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Trihybrid Crosses
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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
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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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
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
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
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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
Step 3: Collect data for the F2 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
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
Analysis of the F2 generation flies will allow us to
map the three genes
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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,
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Parental phenotypes occur most frequently
Double crossover phenotypes occur least frequently
Single crossover phenotypes occur with “intermediate”
frequency
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
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
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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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
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
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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
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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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Parental offspring
Gray body, red eyes
(411 + 61)
Black body, purple eyes
(412 + 60)
Total
Nonparental Offspring
Total
472
Gray body, purple eyes
(30 + 2)
32
472
Black body, red eyes
(28 + 1)
29
944
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61
The map distance between body color and eye color is
61
X 100 = 6.1 map units
Map distance =
944 + 61
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Parental offspring
Gray body, normal wings
(411 + 2)
Black body, vestigial wings
(412 + 1)
Total
Nonparental Offspring
Total
413
Gray body, vestigial wings
(30 + 61)
91
413
Black body, normal wings
(28 + 60)
88
826
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179
The map distance between body color and wing shape is
179
X 100 = 17.8 map units
Map distance =
826 + 179
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Parental offspring
Red eyes, normal wings
(411 + 28)
Purple eyes, vestigial wings
(412 + 30)
Total
Nonparental Offspring
Total
439
Red eyes, vestigial wings
(61 + 1)
62
442
Purple eyes, normal wings
(60 + 2)
62
881
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124
The map distance between eye color and wing shape is
124
X 100 = 12.3 map units
Map distance =
881 + 124
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Step 5: Construct the map
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Based on the map unit calculation the body color and
wing shape genes are farthest apart
 The eye color gene is in the middle
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The data is also consistent with the map being drawn
as vg – pr – b (from left to right)
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In detailed genetic maps, the locations of genes are
mapped relative to the centromere
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To calculate map distance, we have gone
through a method that involved the separation of
data into pairs of genes (see step 4)
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An alternative method does not require this
manipulation
 Rather, the trihybrid data is used directly

This method is described next
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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
= 0.058
1,005
Single crossover
between pr and vg
61 + 60
Double crossover,
between b and pr,
and between
pr and vg
1+2
= 0.120
1,005
= 0.003
1,005
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
To determine the map distance between the genes, we
need to consider both single and double crossovers
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To calculate the distance between b and pr
 Map distance = (0.058 + 0.003) X 100 = 6.1 mu
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To calculate the distance between pr and vg
 Map distance = (0.120 + 0.003) X 100 = 12.3 mu
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To calculate the distance between b and vg
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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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
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
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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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Interference
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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
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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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Interference
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Therefore, we would expect seven or eight offspring to be
produced as a result of a double crossover
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However, the observed number was only three!
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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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Interference (I) is expressed as
 I = 1 – C
 where C is the coefficient of coincidence
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Observed number of double crossovers
C=
Expected number of double crossovers
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C=
3
7.5
= 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
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Since I is positive, this interference is positive interference
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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
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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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