Table Of Key Dates
http://www.accessexcellence.org/RC/AB/BC/Search_for_DNA.html
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1868: Friedrich Miescher isolates nucleic acid from pus cells obtained from discarded bandages
1943: Avery, Macleod and McCarty use bacteria to provide the first evidence that DNA is the bearer of
genetic information
1952: Hershey and Chase show that it is the DNA part of the T2 viral particle, not the protein part, that
enters a host cell, furnishing the genetic information for the replication of the virus
1953: Double helix structure of DNA is revealed by Watson, Crick, Franklin, and Wilkins
1961-65: Genetic code is cracked
1972: First successful DNA cloning experiments are carried out in California
1975: Monoclonal antibodies are produced
1975-77: Rapid methods of sequencing DNA are perfected
1977: The first human gene is cloned
1982: Genetically-engineered insulin is approved for use in diabetics in the USA and UK
1987: First genetically-engineered microorganisms are used in field experiments
Name
Observation / discovery
Lamarck (1809)
proposed theory of evolution based on inheritance of acquired characteristics
Darwin (1859)
On the origin of Species by Means of Natural Selection published
Mendel (1865)
Experiments on the genetics of peas and formulation of his Two Laws
1868: Friedrich Miescher isolates nucleic acid from pus cells obtained from
discarded bandages
Morgan (early 1900s)
pioneered use of Drosophila in genetics experiments and described linkage
Fred Griffith (1920s)
(04AL-I-9)
Found out that a non-lethal strain of bacteria on incubation with a dead lethal
strain would become lethal. This character was inheritable.
Produced evidence suggesting that a chemical ‘ transforming principle’ was
responsible for carrying genetic information
Oswald Avery (1940s)
(04 AL-I-9))

1943: Avery, Macleod and McCarty use bacteria to provide the first
evidence that DNA is the bearer of genetic information
Obtained an extract from the dead lethal bacterial and treated this extract as
follows: http://www.dnaftb.org/dnaftb/17/animation/animation.html
(1) destroyed the protein in it
(2) destroyed the protein and RNA in it
(3) destroyed the protein, RNA and DNA in it
He then incubated a portion of the extract after each treatment with the
non-lethal bacteria and tested the resultant bacteria for their lethal effect.
Chargaff (1950)
Erwin Chargaff discovers regularity in proportions of DNA bases for different
species. In all organisms he studies, the amount of adenine (A) approximately
equals that of thymine (T), and guanine (G) equals cytosine (C).
在 1950 年一位生化學家 E. Chargaff 對不同生物的 DNA 進行化學的分析發現 (by paper
chromatography)，雖然不同來源的 DNA 分子其核甘酸的組成各不相同，但是在這些分子
的化學組成卻有一些共同的原則，這原則就是，不論其組成有多大的不同，但只要是同一
生物來源的 DNA 其中的 adenine 的量永遠與 thymine 的量一樣，同樣的，guanine 與 cytosine
的量是一樣的，稱為 Chargaff’s rule。
Hershey and Chase (1952)
1952: Hershey and Chase show that it is the DNA part of the T2 viral particle,
not the protein part, that enters a host cell, furnishing the genetic information
for the replication of the virus
Showed DNA to be the hereditary material
James Watson and Francis 1953: Double helix structure of DNA is revealed by Watson, Crick, Franklin,
Crick (1953) (04 AL-I-9)) and Wilkins
determined and formulated the detailed molecular structure of DNA
Stahl and Meselson (1959)
described mechanism of semi-conservative replication in DNA
http://www.dnaftb.org/dnaftb/20/concept/index.html
F. Jacob and J. Monod Postulated existence of mRNA in theory on control of protein synthesis
Study of enzyme systems of E. coli
(1961)
 explain the regulation of enzyme production in term of gene regulation.
(Jacob-Monod Hypothesis)
Others

1961-65: Genetic code is cracked

1972: First successful DNA cloning experiments are carried out in
California
1975: Monoclonal antibodies are produced
1975-77: Rapid methods of sequencing DNA are perfected
1977: The first human gene is cloned
1982: Genetically-engineered insulin is approved for use in diabetics in
the USA and UK
1987: First genetically-engineered microorganisms are used in field
experiments
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Adenine
Guanine
Purines
(double-ringed)
Organic bases
Cytosine
Thymine (only in DNA)
Pyrimidines
(single-ringed)
Deoxyribose
Pentose ring
Uracil (only in RNA)
Ribose
Inorganic Phosphate group
Condensation
Nucleotides
Mononucleotides
e.g. ATP (Adenosine triphosphate)
as an energy carrier
Condensation
Dinucleotides
e.g. NAD (Nicotinamide adenine
dinucleotide) as a coenzyme
Polynucleotide single chains
Ribosomal RNA (rRNA)
(Nucleic Acid strands:- DNA / RNA)
Transfer RNA (tRNA)
Messenger RNA (mRNA)
- linear
- backbone with alternating sugar and phosphate
Watson-Crick Model of DNA
- 2 // single-stranded DNA running in opposite
direction
- twisted and form a double helix
(10 base pairs of 3.4 nm per turn)
- backbone of each strand with alternating
sugar (deoxyribose)
and phosphate group
- crossed linked between the complementary bases on the two
strands at regular intervals with hydrogen bonds (A=T ; G≡C)
- bases projecting inwards
DNA as genetic materials
1.
stable complementary nature of the two strands of DNA
 i. makes it suitable for storing genetic information and transmitting the information
from one generation to another
ii. the H bonds between the base pairs tend to drive the double helix to reform
spontaneously after uncoiling in replication and transcription
2.
3.
discrete segment of DNA molecule composed of a particular sequence of nucleotides (gene)
can act as a unit of inheritance of a single characteristic or polypeptide synthesis. (see “Gene
concept” )
genetic information between DNA in different chromosomes and be exchanged (see
“ Meiosis” ) or be changed (see “Mutation”)
 allow variation among different individuals, evolution or forming new species.
DNA Spooling (rough DNA extraction from fresh tissues) -----3 Basic Steps
1.
2.
3.
Break the cell to release DNA and denature cell proteins
The cell and nucleus must be lysed (broken open) to release the DNA. Remove chromosomal proteins.
denature or inactivate DNAase
Protect DNA from degradation by enzymes that will cause shearing.
Precipitate DNA in alcohol
Result:
milk layer
Procedures:
containing DNA at
1. homogenize tissue in a blender
 break cell to release DNA
the middle layer
2. adding detergents
 break cell membranes and alter protein structure
3. adding salt solution
Confirmation:
 denature the cell proteins and sink to the tube bottom
adding Universal indicator
4. heating (60oC water bath)
to denature the DNAase enzymes that cause shearing in DNA
cause the extract turns
5. ice bath
orange yellow in colour
 to slow the enzyme action that degrade DNA
(i.e. acidic)
6. stirring and filtering
7. adding protease
(meat tenderizer has papain, an enzyme that helps clean the protein from the
DNA that can contaminate it. Papaya juice and pineapple juice also contains
this enzyme.)
8. adding iced alcohol to uncoils and precipitate the DNA
Semi-conservative Replication of DNA
Replication by DNA
polymerase
(in 5' to 3' direction)
Features of the Genetic Code (Codon)
by RNA
polymerase
(in 5' to 3'
direction)
Synthesis of non-essential amino acids from other substances (e.g. from
respiratory intermediate via Transamination) or obtain of essential amino
acids from diet
The 1st codon (the start
codon – AUG) is bound with
formylated methionine tRNA
polypeptide is elongated until a stop
codon in the mRNA is read
[HKALE 80 II]
5.
(a)
Briefly explain the meaning of the following terms:
(i)
(ii)
(iii)
(iv)
(v)
gene
allele
chromosome
linkage group
mutation
(10 marks)
(b) Describe how the information carried on DNA is used in protein synthesis.
labelled diagrams to answer the question.)
(If you wish, you may use
(10 marks)
5.
(a)
(b)
(i)
(ii)
A gene is the basic unit of heredity , a number of which form a chromosome.
An allele is one of a pair of genes occupying the same locus on homologous
chromosomes which separate during meiosis.
(iii) A chromosome is a threadlike structure found in the nucleus of eukaryotic cells. A
number constitute the genetic material. They consist of DNA, RNA and protein.
(iv) A linkage group refers to all the gene present on the same chromosome. They do not
show independent assortment.
(v) Mutation involves a change in the amount or structure of DNA in the chromosomes
AND Mutation involves a change in the phenotype of an organism.
2
2
2
2
2
DNA is a double strand/ helix of 4 nucleotide bases with the bases in fixed association i.e. 2
adenine to thymine and vice versa, guanine to cystosine and vice versa
--A—T—G—C----T---A---C---G---The strand begins to unwind and an mRNA chain is produced alonside by complementary
base formation, Transcription, N.B. Uracil replaces thymine.
1
2
mRNA then leaves the nucleus and becomes associated with ribosomes to form polysomes.
The mRNA is coded in triplets and each such group is called a codon.
2
1
1
tRNA associates with these triplets in the anticodon area. This process being known as
translation , thus
1
(20)
And so the amino acids borne at the other end of the tRNA are arranged into a sequence
determined by the original configuration of the bases on the DNA. The amino acids are
then
linked
by
peptide
linkage
to
form
specific
proteins.
[HKALE 89 II]
8.
(a)
The following are codons for some of the amino acids and the start signal for translation found in
messenger RNA:
* The codon AUG serves as a start signal and initiates synthesis of a polypeptide.
Answer the following questions with reference to the synthesis of the peptide shown below:
(i)
Using ONE labelled diagram ONLY, explain the roles of the various components involved
in protein synthesis. In your diagram, you should indicate:
(1) the base sequence of the messenger RNA,
(b)
(2) the site of protein synthesis,
(3) the direction of translation,
(4) the base sequences of the anticodons of the transfer RNAs, and
(5)
the stage at which the amino acids, methionine and histidine, are linked together
while leucine is forming a bond to the growing peptide and cysteine is still in the
pool of free amino acids.
(10 marks)
(ii)
What is the base sequence of the corresponding segment on the DNA molecule ?
(2 marks)
Explain the significance of polysomes in protein synthesis.
(2 marks)
(c)
(i)
(ii)
(iii)
Explain the meaning of the following terms with reference to codons:
universal
degenerate
nonsense triplets
(6 marks)
Total 20 marks
8.
(a)
(i)
correct base sequence of m-RNA
indicate site of protein synthesis - ribosome
indicate correct direction of translation
correct base sequence of anticodons
showing Met and His
showing Leu joining to the growing peptide
showing Cys in the pool of free amino acids
clarity and presentation of diagram
(Diagram should show that:
t-RNAs are amino acid specific such that one end of the t-RNA links up with a
specific amino acid and draws it to the m-RNA on the ribosome. The three unpaired
bases (the anticodon) at the other end link up with the appropriate codon in the
m-RNA molecule. In this way the amino acids are lined up in an order corresponding
to the sequence of codons in the m-RNA.)
(ii)
TAC GTA GAA ACG
2
(b)
Polysomes are the association of a group (5 to 50) of ribosome to a single m-RNA 2
molecule. This enables a large number of polypeptides to be assembled on a single m-RNA
strand in a comparatively short period.
(c)
(i)
'Universal’ means the same code codes for the same amino acid in all kinds of 2
organisms.
(ii)
'Degenerate’ means that a single amino acid may be coded by more than one triplet of 2
bases.
(iii) Nonsense triplets refers to those triplets which do not code for any known amino 2
acid. They sometimes serve as the 'stopping’ code.
[HKALE 98 II]
3.
(a) Explain the features of the genetic code.
(6 marks)
(b) Describe in detail the cellular processes that are necessary in the transfer and decoding of genetic
information for polypeptide synthesis.
(12 marks)
(c) In general, what additional processes are necessary for the formation of the three-dimensional structure
of proteins after polypeptide synthesis ?
(2 marks)
3.
(a) Features of the genetic code
(2 marks on features, 4 marks max. on explanation)
-
triplet code (½)
a group of three nucleotides is necessary to encode for a specific amino acid (1)
½
1
-
universal (½)
same triplet of nucleotides is responsible for coding the same amino acid (½) in all
organisms (½)
½
degenerate (½)
4 different nucleotides to form the triplet (½)64 available codes(½) (i.e. 43 = 64, but
there are only 20+ amino acids (½) some amino acids have, more than 1 code (½)
½
non-overlapping (½)
a nucleotide will only be used once for each translation process / will not be used twice in
adjacent codes in one translation (1)
(accept diagrammatic explanation)
½
-
-
(b) *Transcription(½)in the nucleus (½)
Genetic information on DNA (½). Region of DNA unwinds (½), one strand of the unwound
portion acts as template (½), attracts free ribonucleotides (½)to it through complementary
base-pairing (½):
1
2
1
(max.6)
1
2½
4½
adjacent ribonucleotides join up to form m-RNA (½)
m-RNA leaves the nucleus and attaches to ribosome at rough endoplsmic reticulum (1)
(RER) in the cytoplasm (½)
*Translation (½) at RER
free amino acids in the cytoplasm is activated by each joining to a specific t-RNA (1) using
ATP.
Each t-RNA has a specific anticodon consists of three ribonucleotide bases (1), these pair up
in a complementary manner with the codon bases on the m-RNA (1) bringing the amino acid
with it (½). Adjacent amino acids join by peptide bond (½)to form polypetide. Ribosome
moves along m-RNA (½), amino acids add on one at a time (½).
½
1½
max.8
(½)
1
4
(max.12)
(c) Coiling (½)and folding (½)of the polypeptide due to intramolecular bonds, e.g. H-bond /
disulphide bond (½)formed between different parts of the polypeptide (½).
(2)
Operon
= a group of adjacent genes which act together
(hence located on a particular segment of a
same chromosome)
- mainly consists of two parts:
1. Structural genes
2. Operator gene
Repressor
- protein produced by regulator gene outside an
operon
- binds to operator of the operon to repress the
transcription of the following structural genes
Operator gene
- the gene in front of the structural gene in
the operon
- controls the turning on or off of the
following structural genes
Structural genes
- closely linked genes (same
linkage group)
- responsible for synthesis of
functionally related polypeptides
The Cell Cycle
= is the period from the beginning of one division to the beginning of the next and is customarily
represented in diagrams as a circle.
generation time
= the length of time for a complete cell cycle is termed the
3 main stages:
1.
Interphase / Resting phase
Chromosomes are not visible but only chromatins can be seen after staining:
First growth phase (G1) : cell organelles and biochemicals synthesis
Synthesis phase (S)
: DNA replication
Second growth phase (G2) : centriole duplication, increase energy storage
2. Mitosis
3. Cytokinesis
- animal cells :
constriction of the center of the parent cell from the outside inwards.
- plant cells :
i. fusion of Golgi vesicles
ii. cell plate across the equator from the center outwards
iii.new cell membranes and cellulose is laid down on the plate to form the new cell wall
significance of mitosis and meiosis respectively
Mitosis
Meiosis
gametes for sexual reproduction and increase
genetic stability
- daughter cells are genetically identical to
genetic variation
their parent cell
- segregation of a pair of alleles and independent
 results in genetic stability within
assortment of unlinked genes allows random
population of cells derived from
combination of genetic materials in daughter
parental cells
nuclei
- crossing over of non-sister chromatids in bivalent
allow recombination of genes in a linkage group
- allow fertilization (without double of DNA) of
haploid gametes from different parents
growth
- the number of cells within an organism
increases by mitosis
asexual reproduction and regeneration
[HKALE 96 II]
6.
Compare and contrast the processes of mitosis and meiosis. Discuss the roles and significance of
mitosis and meiosis in the life of flowering plants and mammals, illustrating your answer with
examples where appropriate.
(20 marks)
Ans
6.
Similarity between the processes of mitosis and meiosis:
3
½
Both involve nuclear division (½)
Both involve duplication of DNA/chromosomes (1) and movement of chromosomes along
the spindle fibres. (1)
Both involve spindle formation (1)
Differences between mitosis and meiosis:
Mitosis
Meiosis
2 identical daughter nuclei
4 daughter nuclei are formed which are genetically different
are formed each is
from the parent nucleus and each other (1)
genetically similar to the
parent nucleus (1)
2
only 1 division of the (½)
nucleus, same chromosome
number as the parent nucleus
(½)
2 divisions of the nucleus (½), chromosome number reduced
to half (½).
2
no pairing up of homologues,
linear alignment at the
equator (½).
chromosome splits into
chromatids which move to
opposite poles (1)
homologous chromosomes pair up at the equator in the first
division (1).
1
½
no such events (½)
may involve chiasma formation (½), may involve
crossing-over (½) of genetic material between homologous
chromatids results in variation.
no splitting into chromatids in first division entire
2
chromosome migrates to opposite poles homologues separate
(1).
1
½
Roles and significance of mitosis(a)
Asexual reproduction in flowering plants (½) i.e. vegetative propagation. (not in
mammals), provides for a large number of offspring (½) with the same genetic make-up as
the parent (½) to thrive in the same environment (½).
2
(b)
Cell Multiplication (½) - common to both plants and animals (½), this results in.
1
(1)
6
growth (½)
In flowering plants, mitosis occurs at specific sites (½), the meristems. Apical
meristems (½) (stem and root apices) give rise to elongation in length - in the
longitudinal axis in the plant body (2)
Lateral meristems (½) (vascular cambium and pericycle in root) in dicotyledons (½),
give rise to growth in girth (½) of stem and root, and lateral root (½).
The meristemtic activity at the pericerm (½) gives rise to bark formation (½) which
serves to provide protection (½) to the stem of woody dicotyledons.
In mammals, mitosis takes place in most body cells (½) resulting in increase in size.
(2)
replacement/ repair (½)
4
½
In flowering plants, mitosis at the root tip replaces the root cap (½) which is
constantly wearing away (½). It also forms callus (½) at sites of wound to plug the
opening from microbial and other injury (½).
(c)
In mammals mitosis occurs in body cells e.g. lining of epithelium of alimentary
canal, bone marrow cells. skin epithelium.
Gamete formation in flowering plants (½) from the haploid gametophyte generation.
1
½
Roles and significance of meiosis
(a)
Gamete formation in mammals (½)- producing sperms and ova, each with half the
chromosome number, ensures restoration to diploid at fertilization (½).
1
(b)
Spore formation in flowering plants (½)- these plants exhibit alternation of generations in
their life cycles. 2 types of spores are formed by meiosis, microspores (½) form the anther
sac and megaspore (½) from the ovary of the flower. They give rise to the male
gametophyte (½) and female gametophyte (½) respectively.
2
½
(c)
(Genetic) variation (½) random/ independent assortment (½) at metaphase I, crossing-over
(½) and recombination at syngamy in fertilization (½) result in genetically different progeny
(½) from the parents. These variations may allow for differences in adaptability when
environmental conditions change (1).
3
½
Mendel's First Law (The Law of Segregation)
The characteristic of an organism are determined by internal factors which occur in pairs. Only
one of each pair of such factors can be presented in a single gamete.
回交:
一雜種與其親代中的異性交配
Backcross : cross of a hybrid with one of its
parent
Test cross: a cross between an organism of
unknown genotype and the relevant
homozygous recessive organism.
Dominance : showing their effect in both
homozygous or heterozygous and mask the effect
of another contrasting character
Recessive and only shows its effect when
homozygous.
Mendel's Second Law (Law of Independent Assortment)
each of a pair of contrasted characters may be co-exist (or combine)
randomly with either of another pair
[HKALE 93 II]
1. (a) Distinguish between
(i) codon and anticodon
(ii) continuous and discontinuous variation
(4 marks)
(b) The occurrence of colour blindness and haemophilia in a family is shown by the pedigree below. The
genes for the two traits are found on the X chromosome.
(i)
Deduce the dominance or recessiveness of the gene which leads to
(1) colour blindness
(2) haemophilia
Explain your answer.
(Note : Genetic diagrams will NOT be accepted.)
(6 marks)
(ii) Use appropriate symbols to represent the genotypes of individuals A, B, C, D and E. Define all
the symbols used in your answer.
(7 marks)
(iii) If individual F marries a normal man, what is the probability that her son might be a haemophiliac ?
With the aid of a genetic diagram, explain your answer.
(3 marks)
Total :
20 marks
1.
(a) (i)
= Anticodon is a three-base sequence on a tRNA molecule while
codon is a three-base sequence on a mRNA molecule.
= Anticodon is complementary to codon.
2
(ii) = Continuous variations are variations in a phenotypic character
that are smooth and continuous (e.g. height of adult men).
= Discontinuous variations are distinct or abrupt (e.g. eye colour
of man)
(b) (i) (1) = The gene for colour blindness is recessive.
= It is because daughters of individual A are not colourblind
= while each of them must have inherited an X chromosome
with a colourblind gene from their father. That is, the
colourblind gene must be recessive.
(2) = The gene for haemophilia is recessive.
2
3
3
= It is because a son of individual D is a haemophiliac
= while individual d herself is normal. That is, the son must
have inherited an X chromosome with a recessive
haemophilia gene from his mother.
(ii) Define symbols ;
-Xc as X chromosome with colourblind gene
-Xh as X chromosome with haemophilia gene
-Xhc as X chromosome with both colourblind and haemophila gene
-X and Y as sex chromosomes with the respective normal genes
(N.B. Accept other well-defined symbols, 1/2 mark each)
Genotype for A is XcY
B is XhX
C is XhY
D is XcXh
E is XhXh
(iii)
P:
F1 :
individual F
Xh Xc
Xh X
normal
female
)
)
) (1 mark each)
)
)
x
X hY
haemophilia
male
2
5
normal man
XY
X cX
normal
female
X cY
colourblind
male
 the probability of her son being a haemophiliac is 50 %
2
1
The Chi-Square Test
In any genetic experiment, how can we decide if our data fits any of
the Mendelian ratios we have discussed? A statistical test that can test
out ratios is the Chi-Square or Goodness of Fit test.
Chi-Square Formula
Degrees of freedom (df) = n-1
( where n is the number of classes )
A Chi-Square Table
Probability
Degrees
of
0.9 0.5 0.1 0.05 0.01
Freedom
1
0.02 0.46 2.71 3.84 6.64
2
0.21 1.39 4.61 5.99 9.21
By statistical convention, we use the 0.05 probability level as our
critical value. If the calculated chi-square value is less than the 0 .05
value (i.e. < 7.82 for df=3 with a probability greater than 0.05), we
accept the hypothesis. If the value is greater than the value (i.e. > 7.82
for df=3 with a probability lesser than 0.05), we reject the hypothesis.
3
0.58 2.37 6.25
4
5
1.06
1.61
7.82 11.35
3.36 7.78 9.49 13.28
4.35 9.24 11.07 15.09
Example:
Let's test the following data to determine if it fits a 9:3:3:1 ratio.
Class no.
Observed Values
Expected Values
1
315 Round, Yellow Seed
(9/16)(556) = 312.75 Round, Yellow Seed
2
108 Round, Green Seed
(3/16)(556) = 104.25 Round, Green Seed
3
101 Wrinkled, Yellow
Seed
(3/16)(556) = 104.25 Wrinkled, Yellow
4
32 Wrinkled, Green
-------------- 556 Total Seeds
(1/16)(556) = 34.75
556.00
Wrinkled, Green
Total Seeds
Number of classes (n) = 4
∴ df = n-1 = 4-1 = 3
 Chi-square value = 0.47
According to the Chi-Square table at df = 3 , we see the chi-square value is smaller than 0.58 (i.e. the probability is
greater than 0.90). Threrefore, because the calculated probability (>0.90) is much greater than 0.05,
then we accept the hypothesis that the data fits a 9:3:3:1 ratio.
[HKALE 97 II]
3. (a) What are the process that generate genetic variations in flowering plants ? Explain the mechanisms
involved.
(6 marks)
(b) A diploid plant bears a dominant mutation in one allele of a gene which controls sepal form.
mutation has the effect of changing scale-like sepals to petal-like sepals.
This
(i) As the flowers with petal-like sepals have a better appeal to customers, a plant breeder kept this
mutant plant in a greenhouse allowing it to self-fertilize for several generations. During this period,
the plant breeder made frequent visits to the greenhouse and removed any plant bearing flower buds
with scale-like sepals. Deduce and explain the genotypes of the remaining plants in the F1 and F2
populations and their relative proportions. What will be the long term effect of this breeding
practice ?
(7 marks)
(ii) The original mutant plant was crossed with another plant which produced flowers with red petals
and scale-like sepals. All the F1 plants shows red petals. Yet only half of them produced
petal-like sepals. When these F1 plants were intercrossed, the F2 progeny showed a petal colour
ratio of 3:1 (three red-petal-bearing plants to one white-petal-bearing plants). Without using
genetic diagrams, deduce and explain the dominant petal colour and the genotypes of the two parents.
(7 marks)
3.
(a)
Meiosis (½) in the formation of spores (½) / pollen and embryo sac / male and female
gametophytes. Crossing over (½) at meiosis between homologous chromosomes results in
reciprocal exchange of genetic material (½) , random assortment of chromosomes into
daughter nuclei (½).
Pollination and Fertilization are a random processes (1) resulting in random combination of
genetic make-up from the gametes. (½).
Cross pollination / fertilization involves genes from 2 parents resulting in new combination
of genetic mark-up (1).
Mutation (½) - changes in DNA (½), chromosome structure (½) in all phases in the life
cycle.
2½
1½
1
1½
____
max. 6
(b)
(i)
Let S be the dominant allele for petal -like sepals.
Let s be the recessive allele for scale -like sepals.
For the F1:
mutant / parental genotypes
Ss
x
Ss (½)

F1 genotypes of the remaining plants SS Ss Ss (0.5) ss (removed)
F1 genotypic ratio
1 : 2 (or 1/4 :2/4) (½)
For the F2:
There are two parental types for F2, i.e. SS and Ss, selfing of each will give rise to :
SS x SS (½) Ss x Ss (½)


F2 genotypes
SS (½)
SS, Ss, Ss (½),ss (removed)
of the remaining
plants
SS:Ss=(1/3)x[1/4:2/4]
(1 marks)
Proportion of SS: [ 1/3 x1 +2/3 x 1/4] (0.5) =½
Proportion of Ss : [ 2/3 x 2/4] (0.5) = 1/3
½
½
1
1
1
F2 genotypic ratio : 1/3 (½)
Proportion of SS:Ss = 3:2 (½)
(Deduct ½ mark if answer does not convey idea of removing scale -like sepal plants)
(ii)
1½
1
½
Continual inbreeding and picking of the double recessive genotype will result in the
homozygotes with dominant alleles being more in proportion than the heterzygotes (1) 1
max. 7
Red (½)
The 3:1 red to white petal colour ratio/ the co-existence of red and white petal colour
in F2 indicates that F1 is heterozygous for petal colour (1). As all F1 is red (½)/ 2
white colour cannot express in the presence of red, red petal is the dominant character.
The 3:1 phenotypic petal colour ratio indicates that a single gene is controlling petal
colour (½)
½
Let R be the dominant allele for red petals.
Let r be the recessive allele for white petals.
½
To give rise to heterozygous F1, the 2 parents must be homozygous in their own petal
colour (1) which are different (½) Thus the mutant parent must have white petals (½) 2
since it is given that the other parent has red petals.
For sepal form, the F1 plants show 1:1 ratio (½) indicating that one parent is
homozygous recessive while the other must be heterozygous (½).
1
or
For sepal form , the mutant parent is heterozygous (given ) while the parent with
scale-like sepal must be a double recessive since scale - like sepal is a recessive
character (1)
Genotype of mutant having petaloid sepal is : rrSs (1).
Genotype of parent with red petal and scale-like sepals :RRss (1)
(Deduct ½ mark for not specifying the genotypes belong to which parents)
2
(max.7)
4.
(a) Explain by means of a genetic diagram how a female child could inherit haemophilia.
(8 marks)
Answer
4. (a) (i) Haemophilia is sex-linked/carried on sex-chromosome.
1
H = recessive haemophilia gene (defined)
1+1
parent :
XhY
x
XhX
1+1
haemophiliac male
carrier female
½+½
possible
offspring XhXh
XhX
XhY
XY
1
haemophiliac
carrier
haemophiliac
normal 1
female
female
male
male
Since haemophilia gene is recessive therefore there is a 1:3 chance of a
haemophiliac female being conceived)
(8)
(b)
Explain why it is rare for a haemophiliac female to survive beyond the age of puberty.
(3 marks)
Answer
4. (b)
At puberty menstruation starts. Blood loss from uterus which normally 3
clots, leads to uncontrollable haemorrhage for haemophiliac female.
(3)
(c)
What advice would you give to the mother of the child in part (a) if she were pregnant with a second female foetus ? Give reasons for your advice.
(3 marks)
Answer
4. (c)
(Counsel about chances of survival - advice)
3
Either abortion if in early pregnancy
Attempt birth - warn of frequency transfusions, administration of clotting
factors, lower quality of life, puberty problems
any reasonable suggestions : up to 3 marks
(3)
(d)
Give two ways by which mutations may occur in
(i)
(ii)
genes.
chromosomes.
(4 marks)
Answer
4. (d) (i) Gene mutation
-change in base sequence due to -addition/deletion. substitution etc.
(ii) chromosome mutation
=change in chromosome number
=change in gross structure of chromosome
2
2
(4)
(e)
Suggest how induced mutations in man could be guarded against.
(2 marks)
Answer
4. (e)
=Avoid ingestion/contact/exposure to mutagenic chemicals e.g. suntan oils 2
=Avoid exposure to ionising radiations e.g. X-rays, nuclear events
(2)
an affected female has
all the son be affected
a normal male has all the daughter
be phenotypically normal
Affected boys of normal
mother may have affected gene
from affected grandpa through
their mother
Dwarfism, Huntington’s disease ----- autosomal dominant
Affected progeny should have
at least one affect parent
(i.e. no unexpected affect
progeny form normal parents)
Two affected parents
may have unaffected
progeny.
the phenotype
generally appears
every generation
[HKALE 99 II]
3. (a) Distinguish infectious diseases from inheritable diseases with respect to their causes and modes of
transmission.
(6 marks)
(b) State and explain three features of the human body which are in the first line of defence in preventing the
entry of infectious agents.
(6 marks)
(c) The following pedigree shows the occurrence of a hereditary disease D among the members of a family:
Assuming that disease D is controlled by a single gene, deduce with reasons, why this disease is NOT
sex-linked. (Do not include genetic diagrams in your answer.)
(8 marks)
[HKALE 99 II]
3. Candidates to answer all aspects, i.e., causes and mode of transmission of each type of disease to score full
mark of 6.)
(a) Infectious disease
Inheritable disease
Causes:
Causes (max.3)
-pathogen (1) / any 2, ½ mark
each : bacteria, viruses, fungi
-poor personal hygiene (½) / poor
health / contaminated
environment
- defect in genome / genetic material (½) induced max. 4
(½) or spontaneous (½); gene mutation (½) due to
changes in base sequence (½); chromosome
mutation (½), due to changes in chromosome
number and / or structure (½)
Modes of transmission (max. 3):
Modes of transmission:
- from generation to generation through gametes
(1)
-inhalation of airborne pathogen (1)
-body fluid contact (1) / sexual contact (½) /
blood contact / saliva contact
-vector (½) which includes insects / animals (½)
max. 4
(max. 6)
(b) Any three of the following, 2 marks each:
- dead cells of epidermis of intact skin (1) provide physical barrier (1)
- secretions from sebaceous gland (1) - contains substances that inhibit growth of microbes (1)
- mucus in respiratory tract (1) - traps and stops pathogen from entering the body (1)
- cilia in the respiratory tract (1) - sweep pathogen out of body (1)
- gastric / digestive juice in stomach (1) - strongly acidic, kills ingested pathogen (1) / proteases (½)
- digestive juice (1) - protease (½) kills ingested pathogen (½)
- hair at nasal passage (1) - acts as filter to trap airborne pathogen (1)
(accept other correct alternatives)
2
2
2
2
2
2
2
Deduct 1 mark for incomplete sentences, denote as P = -1
(max.6)
(c) XX is female (½)
XY is male (½)
sex linked means the gene responsible for the disease is on either the X or Y sex chromosome in
human (1)
A mother only transmits X sex chromosome to offspring (½). A father segregates either X or Y sex
chromosome to an offspring (½)
A dominant allele will always express its phenotype in both a male and a female individual (½)
A recessive allele will also express its phenotype in a male individual (½) but it requires
homozygosity for expressing the controlled phenotype in a female individual (½).
1
1
1
1½
max.2
If the disease-causing allele is Y-linked (½),
it is expected that all male progeny of an affected father will be affected individuals (1). However,
from the pedigree, male individual 6 in F1 was not affected although his father (individual 2) was
affected (1). Thus, the disease-causing allele is not Y-linked (½).
Or
it is expected that all affected individuals must be male (1). However, from the pedigree, female
individuals 4and 5 in F1 and female individuals 11 and 12 in F2 were also affected (1).
Thus, the disease-causing allele is not Y-linked (½).
3
If the disease-causing allele is X-linked and recessive (½),
it is expected that all affected females inherit the disease-causing alleles from both father and
mother (1). However, from the pedigree, affected female individuals 11 and 12 in F2 were fathered
by unaffected individual 3 and affected mother 4 (1). Thus, the disease-causing allele is not
recessive X-linked (½).
Or
it is expected that an affected female produces affected male progeny (1).
However, from the pedigree, male individuals 9 and 10 in F2 were unaffected, and they were sons of
the affected mother (individual 4) (1). Thus, the disease-causing allele is not recessive X-linked (½).
If the disease-causing allele is X-linked and dominant (½),
it is expected that an affected son must inherit his disease-causing allele from his mother who is an
affected individual too (1). However, from the pedigree, affected male (individual 7 in F1) was
mothered by unaffected individual 1 (1). Thus, the disease-causing allele is not dominant X-linked
(½).
Or
it is expected that the female progeny of an affected father will be affected (1). However, from the
pedigree, female individuals 15 and 16 in F2 were unaffected although their father (individual 7)
was affected (1). Thus, the disease-causing allele is not dominant X-linked (½).
Thus, from the above deduction, the disease-causing allele is neither X nor Y-linked.
3
3
max.6
(Accept alternative approach that is correct)
(max.8)
Three primary patterns of inheritance:
1.
2.
(Supplementary Summary)
Autosomal Dominant
inheritance (DD or Dd)
e.g. Dwarfism,
Huntington’s disease
Autosomal Recessive inheritance (rr)
e.g. Albinism, Cystic fibrosis, Sickle
cell anaemia and Thalassemia
X-linked recessive inheritance
(XrXr or Xr Y)
e.g. Colour-blindness,
G-6-PD deficiency and Haemophilia
Phenotype in
majority of
progeny
Affected
unaffected
unaffected
Any unexpected
affected
progeny from
two unaffected
parents
No
Yes
Yes (for boys)
(since both parents are dd)
(when both parents are Rr)
 not easy to identify the inheritance
of recessive traits in newborn.because
majority of the affected individuals
have phenotypically normal parents.
(when the mother is carrier XXr) 
inheritance of recessive traits is not easy
to identify because majority of the
affected boys have phenotypically
normal parents.
(usually found in pedigrees with a
high proportion of marriages
between relatives e.g. cousins because
for girls (the father should be affected)
(Most of this type of
affected individuals are
heterozygous for this
characteristic)
the related parents have a greater chance of
producing offspring that is homozygous for a
particular recessive allele than unrelated parents)
3.
4.
5.
6.
Any unaffected
progeny from
two affected
parents
appearance of
affected
phenotype in
every
generation?
If both the
parents are
heterozygote:
among the
siblings of the
affected
individuals, the
ratio of normal
to affected is
Yes
No
No
(when both parents are
Dd)
(since both parents are rr)
(since the parents are either XrXr or
XrY)
Yes
No
No
(appearance of phenotype
generally be shown in
every generation)
(appearance of phenotype may not be
shown in every generation)
(appearance of phenotype may not be
shown in every generation)
Normal : affected =
1:3
Normal : affected = 3:1
Not applicable
(both parents are normal Rr)
since heterozygous situation cannot be
found in male
Is the affected
phenotype
related to sex?
No
No
Yes
(equal chance in both sex)
(equal chance in both sex)
- The trait is in majority exhibited in
male (who are hemizygous)
(both parents are affected
Dd)
provided that the
offspring is in large
number (but it cannot be
achieved in actual case)
provided that the offspring is in large
number (but it cannot be achieved in
actual case)
- an affected male cannot pass on the
trait in the male line, but only through
the female line
(though a normal male has all the
daughter be phenotypically normal,
all daughters (phenotypically normal
carrier or affected individual) of an
affected male will have all the boys
be affected)
- an affected female has all the son be
affected
- normal mother may be carrier to have
normal sons, normal daughter, affected
sons or carrier daughter
Ref: http://www.ndsu.nodak.edu/instruct/mcclean/plsc431/mendel/mendel9.htm
http://web.ukonline.co.uk/webwise/spinneret/genes/pedigr.htm
Human Genetics – Genetic Screening and Counselling of Genetic Diseases
Knowledge for students to apply
in daily life:
1. genes
- the vehicle of heredity
- determine phenotype of
organism
2. DNA
- the genetic material
- found in nucleus of all
eukaryotes
3. Chromosomes
- only visible under
microscope during
metaphase
4. single gene determined
phenotype / character
- inheritance follow
mendelian law
5. mutation
- can occur at
chromosomal level
- or at gene level
Daily Application included:
- understand the basis of genetic diseases
- appreciate the detection or risk estimation of genetic diseases
- gene therapy may be the attempt to treat genetic diseases at the
root.
1. failing of chromosome separation (non-disjunction) during
meiosis cause missed or extra chromosome(s) in gamete.
(knowledge 1, 2 and 5)
 - Down’s syndrome (Trisomy 21)
- Polyploidy plants
(can be induced by colchicines 秋水仙素: blocks spindle
formation and thus prevent anaphase movement)
2. foetus with Down’s syndrome can be identified by observing any
extra 21th chromosomes in nucleus of dividing fetal cells
(karyotype 核型) at particular stage of pregnancy (knowledge 2
and 3)
3. mutation in even one gene can upset the production of important
protein causing a particular genetic disease, (knowledge 1, 2 and
5) e.g.
- G6PD deficiency
- thalassemia
4. estimation of the risk of giving birth to a baby (with genetic
disease, e.g. a thalassemic baby) is possible ((knowledge 4)
5. by introducing selected genes into host cells / replacing the
defective genes with normal one (i.e. gene therapy) is possible
attempt to treat genetic dieases at the root (knowledge 1, 2 and 5)
(see the notes and/or fact sheets provided)
Artificial Selection
Choice of parents:
1. select desirable traits
from natural wild
individuals, or
2.induced
- mutants,
- polyploidy with
even-numbered set
of chromosome, or
- GM
recombinants
Carefully planned
breeding:
- plants:
Artificial pollination
- animals:
Artificial insemination
Controlled further
breeding again and
again. By
- crossing between
progeny or
Via
- inbreeding or
- outbreeding
- backcrossing
change in gene frenquency
in the domesticated
population:
 give rise to stable
genetically distinct line
of new pure breed,
strains and variants
 ensure that future
generations have those
specific desirable traits.
Resulted in an astounding range
of phenotypic variation over
relatively short periods of time.
 - commercial purpose
- experimental tools
- evolution study…etc.
Artificial pollination:
Purposes of cloning
1. For reproduction asexually from a single cell of
an adult
2.Making embryonic stem cells (For therapeutic
purpose as a source of spare parts)
Cloning is the process in
which identical offspring
are formed from a single
cells or tissue.
A clone
is a group of genetically
identical cells or an
individual derived from a
single ancestral cell, tissue
or individual by repeated
asexual divisions.
5 basic steps involved cloning involves
1. Enucleation of the recipient egg
2. Transfer of the donor cell into the recipient
egg
3. Fusion of the cells (activated by a short
electrical pulse.)
4. Culturing the cloned embryo in the incubator
5. Transferring the developing embryo into the
reproductive tract of a surrogate mother
3.For products or tissues or organs from
recombinant DNA (may supply compatible
transplant organs)
Why cloning is used in plant and
animal breeding ?
To maintain desirable traits in
selected plants and animals.
[HKALE 91 II]
1. (a) Explain the meaning of the following terms :
(i) allele
(ii) polyploid
(iii) hybrid vigour
(6 marks)
(b) In a certain species of bird, one gene locus controls general feather colour, with green being dominant to
blue. A second locus controls colour intensity and the alleles at
this locus show incomplete
dominance. The table below shows the six possible phenotypes:
Feather colour
Pale
Green
Blue
Green
Blue
(i)
Colour intensity
Mid
deep green
deep blue
Dark
brownish green
purple
Draw genetic diagrams to illustrate the following crosses :
(1) Homozygous deep green x heterozygous brownish green
(2) Homozygous pale green x deep blue
Include in your diagrams, the genotypes of the parents, zygotes and offspring, the phenotypes of
the offspring, and the phenotype ratio. Define all the symbols you use.
(10 marks)
(ii) Describe the easiest way to determine whether a deep green individual is homozygous or
heterozygous. Explain your answer.
(4 marks)
Total : 20 marks
Answer
1. (a) (i) alleles
2
=one of a pair, =or series of alternative alleles that occur at a given locus in a chromosome;
one contrasting form of a gene.
(ii) polyploidy
2
=a condition where an organism with more than two sets of chromosomes or genomes.
=common in plant species and they often have some advantage e.g. forming larger fruits
etc.
hybrid vigour
2
=unusual growth, strength, and health of hybrids from two less vigorous parents. =This
depends on such alleles for vigour are dominant, and distributed between the 2 parents
(6)
(b) (i) let G be the dominant allele for general feather colour being green
2
g be the recessive allele for general feather colour being blue
i1 be the allele for the pale colour intensity of the feather colour
i2 be the allele for the dark colour intensity of the feather colour.
(N.B. Accept other suitably defined symbols)
(1) P :
GG i1i2
x
Ggi2i2
½x2
Gi1
F1
Gi2
:
G Gi1i2
Deep green
i.e.
Gi2
gi2
Ggi 1i2
Deep green
G Gi2i2
Brownish green
Deep green : Brownish green
1:1
¼x4
¼x4
Ggi2i2
Brownish
green
1
(4)
(2)
½x2
1/3 x 3
½x2
1
(ii) =To cross the deep green individual (G Gi1i2/Ggi1i2) with a purple individual (ggi1i2) :
1
1
(½)
If it is heterozygous :
1
(½)
1
1
½
OR To cross the deep green individual (G Gi1i2/Ggi1i2) with a pale blue individual (ggi1i2) :
1
½
If it is heterozygous :
(20)
Recombinant DNA Technology
Knowledge for students to
apply:
1. gene coding for all
(structural and functional)
proteins
2. genetic codes are essentially
universal
DNA Fringerprinting
Due to
1. random
chromosome
assortment
2. random
crossing over
3. random
mating
4. random
fertilization
5. mutation
Pentadactyl limbs (limb with five digits) are Homologous structures showing Adaptive
radiation due to Divergent evolution
The limitations and accuracy of fossil records
fossil record is very incomplete, patchy, and biased
∵ 1. the remains quickly decay.
2. fossils are best preserved under water in anaerobic conditions, so the fossil record is biased
toward aquatic species, marine types
3. missing links:
4. hard parts of organisms are preserved much more readily than are soft parts, so soft-bodied
invertebrates are represented much less frequently than vertebrate skeletons