PCB 3063General GeneticsLecture Notes Exam 2This set of notes is 16 pages. If you want to find notes for a specific date (ex: 1/27) or topic(codominance), press Ctrl F on your keyboard (Command F for Mac users) and type in what youwant to find. It’ll search the document for whatever specific word/phrase you typed in and saveyou the time of scrolling through everything!--------------------Mon 2/10Cytoplasmic inheritance- extranuclear DNA passed on to offspringExtranuclear- in mitochondria/chloroplastGenome in mitochondria/chloroplast dif than DNA in nucleusMitochondria/chloroplasts thought to be descended from symbiotic bacteriaVariegated-plant w/ 2 dif colors on leaves or branchesChloroplast- green Glucoplast- whiteParent cell has the 2 dif typesParent cell divides, cytoplasm segregatesDaughter cells have one or the other colorEx: 4 o’clock plantWhite female x ___ male = all whiteGreen female x ___ male = all greenVariegated female x ___ male = all variegatedOffspring color only depends on motherMaternal inheritance- only mom’s cytoplasmic info passed onMaternal mitochondrial info passed on to offspringMitochondria get mutations over timeEgg can take mostly mutated, partly mutated or few mutated cellsMaternal effect mutations- nuclear genes, expressed as alternate phenotypeCome from mother’s eggsRequired for embryonic developmentNot influenced by offspring’s own genome at all, only mother’s (from egg, not nucleus)Mutation expressed when mom is homozygous recessive for trait, aaIf she is, ALL offspring are affected (all of her eggs are “bad”)Linkage- 2+ traits correlatedIndependent assortment- nuclear, chromosomes/loci split during cell divisionSegregation- mitochondrial, mitochondria splits (linkage)Ex: Mendel’s peas showed indep assortment, not Mendelian inheritance4 chromosomes with 7 traits#1- 3 loci#2- 2 loci#3- 1 loci#4- 1 loci Showed linkageEx: Punnet experiemntsFlower color + pollen shape in garden peaIn F2, expected 9:3:3:1 ratio, got 15:1:1:3Threw out hypothesis of Mendelian inheritance, turns out it was actually linkageEx: Fruit flyMutations in eye color, wing shape and body colorOriginally thought it was X-linked linkageEye color- red (normal) or purple (mutant)Wing shape- normal or vestigial (mutant)Flies with red eyes tended to have vestigial wingsFlies with purple eyes tended to have normal wingsNot independent assortment but some kind of linkageWildtype- +, normal allele, most common phenotypeEx: normal red and mutant purpleRed = pr+ = +Purple = prEx: Normal/red female x Vestigial/purple maleP: + + v pr - sex chromosome 1 (X)+ + y - sex chromosome 2 (X or Y)F1: all normal/red + + + + v pr for female and y for maleMapping with 2 loci produces 4 dif types of offspringF2: 4 dif types of offspring Genotype Phenotype # of offspring+ + normal/red 1339 Largest #s = parental genotypesv pr vestigial/purple 1195v + vestigial/red 151 Smallest #s = new combinations + pr normal/purple 154 (recombinants)Total 2800Genetic distance- % of new genotypes in F2Genetic distance = (recombinant1 # + recombinant2 #) / total *100%Unit- centimorgan or map unit (mu)1% = 1 centimorgan = 1 muShows when there is no pleiotrophyRecombinant- new combinations of genesPaternal + maternal chromosomes combine in new waysHappens during meiosis when tetrads crossoverResults only revealed in F2 genPhysical distance- distance b/w genes on literal DNA strand, measured in base pairsNot same as % recombinantTester- individual used in test cross Bred to individual with unknown genotypeAlways male, homozygous recessive (no +’s)Recombination happens in female, not maleUse male to measure recombination in femaleCis configuration- +’s line up with +’s Trans configuration- +’s don’t line up+ + + vpr v pr +Mapping- measure connection b/w allelesDetermine map units between two different genesEx: FliesBody color- gray (+) or black (b)Wing type- normal (+) or vestigial (v)P: Gray/vestigial x Black/normal + v x b + Gender doesn’t matter in P gen + v b + Don’t use any Y’sF1: All Gray/normal + v testcross with Tester b v b + (trans) Male b vF2: 4 dif offspring typesPhenotype # of offspringGray/vestigial 42 parentalBlack/normal 42 parentalGray/normal 8 recombinantBlack/vestigial 8 recombinantTotal 100% recombinant = (8+8)/100 * 100% = 16/100 = 16 map units b/w body color and wing typeEx: FliesNormal/vestigial wing type and wildtype/cinnabar (c) eye colorP: Vestigial/wildtype x Normal/cinnabar v + + c v + + cF1: All normal/wildtype v + testcross with Tester v c + c (trans) Male v c (homozygous recessive)F2: Mapping with 2 loci, 4 dif offspring typesPhenotype # of offspringVestigial/wildtype 90 parentalNormal/cinnabar 90 parentalVestigial/cinnabar 10 recombinantNormal/wildtype 10 recombinantTotal 200% recombinant = (10+10)/200 *100% = 20/200 * 100% = 10 map unitsLinkage- no independent assortment, genes are all lined up on the same chromosomeWed 2/12Genes arranged linearly on chromosomeHave specific position/locationEach location can only have 1 gene (any of the one gene’s alleles are possible)Crossover happens in location between 2 gene loci of interestIf not, recombination isn’t noticeableRecombination % always < 50%50% = independent assortment, genes on separate chromosomesOr if it IS recombination, it’s indistinguishable from independent assortment> 50% = impossible, parental genotype frequency > recombinant genotypeEx: MappingCinnabar eye linked to vestigial wing, 10 map units apartVestigial wing linked to black body, 16 map units apartSo cinnabar linked to blackAll 3 loci in linear order on same chromosomeFind gene orderPossibilities: black---(16 mu)---vestigial---(10 mu)---cinnabar (black cinnabar = 26)black---(6 mu)----cinnabar---(10 mu)----vestigial (black vestigial = 16)Find distance between black + cinnabar to determine which order is right P: Black x Cinnabar + b c + If trait not mentioned, assume it’s normal (+) + b c +F1:
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