Chapter 6 Linkage Analysis and Mapping Three …

1 Linkage Analysis and MappingChapter 6 Three point crosses Mapping strategy examples! Mapping human genesThree point crosses Faster and more accurate way to map genes Simultaneous Analysis of Three markers Information on the position of Three genes relative to each other canbe obtained from one mating rather than two independent matings. Example: Drosophila autosomal genes: vg = vestigial wings; vg+ = normal b = black body; b+ = normal body pr = purple eyes; pr+ = normal eyes Cross of pure breeding vestigial winged, black bodied, purple eyedfemale to a pure breeding wild type male: vgvg bb prpr x vg+vg+ b+b+ pr+pr+2/39 Three Factor Testcrosses Result in EightPhenotypic Progeny Classes3/39 Types of Gametes Parental Types Single Crossoverbetween A & B/a & b Single Crossoverbetween B & C/b & C Double CrossoverData Analysis There are eight gametes from the F1 largest number is parental smallest number is double crossover Identify the parental and recombinants Two genes at a time.

2 Compare recombinant to parental Double crossover change (oddball) is guy in the middle Write order of genes The orientation from left to right is purely arbitrary. What the hell are you talking about??????????????????????????????4/39 Data Analysis There are eight gametes from the F1 largest number is parental (P) smallest number is double crossover (DCO) Identify the parental and recombinants Two genes at a time. Compare recombinant to parental Double crossover change (oddball) is guy in the middle Compare DCO to P 2 genes should be the same, one should be the opposite. Check both to check yourself Write order of genes vg - pr - b The orientation from left to right is purely arbitrary.

3 What the hell are you talking about??????????????????????????????5/39v g pr b1779vg b pr1654vg+ b+ pr+ 252vg+ b pr 241vg b+ pr+ 131vg+ b pr+ 118vg b+ pr 13vg b pr+ 9vg+ b+ prvg+ pr+ b+vg pr+ bvg+ pr b+doublecross overchromosomesFind the double recombinant class (the class with the leastnumber of progeny) -- the gene that is different from theparental chromosome in this class is the middle Analysis There are eight gametes from the F1 largest number is parental (P) smallest number is double crossover (DCO) Identify the parental and recombinants Two genes at a time. Compare recombinant to parental Double crossover change (oddball) is guy in the middle Compare DCO to P 2 genes should be the same, one should be the opposite.

4 (the oddball) Check both to check yourself Write order of genes vg - pr - b The orientation from left to right is purely arbitrary. Now determine which numbers go to which genes. Find the numbers where the vg is the oddball and others are same These are numbers for vg to pr region. (252 and 241) What the hell are you talking about?????????????????????????7/39 Analyzing the results of a Three point cross Look at two genes at a time and compare to parental vg and pr vg-pr parentals are: vg - pr - b vg+ - pr+ - b+ vg-pr recombinants are: vg+ - pr - b vg - pr+ - b Numbers that correspond are 252+241+13+9 = = mu41978/39 Analyzing the results of a Three point cross Look at two genes at a time and compare to parental Vg - pr - b Vg+ - pr+ - b+ b-pr recombinants are: Vg+ - b - pr+ Vg - b+- pr Numbers that correspond are 131+118+13+9 = = mu41979/39 Analyzing the results of a Three point cross Look at two genes at a time and compare to parental vg and b vg-b recombinants are.

5 Vg-b+ vg+-b This checks to be sure you have the correct middle gene also. Should be the largest Numbers that correspond are Why count dco twice?252 + 241 + 131 + 118 + 13 + 13 + 9 + 9 4197 X 100 = + 241 + 131 + 1184197X 100 = b dist 252 + 241 + 13 + 9 4197X 100 = pr dist131 + 118 + 13 + 9 4197X 100 = pr is not m. u. !!!!!!WRONG Without dco Double Crossovers Recombination is caused by formation of chiasmata along thechromosome at multiple points. If the distance between two genes is large enough, there canpotentially be multiple chiasmata formation between them; so there could be multiple crossovers.

6 What would happen if there were two crossovers between the twooutside genes (in this case vg and b)? Answer: there would appear to be fewer recombinants betweenthe two genes: it would appear as if the genes are closer; the calculated map distance between these genes will be less recombinationevents occurred inthe interval between a and b. We must count themfor each of the singlecrossover events. They are crossingover events that occuron both sides. Must be used twice incalculationsvg-b + 241 + 131 + 118 + 13 + 9 + 13 +9 4197x 100 = !We need to add the number of doublerecombinants TWICETWICE to our total for theoutside markers:1779vg b pr1654vg+ b+ pr+ 252vg+ b pr 241vg b+ pr+ 131vg+ b pr+ 118vg b+ pr 13vg b pr+ 9vg+ b+ prPDCO15/39 Question Which type of class would you expect toaccount for the lowest frequency?

7 1) Parental2) Single Recombinants3) Double Recombinants4) Middle classDo Genetic and Physical maps correspond? Order of genes in correctly predicted by physical maps Distance between genes is not always similar to physical maps Double, triple, and more crossovers Only 50% recombination frequency observable in a cross Variation across chromosome in rate of recombination Mapping functions compensate for inaccuracies, but are oftenimprecise. In addition, a process called Interference may : The number of double crossovers may be less thanexpected Sometimes the number of observable doublecrossovers is less than expected if the twoexchanges are not independent Occurrence of one crossover reduces likelihood thatanother crossover will occur in adjacent parts of thechromosome Chromosomal interference crossovers do not occur independently Interference is not uniform among chromosomes oreven within a chromosome The product rule allows us to predict the likelihood of a doublecrossover from the individual probabilities of each single crossoverCopyright The McGraw-Hill Companies, Inc.

8 Permission required for reproduction or displayInterferenceExpected(double crossover) =Probability (single crossover)Probability (single crossover ) X = If we analyzed a total of 4197 fly offspring The expected number of double crossover offspring is= 4197 X 33 Observed number is 22 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display Therefore, we would expect 33 offspring to be produced asa result of a double crossover However, the observed number was only 22! This lower-than-expected value is due to a common geneticphenomenon, termed interference The first crossover decreases the probability that a secondcrossover will occur nearby Coefficient of coincidence = ratio between actual or observed dco and expected dco coefficient of coincidence := observed dco / expected dco Interference = 1 coefficient of coincidenceInterferenceMeasuring interference1779vgbpr1654vg+b+pr+ 252vg+ bpr 241vgb+pr+ 131vg+bpr+ 118 vgb+pr 13vgbpr+ 9 vg+b+pr4197p(crossover in region 1) X p(crossover in region 2) =.

9 123 X .064 = .0078from a total of 4197 progeny, we should have seen (expect).0078 x 4197 = or about 33 double recombinants This suggests that a cross over inone gene interval physicallyinhibited a crossover in adjacentregions by = 1 C of CI = 1 -13 + 933= or = 1 Observed # DCOE xpected # DCOA plant heterozygous for Three dominant traits N, T,and U is test crossed, the resulting progeny are asfollows:N U T 2n u T 70N u T 21n u t 4N U t 82n U t 21n U T 13N u t 17 What is the Linkage arrangementof the N, U and T alleles in the parental?

10 Which gene is in the middle?1)N2)U3)T22/39 QuestionWhat is the Linkage arrangementof the N, U and T alleles in the dco? Three Point Cross -- Example where order is not known: start with2 pure breeding strains, F1 test phenotype number+++ 6r++359rs+ 98rsw 4r+w 47+s+ 43+sw351++w 92 Testcross offspringTestcross offspring phenotypenumber +++ 6 r++ 359 rs+ 98 rsw 4 r+w 47 +s+ 43 +sw 351 ++w 92largest class:smallest class:Order: determine an orderwhere it takes two cross-overs togo from parentals to ++359+sw351+++6rsw4srwwrsThe order of Three markers is: s-r-w or w-r-s! Next, sort according to reciprocal products and determine where crossovers occur:F2 products # class + r +359 s + w351 s r + 98 + + w 92 + r w 47 s + + 43 + + + 6 s r w 4parentalparentalSCO(s-r)SCO(s-r)SCO(r-w )SCO(r-w)DCODCOmap distance s-r = map distance r-w = 98+92+6+4 1000= = 20 +43+6+4 1000= = 10 in situ hybridization can also be used tolocalize cloned genes to a particular for a specific geneVisualized with chromosomespecific probesRecent Advances in the FieldHomework ProblemsChapter 6# 19, 20, 21, 22, 23, 24, 27""DONDON T forget to take the online QUIZ!