GATE 2024 Biotechnology Question Paper with Answer Key PDF- Download Here

| Updated On - Oct 14, 2024

GATE 2024 Biotechnology Question Paper PDF is available here. IISc Banglore conducted GATE 2024 Biotechnology exam on February 10 in the Forenoon Session from 9:30 AM to 12:30 PM. Students have to answer 65 questions in GATE 2024 Biotechnology Question Paper carrying a total weightage of 100 marks. 10 questions are from the General Aptitude section and 55 questions are from Engineering Mathematics and Core Discipline.

GATE 2024 Biotechnology Question Paper with Answer Key PDF

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Serial No.	Question	Answer Key	Solution
1	If ‘→’ denotes increasing order of intensity, then the meaning of the words [dry → arid → parched] is analogous to [diet → fast → ________ ]. Which one of the given options is appropriate to fill the blank?	(A) starve	To solve this analogy, let's analyze the progression of intensity in both sequences: For "dry → arid → parched," these words indicate increasing levels of dryness. For "diet → fast → ________," "diet" involves controlled intake of food. "Fast" involves abstaining from food. Thus, the word that fits the pattern of increasing intensity is "starve," representing extreme food deprivation. Therefore, the appropriate word is (A) starve.
2	If two distinct non-zero real variables x and y are such that (x + y) is proportional to (x − y) then the value of x y	(D) is a constant	If (x + y) is proportional to (x − y), it can be expressed as x + y = k(x − y). Solving this leads to a relation where x and y will exhibit a constant product, confirming that xy is a constant.
3	Consider the following sample of numbers: 9, 18, 11, 14, 15, 17, 10, 69, 11, 13. The median of the sample is.	(A) 13.5	Arranging the numbers in ascending order: 9, 10, 11, 11, 13, 14, 15, 17, 18, 69. Since there are 10 numbers, the median is the average of the 5th and 6th numbers: (13 + 14) / 2 = 27 / 2 = 13.5. Thus, the median is 13.5.
4	The number of coins of ₹1, ₹5, and ₹10 denominations that a person has are in the ratio 5:3:13. Of the total amount, the percentage of money in ₹5 coins is.	(C) 10%	Let the number of coins be 5x, 3x, and 13x. Total coins = 21x. Total amount = 1×5x + 5×3x + 10×13x = 150x. Money in ₹5 coins = 15x. Percentage of money in ₹5 coins = (15x / 150x) × 100% = 10%. Thus, the percentage of money in ₹5 coins is 10%.
5	For positive non-zero real variables p and q, if log (p 2 + q 2 ) = log p + log q + 2 log 3 , then, the value of (p⁴+q⁴) /p2q 2 is	(B) 81	Using the equation log(p² + q²) = log(3²)pq, we can simplify to find that p⁴ + q⁴ / p²q² = 81. This follows from manipulating the logarithmic equation to derive a quadratic in terms of p and q.
6	In the given text, the blanks are numbered (i)−(iv). Select the best match for all the blanks. Steve was advised to keep his head (i) before heading (ii) to bat; for, while he had a head (iii) batting, he could only do so with a cool head (iv) his shoulders.	(A) (i) down (ii) down (iii) on (iv) for	The best fit for the blanks is: (i) down (ii) down (iii) on (iv) for, making the sentence coherent.
7	A rectangular paper sheet of dimensions 54 cm × 4 cm is taken. The two longer edges of the sheet are joined together to create a cylindrical tube. A cube whose surface area is equal to the area of the sheet is also taken. Then, the ratio of the volume of the cylindrical tube to the volume of the cube is.	(B) 2/π	The volume of the cylinder (radius = 2 cm, height = 54 cm) is V = πr²h = π(2²)(54) = 432π cm³. The cube has a surface area of 432 cm², so side = 6 cm, volume = 6³ = 216 cm³. Ratio = V_cylinder / V_cube = (432π / 216) = 2/π.
9	A rectangular paper of 20 cm × 8 cm is folded 3 times. Each fold is made along the line of symmetry, which is perpendicular to its long edge. The perimeter of the final folded sheet (in cm) is.	(A) 18	After folding the sheet 3 times, the dimensions are reduced, leading to a final perimeter of 18 cm. The folding process significantly reduces the overall dimensions of the rectangle.
11	In adsorption chromatography, the adsorption of uncharged solute molecules onto a silica-based stationary phase is by ________.	(D) van der Waals forces	In adsorption chromatography, the adsorption occurs primarily through weak intermolecular forces known as van der Waals forces, particularly with a silica-based stationary phase.
12	The transfer function of a process is G(s) = Kp /(Tp S+1), where Kp is the gain and tp is the time constant. This is a ______ process.	(A) first order	The transfer function G(s) = Kpτps + 1 is characteristic of a first-order process, defined by a single time constant and no oscillatory behavior.
13	Which one of the following statements is correct in the context of thermodynamics?	(D) In a closed system, only energy can be transferred across the system boundary and not mass	In a closed system, mass cannot cross the boundary, but energy can be transferred in the form of heat or work, making this statement true.
14	Which one of the following statements is correct about Reynolds Number (NRe) in a stirred tank bioreactor?	(D) In turbulent flow, mixing time is independent of NRe	In turbulent flow, mixing time becomes less dependent on the Reynolds number (NRe) as turbulence ensures efficient mixing throughout the bioreactor, making this statement correct.
15	The relationship that involves the exchange of nutrients between two different species for their mutual growth is called ________.	(D) syntrophism	Syntrophism describes the mutualistic relationship involving nutrient exchange between two species, allowing them to grow together.
16	Mendel’s ‘law of segregation’ applies to the segregation of ________ during gamete formation.	(B) alleles of a gene	Mendel’s law states that during gamete formation, the two alleles for a gene separate so that each gamete carries one allele, thus it applies to alleles of a gene.
17	Co-translational translocation of proteins is observed in ________.	(A) endoplasmic reticulum	Co-translational translocation occurs when the translation of a protein and its translocation across a membrane happen simultaneously, observed in the endoplasmic reticulum (ER).
18	2-mercaptoethanol breaks the ________ covalent bond between light and heavy chains of an immunoglobulin molecule.	(D) S-S	2-mercaptoethanol is a reducing agent that breaks disulfide (S-S) bonds connecting light and heavy chains in immunoglobulins.
19	During normal embryonic development of the mice paw, elimination of cells from the inter-digital space is due to ________.	(C) apoptosis	Apoptosis, or programmed cell death, is responsible for the removal of cells in the inter-digital space, allowing for the proper formation of paws during embryonic development.
20	Which one of the following is the chief source of energy for the body?	(B) carbohydrates	Carbohydrates are the body's primary source of energy, as they are readily metabolized to provide glucose for cellular functions.