Week 2 - Mendelian Genetics Cont.
The genotypes of F1 are on the right
We could use a Punnett Square
This Punnett Square Shows the possible genotypes of the F2 generation
Sum Law
Sum Law of Probability:
The probability of either one or the other of two mutually exclusive events is the sum of their individual probabilities.
P(A or B) = P(A) + P(B)
A real example
You are playing a game with dice
You are almost done with the game.
You win a new car if you roll a 5 or a 6!
What are your chances?
Product Law of Probability
The probability of a joint occurrence of two independent events equals the product of their separate probabilities.
Or the probability of two or more events occurring simultaneously is equal to the product of their individual probabilities.
Product law of probabilities
The probability of a joint occurrence of two independent events equals the product of their separate probabilities.
Chance of rolling two sixes is 1/6 X 1/6 = 1/36
Roll one die Chance of rolling a six is 1/6
Roll two dice What is the chance of rolling 2 sixes?
A a
A AA Aa
a Aa aa
albino
normal pigmentation
If two heterozygous parents (Aa) have 2 children,
what is the probability that the both children will have albinism?
Child 1
Pigmented
Pigmented
Pigmented
Albino
Child 2
Pigmented
Pigmented
Pigmented
Albino
child 1 and
child 2 are
independent events
Real-World Example
Sheep
Important for wool
Many different traits in sheep
Hairy (dominant) vs Wooly (recessive)
White (dominant) vs. Black (recessive)
Brown eyes (dom.) vs. Blue eyes (recessive)
Real World Example
Two villages each have true breeding lines
hairy white with brown eyes HHWWBB
wooly black with blue eyes hhwwbb
A farmer wants a wooly white brown-eyed sheep,
what are the odds shell get one? (shell wait for the F2 generation of course)
The genotypes of F1 are on the right
We could use a Punnett Square
This Punnett Square Shows the possible genotypes of the F2 generation
We could use the Product Law of Probability
These are and events.
wooly and white and brown-eyed so we can use
the Product Law of Probability
hairy white with brown eyes HHWWBB
wooly black with blue eyes
hhwwbb
A farmer wants a wooly white brown-eyed sheep, what are the odds shell get one?
We could use the Product Law of Probability
wooly white brown-eyed
x x = 9/64
New Key Point
You rolled a die
You just got a 6!
What are the odds of rolling a 6
on the next roll?
Practice calculating genotypic and phenotypic ratios
G = green; g = white
P = Powerful; p = weak
C = Creepy; c = nice
Calculate ratios for the following cross:
GgppCc X ggPPCc
Tracking human traits
Same principles apply as with peas
Look at data retrospectively
cant set up controlled crosses
before there were genetic tests, this was the only way to look at data
Analyze family trees in pedigrees
Patterns of inheritance in
humans: pedigree charts
Refer to individuals as e.g. II-2 Figure 3-12
Pedigree for autosomal
recessive trait
example: albinism
Diamond is unknown gender
Figure 3-13
Albinism - recessive trait
Enzyme that produces melanin is defective
recessive trait: heterozygotes are not affected
still have one functioning copy of enzyme
Albinism can happen in many different animals
Pedigree for autosomal
dominant trait
example: hypercholesterolemia
Figure 3-13
Recessive and Dominant human
traits - diseases
Normal variation in humans
caused by single genes(?)
Widows peak
DD dd
Other traits include: earwax, earlobe attachment, freckles, folding hands
S-methyl ester detection
http://udel.edu/~mcdonald/mythintro.html
Pedigrees reveal patterns of
inheritance in humans
Is this likely to be a dominant or a recessive trait?
Probability + Pedigrees
Using the rules of probability that we discussed, can you predict the probability
that a particular child will have a disease?
Product law of probability
Pedigree of a trait
III-5 and III-6 have a child together
What is the probability of that child having the trait?
Dominant or recessive?
What do we know? II 4 and II 5 have an aa daughter (III 4) so they must both have a Aa genotype. III 5 can therefore either be AA (1/3 probability) or Aa (2/3 probability) III 6 is 100% aa. so?
2 analysis
Chi-square analysis
a statistical test commonly used to compare observed data with data we would expect to obtain
taking real-life data and decides if it fits your genetic theory
Chi-square analysis evaluates the influence of chance on data
Working with real-life data
2 analysis
Chi-square analysis
Tests the goodness of fit between observed and expected
tests the null hypothesis
Null Hypothesis
There is no difference between the observed number of phenotypes and the expected number of phenotypes.
Looking for the probability that the true average is the same as the expected average.
If the probability is less than 0.05 then the observed is statistically different than the expected.
Chi squared test for
goodness of fit
Statistical test to check your experimental results against expected ratios
are they significantly different?
e = expected number in a phenotypic class o = observed number in a phenotypic class
= 2 (o-e)2
e ________
Chi-Squared Analysis
Monohybrid cross example from Klug
= 0.53
With this number (0.53), together with degrees of freedom (df),
go to look-up table
df = # phenotypes - 1
Here its 1
= 2 (o-e)2
e ________
Chi-Square look-up table
Figure 3-11
Chi-Square Analysis
Monohybrid cross example from Klug
Example You cross two P1 generations of fruit flies,
long wings and dumpy wings. You believe long is dominant to dumpy. All F1 flies are long winged. Your F2 generation has the following traits:
792 long winged
208 dumpy winged
Can you support the hypothesis that this is a simple case of Mendelian dominance?
= 2 (o-e)2
e ________
Variations on the standard Mendelian model
2 alleles
Complete Dominance
Genes are independent
No genotype is lethal
Nomenclature
In Drosophila, an initial letter or a combination of two-three letters of the name of the mutant is used.
Example: body color
Ebony mutant phenotype is indicated by e.
Normal gray (wild-type) is indicated by e+.
e+/e+: gray homozygote (wild type)
e+/e: gray heterozygote (wild type)
e/e: ebony homozygote (mutant)
OR
+/+: gray homozygote (wild type)
+/e: gray heterozygote (wild type)
e/e: ebony homozygote (mutant)
Nomenclature
If no dominance exists between alleles, italic uppercase italic letters and superscripts are used to denote alternative alleles (R1, R2, CW, CR).
Nomenclature
Many diverse systems of genetic nomenclature are used to identify genes in various organisms.
The symbol used should reflect the function of the gene or even a disorder caused by the gene.
Yeast cdk represents the cyclin dependent kinase gene whose product is involved in cell-cycle regulation.
In bacteria, leu- refers to a mutation that interrupts the biosynthesis of the amino acid leucine, and the wild-type gene is designated leu+.
The symbol dnaA represents a bacterial gene involved in DNA synthesis.
In humans, capital letters are used to name genes: BRCA1, the gene associated with breast cancer.
New Types of Dominance
Mendel saw classic complete dominance
Complete dominance occurs when the phenotype of the heterozygote is
completely indistinguishable from that of the dominant homozygote.
e.g. GGWW and GgWw are both yellow-round
But, other types of dominance exist
Incomplete dominance
Crosses with Snapdragon flowers
Figure 4-1
Hypercholesterolemia: another example
of incomplete dominance
LDL receptor defect
Heterozygotes (+/-)
increased risk of vascular clogging
affected as young adults
Homozygotes (-/-)
have even more risk of vascular disease
affected as children
Co-dominance
Two alleles of a gene can produce distinct, detectable gene products in
heterozygotes: codominance
MN blood group (different glycoproteins)
LM or LN allele
Co-dominance
Two alleles of a gene can produce distinct, detectable gene products in
heterozygotes: codominance
In codominance, the influence of both alleles in a heterozygote is clearly
visible.
Genotype Phenotype
LM LM M
LM LN MN
LN LN N
Multiple alleles of a gene can
exist in a population
ABO blood type Gene for isoagglutinogen: I
3 alleles: IA make A antigen
IB make B antigen
IO make no antigen
Multiple alleles of a gene can
exist in a population
ABO blood type: 3 alleles: IB, IA, IO
Genotype Antigen Phenotype
IAIA A A
IAIO A A
IBIB B B
IBIO B B
IAIB A,B AB
IOIO Neither O
Universal recipient
Universal donor
What kind of dominance is this?
A) incomplete
dominance
B) complete
dominance
C) codominance
Table 4.1
Alterations of Mendelian 3:1 ratio:
Example in agouti mice
Agouti breeds true - homozygous
agouti
Banding pattern in agouti hair All agouti
X
agouti
Figure 4-3
A different mouse strain with yellow fur color
X
yellow
Yellow is a dominant mutation?
agouti
Banding pattern in yellow hair 1/2
yellow 1/2 agouti :
Figure 4-3
Yellow mouse crossed with yellow
mouse gives surprising F1 ratio
What would give rise to a 2:1 ratio??
X
yellow yellow
1/3 agouti : 2/3 yellow
Figure 4-3
Lethal alleles represent
essential genes
Ay allele is recessive for lethality
Ay allele is dominant for yellow fur
Figure 4-3
Altering Mendelian ratios
Epistasis: interaction between
different genes
This can lead to modifications from expected ratios, e.g. 9:3:3:1
Agouti mouse crossed with black
mouse is monohybrid cross
A
A
a a
Aa Aa
Aa Aa
AA agouti
aa black
X
All progeny are agouti
Aa agouti
Epistasis: interaction between
different genes
Separate gene B, determines whether mice are albino
bb mice are albino, regardless of other genes
What happens when we do a dihybrid cross of agouti and albino?
Epistasis: interaction between
different genes
P
F2 ?
AABB agouti
aabb albino
X
AaBb agouti
F1 X
AaBb agouti
Epistasis: interaction between
different genes
F2
Ratio Genotype Phenotype
9/16 A-B- agouti
3/16 A-bb albino
3/16 aaB- black
1/16 aabb albino
Key A- : agouti aa: black bb: albino
Final phenotypic ratio 9/16 agouti 4/16 albino 3/16 black
9:4:3
F1
AaBb agouti
X
AaBb agouti
Modifications to Mendelian ratios
caused by epistasis
Figure 4-7 in Klug Epistasis changes genotype - phenotype connection
Epistasis: another example
Remember ABO blood types:
Gene I has 3 alleles: IA, IB, IO
Epistasis: another example
Remember ABO blood types:
Genotype Antigen Phenotype
IAIA A A
IAIO A A
IBIB B B
IBIO B B
IAIB A,B AB
IOIO Neither O
Confusing pedigree: woman with
O blood type
Bombay phenotype for blood type
=
=
How could this happen?
Another gene plays a role:
Epistasis
Figure 4-2
Epistasis explains Bombay
phenotype
FUT1 gene required to make A or B
H: wild-type version of FUT1 gene
h: mutant version of FUT1 gene
precursor FUT1
A or B antigen X X O phenotype
hh
Epistasis explains Bombay
phenotype
FUT1 gene required to make A or B
Genotypically: B Phenotypically: O
Figure 4-2
Epistasis explains Bombay
phenotype
FUT1 gene (H) required to make A or B
Figure 4-5
Epistasis explains Bombay
phenotype
FUT1 gene (H) required to make A or B
3:6:3:4 ratio
Figure 4-5
Question: If a mother has type A blood and
her son has type O blood, what are the
possible blood types of her sons father?
A) Type O only
B) Types A or O
C) Types B or O
D) Types A, B, or O
E) Any blood type
Question: If a mother has type A blood and
her son has type O blood, what are the
possible blood types of her sons father?
Answer(s): If not considering epistatic effects of other genes:
D. Types A, B, or O
Explanation: The father must carry the IO allele, so he could be IOIO (type O), IAIO (type A), or IBIO (type B).
If considering epistatic effects of FUT1 (H):
E. Any blood type
Explanation: If both mother and father are carriers of the h allele and the son is hh, then the father could have any I blood type, because hh is epistatic to the effects of I (isoagglutinogen)
Epistasis & pleiotropy are
opposites in a way
Epistasis Trait
Gene 1
Gene 2
Gene 3
Gene 4
Pleiotropy
Trait 1
Trait 2
Trait 3
Trait 4
Gene
Pleiotropy example:
Werner syndrome
Recessive allele: defect in DNA helicase causes premature aging
Pleiotropic Effects
- graying and hair loss as teens
- short stature
- thin extremities
- cataracts in their 20s - a change of voice
- osteoporosis, bone deformities
- wrinkled, dry skin
- diabetes
- atherosclerosis
- ankle ulcers
- malignancies
What is two individuals share the same trait, but its caused by
different mutations?
When would this occur and how do you test for it?
Complementation analysis
Allows one to determine if two mutants are defective in the same gene
Screen for mutant phenotypes
Find 2 different mutants with same phenotype
mutant a and mutant b
Are a and b defective in the same gene?
Example: screen for flies without wings
Complementation analysis
comparing wingless mutants
Wingless mutant a
Wingless mutant b
X
P
Wild-type: has wings
F1
What is the result of this test?
A) fail to complement; same gene
B) complement; same gene
C) fail to complement; different genes
D) complement; different genes
Complementation tests
If mutations are in the same gene, the mutants fail to complement
If mutations are in different genes, the mutants complement