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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
5 2014
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Question 1 (DNA and Genomics)
(a) The flow of information in the cell, sometimes called the
Central Dogma of Molecular Biology, primarily involves two classes
of macromolecules.
(i) State the two classes of macromolecules and their cellular
location in eukaryotes. [2]
(ii) Use a simple flow chart to represent the Central Dogma in
the space provided below. Include in the flow chart the processes
by which information flows. [3]
(b) Table 1.1 below shows the relative proportions of the bases
adenine, thymine, guanine and cytosine in DNA from different
organisms.
Table 1.1
Explain the importance of the ratios of A to T and G to C to the
structure of DNA. [2]
(c) During protein synthesis, tRNA and ribosomes are involved in
the formation of a polypeptide chain.
(i) State the exact location in the cell where ribosomal
subunits are made. [1]
(ii) Describe how the correct amino acid is attached to each
tRNA molecule. [3]
[Total: 11 marks]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
5 2014
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Question 2 (DNA and Genomics)
a) (i) Describe the structural features of DNA. [2]
(ii) Outline the main features of DNA replication. [5]
b)
Table 2.2 The genetic code Use the genetic code in Table 2.2 as
shown above to complete the table below. The columns represent
transcriptional and translational alignments. Write down the
directions in terms of 5 and 3 for the nucleic acids.
5 C DNA double
helix
G G G
C A
G
mRNA transcribed
G C A Appropriate
tRNA anticodon
Trp
Amino acids incorporated into
polypeptide [4]
[Total: 11 marks]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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Question 3 (DNA and Genomics) a) The main function of a
chromosome is to act as a template for the synthesis of RNA
molecules, since only in this way does the genetic information
stored in chromosomes become directly useful to the cell. RNA
synthesis is a highly selective process.
Explain what is meant by the term template. [3]
b) Figures 3.1 shows transcription.
(i) Identify structures A to C and describe their function.
[1]
(ii) Describe the function of RNA polymerase. [3]
(iii) State three differences between transcription and DNA
replication. [3] [Total: 10 marks]
Figure 3.1 Transcription
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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Question 4 (Bacteria)
a) An experiment was conducted to determine the identity of
Substance X and Substance Y. Both substances are known to have an
effect on the expression of -galactosidase in Escherichia coli.
Substance X was added after 10 minutes, Substance Y was added after
20 minutes and both substances X and Y were added after 30 minutes.
The results are shown in Fig. 5.1.
Fig. 5.1 a) (i) Suggest an identity for Substance X and
Substance Y
Substance X:
Substance Y:
(ii) With reference to Fig. 5.1, explain how the expression
levels of -galactosidase are affected by Substance X and Substance
Y between 10 minutes to 40 minutes. [5]
Substance X added
Substance Y added
Substance X and
Substance Y added
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
0.0 10.0 20.0 30.0 40.0 50.0 60.0
Amo
un
t of
-ga
lact
osid
ase
(m
g)
Time (min)
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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b) (i) Draw a simple diagram to show all the elements of the trp
operon. [1]
(ii) Explain why it is useful for a bacterial cell to decrease
expression of the trp genes when tryptophan is present. [2]
Table 5.1 below indicates the activity levels of the functional
enzymes E, D, C, B and A in wild type bacterial cells in the
presence and absence of tryptophan (Trp).
Activity level of enzymes/units
Enzyme Trp absent Trp present E 700 0 D 700 0 C 700 0 B 700 0 A
700 0
Table. 5.1
You have managed to obtain several bacterial mutants. Each
mutant is the result of a single base-pair substitution in a single
component of the trp operon. The activity level of functional
enzymes E, D, C, B and A in the bacterial cells having these
individual mutations is shown in Table 5.2.
Activity level of enzymes/units Mutant 1 Mutant 2 Mutant 3
Enzymes Trp absent
Trp present
Trp absent
Trp present
Trp absent
Trp present
E 700 700 700 0 0 0 D 700 700 0 0 0 0 C 700 700 700 0 0 0 B 700
700 700 0 0 0 A 700 700 700 0 0 0
Table. 5.2
c) (i) Explain why it is useful for a bacterial cell to decrease
expression of the trp genes when tryptophan is present. [2]
(ii) A loss-of-function mutation in which component of the trp
operon could explain the phenotype of mutant 3? Explain your
choice. [2]
c) If the phenotype of mutant 3 is caused by a mutation in the
trpR gene, explain how this mutation would affect the structure and
function of the repressor protein. [3]
[Total: 15 marks]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
5 2014
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Question 5 (Bacteria) a) Figure 7.1 shows an electron micrograph
of an E.coli (strain O104:H7) that can enter humans
through ingestion of contaminated food and cause food poisoning
in humans. A serious outbreak of foodborne illness in Germany in
May 2011 was caused by this novel strain of E.coli that had
acquired the genes to produce Shiga toxins.
Figure 7.1 (i) State two structural differences between the
chromosome of this type of microorganism
and a human cell. [2]
(ii) Explain how this bacteria strain of E.coli O104:H7 could
have transferred the genes coding for Shiga toxins to normal E.coli
bacteria in human gut. [4]
b) Figure 7.2 shows a diauxic growth curve of E. coli cultured
in a medium with both glucose and lactose. Growth (!) of bacteria
is measured in terms of optical density (OD) at wavelength of
650nm, concentration of lactose (!) and glucose (") are measured in
millimolar (mM).
Figure 7.2 With reference to Figure 7.2, explain the bacteria
growth rate (i) for the first 6 hours; [3]
(ii) from 6 to 8 hours; [2]
(iii) from 8 hours onwards. [5] [Total: 16 marks]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
5 2014
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Question 6 (Viruses)
a) Explain why viruses are necessarily obligate parasites.
[3]
b) In 1911, Peyton Rous isolated a virus from chickens that
rapidly produced tumours when injected into healthy birds. The
virus was then named the Rous sarcoma virus (RSV). The life cycle
of the virus is shown below in Figure 8.1.
(i) With reference Figure 8.1, describe the common features
shared by RSV and retroviruses e.g. HIV. [3]
(ii) Describe two differences in the mode of entry of influenza
virus and HIV. [4]
[Total: 10 marks]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
5 2014
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Question 7 (Pro and Euk)
a) Describe 2 roles of telomeres. [4]
b) Telomeres can be extended by an enzyme, telomerase.
Telomerase comprises an RNA component which can be used as a
template in the synthesis of telomeric DNA. Figure 4.1 illustrates
the action of telomerase.
Figure 4.1
(i) Suggest how a synthetic single-stranded RNA molecule (not
shown in Fig. 4.1) can be used to inhibit telomerase activity.
[2]
(ii) Suggest how a mutation in DNA can result in increased
telomerase activity in a cell. [2]
c) Figure 4.2 below shows the effect of histone
acetylation/deacetylation on a piece of chromosome.
Figure 4.2
(i) With reference to Figure 4.2, explain the effect of histone
deacetylation on gene expression. [3]
(ii) State and explain another process which can increase the
effect of histone deacetylation on gene expression. [2]
[Total: 9 marks]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
5 2014
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Question 8 (Pro and Euk)
Using the table below, compare how gene regulation at
translational level differs between a bacterium and a eukaryotic
cell. Points of comparison for
translational control Prokaryote Eukaryote
Protecting the RNA and thus the stability of the mRNA
Recognition site for ribosomes
Location of binding for repressor
[Total: 8 marks]
Question 9 (Cell Division)
(a) A root was cut into ten transverse sections at different
distances from the tip. The sections were stained and viewed under
the microscope. The number of cells in mitosis were counted in each
section and the results were used to determine the mitotic
index.
This is calculated as follows:
Fig. 9.1 shows the mitotic index for the ten sections.
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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(i) Using the information in Fig. 9.1, describe how the mitotic
index changes along the length of the root.
[2]
(ii) Explain how the events in the mitotic cell cycle ensure
that all the cells in the root are genetically identical.
[3]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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(b) The diagram shows the main stages in the formation of sperms
in a human testis.
(i) Describe two ways, other than size, in which cells at
anaphase of division A would differ from cells at anaphase in
division B.
[2]
(ii) Explain why meiosis is important for sexual reproduction
[3]
(iii) This man has haemophilia. Haemophilia is a sex-linked
condition. Explain how a child that results from the zygote formed
when sperm P fertilises an ovum may not inherit the haemophilia
allele from its father.
[2]
[Total: 9 marks]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
5 2014
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Question 10 (Cell division)
Fig. 10.1 shows drawings of a cell at various stages in the
mitotic cell cycle.
(a) List the letters shown in Fig. 10.1 in the order in which
these stages occur during a mitotic cell cycle. The first stage has
been entered for you. [1]
(b) Explain what is happening in stage D in Fig.10.1. [2]
(c) Describe and outline what happens to the DNA in the nucleus
during stage A in Fig. 10.1. [3]
(d) State the importance of mitosis in the growth of a
multicellular organism, such as a flowering plant or a mammal.
[1]
[Total: 7 marks]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
5 2014
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Question 11 (Mendelian Genetics)
The figure below shows the pedigrees (family trees) of two
families, A and B, representing the occurrence of colour blindness
in parents and their children.
(a) In individuals A4 and B3 were to produce children, what
proportions of different off spring would you expect to find?
Male colour blind
Male, full colour vision
Female colour blind
Female full colour vision
[1]
(b) Redraw pedigree A only showing all possible genotypes of the
individuals in each generation. ( mark each individual)
[4]
Another human characteristic showing this pattern of inheritance
is haemophilia.
(c) Suggest two reasons why haemophilia is less common than
colour blindness. [2]
The table shows the results of a cross between homozygous
recessive black bodied, bent winged fruit flies (bb, nn) and the
double heterozygote (Bb, Nn).
Phenotype Number Genotype Combination of traits Normal body,
normal wing 83 BbNn Parental Black body, normal wing 71 bbNn
Non-parental Normal body, bent wing 69 Bbnn Non-parental Black
body, bent wing 85 bbnn Parental
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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(d) What is the expected phenotype ratio of the offspring? [1]
Use the chi-squared test to check whether this assumption is
justified.
(e) What is the Null hypothesis for this data? There is no
significant difference* between the observed numbers and the
expected numbers, any difference is due to chance*.
[1]
(f) What is the expected number in each class? [1]
(g) With all working shown clearly, show your calculations for
chi-square below. [2]
Distribution of 2 Degrees of
freedom Probability, p
0.10 0.05 0.02 0.01 0.001 1 2.71 3.84 5.41 6.64 10.83 2 4.61
5.99 7.82 9.21 13.82 3 6.25 7.82 9.84 11.35 16.27 4 7.78 9.49 11.67
13.28 18.47
(h) Explain the conclusion that can be draw from your your
calculated 2 value and the table above.
[2]
[Total: 15 marks]
Question 12 (9700 Nov08/P4/Q8) (Mendelian Genetics) In mice
there are several alleles of the gene that controls the intensity
of pigmentation of the fur. The alleles are listed below in order
of dominance with C as the most dominant.
C = full colour Cch = chinchilla Ch = himalayan Cp = platinum Ca
= albino
The gene for eye colour has two alleles. The allele for black
eyes, B, is dominant, while the allele for red eyes, b, is
recessive. A mouse with full colour and black eyes was crossed with
a himalayan mouse with black eyes. One of the offspring was albino
with red eyes.
Using the symbols above, draw a genetic diagram to show the
genotypes and phenotypes of the offspring of this cross. [6]
[Total: 6 marks]
2 (O E )2
E
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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Question 13 (Mendelian Genetics)
The parental generation consisting of crosses between white-eyed
and normal-eyed forms of fruit fly (Drosophila sp.) and the next
two subsequent generations are shown below. A begins with a cross
between a normal-eyed female and a white-eyed male while B begins
with a reciprocal cross.
(The genotypic ratios obtained in A and B are as follows: first
generation offspring 1:1, second generation offspring 1:1:1:1) (a)
What is meant by reciprocal cross? [1]
(b) Explain the difference in the first generation male obtained
in A and B. [3]
(c) Using appropriate symbols and genetic diagrams, explain the
the cross between the first generation offspring seen in A and B.
[5]
(d) Explain how would you find out which female in the second
generation of A is heterozygous for the eye colour. [2]
[Total: 9 marks]
Parental generation
Cross between first generation offspring respectively
Generation 2
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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Question 14 (Proteins and Enzymes)
Consider the following hypothetical protein (Fig. 14.1) that
spans the membrane of a cell:
Fig. 14.1
(a) (i) What class of macromolecules does the molecules (except
for the protein) that constitute the above membrane belong to?
[1]
(ii) Explain the important features of these molecules that
allow them to form membranes. [2]
(b) Consider the interaction between the side chains of valine
141 and cysteine 136.
(i) Suggest one possible type of interaction between these two
groups. [1]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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(ii) Could changing valine 141 to a different amino acid
strengthen the interaction between the amino acids at position 141
and 136 of this protein? [2]
If no, briefly explain why not. If yes, explain how the
interaction will be weakened.
(c) Amino acids 100 through 129 of this protein are all
alanines, leucines or valines. What interactions occur between the
side chains of these amino acids and the molecules that constitute
the inside of the membrane? [1]
(d) If amino acids 100 through 129 are replaced with serine,
glutamine and asparagines (all three being amino acids that are
hydrophilic), the resulting protein has the same general structure.
Would you then expect that this new protein be still found embedded
in the membrane? Explain your answer. [2]
[Total: 9 marks]
Question 15 (Proteins and Enzymes)
(a) Write down the structural formula of the tripeptide formed
by alanine (R = CH3), glycine (R = H) and serine (R = -CH2OH)
joined together in that order. Indicate and name the bond formed
between these three amino acids.
[2]
(b) Lipase is an enzyme that catalyses the hydrolysis of
triglycerides. It is a soluble globular protein. The function of an
enzyme depends upon the precise structure of its tertiary
structure. Fig. 15.1 represents the structure of an enzyme. The
black strips represent the disulfide bonds which help to stabilize
its tertiary structure.
Fig. 15.1
(i) Describe the nature of the disulfide bonds that help
stabilize the tertiary structure of a protein such as lipase.
[3]
(ii) Name two other types of bonds that help stabilize the
tertiary structure. [1] [Total: 7 marks]
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Raffles Institution Post-Promos Holiday Homework (H2 Biology) Yr
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