Transcript Welcome to the Genetics portion of IB 201!

Welcome to IB 201!
Genetics and Evolution
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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
Course Outline
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
Genetic Data Analysis II
Some simple rules of probability
Sum Rule
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
Genetic Example: Monohybrid Cross
P:
F1:
GG x gg
Gg
Gg x Gg ==>
F2: 1/4 GG: 1/2 Gg: 1/4 gg
What is the probability that the F2
offspring has the dominant phenotype (is
either GG or Gg)?
1/4 GG + 1/2 Gg = 3/4 G-
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 an F2 offspring will have the
dominant phenotype (G-ww or ggW-) for only one of the
two traits?
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
F1: Aa Bb Cc Dd Ee Ff
x
x
aa BB cc dd EE FF
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
Deviations from Mendelism
Lethal Alleles
Epistasis
Unusual sex linkage
Sex influenced inheritance
Genetic Anticipation
Manx Cats
Lethal alleles
F1:
Mm
x
Mm
F1:
1 MM
2 Mm
F2:
1 Lethal: 2 Manx:
1 mm
1 Normal
F2 phenotypic ratio: 2:1 instead of 3:1
Other lethal mutations
Achondroplasia (humans)
Yellow body color (domestic mice)
Curly wings (Drosophila)
Epistasis
Genetic interaction between two (or more)
loci.
One gene modifies the phenotypic effects
of another gene.
Agouti:
wild type
P:
F1:
BB CC
agouti
x
Bb Cc
agouti
bb cc
albino
Simple dominant phenotype?
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
novel phenotype
Epistasis
Normal
dihybrid ratio
is altered from
9:3:3:1 to
9:3:4
C and B
gene have
an epistatic
interaction
Epistasis
Locus 1
BB
Bb
agouti agouti
bb
black
Locus 2
CC
Cc
cc
no effect no effect albino
Biochemical model
C enzyme present ?
Colorless precursor
->
B enzyme present ?
Yes: CC or Cc some melanin
pro du ced
Yes: BB or Bb agouti
No: cc
pro du ced
No:
no melanin
bb
black
CC or Cc: tyrosinase is produced (involved in
production of melanin)
BB or Bb: controls distribution of the pigment
Figure 10.18b
Crosses between pure lines produce novel colors.
Parental
generation
rrYY
X
Yellow
F1 generation
RRyy
Brown
Codominance?
R-YRed
Self-fertilization
F2 generation
R-Y-
rrY-
R-yy
rryy
Red
9/16
Yellow
3/16
Brown
3/16
Green
1/16
Figure 10.18c
Model to explain 9 : 3 : 3 : 1 pattern observed above: Two genes interact to produce pepper color.
yellow
Genotype
R-Y-yy?
---------------->
brown pepper
Colorgreen ----------------->
Explanation of color
chlorophyll Red
rrY-
Yellow
R-yy
Brown
rryy
Y? No chlorophyll
Green
Red red
pigment
+ no chlorophyll
R?
pigment
pigment + no chlorophyll
yellow -------------------> yellow Yellow
------------------>
orange
Red pigment + chlorophyll
R?
red pigment
Yellow pigment + chlorophyll
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
Practice Problem
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;
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?
Other kinds of epistasis
9/16 A-B-
3/16 A-bb
3/16 aaB-
1/16 aabb
Hint: usually given numbers, not fractions
27 agouti; 12 albino; 9 black
28 agouti; 11 albino; 4 black
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
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
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-
4 long
1/16 aa bb
What were the genotypes of the original spherical
parents?
AA bb aaBB
Sex Linkage: mammals, flies
XAXa
Diploid
Adults
Gametes
Heterogametic
Sex
Xa
XA
XA
Y
XA
XAXA
XAY
Xa
XAXa
XaY
XA
Male
Female
XAY
Y
Sex Linkage: birds, butterflies
ZBW
Female
Diploid
Adults
Gametes
Homogametic
Sex
Heterogametic
Sex
ZBZb
Male
W
ZB
ZB
Zb
ZB
ZBZb
ZBZb
W
ZBW
ZbW
ZB
Male
Female
Zb
Y-linked inheritance
Hairy ears
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Male pattern
baldness: what kind
of inheritance?
Sex influenced phenotype
Genotype
bb
bb’
b’b’
Female
Bald
Not bald
Not bald
Male
Bald
Bald
Not bald
Environment-dependent
dependent expression of a
genotype
Siamese or “Himalayan”
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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
Genetic Anticipation
Huntington disease
Fragile-X syndrome
Kennedy disease
Myotonic muscular
dystrophy
Fragile X syndrome
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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.
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.
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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.
Fragile X
Pre-mutation is unstable: maternally-inherited premutation with >100
repeats almost always expands to a full mutation
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.
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.
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)
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