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/
people that have ALREADY picked up
their exams should see Dr. Carr for a re-marking
of p. 2
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
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
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
Approach:
cross parents
observed progeny
compare with expected
Mendel’s First Law
Equal segregation of two members of a gene pair
½ A gametes
Meiosis:
diploid nucleus divides
Aa
½ a gametes
produces haploid nuclei
P(a) = ½
P(A) = ½
Mendel’s Second Law
Independent assortment:
during gamete formation, the segregation of
one gene pair is independent of other gene
pairs.
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
Producing the
F2
F1
YyRr
F2
X
YyRr
1. F1 Gametes produce F2
2. Genotypes
3. Phenotypes
Independent Assortment
Two gene systems:
1. Gametes from dihybrid 4 x 4 = 16
YyRr:
Male gametes
¼ YR
Yr yR yr
¼ YR 1/16YYRR
Female
Yr
F2
gametes
yR
yr
Independent Assortment
2.
F2 Genotypes 3 x 3 = 9
YyRr X YyRr
¼ RR
¼ YY
½ Yy
¼ yy
½ Rr
1/16 YYRR
F2
¼ rr
Independent Assortment
3. F2 Phenotypes 2 x 2 = 4
YyRr X YyRr
¾ Y¼ yy
¾ R- ¼ rr
9/16 R-Y-
F1
YyRr
x
YyRr
YY RR
YY Rr
Yy RR
Yy Rr
Y-R-
YY rr
Yy rr
Y-rr
yy RR
yy Rr
yyRyyrr
yy rr
9 Genotypes
4 phenotypes
Independent Assortment
Any number of independent genes:
Genes
1
2
3
n
Phenotypes
2
4 (2 x2)
8 (2x2x2)
2n
Genotypes
3
9 (3 x 3)
27 (3 x 3 x 3)
3n
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
Human Pedigrees
1. Ambiguous:
Affected
female
Normal
male
2. Unambiguous:
Normal
female
Clues (non sex-linked)
Recessive:
1. individual expressing trait has two
normal parents
2. two affected parents can not have an
unaffected child.
Rare Recessive
Rare = AA
A(AA or Aa)
Cousins
(inbreeding)
Clues
Dominant:
1. every affected person has at least one
affected parent
2. each generation will have affected
individuals
Dominant
Not AA
All genotypes known
Examples
Recessive:
- phenylketonuria (PKU)
- hemophilia (sex linked)
- cystic fibrosis
- albinism
Dominant:
- huntingtons chorea
- brachydactyly (short fingers)
- polydactyly (extra fingers)
- achondroblasia (dwarf)
2n = 46
http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?db=OMIM
Brachydactyly
Bb
bb
Bb short fingers
bb normal
http://omia.angis.org.au/
Online Tutorial:
http://www.biology.arizona.edu/mendelian_genetics/mende
lian_genetics.html
Solving Genetics Problems
1. Don’t panic!
2. Carefully read the problem
3. What information is given? Know
the terms used.
4. What aspect of genetics does the
problem address?
Sex Linked Inheritance
X-linked Dominant
1. affected male ---> all daughters affected
no sons
aa x AY ----> Aa, aY
2. affected female ----> ½ sons, ½ daughters
affected
Aa x aY ----> AY, aY, aa, Aa
*
*
X-Linked Dominant
1.
All daughters
affected, no
sons
2.
1/2 daughters
affected, 1/2
sons affected
X-linked Inheritance
X-linked recessive:
1. more males than females show
recessive phenotype
2. affected female ------> both mother
and father have recessive allele
A a x a Y --------> a a
X-linked Inheritance
X-linked recessive:
3. affected male ----> mother carries allele
A a x AY -----> a Y
carrier
4. affected male -----> no affected offspring
AA x a Y ----> AY, Aa
carrier
X-Linked Recessive
Mother
carrier
Sex Linked Inheritance
(examples)
X linked genes
Humans: - colour blindness
- hemophilia
•
•
More common in males (hemizygous aY)
X linked recessives expressed
Queen
Victoria
(carrier)
QE II
Hemophilic male
Carrier female
X-linked recessive hemophilia
X – linked disease genes
Mendelian Genetics
Topics:
-Transmission of DNA during cell division
Mitosis and Meiosis
- Segregation (Monohybrid)
- Sex linkage
- Inheritance and probability
- Independent Assortment (Dihybrid)
- Mendelian genetics in humans (Pedigree)
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
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
Ch 4 meiosis animation:
http://www.whfreeman.com/mga/
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
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)
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
Linkage maps
Drosophila
Tomato