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/
Weekly Online Quizzes
Marks
Oct. 14 - Oct. 22 Example Quiz 2** for logging in
Oct. 21- Oct. 24 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
Sex-linked Inheritance
Correlation between inheritance
of genes and sex
Drosophila melanogaster
(T. H. Morgan)
White eye
(mutant)
Red eye
(wild)
X
Drosophila
Red Eye
White eye
Cross A
red female
F1
F2
No whiteeyed females
X
white male
all red
red : white
3 : 1
white all male
red 2 : 1 female : male
How to obtain a white-eyed female?
Cross B
white female
F1
X
females
F2
red male
males
females
1
:1
males
: 1
:1
Cross A
Xw+Xw+
w+w+
ww+
F1
ww+
X
X
Xw Y
wY
w+Y
wY
w
w+
w
ww
ww+
Y
wY
w+Y
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
Principle of Segregation
Implications
Answer questions on inheritance:
- mode of inheritance (dominant, recessive
sex-linked)
- paternity
- hybridization
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
Rules of Probability
1. Product rule (AND):
probability of 2 independent events occurring
simultaneously
2. Sum Rule (OR):
probability of either one of two mutually
exclusive events
Probability
1. **Coin toss: P (T) = P(H) = 1/2
P(T, T, T) = P(T) and P(T) and P(T)
= (½)3 = ½ x ½ x ½
2. One die: P (6) or P (5) = 1/6 + 1/6
**http://shazam.econ.ubc.ca/flip/
Self F1
F2
eggs
Aa
X
Aa
equal
Segregation
P(A) = ½
P(a) = ½
sperm
1/2 A
1/2 a
1/2 A 1/4 AA 1/4 Aa
1/2 a 1/4 Aa 1/4 aa
Prob. (AA or Aa) = 1/4 + 2/4 = ¾
Prob. (aa) = ¼
P(AA) = ½ * ½
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
Applying Probability to
Genetics
Dihybrid: RrYy
Hypothesis:
mechanism for putting R or r into a gamete is
independent of the mechanism for putting Y or
y into a gamete
Gametes from Dihybrid
Dihybrid: RrYy (F1)
Principle of segregation during gamete
formation:
Yy -------> P(Y) = P(y) = 1/2
Rr ------->P(R) = P(r) = 1/2
Gametes from dihybrid
RrYy :
Y and R
Y and r
y and R
y and r
probability
1/2 * 1/2 = ¼
1/2 * 1/2 = ¼
1/2 * 1/2 = ¼
1/2 * 1/2 = ¼
4 gamete types
YR
Yr
yR
yr
F1 gametes produce F2
F1
YyRr
X
¼ YR
¼ YR
gametes Yr
yR
yr
YyRr
gametes
Yr yR
yr
1/16 RRYY
F2
Sperm
F2
Fig. 6-7
Eggs
F2
4 Gametes
9 Genotypes
4 Phenotypes
Mendel’s Second Law
Independent assortment:
during gamete formation, the segregation of
one gene pair is independent of other gene
pairs.
(Genes)
Meiosis I
A
Correlation of genes and
Chromosomes during
Meiosis I
a
A
B
A
b
OR
a
b
a
B
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
F1
F2
AaBb
9
3
3
1
X
A-BA-bb
aaBaabb
AaBb
4 phenotypes
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
Independent Assortment
Any number of independent genes:
Genes
1
2
3
n
Phenotypes
2
4
8
2n
Genotypes
3
9
27
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)
http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?db=OMIM
Brachydactyly
Bb
bb
Bb short fingers
bb normal
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)