Genetics WK2PPT

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


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?


different genes

P

F2 ?

AABB agouti

aabb albino

X

AaBb agouti

F1 X

AaBb agouti


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


phenotype

FUT1 gene required to make A or B

Genotypically: B Phenotypically: O

Figure 4-2


phenotype

FUT1 gene (H) required to make A or B

Figure 4-5


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

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Genetics WK2PPT