1. EXTENSIONS OF
MENDELIAN
GENETICS

2. I. INCOMPLETE DOMINANCE
Where a heterozygote has a phenotype intermediate
between the corresponding homozygote phenotypes

3. Incomplete or Partial Dominance involving Flower Color in Snapdragons
R1R1 x R2R2
Red White
R1R2 x R1R2
Pink Pink
R1R R1R2 R2R2
1/4 Red /2 Pink /4 White
genotypic and phenotypic ratios are the same (1:2:1)

4. II. MULTIPLE ALLELES AND CODOMINANCE
codominance - where the heterozygote shows the
phenotype of both alleles equally

5. Multiple Alleles & Codominance
ABO Blood Groups
Discoverer -- Landsteiner ( )
Group A, B, O, AB
Genetic Explanation -- Bernstein (1924)
Series of multiple alleles
IA, 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

6. 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 surface
at the same time
in OO individuals H substance
is not modified

7. Red blood cell antigens
type A
type AB
A antigen A and B antigen
B antigen H antigen
(no A or B antigen)
type B
type O

8. IAIA IAIO Blood Group A 41% A Anti-B IBIB IBIO Blood Group B 10% B
Antibody in
Serum
Genotype
Phenotype
Frequency
Antigen
IAIA
IAIO
Blood Group A
41% A
Anti-B
IBIB
IBIO
Blood Group B
10% B
Anti-A
IOIO
Blood Group O
45%
Neither
Anti-A
Anti-B
IAIB
Blood Group AB 4%
Both
Neither
Note: Caucasian frequencies given. Frequencies differ with population.

9. Legal Applications of Blood Groups
Hospital Cases -- Babies mixed up
Father - O Baby 1 - B
Mother - A Baby 2 - O
IAIA or IAIO x IOIO
IBIB or IBIO impossible, but IOIO possible;
therefore baby #2 belongs to these parents

10.  Caroline is not Charlie Chaplin’s
Blood tests can be used to rule out paternity
Charlie Chaplin Joan Barry
O=IOIO x A=IAIO
Caroline
B= IBIO
 Caroline is not Charlie Chaplin’s
daughter.

11. Rh Blood Group Landsteiner and Weiner (1940)
Rhesus Monkey Blood
Injected into Rabbit
Extract serum from rabbit. Find that rabbit contains
anti-Rh antibodies.
Test in Man Phenotype Genotype
1) 85% Agglutination (+) Rh+ RR or Rr
2) 15% Non-Agglutination (-) Rh- rr

12. Rh Blood Group Alleles Multiple Alleles Weiner R1 R2 R0 Rz r rl rll ry
Closely Linked Loci
Fisher-Race
CDe
cDE
cDe
CDE
cde
Cde
cdE
CdE
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-

13. Rh Incompatibility Rh- x Rh+ rr RR all offspring Rr Affected
(Erythroblastosis fetalis)

14. 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

15. III. Lethal Alleles

16. 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 respect
to coat color but recessive with
respect to embryonic development
genes can be multi-functional

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

18. 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_ agouti
3 A_bb albino
3 aaB_ black
1 aabb albino
b is epistatic to A or a.
B allele required to deposit
pigment; A allele required to
modify pigment pattern
bb aa
precursor black agouti pattern
gene B gene A

19. 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_ white
3 A_bb white
3 aaB_ yellow
1 aabb green
A is epistatic to B or b.
a allele required to make
yellow pigment; b allele
required to convert yellow
to green
A_ B_
white yellow green
gene a (aa) gene b (bb)

20. Blood Group Epistasis Bombay Phenotype
O type woman found to be
incompatible with type O blood
prior to transfusion
based on blood typing of
parents could not be type O
clearly passed on B alleles to
two of her children

21. Red blood cell antigens
type A
type AB
A antigen A and B antigen
B antigen H antigen
(no A or B antigen)
type B
type O

22. Explanation hh IBIO x HH IAIO Hh IAIO or Hh IBIO or Hh IAIB
If genotype is hh conversion does not occur and anti-H is made
in serum….reacts with O-type blood cells which have H antigen
h is epistatic to the A or B alleles

23. V. PLEIOTROPY
condition in which a single mutation in one gene simultaneously affects several characters

24. 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 pituitary
gland required throughout body

25. VI. POLYGENIC INHERITANCE
interactions between two or more genes,
each with an additive effect on the character

26. Kolreuter’s Tobacco Plants

27. Freq Height F1 F2 Tall Population Dwarf Population Intermediate Height
Continuous Variation
Height
Freq. F2
Height

28. Explanation of Kolreuter’s Results
Residual = 2ft
A = B = 1 ft added
a = b = 0 ft added
(non-additive
alleles)
Explanation of Kolreuter’s Results
Parents AABB X aabb
6 feet 2 feet
F AaBb X AaBb F2
AB Ab aB ab AB Ab aB ab
AABB AABb AaBB AaBb
AABb AAbb AaBb Aabb
AaBB AaBb aaBB aaBb
AaBb Aabb aaBb aabb

29. Polygenic Inheritance F2 Results
Genotypes Phenotypes Frequency
AABB 6 feet tall
AABb feet tall
AaBB feet tall
AAbb feet tall
AaBb feet tall
aaBB feet tall
Aabb feet tall
aaBb feet tall
aabb feet tall 1 4 6 4 1

30. 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 with
pure breeding lines.

31. VII. SEX-INFLUENCED TRAITS
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

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

33. Sex-Influenced Trait Inheritance Pattern
BBBB pattern baldness X
BNBNnon- bald
BBBN pattern baldness
BBBN non-bald
B allele is
dominant in
males and
recessive
in females.
N allele is
females and
in males.

34. Pattern Baldness F2 Summary
Genotype Phenotype
Males Females
1 BBBB Bald Bald
2 BBBN Bald Non-Bald
1 BNBN Non-Bald Non-Bald

35. II. 2nd Finger Shorter than 4th
Sex Influenced Trait
II. 2nd Finger Shorter than 4th
SSSS shorter X
SLSL longer
SSSL shorter X
SSSL longer

36. 2nd Finger Shorter than 4th F2 Summary
Genotypes Phenotypes
Males Females
1 SSSS Shorter Shorter
2 SSSL Shorter Longer
1 SLSL Longer Longer

37. VIII. SEX-LIMITED TRAITS
Sex-limited Trait: Autosomal trait expressed in one sex only.

38. Cock and Hen Feathering in Chickens
Hamburgh rooster (cock feathering)
Sebright hen (hen feathering)
cock feathers: curved and pointed
hen feathers: shorter and rounded
phenotype controlled by single gene, h

39. h+h+ hh h+h h+h X cock feathers hen feathers X hen feathers

40. 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