st leonard’s college internal examinations · more athletes have their haemoglobin concentration...
TRANSCRIPT
Students are not permitted to bring mobile phones and / or any other unauthorized electronic devices into the examination room
ST LEONARD’S COLLEGE
Internal Examinations
INTERNATIONAL BACCALAUREATE
STUDENT NAME: ___________________________________
SUBJECT NAME AND LEVEL Year 11
Wednesday 3 June 2015 (Afternoon)
Teacher(s) Responsible
Miss Werba
Mr McKenzie
Reading time: 5min Length of exam(s): 120min
Instructions to candidates
• Do not open this examination paper until instructed to do so • Please use the answer sheet provided for Section A (Multiple Choice
Questions) and the IB Writing booklet for Section C (Extended Response Questions)
Students are not permitted to bring mobile phones and / or any other unauthorized electronic devices into the examination room
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Section A: Multiple Choice Questions [30 marks] For each question, choose the answer you consider to be the best and indicate your choice on the answer sheet provided.
1) The table below shows the level of haemoglobin measured in two different groups of athletes.
Haemoglobin /
grams per 100 cm3 Number of
athletes tested Standard deviation / grams per 100 cm3
Group A 12.6 200 0.8 Group B 11.9 220 3.2
Which of the following statements is correct? A. Results from group B are more accurate because more athletes were tested. B. Results from group B are more reliable because it has a higher standard
deviation. C. More athletes have their haemoglobin concentration close to the mean in group
A than in group B. D. These results indicate that the level of haemoglobin follows a normal distribution.
2) What does the following scatter graph show?
A. No correlation between these variables B. Strong positive correlation between these variables C. Strong negative correlation between these variables D. Weak negative correlation between these variables
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3) If cells of a multicellular organism have the same genes, how can there be many different cell types in a body? a ntiate?
A. Some genes but not others are expressed in each cell type. B. Cells lose some genes as development occurs. C. Genes do not determine the structure of a cell. D. Cells must practice division of labour in order to survive.
Use the following diagram to answer Questions 4 and 5.
4) Process R is an example of
A. exocytosis. B. phagocytosis. C. pinocytosis. D. endocytosis.
5) Organelle X
A. is the site of cellular respiration. B. packages protein molecules for export from the cell. C. absorbs sunlight and produces carbohydrates. D. produces ribosomal RNA.
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6) Which molecule regulates the fluidity of cell membranes?
A. Phospholipid B. Cholesterol C. Glycoprotein D. Peripheral protein
7) The image shows cells in the different stages of mitosis. Which cell is in telophase?
A. AB. BC. CD. D
8) The following diagram is a representation of the cell cycle.
Replication of DNA occurs in section A. one. B. two. C. three. D. four.
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9) The diagrams show three representations of the structure of the same chemical substance
What chemical substance is shown?
A. Ribose B. Glucose C. Fatty acid D. Amino acid
10) Which type of chemical reaction is an example of anabolism?
A. PhotolysisB. CombustionC. HydrolysisD. Condensation.
11) Protein forms part of the structure of
A. polysaccharides. B. transfer RNA. C. phospholipids. D. haemoglobin.
12) What do DNA replication, transcription and translation have in common?
A. Take place in cell nucleus B. Require free nucleotidesC. Catalysed by polymerase D. Complementary base pairing
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13) The diagram shows the translation of a mRNA molecule.
A tRNA molecule with anticodon CAG carries the amino acid phenylalanine. Which codon of mRNA will the tRNA join?
A. CTG B. CAG C. GTC D. GUC
14) What is a feature of shorter wavelength visible radiation?
A. It includes violet light. B. It has less energy per photon than longer wavelengths.C. It is absorbed by greenhouse gases. D. It is reflected by chlorophyll.
15) What is light energy used for in photosynthesis? A. Formation of hydrogen and oxygen
B. Formation of carbon dioxide only C. Formation of ATP and glucose
D. Formation of oxygen only
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16) The fungus Rhytisma grows on the leaves of certain trees, causing a yellow leaf area in which chlorophyll is no longer present. A black, tar-like stain later spreads out. What happens in the leaf when Rhytisma is present? I. An increase in the intake of carbon dioxide II. A reduction in the production of oxygen III. An increase in the loss of water A. I only B. II only C. II and III only D. I, II and III
17) Which of the following is true of cellular respiration? A. Most of the ATP is made in the Krebs Cycle. B. Pyruvate is converted to acetyl CoA under anaerobic conditions. C. No ATP can be formed from the breakdown of glucose under anaerobic
conditions. D. CO2 is a product of lactate fermentation and the Kreb’s Cycle.
18) How does an increase in temperature affect enzyme activity?
Movement of molecules
Chance of collision between enzyme and substrate
A. increases increases B. decreases decreases C. increases decreases D. decreases increases
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19) Examine the following graphs.
From these graphs it is reasonable to infer that at a pH of 4 A. trypsin would be inactive. B. all the pepsin would be denatured. C. all three enzymes would lack a functional active site. D. lysozyme has the highest activity of the three enzymes.
20) What is a gene mutation?
A. Failure of chromosome pairs to separate properly during cell division B. Changes to genes caused by natural selection C. Changes to the nucleotide sequence of the genetic material D. Changes in karyotypes
21) Consider the following diagram of a cellular structure.
This structure is A. visible only using an electron microscope. B. found only in eukaryotic organisms. C. found in all living organisms. D. made up entirely of DNA.
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22) The following karyogram is from a human baby with a genetic defect.
The condition indicated by the karyotype is an example of A. monosomy. B. polyploidy. C. Down syndrome. D. trisomy.
23) Which is a feature of sex-linked genes in humans?
A. Males can only be heterozygous for the gene. B. Females can only be homozygous for the gene. C. Males can be either heterozygous or homozygous for the gene. D. Females can be either heterozygous or homozygous for the gene.
24) A couple, each phenotypically normal, have a child with phenylketonuria, an
autosomal recessive trait. The chance that their second child will have the trait is A. one in four. B. two in three. C. one in three. D. three in four.
25) In humans, neurofibromatosis type 1 (NF-1) is an inherited disorder in which tumours
of the nerve tissue form. NF-1 is an autosomal dominant trait. A male with NF-1 has a child without the trait. The mother of the child does not have the trait. The chance that the next child of these parents has NF-1 is A. zero. B. one in two. C. one in four.
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D. three in four. 26) In the following pedigree, shaded individuals have a particular genetic trait.
The mode of inheritance of the trait is A. X-linked dominant. B. X-linked recessive. C. autosomal recessive. D. autosomal dominant.
27) A woman of blood group A, whose genotype is unknown, and a man of blood group O have a child. Genetically, this is an example of a A. self cross. B. test cross. C. dihybrid cross. D. sex-linked cross.
28) Examine the following pedigree, which shows the phenotype with respect to the ABO gene locus of each individual.
Individuals that would be homozygous at the ABO gene locus include A. ɪ-3. B. ɪɪ-2. C. ɪɪ-6. D. ɪɪɪ-2.
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29) What is amplified using the polymerase chain reaction (PCR)? A. Large amounts of RNA B. Small amounts of DNA C. Small amounts of protein D. Large amounts of polymers
30) Plasmids of bacteria are used to transfer selected genes from one species to another. The process can be represented as follows: bacterial plasmid cut à foreign gene and plasmid mixed à plasmid with inserted foreign gene Enzymes are used to facilitate several of these steps. Which one of the following shows the enzymes required for the fist and last steps of the process?
Cuts plasmid Inserts genes A. restriction enzyme DNA ligase B. restriction enzyme DNA polymerase C. DNA ligase DNA polymerase D. DNA polymerase DNA ligase
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Section B: Short Answer Questions [35 marks] Answer all questions. Write your answers in the boxes provided. 1) Rice (Oryza sativa) is usually intolerant to sustained submergence under water,
although it grows rapidly in height for a few days before dying. This is true for one variety, Oryza sativa japonica. The variety Oryza sativa indica is much more tolerant to submergence. Three genetically modified forms of O. sativa japonica, GMFA, GMFB and GMFC, were made using different fragments of DNA taken from O. sativa indica. The plants were then submerged for a period of 11 days. The heights of all the plants were measured at the beginning and at the end of the submergence period.
[Adapted by permission from Macmillan Publishers Ltd, Xu et al. 2006. “Sub1A is an ethylene-response-factor-like gene
that confers submergence tolerance to rice.” Nature. Vol 442. Pp 705–708. Copyright 2006. http://www.nature.com/ ] a.
(i) State which group of rice plants were the shortest at the beginning of the
experiment. [1]
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(ii) Calculate the percentage change in height for the O. sativa japonica unmodified variety during the submergence period. Show your working. [2]
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b. Explain how the error bars can be used to compare the results for O. sativa
indica. [2]
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c. Deduce the general relationship between the growth of all the japonica varieties
and their stated tolerance level. [1]
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In the same experiment, the researchers hypothesized that the capacity to survive when submerged is related to the presence of three genes very close to each other on rice chromosome number 9; these genes were named Sub1A, Sub1B and Sub1C. The photograph below of part of a gel shows relative amounts of messenger RNA produced from these three genes by the submergence-intolerant variety, O. sativa japonica, and by the submergence-tolerant variety, O. sativa indica, at different times of a submergence period, followed by a recovery period out of water.
[Adapted by permission from Macmillan Publishers Ltd, Xu et al. 2006. “Sub1A is an ethylene-response-factor-like gene
that confers submergence tolerance to rice.” Nature. Vol 442. Pp 705–708. Copyright 2006. http://www.nature.com/]
d.
(i) Determine which gene produced the most mRNA on the first day of the submergence period for variety O. sativa japonica. [1]
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(ii) Outline the difference in mRNA production for the three genes during the submergence period for variety O. sativa indica. [2]
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(iii) Compare the mRNA production for the three genes during the submergence period between the two varieties. [2]
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e. Deduce, using all the data, which gene was used to modify GMFC. [2]
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f. Evaluate, using all the data, how modified varieties of rice could be used to overcome food shortages in some countries. [2]
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(Total marks = 15)
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2) The diagram (not to scale) shows a cell which contains water and sodium ions. This cell is immersed in a salt solution of water and sodium ions.
a. State the mode of transport if:
(i) water moves into the cell. [1]
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(ii) sodium ions move into the cell. [1]
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b. Explain facilitated diffusion. [3]
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c. State the name of the structures formed within a cell by endocytosis. [1]
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(Total marks = 6)
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3) Two foods were measured with a calorimeter to determine the energy in each. Five
trials on potato chips and five trials on walnuts were performed. The results are shown in the table below.
Trial Potato chip / kJ g–1
Walnut / kJ g–1
1 22.4 24.1 2 21.7 23.8 3 21.9 25.2 4 22.0 28.0 5 22.0 27.9
Mean 22.0 missing value Standard deviation 0.1 2.0
a. Calculate the mean energy for the walnut. [1]
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b. Explain how this data shows which food had the greatest variation in its energy content. [2]
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c. Both potato chips and walnuts contain lipids. State one function of lipids. [1]
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(Total marks = 4)
4) Consider the following human karyotype.
a. Name the two types of macromolecules that would be found in the structures in the karyotype. [2]
Macromolecule 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Macromolecule 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
b. Draw two labelled diagrams in the space below to illustrate the general structure of
the monomers of each macromolecule named. [4]
(Total marks = 6)
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5) In Tay Sachs disease (TSD), the enzyme that breaks down glycolipid is faulty due to a genetic mutation. Examine the following two tables.
The amino acid sequence shown in Table Y for a normal individual forms part of the enzyme hexosaminidase A (Hex A).
What is the amino acid sequence for the TSD individual in Table Y? Use the information in Table Y to explain how changing one amino acid in a polypeptide may adversely affect the function of a protein of which the amino acid is part.
TSD amino acid sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Explanation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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(Total marks = 4)
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Section C: Extended Response Questions [20 marks] Answer one question. Up to two additional marks are available for the construction of your answer. Write your answers in the boxes provided.
6) a. Draw a labelled diagram to show the structure of membranes. [5]
b. Explain the importance of surface area to volume ratio as a factor limiting cell
size. [7] c. Outline the process of aerobic respiration. [6]
7) a. Draw a labelled diagram of the molecular structure of DNA including at least
four nucleotides. [5]
b. Explain the process of DNA replication. [7]
c. A small DNA sample found at a crime scene can be used in an investigation. Describe the steps taken in the processing of this small sample of DNA. [6]