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Marking scheme
Biology paper Two July /August 2013
Section A (40 marks)
Answer all questions in this section.
1. In a particular species of tropical beetles, the wings have either red or orange marks. A cross
between red marked beetles with orange marked beetles produces off-springs with yellow
marks only. When the F1 generation off-springs are selfed, they produce F2 generation in
the ratio of 1 red: 2 yellow: 1 orange.
(a) Explain the absence of red and orange marks in the F1 off-springs.
(1 mk)
 Neither the gene is dominant over the other/The gene are co-dominant.
(b) If the gene responsible for red marks is R while the gene responsible for orange marks is
O, use a genetic cross to show how the F2 generation was obtained.
(4 mks)
Parents
Phenotype
Red Marks
Genotype
OR
Gametes
F2 generation
O
OO
Orange
Orange Marks
X
OR
O
R
OR
OR
Yellow
R
RR
Red
(c) Give two examples of sex-linked traits in man.
(2 mks)
 Colour blindness
 Haemophilia
(d) What is non-disjunction?
(1 mk)
 Addition or loss of one or more whole chromosome.
No. of organism in first catch  No. of organisms in sec ond catch
2. (a)Approximate population =
No. of marked organisms recaptured
P
i.e
=
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FM  SC
MR ; 
120 90
;  540 ants; 
20
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(b)
Does not consider migration of organisms into and out of study area.
- Does not consider the effect of paint used in marking on the animals behaviour
- Released animals may not mix freely with the remaining population.
- Marked organism may not have adequate time to mix with the rest.
- Does not consider the effect of weather on the organisms behaviour
(c)
(any 4)
- Quadrat method
- Belt transect method
- Line transect method
3. (a) Villus; Reject Villi
(1mk)
(b) To provide a large surface area for absorption of digested food;
(c) A – Microvillus; Reject – microvillus
B – Lacteal;
D – Arterial; Reject – artery
(d) (i) Succus entericcus / intestinal juice;
(ii) Maltase; Lactase; Sucrase; Peptidase;
(1mk)
(3mks)
(1mk)
[Any two correct] (2mks)
[Total = 8mks]
4. (a)F - PlacentaI; I - Chorion; J – Mucus plug
(b) Give two functions of the part labeled H.
(2 mks)
 Support.
 Shock absorber.
(c) Name two substances that pass through the part labeled G from the foetus to the
mother.
(2 mks)
 Carbon (IV) oxide.
 Nitrogenous waste products.
(d) If the ovaries of a woman are removed during the first four months of pregnancy,
miscarriage is very likely to occur. However, if they are removed after the fourth month,
pregnancy can proceed to full term. Explain.
(1 mk)
 The placenta takes over the role of producing progesterone hormone needed for
maintenance of pregnancy.
5. Form two students from Kabazi Secondary School wanted to investigate a certain process
using potted plants as shown below.
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In set-up A, the potted plant was ring barked but the leaves retained. In set-up B, the potted
plant was ring barked and leaves plucked so that it was always without leaves. The set-ups
were maintained for two months.
(a) What process was being investigated?
(1 mk)
 Translocation.
(b) What observations did the students make in their experiment?
(2 mks)
Set-up 1
 The bark just above the ring barked region swells.
Set-up 2
 The bark just above the ring barked region remains the same.
(c) Give an explanation for the observations in (b) above.
(4 mks)
Set-up 1
 Movement of photosynthetic products being translocated from the leaves are
intercepted at the ring barked region since the phloem has been removed. The
region above the ring barked region therefore swells.
Set-up 2
 Absence of leaves means that no photosynthetic products are being translocated
hence the bark just above the ring barked region remains the same.
(d) Name another method that can be used to investigate the same process.
(1 mk)
 Using aphid stylets.
 Using radioactive carbon 14 (C14)
Section B (40 marks)
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Answer question 6 (compulsory) and either question 7 or 8.
6. The following results were obtained from a study of germination and early growth of maize.
The grains were sown in soil in a greenhouse and at two days intervals. Samples were taken,
oven-dried and weighed.
Time after sowing
0
2
4
6
8
10
12
(days)
Dry mass of
0.002 0.002 0.008 0.016 0.024 0.034 0.035
embryo (g)
(a) Plot a graph of dry mass of embryo against time after sowing.
(6 mks)
0.04
0.035
Dry mass of embryo (g)
0.03
0.025
0.02
0.015
0.01
0.005
0
0
2
4
6
8
10
12
14
Time after sowing (days)
(b) What name is given to the curve you have obtained in (a) above?
(1 mk)
 S-shaped curve/Sigmoid curve
(c) Why is rate of increase low between day one and day three?
(2 mks)
 0-2 – Inhibition (absorption of water)
 2-3 – Germination starts where only few cells have undergone division.
(d) Give reasons for the limited rate of increase between day nine and day eleven. (3 mks)
 Most cells fully differentiated.
 Few cells still diving.
 Environmental factors start limiting.
(e) (i) Name a phylum whose growth does not take the shape of the curve drawn above.
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(1 mk)
 Arthropods. (Rej. Athropods)
(ii) What name is given to the curve exhibited by organisms in the phylum you have
named in (e) (i) above?
(1 mk)
 Intermittent.
(iii) What causes the behavior of the curve mentioned in (e) (ii) above?
(1 mk)
 Ecdysis(moulting) followed by development of exoskeleton.
(f) Apart from temperature, give three external factors necessary for germination. (3 mks)
 Water
 Oxygen.
 Temperature.
(g) Explain what effect would temperature below 100C have on the above seeds? (2 mks)
 Enzymes are deactivated.
 Food reserves are not acted upon hence the seeds will not germinate.


7. The heart is enclosed by a translucent membrane / pericardium which prevents on its over
dilation; the pericardium membrane secretes pericardial fluid; that lubricates the heart; The
heart is covered by a layer of fat; that acts as a shock absorber;
The heart has the cardiac muscles; that contracts and relax rhythmically without
nervous stimulation / myogenic; and does not fatigue throughout life; The cardiac
muscles have intercalated disc between its cells to enable the spread of waves of
contractions throughout whole muscle;
Presence of coronary arteries ; that supply blood with oxygen and nutrients to the heart
tissues / cells; The coronary veins drains blood away from the heart rich in carbon IV
oxide and other metabolic waste from the heart tissues;
Presence of septum; which divides the heart into two halves, thus preventing the mixing
of oxygenated blood and deoxygenated blood;
Presence of Sino atrial node; in the right atrium which acts as a pace maker for the heart
beat; Atrio-ventricular node; receives waves of contractions from SAN and spreads to
the purkinje tissues;
The left ventricle is thicker than the right; to enable it generate a higher pressure
needed to pump blood to the rest of the body;
Presence of atrio-ventricular valves; to prevent the back flow of blood into the
ventricles; The presence of semilunar valves; at the base of pulmonary artery and aorta;
to prevent the back flow of blood to the auricles;
[25 points = max. 20mks]
8. (a) Discuss factors that affect the rate of photosynthesis.
(8 mks)
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(b)






 High light intensity increases the rate of photosynthesis. This continues up to a
certain level when it slows down and finally levels off. Very bright sunshine may
actually damage the plants because of the high amount of ultra-violet light. Low
light intensity slows the process of photosynthesis.
 Increase in carbon (IV) oxide concentration results in a linear increase in the rate of
photosynthesis up to a certain level when it slows down and finally levels off.
Decrease in carbon (IV) oxide concentration slows the process of photosynthesis.
 At low temperature, rate of photosynthesis is slow. This is because enzyme activity is
lowered. A rise in temperature increases the rate of photosynthesis up to about 40 0C
where further increase in temperature decreases rate of photosynthesis and finally
stops. This is because high temperature denatures the enzymes.
Discuss the environmental factors that affect the rate of transpiration.
(12 mks)
High light intensity increases the rate of photosynthesis in the guard cells resulting in the
opening of the stomata. More water vapour is lost to the surrounding atmosphere thus a
higher rate of transpiration.
Increase in atmospheric temperature increases the internal temperature of the leaves.
Latent heat thus a higher rate of transpiration.
In windy conditions, moist air around the leaf is blown away thus increasing the
saturation deficit which increases the diffusion gradient hence higher rate of
transpiration.
Low humidity occurs when the atmosphere is dry and has less water vapour. The
concentration of water vapour in the intercellular air spaces is higher than in the air
surrounding the leaves. This increases the diffusion gradient which in turn increases the
rate of transpiration. High relative humidity occurs when the atmosphere is saturated
with water vapour and the diffusion gradient is lowered thereby reducing the rate of
transpiration.
Low atmospheric pressure at high altitude decreases the saturation of water vapour in
the intercellular air spaces causing an increase in rate of transpiration. Higher
atmospheric pressure at low altitude increases saturation of water vapour in the
intercellular air spaces and reduces diffusion rate causing a decrease in the rate of
transpiration.
When more water is available from the soil, the mesophyll cell walls in the leaves are
moist and avail more water to intercellular spaces. This increases the diffusion gradient
hence a higher rate of transpiration. The guard cells become turgid and stomata remain
open hence more water is lost to the atmosphere.
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