hkcee - biology - 1992 - paper i - a

8/19/2019 HKCEE - Biology - 1992 - Paper I - A

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CE BIO 1992

1. (a) (i) (1)

N . B. No mark for separate food chains

Award 1 mark for each correct feeding relat ionship ( x 10 = 2 ) ------------- 2

Award an addit ional mark for a correct and complete food w eb ----------------

Deduct mark for each incorrect feeding relationship / incorrect

representation in the food web. (No deduction below zero.)

(ii) food chain A ---------------------------------------------------------------------------------------------------- 1

more trophic levels involved / longer food chain ------------------------------------------------------ 1

more energy will be lose during the transfer between successive trophic levels -------------- 1

due to respiration, metabolic activities , death. etc. -------------------------------------------------- 1

(ii) bees -------------------------------------------------------------------------------------------------------------- 1

they help in pollinating the flowers ------------------------------------------------------------------------ 1

(iv) * reptiles --------------------------------------------------------------------------------------------------------- 1

possession of dry scales ------------------------------------------------------------------------------------ 1

Total : 11 marks

1. (a) (i) When asked to construct a food web showing the feeding relations most candidates could

draw the connecting lines linking the organisms. Some, however, forgot to put in the

relevant arrows. A number of candidates put the higher level consumers at the bottom of the

food web which is not the conventional representation of a food web.

(ii) Many candidates tailed to stress that the longer the food chain, the more energy lost

during the transfer between successive trophic levels.

(iv) Many candidates mentioned the presence of scales as a noticeable external feature of

snakes but failed to point out that they are "dry".

Paper I Marking Scheme 1992 Year Report P.1


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1. (b) (i) vessel Y----------------------------------------------------------------------------------------------------------- 1

Its thick / elastic wall -----------------------------------------------------------------------------------------1+

helps resisting the high blood pressure ------------------------------------------------------------------ 1

(ii) vessel X (renal vein) * vena cava * heart * pulmonary artery * lungs

* pulmonary vein * heart * aorta vessel Y (renal artery) ---------------------------------- 4

N.B. 0.5 mark for each correct spelling of the organs in the correct sequence

0.5 mark for the flow chart / direction of blood flow

(iii) vessel X vessel Y

carbon dioxide

concentration

less more ---------------------------------------------1

glucose concentration less more ---------------------------------------------1

Total : 9 marks

1. (b) (i) Many candidates’ descriptions of the function of an artery did not match with the structural

feature stated in the answer. Some candidates thought that the thick and elastic wall of the

artery was to pump or to create a high blood pressure rather than to withstand or resist it.

(ii) Many candidates did not read the question carefully and so drew diagrams of the kidney

and its related blood capillaries instead of a flowchart showing a double circulation of blood

in relation to the kidney. It seems that they had no idea of what a flowchart is.

(iii) A number of candidates were confused about the relative concentrations of carbon dioxide

and glucose in the pulmonary artery when compared with those in the pulmonary vein.


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1. (c) (i) correct axes with labels -------------------------------------------------------------------------------------- 1

correct key ------------------------------------------------------------------------------------------------------ 1

correct plotting of 8 points for each curve (N.B. overall shape must be correct) ---------1,1

N.B. Accept alternative presentations of graph



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would not lead to any estimate of the leaf area.


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2. (a) (i) Tube 3 ----------------------------------------------------------------------------------------------------------- 1

is a passage for sperms / semen and urine ------------------------------------------------------------ 1

(ii) * testis ------------------------------------------------------------------------------------------------------------ 1

(iii) D is for reproduction / production of sperms ----------------------------------------------------------- 1

A is for excretion / osmoregulation ------------------------------------------------------------------------ 1

which is essential for the maintenance of a constant

internal environment / prevents accumulation of toxic materials ---------------------------------- 1

(iv) B --------------------------------------------------------------------------------------------------------------- 1

The elasticity of B enables the temporary storage of urine ----------------------------------------- 1

C --------------------------------------------------------------------------------------------------------------- 1

Erection of C facilitates the transfer of sperms /semen into the female reproduction tract --------------------------------------------------------------- 1

Total : 10 marks

2. (a) (i) Many candidates were confused about the meaning of the terms "urea" and "urine".

(iii) Some candidates were uncertain about the function of the kidney and wrongly thought that

it could produce urea (a function of the liver).


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2. (b) (i) At 1300 hour the transpiration rate is high because

the stomatal pores are widely open ---------------------------------------------------------------------- 1

The resulting high transpiration pull increases the rate of water absorption -------------------- 1

At 2100 hour, the transpiration rate is low because

the stomatal pores are small / closed -------------------------------------------------------------------- 1

This leads to a decrease in the rate of water absorption -------------------------------------------- 1

OR

The stomata are more wide open at 1300 hour than that at 2100 hour ------------------------- 1

Transpiration therefore occurs at a faster rate --------------------------------------------------------- 1

The resulting larger transpiration pull -------------------------------------------------------------------- 1

leads to a higher absorption rate -------------------------------------------------------------------------- 1

(ii) Areas A and B represent either the net amount of water

lost or gained during the specified period of time ----------------------------------------------------- 1 Area B larger than Area A means that there is a net

gain of water over the 24-hour period -------------------------------------------------------------------- 1

The net gain of water is essential for various life

processes e.g. formation of new cells, photosynthesis, etc. ---------------------------------------- 1

(iii) thicker inner cell wall and thinner outer wall / kidney-shaped -------------------------------------- 1

(iv) sweat pores / glands / blood capillaries ----------------------------------------------------------------- 1

prevent overheating ------------------------------------------------------------------------------------------ 1

Total : 10 marks

2. (b) (i) Many candidates misunderstood the question, giving only the changes in amount of water

in the plant from 1300 hours to 2100 hours instead of relating how such changes could be

linked to the relative size of the stomatal pores. Many candidates misread the question as

requiring them to explain changes in transpiration rate with respect to changes in

temperature and light intensity. Only a few candidates correctly pointed out the relationship

between transpiration pull and water absorption. In this regard candidates* concept of

transpiration pull was very vague.

(ii) Many candidates wrongly interpreted areas A and B as the amount of water transpired or

absorbed respectively. These are actually the net gain and net loss of water in the plant

respectively.

(iv) Many candidates gave the function of the sweat pore in the human skin as only one of

temperature regulation without specifying that it is mainly for preventing over heating. Many

incorrectly cited the hair / hair follicle / hair pore / hair gland (rather than the sweat pore or

sweat gland) as the ‘counterpart’ (in human skin) of the plant stoma.


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2. (c) (i) the non-green offspring must be homozygous recessive (gg) ------------------------------------- 1

therefore, they must receive one recessive gene (g) from each parent ------------------------- 1

since both of the parents should have a dominant gene for being green in colour ----------- 1

their genotypes must be heterozygous (Gg) ----------------------------------------------------------- 1

OR

According to Mendel’s Law / Law of Inheritance ------------------------------------------------------ 1

If a cross between two green parents gives rise to

green and non-green offspring in a 3 : 1 ratio ---------------------------------------------------------- 1

(Working of the 3 : 1 ratio) ---------------------------------------------------------------------------------- 1

Then both parents must be heterozygous (Gg) -------------------------------------------------------- 1

(ii)

g

Gg Gg

GG Gg Gg gg

G gG

P

F1

------------------------------------- 1

------------------------------------- 1

Genotypic 1 : 2 : 1 ------------------------------------------- 1

ratio (green) (green) (non-green)

That is, about of the green seedlings have a

heterozygous genotype (Gg) for chlorophyll production i.e. 705 x ----------------------------- 1

= 470 ------------------------------------------------------------------------------------------------------------- 1

(iii) seeds make use of their food reserve for germination ----------------------------------------------- 1

Total : 10 marks

2. (c) (i) Many candidates lacked the competence to deduce the genotypes of parent plants without

using a genetic diagram. This inability was clearly aggravated by the poor standard of

expression in English.

(ii) Some candidates did not use the genetic symbols as given in the question while others

gave the genotypic ratio of 1:2:1 without referring to the genotypes involved. Many

candidates failed to give a logical explanation of how they could deduce the parental

genotypes from what is given in the question.



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3. (a) (i) * chlorophyll ---------------------------------------------------------------------------------------------------- 1

(ii) to remove the residue / green chloroplasts ------------------------------------------------------------- 1

so that results / any colour changes of subsequent food

tests can be easily observed ------------------------------------------------------------------------------- 1

(iii) * cellulose ------------------------------------------------------------------------------------------------------- 1

- it increases the bull of the indigestible material )

- and stimulates the muscular movements along )

the alimentary canal / peristalsis ) any TWO ------------------------------1,1

- to prevent constipation )

(iv) The presence of reducing sugar is shown by the ----------------------------------------------------- 1

formation of orange precipitate in the Benedict’s test ------------------------------------------------ 1

The presence of vitamin C is shown by the decolorization of the DCPIP solution --------- 1+1

N.B. If more than 2 food tests are mentioned, deduct 1 mark for every additional food

substance wrongly deduced. No deduction below 0.

Total : 10 marks

3. (a) (i) The term "chlorophyll" was often misspell. Many candidates mistook "chloroplast" as the

pigment responsible for the green colour of the green pepper.

(iii) Some candidates confused excretion with egestion of undigested food substances.

(iv) Some candidates were not aware that the presence of vitamin C could still be deduced

although the decolorisation of DCPIP was masked by the pale green colour of the original

filtrate. A considerable number of candidates wrongly thought that Benedict’s test was a test

for the presence of glucose. They failed to note that Benedict’s test is only a test for the

presence of reducing sugar.


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3. (b) (i) the iris muscle (circular muscle) contracts -------------------------------------------------------------- 1

to reduce the size of the pupil ------------------------------------------------------------------------------ 1

to prevent too much light entering the eyeballs -------------------------------------------------------- 1

to over stimulate / damage the retina -------------------------------------------------------------------- 1

(ii) the ciliary muscle relaxed ----------------------------------------------------------------------------------- 1

increasing the tension on the suspensory ligaments ------------------------------------------------- 1

the lens became less convex ------------------------------------------------------------------------------ 1

to enable it to focus on a far away object.

(iii)

----------------------------------------- 2

N.B. ---------------------------------------------------------------------------------------- No lens, no mark

No arrows, deduct 1 mark

Light rays not reaching retina, deduct 1 mark

(iv) voluntary -------------------------------------------------------------------------------------------------------- 1

Total : 10 marks

3. (b) (i) Some candidates confused the terms "iris" and "pupil". They failed to realize that the pupil

is an aperture or opening at the centre of the iris which is a circular muscular sheet at the

front of the eye ball. Also quite a few candidates confused the terms: "eye muscles", "ciliary

muscles" and "iris muscles" in their descriptions.

(ii) Many candidates did not realize that in the eye, the relaxation of the ciliary muscles would

lead to an increase in tension of the suspensory ligament. As a result the suspensory

ligament would become tightened or taut but it would not contract on its own.

(iii) Some candidates did not complete the ray diagram fully whilst others forgot to put in the

arrows to indicate the direction of the light rays all the way onto the retina.



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3. (c) (i) 40 - 12 = 26 times / minutes (N.B. no unit, no mark) ----------------------------------------------- 1

(ii) An increase in breathing rate

- provides the body with more oxygen )

- and hence more energy is released )

- by respiration / oxidation of food ) any THREE ------------------------------1,1,1

- It also helps the body to get rid of )

carbon dioxide (product of respiration) )

more quickly )

(iii) Intercostal muscle contracts to raise the rib cage ----------------------------------------------------- 1

Diaphragm muscle contracts to make the diaphragm flattened ----------------------------------- 1

These 2 actions together increase the volume of thoracic cavity --------------------------------- 1

thus decrease the thoracic pressure / pressure inside the lung ----------------------------------- 1The greater atmospheric pressure forces air into the lungs ---------------------------------------- 1

(iv) The amplitude / depth of breath is greater -------------------------------------------------------------- 1

Total : 10 marks

3. (c) This was the most well answered part of the paper indicating that candidates were generally well

prepared for topics related to fundamental human physiological processes like the breathing

mechanism in the human body. In contrast, however, questions set in other parts of the paper on

plant physiological processes like tropism and transpiration led to vague answers which were

often filled with misconceptions.


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4. (a) (i) region C --------------------------------------------------------------------------------------------------------- 1

* stomach ------------------------------------------------------------------------------------------------------- 1

(ii) curve X ---------------------------------------------------------------------------------------------------------- 1

digestion of starch in the mouth cavity ------------------------------------------------------------------- 1

(iii) * pancreatic juice ---------------------------------------------------------------------------------------------- 1

* bile -------------------------------------------------------------------------------------------------------------- 1

* intestinal juice ------------------------------------------------------------------------------------------------ 1

(iv) region D --------------------------------------------------------------------------------------------------------- 1

(v) * large intestine / * colon / * rectum ----------------------------------------------------------------------- 1

absorption of water / temporary storage of faeces ---------------------------------------------------- 1Total : 10 marks

4. (a) (i) Some candidates incorrectly deduced from the graph that digestion of proteins started in the

small intestines.

(iii) Many candidates gave the names of enzymes found in the small intestines rather than the

names of the digestive juices. Some candidates did not realize that "bile" (not "bile juice") is

a digestive juice found in the small intestines.

(v) Some candidates incorrectly believed that the caecum was responsible for reabsorption of

water from the undigested food.


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4. (b) (i) Due to the affect of gravity ---------------------------------------------------------------------------------- 1

there is a higher concentration of auxin accumulated

on the lower side of the horizontally placed seedling ------------------------------------------------ 1

auxin has a growth stimulatory effect -------------------------------------------------------------------- 1

therefore the lower side grows faster than the upper side ------------------------------------------ 1

and the ending curves upwards

(ii) Auxin moves away from light / light destroys or inactivates auxins -------------------------------- 1

resulting in an overall lower auxin concentration in seedling A ------------------------------------ 1

therefore it grows more slowly because of the less stimulatory effect

(iii) The shorter seedling in A ------------------------------------------------------------------------------------ 1

(iv) apply a unilateral light source from the bottom as shown below : ----------------------------2 or 0

Light

stand

N.B. No Light, no mark

Total : 9 marks

4. (b) (i)&(ii) Many candidates could not satisfactorily explain phototropism and geotropism in terms

of the action of plant hormones. It should be cautioned that up to now no one definitive

theory can fully and conclusively explain the mechanism causing phototropism.

Nevertheless, recent findings tend to support the "Light-induced Lateral Transfer of

Auxin Model" as a more plausible explanation of phototropism although the

"Light-induced Destruction or Inhibition of Auxin Model" has not been conclusively

disproved.

(iv) Candidates were generally weak in the provision of experimental designs slightly extended

from the conventional ones on geotropism and phototropism. This was particularly so when

they were asked to consider the joint effects of gravity and light on the growth response of a

plant seedling.



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hkcee - biology - 1992 - paper i - a

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