Trihybridhelper

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TRI HYBRID CROSS BY:Sam Atizado

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Trihybridhelper

  1. 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  b = black body color  b+ = gray body color  pr = purple eye color  pr+ = red eye color  vg = vestigial wings  vg+ = normal wings Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-59
  2. 2.  Step 1: Cross two true-breeding strains that differ with regard to three alleles. Male is homozygousFemale is mutant wildtype for all threefor all three traits traits  The goal in this step is to obtain aF1 individuals that are heterozygous for all three genes Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-60
  3. 3.  Step 2: Perform a testcross by mating F1 female heterozygotes to male flies that are homozygous recessive for all three alleles  During gametogenesis in the heterozygous female F1 flies, crossovers may produce new combinations of the 3 alleles Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-61
  4. 4.  Step 3: Collect data for the F2 generation Phenotype Number of Observed Offspring (males and females) 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 McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-62
  5. 5.  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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-63
  6. 6.  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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-64
  7. 7.  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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-65
  8. 8. Parental offspring Total Nonparental Offspring Total Gray body, red eyes Gray body, purple eyes (411 + 61) (30 + 2) 472 Black body, purple eyes Black body, red eyes 32 (412 + 60) 472 (28 + 1) 944 61 29 The map distance between body color and eye color is 61 Map distance = X 100 = 6.1 map units 944 + 61 5-66
  9. 9. Parental offspring Total Nonparental Offspring Total Gray body, normal wings Gray body, vestigial wings (411 + 2) (30 + 61) 413 Black body, vestigial wings Black body, normal wings 91 (412 + 1) 413 (28 + 60) 826 179 88 The map distance between body color and wing shape is 179 Map distance = X 100 = 17.8 map units 826 + 179 5-67
  10. 10. Parental offspring Total Nonparental Offspring Total Red eyes, normal wings Red eyes, vestigial wings (411 + 28) (61 + 1) 439 Purple eyes, vestigial wings Purple eyes, normal wings 62 (412 + 30) 442 (60 + 2) 881 124 62 The map distance between eye color and wing shape is 124 Map distance = X 100 = 12.3 map units 881 + 124 5-68
  11. 11.  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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-69
  12. 12.  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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-70
  13. 13. Phenotype Number of Observed OffspringGray body, 30purple eyes,vestigial wings Single crossover 30 + 28 = 0.058Black body, 28 between b and pr 1,005red eyes,normal wingsGray body, 61red eyes,vestigial wings Single crossover 61 + 60 = 0.120Black body, 60 between pr and vg 1,005purple eyes,normal wingsGray body, 2purple eyes, Double crossover, 1+2normal wings = 0.003 between b and pr, 1,005Black body, 1 and betweenred eyes, pr and vgvestigial wings Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-71
  14. 14.  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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-72
  15. 15.  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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-73
  16. 16.  The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossoverP (double crossover) = P (single crossover between bP (single crossover between X and pr) pr and vg) = 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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-74
  17. 17. 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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-75
  18. 18.  Interference (I) is expressed as  I = 1 – C  where C is the coefficient of coincidence Observed number of double crossovers  C= Expected number of double crossovers 3  C= = 0.40 7.5 I = 1 – C = 1 – 0.4 = 0.6 or 60%  This means that 60% of the expected number of crossovers did not occur Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-76
  19. 19.  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 Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display 5-77

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