Download pr - vg

Linkage and Recombination
Mendel
Alleles segregate
Alleles for different genes
assort independently
Morgan and Bridges
Genes are on
chromosomes
Bateson et al.
Pea plants have 7 pairs of
chromosomes.
Flies have 4 pairs.
Humans have 23 pairs.
What is the inheritance pattern for two traits
determined by genes on the same chromosome?
Human
X chromosome
DMD
RP3
SCID
XIST
PGK
“Linked” genes are those that do
NOT assort independently.
(They reside on the same
chromosome.)
We will see that...
The corollary isn’t always true!
HPRT
Just because genes are on the
XLP
same chromosome doesn’t
Haemophilia
necessarily mean they are linked.
Fragile X
R-G Colorblindness
Mapping Genes on Chromosomes
Alfred Sturtevant
As a student in
Morgan’s lab,
Sturtevant was the
first person to
correctly interpret
linkage; he created
the first genetic
maps.
Sturtevant in his lab at
Caltech
Segregation of genes on the same chromosome
wild type
pr+ pr+ vg+ vg+
purple eyes
vestigial wings
pr pr vg vg
x
wild type phenotype
pr+ pr vg+ vg
gamete?
gamete?
testcross
[ pr pr vg vg
For I.A., expect: 1
:
phenotypes: pr+ vg+
observed:
1339 :
1
:
pr+ vg
151 :
1
pr vg+
:
]
testcross parent
genotype?
1 ratio
pr vg
154 : 1195
The genes are on the same
Sturtevant’s interpretation: chromosome & the parental
alleles (mostly) co-segregate
wild type
pr+ pr+ vg+ vg+
gamete?
x
parental
purple
eyestype
(same
as gametes
vestigial
wings
made
prthat
pr vg
vg the parent)
wild type phenotype
recombinant
pr+ pr vg+ vg
(non-parental)
type
gamete?
How do we know the genotypes of the gametes?
Testcross parent’s gametes… only recessive alleles present
So testcross progeny phenotype allows us to deduce the
heterozygous parent’s gamete genotype
Recombination… a brief review
cohesins
sister chromatids
(each is a double
helix)
homologues
synaptonemal
complex
Singlestrand
nicks
Holliday
junction
strand
exchange
[see lecture 2 for details]
-No deletions are
caused by
recombination.
-No mutations are
caused by
recombination.
Recombination… a brief review (cont)
Segregation of recombinant chromosomes during Meiosis I
Label the
chromatids
1
2
1
2 3
4
ALWAYS pull
from the
centromeres
3
4
Recombination… a brief review (cont)
Telophase I
Telophase II
Gametes
Parental
Recombinant
Recombinant
Parental
One recombination event: 2 recombinant
and 2 non-recombinant products
Test your Understanding
Which chromatid goes where?
An example with a double crossover:
Draw these chromosomes in anaphase I.
Label all alleles appropriately.
ALWAYS pull from the centromeres
1
2
3
4
Crossing over must occur for
Other types of crossovers
faithful segregation at meiosis I
# xovers
resulting gametes
0
parental
2
parental!
(2 strands)
2
(3 strands)
2
(4 strands)
2 parental
2 non-par.
4 non-par.
Sturtevant’s findings—summary
Genes on the same chromosome can show linkage instead
of independent assortment
Gametes (mostly) have the same allele combinations as
the homologs in the parent
Recombination can give rise to gametes with non-parental
(=recombinant) allele combinations
Two parental types are more abundant and occur at
roughly equal frequency to one another
Non-parental types are less abundant and occur at
roughly equal frequency to each other
Identifying the parental type
Option 1. Know the gametes that made the heterozygous parent
The cross from our
pr+ pr+ vg+ vg+
previous example:
gametes:
define the
parental
type
gametes:
X pr pr vg vg
pr+ vg+
pr+ vg+
pr vg
pr vg
pr+ pr vg+ vg
pr+ vg
A different cross
+ pr+ vg vg
pr
from our example:
pr+ vg
pr vg+
pr vg+
X
pr pr vg+ vg+
dominant alleles
together on the same
chromosome = “cis”
configuration
dominant alleles on
different chromosomes
= “trans” configuration
Identifying the parental type
Option 2. The two most abundant progeny types
(assuming the genes are linked)
Cross:
pr+ pr vg+ vg
x
pr pr vg vg
Progeny:
1287
1204
170
154
What were the gametes that made the heterozygous parent?
pr+ vg
pr vg+
A notation system that defines allele configuration
linkage
=

Configuration
(i.e., cis or trans)
=
???
pr+
vg+
pr
vg
implies
w+
w
;
pr+
pr
separate
chromosomes
Sturtevant’s interpretation of linkage
-Recombination involves the physical exchange
of chromosomal segments between homologs.
-The frequency of recombinant types indicates
the distance between linked genes.
What is the evidence in support
of these hypotheses?
Test of Sturtevant’s hypothesis
The problem: homologous chromosomes look alike…
how to tell if they really exchanged segments?
 Harriet Creighton and Barbara McClintock, maize
 Curt Stern, Drosophila
found genetically marked chromosome with
structurally (visually) distinct homologues
Prediction: Chromosome from genetically recombinant
plants should show structural rearrangement
Creighton/McClintock test Sturtevant’s hypothesis
Structural markers:
Genetic markers:
“knob”
C = colored
c = colorless
translocation
Progeny phenotypes
colored, waxy
x
colorless, starchy
Wx = starchy
wx = waxy
recombinant
genotypes and
chromosomes!
colored, starchy
colorless, waxy
Conclusion: genetic recombination  exchange of
chromosome segments
Mapping genes using recombination
Alfred Sturtevant’s major insight:
If crossovers occur at random
Probability of crossover between
two genes is proportional to the
distance between them
Crossover between A and B much more likely than between
B and D
Map distance… example from last time
Cross:
pr+ pr vg+ vg
x
pr pr vg vg
Progeny:
1287
1204
170
154
What is the map distance between these two genes?
Map distance… example (cont’d)
Step 1. Identify the parental and non-parental types.
parental
1287 types (most
abundant)
170
non-parental
types (least
abundant)
parental =
pr+ vg and pr vg+
non-parental =
pr+ vg+ and pr vg
1204
154
Map distance… example (cont’d)
Step 2. Calculate % recombinant products.
parental
1287 types (most
abundant)
170
% recombinant =
1204
non-parental
types (least
abundant)
(170 + 154)
154
x100
= 11.5%
(170+154+1287+1204)
Map distance = 11.5 map units = 11.5 centiMorgan’s (cM)
Another Example (recessive b mutation)
4615
red
black
4707
purple
tan
307
red
tan
Which are the parental types?
and the recombinant types?
Recombination Frequency =
Number Recombinant Types
Total Number of Progeny
pr b Example:
295
purple
black
Another Example (recessive b mutation)
4615
red
black
4707
purple
tan
307
red
tan
Parental types
pr +
b
pr
295
purple
black
Recombinant types
b+
pr +
b+
pr
b
Recombination Frequency =
Number Recombinant Types
Total Number of Progeny
pr b Example:
307602
+ 295
4615 + 4707
+ 307 + 295
9924
= 0.06 x 100 = 6.0 %
Creating a Genetic Map
Genetic loci:
% Recombinants
+
+
purple - vestigial
- vg)
b+ (pr pr
vg
11.5 %
purple - black (pr - b)
6.0 %
Revealed
vestigial - black (vg - b)
from
16.5 %
other
Alfred Sturtevant’s mental leap:
crosses
b
pr
vg
% recombinants is directly proportional to distance
He drew a genetic map of the second chromosome:
6 cM 11.5 cM
b
pr
vg
Why 17.5cM and not 16.5cM?
Genes very close together…
Low probability of crossover between them
…very few recombinants
“tight linkage”
Genes further apart… more recombinants
Point to ponder:
If you examined a population of meioses, what
is the maximum recombination frequency you
could see between two genes?
Summary
* Crossing-over creates new combinations of traits.
* Two Parental types in ≈ frequencies.
Two Recombinant types ≈ frequencies.
* If genes are linked,
Parental types > recombinant types.
* The frequency of recombinant types
indicates the distance between linked genes.
Practice question
The pedigree shows segregation of two disorders
one is autosomal dominant (A= disease, a = not)
one is autosomal recessive (b = disease, B = not).
I
1
2
Is the gamete that III-1 received from II-2 parental
or non-parental?
BUT FIRST… break down the question:
II
1
2
Talk to your neighbors and devise a systematic,
step-by-step strategy to solve the problem
Step 1.
III
1
= autosomal dominant
Step 2.
Step 3.
= autosomal recessive
= both traits
etc.
Practice question
The pedigree shows segregation of two disorders
one is autosomal dominant (A= disease, a = not)
one is autosomal recessive (b = disease, B = not).
I
1
2
Is the gamete that III-1 received from II-2 parental
or non-parental?
BUT FIRST… break down the question:
II
1
2
Talk to your neighbors and devise a systematic,
step-by-step strategy to solve the problem
Step 1. Figure out all the genotypes!
III
1
= autosomal dominant
= autosomal recessive
= both traits
Step 2. What are the gametes that made
II-2?
Step 3. What is the gamete that II-2 made?
Step 4. Does the gamete that II-2 made
have a different genotype than the
gamete(s) that made him?
The pedigree shows segregation of two disorders
one is autosomal dominant (A= disease, a = not)
one is autosomal recessive (b = disease, B = not).
AB
??
I
1
2
a b Is the gamete that III-1 received from II-2
a b parental or non-parental?
ab gamete
AB
II
aB
ab
1
2
ab
AB
ab
Ab
III
1
Ab
ab
= autosomal dominant
= autosomal recessive
= both traits
Gametes that made II-2=
AB and ab
Gamete that II-2 gave to III-1=
Ab
Quiz Section this week:
yeast tetrad analysis (genetic mapping)
An introduction to the yeast
tetrad analysis terminology. . .
In yeast…
Tetrad of spores… can examine products of individual meioses!
4 haploid spores
diploid
meiosis
2n
1n
Tetrad with only the 2 parental types = “parental ditype” (PD).
Tetrad with only the 2 recomb. types = “non-parental ditype”
non-parental
(NPD).
Tetrad with all four combinations = “tetratype” (T).
Looking at whole tetrads (PD/NPD/T) is informative.
Example
Suppose URA1 and URA2 are on the same chromosome…
Diploid
Spores
Growth on ura plates?
no
yes
no
no
Parental ditype?
Non-parental ditype?
Tetratype?
Exercise
Suppose URA1 and URA2 are on separate chromosomes,
but each tightly linked to their respective centromeres…
what kinds of tetrads (growth on -ura plates) would this diploid
produce?
1+ = URA1
1+
2-
1+ 2-
1and
2-
1- 2-
1- 2+
1+ 2+
1- 2+
1+ 2+
PD
NPD
1- =ura1
2+ =URA2
2- =ura2
Using 2 analysis to explore linkage
P1 cross:
long,
no speck
X
vestigial,
speck
F1 :
All long, no speck
Can these results be
explained by chance
deviation from independent
assortment?
Test cross
Long, no speck: 2929
Vestigial, no speck: 2070
Vestigial, speck: 2921
Long, speck: 2080
Using 2 analysis to explore linkage
Is this really a 1:1:1:1 ratio as we would expect for
independent assortment? Is the deviation from independent
assortment due to chance?
2 analysis: Test the “null” hypothesis—that the observed
deviation from 1:1:1:1 segregation is due to chance
variation.
It
gives
a
precise
Why test the null hypothesis?
expectation (1:1:1:1)
We cannot test directly for linkage, because
(assuming they are linked) we do not know
the map distance separating these genes.
2 analysis
•
2 =
S
(Obs-Exp)2
Exp
Out of 10,000 testcross progeny . . .
observed expected (o-e)2
long no speck
2929
vestigial speck
2921
vestigial no speck
2070
long speck
2080
(o-e)2
e
2
table
P
0.995
0.975
0.900
0.500
0.100
0.050
0.025
0.010
0.005
df
1
2
3
4
5
6
0.000
0.010
0.072
0.207
0.412
0.676
0.000
0.051
0.216
0.484
0.831
1.237
0.016
0.211
0.584
1.064
1.610
2.204
0.455
1.386
2.366
3.357
4.351
5.348
2.706
4.605
6.251
7.779
9.236
10.645
3.841
5.991
7.815
9.488
11.070
12.592
5.024
7.378
9.348
11.143
12.832
14.449
6.635
9.210
11.345
13.277
15.086
16.912
7.879
10.597
12.838
14.860
16.750
18.548
Degrees of freedom?
What is the P value?
What does this P value mean?
NO WAY could these numbers be
due to independent assortment!
Practice question
Test whether the data truly show linkage by doing a Chi-square
analysis. What is the null hypothesis?
speck
P1 cross:
long,
no speck
X
vestigial,
speck
F1 :
All long, no speck
Homework
Draw the chromosomes of the F1 hybrid in prophase of Meiosis I such
that the gametes will produce all four progeny types.
Practice question
What if Sturtevant had analyzed only 100 testcross progeny?
Out of 100 testcross progeny . . .
observed
long no speck
29
vestigial speck
29
vestigial no speck
21
long speck
21
What is the P value?
What does this P value mean?
expected
(o-e)2
(o-e)2
e
2
analysis
The “null” hypothesis:
The testcross data do not
significantly differ from a
• 2 =
Exp
1:1:1:1 ratio.
Out of 100 testcross progeny . . .
2
2
(o-e)
observed expected (o-e)
e
S
(Obs-Exp)2
29
25
42
= 0.64
vestigial speck
29
25
42
= 0.64
vestigial no speck
21
25
42
= 0.64
21
25
42
= 0.64
long no speck
long speck
 = 2.56
2 table
P
0.995
0.975
0.900
0.500
0.100
0.050
0.025
0.010
0.005
df
1
2
3
4
5
6
0.000
0.010
0.072
0.207
0.412
0.676
0.000
0.051
0.216
0.484
0.831
1.237
0.016
0.211
0.584
1.064
1.610
2.204
0.455
1.386
2.366
3.357
4.351
5.348
2.706
4.605
6.251
7.779
9.236
10.645
3.841
5.991
7.815
9.488
11.070
12.592
5.024
7.378
9.348
11.143
12.832
14.449
6.635
9.210
11.345
13.277
15.086
16.912
7.879
10.597
12.838
14.860
16.750
18.548
Degrees of freedom?
What does this P value mean?
Yes, these numbers could be due to independent assortment!
Since Sturtevant DID analyze more flies, however, we are
certain that vg and sp are linked.
Where does speck map?
Genetic loci:
% Recombinants
purple - vestigial (pr - vg)
11.5 %
purple - black (pr - b)
6.0 %
vestigial - black (vg - b)
16.5 %
vestigial - speck (vg - sp)
41.5 %
41.5 cM
sp?
6 cM 11.5 cM
b
pr
41.5 cM
sp?
vg
How do we distinguish these possibilities?
If to the left, fewer recombinants with pr (and b).
If to the right,more recombinants with pr (and b).
P1 cross:
pr+ sp +
pr+ sp+
F1 :
F1 dihybrid
X
pr
pr
X
sp
sp
testcross
parent
Testcross progeny:
2504
red
no speck
2498
purple
speck
Parental types
2501
purple
no speck
2497
red
speck
Recombinant types
Where shall we place speck?
Genetic loci:
% Recombinants
11.5 %
purple - vestigial (pr - vg)
6.0 %
purple - black (pr - b)
16.5 %
vestigial - black (vg - b)
41.5 %
vestigial - speck (vg - sp)
6 cM 10.7 cM
b
pr
41.5 cM
sp
vg
What is the map distance between b and sp? ~58.2 cM