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 chi-squared test will be provided))

Definitions in Genetics

Terms to Define Definition Genetics The study of the inheritance of

genesGene A length of DNA that codes for

a specific polypeptide A gene is found at a gene

locusAllele A variety or specific form of a

gene An alternative form of a gene

Gene locus The position of a gene on a chromosome

Phenotype The physical features observed when the genotype is expressed

The outcome of the interaction between the genotype and the environment

Genotype The combination of all the alleles possessed by an organism

Dominant allele An allele having an effect on the phenotype even when a recessive allele is also present

Codominance The situation when both alleles have an effect on the

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phenotype in a heterozygoteRecessive allele An allele that only has an

effect on the phenotype when dominant alleles are not present

Multiple alleles When there are more than two possible alleles at a gene locus

Homozygous condition 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

Heterozygous condition When the homologous chromosomes have different alleles at the same gene locus

Autosomal linkage Linked genes are present on the same chromosome

The autosomal chromosomes are all the chromosomes except the sex chromosomes

Sex linkage Refers to one (or more) genes located on the sex chromosomes

Most sex linked genes are on the X chromosome

Epistasis 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.

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

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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)

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Possible Genotypes and Phenotypes for CFTR Synthesis in the Human Population

Genotype Description of Genotype

Phenotype

FF Homozygous dominant

Unaffected by cystic fibrosis. CFTR transport proteins are produced

Ff Heterozygous 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 GenotypesFF All FFf 50% F and 50% fff 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

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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

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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 Phenotype IAIA Group AIAIB Group ABIAIO Group AIBIB Group BIBIO Group BIOIO 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 IBIO IAIO

Gamete Genotypes IB IO IA IO

Offspring Genotypes and Phenotypes from a Punnett Square

IB IO

IA IAIB

blood group AB

IAIO

blood group A

IO IBIO

blood group B

IOIO

blood group O

Probability of a child with O blood group being born as the first child is 25% or 0.25

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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

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Table Showing Possible Genotypes and Phenotypes in the Human Population

Genotypes Phenotypes Gamete genotypesHAHA Normal haemoglobin

HSHS Sickle cell anaemia

HAHS 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 HAHS HAHS

Gamete Genotypes HA HS HA HS

Offspring Genotypes and Phenotypes from a Punnett Square

HA HS

HAHAHA

Normal haemoglobinHAHS

Symptomless carrier

HSHAHS

Symptomless carrierHSHS

Sickle cell anaemia

Phenotypic ratio:

25% normal haemoglobin: 50% carrier: 25% sickle cell anaemia

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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 Phenotypes Gamete GenotypesGYPAM GYPAM Blood group M 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 Female Blood group MN 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 MGYPAM GYPAN

Blood Group MNGYPAN

GYPAM GYPAN

Blood Group MN

GYPAN GYPAN

Blood Group NPhenotypic 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.

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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 Gamete Genotypes

CRCRRed/chestnut coats Homozgous dominant

CWCWWhite coats Homozygous dominant

CRCWRed and white hairs described as roan

Heterozygous

Genetic Diagram of Cross between Red Shorthorn Cow and White Shorthorn Bull

Parental phenotypes Cow Bull Red Coat White Coat

Parental genotype CRCR CWCW

Gamete genotypes CR CW

F1 Offspring genotype CRCW

F1 Offspring phenotype Roan coat

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Complete the following:

Genetic Diagram of Cross between Roan Shorthorn Parents

F1 phenotypes Roan Roan

F1 genotypes CRCW CRCW

F1 gamete genotypes CR CW

F2 genotypes and phenotypes from a Punnett Square

CR CW

CR CRCR

Red CoatCRCW

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

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Genetic Cross between Snapdragons with Red Flowers and White Flowers

Genetic Cross between Pink Flowered Snapdragons

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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

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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

XHXH Female Normal blood clotting

XHXh Female Normal blood clotting (a symptomless carrier)

XhXh Female Lethal. Foetus does not develop

XHY Male Normal blood clotting

XhY Male Haemophilia

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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 Complete shading within either shape indicates the phenotypic

presence of a feature such as haemophilia

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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 musclesA Family Pedigree Chart Showing the Inheritance of DMD

Let XD denote the normal allele for dystrophin synthesis and Xd, the mutant allele

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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 XDY XDXd

Gamete Genotypes XD Y XD XD

Punnett Square Showing Offspring Genotypes and Phenotypes

XD Y

XD XD XD

Normal femaleXD YNormal male

Xd XD Xd

Female carrierXd YMale 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

A recessive mutant allele causes red-green colour blindness

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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 normal vision (carrier) Parental genotypes: XRG Y XRGXrg

Gamete genotypes: XRG Xrg Y

Punnett Square showing the offspring genotypes and phenotypes

XRG Xrg

XRGXRG XRG

Female normal red-green vision

XRG Xrg

Female normal vision (carrier)

YXRG YFemale normal red-green vision

Xrg YMale 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 The current specification does not include dihybrid inheritance and you

will not be expected to write out genetic diagrams for these crosses.

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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 alleleSince 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

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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

GR

GGRR

yellow round

GGRr

yellow round

GgRR

yellow round

GgRr

yellow round Gr

GGRr

yellow round

GGrr

yellow wrinkled

GgRr

yellow round

Ggrr

yellow wrinkled gR

GgRR

yellow round

GgRr

yellow round

ggRR

green round

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

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Types of Gene Interaction1) Two genes may work against each other – they are antagonistic.

One gene masks the expression of the other2) Two genes may work together in a complementary way

Antagonistic Epistasis

Two types: recessive epistasis dominant epistasis

Recessive EpistasisThis is where the homozygous presence of two recessive alleles at one gene locus prevents the expression of alleles at a second gene locusThe 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

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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

AB AABB

purple

AABb

purple

AaBB

purple

AaBb

purple Ab

AABb

purple

AAbb

pink

AaBb

purple

Aabb

pink aB

AaBB

purple

AaBb

purple

aaBB

white

aaBb

white ab

AaBb

purple

Aabb

pink

aaBb

white

aabb

whitePhenotypic ratio: 9: 3: 4 purple pink 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 EpistasisThis 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.

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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

DEDDEEwhite

DDEewhite

DdEEwhite

DdEewhite

DeDDEewhite

DDeewhite

DdEewhite

Ddeewhite

dEDdEEwhite

DdEewhite

ddEEyellow

ddEeyellow

deDdEewhite

Ddeewhite

ddEeyellow

ddeegreen

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

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Punnett Square Showing the Genotypes and Phenotypes of the Offspring

IC Ic iC icIC IICC

whiteIICcwhite

IiCCwhite

IiCcwhite

Ic IICcwhite

IIccwhite

IiCcwhite

Iiccwhite

iC IiCCwhite

IiCcwhite

iiCCcoloured

iiCccoloured

ic IiCcwhite

Iiccwhite

iiCccoloured

iiccwhite

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 crCR CCRR

purpleCCRrpurple

CcRRpurple

CcRrpurple

Cr CCRrpurple

CCrrwhite

CcRrpurple

Ccrrwhite

cR CcRRpurple

CcRrpurple

ccRRwhite

ccRrwhite

cr CcRrpurple

Ccrrwhite

ccRrwhite

ccrrwhite

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 abAB AABB

agoutiAABbagouti

AaBBagouti

AaBbagouti

Ab AABbagouti

AAbbwhite

AaBbagouti

Aabbwhite

aB AaBBagouti

AaBbagouti

aaBBblack

aaBbblack

ab AaBbagouti

Aabbwhite

aaBbblack

aabbwhite

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

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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:

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 O E O-E (O-E)2 (O-E)2/Etall 100 95.25 4.75 22.56 0.237short 27 31.75 -4.75 22.56 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.

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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 32Grey body, curled wings 22Ebony body, curled wings 29Ebony body, normal wings 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 104Complete the table on page 33 to determine if there is a significant difference between the observed and expected data

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Class O E O - E (O – E)2 (O – E)2/EGrey and normal

32 26 6 36 1.385

Grey and curled

22 26 -4 16 0.615

Ebony and curled

29 26 3 9 0.346

Ebony and normal

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 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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