Biology 2250
Principles of Genetics
Announcements
Lab 4 Information: B2250 (Innes) webpage
download and print before lab.
Virtual fly: log in and practice
http://biologylab.awlonline.com/
B2250
Readings and Problems
Ch. 4 p. 100 – 112 Prob: 10, 11, 12, 18, 19
Ch. 5 p. 118 – 129 Prob: 1 – 3, 5, 6, 7, 8, 9
Ch. 6 p. 148 – 165 Prob: 1, 2, 3, 10
Ch. 7 p. 188 – 191 (Bacterial conjugation)
Weekly Online Quizzes
Oct. 14 - Oct. 25
Oct. 21- Oct. 25
Oct. 28 – Oct. 31
Nov. 4 – Nov. 7
Nov. 10
(Midterm 2
Marks
Example Quiz 2** for logging in
Quiz 1
2
Quiz 2
2
Quiz 3
2
Quiz 4
2
Thursday Nov. 17)
Questions for practice
1. Gene A and gene B are linked. A test cross produces 10 AaBb
progeny out of a total of 100. The estimated map distance
between gene A and B is: a. 10 b. 20 c. 30 d. 40 e. 50
2. For the pedigree, indicate the most probably mode of inheritance
for the rare trait.
3. For the pedigree, what is the probability that the indicated female
will produce an affected child?
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
Quiz – 2 answers
http://webct.mun.ca:8900/
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
- Independent Assortment
- Mendelian genetics in humans
- Linkage
- Gene mapping
-Gene mapping in other organisms
(fungi, bacteria)
- Extensions to Mendelian Genetics
- Gene mutation
- Chromosome mutation
- Quantitative and population genetics
Linkage Maps
Useful:
- studying recombination (variation)
- study the structure of the genome
- study gene interaction (cis, trans)
- diagnosis (marker gene linked to disease gene)
- constructing particular genetic combinations
Linkage Maps
Mapping two genes:
Test cross: AaBb x aabb
% RF = map distance
3 point test cross:
- multiple crossovers undetected
- underestimate true map distance
Gene Mapping
Requirements:
1. Genetic material from two different individuals
2. Recombination
Examples: fungi
bacteria
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
Independent Assortment
Diploids  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
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.
Tetrad Analysis
Tetrads:
Ordered (linear): map gene to centromere
Sex in Bacteria
E. coli
Haploid
conjugation
Origin of Plasmid
genes from
Lactococcus lactis
Bacteria used to
make cheese and
yogurt
Plasmids: location
of antibiotic
resistant genes
Recombination in Bacteria and
viruses
Human Health:
- antibiotic resistance
- new strains of bacterial and viral
diseases (bird flu)
-horizontal gene transfer (between species)
Linkage: Summary
• Recombination: generates new combinations
(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
Extensions to Mendelian Genetics
Ch. 14 From Gene to Phenotype
Readings: Ch. 14 p. 454 – 473
Problems: Ch. 14: 2, 3, 4, 5, 6, 7
Chapter 1
Genes, environment, organism
Phenotype =
gene + env. + gene x env. + gene x gene
Mendelian Genetics:
Genotype
Phenotype
Dominance ?
G x E interaction
Extensions to Mendelian
Genetics
(Gene  Phenotype)
1. Dominance
2. Multiple alleles
3. Pleiotropy
4. Epistasis (gene interaction)
5. Penetrance and expressivity
Gene interaction
1. Alleles at one gene
Dominance
2. Different genes
Epistasis
1. Dominance
Location of heterozygote between
two homozygotes
1. Complete
2. No dominance
3. Incomplete (partial)
4. Codominance
Homozygotes: A1A1 A2A2
Heterozygote: A1A2
Incomplete Dominance
red
white
pink
Codominance
Human Blood Groups:
Genotype
Phenotype**
AA
A
AB
AB co-dominance
BB
B
** antigen protein on RBC
Codominance
Molecular Markers
Allele
A
B
AB
AA
BB
BB
Heterozygote distinguished from homozygotes
2. Multiple Alleles
(ABO Blood groups - 3 alleles)
Genotype
Phenotype
(6)
(4)
--------------------------------------------OO
O
recessive
AA, AO
A
dominant
BB, BO
B
dominant
AB
AB co-dominant
---------------------------------------------
Multiple alleles
in clover
Test for Allelism
Possibilities:
or
1. alleles for the same gene - all crosses show
mendelian ratios (1:1 3:1 1:2:1)
2. more complex inheritance (> 1 gene)
Example: white, yellow, pink
Cross
white x yellow
white x pink
yellow x pink
F1
yellow
pink
pink
3 alleles: w y
p
6 genotypes: w w y y p p
F2
3:1 yellow : white
3:1 pink : white
3:1 pink : yellow
pw
yw
yp
3. Pleiotropy
(one gene affects > 1 trait)
Example: Mouse
Gene affects:
1. coat colour (
2. survival
AA
Homozygous wildtype
dark
, yellow)
Yellow
Parents
Crosses
A.
x
-----> all
B.
x
---> 1/2
1/2
C.
x
----> 2/3
1/3
Explanation
A. AA
B. AA
x
AA
all AA
x AYA
C. AYA x AYA
½ AYA , ½ AA
¼ AA ½ AYA ¼ AYAY
1
1/3
:
2
2/3
dies
Interpretation
Gene affects both coat colour and
survival
1. AY dominant to A for coat colour
2. AY recessive lethal for survival
Pleiotropy
Genotype
AA
A AY
AY AY
Phenotype
coat colour
survival
dark
dark
yellow
?
live
live
die
G
+
E =
P
Trait 1
Pleiotropy
Gene A
Trait 2
Epistasis
Gene A
Trait
Gene B
Gene interaction
4. Epistasis
(gene interaction)
More than one gene affects a character
One gene pair masks or modifies the
expression of another gene pair
AABB
x aabb ----> AaBb x AaBb ---> F2
F1
Dihybrid
F2
AaBb
Gene A and B
unlinked
x
A- BA- bb
aa Baa bb
AaBb
9/16
3/16
3/16
1/16
4 distinct
phenotypes (2 traits)
(peas: shape, colour)
Epistasis: Gene A and Gene B interact  phenotype of 1 trait
1.
Epistasis
(BbEe X BbEe)
Labrador retriever Coat Colour (B and E genes)
F2 Ratio
9/16
3/16
3/16
1/16
Genotype Phenotype
B- Eblack
B- ee
gold
bb Ebrown
bb ee
gold
Gene E allows colour deposition
Ratio
9/16
4/16
3/16
Epistasis
Allele E
Allele B
Golden
brown
B- ee
bb ee
bb E-
black
B- E-
2.
Epistasis
(AaBb X AaBb)
Example: Flower petal colour
F2 Ratio
9/16
3/16
3/16
1/16
Genotype
A- BA- bb
aa Baa bb
Phenotype
Purple
White
White
White
Ratio
9/16
7/16
Gene B
colourless
(white)
A-bb
aabb
Gene A
colourless
(white)
purple
aaB-
A- B-
5. Penetrance and Expressivity
Phenotype: genotype, genetic background,
and environment
Variable Expression:
Penetrance
Expressivity
Penetrance:
percentage of individuals that show some
degree of expression of a mutant genotype
Example: Polydactyly (P)
extra digits
pp
normal
Pp
PP
10 % normal polydactyly
90 % polydactyly
Expressivity:
degree that a given genotype is expressed
phenotypically
Example: Pp individuals which do express
the extra digits can vary
(a) extra digit on each hand and foot
(b) extra digit on one hand only
(c) complete digit or vestige
Same
genotype
Variable expressivity of
piebald spotting in beagles
Summary
- segregation and independent assortment
can explain a variety of patterns of
genetic variation
- Phenotype = Genotype + Environment
Genetic interaction: genotype, epistasis,
genetic background