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

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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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