Genetics

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 colour
flower colour
Tested for e.g.
blood groups
colour 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 pairs
Genes controlling the same characteristics
occupy identical positions on corresponding
chromosomes
The gene pairs control one characteristic
gene for
eye colour
gene for
nose 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 a
BLACK mouse
The dominant allele is expressed
The recessive allele is masked

Alleles are represented by letters
B 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
B
B
B
b
b
b
b
B
b
meiosis
meiosis
fertilisation
All offspring will
be 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 combinationsFertilisation
sperms
ova
zygotes

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 & Phenotype
genotype: describes the genetic make-up (all
of the genes) of an individual
homozygous dominant
heterozygous
homozygous 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 & Heterozygous
HOMOZYGOUS – 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 breeding
parents)
F1 generation
F2 generation
Purple flowers White flowers

All plants have
purple flowers
Fertilisation
among F1 plants
(F1  F1)
3/4 of plants
have purple flowers
1/4 of plants
have white flowers
Alleleforpurplecolour[100% purpleinF1
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 seed
b 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

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 aa
Gametes:
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 – aa
F1 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

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
Tt
Tt

XX = female
XY = male
X Y
X XX XY
X XX XY
Genotypic ratio - 1 XX : 1XY
Phenotypic 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, Nn
vestigial 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 – bb
Brown eyes Blue eyes
bb
x
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
bb
x
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
bb
x
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 Husband
Bb
Daughter
Bb
X Husband
BB
75% Brown
eyed
25% Blue
eyed
: 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.

Cross1
Parental phenotypes Waltzing x Normal
Parental genotypes _______ _______
Gametes _______ _______
Offspring genotypes _______ _______
Offspring phenotype All normal
Cross2
Parental phenotypes Waltzing x Normal
Parental genotypes _______ _______
Gametes _______ _______
Offspring genotypes _______ _______
Offspring phenotypes Waltzing Normal
50% 50%
Normal – A – AA, Aa
Waltzing – a – aa
aa AA
Aa Aa
a A Aa
aa
aa Aa
Aaaa
aA

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.
WORKING
1. 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 Cow
hh X hh
Offspring
hh
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 Y
X XX XY
X 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
Bb
bb
bb
Bb Bb Bb

Mother Father
Brown
eyes
Blue
eyes
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.
a) How can you tell that brown eyes are dominant? (1)

a) How can you tell that brown eyes are dominant? (1)
All offspring of 3 (blue eyed) and 4 (brown eyed) have
brown eyes.

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

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.

bb
Bb
bb
Bb
bb
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 bb
B B b b
F3 generation: Bb Bb Bb Bb
X

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 bb
B b b b
F3 generation: Bb Bb bb bb
X

A homozygous dominant and a
heterozygous individual have the
same phenotype. How can you
determine their genotype?
Black: B – BB, Bb
Brown: 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

x
bb
50% black : 50% brown
x
100% black
bb
Bb
bbBb
The phenotypic ratios among offspring are different,
depending on the genotype of the unknown parent.

Points to remember: Test Cross
Why 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, 2011
Achondroplastic 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, Dd
d – 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, 2012
Machado-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, Dd
d – 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 Nicholas
Genotype 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 Tania
Genotype 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

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
Rr
RR 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 rrRR
1
red
1
white
2
pink
:
::Phenotype
ratio

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. 82
4)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 Ivory
RR 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 Ivory
RR x rr
F1: Rr
rr
Rr
Rr
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: 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
white
Red & 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. 11
a) 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
sometimes 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. 73
1) 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)
IOIO
ii) 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 IOIO
IBIO
25%
c) What is the blood group phenotype of the
homozygous child? (1) Blood group O

Sex-linked genes are carried on the
sex chromosomes (X chromosome)
autosomes Sex
chromosomes
X X
X Y
Female
carries two
alleles of a gene
Male
carries 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
Females
carry 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
aa
Normal Sick

Question
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

Haemophilia: possible genotypes and
phenotypes
Genotype Phenotype
XHXH Normal female
XHXh Normal female (carrier)
XHY Normal male
XhY Haemophiliac male

Question: pg 63
The 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 B
XH Xh x XH Y
a) What do the X and Y represent?
Sex chromosomes

Person A Person B
XH Xh x XH Y
b) Which person is the male? Give a reason for your
answer.
B Carries a Y chromosome
c) 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 B
XH Xh x XH Y
XH
XH
Xh
Y
XHXH 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

Variation
describes the differences in characteristics
shown by organisms of the same species
 two types of variation:
DISCONTINUOUSCONTINUOUS

Continuous Variation
characteristic 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 effect
2. 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, 2002
Using 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)

Genetics

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