NIOS Class 12 Biology April 2019 Set C (58/OSS/1)

NIOS Biology Solved Paper April 2019 Set C

1 MARK QUESTIONS

1. One of the macronutrients given below plays a key role in food production and normal growth of the plant.

A) Boron

B) Copper

C) Carbon

D) Uranium

2. Which of the following enzyme is called “Joining Enzyme”?

A) Plasmids

B) DNA ligase

C) Protease

D) Amylase

3. Identify the following figure:

A) Sac 

B) Leaf

C) Embryo 

D) Seed Coat

4. Which chemical is involved in the transmission of nerve impulse across a synapse.

A) melanin

B) NaCl

C) Acetyl choline

D) Oxytocin

5. With reference to organisation of life, Rana tigrina comes into which category?

A) Community

B) Species

C) Organism

D) Biome

6. Which of the following bacteria causes cholera?

A) Vibrio cholerae

B) Rhizobium

C) Azotobacter

D) Lactobacillus

7. When the genetically engineered bacteria clean up pollutants from the environment, they are called __________.

A) transgenic animals

B) bio-diesel

C) bio-gas

D) bio-remediation

8. Name the complex tissue which function as a unit to conduct water and minerals upward from root to leaves.

A) Phloem

B) Parenchyma

C) Xylem

D) Sclerenchyma

2 MARKS QUESTIONS

9. What are the factors responsible for population explosion in India?

Ans: 

• Advancement in agriculture
• Advancements in medicine
• Religious and social customs
• Industrialization
• Illiteracy
• Economic reasons
• Desire for a male child

10. Define the term inflorescence. What are the major types of inflorescence in flowering plants.

Ans: Inflorescence is the arrangement of flowers on the floral axis called the peduncle. The two major types of inflorescence are racemose and cymose. 

11. Answer the following:

a) Which phase of growth curve shows rapid and maximum growth?

b) What is the term used for inducing early flowering in plants at low temperature?

c) Name the specially designed equipment used to measure the rate of growth of shoot length of plants.

d) What is the hypothetical plant hormone which is responsible for initiation of flowering in plants?

Ans: 

1. Log phase
2. Vernalisation
3. Auxanometer
4. Florigen

12. Write any four advantages of Bio-diesel.

Ans: Bio-diesel has several advantages; some of them are given below- (any four)

• It is an agriculture-based fuel substitute.
• It can be made from both vegetable oil and animal fats.
• It can be used without major modifications in engines.
• It does not need a separate infrastructure for storage and delivery.
• Handling biodiesel is safer.
• Planting of Jatropha curcus will utilise wasteland in our country.
• Its combustion emits less carbon monoxide, sulphates, unburnt hydrocarbons and particulate matter, thus reducing air pollution.

13. Where does spinal cord is located in human beings. Mention its main function.

Ans: The spinal cord extends from the medulla of the brain downward almost the whole length of the backbone. Its main function is to carry out reflexes below the neck and conduct both sensory impulses and motor responses to and from the brain.

14. Give reasons:

a) Habitat is called an address of the organism.

b) Fins are an adaptations of fish to aquatic life.

Ans: 

1. Habitat is the physical environment in which an organism lives and is thus considered its address.
2. An adaptation is the appearance, behaviour, structure or mode of life of an organism that allows it to survive in a particular environment. The presence of gills and fins is an example of adaptation of fish to an aquatic habitat.

15. What is the importance of balanced diet? Reason out the special diet recommendation to pregnant and lactating women?

Ans: A balanced diet contains all essential nutrients in suitable proportions and amounts to provide necessary energy and keep the body in a healthy state. A pregnant woman has to feed the developing embryo; therefore, she has a special need for extra nutrients. Similarly, nursing mothers (who breastfeed their babies) also need a special diet to take care of their additional requirements of lactation (milk formation). So their diet should contain more protein, calcium and vitamins.

16. Give an example for each of the following:

a) Aerobic bacteria which fixes nitrogen

b) Anaerobic bacteria which fixes nitrogen

c) Cyanobacteria which fixes nitrogen

d) Photosynthetic non-sulphur bacteria which fixes nitrogen

Ans: 

1. Azotobacter
2. Clostridium
3. Anabaena
4. Rhodospirillum

17. Draw a neat diagram of human lung and label the Larynx and trachea parts:

Ans: 

18. A person has protruding eyes, stunted growth, retarded mental growth, delayed puberty and low metabolite rate.

a) What is this condition?

b) How to prevent it?

Ans: 

1. Condition: Cretinism
2. Prevention: Use of iodised table salt and eating seafood, and fish.

19. Observe the diagrams and answer the following:

a) Are these type of stomata found in monocots or dicots?

b) Mention their main function.

Ans:

1. Stomata in dicots.
2. Stomata help in gaseous exchange and allow the loss of water vapour during transpiration.

20. How is the phenotypic ratio of F2-generation in dihybrid cross is different from monohybrid cross?

Ans: Crosses involving plants differing in the inheritance of one contrasting feature only are called monohybrid crosses. Crosses involving two contrasting features are termed dihybrid crosses.

In a monohybrid cross, tallness and dwarfness were involved, TT x tt, 

The F1 was all tall with a genotype of Tt as the gene for tallness is dominant over that for dwarfness. 

In the F2, Tt x Tt, the results will be 

Gametes	T	t
T	TT	Tt
t	Tt	tt

The phenotype of F2 here is 3 Tall: 1 dwarf.

In a dihybrid cross, two contrasting characters are taken into consideration- height and colour of the flower. 

In a cross between a tall red plant with a dwarf red, 

Parents: TTRR x ttrr

Gametes: T, T, R, R x t, t, r, and r

F1: TtRr. 

When F1 is selfed, 

Gametes	TR	Tr	tR	tr
TR	TTRR Tall red	TTRr Tall red	TtRR Tall red	TtRr Tall red
Tr	TTRr Tall red	TTrr Tall white	TTRr Tall red	Ttrr Tall white
tR	TtRR Tall red	TtRr Tall red	ttRR Dwarf red	ttRr Dwarf red
tr	TtRr Tall red	Ttrr Tall white	ttRr Dwarf red	Ttrr Dwarf white

Thus, F2 shows 9 Tall red, 3 Tall white, 3 Dwarf red, and 1 dwarf white.

4 MARKS QUESTIONS

21. Give reasons: [2+2=4]

a) Sino-atrial node is called pacemaker of the heart.

b) Person with blood group ‘O’ called Universal donor.

Ans: 

1. Since the Sinu-atrial Node initiates and regularises the heartbeat, it is also called the pacemaker. The pacemaker is influenced by nerves, hormones, CO₂ and O₂ content of blood, and heat.
2. Blood groups of O type can be given to all groups. It is thus the Universal Donor. This is because there are no antigens in the blood of Group O.

22. Differentiate between ”vascular cambium” and ” cork cambium”.

Ans: Vascular cambium originates as a strip in pericycle cells lying outside the protoxylem and in conjunctive tissue inner to each phloem bundle. Cork cambium (phellogen) also differentiates in the pericycle and gives rise to cork (phellem) towards the periphery and secondary cortex (phelloderm) towards the inside.

23. A man with ‘A’ blood group married a woman with blood group ‘B’. Work out a cross to show possible blood groups that can be expected in their offspring.

Ans: 

Parents: A x B

Genotype: I^AI^A x I^BI^B  

Gametes	IB	IB
IA	IAIB	IAIB
IA	IAIB	IAIB

All their offspring will have the AB blood group. 

24. Write short notes on:

a) Mode of nutrition in Yeast.

b) Sexual recombination in Bacteria.

Ans: 

1. Yeast is saprotrophic. It can directly absorb simple sugar (glucose), but for obtaining sucrose (cane sugar) it gives out the enzyme invertase or sucrase, which breaks down sucrose into simple sugars. The simple sugars are then simply absorbed into the cell.
2. Some bacteria show a primitive mode of sexual reproduction. It is different from sexual reproduction in higher forms. The steps are:
   1. Two conjugating (lie very close for the transfer of genes) bacteria are held together by pili.
   2. A segment of a DNA strand is transferred from one bacterium to another bacterium.

25. Draw a schematic diagram of “Carbon Cycle”.

Ans: 

26. “Genetic code is universal and common for almost all organisms on earth”. Explain.

Ans: Genetic code is universal and common for almost all organisms on earth, as it shows the following characteristics in all organisms. 

• The genetic code is a triplet code, a sequence of 3 bases that stores the information for a particular amino acid. The sequence of codons determines the sequence of amino acids in a protein.
• The genetic code is unambiguous; that is, a particular codon can code for only one amino acid.
• The genetic code is read continuously from beginning to end.
• As there are only 20 amino acids that form the various proteins, there will be more than one codon code for a particular amino acid.
• The genetic code is read on the transcribed mRNA during protein synthesis.
• The AUG codon codes for Methionine and is the initiation codon as it is the first one to be transcribed from a cistron.
• UAA, UAG and UGA are stop codons and anticodons of one of these three codons is present at the end of every cistron to terminate protein synthesis.

6 MARKS QUESTIONS

27. a) What is “Criss-Cross inheritance”?

b) Why is red-green colour blindness more common in males than females?

c) How is colour blindness inherited? Explain with the help of flow chart.

Ans: 

1. The type of inheritance of a recessive sex linked character from father to daughter and then from the daughter to her sons is known as criss-cross inheritance or sex linked or X-linked inheritance.
2. The defective gene for red-green colour blindness is located on the X chromosome. In males, a single defective gene causes the disease, while in females, it requires the presence of two defective genes. This is why this disease is more common in males than in females.
3. In males, the single X chromosome is received from the mother. Hence, a defective gene for colour blindness on the X chromosome of the mother is passed on to the son and expressed as a defect. The daughter who receives the defective gene becomes a carrier.

Parents: XX^C x XY

Gametes: X, X^C, and X, Y

Gametes	X	Y
X	XX Normal daughter	XYNormal son
XC	XCXCarrier daughter	XCYColour blind son

28. a) Name the scientist who discovered C3 cycle.

b) Draw a neat schematic diagram of C3 cycle in plants.

c) Justify your answer why C4 plants are most efficient than C3 plants?

Ans: 

1. Melvin Calvin
2. Calvin Cycle

3. C4 plants are more efficient than C3 plants because C4 plants have no photorespiration and thus there is no loss of additional carbon dioxide, due to breakdown of RuBP to Glycolate and CO₂.

29. a) Write any four salient features of Class-Aves with two examples.

b) Define “Alteration of generation”, with one example.

Ans: 

1. Features of Class Aves (any four)
   1. The body is covered with feathers, and scales are present only on the hind limbs
   2. The body is divisible into three parts: head, neck and trunk.
   3. Jaws with horny beak, no teeth.
   4. Hind limbs with four digits adapted for perching, walking or swimming
   5. Bones with air spaces to make the skeleton light (pneumatic bones).
   6. Forelimbs modified into wings for flight.
2. The alternation between two phases in the life cycle between a gametophytic and a sporophytic generation is called alternation of generations.  Eg. The main body of bryophytes is the gametophyte, and the diploid zygote is the only sporophyte in their lives.

30. a) What is somatic gene therapy?

b) List out the steps involved in ex vivo gene therapy.

c) Name two genetic disorders can be treated by this technique.

Ans: 

1. Somatic gene therapy: Once a normal gene has been cloned, it can be used to correct a genetic defect. Body cells are targeted for genetic transformation (defective gene transformed to normal). This approach helps in the correction of a genetic defect confined to a specific organ or tissue.
2. The steps involved in ex vivo gene therapy are :
   1. Isolating the cells with gene defects from a patient.
   2. Growing the isolated cells in culture.
   3. Altering the genome of the isolated cells with remedial genes.
   4. Selecting, growing and testing the altered cells.
   5. Transplanting or transfusing the altered cells back into the patient
3. Genetic disorders that can be treated by this technique are,
   1. Severe Combined Immuno Deficiency (SCID).
   2. Sickle cell anaemia.
   3. Thalassaemia

Additional Study Materials

• NIOS Class 12 Biology PYQ April 2019 SetA
• NIOS Class 12 Biology PYQ April 2019 Set B