I. Allelic, Genic, and Environmental Interactions
II. Sex Determination and Sex Linkage
III. Linkage
- Overview:
‘Linkage’ is a pattern of correlated inheritance between traits governed by
genes on the same chromosome. Because the genes are part of the same physical
entity, they are inherited together rather than independently.
INDEPENDENT ASSORTMENT (IA)
a
A
B
A
B
A
b
LINKED
AB
ab
b
a
B
a
b
AB
ab
III. Linkage
- Overview:
‘Linkage’ is a pattern of correlated inheritance between traits governed by
genes on the same chromosome. Because the genes are part of the same physical
entity, they are inherited together rather than independently. Only ‘crossing-over’ can
cause them to be inherited in new combinations.
Cross-over
products
III. Linkage
A. ‘Complete’ Linkage
- if genes are immediate neighbors, they are almost never separated by
crossing over and are ‘always’ inherited together. The pattern mimics that of a single
gene.
AABB
aabb
AB
ab
X
AB
ab
III. Linkage
A. ‘Complete’ Linkage
- if genes are immediate neighbors, they are almost never separated by
crossing over and are ‘always’ inherited together. The pattern mimics that of a single
gene.
AABB
aabb
AB
ab
X
ab
AB
Gametes
AB
ab
ab
F1
AB
Double Heterozygote in F1;
no difference in phenotypic
ratios compared to IA
III. Linkage
A. ‘Complete’ Linkage
- if genes are immediate neighbors, they are almost never separated by
crossing over and are ‘always’ inherited together. The pattern mimics that of a single
gene.
ab
ab
1:1 ratio A:a
Test cross
X
AB
ab
1:1 ratio B:b
1:1 ratio AB:ab
ab
Gametes
AB
Phenotype
AaBb
AB – 50%
aabb
Ab – 50%
ab
If it was
I.A., also:
Ab
aB
Aabb
aaBb
Ab
aB
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Crossing over in a region is rare
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Crossing over in a region is rare
- Crossing over events increase as the distance between genes increases
A
B
C
a
b
c
LESS LIKELY
IN HERE
MORE LIKELY IN HERE
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Crossing over in a region is rare
- Crossing over events increase as the distance between genes increases
- So, the frequency of crossing over (‘CO’) gametes can be used as an index of
distance between genes! (Thus, genes can be ‘mapped’ through crosses…)
A
B
C
a
b
c
FEWER ‘CO’
GAMETES:
Ab, aB
MORE ‘CO’
GAMETES:
bC, Bc
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Crossing over in a region is rare
- Crossing over events increase as the distance between genes increases
- So, the frequency of crossing over (‘CO’) gametes can be used as an index of
distance between genes! (Thus, genes can be ‘mapped’ through crosses…)
- How can we measure the frequency of recombinant (‘cross-over’) gametes?
Is there a type of cross where we can ‘see’ the frequency of different types of gametes
produced by the heterozygote as they are expressed as the phenotypes of the
offspring?
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
a
A
b
B
a
a
b
b
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- So, since crossingover is rare (in a particular
region), most of the time it
WON’T occur and the
homologous chromosomes will
be passed to gametes with
these genes in their original
combination…these gametes are
the ‘parental types’ and they
should be the most common
types of gametes produced.
a
A
b
a
B
a
b
A
B
a
b
b
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
- Sometimes, crossing
over WILL occur between these
loci – creating new
combinations of genes…
This produces the ‘recombinant
types’
a
A
b
a
B
a
b
A
B
a
B
A
b
a
b
b
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
a
A
b
a
B
b
a
b
a
As the other parent only
contributed recessive alleles
(whether crossing over occurs or
not)…
a
b
A
B
a
B
A
b
b
TEST CROSS !!!
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
As the other parent only
contributed recessive alleles
(whether crossing over occurs or
not)…
SO: the phenotype of the
offspring is determined by the
gamete received from the
heterozygote…
a
A
b
a
a
B
a
b
A
B
b
b
gamete
genotype
phenotype
ab
aabb
ab
ab
AaBb
AB
a
B
ab
aaBb
aB
A
b
ab
Aabb
Ab
TEST CROSS !!!
III. Linkage
a
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
A
b
a
a
B
b
b
ALTERNATIVES
FREQUENCIES EQUAL TO
PRODUCT OF INDEPENDENT
PROBABILITIES
‘IA’
LINKAGE
a
b
A
B
gamete
genotype
phenotype
ab
aabb
ab
ab
AaBb
AB
LOTS of
PARENTALS
a
B
ab
aaBb
aB
A
b
ab
Aabb
Ab
FEWER
CO’S
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
- Compare the observed
results with what you would expect
if the genes assorted independently
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
- Compare the observed
results with what you would expect
if the genes assorted independently
The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
- Compare the observed
results with what you would expect
if the genes assorted independently
The frequency of ‘AB’ should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28
The frequency of ‘Ab’ should = f(A) x f(B) x N = 55/100 x 49/100 x 100 = 27
The frequency of ‘aB’ should = f(a) x f(B) x N = 45/100 x 51/100 x 100 = 23
The frequency of ‘ab’ should = f(a) x f(b) x N = 45/100 x 49/100 x 100 = 22
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
This is fairly easy to do by creating
a contingency table:
B
b
A
43
12
a
8
37
Col.
Total
Row
Total
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
This is fairly easy to do by creating
a contingency table:
B
b
Row
Total
A
43
12
55
a
8
37
45
Col.
Total
51
49
100
Add across and down…
This gives the totals for each trait
independently.
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Example:
- How do we discriminate
between these two alternatives?
- Conduct a Chi-Square Test of
Independence
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
This is fairly easy to do by creating
a contingency table:
Then, to calculate an expected
value based on independent
assortment (for ‘AB’, for example),
you multiple ‘Row Total’ x ‘Column
Total’ and divide by ‘Grand Total’.
55 x 51 / 100 = 28
B
Exp.
b
Row
Total
A
43
28
12
55
a
8
37
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Offspring
Number
Example:
AB
43
Repeat to calculate the other expected
values… (This is just an easy way to set
it up and do the calculations, but you
should appreciate it is the same as the
product rule:
Ab
12
aB
8
ab
37
F(A) x f(B) x N
55 X 51 x 100 = 55x51
100 100
100
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Compare our observed results with what
we would expect if the genes assort
independently.
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Compare our observed results with what
we would expect if the genes assort
independently.
. If our results are close to the
expectations, then they support the
hypothesis of independence.
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
Compare our observed results with what
we would expect if the genes assort
independently.
. If our results are close to the
expectations, then they support the
hypothesis of independence. If they are
far apart from the expected results, then
they refute that hypothesis and support
the alternative: linkage.
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Typically, we reject the hypothesis of
independent assortment (and accept the
hypothesis of linkage) if our observed results
are so different from expectations that
independently assorting genes would only
produce results as unusual as ours less than
5% of the time…
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
49
100
B
Exp.
b
Exp.
Row
Total
A
43
28
12
27
55
a
8
23
37
22
45
Col.
Total
51
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
We determine this probability with a
Chi-Square Test of Independence.
49
Obs
Exp
(o-e)
(o-e)2/e
AB
43
28
15
8.04
Ab
12
27
-15
8.33
aB
8
23
-15
9.78
ab
37
22
15
10.23
X2 =
36.38
Phenotype
100
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
First, we determine the ‘degrees of
freedom’ = (r-1)(c-1) = 1
B
b
Row
Total
A
43
12
55
a
8
37
45
Col.
Total
51
49
100
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
First, we determine the ‘degrees of
freedom’ = (r-1)(c-1) = 1
Now, we read across the first row in
the table, corresponding to df = 1.
The column headings are the
probability that a number in that
column (diff. between observed and
expected) would occur by chance.
In our case, it is the probability that
our hypothesis of independent
assortment (expected values are
based on that hypothesis) is true.
(and the diff is just due to chance).
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
Note that larger values have a lower
probability of occurring by chance…
This should make sense, and the
value increases as the difference
between observed and expected
values increases.
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
So, for instance, a value of 2.71 will
occur by chance 10% of the time.
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
So, for instance, a value of 2.71 will
occur by chance 10% of the time.
But a value of 6.63 will only occur
1% of the time... (if the hypothesis is
true and this deviation between
observed and expected values is
only due to chance).
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
So, for instance, a value of 2.71 will
occur by chance 10% of the time.
For us, we are interested in the 5%
level. The table value is 3.84.
Our calculated value is much greater
than this; so the chance that
independently assorting genes
would yield our results is WAY LESS
THAN 5%. Our results are REALLY
UNUSUAL for independently
assorting genes.
36.38
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
Our results are REALLY UNUSUAL for
independently assorting genes.
So, either our results are wrong, or
the hypothesis of independent
assortment is wrong. If you did a
good experiment, then you should
have confidence in your results;
reject the hypothesis of IA and
conclude the alternative – the genes
are LINKED.
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Our X2 = 36.38
Our results are REALLY UNUSUAL for
independently assorting genes.
So, either our results are wrong, or
the hypothesis of independent
assortment is wrong. If you did a
good experiment, then you should
have confidence in your results;
reject the hypothesis of IA and
conclude the alternative – the genes
are LINKED.
Of course, there is still that 5%
probability that chance is
responsible for the deviation you
observed, and your conclusion is
WRONG.
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
OK… so we conclude the genes are linked…
NOW WHAT?
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Offspring
Number
OK… so we conclude the genes are linked…
NOW WHAT?
AB
43
Ab
12
We map the genes using the knowledge that
crossing-over is rare, and the frequency of
crossing-over correlates with the distance
between genes.
aB
8
ab
37
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Offspring
Number
OK… so we conclude the genes are linked…
NOW WHAT?
AB
43
Ab
12
1. Crossing-over is rare; so the RARE
COMBINATIONS must be the products of
Crossing-Over. The OTHERS, the MOST
COMMON products, represent the PARENTAL
TYPES.
aB
8
ab
37
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Offspring
Number
OK… so we conclude the genes are linked…
NOW WHAT?
AB
43
Ab
12
1. Crossing-over is rare; so the RARE
COMBINATIONS must be the products of
Crossing-Over. The OTHERS, the MOST
COMMON products, represent the
PARENTAL TYPES.
aB
8
ab
37
A
B
a
b
This tells us the original arrangement of
alleles in the heterozygous parent:
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Offspring
Number
OK… so we conclude the genes are linked…
NOW WHAT?
AB
43
Ab
12
1. Crossing-over is rare; so the RARE
COMBINATIONS must be the products of
Crossing-Over. The OTHERS, the MOST
COMMON products, represent the
PARENTAL TYPES.
aB
8
ab
37
A
B
a
b
This tells us the original arrangement of
alleles in the heterozygous parent:
Segregation without crossing over produces
lots of AB and ab gametes.
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
Offspring
Number
OK… so we conclude the genes are linked…
NOW WHAT?
AB
43
Ab
12
1. Crossing-over is rare; so the RARE
COMBINATIONS must be the products of
Crossing-Over. The OTHERS, the MOST
COMMON products, represent the
PARENTAL TYPES.
aB
8
ab
37
A
B
a
b
2. The frequency of crossing-over is used as an
index of the distance between genes:
The other progeny are the products of
crossing over, and they occurred 20 times
in 100 progeny, for a frequency of 0.2.
Multiply that by 100 to free the decimal,
and this becomes 20 map units
(CentiMorgans).
20 map units
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
So, we used the chi-square test to determine
whether genes are on the chromosome.
Then, we mapped the distance between
genes on the chromosome; all by looking at
the frequency of phenotypes in the offspring
of a test cross.
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
A
B
a
b
20 map units
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
So, we used the chi-square test to determine
whether genes are on the chromosome.
Then, we mapped the distance between
genes on the chromosome; all by looking at
the frequency of phenotypes in the offspring
of a test cross.
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
LIMITATION:
Genes that are 50 map units apart (or more)
will recombine 50% of the time…. Producing
parentals and recombinants in equal
frequency… just like independent assortment
predicts!!!
A
B
a
b
20 map units
III. Linkage
AaBb x aabb
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
So, we used the chi-square test to determine
whether genes are on the chromosome.
Then, we mapped the distance between
genes on the chromosome; all by looking at
the frequency of phenotypes in the offspring
of a test cross.
Offspring
Number
AB
43
Ab
12
aB
8
ab
37
LIMITATION:
Genes that are 50 map units apart (or more)
will recombine 50% of the time…. Producing
parentals and recombinants in equal
frequency… just like independent assortment
predicts!!!
So, genes that are > 50 mu apart assort
independently (even though they are on the
same chromosome).
A
B
a
b
20 map units
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
C. Three-point Mapping
Suppose you conduct two-point mapping and find:
A - C are 23 centiMorgans apart
C – B are 10 centiMorgans apart:
A
C
OR
A
B
C
B
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
C. Three-point Mapping
Three Point Test Cross
AaBbCc x aabbcc
Phenotypic Ratio:
ABC
ABc
Abc
AbC
aBC
aBc
abc
abC
=
=
=
=
=
=
=
=
25
3
42
85
79
39
27
5
III. Linkage
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
C. Three-point Mapping
- combine complementary sets
ABC = 25
abc = 27
52
ABc = 3
abC = 5
= 8
Abc = 42
aBC = 39
81
AbC = 85
aBc = 79
164
Three Point Test Cross
AaBbCc x aabbcc
Phenotypic Ratio:
ABC
ABc
Abc
AbC
aBC
aBc
abc
abC
=
=
=
=
=
=
=
=
25
3
42
85
79
39
27
5
III. Linkage
Three Point Test Cross
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
C. Three-point Mapping
- combine complementary sets
ABC = 25
abc = 27
52
ABc = 3
abC = 5
= 8
Abc = 42
aBC = 39
81
AbC = 85
aBc = 79
164
MOST ABUNDANT ARE
PARENTAL TYPES
AaBbCc x aabbcc
Phenotypic Ratio:
ABC
ABc
Abc
AbC
aBC
aBc
abc
abC
=
=
=
=
=
=
=
=
25
3
42
85
79
39
27
5
III. Linkage
Three Point Test Cross
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
C. Three-point Mapping
- combine complementary sets
ABC = 25
abc = 27
52
ABc = 3
abC = 5
= 8
LEAST ABUNDANT
ARE “DOUBLE
CROSSOVERS”
Abc = 42
aBC = 39
81
AbC = 85
aBc = 79
164
MOST ABUNDANT ARE
PARENTAL TYPES
AaBbCc x aabbcc
Phenotypic Ratio:
ABC
ABc
Abc
AbC
aBC
aBc
abc
abC
=
=
=
=
=
=
=
=
25
3
42
85
79
39
27
5
D
E
F
d
e
f
DCO’s = very rare, and gene in
middle switches chromosomes
III. Linkage
Three Point Test Cross
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
C. Three-point Mapping
- combine complementary sets
ABC = 25
abc = 27
52
ABc = 3
abC = 5
= 8
LEAST ABUNDANT
ARE “DOUBLE
CROSSOVERS”
Abc = 42
aBC = 39
81
AbC = 85
aBc = 79
164
MOST ABUNDANT ARE
PARENTAL TYPES
AaBbCc x aabbcc
Compare Parentals and DCO’s to
determine which gene is in the
middle…
Parentals: AbC
aBc
DCO’s:
abC
ABc
B and C alleles stay together; A
switches and so is in middle.
REMAKE YOUR DRAWING TO
DEPICT ALLELES AND ORDER:
b
A
C
B
a
c
III. Linkage
Three Point Test Cross
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
C. Three-point Mapping
- combine complementary sets
LEAST ABUNDANT
ARE “DOUBLE
CROSSOVERS”
Abc = 42
aBC = 39
81
AbC = 85
aBc = 79
164
So, a “single cross-over” (SCO)
between the B and A locus will
produce the gametes/phenotypes:
BAC and bac – find their frequency:
ABC = 25
abc = 27
52
ABc = 3
abC = 5
= 8
AaBbCc x aabbcc
MOST ABUNDANT ARE
PARENTAL TYPES
52… plus add DCO’s (8) = 60/305
= 0.197 x 100 = 19.7 cM
b
A
C
B
a
c
19.7 cM
III. Linkage
Three Point Test Cross
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
C. Three-point Mapping
AaBbCc x aabbcc
- combine complementary sets
81 + 8 = 89/306 = 0.292 x 100 = 29.2
ABC = 25
abc = 27
52
ABc = 3
abC = 5
= 8
LEAST ABUNDANT
ARE “DOUBLE
CROSSOVERS”
Abc = 42
aBC = 39
81
AbC = 85
aBc = 79
164
MOST ABUNDANT ARE
PARENTAL TYPES
b
A
C
B
a
c
19.7 cM
29.2 cM
III. Linkage
Three Point Test Cross
A. ‘Complete’ Linkage
B. ‘Incomplete’ Linkage
C. Three-point Mapping
- combine complementary sets
We only observed 8.
LEAST ABUNDANT
ARE “DOUBLE
CROSSOVERS”
Abc = 42
aBC = 39
81
AbC = 85
aBc = 79
164
“Coefficient of Coincidence” – are
SCO’s occurring independently? If
so, then DCO’s = product of SCO’s
(0.197) x (0.292) x 305 = 17
ABC = 25
abc = 27
52
ABc = 3
abC = 5
= 8
AaBbCc x aabbcc
MOST ABUNDANT ARE
PARENTAL TYPES
C.O.C. = obs/exp = 8/17 = 0.47
Interference = 1 – c.o.c. = 0.53
b
A
C
B
a
c
19.7 cM
29.2 cM