Extension of Mendel’s Principles (Mendelian Inheritance)

• Simple Mendels experiments established that genes can exist in 2 alternate

forms – called alleles – one dominant, the other recessive.

• An allele is dominant if it has the same phenotypic effect in heterozygotes as

in homozygotes – the genotypes Aa and AA are phenotypically
indistinguishable. → The principle of simple dominance.

• Inheritance patterns that obey the law of segregation and the law of
independent assortment are classified as examples of Mendelian inheritance.

• However, research in the early 20th century showed that genes can exist in
more than two allelic states and each allele can have a different effect on
the phenotype.

• Consider this: Sometimes however, a heterozygote has a phenotype different

from that of either of its associated homozygotes.
• Ratio 9:3:3:1 ( Mendel two-factor cross) – only if two particular pairs of allele
are randomly paired and fully dominance.

• Product of the cross – 9 genotypes and 4 phenotypes (refer to the Punnet

square for Mendel’s dihybrid cross) – ratio = 9:3:3:1

• However, pair of allele – not always freely paired to each other.

- not necessarily fully dominance.

• Gene interactions may occur – resulting the changes of Mendel’s ratio.

• F2 ratios – depending on type of gene interactions.

• Take the flower color of the snapdragon, Antirrhinum majus, for e.g.
Incomplete dominance and codominance

• Incomplete dominance in dihybrid cross can occur at:

a. Both pairs of allele, or
b. One of the two pairs of allele.
- both – will produce different F2 ratio from Mendel’s ratio

• If it occurs at both pair of allele,

- each F2 genotype shows different phenotype.
- 9 phenotypes with the same ratio as genotype ratio. (*** Mendel – 4
phenotypes)
- eg. Human blood group ratio
 Exception of simple dominance:
The phenomena of semidominance
Homozygous red varieties
The red allele – W (WW) when crossed with
The white allele– w homozygous white varieties
(ww) yields heterozygotes
that have pink flowers (Ww).
The allele for red (W) is
therefore said to be
incompletely or partially
dominant or semidominant
over the allele for white color
(w).
Possible reason: the
intensity of the pigmentation
in this species depends on
Genetic basis of flower color
the amount of a product (in
in snapdragons
this case red color) specified
by the color gene.
Polygenic Inheritance

• A polygene (concept) – defined as a gene that individually exerts a slight

effect on a phenotype but, in conjunction with a few or many other genes,
controls a quantitative traits, such as height.

• Polygenic inheritance differs from the classical Mendelian pattern in that a

graded series extends from one parental extreme to the other.

• Polygenic inheritance – a Mendelian concept that is studies by statistical

methods.
Note: codominance = alleles of the same gene are expressed separately to yield different traits in an individual.

a) Similar gene with no codominance

• A trait can be determined by more than one pair of allele (Cc) – nonallelic
gene can have similar function to determine traits.

• Eg. Wheat plant:

- 2 pairs of allele involve in determination of seed colour.
- Cross between dark red seed and white seed colour plants – all
intermediate red colour seed of F1.
- If A and B are genes for red and their alleles a and b for white, F2 genotypes
and phenotypes are as follows:

1/16 AA BB dark red

4/16 AA Bb, Aa BB red
6/16 AA bb, aa BB, Aa Bb intermediate red
4/16 aa Bb, Aa bb pink
1/16 aa bb white

• In this example, no codominance at allele A, a or B, b.

Each gene A or B bring the same amount of red pigment and each gene
increases the red colour of the seed.

• How F2 phenotypes are produced – give one unit to each colour allele.

• Therefore F2 phenotypes have the following colours:

1/6 AA BB dark red 4 unit

4/16 AA Bb, Aa BB red 3 unit
6/16 AA bb, Aa Bb, aa BB intermediate red 2 unit
4/16 aa Bb, Aa bb pink 1 unit
1/16 aa bb white 0 unit
b) Similar gene with codominance at one pair of allele.

• Gene A is codominant to a, but genes B and b has no codominance.

• Therefore, AA phenotype is not similar to Aa phenotype.

• If given one unit to AA (Aa), BB has 2 units (one unit for each B).

• No unit is given to aa and bb.

• Therefore, phenotypes are divided into 4 groups:

3/16 AA BB, Aa BB 3 unit

7/16 AA Bb, Aa Bb, aa BB 2 unit
5/16 AA bb, Aa bb, aa Bb 1 unit
1/16 aa bb 0 unit
 Exception of simple dominance:
The phenomena of codominance

• Another exception of simple dominance: When a heterozygote shows

characteristics found in each of the associated homozygotes – codominance.

• Blood factors in the serum portion of the blood reacts specifically with
antigens.
Homozygotes for M allele – blood type M (produces M antigen).
Homozygotes for N allele – blood type N (produces N antigen).
• Anti-M serum recognizes only the M antigen, anti-N serum recognizes
only the N antigen on the human blood cells → reaction causes blood cells
to clump together or agglutinate.
• Heterozygotes for M and N alleles produce M and N antigens → blood
cells agglutinate when anti-M and anti-N serum are present.
• Because the two alleles appear to contribute independently to the
phenotype of the heterozygotes, they are said to be codominant.
The M antigen allele – M
The N antigen allele – N

Detection of the M and N antigens on blood cells by agglutination

with specific anti-sera.
Codominance (con’t):
(1) Implies that there is an independence in allele
function.
(2) Neither allele is dominant or even partially dominant
over the other.
(3) Therefore, it would be inappropriate to distinguish the
alleles by upper and lower case letters.
(4) Instead, codominant alleles are represented by
superscripts on the symbol for the gene, which in this
case is the letter L – a tribute to Karl Landsteiner, the
discoverer of blood-typing. So, the M allele is LM and the
N allele is LN.
• Eg. 1 – Human blood group ratio (incomplete dominance at both pairs of
allele).

Blood type A, B and AB – is determined by a pair of allele IAIB

Blood type M,N and MN – is determined by a pair allele LM and LN
Ratio produced from the marriage between people with genotype IAIBLMLN;

Phenotype Ratio Genotype Ratio

AM 1/16 AA MM 1/16
A MN 2/16 AA MN 2/16
AN 1/16 AA NN 1/16
AB M 2/16 AB MM 2/16
AB MN 4/16 AB MN 4/16
AB N 2/16 AB NN 2/16
BM 1/16 BB MM 1/16
B MN 2/16 BB MN 2/16
BN 1/16 BB NN 1/16
• Eg. 2 : Height and thickness of tomato plant hair (incomplete dominance at
one pair of allele)

D_ tall h1h1 no hair (bald)

dd dwarf h1h2 thin hair
h2h2 thick hair

Allele D – complete dominance to d.

For the thickness of hair – does not show dominance.
Progeny F2 – genotype and phenotype ratios as following:

D_ h1h1 Tall, no hair 3

D_ h1h2 Tall, thin hair 6
D_ h2h2 Tall, thick hair 3
dd h1h1 Dwarf, no hair 1
dd h1h2 Dwarf, thin hair 2
dd h2h2 Dwarf, thick hair 1
 Multiple Alleles

• Although Mendel discovered that genes exist in two allelic forms,

subsequent discoveries showed that genes can exist in more than two allelic
forms.
• Genes that exist in more than two allelic forms are said to have multiple
alleles.
• A classic example of multiple alleles is the study of human blood types. –
the study is identical with the study of the MN blood types discussed
previously.
• A blood type – A antigen present, B blood-type – B antigen present, AB
blood type – both A and B antigens present, O blood type – neither antigen
present.
• Gene responsible for producing A and B antigens – I. Gene I has three
alleles: IA (specifies the production of A antigen), IB (specifies the production
of B antigen) and i (does not specify an antigen).
• All three alleles are found at appreciable frequencies in human
populations – hence, the I gene is said to be polymorphic (Greek word for
“having many forms”).
• In this system, the IA and IB alleles are codominant – since each is
expressed equally in the IA and IB heterozygotes.
• The i allele is recessive to both IA and IB.
 WORK THIS OUT!

What are the ratios of the blood types of the children of

these couples?

Couple 1 Couple 2 Couple 3

♀ X ♂ ♀ X ♂ ♀ X ♂

IAIB ii IAi ii IBi ii

Couple 4 Couple 5 Couple 6

♀ X ♂ ♀ X ♂ ♀ X ♂

IAIB IAi IAIB IBi IAIB IAIB

 A TEST FOR ALLELISM: MODEL EXPERIMENT INVOLVING
EYE COLOR MUTATIONS IN Drosophila

• What causes a gene to have alleles/multiple alleles?

• Mutation. Mutation causes changes in an existing allele, giving rise to a
new genetic state.
• A simple test to determine the allelic identity of a new mutation – involves
crosses of recessive mutations with recessive mutations of known genes.
(strict rule: only to test recessive mutations! Dominant mutations cannot be
tested in the same way because they exert their effects even if a wild-type
copy of the gene is present).
• Model experiment: Complementation test - involving recessive eye color
mutations in Drosophila.
• This organism has been investigated by geneticists for nearly a century
and many different mutations have been identified.

• Two independently isolated mutations, called cinnabar and scarlet, are

phenotypically indistinguishable – each causing the eyes to be bright red.

• In wild-type (WT) flies, the eyes are dark red.

• We wish to know whether the cinnabar and scarlet are alleles of a single
color-determining gene or if they are mutations in two different genes.
• To know the answer…cross the homozygous mutant strains with each
other to produce hybrid progeny.. E.g: cinnabar x scarlet, cinnabar x
cinnabar-2, cinnabar-2 x scarlet.
If the hybrids have:
• Bright red eyes (mutant phenotype) – cinnabar
and scarlet mutations are alleles of the same
gene
• Dark red eyes (WT phenotype) - cinnabar and
scarlet are not alleles of the same gene but
rather, mutations in two different genes – each
involved in the control of eye pigmentation.
Test of a third mutation called cinnabar-2 for
allelism with the cinnabar and scarlet mutations:
• hybrid combination of cinnabar-2 and cinnabar
has bright red eyes (mutant phenotype) – the
mutations cinnabar and cinnabar-2 are alleles of
the same gene controlling eye pigmentation.
• hybrid combination of cinnabar-2 and scarlet
has dark red eyes (WT phenotype) - the
mutations scarlet and cinnabar-2 are not alleles
of the same gene. Rather the scarlet mutation
defines another color-determining gene.
 Variation Among the Effects of Mutations

• Genes are identified by mutations that alter the phenotype in some way.
E.g a mutation may change the color or shape of the eyes, alter a behavior
may cause even death → tremendous variation among the effects of
individual mutations suggests that each organism carries many different
kinds of genes and that each can mutate in different ways – in nature,
mutations provide the raw material for evolution.
• Visible mutations – mutations that alter some aspect of morphology. Most
visible mutations are recessive but a few are dominant.
• Sterile mutations – mutations that limit reproduction. Some sterile
mutations affect both sexes, but most affect either males or females.
Mutations can be either recessive (impair reproduction slightly) or dominant
(impair reproduction completely).
• Lethal mutations – mutations that interfere with necessary vital functions.
Effect: death.

• Dominant lethals that act early in life are lost one generation after they
occur because the individuals that carry them die. However, dominant
lethals that act later in life, after reproduction, can be passed on the next
generation.

• Recessive lethals may linger a long time in a population because they can
be hidden in heterozygous condition by a wild-type allele – can be detected
by observing unusual segregation ratio in the progeny heterozygous
carriers.

• E.g: The yellow-lethal mutation, AY in mice.

• Lethal genes – cause death of organism (unable to live)

• If the lethal effect is dominant and immediate in expression, all individuals carrying
the gene will die and the gene will be lost.

• There are dominant and recessive lethal genes.

• If the lethal gene is dominant – both types of individuals (homozygote – AA and

heterezygote – AB) will die.

• If the lethal gene is recessive –only homozygote individual will die.

• Some dominant lethals, however, have a delayed effects so that the organism lives
for a time.

• Recessive lethals carried in the heterozygous condition have no effect but may
come to expression when matings between carriers occur.

• Dominant gene can only be inherited if the effect of the gene is shown after
fertilisation.

• Eg. Human; Huntington disease – show the effect in adult individual, nerve
degeneration occur, physical and mental retardation until they die.
A dominant visible mutation –
causes fur to be yellow instead of
gray-brown (the color, also known
as agouti, which is determined by
the allele Ay).
However, the AY mutation is a
recessive lethal, killing AYAY
homozygotes early in their
development.
A cross between AYAy
heterozygotes produces two kinds
of viable progeny: yellow (AYAy)
and gray-brown (AyAy) in a ratio of
2:1. – different from the 3:1 ratio
that would be obtained if AY were
simply a dominant visible
mutation.

Y, the yellow-lethal mutation in mice: a dominant visible that is also a recessive lethal.
• Eg. Lethal recessive kills –at embryo stage – yellow mouse.

- Cross between yellow mouse with another yellow mouse produces yellow
and grey F1 with the ratio of 2:1

Yellow Yellow
P Yy x Yy

Lethal Yellow Grey

F1 YY Yy yy
1/4 2/4 1/4

- Allele Y has two characteristics – act as dominant allele for yellow colour.
- recessive allele for lethal.

- Therefore, all F1 progeny that has YY genotype die at embriyo stage and
ratio between yellow and grey is 2:1.
 Genes Function to Produce Polypeptides

• How do genes affect phenotype? What is it about a gene

that enables it to influence a trait such as eye color, seed
texture, or plant height?
• Genes encode information which is transcribed and
translated into products known as polypeptides. These
polypeptides are the fundamental constituents of proteins. Two
or more polypeptides may combine to form a protein.
• Some proteins, called enzymes, function as catalysts in
biochemical reactions, other form the structural components of
cells and others are responsible in transporting substances
within and between cells.
 Relationship between genes and polypeptides.
• Beadle and Tatum
proposed that each
gene is responsible
for the synthesis of
one polypeptide.
• When a gene is
mutated, its
polypeptide product is
either not made or is
altered in such a way
that its role in the
organism is changed.

• Mutations that eliminate or alter the polypeptide are often

associated with a phenotypic effect – whether the effect is dominant
or recessive depends on the nature of the mutation.
 Factors Affecting Gene Function

1. The environment
➢ A gene must function in the context of both a biological and a
physical environment. The physical environment can affect a
gene’s function.
➢ E.g: Drosophila mutation known as shibire – Japanese word for
“paralysis”. At normal culturing temperature (25C) – shibire flies
are viable and fertile, but are extremely sensitive to sudden shock.
When a shibire culture is shaken, the flies, temporarily paralyzed,
fall to the bottom of the culture. At slightly higher temperature
(29C) – all the flies fall to the bottom and die, even without a
shock. - At 25C, the mutation is viable but at 29C it is lethal →
 the phenotype of the shibire mutation is temperature-sensitive.
➢ Plausible explanation: At 25C the mutant gene makes a partially
functional protein but at 29C, this protein is totally non-functional.
2. Penetrance and Expressivity (Genetic background)
➢ When individuals do not show a trait even though they
have the appropriate genotype, the trait is said to exhibit
incomplete penetrance.
➢ E.g: polydactyly – the presence of extra fingers and toes
in human. – due to a dominant mutation that is
manifested in some of its carriers.
➢ Due to a dominant mutation, P, that is manifested in
some of its carriers.
➢ Incomplete penetrance can be a serious problem in
pedigree analysis because it can lead to incorrect
assignment of genotype.
➢ Expressivity – is used if a trait is not manifested uniformly
among the individuals that show it.
➢ E.g: The dominant Lobe eye mutation in Drosophila. The
phenotype associated with this mutation is extremely variable.
➢ Some heterozygous flies have tiny compound eyes,
whereas others have large, lobulated eyes; between these
extremes, there is a full range of phenotypes.
➢ The Lobe mutation is therefore said to have variable
expressivity.

Both incomplete penetrance and variable expressivity indicate that

the pathway between a genotype and its phenotypes is subject to
considerable modulation. Geneticists know that some of the
modulations are due to environmental factors but some are also
due to genetic background.
Variable expressivity of the Lobe mutation in Drosophila.
3. Gene interactions
• A trait can be influenced by more than one gene.
• Discovered by Bateson and Punnettt from breeding
experiments with chickens.
• Domestic chickens have different comb shapes.
• Discovered that comb type is determined by 2
(independently assorting) genes, R & P - each with 2 alleles.
• Different combinations of alleles from the 2 genes resulted in
different phenotypes - presumably due to interactions between
their products at the biochemical or cellular level.
• Mendel – gene: isolated elements that will produce different
effects – several examples support the idea.

• Scientists – found that genes can interact and produce new

phenotypes.

• A gene interaction indicates that the phenotype is determined

by the contribution of two or more genes.

• According to Mendelian inheritance – a cross involving two

heterozygous individuals for two genes that obey the laws of
segregation and independent assortment – should produce a
9:3:3:1 ratio of phenotypes.

• Deviations from this ratio are a good indication of gene

interactions.
Comb shape in chicken – ratio 9:3:3:1

• The first indication of gene interactions – provided by Bateson and

Punnet (1906) – study comb morphology in chickens.

• The results – produced four distinct phenotypes in a 9:3:3:1 ratio, rather

than the four combinations of two phenotypes in a 9:3:3:1 ratio expected
from a simple Mendelian dihybrid cross.

• Not all gene interactions will produce different F2 ratio (9:3:3:1).

• Eg. Chicken comb shape in cross between Wyandotte and Brahma

chicken.

• F2 ratio = 9:3:3:1, but different phenomena from Mendel.

• Two pairs of different allele – determine the same phenotype.

• When both pairs of allele are dominant (homozygote – RR /

heterozygote – RP) – the allele produces new phenotype.
Comb shapes in chickens of different breeds.

Rose, Wyandottes Pea, Brahmas

Walnut, hybrid Single, Leghorns

from cross
between chicken
with rose and
pea combs
 Wyandotte (rose) – determined
by R gene.

Brahma (pea) – P gene

F1 = Walnut

R gene alone – rose

P gene alone – pea
Combination of R & P – walnut
Combination of recessive
genes, r & p – single

Bateson and Punnett’s

experiment on comb shape in
chickens. The intercross in the
F1 produces four phenotypes,
each highlighted by a different
colour in the Punnet square, in a
© 2003 John Wiley and Sons Publishers
9:3:3:1 ratio
Colour of cow’s hair (Ratio 9:6:1)

Red cow x White cow

F1 Red
F1 x F 1 --- 3 groups of F2 progeny in ratio 9 red: 6 yellowish: 1 white

• Eg. Dihybrid cross with complete dominance at both pairs of gene, but the gene
interaction between both dominance genes will produce the new phenotype.

• Explanation:
Gene A – determine yellowish colour, dominant to allele a (white).
Gene B – also determine yellowish colour, dominant to allele b (white)
Gene interaction between A and B genes – produce red colour

AA BB x aa bb
red white
F1 Aa Bb
red
F2: genotype phenotype
9 A_ B_ Red (gene interactions A & B) 9
3 A_ bb Yellowish (because of A) } =6
3 aa B_ Yellowish (because of B) }
1 aa bb White (because no A & B) 1
4. Epistasis
• When 2 or more alleles influence a trait, an allele of one of
them may have an overriding effect on the phenotype - known
as epistasis.
• E.g. mutations determining Drosophila eye colour.
cinnabar white
X
bright red eye white eye

cinnabar,white

white eye
overrides
white mutation is epistatic to
white cinnabar cinnabar mutation.
• Epistasis : a type of gene interaction which involve masking of a gene
expression (A/a) by another gene (B/b) which is nonallelic gene. (Any gene
that masks the expression of another, nonallelic gene is epistatic to that
gene).

• Gene that masks expression of another gene – epistatic gene.

• Gene that has been masked – hipostatic gene.

• Epistasis should not be confused with dominance. Epistasis is the

interaction between different genes (non-alleles). Dominance is the
interaction between different alleles of the same gene.

Ratio 9:3:4

• Ratio produced when allele aa, in homozygote recessive condition, epistatic

towards B and b genes.

• aa will mask the expression of B or b, although gene a is recessive to A

gene.

• Recessive epistasis – because epistatic gene is a recessive gene (a).

• Eg. Colour of rodent hair (normally is grey)

CC aa x cc AA
Black Albino

F1 Cc Aa
Grey
Self-cross between F1 – produce F2 with ratio:
Genotype Phenotype
9 C_ A_ Grey 9
3 C_ aa Black 3
3 cc A_ Albino}
1 cc aa Albino} 4

For the colour of this rodent;

Gene A – determine grey colour
Gene a – determine black colour
Gene c – in homozygote, epistatic to A and a.
Therefore, phenotypes for cc A_ and cc aa are albino because cc masked the
expression of A_ dan aa, thus produce ratio of 9 grey: 3 black: 4 albino.
Ratio 12:3:1

• Produce with dominant epistasis, gene A epistatics to B and b.

• Eg. Cucurhita pepo. fruit

AA BB x aa bb
white green

F1 Aa Bb
white

F2 (F1 x F1):
Genotype Phenotype
9 A_ B_ White} 12
3 A_ bb White}
3 aa B_ Yellow 3
1 aa bb Green 1
• Gene B – bring yellow colour and dominant to gene b (bring green colour).
• The expression of gene B in A_ B_ class and expression of bb in A_bb
class is masked by gene A.
• B and bb in aa B_ class and aa bb can express their phenotypes (because
there is no gene A).
Students should be able to:

1. Understand gene interactions.

2. Understand epistasis and lethal genes.
3. Know the concept of polygene and polygenic inheritance.
4. Factors effecting gene functions.
Quiz:

1. Assume that “S” represents the dominant trait of having long

leaves, and “s” represents the recessive short-leafed trait in a
plant. In the parental generation, you cross a homozygous
long-leafed plant with a homozygous short-leafed plant. Draw
a Punnet square illustrating this cross, and give the genotypes
and phenotypes of the F1 generation.

2. Using a Punnet square, show the genotypes and phenotypes

of the F2 generation if the F1 plants in question 1 are self-
fertilised.