2019 IJMB biology paper 1

Botany Examination — Questions and Complete Solutions

Instruction: Answer question ONE (1) and any THREE (3) other questions. Question ONE carries 40 marks. The others carry 20 marks each. Use clearly labeled diagrams to illustrate your answers wherever appropriate.


Questions

1. The following data presents the yield (number of seeds produced per pod) in a newly released variety of cowpea from 10 samples in a trial experiment. The experiment was designed to investigate the effect of planting density on yield.

Sample No. T1 T2 T3 T4 T5
1 7 8 10 7 4
2 8 8 8 6 3
3 6 9 11 5 5
4 7 7 7 6 3
5 6 7 9 5 5
6 8 9 9 6 2
7 7 7 10 7 3
8 6 8 8 6 4
9 7 9 7 5 5
10 8 8 11 5 4

i. Calculate the mean number of seeds produced from each treatment.
ii. Present the means in a bar chart.
iii. Based on the means, what preliminary conclusions can you make on the outcome of the experiment?
iv. Calculate the modal yield for each treatment.
v. Calculate the median yield for each treatment.
vi. Calculate the variance of the best treatment.
vii. Calculate the standard deviation of the best treatment.
viii. Calculate the standard error of the best treatment.
ix. Calculate the range of the worst treatment.

2.
a) What are the similarities between algae and fungi?
b) Give the major diagnostic features of each class of the Bryophyta.

3. Briefly explain the following terms:
i. Heterospory
ii. Ecological succession
iii. Natural selection
iv. Alternation of generations

4.
i. Enumerate the economic and ecological importance of algae.
ii. What are the factors that contributed to the success of the pteridophytes as land plants?

5. What are the structural advancements exhibited by the gymnosperms over the pteridophytes?

6. Citing relevant examples, give the different types of root modification in plants.


Solutions

Question 1

Column sums: T1 = 70, T2 = 80, T3 = 90, T4 = 58, T5 = 38

(i) Mean for each treatment

Formula: Mean = Σx / n

  • T1: 70 ÷ 10 = 7.0
  • T2: 80 ÷ 10 = 8.0
  • T3: 90 ÷ 10 = 9.0
  • T4: 58 ÷ 10 = 5.8
  • T5: 38 ÷ 10 = 3.8

(ii) Bar chart

Mean yield by treatment: T1 = 7.0, T2 = 8.0, T3 = 9.0, T4 = 5.8, T5 = 3.8 (Y-axis: mean number of seeds; X-axis: treatment/planting density).

(iii) Preliminary conclusions

Treatment 3 (3 plants per stand) produced the highest mean yield (9.0 seeds per pod). Yield increases from density 1 to density 3, then decreases beyond density 3. Treatment 5 (5 plants per stand) gave the lowest mean yield (3.8), indicating overcrowding reduces yield.

Conclusion: There is an optimal planting density of 3 plants per stand. Beyond this, competition for nutrients, water, and light reduces yield, confirming that planting density significantly affects cowpea yield.

(iv) Mode for each treatment

  • T1: 7 appears 4 times → Mode = 7
  • T2: 8 appears 4 times → Mode = 8
  • T3: 9, 10, and 11 each appear twice → Multimodal (no single mode)
  • T4: 5 appears 4 times → Mode = 5
  • T5: 3 and 5 each appear 3 times → Bimodal (3 and 5)

(v) Median for each treatment

For n = 10, median = average of the 5th and 6th values in ascending order.

  • T1 sorted: 6, 6, 6, 7, 7, 7, 7, 8, 8, 8 → Median = (7+7)/2 = 7.0
  • T2 sorted: 7, 7, 7, 8, 8, 8, 8, 9, 9, 9 → Median = (8+8)/2 = 8.0
  • T3 sorted: 7, 7, 8, 9, 9, 10, 10, 10, 11, 11 → Median = (9+10)/2 = 9.5
  • T4 sorted: 5, 5, 5, 5, 6, 6, 6, 6, 7, 7 → Median = (6+6)/2 = 6.0
  • T5 sorted: 2, 3, 3, 3, 4, 4, 4, 5, 5, 5 → Median = (4+4)/2 = 4.0

(vi) Variance of the best treatment (T3, mean = 9.0)

Formula: Variance (σ²) = Σ(x − x̄)² / n

x x − x̄ (x − x̄)²
10 +1 1
8 −1 1
11 +2 4
7 −2 4
9 0 0
9 0 0
10 +1 1
8 −1 1
7 −2 4
11 +2 4
Σ 20

Variance = 20 / 10 = 2.0

(vii) Standard deviation of the best treatment (T3)

Formula: SD = √Variance

SD = √2.0 = 1.414

(viii) Standard error of the best treatment (T3)

Formula: SE = SD / √n

SE = 1.414 / √10 = 1.414 / 3.162 = 0.447

(ix) Range of the worst treatment (T5, mean = 3.8)

T5 values: 4, 3, 5, 3, 5, 2, 3, 4, 5, 4 — Maximum = 5, Minimum = 2

Range = 5 − 2 = 3


Question 2

a) Similarities between algae and fungi

  • Both are non-vascular (lack xylem and phloem).
  • Both are non-flowering and reproduce by spores.
  • Both lack true roots, stems, and leaves.
  • Both were classified under Thallophyta in older classification systems.
  • Both reproduce sexually and asexually.
  • Both are eukaryotic organisms.
  • Both can be unicellular or multicellular.
  • Both are commonly found in moist or aquatic environments.

b) Major diagnostic features of each class of Bryophyta

Bryophyta is divided into three classes:

1. Hepaticopsida (Liverworts) — e.g., Marchantia

  • Body is a flat, ribbon-like thallus with dorsiventral symmetry.
  • Thallus is lobed (liver-shaped).
  • Rhizoids are unicellular, smooth or tuberculate.
  • Sporophyte is simple with limited independence.
  • No true leaves; some have leaf-like lobes.

2. Anthocerotopsida (Hornworts) — e.g., Anthoceros

  • Thallus is flat and rosette-shaped.
  • Sporophyte is horn-like, growing continuously from a basal meristem.
  • Each cell contains one large chloroplast with a pyrenoid.
  • Stomata present on the sporophyte.
  • Rhizoids are unicellular.

3. Bryopsida (Mosses) — e.g., Funaria, Mnium

  • Plant body is leafy and erect, with stem-like and leaf-like structures.
  • Leaves are spirally arranged on the stem.
  • Rhizoids are multicellular and branched.
  • Sporophyte is well-developed, with a capsule, seta, and foot.
  • Protonema stage present in the life cycle.
  • Distinct operculum and peristome in the capsule.

Question 3 — Brief explanations

i. Heterospory
The production of two distinct types of spores by a plant — microspores (small, male) and megaspores (large, female). Microspores germinate into male gametophytes and megaspores into female gametophytes. Seen in pteridophytes such as Selaginella and Marsilea, and regarded as a forerunner to the seed habit.

ii. Ecological succession
The gradual, sequential change in the composition and structure of a plant community over time in a given area. It begins with pioneer species colonizing bare or disturbed habitats and ends with a stable climax community. It may be primary (on bare rock or new land) or secondary (on previously inhabited land).

iii. Natural selection
The process by which organisms with favorable heritable traits survive and reproduce more successfully than those without such traits in a given environment. Proposed by Charles Darwin, it leads to adaptation over generations, as better-suited organisms pass on their genes, causing gradual evolutionary change.

iv. Alternation of generations
The alternation between a haploid gametophyte generation (producing gametes by mitosis) and a diploid sporophyte generation (producing spores by meiosis) in the life cycle of plants and some algae. The gametophyte is dominant in bryophytes, while the sporophyte is dominant in vascular plants (pteridophytes, gymnosperms, angiosperms).


Question 4

i. Economic and ecological importance of algae

Economic importance:

  • Food: species such as Spirulina and Porphyra (nori) are consumed directly; Chlorella is used as a protein supplement.
  • Agar production: red algae (Gelidium) supply agar for microbiology media and the food industry.
  • Alginates and carrageenan: extracted from brown and red algae for use as emulsifiers and stabilizers in ice cream, cosmetics, and pharmaceuticals.
  • Fertilizers: used as organic manure, especially in coastal areas.
  • Biofuel: microalgae are a source of biodiesel and other biofuels.
  • Iodine and bromine: commercially extracted from marine algae.
  • Diatomite: fossilized diatom shells used in filtration, insulation, and abrasives.

Ecological importance:

  • Oxygen production: phytoplankton produce over half of the world’s atmospheric oxygen through photosynthesis.
  • Primary producers: form the base of aquatic food chains.
  • Carbon fixation: absorb large amounts of CO₂, helping regulate climate.
  • Nitrogen fixation: cyanobacteria fix atmospheric nitrogen, enriching soil and water.
  • Habitat: provide shelter and breeding grounds for aquatic organisms.
  • Bioindicators: sensitive to pollution, used to monitor water quality.

ii. Factors contributing to the success of pteridophytes as land plants

  • Development of true vascular tissue (xylem and phloem) for efficient long-distance transport of water, minerals, and food.
  • Well-developed roots for anchorage and efficient absorption from soil.
  • True stems and leaves (megaphylls), increasing surface area for photosynthesis.
  • Lignified cell walls providing mechanical support for upright growth.
  • Sporangia borne on leaves (sporophylls), enabling efficient spore dispersal on land.
  • Dominant, independent, photosynthetic sporophyte generation, better adapted to terrestrial life.
  • Heterospory in some members, a step toward the seed habit.

Question 5 — Structural advancements of gymnosperms over pteridophytes

  • Seed production: gymnosperms produce seeds (naked, on scales), giving the embryo food reserves and protection, unlike pteridophytes, which only produce spores.
  • Pollen grains: wind-carried to the ovule, removing dependence on water for fertilization.
  • Ovules: naked ovules on megasporophylls develop into seeds.
  • Well-developed wood (secondary xylem) with abundant tracheids, enabling growth into large trees.
  • Cones (strobili): reproductive structures organized into cones for more efficient reproduction.
  • Highly reduced gametophyte, entirely dependent on the sporophyte.
  • Deep root systems for anchorage and water uptake in dry conditions.
  • Needle-like leaves with a thick cuticle, reducing water loss.
  • Siphonogamy: the pollen tube allows sperm to reach the egg without swimming, unlike pteridophytes.

Question 6 — Types of root modification in plants

Root modifications adapt roots to perform functions beyond absorption and anchorage.

  1. Storage roots — store food and water. Taproot examples: Daucus carota (carrot, conical), Beta vulgaris (beet, napiform), Raphanus sativus (radish, fusiform). Tuberous root example: Ipomoea batatas (sweet potato).
  2. Prop roots (pillar roots) — adventitious roots growing from branches down to the soil for support. Example: Ficus benghalensis (banyan tree).
  3. Stilt roots — adventitious roots from the lower nodes of the stem, providing support. Example: Zea mays (maize), Pandanus (screw pine).
  4. Pneumatophores (breathing roots) — negatively geotropic roots that grow upward out of waterlogged soil for gaseous exchange. Example: Avicennia, Rhizophora (mangrove plants).
  5. Climbing/clinging roots (epiphytic roots) — short, adventitious roots that help plants cling to surfaces. Example: Piper betle (betel vine), Monstera.
  6. Haustorial roots (sucking roots) — roots of parasitic plants that penetrate host tissues to absorb water and nutrients. Example: Cuscuta (dodder, total parasite), Striga (partial parasite).
  7. Nitrogen-fixing roots (nodulated roots) — roots with symbiotic nodules containing Rhizobium bacteria that fix atmospheric nitrogen. Example: Glycine max (soybean), Arachis hypogaea (groundnut).
  8. Assimilatory (photosynthetic) roots — green roots that carry out photosynthesis. Example: Tinospora, some orchids (Taeniophyllum).
  9. Floating roots — spongy roots that help aquatic plants float. Example: Jussiaea (water primrose).
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