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by: Charles Kohler


Charles Kohler
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A. Smith

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A. Smith
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This 146 page Class Notes was uploaded by Charles Kohler on Tuesday October 13, 2015. The Class Notes belongs to BIOL 2153 at Louisiana State University taught by A. Smith in Fall. Since its upload, it has received 8 views. For similar materials see /class/222822/biol-2153-louisiana-state-university in Biological Sciences at Louisiana State University.

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Date Created: 10/13/15
Chapter 3 Part II of II Extensions to Mendel Multifactorial Arising from 1nteraction of 2 or more genes 2nteractions between a gene and other factors Two genes can interact to determine one trait I Combined action of 2 genes I Two purebreeding white sweet pea lines I Tan and gray lentils Copyright The McGrawHill Campanies Inc Perm issiun required forrrepmduction or display b A dihybrid cross involving complementary gene action 3 in P Aibb gtlt agiBB Gametes A b a B F1 ail identical QAa 8b X CTAa Bb F2 A B A b a B a b A 5 AABEAA BbAa BBAaB 9 A BApurpla A b Man4A bb A39aBb Aa bb BM bb 7 333 B iwhite a B AaBBAa 353388133 Sb 138 bb a b 43 Rb Aa bb a a Bbaa bb Copyright The McGrawHill Companies Inc Permission required for reproduction or display AA or Aa 83 or Bb Enzyme A Colorless Enzyme B gt precursor gt Pumple pigmeni BB or Bb W No EnzymeB No gt6 colorless Purple precursor pigment 2 AA or A3 bb Enzyme A Colorless No gt precursor gt purple 2 pigment bb No enzyme A No No enzyme B colorless precursor 2 CopyrighwEThe MCGIEWHHJ Companies Inc Permission raquired for repradLmEon 0r display a A dihybrid cross with lentil coal colors What could be P W X 3388 amolecular Ag g explanation F1 all identical Aa Eb Aa Sb F2 A B A b a B a b u 39 39 H quotJ J A 8 AA BBAA BbA as Bbz 9 A B brown J i 91 A3 39 1 V 39 39 r 3 A MD HawI A b AA BbLaA bbAa BbAa bb W I V K 39 3 aa 8 gray x k 1 83 W green 6 B 153 BgAa Bba a B aaB quot W J u U I jg Aa BbMa bb aa 5 ea Db Two genes can interact to determine one trait I Combined action of 2 genes I Two purebreeding white sweet pea lines I Tan and gray lentils I One gene masks another Epistasis I Bombay phenotype I Coat color Copyrighl The McGrawHill Companies Inc Permission required or reproduclion or display b Molecular basis of the Bombay phenotype if H genotype 0 phenotype d Substance H P w A sugar No substance H B sugar 3 hh genotype Bombay phenotype B allele is dominant and determines black b allele is recessive and determines brown If one gene BB and Bb black bb brown 80 where does the yellow come from I as 139quot L a 5 r Capyrigm The McGrawHilh Companies Inc Permission reQuired or reper uclion or display a A dihybrid cross showing recessive epistasis Gametes F1 all identical Black Black Q 3 Es x d 5955 J a BB EE BB Ee Bb EE Bb Ee L 1 mm 1 re quotI 7 V H I a l BB or Bb 3 Default color is essentially colorless EE or Ee E Copynr A rawHlllcnmpames lnc eumssunreamedlaepmmcmuamsmav a A dlhybl ld cross showing recessive epistasis Black Yellow 58 EE bb ee Gameles B E F all identical Black 939 Bb Ee F2 B E black F w Eb E9 Ebee bl Ee Dbee Default color is essentially colorless Copy yr ra quotnmpan as n is m re a A dihybrid cross showing recessive epistasis Black Yellow 58 EE bb ee Gameles B E l F all identical Black Bb E a 5 5 black 3 UL E Drownl Black Labrador Co pam la tea a A dihybrid cross showing recessive epistasis Black Yellow P 58 EE Gameles B E l F all identical Black 5quot Bb Ee Eb E9 Ebee bl Ee Dbee EE or Ee E bb E Brown Chocolate Labrador Default color is essentially colorless Copy vl ra quotnmpan as 55 on lE la 5 o a A dihybrid cross showing recessive epistasis Black Yellow 58 EE bb ee Gameles B E l F all identical Black Bb E a BB or Bb B Eb E9 Ebee DD Ee Dbee ee Yellow Golden Labrador Default color is essentially colorless Two genes can interact to determine one trait I Combined action of 2 genes I Two purebreeding white sweet pea lines I Tan and gray lentils I One gene masks another epistasis I Bombay phenotype I Coat color I Mutation at 1 of 2 or more genes can produce same phenotype A mutant allele at 1 of 2 or more genes yields the same phenotype I eterozygous trait I Deafness in humans may be caused by a mutant allele at one of more than 50 different genes I Albinism I Complementation test I Genetic tool to determine whether a phenotype arises from mutations in the same or separate genes Albinism Copyright The McGrawHill Companies Inc Permission required for reproduction or display a Complementation mutations E E in two different genes AA bb X aa BB i F1 AaBb Genetic mechanism of complemeniation b Noncomplementation mutations in the same gene E E P AA bb x AA bb 1 F1 AAbb Genetic mechanism of noncomplementation The same genotype does not always produce the same phenotype I Phenotype often depends on penetrance andexpressivity I Penetrance percentage of a population with a particular genotype that show the expected phenotype I can be complete 100 or incomplete eg retinoblastoma penetrance is 75 I Expressivity degree or intensity with which a particular genotype is expressed in a phenotype I can be variable or unvarying Phenotype affected by 1 Chance 2 Modifier genes which have subtle secondary effects on a phenotype 3 Environment I Temperature is a common element of the environment that affects phenotype Cooyright E3 The McGrawHiil Companies lnc Permission required for reproduction or display w W 39 quotI v Wing squot m Tip l 39 39t 7 h V t l l Colorless d b precursor u melanin fur Cooler 430049 Colorless lEWF39BrallIFE L temperature 9 33 WSW rw imru lmmlraunl h Melanin Dar fur rigor we arr i 5 lb J Flenee LynniPhoto Researchers inc Coat color in Siamese cats is darker in its extremities legs tails ears etc because of a mutation that renders an enzyme involved in melanin synthesis temperature sensitive Continuous variation can be explained by Mendelian analysis Previous examples of Mendelian inherited traits were discontinuous clear cut phenotypes Continuous traits such as height in humans are determined by segregating alleles of many genes interacting with one another and the environment Continuous traits are called quantitative traits by geneticists and are usually polygenic The more genes that contribute to a trait the greater the number of possible phenotypic classes and the greater the similarity to continuous variation Copynghl The McGrawHIH Campanies Inc Pemussion quuived ior repvoducuon or dispiay Northern European whites African blacks Children of mixed marriages Mating of F1 individuals Amount of dark pigment in skin Two genes can interact to determine one trait I Combined action of 2 genes I Two purebreeding white sweet pea lines I Tan and gray lentils I One gene masks another epistasis I Bombay phenotype I Coat color I Mutation at 1 of 2 or more genes can produce same phenotype One gene can interact with other factors to determine one trait Penetrance and expressivity Modifier genes Environment Chance Monday January 25 2010 Chapter 2 Part II of II Mendel s Breakthrough Assignment Chapter 2 problems 15 79 1115 1820 22 23 26 29 33 Dihybrid cross I Dihybrid an individual that is is heterozygous for two genes l Mendel designed experiments to determine if two genes segregate independently For example yellow round peas YY RR X green wrinkled peas yy rr Gameles F2 F2 contained garental tyges and recombinant type Type Genotype Phenotype Number Phenulypic Parental Y Fab yellow round Hecambinann yy 8 green round Recombinam Y rr yeliow wrinkled Parental yy rr a green wrinkred Ratio of yellow dominanlj 10 green recessive Ratio of mund 39dominant to wrinkled 39recessive 315 108 101 32 ratio 9quot 6 EMS 3 1 6 116 124 or 31 124 or311 i i 1 W l 7 7 l llMEMK Wermw wl ueQeWQETIH During garnete formation different pairs of alleles segregate independently Possible allele combinations in gametes Alleles in Gamete parental cell formation Cl fira pm ix39l a E Jim 7 l H ll eo hzoffdbidlof What is the probability that a pea will be green AND wrinkled We need to use the PRODUCT rule Treat each trait individually Probability ofgreen A Probability of wrinkled A Probability of green and wrinkled A X A 116 Gameles F all identical Q Copynght The McGrawiHill Companies Inc Permissxon required for reproducuon or display Cross 8 YY Rr gtlt Cross C P YyRR x F1 Mendel s work Strength Can make accurate predictions about offspring of complex crosses Limitation Cannot predict phenotype of any one individual Rediscovery of Mendel Mendel s work was unappreciated and remained dormant for 34 years Even Darwin s theories were viewed with skepticism in the late 1800 s because he could not explain the mode of inheritance of variation In 1900 16 years after Mendel died four scientists rediscovered and acknowledged Mendel s work giving birth to the science of gene cs Summary of Mendel39s work I Both parents contribute equally I Inheritance is particulate not blending I There are two copies alleles of each gene associated with a given trait I Gametes contain one copy of the gene I Alleles segregate randomly I Alleles are dominant or recessive I Different pairs of alleles segregate independen y Mendelian inheritance in humans l Most traits in humans are due to the interaction of multiple genes and do not show a simple Mendelian pattern of inheritance I Some traits represent singlegenes Examples are cystic fibrosis and Huntington s disease In humans we must use pedigrees to study inheritance l Pedigree a diagram of a family s relevant genetic features through multiple generations I They help us infer if a trait is from a single gene and if the trait is dominant or recessive Anatomy of a pedigree Comm 2 The McGrawHaii Companies Inc PErmvssin w required tar repioduclmn cw dismay 2 i 3900 Z 23 D20 Mating line Sibship line 1 A Line of descent Generation It i 1 2 Individuai number within generation Huntington s disease A dominant trait 7 i at it i iffi it 37 I j v u H7 77 r r t i m a C 7 4 is rm 7 x 3 t r p7 J 3 7 gt r i i I m m iH i X V J J L t J A V39n i J Cystic fibrosis a recessive trait Friday September 4 2009 Chapter 4 Part of III The Chromosome Theory of Inheritance Chapter 4 problems 112 1517 19 22 26 Evidence that Genes Reside in the Nucleus I 1667 Anton van Leeuwenhoek I Microscopist I Semen contains spermatozoa sperm animals I Hypothesized that sperm enter egg to achieve fertilization I 18541874 confirmation of fertilization through union of eggs and sperm I Recorded frog and sea urchin fertilization using micrographs Evidence that Genes Reside in Chromosomes I 18803 innovations in microscopy and staining techniques identified threadlike structures chromosomes colored bodies I Provided a means to follow movement of chromosomes during cell division Chromosomes contain the genetic material This material needs to be maintained transmitted in accordance with cell division 3 types of cellular reproduction ie division exist binary fission mitosis and meiosis A Eukaryote B Eukaryote C Prokaryote D Eukaryote diploid two haploid haploid single diploid copies of each COPY 0f chromosome chromosome t i haploid haploid haploid diploid 1 Binary fission 2 MItOSIs A Eukaryote 39 d39 Io39d two 3 M cogiesl of each chromosome 11 haploid 1 Binary fission 2 MItOSIs BEukaryote 3 Meiosis hap39 39d haploi 1 Binary fission 239 M ItOSIS C Prokaryote 39 39 haploid sin Ie 3 MeIOSIs copyof 9 chromosome haploid 1 Binary fission 2 Mitosis DEukaryote 3 MeIOSIs diploid diploi Brainstorm Why are there 3 types of cell division P ka ot39c cell Prokaryotic w W I chromosome Doublestranded DNA of DNA I 1 339 I S xquot y r 397 quot quotxxquot 2 39 a iquot x 39 y I N 39 l I I I x I x r39 39 39H s 39 v s 7 Elongation of kc ll Active separation of chromosomes gt Why mitosis and meiosis MitOSiS Meiosis Eukaryote Eukaryote Eukaryote haploid diploid diploid haplid diploid haploid neither Packaging of DNA in a chromosome o oomm MLMEQZCMO 1E aaea 3 E 335 55633E3 Anatomy of a chromosome Copyright The McGrawHill Companies Inc Permission required lor reproduction or display Pair of Homologous Pair of Homologous Metacentric Chromosomes Acrocentric Chromosomes Centromere Centromere 39 Sister Nonsister chromatids chromatids Nonhomologous chromosomes I Homologous chromosomes Homologous chromosomes I Homologous chromosomes homologs contain the same set of genes The genes may carry different alleles I Nonhomologous chromosomes carry completely unrelated sets of genes Mitosis is crucial for development An Overview of Cell Cycle lnterphase G1 8 PHASE amp G2 MITOSIS Cytokinesis New cells start the cycle all over again Mitosis ensures that every cell in an organism carries the same chromosomes I Cell cycle repeating pattern of cell growth and division I Alternates between lnterphase and mitosis I lnterphase period of cell cycle between divisionscells grow and replicate chromosomes I G1 gap phase birth of cell to onset of chromosome replicationcell growth I S synthesis phase duplication of DNA I G2 gap phase end of chromosome replication to onset of mitosis The cell cycle a The cell cycle 1 l39erpl39leS Chromosome replication during 8 phase of cell cycle b Chromosomes replicate during 5 phase U B B A Aall ll ra K quot b b Synthesis of chromosomes Note the formation of sister ch romatids lnterphase I Within nucleus I G1 S and G2 phase cell growth protein synthesis chromosome replication I Outside of nucleus I Formation of microtubules radiating out into cytoplasm crucial for interphase processes I Centrosome organizing center for microtubules located near nuclear envelope I Replicate I Centrioles pair of small darkly stained bodies at center of centrosome in animals not found in plants lnterphase 1 2 Replicate chromosomes Replicate centrosomes l P M A T Ccytokinesis Prophase Metaphase Anaphase Telophase a 1 3 Centrosomes 6 Complete 7 Chromatids 9 Cleavage migrate to the spindle separate furrow forms poles attachment 8 Chromatids 10 Nuclear 4 Chromosomes move to the poles membrane condense and reforms nuclear wall breaks down 11 Spindle apparatus 5 Spindle bers disintegrates a ach to kinetochore Kinetochore microtubules Centromere Centrioles Interphase Prophase Metaphase Anaphase Telophase a J it v A 39 A mutation occurs that affects the process of mitosis Cells with this mutation get stuck in metaphase Which process is the gene that is mutated most likely to affect l PONFDFnerDN Centrosome replication Centriole migration to the poles Chromosome condensation Spindle fiber attachment to kinetochore Completion of spindle attachment Sister chromatid separation Chromatid movement to the poles Cleavage furrow formation is cell of sufficient size Chromosome 39fon eeillfifn39nffnT39s aquot 39 a he been received cemmsome Chromosomes been completely THEN DeCISIon to dupicated9 dupllcate THEN Decision chromosomes and m emer mitosis centrosomes lnte Ongoing protein synthesis nd a Metaphase cell growth Have all chromosomes arrived and Anaphase aligned at the metaphase Telophase and cytokinesis THEN Decision to initiate anaphase Checkpoints help regulate cell cycle What happens if the cycle breaks down One example rapidly dividing cells Karyotyges can be produced by cutting micrograph images of stained chromosomes and arranging them in matched pairs n 5 r 5 B 19 20 71 22 x Gametes vs zygotes Gametes have one set of chromosomes 0 Hagloid Zygotes have two matching sets of chromosomes 0 Digloid 1n the number of chromosomes in a haploid cell Diploid cells 7 r 2n8 2 the number of chromosomes in a diploid cell a E l an X 0 Haploid cells gametes n 4 One Chromosome Pair Determines an lndividual s Sex I Walter Sutton Studied great lubber grasshopper I Observed that spermproducing cells contained 24 chromosomes I 22 were in 11 matched pairs autosomes I The other 2 were unmatched he termed X and Y E chromosomes I Daughter cells contained 11 autosomes and an X or Y After fertilization I Cells with XX were females Cells with XY were males Great lubber grasshopper Brachystoa magna Sex chromosome 3 Provide basis for sex determination a One sex has matching pair o Other sex has one of each type of chromosome Photomicrograph of human X and Y chromosome on K 3 W NE 9 m d Sex determination in humans Children receive only an X chromosome from mother but X or Y from father Sex determination is not the same for all organisms Fruit flies Ratio ofX to autosomes ie the amount of X Y is still important for fertility Complement of sex chromosomes XXX XX XXY XO XY XYY OY Dies Normal Normal Sterile Normal Normal Dies Female Female male male male Drosophlla Nearly Normal Kleinfelter Turner Normal Normal Dies Normal Female male female male or nearly Humans Female sterile sterile normal tall thin webbed male neck Sex determination is not the same for all organisms capynghte rhe McGrewHlll Companies In l ermissinn required inr reproduction Ur diSDlay ABLE 42 Mechanisms of Sex Determination Q oquot Humans an xx XY Drosophila Moths and xx C elegans hermaphrodites X0 in C elegans Birds and 2w zz Butterflies Bees and Diploid Haploid Was 5 Lizards and Cool Warm Alligators temperature temperature Tortoises and Warm Cool Turtles temperature temperature Anemone Fish Older adults Young adults geyse In m l hl39 in pain may 39 Note that the sex of the offspring is determined by the type of gamete re ceived from the parent of the heterogametic sex that is the gender pro ducing two different types of gametes XV X0 or ZW in the table The species highlighted in yeiow have identical chromosomes in the two sexes and sex is determined instead by environmental or other factors Anenome fish bottom row undergo a sex change from male to female as they age in other 7 39 animals 394 quot 394 Change to males as they mature k K M n 5 I 2 u u i 5 3 l I In 3 I I I in all n In If i Meiosis l Chromosomes replicate once prior to Meiosis l Rounds of nuclear division 2 Friday January 22 2010 Chapter 2 Part of II Mendel s Breakthrough Assignment Chapter 2 problems 15 79 1115 1820 22 23 26 29 33 PreMendel I Artificial selection I Definition Purposeful control over mating by selecting parents I Practiced since before recorded history I Domestication of animals I Selective breeding of plants I Breeders could not explain why traits would sometimes disappear and then reappear in subsequent generations Historical theories of inheritance 1 Male parent contributes most features eg homunculus N Hartsoiker 1694 How was this hypothesized Historical theories of inheritance 2 Blended inheritance parental traits become mixed and forever changed in offspring State of genetics in early 1800 s What is inherited How is it inherited What is the role of chance in heredity 1884 Gregor Mendel 1822 1 2 3 4 5 6 Keys to Mendel s experiments Chose garden pea as model 1 Selffertilizing but easy to crossfertilize 2 Produced large number of offspring each generation H Analyzed discrete traits Used purebreeding lines Controlled matings and reciprocal crosses Analyzed large numbers of plants and applied mathematics Mendel was awesome Generation Parental P purebreeding Yellow peas Green peas 0quot pollen 1 Q 9995 All yellow Selffertilization Second filial F2 6022 yellow 2001 green 3 1 Traits have dominant and recessive forms I Disappearance of traits in F1 generation and reappearance in the F2 generation disproves the hypothesis that traits blend I Trait must have two forms alleles that can each breed true I One form must be hidden when plants with each trait are interbred I Trait that appears in F1 is dominant I Trait that is hidden in F1 is recessive Parents contribute equally I Reciprocal crosses showed no difference in progeny e es or eac trait separate urrng gamete formation Gametes pollen or eggs Grows into Iant Gamete lt formation Yerllow pea G from a purebreeding stock Grows into lant Gamele formation yy green pea from a purebreeding stock Two gametes one from each parent unite at random at fertilization Gametes Zygote F1 Hybrid one pollen grain one egg Seed Fertilization develoment Yy yellow pea showing dominant trait Y yellowdetermining allele of pea color gene y greendetermining allele of pea color gene Two alleles for each gene separate segregate during gamete formation and then unite at random one from each parent at fertilization Rules of Probability Mutuallv exclusive events probability of one or another event occurring choosing one excludes the other Sum Rule What IS the probability ofA g B occurring Solution determine probability of each and them together Independent events probability of two events occurring together P What happens in the rst experiment won t affect the roduct Rule probabilities of the second experiment What is the probability thatA m B will occur Solution determine probability of each and multiply them together Probability and Mendel s Results I Cross Yy x Yy pea plants I Chance on sperm uniting with a Y egg I 12 chance of sperm with Y allele I 12 chance of egg with Y allele I Chance on and Y uniting 12 x 12 A I Cross Yy x Yy pea plants I Chance of an offspring being Yy I Two ways of being Yy Yy or yY all possibilities are YY Yy YY or W I Probability of Yy 1A H Self fertilization g0 Sal 11W m 1m w m 5 l 0 Self femllzatlon Selffertilization Genotypes and Phenotypes I Phenotype observable characteristic of an organism I Genotype pair of alleles present in an individual I Homozygous two alleles of gene are the same YY or yy I Heterozygous two alleles of gene are different Yy Genotype for the Seed Color Gene YY Homozygous dominant T F 331338 Yy Heterozygous W ous recessive Phenotype 121 YYYyyy 31 yellow green Test cross reveals unknown genotype Test cross Mating between an individual with the dominant phenotype and an individual with the recessive phenotype all ellow February 10 2010 Chapter 4 part III Exam is this Friday It will cover up through today s lecture chap 4 Wednesday February 1 0 2010 Celebrating Darwin 530 800 pm Science and Humanity Life Sciences Bldg A101 Darwin Day Brought to you by LSU BIOGRADS Comejoin us in celebrating the birthday of Charles Darwin with speakers Dr Bryan Carstens Dr Chris Austin Dr Andrew Whitehead and Dr David Donze Dept of Biological Sciences LSU Refreshments will be provided Milli DARVILV DAY www quot 391ml 1mg FEBRUARY 12 in In r Hirilldwr39v Darwin s birthday Trisomy 21 Down s syndrome ll l ll 3 M ll gt1 ll 1 ll 1 ll l lull lull But I ill I a I I l I I gnug Only other complete trisomies that have led to live births Trisomy 13 Patau s Syndrome Trisomy 18 Edward s Syndrome Trisomy 9 How many cell divisions are in Meiosis 1 One 3 Three 4 Four 5 None How many chromosome replications are in Meiosis 1 One Two Three Four There is one chromosome replication BEFORE Meiosis not IN Meiosis PWN Meiosis Chromosomes replicate once prior to Meiosis Rounds of nuclear division 2 l l 7 2n Chromosomes duplicate xix imieugsl ifaltion K x M 39 39 II a I No guislilzation S In 4 n n Jn Don t let this confuse you Copyright The McGraw Hil l Companies Inc Permission required for reproduction or display ABLE 43 Himsis ani M insis Comparigon Hapiuid and diploid cells can undasrgo mitosis Mitnsla Is needed by S phase chromoma duplicalinn Meiosis Dmurs in germ cells as part 2 he semuai cycle Twn murals n wiaiun meiosis I and meiosis II Only dlplbid sells undergo metoals Maiosis l Meiosis 1II B S 393 J I 1 3 Chromammes duplicate prim Io meiosis I but not J before meiosis II lniarkinasis Gamma romaiiun Meiosis I Meiosis I Meiosis II I Prophase I Prophase 39 LePtOtene I Metaphase I Anaphase II I Diplotene l Telophase II I Diakinesis I Cytokinesis I Metaphase I Anaphase I Telophase I Interkinesis Meiosis Prophase s I A leduc on divismn onphm I Lemmena 1 Chmasomes mm and became vlsmla nun me hlumallds remam mm la 2 Csmrasumas magi la mm lowards opposne mes phna I Zygolene nmamgnus chvommomes emev Synapsls he synaplaneml compax mans Flophlse 39 P54 5 Imam ynansis gt5 complete ross39 aver genetic exchange bewesn ans an hroma ds cl mmo ogcus Dav occurs 13 Meiosis Prophase I continued Crossing over during Prophase produces recombined chromosomes gtc Fgtc How crossing over produces recombined gametes comrith a The McGrawHill companies In emission required my repmuucaon ardisnlay Susrer cmamaua 1 39 Synap onemal comprex sisev cmoman39d 2 Homoragaus chmmosames I ecombr nahan oauras a Leplmene momauds ol each chromosome are nor or 5mm yer visible in the microscope Meiosis Metaphase and Anaphase M aphase I Anaphaapam Talrad5 Mina up alcng the malaphase plate Tm Bammmera does not divide 2 Each chmmusome DE 3 homnl guus pair The chiasmata migrala nh chromaticll attaches In bers 1mm appoalte paras am lands 3 Sister chmmalids anach ta bers 1mm the Lfnmu lngaus chmusnmes move to same pone ppmsite 130135 Meiosis Telophase and Interkinesis Taigphasa i Interkinesis The nuclear envaiapa m inrms i This is similar In iniamhase wiii39i one Fiasuiiani cells haw hali iiia number DI imm ani Biwmi i NC Ehrm mai chmmasnmas each consisting oi iwo dupiicaii m rakes mace sister chmmaiids 2 In some 543mm the chmmcisnmas decandemaa in oiharsiiie1r d0 mail I Meiosis similar to Mitosis Prophase II and Metaphase Meiosis ii An aquamionai divisiun Q Prophase ill Metaphase Ii 1 Chmmnsomes mndense 1 Chmmosornes ailign at the melaphaaae 2 Centriulas move toward the poles Diana 3 The nuciear envelan brEaks dawn at the 2 Sister chrumatids attach tn spindia fibers and cl pmphas e II mi shuwnL 1mm uppaslte pales Meiosis Anaphase II and Telophase Meiosis Cytokenesis Meiosis contributes to genetic diversity in two ways Independent assortment of nonhomologous chromosomes creates different combinations of alleles among chromosomes 2 Crossingover between homologous chromosomes creates different combinations of alleles within each chromosome enpyngmmne MoGrnwHiil ammonia in emmn required nor revmduchan or dismay a Independent assortment b Recombination Orientation Orientation II A B A b x Meta hasel gtltr x p gtalt 9 Prophase l A E A Meiosisl x Telophase l a V g 3 7 56 k lt Meiaphasel a A gt5lt I gtlt Metaphase I i a E 5lt xx gtMeiosis II R A quot 5 Meiaphase 5 a L 9 c quot Telophase H i B J Gametes 539 A q 1 u 5 m A 3 b Gametes I K n I D 5 5 J Gametogenesis involves mitosis and meiosis Mitosis Meiosis Germ cells Gamete precursor Gametes diploid cells diploid haploid Ooqenesis female gametogenesis Oogonia Primam oocyte Ovum egg I m Mitosis Meioiis Germ cells Gamete precursor Gametes diploid cells diploid haploid Clicker How many ova eggs are produced from each female germ cell ImmAmM x CDO IbeMA ZGFO Oogenesis female gametogenesis I Diploid germ cells called oogonia multiply by mitosis to produce primary oocytes I Primary oocytes undergo meiosis l to produce one secondary oocvte and one small polar body which arrests development I Secondary oocyte undergoes meiosis II to produce one ovum and one small polar body I Polar bodies disintegrate leaving one large functional gamete ovum egg Oogonia Primary oocyte Ovum egg I Mitosis I Meiosis III E lllll Q Germ cells Gamete precursor Gametes diploid cells diploid haploid Oogenesis in humans Spermatoqenesis male gametogenesis spermatogonia Primary spermatocyte in il m D i 3 Germ cells Gamete precursor Gametes diploid cells diploid haploid Clicker How many sperm are produced from each male germ cell ICDU IACONA DO ILOONA ZGFO Spermatoqenesis male gametogenesis I Begins in male testis in germ cells called spermatogonia I Mitosis produces diploid primarv spermatocvtes I Meiosis l produces two secondarv spermatocvtes per cell I Meiosis ll produces four equivalent spermatids I Spermatids mature into functional sperm spermatogonia Primary spermatocyte ll u m D i 3 Germ cells Gamete precursor Gametes diploid cells diploid haploid Spermatogenesis in humans Imbalanced chromosomes following Meiosis 1 Nondisiuction I Failure of chromosomes to segregate during meiosis I Results in trisomies trisomy21 XXY etc and monosomies XO etc a The McGrawHHI Companies Inc Permissinn required or repmdumlon or dvsmay I I a Nondigsjunction in an XX fema e 9 White eyed oquot Redeyed w w x M 7 l W Gametes Meiosis Non W Normal disjunction segregation XX 0 l X Y F X Y w w 000 N X X XwaXW wawY dies white 9 0 red XWquot0 QY I steriled dIeS Imbalanced chromosomes following Meiosis 1 Nondisiunction I Failure of chromosomes to segregate during meiosis I Results in trisomies trisomy21 XXY etc and monosomies XO etc 2 Hybrid sterility I Chromosomes can t form perfect pairs results in imbalanced chromosomes in gametes CnpylingTne MnGrawHm campum rm Farmlman nun Var mprcduumn ovd sphy Donkey 39 k 9 Donkey g j I 16 I 23 Donkey 22 XIinked traits XIinked traits Chapter 3 Part Extensions to Mendel singlegene inheritance Chap 3 problems 26 911 15 2325 30 31 34 39 Extensions to Mendel for singlegene inheritance 1 Dominance is not always complete Incomplete Dominance Codominance 2 Genes can have more than two alleles 3 One gene may contribute to multiple traits Incomplete Dominance the progeny have a trait in between those of the parents Lopyngnm I p 5 W a Antirrhinum majus snapdragons b A Punnett square for incomplete dominance A P AA gtlt aa Gametes A a 4i 437139 F1 all identical 9 Aa gtlt o Aa F2 A a A fx x AA Aa a it 1 Aa aa 1 AA lredl 2 Aa mink 1 aa gwhitel Variation occurs due to the additive effect of gene products proteins A a I C 37 O a 3 AA Aa aa A A a A a a I I I l l I C 3 3636 Q7 4E Q 00 0 I Co dominance both traits are visible Copyxighz The McGrawHill Companies Inc Permissicn reguired for reproduction or display a Codominant lentil coat patterns P 0905 x 0000 l J Gametes 03 CD F1 all identical Q 03013 X 0739 0800 3 CD CS 00 CD a 0900 CDCD 1 0909 spotted 2 0900 spotteddotted 1 CDC dotted I2 Cs 0amp6 Ca Another example Blood A B AB RRRRRR od cell A and B A sugar B s u g a r s u g a rs 39 a 5quot 39 J P M i X aa 1 JP R F If Summary Type of Dominance Complete Complete Illncomplete Codominant Copyright The McGrawHill Companies Inc Permission required for reproduction or display AWA A2A2 A IA39 hybrids EXE lX Qgtltv xm A1 is dominant to A2 A2 is recessive to A A2 is dominant to Af AT is recessive to A2 A7 and A2 are incompletely dlomirtant relative to eactn other A1 and Agate codominant relative to each other Do variations on dominance relations negate Mendel s laws I Dominance relations affect phenotype and have no bearing on the segregation of alleles I Gene products control expression of phenotypes differently I Mendel s law of segregation still applies Alleles still segregate randomly I Interpretation of phenotypegenotype relation is more complex Multiple alleles in human blood type LopynghlG M lnc b Codominant blood group alleles Blood Type A Red blood cell A sugar a Genotypes Corresponding Phenotypes AB Types of Molecule on Cell WA A and IA A sugars 0 B B 0 I I 5 B IAIB AB Mutations are the source of new alleles I Novel alleles arise spontaneously in nature due to chance alterations in genetic material I Mutation rate varies from 1 in 10000 to 1 in 1000000 per gamete per generation I Allele freguency is the percentage of the total number of gene copies represented by one allele I Wildtype allele whose frequency is more than 1 I Mutant allele allele whose frequency is less than 1 B advantageous A gtC neutral D disadvantageous Clicker questions advantageousdisadvantageousneutral mutations will become stable and persist Advantageous disappear over time Disadvantageous have a 50 chance of becoming stable have a slight chance of becoming stable Neutral 39 39 Parental Generation F Generatian ereneration ommance senes F pheno panem m crass lrequencles Parenzl gtlt Parentz an pic phenotypes ratio a a z m x a m a 3 2 E m 1 l D on N D m u A gene for seed coat pattern in lentils has 5 alleles Spotted dotted clear marbled1 and marbled2 3 2 E m H m X 2 m 2 a 2 E m l m E m gt m u 0 9 0 v u 9 77 a x a m 2 1 m 1 9 E a m m w o l o a o 2 a a X l m a a o z m a K o a m m m w 3 E E m 3 X E 2 E m 9 N l 3 m a m l g N s m u 0 4 6 3 9 3 av E m 3 X m E a m n a 2 a m 9 g to w b w a g E m 9 X a o m a a m E m 9 in m 1 u s m a a a E m 21 m X a o a m a 3 2 z m 9 T E m u o w 9 9 0 9 0 c edasponeddo edow 339 157 1221 w u 2 a a X n arbled72 gt spoiled dotted gt CleaVI The Maori mao ree Cilia and Flagella Structure 39fnl iquot m 9 Mal cm u u a pa 2 cm 39 WW9 Mlnroluhule A Axoname Pleiotropy sin gl e gen e determi n es Hairlike prujections called a n mill mimequot Cilia line the primary run us to remove microbes and debrls on e n I from the interior of the lungs an d see m i n gly unrelated trait A human example that illustrates many extensions to Mendel Hba 9 uglobin Hb8 9 Bglobin Oxygen molecule Red blood cell Group activity Why has the sickle allele perpetuated 1 Explain the results in this graph 2 What would the graph look like if it represented the population of the United States The correlation of life expectancy and genotype of people in western Kenya A human example that illustrates many extensions to Mendel Copythl The McGraw Hill Companies Inc Perm ission required tor reproducTion or display mEES n i EESSS of NWquot Cam D39seased neiai imZI EZch A A A s s 5 Analysis Hb Hb Hm pr b b Level of Analysis o o livglobin polypeptide r L HbjA and HleS production g 39 V are codominant 0 o o o 0 Red blood cell 9 shape at sea level Sickled cells q quot pre ent 1 Hb is dominant i Hb S is recessive Red blood cell concentration at sea level Lo er w Red blood cell O J quot shape at high altitudes Sickled cells Hb and Hb s U9 LB Qn show incomplete r dominance 3959 I I5 393 I 92 1 lg i 3 I I I Ft Red blood cell concentration at high altitudes Normal Lower Susceptibility J Iquot iquot 1 4 H0133 is dominant to malaria quot39 5 3 t f I tiquot HbIA is recessive Normal l l i SUSGePliUiIiW Resistant Resistant b Group activity Evolution of blood types I Recently 2007 this hypothesis gained wide notoriety people with type 0 blood have better resistance to malaria Knowing that type A is the progenitor type what can you infer about the evolution of the other blood types 0 45 A 40 B 10 AB 5


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