1. 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 5-59 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Trihybrid Crosses b = black body color b + = gray body color pr = purple eye color pr + = red eye color vg = vestigial wings vg + = normal wings

2. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-60 Step 1: Cross two true-breeding strains that differ with regard to three alleles. Female is mutant for all three traits Male is homozygous wildtype for all three traits The goal in this step is to obtain aF1 individuals that are heterozygous for all three genes

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

4. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-62 Step 3: Collect data for the F 2 generation Phenotype Number of Observed Offspring (males and females) Gray body, red eyes, normal wings411 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 wings412

5. 5-63 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Analysis of the F 2 generation flies will allow us to map the three genes The three genes exist as two alleles each Therefore, there are 2 3 = 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

6. 5-64 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 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

7. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-65 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

8. 5-66 Parental offspringTotalNonparental OffspringTotal Gray body, red eyes (411 + 61) 472 Gray body, purple eyes (30 + 2) 32 Black body, purple eyes (412 + 60) 472 Black body, red eyes (28 + 1) 29 944 61 The map distance between body color and eye color is Map distance = 61 944 + 61 X 100= 6.1 map units

9. 5-67 Parental offspringTotalNonparental OffspringTotal Gray body, normal wings (411 + 2) 413 Gray body, vestigial wings (30 + 61) 91 Black body, vestigial wings (412 + 1) 413 Black body, normal wings (28 + 60) 88 826 179 The map distance between body color and wing shape is Map distance = 179 826 + 179 X 100 = 17.8 map units

10. 5-68 Parental offspringTotalNonparental OffspringTotal Red eyes, normal wings (411 + 28) 439 Red eyes, vestigial wings (61 + 1) 62 Purple eyes, vestigial wings (412 + 30) 442 Purple eyes, normal wings (60 + 2) 62 881 124 The map distance between eye color and wing shape is Map distance = 124 881 + 124 X 100 = 12.3 map units

11. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-69 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

12. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-70 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

13. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-71 PhenotypeNumber 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 30 + 28 1,005 = 0.058 Single crossover between b and pr 61 + 60 1,005 = 0.120 Single crossover between pr and vg 1 + 2 1,005 = 0.003 Double crossover, between b and pr, and between pr and vg

14. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-72 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

15. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-73 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

16. The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossover 5-74 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Interference 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

17. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-75 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 Interference

18. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-76 Interference (I) is expressed as I = 1 – C where C is the coefficient of coincidence C = Observed number of double crossovers Expected number of double crossovers 3 7.5 C = I = 1 – C = 1 – 0.4 = 0.6 or 60% This means that 60% of the expected number of crossovers did not occur = 0.40

19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-77 Since I is positive, this interference is positive interference Rarely, the outcome of a testcross yields a negative value for interference This suggests that a 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