Life Sciences
GATE XL discipline:life-sciences PYQs
Build your understanding with focused practice from past exam papers.
Questions 141–150 of 1235. Imported answers require independent review unless marked reviewed.
Question 19
Open questionXL-PNAT1 markNeeds review
Consider the two
values of valine as 2.32 and 9.62. The isoelectric point (pI) of this amino acid is _____. (rounded off to two decimal places)Log in to check the answer
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Open questionXL-PMCQ2 marksNeeds review
A few species are given in Column I. Column II contains the hybrid orbitals used by the central atom of the species for bonding.
The CORRECT match for the species to their central atom hybridization is
(Given: Atomic numbers of B: 5; C: 6; O: 8; F: 9; P: 15; Cl: 17; I: 53)
Column I — Species: i. I₃⁻; ii. PCl₃; iii. BF₃; iv. CO₂
Column II — Hybrid orbitals used by the central atom for bonding: a. sp; b. sp²; c. sp³; d. sp³d
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Open questionXL-PMCQ2 marksNeeds review
For product formation from only one type of reactant (e.g. A → product), the CORRECT match for the order of the reaction (given in Column I) with the half-life expression (given in Column II) is
([A]₀ is the initial concentration and kᵣ is the rate constant)
Column I: Order | Column II: Half-life expression |
|---|---|
i. Zero | P. ln 2 / kᵣ |
ii. First | Q. [A]₀ / 2kᵣ |
iii. Second | R. 1 / kᵣ[A]₀ |
| S. 2kᵣ / [A]₀ |
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Open questionXL-PMSQ2 marksNeeds review
The CORRECT statement(s) for the given reactions is(are)
Reactions:
I) Me–CH(Me)–CHO + i) MeMgBr; ii) H₃O⁺ →
II) MeCHO + i) ⁱPrMgBr; ii) H₃O⁺ →
III) Me–CH(Me)–CO₂Me + i) MeMgBr (excess); ii) H₃O⁺ →
IV) Me₂CO + i) ⁱPrMgBr; ii) H₃O⁺ →
Possible products:
P: Me–C(OH)(Me)–CH₂Me
Q: Me–C(OH)(Me)–CH(Me)₂
R: Me–CH(Me)–CO₂Me

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Open questionXL-PMSQ2 marksNeeds review
Addition of a few drops of concentrated HCl to an aqueous solution of CoCl₂ forms a dark blue complex X.
The CORRECT statement(s) for this reaction is(are)
(Given: Atomic number of Co: 27)
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Open questionXL-PMSQ2 marksNeeds review
The CORRECT statement(s) regarding biomolecules is(are)
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Open questionXL-PNAT2 marksNeeds review
Energy of the transition from
to
for hydrogen atom is E × 10³ cm⁻¹.Given: Rydberg constant for hydrogen: 1.097 × 10⁷ m⁻¹.
Value of E is _____________. (rounded off to two decimal places)
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Open questionXL-PNAT2 marksNeeds review
A non-volatile solute has a molecular weight of 180 g mol⁻¹. Assume that the solute does not associate or dissociate and the boiling-point constant (ebullioscopic constant) of water is 0.51 K kg mol⁻¹.
The amount (in g) of solute added to 500 g of water to elevate the boiling point by 0.153 K is ______ (answer in integer)
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Open questionXL-PNAT2 marksNeeds review
The standard potentials (E⁰) for the Fe³⁺/Fe and Fe³⁺/Fe²⁺ couples are −0.04 V and +0.76 V, respectively.
Given: Faraday constant = 96500 C mol⁻¹.
The value for E⁰ (Fe²⁺/Fe), in V, is _____________. (rounded off to two decimal places)
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Open questionXL-QMCQ1 markNeeds review
Zinc is essential for the function of
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