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香港考試及評核局
HONG KONG EXAMINATIONS AND ASSESSMENT AUTHORITY
2012 年 香 港 中 學 文 憑 考 試
HONG KONG DIPLOMA OF SECONDARY EDUCATION EXAMINATION 2012
生物 試卷一乙
BIOLOGY PAPER 1B
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閱卷員參考之用。閱卷員在完成閱卷工作後，若將本評卷參考
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評卷參考視為標準答案，以致但知硬背死記，活剝生吞。這種
落伍的學習態度，既不符現代教育原則，亦有違考試着重理解
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作，堅守上述原則。
This marking scheme has been prepared by the Hong Kong Examinations and
Assessment Authority for markers’ reference. The Authority has no objection
to markers sharing it, after the completion of marking, with colleagues who are
teaching the subject. However, under no circumstances should it be given to
students because they are likely to regard it as a set of model answers.
Markers/teachers should therefore firmly resist students’ requests for access to
this document. Our examinations emphasise the testing of understanding, the
practical application of knowledge and the use of processing skills. Hence the
use of model answers, or anything else which encourages rote memorisation,
should be considered outmoded and pedagogically unsound. The Authority is
counting on the co-operation of markers/teachers in this regard.
„香 港 考 試 及 評 核 局 保留版權
Hong Kong Examinations and Assessment Authority
All Rights Reserved 2012 ‚
2012-DSE-BIO 1B–1
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Paper 1 Section B
Questions 1 to 5 are common items. Bio and CS Bio share the same MS.
Marks
1. C (1)
A (1)
B (1)
3 marks
2. Method How it works Comment (advantage or disadvantage)
Spraying of Directly kill the larvae / Pros: quickly put mosquito population down
pesticides or adult mosquitoes (1) so in short term (1)
larvicidal oil to that they cannot serve as Cons: mosquitoes may develop resistance to
1,1
mosquito’s habitat vector the pesticides (1) / environmental
contamination when pesticides leak to water
bodies / pesticides are toxic to humans
Clearance of Eradicate the breeding Pros: does not have adverse impacts on
accumulated water places of mosquitoes (1) environment (1)
1,1
in the Cons: it is vritually impossible to clear up all
neighbourhood stagnant water (1)
4 marks
3. (a) ü the cell wall of cell type Q is much thicker than that of cell type P (1) (1)
(¸ accept other answers such as spiral thickening or thickened with ring structures)
(˚ cell type Q being hollow not accepted)
(b) ü when there is ample supply of water (1) (1)
ü cell type P provides turgidity to the plant (1) (1)
ü cell type Q has thickened cell wall (1) (1)
ü which provides rigidity to the plant (1) (1)
5 marks
4. (a) ü Class A (1) (1)
(b) ü the light intensity of the habitat is very low / the habitat is completely dark (1) (1)
ü as organism X does not have eyes to survive in the habitat (1) (1)
(c) ü the key is constructed based on the morphological characteristics of exisiting organism
found (1) / not all the morphological characteristics of the phylum are listed in the key (1)
(d) ü carry out a comparative study about the amino acid sequence of similar proteins / base
sequence of DNA template / mRNA of similar proteins found in organism Y and other
organisms in this phylum (1) (1)
(¸ accept comparison of developmental process / cellular structure / chemical
composition )
ü to establish the phylogenetic relationship between them (1) (1)
6 marks
5. (a) ü 7:00 and 18:00 (1) (1)
(b) ü shorter light period (1), overall rate lower (1) (1,1)
(c) ü the area below the line showing oxygen production rate represents the food production
in 24 hrs (1) (1)
ü whereas the area below the line showing carbon dioxide production rate represents the
food consumption in 24 hrs (1) (1)
ü it is therefore important for food production to be greater than food consumption such
that there is a net amount of food produced (1) (1)
ü as a result, this provide energy for the plant to survive, grow and produce fruits (1) (1)
7 marks
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Marks
Question 6 is a common item which is equivalent to CS Bio question 7.
6. (a) (10 000 000 – 1 000 000) / 10 000 000 x 100% = 90% (Remark: accept -90%)
Method / equation 1 mark (1,1)
Correct answer 1 mark
(b) ü some energy is not obtained by the organisms of higher trophic level as part of the body
of prey is not consumed (1) / part of the food is not digestible or is egested (1) (1)
ü some energy is lost by the organisms at the higher trophic level through excretion (1) /
respiration (1) / in the form of heat (1) (1)
Remarks:
- Any two of the above.
- The words “higher trophic level” are not required. However, if the wrong trophic
level is mentioned, the mark will not be awarded.
- some individuals escape from predation is not accepted.
(c) ü the percentage decrease in energy content is greater (1) (1)
ü as caterpillars mainly feed on leaves of trees, leaving most of the part of the biomass of
tress unconsumed (1) / lignin is not consumed (1)
6 marks
This is an item for Biology only.
7. (a) ü despite the high blood glucose level detected in his blood, his fasting blood insulin level (1)
was lower than that of the healthy person (1)
ü although there is an increase in blood glucose level, the insulin level only shows little (1)
change (1)
ü this shows that Tom failed to produce the normal amount of insulin (1) (1)
ü therefore, Tom suffered from insulin-dependent diabetes (1) / type I diabetes (1)
(b) ü with insufficient insulin, his body cells will not take up extra glucose from the blood
as efficiently as the healthy person (1)
ü as a result, the blood glucose concentration rised to a higher level (1) after glucose
(3)
consumption
ü and remains high for a longer time / decreases slower than the healthy person (1)
Remarks: conversion of glucose to glycogen by insulin is not acceptable
(c) ü by injection of insulin (1) / aerosal spray of insulin applied to nasal cavity (1)
8 marks
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Marks
Question 8 is a common item which is equivalent to CS Bio question 6.
8. (a) ü soluble sugars lower the water potential of the kernels (1) while (1)
ü starch, being insoluble in water, does not affect the water potential of the kernels (1) (1)
ü during development, kernels of sweet corns draw in a larger amount of / more water by
osmosis than those of starch kernels (1) (1)
ü later on, when the kernels dry up, kernels of sweet corn shrinks as a result of lossing
large amount of water (1) (1)
ü as the skin is not elastic (1), thus the kernels of sweet corns become wrinkled (1)
max. 4
(b) (i) ü as both parents are pure-bred plants, the F1 produced are all heterozygous (1) (1)
ü in heterozygous condition, dominant alleles will be expressed while recessive
alleles will be masked (1) (Remark: gene instead of allele is not accepted.) (1)
ü therefore, purple colour and smooth surface are the dominant phenotypes (1) (1)
(Remark: allele instead of phenotype is not accepted.)
If symbols are used in the explanation, the symbols must be defined at the very
beginning.
(ii) ü as the individuals in F1 generation are heterozygous with and the genes for the two (1)
traits are located on different chromosomes, the alleles for the traits are assorted
independently (1)
ü only one type of gametes for the pre-bred corn plant with yellow and wrinkled (1)
kernels (1)
ü and four different types of gametes will be resulted from the F 1 generation (1)
ü after random fertilisation, zygotes with four different combinations of genotypes
will be produced (1) (max. 2)
ü leading to the expressions of all the possible phenotypes (A, B, C and D) in the F2
generation (1) (Remarks: This answer cannot get 1 mark if it is mentioned that the
ratio is 9:3:3:1.)
(c) ü Mendel collected a large amount of experimental evidence (1) (1)
ü and based on his observation on the number of offspring with different phenotypes (1) (1)
ü he worked out a possible explanation for the observation by logical deduction (1) (1)
Remarks:
- The first mark is given to the point that a large number of experiments on different
traits were conducted.
- The second mark is given to the observation of the ratio of different phenotypes.
- The third mark is given to the reasoning of the experimental results.
14 marks
This is an item for Biology only.
9. (a) ü drug X inhibits glycolysis (1) (do not accept more than 1 process) (1)
ü as glycolysis is the first step in the respiratory pathway, the inhibition of glycolysis will
halt the processes that follow, i.e. Krebs cycle and oxidative phosphorylation (1) (1)
ü hence, the overall production of pyruvate, ATP and NADH are greatly reduced, showing
that the whole respiratory pathway was jeopardized (1) (1)
Remarks:
- No mark will be given to bullet point 1 when minor steps in each process instead of
the key process are mentioned.
Alternative answer:
ü Pyruvate is the product of glycolysis (1)
ü As the production of pyruvate is greatly reduced after treating with drug X (1)
ü Glycolysis is inhibited in this case (1)
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(b) ü drug Y inhibits Krebs cycle (1) (do not accept more than 1 process) (1)
ü when the respiratory pathway is halted at Krebs cycle, pyruvate would not be
metabolised (1) (1)
ü but glycolysis still proceeds as usual and produce pyruvate (1), as a result, pyruvate will (1)
accumulate
(c) ü in anaerobic conditions, muscle cells undergoes anaerobic respiration and produce less
ATP (1) and less NADH (1) than aerobic respiration (1,1)
ü at the same time, lactic acid level rises as it is produced (1) as a result of incomplete (1)
oxidation
(d) ü glycolysis: cytoplasm (1) (1)
ü Krebs cycle: mitochondrial matrix (1) (1)
ü oxidative phosphorylation: mitochondrial inner membrane (1) (1)
Remarks:
- if the key processes are not mentioned but the three cellular components are
mentioned in the correct sequence, 3 marks will be given
- if only two or one cellular component(s) are mentioned, no mark will be scored
12 marks
This is an item for Biology only.
10. (a) ü bile salts emulsify fat into droplets (1) (1)
ü such that there is an increase in surface area for the action of the lipase / enzyme (1) (1)
(b) ü as the bile supplementation increased, the fat content of the faeces decreased (1) (1)
ü this indicates increased digestion of fat (1) (1)
Remarks: If absorption instead of digestion is mentioned, no mark will be given
(c) ü to show that the addition of bile supplementation does not adversely affect the growth of
the pigs (1) / indicate the effectiveness of the bile supplementation on promoting piglets’ (1)
growth / effectiveness of fat absorption
(d) Concept for mark award:
ü 1st pt: suitable substrate and correct enzyme used in the experiment
ü 2nd pt: the identification of parameter for meausring the dependent variable
ü 3rd pt: provide expected results
e.g. ü prepare a mixture of lipase, (bile salts) and oil (1) (1)
ü add pH indicator into the mixture / use a data logger with pH sensor/ pH meter to show
the change in pH of the mxiture (1) (1)
ü the faster the drop in the pH of the mixture, the faster the digestion of fat (1) (1)
Accept other reasonable experiments
8 marks
2012-DSE-BIO 1B–5
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Marks
Question 11 is a common item which is equivalent to CS Bio question 8.
11. Differences Significance
ü Pairing of homologous chromosomes ü Such that the daughter cells formed
along the equatorial plane in first division contain the whole set of chromosome /
of meiosis but no such process in mitosis one member of each homologous pairs
(1) (1) after meiosis
ü The pairs of homologous chromosomes ü Random segregation of homologous
segregate into the daughter nuclei during chromosomes results in variations (5)
the first meiotic cell division (1) between gametes formed in meiosis (1)
ü Crossing over may occurs, the exchange
of genetic materials between non-sister
chromatids gives rises to new genetic
combinations (1)
ü Mitosis involves one division only but ü The daughter cells resulted from mitosis
meiosis involves two divisions (1) are genetically identical to the parent cell
(1) which is important for growth of the
organisms (1) / asexual reproduction
ü The daughter cells / gametes formed in
(5)
resulted from meiosis contain half /
haploid the genetic content of the parent
cell (1) such that the amount of genetic
content can be restored after fertilisation
(1)
D = (3) S = max.5 Max. 8
Communication C = max.3
11 marks
Mark award for communication:
Mark Clarity of expression and relevance to the Logical and systematic presentation
question
3 ü Answers are easy to understand. They ü Answers are well structured showing
are fluent showing good command of coherence of thought and
language. organisation of ideas.
ü There is no or little irrelevant material.
2 ü Language used is understandable but there ü Answers are organised, but there is
is some inappropriate use of words. some repetition of ideas.
ü A little relevant material is included, but
does not mar the overall answer.
1 ü Markers have to spend some time and ü Answers are a bit disorganised, but
effort on understanding the answer(s). paragraphing is evident. Repetition
ü Irrelevant material obscures some minor is noticeable.
ideas.
0 ü Language used is incomprehensible. ü Ideas are not coherent and
ü Irrelevant material buries the major ideas systematic. Candidates show no
required by the question. attempt to organise thoughts.
2012-DSE-BIO 1B–6
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