University:
California State University, Northridge
Course:
BIOL 360 | Genetics
Academic year:

2007
Views:

292

Pages:

26
Author:

Rowan Holder

Chapter 6 Linkage Analysis and Mapping Three point crosses • mapping • strategy • examples ! Mapping human genes Three 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 can be 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 eyed female to a pure breeding wild type male: – vgvg bb prpr x vg+vg+ b+b+ pr+pr+ 2/39 Three Factor Testcrosses Result in Eight Phenotypic Progeny Classes 3/39 Types of Gametes • Parental Types • Single Crossover between A & B/a & b • Single Crossover between B & C/b & C • Double Crossover Data 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. – 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. • What the hell are you talking about?????????????????????????????? 5/39 Find the double recombinant class (the class with the least number of progeny) -- the gene that is different from the parental chromosome in this class is the middle gene. 1779 1654 252 241 131 118 13 9 vg b pr vg+ b+ pr+ vg+ b pr vg b+ pr+ vg+ b pr+ vg b+ pr vg b pr+ vg+ b+ pr P DCO vg pr b vg+ pr+ b+ vg pr+ b vg+ pr b+ double cross over chromosomes 6/39 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. (the oddball) Check both to check yourself Write order of genes – The orientation from left to right is purely arbitrary. • vg - pr - b • 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 = 0.123x100 = 12.3 mu 4197 8/39 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 = 0.064x100 = 6.4 mu 4197 9/39 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: – vg-b+ – vg+-b – This checks to be sure you have • the correct middle gene also. • Should be the largest distance…. • Numbers that correspond are 252 + 241 + 131 + 118 + 13 + 13 + 9 + 9 X 100 = 18.7 m.u. 4197 • Why count dco twice? 10/39 WRONG Without dco vg – b dist vg – pr dist b – pr dist 252 + 241 + 131 + 118X 100 = 17.7 m.u. 4197 252 + 241 + 13 + 9 X 100 = 12.3 m.u. 4197 131 + 118 + 13 + 9 4197 X 100 = 6.4 m.u. 17.7 m.u. is not 18.7 m. u. !!!!!! Double Crossovers • Recombination is caused by formation of chiasmata along the chromosome at multiple points. • If the distance between two genes is large enough, there can potentially be multiple chiasmata formation between them; – so there could be multiple crossovers. • What would happen if there were two crossovers between the two outside genes (in this case vg and b)? – Answer: there would appear to be fewer recombinants between the two genes: – it would appear as if the genes are closer; – the calculated map distance between these genes will be less than actual. 12/39 Two recombination events occurred in the interval between a and b. •We must count them for each of the single crossover events. •They are crossing over events that occur on both sides. •Must be used twice in calculations !We need to add the number of double recombinants TWICE to our total for the outside markers: vg-b distance… 1779 1654 252 241 131 118 13 9 vg b pr vg+ b+ pr+ vg+ b pr vg b+ pr+ vg+ b pr+ vg b+ pr vg b pr+ vg+ b+ pr P DCO 252 + 241 + 131 + 118 + 13 + 9 + 13 +9 x 100 = 18.7 m.u. 4197 15/39 Question • Which type of class would you expect to account for the lowest frequency? 1) Parental 2) Single Recombinants 3) Double Recombinants 4) Middle class Do 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 often imprecise. • In addition, a process called Interference may occur. Interference: The number of double crossovers may be less than expected • Sometimes the number of observable double crossovers is less than expected if the two exchanges are not independent – Occurrence of one crossover reduces likelihood that another crossover will occur in adjacent parts of the chromosome – Chromosomal interference – • crossovers do not occur independently – Interference is not uniform among chromosomes or even within a chromosome Interference • The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossover Expected (double crossover) 0.00787 • • Probability Probability = (single crossover) = 0.064 If we analyzed a total of 4197 fly offspring – The expected number – of double crossover offspring is = 4197 X 0.00787 = 33 • Observed number is 22 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display (single crossover ) X X 0.123 Interference • Therefore, we would expect 33 offspring to be produced as a result of a double crossover – However, the observed number was only 22! • This lower-than-expected value is due to a common genetic phenomenon, termed interference – The first crossover decreases the probability that a second crossover will occur nearby Measuring interference • Coefficient of coincidence = – ratio between actual or observed dco and expected dco • coefficient of coincidence : = observed dco / expected dco • Interference = 1 – coefficient of coincidence Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display p(crossover in region 1) X p(crossover in region 2) = .123 X .064 = .0078 from a total of 4197 progeny, we should have seen (expect) .0078 x 4197 = 32.7 or about 33 double recombinants 1779 1654 252 241 131 118 13 9 4197 vg vg+ vg+ vg vg+ vg vg vg+ b b+ b b+ b b+ b b+ pr pr+ pr pr+ pr+ pr pr+ pr I = 1 – C of C I=1– I=1- 13 + 9 33 Observed # DCO Expected # DCO = 0.333 or 33.3% • This suggests that a cross over in one gene interval physically inhibited a crossover in adjacent regions by 33.3%. Question A plant heterozygous for three dominant traits N, T, and U is test crossed, the resulting progeny are as follows: N U n u N u n u N U n U n U N u T 2 T 70 T 21 t 4 t 82 t 21 T 13 t 17 What is the linkage arrangement of the N, U and T alleles in the parental? What is the linkage arrangement of the N, U and T alleles in the dco? Which gene is in the middle? 1) N 2) U 3) T 22/39 Three Point Cross -- Example where order is not known: start with 2 pure breeding strains, F1 test crossed. Testcross offspring progeny phenotype +++ r++ rs+ rsw r+w +s+ +sw ++w number 6 359 98 4 47 43 351 92 Testcross offspring phenotype +++ r++ rs+ rsw r+w +s+ +sw ++w number 6 359 98 4 47 43 351 92 largest class: smallest class: r++ +sw 359 351 +++ 6 rsw 4 Order: determine an order where it takes two cross-overs to go from parentals to DCO. s r w w r s The 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 +r+ s+w sr+ ++w +rw s++ +++ srw # 359 351 98 92 47 43 6 4 class parental parental SCO(s-r) SCO(s-r) SCO(r-w) SCO(r-w) DCO DCO map distance s-r = 98+92+6+4 1000 map distance r-w = 47+43+6+4 1000 10 m.u. s r w 20 m.u. = 0.20 = 20 m.u. = 0.10 = 10 m.u. Recent Advances in the Field Fluorescent in situ hybridization can also be used to localize cloned genes to a particular chromosome. Probe for a specific gene Visualized with chromosome specific probes

Linkage Analysis and Mapping

of 26

Description

Free up your schedule!

Take 5 seconds to unlock