CEE Entrance Examination Model Question Set - 1

Group A: Botany MCQs

1. Sweet Potato is modification of:

A. Stem

B. Flowering axis

C. Root

D. Bud

Correct Answer: C
Explanation: Potato is modification of stem. Sweet potato is modification of root.

2. The fruit that develops from inflorescence is called:

A. Simple fruit

B. Aggregate fruit

C. Composite Fruit

D. None

Correct Answer: C
Explanation: Multiple or composite fruit develops from inflorescence.

3. Axile Placentation is characteristics of:

A. Malvaceae

B. Solanaceae

C. Liliaceae

D. All

Correct Answer: D
Explanation: So → Solanaceae, Ma → Malvaceae, Li → Liliaceae, A → Axile placentation

4. Hybridization followed by polyploidy is called as:

A. Autopolyploidy

B. Allopolyploidy

C. Aneuploidy

D. Euploidy

Correct Answer: B
Explanation: None provided in the document.

5. Tallest Pteridophytes is:

A. Alsophila

B. Cyathea

C. Pteris

D. Dryopteris

Correct Answer: A
Explanation: Largest Pteridophyte: Cyathea, Tallest Pteridophyte: Alsophila, Smallest Pteridophyte: Azolla

6. Chloroplast is found in spores of

A. Riccia and Marchantia

B. Funaria and Equisetum

C. Funaria and Sphagnum

D. Funaria and Anthoceros

Correct Answer: B
Explanation: Chloroplast is found in Funaria and Equisetum.

7. The P/O ratio (No. of ATP produced to the no. of O₂ converted into H₂O) in the first step of glycolysis is:

A. 1/1

B. 1/2

C. 0/0

D. 1/0

Correct Answer: C
Explanation: P/O is 0/0 as neither ATP is formed nor O₂ is released.

8. By weight, each gram of glucose yields biological energy.

A. 9.45 kcal

B. 686 kcal

C. 4.3 kcal

D. 4.1 kcal

Correct Answer: D
Explanation: One mole of glucose (C₆H₁₂O₆ = 180 g) yields 686 kcal and therefore 1 g will yield 686/180 = 3.8 kcal or 15.94 KJ.

9. Loading of phloem is related to:

A. Increase of sugar in phloem

B. Elongation of phloem cell

C. Separation of phloem parenchyma

D. Strengthening of phloem fibres

Correct Answer: A
Explanation: According to mass flow hypothesis, sugar is translocated on mass through phloem along the TP. This increase of sugar in phloem during translocation is called loading of phloem.

10. Guard cell of the stoma is

A. Kidney shaped

B. Convex shaped

C. Longitudinal and cylindrical shaped

D. Dumb bell shaped

Correct Answer: A
Explanation: Guard cell of stoma of normal dicot plant is kidney shaped. Guard cell of stoma of normal monocot plant is dumb-bell shaped.

11. Sugarcane shows high efficiency of CO₂ fixation because it performs

A. Calvin cycle

B. EMP pathway

C. Hatch & Slack pathway

D. TCA cycle

Correct Answer: C
Explanation: C₄ or Hatch & Slack Pathway was 1st observed in sugarcane by Kortschak. Common in a number of tropical plants both monocots & dicots. E.g. Maize, Sugarcane, Sorghum, Millet, Amaranthus etc.

12. Which of the following component of plants have no nucleus?

A. Companion cells

B. Sieve tube

C. Pith

D. Cambium

Correct Answer: B
Explanation: Primary function of sieve tube is to transport sugars & other important molecules necessary for plants. Sieve tube cells are analogous to RBCs of mammals, both living but enucleated at maturity.

13. Jewels of plant kingdom are:

A. bacteria

B. diatoms

C. dinoflagellates

D. none

Correct Answer: B
Explanation: Jewels of plant kingdom: Diatoms, Jewels of pond: Volvox

14. Which one is a true moss?

A. Cord moss

B. Bog or peat moss

C. Hair cap moss

D. All of the above

Correct Answer: D
Explanation: True mosses: Funaria - Cord moss, Polytrichum - Hair cap moss, Sphagnum - Peat or Bog moss

15. Xylem in root is

A. endarch

B. exarch

C. mesarch

D. none

Correct Answer: B
Explanation: Xylem is endarch in stem and exarch in roots.

16. Total number of haploid nuclei in an embryo sac is

A. 4

B. 8

C. 6

D. 3

Correct Answer: B
Explanation: The monosporic embryo sac divides to form 7 celled and 8 nucleated embryo sac & also known as Polygonum type of embryo sac.

17. Rejection of transplanted organ is due to

A. T-cells

B. B-cells

C. Eosinophils

D. Neutrophils

Correct Answer: A
Explanation: It is due to cellular immunity. T-lymphocytes play role in tissue rejection.

18. Tea is a good stimulant due to presence of

A. Morphine

B. Ephedrine

C. Nicotine

D. Caffeine

Correct Answer: D
Explanation: None provided in the document.

19. Tendrils exhibit/twinning of tendrils is due to:

A. Seismonasty

B. Thigmotropism

C. Heliotropism

D. Diageotropism

Correct Answer: B
Explanation: None provided in the document.

20. Gaseous exchange in submerged plant occurs by:

A. Hydathodes

B. Lenticels

C. Stomata

D. General surface

Correct Answer: D
Explanation: Guttation: Loss of water in the form of water drops along with minerals from hydathode.

21. Monoecious plant is

A. Haploid

B. Diploid

C. Both male and female

D. Male or female

Correct Answer: C
Explanation: None provided in the document.

22. Sexual reproduction does not occur in

A. Nostoc

B. Riccia

C. Ulothrix

D. Rhizopus

Correct Answer: A
Explanation: Nostoc is blue green algae also called as cyanobacteria.

23. True alternation of generation is found in:

A. Spirogyra

B. Mucor

C. Selaginella

D. All of these

Correct Answer: C
Explanation: Selaginella is a Pteridophyte.

24. Which group contains fibre yielding plant of economic importance?

A. Gossypium, Crotalaria and Hibiscus

B. Gossypium, Agave and Cassia

C. Gossypium, Cassia and Lycopersicum

D. Gossypium, Brassica and Nicotiana

Correct Answer: A
Explanation: Gossypium hirsutum - cotton: Seed surface (testa) of different species of Gossypium is source of cotton fibres. Crotalaria juncea - Sun hemp: Fibres from phloem and pericycle of stem (bast fibres) are obtained, used for making ropes, mats, etc. Hibiscus cannabinus - Deccan Hemp: (Secondary phloem) of its stem is source of bast fibre.

25. Shrubs have strong wood trunks and they are free supporters. These types of shrubs are found mainly in:

A. Tropical rain forest

B. Temperate forest

C. Alpine habitat

D. None

Correct Answer: A
Explanation: Lianas are woody climbers and twinners; commonly found in tropical rain forest.

26. The total number of base pairs in one turn in B-DNA is:

A. 11

B. 12

C. 10

D. 9

Correct Answer: C
Explanation: B-DNA is the most common DNA. One complete turn of B-DNA helix contains 10 base pairs and the length between successive base pairs is 3.4 Å.

27. In mitochondria which is present

A. Oxysomes

B. Quantasomes

C. Thylakoids

D. Cisternae

Correct Answer: A
Explanation: Oxysome is found in mitochondria, contains ATPase acting as centre of ATP synthesis during oxidative phosphorylation.

28. The marker enzyme of mitochondria is

A. Pyruvate dehydrogenase

B. Aldolase

C. Amylase

D. Succinic dehydrogenase

Correct Answer: D
Explanation: The enzymes that are found only in mitochondria are called marker enzymes of mitochondria. All enzymes of Krebs cycle are found in matrix except succinic dehydrogenase that is found in inner membrane.

29. Enzyme required for removing RNA primer during DNA replication is:

A. Ligase

B. DNA polymerase III

C. DNA polymerase I

D. Primase

Correct Answer: C
Explanation: RNA primers are removed by DNA polymerase I.

30. What will be the codons in m-RNA if the DNA codes are ATG-CAG?

A. UCA-TUA

B. TCA-GTC

C. TAC-GTC

D. UAC-GUC

Correct Answer: D
Explanation: Codons on m-RNA: DNA A → m-RNA U, DNA T → m-RNA A, DNA G → m-RNA C, DNA C → m-RNA G

31. The enzymes of glycolysis are located in the:

A. Cytoplasm

B. Lysosomes

C. Mitochondrion

D. Nucleus

Correct Answer: A
Explanation: Enzymes of glycolysis are present in the cytoplasm.

32. The type of ribosomes found in prokaryotic cells are:

A. 100 s

B. 60 s

C. 80 s

D. 70 s

Correct Answer: D
Explanation: 70 s type of ribosomes is found in prokaryotic cells and in mitochondria and chloroplasts of eukaryotic cells.

33. Transfer of genetic information from a molecule of DNA into mRNA is known as:

A. Transcription

B. Translation

C. Replication

D. Transduction

Correct Answer: A
Explanation: Central dogma indicates unidirectional flow of information from DNA to mRNA by the process of transcription, and then to protein by the process of translation.

34. The most striking example of point mutation is:

A. Thalassemia

B. Sickle-cell anaemia

C. Colour blindness

D. Haemophilia

Correct Answer: B
Explanation: None provided in the document.

35. When the unwanted introns are removed and functional regions (exons) are responsible for coding in transcription. This process is known as

A. Transduction

B. Splicing

C. Inducing

D. Constitutive genes

Correct Answer: B
Explanation: m-RNA formed after transcription is called primary transcript. Primary transcript: Coding part (exon), Non-coding part (intron). Removal of introns from exon is called splicing.

36. Height, skin colour etc are controlled by

A. Polygenic inheritance

B. Mutation

C. Pleiotropism

D. Multiple allelism

Correct Answer: A
Explanation: When regulation of a trait is carried out by genes of more than one locus, it is called polygenic inheritance. E.g. Skin colour, Height, IQ in animals, and Number, size of flowers, fruits and seed, Kernel colour of wheat in plants.

37. In Terai region, the most common disease that appears among the Tharu community is:

A. Phenylketonuria

B. Sickle cell anemia

C. Hemophilia

D. Colour blindness

Correct Answer: B
Explanation: Sickle cell anemia is due to point mutation and is very common among the Tharu community in Terai of Nepal.

38. The most stable ecosystem is

A. Forest

B. Mountain

C. Desert

D. Ocean

Correct Answer: D
Explanation: None provided in the document.

39. Insectivorous plants grow in the soil which is deficient in:

A. Calcium

B. Nitrogen

C. Water

D. Magnesium

Correct Answer: B
Explanation: None provided in the document.