Chapter 6

Linkage & Chromosome Mapping

Chromosome, NOT genes are the unit

of inheritance

Crossing Over -- Prophase I

 Increases genetic variation

 Related to distance
 Chromosome maps

1
Independent Assortment: 2 genes on 2 different
homologous pairs of chromosomes.
Figure 6.1a Page 138

A B

A B
a b

a b

Gametes

A a

B B

A a

2
b b
 Linkage: 2 genes on a single pair of
Homologs: No exchangeFig. 6.1b Page 138

A A a a
B B b b

Sister Sister
Chromatid Chromatid

Non-sister Chromatid
Gametes
A B a b

A B a b 3
 Linkage: 2 genes on a single pair of
Homologs: Single exchange
Fig. 6.1c Page 138

A B

A B
a b

a b
Gametes
A B
Non-Crossover Gamete
a B
Crossover Gamete
A b
Crossover Gamete
a b
Non-Crossover Gamete
4
 Frequency of crossing over for any two
loci on the same chromosome is
proportional to the distance between
them.

 Complete linkage -- Only parental

gametes.

 As distance increases, proportion of

recombinant gametes ⇑ and the
porportion of parental gametes ⇓.

 Frequency of recombinant gemetes

approaches but cannot exceed 50%.

 50% recombinant gametes = 1:1:1:1

(2 parental : 2 recombinant) -- This
would be indistinguishable from 2
non-linked, independently assorting
5
loci.
Incomplete Linkage, Crossing Over &
Genetic Mapping

 Complete Linkage is rare --

Crossing over is common.

 Drosophila geneticists
 Thomas A. Morgan
 Alfred A. Sturtevant

 By examining single trait

crosses, deduced mode of
sex-linked inheritance.

 Puzzling results with 2

sex-linked traits. 6
 y w y+ w+
y w
Yellow body, Wild-type
White eyes

F1 Females -- wild-type

F1 Males -- Expressed both mutant Traits

F2 -- 98.7% Offspring had Parental Phenotype

1.3% either: Yellow body/Wild-type eyes

Or
Wild-type body / White eyes

7
Two Questions:
1. What is the source of gene separation?

 Chiasmata are the source and linked

genes exist in a linear array.

 Morgan proposed the term:

“Crossing-Over”.

2. Why did frequency vary depending on

genes examined?

 Answered by Sturtevant.

 Related to distance and this is

additive.

8
 1. Yellow, White 0.5%
2. White, Minature 34.5%
3. Yellow, Minature 35.4%

1% recombination = 1 map unit =

1 centimorgan
0.5 34.5

y w m

By 1923 Morgan & Sturtevant had shown:

 Linkage & Crossing over not restricted

to sex-linked genes
 In Drosophila, crossing over occurs only
in females
 Substantiated chromosomal theory of 9

inheritance
 Two Pont Mapping

A ---> B = 35
A ---> C = 5
A ---> D = 20
B ---> C = 30
B ---> D = 55
C ---> D = 25

Question: What is the correct order of these

genes and the distance between each pair?

Step I: Find and map the smallest distance.

5
A C
10
STEP II: Continue this process until the
map is complete and all distance add up to
observed recombination frequencies.

20 5 30
D A C B

G = Yellow g = Green
R = Round r = Wrinkled

A plant of unkown genotype & phenotype

is test crossed with a ggrr plant and you
obtain the following progeny:

Yellow, Wrinkled 88
Yellow, Round 12
Green, Wrinkled 8
Green, Round 92 11
Questions:
1. What is the phenotype of the unknown
plant?
2. What is the genotype of the unknown
plant?
3. What is the distance between the
two genes?
4. Cis of Trans configuration?

Yellow, Wrinkled 88 NCO

Yellow, Round 12 SCO
Green, Wrinkled 8 SCO
Green, Round 92 NCO

1. Genotype of unknown plant: Gr/gR

2. Phenotype: Yellow, Round
3. Distance between 2 genes:
(12 + 8) / 200 = 0.1 X 100 = 100 mu
4. Configuration = Trans 12
 Single Cross-Over Event -- Pg. 143

A B A B
gametes A b
A B
a b a B
a b
a b

If only a single exchange -- theoretical limit of

crossing over = 50%

 2 genes on the same chromosome ≥ 50 mu,

a cross over can theoretically be expected
to occur between them in 100% of the
tetrads.

 For several reasons, this limit is never reached

(discussed later). 13
 Multiple Cross Overs -- Pg. 144

A B D
A B D
a b d
a b d

Product Law:

A ---> B = 20% B ---> D = 30%

Double Cross Over = 0.2 X 0.3 = 0.06 =6%

14
 3 Criteria for successful 3-point mapping

1. Individual producing recombinant gametes

must be heterozygous at all 3 loci.

2. Must be able to determine genotype of all

gametes by examining phenotypes of
offspring.

3. Must have a large number of offspring.

3 Autosomal Linked Genes in Maize

bm = Brown mid rib
v virescent seedling
pr = purple aleurone

What genotypes do you need to map these 3 genes?

15
Answer:
G 1 individual heterozygous for all 3 loci
G 1 individual homozygous recessive for all 3 loci.

You don’t know:

G Arrangement of mutant alleles (cis/trans)
G Gene order
G Distances between genes

Offspring Phentoype Observed

+ v bm 230
pr + + 237

pr v + 82
+ + bm 79

+ v + 200
pr + bm 195

pr v bm 44
16
+ + + 42
1109
 Step I: Make an assumption about gene order:

G Assume order to be: pr v bm

Therefore, the cross is:

+ v bm pr v bm
X
pr + + pr v bm

Step II: Determine the correct gene order by

looking at the noncrossover and double crossover
phenotypic classes.

Offspring Phentoype Observed

+ v bm 230 NCO
pr + + 237 NCO
pr v + 82 SCO
+ + bm 79 SCO
+ v + 200 SCO
pr + bm 195 SCO
17
pr v bm 44 DCO
+ + + 42 DCO
This is not the correct gene order so try
another order.

Offspring Phentoype Observed

v + bm 230 NCO
+ pr + 237 NCO

v pr bm 44 DCO
+ + + 42 DCO

This IS the correct gene order! v pr bm

Step III: Calculate map distance between the

3 genes under study.

The distance between pv & v and between v & bm

is equal to the % of all detectable exchanges
occurring between them!

For any two genes, you must consider ALL

appropriate SCO plus the DCO. 18
Now write all genotypes in the correct gene order!

Offspring Phentoype Observed

v + bm 230
+ pr + 237

v pr + 82
+ + bm 79

v + + 200
+ pr bm 195

v pr bm 44
+ + + 42

Distance between v & pr:

(82 + 79 + 44 + 42) / 1109 X 100 = 22.27 mu

Distance between pr & bm:

(200 + 195 + 42 + 44) / 1109 X 100 = 43.37 mu
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Step IV: Construct a chromosomal map

v pr bm
22.27 43.37

Problem: Singed bristles (sn), crossveinless wings

(cv), and vermilion eye color (v) are due to
recessive mutant alleles of 3 X-linked genes in
Drosophila. When a female heterozygous for each
of the three genes was testcrossed with a singed,
crossveinless, vermilion male, the following
progeny were obtained.

20
 singed, crossveinless, vermilion 3
crossveinless, vermilion 392
vermilion 34
crossveinless 61
singed, crossveinless 32
singed, vermilion 65
singed 410
wild-type 3

Question: (a) What is the correct genotype of the

P1 male and female? (b) What is the correct
order of these three genes?, and (C) what are the
map distances between the three genes?

21
sn + + 410 NCO
+ cv v 392 NCO
+ + v 34 SCO
sn cv + 32 SCO
+ cv + 61 SCO
sn + v 65 SCO
sn cv v 3 DCO
+ + + 3 DCO
Check to see if this is the correct order by looking
At NCO and DCO.

The correct gene order is cv sn v

+ sn + 410 NCO
cv + v 392 NCO
+ + v 34 SCO
cv sn + 32 SCO
cv + + 61 SCO
+ sn v 65 SCO
cv sn v 3 DCO
22
+ + + 3 DCO
Calculate distance between each pair of genes
and construct a chromosomal map:

Distance between cv & sn:

(34 + 32 + 3 + 3) / 1000 X 100 = 7.2 mu

Distance between sn & v:

(61 + 65 + 3 + 3) / 1000 X 100 = 13.2 mu

cv sn v
7.2 13.2
23
Interference: Reduction in the number of DCO
events relative to what is expected based on the
map distances.

cv sn v
7.2 13.2
Observed DCOs = 6 / 1000 = 0.006

Expected DCOs = Multiply frequency of

recombination between each pair of genes.

DCO(EXP) = 0.072 X 0.132 = 0.0095

Coefficient of Coincidence (C):

C = Observed (DCO) / Expected (DCO)
C = 0.006 / 0.0095 = 0.632

Interference (I) = 1 - C
I = 1 - 0.632 = 0.368
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I = 1.0 : INTERFERENCE IS COMPLETE and
no DCOs observed.

I = + : POSITIVE INTERFERENCE --
Fewer DCO observed than expected.

I = - : NEGATIVE INTERFERENCE --
More DCO observed than expected.

In eukaryotes, positive interference most

common

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 Why is 50% recombinant gametes the
theoretical maximum?
Multiple strand exchanges Pp. 152
Figure 6.12

A. Two-Strand Double Exchange

B. Three-Strand Double Exchange

A. Four-Strand Double Exchange

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 Problem: The father of Mr. Spock, first officer
of the starship Enterprise, came from planet
Vulcan; Spock’s mother came from Earth. A
vulcan has pointed ears (P) adrenals absent (A),
and a right-sided heart (R). These three loci are
autosomal, and they are linked as shown in the
following map:

P A R
15 MU 20 MU

Questions: If Mr. Spock marries an Earth woman

and there is no (genetic)interference, what
proportion of their children will have:

a. Vulcan phenotypes for all three characters?

b. Earth phenotype for all three characters?
c. Vulcan ears and heart but Earth adrenals?
d. Vulcan ears but Earth heart and adrenals?
27
 PAR/par X par/par

PaR/par (0.15 X 0.20) / 2 = 0.015

pAr/par (0.15 X 0.20) / 2 = 0.015

Par/par (0.15 - 0.03) / 2 = 0.06

pAR/par (0.15 - 0.03) / 2 = 0.06

PAr/par ( 0.20 - 0.03) / 2 = 0.085

paR/par (0.20 - 0.03) / 2 = 0.085

PAR/par (1 - 0.32) / 2 = 0.34

par/par (1 - 0.32) / 2 0.34

a. Vulcan phenotypes for all three characters = 0.34

b. Earth phenotype for all three characters = 0.34

c. Vulcan ears and heart but Earth adrenals = 0.015

d. Vulcan ears but Earth heart and adrenals =280.06

 Sister Chromatid Exchange

 Occurs, but its significance is unclear

 Agents that increase chromosome

damage also increase the frequency
of sister chromatid exchange.

 Frequency of sister chromatid exchange

increases in the human autosomal
recessive “Bloom disease”

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SOMATIC CELL HYBRIDIZATION AND
HUMAN CHROMOSOME MAPS

 Used to determine on which chromosome

a particular trait is located.

 2 cells (mouse and human) can be induced

to form a single hybrid cell with 2
nuclei -- HETEROKARYON.

 Culture heterokaryons

 Nuclei will fuse to form a SYNKARYON.

 After many generations, chromosomes

from one of the two parental species
are gradually lost.
30
Linkage Analysis in Haploid Organisms:

 Vegetative stage = Haploid

 Reproductive cells = Isogametes
 Isogametes from 2 strains (+ and -)
fuse to produce a diploid zygote
 Diploid zygote undergoes meiosis to
re-establish haploid condition
 Haploid products are progenitors of
vegetative stage.

Neurospora -- Meiosis occurs in ASCUS, initial

haploid products are called TETRADS.

 Each cell divides mitotically to produce

8 haploid ascospores
 Ascospores refelct the sequence of
formation

 Tetrad Analysis
Mapping the centromere 31
 Five Combinations

First Division Segregation:

AA++

Second Division Segregation:

A+A+
+A+A
+AA+
A++A

Distance between gene and centromere

1/2 second division segregant asci

Total asci sored 32
Example: 65 1st division segregants
70 2nd division segregants

What is the distance from the gene to the

centromere?

(0.5 X 70) / 135 = 0.259 X 100 = 25.9 mu

Ordered Tetrad Analysis:

 Tedious
 Distinguishes 1st and 2nd division
segregants which is necessary to
map gene in relation to the
centromere.

Unordered Tetrad Analysis:

 Determine if 2 genes are linked
 Map distance between to loci.
33
 Examine 100 tetrads from the following cross:

abX++

TETRAD TYPE P NP T

++ a+ ++
++ a+ a+
GENOTYPES ab +b +b
ab +b ab

#TETRADS 43 43 14

P = PARENTAL DITYPES
NP = NONPARENTAL DITYPES
T = TETRATYPES

 WHEN P = NP, THE TWO GENES ARE NOT

LINKED.

THESE GENES ARE ON SEPARATE

CHROMOSOMES!!!! 34
EXAMPLE 2:

P NP T
64 6 30

Because P ≠ NP, no independent assortment

and hence, 2 genes are linked and can be
mapped.

Exchange Frequency =

[(NP + 1/2T) / Total # Tetrads] X 100

{[(6 + 1/2(30)) / 100} X 100 = 21% or 21 mu

35