Genetics

1. Genetics Instructor: Dr. Jihad Abdallah Topic 9 Linkage and Genetic Mapping

2. Independent assortment • In Mendel’s dihybrid cross, alleles responsible for the two traits assort independently (Law of independent assortment). • This happens when the two genes are on different chromosomes. • In this case we have a ratio of 1:1:1:1 for the F1 gametes (for example, 1RY:1Ry:1rY:1ry)

3. Linkage • In eukaryotic species, each linear chromosome contains a long piece of DNA • A typical chromosome contains many hundred or even a few thousand different genes • Two genes are linked if they occur on the same chromosome (the genes are physically linked to each other). Two loci on different chromosomes are not linked, because they are usually separated by independent assortment. • The term linkage has two related meanings 1. Two or more genes can be located on the same chromosome 2. Genes that are close together tend to be transmitted as a unit

4. Chromosomes are called linkage groups • They contain a group of genes that are linked together • The number of linkage groups is the number of types of chromosomes of the species • For example, in humans • 22 autosomal linkage groups • An X chromosome linkage group • A Y chromosome linkage group • Genes that are far apart on the same chromosome may independently assort from each other • This is due to crossing-over

5. Crossing Over May Produce Recombinant Phenotypes • In diploid eukaryotic species, linkage can be altered during meiosis as a result of crossing over • Crossing over • Occurs during prophase I of meiosis at the bivalent stage • Non-sister chromatids of homologous chromosomes exchange DNA segments

6. The haploid cells contain the same combination of alleles as the original chromosomes The arrangement of linked alleles has not been altered

7. These haploid cells contain a combination of alleles NOT found in the original chromosomes This new combination of alleles is a result of genetic recombination These are termed parental or non-recombinant gametes These are termed nonparental or recombinant gametes

8. Illustration of recombination between two loci A and B:(a) Two pairs of sister chromatids align during meiosis.  A1 and B1 are located on the same chromosome. A2 and B2 are located on a different chromosome. (b) DNA crossover leads to recombination if the chiasma is located between the two loci.  (c) DNA crossover does not lead to recombination if the chiasma is not located between the two loci.

9. The gametes carrying A1B1 or A2B2 are called parental or (non-recombinant) gametes.While the gametes carrying A1B2 or A2B1 are called recombinant gametes because they resulted from recombination during crossing-over. • Recombination is the exchange of alleles between non-sister chromatids due to crossing over and results in new combinations of alleles. • The recombination frequency depends on the distance between the two loci and the position of crossover (the chiasma).  • The closer they are, the less likely the recombination will occur, because recombination occurs only when the chiasma is located between the two loci.

10. Recombination fractions • A recombination event gives parental type (P) and recombinant type (R) offspring • Recombination fraction RF = R/(R+P) • RF is between 0 and 0.5 (0 and 50%) • The closer together the genes are, the smaller is RF • RF = 0.5 for unlinked genes (very far apart or on different chromosomes)

11. Bateson and Punnett Discovered Two Traits That Did Not Assort Independently • In 1905, William Bateson and Reginald Punnett conducted a cross in sweet pea involving two different traits • Flower color and pollen shape • This is a dihybrid cross that is expected to yield a 9:3:3:1 phenotypic ratio in the F2 generation • However, Bateson and Punnett obtained surprising results

12. A much greater proportion of the two types found in the parental generation

13. Genetic Mapping • Genetic mapping (or gene mapping) is to determine the order and distance between genes that are linked to each other on the same chromosome. • Used for construction of genetic linkage maps

14. Genetic maps are useful in many ways: • 1. They allow us to understand the overall complexity and genetic organization of a particular species • 2. They improve our understanding of the evolutionary relationships among different species • 3. They can be used to diagnose, and perhaps, someday to treat inherited human diseases • 4. They can help in predicting the likelihood that a couple will produce children with certain inherited diseases • 5. They provide helpful information for improving agriculturally important strains through selective breeding programs

15. Number of recombinant offspring X 100 Total number of offspring • The units of distance are called map units (mu) or sometimes called centimorgan (cM). • One map distance (1 cM) is equivalent to 1% recombination frequency. • The basis for genetic mapping is that the percentage of recombinant offspring (R) is used to deduce the distance between two genes. • If two genes are far apart, many recombinant offspring will be produced. Map distance =

17. Number of recombinant offspring X100 Total number of offspring 76 + 75 = X 100 542 + 537 + 76 + 75 = 12.3 map units (cM) (Short bristle, Ebony body) x (Normal bristles, Gray body) (se/se) x (se/s+e+) Offspring: Short bristles, ebony body (parental) 542 Normal bristles, gray body (parental) 537 Normal bristles, ebony body (recombinant) 76 Short bristles, gray body (recombinant) 75 Map distance =

18. B. Trihybrid Test Cross Mapping • Data from trihybrid crosses can also yield information about map distance and gene order • The following experiment outlines a common strategy for using trihybrid crosses to map genes • In this example, we will consider fruit flies that differ in body color, eye color and wing shape • b = black body color • b+ = gray body color • pr = purple eye color • pr+ = red eye color • vg = vestigial wings • vg+ = normal wings

19. Step 1: Cross two true-breeding strains that differ with regard to three alleles. black body, purple eye, vestigial wings Gray body, red-eye, normal wings Male is homozygous wildtype for all three traits Female is mutant for all three traits • The goal in this step is to obtain a F1 individuals that are heterozygous for all three genes

20. Step 2: Perform a testcross by mating F1 female heterozygotes to male flies that are homozygous recessive for all three alleles black body, purple eye, vestigial wings Gray body, red-eye, normal wings • During gametogenesis in the heterozygous female F1 flies, crossovers may produce new combinations of the 3 alleles

21. Step 3: Collect data for the F2 generation Double cross-over

22. Analysis of the F2 generation flies will allow us to map the three genes • The three genes exist as two alleles each • Therefore, there are 23 = 8 possible combinations of offspring • If the genes assorted independently, all eight combinations would occur in equal proportions • It is obvious that they are far from equal • In the offspring of crosses involving linked genes, • Parental phenotypes occur most frequently • Double crossover phenotypes occur least frequently • Single crossover phenotypes occur with “intermediate” frequency

23. The combination of traits in the double crossover tells us which gene is in the middle • A double crossover separates the gene in the middle from the other two genes at either end • In the double crossover categories, the recessive purple eye color is separated from the other two recessive alleles • Thus, the gene for eye color lies between the genes for body color and wing shape

24. Step 4: Calculate the map distance between pairs of genes • To do this, one strategy is to regroup the data according to pairs of genes: • From the parental generation, we know that the dominant alleles are linked, as are the recessive alleles • This allows us to group pairs of genes into parental and nonparental combinations • Parentals have a pair of dominant or a pair of recessive alleles • Nonparentals have one dominant and one recessive allele • The regrouped data will allow us to calculate the map distance between the two genes

25. Body color and eye color • The map distance between body color and eye color is Map distance = 61 X 100 = 6.1 map units 944 + 61

26. 179 X 100 826 + 179 Body color, wing shape • The map distance between body color and wing shape is Map distance = = 17.8 map units

27. 124 X 100 881 + 124 • The map distance between eye color and wing shape is Map distance = = 12.3 map units

28. Step 5: Construct the map • Based on the map unit calculation the body color and wing shape genes are farthest apart • The eye color gene is in the middle • The data is also consistent with the map being drawn as vg – pr – b (from left to right) • In detailed genetic maps, the locations of genes are mapped relative to the centromere

29. To calculate map distance, we have gone through a method that involved the separation of data into pairs of genes (see step 4) • An alternative method does not require this manipulation • Rather, the trihybrid data is used directly • This method is described next

30. 30 + 28 1,005 61 + 60 1,005 1 + 2 1,005 Single crossover between b and pr = 0.058 Single crossover between pr and vg = 0.120 Double crossover, between b and pr, and between pr and vg = 0.003

31. To determine the map distance between the genes, we need to consider both single and double crossovers • To calculate the distance between b and pr • Map distance = (0.058 + 0.003) X 100 = 6.1 mu • To calculate the distance between pr and vg • Map distance = (0.120 + 0.003) X 100 = 12.3 mu • To calculate the distance between b and vg • The double crossover frequency needs to be multiplied by two • Because both crossovers are occurring between b and vg • Map distance = (0.058 + 0.120 + 2[0.003]) X 100 = 18.4 mu

32. Alternatively, the distance between b and vg can be obtained by simply adding the map distances between b and pr, and between pr and vg: • Map distance = 6.1 + 12.3 = 18.4 mu • Note that in the first method (grouping in pairs), the distance between b and vg was found to be 17.8 mu. • This slightly lower value was a small underestimate because the first method does not consider the double crossovers in the calculation

33. Interference • The product rule allows us to predict the probability of a double crossover from the individual probabilities of each single crossover P (double crossover) = P (single crossover between b and pr) P (single crossover between pr and vg) X = 0.061 X 0.123 = 0.0075 • Based on a total of 1,005 offspring • The expected number of double crossover offspring is = 1,005 X 0.0075 = 7.5

34. Interference • Therefore, we would expect seven or eight offspring to be produced as a result of a double crossover • However, the observed number was only three! • Two with gray bodies, purple eyes, and normal sings • One with black body, red eyes, and vestigial wings • This lower-than-expected value is due to a common genetic phenomenon, termed positive interference • The first crossover decreases the probability that a second crossover will occur nearby

35. Observed number of double crossovers Expected number of double crossovers 3 7.5 • Interference (I) is expressed as I = 1 – C • where C is the coefficient of coincidence C = C = = 0.40 • I = 1 – C = 1 – 0.4 = 0.6 or 60% This means that 60% of the expected number of crossovers did not occur

36. Since I is positive, this interference is positive interference • Rarely, the outcome of a testcross yields a negative value for interference (the first crossover enhances the rate of a second crossover) • The molecular mechanisms that cause interference are not completely understood • However, most organisms regulate the number of crossovers so that very few occur per chromosome