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OCR A2 F215 GENETICS
Specification:
a) Explain the terms allele, locus, phenotype, genotype, dominant,
codominant and recessive
b) Explain the term linkage
c) Use genetic diagrams to solve problems involving sex linkage and
codominance
d) Describe the interactions between loci (epistasis). Production of
genetic diagrams is not required
e) Predict phenotypic ratios in problems involving epistasis
f) Use the chi-squared (χ2) test to test the significance of the difference
between observed and expected results.(The formula for the chisquared test will be provided))
Definitions in Genetics
Terms to Define
Genetics
Definition
 The study of the inheritance of
genes
 A length of DNA that codes for
a specific polypeptide
 A gene is found at a gene
locus
 A variety or specific form of a
gene
 An alternative form of a gene
 The position of a gene on a
chromosome
 The physical features
observed when the genotype is
expressed
 The outcome of the interaction
between the genotype and the
environment
 The combination of all the
alleles possessed by an
organism
 An allele having an effect on
the phenotype even when a
recessive allele is also present
 The situation when both alleles
have an effect on the
Gene
Allele
Gene locus
Phenotype
Genotype
Dominant allele
Codominance
1
Recessive allele

Multiple alleles

Homozygous condition



Heterozygous condition

Autosomal linkage



Sex linkage


Epistasis

phenotype in a heterozygote
An allele that only has an
effect on the phenotype when
dominant alleles are not
present
When there are more than two
possible alleles at a gene locus
When the same allele is
present on both homologous
chromosomes at a particular
gene locus
If both alleles are dominant,
the individual is homozygous
dominant
If both alleles are recessive,
the individual is homozygous
recessive
When the homologous
chromosomes have different
alleles at the same gene locus
Linked genes are present on
the same chromosome
The autosomal chromosomes
are all the chromosomes
except the sex chromosomes
Refers to one (or more) genes
located on the sex
chromosomes
Most sex linked genes are on
the X chromosome
The situation when the
expression of one gene is
affected by another gene
It is the interaction of genes to
control a feature
Single Gene Inheritance
Example 1 - Cystic fibrosis
Background

Cystic fibrosis is a genetic disease in which abnormally thick mucus is
produced in the lungs and other parts of the body.

Patients are prone to lung infections because the mucus remains in the
airways and is a breeding ground for bacteria.
2

Cystic fibrosis is caused by mutation in a gene that codes for the
production of a protein called CFTR. This protein forms a channel in
the plasma membranes of cells lining the lung airways and the
intestines and allows chloride ions to diffuse from inside the cells to the
outside

The most common mutation in the CFTR gene results in the deletion of
three nucleotides. The resulting protein therefore, has one amino acid
missing in its primary structure

The protein produced is not recognised within the cell and is not
inserted as a transport protein in the cell plasma membranes
Inheritance of Cystic Fibrosis

The faulty allele for CFTR is a recessive allele. The normal allele is
dominant

Represent these alleles by the same letter. Choose a letter that looks
different when written in upper and lower case

It is conventional to use an upper case letter for the dominant allele
and a lower case (same letter) for the recessive allele

Let F represent the dominant allele

Let f represent the recessive allele

Each person has two copies of each gene (within a pair of
homologous chromosomes)
3
4
Possible Genotypes and Phenotypes for CFTR Synthesis in the Human
Population
Genotype
FF
Description of
Genotype
Homozygous
dominant
Ff
Heterozygous
ff
Homozygous
recessive
Phenotype
Unaffected by cystic
fibrosis. CFTR transport
proteins are produced
Unaffected by cystic
fibrosis
Sufferer of cystic fibrosis

When gametes are made by meiosis, the daughter cells only get one
copy of each pair of homologous chromosomes

Therefore the gametes only contain one copy of each gene – only one
allele of the CFTR gene
Parental and Gamete Genotypes of the CFTR Gene
Parental Genotype
FF
Ff
ff

Gamete Genotypes
All F
50% F and 50% f
All f
At fertilisation, any gamete from the father can fertilise any gamete
from the mother
Genetic Diagram

A genetic diagram is a conventional way of showing the relative
chances of a child of a certain genotype or phenotype being born to
parents with particular genotypes/phenotypes

Always include a key with your genetic diagram to indicate the symbols
that relate to each allele

The diagram should always be laid out in a conventional way, as
shown on page 6. The gametes are shown by encircled single letters
representing the presence of one allele

The table showing the fertilisation of gametes is called a Punnett
Square
5
Genetic Diagram to Show the Chances of A Cystic Fibrosis Child being
Born to Parents that are Heterozygous for the CFTR Gene

This genetic diagram indicates that every time the couple have a child,
there is a 25% chance or probability that they will have a child with
cystic fibrosis

An alternative way of expressing this is that the probability of the child
having the genotype ff is 0.25

The probability of these parents having two children (not identical
twins) with cystic fibrosis is (0.25 x 0.25) = 0.0625
Codominance

The situation where both alleles of a genotype have equal effects on
the phenotype of the heterozygote, neither is recessive or dominant
Codominance Example 1 - Inheritance of the ABO Blood Group

Red blood cells contain a glycoprotein in their plasma membranes that
determines the ABO blood group

There are two forms of this protein, known as antigens A and B

The gene determining this blood grouping has three alleles, coding for
antigen A, antigen B or no antigen at all
6

The symbols representing these alleles include the letter I for the gene
locus (I indicates the immunoglobulin glycoprotein), a superscript
distinguishes the three alleles
The three alleles are:
 IA allele for antigen A
 IB allele for antigen B
 IO allele for no antigen
IA and IB are codominant alleles
Both IA and IB are dominant to IO
Possible Genotypes and Phenotypes
Genotype
IAIA
IAIB
IAIO
IBIB
IBIO
IOIO
Phenotype
Group A
Group AB
Group A
Group B
Group B
Group O
A Genetic Diagram to Determine the Chance of a Child with Blood Group
O being born to a Heterozygous Man with Blood Group B and a
Heterozygous Woman with Blood Group A
Parental Phenotypes
Male
Blood group B
IBIO
Parental Genotypes
Gamete Genotypes
Female
Blood Group A
IB
IAIO
IO
IA
IO
Offspring Genotypes and Phenotypes from a Punnett Square
IB
IA
IO
IO
IAIB
IAIO
blood group AB
blood group A
IBIO
IOIO
blood group B
blood group O
Probability of a child with O blood group being born as the first child is 25% or
0.25
7
Codominance Example 2 - Sickle Cell Anaemia
Background

A human genetic disease caused by a gene mutation

The β- polypeptide chains of each haemoglobin molecule differ by one
amino acid. Glutamic acid in the normal chain is substituted by valine in
the mutant polypeptide

This single amino acid difference makes the haemoglobin molecule
insoluble when it is deoxygenated. The abnormal haemoblobin is
crystalline and aggregates into more linear and less globular structures

The abnormal haemoglobin deforms the red blood cells. The cells are
often sickle shaped and cannot easily squeeze through blood
capillaries

After repeated oxygenation-deoxgenation cycles, some red cells are
irreversibly sickled

If sickled cells block capillaries, they reduce blood flow in organs,
particularly bones, heart, lungs and kidneys. These organs suffer
tissue damage
Genotypes and Phenotypes in the Human Population

One gene controls the synthesis of β-polypeptide of haemoglobin

The normal allele is represented by HA

The sickle cell allele is represented by HS
8
Table Showing Possible Genotypes and Phenotypes in the Human
Population
Genotypes
HAHA
Phenotypes
Gamete genotypes
HSHS
Sickle cell anaemia
HAHS
Symptomless carriers of
the sickle cell allele.
Normal haemoglobin
Since the normal and
sickle cell alleles are
codominant, the red blood
cells contain normal and
mutant haemoglobin.
However, the normal
haemoglobin prevents
sickling of the red blood
cells
A Genetic Diagram to Determine the Chance of a Child with Sickle Cell
Anaemia being born to Heterozygous Parents, both Symptomless
Carriers of the Sickle Cell Allele
Parental Phenotypes
Male
Symptomless Carrier
Female
Symptomless Carrier
HAHS
Parental Genotypes
HAHS
HA
Gamete Genotypes
HS
HA
HS
Offspring Genotypes and Phenotypes from a Punnett Square
HA
HS
HAHA
HA
HAHS
Normal haemoglobin
Symptomless carrier
HAHS
HS
HSHS
Symptomless carrier
Sickle cell anaemia
Phenotypic ratio:
25% normal haemoglobin:
50% carrier:
9
25% sickle cell anaemia
Codominance Example 3 – Inheritance of MN Blood Grouping in
Primates
Background

The MN blood grouping system is controlled by 2 codominant alleles at
one gene locus

Allele M codes for the synthesis of glycoprotein antigen M in the
plasma membrane of red blood cells

Allele N codes for the synthesis of glycoprotein antigen N in the plasma
membrane of red blood cells

The alleles are denoted by GYPAM and GYPAN
Genotypes and Phenotypes in the Human Population
Genotypes
GYPAM GYPAM
Phenotypes
Blood group M
Gamete Genotypes
Homozygous dominant
GYPAN GYPAN
Blood group N
Homozygous dominant
GYPAM GYPAN
Blood group MN
Heterozygous
Complete the Genetic Diagram below to determine the Phenotypic Ratio
of Offspring to two Parents with Blood Group MN
Parental Phenotypes
Male
Blood group MN
Female
Blood group MN
Parental Genotypes
GYPAM GYPAN
GYPAM GYPAN
Gamete Genotypes
GYPAM
GYPAN
GYPAM
GYPAN
Offspring Genotypes and Phenotypes from a Punnett Square
GYPAM
GYPAN
GYPAM
GYPAM GYPAM
Blood Group M
GYPAM GYPAN
Blood Group MN
GYPAM GYPAN
Blood Group MN
GYPAN GYPAN
Blood Group N
GYPAN
Phenotypic ratio:
1: 1: 2 There’s an 0.25 probability of the offspring obtaining the Blood
Group M and N and a 0.5 probability of the offspring having a MN Blood
Group.
10
Codominance Example 4 – Coat Colour in Shorthorn Cattle
Background

One of the genes coding for coat colour in shorthorn cattle has 2 alleles

These 2 alleles are denoted by CR coding for red/chestnut hairs and
CW coding for white hairs

CR and CW are codominant alleles. The heterozygous genotype CRCW
produces an animal with a mixture of red and white hairs referred to as
roan
Genotypes and Phenotypes for Coat Colour in the Shorthorn Cattle
Population
Genotype
Phenotype
Red/chestnut coats
Gamete Genotypes
Homozgous dominant
White coats
Homozygous dominant
Red and white hairs
described as roan
Heterozygous
CRCR
CWCW
CRCW
Genetic Diagram of Cross between Red Shorthorn Cow and White
Shorthorn Bull
Parental phenotypes
Cow
Red Coat
Bull
White Coat
Parental genotype
CRCR
CWCW
Gamete genotypes
CR
CW
CRCW
F1 Offspring genotype
F1 Offspring phenotype
Roan coat
11
Complete the following:
Genetic Diagram of Cross between Roan Shorthorn Parents
F1 phenotypes
Roan
Roan
F1 genotypes
CRCW
CRCW
CR CW
F1 gamete genotypes
F2 genotypes and phenotypes from a Punnett Square
CR
CW
CR
CRCR
Red Coat
CRCW
Roan Coat
CW
CRCW
Roan Coat
CWCW
White Coat
F2 Phenotypic Ratio:
1:1:2 where there’s a 0.25 probability of the offspring’s coat will be red
or white and 0.5 probability of the offspring coat is roan.
Codominance Example 5 – Flower colour in Snapdragons (Antirrhinum)
Background

Flower colour in snapdragons is controlled by a single gene with 2
alleles

One allele (denoted by CR) codes for an enzyme that catalyses the
formation of a red pigment in flowers

The other allele (CW) codes for an altered enzyme that lacks this
catalytic activity and does not produce a pigment. Plants with the
genotype CWCW are white

Heterozygous plants (genotype CRCW) with their single allele for red
pigment formation, produce sufficient red pigment to produce pink
flowers
12
Genetic Cross between Snapdragons with Red Flowers and White
Flowers
Genetic Cross between Pink Flowered Snapdragons
13
Sex Determination







Sex is determined by chromosomes rather than genes. The sex
chromosomes are X and Y and they are not homologous. The X
chromosome is longer than the Y chromosome
In humans, females have two X chromosomes (XX). Female
gametes all contain one X chromosome. Human females are
therefore the homogametic sex
Human males have one X and one Y chromosome in their diploid cells
(XY) and produce two types of gamete. 50% of their gametes have an
X chromosome and 50% have a Y chromosome. Human males are
the heterogametic sex
Sex determination is the same in other organisms but the homogametic
and heterogametic sexes may be reversed
In birds, moths, many reptiles and all butterflies, the female is the
heterogametic sex (XY) and the male is homogametic (XX)
Sometimes the Y chromosome is absent. In some insects, the female
has two X chromosomes (XX) whilst the male has just one sex
chromosome (denoted by XO)
Amongst fish and some reptiles, environmental conditions such as
temperature, can affect sex determination
Sex Linkage
 Any gene carried on the X or Y chromosome is described as sex linked
 Very few genes are carried on the Y chromosome in humans. The
SRY gene (sex determining region on the Y chromosome) is only
present in males
 Because it is longer, the X chromosome carries many genes that are
not present on the Y
 The inheritance of the sex linked genes is affected by the person’s
gender - if male or female
 Some of the genes on the X chromosome have recessive alleles that
cause particular abnormal phenotypes
 Since males only have one X chromosome in their body cells, if they
inherit a recessive allele for one of these conditions, they will inherit the
abnormal phenotype
 Therefore, sex linked recessive diseases are much more common
in males than females
Sex Linked Condition Example 1 - Haemophilia
Background

One gene on the X chromosome controls the production of a protein
called factor VIII that is needed for blood clotting

The recessive allele of this factor VIII gene codes for a faulty version of
factor VIII. Blood does not clot properly, a condition called
haemophilia
14

In haemophilia, bleeding occurs into joints and other parts of the body.
It causes great pain and disables the sufferer

Haemophilia can be treated by giving factor VIII throughout life

Since a male only has one X chromosome in his diploid cells, inherited
from his mother, if he has the recessive allele for factor VIII production,
he will be a haemophiliac
Convention for Representing Sex Linked Genes



Note that genes on the X chromosome are represented by superscripts
The normal allele for factor VIII production is represented by XH
The haemophilia allele is represented by Xh
Genotypes and Phenotypes for Factor VIII Gene in the Human
Population
Genotype
Gender
Phenotype for Blood
Clotting
Normal blood clotting
XHXH
Female
XHXh
Female
X hX h
Female
XHY
Male
Normal blood clotting (a
symptomless carrier)
Lethal. Foetus does not
develop
Normal blood clotting
X hY
Male
Haemophilia
15
Genetic Diagram to Show how a Woman Carrier for Haemophilia and a
Man with Normal Blood Clotting can produce a Haemophiliac Son
Pedigree Chart Showing the Inheritance of Haemophilia from Queen
Victoria in Members of various European Royal Families
Background to Pedigree Charts
 Pedigree charts are a useful way of tracing the inheritance of
sex linked diseases such as haemophilia
 A male is represented by a square and a female by a circle
16



Complete shading within either shape indicates the phenotypic
presence of a feature such as haemophilia
A dot within a shape indicates a normal phenotype who carries
the abnormal allele – a carrier of the abnormal allele
The pedigree chart on page 16 shows that only males were
haemophiliacs but their mothers (like Queen Victoria herself)
were symptomless carriers of the haemophilia recessive allele.
The haemophiliac sons inherited the recessive allele on the X
chromosome from their mothers
Sex Linked Condition Example 2 – Duchenne Muscular Dystrophy (DMD)
Background





The X chromosome also carries the DMD gene that codes for the
synthesis of a muscle protein called dystrophin
Dystrophin is a large protein needed for muscle contraction
Mutations of the DMD gene result in no dystrophin synthesis or a much
shorter protein
Boys with DMD develop muscle weakness in childhood and are
wheelchair bound by the age of 10. Death occurs by the early 20’s due
to muscle degeneration particularly involving cardiac and respiratory
skeletal muscles
A Family Pedigree Chart Showing the Inheritance of DMD
Let XD denote the normal allele for dystrophin synthesis and Xd, the
mutant allele
17

Complete the table below to indicate the genotypes of the named
individuals in this pedigree. Each individual may have one or two
genotypes
Individual in the pedigree
Mary
Genotype
Astrid
Jack
Jane
Complete the Genetic Diagram to Show the Probability of Leon and Jane
having Another Child with DMD
Parental Phenotypes
Leon/Male
Normal dystrophin
XDY
Parental Genotypes
XD
Gamete Genotypes
Jane/Female
Normal dystrophin/carrier
XDXd
Y
XD XD
Punnett Square Showing Offspring Genotypes and Phenotypes
XD
Y
XD
XD XD
Normal female
XD Y
Normal male
Xd
XD Xd
Female carrier
Xd Y
Male with DMD
Phenotypic Ratio: 1:1:1:1
Probability of having another child with DMD is: probability of having a
child with DMD is 0.25
Sex Linked Condition Example 3 – Red- Green Colour blindness
Background

The gene controlling normal red-green colour vision is located on the X
chromosome
18


A recessive mutant allele causes red-green colour blindness
Red-green colour blindness is more common in males than in females
since males only need one recessive allele on the X chromosome
inherited from their mother, who carries the recessive allele. Females
need to inherit two recessive alleles, one from a carrier/colour blind
mother and the other from a colour blind father
People with normal red-green colour vision can read the numbers embedded
in the pattern. Those with red-green colour blindness cannot.
Select appropriate symbols to denote the alleles for normal and
abnormal red-green colour vision and complete the genetic diagram to
determine the probability of a carrier mother and normal vision father
producing a red-green colour-blind offspring
Let XRG represent the allele for normal red-green colour vision
Let Xrg represent the abnormal allele
Parental phenotypes: normal red-green colour
Parental genotypes:
XRG Y
Gamete genotypes:
XRG Xrg Y
normal vision (carrier)
XRGXrg
Punnett Square showing the offspring genotypes and phenotypes
XRG
Y
XRG
Xrg
XRG XRG
Female normal redgreen vision
XRG Y
Female normal redgreen vision
XRG Xrg
Female normal vision
(carrier)
Xrg Y
Male with red-green
colour blindness
Phenotype ratio: 1:1:2
Probability of these parents having a red-green colour-blind child: There
is a 0.25 probability of these parents having a male with red-green colour
blindness.
Dihybrid Inheritance
 Dihybrid inheritance is the study of the inheritance of two genes at the
same time
 If these two genes are on different chromosomes they are unlinked
 If these two genes are on the same chromosome they are linked
19

The current specification does not include dihybrid inheritance and you
will not be expected to write out genetic diagrams for these crosses.
However, it is useful to work through an example to improve your
understanding of epistasis that is on the specification
Dihybrid Inheritance of Two Genes that are on Different Chromosomes
One of Gregor Mendel’s genetic studies involved the inheritance of pea seed
colour and shape.
Pea seed colour is either yellow or green
Pea seed shape is either round or wrinkled
Since yellow colour is dominant to green:
G denotes the yellow coloured allele and g denotes the green coloured allele
Since round shape is dominant to wrinkled shape:
R denotes the round allele and r denotes the wrinkled allele
Mendel’s Experiment
1) Mendel crossed pure breeding pea plants producing yellow, round
seeds with pure breeding pea plants producing green, wrinkled
seeds. Pure breeding means that these parents were homozygous for
these two seed features
2) He collected the F1 seeds and observed that they were all yellow and
round
3) He then allowed the F1 plants to self-pollinate and self-fertilise and
collected the F2 generation seeds and observed their features
Genetic Diagram showing Mendel’s Genetic Crosses with Pea Plants
Parental Phenotypes
yellow, round
seeded plants
Parental Genotypes
GGRR
Gamete Genotypes
x
green, wrinkled
seeded plants
ggrr
GR
gr
F1 Genotypes
GgRr
F1 Phenotypes
yellow, round seeds
20
F1 Self Pollination/Self Fertilisation
F1 Phenotypes
yellow, round
Seeded plants
F1 Genotypes
x
yellow, round
seeded plants
GgRr
Gamete Genotypes
GR
GgRr
Gr
gR
gr
GR Gr gR gr
Punnett Square to Show F2 Genotypes and Phenotypes
GR
Gr
gR
gr
GGRR
GGRr
GgRR
GgRr
yellow round
yellow round
yellow round
yellow round
GGRr
GGrr
GgRr
Ggrr
yellow round
yellow wrinkled
yellow round
yellow wrinkled
GgRR
GgRr
ggRR
ggRr
yellow round
yellow round
green round
green round
GgRr
Ggrr
ggRr
ggrr
yellow round
yellow wrinkled
green round
green wrinkled
GR
Gr
gR
gr
F2 Phenotype ratio:
9:
yellow round
3:
yellow wrinkled
3:
green round
1
green wrinkled
This phenotype ratio of 9:3:3:1 is typical in the F2 generation when two
unlinked genes are inherited by self fertilisation of two parents both
heterozygous for two features
Epistasis
 The situation where the expression of one gene is affected by another
gene. It is the interaction of two genes to control a feature. One gene
locus suppresses or masks the expression of another gene locus
Features of Epistasis


It is not inherited
It reduces phenotypic variation because more genes are working
together to influence the gene expression
21
Types of Gene Interaction
1) Two genes may work against each other – they are antagonistic.
One gene masks the expression of the other
2) Two genes may work together in a complementary way
Antagonistic Epistasis
Two types:
 recessive epistasis
 dominant epistasis
Recessive Epistasis
This is where the homozygous presence of two recessive alleles at one gene
locus prevents the expression of alleles at a second gene locus
The homozygous recessive alleles at gene locus 1 are epistatic to the alleles
at gene locus 2.
The alleles at gene locus 2 are hypostatic
Example of Recessive Epistasis – Inheritance of Flower Colour in Salvia




Two gene loci are involved denoted by A/a and B/b
Expression of B at gene locus 2 produces purple flowers.
Expression of bb at gene locus 2 results in pink flowers
Expression of the B/b alleles at gene locus 2 requires at least one
dominant A allele at gene locus 1
The occurrence of homozygous aa alleles at gene locus 1 is epistatic
to both B/b alleles at gene locus 2 and prevents colour expression
controlled by the second gene locus. In the presence of aa at gene
locus 1, the plant produces white flowers, regardless of the alleles at
gene locus 2
Note:

The specification states that students will not be expected to draw
genetic diagrams for epistasis examples in the examination

However, it is useful to work through some examples so that you
can understand how the phenotypic ratios are derived

You are expected to remember the epistatic ratios and which type
of epistasis they are linked to
Genetic Diagram Showing a Cross between Pure Breeding pink
Flowering Salvia and Pure Breeding White Flowered Salvia
Parental Phenotypes
Parental Genotypes
pure breeding
pink flowers
AAbb
pure breeding
white flowers
aaBB
22
Parental Gametes
Ab
aB
F1 Genotype
AaBb
F1 Phenotype
purple flowers
Genetic Diagram Showing an F1 Self-Pollination/Self-Fertilisation
F1 Phenotypes
purple flowers
F1 Genotypes
purple flowers
AaBb
F1 Gametes
AB
AaBb
Ab
aB
ab
AB
Ab
aB
ab
Punnett Square Showing the F2 Genotypes and Phenotypes
AB
AABB
Ab
AABb
aB
ab
AaBB
AaBb
AB
purple
purple
AABb
purple
AAbb
purple
AaBb
Aabb
Ab
purple
pink
AaBB
purple
AaBb
aaBB
pink
aaBb
aB
purple
purple
AaBb
white
Aabb
aaBb
white
aabb
ab
purple
Phenotypic ratio:
pink
9:
purple
white
3:
pink
white
4
white
Remember that a 9:3:4 ratio in the F2 generation with F1 parents both
heterozygous at both gene loci, indicates recessive epistasis
Dominant Epistasis
This is where the presence of a dominant allele (only one is needed) at the
first gene locus masks the expression of the alleles at a second gene locus
Example 1 of Dominant Epistasis – fruit colour in summer squash
 Two gene loci are involved D/d and E/e
 At the second gene locus, the presence of E (only one dominant allele
needed) causes the production of yellow fruit. The presence of ee
causes green fruit
 However, the expression of E or ee requires the presence of dd at the
first gene locus.
23

If there is one dominant D allele at gene locus 1, the squash fruit will be
white
Genetic Diagram Showing the Cross Between two White Coloured Fruited
Plants that are Double Heterozygous at the two Gene Loci
Parental Phenotypes
white fruit
white fruit
Parental Genotypes
DdEe
DdEe
Gamete Genotypes
DE De dE de
DE
De
dE
de
Punnett Square Showing the Genotypes and Phenotypes of the Offspring
DE
DE
De
dE
de
De
dE
de
DDEE
DDEe
DdEE
DdEe
white
white
white
white
DDEe
DDee
DdEe
Ddee
white
white
white
white
DdEE
DdEe
ddEE
ddEe
white
white
yellow
yellow
DdEe
Ddee
ddEe
ddee
white
white
yellow
green
Phenotypic Ratio:
12(white): 3(yellow): 1(green)
Remember that a 12:3:1 phenotypic ratio in the offspring of two parents
that are heterozygous at both gene loci, indicates dominant epistasis
Example 2 of Dominant Epistasis – feather colour in chickens



Two gene loci are involved denoted by I/i (first gene locus)and C/c
(second gene locus)
The presence of only one dominant I allele at the first gene locus
masks the expression at the second gene locus, producing white
chickens
At the second gene locus, only the dominant allele C causes the
production of coloured chickens. Two recessive alleles (cc) at the
second gene locus produces white chickens also
Genetic Diagram to Show the Cross between Two White Chickens,
Heterozygous at both Gene loci
Parental Phenotypes
white
Parental Genotypes
IiCc
Gamete Genotypes
IC
Ic
iC
white
IiCc
ic
24
IC
Ic
iC
ic
Punnett Square Showing the Genotypes and Phenotypes of the
Offspring
IC
Ic
iC
ic
IC
IICC
Ic
IICc
iC
IiCC
ic
IiCc
white
white
white
white
IICc
IIcc
IiCc
Iicc
white
white
white
white
IiCC
IiCc
iiCC
iiCc
white
white
coloured
coloured
IiCc
Iicc
iiCc
iicc
white
white
coloured
white
Phenotypic ratio:
13 (white) : 3 (coloured)
Remember that a 13:3 ratio of offspring from two parental chickens that
are heterozygous at both gene loci, suggests dominant epistasis
Complementary Gene Interaction
Example – white and purple flower colour in sweet peas

Two gene loci are involved C/c and R/r

Purple coloured flowers are only produced if there is at least one
dominant allele at each gene locus (C and R)

The homozygous recessive condition at either gene locus masks the
expression of the dominant allele at the other gene locus

It is suggested that the mechanism of action of these two genes is as
follows:
allele C
allele R
enzyme 1
precursor substance
(colourless)
enzyme 2
 intermediate compound  final pigment
(colourless)
(purple)
Genetic Diagram of a cross between two Heterozygous Parents at both
gene loci
Parental Phenotype
purple flowers
Parental Genotype
Gamete Genotype
purple flowers
CcRr
CR
Cr
CcRr
cR
25
cr
CR
Cr
cR
cr
Punnett Square Showing the Genotypes and Phenotypes of the
Offspring
CR
Cr
cR
cr
CR
CCRR
Cr
CCRr
cR
CcRR
cr
CcRr
purple
purple
purple
purple
CCRr
CCrr
CcRr
Ccrr
purple
white
purple
white
CcRR
CcRr
ccRR
ccRr
purple
purple
white
white
CcRr
Ccrr
ccRr
ccrr
purple
white
white
white
Phenotypic ratio:
9 (purple): 7 (white)
Remember that a 9:7 ratio in the offspring of two parents heterozygous
at both gene loci, suggests complementary epistasis
Other Examples of Epistasis
Coat Colour in Mice

Coat colour may be agouti (alternating bands of melanin pigment
on each hair so that the coat looks grey/brown), black or
albino/white

The gene for agouti has two alleles, A/a. Allele A causes the
banding pattern of colouration called agouti. A mutant allele a
results in uniform black colouration of the hairs. The homozygous
aa condition produces a black haired mouse

A second gene at another gene locus controls the production of the
melanin pigment. This gene is denoted by B/b. For pigment
production by the A/a gene, there must be at least one dominant B
allele at the B/b gene locus

What type of epistasis is this an example of?
Possible Mechanism for Melanin Pigment Production in Mice
allele B
allele A
enzyme 1
enzyme 2
Precursor substance  black melanin pigment  agouti pattern
If the mouse is homozygous recessive at the B/b gene locus, no black
pigment is produced and the mouse is albino/white
26
Genetic Diagram to Determine the Genotypes and Phenotypes of Offspring
from Two Heterozygous Parents
Parental Phenotypes
Parental Genotypes
Gamete Genotypes
agouti coat
agouti coat
AaBb
AB Ab
AaBb
aB ab
AB Ab aB ab
Punnett Square to Show the Offspring Genotypes and Phenotypes
AB
Ab
aB
ab
AB
Ab
aB
ab
AABB
agouti
AABb
agouti
AaBB
agouti
AaBb
agouti
AABb
agouti
AAbb
white
AaBb
agouti
Aabb
white
AaBB
agouti
AaBb
agouti
aaBB
black
aaBb
black
AaBb
agouti
Aabb
white
aaBb
black
aabb
white
Phenotypic ratio: 9 (agouti): 3 (black): 4 (white), suggesting recessive
epistasis
Chi- squared Test

Breeding experiments involve an element of chance

The fusion of gametes at fertilisation is a random event and the
resulting offspring may only approximate to expected ratios

A chi-squared test is a statistical test carried out to determine if the
numbers obtained fit an expected ratio

Chi-squared is given the Greek letter χ2

There are two possible conclusions of the chi-squared test:
1) That there is no significant difference between the observed
and expected data. This means that the numbers obtained fit
the expected ratio
2) That there is a significant difference between the observed
and expected data
Example 1
The height of pea plants is controlled by two alleles at one gene locus.
Since tall is dominant to short, T represents the tall allele
Since short is recessive to tall, t represents the short allele
27
When pure breeding tall and short plants are crossed, the F1 offspring are all
tall, with heterozygous genotype Tt
When two heterozygous plants are crossed, the F2 offspring should show 3
tall plants to every 1 short plant ie a 3:1 ratio of tall:: short
In a breeding experiment, 127 F2 offspring were obtained, 100 tall and 27
short. Does this agree with the 3:1 ratio?
In this experiment, the obtained ratio in the F2 is 100/27 = 3.7: 27/27 = 1
ie 3.7:1.
Is this observed data significantly different from the expected data? To
answer this question, a chi-squared test is carried out as follows:
Χ2 =
∑ (O – E)2 / E
O = observed numbers obtained
E = expected numbers obtained
∑ = sum of
The hypothesis being tested is that the data obtained approximate to a
3:1 ratio
If the data does fit a 3:1 ratio:


the expected numbers are ¾ of 127 = 95.25 tall
and ¼ of 127 = 31.75 short
A table is constructed to calculate the chi-squared value
Class
tall
short
O
100
27
E
95.25
31.75
O-E
4.75
-4.75
(O-E)2
22.56
22.56
(O-E)2/E
0.237
0.710
Χ2 = 0.947
(obtained by adding together the two values for (O-E)2 / E in the table)
To find out what this number indicates, a probability table is used. The
data in the table are critical chi-squared values.
28
Degrees of freedom are calculated by counting the number of classes
minus one. In this case there is one degree of freedom
In biology, the probability level of 0.05 is used, unless you are told otherwise.
If the calculated chi-squared is greater than the critical P value at 0.05 level,
there is a significant difference between observed and expected data.
We are 95% certain that there is a significant difference
In this example, the chi-squared value is less than 3.84, therefore, there is
no significant difference between the observed and expected values at
the P = 0.05 level. The slight differences are due to chance
Question :
In Drosophila, the alleles for ebony body and curled wings are recessive to
the alleles for grey body and normal wings. A heterozygous grey-bodied,
normal winged fly was crossed with an ebony bodied, curled winged fly.
The following offspring were obtained:
Phenotype
Number
Grey body, normal wings
Grey body, curled wings
Ebony body, curled wings
Ebony body, normal wings
32
22
29
21
Are these numbers consistent with the expected 1:1:1:1 ratio?
No though expected no. for all are ¼ of the total no. which is 104
Complete the table on page 33 to determine if there is a significant difference
between the observed and expected data
29
Class
Grey and
normal
Grey and
curled
Ebony and
curled
Ebony and
normal
O
32
E
26
O-E
6
(O – E)2
36
(O – E)2/E
1.385
22
26
-4
16
0.615
29
26
3
9
0.346
21
26
-5
25
0.962
Calculated Χ2 =
3.308
How many degrees of freedom are valid for this comparison of observed
and expected data?
Degrees of freedom = 4-1 = 3.
From the critical chi squared table on page 32, determine the critical chi
squared value for P = 0.05 and compare it to the calculated chi-squared
value to determine if the observed and expected ratios are significantly
different
7.81 was the value in the table. The X2 value is below the value in the
table therefore there is no significant difference.
Null Hypothesis
You may be asked to write out a null hypothesis in chi squared tests.
The null hypothesis is always ‘there is no significant difference between
the observed and expected data’
The object of the chi-squared test is to prove or disprove the null hypothesis
Levels of Significance
As stated previously, P=0.05 is the normal probability level used by biologists.
However, if the calculated chi-squared value is greater than that at P=0.01,
the difference between observed and expected data is even more significant.
It is worthwhile stating this in an answer.
30