BIOLOGY PRACTICAL SOLUTIONS
(1ai)
A – Butterfly
B – Spider
C – Scorpion
D – Grasshopper
E – Tridax fruit
(1aii)
A – Insecta
B – Arachnida
(1aiii)
COMPLETE THE TABLE:
Number of body division:
C – Two
D – Three
Type of mouth part:
C – Chewing
D – Biting and chewing
(1bi)
(i) Possession of exoskeleton
(ii) They have jointed appendages
(iii) They have segmented body
(1bii)
(i)
A – Three distinct body segments – head, thorax, and abdomen.
B – Divided into cephalothorax (combined head and thorax) and abdomen.
(ii)
A – It has six legs.
B – It has eight legs.
(iii)
A – It has two pairs of large, often colorful wings.
B – Absence of wings.
(iv)
A – It has long, slender antennae.
B – Absence of antennae.
(1biii)
A:
(i) It plays a crucial role in the pollination of flowering plants.
(ii) It is aesthetically pleasing and is often a focal point in ecotourism
B:
(PICK ANY TWO)
(i) They can be employed to control unwanted plant species, acting as a form of biological pest control.
(ii) They cause nutrient cycling by breaking down plants, fertilizing soils with their excreta and in turn facilitating plant growth.
(iii) They prevent overgrowth of plants and in turn protecting them for overcrowding and death
(1biv)
I – Pappus
II- Fruit stalk
(1bv)
(i) It is covered in a feathery appendage which helps the fruit to disperse by wind
(ii) It is small in nature
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(2ai)
F – Tilapia fish
G – Tadpole
(2aii)
Pisces
(2aiii)
I – Operculum
II – Mouth
III – Scales
IV – Anal fin
V – Caudal fin
(2aiv)
I – It is used for respiration
II – For locomotion or movement
(2bi)
(i) Possession of scales
(ii) Presence of fins
(2bii)
TABULATE
=SPECIMEN F=
(i) Presence of pectoral fins
(ii) Presence of scales
=SPECIMEN G=
(i) Absence of pectoral fins
(ii) Absence of scales
(2ci)
(i) Possession of mouth for feeding
(ii) Possession of fins for locomotion
(iii) Possession of operculum for respiration
(iv) Possession of scales for temperature regulation
(2cii)
(i) Protein
(ii) Fats and oil
(2ciii)
Frog
(2civ)
Thyroxine
(2cv)
Possession of fins
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(3ai)
I: Funnel
II: Cotton wool
III: Water
(3aii)
X: (60 – 50)cm³ = 10cm³
Y: (60 – 30)cm³ = 30cm³
Z: (60 – 40)cm³ = 20cm³
(3aiii)
X: (10/60) × 100 = 16.67%
Y: (30/60) × 100 = 50%
Z: (20/60) × 100 = 33.3%
(3bi)
X: Sandy soil
Y: Clay soil
Z: Loamy soil
(3bii)
X:
(i) It is fine
(ii) It is coarse
Y:
(i) Sticky when wet
(ii) Least coarse
Z:
(i) Rich in humus
(ii) Dark in colour
(3biii)
Experiment to determine the water holding capacity of soil samples
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