INSTRUCTIONS
- You are required to spend the first 15 minutes of the time allowed for this paper
reading the whole paper before commencing your work. - Answers MUST be written on the spaces provided after each question.
- Candidates may be penalized for recording irrelevant information and incorrect spelling especially of technical terms.
QUESTIONS
-
- You are provided with:
- Specimen P,
- Solution Y,
- DCPIP,
- 0.1 percent Ascorbic acid.
- To 1 cm3 of DCPIP in a test tube, add 0.1 percent solution of Ascorbic acid dropwise; shake the test tube after addition of each drop until color of DCPIP disappears. Record the number of drops used in the table below.
- Repeat the procedure using solution Y and record the number of drops used in the table below.
- Squeeze out the juice from specimen P into a beaker, decant, and discard the residue.To another 1cm3 of DCPIP, add the juice from specimen P drop by drop until the color of DCPIP disappears. Record the number of drops used. Spare the remaining juice of Specimen P for procedure (c) below.
Substance
Number of drops
Ascorbic acid
Solution Y
Specimen P
- From the results obtained in a (i) and (iii) above, calculate the percentage of ascorbic acid in the juice obtained from specimen P. Show your working. (2 marks)
- State two factors that would influence the accuracy of the results (2 marks)
- Suggest the expected results if the juice from specimen P was boiled for 40 min, cooled and added drop by drop to DCPIP solution. (1 mark)
- Explain the expected results in (vi) above (2 marks)
- Between Specimen P and solution Y, which do you recommend for treatment of scurvy.
Give a reason for your answer. (2marks) -
- Using the reagents provided carry out food test to determine other food present in juice of P besides Ascorbic acid. (6marks)
Food substances
Procedure
Observations
Conclusion
Proteins
Reducing sugars
- In which regions of alimentary canal will juice of specimen P be absorbed. (2marks)
- Using the reagents provided carry out food test to determine other food present in juice of P besides Ascorbic acid. (6marks)
- You are provided with:
- You are provided with specimen D and E that are specimens of the same plant
- Using observable features group the specimen into
- Division…………………………………………………………………….…………. (1mark)
Reason………………………………………………………………….………….(1mark) - Class……………………………………………………………………………………(1mark)
Reason…………..………………………………………………………………… (1mark)
- Division…………………………………………………………………….…………. (1mark)
- Press the stem of specimen D hard with a finger repeatedly until the stem is crushed. Record the observation.
- Observation (1mark)
- Name the type of stem in specimen D (1mark)
- Specimen D is the same plant as specimen E. E has been exposed to sunlight for 24hrs.
- Name the phenomenon in exhibited by E (1mark)
- Explain how the phenomenon occurred. (2marks)
- \Name the support tissue in the specimens D and E (1mark)
- Other than the support tissue mentioned in (iii) above, how else do plants with this type of stem obtain mechanical support. (1mark)
- Using observable features group the specimen into
- Below are images of parasite that infest human body. Use it to answer the question that follow.
- Name the parasite and give a reason. (2marks)
- Which part of alimentary canal does the mature parasite infest? (1mark)
- State two effects of the above parasite infestation to humans. (2marks)
- How is the parasite adapted to the parasitic mode of nutrition? (3marks)
- Name any other parasitic worm in human beings (1mark)
CONFIDENTIAL
Each student shall be provided with the following
- Specimen P: A quarter piece of ripe medium size pineapple
- 10ml of Solution Y: Commercially prepared pineapple juice diluted with water at the ratio of 1:2 (e.g. quencher, highlands or any other)
- Specimen D and E: Are from the same Dicotyledonous plant with herbaceous stem and roots ;Specimen D is fresh while E should have been uprooted and exposed to sunlight for at least 24hrs to wilt.
- 10ml measuring cylinder
- A filter paper
- A filter funnel
- 5 test tubes in a rack
- A 50ml empty beaker
- 3 droppers
- Test tube holder
Access to the following: - Benedict’s solution
- 1% copper (II) Sulphate solution with dropper
- 10% Sodium Hydroxide solution with dropper
- 0.01% DCPIP solution with dropper
- 0.1 % Ascorbic acid
- Distilled water in a wash bottle
- Source of heat
Marking scheme
-
- You are provided with:
- Specimen P,
- Solution Y,
- DCPIP,
- 0.1 percent Ascorbic acid.
- To 1 cm3 of DCPIP in a test tube, add 0.1 percent solution of Ascorbic acid dropwise; shake the test tube after addition of each drop until color of DCPIP disappears. Record the number of drops used in the table below.
- Repeat the procedure using solution Y and record the number of drops used in the table below.
- Squeeze out the juice from specimen P into a beaker, decant, and discard the residue.To another 1cm3 of DCPIP, add the juice from specimen P drop by drop until the color of DCPIP disappears. Record the number of drops used. Spare the remaining juice of Specimen P for procedure (c) below.
Substance
Number of drops
Ascorbic acid
1 – 4
Solution Y
10 and above
Specimen P
5-8
(3mks)
NB Number of drops given should be specific not a range
Solution Y to have the most drops, followed by Specimen P then Ascorbic acid to have the least. - From the results obtained in b (i) and b (ii) above, calculate the percentage of ascorbic acid in the juice obtained from specimen P. Show your working. (2 marks)
Number of drops of ascorbic acid x 0.1% 1mk ; correct Ans 1mk
Number of drops of P
Note: use the student’s value from the table - State two factors that would influence the accuracy of the results (2 marks)
Degree of ripening of the fruit/period of storage ;
Size of the drop;
impurity present in the juice - Suggest the expected results if the juice from specimen P was boiled for 40 min, cooled and added drop by drop to DCPIP solution. (1 mark)
Blue colours of DCPIP persist/remains; reject no change, rej discolor
Accept DCPIP did not decolourize - Explain the expected results in (vi) above (2marks).
Boiling destroyed the Ascorbic acid because vitamins are destroyed by extreme high temperature
- Between Specimen P and solution Y, which do you recommend for treatment of scurvy. Reason (2marks)
Specimen P ; requires fewer drops to decolourise DCPIP hence has high concentration of ascorbic acid -
- Using the reagents provided carry out food test to determine other food present in juice of G besides Ascorbic acid. (6mks)
Food substance
Procedure
Observation
Conclusion
Reducing sugar(s);
Reject: Specific sugars e.g. Glucose/ Simple sugars/ Monosaccharide(s)
ü To 1ml/ 2ml or drops of juice of p (in a test tube) add 1ml/ 2ml\drops of Benedict’s solution;
NB food substance and Benedict’s solution used must be of Equal proportion
ü Heat to boil
√Reject: Benedict ^/ benedict’s/ Benedict’s only
Note: √Procedure must be correct, all
the way√Quantities used- ratio of 1:1
Colour changes from Green toYellow to Orange / brown
Accept Final colour only
NB// Procedure must have been marked correctly
Average/
A lot of Reducing sugars present
Conclusion tied to the observation and procedure
Proteins
To the (drops /2ml/ 1ml) of juice of P in a test tube, Add(drops /2ml/ 1ml) Sodium hydroxide followed by drops of Copper (II) sulphate,
Blue colour/any other colour accept but not purple or violet
Protein absent
NB If one commits oneself to volumes, then ratios must be at 1:1 - In which regions of alimentary canal will juice of specimen P be absorbed.(2mks)
Stomach ; Ileum;
- Using the reagents provided carry out food test to determine other food present in juice of G besides Ascorbic acid. (6mks)
- You are provided with:
- You are provided with specimen D and E which are specimens of same plant
- Using observable features only, group the specimen into
- Division :Spermatophyta(1mark) accept Spermaphyta
Reason : Differentiated into roots ,leaves and stem (1mark) - Class : Dicotyledonae (1mark)
Reason…Has Tap root, Leaves with network venation, presence of leaf petiole (1mark)reject characteristics that are not observable
- Division :Spermatophyta(1mark) accept Spermaphyta
- Press the stem of specimen D hard with a finger repeatedly until the stem crush. Record the observation.
- Observation (1mk)
Colourless fluid/ cellsap/ watery substance; reject water alone/ solution - Name the type of stem (1mk)
Herbaceous /Non-woody stem;
- Observation (1mk)
- Specimen D is the same plant as specimen E exposed to sunlight for 24hrs.
- Name the phenomenon in E (1mk)
Wilting; - Explain how the phenomenon occurred.(2mks)
cells in the stem lost water; due to high rate of evaporation/transpiration ; become flaccid; hence the drooping. Max 2mks - Name the support tissue in the plants (1mk)
Parenchyma tissue - Other than the support tissue mentioned in (iii) above, how else do plants with the type of stem obtain mechanical.
Twining around solid objects/climbing stem /Use of tendrils (1mark)
- Name the phenomenon in E (1mk)
- Using observable features only, group the specimen into
- Below are images of parasite that infest human body.Use it to answer the question that follow.
- Name the parasite and give a reason.(2mks)
Ascaris lumbricoides/Round worms
Reason : Tapered on both ends - Which part of alimentary canal does it infest.(1mk)
Small intestines - State two effects of the above parasite infestation to human.(2mks)
Cause irritation of the trachea
Larval stages boars in the lungs causing lung damage and increased chances of infections with other diseases
Leads to malnutrition (as most of the food is consumed by the parasite);
Can cause anaemia (as they feed on blood);
Heavy infestation (where victims develop round stomachs); can lead to blockage of the intestines
The worm may also enter bile duct, pancreatic duct and appendix causing further complications; - How is the parasite adapted to the parasitic mode of nutrition. (3mks)
Has two hosts human beings and pigs to ensure that it has a ready host for survival;
Lay many eggs to increase chances of survival( when some is destroyed);
Eggs have protective shell to survive harsh environmental conditions;
Has a thick elastic cuticle to protect it against the digestive enzymes of the host (enabling it to survive the digestive system);
Has tissues tolerant to low oxygen concentration characteristic of the gut;
Has a muscular pharynx through which it sucks digested food from the host’s intestine into its own gut; - Name any other parasitic worm in human.(1mk)
Bilharzia/Schistosoma ; Tape worm; flukes ; Hook worms reject – ringworms
- Name the parasite and give a reason.(2mks)
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