ACJC_H2_BIO_P2

Name Subject Class Class Candidate Number

2BI

ANGLO-CHINESE JUNIOR COLLEGE Preliminary Examination 2009

BIOLOGY 9747/02 19 AUGUST 2009

HIGHER 2 2 hours Paper 2 Core Paper

Additional Material: Writing Paper

For Examiner's Use

Section A

1

2

3

4

5

6

7

Section B

8 / 9

Total

100

READ THESE INSTRUCTIONS FIRST Write your name, index number and class on this answer booklet. Write in dark blue or black pen. You may use a soft pencil for any diagrams, graphs or rough working. Section A Answer all questions. Section B Answer any one question.

At the end of the examination, circle the number of the Section B question you have answered in the grid opposite. Fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

This question paper consists of 16 printed pages.

[Turn over

2 For Examiner's

Use1 Fig. 1.1 below shows an overview of the synthesis of proteins based on the genetic information contained in the nucleus in a plant cell.

A

B

C

Fig. 1.1 (a) Label processes A and B.

A: B: [1]

(b) In order for process A to occur, the DNA double helix must first be unzipped by enzymes. In

vitro, DNA strands can also be separated by heat denaturation of DNA.

(i) Explain how DNA is denatured by heat.

[1]

ACJC 2009 9747/02/Prelim 09 [Turn over

3 For Examiner's

Use(ii) Two DNA segments are shown below.

CTAGCTGAACGCGCCCGATT TAAGCTTAACGCGACTGATA GATCGACTTGCGCGGGCTAA ATTCGAATTGCGCTGACTAT

Segment 1 Segment 2 Identify the segment that would require a higher temperature for denaturation of the DNA. Explain your answer.

[2]

(c) Process A and DNA replication require different enzymes and substrates. Describe two

other differences between these two processes.

[2] (d) In all cells, there are many types of structure C. State the minimum number of types that

should be present and explain your answer.

[2] (e) Based on Fig. 1.1, explain how protein synthesis in Agrobacterium tumefaciens would differ

from that of a plant.

[2]

[Total: 10]


4 For Examiner's

Use2 The concentration of glucose in human blood is kept within a narrow normal range. All body cells need a continuous supply of glucose. The uptake of glucose into the muscle cells occurs by facilitated diffusion.

(a) Explain why glucose is taken up by facilitated diffusion.

[2] (b) Table 2.1 shows the mean levels of glucose and insulin in two groups of people sampled one

hour after the ingestion of 75 g of glucose. One of the experimental groups consisted of people with diabetes; the other acted as a control.

Table 2.1

group X group Y time after glucose ingestion 0 min 60 min 0 min 60 min

plasma glucose (mmol/L) 5.3 13.0 5.3 7.8

plasma insulin (mmol/L) 68 66 69 380

(i) Using the data in Table 2.1, give two reasons to explain which group (X or Y) included

the people with diabetes.

[3]

(ii) In the diabetic group, the mean plasma glucose concentration remained above 5.3

mmol/L while it returned to 5.3 mmol/L in the control group. Explain how this change occurred.

[1] (iii) A tissue may, over time, lose its responsiveness to insulin, even though insulin

concentration remains unchanged. Suggest a reason for this decrease in responsiveness.

[1]


5 For Examiner's

Use(iv) Streptozotocin is a drug used for treating metastatic cancer of the pancreatic insulin-producing beta cells. Describe the process of metastasis.

[2] To prevent the concentration of glucose in the blood from dropping below the normal range, the pancreas releases another type of hormone, glucagon. A general signal transduction pathway is shown in Fig. 2.1. Some parts of the pathway are labelled.

L Receptor

OUTSIDE OF CELL

1 Plasma membrane

GDP

INSIDE OF CELL

2 Intermediate or relay molecules

3 Activation of

cellular responses

Fig. 2.1 (c) Using glucagon as an example of molecule L, describe what happen at stages 1, 2 and 3 in

a signal transduction pathway.

[3]

[Total: 12]


6 For Examiner's

Use3 Fig. 3.1 shows a segment of the E.coli chromosome that contains the lac operon.

Fig. 3.1

lacI Promoter for lacZ, lacY and lacA

lacZ lacY lacA

lacI promoter Operator

DNA

(a) With reference to Fig. 3.1, describe three differences between the organisation of eukaryotic and prokaryotic genes.

[3] The gene activity of the lac operon in four different E. coli different strains was measured and the results are shown in Table 3.1.

Table 3.1 Key: I+ wild type (i.e. normal) allele for lacI gene I- allele for lacI gene which is not expressed

Z+ wild type (i.e. normal) allele for -galactosidase gene of lac operon Z- allele for -galactosidase gene which is not expressed O+ functional operator of lac operon

Presence () or absence () of -galactosidase activity

E. coli strain Genotype Lactose present Lactose absent

E1 I+ O+ Z+ E2 I O+ Z+ E3 I+ Oc Z+ E4 IS O+ Z+

(b) Account for the activity of -galactosidase in strains E2 and E3 in terms of the expression of

LacI protein and its binding to the operator region.

E2

E3

[2]


7 For Examiner's

Use(c) (i) The activity of -galactosidase in E4 was due to a single-base substitution mutation in the lacI gene, resulting in a drastic change in the LacI protein. Explain how this mutation could have resulted in the change.

[3]

(ii) Based on the results in Table 3.1, state if this mutation in the lacI gene is a gain-of-function or loss-of-function mutation.

[1] Table 3.2 below shows two other strains of E. coli, each containing an F plasmid. The F plasmid includes the fertility factor. All necessary promoters are present. These E. coli strains are partial diploids for a component of the lac operon as indicated in the table below.

Table 3.2

E.coli strain Genotype F1 I O+ Z / F I+

F2 I O+ Z / F O A scientist then mixed two living strains of bacteria, E and F in three different combinations. In each combination, a completely new strain of bacteria was isolated after a few days. The details are shown in Table 3.3. (d) (i) Assuming that no crossing over has taken place, complete the table below for the

genotypes and -galactosidase activity of the new bacterial strains. [2]

Table 3.3

Presence () or absence () of -galactosidase activity E. coli strain

(E) E. coli strain

(F) Genotype of new

strain isolated Lactose present

Lactose absent

I+ O+ Z+ I O+ Z / F O I O+ Z+ I O+ Z- / F I+ IS O+ Z+ I O+ Z / F I+

(ii) Name the process responsible for producing these new strains of E. coli. [1]

[Total: 12]


8 For Examiner's

Use4 mRNAs in a bacterial cell are very unstable, having half-lives of less than a couple of minutes. mRNAs in eukaryotes are relatively more stable, having longer half-lives of not more than 30 minutes. Different mechanisms exist for destroying the eukaryotic mRNAs. One of these is degradation by exonucleases.

(a) Explain why eukaryotic mRNAs are more stable than bacterial mRNAs.

[3] (b) Describe how exonucleases are able to carry out their role in breaking down mRNAs .

[2] (c) Transferrin is the protein that transports iron in the blood stream. It is transported into the cell

by the membrane-bound transferrin receptor (TfR).

(i) Suggest how iron is transported into the cell. [1]

At low intracellular iron concentrations, there are higher amounts of TfR to increase iron intake into the cell. The expression of the TfR gene is mediated by aconitase which is an iron-sensitive RNA-binding protein. This protein binds to the 3-untranslated region (3 UTR) of the TfR mRNA, as shown in Fig. 4.1.

Fig. 4.1

(ii) State the level at which the expression of TfR gene is controlled. [1]


9 For Examiner's

Use(iii) Describe briefly how TfR levels are increased in low intracellular iron concentration.

[3]

(iv) Excessive iron levels in humans are associated with enhanced cancer cell growth. Suggest two ways in which TfRs can be targeted in cancer therapy.

[2]

[Total: 12]


10 For Examiner's

Use5 Pituitary dwarfism is a sex-linked inherited condition in humans in which affected individuals have very short limbs. The allele for pituitary dwarfism is recessive to the allele for normal limbs. The family tree shows part of one affected family.

Fig 5.1

Key: Normal male: Affected male: Normal female: Affected female:

(a) Explain what is meant by 'sex-linked inherited condition'.

[1] (b) With reference to Fig 5.1, identify and explain one piece of evidence to show that the allele

for pituitary dwarfism is recessive to the allele for normal length limbs.

[2] (c) Using appropriate symbols, state the genotype of individuals 10 and 11. Explain how you

derived your answer. (i) Genotype of individual 10:

Reason:

(ii) Genotype of individual 11:

Reason:

[3]


11 For Examiner's

Use(d) Couple 10 and 11 has another son.

(i) State the probability that their second son will show pituitary dwarfism. [1]

(ii) Draw a genetic diagram to explain your answer. [3]

[Total: 10]


12 For Examiner's

Use6 Hexokinase is the first enzyme in the glycolytic pathway. Fig. 6.1 below shows the enzyme alone with the arrow pointing to the active site. Fig. 6.2 shows the enzyme when substrate (not shown in the diagram) is bound to it.

Fig. 6.1 Fig. 6.2

(a) Describe the model by which the substrate binds to hexokinase.

[2] (b) Describe one difference between what you have described in (a) and the binding of a non-

competitive inhibitor to hexokinase.

[1] (c) Explain the role of hexokinase in glycolysis.

[2]


13 For Examiner's

Use(d) Explain the role of glycolysis in aerobic respiration.

[2] (e) Glyceraldehyde-3-phosphate is an intermediate in glycolysis which is also found in the Calvin

cycle. Briefly describe how glyceraldehyde-3-phosphate is formed in the Calvin cycle.

[3] (f) It has been observed that tumour cells have very high rates of glycolysis. Also, in many

tumour cells, hexokinase is bound to the outer membrane of the mitochondrion. Suggest how the association of hexokinase with the mitochondrion can lead to high rates of glycolysis.

[2] (g) The structures of mitochondria and chloroplast share similar adaptations for ATP synthesis.

Describe and explain two of them.

[2]

[Total: 14]


14 For Examiner's

Use7 One population of rats, found on a remote island off the coast of South Australia, have had no genetic contact with mainland rats since they were isolated by rising sea levels at the end of the last glacial period, around 10 000 years ago.

Scientists have taken blood samples from the rats and compared the distribution of unique DNA sequences called microsatellites, which are scattered across the rats' chromosomes. These microsatellites give a measure of the population's genetic diversity, or lack of it. In this case the microsatellite data showed that the island population has low genetic diversity.

The scientists concluded that the island population of rats has been through a genetic bottleneck. (a) Explain the scientists' choice of using microsatellites over genes for the analysis of genetic

diversity of the rat population on the island.

[2] (b) Explain how a genetic bottleneck may lead to a decrease in genetic diversity.

[2] Despite the island rats lack of genetic diversity, the population size has been maintained over many generations. In fact, the rats appear to be thriving. (c) Suggest one reason for the rats' success despite the lack of genetic diversity within the

population.

[1]


15 For Examiner's

UseWarfarin is a poison used to control rat populations. Fig. 7.1 shows changes in the proportion of rats resistant to warfarin in a particular population over a period of about 4 years. High levels of warfarin were used on this population during Year 2 but poisoning stopped at the end of this period. Rats are reproductively mature at an age of three months and can breed about every three weeks.

Fig. 7.1 (d) Explain the process which led to the increase in the percentage of resistant rats during Year

2.

[3] (e) Using the data in Fig. 7.1, explain what can be concluded about the fitness of warfarin-

resistant rats compared to non-resistant rats in an environment without warfarin.

[2]

[Total: 10]



16 For Examiner's

UseSection B

Answer one question. Write your answers in the lined pages provided. Your answers should be illustrated by large, clearly labelled diagrams, where appropriate. Your answers must be in continuous prose, where appropriate. Your answers must be set out in sections (a), (b) etc., as indicated in the question. 8 (a) Explain how meiosis and fertilisation can give rise to genetic variation. [8] (b) Describe how the structure of a myelinated neurone is well-adapted for the transmission of

of an action potential and synaptic transmission. [8] (c) Pufferfish contain a deadly toxin, tetrodotoxin which affects voltage-gated sodium ion

channels. Consuming the toxin may result in paralysis and ultimately death. Suggest how this toxin works. [4]

9 (a) Compare and contrast the ways in which lambda phage and human immunodeficiency

virus (HIV) reproduce themselves. [10]

(b) Using HIV protease as an example, analyse how its structure is related to its function. [6]

(c) Influenza viruses of many different subtypes, e.g. H1N1 and H5N1, have been causing respiratory infections in humans. Explain how these different subtypes come about. [4]

ANGLO-CHINESE JUNIOR COLLEGE

ACJC_H2_BIO_P2

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