Sec.1-Crossing Over..... and Mapping.pptx

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1. MOlecular Genetics II Section 1 Linkage, Recombination ,Crossing over and Chromosome Mapping

2. The source of recombination There are two meiotic processes that cause recombination. (1) Independent assortment of genes of different chromosome pairs. (2) Crossing over between genes on the same chromosome pairs.

3. Independent assortment

4. Law of Independent Assortment the "Second Law” The Law of Independent Assortment states that alleles for separate traits are passed independently of one another i.e from parents to offspring. That is, the biological selection of an allele for one trait has nothing to do with the selection of an allele for any other trait. Mendel found support for this law in his dihybrid cross experiments . In dihybrid crosses, however, he found a 9:3:3:1 ratios. This shows that each of the two alleles is inherited independently from the other.

5. No Linkage: Independent Assortment

6. Linkage without Recombination

7. Linkage with Recombination

8. What is Crossing Over? The exchange of chromosomal segments between two non- sister chromatids. Recall that crossing over occurs in Prophase I of Meiosis I. crossing over provides genetic variation during meiosis. Genetic swapping occurs between paired homologous chromosomes

9. Single Crossovers: Non-crossover (Parental) and Crossover (Recombinant) Gametes

10. The total recombination frequency (RF) It is number of new phenotypes compared to the total phenotypes. It's 25%+25%=50% . This 50% RF is always observed for genes on different chromosome pairs .However, the loci that are very far a part on the same chromosome an also show an RF value of 50%. Ex: AaBb (AB, Ab, aB, ab). map unit: The unit of distance in a genetic map. 1 map unit is equal to 1 percent recombination. 1 RF = 1 map unit = 1 centimorgan (cM)

11. The total recombination frequency (RF) In case of crossing over : The recombination frequency is less than 50% due to the linkage between genes. 1 RF = 1 map unit = 1 centimorgen (CM)

12. Genetic linkage is a term which describes the tendency of certain loci or alleles to be inherited together. Genetic loci on the same chromosome are physically close to one another and tend to stay together during meiosis. Linkage Bateson and Punnett (1905) in peas No independent assortment = No mendelian ratio

13. Morgan (1910) carried out an experiment by which he confirmed the work of Bateson and Punnett He also used another case of linkage between the eye color ,wing shape , and body color on the chromosome

14. Types of Linkage 1-Cis (Coupling ) linkage The 2 dominant alleles tend to be together on the same chromosome ,and 2 recessive alleles tend to be together on the other chromosome

15. % recombination <50% this is due to cis linkage. EX: p Purple ,elongate x red ,round PE/PE x pe / pe G PE pe F1 PE/ pe PE/ pe x pe / pe G PE , Pe,pE , pe pe F2 P-E- ,P- ee , ppE - , ppee 44 6 6 44

16. 2-Trans (repulsion) linkage The dominant allele and recessive allele tend to be together on the same chromosome

17. % recombination <50% this is due to trans linkage. P Purple, round X red elongate Pe/Pe X pE/pE G Pe pE F1 PE/pe PE/pe X pe/pe ( test cross) G P-E-, P-ee, ppE- , ppee 6 44 44 6

18. Morgan determined that the 2 genes of eye color and wing shape are linked due to the deviation of the phenotypic ratio ( Cis,Trans linkage) Red eye W+W+ White eye ww Normal wing m+m+ Miniature wing mm

19. Genetic Mapping The linkage of the genes in a chromosome can be represented in the form of a genetic map, which shows the linear order of the genes along the chromosome with the distances between adjacent genes proportional to the frequency of recombination between them.

20. A genetic map is also called a linkage map or a chromosome map. The concept of genetic mapping was first developed by Morgan's student, Alfred H. Sturtevant , in 1913.

21. Mapping using a trihybrid testcross (Three-point experiments) Three-point experiments is used to determine the loci of three genes in an organism's genome. The two most common phenotypes represent parental gametes. The two least common phenotypes represent double crossover in gamete formation. By comparing the parental and double-crossover phenotypes, the geneticist can determine which gene is located between the others on the chromosome.

22. Chromosome Interference The crossing over in one region prevents the crossing over in the adjacent regions. In higher organisms it has been found that one chiasma formation reduces the probability of another chiasma formation in an immediately adjacent region of the chromosome. The interaction prevents the double crossing over events

23. Coefficient of coincidence (C.O.C) In a three point cross the number of observed double crossovers divided by the number expected based on the observed occurrence of single crossovers. The ratio of the observed number of double recombinants to the expected number (C.O.C) = observed double recombinants  / expected double recombinants  Interference (I) = 1-(C.O.C)

24. Solving a Three-Point Cross Draw a map of these 3 genes (a,b, and d) showing the distances between all pairs of genes, and then calculate the value of interference. Offspring Genotypes Count a + d 510 + b + 498 a + + 3 + b d 1 a b + 61 + + d 59 a b d 35 + + + 37 total 1204 1-What is the order of the genes ? 2-Calculate the recombination frequencies and the interference between the genes ? 3-Draw a genetic map of the genes ?

25. Solving : 1-Step 1: identify parental classes Parentals are the largest groups: Double crossover are the smallest groups. The parental class of offspring is the largest: a + d, and + b +. No crossovers have occurred here: the original parents had these combinations of alleles. The double crossovers class (2CO) is the smallest class: + b d and a + +. The other two pairs are the two single crossover classes. Offspring Ge notypes Count a + d 510 NCO + b + 498 NCO a + + 3 DCO + d b 1 DCO a + b 61 SCO + d + 59 SCO a d b 35 SCO + + + 37 SCO total 1204 Solving a Three-Point Experiment

26. Since we don’t know the orientation of these genes relative to the chromosome as a whole, there are only 3 possible orders. These are based on which gene is in the middle. The genes could be a-- b --d,, or a-- d --b. To determine gene order, set up the parental chromosomes in the F1, then see what the resulting double crossover offspring would look like. If the observed 2CO’s match the expected, then you have the correct gene order. If not, try a different order. Getting Gene Order

27. recombinant offspring(SCO1)= total of recombinant offspring(SCO1 or 2)+ observed double recombinants / Total offspring a—d, so recombinants1 are a b + and + + d. 61 + 59 + 3 + 1 = 124 / Total offspring = 1204. So map distance = 124/1204 * 100 = 10.3 map units. d--b ,so recombinants 2 are + + + and a d b. There are 35 + 37 + 3 + 1 = 76 of them. 76 / 1204 * 100 = 6.3 map units. a—b , so recombinants are a to b . There are 61 + 59 + 35 + 37 +2(3+1) = 200 of them. 200 / 1204 * 100 = 16.6 map units. Gene Distances

28. Interference and Map There were 3 + 1 = 4 observed 2CO’s. Expected 2CO: = ((10.3/100) x (6.3 /100) x1204))= 7.81 Coef. of Coincidence = obs 2CO / exp 2CO = 4 / 7.81 = 0.512 Interference = 1 - C. of C. = 1 - 0.512 = 0.487 a 10.3 d 6.3 b 16.6

29. Steps to Solving 3-Point experiment Problems 1. Determine which pair is the parental class: it is the LARGEST class. 2. Determine which pair is the double crossover (2CO) class: the SMALLEST class. 3. Determine which gene is in the middle. If you compare the parentals with the 2COs, the gene which switched partners is in the middle. If you think two genes have switched sides, it is the other gene that is in the middle. 5. Determine map distances for all three pairs of genes. Count the number of offspring that have had a crossover between the genes of interest, then divide by the total offspring and multiply by 100. 6. Figure the expected double crossovers: (map distance for interval I * map distance for interval II / 100 * total / 100 = exp 2CO. 6. Figure the interference as: I = 1 - (obs 2CO / exp 2CO). The observed 2CO class comes from the data.

30. 580 ct + cv + v 592 ct cv v + 45 ct + cv v 40 ct cv + v + 89 ct cv v 94 ct + cv + v + 3 ct cv + v 5 ct + cv v + 1448 In Drosophila, assuming that you have a cross between Females heterozygous for all three genes were mated with wild type males, and the following progeny were obtained:

31. 1-What is the order of the genes ? 2-Calculate the recombination frequencies and the interference between the genes ? 3-Draw a genetic map of the genes ? Vermilion eyes) v ( ,(+V) red eyes (cv) cross veinless , (+CV) a absence of cross on the wing or vien ct (cut or snipped wing edges) ct (+ smooth wing)

32. In order to draw a genetic map of the gene we need to calculate the recombinants between each two adjacent genes (percentage of crossing events between each two adjacent genes) The recombinants frequency (RF) between v and ct loci are: 89+94+3+5 = 191/ 1448 = 13.2 The recombinants frequency (RF) between ct and cv loci are: 45+40+3+5 = 93/ 1448 = 6.4 The recombinants frequency (RF) between v and cv equal to:

33. (89+94)+ (45+40)+ 2(3+5) = 284/ 1448 = 19.6 Interference = 1- c.o.c =1- observed double recombinants /expected double recombinants  expected double recombinants = % of region I crossing over x % of region II crossing over ×total offspring / 100 = (13.2 × 6.4/100 × 1448) /100 = 0.84 x 1448 / 100 =12 Interference = 1- (8/12) = 33 v 13.2 ct 6.4 cv 19.6

34. Draw a map of these 3 genes (v,w, and z) showing the distances between all pairs of genes, and then calculate the value of interference. Solving a Three-Point Experiment 1-What is the order of the genes ? 2-Calculate the recombination frequencies and the interference between the genes ? 3-Draw a genetic map of the genes ?

35. In corn, c gives a green plant body, while its wildtype allele c + gives a purple plant body. bz (bronze) gives brown seeds, while the wildtype allele bz + gives purple seeds. wx (waxy) gives waxy endosperm in the seeds; wx + gives starchy endosperm. The genes are arranged on the chromosome in the order c-bz-wx. The cross: c bz wx / + + + x c bz wx. Note the +’s are the dominant wildtype alleles of the corresponding gene. Genotype Count c bz wx 318 + + + 324 c bz + 105 + + wx 108 c + + 18 + bz wx 20 c + wx 4 + bz + 3 total 900 1-What is the order of the genes ? 2-Calculate the recombination frequencies and the interference between the genes ? 3-Draw a genetic map of the genes ? Solving a Three-Point Experiment

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