OverviewA) Monohybrid InheritanceB) Dihybrid InheritanceC) The test crossD) Autosomal dominant and autosomal recessive

pedigree charts

E) LinkageF) Sex determinationG) Alleles and their interactions – multiple alleles,

incomplete dominance, codominanceH) Gene interactions – polygenic inheritance,

epistasis

Linkage

Linked genes are situated on the same chromosome

Why do red heads always have freckles?

Genes for red hair & freckles are linked

All genes on a single chromosome :

form a linkage group (abc)

are inherited together as long as crossing-over does not occur

Linked genes:

do not conform to Mendel’s law of independent assortment

fail to produce the expected 9:3:3:1 ratio in a breeding situation involving dihybrid inheritance

in these situations a variety of ratios are produced

Morgan discovered linkage

in the early 1900 when he crossed Drosophila

1891-1945

Linkage in Drosophila

genes for body colour and wing length are linked

Body colour: grey [G] black [g]

Wing length: long wings [L] vestigial (short) wings [l]

Vestigial wings

G gL l

F2 Phenotype ratio expected if GGLL mate with ggll:

Parental genotypes: GGLL x ggllGametes: GL x glF1 genotypes: GgLl

If F1 are allowed to interbreed: GgLl x GgLlExpected F2 phenotypes: 9:3:3:1

However, the F2 showed 3:1

How was a 3:1 ratio obtained?

3 grey body,long wing :

1 black body,vestigial wing

Explanation of 3:1 ratio due to

grey - Gblack - glong wings - Lvestigial wings -l

In practice: this 3:1 ratio is never achieved reason: total linkage is rare[since crossing-over can occur]

In reality: 4 phenotypes are produced

What phenotypes in the offspring did Morgan expect when crossing the following?

GgLl x ggll

G – grey bodyg – black bodyL – long wing l – vestigial wing

Morgan expected a 1:1:1:1 ratio

But he did not get what was expected

GgLl x ggll

Genotype GgLl Ggll ggLl ggll

Phenotype grey, long wing

grey, vestigial

wing

black, long wing

black, vestigial

wing

Actual nos. 965 185 206 944

Expected 25% 25% 25% 25%

Observed 41.5% 8.5% 8.5% 41.5%

GgLl x ggll

Genotype GgLl Ggll ggLl ggll

Phenotype grey, long wing

grey, vestigial

wing

black, long wing

black, vestigial

wing

Observed 41.5% 8.5% 8.5% 41.5%

Like parents

Unlike parents

How did the phenotypes unlike the parents result?

Let us explain how crossing-over may happen during gamete

formation in:

Crossing over produces Recombinant Types

G g

L l

Replication

Crossing over produces Recombinant Types

G g

L l

Replication

G g

L l

G g

L l

Crossing over produces Recombinant Types

G g

L l

Replication Crossing

over

G g

L l

G g

L l

Crossing over produces Recombinant Types

G g

L l

Replication Crossing

over

G g

L l

G g

L l

G

L

g

l

G g

Ll

Recombinants

Grey body,long wing

Grey body,vestigial

wing

Black body,vestigial wing

Black body,long wing

G – grey bodyg – black bodyL – long wing l – vestigial wing Recombinants

Grey body,long wing

Grey body,vestigial

wing

Black body,vestigial wing

Black body,long wing

Recombinants

41.5% grey body, long wing41.5% black body, vestigial wing

8.5% grey body, vestigial wing8.5% black body, long wing

Parental phenotypes

Recombinants

Most breeding experiments involving linkage produce:

1.approximately equal numbers of the parental phenotypes

GreyLong wing

Blackvestigial

BlackLong wing

Greyvestigial

965 944 206 185

2. a significantly smaller number of phenotypes showing new combinations of characteristics also in equal numbers, called recombinants

CROSSING-OVER AND CROSSOVER VALUES (COV)

Recombinant frequency

Recombinant frequency = Number of individuals showing recombination X 100

Number of offspring

parental phenotypes: grey body, long wing 965black body, vestigial wing 944

recombinant phenotypes:black body, long wing 206grey body, vestigial wing 185

206 185Recombinant frequency 100 17%

965 944 206 185

Mutant phenotypes

Short aristae

Black body

Cinnabareyes

Vestigialwings

Brown eyes

Long aristae(appendageson head)

Gray body

Redeyes

Normalwings

Redeyes

Wild-type phenotypes

IIY

I

X IVIII

0 48.5 57.5 67.0 104.5

What did the recombinant frequency or crossover value (COV) demonstrate?

Genes are arranged linearly along the

chromosome.

Question: Getting Started

In Drosophila one gene controls eye colour and another controls body colour. The allele E for red eyes is dominant to e, the allele for purple eyes. The allele B for grey body is dominant to b, the allele for black body.A female Drosophila, heterozygous for both the eye-colour gene and the body-colour gene, was crossed with a male, homozygous recessive for both of these genes. The phenotypes of the offspring are given in the table.

E - red eyes e - purple eyesB - grey bodyb - black body

heterozygous for both the eye-colour gene and the body-colour gene

homozygous recessive for both of these genes

Phenotype Number of offspring

Red eyes, grey bodyPurple eyes, black bodyRed eyes, black bodyPurple eyes, grey body

82791316

a) i) For this cross, write your answer on the lines to give the following information: (3)

Phenotype of male parent x Phenotype of female parentpurple eyes, black body red

eyes, grey body

Genotype of male parent x Genotype of female parenteebb EeBb

Genotype(s) of male x Genotype(s) of female gametes gametes

ii) From the information in (a) (i), explain why there were approximately equal numbers of red-eyed, grey-bodied and purple-eyed, black-bodied offspring. (2)Due to linkage of genes parental phenotypes occur at approximately equal numbers.Phenotype Number of offspring

Red eyes, grey bodyPurple eyes, black bodyRed eyes, black bodyPurple eyes, grey body

82791316

b) i) Explain why there were relatively few red-eyed, black-bodied and purple-eyed, grey-bodied offspring. (2)

These are the recombinants which resulted from crossing over. Crossing over occurs at a low rate.

Phenotype Number of offspring

Red eyes, grey bodyPurple eyes, black bodyRed eyes, black bodyPurple eyes, grey body

82791316

ii) Explain why the number of red-eyed, black-bodied offspring is not exactly equal to the number of purple-eyed, grey-bodied offspring. (1)

Due to random fertilisation of gametes.Phenotype Number of

offspringRed eyes, grey bodyPurple eyes, black bodyRed eyes, black bodyPurple eyes, grey body

82791316

Extent of cross-over is proportional to distance between genes

Most

Least Intermediate

GENE MAPPING

Chromosome maps are constructed by directly converting

the :COV between genes into hypothetical distances along the chromosome

C? C?A B

499

A COV of 4% :between genes

A and B: they are 4 units apart

A COV of 9% :for a pair of genes A and C: they are 9 units apart but does not indicate the linear sequence of genes

consider the following values involving four genes, P, Q, R and S:

P – Q = 24%R – P = 14%R – S = 8%S – P = 6%

consider the following values involving four genes, P, Q, R and S:

P – Q = 24%R – P = 14%R – S = 8%S – P = 6%

consider the following values involving four genes, P, Q, R and S:

P – Q = 24%R – P = 14%R – S = 8%S – P = 6%

consider the following values involving four genes, P, Q, R and S:

P – Q = 24%R – P = 14%R – S = 8%S – P = 6%

A linkage map (or chromosome map) tells the relative distances between

gene loci on a chromosome

Questions: Crossover & Epistasis

In maize the genes for coloured seed and full seed are dominant to the gene for colourless seed and shrunken seed. Pure breeding strains of the double dominant variety were crossed with the double recessive variety and a backcross of the F1 generation produced the following results:

coloured, full seed 380coloured, shrunken seed 14colourless, shrunken seed 386colourless, full seed 10

Calculate the distance in units between the genes for coloured seed and seed shape on the chromosomes. (3)

coloured, full seed 380coloured, shrunken seed 14colourless, shrunken seed 386colourless, full seed 10

Total number of seeds:

380 + 14 + 386 + 10 = 790Number of recombinants:

14 + 10 = 24Recombinant frequency (crossover value): 24 100

3%790

Therefore, distance = 3 units

OverviewA) Monohybrid InheritanceB) Dihybrid InheritanceC) The test crossD) Autosomal dominant and autosomal recessive

pedigree chartsE) Linkage

F) Sex determinationG) Alleles and their interactions – multiple alleles,

incomplete dominance, codominanceH) Gene interactions – polygenic inheritance,

epistasis

Sex Chromosomes (heterosomes)

Females XXHomogametic(one type of gamete, X)

X

Y

X X

Males XYHeterogametic (two types of gametes, X or Y)

Sex Chromosomes in

Drosophila

as in humans

Sex chromosomes in some organisms differ from those in humans

Birds (including poultry), moths & butterflies :

Female

Female

Male

Male

Sex chromosomes in some organisms differ from those in humans

In some insects e.g. grasshopper, the Y chromosome is absent :

XO = male XX = female

Curious Fact:sex is influenced by temperature in

environment in reptiles (turtle, crocodile)

(t <28 °C) (t 28-32 °C) (t >32 °C)

Question

Scientists analysed the:DNA on the Y chromosome and the DNA in the mitochondria of the Swedish

wolves.

Swedish wolves

Finland RussiaXThey concluded that the Swedish wolf population descended from one male wolf from Finland and one female wolf from Russia.

Question

a) Explain why DNA on the Y chromosome helped them to reach this conclusion. (1)Y chromosome is inherited from male.

b) Suggest why DNA in the mitochondria helped them to reach this conclusion. (1)Mitochondria in egg. No mitochondria come from male gamete.

Question: [MAY, 2012]Use your knowledge of biology to describe the significance of the following.All the mitochondria in a human cell are derived from the female parent. (5)

Barr Bodies

Barr Bodies

This held for cats and humans

They thought the spot was a tightly condensed X chromosome

Ewart George Bertram & Murray Llewellyn Barr in 1948

1940’s two Canadian scientists noticed a dark staining mass in the nuclei of cat brain cells

Found these dark staining spots in female somatic cells but not males

The X Chromosomes

One X chromosome: always appears in the

active state

has the normal appearance

If another is present: it is seen in a resting state as a tightly coiled

dark-staining body - the

Where is the Barr body located?

The Barr body is generally located on the periphery of the nucleus.

How can the genotypic sex of a person be determined?

By counting the number of in non-dividing cells

Barr bodies = 0 Barr bodies = 1

How many Barr Bodies?

Barr bodies = 0 Barr bodies = 1 Barr bodies = 3

Tortoiseshell cats

Calico cats

or

Calico cats result from random inactivation of an X chromosome

Zygote

Early embryo

Random inactivation of X chromosome occurs early in

embryonic development

Black inactive

Orange inactive

Male calico cats are rare

What are the sex chromosomes of a

calico male cat?

XXY

Calico manx cat

What is unusual?

X inactivation explains why:

Levels of enzymes or proteins encoded by genes on the X chromosome are the same in both males and females

Even though males have 1 X chromosome and females have 2.

SEX LINKAGE

What is the function of the Y

chromosome?function appears to vary according to

species

In humans: it controls the

differentiation of the testes

In most organisms: it does not carry genes

concerned with sex

Why is the Y chromosome described as genetically inert or genetically empty?

since it carries so few genes

the human Y chromosome contains: only 78 genes the lowest number of known genetic

diseases (44 in total)

Fig. 6Homologous and non-homologous regions of the sex chromosomes.

Homologous portion of the sex chromosomes

Non-homologous portion of the X chromosome

Sex-linkage

affects males and females differently

characteristics determined by genes carried on the non-homologous portion of the X chromosome appear in males even if they are recessive

X Y

Sex-linked Traitsred-green colour

blindnesspremature balding

haemophilia

“3” or “8”??

Question: [MAY, 2006]Suggest explanations for the following observation.

Colour blindness affects 8% of human males but only 0.7% of females. (2)

Normal vision Colour blind vision

Colour-blindness is a sex-linked trait which affects more males than females. The allele responsible for this feature is carried on the X chromosome. As males carry only one such chromosome, chances for the disorder to manifest itself is more than in females, where two X chromosomes are present. A heterozygous female is phenotypically normal, as the recessive allele is not expressed in the presence of the dominant one.

Haemophilia: possible genotypes and phenotypes

Genotype PhenotypeXHXH Normal female

XHXh Normal female (carrier)

XHY Normal male

XhY Haemophiliac male

Parents

XHY XHXhX

What is the probability for these parents to have a sick child?

ParentsParental

Gametes

Y

XHY XHXh

XH XH Xh

X

X

ParentsParental

Gametes

F1 Offspring:

Y

XHY XHXh

XH

Xh

Y

XH

XH XH Xh

X

X

ParentsParental

Gametes

F1 Offspring:

Y

XHXH XHY

XHXh XhY

XHY XHXh

XH

Xh

Y

XH

XH XH Xh

X

25%

X

Using this pedigree, how can you tell that:1. colour blindness is a sex-linked disorder? More males than females are affected.2. colour blindness is due to a recessive allele? Affected parents produce normal offspring.

Eye colour in Drosophila:

Red eye colour:Dominant

is sex-linked

White eye colour:recessive

XR Xr

What is a ‘reciprocal cross’?

A cross, with the phenotype of each sex reversed as compared with the original cross.

Red-eyed White-eyed X X Red-eyedWhite-eyed

XRY XrXr XRXRXrY

Why is a ‘reciprocal cross’ done?

To test the role of parental sex on inheritance pattern.

Red-eyed

male

White-eyed female X X Red-eyed

female

White-eyed male

F1 all red-eyedF1 all males white-eyed: all females red-eyed:

To decide whether trait is sex linked compare offspring phenotypes

XRY XrXr XrY XRXR

XrYXRXr XRXr XRY

Autosomes

If F1 phenotypes are the same from both crosses:

If F1 phenotypes are different both

crosses:

trait is not sex-linkedtrait is sex-linked

N n

Sex chromosomes

N: Normal n: sick

Autosomes

N n

X X

If allele is carried on the

autosomes, males and

females would be normal.

If allele is carried on the sex

chromosomes, males and

females would be different.

Autosomes

n

X Y

What is the phenotypic ratio in the F2 generation if a white-eyed male fly is mated with a red-eyed female? Eye

colour is sex-linked.

Question: [MAY, 2006]

Suggest explanations for the following observation.

In the fruit fly, Drosophila, red eyes are dominant to white eyes. When a red-eyed female is crossed with a white-eyed male, the offspring all have red eyes but when the opposite cross was done (a white-eye male with a red-eyed female) [this should read: a white eyed female and a red eyed male – a mistake in exam paper] all the male offspring had white eyes. (2)

Question: [MAY, 2006]

This is a reciprocal cross carried out to determine if a gene is sex-linked. This result shows that eye colour in Drosophila is sex-linked. A red-eyed female (XRXR) crossed with a white-eyed male (XrY) produce 100% red-eyed offspring (XRXr, XRY). A white-eyed female (XrXr) crossed with a red-eyed male (XRY) produce white-eyed male offspring (XrY).

3. Duchenne muscular dystrophy is a sex-linked inherited condition which causes degeneration of muscle tissue. It is caused by a recessive allele. The diagram shows the inheritance of muscular dystrophy in one family.

a) Give evidence from the diagram which suggests that muscular dystrophy is

i) sex-linked; (1)Only seen in males / not in females;

ii) caused by a recessive allele. (1)Unaffected parents/mother produce a sick child

b) Using the following symbols,XD = an X chromosome carrying the normal alleleXd = an X chromosome carrying the allele for muscular dystrophyY = a Y chromosomegive all the possible genotypes of each of the following persons. [5, 6, 7, 8] (2)

XdY XDY

XDXd

XDXd or XDXD

Xd XD

XDXd

Xd

XD

c) A blood test shows that person 14 is a carrier of muscular dystrophy. Person 15 has recently married person 14 but as yet they have had no children. What is the probability that their first child will be a male who develops muscular dystrophy? (1)

XDY

Person 14: XDXd and Person 15: XDYMale offspring: XDY and XdYThus half of the male offspring develop muscular dystrophyProbability of having male offspring = 0.5Thus probability that their first child will be a male who develops muscular dystrophy: 0.5 x 0.5 = 0.25 / ¼ / 25% / 1 in 4; (NOT ‘1:4’)

5. Fabry’s disease is a sex-linked recessive genetic disorder that causes mental retardation. A study was carried out into the inheritance of this disorder in a family, and the results are shown in the pedigree below.

a) Using the symbol A for the dominant allele and a for the recessive allele, state the genotype of person 2

XAXa (1) XA

Xa

b) Using the evidence from the pedigree, explain why Fabry’s disease is described as a sex-linked recessive genetic disorder. (3)

Only males have the disease. Probably carried on the X chromosome. Male sufferers are produced from unaffected parents. Recessive is only expressed when no dominant allele is present. Females must be carriers. Males have only one locus for the disorder.

c) Explain why person 3 is unaffected but why one of his children (person 5) has Fabry’s disease.

Person Number 3 is the father XAY.Number 5 inherits Y from the father and the recessive allele from the mother who is a carrier. (3)

XA

Xa

XaY

XA

Xa

d) What are the chances of the children of persons 6 and 7 having Fabry’s disease? Give reasons for your answer. (4)

XAY XAXa

Y X a

F2 : No. 6 x No. 7

F2 genotypes: X AYGametes:

F3 genotypes:

F3 phenotypes:

x

x XAX a

X A X A

X AX A

females are all unaffected; half of the males are affected

X AX a X AY X aY

25 % chance of producing a sufferer

3. The phenotypic expression of a dominant gene in Ayrshire cattle is a notch in the tips of the ears. In the pedigree below, square symbols represent male animals, circles represent female animals and solid symbols represent notched individuals.

I

II

III

1 2

1 2 3 4 5

1 2 3 4 5 6

3.1 What is pedigree analysis? (2)The study of phenotypes in closely related families. This is done in order to discover the genotype of each individual.

I

II

III

1 2

1 2 3 4 5

1 2 3 4 5 6

Use the pedigree diagram to determine the probability of notched progeny being produced from each of the following matings. Your working and reasoning must be shown.

3.2 III 1 x III 3 (2)

I

II

III

1 2

1 2 3 4 5

1 2 3 4 5 6

N

n

nn nn

3.3 III 2 x III 3 (2)

I

II

III

1 2

1 2 3 4 5

1 2 3 4 5 6

N

n

nn Nn

3.4 III 3 x III 4 (2)

I

II

III

1 2

1 2 3 4 5

1 2 3 4 5 6

N

n

nn nn

3.5 III 1 x III 5 (2)

I

II

III

1 2

1 2 3 4 5

1 2 3 4 5 6

N

n

nn Nn

3.6 III 2 x III 5 (2)

I

II

III

1 2

1 2 3 4 5

1 2 3 4 5 6

N

n

Nn Nn

HW: GENETICS MATSECNumbers 5, 8,9

Questions: Getting Started

4.In the fruit fly, Drosophila melanogaster the allele r for white eyes is sex-linked and is recessive to the allele R for the normal red eye colour. Sex-linked genes are carried on the X chromosome. In fruit flies, the sex chromosomes are XX in females and XY in males.

When a white-eyed female fly is crossed with a red-eyed male, all of the male offspring have white eyes.

i) In the space below, draw diagrams to represent the chromosomes, and alleles present on them, of the parents in this cross. (3)

Female x MaleXr Xr XR Y

XR – red eyeXr – white eye

ii) State the genotype and phenotype of the female offspring. (2)

Genotype: Phenotype:

XRXr

red eyed

A gene which affects the size of the wings in the fruit flies is not sex-linked (autosomal). The allele for short (dumpy) wings, d, is recessive to the allele, D, for normal wings.A female fly was heterozygous for the genes for eye colour and for wing size. This female was crossed with a white-eyed, short-winged male.

i) Use a genetic diagram to show the expected results of this cross. (4)

D – normal wingd – short wing Parental genotypes: XRXrDd x XrYdd

XRD XRd XrD XrdXrd XRXrDd XRXrdd XrXrDd Xr XrddYd XRYDd XRYdd XrYDd XrYdd

D – normal wingd – short wing Parental genotypes: XRXrDd x XrYdd

XRD XRd XrD Xrd Gametes: Xrd Yd x

ii) Using the columns below, list the genotypes and full phenotypes of the male offspring produced. (2)

Genotypes of male offspring Full phenotypes

ii) Using the columns below, list the genotypes and full phenotypes of the male offspring produced. (2)

Genotypes of male offspring Full phenotypes

XRYDd red eyed, normal wings

XRYdd red eyed, short wings

XrYDd white eyed, normal wings

XrYdd white eyed, short wings

Calico Cats are heterozygous for X-linked alleles determining coat Colour

Orange: XB XB

Black: XbXb

Calico:XBXb

Calico cats have orange &

black patches

What is the probability for offspring to be tortoiseshell (calico) for coat color?

50% Tortoiseshell (calico) cats

Questions: SEX LINKAGE (Calico cats)

1a) Explain what is meant by the term sex linkage. (2)Alleles are located on the sex chromosomes. Most probably, the X chromosome.

b) One of the genes which controls coat colour in cats has

its locus on the X chromosome. This sex-linked gene has two alleles, O and B. Female cats have two chromosomes (XX), male cats have one X and one Y chromosome (XY).In female cats, possible genotypes and phenotypes are shown in the table below.

State two possible genotypes for a male cat and, in each case, indicate the phenotype. (4)

Genotype PhenotypeXOXO Orange

XOXB Orange and black (tortoiseshell)

XBXB Black

Genotype Phenotype

State two possible genotypes for a male cat and, in each case, indicate the phenotype. (4)

Genotype PhenotypeXOXO Orange

XOXB Orange and black (tortoiseshell)

XBXB Black

Genotype Phenotype

XOY Orange

XBY Black

c) An orange and black (tortoiseshell) female cat was mated with a black male cat. Draw a genetic diagram to show the possible genotypes and phenotypes of the kittens resulting from this cross. (5)

Parental phenotypes:

XOXB x XBY

Gametes:

tortoiseshell female

xblack male

F1 genotypes: XOXB XOY XBXB XBY

F1 phenotypes: tortoiseshell female

Black male

Orangemale

BlackFemale

XO XB x XB Y

Parental genotypes:

Calico female with black offspring

2) A sex-linked gene for fur colour on the X chromosome of cats has two alleles. One allele results in orange fur (XO) and the other in black fur (X+).The heterozygous condition (XOX+) results in ‘tortoiseshell’ cats with irregular patches of orange and black fur.

a) Complete the two genetic crosses shown below.black ♀ x orange ♂ Orange ♀ x black ♂

i) Genotypes of parents

ii) Genotypes of kittens ♀ kittens ♂ kittens ♀ kittens ♂ kittens

iii)Phenotypes of kittens

2) A sex-linked gene for fur colour on the X chromosome of cats has two alleles. One allele results in orange fur (XO) and the other in black fur (X+).The heterozygous condition (XOX+) results in ‘tortoiseshell’ cats with irregular patches of orange and black fur.

a) Complete the two genetic crosses shown below.black ♀ x orange ♂ Orange ♀ x black ♂

i) Genotypes of parents

X+ X+ x XOY XO XO x X+Y

ii) Genotypes of kittens

XOX+ X+Y♀ kittens ♂ kittens

XOX+ XOY♀ kittens ♂ kittens

iii)Phenotypes of kittens

tortoiseshell blackfemale male

tortoiseshell orangefemale male

b) Explain why tortoiseshell cats are always female (unless there has been a mutation).Tortoiseshell cats must have two X chromosomes, one being XO, and the other X+. Males have only one X chromosome.

c) Male tortoiseshell cats are very rare and are always sterile with three sex chromosomes. Give these three sex chromosomes with their alleles where appropriate.XOX+Y

OOH OOH, ME ME ME!!

Any Questions?