Trihybrid Crosses n n Data from trihybrid crosses

Trihybrid Crosses n n 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 n In this example, we will consider fruit flies that differ in body color, eye color and wing shape n n n b = black body color b+ = gray body color pr = purple eye color pr+ = red eye color vg = vestigial wings vg+ = normal wings Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -59

n Step 1: Cross two true-breeding strains that differ with regard to three alleles. Female is mutant for all three traits n Male is homozygous wildtype for all three traits The goal in this step is to obtain a. F 1 individuals that are heterozygous for all three genes Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -60

n Step 2: Perform a testcross by mating F 1 female heterozygotes to male flies that are homozygous recessive for all three alleles n During gametogenesis in the heterozygous female F 1 flies, crossovers may produce new combinations of the 3 alleles Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -61

n Step 3: Collect data for the F 2 generation Number of Observed Offspring (males and females) Phenotype Gray body, red eyes, normal wings 411 Gray body, red eyes, vestigial wings 61 Gray body, purple eyes, normal wings 2 Gray body, purple eyes, vestigial wings 30 Black body, red eyes, normal wings 28 Black body, red eyes, vestigial wings 1 Black body, purple eyes, normal wings 60 Black body, purple eyes, vestigial wings 412 Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -62

n Analysis of the F 2 generation flies will allow us to map the three genes n n n 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, n n n Parental phenotypes occur most frequently Double crossover phenotypes occur least frequently Single crossover phenotypes occur with “intermediate” frequency Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -63

n n The combination of traits in the double crossover tells us which gene is in the middle n 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 n Thus, the gene for eye color lies between the genes for body color and wing shape Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -64

n Step 4: Calculate the map distance between pairs of genes n To do this, one strategy is to regroup the data according to pairs of genes n n 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 n n n 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 Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -65

Parental offspring Gray body, red eyes (411 + 61) Black body, purple eyes (412 + 60) Total Nonparental Offspring 472 Gray body, purple eyes (30 + 2) 32 472 Black body, red eyes (28 + 1) 29 944 n Total 61 The map distance between body color and eye color is 61 X 100 = 6. 1 map units Map distance = 944 + 61 5 -66

Parental offspring Total Nonparental Offspring Gray body, normal wings (411 + 2) Black body, vestigial wings (412 + 1) 413 Gray body, vestigial wings (30 + 61) 91 413 Black body, normal wings (28 + 60) 88 826 n Total 179 The map distance between body color and wing shape is Map distance = 179 826 + 179 X 100 = 17. 8 map units 5 -67

Parental offspring Total Nonparental Offspring Red eyes, normal wings (411 + 28) Purple eyes, vestigial wings (412 + 30) 439 Red eyes, vestigial wings (61 + 1) 62 442 Purple eyes, normal wings (60 + 2) 62 881 n Total 124 The map distance between eye color and wing shape is Map distance = 124 881 + 124 X 100 = 12. 3 map units 5 -68

n Step 5: Construct the map n n n Based on the map unit calculation the body color and wing shape genes are farthest apart n 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 Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -69

n n 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 n Rather, the trihybrid data is used directly n This method is described next Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -70

Phenotype Number of Observed Offspring Gray body, purple eyes, vestigial wings 30 Black body, red eyes, normal wings 28 Gray body, red eyes, vestigial wings 61 Black body, purple eyes, normal wings 60 Gray body, purple eyes, normal wings 2 Black body, red eyes, vestigial wings 1 Single crossover between b and pr 30 + 28 1, 005 Single crossover between pr and vg 61 + 60 Double crossover, between b and pr, and between pr and vg 1+2 1, 005 Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display = 0. 058 = 0. 120 = 0. 003 5 -71

n To determine the map distance between the genes, we need to consider both single and double crossovers n n n To calculate the distance between b and pr n Map distance = (0. 058 + 0. 003) X 100 = 6. 1 mu To calculate the distance between pr and vg n Map distance = (0. 120 + 0. 003) X 100 = 12. 3 mu To calculate the distance between b and vg n n The double crossover frequency needs to be multiplied by two n Because both crossovers are occurring between b and vg Map distance = (0. 058 + 0. 120 + 2[0. 003]) X 100 = 18. 4 mu Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -72

n n 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 n 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. n This slightly lower value was a small underestimate because the first method does not consider the double crossovers in the calculation Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -73

Interference n The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossover P (double crossover) = P (single crossover X P (single crossover between b and pr) between pr and vg) = 0. 061 X 0. 123 = 0. 0075 n Based on a total of 1, 005 offspring n The expected number of double crossover offspring is = 1, 005 X 0. 0075 = 7. 5 Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -74

Interference n n 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! n n n 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 n The first crossover decreases the probability that a second crossover will occur nearby Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -75

n n Interference (I) is expressed as n I = 1 – C n where C is the coefficient of coincidence n Observed number of double crossovers C= Expected number of double crossovers n C= 3 7. 5 = 0. 40 I = 1 – C = 1 – 0. 4 = 0. 6 or 60% n This means that 60% of the expected number of crossovers did not occur Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -76

n n n Since I is positive, this interference is positive interference Rarely, the outcome of a testcross yields a negative value for interference n This suggests that a first crossover enhances the rate of a second crossover The molecular mechanisms that cause interference are not completely understood n However, most organisms regulate the number of crossovers so that very few occur per chromosome Copyright ©The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display 5 -77