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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Trihybrid Crosses
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
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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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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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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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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Parental offspring Total Nonparental Offspring Total
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 + 61X 100 = 6.1 map units
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Parental offspring Total Nonparental Offspring Total
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 + 179X 100 = 17.8 map units
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Parental offspring Total Nonparental Offspring Total
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 + 124X 100 = 12.3 map units
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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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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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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
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
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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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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
The product rule allows us to predict the likelihood of a double crossover from the individual probabilities of each single crossover
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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
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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
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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
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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