B2250
Readings and Problems
***Note: Quiz #2: Thurs. Oct. 28
Ch. 4 p. 100 – 112
Ch. 5 p. 118 – 137
Ch. 6 p. 148 – 165
Prob: 10, 11, 12, 18, 19
Prob: 1 – 11, 13
Prob: 1-5, 7, 8, 10, 11, 14
Practice problems:
http://www.mun.ca/biology/dinnes/B2250/B2250.html
(also download Lab. #3 and Print out 3 files)
Genetics in the news
“Junk DNA??”
http://www.jgi.doe.gov/science/highlights/nobrega
1004.html
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
Two Characters
Monohybrid Cross
parents differ for a single character
(single gene ); seed shape
Dihybrid Cross
parents differ for two characteristics
(two genes)
Dihybrid
Two Characters:
1. Seed colour yellow green
Y
y
2. Seed shape Round wrinkled
R
r
4 phenotypes
Dihybrid
P
RRyy
Gametes
F1
X
Ry
rrYY
rY
RrYy
DIHYBRID
F1 Dihybrid ----->F2
F1
RrYy
RrYy
F2
9
3
3
1
Total
315
108
101
32
556
X
RrYy
round, yellow
round, green
wrinkled, yellow
wrinkled, green
Individual Characters
1. Seed shape
round : wrinkled
423 : 133
3: 1
(¾ : ¼)
2. Seed colour
yellow : green
416 : 140
3 : 1
Conclusion
* 3 : 1 monohybrid ratio for each character
* 9 : 3 : 3 : 1 phenotypic ratio a random
combination of 2 independent 3:1 ratios
Two Independent Genes
F2
colour
yellow 3/4
green
1/4
seed shape
3/4
1/4
round
wrinkled
9/16
3/16
3/16
F2
1/16
Phenotypes
Independent Assortment
Test Cross
AaBb
X
gametes
ab
1/4 AB AaBb
1/4 Ab Aabb
4 gamete
types
1/4 aB
aaBb
1/4 ab
aabb
aabb
4 phenotypes
4 genotypes
(Genes)
Meiosis I
A
Correlation of genes and
Chromosomes during
meiosis
a
4 gamete types
A
B
A
b
OR
a
b
a
B
Mendel’s Second Law
Independent assortment:
during gamete formation, the segregation of
one gene pair is independent of other gene
pairs.
Linkage
Chapter 6
- recombination
- linkage maps
Linkage of Genes
- Many more genes than chromosomes
- Some genes must be linked on the same
chromosome; therefore not independent
Independent Assortment
F1
AaBb
X
AaBb
Genotypes
AABB
AaBb
AaBB
F2
9
3
4 phenotypes
3
1
A-BA-bb
aaBaabb
AABb
Aabb, AAbb
aaBb, aaBB
Independent Assortment
Test Cross
AaBb
X
gametes
ab
1/4 AB AaBb
1/4 Ab Aabb
1/4 aB
aaBb
1/4 ab
aabb
aabb
4 phenotypes
4 genotypes
Independent Assortment
Fig 6-6
AB
ab
Ab
aB
Interchromosomal Recombination
Complete Linkage
P
X
A
F1
F1 gametes
B
a
b
A
B
a
b
AaBb
A
B
AB
dihybrid
Parental
Parental
a
b
ab
Recombinant Gametes ?
Crossing over:
- exchange between homologous chromosomes
Crossing over in meiosis I
Meiosis I
- homologous chromosomes pair
- reciprocal exchange between non-sister
chromatids
Crossing over in meiosis I (animation)
Gamete Types
F1
gametes
A
B
a
b
A
a
A
a
B
b
b
B
AaBb
AB
ab
Ab
aB
Parental
Parental
Recomb.
Recomb.
1. Ways to produce dihybrid
P
Cis
A B
A B
X
Note:
Chromatids
omitted
a b
a b
A B
a b
Gametes:
AB
ab
Ab
aB
AaBb
(dihybrid )
P
P
R
R
2. Ways to produce dihybrid
a B
X
a B
AaBb
A b trans
(dihybrid )
a B
Gametes:
P
Ab
P
aB
R
AB
R
ab
P
A b
A b
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
Independent Assortment
Fig 6-6
Interchromosomal
Linkage
Fig 6-11
Intrachromosomal
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)
Summary
Mendelian Genetics:
Monohybrid cross (segregation): Dihybrid Cross (Indep. Assort.):
- ratios (3:1, 1:2:1, 1:1)
- ratios (9:3:3:1, 1:1:1:1)
- dominance, recessive
- linkage (deviation from I.A.)
- autosomal, sex-linked
- recombination
- probability
- linkage maps
- pedigrees
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)
?
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
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
ct 592
+ 45
ct 40
ct 89
+
94
ct
3
+
5
1448
Parental
Parental
Recombinant
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
Recombinant
Parental
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
Recombinant
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
Parental
Recombinant
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
+
cv
v
+
+
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)
Q. Are multiple crossovers
independent ?
Example: v
13.2
ct
6.4 cv
Prob.(single recombinant v---ct) = 0.132
Prob.(single recombinant ct---cv) = 0.064
If independent then:
Prob. (double recombinant)= 0.132 x 0.064
= 0.0084
expected number of doubles = 1448 x 0.0084
= 12
Double Cross Overs
Expect 12
Observe 8
Explanation: a crossover in one region
reduces the probability of a second
crossover in an adjacent region
Interference
Interference (I)
1. Coefficient of coincidence (CC)
Obs # double recombinants
CC =
Exp # double recombinants
2. Interference: I = 1 - CC
= 1 - (8 /12) = 1/3 = 33 %
I = 1 interference complete
I = 0 no interference
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
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:
map units (mu) = m x 50
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
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
dominant 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)
Figure 6-5
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
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
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
MI = 86
a +
a +
MII = 14
a +
a +
+ a
+ a
3 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
+ +
n
2n
meiosis
Unordered Tetrads
Three kinds of unordered tetrads:
a
b
ab
a b+
+
+
ab
a b+
meiosis
a+b+
a+b
+b+
+b
a
a
1. Parental Ditype
2. Nonparental Ditype
3. Tetratype
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 +
a +
+ b
+ 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