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 (
χ 2 ) 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
Gene
Allele
Gene locus
Phenotype
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
Genotype
Dominant allele
Codominance 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
1

Recessive allele
Multiple alleles
Homozygous condition
Heterozygous condition 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
Autosomal linkage
Sex linkage
Epistasis 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 Description of
Genotype
Phenotype
FF
Ff
Homozygous dominant
Heterozygous
Unaffected by cystic fibrosis. CFTR transport proteins are produced
Unaffected by cystic fibrosis ff Homozygous recessive
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 Gamete Genotypes
FF
Ff
All F
50% F and 50% f ff 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:

I A

I B

I O
I A and I B are codominant alleles
Both I A and I B are dominant to I O
Possible Genotypes and Phenotypes
Genotype
I A I A
I A I B
I A I O
I B I B
I B I O
I O I O
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 Female
Blood group B Blood Group A
Parental Genotypes I B I O I A I O
Gamete Genotypes I B I O I A I O
Offspring Genotypes and Phenotypes from a Punnett Square
I B I O
I
I
A
O I
I A I B blood group AB
B I O blood group B
I A I O blood group A
I O I O 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 H A

The sickle cell allele is represented by H S
8

Table Showing Possible Genotypes and Phenotypes in the Human
Population
Genotypes
H A H A
Phenotypes
Normal haemoglobin
Gamete genotypes
H
H
S
A
H
H
S
S
Sickle cell anaemia
Symptomless carriers of the sickle cell allele.
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 Female
Symptomless Carrier Symptomless Carrier
Parental Genotypes
Gamete Genotypes
H
H
A H S
A H S
H
H
A H S
A H
Offspring Genotypes and Phenotypes from a Punnett Square
H A
H A H A
H S
H A
Normal haemoglobin
H A H S
H A H S
H S
Symptomless carrier
Symptomless carrier
H S H S
Sickle cell anaemia
S
Phenotypic ratio:
25% normal haemoglobin: 50% carrier: 25% sickle cell anaemia
9

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 GYPA M and GYPA N
Genotypes and Phenotypes in the Human Population
Genotypes
GYPA M GYPA M
GYPA N GYPA N
GYPA M GYPA N
Phenotypes
Blood group M
Blood group N
Blood group MN
Gamete Genotypes
Homozygous dominant
Homozygous dominant
Heterozygous
Complete the Genetic Diagram below to determine the Phenotypic Ratio of Offspring to two Parents with Blood Group MN
Parental Phenotypes Male Female
Blood group MN Blood group MN
Parental Genotypes GYPA M GYPA N GYPA M GYPA N
Gamete Genotypes GYPA M GYPA N GYPA M GYPA N
Offspring Genotypes and Phenotypes from a Punnett Square
GYPA M GYPA N
GYPA M
GYPA M GYPA M
Blood Group M
GYPA M GYPA N
Blood Group MN
GYPA N
GYPA M GYPA N
Blood Group MN
GYPA N GYPA N
Blood Group N
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 C R coding for red/chestnut hairs and
C W coding for white hairs

C R and C W are codominant alleles. The heterozygous genotype C R C W 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 Gamete Genotypes
Homozgous dominant
C R C R
C W C W
C R C W
Red/chestnut coats
White coats
Red and white hairs described as roan
Homozygous dominant
Heterozygous
Genetic Diagram of Cross between Red Shorthorn Cow and White
Shorthorn Bull
Parental phenotypes Cow
Red Coat
Bull
White Coat
Parental genotype C R C R C W C W
Gamete genotypes C R C W
F
1
Offspring genotype C R C W
F
1
Offspring phenotype Roan coat
11

Complete the following:
Genetic Diagram of Cross between Roan Shorthorn Parents
F1 phenotypes Roan Roan
F1 genotypes C R C W C R C W
F1 gamete genotypes C R C W
F2 genotypes and phenotypes from a Punnett Square
C R C W
C R
C W
C R C R
Red Coat
C R C W
Roan Coat
C R C W
Roan Coat
C W C W
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 C R ) codes for an enzyme that catalyses the formation of a red pigment in flowers

The other allele ( C W ) codes for an altered enzyme that lacks this catalytic activity and does not produce a pigment. Plants with the genotype C W C W are white

Heterozygous plants (genotype C R C W ) 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 X H

The haemophilia allele is represented by X h
Genotypes and Phenotypes for Factor VIII Gene in the Human
Population
Genotype
X H X H
Gender Phenotype for Blood
Clotting
Normal blood clotting
X H X h
Female
Female
X
X h
H
X
Y h Female
Normal blood clotting (a symptomless carrier)
Lethal. Foetus does not develop
Normal blood clotting
X h Y
Male
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 X D denote the normal allele for dystrophin synthesis and X d , 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 Genotype
Mary
Astrid
Jack
Jane
Complete the Genetic Diagram to Show the Probability of Leon and Jane having Another Child with DMD
Parental Phenotypes Leon/Male Jane/Female
Normal dystrophin Normal dystrophin/carrier
Parental Genotypes X D Y X D X d
Gamete Genotypes X D Y X D XD
Punnett Square Showing Offspring Genotypes and Phenotypes
X D Y
X D
X d
X D X D
Normal female
X D X d
Female carrier
X D Y
Normal male
X d 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 X RG represent the allele for normal red-green colour vision
Let X rg represent the abnormal allele
Parental phenotypes: normal red-green colour normal vision (carrier)
Parental genotypes: X RG Y X RG X rg
Gamete genotypes: X RG X rg Y
Punnett Square showing the offspring genotypes and phenotypes
X RG
Y
Phenotype ratio: 1:1:2
X
X
RG
RG
X RG X RG
Female normal redgreen vision
Y
Female normal redgreen vision
X
X rg
RG X rg
Female normal vision
(carrier)
X rg Y
Male with red-green colour blindness
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 x green, wrinkled
seeded plants seeded plants
Parental Genotypes GGRR ggrr
Gamete Genotypes GR gr
F1 Genotypes GgRr
F1 Phenotypes yellow, round seeds
20

F1 Self Pollination/Self Fertilisation
F1 Phenotypes yellow, round x yellow, round
Seeded plants seeded plants
F1 Genotypes GgRr GgRr
Gamete Genotypes GR Gr gR gr GR Gr gR gr
Punnett Square to Show F2 Genotypes and Phenotypes
GR Gr gR
GR
GGRR yellow round
GGRr yellow round
GgRR yellow round
gr
GgRr yellow round
Gr
gR
GGRr yellow round
GgRR yellow round
GGrr yellow wrinkled
GgRr yellow round
GgRr yellow round ggRR green round
Ggrr yellow wrinkled ggRr green round
gr
GgRr yellow round
Ggrr yellow wrinkled ggRr green round ggrr green wrinkled
F2 Phenotype ratio:
9: 3: 3: 1 yellow round yellow wrinkled green round 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 pure breeding pure breeding
pink flowers white flowers
Parental Genotypes AAbb 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 purple flowers
F1 Genotypes AaBb AaBb
F1 Gametes AB Ab aB ab AB Ab aB ab
Punnett Square Showing the F2 Genotypes and Phenotypes
AB
Ab
aB
AB
AABB purple
AABb purple
Ab
AABb purple
AAbb pink
ab
AaBB purple
AaBb purple
AaBb purple
Aabb pink
Phenotypic ratio: 9: 3: 4
purple pink white
aB
AaBB purple
AaBb purple
aaBB white
aaBb white
ab
AaBb purple
Aabb pink
aaBb white
aabb 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
DDEE white
De
DDEe white
dE
DdEE white
de
DdEe white DE
De
dE
DDEe white
DdEE white
DDee white
DdEe white
DdEe white ddEE yellow
Ddee white ddEe yellow
de
DdEe white
Ddee white ddEe yellow ddee 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 white
Parental Genotypes IiCc IiCc
Gamete Genotypes IC Ic iC ic IC Ic iC ic
24

Punnett Square Showing the Genotypes and Phenotypes of the
Offspring
IC Ic iC ic
IC IICC white
IICc white
IiCC white
IiCc white
Ic iC ic
IICc white
IiCC white
IiCc white
IIcc white
IiCc white
Iicc white
IiCc white iiCC coloured iiCc coloured
Iicc white iiCc coloured iicc 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 enzyme 2 precursor substance

intermediate compound

final pigment
(colourless) (colourless) (purple)
Genetic Diagram of a cross between two Heterozygous Parents at both gene loci
Parental Phenotype purple flowers purple flowers
Parental Genotype CcRr CcRr
Gamete Genotype CR Cr cR cr CR Cr cR cr
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Punnett Square Showing the Genotypes and Phenotypes of the
Offspring
CR Cr cR cr
CR
Cr
CCRR purple
CCRr purple
CCRr purple
CCrr white
CcRR purple
CcRr purple
CcRr purple
Ccrr white cR cr
CcRR purple
CcRr purple
CcRr purple
Ccrr white ccRR white ccRr white ccRr white ccrr 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
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Genetic Diagram to Determine the Genotypes and Phenotypes of Offspring from Two Heterozygous Parents
Parental Phenotypes agouti coat agouti coat
Parental Genotypes AaBb AaBb
Gamete Genotypes AB Ab aB ab AB Ab aB ab
Punnett Square to Show the Offspring Genotypes and Phenotypes
AB Ab aB ab
AB
Ab aB
AABB agouti
AABb agouti
AaBB agouti
AABb agouti
AAbb white
AaBb agouti
AaBB agouti
AaBb agouti aaBB black
AaBb agouti
Aabb white aaBb black ab 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
χ 2

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
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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:
Χ 2 = ∑ (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
O
100
E
95.25
O-E
4.75
(O-E) 2
22.56
(O-E)
0.237
2 /E short 27 31.75 -4.75 22.56 0.710
Χ 2 = 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.
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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
32
22
Ebony body, normal wings
29
21
Are these numbers consistent with the expected 1:1:1:1 ratio?
No thou gh 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
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Class
Grey and normal
Grey and curled
Ebony and curled
Ebony and normal
O
32
22
29
21
E
26
26
26
26
O - E
6
-4
3
-5
(O
36
16
9
25
– E) 2 (O – E)
1.385
0.615
0.346
0.962
2 /E
Calculated Χ 2 = 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 X 2 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.
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