non-homolog acrocentric chromosomes. Repair = leads to=20 formation of a small chromosome which is usually lost,=20 and the fusion of the two long arms into a single chromosome. a) Often=20 seen in closely related species. For example, the <SPAN=20 class=3DSpellE>metacentric chromosome 2 of humans resembles in = banding=20 pattern and gene content two acrocentric = chromosomes=20 seen in chimpansees, suggesting a <SPAN=20 class=3DSpellE>robertsonian fusion occurred during primate = evolution. b) Such translocations can also been seen = within=20 species; 14:21 translocations are associated with a heritable form of = Down <SPAN=20 class=3DGramE>Syndrome. c) Reciprocal Translocations and = Cancer: If=20 the breakpoint occurs within a gene (damage at the site of breakage), or = transfers a gene into a region which affects its ability to be = expressed, this=20 can lead to somatic mutations in genes controlling normal growth and = cell=20 proliferation. 1) <SPAN=20 class=3DGramE>formation of hybrid genes with abnormal phenotypes: = in=20 >95% of cases of chronic myelogenous = leukemia, the=20 tumor cells contain a clear reciprocal translocation within the <SPAN=20 class=3DSpellE>abl gene locus that leads to the conversion = of this=20 "proto-oncogene" into an "oncogene". 2) the = movement of the=20 myc gene into the region of the immunglobulin gene cluster produces a change in = its=20 expression pattern which is correlated with a cancer known as <SPAN=20 class=3DSpellE>Burkitt Lymphoma. This activation due to change in = location=20 is an example of position-effect variegation. D. Deletions or = Deficiencies 1. A loss of a small = region of the=20 chromosome may be correlated with abnormalities arising from the = imbalance; most=20 large deletions are inviable. e.g.=20 (46; 5p-) cri du chat syndrome. 2. A loss of a small region may place an = individual at risk if the normal gene copy located on the homolog = chromosome=20 mutates. e.g. Hereditary predisposition to = retinal=20 cancers for individuals deleted for 13q14, the site of the = retinoblastoma (<SPAN=20 class=3DSpellE>Rb) gene locus. (Hartwell, pp. <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:26>634-43</DEL><INS = cite=3Dmailto:MHE dateTime=3D2007-02-08T11:26>702-709</INS>) E. Inversions 1. Once again, if no = genes are=20 damaged at the breakpoint, inversion heterozygotes may=20 only encounter problems relative to chromosome segregation during = meiosis. 2. If crossing over occurs within = inversion, it=20 will produce chromosomes with duplications and deficiencies. a. <SPAN=20 class=3DSpellE>paracentric inversions: centromere=20 not included in the inversion; also produces acentric=20 and dicentric chromosomes. b. <SPAN=20 class=3DSpellE>pericentric inversions: centromere=20 is included in the inversion; recombinant products have <SPAN=20 class=3DSpellE>centromeres but still carry duplications and = deletions. F. Fragile Sites: = X-linked Mental=20 Retardation (Hartwell, <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:26>202-203</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:26>216-217</INS>). = 1. Fragile sites can = be detected=20 by chromosome breakage when cells are cultured under special conditions. = The=20 chromosomal marker is a characteristic of Fragile X syndrome, present in = 1/1000=20 male births and one of the most common forms of mental retardation in = males (~8%=20 of all males with mental retardation). 2. In Fragile X Syndrome the defect is = linked to=20 an unstable replication of a triplet repeat in the FMR-1 gene, located = at the=20 fragile site. An amplification of this sequence from a "<SPAN=20 class=3DSpellE>premutation" leads to the full symptoms of this=20 syndrome. G. Monitoring of = Chromosomal=20 Defects. 1. Amniocentesis. = <SPAN=20 class=3DGramE>Usually performed at ~16 weeks, when fibroblasts can be = collected=20 safely from amniotic fluid. Requires <SPAN=20 class=3DSpellE>mitogenic stimulation of cells prior to = analysis.=20 2. Chorionic = <SPAN=20 class=3DSpellE>Villus Sampling. Can be = performed at ~9th=20 week, from embryonic cells that are already actively dividing.=20 VIII. Departures from = <SPAN=20 class=3DSpellE>Diploidy in Plants. A. Tolerances for <SPAN=20 class=3DSpellE>aneuploidy and polyploidy contribute to speciation = and=20 facilitate genetic manipulations. 1. Plants don't = necessarily need=20 to propagate via sexual reproduction; they can propagate from somatic = tissues=20 (e.g. cuttings). 2. Increases in ploidy=20 may lead to aggricultural benefits (e.g. = increases in=20 plant size). 3. Monoploids can be=20 generated from gametes and used in selection experiments. 4. Chromosome number can be doubled by = <SPAN=20 class=3DSpellE>colchicine treatment. B. <SPAN=20 class=3DSpellE>Polyploid species: (number of complete chromosome = sets >=20 2) 1. <SPAN=20 class=3DSpellE>Autopolyploids: contain multiples of the same = <SPAN=20 class=3DGramE>chromosome set (same species). 2. Allopolyploids: contain <SPAN=20 class=3DSpellE>mutiples of different chromsome=20 sets (different species). 3. Hybridizations may lead to pairing = problems=20 during meiosis. a.=20 triploids are often sterile- difficulties in meiotic segregation. b. a doubling = of the=20 different chromosome sets in allopolyploids can relieve problems during = meiosis:=20 if there are pairing partners for homologs, = sterility=20 often relieved. (e.g. <SPAN=20 class=3DSpellE>Karpechenko, radish and cabbage <SPAN=20 class=3DSpellE>polyploids). <SPAN=20 class=3DGramE>amphidiploid =3D contains the diploid = chromosome=20 complement of both parent species. c. polyploidy = and=20 evolution: origin of common cultivated wheat, an <SPAN=20 class=3DSpellE>allohexaploid. I. Gene Mutation: there is no genetics without = allelic=20 variation (Hartwell, Chapter 7) A. Classes of Mutation = 1. Somatic; <SPAN=20 class=3DSpellE>clonal populations can arise due to mutations = <SPAN=20 class=3DSpellE>occuring during development; e.g. sectors of color = in plant=20 leaves, cancer (somatic mutation of proto-oncogenes=20 [normal cellular genes involved in cellular growth regulation] to <SPAN=20 class=3DSpellE>oncogenes [mutant genes causing abnormal cell = proliferation=20 or metastasis]). 2.Germline; capable=20 of being transmitted to the next generation. B. "Point Mutations": = mutant=20 phenotypes 1. <SPAN=20 class=3DSpellE>Heteromorphs (null, loss-of-function, = gain-of-function) 2. Lethals = 3. Conditional mutants (restrictive vs. = permissive=20 conditions) 4. Biochemical mutants a. <SPAN=20 class=3DSpellE>prototrophs b. <SPAN=20 class=3DSpellE>auxotrophs 5. Resistance mutants = a.=20 antibiotic resistance (e.g. strr) = b. viral = resistance=20 (tonr) C. Selection Systems = and Mutant=20 Detection 1. Advantage of microbes = a.=20 haploid b. fast = generation times=20 c. = inexpensive 2. Eukaryotic systems = (e.g. <SPAN=20 class=3DSpellE>ClB of Muller) D. Nature of Genetic = Change;=20 random mutation/selection or physiological adaptation? (Hartwell, pp. = <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:26>194-196</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:26>210-212</INS>) 1. <SPAN=20 class=3DGramE>tonr mutants: resistance to T1 = <SPAN=20 class=3DSpellE>bacteriophage infection. 2. Delbruck/Luria fluctuation test: = "Mutations of=20 Bacteria from Virus Sensitivity to Virus Resistance": (1943) 3. Lederberg replica plating experiment: = (1952) 4. 1945: Schrodinger's "What is Life?" = published.=20 How do living organisms order themsevles? Do = molecules=20 govern heredity? What is the physical nature of the molecules that = govern=20 heredity? "...likely to involve hitherto unknown 'other laws of = physics,' which,=20 however, once they have been revealed will form just as integral a part = of this=20 science as the former." The influx of individuals trained in physics = provided a=20 basis for the revolution in molecular biology. I. Recombination in Bacteria (Hartwell, Chapter = <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:26>14</DEL><SPAN=20 class=3DmsoIns><INS cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:26>15</INS>)=20 A. How are genes in = bacteria=20 organized? Are there correlates to eukaryotic chromosomes? Is there = linkage? How=20 universal are information storage and retrieval mechanisms in Nature? = B. No genetics without allelic = variation. What=20 phenotypes do we study? II. Mechanisms for Gene Exchange and Recombination = in=20 Bacteria A. Conjugation: = transferred=20 through direct cell-cell contact. B. Transduction: transferred through = viral=20 infection. C. Transformation: Cellular uptake of = DNA from the=20 environment. III. Conjugation A. Lederberg and Tatum = (1946).=20 Using bacteria that were multiply auxotrophic,=20 demonstrated that recombination could occur. B. Davis (1950). The strains used by = Lederberg and=20 Tatum required physical contact for transfer. The = U-tube=20 experiment. C. Hayes (1953). There is a = directionality of=20 transfer. In this process, there is a donor/recipient relationship. = Streptomycin=20 treatment: will allow a donor cell to transfer DNA to a recipient, but = <SPAN=20 class=3DSpellE>str-treated cells will not be capable of = reproducing: 1. Donor (treat c <SPAN=20 class=3DSpellE>str)-->D* (wash out = drug and=20 cross c recipient)--> D* + R=3D recombinants 2. Recipient (treat c str)=20 --->R* (wash out drug and cross to donor)--> D +=20 R*=3D no recombinants. 3. Idea of unidirectional transfer: = D=3D"males";=20 R=3D"females"; crosses =3D "matings" D. Attributed = "maleness" to a=20 specific genetic entity; the "Fertility" Factor. 1. F+ cells would = transfer the=20 genotype of maleness at high frequency. 2. F+ cells would transfer other genotypes = (e.g.=20 prototrophy, phage resistance) at low = frequency. 3. Cavalli = (1950):=20 isolated a peculiar strain of male cell that produced <SPAN=20 class=3DSpellE>prototrophic recombinants at 1000 times the = frequency of=20 normal F+ strains. a. Termed an <SPAN=20 class=3DSpellE>Hfr strain: "high frequency of recombination". = b. The high frequency donor character = pertains=20 only to a limited portion of donor genome. c. Unlike F+ cells, Hfr=20 cells do not transfer male genotype to recipient cells. d. F+ ---> Hfr change=20 attributable to a heritable change in the sex factor's properties. Nature of change? E. Oriented Gene = Transfer and=20 "Time of Entry" Mapping Experiments. 1. Wollman=20 and Jacob: Interrupted mating experiments. Hfr x R=20 cell matings could be conducted under timed = conditions=20 to study the kinetics of gene transfer. 2. Waring = blender=20 experiment: Set up matings, allow <SPAN=20 class=3DSpellE>matings to occur for defined time intervals, = disrupt the=20 matings with a Waring=20 blender, then plate to select recombinants. 3. Time of Entry Mapping: By plotting = frequency of=20 recombinants recovered vs. time, they could extrapolate the "time of = entry",=20 which could be used as an index of distances between genes. 4. Analysis of different <SPAN=20 class=3DSpellE>Hfr strains (isolated from F+ populations by = replica-plate=20 matings on selective media) led to the = conclusion that=20 transfer could occur from different points in the genome in either of = two=20 different directions. How can this be explained by a consistent model? = 5. Both the bacterial chromosome and the F = factor=20 were covalently closed circular molecules. Integration of F factor into = the=20 chromosome DNA led to Hfr. Orientation and = site of the=20 integration event dictated what chromosomal genes were transferred. F. F' factors. 1. Occasionally, F = factors will=20 become unintegrated from bacterial = chromosome in an=20 Hfr cell. This will create an F factor that = can carry=20 a portion of the bacterial genome (F' factor) = . If it=20 conjugates with a recipient cell, that cell receives a chromosomal gene. = 2. F' cells carry two copies of a portion = of the=20 genome (merodiploids). There are also = capable of=20 transferring that gene at high frequency to recipient cells. G. Mapping by = selecting for late=20 markers. 1. Time of entry mapping = is not very=20 high resolution and is not very useful for mapping genes relatively = close to one=20 another. 2. By selecting for a late marker, closely = linked=20 genes can be mapped by examining the recombination frequency. IV. Transduction A. Lytic=20 Cycle: upon entry into cell, virus replicates, packages and <SPAN=20 class=3DSpellE>lysis host. B.Generalized=20 Transduction and Mapping 1. During <SPAN=20 class=3DSpellE>lytic infection, host genetic material may be = packaged. 2. On infection, host genetic material is=20 transferred into new cell, where it may recombine into genome. 3. Since phage particle can only contain = limited=20 amounts of DNA, technique can be used for mapping. <SPAN=20 class=3DGramE>"Co-transduction". C. <SPAN=20 class=3DSpellE>Lysogeny and Specialized Transduction 1. Some phage can = integrate into the=20 host cell genome and replicate along with host chromosome; will not = <SPAN=20 class=3DSpellE>lyse the host cell. 2. Can also leave host chromosome; if = <SPAN=20 class=3DSpellE>unintegration event is imprecise, host genetic = material may=20 be included. 3. These phage can now transfer host genes = into new=20 cell on infection. V. Transformation. A. Certain bacterial = strains=20 incorporate DNA from the environment through their membranes. This DNA = can=20 recombine into the genome. B. To pick up DNA, cells must be in a = "competent"=20 physiological state. C. "Co-transformation" frequencies can = be used to=20 map the location of genes relative to one another. VI. Bacteriophage = Genetics. A. Phage Phenotypes = defined by=20 ability to grow on a particular host, or by plaque morphology. B. Mixed infections between mutant = strains can be=20 used to map genes. I. Biochemical Nature of the Gene (Hartwell, = Chapter 6) A. The "Transforming = Principle".=20 Experiments with mutant strains of Streptococcus = <SPAN=20 class=3DSpellE>pneumoniae (pneumococcus), F.=20 Griffith, 1928. B. Avery, MacLeod and McCarty (1944). = <SPAN=20 class=3DSpellE>Biochemically fractionated <SPAN=20 class=3DSpellE>pneumococcal extracts and tested chemicals for = ability to=20 tranform. DNA =3D transforming principle. = C. Hershey-Chase Experiment (1952). The = material=20 injected into a bacterial cell by a phage programs the phage = reproductive cycle=20 in a lytic infection. That material is = DNA. II. Reluctance to Accept DNA as the Genetic = Material. A. DNA structure = originally=20 considered a monotonous polymer. 1. base=20 composition 2. phosphate-sugar backbone 3. strand = polarity B. Equivalence Rule of = <SPAN=20 class=3DSpellE>Chargaff (1950). III. X-Ray Crystallography A. Three groups = working on=20 developing 3-D reconstruction: Wilkins and Franklin, Watson and Crick, = (at=20 <st1:place w:st=3D"on"><st1:City w:st=3D"on">Cambridge</st1:City>,=20 <st1:country-region = w:st=3D"on">England</st1:country-region></st1:place>) and=20 Pauling (CalTech). B. Franklin produces first high = resolution images.=20 C. Watson and Crick incorporate density = data, base=20 composition and crystallography into their model. 1. Watson-Crick = Structure. <SPAN=20 class=3DGramE>The double helix. 2. Implications of Model. "It has not = escaped our=20 notice that the specific pairing mechanism we have proposed immediately = suggests=20 a possible copying mechanism for the genetic material"- conclusion of = 1953=20 Nature article proposing DNA structure. 3. B-form DNA relative to A-form, = Z-form. IV. DNA Replication. A. Models for DNA = replication.=20 B. Proof of the semi-conservative model: = The <SPAN=20 class=3DSpellE>Meselson-Stahl Experiment (1958). C. Origins of Replication: <SPAN=20 class=3DSpellE>Autoradiographic evidence for bidirectional = replication. V. Synthesis at the Replication Fork. A. Properties of DNA =

From: Subject: 3333outline Date: Wed, 28 Feb 2007 09:32:17 -0600 MIME-Version: 1.0 Content-Type: multipart/related; type="text/html"; boundary="----=_NextPart_000_0000_01C75B1B.566C4810" X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 This is a multi-part message in MIME format. ------=_NextPart_000_0000_01C75B1B.566C4810 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Content-Location: file://C:\Documents and Settings\david\Local Settings\Temporary Internet Files\OLK58E\3333outlineWITHtrackchanges.html 3333outline</TI= TLE> <META http-equiv=3DContent-Type content=3D"text/html; = charset=3Diso-8859-1"> <META content=3DWord.Document name=3DProgId> <META content=3D"MSHTML 6.00.6000.16414" name=3DGENERATOR> <META content=3D"Microsoft Word 11" name=3DOriginator><LINK=20 href=3D"3333outlineWITHtrackchanges_files/filelist.xml" = rel=3DFile-List><LINK=20 href=3D"3333outlineWITHtrackchanges_files/editdata.mso" = rel=3DEdit-Time-Data><o:SmartTagType name=3D"Sn"=20 namespaceuri=3D"urn:schemas:contacts"></o:SmartTagType><o:SmartTagType=20 name=3D"country-region"=20 namespaceuri=3D"urn:schemas-microsoft-com:office:smarttags"></o:SmartTagT= ype><o:SmartTagType=20 name=3D"place"=20 namespaceuri=3D"urn:schemas-microsoft-com:office:smarttags"></o:SmartTagT= ype><o:SmartTagType=20 name=3D"City"=20 namespaceuri=3D"urn:schemas-microsoft-com:office:smarttags"></o:SmartTagT= ype><o:SmartTagType=20 name=3D"State"=20 namespaceuri=3D"urn:schemas-microsoft-com:office:smarttags"></o:SmartTagT= ype> <STYLE>@font-face { font-family: Tahoma; } @page Section1 {size: 8.5in 11.0in; margin: 1.0in 1.25in 1.0in 1.25in; = mso-header-margin: .5in; mso-footer-margin: .5in; mso-paper-source: 0; } P.MsoNormal { FONT-SIZE: 12pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: "Times New Roman"; = mso-style-parent: ""; mso-pagination: widow-orphan; = mso-fareast-font-family: "Times New Roman" } LI.MsoNormal { FONT-SIZE: 12pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: "Times New Roman"; = mso-style-parent: ""; mso-pagination: widow-orphan; = mso-fareast-font-family: "Times New Roman" } DIV.MsoNormal { FONT-SIZE: 12pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: "Times New Roman"; = mso-style-parent: ""; mso-pagination: widow-orphan; = mso-fareast-font-family: "Times New Roman" } A:link { COLOR: blue; TEXT-DECORATION: underline; text-underline: single } SPAN.MsoHyperlink { COLOR: blue; TEXT-DECORATION: underline; text-underline: single } A:visited { COLOR: purple; TEXT-DECORATION: underline; text-underline: single } SPAN.MsoHyperlinkFollowed { COLOR: purple; TEXT-DECORATION: underline; text-underline: single } P { FONT-SIZE: 12pt; MARGIN-LEFT: 0in; MARGIN-RIGHT: 0in; FONT-FAMILY: = "Times New Roman"; mso-pagination: widow-orphan; = mso-fareast-font-family: "Times New Roman"; mso-margin-top-alt: auto; = mso-margin-bottom-alt: auto } P.MsoAcetate { FONT-SIZE: 8pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: Tahoma; = mso-pagination: widow-orphan; mso-fareast-font-family: "Times New = Roman"; mso-style-noshow: yes } LI.MsoAcetate { FONT-SIZE: 8pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: Tahoma; = mso-pagination: widow-orphan; mso-fareast-font-family: "Times New = Roman"; mso-style-noshow: yes } DIV.MsoAcetate { FONT-SIZE: 8pt; MARGIN: 0in 0in 0pt; FONT-FAMILY: Tahoma; = mso-pagination: widow-orphan; mso-fareast-font-family: "Times New = Roman"; mso-style-noshow: yes } SPAN.msoIns { COLOR: teal; TEXT-DECORATION: underline; text-underline: single; = mso-style-type: export-only; mso-style-name: "" } SPAN.msoDel { COLOR: red; TEXT-DECORATION: line-through; mso-style-type: export-only; = mso-style-name: "" } SPAN.msoChangeProp { mso-style-type: export-only; mso-style-name: "" } SPAN.SpellE { mso-style-name: ""; mso-spl-e: yes } SPAN.GramE { mso-style-name: ""; mso-gram-e: yes } DIV.Section1 { page: Section1 } OL { MARGIN-BOTTOM: 0in } UL { MARGIN-BOTTOM: 0in } </STYLE>  <META content=3DC:\MSOFFICE\OFFICE\html.dot name=3DTemplate></HEAD> <BODY lang=3DEN-US style=3D"tab-interval: .5in" vLink=3Dpurple = link=3Dblue> <DIV class=3DSection1> <SPAN=20 style=3D"FONT-SIZE: 13.5pt">This outline is designed to provide you with = a general=20 summary of the topics that will be discussed in class. It is not = designed to=20 substitute for your own in-class notes, or for the detailed discussions = of these=20 topics in the assigned readings. Interspersed within this outline are = references=20 to readings in the Hartwell text, where this subject material can be=20 found. = I<SPAN=20 style=3D"FONT-SIZE: 10pt">. Course Description A. Text B. Grading C. Course Schedule II. =20 What is a Gene? A. It must be = transmitted between=20 generations and each individual must have a physical copy of this = material. B. It must provide information to its = carriers with=20 respect to biological structure and function. III. =20 Historical Perspectives:PreMendelian = concepts=20 of inheritance A. Spontaneous = generation (till=20 1861!) B. Concepts emphasizing continuity between = generations 1. Early Greeks (~500 = B.C.) a. Pangenesis: somatic = tissue=20 contributes to heritable information b. Heritable contributions of males and = females=20 differ. 2. <SPAN=20 class=3DSpellE>Preformationism (1600's) 3. Studies of early embryogenesis and = the concept=20 of epigenesis (~1750-1880) 4. Charles Darwin's concepts of = inheritance (1859)=20 a. <SPAN=20 class=3DSpellE>gemmule theory b. the = observation of=20 continuous variation across the biological spectrum and an emphasis on = blending=20 of traits in offspring. C. <SPAN=20 class=3DSpellE>Gregor Mendel: a contemporary of Charles Darwin = (1822-1884)=20 1. Entered monastery and = was=20 educated at <st1:City w:st=3D"on"><st1:place=20 w:st=3D"on">Vienna</st1:place></st1:City>, Brunn. 2. Started experiments in peas in 1854 = 3. Published experiments in 1866 in a = journal=20 distributed to over 100 libraries in <st1:place = w:st=3D"on">Europe</st1:place>;=20 corresponded with major biologists of his day. 4. Assumed administrative responsibilities = as abbot=20 (1868); died 1884 in scientific obscurity. IV. = Principles of=20 Transmission Genetics: Mendel's experiments in Peas. A. His experimental = organism, the=20 common garden pea, very useful for genetic analysis. 1. Chose characteristics = that were=20 unambiguous. 2. Relatively quick reproductive cycle; = large=20 numbers of progeny. 3. Organism could be either = self-pollinated or <SPAN=20 class=3DSpellE>outcrossed (hybridized). 4. Took a quantitative approach to data=20 collection. B. The monohybrid = <SPAN=20 class=3DGramE>cross: principles of unit inheritance and = segregation. 1. If two plants with = differing=20 traits were crossed (P generation) the next generation (F1) = always=20 gave rise to plants displaying only one parental character. If=20 the F1 plants are now allowed to self-fertilize, the other = parental=20 character reappears in the next generation (F2), representing = 25% of=20 the offspring. 2. Generated a hypothesis consistent with = his=20 results. a. Traits could be = represented as=20 discrete, particulate entities or factors (represented by symbol). b. Factors appear to be paired, although = some=20 traits are not observed in hybrids even though the factor associated = with their=20 expression may be present in the hybrid (concept of dominant and = recessive=20 traits). c. Factors will randomly segregate = during=20 production of gametes; one factor is received from each parent; = restoration of=20 paired factors upon fertilization. C. The <SPAN=20 class=3DSpellE>dihybrid cross: <SPAN=20 class=3DSpellE>prinicple of independent assortment 1. Expanded his = observations,=20 watching the segregation of two traits simultaneously in the same cross. = 2. Results confirmed earlier observations. = a. Two traits observed = together=20 will partition independently of one another. b. observed = F2=20 ratios (9:3:3:1) were two independent 3:1 ratios. D. The testcross = (backcross):=20 Supported his basic hypothesis. 1. Backcross of = F1=20 monohybrid to single recessive parent produced expected 1:1 ratios. 2. Backcross of F1 <SPAN=20 class=3DSpellE>dihybrid to double recessive parent produced = expected=20 1:1:1:1 ratios. E. Basic Terminology = 1. Merkmal=20 (Mendel's German term) =3D factor =3D gene. 2. alleles: = variant genes=20 for any given trait; capital letter (A) represents dominant; small = letter (a)=20 represents recessive. 3. homozygote: = identical=20 pair at a given locus; AA or aa. (<SPAN=20 class=3DGramE>locus =3D location on a chromosome) 4. heterozygote: a <SPAN=20 class=3DSpellE>nonidentical pair at a given locus; <SPAN=20 class=3DSpellE>Aa. 5. genotype: = the genetic=20 composition or complement (for any given gene pair or locus) 6. phenotype: = the physical=20 appearance observed based on expression of the genotype. V. 19th Century Advances = in Cell=20 Biology: Indentifying the Physical Basis for = Mendel's=20 Laws. A. Nucleus and = discovery of cell=20 structure. (Brown, 1833) B. Fertilization and pronuclear fusion. = (<SPAN=20 class=3DSpellE>Hertwig et al. ~1875) C. Mitosis (Fleming et al. ~1880) D. Germplasm = theory=20 (Weismann, ~1880) E. Meiosis (Van Beneden,=20 Boveri et al. 1880s-early 1900s) F. Rediscovery of Mendel's Laws (<SPAN=20 class=3DSpellE>deVries, Correns, von = <SPAN=20 class=3DSpellE>Tschermak; 1900) VI. The Sutton-<SPAN=20 class=3DSpellE>Boveri Hypothesis (1903) A. Both chromosomes = and <SPAN=20 class=3DSpellE>Mendelian factors are found in pairs; of both = maternal and=20 paternal origin. B. Chromosomes of each pair separate = during=20 meiosis, as do Mendelian factors. C. On fertilization, the "<SPAN=20 class=3DSpellE>doubleness" or the chromosome complement is = restored, as are=20 Mendelian factors. VII. Segregation, = Independent=20 Assortment and Probability. A. Probability =3D = number of=20 anticipated events/total number of possibilities; e.g. the = probability of=20 pulling an ace out of a straight deck of cards: P(ace) =3D 4/52 or 1/13. = B. Product Rule: For independent = events,=20 the probability of several events occurring is the product of = their=20 individual probabilities; P(A and B) = =3D P(A) *=20 P(B). e.g. P (ace = of spades) =3D=20 P (ace) * P (spade) =3D 1/13 * 1/4 =3D 1/52 C. Sum Rule: For mutually exclusive = events,=20 the probability represents the sum of the individual = probabilities. P (A=20 or B) =3D P(A) + P(B). e.g. P (ace = or king) =3D=20 1/13 + 1/13 =3D 2/13 D. For <SPAN=20 class=3DSpellE>Mendelian patterns of inheritance, the events that = occur=20 during the process of meiosis and fertilization (segregation, = independent=20 assortment) can be expressed in probabilistic terms. 1. In a monohybrid = cross, the=20 fraction of progeny with a given phenotype follows the sum rule: <SPAN=20 class=3DGramE>P(AA or Aa) =3D 1/4 +=20 1/2 =3D 3/4; P(A_ or aa) =3D = 3/4 + 1/4. 2. Assuming independent assortment, a = <SPAN=20 class=3DSpellE>dihybrid cross would be expected to yield a = 9:3:3:1=20 phenotypic ratio- applying the product rule to two independent events. = P( A_ and=20 B_) =3D P(A_) * P(B_) =3D 9/16 A_B_ P (A_ and bb) =3D P(A_) *=20 P(bb) =3D 3/16 A_bb P (aa and = B_) =3D <SPAN=20 class=3DGramE>P(aa) * P(B_) =3D 3/16 = <SPAN=20 class=3DSpellE>aaB_ P (aa and = bb) =3D <SPAN=20 class=3DGramE>P(aa) * P(bb) =3D 1/16 = <SPAN=20 class=3DSpellE>aabb 3. In the cross <SPAN=20 class=3DSpellE>AaBbCc x <SPAN=20 class=3DSpellE>AaBbcc, what is the probability of = producing progeny=20 with the same phenotype as the first parent? (ans.=20 9/32) I. Documenting Patterns = of=20 Transmission for Single Gene Traits in Humans. A. Importance of Familiy Histories for Human Genetic Analysis B. Pedigree Symbols and Guidelines C. Pedigree Patterns for <SPAN=20 class=3DSpellE>Autosomal Dominant Inheritance. 1. Trait appears in = every generation=20 (exceptions; new mutation or incomplete dominance). 2. Trait is passed from affected person to = 1/2 of=20 progeny, regardless of sex. 3. Unaffected members will not pass on the = trait. D. Pedigree Patterns = for <SPAN=20 class=3DSpellE>Autosomal Recessive Inheritance. 1.Trait=20 appears to skip generations, often observed in sibs but not parents. = 2. On the average, the trait will show up = in 25% of=20 the sibs of an affected proband, <SPAN=20 class=3DSpellE>regarless of sex. 3. The parents may be consanguineous. E. Probability and = Risk=20 Assessment. <IMG id=3D_x0000_i1025 height=3D245=20 src=3D"http://www.ou.edu/cas/zoology/Courses/3333/Image3.gif" = width=3D192> 1. From pedigree, trait = appears=20 inherited as an autosomal recessive. 2. From phenotype of II-3, parents are = obligate=20 heterozygotes. 3. If parents are obligate <SPAN=20 class=3DSpellE>heterozygotes, III-2 and III-3 have a 2/3 = probability of=20 being heterozygotes. 4. If parents of IV-1 and IV-2 are <SPAN=20 class=3DSpellE>heterozygotes, IV-1 and IV-2 each stand a 1/2 = probability of=20 inheriting disease gene allele from their parents. 5. If they are <SPAN=20 class=3DSpellE>heterozygotes, they have a 1/4 probability of = having an=20 affected child. 6. The risk to IV-1 and IV-2 of having an = affected=20 child is therefore: 2/3*2/3*1/2*1/2*1/4 =3D = 1/36 II. Evidence for the = Chromosomal=20 Theory of Inheritance A. Sex Chromosome = Dimorphism: homo-=20 and heterogametic sexes. B. The white locus in <SPAN=20 class=3DSpellE>Drosphila: Thomas Hunt Morgan (1910). 1. Characteristic = segregation=20 pattern: white female x normal male shows criss-cross=20 inheritance. 2. Calvin Bridges: Correlation of = exceptional=20 progeny with nondisjunction during = meiosis. C. Sex Determination: = Nature of the=20 "Primary" Switch; mechanisms vary. (See Hartwell, = pp. <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:20>605-610</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:20>669, 672-677</INS> = for detailed=20 discussion of molecular mechanisms in Drosophila.) 1. Drosophila: = X<SPAN=20 class=3DGramE>:A ratio. <P=20 style=3D"MARGIN-LEFT: 2.5in; mso-list: none; mso-list-ins: MHE = 20070208T1120">2.=20 Mammals: TDF/SRY gene on Y chromosome. (See Hartwell, pp. <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:20>653</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:20>720</INS><SPAN=20 style=3D"BACKGROUND: windowtext; mso-highlight: windowtext"><SPAN=20 style=3D"BACKGROUND: yellow; mso-highlight: yellow">)<INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:20> Please note: part of this = discussion=20 was deleted from this chapter in the 3/e. =20 Please check</INS><INS=20 cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:20><o:p></o:p></INS> <o:p> </o:p> III. Patterns of = Sex-Linked=20 Expression in Humans. A. X-linked Recessive = Inheritance.=20 1. Incidence much = higher in males=20 than in females. 2. Affected male will give rise to = daughters who=20 are carriers. On average, 1/2 of a carrier female's sons will be = affected; half=20 of her daughters will also be carriers.. 3. Never father to son transmission. B. X-linked Dominant = Inheritance.=20 1. Affected males will = have no=20 affected sons and no normal daughters. 2. Transmission by females is not = distinguishable=20 from an autosomal dominant. 3. Since most females for <SPAN=20 class=3DSpellE>autosomal dominants are heterozygous and most = males will be=20 hemizygous, the trait is often more severe = in=20 males. C. <SPAN=20 class=3DSpellE>Holandric Inheritance. 1. Relatively rare in = human=20 populations. Y chromosome, relative to X, is gene poor (constitutive=20 heterochromatin). 2. TDF/SRY gene. D. Sex-limited and = Sex-Influenced=20 Traits. 1. Sex-limited; e.g. = precocious=20 puberty. 2. Sex-influenced; e.g. pattern = baldness. VI. Probability = Revisited. <SPAN=20 class=3DGramE>(N.B. See Chapter 5 of Thompson; this is not = formally=20 discussed in Hartwell) A. Use of the Binomial = Expansion.=20 1. In a three child = family, what=20 is the probability of two boys and 1 girl? P(boy) =3D = p=3D 1/2; P(girl)=20 =3D q=3D 1/2; n =3D number of events (births) =3D 3. All potential possiblities are represented by the binomial = expansion, with=20 the specific case in question emphasized: (p + q)3 =3D 1=20 or p3 <BLINK>+3p2q </BLINK>+ = 3pq2 +=20 q3 =3D 1; 3p2q =3D 3(1/2)(1/2)(1/2) =3D 3/8 = For any given term in the expansion = (e.g. in a=20 three child family, the odds of two boys and a girl) is also represented = by the=20 equation: <IMG id=3D_x0000_i1026 height=3D41=20 src=3D"http://www.ou.edu/cas/zoology/Courses/3333/Image4.gif" = width=3D58> where n =3D = number of=20 children =3D 3, P(boy) =3D p =3D 1/2; P(girl) =3D q =3D 1/2; s =3D = number of boys=3D 2; t =3D=20 number of girls =3D 1. Note that p + q =3D 1; s + t =3D n. n!/<SPAN=20 class=3DSpellE>s!t! =3D (3*2*1)/(2*1)(1) =3D 3; and=20 <SPAN=20 class=3DGramE>psqt =3D = (1/2) (1/2)=20 (1/2) =3D 1/8; therefore; = 3* 1/8 =3D 3/8 <P class=3DMsoNormal=20 style=3D"MARGIN-LEFT: 1.5in; TEXT-INDENT: -0.25in; mso-margin-top-alt: = auto; mso-margin-bottom-alt: auto; mso-list: l0 level3 lfo1; = mso-list-change: '%3:1:0:.' MHE 20070208T1119; tab-stops: list = 1.5in"><![if !supportLists]><SPAN=20 style=3D"mso-list: Ignore">1.<SPAN=20 style=3D"FONT: 7pt 'Times New Roman'"> =20 <![endif]>The binomial expansion is a special case of the = more=20 general multinomial distribution, where the model can be expanded to = include=20 more than two alternatives: <IMG id=3D_x0000_i1027 = height=3D41=20 src=3D"http://www.ou.edu/cas/zoology/Courses/3333/Image5.gif" = width=3D186> B. Testing Experimental = Data for=20 Conformity to an Experimental Hypothesis: the Chi Square Test. (See = Thompson,=20 ch. 5; discussed in Hartwell with regard to = linkage=20 analysis, pp. <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:21>117-120</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:21>127-131</INS>) 1. Using Mendel's data = set as an=20 example of observed vs. experimental data. 2. Concept of Degrees of Freedom; for = genetic=20 tests like these, the degree of freedom is one less than the number of = classes.=20 3. The chi square table and assessment = of the=20 probability the differences between the observed and expected values are = due to=20 chance. 4. The 0.05 criterion for level of=20 significance. I. Dominance = Relationships A. Complete Dominance = 1. Mendel's pea = experiments. 2. A. Garrod:=20 (1902-1909) a. Demonstrated that = human disorders=20 (e.g. alcaptonuria) were transmitted in = human=20 pedigrees according to Mendelian laws of = inheritance=20 (e.g. alcaptonuria shows a=20 autosomal recessive pattern of inheritance.) = b. Postulated that the reason for the = disease was a=20 metabolic defect that was genetically inherited- that genes are = ultimately=20 responsible for making enzymes and the genetic defect in <SPAN=20 class=3DSpellE>alcaptonuria was related to the absence of = functional enzyme=20 ("inborn error of metabolism"). Like Mendel, this postulate was ahead of = its=20 time, and not rigorously tested until the 1030s-40s. B. Incomplete Dominance = 1. Transmission = genetics could=20 demonstrate that some heterozygotes = displayed a=20 phenotype different from either parent, although pattern of segregation = followed=20 Mendel's laws. e.g. flower color in 4 o'clock = plants.=20 Pure breeding red and white individuals produce pink F1s. = Pink=20 F1 x Pink F1 produce 1 red: 2 pink: 1 white = offspring; red=20 and white breed true; pink phenotype due to heterozygote state. 2. If phenotype, in certain instances, = is=20 reflected by level of enzymatic activity, then some phenotypes may show = a=20 threshold effect (e.g. alcaptonuria) where = any enzyme=20 activity above a certain level confers a wild type phenotype, and some = may show=20 a saturation effect, where incremental amounts of the enzyme may lead a = gradient=20 of phenotypes (e.g. sugar deposition in some plant seeds). C. Multiple Alleles and = <SPAN=20 class=3DSpellE>codominance 1. Although any = individual in a=20 population may contain only two allelic variants (<SPAN=20 class=3DGramE>Aa), within the population there may be may = different=20 variants (A, a, A1, A2, A3 etc.) 2. Allelic variation can be detected = through a=20 number of means: a.=20 electrophoresis (protein level) b. nucleic acid = analysis=20 (DNA level) 3. <SPAN=20 class=3DGramE>examples: beta globin,=20 alpha1-anti-trypsin. 4. <SPAN=20 class=3DGramE>codominance: the heterozygote exhibits a = phenotype=20 based on the expression of both alleles. e.g. = ABO blood=20 group locus. 5. Molecular discussion of beta <SPAN=20 class=3DSpellE>globin variants and ABO blood group locus: = (Hartwell, <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:21>277-78; 308-310;=20 46-7; 56-7</DEL><INS cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:21>301-302; 336-339; 48-49; = 60-61</INS>) II. Deviation from = <SPAN=20 class=3DSpellE>Mendelian Ratios due to Single Gene, Polygenic, or = Environmental Interactions. A. = lethal=20 genes. e.g. yellow allele in mice = leads to 2:1=20 ratio if two yellow animals are crossed. B. genetic = heterogeneity.=20 e.g. albinism can be caused by a defect at = more than=20 one genetic locus. C. <SPAN=20 class=3DSpellE>phenocopy. e.g. = <SPAN=20 class=3DSpellE>kwashiorkhor- environmental factors mimic genetic = disorder=20 D. Variable Expressivity and <SPAN=20 class=3DSpellE>Penetrance. 1. Variable = Expression: single=20 gene effects can be variable in severity of expression and <SPAN=20 class=3DSpellE>pleiotropic (have several different phenotypic = effects).=20 e.g. anonychia = and=20 neurofibromatosis. 2. Penetrance: the=20 frequency with which a gene manefests itself = in the=20 appropriate genotype: some genes are incompletely <SPAN=20 class=3DSpellE>penetrant (do not express the appropriate = phenotype due to=20 gene or environmental modifiers). III. Interactions of = Genes with=20 other Genes (epistatic interactions). A. 4 distinct phenotypes = (comb shape=20 in chickens). B. complementary gene action (<SPAN=20 class=3DSpellE>anthocyanin pigment formation in pea flowers) C. recessive epistasis=20 (albinism in mice) D. recessive suppression of a recessive = phenotype in=20 Drosophila; purple eye color E. duplicate genes (fruit shape in <SPAN=20 class=3DSpellE>shepard's purse) I. Linkage and = Recombination A. Two genes close = together on the=20 same chromosome will not assort independently; they will be transmitted = to the=20 same gamete >50% of the time since they are attached to a common = <SPAN=20 class=3DSpellE>centromere. B. In heterozygotes,=20 linked genes may be arrayed in two possible configurations (phasing=20 relationships). 1. <SPAN=20 class=3DGramE>cis (or coupling): both dominant loci on = same=20 chromosome and both recessive loci on the same chromosome: (AB/<SPAN=20 class=3DSpellE>ab). 2. trans (or = repulsion):=20 A dominant gene at one locus associated on the same chromosome with a = recessive=20 gene at a second locus: (Ab/aB). C. Linkage can be = followed in a=20 testcross. II. Use of testcrosses = to=20 demonstrate linkage (e.g. T.H. Morgan). A. e.g. The=20 white and minature genes on = the X=20 chromosome of Drosophila. w=20 m/ + + x w m/Y or The white=20 and yellow genes on the X chromosome of Drosophila. = y +/+ w =20 x y=20 w/Y B. Parental types = recovered in=20 greater frequency than on basis of random assortment. 1. Results deviated = from expected=20 1:1:1:1 ratio; non-recombinant (parental) classes recovered in greater = frequency=20 than the recombinant (non-parental) classes. Hypothesized that=20 the non-parental gene combinations arose from physical exchange of = chromosomal=20 material during meiosis in the heterozygote. 2. The frequency of recombination is a = function of=20 how close two loci are linked. 3. The crossover frequency can be = directly related=20 to physical distances between genes on the chromosome (A.H. Sturtevant, = 1911).=20 a) the=20 farther apart two loci are located from one another, the more likely a = crossover=20 will occur between them. b) 1% recombination in gametes of = heterozygote=20 represents 1 map unit ( m.u.=20 or CentiMorgan, cM). c) for = white and=20 miniature, distance is large (~37.6 m.u.); for=20 yellow and white, distance between genes in much smaller (~1.4 <SPAN=20 class=3DSpellE>m.u.); d) the closer = the loci, the=20 greater the number of progeny that must be examined to detect a = crossover=20 event. 4. Conclusions from = Linkage=20 Experiments. a) Genes are linked = through all=20 generations, no random mixing. b) Genes are arrayed in a specific linear = order=20 along the length of a chromosome, which is a species characteristic. = c) Homologs = carry the same=20 array of gen loci, but not necessarily the = same=20 alleles. During meiosis, homologs could = exchange parts=20 by crossing over. Chiasmata represent = crossover points=20 between homologs (Janssens,=20 1909). III. Three point = testcrosses. A. = <SPAN=20 class=3DSpellE>eg. Corn: Mapping the genes for the <SPAN=20 class=3DSpellE>rececessive traits lazy growth, glossy leaf, and = sugary=20 endosperm. Progeny <SPAN=20 class=3DGramE>From a Three Point Testcross in Corn <o:p> </o:p> <DIV align=3Dcenter> <TABLE class=3DMsoNormalTable style=3D"BACKGROUND: #80ff80; = mso-cellspacing: 0in"=20 cellSpacing=3D0 cellPadding=3D0 border=3D0> <TBODY> <TR style=3D"mso-yfti-irow: 0; mso-yfti-firstrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"=20 colSpan=3D3> <o:p> </o:p></TD></TR> <TR style=3D"mso-yfti-irow: 1"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Phenotype of testcross = progeny</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Genotype of Gamete from Hybrid = Parent</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Number</TD></TR> <TR style=3D"mso-yfti-irow: 2"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> normal (<SPAN=20 class=3DSpellE>wildtype)</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Lz Gl=20 Su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 286</TD></TR> <TR style=3D"mso-yfti-irow: 3"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> lazy</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lz Gl=20 Su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 33</TD></TR> <TR style=3D"mso-yfti-irow: 4"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> glossy</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Lz gl=20 Su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 59</TD></TR> <TR style=3D"mso-yfti-irow: 5"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> sugary</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Lz Gl = <SPAN=20 class=3DSpellE>su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 4</TD></TR> <TR style=3D"mso-yfti-irow: 6"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lazy,glossy</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lz gl=20 Su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2</TD></TR> <TR style=3D"mso-yfti-irow: 7"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> lazy,=20 sugary</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lz Gl = <SPAN=20 class=3DSpellE>su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 44</TD></TR> <TR style=3D"mso-yfti-irow: 8"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>glossy,sugary</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Lz gl = <SPAN=20 class=3DSpellE>su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 40</TD></TR> <TR style=3D"mso-yfti-irow: 9; mso-yfti-lastrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> lazy, <SPAN=20 class=3DSpellE>glossy,sugary</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>lz gl = <SPAN=20 class=3DSpellE>su</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 272</TD></TR></TBODY></TABLE></DIV> B. Review of data set = shows=20 deviation from 1:1:1:1:1:1:1:1 ratio that would be expected for = independent=20 assortment; instead, two classes are in very high frequency, two classes = are in=20 very low frequency and 4 classes are recovered in intermediate = frequencies. C. The parental class represents the most = frequent=20 phenotypic classes amongst the progeny. This allows the determination of = the=20 orientation of the dominant and recessive alleles in the heterozygous = parent.=20 D. The double recombinant classes = represent the=20 least frequent phenotypic classes amongst the progeny. The 4 = intermediate=20 classes represent the single recombinant classes in each interval. E. Comparison of parental and double = recombinant=20 classes allows the determination of which gene is in the middle; from = the data=20 set given in class, the phasing relationship and order of the genes in = the=20 heterozygous parent was: <SPAN=20 class=3DSpellE>Lz = Su =20 Gl/lz su <SPAN=20 class=3DSpellE>gl F. Once gene array in = the parental=20 heterozygote is determined, identify the phenotypic classes in the = progeny that=20 represent the single and double crossovers which occur within each gene=20 interval: RF =3D (#SCO + #DCO)/total progeny for each gene interval. = <SPAN=20 class=3DSpellE>eg. for=20 the example given in class, The gene interval between <SPAN=20 class=3DSpellE>Lz and Su could be calculated as : = RF =3D ([33 +=20 40] + [4 + 2])/740 =3D 0.107 =3D 10.7 m.u. = E. Interference. 1. A crossover in one = area may=20 interfere with crossing over in an adjacent area of the chromosome. Male Progeny=20 from the mating of Drosophila melanogaster=20 females heterozygous for three X-linked Genes with <SPAN=20 class=3DSpellE>Wildtype Males <o:p> </o:p> <DIV align=3Dcenter> <TABLE class=3DMsoNormalTable style=3D"BACKGROUND: #ffff80; = mso-cellspacing: 0in"=20 cellSpacing=3D0 cellPadding=3D0 border=3D0> <TBODY> <TR style=3D"mso-yfti-irow: 0; mso-yfti-firstrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"=20 colSpan=3D3> <o:p> </o:p></TD></TR> <TR style=3D"mso-yfti-irow: 1"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Phenotype</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Genotype of=20 female gamete</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> number</TD></TR> <TR style=3D"mso-yfti-irow: 2"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>wildtype</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> + +=20 +</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 5</TD></TR> <TR style=3D"mso-yfti-irow: 3"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>crossveinless</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>cv + +</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2207</TD></TR> <TR style=3D"mso-yfti-irow: 4"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> echinus</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> + <SPAN=20 class=3DSpellE>ec +</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 217</TD></TR> <TR style=3D"mso-yfti-irow: 5"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> cut</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> + +=20 ct</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 265</TD></TR> <TR style=3D"mso-yfti-irow: 6"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>crossveinless, echinus</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>cv ec = +</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 273</TD></TR> <TR style=3D"mso-yfti-irow: 7"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>crossveinless,cut</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>cv + ct</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 223</TD></TR> <TR style=3D"mso-yfti-irow: 8"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>echinus,cut</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> + <SPAN=20 class=3DSpellE>ec ct</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2125</TD></TR> <TR style=3D"mso-yfti-irow: 9; mso-yfti-lastrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>crossveinless,echinus,cut</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>cv ec=20 ct</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 3</TD></TR></TBODY></TABLE></DIV> 2. This may be = observed in three=20 point testcrosses, where the number of expected double crossovers that = are=20 actually recovered is less than the number that are=20 expected. e.g. for the example taken from a=20 Drosophila cross in class (see above data set), the calculated = map=20 distances between the echinus, crossveinless=20 and cut loci were: <SPAN=20 class=3DGramE>ec-------10.3 <SPAN=20 class=3DSpellE>m.u.--------<SPAN=20 class=3DSpellE>cv------8.4 <SPAN=20 class=3DSpellE>m.u.------ct and out of = 5318 progeny,=20 8 double crossovers were actually observed. The expected number of = double=20 crossovers was therefore: (0.103)(0.084) =3D (0.0086<SPAN=20 class=3DGramE>)(5318) =3D 46 expected double crossovers. coefficient = of=20 coincidence =3D observed double crossovers/expected double crossovers = =3D 0.174 Interference =3D 1- coefficient of = coincidence =3D=20 0.826 IV. Ordered tetrads in = fungi. A. Some fungi produce = meiotic=20 products, haploid spores, in a saclike structure (<SPAN=20 class=3DSpellE>ascus). All meiotic products from the same = division cycle=20 can be recovered; in some species, each member of the tetrad giving rise = to=20 these spores are arrayed in the order of = their=20 segregation. Likewise, in some species, a mitotic division follows = meiosis=20 giving rise to two spore pairs for each tetrad. B. Other advantages to genetic analysis. = 1. Organism is haploid = for majority=20 of life cycle. 2, They produce = large=20 numbers of spores- rare events are easily detected. C. Life cycle of Ascomycetes: e.g. <SPAN=20 class=3DSpellE>Neurospora crassa,=20 bread mold. 1. Haploid spores = generate <SPAN=20 class=3DSpellE>hyphal structures. 2. Haploid nuclei of opposite mating types = can=20 undergo nuclear fusion and meiosis in fruiting bodies (<SPAN=20 class=3DSpellE>perithecia). 3. The zygote is the only diploid stage; = this=20 transient diploid state undergoes meiosis and then mitosis; 8 <SPAN=20 class=3DSpellE>ascospores are produced. 4. The meiotic products (tetrads) are = maintained in=20 a linear order. V. Mapping by Ordered = Tetrad=20 Analysis. All meiotic products are recovered in the same structure in an = ordered=20 array. This permits the mapping of genes relative to the <SPAN=20 class=3DSpellE>centromere. A. Calculating a Map = Distance=20 Between the Gene and <SPAN=20 class=3DSpellE>Centromere. 1. If no recombination = occurs, you=20 will expect to see a 1st division segregation pattern: a 4:4=20 array of spores relative to phenotype. 2. If a single crossover occurs between = the gene and=20 the centromere, you will expect to see a 2nd = division=20 segregation pattern: a 2:2:2:2 or a 2:4:2 array (fig 5-<DEL=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:22>23 </DEL><SPAN=20 class=3DmsoIns><INS cite=3Dmailto:MHE dateTime=3D2007-02-08T11:22>21 = </INS>in=20 Hartwell). 3. The percentage of 2nd division = segregation=20 patterns is related to the distances between the gene and <SPAN=20 class=3DSpellE>centromere. 1/2(asci with=20 2nd division segregation pattern) X 100 =3D map distance=20 total number of = <SPAN=20 class=3DSpellE>asci Remember that you a scoring <SPAN=20 class=3DSpellE>asci here, and that the "1/2" in the above = equation corrects=20 for the fact that only 1/2 of the chromosomes arising from that meiosis = will be=20 recombinant. B. Gene Linkages. = Suppose two loci=20 are present in the same cross. In the example given in class, the cross = AB x=20 ab yields the following data set: Hypothetical Data Set from the <SPAN=20 class=3DSpellE>Neuospora Cross AB x <SPAN=20 class=3DSpellE>ab <o:p> </o:p> <DIV align=3Dcenter> <TABLE class=3DMsoNormalTable style=3D"BACKGROUND: aqua; = mso-cellspacing: 0in"=20 cellSpacing=3D0 cellPadding=3D0 border=3D0> <TBODY> <TR style=3D"mso-yfti-irow: 0; mso-yfti-firstrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"=20 colSpan=3D5> <o:p> </o:p></TD></TR> <TR style=3D"mso-yfti-irow: 1"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <o:p> </o:p> <o:p> </o:p></TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Ascus = Type</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Ascus = Type</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Ascus = Type</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Ascus = Type</TD></TR> <TR style=3D"mso-yfti-irow: 2"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Spore=20 pair</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 1</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 3</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 4</TD></TR> <TR style=3D"mso-yfti-irow: 3"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> </TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> (PD)</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> (TT)</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> (PD)</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> (NPD)</TD></TR> <TR style=3D"mso-yfti-irow: 4"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 1</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>aB</TD></TR> <TR style=3D"mso-yfti-irow: 5"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>aB</TD></TR> <TR style=3D"mso-yfti-irow: 6"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 3</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>aB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> AB</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Ab</TD></TR> <TR style=3D"mso-yfti-irow: 7"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 4</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>ab</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> <SPAN=20 class=3DSpellE>Ab</TD></TR> <TR style=3D"mso-yfti-irow: 8; mso-yfti-lastrow: yes"> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> Total number</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 108</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 60</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 30</TD> <TD=20 style=3D"PADDING-RIGHT: 0.75pt; PADDING-LEFT: 0.75pt; = PADDING-BOTTOM: 0.75pt; PADDING-TOP: 0.75pt"> 2</TD></TR></TBODY></TABLE></DIV> 1. 3 types of tetrads = are produced:=20 a) <SPAN=20 class=3DGramE>parental ditype (PD): = <st1:place=20 w:st=3D"on"><st1:City w:st=3D"on">AB</st1:City> <st1:State = w:st=3D"on"><SPAN=20 class=3DSpellE>AB</st1:State></st1:place> <SPAN=20 class=3DSpellE>ab ab b) <SPAN=20 class=3DGramE>nonparental ditype (NPD):=20 aB aB = <SPAN=20 class=3DSpellE>Ab Ab c) <SPAN=20 class=3DGramE>tetratype (TT): AB Ab=20 aB ab 2. If genes are = unlinked, then the=20 A and B loci might be expected to assort independently; = therefore=20 PD =3D NPD (approximately the same numbers). 3. If the two genes are linked, then the = only way to=20 generate a NPD would be through a 4 strand double crossover; and = PD>> NPD.=20 In the above data set, the A and B loci are linked. 4. Centromere = order can be=20 determined by examining frequency of classes showing 1st and second = degree=20 segregation patterns in relation to frequency of NPD classes. For = example, the=20 type 3 ascus pattern above is informative; = both the=20 A and B loci show a second degree segregation pattern, = indicating=20 a cross over between both genes and the centromere. If=20 genes were on opposite sides of the centromere, you=20 would need a double crossover to generate this class. Its frequency = relative to=20 the other classes indicates that it is a single crossover class, and = that=20 A is closer to the centromere than = B:=20 centromere--------A----------B. 5. Recombination between <SPAN=20 class=3DSpellE>centromere and B: 1/2 (60 + 30) X 100 =3D 22.5 map = units=20 200 6. Recombination between <SPAN=20 class=3DSpellE>centromere and A: 1/2 (30) X 100 =3D 7.5 map units=20 200 The deduced difference between the genes = is=20 therefore 15 map units. 7. The above calculation will lead to an=20 underestimate of distance, since the above technique is not accurately=20 estimating the number of double crossovers. The number of double = crossovers can=20 be estimated, however from the NPD class (presented in class: diagram = found in=20 ZAP room notes). From this class, we can estimate the total number of = single and=20 double crossovers occurring in the interval between A and = B: DCO =3D 4 NPD T =3D SCO + 1/2 DCO; therefore SCO =3D T - 2 NPD RF =3D 1/2 SCO + DCO RF =3D 1/2[TT-2NPD] + 4NPD=20 &nb= sp; =20 total asci for the above = data set:=20 RF =3D 1/2(60-4) + 8=20 &nb= sp; =20 200 =3D 36/200 =3D = 0.18 =3D 18 map=20 units between the A and B loci. (note our=20 previous estimate was 15 map units, an underestimate). VII. Mitotic = Recombination. A. Although much rarer = than meiotic=20 recombination during gametogenesis, = recombination=20 events can be detected in somatic tissue. A classic example is the = formation of=20 twin spots during Drosophila development (Stern, 1936). = B. Twin spots have proven useful in the = analysis of=20 the developmental effects of mutations which may otherwise cause = lethality if=20 inherited in homozygous form in the zygote. VIII. Human Gene = Linkage.=20 (Hartwell, pp. <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:22>392-33</DEL><INS = cite=3Dmailto:MHE dateTime=3D2007-02-08T11:22>416-417</INS>; = Thompson,=20 Chapters 10-12) A. Information from = Traditional=20 Linkage Data. It is obviously difficult to generate large numbers of = progeny and=20 perform selective matings in humans. = Nevertheless,=20 information on linkage can be obtained from pedigree data using = statistical=20 techniques. B. Lod Scores. = <SPAN=20 class=3DSpellE>Lod score analysis is a statistical technique that = permits=20 linkage relationships to be derived based on relative probabilities of = linkage=20 given the segregation patterns observed in pedigrees. The statistical = data=20 handling is beyond the scope of this course; for those who are = interested, I can=20 provide you with a more detailed description of the analysis method. = Briefly,=20 linkage data can be represented as a likelihood ratio, which represents = the=20 probability that two alleles are linked at some given recombination = frequency=20 (q = ), divided=20 by the probability that the alleles assort independently (RF =3D 0.5). = The data is=20 usually presented as the log of the likelihood ratios (<SPAN=20 class=3DSpellE>Lod Score): Zq =3D log=20 10 (Probability of family for <SPAN=20 style=3D"FONT-FAMILY: Symbol">q =3D 0.01, 0.02, etc.)=20 &nb= sp; =20 (Probability of family for q = =3D 0.50)=20 Lod scores = from individual=20 experiments can be calculated and are additive; it is possible to arrive = at=20 linkage probabilities without knowledge of the phasing relationships in = the=20 heterozygous parent. A likelihood ratio ratio of=20 1000:1 (Lod score =3D 3) is usually = considered "proof"=20 of linkage. IX. Assigning Genes to = Chromosomes=20 Using Somatic Cell Genetics. (see Thompson, = chapter 10)=20 A. Production of Hybrid = Cells.=20 Usually a rodent tumor cell line is fused with human fibroblast cells. = Cell=20 fusion is promoted by either viral infection or chemicals (polyethylene = glycol;=20 PEG). 1. HAT Selection = System. Used to=20 enrich for hybrid cells and eliminate parental cells and same-parent = fusions=20 from the culture. Growth medium contains a chemical (<SPAN=20 class=3DSpellE>aminopterin) to shut down the major pathway for = nucleotide=20 precursor synthesis- forcing the use of a salvage pathway to produce the = subunits needed for DNA. The "salvage" molecules hypoxanthine and <SPAN=20 class=3DSpellE>thymidine are also present in the growth medium = and can be=20 incorporated into DNA, provided the cells have two salvage pathway = enzymes-=20 HGPRT (hypoxanthine-guanine phosphoribosyltransferase)=20 and TK (thymidine kinase)=20 necessary to convert them into the appropriate DNA precursors. By using = parental=20 lines deficient (i.e. mutant) for one or the other of the two enzyme = activities,=20 (e.g. human <SPAN=20 style=3D"FONT-SIZE: 10pt">HGPRT-/TK+; = rodent /<SPAN=20 style=3D"FONT-SIZE: 10pt">HGPRT+/TK-; = hybrid <SPAN=20 style=3D"FONT-SIZE: 10pt">HGPRT+/TK+) one = can select=20 for the rare cell fusions that will produce hybrid cells = (complementation). 2. The hybrid cells will eventually = undergo=20 nuclear fusion to form heterokaryons. This = process=20 will lead to a spontaneous loss of human chromosomes from the hybrid = cells. The=20 chromosomes will be lost at random, giving rise to colonies which will = contain=20 different portions of the human chromosome complement. 3. The presence or absence of a human = gene product=20 can then we assayed in the hybrid cells. Only those hybrid cells = containing the=20 human gene (and the chromosome on which it resides) would be expected to = produce=20 the human product. Concordance and discordance between the presence or = absence=20 of the human gene product (enzyme or protein) and the human chromosomes = a hybrid=20 cell line carries establishes what chromosome the gene resides on (synteny). 4. Once synteny is=20 established, the cell lines carrying fragments of the <SPAN=20 class=3DSpellE>chromsome in question can also be assayed. 5. The above technique requires that we = be able to=20 identify individual human chromosomes. It also assumes that we can = identify a=20 human gene product in the hybrid cell. We can do both. Indeed, the = nucleic acid=20 sequence of the gene itself is sufficiently different to distinguish = mouse from=20 human genes, allowing DNA gene probes to be used for mapping = purposes. X. Aberrant <SPAN=20 class=3DSpellE>Asci Patterns, Gene Conversion and Recombination = at the=20 Molecular Level.(Hartwell, <DEL=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:22>178-186</DEL><SPAN=20 class=3DmsoIns><INS cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:22>191-200</INS>) A. Aberrant <SPAN=20 class=3DSpellE>asci patterns (e.g. 3:1:1:3; 6:2; 5:3) have led to = insights=20 into the molecular nature of recombination. The recovery of products of = a=20 meiotic division in a heterozygote at some ratio other than the expected = 1a:1A=20 resulting from reciprocal recombination is known as gene = conversion.=20 Because we can trap all the meiotic products of the same division cycle, = and=20 because the mitotic division following meiosis allows us to analyze the=20 individual DNA molecules that underwent recombination in each tetrad, we = have=20 been able to develop models of how recombination occurs. Many of these = models=20 have common features which can explain the origin of aberrant <SPAN=20 class=3DSpellE>asci during fungal meiosis. These models suggest a = process=20 of DNA nicking, strand invasion, heteroduplex=20 formation and DNA repair occurring at the recombination site. = Aspects of=20 these models have been verified by observing recombination in other = model=20 systems (e.g. bacteria). The Holliday Model is one such model = developed=20 to explain gene conversion. Features of the model and how it can = explain an=20 aberrant ascus pattern are covered in pp. = <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>178-186</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:23>191-200</INS> of=20 Hartwell. I. Overview of Mutation = Section A. Mutation: there is = no genetics=20 without variation B. Gene or "Point" Mutation: allelic = variations=20 occuring within a single gene locus, usually = resulting=20 in a change or loss of a small number of nucleotides. (Hartwell, = Chapter=20 7) C. Chromosomal Mutation: a change in a = segment,=20 chromosome or set of chromosomes, which may, or may not include a change = within=20 a single gene locus. 1. Aberrations due to = chromosome=20 structure (Hartwell, Chapter <DEL cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:23>13</DEL><INS = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>14</INS>) 2. Aberrations due to chromosome number = (Hartwell,=20 Chapter <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>13</DEL><INS = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>14</INS>) D. We will cover this = section in=20 the order given below, first discussing chromosome structure and changes = that=20 can occur at the chromosome level. We will return to a discussion of = mutation at=20 the allele level prior to a discussion of bacterial genetics, and = emphasize the=20 important role bacterial genetics had in laying the groundwork for = molecular=20 genetics- the study of information transfer at the molecular level. II. Techniques Used to = Visualize=20 Chromosome Structure: Karyotype analysis = (Hartwell,=20 pp. <DEL cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:23>81;=20 422-24</DEL><INS cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:23>84; 472-474</INS>) A. Chromosomes are a = complex of=20 primarily DNA and protein (chromatin). <SPAN=20 class=3DGramE>euchromatin: lightly staining, genetically = active;=20 heterochromatin, darkly staining, genetically inactive. <SPAN=20 class=3DGramE>(Hartwell, <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:24>422-24</DEL><INS = cite=3Dmailto:MHE dateTime=3D2007-02-08T11:24>472-474</INS>) for = definitions=20 and description). B. During interphase,=20 chromosome structure is not resolved due to <SPAN=20 class=3DSpellE>decondensed state of chromatin. C. DNA becomes highly compacted during = process of=20 mitosis; chromosome structure can be visualized. D. In humans, a cell population may need = to be=20 stimulated to start mitosis before analysis can be performed (e.g. white = blood=20 cells (lymphocytes) from peripheral blood; or fetal fibroblasts (<SPAN=20 class=3DSpellE>amniocytes) taken from amniotic fluid). 1. <SPAN=20 class=3DSpellE>Interphase cells are induced to enter mitosis by = stimulation=20 with a mitogen (e.g. <SPAN=20 class=3DSpellE>phytohemagluttinin). 2. Cell may be stopped in metaphase by = disrupting=20 spindle fiber formation (tubulin = polymerization) with=20 the chemical colchicine. 3. The cells must be prepared to = preserve=20 chromosome structure and specifically disrupted to spread the = chromosomes. III. Basic Chromosome = Morphology=20 A. Size B. Centromere = position=20 (metacentric, <SPAN=20 class=3DSpellE>submetacentric, acrocentric, <SPAN=20 class=3DSpellE>telocentric). C. Other distinguishing features such as = nucleolus=20 organizer regions (secondary constrictions, site of ribosomal RNA = genes. IV. Differential = Banding=20 Techniques. A. It is possible to = identify every=20 human chromosome on the basis of a distinctive banding pattern revealed = by=20 partially disrupting chromosome and staining with specific dyes. B. Several Methodologies: require = treatments with a=20 variety of agents that will partially disrupt chromatin structure = (acids, bases,=20 proteolytic enzymes, heating). C. Most common techniques: 1. G Banding: standard = for=20 comparisons (Paris Convention, 1971). 2. Q Banding: fluorescent staining: = similar to G=20 Band pattern. 3. R Banding: banding pattern appears = reversed=20 from G Band pattern. 4. C Banding: will stain only the = heterochromatin=20 surrounding the centromere. D. Normal Standard of = <SPAN=20 class=3DGramE>Resolution : ~ 300-500 bands at metaphase; ~1200 = bands at=20 prometaphase, when chromatin is less = condensed. (The=20 most recent estimates of the number of genes within human genome <SPAN=20 class=3DGramE>is on the order of 25,000 (revised since Hartwell = was printed=20 in 2004 and still not a certainty); so this level of cytological = resolution is=20 obviously limited in its ability to detect structural defects that may = exist in=20 chromosome.) E. Convention for designating sites on = chromosomes.=20 Sites are designated by first giving the chromosome number, followed by = the=20 chromosome arm (p=3D "petite" or short arm; q =3D long arm). Chromosome = regions are=20 identified on the basis of "landmark bands" which define <SPAN=20 class=3DSpellE>chromsome regions; these regions are numbered away = from the=20 centromere. The band found within a region = is given=20 last; the individual bands are also numbered away from the <SPAN=20 class=3DSpellE>centromere:<IMG id=3D_x0000_i1028 height=3D259=20 src=3D"http://www.ou.edu/cas/zoology/Courses/3333/Image7.gif" = width=3D142> e.g. In the = above example,=20 the arrow would be located at 8q22 in the human chromosome = complement. V. Morphology of Human = Sex=20 Chromosomes. A. Mammalian Y = Chromosomes.=20 (Hartwell, pp. <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:24>79-82; 103-4</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:24>82-86; = 110-112</INS>) 1. Apparently does not = contain=20 many genes, other than those involved in human sex determination, or = sperm=20 fertility factors. 2. The chromosome contains much = chromatin that=20 appears genetically inert (consitutive=20 heterochromatin). There is much population polymorphism in the size of = the long=20 arm of the Y. 3. Pairs with the X chromosome during = meiosis=20 along a small region that is homologous to the X (<SPAN=20 class=3DSpellE>pseudoautosomal region). 4. TDF gene serves as a major = developmental switch=20 for determining maleness. B. = Mammalian X=20 Chromosomes. (Hartwell, pp. <DEL = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:24>79-82; 431-32</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:24>82-86; 4</INS><INS cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:25>81-<SPAN=20 class=3DGramE>482;) </INS> 1. Females have two = copies of the=20 X; males have only one; Is there a dosage = compensation=20 mechanism that operates to control and normalize levels of X chromosome = gene=20 products? 2. Interphase cells of=20 females show a dark body (Barr body) usually associated with the = periphery of=20 nucleus; males have no Barr body; individuals with extra X chromosomes = show=20 extra Barr bodies, such that the number of Barr bodies observed are 1 = less that=20 the total number of X chromosomes. 3. <st1:place = w:st=3D"on">Lyon</st1:place>=20 Hypothesis: In a normal female (46; XX), one X chromosome becomes = functionally=20 inactive early in embryogenesis (facultative heterochromatin). The = inactivation=20 process is random, and the chromosome that is inactivated remains so in = all=20 progeny cells. Since inactivation is a random event, all females are=20 functionally mosaic; some cells contain one inactivated X, some cells = the other.=20 e.g. X linked coat color genes in mice; anhydrotic dysplasia = in=20 humans. C. Genomic Imprinting. = (Hartwell,=20 pp. <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>596-602</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:25>660-664</INS>) 1. In mammals, a small = number of=20 genes are passed on from the parent to the zygote in an inactive form; = embryos=20 therefore inherit only one active gene. 2. Some of = these genes=20 are inactivated during spermatogenesis; some during <SPAN=20 class=3DSpellE>oogenesis. 3. If one parent passes a wild-type = inactive gene=20 and the other parent passes a mutant allele, the resultant individual = will show=20 a mutant phenotype. e.g. <SPAN=20 class=3DSpellE>Praeder-Willi Syndrome (maternal gene inactive); = <SPAN=20 class=3DSpellE>Angelman Syndrome (paternal gene inactive). 4. Why would one gene be specifically = inactivated=20 by a parent? One theory suggests that imprinting is related to = competition=20 between embryo and mother for resources. According to this theory, genes = that=20 are passed in an active form by males would lead to increased growth = during=20 embryogenesis; genes that are passed in an active form by females would = tend to=20 retard embryonic growth. VI. Departures from = <SPAN=20 class=3DSpellE>Diploidy: Heritable Aberrations in Chromosome = Number;=20 Humans. (Hartwell, Chapter <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>13</DEL><INS = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>14</INS>) A. <SPAN=20 class=3DSpellE>Aneuploidy: a chromosome number that is not an = exact=20 multiple of the haploid chromosome number. <SPAN=20 class=3DSpellE>Aneuploidies usually arise from de novo errors = (<SPAN=20 class=3DSpellE>nondisjunction) during cell division. B. In humans, aneuploidy=20 involving sex chromosomes are usually the least severe of the possible = <SPAN=20 class=3DSpellE>aneuploidies: 1. Turner Syndrome = (45, X) 2. Klinefelter Syndrome=20 (47, XXY). 3. Multiple Y males (47, XYY). 4. Poly X females (47, XXX; 48, XXXX, = etc.) C. In humans, <SPAN=20 class=3DSpellE>aneuploidy involving autosomes is=20 usually fatal; all autosomal <SPAN=20 class=3DSpellE>monoploidy is fatal; most <SPAN=20 class=3DSpellE>trisomies are fatal with the exception of <SPAN=20 class=3DSpellE>Patau, Edwards and Down Syndrome. 1. Patau=20 Syndrome (47, +13): seen approximately 1/4000 births; severe mental = retardation=20 and physiological defects. Usually fatal in early=20 infancy. 2. Edwards Syndrome (47, +18): seen in=20 approximately 1/7000 births; severe mental retardation and other = physiological=20 defects. Usually fatal in early infancy. 3. Down = Syndrome (47,=20 +21): seen in approximately 1/800 births; physical signs in infants are = lowered=20 muscle tone, protruding tongue, characteristic folds at eye; spots on = iris,=20 characteristic fingerprints (dermatoglypics). Children=20 are at risk for increased cardiac defects, severe mental retardation; = older=20 individuals will develop early onset Alzheimer disease. 4. There is a pronounced increase in = risk for=20 chromosomal aneuplodies in older women. VII. Structural = Chromosomal=20 Aberrations (Hartwell, Chapter <DEL = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>13</DEL><INS = cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:25>14</INS>) A. Factors Generating = Chromosomal=20 Damage (clastogenic agents) 1. <SPAN=20 class=3DGramE>ionizing radiation. 2. some viral = infections.=20 3. chemicals. B. Breakage Events. 1. Telomeres <SPAN=20 class=3DGramE>represents "proper" chromosomal ends; they are = specialized=20 structures that with a highly conserved DNA sequence which marks the = natural end=20 of a chromosome. 2. If a break is introduced, new "ends" = are not=20 properly terminated and will be subject to DNA repair mechanisms. The = joining of=20 inappropriate ends can lead to structural abnormalities. C. Reciprocal = Translocations: a=20 reciprocal exchange between non-homologous chromosomes. 1. If no genes are = damaged at the=20 site of breakage, then individuals heterozygous for the reciprocal = translocation=20 (i.e. carrying two translocation chromosomes and two normal chromosomes) = may=20 show no overt abnormal phenotype. 2. Pairing problems during meiosis, = however, will=20 lead to the formation of tetravalents; = abnormal=20 segregation from this structure (adjacent segregation) will lead to = unbalanced=20 gametes carrying duplications and deficiencies for certain genes. The = result is=20 semisterility in the translocation = heterozygote. 3. Robertsonian=20 Translocations: a specific type of reciprocal translocation, where = breaks are=20 introduced to opposite sides of the centromeres of two=20 non-homolog acrocentric chromosomes. Repair = leads to=20 formation of a small chromosome which is usually lost,=20 and the fusion of the two long arms into a single chromosome. a) Often=20 seen in closely related species. For example, the <SPAN=20 class=3DSpellE>metacentric chromosome 2 of humans resembles in = banding=20 pattern and gene content two acrocentric = chromosomes=20 seen in chimpansees, suggesting a <SPAN=20 class=3DSpellE>robertsonian fusion occurred during primate = evolution. b) Such translocations can also been seen = within=20 species; 14:21 translocations are associated with a heritable form of = Down <SPAN=20 class=3DGramE>Syndrome. c) Reciprocal Translocations and = Cancer: If=20 the breakpoint occurs within a gene (damage at the site of breakage), or = transfers a gene into a region which affects its ability to be = expressed, this=20 can lead to somatic mutations in genes controlling normal growth and = cell=20 proliferation. 1) <SPAN=20 class=3DGramE>formation of hybrid genes with abnormal phenotypes: = in=20 >95% of cases of chronic myelogenous = leukemia, the=20 tumor cells contain a clear reciprocal translocation within the <SPAN=20 class=3DSpellE>abl gene locus that leads to the conversion = of this=20 "proto-oncogene" into an "oncogene". 2) the = movement of the=20 myc gene into the region of the immunglobulin gene cluster produces a change in = its=20 expression pattern which is correlated with a cancer known as <SPAN=20 class=3DSpellE>Burkitt Lymphoma. This activation due to change in = location=20 is an example of position-effect variegation. D. Deletions or = Deficiencies 1. A loss of a small = region of the=20 chromosome may be correlated with abnormalities arising from the = imbalance; most=20 large deletions are inviable. e.g.=20 (46; 5p-) cri du chat syndrome. 2. A loss of a small region may place an = individual at risk if the normal gene copy located on the homolog = chromosome=20 mutates. e.g. Hereditary predisposition to = retinal=20 cancers for individuals deleted for 13q14, the site of the = retinoblastoma (<SPAN=20 class=3DSpellE>Rb) gene locus. (Hartwell, pp. <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:26>634-43</DEL><INS = cite=3Dmailto:MHE dateTime=3D2007-02-08T11:26>702-709</INS>) E. Inversions 1. Once again, if no = genes are=20 damaged at the breakpoint, inversion heterozygotes may=20 only encounter problems relative to chromosome segregation during = meiosis. 2. If crossing over occurs within = inversion, it=20 will produce chromosomes with duplications and deficiencies. a. <SPAN=20 class=3DSpellE>paracentric inversions: centromere=20 not included in the inversion; also produces acentric=20 and dicentric chromosomes. b. <SPAN=20 class=3DSpellE>pericentric inversions: centromere=20 is included in the inversion; recombinant products have <SPAN=20 class=3DSpellE>centromeres but still carry duplications and = deletions. F. Fragile Sites: = X-linked Mental=20 Retardation (Hartwell, <DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:26>202-203</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:26>216-217</INS>). = 1. Fragile sites can = be detected=20 by chromosome breakage when cells are cultured under special conditions. = The=20 chromosomal marker is a characteristic of Fragile X syndrome, present in = 1/1000=20 male births and one of the most common forms of mental retardation in = males (~8%=20 of all males with mental retardation). 2. In Fragile X Syndrome the defect is = linked to=20 an unstable replication of a triplet repeat in the FMR-1 gene, located = at the=20 fragile site. An amplification of this sequence from a "<SPAN=20 class=3DSpellE>premutation" leads to the full symptoms of this=20 syndrome. G. Monitoring of = Chromosomal=20 Defects. 1. Amniocentesis. = <SPAN=20 class=3DGramE>Usually performed at ~16 weeks, when fibroblasts can be = collected=20 safely from amniotic fluid. Requires <SPAN=20 class=3DSpellE>mitogenic stimulation of cells prior to = analysis.=20 2. Chorionic = <SPAN=20 class=3DSpellE>Villus Sampling. Can be = performed at ~9th=20 week, from embryonic cells that are already actively dividing.=20 VIII. Departures from = <SPAN=20 class=3DSpellE>Diploidy in Plants. A. Tolerances for <SPAN=20 class=3DSpellE>aneuploidy and polyploidy contribute to speciation = and=20 facilitate genetic manipulations. 1. Plants don't = necessarily need=20 to propagate via sexual reproduction; they can propagate from somatic = tissues=20 (e.g. cuttings). 2. Increases in ploidy=20 may lead to aggricultural benefits (e.g. = increases in=20 plant size). 3. Monoploids can be=20 generated from gametes and used in selection experiments. 4. Chromosome number can be doubled by = <SPAN=20 class=3DSpellE>colchicine treatment. B. <SPAN=20 class=3DSpellE>Polyploid species: (number of complete chromosome = sets >=20 2) 1. <SPAN=20 class=3DSpellE>Autopolyploids: contain multiples of the same = <SPAN=20 class=3DGramE>chromosome set (same species). 2. Allopolyploids: contain <SPAN=20 class=3DSpellE>mutiples of different chromsome=20 sets (different species). 3. Hybridizations may lead to pairing = problems=20 during meiosis. a.=20 triploids are often sterile- difficulties in meiotic segregation. b. a doubling = of the=20 different chromosome sets in allopolyploids can relieve problems during = meiosis:=20 if there are pairing partners for homologs, = sterility=20 often relieved. (e.g. <SPAN=20 class=3DSpellE>Karpechenko, radish and cabbage <SPAN=20 class=3DSpellE>polyploids). <SPAN=20 class=3DGramE>amphidiploid =3D contains the diploid = chromosome=20 complement of both parent species. c. polyploidy = and=20 evolution: origin of common cultivated wheat, an <SPAN=20 class=3DSpellE>allohexaploid. I. Gene Mutation: there is no genetics without = allelic=20 variation (Hartwell, Chapter 7) A. Classes of Mutation = 1. Somatic; <SPAN=20 class=3DSpellE>clonal populations can arise due to mutations = <SPAN=20 class=3DSpellE>occuring during development; e.g. sectors of color = in plant=20 leaves, cancer (somatic mutation of proto-oncogenes=20 [normal cellular genes involved in cellular growth regulation] to <SPAN=20 class=3DSpellE>oncogenes [mutant genes causing abnormal cell = proliferation=20 or metastasis]). 2.Germline; capable=20 of being transmitted to the next generation. B. "Point Mutations": = mutant=20 phenotypes 1. <SPAN=20 class=3DSpellE>Heteromorphs (null, loss-of-function, = gain-of-function) 2. Lethals = 3. Conditional mutants (restrictive vs. = permissive=20 conditions) 4. Biochemical mutants a. <SPAN=20 class=3DSpellE>prototrophs b. <SPAN=20 class=3DSpellE>auxotrophs 5. Resistance mutants = a.=20 antibiotic resistance (e.g. strr) = b. viral = resistance=20 (tonr) C. Selection Systems = and Mutant=20 Detection 1. Advantage of microbes = a.=20 haploid b. fast = generation times=20 c. = inexpensive 2. Eukaryotic systems = (e.g. <SPAN=20 class=3DSpellE>ClB of Muller) D. Nature of Genetic = Change;=20 random mutation/selection or physiological adaptation? (Hartwell, pp. = <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE=20 dateTime=3D2007-02-08T11:26>194-196</DEL><INS=20 cite=3Dmailto:MHE dateTime=3D2007-02-08T11:26>210-212</INS>) 1. <SPAN=20 class=3DGramE>tonr mutants: resistance to T1 = <SPAN=20 class=3DSpellE>bacteriophage infection. 2. Delbruck/Luria fluctuation test: = "Mutations of=20 Bacteria from Virus Sensitivity to Virus Resistance": (1943) 3. Lederberg replica plating experiment: = (1952) 4. 1945: Schrodinger's "What is Life?" = published.=20 How do living organisms order themsevles? Do = molecules=20 govern heredity? What is the physical nature of the molecules that = govern=20 heredity? "...likely to involve hitherto unknown 'other laws of = physics,' which,=20 however, once they have been revealed will form just as integral a part = of this=20 science as the former." The influx of individuals trained in physics = provided a=20 basis for the revolution in molecular biology. I. Recombination in Bacteria (Hartwell, Chapter = <SPAN=20 class=3DmsoDel><DEL cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:26>14</DEL><SPAN=20 class=3DmsoIns><INS cite=3Dmailto:MHE = dateTime=3D2007-02-08T11:26>15</INS>)=20 A. How are genes in = bacteria=20 organized? Are there correlates to eukaryotic chromosomes? Is there = linkage? How=20 universal are information storage and retrieval mechanisms in Nature? = B. No genetics without allelic = variation. What=20 phenotypes do we study? II. Mechanisms for Gene Exchange and Recombination = in=20 Bacteria A. Conjugation: = transferred=20 through direct cell-cell contact. B. Transduction: transferred through = viral=20 infection. C. Transformation: Cellular uptake of = DNA from the=20 environment. III. Conjugation A. Lederberg and Tatum = (1946).=20 Using bacteria that were multiply auxotrophic,=20 demonstrated that recombination could occur. B. Davis (1950). The strains used by = Lederberg and=20 Tatum required physical contact for transfer. The = U-tube=20 experiment. C. Hayes (1953). There is a = directionality of=20 transfer. In this process, there is a donor/recipient relationship. = Streptomycin=20 treatment: will allow a donor cell to transfer DNA to a recipient, but = <SPAN=20 class=3DSpellE>str-treated cells will not be capable of = reproducing: 1. Donor (treat c <SPAN=20 class=3DSpellE>str)-->D* (wash out = drug and=20 cross c recipient)--> D* + R=3D recombinants 2. Recipient (treat c str)=20 --->R* (wash out drug and cross to donor)--> D +=20 R*=3D no recombinants. 3. Idea of unidirectional transfer: = D=3D"males";=20 R=3D"females"; crosses =3D "matings" D. Attributed = "maleness" to a=20 specific genetic entity; the "Fertility" Factor. 1. F+ cells would = transfer the=20 genotype of maleness at high frequency. 2. F+ cells would transfer other genotypes = (e.g.=20 prototrophy, phage resistance) at low = frequency. 3. Cavalli = (1950):=20 isolated a peculiar strain of male cell that produced <SPAN=20 class=3DSpellE>prototrophic recombinants at 1000 times the = frequency of=20 normal F+ strains. a. Termed an <SPAN=20 class=3DSpellE>Hfr strain: "high frequency of recombination". = b. The high frequency donor character = pertains=20 only to a limited portion of donor genome. c. Unlike F+ cells, Hfr=20 cells do not transfer male genotype to recipient cells. d. F+ ---> Hfr change=20 attributable to a heritable change in the sex factor's properties. Nature of change? E. Oriented Gene = Transfer and=20 "Time of Entry" Mapping Experiments. 1. Wollman=20 and Jacob: Interrupted mating experiments. Hfr x R=20 cell matings could be conducted under timed = conditions=20 to study the kinetics of gene transfer. 2. Waring = blender=20 experiment: Set up matings, allow <SPAN=20 class=3DSpellE>matings to occur for defined time intervals, = disrupt the=20 matings with a Waring=20 blender, then plate to select recombinants. 3. Time of Entry Mapping: By plotting = frequency of=20 recombinants recovered vs. time, they could extrapolate the "time of = entry",=20 which could be used as an index of distances between genes. 4. Analysis of different <SPAN=20 class=3DSpellE>Hfr strains (isolated from F+ populations by = replica-plate=20 matings on selective media) led to the = conclusion that=20 transfer could occur from different points in the genome in either of = two=20 different directions. How can this be explained by a consistent model? = 5. Both the bacterial chromosome and the F = factor=20 were covalently closed circular molecules. Integration of F factor into = the=20 chromosome DNA led to Hfr. Orientation and = site of the=20 integration event dictated what chromosomal genes were transferred. F. F' factors. 1. Occasionally, F = factors will=20 become unintegrated from bacterial = chromosome in an=20 Hfr cell. This will create an F factor that = can carry=20 a portion of the bacterial genome (F' factor) = . If it=20 conjugates with a recipient cell, that cell receives a chromosomal gene. = 2. F' cells carry two copies of a portion = of the=20 genome (merodiploids). There are also = capable of=20 transferring that gene at high frequency to recipient cells. G. Mapping by = selecting for late=20 markers. 1. Time of entry mapping = is not very=20 high resolution and is not very useful for mapping genes relatively = close to one=20 another. 2. By selecting for a late marker, closely = linked=20 genes can be mapped by examining the recombination frequency. IV. Transduction A. Lytic=20 Cycle: upon entry into cell, virus replicates, packages and <SPAN=20 class=3DSpellE>lysis host. B.Generalized=20 Transduction and Mapping 1. During <SPAN=20 class=3DSpellE>lytic infection, host genetic material may be = packaged. 2. On infection, host genetic material is=20 transferred into new cell, where it may recombine into genome. 3. Since phage particle can only contain = limited=20 amounts of DNA, technique can be used for mapping. <SPAN=20 class=3DGramE>"Co-transduction". C. <SPAN=20 class=3DSpellE>Lysogeny and Specialized Transduction 1. Some phage can = integrate into the=20 host cell genome and replicate along with host chromosome; will not = <SPAN=20 class=3DSpellE>lyse the host cell. 2. Can also leave host chromosome; if = <SPAN=20 class=3DSpellE>unintegration event is imprecise, host genetic = material may=20 be included. 3. These phage can now transfer host genes = into new=20 cell on infection. V. Transformation. A. Certain bacterial = strains=20 incorporate DNA from the environment through their membranes. This DNA = can=20 recombine into the genome. B. To pick up DNA, cells must be in a = "competent"=20 physiological state. C. "Co-transformation" frequencies can = be used to=20 map the location of genes relative to one another. VI. Bacteriophage = Genetics. A. Phage Phenotypes = defined by=20 ability to grow on a particular host, or by plaque morphology. B. Mixed infections between mutant = strains can be=20 used to map genes. I. Biochemical Nature of the Gene (Hartwell, = Chapter 6) A. The "Transforming = Principle".=20 Experiments with mutant strains of Streptococcus = <SPAN=20 class=3DSpellE>pneumoniae (pneumococcus), F.=20 Griffith, 1928. B. Avery, MacLeod and McCarty (1944). = <SPAN=20 class=3DSpellE>Biochemically fractionated <SPAN=20 class=3DSpellE>pneumococcal extracts and tested chemicals for = ability to=20 tranform. DNA =3D transforming principle. = C. Hershey-Chase Experiment (1952). The = material=20 injected into a bacterial cell by a phage programs the phage = reproductive cycle=20 in a lytic infection. That material is = DNA. II. Reluctance to Accept DNA as the Genetic = Material. A. DNA structure = originally=20 considered a monotonous polymer. 1. base=20 composition 2. phosphate-sugar backbone 3. strand = polarity B. Equivalence Rule of = <SPAN=20 class=3DSpellE>Chargaff (1950). III. X-Ray Crystallography A. Three groups = working on=20 developing 3-D reconstruction: Wilkins and Franklin, Watson and Crick, = (at=20 <st1:place w:st=3D"on"><st1:City w:st=3D"on">Cambridge</st1:City>,=20 <st1:country-region = w:st=3D"on">England</st1:country-region></st1:place>) and=20 Pauling (CalTech). B. Franklin produces first high = resolution images.=20 C. Watson and Crick incorporate density = data, base=20 composition and crystallography into their model. 1. Watson-Crick = Structure. <SPAN=20 class=3DGramE>The double helix. 2. Implications of Model. "It has not = escaped our=20 notice that the specific pairing mechanism we have proposed immediately = suggests=20 a possible copying mechanism for the genetic material"- conclusion of = 1953=20 Nature article proposing DNA structure. 3. B-form DNA relative to A-form, = Z-form. IV. DNA Replication. A. Models for DNA = replication.=20 B. Proof of the semi-conservative model: = The <SPAN=20 class=3DSpellE>Meselson-Stahl Experiment (1958). C. Origins of Replication: <SPAN=20 class=3DSpellE>Autoradiographic evidence for bidirectional = replication. V. Synthesis at the Replication Fork. A. Properties of DNA =