Multiple Allelism

1. Dr. Thirunahari Ugandhar
Associate Professor of Botany
Department of Botany
Kakatiya Govt. College (A), Hanamkonda

3. Multiple Allelism – Definition and Explanation
• Definition:
Multiple allelism refers to the presence of more
than two alternative forms (alleles) of a gene
that occupy the same locus (position) on a
homologous pair of chromosomes, although only
two alleles can be present in a diploid individual
(one on each chromosome).

4. Example in Humans:
ABO blood group system is a classic
example.
• Gene I has three alleles:
• IAI^AIA – produces A antigen
• IBI^BIB – produces B antigen
• iii – does not produce any antigen
• These combine to form four blood types: A, B, AB,
and O.
Genotypic Combinations:
With three alleles, the possible
combinations are:
• IAIAI^A I^AIAIA, IAiI^A iIAi, IBIBI^B I^BIBIB, IBiI^B
iIBi, IAIBI^A I^BIAIB, iii iii

5. Failure of Dominance – Definition and Explanation
• Definition:
Failure of dominance refers to a genetic condition
where the dominant allele does not completely
mask the effect of the recessive allele in a
heterozygous individual. As a result, the phenotype is
an intermediate or a blend between the dominant
and recessive traits.
Key Concepts:
1.Also Known As:
1.. Incomplete dominance or partial dominance.
2.Contrast to Complete Dominance:
1. In complete dominance, the dominant allele completely
overrides the effect of the recessive allele (e.g., tall ×
dwarf tall).
→
2. In failure of dominance, neither allele is completely
dominant, and both contribute to the phenotype.

6. Example:
Snapdragon Flower Color (Antirrhinum majus):
• Red flower (RR) × White flower (rr) Pink flower (Rr)
→
• Phenotype of F1 generation is intermediate,
showing incomplete dominance, not red
(dominant) or white (recessive).
Genotypic and Phenotypic Ratios in F2 Generation:
• Cross: Rr × Rr
• Genotype Ratio: 1 RR: 2 Rr: 1 rr
• Phenotype Ratio: 1 Red: 2 Pink: 1 White
Failure of dominance shows that dominance is not
always complete. It provides important evidence
that inheritance patterns can be more complex than
simple Mendelian laws.

9. • A given phenotypic trait of
an individual depends on
a single pair of genes,
each of which occupies a
specific position called
the locus on a
homologous chromosome.
• When any of the three or
more allelic forms of a
gene occupy the same
locus in a given pair
of homologous
chromosomes, they are
said to be called multiple
alleles.

10. Characteristics of multiple
alleles
• Multiple alleles of a series always
occupy the same locus in the
homologous chromosome.
Therefore, no crossing over occurs
within the alleles of a series.
• Multiple alleles are always
responsible for the same character.
• The wild-type alleles of a series
exhibit dominant character,
whereas mutant types will
influence dominance or an
intermediate phenotypic effect.
• When any two of the mutant
multiple alleles are crossed, the
phenotype is always mutant type
and not the wild type

11. Wild type
Mutant type
Amorphs
Hypomorphs
Neomorphs
Isoalleles
Unstable alleles

12. Self-sterility in Nicotiana
• In plants, multiple alleles have
been reported in association
with self-sterility or self-
incompatibility.
• Self-sterility means that the
pollen from a plant is unable to
germinate on its stigma and will
not be able to bring about
fertilization in the ovules of the
same plant.
• East (1925) observed multiple
alleles in Nicotiana which are
responsible for self-
incompatibility or self-sterility.
• The gene for self-incompatibility
can be designated as S, which
has allelic series S1, S2, S3, S4 and
S5

14. • The cross-fertilizing tobacco plants
were not always homozygous as S1S1
or S2S2, but all plants were
heterozygous as S1S2, S3S4, S5S6.
• When crosses were made between
different S1S2 plants, the pollen tube
did not develop normally. But
effective pollen tube development
was observed when crossing was
made with other than S1S2 for
example S3S4.

16. • When crosses were made
between seed parents
with S1S2 and pollen
parents with S2S3, two
kinds of pollen tubes were
distinguished.
• Pollen grains carrying S2
were not effective, but the
pollen grains carrying S3
were capable of
fertilization.
• Thus, from the cross
S1S2XS3S4, all the pollens
were effective and four
kinds of progeny resulted:
S1S3, S1S4, S2S3 and S2S4.

17. Coat colour in rabbit
As an example, let's
consider a gene that
specifies coat color in
rabbits, called the C gene.
The C gene comes in four
common alleles: C, cch, ch
and c:A CC, rabbit has black
or brown fur A cchech,
⚫
rabbit has chinchilla
coloration (grayish fur).A
chch rabbit has Himalayan
(color-point) patterning,
with a white body and dark
ears, face, feet, and tailA cc

18. ⚫ Multiple alleles makes for many possible dominance relationships. In
this case, the black C allele is completely dominant to all the others; the
chinchilla c^{ch}cchc, start superscript, c, h, end superscript allele is
incompletely dominant to the Himalayan c^hchc, start superscript, h, end
superscript and albino ccc alleles; and the Himalayan c^hchc, start
superscript, h, end superscript allele is completely dominant to the
albino ccc allele.
⚫ Rabbit breeders figured out these relationships by crossing different
rabbits of different genotypes and observing the phenotypes of the
heterozygous kits (baby bunnies).

19. Human Blood Type
⚫An excellent example of multiple allele
inheritance is human blood type. Blood type
exists as four possible phenotypes: A, B, AB, &
O.There are 3 alleles for the gene that
determines blood type.
⚫The alleles are as follows:
ALLELE
IA
⚫IB
⚫I
CODES
FOR
Type "A"
Blood
Type "B"
Blood Type

20. ⚫Notice that, according to the symbols used in
the table above, that the allele for "O" (i) is
recessive to the alleles for "A" & "B".
⚫With three alleles we have a higher
number of possible combinations in
creating a genotype.
⚫ GENOTYPE
S
IAIA
Iai
RESULTING
PHENOTYPES
Type
A
Type
A
IBIB
Ibi
Type B
Type
B
⚫
IAIB Type
AB
⚫ii Type
O

21. Extensions to Mendelian
Genetics
 Gene Interactions

22. Gene Interactions – Extensions
to Mendelian Genetics
22
• 2 different genes can also act together to modify
a phenotype:
• 2 genes 1 phenotype (Additive Gene Action)
Complementation (complementary gene action)
• Epistasis (recessive and dominant)
Just as
different
alleles of 1
gene can
interact in
complex
ways,
Redundancy

23. Multifactorial Inheritance
23
Vast majority of traits are determined by multiple factors:
• genetic as well as environmental.
Gene interactions between two or more genes
• Example: Lentil Seed color.
F1 all same, F2: 4 different phenotypes
F2 phenotypic ratio is 9:3:3:1
• (same as F2 dihybrids in Mendel’s original crosses).
Difference:
• in original crosses: 2 independent traits/phenotypes=2 independent
genes;
• Seed color and seed shape
• here: multiple phenotypes of 1 trait=2 independent genes
• Seed color only.

24. You can tell this
genotype is
caused by more
than one gene :
•because there
are 4 phenotypes
not 3 in F2
•1 gene F2
would have 3
phenotypes
1:2:1 ratio
(Additive Gene
Action)

25. F2 phenotypes
– Tan:
– Gray:
A_bb
aaB_
– Green:
aabb
Dominance Relationships:
• Tan is dominant to green
• Gray is dominant to green
• Brown is dominant to gray, green and tan.
• Tan and Gray are incompletely dominant, giving rise
to brown.
Genotypic classes:
• Brown: A_B_

26. 6
Complementary Gene
Action
The must have a dominant allele in both genes to result
in the purple flower phenotype
Each genotypic class may not always dictate a unique phenotype
A pair of genes can often work together to create a specific phenotype. We call this
complementary interaction.
With this type of interaction we see 2 different phenotypes instead of the 4 seen in
2 genes 1 phenotype
Two or more genotypic classes may display an identical phenotype.
• – Example: Two lines of pure breeding white flowered pea plants falling
into different genotypic classes: AAbb & aaBB

27. 27

28. 28

29. Epistasis
29
One gene’s allele masks the phenotype of the
other gene’s alleles.
Four genotypic classes produce fewer than four
phenotypes.
Different types of epistasis:
Recessive epistasis: when the recessive allele of
one gene masks the effects of either allele of the
second gene.
Dominant epistasis: when the dominant allele of
one gene masks the effects of either allele of the
second gene.

30. Recessiv
e
Epistasis
30
Phenotypic ratios are
9:3:4 in F2.
Example 2: ABO blood
groups: Bombay
phenotype.
Example 1: Coat color of
Labrador retriever

31. X
P
F1
Coat-Color Inheritance in
Labrador Retrievers
blac
k
11
golde
n
blac
k

32. Recessive Epistasis:
a recessive mutation in one gene masks the phenotypic effects of
another
F1
X
F2
Appears like
incomplete
dominance
because some of
the progeny look
like neither parent,
but the ratio is
wrong.
9 : 3 : 4 12

33. 13
(9 B-E-: 3 bbE-: 3 B-ee: 1 bbee)
x
F2
BE
Be
bE
be
AACC AACc AacCC AacCc
AACc AAcc AacCc Aaccc
AacCC AacCc acacCC acacCc
AACc Aaccc acacCc acaccc
BbEe
BbEe
BE Be bE be
BBEE BBEe BbEE BbEe
BBEe
BbEE BbEe
BbEe
BBee Bbe
e
bbEe
bbEE
BbEe Bbee bbEe bbee
9 black: 3 brown: 4 golden
Dihybrid
Cross:

34. 14
Molecular Explanation
Pigment production (B) and subsequent incorporation (E)
into the hair shaft are controlled by two separate genes. To
be black, both genes must function. Mutations in B (b) lead
to brown pigment. Mutations in E (e) lead to no pigment in
coat.
gene B
gene E
3
4
9

35. 15
Recessive Epistasis
alleles of another
Two genes involved in coat color determination.
Gene B determines whether black (B) or brown (bb) pigment is produced.
Gene E determines if pigment is deposited in hair
• – golden retrievers (ee) make either black (B-) or brown (bb) pigment (look at noses)… but not in fur
The recessive allele is epistatic to (stands over) other genes when homozygous -- hence the
name “recessive epistasis”
Phenotypes do not segregate according to Mendelian ratios (the phenotypic ratios are
modified Mendelian ratios).
epistasis - (Greek, to stand upon or stop) the differential phenotypic expression of a genotype
at one locus caused by the genotype at another, non allelic, locus. A mutation that exerts its
expression by canceling the expression of the

36. Dominant Epistasis
• caused by the dominant allele of one
gene, masking the action of either allele
of the other gene.
• Ratio is 12:3:1 instaed f 9:3:3:1
• Example: Summer Squash

38. 18
Petal color in
snapdragons
-
if Mendel had
used snap
dragons for his
experiments,
he wouldn’t
be famous!
X
F1
F2 15/16 red; 1/16
white
P
AABB
A-B-
aab
b
Redundancy:
Duplicate Genes

39. AACC AACc AacCC
AACc AAcc AacCc
AacCC AacCc acacCC
AACc Aaccc acacCc
acaccc
AB Ab aB ab
AABB AABb AaBB AaBb
AABb AAbb AaBb Aabb
AaBB AaBb aaBB aaBb
AaBb Aabb aaBb aabb
A-B- A-B-
X
AB
Ab
aB
ab
15/16 A-B-
red;
1/16 aabb

40. • If one gene is involved in the
trait, then the monohybrid
phenotypic ratio is:
• 3:1 or 1:2:1 or 2:1
• If two genes are involved in the
trait, then the dihybrid
phenotypic ratio is:
• 9:3:3:1 or some permutation
(9:4:3 or 9:7 or 12:3:1)
Hints for figuring out gene
interactions:
Look at the F2 phenotypic
ratios!!

41. 21
Hints for figuring out gene
interactions:
• 2 Genes 1 Phenotype (Additive Gene Action): You can tell
this genotype is caused by more than one gene because
there are 4 phenotypes not 3 in F2 (9:3:3:1)
– 1 gene F2 would have 3 phenotypes 1:2:1 ratio
• Complementary Gene Action: one good copy of each
gene is needed for expression of the final phenotype
– 9:7 ratio
• Epistasis: one gene can mask the effect of another
gene
– 9:3:4 ratio for recessive epistasis
– 12:3:1 ratio for dominant epistasis
• Duplicate genes: only double mutant has mutant
phenotype
– 15:1 ratio

42. 2
variations on Mendelian
inheritance
Gene
interaction
Additive
Complementary
Recessive
Epistasis
Dominant
Epistasis
One dominant a lele
from either of twogenes
needed for phenotype
9 3 3 1 15:1
Duplicate
Genes
Inheritanc
e pattern
A-/B- A-/bb aa/B- aabb ratio
Each genotype
results in a unique
phenotype
9 3 3 1 9:3:3:1
At least one
dominant alele
from each of two
genes needed for
phenotype
9 3 3 1 9:7
Homozyous recessive
genotype atone locus
masks expression at
second locus
9 3 3 1 9:3:4
Dominant allele at one
locus masks
expression at second
locus
9 3 3 1 12:3:1
2

43. 23
true breeding brown dogs X true breeding white
dogs F1 = all white
F2 = 118
white
32 black
10 brown
 Find the genotypes of the dogs in each
class: What is the ratio?
12
3
1
How many
genes?
2
What is the ratio of white to colored dogs? 12:4 =
3:1
This means that white is dominant to colored so let’s
call one gene: W= white w=colored
Sample
Problem

44. 24
What is the ratio of black to brown
dogs?
3 : 1
So black must be dominant to brown. So we will call
the second gene: B=black and b=brown
What class of dogs are the double recessive
homozygotes and what is their genotype?
Brown - wwbb
What is the genotype of the black
dogs? Must be wwB-
What are the genotypes of the white
dogs?
W_ B_ and W_bb  This is an example of
dominant
epistasis
(white).
F2 = 118 white
32 black
10 brown

45. Same Genotype may produce
different Phenotypes
• Penetrance: Genotype does not necessarily define
phenotype. The proportion of individuals with a given genotype
express the phenotype determines penetrance.
• 100% penetrance = all individuals show phenotype.
• 50% penetrance = half the individuals show phenotype.
– Example: retinoblastoma: only 75% individuals affected.
• Expressivity: the degree or intensity with which a particular
genotype is expressed in a phenotype in a given individual.
– Retinoblastoma: some have both eyes affected,
some only one.

46. Modifier Effects
46
Modifying environment: The
environment may influence the effect
of a genotype on the phenotype.
E.G.: Siamese cats: temperature dependent
color of coat. Color shows up only in
extremities, where the temp is lower (enzyme
for pigment formation is active only at lower
temp.)
Modifier Genes: they have a subtle,
secondary effect which alters the
phenotypes produced by the primary
genes.
E.G. Tail length in mice. The mutant allele t
causes a shortening of the tail. Not all short
tails are of the same length: another gene
affects the actual length. (Variable
expressivity).

47. Modifying environment:
The environmental influence
of a genotype on the
phenotype= phenocopy