Download Genetics --- introduction

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
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 – Oct. 31 Quiz 2
2
Nov. 4
Quiz 3
2
Nov. 10
Quiz 4
2
Mendelian Genetics
Topics:

-Transmission of DNA during cell division




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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
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
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
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 (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
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
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
Practice Questions:
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?
Practice Questions:
1. Mode of inheritance: every generation; father to daughter: sex-linked dominant
2. Probability that the indicated female will produce an affected child?
Aa
x
aY  ¼ Aa ¼ aa
¼ AY ¼ aY
Prob. = 1/2
Quiz 2 Questions
Quiz 2 Answers:
http://webct.mun.ca:8900/
Using Linkage to Hunt for
Human Disease Genes
Basic approach:
1. Collect pedigree information on disease
2. Collect blood samples from individuals
3. Correlate genetic marker information with
disease
4. Use recombinants to map gene and marker
Huntington’s Disease
Huntington’s Disease
Not sex-linked
Autosomal Dominant
Huntington’s Disease
Linked marker
Huntington’s Disease
As a result of the gene discovery, a direct genetic
test for Huntington's disease has replaced the
indirect linkage marker test.
While the Huntington's disease gene discovery
alters the technical aspects of predictive testing for
Huntington's disease, there is still no cure for
Huntington's disease and no available treatment to
delay its onset or to slow, stop or reverse the
disease's relentless progression.
Hunting for Human Disease
Genes
Newfoundland Population
Small founding population – high freq. of allele
Isolated – little gene flow
Inbreeding – increased chance of “aa”
Good pedigree records
Rare Recessive
a
Rare = AA
A(AA or Aa)
Cousins
(inbreeding)
Rare
autosomal
recessive
BBS1
Chr - 11
B10
Aa
Aa
Chromosome 11
aa aa
Aa
AA Aa
Aa
AA 3 (2.5)
Aa AA Aa AA aa
Aa 5 (5)
aa 2 (2.5)
aa
Genetic
Diseases
dominant
Gene Discovery
Genetic marker and linkage analysis
Narrow location of gene (chromosome and region)
Genome sequencing  identify gene
Genetic counseling, gene therapy??
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
mutant
v
vermillion
cv
crossveinless
ct
3. Wing
cut wing
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
Three point test cross
Observations:
13.2
map v
6.4
ct
cv
18.5
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
8 gamete types
(correct order)
F1 v/+ ct/+ cv/+
ct - cv
v
+
v
+
v
+
v
+
+
ct
+
ct
ct
+
ct
+
Parental = non crossover
+ 580
cv 592
cv 45
+
40
cv 89
+
94
cv
3
cv
5
1448
Parental
Parental
(most frequent)
Single cross over
Single cross over
Double cross over
8 gamete types
(correct order)
F1 v/+ ct/+ cv/+
v - ct
v
+
v
+
v
+
v
+
+
ct
+
ct
ct
+
ct
+
Parental = non crossover
+ 580
cv 592
cv 45
+
40
cv 89
+
94
cv
3
cv
5
1448
Parental
Parental
(most frequent)
Single cross over
Single cross over
Double cross over
8 gamete types
(correct order)
F1 v/+ ct/+ cv/+
v - ct - cv
v
+
v
+
v
+
v
+
+
ct
+
ct
ct
+
ct
+
Parental = non crossover
+ 580
cv 592
cv 45
+
40
cv 89
+
94
cv
3
cv
5
1448
Parental
Parental
(most frequent)
Single cross over
Single cross over
Double cross over
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