bft1033 6 linkage genes_print
TRANSCRIPT
(GENE MAPPING)
INTRODUCTION
Most chromosomes consist of very large numbers of genes
Genes that are part of the same chromosome are said linked
These genes demonstrate linkage in genetic crosses
During meiosis, they are not free to undergo independent assortment They are transmitted as a unit
Crossover results in reshuffling or recombination of alleles between homolog.
Introduction
No crossover: two genetically different gametes are formed Each gamete receive the alleles present on one
homolog or the other
Illustrate complete linkage
Produce parental or noncrossover gametes
Crossover: produce four types of gametes Two parental gametes
Two recombinant or crossover gametes
Introduction
The Linkage Ratio
Complete linkage in Drosophila melanogaster: Mutants: brown (bw) eye and heavy (hv) wing vein
Normal alleles: bw+ (red eye) and hv+ (thin wing vein)
Cross between brown eye and thin vein with red eyes and heavy vein
P:
F1:
bwhv bw hv
bwhv bw hv
red, thin
Brown-thin Red-heavy
bwhv
bw hv
The Linkage Ratio
When the F1 generation is interbred, the F2 generation will be produced in a 1:2:1 phenotypic and genotypic ratio.
When the F1 is tescrossed, it will produce a 1:1 ratio of brown thin and red heavy
The Linkage Ratio
The Linkage Ratio
Morgan crossed yellow bodied white eyed female and wild type male
P: yellow, white ♀ X wild-type ♂
F1: ♀: wild type
♂: expressed both mutant traits
F2: 98.7% parental types (gray bodied, red eyed)
1.3% either: yellow bodied with red eyed , or
gray bodied with white eyed
Crossover and Gene Distance
Morgan made crosses involving other X-linkage genes
P: White-eye, miniature wing ♀ X wild type ♂
F1: even more puzzling
F2: phenotypes differed
62.8%: parental types
37.2%: either: white eyed or
miniature wing
Crossover and Gene Distance
Crossover and Gene Distance
Morgan crossed yellow bodied white eyed female and wild type male:
yw/yw Xy+w+Y
Morgan made crosses involving other X-linkage genes:White-eye, miniature wing ♀X wild type ♂
Morgan postulated that exchange occurred between the mutant genes on the two X chromosomes of the F1 females
Lead to 1.3 and 37.2 recombinant gametes
The closer two gene are, the less likely genetic exchange will occur between them
Morgan proposed the term crossing over to describe the physical exchange leading to recombination.
Crossover and Gene Distance
Crossover and Gene Distance
Crossover and Gene Distance
Concept of a genetic map
Two arrangements of alleles exist for an individuals heterozygous at two loci:
Cross-over of cis results in trans and vice versa
Frequency of recombinants (%) is a characteristic of each gene pair, regardless of cis or trans arrangements
‘cis’ or coupling
w+ m+
w m
‘trans’ or repulsion
w+ m
w m+
In cross A Parental types (yellow-white, wild type): 98.7%
Recombinant types (white, yellow): 1.3%
Distance between genes: 1.3 mu
In cross B Parental types (white-miniature, wild type): 62.8%
Recombinant types (white, miniature): 37.2%
Distance between genes: 37.2 mu
Concept of a genetic map
Cross-over is more likely to occur between distant genes than close genes
Calculating Recombination frequency
Sturtevant (1913) recognized that recombination frequencies could be used to create a map
1% cross-over rate = 1 map unit (mu) or centiMorgan (cM)
Map units (mu) and centiMorgans (cM) are relative measures.
# of recombinant progenyRecombination frequency X 100%
total # of progeny
Example
The test cross Ab/aB x ab/ab is performed. The following numbers of progeny of each genotype are obtained: 87 AaBb, 409 Aabb, 390 aaBb, 114 aabb. What is the approximate distance (in map units)
between the two genes in question?
RF = (87 + 114)/(87 + 409 + 390 + 114) x 100%
= 201/1000 x 100%
= 20.1%
So the distance between the two genes is 20.1 cM
# of recombinant progenyRecombination frequency X 100%
total # of progeny
First genetic map was for Drosophila:3 sex-linked genes
w = white-eyes
m =miniature wings
y = yellow body
Recombination frequencies:w x y = 0.5%
w x m = 34.5%
m x y = 35.4%
0.5 34.5
35.5
Single Crossover
Single crossover
Double Crossover
Double crossover
Three-Point Mapping
The genotype of the organism producing the crossover gametes must be heterozygous at all loci
The cross must be constructed so that genotypes of all gametes can be accurately determined by observing the phenotypes of the resulting offspring
A sufficient of number of offspring must be produced in the mapping experiment to recover a representative sample of all crossover.
Three-Point Mapping
Males hemizygous for all three wild type alleles are crossed to female with three mutant traits (yellow body, white eyes, and echinus eye shape)
F1 consists of females heterozygous at all loci and males hemizygous for all three mutant alleles
When the F1 is intercrossed to produce F2, it produces 8 different classes: Two classes of parental types (the biggest
proportion)
Two classes from single crossover in region I
Two classes from single crossover in region II
Two classes from double crossover (the smallest proportion).
Determining Gene Sequence
Phenotypes
white, echinus
yellow
yellow, white
echinus
yellow, echinus
white
A Mapping Problem in Maize
In maize, the recessive mutant genes: bm (brown midrib), v (virescent seedling), and pr (purple aleurone)
are linked on chromosome 5
A female plant is heterozygous for all three traits is crossed with a male homozygous for all three mutant alleles
F1 data: [+ v bm] 230
[pr + +] 237
[+ + bm] 82
[pr v +] 79
What is the correct sequence of genes?
What is the distance between each pairs of gene?
[+ v +] 200
[pr + bm] 195
[pr v bm] 44
[+ + +] 42
The Five Steps to Solve the Problem
1. Determine the parental and dco types
2. Examine the gene in the middle
3. Re-order the genes (if necessary)
4. Examine sco in region I and II
5. Calculate the distance
A Mapping Problem in Maize
The Five Steps to Solve the Problem
1. Determine the parental and dco types
2. Examine the gene in the middle
3. Re-order the genes (if necessary)
4. Examine sco in region I and II
5. Calculate the distance
A Mapping Problem in Maize
Determine the parental and dco types
The parental types are the biggest number, and dco types are the smallest [+ v bm] 230
[pr + +] 237
[+ + bm] 82
[pr v +] 79
[+ v +] 200
[pr + bm] 195
[pr v bm] 44
[+ + +] 42
Determine the parental and dco types
The parental types are the biggest number, and dco types are the smallest [+ v bm] 230 parental type
[pr + +] 237 parental type
[+ + bm] 82
[pr v +] 79
[+ v +] 200
[pr + bm] 195
[pr v bm] 44 dco type
[+ + +] 42 dco type
The Five Steps to Solve the Problem
1. Determine the parental and dco types
2. Examine the gene in the middle
3. Re-order the genes (if necessary)
4. Examine sco in region I and II
5. Calculate the distance
A Mapping Problem in Maize
Examine the gene in the middle
[+ v bm] 230 parental type
[pr + +] 237 parental type
[pr v bm] 44 dco type
[+ + +] 42 dco type
v
v
v
vv
vbm
bm
bmbm
+
bm
+
+
+
+
+
+
+
+
pr
+
pr
pr
pr
pr
pr
bm+ +
+
++
+
++
The Five Steps to Solve the Problem
1. Determine the parental and dco types
2. Examine the gene in the middle
3. Re-order the genes (if necessary)
4. Examine sco in region I and II
5. Calculate the distance
A Mapping Problem in Maize
Re-order the genes (if necessary)
Temporary Order
[+ v bm] 230 [pr + +] 237 [+ + bm] 82 [pr v +] 79 [+ v +] 200 [pr + bm] 195 [pr v bm] 44 [+ + +] 42
Correct Order
v + bm 230 + pr + 237 + + bm 82 v pr + 79 v + + 200 + pr bm 195 v pr bm 44 + + + 42
The Five Steps to Solve the Problem
1. Determine the parental and dco types
2. Examine the gene in the middle
3. Re-order the genes (if necessary)
4. Examine sco in region I and II
5. Calculate the distance
A Mapping Problem in Maize
Examine sco in region I and II
v
+
bm
+
+
pr
v
+
bm
+
+
pr
v
+
bm
+
+
pr
v bm+
++ pr
v +pr
bm+ +
v ++
bm+ pr
I
II
v + bm 230 parental type
+ pr + 237 parental type
v pr + 79 scoI type
+ + bm 82 scoI type
v + + 200 scoII type
+ pr bm 195 scoII type
The Five Steps to Solve the Problem
1. Determine the parental and dco types
2. Examine the gene in the middle
3. Re-order the genes (if necessary)
4. Examine sco in region I and II
5. Calculate the distance
A Mapping Problem in Maize
Calculate the distance
The formula to calculate the distance between two genes:
In region I =
In regio II =
100xTotal
dcoscoI
100xTotal
dcoscoII
v + bm 230 parental type
+ pr + 237 parental type
+ + bm 82 scoI type
v pr + 79 scoI type
v + + 200 scoII type
+ pr bm 195 scoII type
v pr bm 44 dco type
+ + + 42 dco type
Total 1109Calculate the distance
Distance between v-pr =
Distance between pr-bm =
Distance between v-bm =
And the map is
cM 22.27 100x
1109
42 44 79 82
cM 43.37 100x
1109
42 44 195 200
Calculate the distance
cM 65.64 43.37 22.27
v + bm 230 parental type
+ pr + 237 parental type
+ + bm 82 scoI type
v pr + 79 scoI type
v + + 200 scoII type
+ pr bm 195 scoII type
v pr bm 44 dco type
+ + + 42 dco type
v+/v pr+/pr bm+/bm
22.27 cM 43.37 cM
Interference and Coincidence
Interference: a crossover at one spot on a chromosome decreases the likelihood of a crossover in a nearby spot
where c: coefficient of coincidence
obs dco: observed data
exp dco: expected dco = sco I x sco II
obsdco
expdcoc
I = 1 – c
From the data:
obs dco = 86/1109 = 0.0775
exp dco = 0.2227 x 0.4337 = 0.0966
c = 0.0775/0.0966 = 0.80
I = 1 – 0.80 = 0.20
v+/v pr+/pr bm+/bm
22.27 cM 43.37 cM
v + bm 230 parental type
+ pr + 237 parental type
+ + bm 82 scoI type
v pr + 79 scoI type
v + + 200 scoII type
+ pr bm 195 scoII type
v pr bm 44 dco type
+ + + 42 dco type
Interference and Coincidence
Any questions?Thank you
