Course Outline
IB 201: Review Question
A phenotype ratio of 9:3:3:1 in the
offspring of a mating between two
individuals that are heterozygous
for two traits occurs when:
A. the genes reside on the same
chromosome
B. each gene contains two
mutations
C. the gene pairs assort
independently during meiosis
D. only recessive traits are scored
E. none of the above
Genetic Data Analysis:
Probability & Statistics
Deviations from Mendelism:
Epistasis; Unusual Modes of Inheritance
Chromosomal Inheritance:
Chromosomal Abnormalities; Sex Determination
Mapping:
Gene and Genome Mapping
Traits Affected by Genes & Environment
Quantitative Traits
Genes in Populations
Genetic Mechanisms of Evolution; Population Genetics of Disease
and Disease resistance
Sum Rule
Genetic Data Analysis II
Some simple rules of probability
The combined probability of two
events that are mutually exclusive is
the sum of the individual
probabilities. Clue: look for “or”
Q: What’s the probability of rolling a ‘five’
or a ‘six’ on one six-sided die?
A: 1/6 + 1/6 = 1/3
1
Genetic Example: Monohybrid Cross
P:
F1:
GG x gg
Gg
Gg x Gg ==>
F2: 1/4 GG: 1/2 Gg: 1/4 gg
Genetic Example 2: Dihybrid Cross
P: GG ww x gg WW
F1:
Gg Ww
Gg Ww x Gg Ww ==>
F2: 9/16 G-W- 3/16 G-ww 3/16 ggW- 1/16 ggww
What is the probability that the F2
offspring has the dominant phenotype (is
either GG or Gg)?
What is the probability that an F2 offspring will have the
dominant phenotype (G-ww or ggW-) for only one of the
two traits?
1/4 GG + 1/2 Gg = 3/4 G-
3/16 G-ww + 3/16 ggW- = 6/16=3/8
Product Rule
The probability of several
independent events is the product
of the individual probabilities.
Two events are independent if the
occurrence of the first event has no
effect on the probability of the second
event. Clue: look for “and”.
Q: You roll two dice. What’s the
probability of getting a ‘two’ on the
first one and a ‘five’ on the second
one.
A: 1/6 * 1/6 = 1/36
Genetic example of product rule
P: AA bb CC DD ee ff
x
aa BB cc dd EE FF
F1: Aa Bb Cc Dd Ee Ff
x
Aa Bb Cc Dd Ee Ff
Q: What proportion of F2 progeny will be
AA bb Cc DD ee Ff ?
A: 1/4 * 1/4 * 1/2 * 1/4 * 1/4 * 1/2 = 1/1024
2
Deviations from Mendelism
Manx Cats
Lethal Alleles
Epistasis
Unusual sex linkage
Sex influenced inheritance
Genetic Anticipation
Lethal alleles
F1:
Mm
x
Other lethal mutations
Achondroplasia (humans)
Mm
F1:
1 MM
2 Mm
F2:
1 Lethal: 2 Manx:
1 mm
1 Normal
Yellow body color (domestic mice)
Curly wings (Drosophila)
F2 phenotypic ratio: 2:1 instead of 3:1
3
Epistasis
Agouti:
wild type
Genetic interaction between two (or more)
loci.
One gene modifies the phenotypic effects
of another gene.
P:
F1:
BB CC
agouti
x
Bb Cc
agouti
bb cc
albino
Simple dominant phenotype?
Epistasis
Normal
dihybrid ratio
is altered from
9:3:3:1 to
9:3:4
F2: 9/16 B- C- 3/16 bb C- 3/16 B- cc 1/16 bb cc
albino
agouti
albino
black
F2 Phen. ratio: 9 agouti : 3 black : 4 albino
C and B
gene have
an epistatic
interaction
novel phenotype
4
Biochemical model
Epistasis
C enzyme present?
Colorless precursor ->
Locus 1
BB
Bb
agouti agouti
bb
black
Locus 2
CC
Cc
cc
no effect no effect albino
B
enzyme present?
Yes: CC or Cc some melanin
produced
Yes: BB or Bb agouti
No: cc
no melanin
produced
No: bb
black
CC or Cc: tyrosinase is produced (involved in
production of melanin)
BB or Bb: controls distribution of the pigment
Figure 10.18b
Figure 10.18c
Crosses between pure lines produce novel colors.
Model to explain 9 : 3 : 3 : 1 pattern observed above: Two genes interact to produce pepper color.
Parental
generation
rrYY
X
Yellow
yellow
Genotype
RRyy
R-Y-yy?
Brown
---------------->
brown pepper
Colorgreen ----------------->
Explanation of color
chlorophyll Red
rrY-
R?
pigment
Red red
pigment
+ no chlorophyll
Yellow
pigment + no chlorophyll
yellow -------------------> yellow Yellow
------------------>
orange
F1 generation
Codominance?
R-Y-
R-yy
Brown
rryy
Y? No chlorophyll
Green
Red pigment + chlorophyll
R?
red pigment
Yellow pigment + chlorophyll
Red
Self-fertilization
F2 generation
R-YRed
9/16
rrYYellow
3/16
R-yy
rryy
Brown
3/16
Green
1/16
Gene 1
Gene 2
R = Red
Y = Absence of green (no chlorophyll)
r = Yellow
y = Presence of green (+ chlorophyll)
(-) = R or r
(-) = Y or y
5
Practice Problem
Other kinds of epistasis
9/16 A-B-
3/16 A-bb
3/16 aaB-
1/16 aabb
In Labrador retrievers, coat color is controlled by two
loci each with two alleles B,b and E,e respectively.
When pure breeding Black labs with genotype BB EE
are crossed with pure breeding yellow labs of
genotype bb ee the resulting F1 offspring are black.
F1 offspring are crossed (Bb Ee x Bb Ee). Puppies
appear in the ratio:
9/16 black;
3/16 chocolate;
chocolate;
9/16 B- EB- E-
3/16 B- ee
B- ee
4/16=1/4 yellow.
3/16 bb E- 1/16 eebb
bb E- and bb ee
What genotypes correspond to these three
phenotypes?
Practice Problem
In the summer squash (Cucurbita pepo) fruit shape is
determined by two genes. Two different true-breeding
spherical types were crossed. The F1's were all disk,
and the F2's segregated 35 disk, 25 spherical and 4
long. Explain these results.
What’s the first step?
Notice novel phenotype: disk, long.
What’s the next step?
Notice there are three F2 phenotypes. What kind of
inheritance will give three F2 phenotypes?
Genetic Model?
Incomplete, codominance
Epistasis
Expected F2 ratio?
1:2:1
Variation on 9:3:3:1
Hint: usually given numbers, not fractions
27 agouti; 12 albino; 9 black
28 agouti; 11 albino; 4 black
Practice Problem, cont.
In the summer squash (Cucurbita pepo) spherical fruit is
recessive to disk, True-breeding spherical types from different
geographic regions were crossed. The F1's were disk, and the
F2's segregated 35 disk, 25 spherical and 4 long. Explain
these results.
Are the phenotypic ratios closer 1:2:1 or to a
variant of 9:3:3:1 ?
If phenotypic ratios closer to a variant of 9:3:3:1,
then what variant is it?
Total # of individuals = 35 + 25 + 4 = 64
64/16 = 4
9*4 = 36
6*4 = 24
1*4 = 4
Phenotypic ratio close to 9:6:1
6
Practice Problem, cont.
In the summer squash (Cucurbita pepo) spherical fruit is
recessive to disk, True-breeding spherical types from different
geographic regions were crossed. The F1's were disk, and the
F2's segregated 35 disk, 25 spherical and 4 long. Explain
these results.
If phenotypic ratios are close to 9:6:1, then what are
the genotypes associated with each phenotype?
35 disk
9/16 A- B-
25 spherical
3/16 A- bb + 3/16 aa B-
Sex Linkage: mammals, flies
XAXa
Diploid
Adults
Gametes
Xa
XA
XA
Y
XA
XAXA
XAY
Xa
XAXa
XaY
XA
4 long
1/16 aa bb
What were the genotypes of the original spherical
parents?
Male
Female
AA bb aaBB
Heterogametic
Sex
XAY
Y
Sex Linkage: birds, butterflies
Diploid
Adults
Gametes
Heterogametic
Sex
ZBW
ZBZb
Female
Male
W
ZB
ZB
Zb
ZB
ZBZb
ZBZb
W
ZBW
ZbW
ZB
Male
Female
Homogametic
Sex
Zb
7
Hairy ears
Y-linked inheritance
Sex influenced phenotype
Male pattern
baldness: what kind
of inheritance?
Genotype
bb
bb’
b’b’
Female
Bald
Not bald
Not bald
Male
Bald
Bald
Not bald
8
Siamese or “Himalayan”
Environment-dependent
dependent expression of a
genotype
Different allele of the C locus that causes albinism.
Temperature sensitive.
Phenotypes are not always a
direct reflection of genotypes
Temperature-sensitive alleles: Siamese
color pattern
Nutritional effects: phenylketonuria
Genetic anticipation: several genetic
diseases
Phenylketonuria
Nutritional defect: can’t metabolize
phenylalanine.
Can lead to severe physical and mental
disorders in children, but only if they
consume phenylalanine.
Disease phenotype can be avoided by
eliminating phenylalanine from the diet
9
Genetic Anticipation
Fragile X syndrome
Symptoms: delayed
development & mental
retardation. More severe
in males than females
Caused by expansion of
triplet repeat (CGG) in a
gene on the long arm of
the X chromosome
Named for breakage of X
chromosome in cell
preparations.
Huntington disease
Fragile-X syndrome
Kennedy disease
Myotonic muscular
dystrophy
Fragile X
Normal range: 7-52 (average=30)
“Pre-mutation”: 60-200 repeats
Full Mutation: > 230-1000s. DNA
becomes abnormally methylated,
promoter is inactivated, and gene
silenced.
Pre-mutation is unstable: maternallyinherited premutation with >100
repeats almost always expands to a full
mutation
Genetic Anticipation: Fragile X
Most common kind of inherited
mental retardation.
Named for “fragile site”
Due to expansion of 3-base pair
repeat (CGG) in a gene near the
tip of the long arm of X
chromosome.
10
Fragile X
Pre-mutation is unstable: maternally-inherited premutation with >100
repeats almost always expands to a full mutation
Huntington Disease
Autosomal dominant lethal (chromosome 4)
Progressive neurological deterioration
First symptoms appear after reproductive age
One of 8 known neurodegenerative diseases caused by expansion of (CAG)
repeats
All show inverse correlation with age of onset and number of repeats.
Genetic Anticipation
causes subsequent
generations in a family to
be more severely affected
by a disease. It does this
by increasing the number
of triplet repeats in the
fragile area of the X
chromosome through the
generations.
Huntington Disease
Autosomal dominant lethal (chromosome 4)
Progressive neurological deterioration
First symptoms appear after reproductive age
One of 8 known neurodegenerative diseases caused by expansion of (CAG)
repeats
All show inverse correlation with age of onset and number of repeats.
11
Huntington Disease
Autosomal dominant lethal (chromosome 4)
Progressive neurological deterioration
First symptoms appear after reproductive age
One of 8 known neurodegenerative diseases caused by expansion of (CAG)
repeats
All show inverse correlation with age of onset and number of repeats.
Which is the pedigree of
autosomal dominant (like HD)
12