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Linkage and Recombination
Genes are on chromosomes
Alleles segregate
Alleles for different genes assort independently
Pea plants have 7 pairs of chromosomes. Flies have 4 pairs. Humans have 23 pairs.
Mendel
Morgan and Bridges
Bateson et al.
What is the inheritance pattern for two traits determined by genes on the same chromosome?
“Linked” genes are those that do NOT assort independently.(They reside on the same chromosome.)
We will see that...
DMD
SCIDXIST
XLP
Fragile XHaemophilia
R-G Colorblindness
Human X chromosome
RP3
HPRT
PGKThe corollary isn’t always true!Just because genes are on the same chromosome doesn’t necessarily mean they are linked.
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
Alfred Sturtevant
Mapping Genes on Chromosomes
Segregation of genes on the same chromosome
wild typepr+ pr+ vg+
vg+
purple eyesvestigial wings
pr pr vg vg
gamete? gamete?wild type
phenotypepr+ pr vg+ vg
x
testcross
[ ]
testcross parent genotype?
pr+ vg+
pr+ vg
pr vg+ pr vgphenotypes:
For I.A., expect: 1 : 1 : 1 : 1 ratio
observed: 1339 : 151 : 154 : 1195
pr pr vg vg
Sturtevant’s interpretation:
parental type(same as gametes that made the parent)mostl
y
recombinant (non-parental) type
occasional
crossover
wild typepr+ pr+ vg+
vg+
purple eyesvestigial wings
pr pr vg vg
gamete? gamete?wild type
phenotypepr+ pr vg+ vg
x
The genes are on the same chromosome & the parental alleles (mostly) co-segregate
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
sister chromatids(each is a double helix)
homologues
cohesins
synaptonemal complex
Recombination… a brief review
[see lecture 2 for details]
Single-strand nicks
strand exchange
Holliday junction -No deletions are
caused by recombination.-No mutations are caused by recombination.
Recombination… a brief review (cont)
ALWAYS pull from the centromeres
Segregation of recombinant chromosomes during Meiosis I
3
4
pr +
prvg
vg
1
2
prpr
+
vg +
vg +
1
4
32
Label the chromatids
Telophase I Gametes
Parental
Parental
Recombinant
Recombinant
Telophase II
One recombination event: 2 recombinant and 2 non-recombinant products
Recombination… a brief review (cont)
Which chromatid goes where?
An example with a double crossover:
Test your Understanding
Draw these chromosomes in anaphase I.
Label all alleles appropriately.
ALWAYS pull from the centromeres
3
4
pr +
pr
vg
vg+
1
2
prpr
+
vg +
vg
Other types of crossovers
# xovers resulting gametes
0 parental
parental!2(2 strands)
2 parental2 non-par.
2(3 strands)
4 non-par.
2(4 strands)
Crossing over must occur for faithful segregation at meiosis I
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
pr+ pr+ vg+ vg+
pr pr vg vg
pr+ vg+ pr vggametes:
pr+ pr vg+ vg
pr+ vg pr
vg+gametes:
dominant alleles together on the same chromosome = “cis” configuration
dominant alleles on different chromosomes = “trans” configuration
define the parental type
X
pr+ pr+ vg vg pr pr vg+ vg+X
pr+ vg+
pr vgpr+ vgpr
vg+
The cross from our previous example:
A different cross from our example:
Identifying the parental type
Option 2. The two most abundant progeny types
(assuming the genes are linked)
1287 1204
170 154
Cross: pr+ pr vg+ vg x pr pr vg vg
Progeny:
What were the gametes that made the heterozygous parent? pr+ vg pr
vg+
=
=
pr+ prvg+ vg implie
s
Doesn’t make se
nse!???
A notation system that defines allele configuration
linkage
Configuration(i.e., cis or trans)
separate chromosomes
pr+
pr;
w+
w
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
Harriet Creighton and Barbara McClintock, maize Curt Stern, Drosophila
The problem: homologous chromosomes look alike… how to tell if they really exchanged segments?
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
C = colored Wx = starchyc = colorless wx = waxy“knob
”translocation
x
Progeny phenotypescolored, waxy
colorless, starchy
colored, starchy
colorless, waxy
Conclusion: genetic recombination exchange of chromosome segments
Structural markers: Genetic markers:
recombinant genotypes and chromosomes!
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
1287 1204
170 154
Cross: pr+ pr vg+ vg x pr pr vg vg
Progeny:
What is the map distance between these two genes?
non-parental types (least abundant)
parental types (most abundant)
1287 1204
170 154
Step 1. Identify the parental and non-parental types.
parental =
non-parental =
pr+ vg and pr vg+
pr+ vg+ and pr vg
Map distance… example (cont’d)
non-parental types (least abundant)
parental types (most abundant)
1287 1204
170 154
Step 2. Calculate % recombinant products.
% recombinant =
(170 + 154)
(170+154+1287+1204)
x100= 11.5%
Map distance =
11.5 map units = 11.5 centiMorgan’s (cM)
Map distance… example (cont’d)
Number Recombinant Types
Total Number of Progeny
Another Example (recessive b mutation)
Which are the parental types? and the recombinant types?
Recombination Frequency =
4615
red blac
k
4707purple
tan
307red tan
295purpl
e black
pr b Example:
Recombination Frequency =
Total Number of Progeny
pr b Example:307 + 295
4615 + 4707 + 307 + 295= 0.06
Parental types Recombinant types
pr + b pr b + pr + b + pr b
= 6.0 %x 100602
9924
4615
red blac
k
4707purple
tan
307red tan
295purpl
e black
Number Recombinant Types
Another Example (recessive b mutation)
Genetic loci: % Recombinants
Alfred Sturtevant’s mental leap:
11.5 %
% recombinants is directly proportional to distance
purple - vestigial (pr - vg)
6.0 %purple - black (pr - b)16.5 %vestigial - black (vg - b)
He drew a genetic map of the second chromosome:
vgpr
11.5 cM
b
6 cM
Creating a Genetic Map
Revealed from other
crosses
Why 17.5cM and not 16.5cM?
b+
b vg
vg+
pr
pr+
Point to ponder:If you examined a population of meioses, what is the maximum recombination frequency you could see between two genes?
Genes very close together…
Low probability of crossover between them
…very few recombinants
“tight linkage”
Genes further apart… more recombinants
* Crossing-over creates new combinations of traits.
* The frequency of recombinant types indicates the distance between linked
genes.
* Two Parental types in ≈ frequencies.Two Recombinant types ≈ frequencies.
* If genes are linked,Parental types > recombinant
types.
Summary
Practice questionThe pedigree shows segregation of two disorders
one is autosomal dominant (A= disease, a = not) one is autosomal recessive (b = disease, B = not).
I
II
III
1 2
1 2
1
= autosomal dominant= autosomal recessive= both traits
Is the gamete that III-1 received from II-2 parental or non-parental?
BUT FIRST… break down the question:
Talk to your neighbors and devise a systematic, step-by-step strategy to solve the problem
Step 1.
Step 2.
Step 3.
etc.
The pedigree shows segregation of two disordersone is autosomal dominant (A= disease, a = not)
one is autosomal recessive (b = disease, B = not).
I
II
III
1 2
1 2
1
= autosomal dominant= autosomal recessive= both traits
Is the gamete that III-1 received from II-2 parental or non-parental?
BUT FIRST… break down the question:
Talk to your neighbors and devise a systematic, step-by-step strategy to solve the problem
Step 1. Figure out all the genotypes!
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?
Practice question
I
II
III
1 2
1 2
1
Is the gamete that III-1 received from II-2 parental or non-parental?
a ba b
A B
? ?
AB
A B
a b
A b
a b
a B
a b
ab Ab
ab gamete
Gametes that made II-2=
Gamete that II-2 gave to III-1=
AB and ab
Ab= autosomal dominant= autosomal recessive= both traits
The pedigree shows segregation of two disordersone is autosomal dominant (A= disease, a = not)
one is autosomal recessive (b = disease, B = not).
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!
Tetrad with only the 2 parental types = “parental ditype” (PD).
Tetrad with only the 2 recomb. types = “non-parental ditype” (NPD).Tetrad with all four combinations = “tetratype” (T).
Looking at whole tetrads (PD/NPD/T) is informative.
non-parental
4 haploid spores
1n
meiosis2n
diploid
Example
Suppose URA1 and URA2 are on the same chromosome…
Diploid SporesGrowth on -ura plates?
no
no
no
yes
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- =ura1
2+ =URA2
2- =ura2
1+ 2-
1+ 2-
1- 2+
1- 2+
1- 2-
1- 2-
1+ 2+
1+ 2+
and
PD NPD
Using 2 analysis to explore linkage
P1 cross:
F1 :All long, no speck
Xlong, no speck
vestigial, speck
Test cross
Long, no speck: 2929Vestigial, speck: 2921
Vestigial, no speck: 2070Long, speck: 2080
Can these results be explained by chance deviation from independent assortment?
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.
Why test the null hypothesis?It gives a precise 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.
Using 2 analysis to explore linkage
Out of 10,000 testcross progeny . . .expectedobserved
long no speck 2929
(o-e)2 (o-e)2
e
(Obs-Exp)2
Exp2 = •
2 analysis
vestigial speck 2921vestigial no speck 2070long speck 2080
P
df123456
0.995
0.0000.0100.0720.2070.4120.676
0.975
0.0000.0510.2160.4840.8311.237
0.900
0.0160.2110.5841.0641.6102.204
0.500
0.4551.3862.3663.3574.3515.348
0.100
2.7064.6056.2517.7799.236
10.645
0.050
3.8415.9917.8159.488
11.07012.592
0.025
5.0247.3789.348
11.14312.83214.449
0.010
6.6359.210
11.34513.27715.08616.912
0.005
7.87910.59712.83814.86016.75018.548
2 table
What does this P value mean?REJECT
NO WAY could these numbers be due to
independent assortment!
What is the P value?
Degrees of freedom?
P1 cross:
F1 :
All long, no speck
Xlong, no speck
vestigial, speck
Test whether the data truly show linkage by doing a Chi-square analysis. What is the null hypothesis?
speck
Draw the chromosomes of the F1 hybrid in prophase of Meiosis I such that the gametes will produce all four progeny types.
Homework
Practice question
Out of 100 testcross progeny . . .
expectedobserved
long no speck 29
(o-e)2 (o-e)2
e
vestigial speck 29
vestigial no speck 21
long speck 21
What does this P value mean?
What is the P value?
What if Sturtevant had analyzed only 100 testcross progeny?
Practice question
Out of 100 testcross progeny . . .expectedobserved
long no speck 29
(o-e)2 (o-e)2
e
= 2.56
(Obs-Exp)2
Exp2 = •
vestigial speck 29vestigial no speck 21long speck 21
25
25
25
25
42
42
42
42
= 0.64
= 0.64
= 0.64
= 0.64
2 analysis The testcross data do
not significantly differ from a 1:1:1:1 ratio.
The “null” hypothesis:
P
df123456
0.995
0.0000.0100.0720.2070.4120.676
0.975
0.0000.0510.2160.4840.8311.237
0.900
0.0160.2110.5841.0641.6102.204
0.500
0.4551.3862.3663.3574.3515.348
0.100
2.7064.6056.2517.7799.23610.645
0.050
3.8415.9917.8159.48811.07012.592
0.025
5.0247.3789.34811.14312.83214.449
0.010
6.6359.21011.34513.27715.08616.912
0.005
7.87910.59712.83814.86016.75018.548
2 table
What does this P value mean?
Can’t Reject
Yes, these numbers could be due to independent assortment!
Degrees of freedom?
Since Sturtevant DID analyze more flies, however, we are certain that vg and sp are linked.
41.5 %
Genetic loci: % Recombinants11.5 %purple - vestigial (pr - vg)
vestigial - speck (vg - sp)
6.0 %purple - black (pr - b)16.5 %vestigial - black (vg - b)
vgpr
11.5 cM sp?41.5 cM
b
6 cMsp?41.5 cM
If to the left, fewer recombinants with pr (and b). If to the right,more recombinants with pr (and b).
Where does speck map?
How do we distinguish these possibilities?
P1 cross:
X
2504
F1 :F1 dihybrid
2498 2501 2497
Parental types Recombinant types
testcross parent
X
Testcross progeny:
redno speck
purplespeck
purpleno speck
redspeck
pr+ sp +
pr+ sp+
pr sppr sp
Genetic loci: % Recombinants11.5 %
41.5 %
purple - vestigial (pr - vg)
vestigial - speck (vg - sp)
6.0 %purple - black (pr - b)16.5 %vestigial - black (vg - b)
vgpr
10.7 cM sp41.5 cM
b
6 cM
Where shall we place speck?
What is the map distance between b and sp?~58.2 cM