Biology 2250
Principles of Genetics
Announcements
Lab 3 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
Ch. 5 p. 118 – 129
Ch. 6 p. 148 – 165
Prob: 10, 11, 12, 18, 19
Prob: 1 – 3, 5, 6, 7, 8, 9
Prob: 1, 2, 3, 10
Weekly Online Quizzes
Marks
Oct. 14 - Oct. 25 Example Quiz 2** for logging in
Oct. 21- Oct. 25 Quiz 1
2
Oct. 28
Quiz 2
2
Nov. 4
Quiz 3
2
Nov. 10
Quiz 4
2
Weekly Online Quizzes
Results
Example quiz:
Quiz 1:
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
- Tetrad Analysis (mapping in fungi)
- Extensions to Mendelian Genetics
- Gene mutation
- Chromosome mutation
- Quantitative and population genetics
Mendelian Inheritance
Determining mode of inheritance:
- single gene or more complicated
- recessive
or dominant
- sex linked or autosomal
- probability
Approach:
cross parents
observed progeny
compare with expected
Mendelian Genetics
in Humans
Determining mode of inheritance
Problems:
1. long generation time
2. can not control mating
Alternative:
* information from matings that have
already occurred “Pedigree”
Human Pedigrees
Pedigree analysis:
•
•
•
•
trace inheritance of disease or condition
provide clues for mode of inheritance
(dominant vs. recessive)
(autosomal vs. sex linked)
however, some pedigrees ambiguous
determine probability
Mendel’s Second Law
Independent assortment:
during gamete formation, the segregation of
one gene pair is independent of other gene
pairs.
Genes independent because they are on
different chromosomes
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
Inferred F1
gamete types
Fig 6-6
AB
ab
Ab
aB
Interchromosomal Recombination
(Genes)
Meiosis I
A
Correlation of genes and
Chromosomes during
meiosis
a
4 gamete types
A
B
A
b
OR
a
b
a
B
Linkage of Genes
- Many more genes than chromosomes
- Some genes must be linked on the same
chromosome; therefore not independent
Complete Linkage
P
X
A
F1
B
a
A
B
a
b
b
AaBb
dihybrid
AB
F1 gametes
A
B
AB
ab
a
b
ab
parental
Recombinant Gametes ?
Crossing over:
- exchange between homologous chromosomes
Crossing over in meiosis I
Meiosis I
- homologous chromosomes pair
- reciprocal exchange between non-sister
chromatids
Ch 4 meiosis animation:
http://www.whfreeman.com/mga/
Crossing over in meiosis I (animation)
Gamete Types
F1
A
B
X
gametes
a
b
A
a
A
a
B
b
b
B
AaBb
AB
ab
Ab
aB
Parental
Parental
Recomb.
Recomb.
Two Ways to produce dihybrid
1
AABB
A B
A B
2
AAbb
A b
A b
aabb  AaBb
x
X
a b
a b
aaBB  AaBb
x
X
a B
a B
Note:
Chromatids
omitted
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
P
AaBb
(dihybrid )
P
P
R
R
A b
A b
a B
X
a B
A b trans
a B
Gametes:
Ab
aB
AB
ab
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
Alfred Sturtevant (1913)
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
Linkage maps
Tomato
Drosophila
Linkage group = chromosome
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
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
1
2
3
Three genes
Wild (+)
1. Eye
colour
2.Wing
3. Wing
mutant
v
cv
ct
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
Parental = non crossover
+ 580
ct 592
+ 45
ct 40
ct 89
+
94
ct
3
+
5
1448
Parental
Parental
(most frequent)
Recombinant
8 gamete types
Examine
two genes
at a time
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
Correct gene order
+
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)