genetics

INHERITANCE

is the study of heredity

is the process in which traits are passed from parents to offspring

Characters or Traits are resemblances or differences which can be:

Seen e.g. eye colourflower colour

Tested for e.g. blood groupscolour blindness

Gregor Mendel

(1822-1884)

Austrian monk who formulated fundamental laws of heredity in early 1860s

Experimented with peas

Over seven years, he made crosses with 24,034 plants

Let’s revise:Genes: control the characteristics of living organisms are carried on the chromosomes

Chromosomes are in pairs, one from each parent

Genes are in pairsGenes controlling the same characteristics

occupy identical positions on corresponding chromosomes

The gene pairs control one characteristic

gene for eye colour

gene fornose shape

gene for making insulin

The genes of a corresponding pair are called alleles

Homologous chromosomes have the same length and carry the same gene sequences

Alleles are alternative forms of the same gene

Gene

Let’s take coat colour in mice as an example

Mice can be: Black

of the gene pair which help determine coat colour in mice:

one allele might try to produce black fur

and its partner might try to produce brown fur

Brown

The allele for black fur is dominant to the allele for brown fur

This combination of alleles gives aBLACK mouse

The dominant allele is expressedThe recessive allele is masked

Alleles are represented by lettersB and b are alleles of the gene for coat colour the allele for black fur is given the letter B the allele for brown fur is given the letter b the alleles must have the same letter but the

dominant allele is always in capitals

Black mouse (B – dominant allele)

Brown mouse (b – recessive allele)

Possible combinations of alleles

A black mouse (BB) is crossed with a brown one (bb). What will the offspring look like?

B

B

b

b

B

b

PURE-BREEDING organism – both alleles are the same [BB and bb]

B

BB

B

b

b

b

b

B

b

meiosis

meiosis

fertilisation

All offspring willbe black (Bb)

sperm mother cell

ovum mother cell

zygote

If two of the offspring (Bb) thus produced are mated,

what type of offspring would result?

Bb Bb

x

?

B

b

B

b

B

b

B

b

B

B

B

b

B

b

b

b

BB

Bb

Bb

bb

sperm mother cell

ovum mother cell

meiosis

Possible combinationsFertilisationsperms

ovazygotes

FIRST FILIAL GENERATION (F1) the offspring produced by a

parental generation

xParents:

SECOND FILIAL GENERATION (F2)

offspring of the F1

When two F1 offspring mate, they produce the

F2

Genotype & Phenotypegenotype: describes the genetic make-up (all

of the genes) of an individual

homozygous dominantheterozygoushomozygous recessive

phenotype: outward appearance of an individual

Which of the offspring is/are: a carrier:The heterozygous ones.

phenotypically normal:

Nn. Persons look normal but carry the defective allele.

Normal - N – NN, Nn Sick – n – nn

Homozygous & HeterozygousHOMOZYGOUS – alleles on corresponding

positions of homologous chromosomes are identical e.g. BB or bb

HETEROZYGOUS – pairs of different alleles are present on corresponding positions of homologous chromosomes e.g. Bb

Which is the dominant allele?

Parents(true breedingparents)

F1 generation

F2 generation

Purple flowers White flowers

All plants havepurple flowers

Fertilisationamong F1 plants

(F1 F1)

3/4 of plantshave purple flowers

1/4 of plantshave white flowers

Allele for purple colour [100% purple in F1 generation]

Let us become familiar with terms learned

R represent round seed r represent wrinkled seed

Round

What is the:a) phenotype of a homozygous dominant plant?

b) genotype of a homozygous dominant plant?

c) genotype of a heterozygous plant?RR

Rr

B represent yellow seedb represent green seed

What is the:a) dominant allele for seed colour?

b) genotype of a homozygous recessive plant?

B

bb

c) genotype of a true breeding plant that produces green seeds?bb

GENETIC CROSSES

Let A represent the allele for purple flower colour and a for white colour. A pure breeding purple and a pure breeding white flower are crossed.What will the phenotype and genotype ratios be in the F1 generation?

purple - A – AA, Aa white – a – aaParents: Purple x White

AA x aaGametes:

F1 generation: Aa Aa Aa Aa

A A ax a

F1 Phenotype: 100% purple F1 Genotype: 100% heterozygous

The cross does NOT mean that FOUR offspring are produced. It shows PROBABILITY.Cross can be shown as:

Parents: Purple x White AA x aa

Gametes:

F1 generation: Aa

A x a

F1 Phenotype: 100% purple F1 Genotype: 100% heterozygous

IMPORTANT!!

Self-pollination occurs in one of the F1 plants. What will the phenotype and genotype ratios be in the F2 generation? purple – A – AA, Aa

white – a – aaF1 generation: Purple x Purple

Aa x Aa

Gametes:

F2 generation:

xA a aA

AA Aa Aa aa

Phenotype- 3 purple : 1 white OR 75% purple: 25% white Genotype- 1 AA : 2 Aa : 1 aa

Remember: when both parents are heterozygous, they produce offspring in such a

ratio

Pp Pp

x

Parents

Offspring

THE PUNNETT SQUARE METHOD

Reginald Punnett (1875-1967)

In 1902, created the Punnett Square - a chart which helped to determine the probable results of a genetic cross

T t

T TT Tt

t Tt tt

Male gametes

Female gametes

TtTt

XX = femaleXY = male

  X Y

X XX XY

X XX XY

Genotypic ratio - 1 XX : 1XYPhenotypic ratio - 1 female: 1 male

Sex Determination

GENETICS Questions [pg.65]1. The diagram shows a genetic cross between two

pure bred mice. Black coat (B) is dominant to white coat (b).

a) Complete the diagram by giving the:genotype of the white parent. phenotype of the offspring.

bb

Black

b) Two of the offspring were mated.(i) Complete the Punnett square to show this cross.

Parents: Bb × Bb

ii) Draw a circle round the homozygous recessive offspring.

iii) Give the expected ratio of black mice to white mice.3 black : 1 white

bb

bB

BB

Bb

Bb

2) Drosophila is a fruit fly often used in genetic research. The diagram below shows two different strains of Drosophila.

a) Vestigial wing is caused by a mutation. For wings, the normal allele is represented by the symbol N and the allele for vestigial wings is represented by n.

In a breeding experiment, pure bred normal-winged male flies were mated with pure bred vestigial-winged female flies. All the offspring of this cross had normal wings.State the genotype of the male parents.

pure bred normal-winged flies

normal wings – N – NN, Nnvestigial wings – n - nn

a) State the genotype of the male parents.

pure bred vestigial-winged flies

F1: 100% normal wings

x

NN

nn

Nn

b) State the genotype of the female parents.

c) State the genotype of the offspring.

d) In a second experiment, flies with the genotype Nn (for wings) were allowed to mate together.

i) Use a genetic diagram to show this cross.

Parents: Nn × Nn

N n nN

NN Nn Nn nnF2 generation:

Gametes:

ii) Give the expected ratio of the genotypes of the offspring of this cross.

iii) Give the expected ratio of the phenotypes of the

offspring of this cross.

1 NN : 2 Nn: 1 nn

3 normal wings : 1 vestigial

×

3) In humans the allele for brown eyes (B) is dominant to the allele for blue eyes (b). A brown-eyed man and a blue-eyed woman have a daughter with brown eyes and a blue-eyed son.

Write down the genotype for eye colour of:i) father ii) mother iii) daughter iv) son

Brown eyes - B – BB, Bb Blue eyes – b – bbBrown eyes Blue eyes

bb

x

Brown eyes Blue eyes bb

B_

B_

3) In humans the allele for brown eyes (B) is dominant to the allele for blue eyes (b). A brown-eyed man and a blue-eyed woman have a daughter with brown eyes and a blue-eyed son.

Write down the genotype for eye colour of:i) father ii) mother iii) daughter iv) son

Brown eyes - B – BB, Bb Blue eyes – b – bbBrown eyes Blue eyes

bb

x

Brown eyes Blue eyes bb

Bb

Bb

b) What is the phenotype for eye colour for the mother?

c) Which of the family are homozygous for eye colour?

Blue eyes

Brown eyes Blue eyes bb

x

Brown eyes Blue eyes bb

Bb

Bb

Mother and son

d) If the daughter were to marry a brown-eyed man, explain why the eye colour of her children would depend on her husband’s genotype.

Daughter Bb

X HusbandBb

Daughter Bb

X HusbandBB

75% Brown eyed

25% Blueeyed

: 100% Brown eyed

4) ‘Waltzing’ mice spin round and appear to chase their tails instead of crawling normally. This behaviour is controlled by a gene with two alleles.

Using these mice, the following two crosses were made. In Cross 1 waltzing mice were crossed with normal mice and a large population of normal mice was obtained. In Cross 2 normal mice from the offspring of Cross 1 were crossed with waltzing mice. Half of the resulting population were waltzing mice and the other half were normal. Use A for the dominant allele and a for the recessive allele, write the genotypes and phenotypes of the stages in the two crosses shown below.

Cross1Parental phenotypes Waltzing x NormalParental genotypes _______ _______Gametes _______ _______Offspring genotypes _______ _______Offspring phenotype All normal Cross2Parental phenotypes Waltzing x NormalParental genotypes _______ _______Gametes _______ _______Offspring genotypes _______ _______Offspring phenotypesWaltzing Normal 50% 50%

Normal – A – AA, Aa Waltzing – a – aa

aa AA

Aa Aa

a A Aa

aaaa Aa

AaaaaA

5a) In cattle, the presence of horns is controlled by a single pair of alleles. Animals with horns have a homozygous recessive genotype. The diagrams below show two cows, C1 and C2.

A bull is crossed with the two cows, C1 and C2.

C1 was without horns and produced a calf which grew horns.C2 had horns and produced a calf which did not grow horns.i) Using the symbol H to represent the dominant allele and h

the recessive allele, write down the following.1. The genotype of cow C1.

2. The genotype of the calf produced by cow C2.

3. The genotype of the bull. 4. The phenotype of the bull.

WORKING1. First write the crossings given in the following format to

help you find genotype of each animal:

Bull X C1 Bull X C2 (no horns) (horns)

(horns)_ _

(no horns)_ _

_ _ _ _ _ _ _ _

2. Next find out what H and h refer to. In problem it is given that ‘animals with horns have a homozygous recessive genotype’.

Bull X C2 (horns)

(no horns)_ _

Hh

Bull X C1 (no horns)

(horns)_ _

_ _ _ _ _ _

No horns – H – HH, Hh Horns – h – hh

3. Write genotype of animals with horns & then continue for the others.

_ _

hh

hhHh Hh

Hh

i) Using the symbol H to represent the dominant allele and h the recessive allele, write down the following.1. The genotype of cow C1.

2. The genotype of the calf produced by cow C2.

3. The genotype of the bull.

4. The phenotype of the bull.

Hh

Hh

Hh

No horns

ii) A bull had its horns removed by a farmer to make it less dangerous. In a further cross this bull was mated with a cow with horns. Describe the appearance of calves from this cross and give reasons for your answer. (3)

No horns – H – HH, Hh Horns – h – hh

Bull X Cowhh X hh

Offspringhh

All calves have horns. If a bull (hh) mates with a cow (hh), all offspring are hh. Cutting the horns does not change the genotype of the bull.

c) Explain why the offspring of a single Amoeba may differ from one another. (2)

Due to changes in their DNA i.e. mutations.

  X YX XX XYX XX XY

d) Explain why in humans the number of female babies born is approximately equal to the number of male babies born. (4)

All gametes of a female carry an X chromosome. Half of the gametes produced by a male carry an X chromosome and the other half carry a Y chromosome.

6) In humans the allele of the gene for brown eyes, B, is dominant to the allele of the gene for blue eyes, b.A brown-eyed woman married a blue-eyed man. They had four children, three with brown eyes and one with blue eyes.

Mother Father

Brown eyes

a) i) Complete the diagram below by writing in each circle, the genotype for eye colour of each member of the family.

Blue eyes

Brown eyes – B – BB, Bb Blue eyes – b – bb

Bb

bb

bb

Bb Bb Bb

Mother Father

Brown eyes

Blue eyes

Brown eyes – B – BB, Bb Blue eyes – b – bb

Bb

bb

bb

Bb Bb Bb

ii) How many different genotypes for eye colour exist in the family?

iii) How many different phenotypes for eye colour exist in the family?

iv) How many members of the family are heterozygous for eye colour?

4 [those who are Bb]

2 [Bb and bb]

2 [Brown and blue]

b) i) If the couple were to produce another child with brown eyes, what would be its genotype?

ii) What is the probability of the next child

having brown eyes?

Bb

50%

Parents: Bb x bb

B bGametes: b bx

F1 generation: Bb Bb bbbb

7) Brown eyes (B) and blue eyes (b) are two different alleles of the gene which determine eye colour. The diagram below shows the eye colours of the members of one family numbered 1 to 14.

Brown eyes – B – BB, Bb Blue eyes – b – bb

a) How can you tell that brown eyes are dominant? (1)

Brown eyes – B – BB, Bb Blue eyes – b – bb

a) How can you tell that brown eyes are dominant? (1)

All offspring of 3 (blue eyed) and 4 (brown eyed) have brown eyes.

Brown eyes – B – BB, Bb Blue eyes – b – bb

b) What are the eye colour phenotypes of the following? Person 1, Person 10 and Person 12

Person 1: brown ; Person 10: brown; Person 12: blue

Brown eyes – B – BB, Bb Blue eyes – b – bb

bb bb

bb

c) i) Which people in the diagram above are definitely homozygous for eye colour? (3)

Those with blue eyes i.e. persons 3, 12 and 13.

Brown eyes – B – BB, Bb Blue eyes – b – bb

bb

Bb

bb

Bbbb

Bb/BB

ii) Which people in the diagram above are definitely heterozygous for eye colour? (3)Persons: 8, 9, 10, 11, 6 and 7.

Bb Bb Bb Bb

100%

Gametes:

d) i) If person 14 is homozygous for eye colour, what would be the chance of a child born to 13 and 14 being brown eyed? (1)

F2 generation: Person 14 X Person 13

BB X bbB B b b

F3 generation: Bb Bb Bb BbX

50%

Gametes:

d) ii) If person 14 is heterozygous for eye colour, what would be the chance of a child born to 13 and 14 being brown eyed? (1)

F2 generation : Person 14 X Person 13

Bb X bbB b b b

F3 generation: Bb Bb bb bbX

TEST CROSS

A homozygous dominant and a heterozygous individual have the same phenotype. How can you

determine their genotype? Black: B – BB, BbBrown: b – bb

bb

Carry out a test cross.

x

bb

Look at phenotypic ratio of the offspring to determine genotype of parent.

A test cross is done by crossing an organism of unknown genotype with a

homozygous recessive organism

xbb

50% black : 50% brown

x

100% black

bb

BbbbBb

The phenotypic ratios among offspring are different, depending on the genotype of the unknown parent.

Points to remember: Test CrossWhy is it done?

How is it done?

How is a conclusion drawn?

To find if an organism is homozygous dominant or

heterozygous.

Organism of unknown genotype is mated with a

homozygous recessive one.

Question: MAY, 2011Achondroplastic dwarfism is a genetic condition that affects the long bones of the body which do not grow to normal size. This condition is a result of an autosomal dominant trait (represented by D; while d represents the recessive allele). The inheritance of the condition is shown in the following family tree.

a) Write the genotypes of:i) Paul:ii) Sue:

D – sick – DD, Ddd – normal – dd

Dd

dd

b) Ian and Elaine are expecting their second child. They would like to know whether their second child will be affected or not. Draw a genetic diagram to work out the percentage chance of having an unaffected child. (4)

Percentage chance of having an unaffected

child (dd) is 50%.

Dd dd

Question: MAY, 2012Machado-Joseph Disease (MJD) is an autosomal dominant disorder, represented by D, that is characterised by slow progressive clumsiness in the arms and legs. The following diagram shows the pattern of inheritance of this disease in a particular family.

a) Write the genotypes of:i) Peter: ii) Anton: iii) Mario:

D – sick – DD, Ddd – normal – dd

Dd

Dd

Dd

dd dd

dd

dd dd

dd dd

b) Christine, the only female in the diagram is affected by MJD, and her husband Nicholas are expecting their first child. Nicholas is also affected by MJD. The genetic counsellor informed the couple that there is only 25% chance of having a child not affected with the disorder. Work out a genetic diagram to confirm the counsellor’s prediction. (4)

Parents: Christine NicholasGenotype of parents:Gametes:F1 generation:

Explanation:

Parents:Christine Nicholas

Genotype of parents:

Dd x Dd

Gametes: x

F1 generation:

DD Dd Dd dd

Explanation: If father was DD, all offspring would have been normal. For ¼ i.e. 25% to be normal (dd), father must have been Dd.

D Dd d

c) Mario and his wife Tania are expecting their third child. Their first two children, Sam and Jack, are both affected by the disorder. The genetic counsellor informed the couple that due to Tania’s genotype, all their future children will be affected with the disorder. Work out a genetic diagram to confirm the counsellor’s prediction. (4)

Parents: Mario TaniaGenotype of parents:Gametes:F1 generation:

Explanation:

Parents:Mario Tania

Genotype of parents:

dd x DD

Gametes: x

F1 generation:

Dd Dd Dd Dd

Explanation: If mother was Dd, 50% of her offspring would have been affected. For all her offspring to be affected, she must have been DD.

d Dd D

INCOMPLETE DOMINANCE

neither allele is dominant the heterozygote shows an intermediate

phenotype

Incomplete dominance is a pattern of inheritance where:

Red White

Pink

R allele: is partially dominant

Compare ‘complete’ with ‘incomplete’ dominance

Complete dominance Incomplete dominance

The dominant allele completely masks the

recessive one

Neither allele is dominant

RR rr

RrRR Rrrr

Alleles which show Incomplete dominance are sometimes written

differently:

Red White

Pink

Suppose a pink flower is self pollinated, what phenotypes and in what ratios would the

offspring be?

Rr Rr

Rr Rr rrRR1

red1

white2

pink

:

::Phenotyperatio

F2 Phenotypic ratio:1 white : 2 pink :1 red

In the F2: phenotypic and genotypic ratios

are the same

F2 Genotypic ratio:1 rr : 2 Rr :1 RR

Pg. 824)A pure breeding strain of red flowered

snapdragon plants was cross-pollinated with a pure breeding variety of ivory flowered plants. When the seeds obtained from this cross were sown and grown, all the first filial (F1) generation plants had pink flowers.

Alleles producing red flowers are the same.

red x ivory pink

a) Why do pure breeding red flowers keep on producing plants having the same red flower colour, generation after generation, if they carry out self-pollination? (2)

b) Using appropriate symbols, give the genotype of the parental:

i) red flowered plants:ii) ivory flowered plants: c) The appearance of pink flowered plants is an

example of

RR

rr

incomplete dominance

d) Explain why all the first filial (F1) generation plants:

i) Did not have either red or ivory coloured flowers like the parental plants.

Parents: Red x IvoryRR x rr

F1: Rr

Genotype is heterozygous. An intermediate phenotype results due to incomplete dominance .

ii) Had pink flowers. (2)

A red flower is RR and an ivory one is rr, but the F1 plants are Rr. The R allele is partially dominant and results in a different phenotype from that of parents.

e) Two of the first filial (F1) generation plants were cross-pollinated. Use the Punnett Square drawn below to work out the genotypes of the second filial (F2) generation plants. (2)

Parents: Red x IvoryRR x rr

F1: Rr

rr

RrRr

R

R

r

r

RR Rr

Rr

f) Give the ratio of the phenotypes obtained from the cross in (e) above. (2)

1 red : 2 pink : 1 ivory

CODOMINANCE

Codominance: a condition in which both alleles of a gene pair in a heterozygote

are fully expressed, with neither one being dominant or recessive to the other

e.g. in white clover leaves , in the heterozygote, both the chevron and the patch pattern appear together

Chevron pattern VCVC

Patch pattern VPVP

Alleles are shown by a superscript in

codominance

Codominance: coat colour in cattle

Red [RR] White [R’R’]  Roan [RR’]  All hairs are

red All hairs are

whiteRed & white hairs occur together:

both alleles are expressed

Roan cow 

Parents: RR' X RR'

  R R'

R RR RR'

R' RR' R'R'

Genotypic ratio: 1 RR: 2 RR': 1R'R'

Phenotypic ratio: 1 red: 2 roan: 1 white

What genotype and phenotype ratios occur in the offspring from the cross:

Question: pg. 72 No. 11a) Certain varieties of cattle can exist in three

colours: red, white and roan. When a red bull is mated with a white cow the calves have a mixture of red and white hairs, giving them an overall colour called roan. These roan calves are different in colour from both parents.

i) State the type of dominance shown by colour in

these cattle.

Codominance

ii) Using the symbols CR for the allele for red hair, and CW for the allele for white hair, state the genotypes of the red bull and the white cow. Red bull: White cow:CR CR CW CW

iii) Give the genotypes of the gametes produced by each parent.

Gametes from red bull: Gametes from white cow:

iv) Give the genotype of the offspring from a cross

between a red bull and a white cow.

CR

CW

CR CW

b) By means of a genetic diagram, show the results of a cross between a roan bull and a roan cow. Your diagram should show the genotypes of the parents, the gametes they produce and the genotypes and phenotypes of all the possible offspring.

Parents: CRCW X CRCW

CR CW

CW CWCW

CRCW

CRCW

CRCRCR 1 Red : 2 Roan : 1 White

CRCW

CRCW Offspring phenotypes:

BLOOD GROUPS

BLOOD GROUPSsometimes a characteristic is controlled by

more than two alleles e.g. three  alleles control human blood:

A, B and O a person has two out of three alleles

BLOOD GROUP (phenotype)

GENOTYPE

A IA IA or IA IO

B IB IB or IB IO

AB IA IB

O Io Io

The alleles for groups A & B are CODOMINANT

Questions: pg. 731) The diagram below shows part of a family tree.

The letters show the blood group of each member of the family.

a) i) State the number of one person in the family tree who must be homozygous for blood group genotype. (1) 6

ii) Give the genotype of this person. (1) IoIo

b) i) State the blood group genotype and phenotype of person 5. (2)

IoIo

ii) Explain how you worked this out. (2)

Genotype: Phenotype: IAIo

IAIB

IBIo

IBIo

Blood group B

Person 6 is IoIo and must have obtained one Io allele from each parent. Person 7 is IAIB and got IB from person 5 and IA from person 4.

c) i) State the genotype of person 1. (1)

IOIOii) Explain how you worked

this out. (2)

IAIO

IAIB

IBIO

IAIO

Person 4 is IAIO and must have obtained IO allele from person 1 as person 2 has the genotype IAIB . Person 4 is blood group A and so the other allele must be IA.

IAIB

IAIO or IAIA

IAIO

2) Two parents, one with blood group A and the other with blood group B, have a child whose genotype is homozygous.

a) Complete the diagram below to show how this can happen. (5)

IoIo

Io Io

IAIo IBIo

IBIA

b) What is the chance of these parents producing a homozygous child? (1)

Parents: IAIO x IBIO

Gametes: IA

F1 generation:

xIO IB IO

IAIOIAIB IOIOIBIO

25%

c) What is the blood group phenotype of the homozygous child? (1) Blood group O

SEX-LINKAGE

Sex-linked genes are carried on the sex chromosomes (X chromosome)

autosomes Sex chromosomes

X X

X Y

Femalecarries two

alleles of a gene

Malecarries one

allele of a gene

Sex-Linkage

red-green colour blindness haemophilia

“3” or “8”??

is the appearance of a certain character in ONE sex, namely the male

examples of sex linked traits:

Males are more likely to suffer from sex-linked diseases

Normal

A

Femalescarry two allele of a gene. If

one allele is defective, female is still normal as effect is

masked by the normal allele.

A Normal: A Sick: a

SickPhenotypically normal / carrier

A

A

a

a

a a

Normal Sick

QuestionSuggest explanations for the following observation.

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

Normal vision Colour blind vision

Haemophilia: possible genotypes and phenotypes

Genotype PhenotypeXHXH Normal female

XHXh Normal female (carrier)

XHY Normal male

XhY Haemophiliac male

Question: pg 63The diagram shows a cross between a person who is a carrier of the disorder haemophilia and a person who has normal blood.Let H represent the allele for normal blood and h the allele for haemophiliac blood.

Person A Person BXH Xh x XH Y

a) What do the X and Y represent?

Sex chromosomes

Person A Person BXH Xh x XH Y

b) Which person is the male? Give a reason for your answer.

B Carries a Y chromosomec) Explain why the Y chromosome does not carry

either an H or h allele.

The Y chromosome is short.

d) Explain the term carrier. (2)

A person who looks normal but carries the defective allele. Has a heterozygous genotype.

e) Complete the Punnett square to show the children which could be born in this cross.

Person A Person BXH Xh x XH Y

XH

XH

Xh

YXHXH XHXh

XHY XhY

f) Give the probability of the first child being male.

50%

g) Give the phenotypic ratio of haemophiliac to normal children. normal – H

sick – h

1 haemophiliac : 3 normal

h) Describe what happens during fertilisation. The nucleus of the sperm and ovum fuse.

i) Explain why human reproduction has both fertilisation and cell division by meiosis in each generation.

Meiosis is important to form gametes having half the number of chromosomes so that after fertilisation, the full set of chromosomes is restored.

fertilisation

VARIATIONWITHIN A

POPULATION

Variationdescribes the differences in characteristics

shown by organisms of the same species two types of variation:

DISCONTINUOUSCONTINUOUS

Continuous Variationcharacteristic varies amongst the members

of a species in a smooth continuous way from one extreme to the other

Continuous Variation: Examples:

Mass Height Intelligence Colour of organs & organisms

Continuous variation is affected by:1. many genes of small effect2. AND the environment

The environment CAN change these variations

Sun bathing causes tanning of skin.

Eating carrots makes human skin and feathers

in birds turn orange.

Discontinuous Variation Produces:

individuals with clear-cut differences with no intermediates between them

examples: blood group in humans sex tongue rolling

Either OR

traits

The environment cannot change discontinuous type of variations

No matter how much you stay in the sun or what you eat – will not change your blood

group!!

Question: SEP, 2002Using examples, distinguish between:i) inherited and

non-inherited variations; (2, 2)

Inherited traits are passed on from parents to offspring due to information in genes e.g. colour of the eyes or shape of nose.

Non-inherited traits are not passed on e.g. pierced ears or a tattoo.

Using examples, distinguish between:ii) continuous and discontinuous variations; (3, 3)

Continuous variation results when a characteristic varies amongst the members of a species in a smooth continuous way from one extreme to the other e.g. colour of the skin, weight and height of people.Discontinuous variation produces individuals with clear-cut differences with no intermediates between them

e.g. blood groups in humans , long or short wings in fruit flies, normal or sick persons.

Question: SEP, 2005 [pg. 83]Look at the drawings of the two men.List two features shown in the diagram, that are inherited and three features that are not inherited. (5)