1!

product rule - if two events are independent, the probability of both occurring is the product of each of them occurring. !p(a and b) = p(a) x p(b).!p(RRYY) = p(RY) p(RY) = (1/4)(1/4) = 1/16!

sum rule - for two exclusive events, the probability for either occurring is the sum of them both. !p(a or b) = p(a) + p(b).!p(RRyy or Rryy) = 1/16 + 2/16 = 3/16!

Simple statistics used in genetics.!

P !RR yy PP x rr YY pp!

F1 ! !Rr Yy Pp (round, yellow, purple)!

If Rr Yy Pp were selfed, what proportion of the progeny would be

homozygous recessive for all three genes?!

1." 1/8!

2." 1/16!

3." 3/16!

4." 1/64!

5." 3/64!

6." 9/64!

Clicker Q 5: Multiple Genes!

P !RR yy PP x rr YY pp!

F1 ! !Rr Yy Pp (round, yellow, purple)!

If Rr Yy Pp were selfed, what proportion of the progeny would have a

dominant phenotype for R and Y, but recessive for P?!

1." 1/8!

2." 1/16!

3." 3/16!

4." 1/64!

5." 3/64!

6." 9/64!

Clicker Q 6: Multiple Genes!

P !RR yy PP x rr YY pp!

F1 ! !Rr Yy Pp (round, yellow, purple)!

1." 1/8!

2." 1/16!

3." 3/16!

4." 1/64!

5." 3/64!

6." 9/64!

If Rr Yy Pp were selfed, what proportion of the progeny would have a

dominant phenotype for one of the genes, but be recessive for the other

two genes?!

Clicker Q 7: Multiple Genes!

P !RR yy PP x rr YY pp!

F1 ! !Rr Yy Pp (round, yellow, purple)!

1." 1/8!

2." 1/16!

3." 3/16!

4." 1/64!

5." 3/64!

6." 9/64!

If Rr Yy Pp were selfed, what proportion of the progeny would breed

true?!

Clicker Q 8: Multiple Genes!

Most traits are affected by more than one gene and cannot be analyzed using simple Mendelian genetics.!

Many single gene traits that can be followed cause rare diseases, rather than common phenotypes.!

Human genetics is not straight forward:!No pure breeding humans.!Generation time is long.!No controlled matings.!Siblings rarely mate to generate F2.!Small family size.!

How can one study human genes that cause human diseases?!

Geneticists make use of the variation that already exists in the population.!

Pedigree analysis. Rare traits can be studied by carefully analyzing families that are affected. In other words, the trait may

be rare in the population but common in a particular family.!

Mendelian Analysis in Humans!

Female!

Male!

Unknown Sex!

Deceased!

Consanguineous!

mating!

Individual with a trait!

Heterozygous!

Number of children !

of indicated sex!

4!

I!

II!1 2 3 4 5!

2!III!

Carrier (heterozygous)!

Pedigree of a rare

recessive phenotype. !

A is dominant allele.!

a is recessive allele.!

a is rare.!

Cystic Fibrosis!

- Afflicted individuals often die by the age 20.

- Prevalent in Northern European populations (carrier frequency can be as high as 1/20)

- Gene encodes chloride transport channel protein

- May confer resistance to cholera toxin

Cystic fibrosis is a rare recessive disorder. Jack whose brother has the disease

marries Jill whose aunt has the disease. !

If they have one child, what is the probability that the child will have cystic

fibrosis?!

A. ! 2/3 x 2/3 x 1/2 x 1/4 = 1/18!

B.! 1/2 x 2/3 x 1/2 x 1/4 = 1/24!

C. ! 1/2 x 1/2 x 1/2 x 1/4 = 1/32!

D. ! 2/3 x 2/3 x 2/3 x 1/4 = 2/27!

E.! None of above.!

Clicker Q 9: Pedigree Analysis in Humans!

Cystic fibrosis is a rare recessive disorder. Jack whose brother has the disease

marries Jill whose aunt has the disease. !

b) If their first child has cystic fibrosis, what is the probability that their

second child will have cystic fibrosis?!

A. ! 2/3 x 2/3 x 1/2 x 1/4 = 1/18!

B.! 1/2 x 2/3 x 1/2 x 1/4 = 1/24!

C. ! 1/2 x 1/2 x 1/2 x 1/4 = 1/32!

D. ! 2/3 x 2/3 x 2/3 x 1/4 = 2/27!

E.! None of above.!

?!

Clicker Q 10: Pedigree Analysis in Humans!

Cystic fibrosis is a rare recessive disorder. Jack whose brother has the disease

marries Jill whose aunt has the disease. !

c) If their first and second child has cystic fibrosis, what is the probability

that their third child will have cystic fibrosis?!

d) If their first child has cystic fibrosis, what is the probability that of their

next three children, two will be carriers and the other affected?!

?!

Pedigree Analysis in Humans!

Incomplete (or partial) dominance: Phenotype of heterozygous hybrid is intermediate between homozygous parents.

Incomplete Dominance!

Extension on Mendel!

•"Genotypic ratio is 1:2:1 and so is the phenotypic ratio. •"Mendel’s laws are still valid at the genotypic level.

Extension on Mendel: Incomplete Dominance!

AA Aa aa

Pigment forming enzyme is present in limiting quantities, so having more makes the flowers more red.

Molecular Explanation of one example of incomplete

dominance!

Precursor Red pigment

Red pigment forming enzyme A = Dominant allele makes a normal enzyme a = Recessive allele makes a defective one

phenotypes of both alleles are observed in the heterozygotes

Extension on Mendel: Co-Dominant Alleles!

~ Complete dominance!

~ Codominance!

~ Incomplete dominance!

Multiple alleles!Coat color in guinea pigs!CC = black ! ! !Ccd = black!ckck = sepia ! ! !ckca = sepia!cdcd = cream! ! !!

caca = albino!

Coat color in guinea pigs have a hierarchy of dominance:!C > ck > cd > ca!

What is the color (phenotype) of a cdca guinea pig.!A."Black!B."Sepia!C."Cream!D."Albino!E."Can#t tell.!

Extension on Mendel: Multiple Alleles!

Clicker Q 11: Multiple Alleles!

Multiple alleles!Coat color in guinea pigs!CC = black ! ! !Ccd = black!ckck = sepia ! ! !ckca = sepia!cdcd = cream! ! !!

caca = albino!Hierarchy of dominance:!C > ck > cd > ca!

What ratio of color (phenotype) is the F1 guinea pig progeny?!A." 2 Black: 1 sepia: 1cream!B." 3 Black: 1 sepia!C." 2 Black: 2 cream!D." 1 Black: 1 sepia: 1 cream: 1 albino!E." Can#t tell or don#t know.!

Ccd! ckca!

?!

Clicker Q 12: Multiple Alleles!

Multiple alleles!Coat color in guinea pigs!CC = black ! ! !Ccd = black!ckck = sepia ! ! !ckca = sepia!cdcd = cream! ! !!

caca = albino!Hierarchy of dominance:!C > ck > cd > ca!

What are the genotypes of the parents?!A." CC x ckck!

B." Cca x ckck!

C." CC x ckca!

D." Cca x ckca!

E." Can#t tell or don#t know.!

Black x sepia!

2 black + 1 sepia + 1 albino!

Clicker Q 13: Multiple Alleles!

Multiple alleles and Incomplete Dominance!ABO Blood Group!

There are three alleles: IA IB and i!IA and IB are co-dominant; i is recessive.!

Phenotype !Genotype!O ! !ii!A ! !IAIA or IAi!B ! !IBIB or IBi!AB ! !IAIB!

AB! AB!

B!A!

?!

What are the possible phenotypes of the

child?!

ABO Blood Group!

Multiple alleles and Incomplete Dominance!ABO Blood Group!

Phenotype !Genotype!O ! !ii!A ! !IAIA or IAi!B ! !IBIB or IBi!AB ! !IAIB!

There are 4 children who have the following blood types; A, B, O, AB!

These children belong to 4 sets of parents who are AB x O; A x O; A x AB and O x O.!

As best as you can, match up the children with their parents.!

! !#1 ! !#2 ! !#3 ! !#4!

AB x O ! !AB ! !AB ! !B ! !AB or B!

A x O ! !O ! !A ! !A ! !A!

A x AB ! !B ! !B ! !AB ! !AB or B!

O x O ! !A ! !O ! !O ! !O!

Clicker Q 14: ABO Blood Group!

Genes Affecting the same character!P !white x white!

F1 ! purple!

F2 !9 purple : 7 white!

These genes have complementary action. !

Complementation: 2 or more genes necessary to give a function. If they complement they are different genes. If they fail to complement they are the same gene.!

Extension on Mendel: Complementation!

The complementation test.!If two mutants

complement, they

are in different

genes.!

If two mutants fail

to complement,

they are in the

same gene.!

Genes Affecting the same character!P albino x black!

F1 agouti!

F2 9 agouti : 3 black : 4 albino!

Epistasis - one gene interferes with or masks another gene.!

Extension on Mendel: Epistasis!

Genes Affecting the same character!Shepherds Purse! Round fruit x Narrow fruit!

! !all round!

15 round : 1 narrow!

Extension on Mendel: Redundant Genes!

Redundant genes are usually caused by a gene duplication.!

Lethal Alleles - All biological functions are under the control of genes, some genes are essential for viability.!Yellow gene of mice.!

!Y/y x y/y!

!y/y and Y/y in 1:1 ratio!

!Y/y x Y/y!

!2 yellow : 1 normal (expect 3:1)!

backcross yellow and they look like heterozygote.!Look at pregnant females and find 1/4 embryos are dead.!Y is dominant for yellow hair color!Y is a recessive lethal (Y/Y individuals die).!

Extension on Mendel: Lethal Alleles!

You isolate a recessive lethal allele of the dead gene in Drosophila

melanogaster. This imaginary gene maps on the X chromosome.

A female heterozygous for dead (d+/d) is phenotypically wild type but males carrying the recessive allele (d/Y) are never found, the

d allele has a recessive lethal phenotype. A female heterozygous

for dead is mated with a wild type male.!

What is the ratio of males to females among the progeny of the above cross?!

A. 3 males : 1 female!

B. 2 males : 1 female!

C. 1 male : 1 female!

D. 1 male : 2 female!E. 1 male : 3 female!

Clicker Q 15: Lethal Alleles!

PENETRANCE = % with expected phenotype for a given genotype.!Examples of 100% penetrance: Mendal#s Peas, ABO!

Incomplete penetrance (<100%)!Retinoblastoma (Autosomal Dominant).!

Huntington#s Disease. “Woody Guthrie” Disease. Autosomal dominant.!Neurological defects with old age (40s to 80s).!

Extension on Mendel: Penetrance!

EXPRESSIVITY = variation in intensity of phenotype for given genotype.!

phenylketonurea (PKU) - rare recessive allele in humans. PKU individuals cannot convert phenylalanine to tyrosine. Severe mental retardation.!

Extension on Mendel: Expressivity!

!" Phenotype controlled by many genes with many alleles.!

!" Produce a continuum of phenotypes. !

Connecticut state university!

Crow, 1997 Genetics 147:1!

These are called continuous traits, or quantitative traits, because they vary over

a wide range of values.!

All of the mechanisms that we discussed contribute to continuous variation.

Complex dominance relationships, multiple alleles, and multiple genes

interacting in complex ways determine the genetic diversity of life. !