TRI HYBRID CROSS BY:Sam Atizado

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
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2.  Step 1: Cross two true-breeding strains that differ
with regard to three alleles.
Male is homozygous
Female is mutant
wildtype for all three
for all three traits
traits
 The goal in this step is to obtain aF1 individuals that
are heterozygous for all three genes
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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13. Phenotype Number of
Observed
Offspring
Gray body, 30
purple eyes,
vestigial wings Single crossover 30 + 28
= 0.058
Black body, 28 between b and pr 1,005
red eyes,
normal wings
Gray body, 61
red eyes,
vestigial wings Single crossover 61 + 60
= 0.120
Black body, 60 between pr and vg 1,005
purple eyes,
normal wings
Gray body, 2
purple eyes, Double crossover, 1+2
normal wings = 0.003
between b and pr, 1,005
Black body, 1 and between
red eyes, pr and vg
vestigial wings
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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
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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
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16.  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 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
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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
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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
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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
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