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Transcript of 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)