Download Gene Mapping using 3 Point Test Crosses: Outlined below are the

Gene Mapping using 3 Point Test Crosses:
Outlined below are the steps necessary for correctly analysing three-point test cross data, along with an
example of a problem. Examples of other problems can be found in your text.
Mapping Data: Data is derived from an F1 individual backcrossed to an individual that was
homozygous recessive a/a; s/s; r/r. Outline the genotype of the F1 used in the cross, and the linkage,
along with map distances where applicable of these three genes.
+++
a++
+s+
++r
as+
a+r
+sr
asr
58
1140
2
110
70
0
1078
43
1) Look at the data and determine which groups belong to the following three classes.
NCO = Nonrecombinant Class (Classes of gametes where no recombination has occurred)
SCO = Single cross over classes. Usually these can be divided into groups of two with roughly
equal numbers in each of the two classes in a group.
DCO = Double cross over classes. Always tends to be the smallest class with the very few
numbers in it.
2) In almost all cases (and unless clearly specified), the data will be derived from an F1 individual that
has been backcrossed (test-cross), to a homozygous, recessive individual. Therefore the two groups
in the NCO class represent the two chromosomes in the F1 parent. These should then be drawn out,
such as below.
F1
a
+
+
s
+
r
1140
1078
3) Determine which two groups belong in the DCO class. This information can then be used to
determine the gene order. The gene order is determined by predicting the events resulting from a
double cross over event. This is shown below. Try all three combinations, and make sure that you
have the correct gene order. Two orders should not work, the other should work.
4) Once the gene order has been established, use your map of the chromosomes crossing over during
meiosis in the F1 to predict the effect of a SCO event between A and S or the left and then centre
gene. The map distance is: MU = Sum of both classes of progeny from a SCO + the sum of both
classes from the DCO event.
5) The Co-efficient of coincidence (COC).
COC = Observed frequency of DCO’s/Expected frequency of DCO’s, which should be between
0 and 1. Interference is defined as 1 – c.o.c. Thus an interference of 1 means one CO event
completely prevents a second event.
Example 1:
Type of gamete
Parental
Parental
SCO
SCO
SCO
SCO
DCO
DCO
Gene Order a + +
+ s r
genotype
a++
+sr
+++
++r
as+
asr
+s+
a+r
# of progeny
1140
1078
58
110
70
43
2
0
or + a + or
s + r
+ r +
a + s
DCO = + s + and a + r. Model these.
Therefore gene order = a r s
Map Distance a - r = 70 + 110 + 2 + 0 = 182/2501 = 7.27 MU
Map Distance r - s = 58 + 43 + 2 + 0 = 103/2501
= 4.11
Interference: Predicted DCO’s = (0.0411 * 0.0727) = 0.00299 x 2501 = 7.48
COC = 2/7.48 = 0.267
Interference = 1 – 0.267 or 0.733
Example 2:
A true breeding Drosophila with vermilion eyes is crossed to a fly that is true breeding for the
crossveinless, cut phenotype. cv/cv t/t
(crossveinless, cut phenotype) cross to
v/v cv+/cv+ t+/t+ (vermilion)
v+ cv t
v+ cv t
x
v cv+ t+
v cv+ t+
P1
v cv+ t+
v+ cv t
F1
2) Cross F1 triple heterozygote to triply recessive stock (v cv t)
Progeny: 8 gametic types are possible: | v-cv | v-t | cv-t |
Parental
v
cv+ t+
v+
cv t
Recombinants
v cv t+
31
v+ cv+ t
28
Total
400
409
1000
v cv t
v+ cv+ t+
61
65
v cv+ t
v+ cv t+
2
0
3 Point Test Cross
Double Recombinants
Model 1 ;
F1 :
v cv+ t+
v+ cv t
Reject
t --------------------- cv -------------- v
F1
t cv v / t+ v+ cv+
Model 3 ;
v t+ cv+
v+ t cv
F1 :
2
4
t --------------------- v -------------- cv
t v+ cv/ t+ v cv+
Model 2 ;
F1 :
v cv+ t
v+ cv t+
Reject
cv --------------------- t -------------- v
F1
cv+ t v / t+ v+ cv
Accept
Example 3: In corn, a triple heterozygote was obtained carrying mutant alleles s (shrunken), w (white
aleurone) and x (waxy), all paired with their normal wild type alleles. This triple heterozygote was
testcrossed and the progeny contained the following.
116
4
2538
601
626
2708
2
113
Shrunken, white
wild type
shrunken
shrunken, waxy
white
white, waxy
shrunken, white, waxy
waxy
a) Determine if any of these loci are linked and if so, show map distances.
b) Show the allele arrangement on the chromosomes of the triple resistant heterozygotes used in
the testcross.
c) Calculate interference, if possible.
Steps:
1)
2)
3)
4)
5)
What are the parental and DCO gametes.
What is the order of genes on the chromosome.
What does the parental chromosomes look like.
Calculate your map distances.
Calculate your co-efficient of coincidence.
Create gene designations.
s/s+; w/w+; x/x+
x
s/s; w/w; x/x
Gametes from F1
Parental
s; w+; x+
s+; w; x
2538
2708
DCO; From this order, lets look at the order of the genes on the chromosomes.
s; w; x (2)
and
s+; w+; x+ (4)
Deriving Linkage Distance and Gene Order From Three-Point Crosses
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 F1 gamete
composition.
Genotype Observed
Type of Gamete
ABC
390
Parental
abc
374
Parental
AbC
27
Single-crossover between genes C and B
aBc
30
Single-crossover between genes C and B
ABc
5
Double-crossover
abC
8
Double-crossover
Abc
81
Single-crossover between genes A and C
aBC
85
Single-crossover between genes A 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 genotypes. 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 doublecrossover. The double-crossover gametes are always in the lowest frequency. From the
table the ABc and 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
A and 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 C and 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 genes ct and cv
v+ cv+ ct
40
Single-crossover between genes ct and cv
v cv ct
89
Single-crossover between genes v and ct
v+ cv+ ct+
94
Single-crossover between genes v and 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 parental 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 genotypes As we mentioned above, the least frequent genotypes are the doublecrossover genotypes. These genotypes 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, the ct 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.
•
v - ct distance caluculation. This distance is derived as follows:
100*((89+94+3+5)/1448) = 13.2 cM
•
ct - cv distance calculation. This distance is derived as follows:
100*((45+40+3+5)/1448) = 6.4 cM
Step 4. Draw the map.
Three-point crosses also allows one to measure interference (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 ct andcv 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 cv data, 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.