Uhs Mdcat 2019 — Solved Past Paper with Answers

Q1. The culpable child _ some words to her mother for pardoning his delinquency.

• A. Mumbled✓
• B. Showy
• C. Rude
• D. Crazy

Explanation: 'Mumbled' is to say something indistinctly and quietly, making it difficult for others to hear. So, the correct sentence is, "The culpable child mumbled some words to her mother for pardoning his delinquency."

Q2. She was feeling _ even after five hours of the surgery.

• A. Mumbled
• B. Haggard✓
• C. Pally
• D. Grope

Explanation: 'Haggard' means looking exhausted and unwell, especially from fatigue, worry, or suffering. So, the correct sentence is, "She was feeling haggard even after five hours of the surgery."

Q3. The parents were stunned when they saw that children had created complete _ in the bedroom.

• A. Knack
• B. Groggy
• C. Dank
• D. Mayhem✓

Explanation: 'Mayhem' means violent or extreme disorder; chaos. So, the correct sentence is, "The parents were stunned when they saw that children had created complete mayhem in the bedroom."

Q4. I _ caution in interpreting these results.

• A. Urge✓
• B. Usher
• C. Usurp
• D. Uproot

Explanation: As a noun, 'urge' means a desire. As a verb, it means to strongly encourage. 'Urge' is related to the word, 'urgent' or 'pressing'. An 'urge' is a pressing want, one that is almost a compulsion, like when you're so frustrated, you have the urge to scream. If you urge someone to do something, you feel strongly about it.

Q5. They sometimes feel a _ for the mountains and the sea.

• A. Yearning✓
• B. Yapping
• C. Yelling
• D. Yielding

Explanation: 'Yearning' is a feeling of intense longing for something. Phrase: "He felt a yearning for the mountains".

Q6. The new teacher showed no _ about hitting the students.

• A. Quakes
• B. Qualms✓
• C. Quarrel
• D. Quotation

Explanation: 'Qualms' is an uneasy feeling of doubt, worry, or fear, especially about one's own conduct; a misgiving. Phrase: "Military regimes generally have no qualms about controlling the press." “Some people have no qualms about correcting other people's grammar."

Q7. The accident happened due to the driver’s _ .

• A. Negligence✓
• B. Reluctance
• C. Regret
• D. Quotation

Explanation: 'Negligence' is the failure to take proper care over something. Phrase: "His injury was due to the negligence of his employers."

Q8. In the following question, four alternative sentences are given. Choose the CORRECT option.

• A. I was been to America for a medical check up.
• B. I had being to America for a medical check up.
• C. I have been to America for a medical check up.✓
• D. I has been to America for a medical check up.

Explanation: This is an example of a Past Perfect tense sentence due to the presence of 'have' and 'been' implies an incident, that occurred in the past. The best way to solve this question is to read each option in your head and then decide which one sounds the best and fits in well.

Q9. In the following question, four alternative sentences are given. Choose the CORRECT option.

• A. After breaking the glass, Ruby said "Please don't tell on me".
• B. After breaking the glass Ruby said: "Please don't tell on me."
• C. After breaking the glass, Ruby said: "Please don't tell on me."✓
• D. After breaking the glass Ruby said: please don't tell on me.

Explanation: The sentence 'after breaking the glass' is another idea in the sentence hence a comma should be added between these two sentences. The rule of punctuation in quotation marks is that the full stop or comma should come within the quotation marks. 'Don't' is a contraction, which stands for 'do not'.

Q10. In the following question, a sentence is given with blanks to be filled in with an appropriate word. Four alternatives are suggested. Choose the correct alternative, out of the four as your answer.Ravens appear to behave _, actively helping one another to find food.

• A. Mysteriously
• B. Warily
• C. Aggressively
• D. Defensively✓

Explanation: Option D is the correct answer because this sentence asks you to look for a word that describes how the ravens behave. The information after the comma restates and defines the meaning of the missing word.

Q11. The prince abdicated the crown and returned to his castle. Abdicated means:

• A. Gave up✓
• B. Sold
• C. Reinvested into
• D. Auctioned

Explanation: 'The word "abdicated" means to renounce or give up a high office or responsibility, such as a monarch giving up their crown.

Q12. SPOT THE ERROR: In the following sentence, some segments of the sentence are underlined. Your task is to identify that underlined segment of the sentence, which contains the mistake that needs to be corrected.

• A. Option A
• B. Option B✓
• C. Option C
• D. Option D
• E. No error

Explanation: Correct is option B as 'social and economic progress, good governance' are the main verbs which are plural so 'take' should be used.

Q13. SPOT THE ERROR: In the following sentences, some segments of the sentence are underlined. Your task is to identify that underlined segment of the sentence, which contains the mistake that needs to be corrected.

• A. Let's
• B. The
• C. is
• D. to✓

Explanation: The incorrect preposition 'of' is used here. It should be replaced by 'for'. 'The bus is leaving for the last stop.'

Q14. Ester linkages occur in:

• A. Nucleic acids✓
• B. Nucleus
• C. Carbohydrates
• D. Proteins

Explanation: DNA and RNA are the chain of polynucleotides, that are joined by ester linkage known as 'phosphodiester bonds'. The two nucleotides, within a nucleotide chain, are linked by a phosphodiester bond. It is formed by the linkage between the 3' carbon atom of one sugar molecule and the 5' carbon atom of another, with the phosphate group acting as a bridge between the 2 carbons.So, the correct answer is option A.

Q15. The finger like infoldings, which are formed by inner membrane of mitochondria, are called:

• A. Ribosomes
• B. Matrix
• C. Porin
• D. Cristae✓

Explanation: Factual recall; 'Ribosomes' are organelles, that serve as sites of protein synthesis. 'Matrix' refers to the space confined by the inner mitochondrial membrane within the mitochondria. 'Porin' refers to a transport protein that enables the passive diffusion of certain molecules across a membrane. Cristea are finger-like infoldings, which are formed by the inner membrane of the mitochondria.

Q16. The main neurotransmitter for synapses is _, which lie outside the central nervous system.

• A. Acetylcholine✓
• B. Acetaldehyde
• C. Choline
• D. Phosphatidylcholine

Explanation: Acetylcholine is a neurotransmitter that is used in transmission across synapses within both, the central nervous system and peripheral nervous system (i.e. outside the central nervous system). Acetaldehyde (B) is an aldehyde with two carbon atoms and is not used as a neurotransmitter. Choline (C) is a constituent of the neurotransmitter, 'Acetylcholine' but, by itself, is not a neurotransmitter. Phosphatidylcholine (D) contains a choline subunit, similar to acetylcholine, but itself is not a neurotransmitter. It is actually a constituent of cell membranes as well as of surfactant, which is present in alveoli.

Q17. The structure present in a eukaryotic cell but absent in prokaryotic cells is:

• A. Nucleus✓
• B. DNA
• C. Ribosomes
• D. Cell surface membrane

Explanation: :The term ' prokaryotes' is derived from the Greek word ' pro meaning before' and ' karyon meaning nucleus'. Prokaryotic cells are cells that do not have a true nucleus and membrane-bound organelles.

Q18. If water has a high latent heat of vaporization, how could this property of water be helpful to plants and animals?

• A. With the release of large amount of water, a small amount of heat loss can take place.
• B. With the release of small amount of water vapours, a great amount of heat loss can take place.✓
• C. It will keep their temperature very high.
• D. No cooling effect, with the release of even a large amount of water vapours.

Explanation: As per the formula Q=mL, since water has a high latent heat of vaporization (L), for a small mass or amount of water (m), a large amount of energy (Q) will be released which results in a large amount of heat loss, enabling the organism to remain cool via sweating (animals) or transpiration (plants). Option A describes this effect but in reverse, option C states the opposite effect as this process prevents their temperature from becoming too high, and option D is false as a cooling effect does indeed occur.

Q19. Which cell organelle is responsible for protein modification?

• A. Chloroplast
• B. Golgi body✓
• C. Ribosome
• D. Mitochondrion

Explanation: The 'Golgi apparatus' is a cellular organelle responsible for the modification and trafficking of proteins to other organelles such as the lysosome, the digestive organelle of the cell.

Q20. Nowadays every newborn gets regular shots of vaccine for polio. It contains _ for polio to make a child immune against this disease.

• A. Antibiotics
• B. Antibodies
• C. Antigens✓
• D. Antisera

Explanation: A vaccine elicits an immune response by the host so that the host is better equipped to fight against the invading organism, in any potential future instance, via the production of memory cells. It may do so by containing the antigens of the invading organism (in this case, poliovirus) so that the host’s immune system produces memory cells against the specific organism, thereby conferring active immunity against the disease. Antibiotics (option A) are not present in vaccines; they are drugs used to treat bacterial illnesses. Antibodies (option B) are not present in vaccines; they are proteins, produced by plasma cells, to ward off infection. They are produced by the immune system to fight an active infection, but they are not used in vaccines since they do not confer active immunity as the host does not produce memory cells against the invading organism. Antisera (option D) is a blood serum preparation that contains antibodies against a certain antigen or disease; they are used as a treatment for certain conditions and do not confer active immunity to the individual.

Q21. Change in frequency of alleles that occurs by chance is called as:

• A. Genetic drift✓
• B. Mutation
• C. Migration
• D. Natural selection

Explanation: The question describes the definition of a genetic drift, which is a change in the frequency of alleles that occurs purely by chance. A mutation (option B) describes a change in DNA sequence. Migration (option C) describes the influx or the efflux of individuals into or out of a population, respectively. Natural selection (D) can be confused with genetic drift as both are mechanisms for evolution, but the main difference lies in the fact that in natural selection, allele frequencies change over time not by chance, but due to certain alleles conferring a selective advantage to some individuals over the other (hence, it is not driven by chance, but driven by factors that are useful to the individual and ensures their survival).

Q22. Lipids contain double amount of energy as compared to the same amount of carbohydrates due to the presence of:

• A. Higher proportion of C-H✓
• B. Higher proportion of C-O
• C. Higher proportion of Oxygen
• D. Low proportion of C-H bonds

Explanation: C-H bond is a stronger bond than the C-O bond, and since lipids contain a much higher proportion of C-H bonds than carbohydrates (because lipids contain much less oxygen per mole than carbohydrates), lipids contain double the amount of energy than carbohydrates. The remaining three options (B,C,D) are incorrect as they explain the traits of lipids in reverse.

Q23. The phase of mitosis in which sister chromatids move towards opposite poles:

• A. Anaphase✓
• B. Metaphase
• C. Prophase
• D. Telophase

Explanation: Anaphase is the most critical phase of the mitosis, which ensures equal distribution of chromatids in the daughter cells. The kinetochore fibers of spindle contract towards their respective poles, at the same time polar microtubules elongates exert force and sister chromatids are separated from centromere. As a result, half sister chromatids travel towards each pole. In metaphase, chromosomes line at the equator of the spindle. In prophase, chromosomes condense and the nuclear envelope disintegrates. In telophase, the sister chromatids have reached their respective poles and the two daughter nuclei form.

Q24. Starch is present in tubers, fruits, and grains but absent in animal cells. Instead, animals have a substance stored in liver and muscles known as:

• A. Galactose
• B. Glucagon
• C. Glycogen✓
• D. Glucose

Explanation: Glycogen is the storage form of carbohydrate found in animals. Galactose (A) is a monosaccharide, Glucagon (B) is a hormone, Glucose (D) is a monosaccharide and the monomer of glycogen and starch.

Q25. Thin filaments of muscles contain _ chains of actin molecules.

• A. Two✓
• B. One
• C. Three
• D. Four

Explanation: The actin molecules contain active sites, to which myosin heads will bind, during contraction. The thin filaments contain two intertwined bands of actin monomers. Alongside, they possess regulatory proteins called tropomyosin and troponin, which regulate the interaction of actin and myosin.

Q26. The reflex action is the phenomena which only involves:

• A. Brain. receptors, spinal cord
• B. Receptors, neurons, brain
• C. Receptors, effectors and spinal cord✓
• D. Receptors and effectors

Explanation: The simplest arrangement of a reflex are consists of the receptors, an interneuron, and an effector; together these units from a finctional reflex group. Brain is, however, excluded in the reflex arc.

Q27. Complementary DNA molecule is:

• A. DNA from mRNA✓
• B. An artificial DNA
• C. Single stranded DNA
• D. A small segment of chromosomal DNA

Explanation: A complementary DNA molecule or cDNA is, by definition, a DNA copy of a messenger RNA or mRNA molecule, produced by reverse transcriptase.

Q28. An example of competitive inhibition of an enzyme is the inhibition of:

• A. Succinic dehydrogenase by malonic acid.✓
• B. Cytochrome oxidase by cyanide.
• C. Hexokinase by glucose.
• D. Carbonic anhydrase by carbon dioxide.

Explanation: The inhibition of succinic dehydrogenase by malonic acid is an example of competitive inhibition. Succinic dehydrogenase is an enzyme involved in the citric acid cycle (also known as the Krebs cycle) and plays a role in the conversion of succinate to fumarate. Malonic acid serves as a competitive inhibitor in this context. It structurally resembles the substrate (succinate) and competes with it for binding to the active site of succinic dehydrogenase. Malonic acid binds to the active site of the enzyme, preventing the normal substrate (succinate) from binding and undergoing the catalytic reaction. The presence of malonic acid decreases the rate of the succinic dehydrogenase-catalyzed reaction by interfering with the binding of succinate.

Q29. Meselson and Stahl transferred few bacteria grown in N15 medium to N14 medium for replicating their DNA. What would be the result after two rounds of replication?

• A. 100% heavy duplex
• B. 100% hybrid duplex
• C. 50% hybrid duplex and 50% heavy duplex
• D. 50% hybrid duplex and 50% light duplex✓

Explanation: After two rounds of replication, half of the duplexes will be hybrid i.e. contain one N15 strand and one N14 strand, while the other half will contain only N14 or light strands.

Q30. In an action potential, the permeability of sodium ions in the neurons increases due to:

• A. Repolarization
• B. Sodium ions forming an ionic bonding
• C. The opening of sodium channels/gates✓
• D. The action of the acetylcholinesterase enzyme

Explanation: Option A mentions repolarization, which is the stage where there is a large influx of potassium ions out from the cell, during which sodium ions do not enter the cell. Option B mentions the concept of ionic bonding which is irrelevant to the scenario. Option C is correct, as when sodium voltage-gated channels open, only then does the permeability of sodium ions across the membrane increases, resulting in their influx into the cell. Option D describes an enzyme, that is important in recycling ACh across synapses; it is not relevant to the concept of sodium permeability during action potentials.

Q31. In which situation are genes, are not assorted independently, during meiosis in a chromosome?

• A. When genes are linked and their loci are close to each other✓
• B. When some genes have mutated on the chromosome
• C. When there are too many Genes on a chromosome
• D. When genes are not linked and their loci are far apart

Explanation: According to the law of independent assortment, the alleles of two or more different genes get sorted into gametes independently of one another, i.e. the allele, a gamete receives for one gene, does not influence the allele received for another gene. Hence, two different traits or genes are inherited separately or independent of each other. For genes to NOT assort independently, this implies that they are inherited TOGETHER, i.e. are linked. This is more likely to occur when the positions of the genes (i.e. their loci) are close to each other on one chromosome. This scenario is only described by option A.

Q32. During spermatogenesis, the _, which are haploid cells eventually mature into spermatozoa/mature sperms.

• A. Spermatogonia
• B. Primary spermatocytes
• C. Spermatids✓
• D. Secondary spermatocytes

Explanation: During spermatogenesis, the spermatids, which are haploid cells, eventually mature into spermatozoa or mature sperm. Spermatogenesis is the process by which male germ cells (spermatogonia) undergo a series of divisions and differentiations to ultimately form functional sperm cells.

Q33. Site of mRNA translation:

• A. Lysosomes
• B. Ribosomes✓
• C. Cisternae
• D. Golgi body

Explanation: Factual recall; Option A mentions lysosomes, which contain hydrolytic enzymes that digest old organelles or phagocytosed particles. Option B mentions ribosomes; messenger RNA (mRNA) is decoded in a ribosome, outside the nucleus, to produce a specific amino acid chain or polypeptide. Option C, cisternae, describes the sac-like tubules that make up the endoplasmic reticulum. Option D, Golgi body, is involved in packaging of already synthesized proteins and their subsequent transport or secretion.

Q34. The photosynthetic pigments of plants are arranged as clusters in thylakoid membranes. The reaction centers of these clusters consist of _ molecules.

• A. Chlorophyll✓
• B. ATP
• C. Carotenoids
• D. Glucose

Explanation: The reaction centers of photosynthetic pigment clusters in thylakoid membranes consist of chlorophyll molecules. Chlorophyll is the primary pigment responsible for capturing light energy during the light-dependent reactions of photosynthesis. In the thylakoid membrane, chlorophyll molecules are organized into complexes known as photosystems.

Q35. Acetylcholine and noradrenaline are two types of _ used in our nervous system.

• A. Enzymes
• B. Channel and carrier proteins in the cell membrane of a neuron
• C. Neurotransmitters✓
• D. Hormones

Explanation: Factual recall: Both of these substances are neurotransmitters and not any of the things, mentioned in the other options.

Q36. The type of energy, reduced by the enzymes, for biological reactions to occur is called the:

• A. Light Energy
• B. Heat energy
• C. Active energy
• D. Activation energy✓

Explanation: The type of energy reduced by enzymes to facilitate biological reactions is referred to as "activation energy." Enzymes play a crucial role in biological systems by lowering the activation energy required for a chemical reaction to occur. Activation energy is the energy input needed to initiate a chemical reaction by overcoming the energy barrier associated with the conversion of reactants into products. Enzymes accelerate reactions by providing an alternative pathway with a lower activation energy, making it easier for the reaction to proceed.

Q37. The prokaryotes possess ribosome with larger subunit of:

• A. 70S
• B. 50S✓
• C. 60S
• D. 40S

Explanation: Prokaryotes contain 70S ribosomes, which themselves comprise of two further subunits; a larger 50S subunit and a smaller 30S subunit. 80S ribosomes are present in eukaryotes.

Q38. Homozygous means:

• A. Alleles in an organism
• B. Two different alleles of a gene
• C. Having two identical genes
• D. Having two identical alleles of a gene✓

Explanation: Homozygous refers to a genotype where, for a given trait or gene, the organism possesses two alleles of the same type (i.e. both alleles are dominant or both alleles are recessive). Option D fits this description. Option A merely mentions alleles in an organism, whereas homozygous refers to the allelic combination of a particular trait/gene. Option B refers to the heterozygous state, not homozygous. Option C refers to having two identical genes. Homozygous (and even heterozygous) terms refer to the allelic combination the organism possesses of one particular gene. It is important to remember that an allele is a variant of one particular gene so, one gene may have two or more variants or alleles.

Q39. Most proteins are made up of:

• A. 10 types of amino acids
• B. 20 types of amino acids✓
• C. 170 types Of amino acids
• D. 16 types of amino acids

Explanation: There are a total of 20 naturally-occurring amino acids, which are found in proteins, which consist of long chains of these amino acids. Option B mentions 20 amino acids so, it is the correct answer, while the other options mention numbers aside from 20.

Q40. The route of urine excretion from kidney to outside of body is:

• A. Kidney → ureter → urinary bladder → urethra✓
• B. Urinary bladder → kidney → ureter → urethra
• C. Kidney → urethra → urinary bladder → ureter
• D. Kidney → ureter → urethra → urinary bladder

Explanation: Option A describes the correct route of urine excretion. The kidney filters blood and forms urine. This urine is to be excreted out of the body, and from the kidney, it drains into the ureter. Each kidney drains into one ureter, which is a long muscular tube, that leads into the urinary bladder, which is a sac-like organ that collects urine from the kidney. When a certain amount of urine has been collected, it is urinated out by a thin tube known as the urethra.All of the remaining options describe jumbled up/incorrect routes of urine excretion.

Q41. In genetics, the term locus refers to the _ of the gene on the chromosome.

• A. Position✓
• B. Frequency
• C. Copy
• D. Inversion

Explanation: The locus describes the position of the gene on a chromosome.

Q42. Glycolysis takes place in the _ of cells.

• A. Nucleus
• B. Mitochondria
• C. Golgi complex
• D. Cytoplasm✓

Explanation: Glycolysis, the first step of aerobic respiration, which results in the formation of two molecules of pyruvate from one molecule of glucose, occurs in the 'cytoplasm' of the cell.

Q43. If a carrier haemophilic female (XHXh) is married to a haemophilic male (XhY). What will be the ratio of the presence of haemophilia in the children? Select the best answer from the given options.

• A. 100% of all females and males will be hemophilic
• B. Females have a 50% chance of getting haemophilia and males will be 100% hemophilic
• C. Carrier female 25%, hemophilic female 25%, normal male 25% and 25% hemophilic male✓
• D. Females and males both have 50% chance of getting hemophilia

Explanation: As per the cross given above, there are four possible genotypes and hence, four possible phenotypes. These are; carrier daughter, hemophiliac daughter, normal son, and hemophiliac son, and all four have an equal probability or likelihood of occurring, which is 25 percent. Option C describes this scenario correctly so, it is the correct option. Option A describes a scenario in which the mother is a hemophiliac (i.e. has a genotype of XhXh) and the father is also a hemophiliac (genotype XhY), which results in all the children being hemophiliacs. Option B describes an impossible scenario, as in order for all the males to be hemophiliac, the mother must have a genotype of XhXh, but in this scenario, the daughters will either be all hemophiliacs or all carriers (father's genotype being either XhY or XHY, respectively). Hence, for all males to be hemophiliacs, it is impossible for the females to have a 50% chance of being hemophiliacs. Option D also describes the given scenario, because if you look at the figure above, males will either be normal or hemophiliac, and females will either be carriers or hemophiliac, as such, each gender has a 50% chance of being hemophiliacs. However, option C better describes this scenario as it specifically lists the probabilities of all four possible phenotypes. (NOTE: The question asks the candidate to select the BEST possible answer from the given options).

Q44. A disease caused by gradual breakdown of the thin walls of alveoli is:

• A. Tuberculosis
• B. Asthma
• C. Emphysema✓
• D. Bronchitis

Explanation: Option A, tuberculosis, is a bacterial infection of the lungs, characterized by persistent cough, fever, and weight loss. This involves granuloma formation and cavitation of the lung tissue so, it is not the disease described in the question. Option B, asthma, is the thickening of the bronchioles due to an excessive allergic reaction upon exposure to certain agents and irritants (allergens). Option C, emphysema, is the correct answer, as it involves the breakdown of the alveolar elastic tissue resulting in the breakdown of alveolar walls, leading to the formation of large pockets within the lungs. Option D, bronchitis, is a chronic disease, that involves excessive mucus production in response to prolonged smoking.

Q45. Substances responsible for increasing the set point of the hypothalamus are called:

• A. Androgens
• B. Pyrogens✓
• C. Pepsin
• D. Prions

Explanation: Substances responsible for increasing the set point of the hypothalamus are called pyrogens. Pyrogens are substances that induce a fever response in the body by raising the set point of the hypothalamus, which is the body's internal temperature-regulating center. When pyrogens are present, they signal the hypothalamus to increase the set point, resulting in the body generating and conserving heat to reach the new higher temperature. Pyrogens can be endogenous (produced within the body) or exogenous (coming from external sources). Examples of endogenous pyrogens include cytokines such as interleukin-1 (IL-1) and interleukin-6 (IL-6), which are released during an immune response. Exogenous pyrogens can include bacterial toxins or certain drugs. By increasing the set point of the hypothalamus, pyrogens contribute to the development of a fever, which is a protective response that helps the body fight off infections and promote immune function.

Q46. DNA polymerase enzyme for PCR is isolated from bacteria, Thermus aquaticus, because:

• A. It can work at high speed
• B. It can withstand high denaturation temperature✓
• C. It can withstand low denaturation temperature
• D. It can be used again and again

Explanation: 'Thermos aquaticus' is a heat-tolerant bacterium and its enzymes, including DNA polymerase (or Taq polymerase), are resistant to high temperatures. In PCR, a high temperature is used (denaturation temperature) to denature the template DNA or separate its strands. The fact that this DNA polymerase can withstand the high temperatures, used in PCR, enables it to work efficiently without denaturing itself, thereby enabling the reaction to occur, adequately. Hence, option B is the correct answer. Option A is not exclusive to this enzyme, as most enzymes can work at a high speed and increase the rate of reaction. Option C is technically correct as if this DNA polymerase can withstand high temperature, it can obviously withstand low temperature as well, but since PCR involves a high temperature, this trait is not relevant to the scenario. Option D is a feature shared by all enzymes as they are recyclable, not exclusive to this DNA polymerase.

Q47. Which of the following photosystems is involved in cyclic photophosphorylation?

• A. PS II
• B. PS I✓
• C. PS I and PS II
• D. PS III

Explanation: Cyclic photophosphorylation (which is depicted by the yellow arrows below) employs the use of only photosystem I (PS1) and not PSII, which is involved in non cyclic photophosphorylation.

Q48. Which hormonal pair would maintain the endometrium and make it receptive for implantation of embryo?

• A. Luteinizing Hormone and Progesterone
• B. Estrogen and Progesterone✓
• C. Estrogen and Follicle Stimulating Hormone
• D. Luteinizing Hormone and Follicle Stimulating Hormone

Explanation: The hormonal pair that plays a crucial role in maintaining the endometrium (the lining of the uterus) and making it receptive for the implantation of an embryo consists of progesterone and estrogen. During the menstrual cycle, estrogen is primarily produced by the developing ovarian follicles. In the early phase of the menstrual cycle (follicular phase), rising estrogen levels stimulate the growth and proliferation of the endometrial tissue. Estrogen also promotes the development of blood vessels and glands in the endometrium. Following ovulation, the ruptured follicle transforms into a structure called the corpus luteum, which secretes progesterone. Progesterone acts in conjunction with estrogen to further prepare the endometrium for possible implantation. It causes the glands in the endometrium to become more secretory, increases the thickness of the uterine lining, and helps create a receptive environment for a potential embryo.

Q49. The thick filaments, in a myofibril of muscles, are made of _.

• A. Myoglobin
• B. Myosin✓
• C. Actin
• D. Haemoglobin

Explanation: Option A, myoglobin, is a heme-containing protein found in muscles, that binds to oxygen, which it supplies to muscle tissue. Option B, myosin, is the correct answer as it forms thick filaments in a myofibril. Option C, actin, forms thin filaments in a myofibril. Option D, haemoglobin, is the heme-containing protein found in red blood cells and serves to carry oxygen within the blood.

Q50. If sequence in DNA is CCCTAGAG, then what would be the sequence in messenger RNA after transcription?

• A. GGGAUCUC✓
• B. GGGATCTC
• C. GGAAUCUC
• D. GGGGTCTC

Explanation: C pairs with G, A pairs with T in DNA while A pairs with U in RNA. The question asks for the corresponding mRNA sequence, after transcription of the DNA sequence. The C bases will correspond to G bases on the mRNA, T will correspond to A, G will correspond to C, whereas A will correspond to U and not T because the transcript is made up of RNA and not DNA, hence U takes the place of T. Option A describes the correct transcript: C C C T A G A G G G G A U C U C

Q51. In chemiosmosis, the proton (H+) moves from _.

• A. Stroma to Lumen
• B. Cytoplasm to Stroma
• C. Lumen to Stroma✓
• D. Stroma to cytoplasm

Explanation: Chemiosmosis is the process, whereby protons move down their electrochemical gradient i.e. from a region of their higher concentration to lower concentration. During this process, they move through ATP synthase, and their movement results in the coupling of ADP and inorganic phosphate to from ATP. In chloroplasts, the concentration of protons is greater in the thylakoid lumen relative to the stroma, hence protons move across the semi-permeable membrane, from the lumen into the stroma.

Q52. Microtubule subunits (for spindle fibers) are synthesized in _ phase.

• A. G1
• B. M
• C. G2✓
• D. S

Explanation: Microtubule subunits are tubulin proteins, which are formed prior to the initiation of mitosis. They are formed during the G2 phase, hence option C is the correct answer. G1 phase, involves normal growth and protein synthesis of the cell, leading up to the DNA replication phase (S). M is mitosis in which nuclear division occurs and it consists of prophase, metaphase, anaphase, and telophase. Prophase involves the assembly of microtubules into the spindle fibers, but the subunits of the microtubules, which are tubulin proteins, are formed prior to this stage, i.e. G2. S phase involves DNA replication.

Q53. How many molecules of ATP would be utilized for phosphorylation of one glucose molecule during glycolysis?

• A. Three
• B. Two✓
• C. Four
• D. One

Explanation: Two ATP molecules are used to phosphorylate glucose; the first is used to convert glucose into glucose-6-phosphate, while the second is used to convert fructose-6-phosphate into fructose-1,6-bisphosphate.

Q54. The function of calcium ions in muscle contraction is to:

• A. Bind to troponin molecule and cause them to move✓
• B. Polarize visible light
• C. Aid in the transmission of nerve impulse
• D. Bind to tropomyosin molecule and cause them to form cross bridges

Explanation: Calcium ions play a critical role in muscle contraction by binding to the troponin molecules present on the thin filaments. When calcium ions bind to troponin, it induces a conformational change in the troponin-tropomyosin complex. This change exposes the myosin-binding sites on the actin filaments, allowing the myosin heads to form cross-bridges and initiate muscle contraction.

Q55. According to the theory of natural selection, organisms produce:

• A. More offspring than supported✓
• B. Less offspring than supported
• C. Offspring according to the resources available
• D. Offspring to create resources

Explanation: According to theory of natural selection, those organisms whose inherited characterstics fit them best to their enviroment are likely to leave more offsprings than the less fit individuals.

Q56. A person was married to his cousin and both are heterozygous for Sickle Cell anemia. Among their four kids, what will be the proportion of homozygotes?

• A. 75%
• B. 50%✓
• C. 100%
• D. 25%

Explanation: As can be seen from the cross below, 50 percent of the offspring are heterozygotes, i.e. 50 percent possess the sickle cell trait. 25 percent are normal i.e. homozygous for HbN, while 25 percent have sickle cell anemia, i.e. homozygous for HbS. In total, 50 percent are homozygous (25 percent plus 25 percent) so, option B is correct.

Q57. The major function of basophils is to:

• A. Destroy small particles by phagocytosis
• B. Release heparin to prevent blood clotting✓
• C. Transport oxygen
• D. Inactivate inflammation producing substances

Explanation: Basophils have a variety of functions, including mediating inflammatory and allergic responses as they contain histamine, in addition to preventing blood clotting as they contain heparin, which is a blood thinner/anticoagulant.

Q58. Which enzyme is administered to the patients of Severe Combined Immunodeficiency Disease (SCID)?

• A. Adenosine Deaminase (ADA)✓
• B. Pancreatic Enzyme
• C. ß-galactosidase
• D. ß-lactamase

Explanation: 'SCID' is due to Adenosine Deaminase (ADA) deficiency so, it is administered to overcome the deficiency. Hence, option A is correct. Pancreatic enzymes (B) partake in digestion. Beta-galactosidase (C) breaks down lactose into glucose and galactose in certain organisms. Beta-lactamase (D) is an enzyme produced by bacteria, that confers their resistance against beta-lactam drugs, such as penicillin, as it breaks down the beta-lactam ring present in these drugs.

Q59. Given below, is the diagram of a nephron without a vascular supply.What is the name of Part C?

• A. Collecting tubule
• B. Proximal tubule
• C. Distal tubule✓
• D. Loop of Henle

Explanation: Part A is the Bowman's capsulePart B is the Descending limb of the loop of henle.Part C is the Distal tubule so, option C is the correct answer.Part D is the Collecting duct.

Q60. Inside ovary, primary oocyte divides through first meiotic division, forming two haploid cells, the secondary oocyte and:

• A. Ovum
• B. Oogonium
• C. Follicle cell
• D. Polar body✓

Explanation: Inside the ovary, during the process of oogenesis, the primary oocyte undergoes the first meiotic division, resulting in the formation of two haploid cells. These cells are the secondary oocyte and a smaller cell called the first polar body. The first polar body is a smaller cell that contains a portion of the genetic material but minimal cytoplasm. The polar bodies typically do not play a direct role in fertilization and are eventually broken down or reabsorbed.

Q61. Genetic engineering has been successfully used to produce?

• A. Animals, with inferior traits
• B. Transgenic mice for testing the safety of the polio vaccine before use in humans✓
• C. Animals, that have the same genetic constitution
• D. None of the above

Explanation: Transgenic mice are being formed for use in testing the safety of vaccines before they are used on human beings. Transgenic mice are being used to test the safety of the polio vaccine. Many transgenic animals are developed to increase our understanding of how genes contribute to the development of disease so, that investigation of new treatments for diseases is made possible. Now transgenic models exist for many human diseases such as cancer, cystic fibrosis, rheumatoid arthritis, Alzheimer's disease, hemophilia, thalassemia, etc. Therefore, the correct answer is option B.

Q62. In plants, which sugar is transported from source to sink through sieve tubes?

• A. Glucose
• B. Sucrose✓
• C. Fructose
• D. None of the above

Explanation: Factual recall; Glucose is the monosaccharide, that makes up the disaccharide sucrose along with fructose, whereas sucrose is the form, in which sugar is transported from source to sink. Sucrose is broken down to its constituents and then used by tissue cells.

Q63. Which of the following hormone stimulates the ovulation from the follicle into oviduct?

• A. Estrogen
• B. Progesterone
• C. Luteinizing hormone✓
• D. Follicle stimulating hormone

Explanation: Option A, estrogen, is the hormone that causes the development and thickening of the endometrial lining.Option B, progesterone, is the hormone, secreted by the corpus luteum, to maintain and thicken the endometrial lining, making it ready for implantation.Option C, LH, is the main trigger of ovulation, as a surge in LH causes the release of the secondary oocyte from the Graafian follicle into the oviduct. Hence, it is the correct option. Option D, FSH, is responsible for the development of follicles in the ovary.

Q64. Which one is an example of a Nucleotide?

• A. Adenosine
• B. NAD
• C. ATP✓
• D. Guanine

Explanation: ATP is a nucleotide, as it contains a pentose sugar, an adenine base, and three phosphate groups. A nucleotide may have one or more phosphate groups.

Q65. Four plants are present in different environmental conditions. Plant A is present in a warm climate with continuous rainfall, plant B is present in a cool forest, plant C is present in a warm climate with little breeze, while plant D is present in a warm climate with high wind speed. Which one of the above plants will have the highest rate of transpiration?

• A. Plant B
• B. Plant D✓
• C. Plant A
• D. Plant C

Explanation: When it comes to questions like these, do not confuse the option letter with the letter of the plant itself. The factors which affect the rate of transpiration are: temperature, as at a higher temperature, more water vapour is given up by the leaves of the plant; humidity, as more humid conditions decrease the concentration gradient of water vapour between the air spaces of the leaves and the atmosphere so, the rate of transpiration decreases; wind speed, as a greater wind speed moves away more of the water vapour collected around the leaves and stomata, thereby increasing the concentration gradient between the air spaces of the leaves and the atmosphere. Plant A has a warm atmosphere but has continuous rainfall. Rainfall would imply a humid atmosphere so, the rate of transpiration is not very high. Plant B has a cool atmosphere so, the rate is low. Plant C is present in a warm climate but the breeze is slow so, when wind speed is low, more water vapour accumulates around the leaves of the plant thereby reducing the concentration gradient between the air spaces and the atmosphere. Hence, the rate of transpiration is not very high. Plant D is present in a warm climate in addition to having a high wind speed so, both factors contribute to a high rate of transpiration. Hence, the option which mentions plant D is the correct answer, which is option B.

Q66. The protein coat of virus is called:

• A. Capsid✓
• B. Cosmid
• C. Capsomere
• D. Chromophore

Explanation: The outer protein coat of a virus is known as 'the capsid'. It consists of several oligomeric structural subunits made of a protein called protomers, which cover the nucleic acid, present in the virus, and protects it when the virus inserts itself into the host. The capsid proteins also help the virus attach and penetrate the host cells, thereby infecting the host. Hence, option A is correct.

Q67. If stimulation is above _, impulses travel to the brain along the sensory neuron.

• A. Action Potential
• B. Resting Potential
• C. Threshold✓
• D. Recovery Period

Explanation: The threshold is the minimum level of stimulation required for a neuron to generate an action potential. If the stimulation is above the threshold, impulses can travel to the brain along sensory neurons, initiating the transmission of sensory information.

Q68. The covalent bond or bridge between two monosaccharides to form a disaccharide is called a:

• A. Carboxyl bond
• B. Hydrogen bond
• C. Hydroxyl bond
• D. Glycosidic bond✓

Explanation: Hydrogen bond: Hydrogen bonds are weak electrostatic attractions between a hydrogen atom in a polar covalent bond and an electronegative atom (such as oxygen, nitrogen, or fluorine) in another molecule. While hydrogen bonding can play a role in the interactions between monosaccharides or other biomolecules, it is not the bond that joins two monosaccharides to form a disaccharide. Hydroxyl bond: A hydroxyl bond is not a commonly used term in chemistry. The correct term to describe the bond between two monosaccharides to form a disaccharide is a glycosidic bond. Glycosidic bond: The correct option is glycosidic bond. A glycosidic bond is a covalent bond that forms when a hydroxyl group (-OH) on one monosaccharide reacts with the anomeric carbon (the carbon containing the carbonyl group, usually the first carbon) of another monosaccharide. This reaction involves the elimination of a water molecule (H2O), resulting in the formation of a glycosidic linkage between the two monosaccharides. Disaccharides, such as maltose, lactose, and sucrose, are formed by glycosidic bonds between two monosaccharides.In summary, option d) Glycosidic bond is the correct option. A glycosidic bond is the covalent bond between two monosaccharides that forms a disaccharide or other types of oligosaccharides.

Q69. The structure of a fibrous protein comprises of polypeptide chains in the form of:

• A. Cluster
• B. Spherical or curried up ball
• C. Long strands or fibrils✓
• D. Flat uncoiled chains

Explanation: The options describe different conformations of polypeptide chains. Option A describes a formation, not characteristic of fibrous proteins.Option B describes the conformation of globular proteins.Option C describes that of fibrous proteins.Option D most probably is indicative of beta pleated sheets or another structure, not characteristic of fibrous protein.

Q70. Taxonomy includes the arrangement of organisms into different taxa. Which of the following represents the correct hierarchy of various taxa of classification?

• A. Species, genus, family, order, class, phylum✓
• B. Order, family, class, phylum, kingdom
• C. Species, genus, family, class, order, phylum
• D. Species, genus, order, family, class, phylum

Explanation: The correct hierarchy of various taxa in biological classification is:Species, genus, family, order, class, phylumExplanation:1. Species: The fundamental unit of classification, consisting of organisms that can interbreed and produce fertile offspring.2. Genus: A group of closely related species.3. Family: A group of related genera.4. Order: A group of related families.5. Class: A group of related orders.6. Phylum: A group of related classes.This hierarchical arrangement helps organize and categorize living organisms based on their evolutionary relationships. Mnemonics, that one can use to memorize the taxonomic hierarchy, which is Domain, Kingdom, Phylum, Class, Order, Family, Genus, and Species, in that specific order are as follows:-“Dear King Philip Came Over For Good Soup" (the first letter of each word corresponds to a taxa)- Kingdom Phylum went to a Class and Ordered his Family to look for a Genius(Genus) who is a rare Species. The diagram, given below, illustrates the hierarchy.

Q71. The Plasmid pBR322 has antibiotic resistance genes for:

• A. Streptomycin
• B. Ampicillin and Tetracycline✓
• C. Tetracycline and doxycycline
• D. Ampicillin

Explanation: Factual recall;pBR322 is base pairs in length and has two antibiotic resistance genes - the gene 'bla', encoding the ampicillin resistance (AmpR) protein, and the gene 'tetA', encoding the tetracycline resistance (TetR) protein.

Q72. Which of the following blood vessels contain semilunar valves?

• A. Arteries
• B. Capillaries
• C. Veins✓
• D. Arterioles

Explanation: Valves in veins are one-way(semi lunar) flaps that help prevent blood from flowing backward in the circulatory system. They play a crucial role in maintaining blood flow towards the heart and preventing the backflow of blood. This mechanism is particularly important in the legs, where gravity can make it harder for blood to return to the heart, leading to conditions like varicose veins when the valves don't function properly.

Q73. The main nitrogenous excretory product of humans is:

• A. Ammonia
• B. Urea✓
• C. Ammonium
• D. Uric acid

Explanation: Option A, Ammonia, is the primary excretory product in certain animals such as fishes, aquatic amphibians, aquatic insects, tadpoles, etc. as ammonia is extremely soluble in nature and excreted via simple diffusion in aquatic animals. It is not the main nitrogenous excretory product in humans, as due to its high solubility, it would need a very large amount of water to be excreted, which would be unfavorable for humans.Option B, Urea, is the correct answer. Urea is relatively less toxic and requires less water than ammonia to excrete so, it is the main nitrogenous excretory product in humans as it best suits the organism’s condition and needs.Option C, Ammonium, is a derivative of ammonia or its ionic form so, the same concept, associated with ammonia, applies here.Option D, Uric Acid, is the chief excretory product in organisms such as birds, cockroaches, snakes, lizards, and other terrestrial insects, which excrete it in a solid or semisolid form. Uric Acid is the least soluble of the waste products so, it requires less water. Humans excrete a small portion of this because if a large amount of it were to accumulate, it would precipitate and form uric acid stones, thereby damaging the urinary system.

Q74. Water and Minerals move down their concentration gradient through plasmodesmata to cells of cortex, endodermis, pericycle, and then to sap in the xylem cells. This is also known as the:

• A. Vacuolar pathway
• B. Apoplastic pathway
• C. Symplast Pathway✓
• D. Mineral absorption Pathway

Explanation: The symplast pathway, involves movement of water from one cell to the other by plasmodesmata, hence water moves from cytoplasm of one cell to the cytoplasm of the other cell via the plasmodesmata, connecting them. This scenario is mentioned in the question, hence C is the correct answer.

Q75. If 15 um size object is observed under light microscope, using 5X eyepiece and 10X objective, its magnified image size will be:

• A. 750 um✓
• B. 50 um
• C. 500 um
• D. 250 um

Explanation: Recall the formula for magnification, where Image Size (I)= Actual Size (A) x Magnification (M), or I=A x M. The magnification is the product of the eyepiece lens and objective lens’ respective magnifications so; M= 5 x 10 = 50. The question asks for the size of the image so; I=A x M = 15 x 50 = 750 um.

Q76. A person got an infection, he became ill but then he survived. What do you think which of immunity he would have developed?

• A. Active immunity
• B. Artificially induced active immunity
• C. Passive immunity
• D. Naturally induced active immunity✓

Explanation: Naturally induced active immunity occurs when the body's immune system responds to an actual infection by pathogens or exposure to antigens in the environment. In this process, the immune system recognizes the invading pathogens, such as bacteria or viruses, and mounts a specific and targeted immune response. During the course of the infection, the immune system produces antibodies and memory cells. Antibodies help in the elimination of the pathogens, while memory cells "remember" the specific antigens, providing the basis for a faster and more robust immune response upon subsequent exposure to the same pathogen. This form of immunity is a natural consequence of the body's encounters with various infectious agents throughout life.

Q77. The nitrogen-containing bases in nucleotide are of two types; Purines and Pyrimidines. The purines bases are:

• A. Adenine, Guanine, and Cytosine
• B. Guanine and Cytosine
• C. Adenine and Guanine✓
• D. Adenine and Thymine

Explanation: Purine bases are adenine and guanine, while pyrimidine bases are thymine, cytosine, and uracil. Option C is the only option that mentions only purine bases, while the others mention a combination of purine and pyrimidine bases.

Q78. The process in which a complementary copy of the code from a gene is produced by RNA Polymerase in the nucleus:

• A. Transcription✓
• B. Translation
• C. Proofreading
• D. DNA Replication

Explanation: Transcription, is when RNA polymerase makes a copy of a gene on a template DNA strand in the nucleus; this copy is messenger RNA.

Q79. Among the following, which cellular organelle contains circular DNA, similar to those found in bacteria?

• A. Lysosome
• B. Nucleus
• C. Nucleosome
• D. Mitochondria✓

Explanation: Chloroplasts and mitochondria are the two organelles, that contain circular DNA similar, to those found in prokaryotic cells, along with 70S ribosomes, which are also similar to the ones in prokaryotes, such as in bacteria. This can be traced back to the endosymbiotic theory, where mitochondria and chloroplasts are understood to be ancient prokaryotes.

Q80. The following flowchart depicts the steps of the Calvin Cycle. Which option, according to you, fits in as the correct answer of the missing step?

• A. Pyruvate
• B. Hydrogenase
• C. Ribulose bisphosphate✓
• D. Oxaloacetate

Explanation: The first step involves the fixing of carbon (derived from CO2) into molecules of Ribulose Bisphosphate (RuBP) via the action of the enzyme 'rubisco', resulting in the formation of 3-phosphoglycerate. This is further processed into 1,3-BPG and eventually, G-3-P, before the cycle starts again. The chart above only has RuBP missing.

Q81. Given figure shows:

• A. Structure of a Lenticel✓
• B. Hydathode showing gaseous vapour exchange
• C. Fungus reproducing by spore formation
• D. Algae reproducing by spore formation

Explanation: The figure shows the structure of 'Lenticel', which facilitates gaseous exchange and transpiration. These are aerating pores in the bark of woody trees. These are surrounded by loosely arranged thin walled complementary cells, enclosing intercellular spaces for gaseous exchange.

Q82. Xerophytes have small, thick leaves to:

• A. Help them float on water
• B. Limit water loss, by increasing the surface area
• C. Help them survive in salty environment
• D. Limit water loss, by reducing the surface area✓

Explanation: Xerophytes are plants that are adapted to survive in environments, with very little water. Option A describes a feature of a hydrophyte, that grows on water.Option B describes the correct feature, which is limiting the loss of water, but this is not done by 'increasing the surface area'. Water loss is limited by 'decreasing the surface area of the leaves'.Option C describes a feature of halophytes, which grow in conditions of high salt content in the soil or water.Option D describes the correct feature along with the correct adaptation, i.e. limiting water loss by 'reducing the surface area' of the leaves, as xerophytes need to conserve water and they do this by having small leaves, with a thick waxy coating and sunken stomata.

Q83. Passive processes for the movement of molecules across cell surface membrane are:

• A. Facilitated diffusion and osmosis✓
• B. Diffusion and exocytosis
• C. Pinocytosis and facilitated diffusion
• D. Osmosis and phagocytosis

Explanation: Osmosis, diffusion and facilitated diffusion are passive processes of passage of molecules across the cell membrane but exocytosis and endocytosis are active processes of molecules across the cell.

Q84. During the G2 phase:

• A. Chromosome number is duplicated.
• B. The chromosomes are left with only one chromatid.
• C. Energy is stored for chromosome movement and mitotic specific proteins (Tubulin) are produced.✓
• D. Specific enzymes are synthesized and DNA base units are accumulated.

Explanation: Option A occurs during the S phase, where DNA is replicated. Option B occurs after the M phase, when the cell has divided. Option C is 'G2', which occurs right before mitosis. Steps are taken to prepare the cell for cell division, and they include storing energy for chromosome movement and production of tubulin proteins, which are microtubule subunits that will be used in the formation of spindle fibers in the M phase. Option D is G1 phase, where the cell prepares for DNA replication so, enzymes, such as DNA polymerase, are synthesized along with nucleotides that will be used to form new DNA.

Q85. During inspiration, the space inside the chest cavity is increased due to:

• A. The relaxation of the muscles of the diaphragm
• B. Relaxation of the external intercostal muscles
• C. Increased pressure
• D. The contraction of the muscles of the diaphragm✓

Explanation: Option A occurs during expiration, as the diaphragm’s muscles relax, causing the chest cavity space to decrease and the air is pushed out.Option B also occurs during expiration, as in inspiration, the external intercostal muscles contract while the inner ones relax, and in expiration, the external intercostal muscles relax and inner ones contract.Option C occurs during expiration, as the volume of the chest cavity decreases, causing an increase in pressure which drives the movement of air back into the atmosphere. The reverse process occurs in inspiration.Option D occurs during inspiration, as the diaphragm muscles contract, causing the diaphragm to flatten. This increases the volume of the chest cavity (along with the external intercostal muscles, contracting, and inner ones, relaxing, resulting in the ribcage, moving upwards and outwards), which causes a decrease in pressure, resulting in the movement of air into the lungs from the atmosphere.

Q86. A student of chemical engineering mistakenly engulfed the toxic compound "A" which was a potent inhibitor of a certain enzyme. He was immediately brought to a hospital where the doctor injected, intravenously, substrate "B" to minimize the toxic effect of Compound A. His life was saved from serious damage. The treatment method shows that compound A was a _ inhibitor.

• A. Non competitive
• B. Irreversible
• C. Temperature sensitive
• D. Competitive reversible✓

Explanation: 'Compound A' caused a toxic effect by inhibiting the action of a certain enzyme. 'Substrate B' was injected, which minimized the toxic effect of compound A, and this toxic effect was brought about by inhibiting the enzyme. The addition of a larger quantity of substrate allowed the enzyme to work to its normal capacity once again, as the inhibitory or toxic effect of compound A was minimized. Competitive reversible inhibitors compete with the substrate for the active site and if there is a greater amount of the substrate, the active site is more likely to bind to the substrate, rather than the inhibitor so, the activity of the enzyme is restored. This is the exact scenario described in the question, as the inhibitory effect was reversed upon addition of more substrate. Hence, the correct option is D. Option A, non-competitive inhibitors, decrease the Vmax of the enzyme so, addition of more substrate will not change or restore the enzyme's effectiveness. Option B, irreversible inhibitors, would have an effect that cannot be reversed under any circumstances so, addition of substrate would have no effect. Option C, temperature sensitive inhibitor has no relevance to the given scenario as temperature has not been mentioned as a factor in the question.

Q87. Which is an example of a Disaccharide?

• A. Starch
• B. Lactose✓
• C. Fructose
• D. Glycogen

Explanation: Option A is a polysaccharide, consisting of glucose subunits joined together to form a large chain. (starch and glycogen are polysaccharides)Option B is a disaccharide, consisting of two monosaccharides, glucose and galactose, joined together. (sucrose, maltose, and lactose are all disaccharides)Option C is a monosaccharide (fructose, glucose, and galactose are monosaccharides).Option D is a polysaccharide.

Q88. In glycine, R is _.

• A. Ethane
• B. Fatty acid
• C. H atom✓
• D. Competitive reversible

Explanation: Glycine is the simplest amino acid, with its R group (the side chain of an amino acid) consisting of just a single hydrogen atom. Hence, option C is correct.

Q89. Sequence of amino acids, in a polypeptide chain of protein molecule, corresponds to the sequence of nucleotides on mRNA for that protein. If the reading frame of mRNA for a human protein is 993 nucleotides, including a stop codon at the end, how many amino acids would be incorporated in the polypeptide chain?

• A. 93
• B. 330✓
• C. 331
• D. 993

Explanation: Codon consists of three nucleotides, and each codon corresponds to a single amino acid. A stop codon, however, does NOT code for an amino acid; they signal the end or termination of the polypeptide chain. Hence, out of the 993 nucleotides, the last three correspond to a stop codon and hence 990 nucleotides are to be considered. 990 nucleotides correspond to 330 codons (990 divided by 3), and since each codon codes for a single amino acid, 330 codons will code for 330 amino acids, hence, the answer is B.

Q90. Blood group AB is an example of _ .

• A. Complete dominance
• B. Recessive alleles
• C. Co-dominance✓
• D. Incomplete dominance

Explanation: Option A describes one allele being completely dominant over the other allele i.e. one allele is dominant while the other is recessive so, in a heterozygous condition, only the dominant allele is expressed as a trait. However, in the AB blood group, BOTH A and B antigens are expressed so, that one allele is not completely dominant over the other. Option B describes alleles that are only expressed in the phenotype in homozygous conditions; they will not be expressed if a dominant allele is present in heterozygous conditions. One allele is recessive over the other, but in this case, both traits are expressed. Option C is the correct answer, as it describes the scenario in which both alleles are equally expressed in the phenotype. Since both antigens are expressed individually, this means that both alleles are equally expressed. Option D describes a scenario in which the alleles are expressed in such a way that the phenotype is actually a mixture of intermediate expression of both alleles. The individual alleles are not expressed completely. For example, alleles of black and white fur color are not expressed as black and white spots, they are expressed as grey fur color, i.e. an intermediate or mixture of the two traits.

Q91. As a result of replication, parental DNA would become completely dispersed and that each strand of all the daughter molecules would be a mixture of old and new DNA. This is called as:

• A. Dispersive idea✓
• B. Conservative idea
• C. Disruptive idea
• D. Semi-conservative idea

Explanation: It is 'dispersive replication', which involves each daughter strand containing both parental and new DNA, as the old DNA is broken up and randomly dispersed into each daughter strand.

Q92. Smooth endoplasmic reticulum is responsible for the metabolism of:

• A. Carbohydrates✓
• B. Nucleic acids
• C. Proteins
• D. Lipids

Explanation: Mitochondria are responsible for the carbohydrate metabolis. To convert carbohydrates into CO2 , H20 and generate ATPs through cellular respiraton. SER also metabolis carbohydrates

Q93. Among followings, _ enzyme is naturally found in human immunodeficiency virus (HIV).

• A. Ligase
• B. Reverse Transcriptase✓
• C. RNA Polymerase
• D. DNA Polymerase

Explanation: HIV is a retrovirus. Its genome consists of RNA molecules, and it possesses a 'reverse transcriptase enzyme', which allows the virus to use its RNA as a template to construct copies of DNA, which it uses to incorporate into the genome of the host cell. The remaining enzymes (ligase, RNA, and DNA polymerase) are present in the host cell, which HIV uses for its own proliferation upon infecting the cell. Hence, the correct option is B.

Q94. The average atomic mass of Boron is 10.8. It has two isotopes of masses 10 and 11 respectively. What is the percentage of isotope with the average mass of 10?

• A. 20%✓
• B. 60%
• C. 80%
• D. 50%

Explanation: The formula for calculating the average atomic mass is[ (Mass of one isotope x percentage abundance) + (mass of other isotope x percentage abundance) divided by 100]. In this question, aside from using the formula for each option which would be a bit time consuming, you can take clues from the question itself. The average mass of Boron is 10.8, which is closer to 11 than 10. As such, you can conclude that the abundance of 10 isotope is lesser than the abundance of the 11 isotope, as such, its abundance MUST be less than 50%. The only option with an abundance less than 50 percent is option A, 20 percent. If you check this with the formula, the answer comes out to be correct: [(10 x 20) + (11 x 80) divided by 100] is 10.8.

Q95. Which one of the following compounds is an additional polymer?

• A. Polyvinyl chloride✓
• B. Nylon
• C. Carbohydrate
• D. Polyester

Explanation: An 'addition polymer' is a polymer which consists of monomers, that add on to each other or combine without the formation of a secondary product, whereas in 'condensation polymers', the functional groups of monomers condense, such that a secondary product, such as H2O, is also formed. Out of all these options, only option A is an addition polymer, whereas all three others are condensation polymers.

Q96. Which of the following compounds will give a secondary alcohol after reaction with NaBH4?

• A. CH3COCH3✓
• B. CH3COOCH3
• C. CH3CH2CHO
• D. CH3CH2COOH

Explanation: Option A is a ketone, and its reduction (NaBH4 is a reducing agent) results in a secondary alcohol. This is because the carbonyl functional group is present in the middle of the chain so, the hydroxyl group is formed in the same place after reduction, giving a secondary alcohol. Option B is an ester, which does not undergo reduction with NaBH4.Option C is an aldehyde, that forms a primary alcohol, upon reduction, as its carbonyl group is towards the terminal end of the chain, thereby forming a terminal hydroxyl group. Option D is a carboxylic acid, which does not undergo reduction with NaBH4.

Q97. Select the reagent X from the following choices for this conversion:

• A. Acidified Potassium hydroxide
• B. Acidified Potassium dichromate (VI)✓
• C. Acidified Phosphoric acid
• D. Acidified Oxalic acid

Explanation: The reaction above is oxidation of a secondary alcohol into a ketone. As such, an oxidizing agent must be chosen for reagent X. Among the options, 'Potassium dichromate (VI)' is an oxidizing agent so, the correct option is B. Another answer could be 'potassium manganate (VII)', but it is not mentioned in any of the options. The other options do not oxidize alcohols.

Q98. The pH of 10-2 M aqueous solution of sodium hydroxide is:

• A. 14
• B. 10
• C. 12✓
• D. 13

Explanation: The concentration of OH- in the 10-2 M solution of NaOH would be [10-2]. The pOH of this solution is -log (10-2) which is 2, and the pH is (14 - pOH), which equals 12.

Q99. Ketones can be made by oxidation of:

• A. Aldehydes
• B. Primary Alcohols
• C. Secondary Alcohols✓
• D. Tertiary Alcohols

Explanation: Option A; aldehydes oxidize into carboxylic acids. Option B; primary alcohols oxidize into aldehydes and then to carboxylic acids. Option C; secondary alcohols oxidize into ketones. Option D; tertiary alcohols are resistant to oxidation.

Q100. In the following reaction sequence, product D will be:

• A. 1 - propanol
• B. Propanoic acid
• C. 2 - propanol✓
• D. Mixture of methanol and ethanol

Explanation: The above reaction scheme describes dehydrohalogenation of the alkyl halide into an alkene, followed by hydration of the alkene. The first step involves the reaction of propyl bromide with alcoholic KOH, which will result in elimination or dehydrohalogenation, whereby the halogen atom and adjacent carbon atom's hydrogen atom is removed, resulting in the formation of a double bond between the two carbons, forming propane, which is product C. Propene is acted upon by aqueous H2SO4, which implies hydration or addition of H2O, resulting in the formation of two products; the major product which is 2-propanol and the minor product which is 1-propanol so, the major product is the preferred option, i.e. option C.

Q101. Copper is a typical transition metal. Its atomic number is 29. In which oxidation state does it have a partially filled orbital in d-subshell?

• A. Cu
• B. Cu+
• C. Cu-
• D. Cu2+✓

Explanation: Electronic configuration of the options are as follows: Option A, Cu: 1s2 2s2 2p6 3s2 3p6 3d10 4s1, so its d subshell is filled Option B, Cu+: 1s2 2s2 2p6 3s2 3p6 3d10, so its d subshell is filled Option C, Cu-: 1s2 2s2 2p6 3s2 3p6 3d10 4s2, so its d subshell is filled. Option D, Cu2+: 1s2 2s2 2p6 3s2 3p6 3d9, so its d subshell is not filled completely, one orbital is partially filled, so it is the correct answer.

Q102. The decomposition of phosphorus pentachloride in the presence of moisture takes place by the following mechanism. The rate equation for this reaction will be:

• A. Rate = k [POCl3][H2O]3
• B. Rate = k [PCl5][H2O]✓
• C. Rate = k [PCl5][H2O]4
• D. Rate = [PCl5][H2O]

Explanation: For the rate equation, the reactants of ONLY the rate-determining step, which is the slow step, are taken into account in a multi-step reaction, NOT the overall equation or the fast step. Since the slow step here consists of one mole of PCL5 and one mole of H2O in the reactants, the rate equation will contain these two substances, each raised to the power 1, which is reflected in option B only. Option A takes into account the fast step only. Option C takes into account the overall reaction. Option D does not have the rate constant k mentioned in the question.

Q103. Which type of reaction takes place when a carbonyl compound is treated with a mixture of NaCN and an acid?

• A. Electrophilic addition reaction
• B. Substitution reaction
• C. Nucleophilic addition reaction✓
• D. Displacement reaction

Explanation: The given scenario is an example of a nucleophilic addition reaction, as the reactant, CN- is a nucleophile which adds across the carbonyl bond. Option A describes reactions undergone by alkenes, where an electrophile is added across the carbon-carbon double bond. Option B describes a reaction where one group of atoms is replaced by another group of atoms, but this is not the case here. Option D describes a reaction in which one element is replaced by another element in a compound.

Q104. An intermolecular force of attraction X is relatively stronger than the other intermolecular forces. It stabilizes alpha-helix and beta-pleated sheets of proteins. The double helical structure of DNA is also stabilized by this force of attraction. Identify X:

• A. Van Der Waals Forces
• B. Hydrogen bonding✓
• C. Ionic interactions
• D. Dipole dipole attraction

Explanation: Option B is correct, as H-bonding stabilizes DNA structure as it is between bases in a complementary base pair as well as it stabilizes the secondary structures, such as alpha-helix and beta-pleated sheet. It is also stronger than the other intermolecular attraction, which is Van der Waals forces.

Q105. Which of the following molecule shows cis-trans isomers?

• A. C2H4
• B. C2H2Cl4
• C. C2H2Br2✓
• D. C2HCl3

Explanation: For cis-trans isomerism, the rotation must be restricted in the molecule (so, cannot occur in compounds that only have C-C single bonds) and there must be two non-identical groups on each double-bonded carbon atom. Option A does not fulfill this requirement, as each double-bonded carbon atom has two identical groups i.e. each C has two H atoms attached to it. Option B does not fulfill these requirements as it is a halogenoalkane, with no C-C double bonds so, rotation is not restricted in the molecule. Option C fulfills these requirements, as not only is rotation restricted as it has a C-C double bond, each carbon atom has two non-identical atoms, i.e. each C has one Hydrogen atom and one Bromine atom so, this option is correct. Option D does not fulfill the requirements as well as even though one double-bonded carbon atom has two non-identical atoms i.e. one H atom and one Cl atom, the other carbon has two identical Cl atoms.

Q106. Modern periodic table is arranged in ascending order of?

• A. Nucleon number
• B. Atomic mass
• C. Proton number✓
• D. Mass number

Explanation: Elements of the periodic table are arranged in ascending order of their atomic number or proton number (which corresponds to the number of protons in the nucleus). Nucleon number, mass number, or atomic mass (all terms used to describe the same thing i.e. relative atomic mass) may have been used in earlier periodic tables.

Q107. Disposable cups are made of a polymer 'polystyrene'. Polystyrene is:

• A. A polyamide
• B. A condensation polymer
• C. An addition polymer✓
• D. A polyester

Explanation: As you can see in the above image, polystyrene is an addition polymer, where the monomers (in this, a styrene molecule) are simply linked together to form a long chain, without the formation of a secondary product so, the answer is C. Option A describes polyamide, which is a condensation polymer formed by monomers, linked by amide bonds. Option B describes condensation polymers, where the functional groups of monomers are linked together, forming a secondary product, such as H2O, in the process. Option D describes a polyester, which is a condensation polymer, formed by monomers linked by ester linkages.

Q108. Which of the following element is not present in halogens?

• A. I
• B. Cl
• C. Fe✓
• D. F

Explanation: Halogens are elements present in group 7A of the periodic table and include, Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), and Astatine (At). Among the given options, only option C is NOT a halogen, it is Iron (Fe), a transition element.

Q109. Chlorofluorocarbons (CFCs) are important compounds that are used as refrigerants but these are also responsible for Ozone layer depletion. If a Chlorofluorocarbon CFCl3 is present in stratosphere, which of it’s reaction intermediates are actually responsible for the breakdown of the ozone molecule?

• A. CFCl2* and Cl*
• B. Cl* and ClO*✓
• C. CFCl2* and ClO*
• D. CFCl2 and CFCl3

Explanation: Ozone breakdown is mediated by free radicals derived from CFCs. Free radicals have unpaired electrons and are highly unstable, thereby breaking down ozone into oxygen. As per the following reaction mechanism, you can see that chlorine radicals are the chief substance involved in ozone depletion, as it catalyses the breakdown of ozone into oxygen and gets regenerated at the end of the reaction via forming a ClO* intermediate, enabling it to break down more ozone molecules.As such, the option, containing Cl* and ClO*, is correct, which is option B.Option A contains a carbon or alkyl containing radical, which is not as reactive as chloride radical due to the presence of a stabilising carbon/ alkyl group, and does not mention ClO*. Option C does not mention the chloride radical.Option D merely mentions examples of CFCs.

Q110. Which of the equations shows ‘twice’ the enthalpy change of neutralization as the following equation: HCl + NaOH → NaCl + H2O

• A. MgCO3 + 2HCl → MgCl2 + CO2 + H2O
• B. KOH + HCl → KCl + H2P
• C. NH4Cl + NaOH → NaCl + H2O + NH3
• D. H2SO4 + Mg(OH)2 → MgSO4 + 2H2O✓

Explanation: The given reaction depicts a neutralization reaction, where one mole of acid is neutralized by one mole of alkali (as indicated by 1 mole of HCl reacting with 1 mole of NaOH). For a reaction to have twice the enthalpy change as this reaction, it should involve neutralization of TWO moles of acid with TWO moles of alkali. Such an option is D, which has two moles of acid (which are present in H2SO4, as one molecule of H2SO4 has two molecules of hydrogen ions) reacting with two moles of alkali (Mg(OH)2 has two moles of hydroxide ions). Option B has the same enthalpy change of neutralisation as the reaction in the question as both involve one mole of H+ reacting with one mole of OH-. Options A and C are not reactions that fit the description of enthalpy change of neutralization, as they form gases in the products aside from a salt and water (for enthalpy change of neutralisation, an acid and a base react to form a salt and water only) so, their enthalpies cannot be compared to the given reaction.

Q111. Which two elements are isotopes?

• A. 16X8 and 16Y8
• B. 14X8 and 15Y8✓
• C. 18X9 and 20Y10
• D. 12X6 and 12Y7

Explanation: In option A, both elements are the exact same, as they have same proton number (8) and same mass number (16) so, they are not isotopes (which vary in mass number but have the same proton number). Option B describes isotopes, as they have the same proton number (8), while having differing mass numbers. Option C describes two different elements entirely as they have varying proton numbers. Option D also describes two different elements due to varying proton numbers.

Q112. Carboxylic acids can be reduced to corresponding alcohols. Which of the following reagent can be used for this purpose?

• A. KMnO4
• B. LiAlH4✓
• C. K2Cr2O4
• D. H2SO4

Explanation: For this reaction to take place, you need the proper reducing agent, which is only LiAlH4 i.e. option B. (NOTE: NaBH4 is not a strong enough reducing agent to reduce carboxylic acids, but LiAlH4 is.)Options A, C, and D; all describe OXIDISING agents.

Q113. Which enthalpy change is relevant in the following process Na(s) → Na (g) ΔH = +107

• A. Enthalpy of atomization✓
• B. Enthalpy of fusion
• C. Enthalpy of vaporization
• D. Enthalpy of formation

Explanation: Option A, enthalpy of atomization, is the energy required to completely separate all the atoms in a chemical substance. The separated atoms are in a gaseous state and get separated from the corresponding substance in its standard state. Sodium is solid in its standard state, and after atomization, the atoms are in their gaseous state. Hence, this is the correct answer. Option B is the enthalpy change when a solid substance melts into a liquid. Here, the state change is from a solid to a gas, no liquid state involved. Option C is the enthalpy change when a liquid is vaporized into a gas. Here, a solid is atomized to its gaseous state so, vaporization does not apply here. Option D is the enthalpy change when one mole of a substance is formed from its constituent elements. This does not describe the given reaction.

Q114. Ionization energy decreases down the group, from top to bottom, due to:

• A. Increase in atomic mass
• B. Increase in shielding effect of the intervening electrons✓
• C. Increase in proton number
• D. Decrease in atomic size

Explanation: Ionization energy is the energy required to remove an electron from the outermost shell of the atom. The factors that affect activation energy are: An increase in proton number, as more protons cause a greater electrostatic force of attraction between the nucleus and the outermost electrons, requiring more energy to overcome. A decrease in atomic size, as the force of attraction is inversely proportional to distance so, the closer the atoms are to the nucleus, the greater would be the force of attraction, which requires greater energy to overcome. The shielding effect describes the reduction in effective nuclear force of attraction between the nucleus and outer electrons due to the repulsion exerted by the inner shells of the atom. As we go down the group, the proton number increases, which would theoretically cause ionization energy to increase, but as we go down the group, the number of shells of each atom increases, successively, resulting in a greater shielding effect which reduces the effective force of attraction between the nucleus and the valence electrons. Hence, ionization energy decreases due to an increase in shielding effect down the group, i.e. option B. Option A does not have a direct impact on ionization energy, albeit atomic mass is accompanied by atomic number, closely in most atoms. Option C is a correct observation down the group, but the ionization energy still decreases even though the atomic number increases. Option D is an incorrect statement as atomic size increases down the group.

Q115. The Ka values of HCI, CH3COOH, HF, and H2SO4 are 107, 1.85 x 10-5, 6.7 x 10-5 and 102 respectively. The decreasing order of acidic strength is:

• A. CH3COOH > HF > H2SO4 > HCI
• B. HCI > H2SO4 > HF > CH3COOH✓
• C. HCl > CH3COOH > HF > H2SO4
• D. HCl > HF > H2SO4 > CH3COOH

Explanation: Ka is a measure of ionization of the acid, thereby a greater Ka corresponds to greater acidic strength. The question asks for DECREASING order of acidic strength so, the acids are to be listed from strongest to weakest. HCl is the strongest with the greatest Ka (107), followed by H2SO4 (102), followed by HF (6.7 x 10-5), and then CH3COOH which is the weakest having the lowest Ka (1.85 x 10-5). Option B has this correctly laid out so, it is the answer. Option A has the order in reverse, as it lists the acid down in increasing order of acidic strength. Options, C and D, have them listed incorrectly, as C has CH3COOH before the rest while D has HF mentioned before H2SO4.

Q116. In contact process, to which substance are adequate quantities of water added to convert it to sulphuric acid?

• A. H2S2O7✓
• B. HSO4-
• C. SO3
• D. S

Explanation: In the contact process, sulfur dioxide (formed by roasting sulfur in oxygen) is reacted with oxygen to form sulfur trioxide, which is mixed with H2SO4 to form oleum or fuming sulfuric acid, with the formula of H2S2O7. This is then mixed with adequate quantities of water to form aqueous sulfuric acid, so the answer is A. Option B is hydrogen sulfate, the species formed when H2SO4 donates one proton. This is not involved in the contact process. Option C is SO3, which can react with water to directly form H2SO4, but the reaction is so violent and inefficient that it can not be a practical source of sulfuric acid so, SO3 is reacted with H2SO4 to form oleum. Option D is elemental sulfur, which is used to make SO2 upon reacting with oxygen in the contact process.

Q117. For the following equilibrium reaction, when the forward reaction is exothermic; an increase in temperature shifts the equilibrium position towards the left. What will occur in that scenario? 2SO2 + O2 ⇌ 2SO3

• A. The concentrations of SO2 and O2 increases and concentration of SO3 decreases as the temperature increases.✓
• B. The concentrations of SO3, SO2, and O2 increase as the temperature increases.
• C. The concentrations of SO2 and O2 decreases and concentration of SO3 increases as the temperature increases.
• D. The concentrations of SO2 and O2 and SO3 remains the same as the temperature increases.

Explanation: Since the equilibrium of the reaction shifts to the left, this means that the left hand side of the reaction is favoured. Increasing the reactants (LHS) and decreasing the products (RHS). Le Chatelier's principle will be used.

Q118. Which one of the following molecules has sp3 hybridization?

• A. CH4✓
• B. C2H4
• C. CO2
• D. C2H2

Explanation: CH4 has sp3 hybridization which enables all four bonds to be of the same length and strength. Methane molecule is formed by the overlap of sp3 hybrid orbitals of carbon with 1s orbitals of four hydrogen atoms separately to form four sigma bonds.

Q119. During stoichiometric calculations, which of the following laws must be followed?

• A. Dalton's law
• B. Avogadro's law
• C. Law of conservation of mass✓
• D. Law of conservation of energy

Explanation: Option A describes the law of partial pressures, which states that the total pressure by a mix of gases is equal to the sum of partial pressures of each of the constituent gases. Option B describes the law which states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of molecules. (NOTE: Do not confuse this with Avogadro's number and the concept of a mole). Option C describes the law which states that matter can neither be created nor destroyed so, the mass of every element, present in the products of a chemical reaction, must be equal to the mass of each and every element present in the reaction, on the reactant side. This is the basis of balancing a chemical equation, without which stoichiometric calculations cannot be carried out so, it is the correct answer. Option D describes the law which states that energy can neither be created nor destroyed; it can only be transferred from one form to the other. This has more relevance in energy conversion and work or power calculations rather than stoichiometry.

Q120. Nitriles (RCN), on hydrolysis, in the presence of a mineral acid yield:

• A. Ethers
• B. Carboxylic acids✓
• C. Aldehydes
• D. Alcohols

Explanation: As can be seen from the reaction mechanism below, nitrile hydrolysis yields carboxylic acids and the liberation of an ammonium/ammonia molecule.

Q121. Alkenes undergo:

• A. Electrophilic Addition✓
• B. Electrophilic substitution
• C. Nucleophilic substitution
• D. Nucleophilic addition

Explanation: Alkenes have a pi bond between carbon atoms, which is an area of electron density i.e. a region of great negative charge, which attracts positively charged species, i.e. electrophiles, which add across the double bond without the removal of any atom/group of atoms. Hence, alkenes undergo electrophilic addition, option A. The remaining options describe reactions that are undergone by other organic compounds. Option B, electrophilic substitution is undergone by benzene. Option C, nucleophilic substitution is undergone by alkyl halides. Option D, nucleophilic addition is undergone by aldehydes and ketones.

Q122. Treatment of ethene with cold sulphuric acid, followed by reaction with boiling water, yields:

• A. Ethane
• B. Ethanal
• C. Ethanol✓
• D. Ethyne

Explanation: The scenario describes the overall reaction of the addition of OH or hydration of ethene. Upon reacting with sulfuric acid, it forms an intermediate, which is ethyl hydrogen sulfate, and upon boiling with water, it forms ethanol as sulfuric acid is regenerated. The answer is C. Option A, ethane is formed upon hydrogenation of ethene. Option B, ethanal is formed upon oxidation of the ethanol, which is formed. Optio D, ethyne is formed upon treatment of ethene with Bromine, followed by heating in the presence of alcoholic KOH.

Q123. How many moles of Calcium Carbonate are present in 1.75 kg of Calcium Carbonate? (Ar of Ca = 40, Ar of C = 12, Ar of O = 16)

• A. 1.75 mol
• B. 1750 mol
• C. 0.0175 mol
• D. 17.5 mol✓

Explanation: The molar mass of CaCO3 (40 + 12 + 16 + 16 + 16) is 100g per mole. If one mole equals 100g, than 1750g (1.75 kg x 1000) equals 1750/100, which is 17.5 mol so, D is the answer.

Q124. Which balanced chemical equation show the formation of ethanoyl chloride using thionyl chloride?

• A. HCOOH + SOCl2 → HCOCI + SO2 + HCI
• B. CH3CH2COOH + 2SOCl → CH3CH2COCl + SO2 + HCI
• C. CH3CH2COOH + 2SOCl → CH3CH2COC1 + SO3 + HCI
• D. CH3COOH + SOCl2 → CH3COCl + SO2 + HCI✓

Explanation: Thionyl chloride is SOCl2 and it reacts with carboxylic acids to form acyl chlorides.Options B and C do not involve thionyl chloride, it mentions SOCl2 instead of SOCl2 so, they are incorrect.Options A and D are both valid reactions involving thionyl chloride. However, the question asks for the formation of ethanoyl chloride. Option A shows the formation of METHANOYL Chloride, whereby D shows ethanoyl chloride formation so, D is the correct answer.

Q125. All the collisions between the particles of gases are elastic in nature. What is meant by "Elastic Collisions"?

• A. No change in the total kinetic energy✓
• B. The velocity of the molecules changes
• C. No change in potential energy during the collisions
• D. No change in mass during the collisions

Explanation: By definition, an 'Elastic collision' describes a collision whereby the total kinetic energy of the molecules is unchanged so, option A is correct. Option B, velocity may or may not change in elastic collisions, and the velocity also changes in inelastic conditions as well. Option C, potential energy is not usually used in the context of elastic collisions. Option D is a correct statement but it holds for all types of collisions and is not exclusive to elastic collisions.

Q126. Which of the following reagent is required for preparation of acyl chloride (CH3COCl) from ethanoic aicd?

• A. PCl5✓
• B. POCl3
• C. CH3Cl
• D. HCl

Explanation: Production of acyl chloride from a carboxylic acid can be used using the following reagents, PCl3, PCl5, SOCl2 (thionyl chloride). Only option A mentions one of the three, which is PCl5 so, it is the correct answer.

Q127. Identification tests for functional groups of organic compounds are associated with specific observations. Tollen’s reagent is an ammoniacal silver nitrate solution, which is used for the identification of a functional group X with an observation O. Identify X and O.

• A. X = Aldehyde O = Silver mirror✓
• B. X = Ketone, O = Grey precipitate
• C. X = Ketone, O = Silver mirror
• D. X = Aldehyde O = Red precipitate

Explanation: Tollens reagent test is also known as the silver mirror test. It cannot react with ketones but it reacts with aldehydes, thereby identifying the aldehyde functional group. Upon reacting, the silver precipitates as a silver mirror which coats the bottom of the test tube. Hence, option A is the correct answer as it is the only one with both correct functional group and observation.

Q128. Free Nitrogen and Oxygen are present in atmosphere but they do not react with each other under normal conditions because:

• A. Nitrogen requires a catalyst
• B. Oxygen is very inactive
• C. Oxygen is found in less concentration
• D. Nitrogen is highly inactive gas✓

Explanation: Option A is a true statement, but it does not, by itself, explain why nitrogen does not react under normal conditions; Nitrogen is highly inactive so, it requires a catalyst because of its inactivity/inertness.Option B is incorrect as oxygen is quite reactive, taking part in numerous reactions such as combustion reactions.Option C is a true statement because it is found in less concentration than nitrogen, but that should not, in itself, prevent a reaction from taking place under normal conditions.Option D is correct. Nitrogen is nonpolar and has a triple bond, which requires an enormous amount of energy to overcome so, under normal conditions, it cannot react with oxygen; it requires a catalyst and heat energy to overcome its natural inertness.

Q129. CFCs are organic compounds, that are derivatives of saturated hydrocarbons. They have high bond dissociation values therefore they are inert and non-toxic for living organisms.The word CFCs stands for:

• A. Chlorofluorocarbons✓
• B. Carbonfluorochlorines
• C. Clorofluorocarbonos
• D. Chlorofluoridecarbons

Explanation: Factual recall.

Q130. According to Watson and Crick’s model of DNA, the DNA molecule consists of a double helix. What type of forces are responsible to keep two strands of DNA together?

• A. Van der Waals forces
• B. Ionic bonding
• C. Hydrogen bonding✓
• D. Dipole-induced dipole forces

Explanation: The bases in each complementary base pair have hydrogen bonds between them, which keeps the two strands together and stable. 'A' and 'T' have 2 hydrogen bonds between them, while 'G' and 'C' have three hydrogen bonds between them. The answer is C, as the other options do not mention hydrogen bonding.

Q131. Nitrogen has an atomic number of 7.Which of the following electronic configurations is of a nitrogen atom in the ground state?

• A. 1s2, 2s2, 2px1, 2py1, 2pz1✓
• B. 1s2, 2s2, 2px2, 2pz1
• C. 1s2, 2s2, 2px2, 2py1
• D. 1s2, 2s2, 2py2, 2pz1

Explanation: The electronic configuration of nitrogen is as follows: 1s2, 2s2, 2p3. The 'Auf-bau principle' states that electrons fill lower-energy atomic orbitals before filling higher-energy ones. Moreover, we have the 'Hund's Rule' stating that when electrons occupy degenerate orbitals (i.e. of the same energy level), they must first occupy the empty orbitals before double occupying them.

Q132. Which of the following substances exhibits strong hydrogen bonding?

• A. H2S
• B. SiH4
• C. NH3✓
• D. HI

Explanation: Hydrogen bonding occurs in those substances in which a highly electronegative atom (such as Fluorine (F), Oxygen (O) or Nitrogen (N)) is directly bonded to Hydrogen. The presence of one of these three atoms is a REQUIREMENT for hydrogen bonding to occur. Only option C mentions a substance which has one of these three atoms, i.e. Nitrogen in NH3, while the other options mention other atoms aside from the three highly electronegative species mentioned previously.

Q133. Aqueous solutions of Iodine and Sodium hydroxide were mixed in a round bottom flask at 70 C. Following chemical reaction was carried out. 3I2 + 6 NaOH → NaIO3 + NaI + H2OThis reaction is termed as:

• A. Redox reaction✓
• B. Precipitation reaction
• C. Substitution reaction
• D. Free radical reaction

Explanation: To determine whether such a reaction is a redox reaction, the oxidation states of the reactants and products is compared. The oxidation state of Na remains +1 throughout, as does the oxidation states of Hydrogen (+1) and Oxygen (-2). Iodine undergoes a change in oxidation state. Its oxidation state is 0, in its molecular form, on the reactant side. In NaI, iodine's oxidation state is -1 so, Iodine has undergone reduction. In NaIO3, the oxidation states of Na remain +1 and oxygen's also remains -2 so, we can determine that the oxidation state of iodine is +5 in this reaction. (NOTE: For oxidation numbers, the total oxidation state of a compound is zero so, Iodine's oxidation number must be +5 to ensure NaIO3’s oxidation state is zero). Since Iodine underwent oxidation AND reduction, we can conclude that this is a redox reaction (disproportionation to be precise, as the same species has undergone oxidation and reduction) so, option A is correct. Option B is incorrect, as the two salts produced are soluble in water. Option C does not describe this reaction. Option D is incorrect as no free radicals are produced during this reaction.

Q134. Which of the following bond is responsible for joining the amino acids in proteins?

• A. Peptide Bond✓
• B. Ionic Bond
• C. Metallic Bond
• D. Disulfide Bond

Explanation: Option A is correct, as within a protein chain, each amino acid is linked to each other by a peptide bond which is a covalent bond.Options B and D describe those bonds that hold proteins in their complex three dimensional structures, but the question asks for the bond that holds individual amino acids within proteins.Option C describes a metallic bond which is only found in metals, and not in proteins as it is molecular in nature.

Q135. Which of the following is the electronic configuration of Cr?

• A. [Ar] 3d5 4s2
• B. [Ar] 3d5 4s1✓
• C. [Ar] 3d4 4s2
• D. [Ar] 3d6 4s0

Explanation: Chromium (Cr) has atomic number 24. Theoretically, the electronic configuration of Cr should be 1s2 2s2 2p6 3s2 3p6 3d4 4s2 (NOTE: [Ar] symbol denotes the electronic configuration for 1s2 2s2 2p6 3s3 3p6). However, in reality, one electron from 4s jumps to the 3d subshell so, that the configuration is [Ar] 3d5 4s1. This is because a half-filled d subshell, which is 3d5, is much more stable than 3d4, and atoms favor stability. Hence, option B is the correct answer.

Q136. The number of moles of water in 1 kg ice are:

• A. 50 moles
• B. 100 moles
• C. 1000 moles
• D. 55.5 moles✓

Explanation: 1 mole of water corresponds to 18g. So, to calculate the number of moles of water in 1kg or 1000g of ice (NOTE: The state of the substance does not have an effect in such stoichiometric calculations), 1000 is divided by 18, which comes out to 55.5 moles.

Q137. Which of the following set constitutes of all the molecules and ions of non-planar geometry?

• A. PH4+, NH3, SO3, Benzene
• B. CH4, NH4 +, MnO4- , NF3✓
• C. Ethene, H2O, BeCl2, H2S
• D. SO2, C2H4, BF3, NO3-

Explanation: The question asks for species that have NON-PLANAR geometry so, those structures which are planar are to be eliminated. Option A includes benzene and SO3, which are planar molecules as all carbons of benzene have trigonal planar geometry around them, while SO3 is a trigonal planar molecule as well (due to the presence of three bonding pairs of electrons around them). PH4+ and NH3 are not planar however, PH4+ is tetrahedral (its shape corresponds to ammonium ion as P and N are in the same group), while NH3 has a trigonal pyramidal shape (3 bonding pairs and one lone pair around N). Option B is correct, as all four substances have non-planar structures. CH4 and NH4+ have tetrahedral structures due to four pairs of bonding electrons around the central atom, MnO4- also has tetrahedral geometry (see pictures below), while NF3 has a trigonal pyramidal shape (compare it to NH3 structure as both have three single bonds with a lone pair of electrons). Option C is also incorrect, as ethene and BeCl2 are planar molecules. Ethene is trigonal planar, while BeCl2 is linear. H2O and H2S are non planar however, with both having the same bent structure with two bonding pairs of electrons and two lone pairs (the two structures are similar as both O and S belong to group 6). Option D is incorrect due to the presence of planar C2H4 (ethene), NO3-, and BF3, as they both are trigonal planar. SO2 only is the non-planar bent substance in this list.

Q138. The given diagram shows the enthalpy changes during a chemical reaction. What type of reaction is this?

• A. An endothermic reaction
• B. An exothermic reaction✓
• C. An isothermic process
• D. A non-spontaneous process

Explanation: This diagram shows a reaction in which the reactants are at a greater energy level than the products, indicating that energy is released during the reaction, i.e. change in enthalpy or (H2-H1) would be negative, indicating that it is an exothermic reaction so, option B is correct. Option A describes an endothermic reaction which would have a diagram depicting products, at a higher energy level than the reactants. Option C indicates a process in which the heat energy of the system does not change, but the diagram itself shows a change in enthalpy. Option D cannot be determined from a mere energy diagram as shown.

Q139. What is the measure of activation energy in an endothermic reaction?

• A. The energy of activation of backward reaction is less than that of forward reaction✓
• B. The energy of activation of backward reaction is more that that of forward reaction
• C. The energy of activation of forward reaction is less than that of backward reaction
• D. The energy of activation of forward - backward reaction is same

Explanation: In an endothermic reaction, the activation energy of a forward reaction will always be greater than that of the reverse reaction as can be seen in the diagram given above so, the answer is A which is the activation energy of the reverse reaction being less than that of the forward reaction. Options B and C describe the conditions of an exothermic reaction. Option D would describe a reaction in which the products and reactants are at the same energy level.

Q140. Maleic and Fumaric acid, both have chemical C4H4O4. The structure of these acids is shown below. What type of isomers are these?

• A. Cis-trans isomers✓
• B. Structural isomers
• C. Metamers
• D. Position Isomers

Explanation: Option A describes isomers in which one isomer has the same groups of atoms on one side of the double bond (cis), while the other isomer (trans) would have different groups of atoms on each side of the double bond. Maleic acid is the cis-isomer, while fumaric acid is the trans-isomer. Option B does not describe these compounds as these isomers are stereoisomers, having the same arrangements of atoms, structurally. Option C, metamers, describes compounds that have differing alkyl groups on each side of the functional group. As you can see from their structures, both compounds have the same groups on either side of the C=C. Option D also is incorrect as positional isomers have different locations of the functional groups, but both compounds have the double bond located at the exact same location.

Q141. Amino acids are bi-functional compounds, with a general formula NH2CH(R)CO2H. A tripeptide is formed between Alanine (ala), Glycine (gly), and lysine (lys). There is no repetition of amino acid in this tri-peptide; suggest how many tri-peptides are possible?

• A. 9
• B. 6✓
• C. 3
• D. 12

Explanation: For such questions, try to make all possible combinations of amino acids. They include:ala-gly-lys gly-ala-lys lys-ala-glyala-lys-gly gly-lys-ala lys-gly-alaThus, there are six possible amino acids that can be formed without repetition. If repetition was allowed, then the answer would be 27 (3 x 3 x 3).NOTE: Tripeptides which can be written in reverse such as ala-lys-gly and gly-lys-ala are different and distinct from each other so do not eliminate such examples or count them as one instead of two, because in such tripeptides the C-terminal and N-terminals vary.

Q142. If the energy of activation of a chemical reaction is very low, the rate of that chemical reaction is observed to be very high because?

• A. Concentration of the reactants becomes irrelevant
• B. Reaction proceeds without any transition state
• C. Number of efficient or fruitful collisions increase✓
• D. Molecules of the reactants move slowly

Explanation: Option A is incorrect because the concentration of reactants, still, is a factor regardless of other factors such as low activation energy. (Same goes for all other factors that affect the rate of reaction) Option B is incorrect, as the transition state is always present regardless of activation energy. Option C is correct, as low activation energy means that many more reactant particles have the minimum energy required to produce products so, with a greater number of particles possessing energy equal to or greater than the activation energy collide, they produce products. Hence, more fruitful collisions occur. Option D is incorrect, as reactant particles move slowly if there is less heat energy applied, which was not mentioned in the question.

Q143. Which of the following compounds is responsible for the depletion of ozone layer?

• A. Chlorofluorocarbons✓
• B. Carbon tetrachlorie
• C. Methane
• D. Hydroflorocarbons

Explanation: Factual recall.

Q144. Which product is obtained by the hydrolysis of 1-chlorobutane with the aqueous sodium hydroxide?

• A. 1-butanal
• B. 1-butanol✓
• C. 1-butene
• D. Butanone

Explanation: The scenario describes nucleophilic substitution of 1-chlorobutane, where -OH serves as the nucleophile, as sodium hydroxide is aqueous instead of alcoholic. As such, alcohol is produced as -OH substitutes the Cl atom so, option B is correct.Option A is incorrect as alkyl halides cannot be directly converted to aldehydes.Option C, an alkene, would be produced if the sodium hydroxide was in an alcohol solvent or if the reactant was alcoholic KOH instead of aqueous. This is because, in these conditions, an elimination reaction occurs instead of a substitution reaction, producing an alkene upon the elimination of the halogen and adjacent carbons hydrogen atom.Option D is incorrect as ketones cannot be formed directly from alkyl halides.

Q145. Oxidation number of particular element can be directly or indirectly inferred from its:

• A. Group number✓
• B. Atomic size
• C. Atomic mass
• D. Physical state

Explanation: Generally, groups 1, 2, 3, and 4 have oxidation numbers of +1, +2, +3, and +/-4. And groups 7, 6, and 5 have oxidation numbers -1, -2, and -3. This is the general trend among the groups; however, there are many exceptions to this.

Q146. Which of the following reactions is used for the production of alcohols on industrial scale?

• A. Hydrogenation of alkenes
• B. Hydration of alkenes✓
• C. Hydrohalogenation of alkenes
• D. Hydroxylation of alkenes

Explanation: Option A describes a reaction that produces alkanes from alkenes. Option B describes a reaction that produces alcohol from alkenes so, it is correct. Option C describes a reaction that produces alkyl halides from alkenes. Option D describes a reaction that produces glycol from alkenes, which is a type of alcohol but not the type that is produced on an industrial scale for commercial use.

Q147. Solution contains 85.5g of sucrose (C12H22O11) in 250 cm3. What is its molarity?

• A. 1 M✓
• B. 0.5 M
• C. 0.25 M
• D. 2 M

Explanation: Molarity is moles of a substance divided by the volume in liters or cubic decimeters. For moles, divide the given mass with the molar mass of sucrose, which is (12x12+22+11x16) 342 grams per mol. Number of moles is 85.5/342 which comes out to 0.25 mol. To convert 250 cubic cm to liters, divide by 1000, so this comes out to 0.25 liters. Molarity = 0.25/0.25 = 1M so, answer is A.

Q148. Which derivative of benzene shows maximum reactivity in electrophilic substitution reactions?

• A. Benzoic acid
• B. Methyl benzene✓
• C. Benzaldehyde
• D. Nitrobenzene

Explanation: Activating groups increase the reactivity of electrophilic substitutions, whereas deactivating groups decrease it. Activating groups are (from most activating to least): -NH2, -NR2 > -OH, -OR> -NHCOR> -CH3 and other alkyl groups, where R is an alkyl group. Deactivating groups are (from most deactivating to least): -NO2, -CF3> -COR (COR can be both carboxylic acids or aldehydes), -CN, -CO2R, -SO3H > Halogens. Among the options, only B describes benzene with an activating group so, it is correct. All other options describe deactivating groups as one can see from the lists mentioned.

Q149. The names of functional groups in the following compound X are:

• A. Secondary alcohol, nitrile, and phenol ring
• B. Secondary alcohol, amine, and benzene ring
• C. Secondary alcohol, nitrile, and aryl ring✓
• D. Primary alcohol, nitrile, and benzene ring

Explanation: The functional groups are: secondary alcohol, nitrile, and benzene/aryl ring. Option A incorrectly mentions phenol as there is no hydroxyl group directly bonded to the benzene ring. Option B incorrectly mentions amine, as for an amine, the nitrogen atom must be bonded to at least three other species/atoms, but in this case, nitrogen is directly bonded to one carbon atom, indicating a C-N triple bond which is characteristic of nitriles. Option C correctly lists all three functional groups. Option D incorrectly mentions primary alcohol, as for the alcohol group to be primary it must have been terminal/present at the end of the alkyl group, which is not the case here as -OH group is mentioned in the structural formula as being in the middle of the alkyl group.

Q150. In the second period of elements, although oxygen lies next to nitrogen yet its ionization first energy is lower than that of nitrogen because?

• A. Oxygen is paramagnetic in character.
• B. Nuclear charger of oxygen is greater than nitrogen.
• C. Oxygen has higher electron affinity.
• D. In oxygen, there exists repulsion between pair of electrons present in the same orbital of valence shell.✓

Explanation: Option A is factually correct but has no bearing on ionization energy. Option B is factually correct as oxygen has one more proton in its nucleus than nitrogen, however, because of this, the ionization energy must increase, but in reality, the ionization energy of oxygen is lower than that of nitrogen. Option C is factually correct, but has no bearing on activation energy. Option D is correct because oxygen has 6 valence electrons compared to Nitrogen's 5 valence electrons so, the electronic configuration of the last shell of O is 2s2 2p4 compared with nitrogen's 2s2 2p3. In N, the p subshells have all three orbitals, singly filled with electrons whereby in oxygen, one p orbital has a pair of electrons. In this orbital, the two electrons exert repulsion which will lower the ionization energy due to a decrease in the effective nuclear force of attraction between the outermost electron and nucleus in the oxygen atom.

Q151. The structure of Xenon trioxide is shown below: With reference to the valence shell electron pair repulsion theory (VSEPR), the shape of XeO3 is:

• A. Trigonal planar.
• B. Tetrahedral.
• C. Bent(or angular).
• D. Trigonal pyramidal.✓

Explanation: To determine the shape of the molecule, you must find the number of bonding pairs of electrons and lone pairs of electrons. Pi bonds are to be excluded. The molecule has three single bonds which correspond to three bonding pairs of electrons, and one lone pair of electrons, resulting in a structure similar to that of ammonia, which is trigonal pyramidal so, the answer is D. Option A would occur in a molecule with only three bonding pairs of electrons. Option B would occur in a molecule with four bonding pairs of electrons. Option C would occur in a molecule with either two bonding pairs with one lone pair or with two bonding pairs and two lone pairs of electrons.

Q152. Which of the following will give a positive test with Tollen’s reagent?

• A. Tertiary alcohols
• B. Carboxylic acids
• C. Aldehydes✓
• D. Ketones

Explanation: Only aldehydes can be easily oxidised by mild oxidising agents such as the Tollen reagent, Fehling solution and Benedict solution.

Q153. Keeping in view the values of standard reduction potential given above, which one of the following would you select as a feasible redox chemical reaction?

• A. Cl- + I2 -> Cl2 + 2I-
• B. Cu + Zn2+ -> Cu2+ + Zn
• C. Mg + 2H + -> Mg2+ + H2✓
• D. 2Au + 6H+ -> 2Au3+ + 3 H2

Explanation: For a redox reaction to be feasible, the emf of it must be positive. (NOTE: When calculating emf, you take the voltage of each half equation according to the scenario, i.e. one will exhibit the same order as those shown above so, the E value is directly taken, but for some, you have to reverse the sign of the E value since the reaction, itself, is reversed). In option A, the emf is = -1.36 + 0.54 = -0.82, so NOT FEASIBLE. In option B, the emf is = -0.34 -0,76 = -1.10, so NOT FEASIBLE. In option C, the emf is = 2.37 + 0.00V = 2.37 so, it IS FEASIBLE; the answer is C. In option D, the emf is = -1.50 + 0 = -1.54 (the coefficient of the reaction is not taken into account) so, it is NOT FEASIBLE.

Q154. The range of the projectile depends upon the velocity of the projection and the angle of the projection i.e. 45. For a fixed velocity, when the angle of projection is larger than 45. Which of the following is correct?

• A. The height attained by the projectile will be more but the range is less.✓
• B. Both the height and the range attained by the projectile will be more.
• C. The height attained by the projectile will be less but the range is more.
• D. Both the height and the range attained by the projectile will be less.

Explanation: In the formula for range, it is maximum when the value of sin(2θ) is maximum. This value occurs when 2θ is equal to 90 therefore θ = 45. Beyond θ = 45, the Range decreases. The height increases beyond θ = 45 because it is calculated using Sin(θ).

Q155. The wavelength of the electromagnetic wave having frequency of 3 kHz will be?

• A. 120 km
• B. 100 km✓
• C. 80 km
• D. 140 km

Explanation: λ = v / fInputting speed of light or 3 x 108 m/s as v and 3 x 103 Hz as f gives us 1 x 105 m or 100 km.

Q156. An automobile is moving forwards with uniform velocity due to the force exerted by its engine. If the force is doubled with the velocity remaining constant, what happens to its total power?

• A. It is squared
• B. It is halved
• C. It is doubled✓
• D. It does not change

Explanation: Power = force * velocityWith velocity kept constant, Power is directly proportional to force. Therefore, doubling force would double power.

Q157. Which of the following statement shows that no work is done?

• A. Pushing a car to start it moving
• B. Lifting the weights.
• C. The moon Orbiting the earth.✓
• D. Writing an essay on a page.

Explanation: When the moon revolves around the earth, the displacement is normal to the direction of force on the moon. Therefore no work is done by the moon.Work is the dot product of force and displacement.W=F.dcos theta

Q158. If two objects of equal masses 'm' are moving towards each other with the same speeds 'v' then what will be the total final momentum after elastic head-on collision?

• A. mv kg m/s
• B. - mv kg's
• C. 2 mv kg's
• D. 0 kg m/s✓

Explanation: According to the law of conservation of momentum, total initial momentum of a closed system is equal to total final momentum of a closed system. Since two equal masses are moving in opposite direction, computing vector sum of total initial momentum gives us zero:m * (+v) + m * (-v) = 0Therefore, total final momentum is also equal to 0

Q159. In double slit experiment, the fringe spacing of the diffracted rays increases when:

• A. The wavelength of the diffracted rays increases✓
• B. The distance between the slits increases
• C. The distance from midpoints of the slits to the central of the fringe on the screen increases
• D. The distance between the screen and the slits decreases

Explanation: Increasing the wavelength of the light source or distance between the slit and the screen would increase the fringe separation. Decreasing slit separation increases the fringe spacing.

Q160. What is the quark composition of a Proton?

• A. Two up quarks and one down quark✓
• B. Two up quarks and one strange quark
• C. One up quark and two strange quarks
• D. Two down quarks and one up quark

Explanation: The quark composition of a proton consists of two "up" quarks (u) and one "down" quark (d).

Q161. The sum of all forms of molecular energies (kinetic and potential) of a substance is termed as:

• A. Elastic energy
• B. Absolute energy
• C. Internal energy✓
• D. Heat energy

Explanation: Elastic energy is a form of potential energy. Absolute energy does not exist. By definition, Internal energy is the sum of total kinetic and potential energy as defined in 1st law of thermodynamics.

Q162. Molecules of a gas at constant pressure for a fixed amount of gas have average kinetic energy X. Increasing temperature from 27 oC to 327 oC, the average K.E. of molecules will become:

• A. 20X
• B. 300X
• C. 200X
• D. 2X✓

Explanation: Total Kinetic Energy = 3/2 * Boltzmann's Constant (k) * temperature in Kelvins27C + 273 = 300K327 + 273 = 600KAs rest of quantities are kept constant increasing temperature from 300K to 600K will double the average kinetic energy of the system.

Q163. Kirchhoff's first law/rule corresponds to:

• A. Law of conservation of energy
• B. Law Of conservation Of momentum
• C. Law of conservation of charge✓
• D. Law of conservation of mass

Explanation: Kirchhoff's first law states that the algebraic sum of current flowing into a junction is zero. This rule corresponds to conservation of charge which states that total charge in a closed system remains constant because current is the rate of flow of charge.

Q164. Minimum energy required to eject an electron from metal surface is called:

• A. Stopping potential
• B. Electromotive force
• C. Work function✓
• D. Threshold frequency

Explanation: The work function is the minimum energy needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface. Stopping potential is defined as the potential required to stop ejection of electrons from a metal surface when an incident beam of energy greater than the work potential of metal is directed on it. Threshold frequency is the minimum frequency of incident light that can cause emission of electrons from the metal surface.

Q165. If we change the magnetic flux linking a coil by rotating the coil in a constant magnetic field, the rate of change of this flux is:

• A. Proportional to the emf produced in it✓
• B. Proportional to the material Of the coil
• C. Proportional to the resistance of the coil
• D. Proportional to the change in magnetic field

Explanation: Explanation is given below.

Q166. The area under the extension-load graph Of an elastic material whose elastic limit has not been exceeded gives its:

• A. Stress
• B. Strain energy✓
• C. Strain
• D. Young modulus

Explanation: Strain energy is the energy stored by an object undergoing deformation. Area under the extension-load graph represents work done to cause the deformation. An object which has not reached its elastic limit, the work done during stretching is recovered when it is unloaded.

Q167. What will be the expression for the observed frequency f0 , if the source is moving towards the stationary observer?

• A. Option A
• B. Option B
• C. Option C
• D. Option D✓

Explanation: The following is the solution:

Q168. The direction of current through the load resistance of a full-wave rectification circuit:

• A. remains constant✓
• B. inverts for positive cycle
• C. changes for every cycle
• D. inverts for negative cycle

Explanation: Rectification changes alternate current to direct current where current remains positive throughout the cycle.

Q169. A negligible small current between input terminals of the operational amplifier is because of:

• A. High output resistance
• B. Low output resistance
• C. High input resistance✓
• D. Low input resistance

Explanation: Ideal op-amps have zero input current. A small input current passes because of very high input impedence.

Q170. Heavy nucleus of atoms go through fission so that they can:

• A. absorb low amount of energy
• B. absorb high amount of energy
• C. reduce their binding energy nucleon
• D. increase their binding energy per nucleon✓

Explanation: Maximum binding energy is achieved around Z = 50 (Iron is close to peak at Z = 56). Heavy nuclei undergo fission to decrease nuclear size and increase binding energy per nucleon.

Q171. A wire has a spring constant of 5 x 104 N m-1. It is stretched by a force to the extension of 1.4 mm. Calculate the strain energy stored in the wire.

• A. 4.9 x 10-5 J
• B. 4.9 x 10-2 J✓
• C. 4.9 J
• D. 4.9 x 10-3 J

Explanation: E=0.5kx^2E=0.5(50000)(0.0014)2 =0.049 J

Q172. An object is moving along a circular path of radius 4m. What will be its angular displacement if it moves 14 m on this circular path?

• A. 5.0 radians
• B. 3.5 radians✓
• C. 4.5 radians
• D. 5.5 radians

Explanation: Arc length = Radius x Angular displacementS = rθKeeping Arc length as 14m and radius as 4m, we get angular displacement as 3.5 radians.

Q173. A copper wire has length L and cross-sectional area A. Its resistance is R. If we halved the length and halved the diameter of wire, then what will be the resistance of this wire?

• A. 4R
• B. 3R
• C. 2R✓
• D. R

Explanation: When length is halved, resistance is halved.When diameter is halved, area is divided by 4 (square relationship) hence, resistance is quadrupled. Therefore, (4/2) x R = 2R

Q174. If a conductor of length 7m is placed in a magnetic field of strength 0.3 T carrying current 1 A, parallel to the field. What will be the force, acting on it, due to this magnetic field?

• A. 7 N
• B. 0 N✓
• C. 3.1 N
• D. 2.1 N

Explanation: When the conductor is placed parallel to magnetic field, no force will act on it.

Q175. In relation, which quantity is represented by λ?

• A. Decay constant✓
• B. Half life
• C. Activity
• D. Wavelength

Explanation: Explanation is given below.T1/2 represents the 'half-life' with 'T', implying the quantity has something to do with time, and '1/2' representing 'half'.

Q176. Work done, due to centripetal force, for circular motion will be:

• A. Reduced
• B. Zero✓
• C. Maximum
• D. Half

Explanation: The displacement of an object after one complete revolution is 0 as the object returns to its original position. The beginning and final locations of a body that travels a distance and returns to its original place are the same. The displacement is zero in this situation, but the distance traveled is not.As work done = force x displacement, given the displacement is 0, work done is also 0.

Q177. Electric field strength of a point charge is E and electric potential is V, at a distance 'r' from the point charge. What is the electric potential at a point for the same point charge where electric field strength is E/4?

• A. 4V
• B. 2V✓
• C. V/2
• D. V/4

Q178. Light, emitted by a single source, passes through 2 narrow slits 1.00 mm apart. The interference pattern is observed on a screen 200 cm away and the separation between the centers of adjacent bright fringes is 2.00 mm. What would be the wavelength of the light?(u= micro)

• A. 2 pm
• B. 1 um✓
• C. 2 um
• D. 1 nm

Explanation: The explanation is given below:Wavelength = slit separation x firing separation/Distance between slits and screenAll the values, of distance, would be converted to meters (m).Slit separation = 1 x 10-3mFringe separation = 2 x 10-3mDistance between slit and screen = 200 x 10-2m

Q179. Calculate the energy of a photon, with a frequency of 3.0 x 1018 Hz.(h = 6.63 x 10-34)

• A. 19.89 x 10-16 J✓
• B. 11.89 x 10-16 J
• C. 1.89 x 10-16 J
• D. 19.89 x 10-18 J

Explanation: The correct option is A. E=hf = 6.63×10-34 × 3×10¹⁸ = 19.89×10-16 J

Q180. The path difference for the destructive interference can be written as:

• A. Δs = n λ
• B. Δs = (n+ 1/3) λ/2
• C. Δs = (2n+ 1) λ/2✓
• D. Δs = (2n) λ

Explanation: Explanation is given below.

Q181. Calculate the rate at which energy is transferred by a 220 V mains supply, which provides a current of 0.1 A to an LED?

• A. 2.2 W
• B. 2.2 kW
• C. 22 kW
• D. 22 W✓

Explanation: Power = Voltage x Current; 220 x 0.1 = 22W

Q182. An alternating voltage V (in volts) is represented by the equation:V = 300 sin(100πt) What is the value of “f” for this voltage?

• A. 100 Hz
• B. 50 Hz✓
• C. 25 H
• D. 200 Hz

Explanation: Voltage = Maximum Voltage x sin (2πft) Therefore, 2f = 100 and frequency = 50 Hz

Q183. A particle, carrying a charge of 5e, falls through a potential difference of 25 V. What would be energy acquired by the particle in ‘J’.

• A. 1.6 x 10-19 J
• B. 125 J
• C. 125 x 10-19 J
• D. 125 x 1.6 x 10-19 J✓

Explanation: Electric Potential Energy = Electric Potential x Charge Charge = 5 x 1.6 x 10-19 C Electric Potential Energy = 25V x (5 x 1.6 x 10-19) C This could be manipulated to give us (125 x 1.6 x 10-19) J, which is Option D.

Q184. The unit of magnetic flux density is the tesla, ‘T’. It can also be expressed as:

• A. 1 N A-1 m
• B. 1 N-1 A-1 m
• C. 1 N-1 A-1 m-1
• D. 1 N A-1 m-1✓

Explanation: Explanation is given below.

Q185. If we give a direct current to the transformer’s primary coil, then there will be:

• A. No emf produced in the secondary coil✓
• B. More emf in the secondary coil
• C. Equal emf produced in the secondary coil
• D. Less emf produced in the secondary coil

Explanation: A direct current will produce no varying magnetic field and therefore, no emf will be generated in the secondary coil.

Q186. Percentage uncertainty in length and width of a rectangle is 2% and 3%. The total uncertainty in the area of that rectangle is:

• A. 1.5%
• B. 5%✓
• C. 1%
• D. 6%

Explanation: The following is the solution:Area = length x widthRelative uncertainties are added, when two quantities are multiplied.For addition or subtraction: The uncertainties add up directly.For multiplication or division: The relative uncertainties (expressed as a fraction or percentage) add up directly. To obtain the absolute uncertainty, multiply the combined relative uncertainty by the result of the multiplication or division.For powers or roots: Multiply the absolute uncertainty by the power or root exponent.

Q187. Electric field strength at a point between oppositely charged plates is 'E'. If the distance between plates is reduced to half, what will be the new value of electric intensity?

• A. 4E
• B. E/2
• C. E/4
• D. 2E✓

Explanation: Between two parallel plates, the electric field strength is given by: E = V/d If the distance is halved, electric field strength is doubled since both are inversely related.

Q188. The horizontal component of Earth magnetic flux density is 1.8 x 10-6 T. The current in a horizontal cable is 160 A. Calculate the maximum force per unit length?

• A. 2.88 x 10-6 N/m
• B. 2.88 x 10-4 N/m✓
• C. 2.88 x 10-2 N/m
• D. 2.88 x 10-8 N/m

Explanation: Force/Length = Magnetic Flux density x Current x sin(θ) For maximum force, sin(θ) = 1. Putting in the values gives us the answer.

Q189. The value and units of the Planck constant ‘h’ can be expressed as:

• A. 3.63 x 10-34 Js
• B. 6.63 x 10-34 Js-1
• C. 6.63 x 10-34 Js✓
• D. 6.63 x 10-43 Js

Explanation: Planck's Constant tells us the relationship between energy and frequency of light.

Q190. The diameter of a wire is measured by using a micrometer screw gauge with least count of 0.01 mm. Which of the following readings will be correct?

• A. 0.20 cm
• B. 0.002 m
• C. 2.00 mm✓
• D. 0.2 cm

Explanation: The following is the solution:Since least count is 0.01 mm, therefore, diameter must be expressed in multiples of 0.01.2.00 mm is 200 times 0.01 mm therefore, (C).

Q191. When the length of a simple pendulum is doubled, the ratio of its new time period to old time period is:

• A. 3√2
• B. 1√2
• C. 2√2
• D. √2✓

Explanation: The formula for time period is: T = 2π x √L/gUsing this formula, we see that doubling length results in multiplying time period with √2.

Q192. For projectile motion, in the absence of air resistance:

• A. Horizontal acceleration is zero✓
• B. Vertical acceleration is zero
• C. Horizontal force is constant
• D. Vertical speed is constant

Explanation: Air resistance acts on horizontal component, mainly. In the absence of air resistance, there will be no horizontal acceleration. Vertical acceleration is due to gravity, hence options B and D are wrong. In the absence of air resistance, no horizontal force will be applied and thus, option C is wrong.

Q193. In simple harmonic motion, acceleration will be maximum, when the object is at:

• A. Maximum displacement from the mean position✓
• B. Centre position
• C. Half of the maximum displacement from mean position
• D. Mean position

Explanation: When displacement is maximum, acceleration will be maximum. The displacement and acceleration are related by the equation, a= -w2x, where 'a' (acceleration) and 'x' (displacement) are proportional.

Q194. Choose the correct option

• A. The teacher made all the students rewrite their papers, because the first drafts were not acceptable.✓
• B. The teacher made all the students to rewrite their papers, because the first drafts were not acceptable.
• C. The teacher made all the students rewrite their papers, because the first drafts was not acceptable.
• D. The teacher made all the students rewrite their papers, because first drafts were not acceptable.

Explanation: This sentence is correct. "Made" is correctly followed by the base form of the verb "rewrite." "Were" is the correct verb to agree with the plural subject "drafts."

Q195. When I was young, I — run very fast.

• A. Can
• B. Could✓
• C. Could have
• D. Must

Explanation: This is the correct choice because it indicates past ability or capability. It fits well with the past tense "was" in the sentence, indicating that in the past (when the speaker was young), they had the ability to run very fast.

Q196. She has a driving license. She — drive the car on the motorway.

• A. Might
• B. May
• C. Can✓
• D. Could have

Explanation: This is the correct answer because it indicates present ability or capability. Since she has a driving license, she is able to drive the car on the motorway.

Q197. — you turn down the music a bit, please?

• A. Might
• B. Can✓
• C. Could
• D. Would

Explanation: This is the correct answer because it is used to ask for permission or to make a polite request. In this context, it is asking if it's possible for someone to lower the music volume.

Q198. You — reduce the speed. There is a speed limit.

• A. Need to
• B. Ought to
• C. Must✓
• D. will

Explanation: This is the correct answer because it expresses a strong necessity or obligation. It indicates that reducing the speed is mandatory because there is a speed limit in place.

Q199. The old garage suddenly collapsed one day. Luckily, the owner wasn’t there or he — received serious injuries.

• A. Might as well
• B. Could have✓
• C. Have to
• D. Would

Explanation: This is the correct answer because it expresses a past possibility or hypothetical situation. It suggests that if the owner had been there, it was possible (but did not actually happen) that he could have received serious injuries due to the collapsing garage.

Q200. You — drive fast. We have plenty of time.

• A. Will not
• B. Shall not
• C. Must not
• D. Need not✓

Explanation: This is the correct answer because it means it is not necessary to drive fast. The sentence implies that driving fast is optional because there is plenty of time available.