By adding a third gene, we now have several different types of crossing over
products that can be obtained. The following figure shows the different
recombinant products that are possible.
Now if we were to perform a testcross with F1, we would expect a
1:1:1:1:1:1:1:1 ratio. As with the two-point analyzes described above, deviation
from this expected ratio indicates that linkage is occurring. The best way to
become familiar with the analysis of three-point test cross data is to go
through an example. We will use the arbitrary example of genes A, B,
and C. We
first make a cross between individuals that are AABBCC and aabbcc.
Next the F1 is testcrossed
to an individual that is aabbcc.
We will use the following data to determine the gene order and linkage
distances. As with the two-point data, we will consider the F1gamete
composition.
Genotype |
Observed |
Type of Gamete |
ABC |
390
|
Parental |
abc |
374
|
Parental |
AbC |
27
|
Single-crossover between
genesC and B |
aBc |
30
|
Single-crossover between
genesC and B |
ABc |
5
|
Double-crossover |
abC |
8
|
Double-crossover |
Abc |
81
|
Single-crossover between
genesA and C |
aBC |
85
|
Single-crossover between
genesA and C |
Total |
1000
|
|
The best way to solve these problems is to develop a systematic approach. First,
determine which of the the genotypes are the parental gentoypes. The
genotypes found most frequently are the parental genotypes.From the table it
is clear that the ABC and abc genotypes
were the parental genotypes.
Next we need to determine the order of the genes. Once we have determined the
parental genotypes, we use that information along with the information obtained
from the double-crossover. The
double-crossover gametes are always in the lowest frequency. From
the table the ABcand abC genotypes
are in the lowest frequency. The
next important point is that a double-crossover event moves the middle allele
from one sister chromatid to the other. This
effectively places the non-parental allele of the middle gene onto a chromosome
with the parental alleles of the two flanking genes. We can see from the table
that the C gene
must be in the middle because the recessive c allele
is now on the same chromosome as the Aand B alleles,
and the dominant C allele
is on the same chromosome as the recessive a and b alleles.
Now that we know the gene order is ACB,
we can go about determining the linkage distances between A and C,
and Cand B.
The linkage distance is calculated by dividing the total number of recombinant
gametes into the total number of gametes. This is the same approach we used with
the two-point analyses that we performed earlier. What is different is that we
must now also consider the double-crossover events. For these calculations we
include those double-crossovers in the calculations of both interval distances.
So the distance between genes A and C is
17.9 cM [100*((81+85+5+8)/1000)], and the distance between C and B is
7.0 cM [100*((27+30+5+8)/1000)].
Now let's try a problem from Drosophila,
by applying the principles we used in the above example. The following table
gives the results we will analyze.
Genotype |
Observed |
Type of Gamete |
v cv+ ct+ |
580
|
Parental |
v+ cv
ct |
592
|
Parental |
v cv ct+ |
45
|
Single-crossover between
genesct and cv |
v+ cv+ ct |
40
|
Single-crossover between
genesct and cv |
v cv ct |
89
|
Single-crossover between
genes vand ct |
v+ cv+ ct+ |
94
|
Single-crossover between
genes vand ct |
v cv+ ct |
3
|
Double-crossover |
v+ cv
ct+ |
5
|
Double-crossover |
Total |
1448
|
|
Step 1: Determine the parental genotypes.
The most abundant genotypes are the partenal types. These genotypes are v
cv+ ct+ and v+ cv
ct. What is different from our first three-point cross is that one parent
did not contain all of the dominant alleles and the other all of the recessive
alleles.
Step 2: Determine the gene order
To determine the gene order, we need the parental genotypes as well as the
double crossover geneotypes As we mentioned above, the least frequent genotypes
are the double-crossover geneotypes. These geneotypes are v
cv+ct and v+ cv
ct+. From this information we can determine the order by asking
the question: In the double-crossover genotypes, which parental allele is not
associated with the two parental alleles it was associated with in the original
parental cross. From the first double crossover, v
cv+ ct, thect allele
is associated with the v and cv+ alleles,
two alleles it was not associated with in the original cross. Therefore, ct is
in the middle, and the gene order is v
ct cv.
Step 3: Determing the linkage distances.
Step 4. Draw the map.
Three-point crosses also allows one to measureinterference (I)
among crossover events within a given region of a chromosome. Specifically, the
amount of double crossover gives an indication if interference occurs. The
concept is that given specific recombination rates in two adjacent chromosomal
intervals, the rate of double-crossovers in this region should be equal to the
product of the single crossovers. In the v
ct cv example
described above, the recombination frequency was 0.132 between genes v and ct,
and the recombination frequency between ctandcv was
0.064. Therefore, we would expect 0.84% [100*(0.132 x 0.64)] double
recombinants. With a sample size of 1448, this would amount to 12 double
recombinants. We actually only detected 8.
To measure interference, we first calculate the coefficient
of coincidence (c.o.c.)
which is the ratio of observed to expected double recombinants. Interference is
then calculated as 1 - c.o.c. The formula is as follows:
For the v ct cvdata,
the interference value is 33% [100*(8/12)].
Most often, interference values fall between 0 and 1. Values less than one
indicate that interference is occurring in this region of the chromosome.