Where a heterozygote has a phenotype intermediate between the corresponding homozygote phenotypes

Incomplete or Partial Dominance involving Flower Color in Snapdragons

R1R1 x R2R2 Red White

R1R2 x R1R2

Pink Pink

R1R1 R1R2 R2R2

1/4 Red 1/2 Pink 1/4 White

• genotypic and phenotypic ratios are the same (1:2:1)

codominance - where the heterozygote shows the phenotype of both alleles equally

Multiple Alleles & Codominance

ABO Blood Groups

Discoverer -- Landsteiner (1900-1901)Group A, B, O, AB

Genetic Explanation -- Bernstein (1924)Series of multiple allelesIA, IB, IO (i)

…O recessive, A and B alleles codominant

• the A and B alleles make a different version of the same protein with a slightly different function

Molecular basis of ABO alleles

• fucose added to H-substance precursor

• product of A allele adds N-acetylgalactosamine

• product of B allele adds galactose

• both molecules (A and B) can be expressed on cell surfaceat the same time

• in OO individuals H substance is not modified

type A

type B type O

type AB

Red blood cell antigens

A antigen A and B antigen

B antigen H antigen(no A or B antigen)

IAIA

IAIO

Blood GroupA 41% A Anti-B

IBIB

IBIO

Blood GroupB 10% B Anti-A

IOIO

Blood GroupO 45% Neither

Anti-AAnti-B

IAIB

Blood Group AB 4% Both Neither

Genotype Phenotype Frequency Antigen

Antibodyin

Serum

Note: Caucasian frequencies given. Frequencies differ with population.

Legal Applications of Blood Groups

Hospital Cases -- Babies mixed up

Father - O Baby 1 - BMother - A Baby 2 - O

IAIA or IAIO x IOIO

IBIB or IBIO impossible, but IOIO possible;therefore baby #2 belongs to these parents

Blood tests can be used to rule out paternity

Charlie Chaplin Joan BarryO=IOIO x A=IAIO

Caroline B= IBIO

Caroline is not Charlie Chaplin’sdaughter.

Rh Blood Group

Landsteiner and Weiner (1940)

Rhesus Monkey BloodInjected into Rabbit

Extract serum from rabbit. Find that rabbit containsanti-Rh antibodies.

Test in Man Phenotype Genotype

1) 85% Agglutination (+) Rh+ RR or Rr2) 15% Non-Agglutination (-) Rh- rr

Rh Blood Group Alleles

Multiple AllelesWeiner

R1

R2

R0

Rz

rrl

rll

ry

Closely Linked LociFisher-Race

CDecDEcDeCDEcdeCdecdECdE

• Weiner - One complex locus, 8 different alleles

• Fisher-Race - Three separate loci; simple alleles, D allele most

important; D = Rh+; dd = Rh-

• C and E loci may effect expression of Rh phenotype

Rh+

Rh-

Rh Incompatibility

Rh- x Rh+

rr RR

Affected(Erythroblastosis fetalis)

all offspring Rr

Explanation for Rh Incompatibility• Mother Rh-, lacks Rh antigen on blood cell surfaces

• Seepage of blood from 1st born into mother’s blood stream from trauma of birth

• Mother’s immune system interprets entering Rh antigens as foreign protein to be eliminated

• Offspring heterozygous Rr; produce Rh antigen on red blood cell surface

• Mother builds up antibody against Rh antigen after birth of 1st child; takes time to build antibody; 1st child spared

• Rh antibody is IgG class and can pass across placenta and enter blood stream of fetus 2, 3 & 4

• Mother’s antibodies destroy fetus red blood cells; results in erythroblastosis fetalis

• Expected frequency is 10% of all pregnancies but observed at frequency of 1/200….C and E loci may modify immune response

• mice can have agouti (silver gray) or yellow coat color

• yellow allele dominant to agouti allele

• homozygotes for the yellow allele (AY) do not survive embryonic development

• F1 phenotypic ratio distorted (2:1)due to presence of lethal allele

• yellow allele dominant with respectto coat color but recessive with respect to embryonic development

• genes can be multi-functional

An interaction between genes such that a mutation in one gene masks the expression of the phenotype of another gene.

recessive epistasis

mouse coat color A_ = agouti aa = black

In a cross thought to be Aa x Aa see 9 agouti : 3 black : 4 albino.

Modified dihybrid ratio implies 2 genes involved (9:4:3 is typical recessive epistatic ratio if 2 genes determine character)

Can explain results if cross is actually AaBb x AaBb

9 A_B_ agouti3 A_bb albino3 aaB_ black1 aabb albino

gene B gene A

b is epistatic to A or a.

bb aa

• B allele required to depositpigment; A allele required to modify pigment pattern

precursor black agouti pattern

white yellow green

dominant epistasis

squash color A_ = white aa = yellow

In a cross thought to be Aa x Aa see 12 white : 3 yellow : 1 green.

Modified dihybrid ratio implies 2 genes involved (12:3:1 is typical dominant epistatic ratio if 2 genes determine character)

Can explain results if cross is actually AaBb x AaBb

9 A_B_ white3 A_bb white3 aaB_ yellow1 aabb green

gene a (aa) gene b (bb)

A is epistatic to B or b.

A_ B_

•a allele required to make yellow pigment; b allele required to convert yellow to green

• O type woman found to be incompatible with type O bloodprior to transfusion

• based on blood typing of parents could not be type O

• clearly passed on B alleles to two of her children

type A

type B type O

type AB

Red blood cell antigens

A antigen A and B antigen

B antigen H antigen(no A or B antigen)

hh IBIO x HH IAIO

Hh IAIO or Hh IBIO or Hh IAIB

Explanation

If genotype is hh conversion does not occur and anti-H is madein serum….reacts with O-type blood cells which have H antigen

h is epistatic to the A or B alleles

condition in which a single mutation in one gene simultaneously affects several characters

Examples of pleiotropy

White eye gene in Drosophila• eye color white• flight muscles defective

Albinism in humans• lack of pigment• eye sight can be affected• hearing can be affected

Hormones produced in pituitarygland required throughout body

interactions between two or more genes, each with an additive effect on the character

Kolreuter’s Tobacco Plants

Dwarf Population

Tall Population

Height

Freq

Intermediate Height

F1

Height

Freq.

Freq.

Height

F2

Continuous Variation

Explanation of Kolreuter’s Results

Parents AABB X aabb

6 feet 2 feet

F1 AaBb X AaBb

AABB AABb AaBB AaBb

AABb AAbb AaBb Aabb

AaBB AaBb aaBB aaBb

AaBb Aabb aaBb aabb

AB

Ab

aB

ab

F2 AB Ab aB ab

•Residual = 2ft

•A = B = 1 ft added

•a = b = 0 ft added (non-additive alleles)

Polygenic Inheritance F2 Results

Genotypes Phenotypes Frequency

AABB 6 feet tall 1

AABb 5 feet tall 2

AaBB 5 feet tall 2

AAbb 4 feet tall 1

AaBb 4 feet tall 4

aaBB 4 feet tall 1

Aabb 3 feet tall 2

aaBb 3 feet tall 2

aabb 2 feet tall 1

1

4

6

4

1

Calculating the number of genes involved in determining a polygenic character.

The number of possible different F2 phenotypes = 2n+1

n = the number of heterozygous gene pairs determining the character

We know that all genes pairs in the F1 cross are heterozygous because we start the parental cross withpure breeding lines.

sex chromosome - a chromosome whose presence or absence is correlated with the sex of the bearer,

or, a chromosome that plays a role in sex determination

autosome - any chromosome that is not a sex chromosome

I. Pattern Baldness

Sex-influenced trait: Autosomal trait expressed as a dominant in one sex and as a recessive in the other.

Sex-Influenced Trait Inheritance Pattern

BBBB pattern

baldness

X BNBN

non- bald

BBBN pattern

baldness

BBBN non-bald

X

B allele is dominant in males andrecessive in females.

N allele is dominant in females andrecessive in males.

Pattern Baldness F2 Summary

Genotype Phenotype

Males Females

1 BBBB Bald Bald

2 BBBN Bald Non-Bald

1 BNBN Non-Bald Non-Bald

Sex Influenced Trait

II. 2nd Finger Shorter than 4th

SSSS shorter

X SLSL

longer

SSSL shorter

SSSL longer

X

2nd Finger Shorter than 4th F2 Summary

Genotypes Phenotypes

Males Females

1 SSSS Shorter Shorter

2 SSSL Shorter Longer

1 SLSL Longer Longer

Sex-limited Trait: Autosomal trait expressed in one sex only.

Cock and Hen Feathering in Chickens

Hamburgh rooster (cock feathering)

Sebright hen (hen feathering)

cock feathers: curved and pointedhen feathers: shorter and rounded

phenotype controlled by single gene, h

hh X h+h+

h+h h+hX

cock feathers

hen feathers

hen feathers

hen feathers

Feathering in Chickens F2 Summary

Genotype Phenotype

Males Females

1 h+h+ hen hen

2 h+h hen hen

1 hh cock hen

Leghorn chickens all hh - males always different Sebright bantams all h+ h+ - males/females have same feathers