Kerala Plus One Zoology Previous Year Question Paper 2018
Time: 1 Hours
Cool off time : 10 Minutes
Maximum : 30 Scores
General Instructions to candidates
- There is a ‘cool off time’ of 10 minutes each for Botany and Zoology in addition to the writing time of 1 hour each. Further there is a ‘ 5 minutes’ ‘preparatory time’ at the end of the Botany Exami¬nation and before the commencement of the Zoology Examination.
- Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
- Read the instructions carefully.
- Read questions carefully before you answering.
- Calculations, figures and graphs should be shown in the answer sheet itself.
- Malayalam version of the questions is also provied.
- Give equations wherever necessary.
- Electronic devices except non programmable calculators are not allowed in the Examination Hall.
Answer all questions from question numbers 1 to 3. Each carry one score. (3 × 1 = 3)
Rearrange the following taxonomic categories in the correct sequence.
Note the relationship in the first pair and then complete the second pair
Osteichthyes : Cycloid scales
Chondrichthyes : ……………….
The innermost layer of the eyeball is ……………….
Answer any nine from question numbers 4 to 14. Each carries two scores.
The molecular structure of 2 amino acids are given below.
Arrange the following terms under two headings based on symmetry.
a. Expand GFR.
b. Even though GFR in a healthy person is 180 litres per day, the amount of urine released per day is only about 1.5 litres. Give a reason.
Select the bones of the leg from the given list of bones.
Name the following :
a. The antibacterial enzyme present in the saliva of man which helps in prevention of infection.
b. The digestive enzyme present in salvia.
Observe the diagram showing the alimen¬tary canal of cockroach. Name the parts labelled A,B,C and D.
Diagrammatic representation of a standard ECG is given below :
a. What does the QRS complex denotes ?
b. Mention the clinical significance of ECG.
Distinguish between the following :
Complete the given flow chart.
Name the hormones whose deficiency is responsible for the following :
b. Diabetes mellitus
Identify the following tissues :
a. Tissue that stores fat
b. Tissue that connects bones together
c. Tissue that connects bones to muscles
d. Tissue that conducts impulses
Answer any three from question numbers 15 to 18. Each carries three scores.
Match the terms in column A with those in columms B and C
a. Complete the diagrammatic representation showing the nature of enzyme action :
b. List out any two factors affecting enzyme activity.
c. Based on the reaction formulae given below, identify the classes of the enzymes.
Protein digestion by proteolytic enzymes is given below :
a. Name the enzymes marked as A and B.
b. Identify the gland which secrete these enzymes.
c. Write the inactive form of the enzymes A and B.
a. A table showing examples of vertebrates given below. But some of the examples are wrongly given. Identify and rearrange it
b. Which of the above mentioned class is characterized by the presence of pneumatic bones?
Kingdom ← Phylum ← Class ← Order ← Family ← Genus ← Species
Radial symmetry – Ctenophores, Coelen- terates.
Bilateral symmetry – Arthropods, Molluscs
a. Glomerular Filtration Rate
b. Due to tubular reabsorption. Nearly 99% of the filtrate is reabsorbed by the renal tubules.
Femur, Tibia, Fibula, Tarsals
b. Salivary amylase / amylase
A – Crop
B – Hepatic caecae
C – Malpighian tubules
D – Colon / Hindgut
a. Depolarisation of the ventricles or ven tricular contraction.
b.Any deviation from the normal shape indicates a possible abnormality or dis-ease, heart disease, defective heart functioning and chance of heart attack.
a. IRV (Respiratory Reserve Volume) :
Additional volume of air that can be inspired forcibly after a normal inspiration. This averages 2500mL – 3000mL.
ERV (Expiratory Reserve Volume) :
Additional volume of air that can be expired forcibly after a normal expira¬tion. This averages 1100 mL – 1200mL.
b. IC(inspiratory Capacity)
Total volume of air, a person can inspire after a normal expiration/ relevant volume. Which is equal to (TV + IRV).
Total volume of air, a person can expire after a normal inspiration / relevant volume. Which is equal to TV + ERV.
a. Growth hormone
a. Adipose tissue
d. Neural / Nervous tissue
a. (a) EP / Enzyme Product Complex, (b) Product / P
b. i. Temperature
c. i. Oxidoreductases/dehydrogenases.
a. A – Chymotrypsin
B – Carboxy peptidase
c. A – Chymotrypsinogen
B – Proarboxypeptidases
b. Class – Aves / Birds