3. Genetic Linkage and Mapping
• Thomas Hunt Morgan
won the Nobel Prize
for work establishing
the chromosome
theory of inheritance
and also for his role
identifying and
explaining genetic
linkage and
recombination
• He applied linkage
and recombination
to genetic mapping
3
4. What we have learned…
All genes have been on different chromosomes
Allows “Independent Segregation”
1:1:1:1
gamete
ratio
6. Meiosis I
Anaphase
MetaphaseI I
A A a a
A A a a
or
BBb b
What will happen if two
genes are on the same
chromosome?
b b BB
AB or ab gametes
Ab or aB gametes
Over all, 1 AB : 1 Ab : 1 aB : 1 ab
7. Meiosis – Two Genes on Same Chromosome
Anaphase
Prophase II and
Metaphase I
A A
a a
BB
b b
Usually only get AB or ab gamete (1:1 ratio)
“Parental-type gametes”
Will probably get a low frequency of Ab or aB
“Nonparental-type (recombinant-type) gametes”
How?
10. The Discovery of Genetic Linkage
• William Bateson and Reginald
Punnett, in a series of crosses
with sweet peas, found
evidence of genetic linkage
• They crossed pure-breeding
purple-flowered, long-pollen
plants to white-flowered, roundpollen plants; the purple, longpollen F1 were interbred to
produce F2
• The expected 9:3:3:1 ratio was
not observed
11.
12. Discovery of Genetic Linkage
Pea flower color: Purple (P) is dominant to red (pp)
Pea pollen length: Long (L) is dominant to short (ll)
F1:
PPLL
ppll
Purple long
Parents:
Red short
All PpLl, (Purple, long)
9
Purple, short:
3
21
Red, long:
3
21
Red, short:
F2:
Expected
Purple, long:
Observed
284
1
55
How do you test if an observed ratio is significantly different from expected?
13. Chi Square Analysis
Class
observed expected
O-E
(O-E)2
(O-E)2/ E
Purple, long
284
9/16 x 381
214.3
69.7
4858
22.6
Purple, short
21
3/16 x 381 -50.4
71.4
2540
35.6
Red, long
21
3/16 x 381 -50.4
71.4
2540
35.6
Red, short
55
1/16 x 381
23.8
973
40.9
31.2
381
Σ = 134.7
df = n-1 = 4-1 = 3
2
P < .01
Chi Square Probability Values
DF
3
P=
.99
.95
.90
.75
.50
.25
.10
.05
.01
0.114
0.351
0.584
1.212
2.365
4.108
6.251
7.814
11.34
Conclusion?
SIGNIFICANTLY DIFFERENT! Traits
are not sorting independently!
14. Discovery of Linkage
Pea flower color: Purple (P) is dominant to red (pp)
Pea pollen length: Long (L) is dominant to short (ll)
PPLL
ppll
Purple long
Red short
Parents:
All PpLl, (Purple, long)
F1:
F2:
Purple, long:
Expected
215
Observed
284 Significantly more!
21
Significantly less!
Purple, short:
71
Red, long:
71
21
Significantly less!
Red, short:
24
55
Significantly more!
P and L genes appear to be “linked” together
15. What Bateson and Punnett Concluded
• Observation: MORE parental
phenotypes, LESS nonparental types
• Conclusion: Bateson and Punnett
suggested that an unknown
mechanism kept the two parental
gamete combinations together,
which they called “coupling”
16. 5.1 Linked Genes Do Not Assort Independently
• Genes located on the same chromosome are called syntenic
genes
• Syntenic genes so close together that their alleles cannot
assort independently are called linked genes
• Genetic linkage can be quantified to map the positions of
genes on chromosomes
genemol.org
17. Recombination and Syntenic Genes
• Alleles of syntenic genes can
be reshuffled when crossing
over occurs between homologs
to produce recombinant
chromosomes
• Homologs that do not reshuffle
alleles under study are called
parental chromosomes or
nonrecombinant
chromosomes
• Genetic linkage mapping
plots the positions of genes on
chromosomes
http://viirulentscience.wordpress.com/
18.
19.
20. Complete Genetic Linkage
• Complete genetic linkage
is observed when no
crossing over occurs
between linked genes;
only parental gametes are
formed
• Some organisms exhibit
complete linkage, e.g.,
Drosophila males have no
crossing over
• The biological basis for
this is unknown
Parental gametes
21.
22. Incomplete Genetic
Linkage
• Incomplete genetic
linkage is much more
common than
complete linkage; in
this case a mixture of
parental and
nonparental gametes
are produced
23. Calculating Recombination Frequency
• Recombination frequency, expressed as r, is
calculated as
• Recombination frequency is likely a reflection of the
physical distance between two genes
r=
# or recombinant animals
total number of animals
24. Correlation Between Recombination
Frequency and Gene Distance
• Crossing over occurs at a
higher rate between genes
that are farther apart, and a
lower rate between genes
that are closer together
Smaller r
Bigger r
25. Morgan’s Crosses
• Morgan studied the white
(eye color) and miniature
(wing size) genes
• He crossed females
pure-breeding for white
eyes and small wings
(wm/wm) to males that
were wild type for both
(w m /Y)
• The F1 were w m /wm
females and wm/Y males
http://bioweb.wku.edu/courses/biol114/vfly1.asp
26. Morgan’s F1
F1 Results
• Morgan interbred the
w+m+/wm females and
wm/Y males
• A 1:1:1:1 ratio was
predicted based on the
assumption of
independent
assortment of the
genes
• However, Morgan
observed many more
parental types than
recombinant
types, suggesting that
the genes in question
were found on the
same chromosome, in
this case the X
Less than 50%? →
27. Chi-Square Analysis of Morgan’s w,
m Crosses
•
• There are 3 degrees of freedom in this case, and
the p value is p 0.005
• Significant??
Yes! Thus the genes are linked!
29. Detecting Autosomal Genetic Linkage
Through Test-Cross Analysis
• Morgan realized that linkage
of autosomal genes in
Drosophila could be
interpreted using a twopoint test-cross analysis
• In a test cross, the
homozygous recessive
parent contributes only
recessive alleles
• Thus, the alleles contributed
by just the dihybrid parent
can be examined
30. Crosses with vg and pr
• Morgan crossed flies with
purple eyes and vestigial
(short) wings to wild type
and obtained wild-type F1
• The F1 females were then
crossed to males that had
purple eyes and vestigial
wings
• The alleles in the
female gametes in
this cross determined
the phenotype in
each of the progeny
• Offspring produced did not
fit the 1:1:1:1 expected
ratio
31.
32. Important Conclusions from All of Morgan’s
Crosses
1. Genetic linkage is a physical
relationship between genes located
near one another on a chromosome
2. Recombination occurs between
linked genes less than 50% of the
time, and greater than 50% of the
gametes contain parental allele
combinations
3. Recombination frequency varies
among linked genes in proportion to
the distance between them
33. 5.2 Genetic Linkage Mapping Is Based on
Recombination Frequency Between Genes
• Morgan recognized that with linked genes, more parental than
recombinant progeny occurred and that the recombinant frequency varied
among gene pairs
• Morgan suggested that closer proximity of genes produced a
correspondingly higher frequency of parental allele combinations
• Therefore, we can use r values to make a map!
34. The First Genetic Linkage Map
• Morgan’s
student, Alfred
Sturtevant, realize
d that the
variations in
recombination
frequency could
be used to
determine genetic
maps for genes
• He used the
results of several
experiments to
create a genetic
map for five Xlinked genes
http://www.ncbi.nlm.nih.gov
35.
36. General logic for generating the map:
1. Of the genes tested, the pair 3. M is close to v, but more distant from
with the smallest
w, so y-w-v-m
recombination frequency
4. R is very far away from w, and fairly
must be the closest in
distant from v. This suggests that r is
difference (y & w)
on the opposite end of the map, so y2. V is more distant from y than
w-v-m-r
w, suggesting y-w-v
You will try this in your homework!
37. Map Units
• Recombination frequencies between two genes can
be converted into units of physical distance, map
units (m.u.)
• A map unit is also called a centiMorgan (cM)
• By convention, 1% recombination
1 m.u. or 1 cM
http://www.tutorvista.com
38. Example Mapping Problem
A homozygous pea plant with purple flowers and long
pollen (PPLL) is crossed with a second inbred line with
red flowers and short pollen (ppll)
How would you show their genotypes using new method?
PL
pl
PL
pl
What would the F1 look like?
PL
pl
39. Example Mapping Problem, Cont’
An F1 plant is testcrossed and the following progeny
were observed:
PL
pl
pl
pl
Purple, long
39
Calculate the map
distance between
the P and L genes.
Purple, short
9
Step 1: Write genotypes of parents
Red, long
10
Step 2: Write genotypes of kids
Red, short
42
# of Progeny
40. Example Mapping Problem, Cont’
PL
pl
pl
pl
# of Progeny
Note: We are only looking
at recombination in the
heterozygous parent!
Purple, long
39
From mom?
PL
From dad?
pl
Purple, short
9
Pl
pl
Red, long
10
Red, short
42
pL
pl
pl
pl
Which are the result of a parental-type gamete?
Which are the result of a recombinant-type gamete?
41. Example Mapping Problem, Cont’
# of Progeny
Purple, long
39
Purple, short
9
Red, long
10
Red, short
42
These 19 progeny were the result of a recombination
between the “P” and “L” genes.
# Recombinants
Total # of progeny
What is the map distance?
Recombination % =
X 100
= [(9 + 10) / 100] x 100 = 19%
The P and L genes are 19 cM apart
42. Final Step: Draw the Map
You can then draw a map showing the distance between
the two genes
19 cM
P
L
This is Two Point Linkage Analyses
