Q38. Let X, Y and Z be three amino acids which are encoded by UCA=X, ACU=Y and GUC=Z. What will be the sequence of nucleotide in DNA coding/sense strand that comprises information for following polypeptide: Polypeptide= Y-X-Y-Y-X-Z-Z-X-Z-Y a) 5′-ACUUCAACUACUUCAGUCGUCUCAGUCACU-3′ b) 5′-ACTTCAACTACTTCAGTCGTCTCAGTCACT-3′ c) 3′-TGAAGTTGATGAAGTCAGCAGAGTCAGTGA-5′ d) 3′-UGAAGUUGAUGAAGUCAGCAGAGUCAGUGA-5′
- Translate the polypeptide into codons, using UCA = X, ACU = Y and GUC = Z. Y-X-Y-Y-X-Z-Z-X-Z-Y gives ACU UCA ACU ACU UCA GUC GUC UCA GUC ACU.
- The mRNA is therefore 5′-ACUUCAACUACUUCAGUCGUCUCAGUCACU-3′.
- The coding (sense) strand has the same sequence as the mRNA, except that T replaces U. Its sequence is 5′-ACTTCAACTACTTCAGTCGTCTCAGTCACT-3′.
- (a) contains U, and DNA has no uracil.
- (c) is the template (antisense) strand, written 3′ to 5′ and complementary to the sense strand.
- (d) contains U and is also written 3′ to 5′, so it is not DNA.
EXPLAIN C
| Sense (coding) strand | 5′-ACTTCAACTACTTCAGTCGTCTCAGTCACT-3′ |
| Template (antisense) strand | 3′-TGAAGTTGATGAAGTCAGCAGAGTCAGTGA-5′ |
- The question asks for the coding/sense strand. This is the strand whose sequence matches the mRNA, with T in place of U.
- RNA polymerase reads the template strand 3′→5′. It builds the mRNA 5′→3′, so the mRNA is complementary to the template strand and identical to the coding strand.
- Transcribing (c) gives 5′-ACUUCAACUACUUCAGUCGUCUCAGUCACU-3′. That mRNA does encode Y-X-Y-Y-X-Z-Z-X-Z-Y, so (c) does carry the information. However, it carries it as the template, not as the coding strand.
Q43. When the muscle is at rest, is disposed in such a way that it covers the sites on the chain where the head of becomes attached. a) tropomyosin : actin : myosin b) myosin : actin : tropomyosin c) actin : tropomyosin : myosin d) actin : myosin : tropomyosin
- Tropomyosin is a long, thin protein that lies along the groove of the actin filament.
- At rest, it blocks the myosin-binding sites on actin. The troponin complex holds it in this position.
- When a nerve impulse arrives, calcium is released from the sarcoplasmic reticulum. Ca²⁺ binds troponin C, and the troponin-tropomyosin complex shifts to uncover the binding sites. The myosin heads can then attach and form cross-bridges, so contraction begins.
Q44. If green is a dominant and yellow is a recessive trait then what would be the ratio of homozygotes in F2 generation? a) 0% b) 25% c) 50% d) 75%
- P generation: GG (green) × gg (yellow).
- F1: all Gg (green).
- F2 (Gg × Gg): 1 GG : 2 Gg : 1 gg.
Q45. In human respiratory system, bronchioles have: a) Cartilage rings b) Cartilage plates c) Circular smooth muscles d) Longitudinal smooth muscles
- Bronchioles have no cartilage. Cartilage rings (trachea, primary bronchi) and cartilage plates (secondary and tertiary bronchi) are present only in the larger airways. This rules out (a) and (b).
- The bronchiolar wall has a layer of circularly arranged smooth muscle. It holds the airway open without cartilage, and contraction or relaxation changes the airway diameter. This is why bronchoconstriction in asthma narrows these airways.
- The epithelium thins to simple columnar or cuboidal, and goblet cells and glands are lost. Terminal bronchioles lead into respiratory bronchioles and then alveoli.
Q50. Over time, many dihybrid cross experiments of autosomally-linked genes in sweet pea plants (Lathyrus odoratus) have been carried out by reputed scientists. Which of the following best explains the presence of some non-parental phenotypes in the offspring of such crosses? a) Crossing over has occurred, leading to the production of recombinant plants b) The scientists made errors such as not cleaning the small paint brush that they used to transfer pollen grains between plants c) Independent assortment had occurred, so unusual phenotypes are to be expected d) Only plants of one gender exhibit parental phenotypes due to linkage
- Linked genes lie close together on the same chromosome and tend to be inherited together. This produces mostly parental phenotypes.
- During prophase I of meiosis, crossing over between non-sister chromatids of homologous chromosomes can separate the linked alleles. This produces recombinant gametes, which give the non-parental phenotypes in the offspring.
- The recombinants are a minority, because the closer two genes are, the less often a crossover falls between them. The recombination frequency reflects the distance between the genes. This is the basis of linkage mapping.
- The sweet pea flower colour and pollen shape genes, studied by Bateson and Punnett, were an early example of linkage.
- (b) is not a scientific explanation, since the results were consistent across many experiments.
- (c) is wrong because independent assortment applies to unlinked genes. Linked genes deviate from it, which is why the ratio is not 9:3:3:1.
- (d) is wrong because sweet pea flowers are hermaphroditic, and linkage does not depend on gender.
Q65. What unique characteristic is typically found in sensory neurons of spinal nerves? a) A single axon b) A single dendron c) Multiple dendrites d) Multiple nodes of Ranvier
- Sensory neurons of spinal nerves have their cell bodies in the dorsal root ganglion. They are pseudounipolar (unipolar) neurons.
- Only one process leaves the cell body. It then divides in a T shape into a peripheral branch and a central branch. The peripheral branch carries impulses from the receptor, and the central branch enters the spinal cord through the dorsal root.
- Textbooks describe the unipolar type as a cell body with one axon only, with no separate dendrites. That is why "a single axon" is the best answer.
- (b) "Single dendron" is not the standard description, even though the peripheral branch functions like a dendrite.
- (c) Multiple dendrites describe multipolar neurons, such as motor neurons and interneurons.
- (d) Multiple nodes of Ranvier are found on any myelinated axon, so they are not unique to these neurons.
Q85. The solubility of AgBr is 7.1 X 10⁻⁷ M. This dissolved AgBr produces equal moles of Ag+ and Br - , then the solubility product of this equilibrium reaction would be: a) 5.041 X 10⁻¹³ b) 4.941 X 10⁺¹³ c) 7.1 X 10⁻⁷ d) 14.082 X 10⁻¹⁴
- (b) has a positive exponent, which is impossible for a sparingly soluble salt.
- (c) is just the solubility, not its square.
- (d) doubles s² instead of squaring it. 14.082 × 10⁻¹⁴ is 2 × 7.041 × 10⁻¹⁴, which is not s².
Q89. Acetic acid is converted into acetyl chloride using which reagent? a) thionyl chloride b) chlorine c) sodium chloride d) phosphorous trioxide
- (b) Chlorine (Cl₂) does not replace the OH group of the acid. It chlorinates the alpha carbon instead (Hell-Volhard-Zelinsky type), which gives chloroacetic acid.
- (c) Sodium chloride is a salt and is not a chlorinating agent.
- (d) Phosphorus trioxide is not a standard reagent for this conversion. The usual phosphorus reagents are PCl₅ and PCl₃. PCl₃ gives acetyl chloride with H₃PO₃ as the by-product.
Q92. The value of ideal gas constant in dm³.torr. K⁻¹.mol⁻¹ is: a) 0.0821 b) 1.98722 c) 8.3143 d) 62.364
- (a) 0.0821 is in dm³·atm·K⁻¹·mol⁻¹.
- (b) 1.98722 is in cal·K⁻¹·mol⁻¹.
- (c) 8.3143 is in J·K⁻¹·mol⁻¹.
Q93. How many unpaired electrons are there in Fe⁺⁺(Z=26)? a) 6 b) 4 c) 3 d) 2
- Fe (Z = 26) has the configuration [Ar] 3d⁶ 4s².
- Fe²⁺ forms by losing the two 4s electrons first, giving [Ar] 3d⁶.
- The six 3d electrons fill the five d orbitals following Hund's rule. One orbital holds a pair (↑↓) and the other four hold one electron each (↑ ↑ ↑ ↑).
Q95. A thermodynamic system goes through a cyclic process which absorbs 1200 J heat and ejects 700 J heat. Which statement correctly describes this? a) The ∆E of a system increases by 500 J b) The ∆E of a system decreases by 500 J c) The system performs network of 500 J d) The surroundings perform network of 500 J Q96. Which of the following elements from Period 3 of Periodic table, has the largest ionic radius when forms a stable ion? a) Sodium b) Silicon c) Phosphorous d) Chlorine Q97. Which type of interaction is primarily responsible for holding molecules together in noble gases? a) Covalent bond b) Electrostatic forces of attraction c) Hydrogen bonding d) London dispersion forces Q98. The reaction between ethyl chloride and dry silver oxide on heating yields: a) dimethyl ether. b) diethyl ether c) ethanoic acid. d) ethanal.
- The process is cyclic, so the system returns to its initial state and ΔE = 0. This rules out (a) and (b).
- Net heat absorbed is q = 1200 − 700 = 500 J.
- From the first law, ΔE = q + w = 0, so w = −500 J. The sign convention here gives w as work done on the system, so the system does 500 J of work on the surroundings. This is how a heat engine behaves.
- The stable ions are Na⁺, Si⁴⁺ (or Si⁴⁻), P³⁻ and Cl⁻.
- Na⁺ and Si⁴⁺ have lost their outer shell, so they are small cations (about 102 pm and 40 pm).
- P³⁻ and Cl⁻ are isoelectronic with Ar (18 electrons). The one with the lower nuclear charge has the larger radius, because the electrons are held less tightly.
- P³⁻ (Z = 15) is larger than Cl⁻ (Z = 17), at about 212 pm against 181 pm.
Q102. Carboxylic acid can be formed by reaction of Grignard reagent with: a) Carbonate b) Carbon dioxide c) Aldehyde d) Carbon mono oxide
- (a) Carbonate is not a useful electrophile for this reaction.
- (c) An aldehyde gives a secondary alcohol (formaldehyde gives a primary alcohol).
- (d) Carbon monoxide is not a standard reagent for this reaction.
Q109. Rate of reaction in E₁ depends on the: a) R-X only b) Base only c) Nucleophile only d) Both base and R-X
- Slow step (rate-determining): the leaving group departs from R-X and forms a carbocation.
- Fast step: a base removes a β-hydrogen and the alkene forms.
Q111. Grignard's reagent is treated with methanal to produce: a) Phenol b) Primary alcohol c) Propanoic acid d) Ketone Q112. Probability of finding an electron is maximum in: a) Orbit b) Orbital c) Degenerate orbit d) Nodes
- (a) An orbit is a fixed circular path from Bohr's model. Quantum mechanics (the Heisenberg uncertainty principle) does not support it.
- (c) Degenerate orbitals are orbitals of equal energy. "Degenerate orbit" is not a meaningful term here.
- (d) Nodes are regions where the probability of finding the electron is zero.
Q121. The oxidation number of both nitrogen atoms in NH₄NO₃ is/are: a) +4 and -4 b) -3 and +5 c) -4 and +6 d) -4 and +5
- NH₄⁺: H is +1, so x + 4(+1) = +1, which gives x = -3.
- NO₃⁻: O is -2, so x + 3(-2) = -1, which gives x = +5.
Q132. If the speed of a standing wave in a string is 2 m/s and its wavelength is 0.25 m, then the frequency of the standing wave will be: a) 0.125 Hz. b) 2.0 Hz. c) 4.0 Hz. d) 8.0 Hz. Q133. The distance traveled by a wave in unit time is termed as: a) Wavelength b) Amplitude c) Displacement d) Wave speed
Q137. When a car is moving with constant acceleration, the shape of velocity time graph is: a) Parabola b) Incline curve c) Straight line d) Decline curve
- A parabola is the shape of the position-time graph under constant acceleration, since s = ut + ½at².
- A curved line on a velocity-time graph means the acceleration is changing.
Q143. A conducting coil with its plane perpendicular to a uniform magnetic field is moved at constant speed in the field region, the induced emf in the coil is: a) zero b) constant c) increasing d) decreasing
- The field is uniform, so B is the same everywhere in the region.
- The coil is perpendicular to the field, so θ = 0 and cosθ = 1.
- Moving the coil at constant speed inside the field region does not change B, A or θ. The flux stays constant and dΦ/dt = 0.
But the key says it's constant. Constant.
- The field is uniform and the coil's plane stays perpendicular to it, so the flux is Φ = BA throughout the motion.
- Moving the coil at constant speed inside the field region changes neither B nor A, so Φ stays constant.
- Faraday's law gives ε = −dΦ/dt = 0. Nothing is being cut at a net rate. In a rectangular coil, the motional emfs on opposite sides are equal and opposite, so they cancel around the loop.
- The key may be mistaken.
- The question writer may have been thinking of the flux being constant, or of the motional emf on each side being constant (vBl). That does not give a net emf in the loop.
- If the coil were entering or leaving the field region, the emf would be constant (for a uniform speed) while the flux changes. The question says the coil is moved in the field region, which describes a coil fully inside the field.
Q144. The work done between two points per unit positive charge against the electric field due to a source charge is called as: a) Electric Power b) Electric Potential difference c) Electric flux d) Electric field Intensity
- (a) Electric power is the rate of energy transfer (P = VI), in watts.
- (c) Electric flux is the number of field lines through a surface (Φ = E·A).
- (d) Electric field intensity is the force per unit positive charge (E = F/q₀), in N/C. It is force, not work.
Q148. Two identical masses are placed at heights h and 2h, the ratio of gravitational potential energies will be: a) 1:2 b) 2:1 c) equal d) dependent on mass
Q152. In an isothermal process internal energy of the system: a) Decreases b) Increases c) Remain the same d) Becomes zero
Q153. A gas expands adiabatically and does 10 J of work then the change in internal energy is: a) 10J b) -10J c) 20J d) 0J
Q158. If 10 waves pass through a denser medium in 1 second with a speed of 10 m/sec, then the wavelength of the wave will be: a) 0.1 m b) 1 m c) 10 m d) 100 m
Q162. What is meant by random nature of radioactive decay of a substance? a) We cannot control the decay rate. b) Probability of decay of every unstable nucleus is constant. c) We can locate every next decaying nuclei. d) We can speed up the decay process.
- We cannot predict when a particular nucleus will decay. Each unstable nucleus has the same fixed probability of decaying in a given time interval, and this probability is the decay constant λ.
- Because of this, the number of decays in a large sample follows N = N₀e^(−λt), and the activity falls exponentially.
- The probability does not depend on the nucleus's age or history, or on its surroundings.
- (a) is true but it is a consequence, not the meaning of randomness. Temperature, pressure and chemical state do not change the decay rate.
- (c) is false because we cannot locate or predict which nucleus decays next.
- (d) is false because decay cannot be sped up by ordinary physical or chemical means.
Q166. The captain stood at the head of the table during the meeting. What does "head" most likely mean? a) A body part b) A leader or position of authority c) The front part of something d) A senior person
- (a) A body part is the literal meaning. The captain isn't standing at a person's head.
- (b) Leader or position of authority could be argued, because the head of a table is often where the leader sits. But the sentence says where the captain stood, and "the head of the table" is a fixed phrase for the end of the table.
- (d) A senior person doesn't fit, because the head here belongs to the table and not to the captain.
Can you suggest some book or page number of a specific book for the evidence of the above question of English? 105. إنفيايسي. الـ SBI SE. SBI SE. SBI SE. E. Tchau.
"head of the table" definition dictionary the place at the end of a table
- Reverso Dictionary lists "head of the table" as a noun meaning a seating position at the end of a table. Its examples are "He sat at the head of the table during dinner" and "The host always sits at the head of the table."
- WordReference Forums has a discussion thread that describes the phrase as the most important seat, or the person in it, at the "top" end of a long table. This is a forum thread, not an authoritative reference.
- Merriam-Webster defines "head table" as the table where the most important people sit during a formal meal. This is a related term, not the exact phrase.
- Oxford Advanced Learner's Dictionary
- Cambridge Dictionary
- Merriam-Webster
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