1. Linkage

2. What is Linkage? Linkage is defined genetically: the failure of two genes to assort independently. Linkage occurs when two genes are close to each other on the same chromosome. However, two genes on the same chromosome are called syntenic. Linked genes are syntenic, but syntenic genes are not always linked. Genes far apart on the same chromosome assort independently: they are not linked. Linkage is based on the frequency of crossing over between the two genes. Crossing over occurs in prophase of meiosis 1, where homologous chromosomes break at identical locations and rejoin with each other.

3. Discovery of Linkage In 1900, Mendel’s work was re-discovered, and scientists were testing his theories with as many different genes and organisms as possible. William Bateson and R.C. Punnett were working with several traits in sweet peas, notably a gene for purple (P) vs. red (p) flowers, and a gene for long pollen grains (L) vs. round pollen grains (l).

4. Bateson and Punnett’s Results PP LL x pp ll selfed F1: Pp Ll F2 results in table Very significant deviation from expected Mendelian ratio: chi-square = 97.4, with 3 d.f. Critical chi square value = 7.815. The null hypothesis for chi square test with 2 genes is that the genes assort independently. These genes do not assort independently. phen otype obsexp ratio exp num P_ L_2849/16215 P_ ll213/1671 pp L_ 213/1671 pp ll551/1624

5. B+P Genes in a Test Cross Purpose of a test cross: the offspring phenotypes appear in the same ratio as the gametes in the parent being tested. Here, we want to see how many gametes are in the original parental configuration (PL or pl) and how many are in the recombinant configuration (Pl or pL). The parental types have the same combination of alleles that were in the original parents, and the recombinant types have a combination of the mother’s and father’s alleles. Original parents: PP LL x pp ll F1 test cross: Pp Ll x pp ll Pheno- type obs purple long 392 purple round 116 red long 127 red round 365 total1000

6. More Test Cross Parentals: 392 PL + 365 pl = 757. 757/1000 total offspring = 75.7% parental Recombinant: 116 Pl + 127 pL = 243. 243 /1000 = 24.3% recombinant. If the genes were unlinked, 50% would be recombinant. These genes are linked, with 24.3% recombination between the P gene and the L gene. If the genes were right on top of each other, that is, the two phenotypes were both caused by the same gene (pleiotropy), then there would be 0% recombination between them. The percentage of recombinants is always between 0% and 50%, and the percentage of parentals is always between 50% and 100%.

7. Better Symbolism We have been following Mendel’s tradition in writing the two alleles for each gene together, as in PP LL x pp ll. Now we need to start paying attention to the fact that genes are on chromosomes. If one parent contributes a P L chromosome and the other parent contributes a p l chromosome, we write the heterozygote as PL/pl. Homozygotes (for all genes on that chromosome) are written without the slash: the pp ll homozygote used in the test cross is written as p l.

8. Coupling vs. Repulsion The original test cross we did was PL/pl x p l. Among the offspring, PL and pl were parental types, and pL and Pl were the recombinant types. There was 24.3% recombination between the genes. The condition of having the dominant alleles for both genes on the same parental chromosome, with both recessives on the other parental chromosome, is called “coupling”: the P and L genes are “in coupling phase”. The opposite condition, having one dominant and one recessive on each parental chromosome, is called “repulsion”. Thus, if the original parents were P l x p L, their offspring would have the genes in repulsion phase: Pl / pL.

9. Test Cross in Repulsion Now do the test cross in repulsion: Pl / pL x p l Here, the parental types are P l and p L, and the recombinant types are P L and p l. The numbers of offspring in each type are quite different from the originals. However, the percentage of recombinants is the same: 24.3%. 123 P L + 120 p l = 243 recombinant offspring. 243/ 1000 total offspring = 24.3 % The percentage of recombination depends on the distance between the genes on the chromosome, and NOT on which alleles are on which chromosome. phenotypeobs P L123 P l372 p L385 p l120 total1000

10. Process of Recombination From an evolutionary point of view, the purpose of sex is to re-shuffle the combinations of alleles so the offspring receive a different set of alleles than their parents had. Natural selection then causes offspring with good combinations to survive and reproduce, while offspring with bad combinations don’t pass them on. Genes are on chromosomes. Meiosis is a mechanism for re-shuffling the chromosomes: each gamete gets a mixture of paternal and maternal chromosomes. However, chromosomes are long and contain many genes. To get individual genes re-shuffled, there needs to be a mechanism of recombining genes that are on the same chromosome. This mechanism is called “crossing over.

11. More Recombination Crossing over occurs in prophase of meiosis 1, when the homologous chromosomes “synapse”, which means to pair closely with each other. DNA strands from the two chromosomes are matched with each other. During synapsis, an enzyme, “recombinase”, attaches to each chromosome at several randomly chosen points. The recombinase breaks both DNA molecules at the same point, and re-attaches them to opposite partners. The result of crossing over can be seen in the microscope as prophase continues, as X-shaped structures linking the homologues. The genetic consequence of crossing over is that each chromosome that goes into a gamete is a combination of maternal and paternal chromosomes.

12. Recombination Process

13. The Process of Recombination, animated version http://www.youtube.com/watch?v=BhJf9MHHmc4 Or, maybe this animated GIF image will work

14. Linkage Mapping Each gene is found at a fixed position on a particular chromosome. Making a map of their locations allows us to identify and study them better. In modern times, we can use the locations to clone the genes so we can better understand what they do and why they cause genetic diseases when mutated. The basis of linkage mapping is that since crossing over occurs at random locations, the closer two genes are to each other, the less likely it is that a crossover will occur between them. Thus, the percentage of gametes that had a crossover between two genes is a measure of how far apart those two genes are. As pointed out by T. H. Morgan and Alfred Sturtevant, who produced the first Drosophila gene map in 1913. Morgan was the founder of Drosophila genetics, and in his honor a recombination map unit is called a centiMorgan (cM). A map unit, or centiMorgan, is equal to crossing over between 2 genes in 1% of the gametes.

15. Three Point Cross The easiest way to map genes is to compare them in groups of 3. This allows both the distances between them and their order to be determined. Further genes can be added to the map by using overlapping groups of 3. Mapping is usually done in test crosses. One parent is heterozygous for two versions of the chromosome being mapped, and the other parent is homozygous for the recessive mutants being mapped. Recombination in the heterozygous parent gives different combinations of alleles, which are counted. Note that recombination also occurs in the homozygous recessive parent, but it has no effect on the alleles in the offspring because it is homozygous.

16. Data 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. phenotypecount c bz wx318 + + +324 c bz +105 + + wx108 c + +18 + bz wx20 c + wx4 + bz +3 total900

17. Notes on the Data Genes are arranged in reciprocal pairs: each pair has 1 copy of the mutant allele and the wildtype allele for each gene. The counts are roughly equal for reciprocal pairs, because they are both products of the same crossing over events. Parentals are the largest groups: c bz wx and + + +. Double crossovers, one between c and bz and another between bz and wx, are the smallest groups.

18. Calculating Map Distances Basic process: determine the percentage of offspring that had a crossover between each pair of genes. 1. Examine c and bz first. Parental configuration was c bz and + +. Therefore, the recombinant configurations are c + and + bz. Count recombinants, ignoring the other gene (wx): 18 c + +, 20 + bz wx, 4 c + wx, 3 + bz +. Total is 45 recombinants out of 900 total offspring. 45 / 900 = 0.05. Need to multiply by 100 to get percentage: 0.05 x 100 = 5.0 map units between c and bz.

19. More Calculating Map Distances 2. Next examine bz and wx. Parentals are bz wx and + +, so recombinants are bz + and + wx. Ignoring c, the count of recombinants is: 105 c bz +, 108 + + wx, 4 c + wx, 3 + bz +. Total = 220 recombinants. 220 / 900 = 0.244. 0.244 x 100 = 24.4 map units between bz and wx. 3. Now do c and wx. Parentals are c wx and + +, so recombinant offspring are c + and + wx. Ignoring bz, the recombinants are 105 c bz +, 108 + + wx, 18 c + +, 20 + bz wx. Total = 251. 251/ 900 x 100 = 27.9 map units.

20. Map of c, bz, and wx All 3 genes are in the proper order, and all 3 distances between pairs of genes are shown. Note that distances don’t add up. This is due to double crossovers, which we will discuss next.

21. Double Crossovers and Mapping A double crossover is two crossovers both occurring between the two genes being examined. The first crossover changes the parental configuration of alleles to the recombinant configuration. The second crossover changes the recombinant configuration back to the parental. The net result is that the genes are in the parental configuration, same as if no crossovers had occurred. Thus, any even number of crossovers is the same as 0 crossovers, and any odd number is the same as 1 crossover. Since you only see the offspring and not the actual crossovers, it is very easy to undercount the number that occurred Consider the c bz wx cross. If you were just looking at c and wx, and hadn’t examined bz, the c + wx and + bz + offspring would be parental and count as 0 crossovers. The only way you know that 2 crossovers occurred is by examining bz. Perhaps other crossovers also occurred that we didn’t detect, since we didn’t examine any other genes in between bz and wx. This is the main reason why the sum of the c--bz and bz--wx distances didn’t add up: double crossovers were counted as parentals.

22. Mapping Function The further apart 2 genes are, the more likely it is that undetected double crossovers will occur between them. For this reason, gene maps are created using short intervals. One consequence: the maximum percentage of recombinant offspring is 50%, but many chromosomes are several hundred map units long. For instance, the human chromosome 13 is 125 map units long. Genes on opposite ends of this chromosome would have a 50% frequency of recombinant offspring. Mapping function: number of actual crossovers on x-axis, frequency of recombinant offspring on y-axis. In general, –for less than 20% recombinants, the number of recombinants is roughly equal to the number of crossovers; –for 20-50% recombinants, the number of recombinants is significantly less than the number of crossovers. –for 50% recombinants, the number of crossovers is not determinable.

23. Interference There is a second issue with double crossovers: interference. Interference is the inability of 2 crossovers to occur very close to each other. Think of the chromosome as a thick rope: it is impossible to bend it too tightly. It is possible to measure the amount of interference, by comparing the actual number of double crossovers to the number that you would expect based on the number of single crossovers that occurred.

24. Measuring Interference Consider our c--bz--wx example. The c--bz interval was 5.0 map units, or 5% recombinants between those 2 genes. The probability of a crossover between c and bz is 0.05, which you get by dividing the map distance by 100 to put it on a 0-1 scale instead of a 0-100 scale. The bz--wx interval was 24.4 map units, or 24.4% recombinants, or a 0.244 probability. If a crossover between c and bz had no effect on the possibility of a crossover between bz and wx (i.e. no interference), then a chance of a c--bz crossover AND and bz--wx crossover would be the product of their individual probabilities. That is, the expected frequency of double crossovers would be 0.05 * 0.244 = 0.0122. Since there were 900 offspring, the expected number of double crossover offspring would be 10.98. The observed number of double crossovers was 7. The coefficient of coincidence is the ratio of observed double crossovers to expected: 7 / 10.98 = 0.638. The interference is 1 minus the coefficient of coincidence, = 1 - 0.638 = 0.362. This means that about 36% of the expected double crossovers are not occurring due to interference.

25. Interference Formulas

26. Gene Order Another problem that can arise is that when mapping genes you usually don’t know their order. You need to infer the order from the data. Also, not all the alleles are necessarily going to be in coupling. Sometimes some alleles are in repulsion. Getting the gene order is a matter of comparing the alleles present on the parental chromosomes to those on the double crossover chromosomes.

27. Example Note that the offspring counts are arranged in reciprocal pairs. Each member of the pair has the opposite alleles, and the counts for the two members of the pair are approximately equal. 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 phenotype count a + d510 + b +498 a + +3 + b d1 a b +61 + + d59 a b d35 + + +37 total1204

28. Getting Gene Order Since we don’t know (or care about) 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, b--a--d, or a--d--b. Note that a--b--d is identical to d--b--a, etc. The order in which the genes are listed in the offspring counts has nothing to do with their order on the chromosome!!! The gene order is unknown at this point. 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.

29. Gene Order 1. Try a--b--d. The F1 chromosomes would be a + d/ + b +. A double crossover would give a b d and + + + 2CO offspring. This does not match the observed a + + and + b d. 2. Try b--a--d. The F1 chromosomes are + a d / b + +. A double crossover would give b a + and + + d 2CO offspring. This does not match the observed b + d and + a +. 3. Try a--d--b. The F1 chromosomes are a d + and + + b. A double crossover gives a + + and + d b, which matches the observed 2CO offspring. Therefore, a--d--b is the correct order; d is in the middle.

30. Gene Distances a--d. Parentals are a d and + +, so recombinants are a + and + d. There are 61 + 59 + 3 + 1 = 124 recombinant offspring. Total offspring = 1204. So map distance = 124/1204 * 100 = 10.3 map units. d--b. Parentals are d + and + b, so recombinants are d b and + +. There are 35 + 37 + 3 + 1 = 76 of them. 76 / 1204 * 100 = 6.3 map units. a--b. Parentals are a + and + b, so recombinants are a b and + +. There are 61 + 59 + 35 + 37 = 192 of them. 192 / 1204 * 100 = 15.9 map units.

31. Interference and Map There were 3 + 1 = 4 observed 2CO’s. Expected 2CO: = (10.3/100) * (6.3/100) * 1204 = 7.81 Coef. of Coincidence = obs 2CO / exp 2CO = 4 / 7.81 = 0.512 Interference = 1 - C. of C. = 1 - 0.512 = 0.487

32. Steps to Solving 3-Point Cross Problems 1. Organize the data into reciprocal pairs. For example, a + c and + b + are a reciprocal pair, and so are a b c and + + +. The number of offspring for each member of a pair will be similar. 2. Determine which pair is the parental class: it is the LARGEST class. 3. Determine which pair is the double crossover (2CO) class: the SMALLEST class. 4. 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 / 100 ) * (map distance for interval II / 100 ) * (total offspring) = exp 2CO. 6. Figure the interference as: I = 1 - (obs 2CO / exp 2CO). The observed 2CO class comes from the data.

33. One More Example Here are some test cross data for a 3 point cross. Determine the order of the genes, make a map showing all map distances, and determine the interference value. pr = purple eyes bl = black body dp = dumpy bristles Answer will be given in class.

34. Data phenotypecount wild type756 purple59 dumpy414 black21 purple dumpy23 purple black417 dumpy black60 purple dumpy black750 total2500

35. More Realistic Data phenotypecount wild type60 purple756 dumpy23 black417 purple dumpy414 purple black21 dumpy black750 purple dumpy black59 total2500