BIO 305 1st Edition Lecture 12 Outline of Last Lecture I. Quantitative Genetics and Additive EffectsII. Normal Distribution and 6 ParametersIII. HeritabilityA. Broad-Sense HeritabilityB. Narrow-Sense HeritabilityOutline of Current Lecture I. QTL (Quantitative Trait Loci) Mapping and Association MappingII. Population GeneticsA. Allele FrequencyB. The Hardy Weinberg LawIV. Violations of the Hardy Weinberg Law (Part !.)A. Population StructureB. InbreedingV. Vocabulary and Sample QuestionsCurrent LectureI. QTL MappingKnown Pedigrees: Suppose we have a strain of drosophila that is resistant to DTD vs. a strain that is resistant to DTDWe want flies that carry part of the yellow genome and part of the blue genomeEx: Beefmaster vs. Sungold TomatoesIn general, we can also do QTL mapping from a backcross: say we have two types of tomatoes and want to know the gene controlling size. The F1 will have two G genes and two S genes and you need to relate genotype and size. Key Point: Distinct distributions for genotypic classes at marker locus signal the location of a QTLnear the markerThe parental line = 230 gramsThere are 5 lines, each representing a marker that we have to look at: which marker has information about size?M1: Some have BB, some have B/S; we find that both sides are very similarconclusion: BB or BS at this locus will have no effect on sizeM3: The cut-off LOD score = 3, thus the peak at M2 is significantconclusion: There is at least one gene near the M3 chromosome that contributes to the size of the tomatoLod scores also provide statistical evidence for QTLOther Examples of QTL mapping:Ex: Tomato Fruit Size fw2.2Note that size is a quantitative trait so it is likely that more than one gene is involvedHow do we prove which gene contributes to size?Tomato fruit size – note that this is quantitive so more than one gene probably involved, how dowe prove that gene contributes?Experiment: (-) was the control and was large(+) was mutant When fw2.2 is inserted, saw that mutants were 30% smallerThis provided transgenic proof: the gene encodes a protein controlling # carpel cellsQTL mapping provided means to identify genes contributing to quantitative variationSuppose we genotype at each SNP loci and we know phenotypeAssociation MappingMapping QTL without PedigreesUsing the marker to infer:red means ‘equivalent’ while each side means ‘linkage disequivalent’ because their distinguishability Each small diamond tells you linkage equivalency between adjacent SNPSTry to statistically correlate genotype at each position with a trait:Which SNP has some statistical correlation with diabetes?Answer: From looking at the data, if you have G you tend to be NO, A = YES; heterozygote can be eitherAssociation mapping can be used to find genes for a quantitative trait or even identify genes for disease susceptibility:II. Population GeneticsA. Allele FrequencyHow do we estimate allele frequency in a population?Ex: MN blood group, where MN is co-dominantM type = 600; N type = 300; MN type = 100 individualsFrequency of the M allele = (600*2+100)/2000 = 0.65Frequency of the N allele = (300*2+100)/2000 = 0.35Ex: M type = 650; N type = 350; MN type = 0 individuals Frequency of M allele = (650*2+0)/2000 = 0.65Frequency of N allele = (350*2+0)/2000 = 0.35Ex: Chemokine receptor CCR5HIV uses CCR5 as a co-receptorwild-type allele and Δ32 allele‘Δ32’ indicates a 32bp deletion, which makes the gene nonfunctional+/+: 79+/ Δ32: 20Δ32/ Δ32: 1Frequency of + allele = (79*2+20)/200 = 0.89Frequency of + allele = (1*2+20)/200 = 0.11There is variation among different human population in susceptibility to HIVcorresponds to variation of allele frequency of CCR5-Δ32 in EuropeYou can estimate allele frequencies from genotype frequencies; you cannot get genotype frequencies from allele frequenciesSpecial exception: Hardy-Weinberg LawB. The Hardy Weinberg LawWhat six conditions must be met?1) No selection2) No migration3) Random mating4) No mutation5) Infinite population size6) Equal allele frequencies in males and femalesAllele frequencies are not shared; after one generation of random mating, can compute genotype from alleles:Demonstration:The demonstration of the HW Law is similar to the Punnett Square. When males have the same allele frequency as females, among sperm: the fraction of ‘A’ sperm is P, and ‘a’ sperm is Qafter random mating, p2 + 2pq + q2 = 1, phenotype frequencies after one generation of matingWhat are the 3 consequences of a condition where HW applies?1) Genotype frequency can easily be calculated from allele frequency (p2 + 2pq + q2 = 1)2) A certain genetic variation can be maintained in a population3) No evolution occurs when all 6 conditions are metEx: Cross of a tall plant (DD) with a short plant (dd) received 36 tall plants and 64 short plants. How do we estimate allele and genotype frequencies, with the assumption that HW applies? We know dd = q2 = 64/100 (dwarf over total), thus q = 0.8p = 1 – q = 1 - 0.8 = 0.2Now can find all phenotype frequencies: DD = p2, Dd = 2pq and dd = q2Testing HW Equilibrium with Statistics:Different df in this case = independent observations - independent parameters – 1Back to our example of CCR5: Now calculate expected valuesFirst, calculate expected number of +/+ = 100 * (.89*.89) = 79.2Then, +/ Δ32 = 100(2pq) = 100 * (2*.89*.11) = 19.6And, Δ32/ Δ32 = 100(.11*.11) = 1.2Using observed values:+/+: 79+/ Δ32: 20Δ32/ Δ32: 1 UseDo a chi square test, df= 3-1-1 = 1, and you will find that p > 0.5Estimating Heterozygote Frequency:Ex: Cystic fibrosis, an autosomal recessive disease, occurs in 1/2500 of North Europeans.How do we estimate the frequency of carriers?We know aa = q2 = 1/2500, thus q = 1/50 = 0.02p = 1 – q = 0.98Now can find heterozygote frequency 2pq = 2*0.98*0.02 = 0.0392, 1 in 25 peopleGenotype Frequency under Hardy-WeinbergStudy the amount of frequency of AA, Aa, and aa in relation to each other:III. Violation of the Hardy Weinberg Law1st: Mating Bias – violates random mating lawtwo types of effects: Population structure and InbreedingA. Population Structurepopulation structure is a type of nonrandom mating that leads to a deficiency in heterozygosity and excess in homozygositybut doesn’t necessarily mean allele frequencies are affectedhomozygosity + heterozygosity = 1Ex: A barrier separates two populations; when merged, you see a higher deficiency in Aa than expected: B. InbreedingAlso leads to a