Category: IIT JAM Biotechnology Previous Year Question Paper 2017 Video Solution

Category: IIT JAM Biotechnology Previous Year Question Paper 2017 Video Solution

IIT JAM Biotechnology Previous Year Question Paper 2017 – Video & Text Solutions (Complete Analysis)

Preparing for IIT JAM Biotechnology (BT) without solving previous year question papers is one of the biggest strategic mistakes aspirants make. Among all PYQs, the IIT JAM Biotechnology 2017 question paper holds special importance because it clearly reflects the shift towards numerical analysis, application-based biology, and biotechnology techniques.

In this article, Let’s Talk Academy provides a complete guide to IIT JAM BT 2017 PYQ, including video solutions, text-based explanations, exam relevance, and preparation strategy.

About IIT JAM Biotechnology Exam

The IIT JAM (Joint Admission Test for M.Sc.) is a national-level entrance examination conducted by IITs for admission into postgraduate science programs. The Biotechnology (BT) paper evaluates a candidate’s understanding of core life science subjects combined with biotechnology applications.

Why IIT JAM Biotechnology PYQ 2017 Is Extremely Important?

The 2017 Biotechnology paper is considered a benchmark paper because:

  • NAT questions became more calculation-intensive

  • Increased focus on biochemical numericals

  • Strong weightage to molecular biology and genetics

  • Technique-based questions tested conceptual clarity, not memory

For this reason, IIT JAM BT 2017 video solutions are highly searched and extremely valuable for aspirants.

IIT JAM Biotechnology 2017 Exam Pattern (PYQ Based)

Section Question Type Nature
Section A MCQs Concept + logic
Section B MSQs Multiple correct
Section C NAT Numerical & analytical

📌 Section C alone decided rank in IIT JAM 2017, as observed from toppers’ analysis.

Subject-wise Breakdown of IIT JAM Biotechnology 2017 PYQs

1. Molecular Biology & Genetics

  • Replication and transcription mechanisms

  • Gene regulation models

  • Mendelian ratios with logic

  • Mutation and repair concepts

✔️ Questions were indirect but conceptually deep, making video explanations crucial.

2. Biochemistry

  • Enzyme kinetics numericals

  • Protein structure-based logic

  • Metabolic pathway calculations

✔️ One of the highest scoring sections if PYQs are practiced properly.

3. Cell Biology

  • Cell cycle regulation

  • Membrane transport

  • Signaling pathways

4. Biotechnology & Techniques

  • PCR efficiency-based questions

  • Cloning and vectors

  • Experimental interpretation

✔️ These questions cannot be solved by memorization—video solutions help visualize the logic.

IIT JAM Biotechnology 2017 Video Solution – Why It Matters

Many students read answers but fail to understand the thinking process. This is where IIT JAM BT 2017 video solutions become essential.

Benefits of Video Solutions:

  • Step-by-step problem-solving approach

  • Visualization of numericals and techniques

  • Concept linking with syllabus

  • Common mistake identification

At Let’s Talk Academy, video solutions are designed to teach the approach, not just the answer.

IIT JAM Biotechnology 2017 Text (Written) Solutions – When to Use Them?

Text solutions are useful when:

  • Revising concepts quickly

  • Practicing again before the exam

  • Making short notes

  • Cross-checking logic

👉 Best practice: Use video solution first, then revise with text solution.

How to Prepare IIT JAM Using 2017 PYQs? (Expert Strategy)

Step 1: Attempt the Paper as a Mock Test

Solve the IIT JAM BT 2017 paper in 3 hours without help.

Step 2: Analyze with Video Solutions

Watch topic-wise video solutions and note:

  • Concept used

  • Formula applied

  • Reason for wrong options

Step 3: Convert PYQs into Concepts

Each question should become a revision point, not just a solved problem.

Common Errors Students Make with IIT JAM 2017 PYQs

  • Watching video solutions without attempting first

  • Ignoring numerical derivations

  • Skipping biotechnology technique questions

  • Not revising solved PYQs

Let’s Talk Academy addresses these gaps with structured PYQ discussion classes.

Why Choose Let’s Talk Academy for IIT JAM Biotechnology PYQ Solutions?

Let’s Talk Academy is trusted by Life Science aspirants for its PYQ-focused, result-driven teaching approach.

What You Get:

  • IIT JAM Biotechnology 2017–latest PYQ video solutions

  • Detailed text-based explanations

  • Concept mapping from PYQs

  • Exam-level mock tests

  • Guidance aligned with CSIR NET + IIT JAM synergy

Final Conclusion

The IIT JAM Biotechnology previous year question paper 2017 is not just a practice set—it is a conceptual blueprint of how IIT JAM frames Biotechnology questions. When combined with expert video solutions and structured text explanations, it becomes one of the most powerful tools for rank improvement.

If your goal is IIT JAM Biotechnology selection, start with PYQs like 2017, analyze them deeply, and prepare smartly with Let’s Talk Academy.

Q.60 In a mass spectrometer, a deuteron with kinetic energy 17 MeV enters a uniform magnetic field of 2.4 T with its velocity perpendicular to the field. The deuteron moves in a circular path in the magnetic field. The radius of its path in the magnetic field (correct to two decimal places) is _______ cm. [mass of deuteron is 3.34 × 10⁻²⁷ kg, 1 MeV = 1.6 × 10⁻¹³ J and e = 1.6 × 10⁻¹⁹ C]

Deuteron Radius in Mass Spectrometer

Q.59 An X-ray tube operates at 30 kV. If one electron converts 10% of its energy into a photon at first collision, then the wavelength of the photon (correct to two decimal places) is _______ Å. [h = 4.14 × 10⁻¹⁵ eV·s, c = 3 × 10⁸ m s⁻¹ and e = 1.6 × 10⁻¹⁹ C]

X-Ray Tube Wavelength Calculation

Q.58 Using the letters in the word TRICK a new word containing five distinct letters is formed such that T appears in the middle. The number of distinct arrangements is _______.

Using word TRICK has five distinct letters

Q.57 If a and b are unit vectors and the angle between them is π/3, then the magnitude of a − b is _______.

Magnitude of a – b for Unit Vectors at π/3 Angle

Q.56 In acidic solution, permanganate ion is reduced by ferrous ion. The number of electrons involved in the reduction of permanganate ion is _______.

Permanganate Ion Reduction by Ferrous Ion in Acidic Solution

Q.55 In 2 N H₂SO₄, an organic compound shows fluorescence with quantum yield, ϕ = 0.42 and fluorescence rate constant, kf = 5.25 × 10⁷ s⁻¹. The observed fluorescence lifetime of it under the same conditions (correct to 1 decimal place) is _______ ns.

Fluorescence Lifetime Calculation in 2N H₂SO₄

Q.54 In bacteria, a ribosome synthesizes a protein containing 300 amino acids from mRNA in 20 seconds. If the average lifetime of mRNA is 2 minutes, the number of ribosomes that can translate a single mRNA containing 1350 nucleotides is _______.

How Many Ribosomes Translate Single Bacterial mRNA

Q.53 The relationship between log(MW) [where MW – molecular weight in kDa] of a mixture of protein standards and their retention factors (Rf) obtained from native-PAGE is log₁₀(MW) = −2Rf + 3. If the measured retention factor for a protein with 180 amino acids is 0.5, then the number of identical monomers in the protein is _______.

Calculating Protein Oligomer Number from Native-PAGE Rf and Amino Acids

Q.52 The Kₘ and kcat of an enzyme are 4 mM and 0.1 μM h⁻¹ respectively. In the presence of 1.5 mM inhibitor, the Kₘ and kcat of the enzyme are 6 mM and 0.1 μM h⁻¹, respectively. The value of inhibition constant, Ki (correct to 1 decimal place) is _______ mM.

Enzyme Inhibition Ki Calculation

Q.51 The standard oxidation potentials for oxidation of NADH and H₂O are +0.315 V and −0.815 V, respectively. The standard free energy for oxidation of 1 mole of NADH by oxygen under standard conditions (correct to 1 decimal place) is _______ kJ. [Faraday constant is 96500 C mol⁻¹]

Standard Free Energy Change for NADH Oxidation by Oxygen

Q.50 Two sources P and Q produce electromagnetic waves with wavelengths Z and 2Z, respectively. Source P ejects a photon with a maximum kinetic energy of 4.0 eV from a metal with work function 2.0 eV. The maximum kinetic energy (eV) of a photon ejected by source Q from the same metal is ________.

Photoelectric Effect

Q.49 One gram of radioactive nuclei with a half‑life of 300 days is kept in an open container. The weight of nuclei remaining after 900 days (correct to 1 decimal place) is ________ mg.

Radioactive Decay Half-Life Calculation

Q.48 ∫₀¹ z dx + ∫₁² (2 − x) dx = ________.

How to Evaluate ∫01 2x dx + ∫12 (2 − x) dx

Q.47 The positive root of the equation x³ + x² − 2 = 0 is ________.

Positive Root of x³ + x² – 2 = 0

Q.46 The [H⁺] of 0.1 N acetic acid solution is 1.33 × 10⁻³. The pH of the solution (correct to two decimal places) is ________.

Calculate pH of 0.1 N Acetic Acid with [H⁺] 1.33 × 10⁻³ M

Q45. Total number of singlets observed in the H NMR of the following compound is

Total Number of Singlets in the 11H NMR of 2‑Methoxy‑3‑Hydroxybenzaldehyde Explained

Q.44 A linear DNA contains five restriction sites for EcoRI and three restriction sites for BamHI. The number of fragments that will be generated after digestion with EcoRI is ______.

 Linear DNA EcoRI Restriction Sites

Q.43 The Kₘ and Vmax of lactate dehydrogenase for conversion of pyruvate to lactate are 1 mM and 5 pM s⁻¹, respectively. At 0.25 mM pyruvate, the velocity of the reaction catalyzed by lactate dehydrogenase is ______ pM s⁻¹.

Lactate Dehydrogenase Km and Vmax Problem

Q.42 The population of a bacterial culture increases from one thousand to one billion in five hours. The doubling time of the culture (correct to 1 decimal place) is ______ min. ​

Bacterial Culture Doubling Time

Q.41 Antigen and antibody interaction is shown by the following scheme Ab + Ag ⇌ Ab–Ag Where Ab is antibody, Ag is antigen and Ab–Ag is antigen–antibody complex. The values of k₁ and k₋₁ are 5 × 10⁸ pM⁻¹ s⁻¹ and 2 × 10⁷ s⁻¹, respectively. The dissociation constant for the complex is ______ pM.

Antigen–antibody interaction

Q.40 The electric field and capacitance of a capacitor in the absence of dielectric material are E and C, respectively. When the capacitor is filled with a dielectric material, the electric field and capacitance of the capacitor becomes E′ and C′, respectively. Which of the following is (are) correct? (A) E′ > E and C′ = C (B) E′ < E and C′ > C (C) E′ = E and C′ > C (D) E′ > E and C′ < C

Effect of Dielectric on Capacitor Electric Field and Capacitance

Q.39 Kinetic theory of an ideal gas is based upon the following assumption(s): (A) Gases are made of molecules with negligible volume (B) The gaseous molecules do not possess kinetic energy (C) The molecules are in constant random motion (D) Intermolecular forces of attraction are negligible ​

 Kinetic Theory of Ideal Gas Assumptions

Q.38 Which of the following point(s) lies (lie) on the plane 2x + 3y + z = 6? (A) (0, 0, 6) (B) (0, 2, 0) (C) (1, 1, 1) (D) (3, 0, 0)

Points on Plane 2x + 3y + z = 6

Q.37 Fehling’s solution (A) contains a copper complex of tartaric acid (B) forms a brick‑red precipitate with glucose (C) gives a purple precipitate with aldehydes (D) is used as a test reagent for reducing sugars

Fehling’s Solution

Q.36 The elements with atomic numbers 19, 37 and 55 (A) form cubic chloride salts with the coordination number of cation being 6 (B) form ionic fluorides with general formula MF (C) have lowest density of solids in their respective periods (D) have lowest ionization energy in their respective periods

Elements Atomic Numbers 19, 37, 55

Q.35 Bacterial plasmid genes of non‑chromosomal origin are associated with (A) providing resistance against antibacterial agents (B) the degradation of toxic materials (C) the production of certain toxins (D) the transfer of genetic material from one cell to another cell

Bacterial Plasmid Genes Non-Chromosomal Origin

Q.34 The ΔG°′ for homolactic fermentation converting glucose to lactate is −196 kJ mol⁻¹. If ΔG°′ for the formation of ATP is +30.5 kJ mol⁻¹, then (A) homolactic fermentation is 31% energy efficient (B) the efficiency of energy conservation is 69% (C) the energy stored in the form of ATP is 31% (D) the process results in the loss of 31% of energy

Homolactic Fermentation Energy Efficiency

Q.33 Which of the following organs are correctly paired with their function? (A) Large intestine — Protein digestion (B) Oral cavity — Starch digestion (C) Pancreas — Bile production (D) Small intestine — Fat digestion

Organs Correctly Paired with Digestion Functions

Q.32 Which of the following is (are) INCORRECT in the regulation of the trp operon? (A) It is an example of a negatively controlled repressible operon. (B) The amino acid Trp inactivates the repressor. (C) The amino acid Trp induces the operon. (D) The repressor binds to the operator in the presence of amino acid Trp.

Trp Operon Regulation

Q.31 The cells involved in allergic reactions and containing surface receptors of IgE antibodies and histamine are (A) Basophils (B) Mast cells (C) Monocytes (D) Neutrophils

Cells Involved in Allergic Reactions with IgE Receptors and Histamine

Q.30 A body XX of mass MM moving with velocity vv hits a stationary body YY of mass mm. If M≫mM≫m and XX moves with the velocity v′v′, then the velocity of YY after an elastic collision is (A) 2v (B) v+v′ (C) v−v′ (D) 2v′

Elastic Collision Heavy Mass M >> m

Q.29 Two inductors P P and Q Q having inductance ratio 1 : 2 1:2 are connected in parallel in an electric circuit. The energy stored in the inductors P P and Q Q are in the ratio (A) 1 2 : 4 2 1 :4 (B) 1 : 2 1:2 (C) 2 : 1 2:1 (D) 4 : 1 4:1

Energy Stored in Parallel Inductors Ratio 1:2 Explained

Dimensions of P in Particle Position Equation y=Pt+Q

Q.27 If P = { 1 , 2 , − 1 , 3 } P={1,2,−1,3}, Q = { 0 , 4 , 1 , 3 } Q={0,4,1,3} and R = { 1 , 6 , 7 } R={1,6,7}, then P ∖ ( Q ∪ R ) = P∖(Q∪R)= (A) { 1 , 2 } {1,2} (B) { 4 , 3 } {4,3} (C) { 2 , 1 } {2,1} (D) { 2 , 3 } {2,3}

Solving P ∖ (Q ∪ R) Set Difference

Q.26 If one of the diameters of a circle has end points ( 2 , 0 ) (2,0) and ( 4 , 0 ) (4,0), then the equation of that circle is (A) x 2 − 3 x + y 2 + 5 = 0 x 2 −3x+y 2 +5=0 (B) x 2 − 4 x + y 2 + 6 = 0 x 2 −4x+y 2 +6=0 (C) x 2 − 5 x + y 2 + 7 = 0 x 2 −5x+y 2 +7=0 (D) x 2 − 6 x + y 2 + 8 = 0 x 2 −6x+y 2 +8=0

Equation of Circle with Diameter Endpoints (2,0) and (4,0) – MCQ Solved

Q.25 Two dice are thrown simultaneously. The probability that the sum of the numbers obtained is divisible by 7 is (A) 1/t6 (B) "36 (C) 0 (D) 1/18

Probability Two Dice Sum Divisible by 7

Q.24 If u ( x ) u(x) and r ( x ) r(x) are differentiable at x = 0 x=0, and if u ( 0 ) = 5 u(0)=5, u ′ ( 0 ) = − 3 u ′ (0)=−3, r ( 0 ) = − 1 r(0)=−1 and r ′ ( 0 ) = 2 r ′ (0)=2, then the value of − g ( u r + t 2 ) −g(ur+t 2 ) at x = 0 x=0 is (A) − 20 −20 (B) − 7 −7 (C) 6 6 (D) 13 13

How to Find the Value of −(ur+u)′ at x=0 When u(x) and r(x) Are Differentiable

Q.23 In metal–carbonyl complexes, the π-back bonding is (A) p π – d π pπ–dπ type (B) d π – d π dπ–dπ type (C) d π – p π dπ–pπ type (D) d π – s π dπ–sπ type

π-Back Bonding in Metal-Carbonyl Complexes

Q.22 For an autocatalytic second order reaction R → P, the rate law is [where v is rate of the reaction and k is the rate constant] (A) v = k [R] (B) v = 1 [R][ ] 1 (C) v = k [R]² (D) v = k [P]²

Autocatalytic Second Order Reaction R→P

Q21. The reaction that produces o-bromophenol as the major product is

Reaction that Produces o‑Bromophenol as Major Product

Q20. The major product formed in the following reaction is

Major Product in Acid‑Catalyzed Cyclization of γ‑Hydroxycarboxylic Acid

Q.19 The correct sequence of the following events in the human female reproductive cycle is I: Secretion of FSH II: Growth of corpus luteum III: Growth of follicle and oogenesis IV: Ovulation V: Sudden increase in the levels of LH (A) I, II, IV, V, III (B) II, I, III, IV, V (C) I, III, V, IV, II (D) I, V, III, IV, II

Human Female Reproductive Cycle Sequence

Q.18 During eukaryotic cell division, the amount of DNA doubles (A) between prophase and anaphase of mitosis (B) between prophase I and prophase II of meiosis (C) between the G1 and G2 phases of the cell cycle (D) during the M phase of the cell cycle

When Does DNA Double in Eukaryotic Cell Division?

Nitrogenase of Diazotrophs

Q.16 Match the entries in Group I with that in Group II Group I P) Nucleolus Q) Spherosomes R) Peroxisomes S) Plasmodesmata Group II lipid storage breakdown of fatty acids transport of macromolecules RNA synthesis (A) P‑4, Q‑3, R‑1, S‑2 (B) P‑4, Q‑1, R‑2, S‑3 (C) P‑2, Q‑1, R‑3, S‑4 (D) P‑1, Q‑3, R‑4, S‑2

Cell Organelles and Their Functions

Q.15 Which of the following is NOT an example of an adaptive defense mechanism against predation? (A) Bright colors of bird‑pollinated flower (B) Insect that resembles a stick (C) Nicotine in the tobacco plant (D) Spines on porcupine

Adaptive Defense Mechanisms Against Predation

Q.14 The natural geographical distribution of kangaroos is limited to the Australian continent because (A) abiotic factors determine the distribution (B) dispersal is limited by accessibility to other continents (C) kangaroos have not selected habitats in other continents (D) predators limit their distribution in other continents

Why Kangaroos Natural Geographical Distribution Limited to Australian Continent

Q.13 If an aldol cleavage of glucose-6-phosphate occurs in glycolysis, it will result in (A) Products of equal carbon chain length (B) Products of unequal carbon chain length (C) Removal of phosphate group (D) Three C₃ compounds

Aldol Cleavage of Glucose-6-Phosphate in Glycolysis

Q.12 Match the entries in Group I with that in Group II Group I Pectinase Xylanase Cellulase Enzymes Group II P) In paper and pulp processing Q) Delignification R) Gluten complex reduction S) Improve mineral availability T) Phosphorolization (A) P-4, Q-1, R-3, S-5 (B) P-4, Q-1, R-2, S-3 (C) P-1, Q-3, R-4, S-2 (D) P-5, Q-1, R-2, S-3

Industrial applications of phytase, xylanase, laccase and bromelain

Q.11 Which of the following is true of protein synthesis ONLY in prokaryotes? (A) Translation and transcription are coupled (B) The codon AUG codes for the start signal (C) The tRNA anticodon can bind to two or more different codons (D) The functional ribosomes contain two subunits constructed of proteins and RNA

 Protein Synthesis Only in Prokaryotes

Q.10 The refractive index of diamond is 2.419. If the speed of light in vacuum is 3 · 10⁸ m s⁻¹, then the speed of light in diamond is (A) 1.240 · 10⁸ m s⁻¹ (B) 1.352 · 10⁸ m s⁻¹ (C) 1.521 · 10⁸ m s⁻¹ (D) 2.433 · 10⁸ m s⁻¹

Speed of Light in Diamond with Refractive Index 2.419 Calculation

Q.9 The average energy of a diatomic gaseous molecule at temperature T is kBT where kB is Boltzmann’s constant. The average energy of this molecule per degree of freedom is (A) ½ kBT (B) 2⁄3 kBT (C) kBT (D) 2⁄5 kBT

Average Energy of Diatomic Gaseous Molecule per Degree of Freedom Explained

Q.8 x y 0 0 0 If pq R = 0 0 0 , then the order of R is u v 0 0 i (A) 2 · 3 (B) 3 · 2 (C) 2 2 (D) 3 · 3

Order of a Relation Matrix

Q.7 cos(8(x + y)) = (A) cos8(x) cos8(y) − sin(x) sin(y) (B) cos8(x) cos8(3x) + sin(x) sin(y) (C) cos8(x) sin(3x) = sin(3x) cos8(x) (D) cos8(x) sin(3x) + sin(3x) cos8(x)

Solve cos(8(x + y)) Trigonometric Identity

Q.6 The result of an electrophoretic separation of a mixture of amino acids X, Y and Z at pH = 5.0 is represented as (Given the isoelectric points of X, Y and Z are 9.87, 3.22 and 5.43, respectively.)

Electrophoretic Separation of Amino Acids X, Y and Z at pH 5.0

Q.5 Among CH4, H2O, NH3 and PH3, the molecule having the smallest percent s character for the covalent bond (X—H) between the central element (X = C, O, N or P) and hydrogen is (A) CH4 (B) H2 (C) NH3 (D) PH3

Smallest Percent S-Character in CH4 H2O NH3 PH3

Q.4 The role of enzyme E synthesized by phage X174 during host infection is to (A) block peptidoglycan synthesis (B) enhance synthesis of viral +RNA (C) inhibit lipid metabolism (D) stimulate dsDNA replication

Role of Enzyme E in Phage φX174 Host Infection

Q.3 Which of the following statements is correct? (A) Gram negative bacteria are colored purple after Gram staining (B) Gram negative bacteria are commonly more resistant to antibiotics than Gram positive bacteria (C) Gram negative bacteria cell wall consists of a thick layer of peptidoglycan outside the plasma membrane (D) Cell wall of Gram negative bacteria does not contain an outer membrane

Gram Negative Bacteria

Q2. Which of the following is NOT involved in eukaryotic translation? (A) Ribosome             (B) Spliceosome         (C) mRNA       (D) tRNA

Eukaryotic Translation Components

The antigen binding site of an antibody is present? (A)at the constant region (B) at the C-terminal (C) at the variable region                              (D) between the constant and the variable region

Antigen Binding Site of Antibody

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