Biology Paper 1 Questions And Answers - Form 4 Term 2 Opener 2021

INSTRUCTIONS TO CANDIDATES

Answer All questions

What is meant by the term sex linkage. (1mk)
Part of one strand of DNA molecule was found to have the following sequence
G-C-C- G – A – T- T – T – A – C – G – G
What is the sequence
1. of the complimentary DNA strand? (1mk)
2. On a m-RNA strand copied from this DNA portion? (1mk)
State three regions ion a plant where the end products of photosynthesis are translocated to? (3mks )
With reference to circulatory system only give two reasons why birds and mammals are more active compared to other organisms? (2mks)
1. What three characteristics are used to divide the phylum arthopoda into classes? (3mks)
2. The diagram below shows an organisms from a division in Kingdom plantae. Study it and answer the questions that follow.
  1. Identify the division from which the plant was obtained. (1mk)
  2. Name the parts labelled X and Z (2mks )
What is the relationship between a genus and a species? (1mk)
A drawing of 3 cm was made of a giant spider whose actual length was 7cm. calculate the magnification of the drawing? (3mks)
Explain why osmosis is described as a special type of diffusion? (1mk)
The following table shows the estimated number of organisms recorded in a dam.

Organisms Number

Small fish 3500

Microscopic algae 12000

Crocodiles 100

Large fish 950

Mosquito larvae 8900
1. Construct a possible food chain for the dam? (1mk)
2. Construct a pyramid of numbers for the given data? (1mk)
3. Explain the shape of pyramid obtained? (2mks)
1. Explain why leaves of most plants are thin and broad. (2mks)
2. State the function of the following enzymes during digestion in the stomach?
  1. Pepsin (1mk) ………………………………………………………………………………………………….
  2. Renin (1mk) ………………………………………………………………………………………………….
Explain the following:
1. Respiratory surface must be moist? (1mk) ………………………………………………………………………………………………….
2. Respiratory surface must be thin (1mk) ………………………………………………………………………………………………….
3. Palisade cells are cylindrical shaped and arranged with long axis perpendicular to the leaf surface. (1mk)
Germinating beans seeds were placed in the clinostat as shown in the diagram and left for three days. (a)
1. What is a clinostat. (1mk) ………………………………………………………………………………………………….
2. The Clinostat was switched on and left to run for the three days. Suggest the direction the seedling will be facing at the end of the three days. (1mk)
3. Give a reasons for your answer in (b) above? (1mk)
Explain why the body temperature of a healthy person rises slightly during humid days? (2mks)
Nocturnal animals such as owl are capable of seeing fairly at night. What two retinal adaptations have made this possible? (2mks)
State the function of the following organelles:
1. Granulated Endoplasnic reticulum (1mk) ………………………………………………………………………………………………….
2. Nucleolus (1mk) ………………………………………………………………………………………………….
State three gaseous exchange sites in plants? (3mks)
The diagram below shows an apparatus used during collection of specimen or biological study.
1. Identify the apparatus? (1mk) ………………………………………………………………………………………………….
2. What is the use of the apparatus named above? (1mk)
List three limitations of fossil records as an evidence of organic evolution? (3mks)
Distinguish between enzyme co-factors and co-enzymes? (2mks)
Give two reasons for the rapid growth during the exponential phase of growth curve? (2mks)
Give two reasons why Carolus Linneaus preferred the use of latin language in the scientific naming of living organisms. (2mks)
State three roles played by active transport in living organisms. (3mks)
List three factors affecting the rate of respiration? (3mks)
Study the diagram below and answer the questions that follow.
1. Identify the cell (1mk)
2. Label the parts X,Y and W (3mks)
The diagram below shows a bone of the hind limb. Study it and answer the questions that follow.
1. Name the bone (1mk) ………………………………………………………………………………………………….
2. Name the parts labelled Q and R (2mks)
3. Name the structure that articlualtes with the part labelled Z and the joint formed? (2mks)
  Structure ………………………………………………………………………………………………
  Joint …………………………………………………………………………………………………… .
List two functions of inter-vertebral discs between two adjacent vertebrae. (2mks)
Explain why it is becoming more difficult to treat malaria using chloroquine? (4mks)
State four ways by which the ileum is adapted fro absorption of food materials? (4mks)
Name two processes that contribute to variation during gamete formation? (2mks)

MARKING SCHEME

These are genes located on the sex chromosomes and are transmitted together with those determine sex. 1mk
1. C – G – G – C – T – A – A – A – T – G – C – C 1mk
2. C – G – G – C – U – A – A – A – U – G – C – C 1mk
- The growing and developing regions such as shoots, leaves, flowers, fruits and roots
- Storage organs or tissues such as tubers, corns, bulbs, rhizomes and seeds.
- The secretors organs such as nectar glands in some insect pollinated plants such as bananas
  3×1= 3mks
1. Deoxygenated and oxygenated blood do not mix
2. Blood is at a higher pressure once the heart pumps it twice.
  2×1= 2mks
1. - Number of limbs
  - Presence and number of antennae
  - Number of body parts
    3×1= 3mks
2. 1. Bryophyta
  2. X – Seta
    Z – Rhizoid rej Rhizoids.
A species is a subset of genus i.e. one genus contains several species. 1×1= 1mk
Magnification = length of drawing √1
length of real object
= 3cm = 0.429 √1
7cm
X 0.43√1
3×1= 3mks
Because it involves movement of solvent (water) molecules from their region of high concentration to region of low concentration across a semi permeable membrane. 1×1= 1mk
1. Microscopic algae → mosquito larvae → small fish → large fish → crocodile NB: mark as a whole
  1×1= 1mk
2. 1×1= 1mk
3. 1. Body size of the organism increase at each trophic level from the base as their numbers decrease.
  2. At each trophic level much of the energy obtained is lost in respiration thus fewer organisms can be supported at the succeeding level. 2mks
1. 1. Thin
    - To reduce distance for diffusion of gases. 1×1= 1mk
    - To reduce distance for sunlight to reach the photosynthetic cells. 1×1= 1mk
  2. Broad – To provide large surface area for maximum light absorption. 1×1= 1mk
2. Pepsin – Breaks down proteins into peptides. 1×1= 1mk
  Renin – Digests protein caseinogens in milk to casein (curd) 1×1= 1mk
1. Moist o dissolve the diffusing gases across the respiratory surface. 2mks
2. Thin to reduce distance covered by diffusing gases i.e. for the gases to diffuse through short distance. 1mk
3. Many palisade cells in a small area to enable them receive maximum sunlight. 1×1= 1mk
1. A clinostat is a device which slowly rotates a plant to nullify the effect of unidirectional stimulus.
  1×1= 1mk
2. Horizontal direction as in diagram. 1mk
3. The clinostat will nullify the effect of gravity on the seedling hence the seedling continues to grow horizontally. 1×1= 1mk
This is because during humid day, there is low rate of sweating; since less water is lost from the body surface, leading to less heat loss through sweat hence body temperature tend to rise slightly.
1×2= 2mks
- High concentration of rods in the retina of their eyes.
- There is more rods than cones in the retina.
  1x2 = 2mks
1. Helps in the transport of proteins. 1mk
2. Manufacture of ribosomes 1mk
- Stomata;
- Lenticels of woody plants
- Cuticles
  3×1= 3mks
1. Pitfall trap
2. For catching crawling animals.
- Distortion of parts of fossils during sedimentation hence can give wrong impression of the structure; -
- There was several missing links of fossils records as some parts or whole organism decomposed, some scavenged upon and conditions may not be conducive for fossilization (O.W.T.T.E).
- Destruction of fossils by geological activities like earthquakes, faulting and mass movement.
  3×1= 3mks
Enzymes cofactors are non-proteinous substances which activate enzymes; while co-enzymes are organic non protein molecules that work is association with particular enzymes. (Mark as a whole) 2mks
- Cells have adjusted to the new environment
- Food and other factors are not limiting hence no competition for resources.
- Rate of cell increase is higher than cell death.
- There is an increase in the number of cells dividing
  First two 2×1= 2mks
- Latin language was widely spoke and used by scientists during his time;
- Local names used previously could not be understood by everyone thus Latin language enhanced scientific communication worldwide. 2×1= 2mks
- Excretion of waste products from the body cells.
- Absorption of digested food from alimentary canal of animals in the blood stream.
- Absorption of some minerals salt from soil by plant roots.
- Accumulation of substances into the body to offset osmotic imbalance in arid and saline environment. - Reabsorption of sugars and some salts by the kidney. 3×1= 3mks
- Oxygen concentration
- Presence or absence of hormones
- Substrate concentration
- Surface area to volume ratio/body size of an organism. 3×1= 3mks
1. Mature human ovum 1mk
2. X – follicle cell
  Y – viteline membrane 1mk
  W – Plasma membrane 1mk
1. Femurs
2. Q – neck
  R – Shaft
3. Structure - patella
  Joints – hinge joint
- Act as a cushion that absorbs shock thus reducing friction.
- Allows for a certain degree of movement between the vertebrae in the vertebral column.
  2×1= 2mks
Some malaria plasmodium developed resistance; to chloroquine drug; through mutation; those resistant individuals transmit the characteristic to their offspring through reproduction thus establishing a new population of resistant forms. 4mks
- It is long to provide large surface area for absorption.
- It is numerous to bring digested food into close contact with walls of the ileum for easier absorption. -
- Highly coiled to slow down movement of food, allowing more time for absorption.
- Higher surface has large number of villi and micro-villi which increase the surface area for absorption of end products of digestion.
- Presence of thin layer of cells through which digested food diffuses.
- Presence of tense network of blood capillaries in villi into which nutrients are absorbed.
- Presence of lacteals in the villi for absorption of fatty acids and glycerol.
  4×1= 4mks
- Independent assortment
- Crossing over
  2×1= 2mks