Mye Sec 3 Marking Scheme

Section A

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Multiple-Choice Questions (30 marks)

Choose the most suitable answer and write the corresponding letter (A, B, C or D) in the OTAS answer sheet provided. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30.

D A C D B A B C C B D B C C D C A D B A C B D A C D D B A A

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Section B

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Short-Structured Questions (60 marks)

Answer the following questions in the spaces provided. 31.

Fig 1.1 shows an aphid on a longitudinal section of a plant shoot.

Fig 1.1 (c)(i) [1] (b) lignin pattern on xylem wall [1]

(a)

Name the parts A, B and C of the stem and state their functions.

[3]

A: Phloem: - transports manufactured food from the leaf to other parts of the plant [1]; B: Cambium: - cell division to eventually differentiate into new phloem and xylem [1]; C: Xylem: - Transports water and mineral salts from the roots to other parts of the plant [1]. 3 Correct Labels + wrong function(s) Award [1] only; no mks for wrong label (b)

Draw on Fig 1.1 a structural adaptation that helps provide mechanical support to the plant. [1]

(c)

The aphid is an insect that is often found on young shoots during the day. (i)

Complete the drawing of the aphid in Fig 1.1 by inserting its mouthpart. Mark clearly the tissue from which the aphid obtains its food. [1]

(ii)

Suggest why aphids are usually found on young shoot.

[1]

Less woody, hence easier for the mouthpart to penetrate / Young shoots are smaller; hence the vascular bundles are nearer to the surface [1]. (iii) Explain why aphids are usually found only during the day.

[1]

Photosynthesis takes place during the day, resulting in production of sugars (glucose), which are carried by the phloem / Manufactured food is being constantly removed from the leaves during the day [1].

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(d)

Briefly explain why an infestation of aphids will kill a plant.

[1]

Food is being removed in such large quantities by the aphid that the plant does not have enough for its own growth [1]. (e) How does leaving the stylet attached to the shoot help scientists?

[1]

Helps in identification of contents of the phloem and gives an indication of the flow in the phloem [1]. [Total: 9] 32.

Fig. 2.1 shows the concentration of glucose in blood extracted from the hepatic vein over a period of time.

Fig 2.1

(b) Hepatic portal vein [1]

(a)

C

Explain what is happening at the following points.

[3]

A: Liver is converting excess glucose into glycogen for storage [1]; B: Glucose is being exported to hepatic vein [1]; C: Level of glucose is returning to normal [1]. (b)

Draw and indicate, on Fig 2.1, the glucose concentration in the hepatic portal vein. [1]

(c)

Describe how the hepatic portal vein is different from the hepatic vein.

[2]

Hepatic portal vein leads from the capillaries in the small intestine to the liver [1]; Hepatic vein leads from the liver to the main blood circulation [1].

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(d)

Describe another difference between the blood in the hepatic portal vein and the hepatic vein. Explain what causes this difference. [2] Hepatic portal vein contains more amino acids resulting from absorption through the capillaries of the villus in the ileum whereas hepatic vein may contain more urea [1]; Excess amino acids are deaminated by the liver, resulting in formation of urea, which will be carried by the hepatic vein and excreted by the kidneys [1]. [Total: 8]

33.

Fig 3.1 shows an experimental setup. A water plant was submerged in water enriched with carbon dioxide and exposed to light placed at a fixed distance. The gas produced in five minutes was collected. The results of the experiment were plotted on the graph shown in Fig 3.2.

Fig. 3.1

Fig. 3.2

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(a)

Explain why a gas was produced.

[1]

Photosynthesis is taking place [1]. (b)

What gas was being collected?

[1]

Oxygen [1]. (c)

What was the aim of the experiment?

[1]

To investigate how the process of photosynthesis is affected by light intensity [1]. (d)

Describe the effect of life intensity at the following parts of the graph.

[4]

Part 1: Light is less than 10 cm from the plant, thus water is heated up by the bulb [1]; Plant cannot photosynthesize well as enzymes involved in photosynthesis are being denatured [1]; Part 2: At this temperature, photosynthesis is at its most efficient [1]; Part 3: As light intensity decreases, photosynthesis also decreases [1]. (e)

Explain why light have the effect described in (d)?

[3]

Light energy is needed in photosynthesis mechanism when sugars are produced [1]; During the first phase of photosynthesis, oxygen is produced [1]; When less light is present, the light dependent phase of photosynthesis will be less efficient so less oxygen is produced [1] (f)

When the experiment was repeated without changing any materials, the set of results shown by the dotted line in Fig 3.2 was obtained. What was the most likely cause of the difference in results? [1] The amount of carbon dioxide dissolved in water could have been depleted by the photosynthesizing plant of the earlier experiment [1]. [Total: 11]

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34.

Fig. 4.1 shows a cross-section of a leaf. Light [1]

Chloroplast containing chlorophyll [1] Water [1]

Fig. 4.1 Carbon dioxide [1]

(a)

Complete the table below by naming the labeled parts and stating the adaptations that allows each part to fulfill its functions. [8] Part A B

F G

Name Cuticle + Upper Epidermal cell + Palisade mesophyll cell + Spongy mesophyll cell + Intercellular air spaces + Stoma + Guard cell +

H

Veins +

C D E

Adaptations Waxy to prevent excess loss of water [1] No chloroplasts present to allow maximum penetration of light [1] Many chloroplasts to trap light; max surface area available for photosynthesis [1] Irregular cells covered with film of water to provide a large SA over which gases can dissolve [1] Large air spaces to allow gases to diffuse into and out of the leaf [1] Allow movement of gases into and out of leaf [1] Uneven thickening of cell wall to control size of stomata / Presence of chloroplasts [1] Bring water and mineral salts near every cell of the leaf, and manufactured food to other parts of plant [1]

Every 2 correct labels, award [1]. No mks awarded for wrong label though function is correct.

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(b)

On Fig 4.1, indicate with the help of arrows, where the shaded cell obtains the substances for photosynthesis. [4] [Total: 12]

35.

In an experiment, a mixture of amylase and a 1% starch solution were placed in selectively permeable tubing. The tubing was sealed at both ends before being placed in a water bath of known temperature. The water surrounding the tubing was tested at various time intervals for the presence of reducing sugars. The experimental setup was performed at different temperatures and the results were recorded in Fig 5.1.

Water bath temperature (°C) Time at which reducing sugar was first detected in the water (min)

0

10

20

30

40

50

55

60

20

10

5

4

4

9

15

>40

Fig. 5.1 (a)

Using the readings in Fig 5.1, draw a graph to show the relationship between the temperature of the water bath and the time taken for the reducing sugar to be first detected. [3] Correct axes labeled with units [1];

Time / min

Points plotted accurately [1]; Appropriate smooth curve / best straight line drawn [1].

Temperature / °C

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(b)

Explain how reducing sugars can appear in the water bath when starch was originally placed in the tubing.

[3]

Amylase is an enzyme which breaks down starch into maltose [1]; Maltose have reducing properties [1]; It diffuse out of the tubing into the surrounding water where it can be detected using Benedict’s reagent [1]. (c)

At which temperature was the rate of reaction fastest?

[1]

35° C [1] (d)

Compare the rates of reaction at 0°C and 10°C. Explain your observations.

[1]

Rate of reaction at 10°C is double that at 0°C. This means that the enzyme is less active at lower temperatures [1]. (e)

Compare the rates of reaction at 50°C and 60°C. Explain your observations.

[2]

Rate of reaction at 60°C cannot be measured [1]; The enzyme has probably been denatured by heat, so it cannot function [1]. [Total: 10] 36.

(a)

In what way is the transport of water assisted by the fact that the xylem vessels are dead. [2] Vessels are dead / continuous, cell contents such as cytoplasm or organelles are absent [1]; This allows the unrestricted flow of water through the vessel [1].

(b)

If xylem vessels are non-living, how could you explain the observation that the stems of woody plants increase in diameter? [2] New xylem tissue, which is made up of living cells, is being added all the time [1]; Old tissue that is made of non-living cells is not lost from the plant. As a result, the diameter of the stem increases [1]. [Total: 4]

37.

(a)

Explain why it is important for root hairs of the plant to be long.

[1]

To provide a large surface area for absorption of water and mineral salts from the soil [1]. (b)

Root hairs are very delicate structures that can be easily damaged. How does this explain why, when re-potting the plant, the roots must be disturbed as little as possible in order for the plant to survive? [2] If too many roots are damaged, the ability of the plant to absorb water is reduced [1]; Water is necessary for growth so a shortage of water would result in death of the plant [1]. [Total: 3]

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38.

Normal human red blood cells are biconcave in shape. They are completely filled with haemoglobin, a red pigment that combines with oxygen. As oxygenated red blood cells squeeze along the narrow capillaries, oxygen moves out of them due to a higher level of oxygen in the red blood cell than in the plasma, through the capillary walls and into the surrounding tissues. (a)

Why is the biconcave shape better than a spherical one in allowing the cell to absorb oxygen? [2] Ensures minimum diffusion distance / also gives cell more flexibility as it can be bent over as it passes through the narrowest capillaries [1]; there is a higher surface area: volume ratio [1].

(b)

What impact would the presence of a nucleus have on the oxygen-carrying capacity of the cell?

[2]

Less haemoglobin molecules for oxygen transport [1]; Nucleus occupies space in the cell [1]. [Total: 4] Section C

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Essay Questions (10 marks)

Answer the following question in the writing papers provided. 39.

(a)

Describe the advantage to a herbivore, such as a cow, of chewing its food.

[2]

Chewing increases the surface area for enzyme action on the masticated food [1]; Softens the cellulose / food and enables the enzymes to work on the vegetation easily [1]. (b)

Describe (i)

how a molecule of digested food from the gut enters the bloodstream, [4]

A molecule of digested food eg. Glucose is broken down from carbohydrates in the mouth and stomach pass along the gut by peristalsis until they reach the small intestine [1]; The walls of the intestine are made up of numerous finger-like villi, which project into the intestinal cavity [1]; The villi are richly supplied with blood and lymphatic vessels to carry away the digested food substances [1]; The glucose molecules is absorbed by the villi and the glucose passes through the walls of the blood vessels and into the bloodstream [1]

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(ii)

the pathway by which it eventually reaches cells in the hand.

[4]

Blood leaving the small intestine with this molecule of glucose enters the hepatic portal vein [1]; This passes through the liver and via the hepatic vein and then towards the heart through the vena cava [1]; The heart pumps this blood to the lungs which then, after that, flows back to the heart through the pulmonary vein [1]; The glucose in the blood will flow through the aorta, which branches off into arteries and then reaches the tissues in the hand [1]. [Total: 10]

~~~END OF PAPER~~~

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