1. (GENE MAPPING)

2.  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

3.  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

4. Introduction

5.  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-heavy
Brown-thin
bwhv
bw hv
red, thin
The Linkage Ratio

6. The Linkage Ratio

7.  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

8. The Linkage Ratio

9. 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

10. Morgan made crosses involving other X-linkage
genes
P: White-eye, miniature wing ♀
F1:
F2:
X wild type ♂
even more puzzling
phenotypes differed
62.8%: parental types
37.2%: either: white eyed or
miniature wing
Crossover and Gene Distance

11. Morgan
crossed
yellow
bodied
white eyed
female and
wild type
male:
yw/yw X
y+w+Y
Crossover and Gene Distance
Morgan
made
crosses
involving
other Xlinkage
genes:
Whiteeye,
miniature
wing ♀
X
wild type
♂

12.  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

13. Crossover and Gene Distance

14. Crossover and Gene Distance

15.  Two arrangements of alleles exist for an individuals
heterozygous at two loci:
‘cis’ or coupling
‘trans’ or repulsion
w + m+
w+ m
w
w
m
m+
 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
Concept of a genetic map

16.  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

17.  Cross-over is more
likely to occur
between distant
genes than close
genes

18.  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 progeny
Recombination frequency 
X 100%
total # of progeny
Calculating Recombination frequency

19.  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?
Recombination frequency 
# of recombinant progeny
X 100%
total # of progeny
 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
Example

20.  First genetic map was for Drosophila:
3 sex-linked genes
w = white-eyes
m =miniature wings
y = yellow body
 Recombination frequencies:
wxy
wxm
mxy
= 0.5%
= 34.5%
= 35.4%
0.5
34.5
35.5

21.  Single crossover
Single Crossover

22.  Double crossover
Double Crossover

23.  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

24.  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
Three-Point Mapping

25.  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).

27. Phenotypes
white, echinus
yellow
yellow, white
echinus
yellow, echinus
white
Determining Gene Sequence

28.  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




[+ v +]
200
[pr + bm]
195
[pr v bm]
44
[+ + +]
42
 What is the correct sequence of genes?
 What is the distance between each pairs of gene?
A Mapping Problem in Maize

29.  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

30.  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

31.  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

32.  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
Determine the parental and dco types

33.  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

34. +
v
bm
+
+
bm
pr
+
+
pr
v
+
+
bm
v
+
+
v
pr
+
+
v
+
bm
pr
v
bm
pr
+
bm
+
pr
+
+
+
+




[+ v bm]
[pr + +]
[pr v bm]
[+ + +]
230  parental type
237  parental type
44  dco type
42  dco type
Examine the gene in the middle

35.  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

36. Temporary Order








[+ v bm]
[pr + +]
[+ + bm]
[pr v +]
[+ v +]
[pr + bm]
[pr v bm]
[+ + +]
Correct Order
230
237
82
79
200
195
44
42








v
+
+
v
v
+
v
+
+ bm
pr +
+ bm
pr +
++
pr bm
pr bm
++
Re-order the genes (if necessary)
230
237
82
79
200
195
44
42

37.  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

38. v
+
pr
+
+
pr
+
pr
+
bm
+
bm
v
v
+
I
v
+
bm
+
pr
+
II
+
v
v
+
pr
+
+
bm
+
bm
+
+
+
pr
bm
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
Examine sco in region I and II

39.  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

40.  The formula to calculate the distance between
two genes:
 In region I =
scoI dco
x 100
Total
 In regio II =
scoII dco
x 100
Total
Calculate the distance

41.  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
Calculate the1109
distance

42.  Distance between v-pr =
82  79  44  42
x 100  22.27 cM
1109
 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
 Distance between pr-bm =
 v pr bm
44  dco type
 + ++
42  dco type
200  195  44  42
x 100  43.37 cM
1109
 Distance between v-bm =
22.27  43.37  65.64 cM
 And the map is
22.27 cM
v+/v
Calculate the distance
43.37 cM
pr+/pr
bm+/bm

43.  Interference: a crossover at one spot on a
chromosome decreases the likelihood of a crossover
in a nearby spot
I=1–c
 where c: coefficient of coincidence
obs dco
c
exp dco
 obs dco: observed data
 exp dco: expected dco = sco I x sco II
Interference and Coincidence

44.  From the data:
22.27 cM
v+/v
43.37 cM
pr+/pr
bm+/bm
 obs dco = 86/1109 = 0.0775
 exp dco = 0.2227 x 0.4337 = 0.0966
 v + bm
 c = 0.0775/0.0966 = 0.80
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
 I = 1 – 0.80 = 0.20
 v pr bm
44  dco type
 + ++
42  dco type
Interference and Coincidence

45. Any questions?
Thank you