BIOB 270 1st Edition Lecture 12Outline of Last Lecture Stats, Hypothesis Testing, and LinkageI. Probability: a. Product Ruleb. Sum Rulec. Conditional ProbabiltiyII. Genes on Same Chromosome vs. Genes on Different Chromosomes after MeiosisIII. Hypothesis TestingChi-Square goodness of fit testOutline of Current Lecture LinkageI. Linkage in Sweet Peas ExampleII. Test CrossesIII. Crossing OverIV. Recombination RateV. Linkage TermsVI. Gene Order DeterminationVII. Double-Stranded RecombinationVIII. 3 Types of Genome Maps These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.Current LectureLinkageI. Linkage in Sweet Peas Example:Sweet Pea combinations: Red Flower (R), Long Pollen Grains (L), White Flower (r), Short Pollen Grains (l)P: RL x rlF1:RL x rl - shows that Red flower with long grains are both dominantF2(expected): RL x Rl x rL x rl – 9:3:3:1 ratioExp (451.7) (150.1) (150.1) (50.2) Observed: (9/16) x 803 = (451.7) (3/16) x 803 = (150.1) (1/16) x 803 = (50.2)x^2= 37.97 + 102.60 + 105.93 + 285.90 = 532.40Df=3 Chi-Table shows 7.82 as chi-square valueIf the null hypothesis (HO 9:3:3:1) is correct, we expect a chi-square value greater than 7.82 (df=3) less than 5% of the time.532.40>>>>7.82Here, we reject the null hypothesis because the chi-square value is greater than 7.82. The phenotypes do not follow 9:3:3:1Explanation:The R and L loci are on the same chromosome and they are genetically linked.-this means that we see the recombinant combination of (Rl and rL) very little and the parental combinations (RL and rl) are much more common…it’s not a 1:1:1:1 ratio.II. Test Crosses: a simple way to estimate recombination1. Step one: A heterozygote for two linked genes is testcrossed to a recessive homozygote:(RL and rl) x (rl and rl)2. Step Two: In the progeny, recombinants are less than 50% of the total:(450 RL, rl) (42 Rl, rl) (38 rL, rl) (470 rl, rl) =80 recombinants total, 920 parental total3. Step Three: Find the frequency of recombinationFrequency of recombinants= recombinants total/ (recombinants total+ parental total)=80/(80+920)= 0.08 **this low frequency of recombination indicates that the genes are rather tightly packedIII. Crossing Over: the basis of recombination between syntenic genes- AB, AB x ab,ab => cross over at chromatids AB and ab (chiasmata- genetic exchange)= 2 nonrecombinant (parental) chromosomes AB and ab2 recombinant chromosomes Ab and aB*this occurs during pachytene of Prophase I- If no crossing over occurs, the alleles remain in original (parental) combinationsIV. Recombination Rate: (r) is a measure of genetic distance. Genes that are closer together are less likely to have crossovers occur between them.- 1% recombination =- 1 map unit (mu) =- 1 centiMorgan (cM)- r= # of recombinant gametes/ total # of gametesDrosophila Example:Crossing over does not occur in male DrosophilasWildtype (+) is dominatvg+=long wings vg= vestigial wingsv+=gray body b=black bodyP: vg+b+, vg+b+(male) x vgb, vgb(female)F1: vgb, vg+b+(male) x vgb, vgb(female)F2: vg+b+, vgb vgb+,vgb vg+b, vgb vgb,vgb180 recombinants, 820 parentalsFrequency of recombination= 180/1000=0.18Clicker question: If vg & b are linked, which female gametes will be in excess? (a) + + and vg b (b) + + and + b answer: a. ++ and vg b(c) + + and vg + (d) v + and + bV. Linkage Terms:- Synteny – co-localization of genes on same chromosome in an individual or species.- Syntenic group – a group or block of genes occurring in the same order in different species- Genes on same chromosome may be genetically linked as well. - Genetic linkage map - The linear order of linked genes can be determined by examining variation in the strength of linkage. - Most genomes have multiple groups of linked genes (linkage groups) that often correspond to different chromosomes.VI. Gene Order Determination- Can be determined on a chromosome using observed recombination rates between all pairwise combinations- When loci are far apart there is a high probability of double crossovers happening between them- Can use genetic maps to help determine where genes are relative to each other, but genetic map distances are NOT additive because of double crossoversVII. Double-Stranded Recombination- Increase probability of double crossovers when farther apart- Double crossover: 2 strands cross over at two different spotso 4 chromatids= 100% recombinantso 3 chromatids= 50% recombinants o 2 chromatids=0 recombinants*overall average=50% for all possibilities, so r=0.50o Loci on the same chromosome are syntenic.Loci that show less than 50% recombinationare genetically linked.VIII. 3 Types of Genome Maps:1. Genetic or Linkage map: linear arrangement of genes based on the analysis of recombination (crossing over) in crosses or pedigrees. 2. Cytological map: chromosome features based on microscopic image of chromosomes.3. Physical (sequence) map: Genome map based on determining the linear sequence of DNA. (more on this
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