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Mendelian Genetics
Topics:

-Transmission of DNA during cell division


Mitosis and Meiosis
- Segregation
- Sex linkage (problem: how to get a white-eyed female)



- Inheritance and probability
- Mendelian genetics in humans
- Independent Assortment
- Linkage
- Gene mapping
- 3 point test cross
- Tetrad Analysis (mapping
in fungi)
- Extensions to Mendelian Genetics
- Gene mutation
- Chromosome mutation
- Quantitative and population genetics
Linkage
Chapter 6
- recombination
- linkage maps
Ch. 6 p. 148 – 165
Prob: 1-5, 7, 8, 10, 11, 14
Linkage of Genes
- Many more genes than chromosomes
- Some genes must be linked on the same
chromosome; therefore not independent
Independent Assortment
Fig 6-6
Interchromosomal
Linkage
Fig 6-11
Intrachromosomal
Two ways to produce dihybrid
A B
a b
X
A B
a b
cis A B
a b
Gametes:
AB
ab
Ab
aB
P
AaBb
(dihybrid )
P
P
R
R
A b
A b
a B
X
a B
A b trans
a B
Ab
aB
AB
ab
Example
Test Cross
How to distinguish:
Parental high freq.
Recombinant
low freq.
AaBb
AB
Ab
aB
ab
X
ab
AaBb
Aabb
aaBb
aabb
aabb
Exp.
25
25
25
25
100
Obs.
10 R
40 P
40 P
10 R
100
Example (cont.)
Gametes:
AB R
Ab P
aB P
ab R
Therefore dihybrid:
A
a
b (trans)
B
Linkage Maps
Genes close together on same chromosome:
- smaller chance of crossovers
between them
- fewer recombinants
Therefore:
percentage recombination can be
used to generate a linkage map
Linkage maps
A
a
C
c
B
b
D
d
large # of recomb.
small number of recombinants
Linkage maps
example
Testcross progeny:
P AaBb 2146
R Aabb
43
65
R
aaBb
22 4513 = 1.4 % RF
P
aabb 2302
Total 4513
1.4 map units
A
1.4 mu
B
Additivity of map distances
separate maps
A
B
A
7
combine maps
C
2
A
2
B
7
or
A
C
2
C
B
5
Locus
(pl. loci)
Linkage
Deviations from independent assortment
Dihybrid gametes
2 parent (noncrossover) common
2 recombinant (crossover) rare
% recombinants a function of distance between
genes
% RF = map distance
Gametes
Number of Genes
Number of Different
Gametes
monohybrid 1 (Aa)
2
dihybrid
2 (AaBb)
4
trihybrid
3 (AaBbCc)
8
Three Point Test Cross
Trihybrid
AaBbCc
ABC
ABc
AbC
Abc
aBC
aBc
abC
abc
X
aabbcc
abc
8 gamete types
Three Point Test Cross
Trihybrid Gametes
C
ABC
c
ABc
C
AbC
c
Abc
B
A
b
a
Three Point Test Cross
Trihybrid
AaBbCc 3 genes:
Possibilities:
1. All unlinked
2. Two linked; one unlinked
3. Three linked
- order ? A---B---C
B---C---A
B---A---C
Three Point Test Cross
Three recessive mutants of
Drosophila:
P +/+ cv/cv ct/ct
+, v vermilion eyes
+, cv crossveinless
+, ct cut wing
X
v/v +/+ +/+
Three Point Test Cross
P +/+ cv/cv ct/ct
Gametes
F1 trihybrid
+ cv ct
x
v/v +/+ +/+
v + +
v/+ cv/+ ct/+
Three Point Test Cross
F1 v/+ cv/+ ct/+
8 gamete types
x
v/v cv/cv ct/ct
v cv ct
one gamete type
8 gamete types
F1 v/+ cv/+ ct/+
v
+
+ cv
v cv
+ +
v cv
+ +
v +
+ cv
+ 580
Parental (most frequent)
ct 592
+ 45
ct 40
ct 89
Recombinant
+
94
ct
3
+
5
1448
8 gamete types
F1 v/+ cv/+ ct/+
v
+
+ cv
v cv
+ +
v cv
+ +
v +
+ cv
+ 580
ct 592
+ 45
ct 40
ct 89
+
94
ct
3
+
5
1448
Parental
Recombinant
268
Recombinant
Parental
268
1448
= 18.5 %
8 gamete types
F1 v/+ cv/+ ct/+
v
+
+ cv
v cv
+ +
v cv
+ +
v +
+ cv
+ 580
ct 592
+ 45
ct 40
ct 89
+
94
ct
3
+
5
1448
Parental
Parental
Recombinant
191
Recombinant
191
1448
= 13.2 %
8 gamete types
F1 v/+ cv/+ ct/+
v
+
+ cv
v cv
+ +
v cv
+ +
v +
+ cv
+ 580
ct 592
+ 45
ct 40
ct 89
+
94
ct
3
+
5
1448
Parental
Recombinant
93
Parental
Recombinant
93
1448
= 6.4 %
Calculate Recombination Fraction
1.
v - cv
2.
v - ct
3. ct - cv
R v cv
R + +
R + +
R v ct
R ct +
R + cv
45 + 89
40 + 94
268 / 1448 = 18.5 %
94 + 5
89 + 3
191/1448
= 13.2 %
40 + 3
45 + 5
93/1448
=
6.4 %
Three point test cross
Observations:
all 3 RF < 50 %
3 genes on same chromosome
v-----cv largest distance ct in middle
map v-------ct-------cv = cv-------ct-------v
13.2 + 6.4 = 19.6 > 18.5 !! Why ?
Three Point Test Cross
P +/+ ct/ct cv/cv
gametes
F1 trihybrid
+ ct
x
v/v +/+ +/+
cv
v
+
v +
+
ct
+
cv
+
Three Point Test Cross
Double crossover class rarest:
v---cv
P v
P +
R
R
v
+
X
+
ct
ct
+
X
+
cv
v
+
+
cv
v
+
X
+
X
cv
+
cv
Three Point test cross
1. Double crossovers not counted in v--cv RF
2. Double crossovers generate P types (with
respect to v--cv)
3. Double crossovers not detected as
recombinants
Consequence:
underestimate of v----cv map distance
Greater distance of genes  greater error
Double recombinant class:
(3 + 5) x 2 = 16
268 + 16 = 284
284/1448 = 19.6
NOTE: double crossovers detected
because of middle gene (ct)
Mapping Function
Genes close together on chromosome
-RF good estimate of map distance
Genes far apart on chromosome
- RF underestimates true map distance due
to undetected multiple crossovers
Mapping Function
m = avg. # crossovers per meiosis
(linear with true map distance)
if m = 1 (1 cross over for every meiosis)
then 50 % recombinants produced
Therefore:
Gamete Types
map units (mu) = m x 50
F1
gametes
A
B
a
b
AaBb
A
a
A
a
B
b
b
B
AB
ab
Ab
aB
Parental
Parental
Recomb.
Recomb.
Recomb.
Recomb.
Mapping Function
Mapping function:
- relates RF to true map distance
(better estimate for genes separated by
large distances)
m = -ln (1 - 2RF)
mu = m x 50
Mapping function
Mapping Function
Observed RF (%)
60
50
40
30
m = -ln(1 - 2RF)
20
10
0
0
50
100
150
True Map Distance (m x 50)
200
Mapping Function
example
1. RF = 18.5 %
m = 0.46
mu = 23.1
2. RF = 6.4 %
m = 0.137
mu = 6.8
Summary:
- short distances: use RF
- long distances: use mapping function
Linkage
Other Points:
1. No crossing over in male Drosophila
male: AaBb A B  gametes AB, ab
a b
use female dihybrid: AaBb x aabb
O
O
Linkage
2. Linkage of genes on the X chromosome:
AaBb x --Y
O
O
Male progeny:
AB Y
Ab Y
male progeny direct
aB Y
measure of female meiotic
ab Y
products
Fungal Genetics
Fungi:
important organisms in the ecosystem
- decomposers
- pathogens
important for humans
- food
- pathogens
(Biology 4040 – Mycology)
Fun Facts About Fungi
http://www.herbarium.usu.edu/fungi/funfacts/factindx.htm
Fungi
Neurospora crassa
(bread mold)
Morphological mutants
Biochemical mutants (one gene, one enzyme)
Linkage Map
Neurospora crassa Linkage group I
Fungus Life Cycle
vegetative stage haploid
+, - mating types
brief diploid stage  meiosis
n
n
+
spores
+
meiosis
n
-
2n
n
Gamete Pool
Gametes: Products of many meioses
all pooled together
A B
a b
AB AB ab ab AB ab
P AB
ab ab AB ab Ab AB
Gamete
P ab
AB aB ab ab AB AB
pool
R aB
ab AB AB ab
R Ab
Tetrad Analysis
Some Fungi and algae: 4 products of a single
meiosis can be recovered
Advantages:
1. haploid organism - no dominance
2. examine a single meiosis - test cross not needed
3. small, easy to culture
4. Tetrad Analysis - map gene to centromere
Ascus with ascospores
Tetrad Analysis
Types of Tetrads:
1. Unordered - 4 products mixed together
2. Ordered (linear) - 4 products lined up, each
haploid nucleus can be traced
back through meiosis
3. Octads - mitotic division after meiosis
8 products (2 x 4)
Linear Tetrad Analysis
Life Cycle:
+ = a+
a
a
a
+
+
a
+
a
Meiosis
+
Diploid
Haploid
Mating: a
n
+
x
+  a /+
n
2n
4 haploid
products
Linear Tetrad Analysis
a
a
a
a
+
+
+
+
8 haploid
spores
mitosis
a
a
+
+
4 haploid
products
(Octad)
Linear Tetrad Analysis
Two types of asci:
1. no crossover----> first division segregation (MI)
2. crossover between
gene and centromere-----> second division
segregation (MII)
Mapping gene to centromere
First Division
a
a
+
+
No
Crossover
a
a
a
a
+
+
+
+
First division segregation
meiosis
A
A
A
A
a
a
a
a
Mapping gene to centromere
Second division
a
a
a
+
a
+
+
a
+
a
+
+
crossover
Second division segregation
A
A
a**
a**
A**
A**
a
a
**
recombinant
st
1
and
nd
2
Division segregation
First Division
a
a
+
+
No
Crossover
a
a
a
a
+
+
+
+
Second division
a
a
a
+
a
+
+
a
+
a
+
+
Crossover
Mapping gene to centromere
I
a
a
a
a
+
+
+
+
43
+
+
+
+
a
a
a
a
43
II
a
a
+
+
a
a
+
+
3
+
+
a
a
+
+
a
a
4
+
+
a
a
a
a
+
+
3
a
a
MI = 86
+
+
MII = 14
+
+
a
a
4 Total = 100
Mapping gene to centromere
MI = 86
MII = 14
14/100 = 14 % of meioses showed a crossover
½ of the crossover products
recombinant
RF = ½ x 14 % = 7 %
a
7 m.u.
Unordered Tetrad Analysis
1. still products of a single meiosis
2. can not map gene to centromere
3. linear tetrads can be analyzed as unordered
4. map distance between linked genes
a
b
n
X
a+ b+
a b
a + b+
n
2n
meiosis
Unordered Tetrads
Three kinds of unordered tetrads:
a
a+
b
b+
meiosis
1. Parental Ditype
2. Nonparental Ditype
3. Tetratype
ab
ab
a+b+
a+b+
a b+
a b+
a+b
a+b
PD
NPD
ab
a b+
a+b
a+b+
T
PD
ab
ab
a + b+
a + b+
T
NPD
ab
a b+
a b+
a b+
a + b+
a+ b
a+ b
a+ b
Unlinked genes PD = NPD
a
b
X
+ +
ab
ab
a b
+ +
a/+, b/+
meiosis
a +
+ b
PD
a+ b+
a+ b+
a b+
a b+
a+ b
a+ b
NPD
Unordered Tetrads
Unlinked Genes: PD = NPD
Linked Genes: PD >> NPD
NPD----> all products recombinant
T--------> ½ products recombinant
PD-----> all parental type
PD = 58
RF = ½ T + NPD
T = 40
T + NPD + PD
NPD = 2
RF = 0.22
22 m.u.
Tetrad Analysis
Types of Tetrads:
1. Ordered (linear): map gene to centromere
2. Unordered:
map genes
Linkage: Summary
• Recombination: generates variation
(inter and intrachromosomal)
• Genetic maps:
- genes linked on the same chromosome
- location of new genes relative to genes
already mapped
Linkage: Summary
• Hunting for genes (Human Diseases)
- genetic markers: DNA variation
- co-inheritance with diseases using pedigree
information
- recombinants used to estimate linkage
- MUN Medical Genetics
Linkage vs. Association
• Association - test if a disease and a marker allele
show correlated occurrence in a population
• Linkage – test if disease and a marker allele
show correlated transmission within a pedigree
Linkage
Association
3,4
1,3
3,3
1,3
2,2
1,4
3,4
2,3
2,4
4,4
4,4
1,1
1,4
2,4 1,2 3,4
1,2
3,3
1,3
1,4 3,4 3,4
1,4
1,4
Genetic linkage in humans
Nail- patella
syndrome
Rare disease
inherited as a
dominant
Genetic linkage in humans
ABO Blood group marker:
Alleles: A, B, O
Genotypes: OO, AO, BO, AB
Two Genes: N-P syndrome; Blood Group
Evidence for linkage
Genetic linkage in humans
http://www.ndsu.nodak.edu/instruct/mcclean/plsc431/link
age/linkage5.htm