Biology Book 4
Public examination questions
Suggested Answers
22 Basic Genetics
A.
Multiple choice
1.
A
2.
D
3.
C
4.
D
5.
C
6.
C
7.
A
8.
C
9.
D
10.
C
(1 mark each)
(Total: 10 marks)
B.
Short questions
1.
(a)
chromosome
(b)
heterozygote
(c)
dominant
(d)
meiotic cell division
(e)
gamete
(f)
diploid
(1 mark each)
(Total: 6 marks)
2.
Main concept:
-
definition of symbols showing recessiveness of the allele for colour-blindness (1 mark)
-
genotypes of parents showing sex linkage for colour-blindness
(1 mark)
-
genetic diagram correctly / properly laid out
(1 mark)
-
genotypes and phenotypes for boys and girls
(1 mark)
-
showing probability – 50% colour-blind for both boys and girls
(1 mark)
1
e.g.
-
definition of genetic symbols
(1 mark)
~ let b be the recessive allele for colour blindness
~ let B be the dominant allele for normal vision
Father
Mother
XbY
Parental genotype
XBXb
x
Gamete types
Xb
Genotypes of offspring
XBXb
XbXb
XBY
XbY
Phenotypes of offspring
normal
colour-blind
normal
colour-blind
Probability
XB
(1 mark)
Y
Xb
girls
boys
50%
50%
(1 mark)
(1 mark)
(1 mark)
(Total: 5 marks)
C.
Structured questions
1.
(i)
Any two of the following: (1 mark each)
(2 marks)
-
Four daughter cells are formed from a single parent cell.
-
The two members of a pair of homologous chromosomes are separated; each
goes to a different daughter cell.
(ii)
-
Each daughter cell contains the haploid number of chromosomes.
(1)
Type A gamete has both members of the homologous pair, while type B gamete
has none of that homologous pair.
(1 mark)
(2)
Down syndrome (Accept other correct answers)
(1 mark)
(3)
Any one set of the following: (3 marks each)
(3 marks)
The X chromosome carries some vital genes
that are essential to the survival of the zygote and its subsequent development.
These genes are absent in the Y chromosome.
OR
The X chromosome carries more genes than the Y chromosome.
Absence of the X chromosome will result in the loss of some genes
2
That may be essential to the survival of the zygote and its subsequent
development.
(Accept other reasonable answers)
(Total: 7 marks)
2.
(a)
Yes.
Any one set of the following: (3 marks each)
(3 marks)
To produce offspring with different phenotypes,
There must be two different combinations of gametes.
Therefore, either one of the parents must be heterozygous, producing two types of
gametes carrying different alleles.
OR
The parents are of different phenotypes, hence, one of the parents must be homozygous
recessive.
To produce offspring with different phenotypes,
the other parent must be heterozygous, producing two types of gametes carrying
different alleles.
(b)
Individual 1 possesses straight little fingers, she must be homozygous recessive
(1 mark)
and pass an allele for straight little fingers to individual 4.
(1 mark)
Individual 4 possesses bent little fingers, she must have at least one allele for bent little
(c)
fingers.
(1 mark)
Hence, individual 4 is heterozygous.
(1 mark)
Effective Communication
(1 mark)
Define symbol:
Let B be the allele for bent little fingers
and b be the allele for straight little fingers.
Individual 5
Parent
bb
Gamete
b
Offspring
x
Individual 6
Bb
B
(1 mark)
b
Bb
bb
(bent little finger)
(straight little finger)
(1 mark)
(1 mark)
Deduct 1 mark for wrong format, e.g. no identification of parent, gamete or offspring,
or no line indicating the combination of gamete.
3
The probability for their child to have straight little finger is 1/2 or 0.5 or 50%. (1 mark)
(Total: 12+1 marks)
3.
(i)
Heterozygous /Aa
(1 mark)
Individual 3 has normal vision, so he must be homozygous recessive.
(1 mark)
He must receive one recessive allele from individual 1.
(1 mark)
As individual 1 has astigmatism, she must have an allele for astigmatism
(1 mark)
So she must be heterozygous.
Effective Communication
(1 mark)
(ii)
Individual 4
Parent
Husband
Aa
Gamete
A
Offspring
AA
x
Aa
a
(1 mark)
A
Aa
Aa
with astigmatism
a
(1 mark)
aa
with normal
vision
The probability of the second child to have normal vision is 1/4.
(1 mark)
(1 mark)
Deduct 1 mark if leave out parent / gamete / offspring.
(Total: 8+1 marks)
4
23 Molecular Genetics
A.
Multiple choice
1.
D
2.
C
3.
C
4.
C
5.
A
(1 mark each)
(Total: 5 marks)
B.
Short questions
1.
(i)
CGCAAGAGGUCU
(ii)
Concept for mark award:
(1 mark)
- fate of mRNA after transcription
(1 mark)
- amino acids are carried by tRNA
(1 mark)
- complementary pairing between anticodons of tRNA and codons of mRNA (1 mark)
- attachment of tRNA to ribosomes in sequence
(1 mark)
- formation of peptide bond
(1 mark)
e.g.
The mRNA transcribed moves out to the cytoplasm
(1 mark)
and attaches onto rRNA / the ribosomes.
(1 mark)
Different tRNA molecules carry different amino acids
(1 mark)
Aminoacyl-tRNA molecules / amino acid-tRNA complex with anticodons
complementary to the codons of the mRNA
(1 mark)
will attach to the ribosome in sequence and a peptide bond will be formed between
adjacent amino acids, thus forming the polypeptide.
(1 mark)
(Total: 6 marks)
2.
(a)
Any one of the following:
(1 mark)
- ionising radiation
- high energy radiation
- high energy particles
- example of radiation
- named mutagenic agent
(b)
For methionine,
(1/2 mark)
substitution (always) gives different amino acids,
(1/2 mark)
e.g. substitution of C gives isoleucine.
(1/2 mark)
5
For glycine or isoleucine,
(1/2 mark)
substitution of either of first two bases gives different amino acid,
(1/2 mark)
e.g. in glycine, substitution of third base still codes for glycine.
(1/2 mark)
(Total: 4 marks)
C.
Structured questions
1.
(i)
mRNA of S: UAUUCGUACUGUAAU
(2 marks)
(ii)
Amino acid sequence of S: Tyr – Ser – Tyr – Cys - Asn
(2 marks)
(iii) R has nucleotide AAG instead of ATG in the third triplet of nucleotides. OR
The 8th base T is replaced by A.
(1 mark)
The third amino acid changes from Tyr to Phe / replaced by Phe in the encoded peptide.
(2 marks)
(Total: 7 marks)
2.
(a)
(b)
For AGC: AGC
(1 mark)
For UUC: TTC
(1 mark)
The anticodon of tRNA is complementary to codon/reads message on mRNA. (1 mark)
Specific amino acid is carried/transferred to ribosome through tRNA.
(1 mark)
It ensures correct sequence of amino acids along polypeptide formed.
(1 mark)
(c)
(Met)
Phe
Gln
Gln
Lys
Gln
Phe
(2 marks)
(Total: 7 marks)
3.
(a)
Sequence of bases in gene/on DNA determines the amino acid sequence.
(1 mark)
DNA is transcribed to form mRNA / Triplet DNA is transcribed into codons, (1 mark)
with complementary base-pairing (e.g. A to U, T to A, C to G).
(1 mark)
Each amino acid is attached to a specific tRNA,
(1 mark)
with the tRNA having anticodon to bind to mRNA’s codon.
(1 mark)
Ribosome is composed of rRNA and it is the site for translation, forming a polypeptide
chain.
(b)
(i)
(ii)
(c)
(1 mark)
X is phosphate.
(1/2
Y is deoxyribose.
(1/2 mark)
hydrogen bond
AZT binds to adenine/ A (on single DNA strand).
mark)
(1 mark)
(1 mark)
However, it lacks phosphates/ OH group and is unable to bind to another nucleotide.
(1 mark)
So, HIV is unable to form new strand of DNA.
(Total: 10 marks)
6
D.
Essay
1.
Structure of DNA
(Max. 5 marks)
- DNA is a macromolecule composed of repeating monomers called nucleotides.
- A nucleotide is formed by condensation of a five-carbon sugar, a phosphate group and a
nitrogenase base (adenine, guanine, thymine and cytosine).
- DNA is formed by condensation of a sequence of nucleotides in which the phosphate of a
nucleotide links with the deoxyribose of another nucleotide.
- DNA exists as a double helix in which two polynucleotide chains coil in a spiral.
- The two polynucleotide chains are bound by hydrogen bonds between their complementary
bases. The bases pair up according to the following rules: adenine pairs with thymine;
guanine pairs with cytosine.
- The sequence of one chain determines that of the other, making the two chains
complementary. This complementary nature makes the repair of DNA possible when one
strand is damaged, as the other can act as a template for repairing.
Information stored in DNA
(Max. 3 marks)
- The genetic information stored in a DNA is determined by its base sequence.
- The piece of DNA which codes for a protein is called a gene.
- As DNA produce protein molecules, the sequence of nucleotide bases in DNA forms the
code for the amino acid sequence of a protein.
- The genetic code is a triplet code, i.e. one codon consists of three bases and codes for one
specific amino acid. The triplet codon enables the four types of nucleotides to code for the
20 amino acids found in our bodies. But clearly 4 x 4 x 4 =64, there should be 64 kinds of
amino acids. Therefore, we can conclude that the code is degenerate: one amino acid can
be coded for by more than one codon.
- The code is universal, i.e. all living organisms contain the same 20 amino acids and the
same five bases.
How DNA controls cellular activities
(Max. 7 marks)
Self-replication
- The DNA unwinds by breaking the hydrogen bonds between complementary bases. One of
the single-stranded DNA acts as the template, free nucleotides have complementary bases
attached to the template by base pairing.
- A new piece of DNA is formed and can be passed to the next generation of the same
species.
- The phosphate group of the nucleotides enables it to bind to a chromatin protein, so that it
will coil up and become more stable, and the DNA molecules can be packed in a nucleus.
7
Protein synthesis
- The sequence of triplet codes, codons, determines the sequence of amino acids.
- As the proteins produced may form enzymes or hormones that control various
physiological functions of our body, the cellular activities are, in fact, controlled by the
codon.
- First, transcription occurs in the nucleus and in the process, the genetic information on a
DNA is converted into the complementary base sequence of an mRNA (messenger RNA).
- One strand of the unwound portion of the DNA acts as a template on which free
nucleotides attach by complementary base-pairing.
- In RNA, uracil instead of thymine in DNA pairs with adenine.
- Adjacent nucleotides join together and form an mRNA. This process requires energy.
- The mRNA leaves the nucleus through nuclear pores, and it attaches to a ribosome which is
made of rRNA at rough endoplasmic reticulum in the cytoplasm.
- Translation in which the base sequence of the mRNA is converted into a sequence of
amino acids in a polypeptide chain then takes place.
- The free amino acids in the cytoplasm are activated by joining a specific tRNA (transfer
RNA) with the use of ATP.
- Each tRNA has a specific anticodon which is made up of three nucleotide bases and can
pair up in a complementary manner with the codon bases of the mRNA by forming
hydrogen bonds.
- Adjacent amino acids are joined by a peptide bond.
- As the ribosome moves along the mRNA, an amino acid is added at a time for each codon.
- When the terminal codon is reached, translation stops and the polypeptide chain leaves the
ribosome.
- A three-dimensional structure is resulted from coiling up and folding of the polypeptide
due to the formation of hydrogen bonds between different parts.
- The polypeptide then becomes a structural protein that composes our bodies, or a
functional protein such as enzyme for the synthesis or degradation of other molecules, or a
hormone for regulating the metabolism of other tissues, thus controlling our cellular
activities.
Effective Communication:
•
Clear, logical and systematic presentation
(3 marks)
•
Relevance to the questions
(2 marks)
(Total: 20 marks)
8
24 Biodiversity
A.
Multiple choice
1.
C
2.
B
3.
C
4.
A
5.
A
(1 mark each)
(Total: 5 marks)
B.
Short questions
1.
(a)
(b)
Smaller groups are put within big groups / Organisms are categorised in a hierarchical
way.
(1 mark)
It is based on similarities / features in common / with named example,
(1 mark)
as well as evolutionary relationships / common ancestry.
(1 mark)
(i), (ii)
Taxonomic
Name
Sequence
Kingdom
Animalia
1
Genus
Drosophila
6
Species
melanogaster
7
Phylum
Arthropoda
2
Class
Insecta
3
Order
Diptera
4
Family
Drosophilidae
5
group
Correct answer for first two columns
(1 mark)
Correct sequence
(1 mark)
(Total: 5 marks)
2.
(a)
virus
(1 mark)
(b)
prokaryotes
(1 mark)
(c)
fungi / protists
(1 mark)
(d)
Protists / fungi
(1 mark)
(Total: 4 marks)
9
3.
1
Key sequence for
Animal
Scientific name
B
Paphia euglypta
1b → 4a → 5b
C
Ostrea nigromarginata
1b → 4b
D
Cypraea vitellus
1b → 2b → 3b
identification
/2 mark each for the correct scientific name
(11/2 marks)
1 mark each for the correct key sequence for identification
(3 mark)
1
1
/2 mark for underlining the scientific names of all specimens
( /2 mark)
(Total: 5 marks)
10
25 Evolution
A.
Multiple choice
1.
C
2.
A
3.
C
4.
B
5.
D
6.
A
(1 mark each)
(Total: 6 marks)
B.
Short questions
1.
(a)
Any two of the following: (1 mark each)
(2 marks)
- to obtain information for comparing the anatomy of organisms
- to obtain information for comparing the morphology of organisms
- to obtain information about the time of existence of organisms / chronological order
of existence of organisms
- to find the missing links of existing organisms
(b)
Any two of the following: (1 mark each)
(2 marks)
- There are missing links in the fossil records that indicate the intermediate forms
between related groups of organisms.
- Most fossils are incomplete and may be damaged, thus giving an incomplete record
of organisms in the past.
- Some organisms may not be fossilised as they may have a soft body.
(Total: 4 marks)
C.
Structured questions
1.
(i)
Any one set of the following: (2 mark each)
(2 marks)
Some mosquitoes survived after the spraying of the insecticide.
They would reproduce to give many offspring
and thus the population would rise again.
OR
Many eggs were laid before the death of the mosquitoes.
The eggs hatched
and thus the population would rise again.
11
(ii)
Any two sets of the following: (2 mark each)
(4 marks)
Mutation
It results in the change of the gene(s) controlling sensitivity / resistance to the
insecticide.
OR
Independent assortment of homologous chromosomes occurs during meiosis
leading to the formation of gametes with different alleles for insecticide-resistance.
OR
Fertilisation
It results in a random combination of alleles for insecticide-resistance in the zygote.
(iii) There are genetic variations among the mosquitoes in their resistance against
insecticide.
(1 mark)
Those that are resistant have a higher chance of survival in the presence of the
insecticide.
(1 mark)
They have a greater chance of reproduction / producing more offspring.
(1 mark)
Thus the proportion of the insecticide-resistant mosquitoes increases in the subsequent
generations.
(1 mark)
Hence, the insecticide becomes less effective in killing the mosquitoes.
Effective Communication
(1 mark)
(Total: 10+1 marks)
2.
(a)
(b)
If the two population can breed together
(1 mark)
and produce fertile offspring, then they belong to same species.
(1 mark)
There is geographical isolation of the two populations.
(1 mark)
Variations are already present between populations due to mutations.
(1 mark)
Different environmental conditions select different features and hence different alleles,
(c)
so there is different frequency of alleles in the two population.
(1 mark)
Finally, there is no interbreeding between the two populations.
(1 mark)
Individuals will select mate dependent on colour pattern.
(1 mark)
This can prevent interbreeding.
(1 mark)
(Total: 8 marks)
12