BIO207 ALL TUTORIAL.pdf

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Department
Biology
Course
BIO207H5
Professor
Fiona Rawle
Semester
Winter

Description
TUT1 January-17-14 2:23 PM Tutorials: focused on problems Problem sets: due at the beginning of the class SHOW WORK FOR PROBLEM SETS CHAPTER2 Genetics: science that deals w/ the structure & function of genes & their transmission from 1 generation to the next • Is the science of heredity • Variation w/in & btw species Branches of Genetics: Transmission genetics Molecular genetics Population genetics Quantitative genetics Chapter 2: Mendelian genetics Terminology: Allele Character(trait) Cross Diploid(2n) F1 generation (1st filial generation) F2 generation (2nd filial generation) Gamete: mature reproductive cells specialized for sexual fusion; haploid cells Gene: factors; determinant of a characteristic of an organism Haploid (n) Locus (loci) True-breeding/ pure-breeding Zygote Genotype - Genetic constitution of an organism - Homozygous(SS or ss) - Heterozygous(Ss) - Dominant (SS or Ss) - Recessive(ss) Phenotype - observable/physicalset of traits (structural or functional) Monohybrid cross:cross b/w organisms that differ in a single trait - Yield genotypic ratio: 1:2:1(SS:Ss:ss) Phenotypicratio: 3:1 (smooth:wrinkled) Mendel's Principle #1: Segregation Confirm Principle #: Test-cross (**cross with homozygousrecessive**):cross of an individual of unknown genotype (usually showing dominant phenotype) with a homozygousrecessive individual to determine the unknown genotype - S__ x ss (smoothx wrinkled) - If offspring are all smooth,then unknown parent's genotype is SS - If offspring show ratio of 1 smooth:1 wrinkled, then unknown parent genotype is Ss Dihybrid crosses and principle of independent assortment Dihybrid cross - cross between organisms for 2 pairs of alleles at 2 different loci Mendel's law of independent assortment If crossed with itself, always 9:3:3:1 Trihybrid crosses: 27:9:9:9:3:3:3:1 Principlesof Probability:ratio of the number of times a particular event is expected • Product rule ("and") - take probability of 2 independent events occurring simultaneously • Sum rule ("or") - probability of either one or the other of 2 mutually exclusive events occurring at the same time Conditional probability Binomial probability - involve predicting the likelihood of a series of events (for which there are two outcomeseach time) p(success) and q(failure) Expands the binomial expression to reflect the # of outcomecombos& probability of each ○ Use 2 variable: ○ p = frequency of 1 outcome ○ q = frequency of other outcome ○ Exponents: # of events ○ Use Pascal's triangle Take note of all diseased and what type of disorder it is and how it's transmitted(dominant, recessive,somatic, etc.) Practiceproblems: 1. In maize, a dominant allele A is necessary seed color, as opposed to colorless (a). Another gene has a recessive allele (w) that results in waxy starch, as opposed to normal starch (W). The two genes segregate independently. An AaWW plant is test-crossed. What are the genotypes of offspring? ○ AaWw, aaWw ○ Ans: a plant that is genotypically AaWW has 2 types of gametes; AW & aW (notice that W is a symbolfor one allele, not 2) ○ In a testcross,this plant is crossed to one homozygousfor recessive alleles at both the colour gene & the waxy gene (aaww) The gametes of this plant are all aw ○ Genetic Cross: aaww x AaWW 2. In cattle, the polled (hornless) (P) is dominant over the horned (p) phenotype. Another gene recessive(c) that results in white cattle, as opposed to spotted cattle (C). A hornless and spotted cow (heterozygousfor both alleles) is test-crossed.What proportion of the offspring will be horned and spotted? (both recessive) ○ 1/4 3. In jimsonweed,purple (P) flowers are dominant to white (p). Self-fertilization of a particular purple-flowered jimsonweedproduces 68 purple-flowered & 22 white-floweredprogeny. What proportion of the flowers will be true breed? ○ ~33% 4. What is probability of a cross between an individual of genotype AaBBccDdee and an individual genotype AAbbCcDdee producing an offspring of genotype AaBbccDDee? ○ ½ x 1 x ½ x ¼ x 1 = 1/16 5. In mimes, a rare autosomal recessivemutation of a particular gene results in an diamond shape around the eyes (dd) rather than the normal lash lines (Dd or DD). A dominant mutation on another autosomalgene results in plain black shirts (BB or Bb) as opposed to the usual stripes (bb). A pure breeding diamond-eyed,black-shirt mime is crossed with a lash-line eyed, striped shirt mime(pure-breeding). What are the possible genotypes and phenotypes of the offspring in the F2 generation? ○ Answer: ddBB x DDbb ○ F2 generation: 100% DdBb (all lash-line, black-shirt) Intercross the F1 generation, what are the phenotypic ratios? ○ Answer: 9:3:3:1 --always this outcomefor DdBb x DdBb ○ D-B- (lash line & black shirt): 9/16 ○ D-bb (lash line & striped shirt): 3/16 ○ ddB-: (diamond-eye & black shirt): 3/16 ○ ddbb (diamond-eye & striped shirt): 1/16 Show your work to get marks Pedigree analysis - Tracks phenotype of a specified human trait over several generations - The affected individual through whom the pedigree is discoveredcalled the proband Know characteristics of inheritance for traits/diseases: - Autosomaldominant - Autosomalrecessive - X-linked recessive - X-linked dominant Review: Chi Square test 1. A man whose father expresses a recessivetrait marries a woman who has a brother expressing the same trait. What is the probability that the two individuals will produce a child expressing that recessive trait? Hint: begin by doing a pedigree ○ 1/6 2. Bob and alice are married and thinking of having children. Alice has a brother with galactosemia(a rare autosomalrecessivedisease). Bob's aunt also has galactosemia • Draw pedigree for this family. Identify the genotypes of all individuals in the pedigree. • Determinethe probability that Bob and Alice will have a child that has galactosemia. ○ p(child is gg) = p(Alice is Gg)p(Bob is Gg)p(both pass g to child) 2/3 (2/3)(1/2) (1/2)(1/2) ○ 4/72= 1/18 • Determineprobability that Bob and Alice will have a child that is a carrier for the recessive allele p(child is Gg) = p(Alice is Gg)p(Bob is GG)p(Alice passes g) + p(Alice is GG)p(Bob is Gg)p(Bob passes g) etc. • Assume homozygousnormal unless otherwise indicated when someoneis marrying into a family 3. Gaucher disease is caused by a chronic enzyme deficiency. A man has a sister affected with the 3. Gaucher disease is caused by a chronic enzyme deficiency. A man has a sister affected with the disease. His parents, grandparents. And three siblings are not affected. Discussions with his wife's family reveal that the disease is not likely present in her family, although somerelatives recall that the brother of the wife's paternal grandmother suffered from a similar disease. Determinethe higher p that, if this couple have a child, the child will be affected (what is the chance of the worst-case scenario occurring?) ○ 4/144= 1/36 TUT2 January-24-14 2:09 PM NOTE that students should be able to do ALL of the problems at the end of Chapter 2 but the following may be helpful preparation for Problem Set 1. *Brief review terminology& expected Mendelian ratios (e.g, #8, 16,17) *Determiningparental genotypesbased on offspring phenotypes(e.g., #29) *Basic Probability (e.g., #35) CHAPTER3&4 Chi Square test checks for goodness of fit between the expected and observedtraits Due to chance Example: SsYy x ssyy - observed progeny data from dihybrid cross (total 568) - 154 smooth;yellow - 124 smooth;green - 144 wrinkled; yellow - 146 wrinkled; green 1. Expect proportion of each phenotypic class to be 1/4 ○ If n = 568, we expect that each class will contain 1/4 of the total progeny or... ○ 568 x 1/4 = 142 progeny expected in each phenotypic class 2. Next, subtract each observed number by its corresponding number (o-e) 3. Let d = (o-e), square each d 4. Divide each d by its corresponding expected values 5. Add all these numbers 6. Determineyour degrees of freedom - n of classes you are looking at (4 phenotypic classes) ○ Degrees of freedom is n-1 Use chi-square distribution table .05 use as tolerance - anything above we fail to reject null hypothesis - Anything below reject null hypothesis Chi Square test cannot tell us that hypothesis is corrector incorrect Epistasis One gene is going to mask another gene No new phenotypes produced We can have different types Recessiveepistasis: 9:3:4 Dominant epistasis: 12:3:1 Complementarygene interaction/ duplicate recessiveepistasis: 9:7 Duplicate (gene interaction)dominant epistasis: 15:1 Dominant suppression: 13:3 Problem: Barranca Mesa Flying Lizard. The results suggested that the wing shape for the lizard is due to a recessiveepistatic system (9:3:4).To test this, the researcher performed a dihybrid cross, where he crossed two breeding lines of lizard, one with dumpy, and second with no wings. The resulting F1 progeny were all long-winged, the F1 progeny were intercrossed and the following F2 progeny were produced: 1025no wing; 845 dumpy; 2465 long-winged Carry 3 decimal places Chapter 3: chromosomal basis of inheritance *Read non-disjunction, lethality *Read non-disjunction, lethality Sex chromosomes Autosomes Heterogametic- not always male Homogametic- same sex chromosome,not always female Hemizygous Y-chromosome- testis determining factor ○ Y present - male ○ Y absent - female XX males - a small fragment of Y is translocated to an X XY females - have a deletion of the same region of Y In Drosophila.. X-chromosome:autosomebalance system Y chromosomehas no effect on sex determination (opposite from humans) XXY-female XO-sterile male In birds and moths..ZW-system ZZ-males ZW-females Sample question: in particular species of moth, body colour is determined by the d+ and d alleles. The wild-type, dark allele (d+) is dominant to the light (d) allele. A dark female moth is crossed to a dark male (heterozygous) a. What are the genotypes of the parentals? b. What are the phenotypic proportionsof the F1 moths? c. If a light F1 female moth is mated with the father, what are the phenotypic proportionsof the F2 moths? Drosophila gene symbols '+' = wild type - most prevalent in nature Dominant and recessiveis symbolizedby upper-case and lower-case If genes were located in the X-chromosome ○ Male v+/YOR X Y v+ v+ v+ ○ Female: v+/v+or X X Sample question: True-breeding red-eyed female and a true-breeding white-eyed male - Red is wild-type (w+) to white (mutant; w) - Gene is on X-chromosome a. What are the gametes produced by each parent? b. What phenotypes and genotypes F1 generation? The proportions? c. F1 generation are intercrossed,what phenotypes and genotypes are observed in F2? The proportions? d. Performa reciprocal cross. Do we still observe the same phenotypic and genotypic ratio? ○ We do not obeserve,2 different ratios for the 2 sexes e. What would we expect if the gene was autosomal ○ ALWAYS the same if the genes are located on AUTOSOMES Problem: You are genetic counsellor, and a man and woman have come to you to discus their family history and asses their risk of having a child affected with Tay-Sachs or haemophilia A. the couple indicates that the woman's maternal grandfather had haemophilia A. no other people in her family are known to have had haemophilia. The mans' paternal grandfather's brother had Tay-Sachs, and the woman's paternal uncle had Tay-Sachs. No other people in the family are known to have had Tay-Sachs. Assume that these mutations are very rare in this population. Tay-Sachs is an autosomalrecessive Assume that these mutations are very rare in this population. Tay-Sachs is an autosomalrecessive disorder, and haemophilia A is an x-linked disorder. a. Draw pedigree b. P child has Tay-Sachs? 1/72 c. P carrier for Tay-Sachs? 8/36 d. P boy with haemophilia A? 1/8 e. P girl carrier for haemophilia A? 1/8 TUT3 January-31-14 2:11 PM CH 3,4 (continued) & gene mapping Review: CH2 must do - 8, 16-18,23, 24, 28, 29, 33-35 CH3 must do - 12, 13, 14, 23 CH4 must do - 18, 19, 30, 31 (ABO system) CH3: Non-disjunction and Drosophila Practise Question CH4: Lethal alleles, Gene interactions (continued) Start CH5: Introduction, Two and three point testcrosses Sections 5.1-5.2 Non-disjunction of X chromosomes - Fail to segregate properly - Meiosis I, homologouschromosomesfail to segregate - Meiosis II (and Mitosis), sister chromatids fail to segregate - Can occur with autosomesor sex chromosomes - In humans, resulting progeny (XXX; XO female) (XXY male) - Drosophila, XXX and YO offspring die! - In fruit flies, XO male, XXY female Non-disjunction question: - In drosophila, vestigial (partly formed) wings (vg) are recessiveto normal long wings (wild- type), and the gene for this trait is autosomal.The gene for the white eye trait (w) is on the X chromosome(red eyes are wild-type). Suppose a homozygouswhite-eyed, long-winged female fly is crossed to a homozygousred-eyed, vestigial winged male ○ Show genotypes of the parentals. ○ What re the genotypesand phenotypic proportionsof F1 ○ Assume that an F1 female undergoes non-disjunction of the sex chromosomesin meiosis I when producing her all eggs. What will the surviving progeny of a cross between this female and an F1 male look like? Lethal allele - Affects an essential gene (mutated genes that result in a lethal phenotypes) - There are dominant and recessive lethal genes - Mice (two copies of lethal allele dies) - example of recessive lethal Question: you know that gene F codes for fur colour in bunnies. Bunnies that are hetero (Ff) have white fur. Bunnies that are homo for dominant (FF) are brown. ff condition is lethal. - Probability of cross between two white bunnies producing a brown bunny? 2/3 Ff white and 1/3 FF brown - What can we conclude about the fur allele? Allele F is recessive lethal Two mice ratty a lethal allele which has dominant effect of colour, making them yellow rather than the wild type brown colour. The mice mate multiple times and 100 pups are produced. How many pups are expected to be yellow? ~66% a+ b+ c+ d+ = Independently assorting controlling black pigment. The alternative alleles that give abnormal functioning of these genes are a, b, c, d. black individual of genotype a+/a+, b+/b+, c+/c+, d+/d+ is crossed with a colourless individual of genotype a/a, b/b, c/c, d/d to produce all black F1. Two F1 progeny are crossed. Two F1 progeny are crossed. - What proportion of F2 progeny are colorless?37/64 - What proportion of F2 progeny are brown? 27/256 Black can be produced only if both the red pigments are present and thus the c+ convertsboth pigments together into black. - What proportion of the F2 progeny is colorless? - What proportion of the F2 progeny is black? - What proportion of the F2 progeny is red? Genes A, B, and C are independently assorting and control the production of a black pigment Suppose instead that a diff pathway is utilized. In it, the C allele produces an inhibitor that prevents the formation of black by destroying the ability of B to carry out its function. A colorlessA/A B/B C/C indv is crossed with a/a b/b c/c, a colorless giving colorless F1. F1 are selfed. What is the ratio if colorless to black in the F2? Only colorlessand black phenotypes are observed - For this pathway to produce black, ind must have A and B functions BUT NOT the inhibitor function provided by C - A/- B/- c/c (black) = 3/4 x 3/4 x 1/4 = 0.1406 - Colorless = 55/64 TUT4 February-07-14 2:10 PM Please note that you are responsible for ALL end-of-chapter problems for Chapters 3 and 4. However,the following is a list of concepts and textbookquestions may be helpful preparation for Problem Set 2. Chapter 3: *ZW sex determination(e.g., #23) *Pedigree Analysis with sex-linked traits (e.g., #13) *Determiningparental genotypesw/ sex-linked traits (e.g., #14) -students are responsible for ALL end-of-chapter problems. Chapter 4: *Chi-Squared Analysis (e.g., #30) *Genetic Pathways (e.g., #18) *Reviewgene interactions,epistasis, and modified Mendelian ratios (…lotsof possible questions; also see table 4.19 on page 127) -students are responsible for ALL end-chapter problems. Ch5 - Gene Mapping (Part I) Questions do not do: 18, 20, 28, 29, 31 Today: • Emphasis on Sections 5.1 to 5.3 • Introduction - "Lingo" for gene mapping • 2 & 3 point test crosses • Chi-Square analysis - genetic linkage Gene mapping - Several genes are located on same chromosomes= linked genes - Two parents w/ different alleles are crossed (e.g. white eyes/ mini wings x wild-type eyes and wings) - Parental combination of alleles = parentals (white eyes/ mini wings x wild-type eyes and wings) - Non-parental combinationof alleles = recombinants(white eyes/ wild-type wings x wild-type eyes and mini wings) Fundamentals to genetic mapping - Linked genes are always SYNTENIC and are always located near one another on a chromosome - Genetic linkage leads to more frequent parental phenotypic classes than recombinant - CO is less likely to occur when linked genes are closer together on a chromosomethan b/w genes farther apart on the chromosome ○ Frequency of CO is roughly proportionateto the distance b/w genes -> relationship that allows for genes to be mapped! 1. % recombinants below 50% suggest genes are located on same chromosome(Linked genes) 2. % recombinants above 50%, genes are on different chromosomes(genes are not linked) 3. Max recombinantis 50% Concept of genetic mapping - In heterozygoteof two linked genes ○ Two arrangements of alleles: ○ a b /ab or a b/ ab (both have same phenotypes) Coupling (CIS) configuration: when wild type alleles are on one chromosomeand mutant ○ Coupling (CIS) configuration: when wild type alleles are on one chromosomeand mutant alleles are on another chromosome ○ Repulsion (TRANS) configuration: when one wild type and one mutant allele are on each chromosome Genetic distances - Can use recombinant frequencies as a quantitative measure of the genetic distances b/c 2 genes are on a genetic map - mu (map units) or cM (centimorgan) - 1 mu = 1 cM = 1% recombinationfrequency - Be careful in using cM, just use mu to be safe - Crude approximations, but good enough for genetic instructions! Chi-square and gene linkage - Analyze testcrossdata and determine whether a deviation is "significant" - H oNull hypothesis) - genes are not linked ○ Used because it is not possible to predict phenotype frequencies produced by linked genes ○ If 2 genes are not linked, a testcrossshould be 1:1 ratio of parentals: recombinant  Genes assort independently in meiosis - Expected values - based in the assumption that the genes in question are assorting independently Sample chi-square test - linkage - In drosophila, b is recessiveautosomal mutation that results in black body color, and vg is a recessiveautosomal mutation that results in vestigial (short, crumpled wings). Wild-type flies have grey bodies and ling, non-crumpled wings (normal) - Cross: true-breeding black, normal wings x true-breeding grey, vestigial wings - P: b/b vg vg x b b vg/vg + + - F1: all progeny flies -> grey with normal wings (b /b vg /vg) - Test cross F1 female x black body, vestigial wing (b/b vg/vg) - Males cannot be used for these studies (they do not go over crossing over, only female does!) - Testcross:F1 female x black body, vestigial wing (b/b vg/vg) - Testcrossprogeny (3236total) ○ Grey, normal wings - 283 (non-parental or recombinant) ○ Grey, vestigial wings - 1294 (parental) ○ Black, normal wings -1418(parental) ○ Black, vestigial wings -241 (non-parental or recombinant) Phenotype Observed Expected (O-E) /E2 Grey, normal wings 283 809 341.998 Grey, vestigial wings 1294 809 290.760 Black, normal wings 1418 809 458.443 Black, vestigial wings 241 809 398.793 - Chi-square = 1489.994;df = 3; reject null hypothesis - Is it likely that the observed numbers are the result of 2 independent assorting genes? NO, since p value is less than 0.001. There is a significant deviation. - Alternate hypothesis explain our observed data ○ Genes are linked! ○ Genes are (241+283)/3236x 100% = ~16% or ~16 mu apart Two point test crosses - What have we seen up to this point - Remember:a b /a b x a b/a b cross + + - Results in the following two point test cross: a b /ab x a b/a b - Results in the following two point test cross: a b /ab x a b/a b - Frequency of recombinantsgives an estimateof genetic distance in map units ○ rf= (number of recombinants/numberof testcross progeny) x 100% - Number of phenotypes of recombinants changes when arrangement of F1 (cis or trans) changes ○ But the number of recombinantsdoes not change - There is large number of frequency of parentals if there is linkage - This method can be considered accurate as long as the genes are relatively close together - This method is not accurate when genes are far apart (figure from textbook) Three-point testcrosses - Used to determine linkage and rf's of three genes - Three-point testcross: ○ a b c /a b c x a b c/a b c - There is 8 phenotypic classes (see figure from textbook) p = purple p = yellow (WT) r = round r = elongated j = juicy + j = dry p r j + 179 parental prj 173 parental p rj 46 recombinant + + pr j 52 recombinant + + p rj 22 recombinant pr j 22 recombinant p r j 4 recombinant, double crossover(fewestprogeny) prj+ 2 recombinant, double crossover(fewestprogeny) Steps in genetic mapping 1. Classify phenotypic classes in progeny: of the 8 classes, 2 are parentals, 4 are recombinants. ○ Total progeny = 500 2. In recombinants,determine which arose due to a double crossover ○ Double crossoversare rare; hence, the recombinants associated with the fewest progeny arose due to double crossovers! 3. Determineorder of genes: after a double crossover,only one of the genes will be exchanged between the homologouschromosomes,leaving the other two unchanged (i.e. parental arrangement) ○ We see that only the j gene is changed. While the other two remain in their parental arrangement ○ J is in the middle. Therefore,the order of the genes: p j r or r j p 4. Determinewhich SCO progeny is produced in the chromosomes ○ First single crossover  Results: p jr = 46 (yellow, round, juicy fruit) + + pr j = 52 (purple, elongated, juicy fruit) ○ Second single crossover  Results: p jr = 4 + pj r = 2 5. Calculate the recombinant frequency for two genes at a time Calculate rf between p and j: ○ Calculate rf between p and j: Single crossoverrecombinantsp and j = 46+52= 98 Double crossoverrecombinants p and j = 4+2 = 6 Rf = (single crossoverprogeny + double crossoverprogeny / total progeny) = 20.8 mu ○ Calculate rf between r and j: Single crossoverrecombinantsr and j = 44 Double crossoverrecombinants r and j = 6 Rf = (single crossoverprogeny + double crossoverprogeny / total progeny) = 10.0 mu ○ Combine data: Single r and j = 44 Double r and j = 6 Single p and j = 98 Double p and j = 6 Rf = 30.8 mu 6. Draw you genetic map! Rememberyour distances and units! 7. Calculate coefficientof Interference and coincidence (if necessary) - If each single crossoveris considered an independent event. Then we can say that the double crossover(2 single crossoversoccurring simultaneously)probability is the product of each single crossover - Prob/freq for 1st single crossover= 0.208, 2nd single cross = 0.10 - Expect the double crossoverto be = Product rule -> 0.208 x 0.10 = 0.0208 - Observed is 6/500= 1.2% . Expected is higher than observedbecause of interference. - Interference = one single crossoverinterfering with another single crossover ○ I = 1 - coefficient of coincidence - Coefficient of coincidence = (observed/expectedfreq) = 0.012 / 0.0208 ○ Coeff of 1 = DCO occurred that were expected on the basis of 2 independent events (no interference) ○ Coeff of 0 = none of the expected DCO occurred (total interference,with one CO completelyinterfering a second crossoverin that region) Generating your genetic maps: summaryof the 7 steps to creating your genetic map are 1. Examine testcrossprogeny numbers and indicate which are parentals and recombinants. Determinegene confirmation (i.e. cis or trans)? 2. Indicate your double crossoverprogeny (look for the lowestnumber of progeny) 3. Determinethe order of your genes via DCO 4. Determinewhich SCO progeny is produced from CO between regions 1 & 2 5. Calculate your recombination frequency for each region 6. Draw your final genetic map 7. Calculate coefficientof coincidence and interference(if necessary or required by question) TUT5 February-21-14 12:19 PM Finish Chapter 5 (up to 5.3) Studentsare responsible for the following end-of-chapter questionsfor Chapter 5: 1-17, 19, 21-28, 30-32. ...However, some questions/concepts that may be helpful preparation for Problem Set 3 include: Three-pointtest crosses (e.g., #15, #16, #23, 26) Basic recombination (leading to more challenging recombination)(e.g., #17) X2 with linkage (e.g., #13, #32) TAs may also: -Revisit complementation analysis (Chapter 4, Section 4.4). -Start Chapter 10. CH 5 - Gene Mapping (Part II) Questions: 1. In soy bean plants, there are three linked genes coding for pod color, pod texture, and pod size. The wild-type phenotypes are green, smooth, and big. The mutant phenotypesare turquoise(t), wrinkled(w), petite(p).A triple heterozygote is test-crossed producing the ff. progeny types: F2genotype Number a.) Recombinantor parental? b.) None, SCO, or DCO types? w++ 476 P none ++p 375 R SCO (t-w) +tp 496 P none wtp 88 R DCO wt+ 385 R SCO (t-w) w+p 204 R SCO (w-p) +t+ 249 R SCO (w-p) +++ 66 R DCO c.) What is the gene order? twp or pwt Solution: Parents: DCO w++ +++ +tp wtp 'w' is middle gene d.) draw the genetic map showing the correct order and relative distances (including units) between the genes Rf(t-w) = Rf(w-p) = t---39.08mu----w---25.95mu----p p---25.95mu---w---39.08mu----t e.) Coefficient of coincidence and interference values C.O.C= observed DCO/expected DCO = 0.6492 I = 1- 0.6492 = 0.3508 f.) What does the above data tell you? Most of the DCO are occurring 35.07% 2. Three mutant phenotypesare observed: brown kernels (b); wide leaves (w); short (s). The wild-type (+) phenotypes are yellow, narrow leaves and tall, respectively. All mutant alleles are recessive to wild-type. A triple heterozygote corn plant is testcrossed to produce the ff. progeny types: heterozygote corn plant is testcrossed to produce the ff. progeny types: F2phenotype Number a.) F 2enotype b.) Recombinantor parental? c.) None, SCO, or DCO types? Brown, short 18 bs+ R DCO brown 279 b++ P None short 82 +s+ R SCO (s-w) wide 12 ++w R DCO Wild-type 127 +++ R SCO(b-s) Wide, short 251 +sw P None Brown, wide, short 143 bsw R SCO(b-s) Brown, wide 88 b+w R SCO(s-w) d.) What is the gene order? bsw or wsb Solution: Parentals DCO b++ bs+ +sw ++w "s" is the middle gene Rf=(b-s) = 30.0mu s-w = 20.0mu b--30.0mu---s---20.0mu----w COC=0.50 I=0.50 or 50% What does our interference value indicate? Most of our expected DCO are occurring although there is 50% interference indicates that 50% of DCO to occur are not occurring due to in of DCO event w/ 3. A cross was made between an aaccFF female and an AACCff male. The1F males were testcrossed and produced the ff: F1testcross -> AaCcFf x aaccff F2progeny Phenotype Observed ACf 394 aCF 402 acf 90 aCf 102 AcF 98 ACF 110 Acf 398 acF 406 Assume A, B, C are dominant phenotypes to a, b and c. a. Which pair of genes is linked> A and C, A and F, F and C? - If a pair of genes is linked, then we should expect to see more parental types than recombinant types. That is parental recombination would be greater than 50% and rec would be less than 50% - If a pair of genes are not linked, then we should to see equal numbersof parental to recombinanttypes (i.e. independentassortment -> 1:1:1:1) A and C C and F A and F A C -> 394 + 110 = 504; C F -> 402 + 110 = 512; A F -> 98 + 110 = 208 ; 208/2000=0.104 = ~ 504/2000=0.252 = ~25% 512/2000=0.256 = ~25% 10% << 25% (recombinant) a c = ~25% cf= ~25% af=~10% A c= ~25% Cf= ~25% Af=~40% (parental) A c= ~25% Cf= ~25% Af=~40% (parental) a C= ~25% cF= ~25% aF=~40% (parental) C and F are unlinked Therefore, A and F are linked and are on the same chromo
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