GATE XL 2023 Question Paper with Answer Key

GATE XL 2023 question paper with answer key: all 122 questions

All 122 questions from the GATE 2023 Life Sciences (XL) paper, typed out in full, with the official final answer key and every diagram: General Aptitude, Chemistry and all five optional sections.

122questions across all sections
65you answer: GA, XL-P and two sections
100marks, 3 hours
30diagrams, as set in the exam

GATE 2023 Life Sciences (XL) was organised by IIT Kanpur. Every candidate answered General Aptitude (GA) and XL-P Chemistry, then chose two of the five optional sections: XL-Q Biochemistry, XL-R Botany, XL-S Microbiology, XL-T Zoology and XL-U Food Technology. This page has all of them, in the order they were set, so you can practise the two you plan to choose and look at the rest. Each question is tagged with its topic. Tap Show answer under any question to check the official answer.

How to use this pageTry each question before you open its answer. Answers come from the organising institute’s final answer key. For MSQ questions every correct option must be chosen; numerical answers are accepted anywhere in the official range.
Take it as a timed examThe same paper is available as a free, timed test with GATE marking and an instant score: GATE XL 2023 in live exam mode (log in to save your score).
Want full-length practice?The GATE-XL Life Sciences Full Test Series gives you full-length mocks on the GATE pattern, with MCQ, MSQ and numerical questions and GATE marking. See the GATE-XL Life Sciences Full Test Series.

Marking scheme

SectionQuestionsMarksYou answer
GA: General Aptitude10 (5 × 1 mark, 5 × 2 marks)15Compulsory
XL-P: Chemistry17 (9 × 1 mark, 8 × 2 marks)25Compulsory
XL-Q to XL-U: five optional sections19 each (8 × 1 mark, 11 × 2 marks)30 eachAny two
Total you answer651003 hours
Negative markingOnly MCQs are penalised: a wrong 1-mark MCQ loses 1/3 mark and a wrong 2-mark MCQ loses 2/3 mark. MSQ and numerical answer questions have no negative marking, and there is no partial credit on MSQ.

Which topics this paper tested

We tagged every question with its GATE XL syllabus topic. The table shows the three biggest topics in each section; each section below opens with a chart of all its topics.

SectionQuestionsMost-tested topics (questions)
GA General Aptitude10Quantitative Aptitude (4), Verbal Aptitude (3), Spatial Aptitude (2)
XL-P Chemistry17Organic Chemistry (4), Chemical Equilibria (2), Biomolecule Chemistry (2)
XL-Q Biochemistry19Biomolecules (6), Metabolism & Bioenergetics (4), Biochemical Techniques (3)
XL-R Botany19Genetics & Genomics (5), Plant Physiology & Biochemistry (4), Plant Anatomy (3)
XL-S Microbiology19Microbial Growth (4), Microbial Metabolism (3), Prokaryotic Cells (3)
XL-T Zoology19Biochemistry & Molecular Biology (6), Developmental Biology (2), Parasitology & Immunology (2)
XL-U Food Technology19Food Chemistry & Nutrition (10), Food Engineering (5), Food Products Technology (2)

Jump to a section

GA: General Aptitude (Q1–10)

10 questions (5 of 1 mark, 5 of 2 marks). Compulsory for every XL candidate.

GATE XL 2023, GA General Aptitude: questions by topic
10 questions, tagged by Lesson Basket.
Quantitative Aptitude40%
Verbal Aptitude30%
Spatial Aptitude20%
Analytical Aptitude10%
Topic tags: Lesson Basket. See also our GATE-XL weightage analysis.
Q1Verbal AptitudeGA Q.1 · MCQ · 1 mark, −1/3 for a wrong answer

The village was nestled in a green spot, _______ the ocean and the hills.

  1. Athrough
  2. Bin
  3. Cat
  4. Dbetween
Show answer
Answer: (D) between
Q2Verbal AptitudeGA Q.2 · MCQ · 1 mark, −1/3 for a wrong answer

Disagree : Protest : : Agree : _______ (By word meaning)

  1. ARefuse
  2. BPretext
  3. CRecommend
  4. DRefute
Show answer
Answer: (C) Recommend
Q3Quantitative AptitudeGA Q.3 · MCQ · 1 mark, −1/3 for a wrong answer

A ‘frabjous’ number is defined as a 3 digit number with all digits odd, and no two adjacent digits being the same. For example, 137 is a frabjous number, while 133 is not. How many such frabjous numbers exist?

  1. A125
  2. B720
  3. C60
  4. D80
Show answer
Answer: (D) 80
Q4Quantitative AptitudeGA Q.4 · MCQ · 1 mark, −1/3 for a wrong answer

Which one among the following statements must be TRUE about the mean and the median of the scores of all candidates appearing for GATE 2023?

  1. AThe median is at least as large as the mean.
  2. BThe mean is at least as large as the median.
  3. CAt most half the candidates have a score that is larger than the median.
  4. DAt most half the candidates have a score that is larger than the mean.
Show answer
Answer: (C) At most half the candidates have a score that is larger than the median.
Q5Spatial AptitudeGA Q.5 · MCQ · 1 mark, −1/3 for a wrong answer

In the given diagram, ovals are marked at different heights (h) of a hill. Which one of the following options P, Q, R, and S depicts the top view of the hill?

GATE XL 2023 Q5 figure: In the given diagram, ovals are marked at different heights (h) of a hill. Which one of the following options
  1. AP
  2. BQ
  3. CR
  4. DS
Show answer
Answer: (B) Q
Q6Analytical AptitudeGA Q.6 · MCQ · 2 marks, −2/3 for a wrong answer

Residency is a famous housing complex with many well-established individuals among its residents. A recent survey conducted among the residents of the complex revealed that all of those residents who are well established in their respective fields happen to be academicians. The survey also revealed that most of these academicians are authors of some best-selling books. Based only on the information provided above, which one of the following statements can be logically inferred with certainty?

  1. ASome residents of the complex who are well established in their fields are also authors of some best-selling books.
  2. BAll academicians residing in the complex are well established in their fields.
  3. CSome authors of best-selling books are residents of the complex who are well established in their fields.
  4. DSome academicians residing in the complex are well established in their fields.
Show answer
Marks to all. The organising institute awarded this question’s marks to every candidate in the final answer key, so it has no single correct answer.
Q7Quantitative AptitudeGA Q.7 · MCQ · 2 marks, −2/3 for a wrong answer

Ankita has to climb 5 stairs starting at the ground, while respecting the following rules:

1. At any stage, Ankita can move either one or two stairs up.
2. At any stage, Ankita cannot move to a lower step.

Let F(N) denote the number of possible ways in which Ankita can reach the Nth stair. For example, F(1) = 1, F(2) = 2, F(3) = 3. The value of F(5) is _______.

  1. A8
  2. B7
  3. C6
  4. D5
Show answer
Answer: (A) 8
Q8Verbal AptitudeGA Q.8 · MCQ · 2 marks, −2/3 for a wrong answer

The information contained in DNA is used to synthesize proteins that are necessary for the functioning of life. DNA is composed of four nucleotides: Adenine (A), Thymine (T), Cytosine (C), and Guanine (G). The information contained in DNA can then be thought of as a sequence of these four nucleotides: A, T, C, and G. DNA has coding and non-coding regions. Coding regions—where the sequence of these nucleotides are read in groups of three to produce individual amino acids—constitute only about 2% of human DNA. For example, the triplet of nucleotides CCG codes for the amino acid glycine, while the triplet GGA codes for the amino acid proline. Multiple amino acids are then assembled to form a protein. Based only on the information provided above, which of the following statements can be logically inferred with certainty? (i) The majority of human DNA has no role in the synthesis of proteins. (ii) The function of about 98% of human DNA is not understood.

  1. Aonly (i)
  2. Bonly (ii)
  3. Cboth (i) and (ii)
  4. Dneither (i) nor (ii)
Show answer
Answer: (D) neither (i) nor (ii)
Q9Quantitative AptitudeGA Q.9 · MCQ · 2 marks, −2/3 for a wrong answer

Which one of the given figures P, Q, R and S represents the graph of the following function?

f(x) = | |x + 2| − |x − 1| |

GATE XL 2023 Q9 figure: Which one of the given figures P, Q, R and S represents the graph of the following function?
  1. AP
  2. BQ
  3. CR
  4. DS
Show answer
Answer: (A) P
Q10Spatial AptitudeGA Q.10 · MCQ · 2 marks, −2/3 for a wrong answer

An opaque cylinder (shown below) is suspended in the path of a parallel beam of light, such that its shadow is cast on a screen oriented perpendicular to the direction of the light beam. The cylinder can be reoriented in any direction within the light beam. Under these conditions, which one of the shadows P, Q, R, and S is NOT possible?

GATE XL 2023 Q10 figure: An opaque cylinder (shown below) is suspended in the path of a parallel beam of light, such that its shadow is
  1. AP
  2. BQ
  3. CR
  4. DS
Show answer
Answer: (D) S

XL-P: Chemistry (Q11–27)

17 questions (9 of 1 mark, 8 of 2 marks). Compulsory for every XL candidate.

GATE XL 2023, XL-P Chemistry: questions by topic
17 questions, tagged by Lesson Basket.
Organic Chemistry23.5%
Chemical Equilibria11.8%
Biomolecule Chemistry11.8%
s, p & d Block Elements11.8%
Thermodynamics11.8%
Atomic Structure & Periodicity11.8%
Structure & Bonding11.8%
Electrochemistry5.9%
Topic tags: Lesson Basket. See also our GATE-XL weightage analysis.
Q11Chemical EquilibriaXL-P Q.1 · MCQ · 1 mark, −1/3 for a wrong answer

Which one among the following mixtures gives a buffer solution in water?

  1. ACH3COOH + CH3COONa
  2. BCH3COOH + NaCl
  3. CNaOH + NaCl
  4. DNaOH + CH3COONa
Show answer
Answer: (A) CH3COOH + CH3COONa
Q12Biomolecule ChemistryXL-P Q.2 · MCQ · 1 mark, −1/3 for a wrong answer

What is the major product formed in the given reaction?

GATE XL 2023 Q12 figure: What is the major product formed in the given reaction?
  1. AGATE XL 2023 Q12, option A
  2. BGATE XL 2023 Q12, option B
  3. CGATE XL 2023 Q12, option C
  4. DGATE XL 2023 Q12, option D
Show answer
Answer: (C)
Q13s, p & d Block ElementsXL-P Q.3 · MCQ · 1 mark, −1/3 for a wrong answer

The CORRECT order of stability of the given metal oxides is

  1. ALiO2 > NaO2 > KO2 > RbO2
  2. BLiO2 < NaO2 < KO2 < RbO2
  3. CLiO2 < NaO2 > KO2 > RbO2
  4. DLiO2 > NaO2 < KO2 < RbO2
Show answer
Answer: (B) LiO2 < NaO2 < KO2 < RbO2
Q14ThermodynamicsXL-P Q.4 · MSQ (one or more correct) · 1 mark, no negative marking

Which of the following is/are CORRECT when two single complementary strands of DNA come together to form a double helix at a given temperature?

(ΔS and ΔH are changes in entropy and enthalpy of the process, respectively.)

  1. AΔS > 0
  2. BΔS < 0
  3. CΔH > 0
  4. DΔH < 0
Show answer
Answer (select all of these): (B) ΔS < 0 and (D) ΔH < 0
Q15Organic ChemistryXL-P Q.5 · MSQ (one or more correct) · 1 mark, no negative marking

Suitable reagent(s) to bring about the conversion of P to Q in good yield is/are

GATE XL 2023 Q15 figure: Suitable reagent(s) to bring about the conversion of P to Q in good yield is/are
  1. AGATE XL 2023 Q15, option A
  2. BGATE XL 2023 Q15, option B
  3. CGATE XL 2023 Q15, option C
  4. DGATE XL 2023 Q15, option D
Show answer
Answer (select all of these): (B) and (D)
Q16Atomic Structure & PeriodicityXL-P Q.6 · MSQ (one or more correct) · 1 mark, no negative marking

Choose the CORRECT trend(s) of the first ionization energies among the following. (Given: Atomic numbers C: 6; N: 7; O: 8; F: 9; Si: 14; P: 15; S: 16; Cl: 17)

  1. AC < N > O < F
  2. BSi < P > S < Cl
  3. CC < N < O < F
  4. DSi < P < S < Cl
Show answer
Answer (select all of these): (A) C < N > O < F and (B) Si < P > S < Cl
Q17Chemical EquilibriaXL-P Q.7 · Numerical answer · 1 mark, no negative marking

The depression of freezing point of water (in K) for 0.1 molal solutions of NaCl and Na2SO4 are ΔT1 and ΔT2, respectively. Assuming the solutions to be ideal, the ratio ΔT1/ΔT2 is _____________ (rounded off to two decimal places).

Type a number as your answer.

Show answer
Answer: any value from 0.66 to 0.68 (numerical answer, as in the official key).
Q18Organic ChemistryXL-P Q.8 · Numerical answer · 1 mark, no negative marking

Considering cyclobutane to be planar, the number of planes of symmetry in the following compound is _____________ (in integer).

GATE XL 2023 Q18 figure: Considering cyclobutane to be planar, the number of planes of symmetry in the following compound is __________

Type a number as your answer.

Show answer
Answer: 1 (numerical answer, as in the official key).
Q19Structure & BondingXL-P Q.9 · Numerical answer · 1 mark, no negative marking

The dipole moment (μ) of BrF is 1.42 D and the bond length is 176 pm. The atomic charge distribution (q) in the molecule is _____ (rounded off to two decimal places).

(Given: 1 D = 3.34 × 10−30 C m; the factor e (electronic charge) = 1.60 × 10−19 C)

Type a number as your answer.

Show answer
Answer: any value from 0.16 to 0.18 (numerical answer, as in the official key).
Q20ThermodynamicsXL-P Q.10 · MCQ · 2 marks, −2/3 for a wrong answer

Consider two different paths in which the volume of an ideal gas doubles isothermally:

i) Reversible expansion (work done = wrev)
ii) Irreversible expansion, with the external pressure equal to the final pressure of the gas (work done = wirrev)

Here, wrev/wirrev = ________________.

  1. A2 ln 2
  2. B½ ln 2
  3. C½ ln ½
  4. D2 ln ½
Show answer
Answer: (A) 2 ln 2
Q21Biomolecule ChemistryXL-P Q.11 · MCQ · 2 marks, −2/3 for a wrong answer

A mixture of four peptides, PKKRK, RGERV, RYRGV and LVVYP, is loaded onto an ion-exchange column at pH = 7.2. If carboxymethyl (CM) cellulose is used as the stationary phase of this column, then which peptide elutes first?

Given:

GATE XL 2023 Q21 figure: A mixture of four peptides, PKKRK, RGERV, RYRGV and LVVYP, is loaded onto an ion-exchange column at pH = 7.2.
  1. APKKRK
  2. BRGERV
  3. CRYRGV
  4. DLVVYP
Show answer
Answer: (D) LVVYP
Q22s, p & d Block ElementsXL-P Q.12 · MCQ · 2 marks, −2/3 for a wrong answer

Match the coordination complexes given in Column I with the most appropriate properties in Column II.

(Given: Atomic numbers of Mn: 25; Co: 27; Ni: 28)

Column I
Coordination complexes
Column II
Properties
E. [Mn(H2O)6]2+1. 5.92 Bohr Magneton (BM)
F. [CoF6]3−2. CFSE = 0.4 Δo
G. [NiCl4]2−3. Metal ion hybridisation is sp3
H. [Ni(CN)4]2−4. Diamagnetic
  1. AE-1, F-2, G-3, H-4
  2. BE-2, F-1, G-4, H-3
  3. CE-4, F-2, G-1, H-3
  4. DE-1, F-4, G-3, H-2
Show answer
Answer: (A) E-1, F-2, G-3, H-4
Q23Organic ChemistryXL-P Q.13 · MSQ (one or more correct) · 2 marks, no negative marking

Compounds P and Q undergo E2 elimination with reaction rate constants of k1 and k2, respectively, as shown below. Which is/are the CORRECT option(s)?

GATE XL 2023 Q23 figure: Compounds P and Q undergo E2 elimination with reaction rate constants of k 1 and k 2 , respectively, as shown
  1. Ak1 > k2
  2. Bk2 > k1
  3. CMost stable conformer of P gives the product
  4. DMost stable conformer of Q gives the product
Show answer
Answer (select all of these): (A) k1 > k2 and (C) Most stable conformer of P gives the product
Q24Structure & BondingXL-P Q.14 · MSQ (one or more correct) · 2 marks, no negative marking

According to Hard-Soft Acid-Base (HSAB) principle, the CORRECT option(s) for the solubility trend in water is/are

  1. AAgF > AgCl > AgBr > AgI
  2. BLiBr > LiCl > LiF
  3. CAgF < AgCl < AgBr < AgI
  4. DLiBr < LiCl < LiF
Show answer
Answer (select all of these): (A) AgF > AgCl > AgBr > AgI and (B) LiBr > LiCl > LiF
Q25Organic ChemistryXL-P Q.15 · MSQ (one or more correct) · 2 marks, no negative marking

Compound X gives alcohol P as the major product for the reaction shown below. Suitable option(s) for X is/are

GATE XL 2023 Q25 figure: Compound X gives alcohol P as the major product for the reaction shown below. Suitable option(s) for X is/are
  1. AGATE XL 2023 Q25, option A
  2. BGATE XL 2023 Q25, option B
  3. CGATE XL 2023 Q25, option C
  4. DGATE XL 2023 Q25, option D
Show answer
Answer (select all of these): (B) and (C)
Q26Atomic Structure & PeriodicityXL-P Q.16 · Numerical answer · 2 marks, no negative marking

The number of radial node(s) for the valence orbital of U(III) ion is _____ (in integer).

(Given: Atomic number of U is 92)

Type a number as your answer.

Show answer
Answer: 1 (numerical answer, as in the official key).
Q27ElectrochemistryXL-P Q.17 · Numerical answer · 2 marks, no negative marking

E° = 1.10 V for the following cell reaction:

Zn(s) + Cu2+(aq) ⇌ Zn2+(aq) + Cu(s)

For this reaction, the equilibrium constant is y × 1037 at 298 K. The value of y is _____ (rounded off to two decimal places).

(Given: F = 96485 C mol−1, R = 8.314 J K−1 mol−1)

Type a number as your answer.

Show answer
Answer: any value from 1.58 to 1.66 (numerical answer, as in the official key).
Two sections to go. Now test yourself under time
The GATE-XL Life Sciences Test Series gives you full-length mocks on the GATE pattern, with MCQ, MSQ and numerical questions.

XL-Q: Biochemistry (Q28–46)

19 questions (8 of 1 mark, 11 of 2 marks). Optional: answer it only if you pick Biochemistry. From GATE 2027 this section is called XL1.

GATE XL 2023, XL-Q Biochemistry: questions by topic
19 questions, tagged by Lesson Basket.
Biomolecules31.6%
Metabolism & Bioenergetics21.1%
Biochemical Techniques15.8%
Immunology15.8%
Cell Biology10.5%
Enzymes5.3%
Topic tags: Lesson Basket. See also our GATE-XL weightage analysis.
Q28Metabolism & BioenergeticsXL-Q Q.1 · MCQ · 1 mark, −1/3 for a wrong answer

Determine the correctness or otherwise of the following Assertion [a] and the Reason [r].

Assertion [a]: On a per carbon basis, palmitic acid yields more ATP than glucose.
Reason [r]: Carbons in palmitic acid are more reduced than those in glucose.

  1. ABoth [a] and [r] are true and [r] is the correct reason for [a]
  2. BBoth [a] and [r] are true but [r] is not the correct reason for [a]
  3. C[a] is true but [r] is false
  4. D[a] is false but [r] is true
Show answer
Answer: (A) Both [a] and [r] are true and [r] is the correct reason for [a]
Q29Biochemical TechniquesXL-Q Q.2 · MCQ · 1 mark, −1/3 for a wrong answer

When cell components are fractionated by sedimentation, the correct order (from lower to higher gravitational force, g) in which the components get separated is _______.

  1. Anuclei, mitochondria, microsomes, and ribosomes
  2. Bmicrosomes, mitochondria, ribosomes, and nuclei
  3. Cnuclei, ribosomes, mitochondria, and microsomes
  4. Dribosomes, microsomes, mitochondria, and nuclei
Show answer
Answer: (A) nuclei, mitochondria, microsomes, and ribosomes
Q30ImmunologyXL-Q Q.3 · MCQ · 1 mark, −1/3 for a wrong answer

In a population, the probability of a susceptible individual getting infected with SARS-CoV-2 is low when a majority of individuals in the population becomes immune to this virus. This phenomenon is known as _______.

  1. Ainnate immunity
  2. Badaptive immunity
  3. Cactive immunity
  4. Dherd immunity
Show answer
Answer: (D) herd immunity
Q31Metabolism & BioenergeticsXL-Q Q.4 · MSQ (one or more correct) · 1 mark, no negative marking

Given below are four reactions of the glycolytic pathway catalyzed by the enzymes E1, E2, E3, and E4, as indicated. Which of these enzymes is/are NOT part of the gluconeogenesis pathway?

GATE XL 2023 Q31 figure: Given below are four reactions of the glycolytic pathway catalyzed by the enzymes E1, E2, E3, and E4, as indic
  1. AE1
  2. BE2
  3. CE3
  4. DE4
Show answer
Answer (select all of these): (A) E1 and (D) E4
Q32Cell BiologyXL-Q Q.5 · MSQ (one or more correct) · 1 mark, no negative marking

Which of the following molecules is/are second messenger(s) produced by the phosphoinositide signaling cascade?

  1. APhosphatidylinositol 4,5-bisphosphate
  2. BInositol 1,4,5-triphosphate
  3. CInositol 1,3,5-triphosphate
  4. DDiacylglycerol
Show answer
Answer (select all of these): (B) Inositol 1,4,5-triphosphate and (D) Diacylglycerol
Q33BiomoleculesXL-Q Q.6 · Numerical answer · 1 mark, no negative marking

A protein has seven cysteine residues. The maximum number of disulfide bonds of different combinations that can possibly be formed by these seven cysteine residues is _______ (in integer).

Type a number as your answer.

Show answer
Answer: 48 (numerical answer, as in the official key).
Q34BiomoleculesXL-Q Q.7 · Numerical answer · 1 mark, no negative marking

A lyophilized sample of 20 nanomoles of an oligonucleotide is dissolved in water and the volume of the solution is made up to 200 μL. The concentration (in μM) of the oligonucleotide in this solution is ______ (in integer).

Type a number as your answer.

Show answer
Answer: 100 (numerical answer, as in the official key).
Q35BiomoleculesXL-Q Q.8 · Numerical answer · 1 mark, no negative marking

DNA in a 1 cm long chromatin contains 5 × 109 base pairs. The fold compaction of this DNA within the chromatin is _______ (in integer).

Type a number as your answer.

Show answer
Answer: 170 (numerical answer, as in the official key).
Q36Metabolism & BioenergeticsXL-Q Q.9 · MCQ · 2 marks, −2/3 for a wrong answer

Intracellular concentrations of ATP, ADP, and inorganic phosphate in four cell types are given below. Which one of these cell types has the most negative ΔG for ATP hydrolysis?

Cell typeATP (mM)ADP (mM)Inorganic phosphate (mM)
L3.01.85.0
K3.91.33.0
B2.70.72.7
M7.20.98.0
  1. AL
  2. BK
  3. CB
  4. DM
Show answer
Answer: (C) B
Q37BiomoleculesXL-Q Q.10 · MCQ · 2 marks, −2/3 for a wrong answer

Which one of the following amino acids has more than two acid-base groups?

  1. AAlanine
  2. BLeucine
  3. CPhenylalanine
  4. DTyrosine
Show answer
Answer: (D) Tyrosine
Q38EnzymesXL-Q Q.11 · MCQ · 2 marks, −2/3 for a wrong answer

Enzyme activity profiles as a function of time in the absence or presence of different types of feedback mechanisms are shown in the figure below. Match the following feedback mechanisms with the corresponding profiles in the figure.

(p) No feedback mechanism
(q) Negative feedback mechanism with short delay
(r) Negative feedback mechanism with long delay
(s) Positive feedback mechanism

GATE XL 2023 Q38 figure: Enzyme activity profiles as a function of time in the absence or presence of different types of feedback mecha
  1. A(p) – (i); (q) – (ii); (r) – (iii); (s) – (iv)
  2. B(p) – (iv); (q) – (iii); (r) – (ii); (s) – (i)
  3. C(p) – (iv); (q) – (ii); (r) – (iii); (s) – (i)
  4. D(p) – (i); (q) – (iii); (r) – (ii); (s) – (iv)
Show answer
Answer: (B) (p) – (iv); (q) – (iii); (r) – (ii); (s) – (i)
Q39Biochemical TechniquesXL-Q Q.12 · MCQ · 2 marks, −2/3 for a wrong answer

A linear DNA fragment of 5 kilobase (kb) when completely digested with EcoRI produces 2.5 kb, 1.5 kb, and 1 kb fragments. Complete digestion of the same 5 kb fragment with XbaI produces 3.5 kb and 1.5 kb fragments. Which one of the following sets of fragments will be obtained if the 5 kb fragment is fully digested with EcoRI and XbaI simultaneously?

  1. A3 kb and 2 kb
  2. B2 kb and 1 kb
  3. C2 kb, 1.5 kb, 1 kb, and 0.5 kb
  4. D2.5 kb, 1.5 kb, 0.75 kb, and 0.25 kb
Show answer
Answer: (C) 2 kb, 1.5 kb, 1 kb, and 0.5 kb
Q40ImmunologyXL-Q Q.13 · MCQ · 2 marks, −2/3 for a wrong answer

Match the cell types listed in Group I with associated processes listed in Group II.

Group IGroup II
(p) NK cells(i) Antibody production
(q) B cells(ii) First cells to be recruited at the site of infection
(r) Mast cells(iii) Antibody-dependent cell-mediated cytotoxicity
(s) Neutrophils(iv) Histamine production
  1. A(p) – (ii); (q) – (i); (r) – (iii); (s) – (iv)
  2. B(p) – (ii); (q) – (i); (r) – (iv); (s) – (iii)
  3. C(p) – (iii); (q) – (i); (r) – (iv); (s) – (ii)
  4. D(p) – (iii); (q) – (ii); (r) – (i); (s) – (iv)
Show answer
Answer: (C) (p) – (iii); (q) – (i); (r) – (iv); (s) – (ii)
Q41BiomoleculesXL-Q Q.14 · MSQ (one or more correct) · 2 marks, no negative marking

Four statements about lipids are given below as options. Choose the statement(s) which is/are CORRECT.

  1. ACholesterol is amphipathic
  2. BSelf-assembly of phospholipids in water is due to hydrophobic effect
  3. CThe temperature at which the gel phase changes to liquid crystalline phase increases with an increase in the degree of unsaturation of fatty acyl tails
  4. DThe choline head group of lipids is positively charged
Show answer
Answer (select all of these): (A) Cholesterol is amphipathic and (B) Self-assembly of phospholipids in water is due to hydrophobic effect and (D) The choline head group of lipids is positively charged
Q42Biochemical TechniquesXL-Q Q.15 · MSQ (one or more correct) · 2 marks, no negative marking

Which of the following technique(s) can be used to separate proteins according to their molecular weights from a mixture of proteins?

  1. AIon exchange chromatography
  2. BSize exclusion chromatography
  3. CSodium dodecylsulfate – polyacrylamide gel electrophoresis (SDS-PAGE)
  4. DSucrose density gradient centrifugation
Show answer
Answer (select all of these): (B) Size exclusion chromatography and (C) Sodium dodecylsulfate – polyacrylamide gel electrophoresis (SDS-PAGE) and (D) Sucrose density gradient centrifugation
Q43ImmunologyXL-Q Q.16 · MSQ (one or more correct) · 2 marks, no negative marking

B cells produce two forms of an immunoglobulin: (i) membrane-bound form, known as B cell receptor (BCR) and (ii) soluble form, known as antibody. Which of the following statements is/are CORRECT about BCR and antibody produced by the same B cell?

  1. ABCR and antibody have identical antigen binding site
  2. BBCR and antibody recognize different epitopes
  3. CBCR and antibody are encoded by the same gene
  4. DBCR and antibody are formed by differential splicing
Show answer
Answer (select all of these): (A) BCR and antibody have identical antigen binding site and (C) BCR and antibody are encoded by the same gene and (D) BCR and antibody are formed by differential splicing
Q44BiomoleculesXL-Q Q.17 · Numerical answer · 2 marks, no negative marking

A 100 ml solution of pH 10 was well-mixed with a 100 ml solution of pH 4. The pH of the resultant 200 ml solution is ___________ (rounded off to two decimal places).

Type a number as your answer.

Show answer
Answer: any value from 4.29 to 4.31 (numerical answer, as in the official key).
Q45Metabolism & BioenergeticsXL-Q Q.18 · Numerical answer · 2 marks, no negative marking

An organism uses only the glycerophosphate shunt pathway to transport cytosolic NADH to mitochondria. For every two electrons transported, complex I, complex III, and complex IV of the electron transport chain in this organism transport 2.5, 1.5, and 2.0 protons (H+), respectively. The H+ to ATP ratio of FOF1-ATPase of this organism is 4.0. Terminal electron acceptor is oxygen. The number of ATP molecules synthesized by oxidizing NADH from glycolysis is __________ (rounded off to two decimal places).

Type a number as your answer.

Show answer
Answer: any value from 1.74 to 1.76 (numerical answer, as in the official key).
Q46Cell BiologyXL-Q Q.19 · Numerical answer · 2 marks, no negative marking

If the extracellular concentration of sodium ion (Na+) is ten times more than its intracellular concentration, then the sodium equilibrium potential at 20 ºC in mV is ________________ (rounded off to two decimal places). Assume that the membrane is permeable only to Na+ ions. [Use R = 1.987 cal deg−1 mol−1 and F = 23062 cal mol−1 V−1]

Type a number as your answer.

Show answer
Answer: any value from 57.80 to 58.80 (numerical answer, as in the official key).

XL-R: Botany (Q47–65)

19 questions (8 of 1 mark, 11 of 2 marks). Optional: answer it only if you pick Botany. From GATE 2027 this section is called XL2.

GATE XL 2023, XL-R Botany: questions by topic
19 questions, tagged by Lesson Basket.
Genetics & Genomics26.3%
Plant Physiology & Biochemistry21.1%
Plant Anatomy15.8%
Morphogenesis & Development15.8%
Plant Pathology10.5%
Economic Botany5.3%
Plant Breeding & Genetic Modification5.3%
Topic tags: Lesson Basket. See also our GATE-XL weightage analysis.
Q47Plant AnatomyXL-R Q.1 · MCQ · 1 mark, −1/3 for a wrong answer

Which one of the following statements on Casparian strips is correct?

  1. ACasparian strips are specific to vascular plants found in epidermal cells.
  2. BCasparian strips are modifications mostly found in shoot tissue.
  3. CCasparian strips act as a cellular barrier to allow selective nutrient uptake and exclusion of pathogens.
  4. DCasparian strips are common in root endodermal cells of non-vascular plants.
Show answer
Answer: (C) Casparian strips act as a cellular barrier to allow selective nutrient uptake and exclusion of pathogens.
Q48Plant Physiology & BiochemistryXL-R Q.2 · MCQ · 1 mark, −1/3 for a wrong answer

Rotenone is a chemical often used to kill insect pests on crop plants and fishes in lakes. Rotenone acts by inhibiting electron transport from the NADH dehydrogenase enzyme in Complex I to ubiquinone in the mitochondrial electron transport chain. Which one of the following explains why plants can tolerate rotenone application?

  1. AThe Complex I in plants is resistant to rotenone.
  2. BPlants inactivate rotenone by enzymatic degradation.
  3. CPlants have specific channels that efflux rotenone out of the cell.
  4. DPlants have additional NAD(P)H dehydrogenases that are resistant to rotenone.
Show answer
Answer: (D) Plants have additional NAD(P)H dehydrogenases that are resistant to rotenone.
Q49Plant PathologyXL-R Q.3 · MCQ · 1 mark, −1/3 for a wrong answer

Although Pseudomonas syringae infection in plants is actively inhibited by the endogenous salicylic acid (SA) of host origin, a successful infection is still established because the bacterium secretes coronatine, an effector molecule. Which one of the following best describes the mode of action of coronatine?

  1. ACoronatine inhibits SA biosynthesis.
  2. BCoronatine promotes the biosynthesis of jasmonic acid (JA), and JA signaling in turn inhibits SA response.
  3. CCoronatine is a structural analogue of SA, which binds to the SA receptor and inhibits its function.
  4. DCoronatine is a structural analogue of jasmonic acid (JA), which activates JA signaling to inhibit SA response.
Show answer
Answer: (D) Coronatine is a structural analogue of jasmonic acid (JA), which activates JA signaling to inhibit SA response.
Q50Morphogenesis & DevelopmentXL-R Q.4 · MCQ · 1 mark, −1/3 for a wrong answer

The schematic depicts an unexpanded plant cell within a hypocotyl with the arrangement of cellulose microfibrils marked on its cell wall.

GATE XL 2023 Q50 figure: The schematic depicts an unexpanded plant cell within a hypocotyl with the arrangement of cellulose microfibri

Which one of the following shapes would most likely result from the expansion of this cell if the pattern of the cellulose fibrils does not change?

  1. AGATE XL 2023 Q50, option A
  2. BGATE XL 2023 Q50, option B
  3. CGATE XL 2023 Q50, option C
  4. DGATE XL 2023 Q50, option D
Show answer
Answer: (A)
Q51Morphogenesis & DevelopmentXL-R Q.5 · MSQ (one or more correct) · 1 mark, no negative marking

Which one or more of the following statements is/are NOT CORRECT with respect to pollen development in angiosperm?

  1. ATapetal cell wall in all angiosperms breaks down to release the cytoplasmic content.
  2. BTapetal cell wall in all angiosperms remains intact.
  3. CTapetal cell wall breaks down in some angiosperm species, whereas it remains intact in others.
  4. DWithin an angiosperm species, the tapetal cell wall breaks down in some individuals and not in others.
Show answer
Answer (select all of these): (A) Tapetal cell wall in all angiosperms breaks down to release the cytoplasmic content. and (B) Tapetal cell wall in all angiosperms remains intact. and (D) Within an angiosperm species, the tapetal cell wall breaks down in some individuals and not in others.
Q52Plant Physiology & BiochemistryXL-R Q.6 · MSQ (one or more correct) · 1 mark, no negative marking

Regulation of phosphoenolpyruvate carboxylase (PEPCase) governs CO2 fixation in both C4 and CAM (crassulacean acid metabolism) plants. Which one or more of the following statements with respect to PEPCase activity is/are CORRECT?

  1. APEPCase in C4 plants is inactivated by dephosphorylation during the day.
  2. BPEPCase in CAM plants is inactivated by dephosphorylation during the day.
  3. CPEPCase in C4 plants is inactivated by dephosphorylation at night.
  4. DPEPCase in CAM plants is inactivated by dephosphorylation at night.
Show answer
Answer (select all of these): (B) PEPCase in CAM plants is inactivated by dephosphorylation during the day. and (C) PEPCase in C4 plants is inactivated by dephosphorylation at night.
Q53Plant Physiology & BiochemistryXL-R Q.7 · MSQ (one or more correct) · 1 mark, no negative marking

Which one or more processes listed below DOES NOT/DO NOT produce carbon dioxide during fermentation?

  1. ABrewing wine using yeast.
  2. BBaking bread using yeast.
  3. CMaking yogurt using lactobacillus.
  4. DMaking cheese using fungus.
Show answer
Answer (select all of these): (C) Making yogurt using lactobacillus. and (D) Making cheese using fungus.
Q54Genetics & GenomicsXL-R Q.8 · Numerical answer · 1 mark, no negative marking

The ovule of a diploid species with 2n = 8 undergoes double fertilization. If the pollen is contributed by an individual with meiotic nondisjunction, the chromosome number of the zygote will be _________.

Type a number as your answer.

Show answer
Answer: 12 (numerical answer, as in the official key).
Q55Genetics & GenomicsXL-R Q.9 · MCQ · 2 marks, −2/3 for a wrong answer

Match the tasks given in Group I with the associated techniques conventionally used as listed in Group II.

Group IGroup II
P. Ploidy analysis1. RNA sequencing
Q. Profiling DNA methylation2. Exome sequencing
R. Identifying non-coding RNAs3. Fluorescence in situ hybridization
S. Identifying SNPs4. Bisulfite sequencing
T. Satellite DNA isolation5. Density-gradient centrifugation
  1. AP-2; Q-1; R-3; S-4; T-5
  2. BP-3; Q-4; R-1; S-2; T-5
  3. CP-5; Q-4; R-1; S-2; T-3
  4. DP-3; Q-5; R-1; S-2; T-4
Show answer
Answer: (B) P-3; Q-4; R-1; S-2; T-5
Q56Plant AnatomyXL-R Q.10 · MCQ · 2 marks, −2/3 for a wrong answer

Periderm is a protective tissue found in stems and roots of gymnosperm and woody dicotyledons. It contributes to the increased thickness by secondary growth. Match the peridermal components given in Group I with the cell/tissue types given in Group II.

Group IGroup II
(P) Phelloids(1) Tissue resembling cortical parenchyma
(Q) Phellogen(2) Cork cambium
(R) Phellem(3) Cork-like cells
(S) Phelloderm(4) Cork
  1. AP-4; Q-3; R-1; S-2
  2. BP-3; Q-2; R-4; S-1
  3. CP-2; Q-1; R-3; S-4
  4. DP-4; Q-1; R-3; S-2
Show answer
Answer: (B) P-3; Q-2; R-4; S-1
Q57Plant PathologyXL-R Q.11 · MCQ · 2 marks, −2/3 for a wrong answer

Match the following rice diseases in Group I with their causal agents in Group II.

Group IGroup II
(P) Blast(1) Sclerophthora macrospora
(Q) False smut(2) Rhizoctonia solani
(R) Sheath blight(3) Ustilaginoidea virens
(S) Downy mildew(4) Puccinia graminis
(5) Magnaporthe grisea
  1. AP-5; Q-3; R-2; S-1
  2. BP-4; Q-2; R-5; S-3
  3. CP-4; Q-5; R-3; S-1
  4. DP-5; Q-4; R-1; S-2
Show answer
Answer: (A) P-5; Q-3; R-2; S-1
Q58Plant AnatomyXL-R Q.12 · MCQ · 2 marks, −2/3 for a wrong answer

Central vascular cylinder or stele consists of the primary vascular system (xylem and phloem) and the associate fundamental tissue. Match the schematics of stele in Group I (xylem shown in green, and phloem shown as a hatched block) with their respective types in Group II.

GATE XL 2023 Q58 figure: Central vascular cylinder or stele consists of the primary vascular system (xylem and phloem) and the associat
  1. AP-2; Q-4; R-1; S-3
  2. BP-5; Q-1; R-4; S-2
  3. CP-5; Q-3; R-1; S-2
  4. DP-3; Q-4; R-2; S-5
Show answer
Answer: (A) P-2; Q-4; R-1; S-3
Q59Morphogenesis & DevelopmentXL-R Q.13 · MCQ · 2 marks, −2/3 for a wrong answer

Consider the following four experimental observations (i, ii, iii, iv) on the effect of the FT gene on flowering transition in the shoot apical meristem (SAM) of Arabidopsis thaliana.

i) The FT promoter is active in leaves alone.
ii) The ft null mutation causes delayed flowering transition of the SAM.
iii) Expressing a recombinant FT protein fused to nuclear localization signal sequence under the endogenous promoter does not rescue the delayed-flowering phenotype of the ft null mutant.
iv) Downregulation of FT transcript in the SAM by RNA interference in the wild-type background does not alter flowering transition.

Which one of the following conclusions best explains the above observations?

  1. AFT protein resident in leaves causes flowering transition of the SAM.
  2. BFT transcript moves from leaves to the meristem and promotes flowering.
  3. CFT protein moves from leaves to the SAM and promotes flowering.
  4. DBoth FT transcript and FT protein are required in the SAM to promote flowering.
Show answer
Answer: (C) FT protein moves from leaves to the SAM and promotes flowering.
Q60Economic BotanyXL-R Q.14 · MCQ · 2 marks, −2/3 for a wrong answer

Which one of the options given correctly matches the alkaloids in Group I with their source plants in Group II?

Group IGroup II
P. Cocaine1. Cocoa
Q. Caffeine2. Nightshade
R. Morphine3. Coca
S. Atropine4. Poppy
  1. AP-3; Q-1; R-4; S-2
  2. BP-1; Q-3; R-4; S-2
  3. CP-2; Q-1; R-3; S-4
  4. DP-4; Q-2; R-1; S-3
Show answer
Answer: (A) P-3; Q-1; R-4; S-2
Q61Genetics & GenomicsXL-R Q.15 · MSQ (one or more correct) · 2 marks, no negative marking

A drought tolerant rice genotype was found to be associated with a missense mutation in the gene A. Which one or more of the following experiments is/are appropriate to validate whether the mutation in A is the causal factor for drought tolerance?

  1. AIntroduce the same mutation in a drought sensitive rice genotype and test if it becomes drought tolerant.
  2. BDelete the wild-type A in drought sensitive plant and test if it becomes drought tolerant.
  3. CDetermine the stability of the protein encoded by the wild-type and the mutant forms of A.
  4. DRepair the mutation in the drought tolerant rice genotype and test if it becomes drought sensitive.
Show answer
Answer (select all of these): (A) Introduce the same mutation in a drought sensitive rice genotype and test if it becomes drought tolerant. and (D) Repair the mutation in the drought tolerant rice genotype and test if it becomes drought sensitive.
Q62Plant Physiology & BiochemistryXL-R Q.16 · MSQ (one or more correct) · 2 marks, no negative marking

Blue light can directly induce opening of stomata. Blue light also triggers photosynthesis in the guard cells, which indirectly induces stomatal opening. Which one or more of the following experimental approaches would test the direct effect of blue light on stomatal opening?

  1. AApplication of low photon fluxes of red light followed by high fluence rate of blue light.
  2. BApplication of high fluence rates of red light followed by low photon fluxes of blue light.
  3. CApplication of high fluence rates of blue light followed by high photon fluxes of red light.
  4. DInhibition of photosynthetic electron transport by dichlorophenyldimethylurea (DCMU).
Show answer
Answer (select all of these): (B) Application of high fluence rates of red light followed by low photon fluxes of blue light. and (D) Inhibition of photosynthetic electron transport by dichlorophenyldimethylurea (DCMU).
Q63Genetics & GenomicsXL-R Q.17 · Numerical answer · 2 marks, no negative marking

In a diploid angiosperm species, flower colour is regulated by the R gene. RR and Rr genotypes produce red flowers, whereas the rr genotype produces white flowers. If two individual plants are randomly selected from a large segregating population of a genetic cross between RR and rr parents, the probability of both the plants producing red flowers is _______. (Rounded off to two decimal places)

Type a number as your answer.

Show answer
Answer: 0.56 (numerical answer, as in the official key).
Q64Plant Breeding & Genetic ModificationXL-R Q.18 · Numerical answer · 2 marks, no negative marking

A cytoplasmic male-sterile female plant with the restorer (nuclear) genotype rr is crossed to a male-fertile male plant with the genotype RR. Both RR and Rr can restore the fertility, whereas rr cannot. When an F1 female plant with Rr genotype was test-crossed to a male-fertile male plant with the rr genotype, the percentage of the population that is male fertile would be _______ %. (Answer in integer)

Type a number as your answer.

Show answer
Answer: 50 (numerical answer, as in the official key).
Q65Genetics & GenomicsXL-R Q.19 · Numerical answer · 2 marks, no negative marking

The frequencies for autosomal alleles A and a are p = 0.5 and q = 0.5, respectively, where A is dominant over a. Under the assumption of random mating, the mating frequency among dominant parents is _________. (Rounded off to two decimal places)

Type a number as your answer.

Show answer
Answer: any value from 0.5 to 0.6 (numerical answer, as in the official key).

XL-S: Microbiology (Q66–84)

19 questions (8 of 1 mark, 11 of 2 marks). Optional: answer it only if you pick Microbiology. From GATE 2027 this section is called XL3.

GATE XL 2023, XL-S Microbiology: questions by topic
19 questions, tagged by Lesson Basket.
Microbial Growth21.1%
Microbial Metabolism15.8%
Prokaryotic Cells15.8%
Microbial Genetics15.8%
Microbial Taxonomy & Diversity10.5%
Microbial Pathogenesis & Immunology10.5%
Microbial Ecology & Applied Microbiology5.3%
Historical Perspective5.3%
Topic tags: Lesson Basket. See also our GATE-XL weightage analysis.
Q66Microbial Taxonomy & DiversityXL-S Q.1 · MCQ · 1 mark, −1/3 for a wrong answer

Monkey pox is caused by a

  1. Adouble-stranded DNA virus
  2. Bsingle-stranded DNA virus
  3. Cdouble-stranded RNA virus
  4. Dsingle-stranded RNA virus
Show answer
Answer: (A) double-stranded DNA virus
Q67Microbial MetabolismXL-S Q.2 · MCQ · 1 mark, −1/3 for a wrong answer

Which one of the following converts sulfate to hydrogen sulfide?

  1. ABeggiatoa
  2. BDesulfovibrio
  3. CThiobacillus
  4. DThiothrix
Show answer
Answer: (B) Desulfovibrio
Q68Prokaryotic CellsXL-S Q.3 · MCQ · 1 mark, −1/3 for a wrong answer

Which one of the statements about bacterial flagella is correct?

  1. AFlagella varies in length ranging from 0.5 to 2 µm.
  2. BFlagella are adjacent fibrils with regular patterns.
  3. CFlagella helps in conjugation.
  4. DFlagella originates from basal body.
Show answer
Answer: (D) Flagella originates from basal body.
Q69Microbial Ecology & Applied MicrobiologyXL-S Q.4 · MCQ · 1 mark, −1/3 for a wrong answer

Microbial plastics are made from

  1. Apolyhydroxyalkanoates
  2. Bpolystyrene
  3. Cpolyurethane
  4. Dpolyvinyl chloride
Show answer
Answer: (A) polyhydroxyalkanoates
Q70Microbial MetabolismXL-S Q.5 · MCQ · 1 mark, −1/3 for a wrong answer

The correct sequence of metabolic intermediates in Krebs cycle is

  1. Aα-ketoglutarate → fumarate → succinate → malate
  2. Bfumarate → malate → succinate → α-ketoglutarate
  3. Cα-ketoglutarate → succinate → fumarate → malate
  4. Dsuccinate → α-ketoglutarate → malate → fumarate
Show answer
Answer: (C) α-ketoglutarate → succinate → fumarate → malate
Q71Microbial GeneticsXL-S Q.6 · MCQ · 1 mark, −1/3 for a wrong answer

Catabolite repression in bacteria is regulated by the concentration of

  1. Aamino acids
  2. Bglucose
  3. Cmessenger RNA
  4. Dlactose
Show answer
Answer: (B) glucose
Q72Historical PerspectiveXL-S Q.7 · MCQ · 1 mark, −1/3 for a wrong answer

Phagocytosis was first described by

  1. AElie Metchnikoff
  2. BRobert Hooke
  3. CRobert Koch
  4. DPaul Ehrlich
Show answer
Answer: (A) Elie Metchnikoff
Q73Microbial GrowthXL-S Q.8 · MCQ · 1 mark, −1/3 for a wrong answer

Which one of the following statements about batch culture of microbes is NOT correct?

  1. ACells from stationary phase will show longer lag phase when inoculated in fresh growth medium compared to those collected from exponential phase.
  2. BDeath phase of culture is often exponential in nature.
  3. CStationary phase is the cryptic growth phase.
  4. DThe rate of generation of new cells during exponential growth phase is constant.
Show answer
Answer: (D) The rate of generation of new cells during exponential growth phase is constant.
Q74Microbial Pathogenesis & ImmunologyXL-S Q.9 · MCQ · 2 marks, −2/3 for a wrong answer

Match the test in Group I with its application in Group II

Group IGroup II
P. Oakley-Fulthorpe test1. IgM detection
Q. Limulus amoebocyte lysate test2. Determining antigen-antibody specificity
R. Weil-Felix reaction test3. Endotoxin detection
S. Complement-fixation test4. Rickettsial infection diagnosis
  1. AP-2, Q-3, R-4, S-1
  2. BP-2, Q-1, R-4, S-3
  3. CP-3, Q-1, R-2, S-4
  4. DP-4, Q-3, R-2, S-1
Show answer
Answer: (A) P-2, Q-3, R-4, S-1
Q75Microbial GrowthXL-S Q.10 · MCQ · 2 marks, −2/3 for a wrong answer

Which one of the following is NOT correct about antibiotic resistance mechanism in microbes?

  1. AMycoplasma is naturally resistant to penicillins due to presence of R plasmid.
  2. BGram-negative bacteria are impermeable to penicillin G.
  3. Cβ-lactamases of bacteria can cleave penicillins.
  4. DSelective microbes can efflux penicillins entering the cell and develop resistance.
Show answer
Answer: (A) Mycoplasma is naturally resistant to penicillins due to presence of R plasmid.
Q76Microbial MetabolismXL-S Q.11 · MCQ · 2 marks, −2/3 for a wrong answer

A suspension of photosynthetic green algae was illuminated in the presence of 14CO2 for few seconds. The first metabolite in the Calvin cycle to be radiolabeled will be

  1. Aglyceraldehyde
  2. B1,3-bisphosphoglycerate
  3. C3-phosphoglycerate
  4. Dribulose 1,5-bisphosphate
Show answer
Answer: (C) 3-phosphoglycerate
Q77Prokaryotic CellsXL-S Q.12 · MCQ · 2 marks, −2/3 for a wrong answer

Determine the correctness or otherwise of the following Assertion [a] and the Reason [r].

Assertion [a]: Endospore can survive heat that would rapidly kill vegetative cells of the same species.
Reason [r]: In endospore, the protoplasm is reduced to minimum volume as a result of the accumulation of calcium-dipicolinic acid complexes and small acid-soluble spore proteins, which forms a cytoplasmic gel and a thick cortex.

  1. ABoth [a] and [r] are true and [r] is the correct reason for [a]
  2. BBoth [a] and [r] are true and [r] is not the correct reason for [a]
  3. CBoth [a] and [r] are false
  4. D[a] is true but [r] is false
Show answer
Answer: (A) Both [a] and [r] are true and [r] is the correct reason for [a]
Q78Microbial GeneticsXL-S Q.13 · MCQ · 2 marks, −2/3 for a wrong answer

Which one of the following conjugations will result in formation of merodiploids?

  1. AF+ donor × F− recipient
  2. BHfr donor × F− recipient
  3. CF′ donor × F− recipient
  4. DF+ donor × Hfr recipient
Show answer
Answer: (C) F′ donor × F− recipient
Q79Microbial Taxonomy & DiversityXL-S Q.14 · MSQ (one or more correct) · 2 marks, no negative marking

Which of the following genus is/are a spirochete(s)?

  1. ABorrelia
  2. BLeptospira
  3. CSpirulina
  4. DTreponema
Show answer
Answer (select all of these): (A) Borrelia and (B) Leptospira and (D) Treponema
Q80Prokaryotic CellsXL-S Q.15 · MSQ (one or more correct) · 2 marks, no negative marking

Which of the following is/are non-membrane bound inclusion bodies?

  1. ACarboxysomes
  2. BCyanophycin granules
  3. CPoly-β-hydroxybutyrate granules
  4. DPolyphosphate granules
Show answer
Answer (select all of these): (B) Cyanophycin granules and (D) Polyphosphate granules
Q81Microbial GrowthXL-S Q.16 · MSQ (one or more correct) · 2 marks, no negative marking

Which of the following antibiotics is/are isolated from Streptomyces spp.?

  1. AGentamicin
  2. BNystatin
  3. CPolymyxins
  4. DTetracyclines
Show answer
Answer (select all of these): (B) Nystatin and (D) Tetracyclines
Q82Microbial Pathogenesis & ImmunologyXL-S Q.17 · MSQ (one or more correct) · 2 marks, no negative marking

Which of the following statements about the primary and secondary adaptive immune responses to an antigen is/are correct?

  1. AIgM antibodies appear first in response to the initial exposure of the antigen.
  2. BMajority of the antibodies produced in response to the second exposure of the same antigen are IgM isotype.
  3. CSecond exposure of the same antigen stimulates production of memory cells.
  4. DPrimary antibody response has shorter lag phase than secondary antibody response.
Show answer
Answer (select all of these): (A) IgM antibodies appear first in response to the initial exposure of the antigen. and (C) Second exposure of the same antigen stimulates production of memory cells.
Q83Microbial GeneticsXL-S Q.18 · MSQ (one or more correct) · 2 marks, no negative marking

The spontaneous, and induced mutations in bacteria can be distinguished by

  1. Afluctuation test
  2. Breplica plating
  3. Cdisc diffusion test
  4. Duse-dilution test
Show answer
Answer (select all of these): (A) fluctuation test and (B) replica plating
Q84Microbial GrowthXL-S Q.19 · Numerical answer · 2 marks, no negative marking

During the exponential growth, it took 6 hours for the population of bacterial cells to increase from 2.5 × 106 to 5 × 108. The generation time of the bacterium, rounded off to the nearest integer, is ________ minutes.

Type a number as your answer.

Show answer
Answer: any value from 46 to 48 (numerical answer, as in the official key).

XL-T: Zoology (Q85–103)

19 questions (8 of 1 mark, 11 of 2 marks). Optional: answer it only if you pick Zoology. From GATE 2027 this section is called XL4.

GATE XL 2023, XL-T Zoology: questions by topic
19 questions, tagged by Lesson Basket.
Biochemistry & Molecular Biology31.6%
Developmental Biology10.5%
Parasitology & Immunology10.5%
Cell Biology10.5%
Genetics10.5%
Animal Diversity5.3%
Animal Behaviour5.3%
Ecology5.3%
Evolution5.3%
Animal Anatomy & Physiology5.3%
Topic tags: Lesson Basket. See also our GATE-XL weightage analysis.
Q85Animal DiversityXL-T Q.1 · MCQ · 1 mark, −1/3 for a wrong answer

Which one of the following animals has “Book Lungs” as a respiratory organ?

  1. AEarthworm
  2. BScorpion
  3. COctopus
  4. DStarfish
Show answer
Answer: (B) Scorpion
Q86Animal BehaviourXL-T Q.2 · MCQ · 1 mark, −1/3 for a wrong answer

Which one of the following describes the “innate behavior” of an animal?

  1. AA behavior that is triggered due to the change in environment.
  2. BA behavior that is trained by the parents.
  3. CA behavior that is determined by heredity.
  4. DA behavior that is learnt by “hit and trial” approach.
Show answer
Answer: (C) A behavior that is determined by heredity.
Q87EcologyXL-T Q.3 · MCQ · 1 mark, −1/3 for a wrong answer

Which one of the following represents a true “Ecological population” ?

  1. AA pitcher plant and a trapped fly in it
  2. BAll animals that live near each other in a national park
  3. CThe leeches and the flatworms that live in a forest
  4. DAll the lions in a reserve forest
Show answer
Answer: (D) All the lions in a reserve forest
Q88Developmental BiologyXL-T Q.4 · MCQ · 1 mark, −1/3 for a wrong answer

Which of the following animals show “Bottle cells” during the gastrulation stage of development?

  1. ASnails
  2. BAmphibians
  3. CBirds
  4. DMammals
Show answer
Answer: (B) Amphibians
Q89Biochemistry & Molecular BiologyXL-T Q.5 · MCQ · 1 mark, −1/3 for a wrong answer

The organisms that obtain energy from inorganic compounds are known as

  1. AAutotrophs
  2. BOrganotrophs
  3. CLithotrophs
  4. DPhototrophs
Show answer
Answer: (C) Lithotrophs
Q90Parasitology & ImmunologyXL-T Q.6 · MSQ (one or more correct) · 1 mark, no negative marking

Which of the following is/are the causative agent(s) of Filariasis?

  1. AWuchereria bancrofti
  2. BLeishmania donovani
  3. CBrugia malayi
  4. DTrypanosoma gambiense
Show answer
Answer (select all of these): (A) Wuchereria bancrofti and (C) Brugia malayi
Q91EvolutionXL-T Q.7 · Numerical answer · 1 mark, no negative marking

In a population of 1000 wild dogs in a grassland, 360 and 480 dogs had black body colour with genotypes BB and Bb, respectively. In the same population, remaining dogs were white in colour with a genotype of bb. Based on this data, the frequency of allele “b” in the population is ______ (round off to one decimal place).

Type a number as your answer.

Show answer
Answer: 0.4 (numerical answer, as in the official key).
Q92Cell BiologyXL-T Q.8 · Numerical answer · 1 mark, no negative marking

A mature rat sperm cell has 2.5 µg of genomic DNA that is equivalent of a haploid genome. Compared to this sperm cell, the amount of genomic DNA (in µg) in a somatic cell, which is in the G2 phase of cell cycle, will be _____ (in integer).

Type a number as your answer.

Show answer
Answer: 10 (numerical answer, as in the official key).
Q93Developmental BiologyXL-T Q.9 · MCQ · 2 marks, −2/3 for a wrong answer

In an experiment, excess amount of bicod mRNA (more than wild-type expression level) was injected into the posterior pole of a wild-type Drosophila embryo at pre-blastodermal stage. Out of the following options, which one represents the best expected phenotype in the resulted developing embryo?

  1. ANormal embryo with head structure at anterior and tail structure at posterior pole
  2. BHead structure only at posterior pole of the embryo
  3. CTail structure at anterior and head structure at posterior poles of the embryo
  4. DHead structure at both anterior and posterior poles of the embryo
Show answer
Answer: (D) Head structure at both anterior and posterior poles of the embryo
Q94Animal Anatomy & PhysiologyXL-T Q.10 · MCQ · 2 marks, −2/3 for a wrong answer

Match the hormones/precursors listed in Column I with their chemical type in Column II and the tissue of origin listed in Column III

Column IColumn IIColumn III
P. Glucagon(i) Tryptophan derivativea. Anterior pituitary
Q. Pregnenolone(ii) Peptideb. Pineal
R. FSH(iii) Steroidc. Adrenal
S. Melatonin(iv) Glycoproteind. Pancreas
  1. AP – (ii) – d ; Q – (iii) – c ; R – (iv) – a ; S – (i) – b
  2. BP – (ii) – d ; Q – (iv) – a ; R – (i) – c ; S – (iii) – b
  3. CP – (i) – c ; Q – (ii) – b ; R – (iv) – d ; S – (iii) – a
  4. DP – (iv) – a ; Q – (i) – d ; R – (ii) – b ; S – (iii) – c
Show answer
Answer: (A) P – (ii) – d ; Q – (iii) – c ; R – (iv) – a ; S – (i) – b
Q95GeneticsXL-T Q.11 · MCQ · 2 marks, −2/3 for a wrong answer

Match the syndromes listed in Column I with the cause/symptoms listed in Column II

Column IColumn II
P. Prader-Willi syndrome(i) a collection of signs and symptoms due to prolonged exposure to corticosteroids like cortisol
Q. Down syndrome(ii) a syndrome of inadequate reabsorption in the proximal renal tubule of the kidney
R. Cushing syndrome(iii) a genetic disorder usually caused by the deletion of a part of chromosome 15
S. Turner syndrome(iv) a genetic disorder caused by the presence of all or part of a third copy of chromosome 21
T. Fanconi syndrome(v) a genetic condition in which a female has partially or completely missing an X chromosome
  1. AP – (ii) ; Q – (v) ; R – (iv) ; S – (iii) ; T – (i)
  2. BP – (iv) ; Q – (iii) ; R – (i) ; S – (ii) ; T – (v)
  3. CP – (iii) ; Q – (iv) ; R – (i) ; S – (v) ; T – (ii)
  4. DP – (v) ; Q – (iv) ; R – (ii) ; S – (i) ; T – (iii)
Show answer
Answer: (C) P – (iii) ; Q – (iv) ; R – (i) ; S – (v) ; T – (ii)
Q96Parasitology & ImmunologyXL-T Q.12 · MCQ · 2 marks, −2/3 for a wrong answer

Match the immunological statements in Column I with the appropriate descriptions from Column II

Column IColumn II
P. Active acquired immunity(i) Complement proteins and interferons
Q. First line of defense(ii) Direct contact with pathogens that enter the body
R. Passive natural immunity(iii) Surface barriers
S. Second line of defense(iv) Antibodies pass through placenta
  1. AP – (ii) ; Q – (iii) ; R – (iv) ; S – (i)
  2. BP – (iv) ; Q – (iii) ; R – (i) ; S – (ii)
  3. CP – (iv) ; Q – (ii) ; R – (iii) ; S – (i)
  4. DP – (i) ; Q – (iv) ; R – (iii) ; S – (ii)
Show answer
Answer: (A) P – (ii) ; Q – (iii) ; R – (iv) ; S – (i)
Q97Biochemistry & Molecular BiologyXL-T Q.13 · MCQ · 2 marks, −2/3 for a wrong answer

Match the standard/stated cofactors in Column I with their respective enzymes in Column II

Column IColumn II
P. Cu2+(i) Dinitrogenase
Q. Se(ii) Cytochrome oxidase
R. Ni2+(iii) Pyruvate kinase
S. K+(iv) Glutathione peroxidase
T. Mo(v) Urease
  1. AP – (v) ; Q – (ii) ; R – (iv) ; S – (i) ; T – (iii)
  2. BP – (ii) ; Q – (iv) ; R – (v) ; S – (iii) ; T – (i)
  3. CP – (iv) ; Q – (ii) ; R – (iii) ; S – (i) ; T – (v)
  4. DP – (ii) ; Q – (i) ; R – (iii) ; S – (iv) ; T – (v)
Show answer
Answer: (B) P – (ii) ; Q – (iv) ; R – (v) ; S – (iii) ; T – (i)
Q98Biochemistry & Molecular BiologyXL-T Q.14 · MSQ (one or more correct) · 2 marks, no negative marking

The presence of excess glucose has been known to prevent the induction of lac operon as well as other operon controlling enzymes involved in carbohydrate metabolism in E. coli. Which of the following processes define(s) the phenomenon?

  1. ACatabolite repression
  2. BAttenuation
  3. CGlucose effect
  4. DFeedback inhibition
Show answer
Answer (select all of these): (A) Catabolite repression and (C) Glucose effect
Q99Biochemistry & Molecular BiologyXL-T Q.15 · MSQ (one or more correct) · 2 marks, no negative marking

Which of the following techniques is/are used for determining the three-dimensional structure of proteins?

  1. ACryo-electron Microscopy
  2. BCircular Dichroism
  3. CNuclear Magnetic Resonance Spectroscopy
  4. DX-ray Diffraction
Show answer
Answer (select all of these): (A) Cryo-electron Microscopy and (C) Nuclear Magnetic Resonance Spectroscopy and (D) X-ray Diffraction
Q100Biochemistry & Molecular BiologyXL-T Q.16 · MSQ (one or more correct) · 2 marks, no negative marking

Among the following statements, which is/are TRUE regarding the replication of DNA?

  1. AReplication is bidirectional and conservative in nature.
  2. BReplication in eukaryotes takes place at multiple Ori sites simultaneously.
  3. CBoth the strands replicate in discontinuous manner.
  4. DOne strand replicates in continuous while the other replicates in discontinuous manner.
Show answer
Answer (select all of these): (B) Replication in eukaryotes takes place at multiple Ori sites simultaneously. and (D) One strand replicates in continuous while the other replicates in discontinuous manner.
Q101Cell BiologyXL-T Q.17 · MSQ (one or more correct) · 2 marks, no negative marking

Which of the following statements is/are TRUE for Colchicine?

  1. AIt binds to tubulin molecule and disrupts the assembly/polymerization of microtubule.
  2. BIt inhibits crossover of chromosomes during meiosis.
  3. CIt inhibits chromosome condensation during Prophase.
  4. DIt blocks mitotic cells in Metaphase.
Show answer
Answer (select all of these): (A) It binds to tubulin molecule and disrupts the assembly/polymerization of microtubule. and (D) It blocks mitotic cells in Metaphase.
Q102GeneticsXL-T Q.18 · Numerical answer · 2 marks, no negative marking

Wild-type Drosophila females having three linked genes (AABBCC) were crossed with triple recessive mutant (aabbcc) males. The F1 female progenies (AaBbCc) were back crossed with the triple negative mutant (aabbcc) males. The cross resulted in following number of progenies in F2:

AaBbCc241
Aabbcc112
aaBbCc103
aabbcc252
aaBbcc17
aabbCc134
AabbCc14
AaBbcc127
Total1000

The order of genes as determined from the above data was found to be “ABC” (note that the order is equivalent to “CBA” and the order outside the markers are arbitrary). The recombination map distance (in centi Morgan) between “A to C” is ______ (round off to one decimal place).

Type a number as your answer.

Show answer
Answer: 53.8 (numerical answer, as in the official key).
Q103Biochemistry & Molecular BiologyXL-T Q.19 · Numerical answer · 2 marks, no negative marking

The length of a double helical DNA molecule is 13.6 km. If the DNA double helix weighs 1 × 10−18 g per 1000 nucleotide pairs and rise per base pair is 3.4 Å, then weight of the double helical DNA molecule (in nanogram) will be ______ (in integer).

Type a number as your answer.

Show answer
Answer: 40 (numerical answer, as in the official key).

XL-U: Food Technology (Q104–122)

19 questions (8 of 1 mark, 11 of 2 marks). Optional: answer it only if you pick Food Technology. From GATE 2027 this section is called XL5.

GATE XL 2023, XL-U Food Technology: questions by topic
19 questions, tagged by Lesson Basket.
Food Chemistry & Nutrition52.6%
Food Engineering26.3%
Food Products Technology10.5%
Food Microbiology10.5%
Topic tags: Lesson Basket. See also our GATE-XL weightage analysis.
Q104Food Chemistry & NutritionXL-U Q.1 · MCQ · 1 mark, −1/3 for a wrong answer

Choose the correct group of fat soluble vitamins

  1. ACholecalciferol, α-Tocopherol, Menadione
  2. BThiamine, Cholecalciferol, α-Tocopherol
  3. CNiacin, α-Tocopherol, Menadione
  4. DBiotin, Thiamin, Niacin
Show answer
Answer: (A) Cholecalciferol, α-Tocopherol, Menadione
Q105Food Chemistry & NutritionXL-U Q.2 · MCQ · 1 mark, −1/3 for a wrong answer

The synthesis of thyroxine T4 in human body requires

  1. ASelenium
  2. BIodine
  3. CIron
  4. DZinc
Show answer
Answer: (B) Iodine
Q106Food Chemistry & NutritionXL-U Q.3 · MCQ · 1 mark, −1/3 for a wrong answer

Which among the followings is NOT an essential amino acid?

  1. AL-Phenylalanine
  2. BL-Valine
  3. CL-Lysine
  4. DL-Arginine
Show answer
Answer: (D) L-Arginine
Q107Food EngineeringXL-U Q.4 · MCQ · 1 mark, −1/3 for a wrong answer

The time required for stipulated destruction of a microbial population at a given temperature is

  1. AD-value
  2. BF-value
  3. Cz-value
  4. DQ10 value
Show answer
Answer: (B) F-value
Q108Food Chemistry & NutritionXL-U Q.5 · MCQ · 1 mark, −1/3 for a wrong answer

Which among the following statements is NOT correct?

  1. ACod fish is a major source of ω-3 fatty acids.
  2. BBeetroot is a good source of β-carotene.
  3. CApple is a good source of vitamin B12.
  4. DFresh sugarcane juice is a good source of polyphenol oxidase.
Show answer
Answer: (C) Apple is a good source of vitamin B12.
Q109Food EngineeringXL-U Q.6 · Numerical answer · 1 mark, no negative marking

Calculate the efficiency in percent (rounded off to 1 decimal place) of an oil expeller which yields 37 kg oil containing 5% solid impurities from 100 kg mustard seeds. The oil content of the mustard seed is 38%.

Type a number as your answer.

Show answer
Answer: any value from 92.0 to 93.0 (numerical answer, as in the official key).
Q110Food Chemistry & NutritionXL-U Q.7 · Numerical answer · 1 mark, no negative marking

Orange juice is packaged aseptically and stored under ambient conditions. The degradation of vitamin C in the juice occurs during storage and it follows first order reaction kinetics. The degradation rate constant is 5.2×10-3 day-1. The half-life of vitamin C in days is ____________ (in integer).

Type a number as your answer.

Show answer
Answer: any value from 132 to 134 (numerical answer, as in the official key).
Q111Food Products TechnologyXL-U Q.8 · Numerical answer · 1 mark, no negative marking

The weight of 10 kg dried cauliflower containing 5% moisture (wet basis) after rehydration is 60 kg. If the fresh cauliflower contained 87% moisture (wet basis), calculate the coefficient of rehydration (rounded off to 2 decimal places).

Type a number as your answer.

Show answer
Answer: any value from 0.75 to 0.90 (numerical answer, as in the official key).
Q112Food MicrobiologyXL-U Q.9 · MCQ · 2 marks, −2/3 for a wrong answer

Some of the industrial products are produced by fermentation processes. Identify the correct pair of product and fermentative microorganism.

  1. AVinegar – Acetobacter aceti
  2. BCitric acid – Enterbacter aerogenes
  3. CEthanol – Saccharomyces cerevisiae
  4. DL-Lysine – Aspergillus niger
Show answer
Marks to all. The organising institute awarded this question’s marks to every candidate in the final answer key, so it has no single correct answer.
Q113Food Chemistry & NutritionXL-U Q.10 · MCQ · 2 marks, −2/3 for a wrong answer

Choose the correct statement(s) about the enzyme and its application in food processing reaction.

  1. AChymosin is widely used in cheese manufacturing.
  2. BThermolysin is used in the synthesis of Aspartame.
  3. Cβ-Galactosidase catalyzes the hydrolysis of galactose.
  4. DLipase is used for restructuring of acyl glycerol.
Show answer
Marks to all. The organising institute awarded this question’s marks to every candidate in the final answer key, so it has no single correct answer.
Q114Food MicrobiologyXL-U Q.11 · MCQ · 2 marks, −2/3 for a wrong answer

Identify the Gram +ve bacteria responsible for causing food borne diseases among the followings

  1. ACampylobacter jejuni
  2. BClostridium botulinum
  3. CVibrio cholerae
  4. DSalmonella typhi
Show answer
Answer: (B) Clostridium botulinum
Q115Food EngineeringXL-U Q.12 · MCQ · 2 marks, −2/3 for a wrong answer

Extrusion cooking is accomplished in four different stages, which are indicated as I, II, III and IV in the figure given below. Choose the correct option representing the name of each stage.

GATE XL 2023 Q115 figure: Extrusion cooking is accomplished in four different stages, which are indicated as I, II, III and IV in the fi
  1. AI – Feeding, II – Cooking, III – Kneading, IV – Expansion
  2. BI – Kneading, II – Feeding, III – Cooking, IV – Expansion
  3. CI – Feeding, II – Kneading, III – Cooking, IV – Expansion
  4. DI – Cooking, II – Kneading, III – Feeding, IV – Expansion
Show answer
Answer: (C) I – Feeding, II – Kneading, III – Cooking, IV – Expansion
Q116Food Chemistry & NutritionXL-U Q.13 · MCQ · 2 marks, −2/3 for a wrong answer

Match the method/ value used for measuring lipid characteristics in Column I with the corresponding properties indicated by them, in Column II.

Column IColumn II
P. Thiobarbituric acid test1. Induction time
Q. Rancimat method2. Degree of unsaturation
R. Peroxide value3. Carbonyl content
S. Iodine value4. Hydroperoxide content
  1. AP-3, Q-1, R-4, S-2
  2. BP-1, Q-3, R-4, S-2
  3. CP-3, Q-1, R-2, S-4
  4. DP-3, Q-4, R-1, S-2
Show answer
Answer: (A) P-3, Q-1, R-4, S-2
Q117Food Products TechnologyXL-U Q.14 · MCQ · 2 marks, −2/3 for a wrong answer

Match the peeling technique in Column I with the vegetable, for which it is used in industry, given in Column II.

Column IColumn II
P. Knife peeling1. Brinjal
Q. Abrasion peeling2. Tomato
R. Flame peeling3. Potato
S. Flash peeling4. Cucumber
  1. AP-3, Q-4, R-1, S-2
  2. BP-4, Q-1, R-3, S-2
  3. CP-4, Q-3, R-2, S-1
  4. DP-4, Q-3, R-1, S-2
Show answer
Answer: (D) P-4, Q-3, R-1, S-2
Q118Food Chemistry & NutritionXL-U Q.15 · MCQ · 2 marks, −2/3 for a wrong answer

Match the process in Column I with the related food component in Column II.

Column IColumn II
P. Caramelization1. Lipid
Q. Denaturation2. Sugar
R. Oxidation3. Pigment
S. Bleaching4. Enzyme
  1. AP-2, Q-4, R-1, S-3
  2. BP-2, Q-1, R-4, S-3
  3. CP-1, Q-3, R-2, S-4
  4. DP-2, Q-1, R-3, S-4
Show answer
Answer: (A) P-2, Q-4, R-1, S-3
Q119Food Chemistry & NutritionXL-U Q.16 · MSQ (one or more correct) · 2 marks, no negative marking

Identify the correct statement(s) related to grain polysaccharides among the followings.

  1. ADextrin are a group of low molecular weight polysaccharides produced by dry hydrolysis of starch.
  2. BAmylose is a linear polymer of D-glucose units joined by α (1→6) glycoside linkages.
  3. CAmylopectin is a branched chain polymer of D-galactose monomer units.
  4. DRetrogradation is a process of reassociation of amylose and formation of crystalline structure by gelatinized starch upon cooling.
Show answer
Answer (select all of these): (A) Dextrin are a group of low molecular weight polysaccharides produced by dry hydrolysis of starch. and (D) Retrogradation is a process of reassociation of amylose and formation of crystalline structure by gelatinized starch upon cooling.
Q120Food Chemistry & NutritionXL-U Q.17 · Numerical answer · 2 marks, no negative marking

A sample of glucose isomerase enzyme converts 15 µmoles of substrate glucose into product fructose min-1 mL-1 under standard assay conditions. The enzyme activity of the glucose isomerase in International Unit (IU) is _______ (in integer).

Type a number as your answer.

Show answer
Answer: 15 (numerical answer, as in the official key).
Q121Food EngineeringXL-U Q.18 · Numerical answer · 2 marks, no negative marking

If D10 for Salmonella in egg yolk is 0.75 kGy, calculate the radiation dose in kGy (rounded off to 2 decimal places) required for reducing the Salmonella count in egg yolk by 8 log cycles.

Type a number as your answer.

Show answer
Answer: any value from 13.00 to 14.50 (numerical answer, as in the official key).
Q122Food EngineeringXL-U Q.19 · Numerical answer · 2 marks, no negative marking

The average moisture binding energy of a textured protein product (TPP) at 8% moisture content (dry basis) is 3200 cal mol−1. If the water activity of the TPP at the above moisture content is 0.30 at 30 ℃, the water activity of the sample at 45 ℃ is __________ (rounded off to 2 decimal places). The value of Gas constant R = 1.987 cal mol−1 K−1.

Type a number as your answer.

Show answer
Answer: any value from 0.32 to 0.41 (numerical answer, as in the official key).

Changes in the final answer key

Marks to all: Q6, Q112, Q113. Every other question has the answer shown.

How to revise from this paper

  1. Pick your two optional sections firstAttempt GA, XL-P and the two sections you plan to choose, 65 questions in one 3-hour sitting.
  2. Score it the GATE wayAdd 1 or 2 marks per correct answer and subtract 1/3 or 2/3 only for wrong MCQs. MSQ must be fully correct to score.
  3. Sort your mistakes by topicUse the topic tag on each question. Two or three misses in the same topic show where to revise first.
  4. Repeat with another yearTake a different year a week later and compare your topic-wise mistakes.

Keep going

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Source: GATE 2023 Life Sciences (XL) question paper and final answer key, published by the organising institute (IIT Kanpur). Question text is reproduced for practice; any transcription error is ours, and the official documents are the authority.
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