👉 Download Biology Question Paper PDF - SET 6
PART A: BOTANY
I. Answer any 3 questions from 1 – 5. Each carries 1 score. (3 x 1 = 3)
✅ Seed Dormancy
✅ Palindromic sequence
✅ Carrying capacity.
✅ Bacillus thuringiensis
✅ Grazing food chain (GFC)
II. Answer any 9 questions from 6 – 16. Each carries 2 scores. (9 x 2 = 18)
a. Give name of one such genetically modified pest resistant crop.
b. Which gene is used for its production?
c. Write about its mode of action.
✅ Answer
(a) Bt cotton.(b) Cry genes from Bacillus thuringiensis.
(c) When an insect ingests the Bt toxin, it becomes active due to alkaline pH of the gut which solubilise the crystals. Toxin binds to surface of mid-gut epithelial cells creating pores. It causes cell swelling and lysis and death of the insect.
✅ Answer
Characters of insect pollinated flower:a. Large, colourful, fragrant and rich in nectar. Nectar & pollen grains are the floral rewards for pollination.
b. Small flowers form inflorescence to make them visible.
c. The flowers pollinated by flies and beetles secrete foul odours to attract these animals.
d. The pollen grains are generally sticky.
✅ Answer
E = Escherichia (Genus name of bacteria)Co = Coli (Species name of bacteria)
R = Strain
I = First extracted enzyme (Order of extraction).
b. What does K stand for?
✅ Answer
(a) Exponential (Geometric) growth curve and Logistic (Sigmoid) growth curve.(b) Carrying capacity.
Grass – 5,842,000
Wolf – 28
Birds – 215
a. Draw a pyramid of numbers showing various trophic levels.
b. Explain trophic level.
✅ Answer
Insulin consists of two short polypeptide chains such as chain A and chain B. They are linked by disulphide bridges.Two DNA sequences corresponding to A and B, chains of human insulin are prepared and inserted into E. coli plasmids to produce insulin chains. Chains A and B are extracted and combined by creating disulfide bonds to form human insulin.
a. rice pollen b. rice endosperm
✅ Answer
a. Rice Pollen: The rice pollen is a haploid cell, so it contains half the number of chromosomes as the meiocyte. Therefore, it has 12 chromosomes.b. Rice Endosperm: The rice endosperm is a triploid cell, so it contains one and a half times the number of chromosomes as the meiocyte. Therefore, it has 36 chromosomes.
✅ Answer
Treating the cells with specific concentration of a divalent cation such as calcium.Incubating the cells with recombinant DNA on ice. Place them briefly at 42°C (heat shock). Again, place them back on ice.
a. Define GPP and NPP.
b. How can we relate GPP and NPP?
✅ Answer
(a) Gross primary productivity (GPP): It is the rate of production of organic matter during photosynthesis. A considerable amount of GPP is used by plants in respiration.Net primary productivity (NPP): It is the available biomass for the consumption to heterotrophs (herbivores & decomposers).
(b) NPP is the Gross primary productivity minus respiration losses (R).
✅ Answer
The substance is Humus.Peculiarities:
It is highly resistant to microbial action and decomposition.
Serves as a reservoir of nutrients.
Release inorganic nutrients.
✅ Answer
Natality: It is the number of births in a population during a given period.Mortality: It is the number of deaths in a population during a given period.
Natality is indicated by the letter B.
Mortality is indicated by the letter D.
III. Answer any 3 questions from 17 – 20. Each carries 3 scores. (3 x 3 = 9)
b. Explain how parasitism differs from predation.
c. Give the significance of species interaction.
✅ Answer
(a) i. Mutualismii. Competition
iii. Commensalism
iv. Amensalism
(b) Predation: A predator kills and consumes its prey.
Parasitism: A parasite lives off a host, causing harm but not immediate death.
(c) It is essential to form a biological community, for energy flow and nutrient cycling.
✅ Answer
It is a diagrammatic representation of the mature embryo sac.A= Antipodals
B= Polar nuclei
C= Egg
D= Synergids
E= Filiform apparatus
3’ – CTTAAG – 5’
a. Give salient features of this sequence.
b. Write name of enzymes which recognize such sequences.
c. Elaborate importance of this enzyme in Genetic engineering.
✅ Answer
(a) Palindromic sequence(b) EcoRI Restriction endonuclease
(c) This enzyme cuts at specific nucleotide sequence to get sticky ends. The same restriction enzyme is used to cut both foreign DNA and cloning vector. This helps to create recombinant DNA.
PART B: ZOOLOGY
I. Answer any 3 questions from 1 – 5. Each carries 1 score. (3 x 1 = 3)
a. Gastrula b. Morula
c. Zygote d. Blastocyst
✅ d. Blastocyst
a. AAA b. AUG
c. GUA d. UGA
✅ b. AUG
a. HIV b. Haemophilia
c. Genital warts d. Hepatitis B
✅ Haemophilia. It is a genetic disorder. Others are STIs.
✅ Paul Ehrlich.
a. MALT b. NACO
✅ a. Mucosa Associated Lymphoid Tissues.
b. National AIDS Control Organisation.
II. Answer any 9 questions from 6 – 16. Each carries 2 scores. (9 x 2 = 18)
a) Expand HCG.
b) What does the presence of HCG indicate?
c) Which is the source of HCG?
✅ Answer
a) Human Chorionic Gonadotropinb) She is pregnant.
c) Placenta.
a. Suggest the contraceptive action of CuT.
b. Name the hormone releasing IUDs.
✅ Answer
a) Cu ions suppress motility and fertilising capacity of sperms.b) Progestasert, LNG-20.
a. Name the test used.
b. Procedure of the test is given below. Complete it.
i. Isolation of DNA
ii. DNA is cut using restriction endonuclease
iii. …………………
iv. …………………
v. Hybridisation using VNTR probe
vi. ………………….
✅ Answer
a) DNA fingerprinting.b) iii. Separation of DNA fragments by gel electrophoresis.
iv. Transfer of separated DNA fragments into nitrocellulose or nylon membrane.
vi. Detection of DNA using autoradiography.
b. How many histone molecules are present in the histone core?
c. Distinguish Euchromatin and Heterochromatin.
✅ Answer
a) Nucleosome.b) 8
c) Euchromatin: Loosely packed, light stained and transcriptionally active chromatin.
Heterochromatin: Densely packed, dark stained and transcriptionally inactive chromatin.
✅ Answer
Darwinian variations: These are minor, slow and directional. It results in gradual evolution.Mutational variations: These are sudden, random and directional. Speciation is by saltation.
a. Double helical model of DNA was proposed by Jacob and Monod.
b. Sugar present in RNA is Ribose.
c. Introns are the coding sequences of a eukaryotic gene.
d. DNA replication occurs by semi-conservative method.
✅ Answer
a) False. Double helical model of DNA was proposed by Watson and Crick.b) True.
c) False. Exons are the coding sequences of a eukaryotic gene.
d) True.
b. State that theory and name the scientists who proposed it.
✅ Answer
a) Theory of Chemical Evolution.b) It states that the first form of life was originated from non-living inorganic & organic molecules.
This theory was proposed by Oparin and Haldane.
“Allele frequencies in a population are stable and is constant from generation to generation.”
a. Name the principle mentioned here.
b. Mention any three factors that affect the equilibrium.
✅ Answer
a) Hardy-Weinberg Principle.b) Gene migration, genetic drift, Mutation, Natural selection etc.
a. How do the bacteria respond to the above situation at genetic level?
b. If lactose is removed from the medium, what will happen?
✅ Answer
a) In E. coli, lac operon will be switched on. Lactose (inducer) binds with repressor protein. So repressor protein cannot bind to operator region. It induces the RNA polymerase to bind with promoter. Then transcription starts.b) If lactose is removed, lac operon remains switched off. The regulator gene synthesizes mRNA to produce repressor protein. This protein binds to the operator region and blocks RNA polymerase movement. So the structural genes are not expressed.
(b) A person has earlier symptoms of STDs. What will happen if he/she does not consult a doctor? (Mention any two consequences).
✅ Answer
a) Sexually Transmitted Diseases.b) Pelvic Inflammatory Diseases (PID), Infertility, Ectopic Pregnancy, abortions, still births, cancer of the reproductive tract etc.
III. Answer any 3 questions from 17 – 20. Each carries 3 scores. (3 x 3 = 9)
a. What are these?
b. Mention any 4 ill-effects of alcoholism.
✅ Answer
a) Cannabinoids are a group of drugs obtained from Cannabis sativa. E.g., marijuana, hashish, ganja.b) • Reckless behaviour, vandalism and violence.
• Coma and death.
• Damage of nervous system and liver cirrhosis.
• Mental and social distress to family and friends.
b. Name the enzyme involved in this process.
c. Explain the three major steps in this process.
✅ Answer
a) Transcription.b) RNA polymerase.
c) Initiation, Elongation, Termination.
• Initiation: Here, RNA polymerase binds at the promoter site of DNA. An initiation factor (σ factor) present in RNA polymerase initiates the RNA synthesis.
• Elongation: RNA chain is synthesized in 5’-3’ direction. Ribonucleoside triphosphates are added.
• Termination: A termination factor (ρ factor) binds to the RNA polymerase and terminates the transcription.
✅ Answer
Three arguments of the reasons for biodiversity conservation:a. Narrowly utilitarian arguments: Human derive economic benefits from nature.
b. Broadly utilitarian arguments: Biodiversity has ecosystem services. E.g. production of O2, Pollination, Aesthetic pleasures.
c. Ethical arguments: Every species has an intrinsic value. We have to care them.
(b) Categorise the given birth control methods into two groups under proper headings.