BIOL 2416 GENETICS
Dr. A. Tineke Berends
TWO-POINT TEST CROSSES
WHY?
1.
2.
WHAT?
AUTOSOMAL
RECESSIVE
double heterozygote
X
homozygous
recessive
dominant =
+
symbolized by a
+
and b = wild-type
a+b+/ab x
ab/ab
To determine if 2 genes are linked
To estimate the map distance between 2 linked genes (more
accurate if genes are close together and when large numbers of
progeny are scored)
AUTOSOMAL
DOMINANT
double heterozygote
X
homozygous
recessive
mutant alleles are
dominant, so
recessive alleles =
+
wild type = A and
B+
AB/A+B+ x
A+B+/A+B+
X-LINKED
RECESSIVE
female double
heterozygote
X
male hemizygous
for wild-type alleles
+
dominant = a and
b+ = wild-type
X-LINKED
DOMINANT
female double
heterozygote
X
Male hemizygous
for wild-type alleles
recessive = wild+
+
type = A and B
a+b+/ab x
ab/Y
AB/A+B+ x
A+B+/Y
NOTE: The baby phenotypes will depend on how the linked alleles are arranged in the
heterozygous parent! In the above examples, both dominant alleles are arranged on one
chromosome, while the recessive alleles are arranged on the other chromosome (called
CIS coupling). But you could also put one dominant and one recessive on each
chromosome (e.g. a+b/ab+; called TRANS coupling).
HOW?
e.g. for the 2-PT test cross (using cis-coupling of an autosomal recessive trait):
P genotypes:
P phenotypes:
a+b+/ab
a+b+
x
ab/ab
ab
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Note we are using the allele symbols to represent PHENOtypes. From this cross, you
might get the following progeny phenotypes (baby flies):
Baby phenotype
+ +
ab
ab
a+b
ab+
Looks like?
Look like 1st parent
Observed # of babies
965
Look like 2nd parent
Recombinant type
944
206
Recombinant type
185
Total = 2300 babies
1.
HOW CAN YOU TELL FROM THESE NUMBERS THAT THE A AND B
GENE ARE LINKED ON THE SAME CHROMOSOME?
IF the genes were NOT linked (carried on different chromosomes), the heterozygous
parent could form the following gametes in EQUAL proportions (allowed to FOIL to find
gametes):
+ +
+
+
1 a b : 1 ab : 1 a b : 1 ab
That means you would EXPECT to see the following numbers (based on the same total #
of actual babies) in the offspring:
Baby phenotype
Looks like?
Observed # of
babies
a+b+
ab
a+b
ab+
Look like 1st parent
965
Expected # of
babies IF genes are
NOT linked
2300/4 = 575
Look like 2nd parent
Recombinant type
944
206
2300/4 = 575
2300/4 = 575
Recombinant type
185
2300/4 = 575
Total = 2300 babies
Total = 2300 babies
The fact that the actual number pattern does NOT match the expected number pattern for
unlinked genes, means the genes are NOT carried on different chromosomes. This
implies we are dealing with LINKED genes.
But does the actual number pattern match the case where we have two genes that stay
linked all the time? If that were the case, the heterozygous (cis) parent would only be
able to make the following gametes, since NO FOILING is allowed:
+ +
1 a b : 1 ab
2
Then we would expect to see the following:
Baby phenotype
Looks like?
Observed # of
babies
a+b+
ab
a+b
ab+
Look like 1st parent
965
Expected # of
babies IF genes are
linked ALL THE
TIME (no
crossover at all)
2300/2 = 1150
Look like 2 parent
Recombinant type
944
206
2300/0 = 1150
0
Recombinant type
185
0
Total = 2300 babies
Total = 2300 babies
nd
Obviously, the actual number pattern does not match the “totally linked” scenario either.
We seem to have a pattern that falls somewhere in between the UNLINKED and
TOTALLY LINKED scenarios. The only way to explain this is to assume the two genes
ARE LINKED, but are sometimes “unlinked” by a single crossover event between them.
2. NOW THAT WE KNOW THE TWO GENES ARE LINKED, HOW FAR
APART ARE THEY ON THE SAME CHROMOSOME?
Assuming that the further the 2 genes are apart, the greater the chance that crossover will
occur between them, we can use the (single) crossover frequency as an estimate of map
distance between the genes.
Since the only way you can get a recombinant type baby is by using crossover gametes in
the heterozygous parent (the ones you can only get by foiling), we can use the numbers
(%) of recombinant babies to estimate map distances:
(# of observed recombinant babies / total # of babies)(100%)
= % crossover events
= approximate map distance between the 2 genes (in map units or cM)
So for the above example:
(206 + 185)/2300 x 100%
= 17 % crossover
= genes
a+ and b+ are 17 map units or cM apart
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2-POINT TEST CROSS METHOD RECAP:
1.
Determine the genotypes of the test cross parents (this will depend on the
inheritance pattern of the trait; we will mostly use autosomal recessive; one
parent is heterozygous, the other is homozygous recessive).
2.
Use allele symbols to represent the phenotypes of the test cross parents (the
heterozygous parent will be dominant-dominant, the homozygous recessive
parent will be recessive-recessive).
3.
Look at the list of observed phenotypes in the babies. What kind of pattern do
you see?
OBSERVED PATTERN
1:1:1:1
1:1
More parental types than recombinant
types.
CONCLUSION
The two genes are NOT linked.
The two genes are linked and STAY linked
ALL of the time.
The two genes are linked, but there is some
crossover between them.
4.
If desired, do a Chi-square test, comparing the observed pattern with the
1:1:1:1 pattern (you will have to calculate expected progeny numbers using
the same total number of baby flies). You will either “reject” or “fail to
reject” the hypothesized 1:1:1:1 pattern.
5.
If you have linked genes with crossover:
(# observed recombinants/# observed total progeny)/100% = % crossover =
map distance in map units or cM between the genes.
4