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PART B

  • Acridine orange is a type of Intercalating agent. Intercalating agents can insert themselves between the bases in the DNA molecule, causing the DNA to become distorted. This can lead to errors during DNA replication and transcription, which can inhibit cell growth and division. This is why the presence of acridine orange in the medium would inhibit the growth and division of the bacteria.

  • Thiobacillus ferroxidans is used in mining operations because it oxidizes iron-sulfide minerals, particularly pyrite (FeS2), converting them into soluble sulfate compounds. This process, known as bioleaching or biomining, helps to recover valuable metals like copper, gold, and nickel from the extracted metal sulfides.
  • Actually, Thiobacillus ferrooxidans can also oxidize sulfur. This bacterium is capable of oxidizing both iron and sulfur to derive energy. In the context of mining, the oxidation of sulfur can help to break down sulfide minerals, which are common in ore deposits. So, while the answer to the original question was (c) iron
  • However, this oxidation process is intricately linked to the oxidation of sulfur in iron-sulfide minerals, particularly pyrite (FeS2). While Thiobacillus ferroxidans directly oxidizes ferrous iron, it utilizes the energy released from this reaction to indirectly oxidize sulfur within the pyrite molecule. This complex process ultimately converts the pyrite into soluble sulfate compounds and frees the iron (as ferric ions) and other valuable metals trapped within the mineral.
  • Viruses often have glycoproteins on their surface, which are proteins with attached sugar groups. These glycoproteins can interact with specific receptors on the surface of red blood cells, causing the cells to clump together or agglutinate.
  • This is a common method used by viruses to infect host cells.
  • The other options, nucleic acids, capsomeres, and lipid molecules, are also components of viruses, but they do not directly cause agglutination of red blood cells.
  • 1. Gram-negative: E. coli does not retain the violet stain in the Gram staining method, but instead takes up the red counterstain.
  • 2. Rod-shaped: E. coli cells are elongated and cylindrical, resembling the shape of a rod.
  • 3. Peritrichously flagellated: E. coli has multiple flagella distributed all over its surface, allowing it to move in various directions.
  • 4. Non-endospore forming: E. coli does not form endospores, which are a type of dormant stage that some bacteria can enter when conditions are unfavorable.
  • 5. Facultative anaerobe: E. coli can survive in both oxygen-rich and oxygen-poor environments.
  • 6. Mesophilic: E. coli prefers moderate temperatures, with an optimal growth temperature of around 37°C, which is the normal body temperature of humans.
  • 7. Found in the intestines: E. coli is a normal part of the gut flora in humans and other warm-blooded animals.
  • 8. Some strains can be pathogenic: While many strains of E. coli are harmless, some can cause food poisoning, urinary tract infections, and other diseases.
  • In microbial fermentation, pyruvate acts as the electron acceptor because it is the end product of glycolysis.
  • During fermentation, cells need to regenerate NAD+ from NADH so that glycolysis can continue. This is done by reducing pyruvate into various end products depending on the type of fermentation. For example, in lactic acid fermentation, pyruvate is reduced to lactic acid. In alcohol fermentation, pyruvate is first decarboxylated to acetaldehyde, which is then reduced to ethanol. In both cases, pyruvate or its derivative acts as the electron acceptor.

The two most common types are lactic acid fermentation and alcohol fermentation.

1. Lactic Acid Fermentation: This type of fermentation occurs in certain bacteria and animal cells (like muscle cells under strenuous activity). In this process, pyruvate from glycolysis is reduced to lactic acid by NADH, which is oxidized to NAD+. This allows glycolysis to continue producing ATP in the absence of oxygen.

2. Alcohol Fermentation: This type of fermentation occurs in yeast and some types of bacteria. Pyruvate is first converted to acetaldehyde and releases CO2. The acetaldehyde is then reduced by NADH to ethanol, regenerating NAD+ for glycolysis.

  • Peptidoglycan, a polymer that forms a mesh-like layer outside the plasma membrane of bacteria, is composed of alternating units of N-acetylglucosamine (NAG) and N-acetylmuramic acid (NAM).
  • These are linked together in a chain by beta (1,4) glycosidic bonds. N-acetyl neuraminic acid, also known as sialic acid, is not a component of peptidoglycan.
  • Instead, it is often found on the surfaces of animal cells, particularly in the brain and other nerve tissue.
  • Reverse phase chromatography is the best method to separate two proteins of similar molecular weights and the same net charge but differing in their amino acid composition because it separates molecules based on their hydrophobicity, or their affinity for water.
  • The amino acid composition of a protein greatly influences its hydrophobicity. Therefore, even if two proteins have similar weights and charges, if their amino acid compositions are different, their hydrophobicities will likely be different as well, allowing them to be separated by reverse phase chromatography.
  • Gel filtration chromatography, also known as size exclusion chromatography, separates proteins based on their size, not their amino acid composition.
  • In this method, smaller molecules enter the pores of the gel and take longer to elute, while larger molecules are excluded from the pores and elute more quickly.
  • Since the two proteins in your question have similar molecular weights (and thus similar sizes), gel filtration chromatography would not effectively separate them. On the other hand, reverse phase chromatography, which separates based on hydrophobicity, can differentiate between the two proteins due to their different amino acid compositions.
  • The dihedral angles φ (phi) and ψ (psi) are used in the Ramachandran plot to describe the rotation of the polypeptide backbone around the bonds between the nitrogen (N) and the alpha carbon (Cα), and the alpha carbon (Cα) and the carbonyl carbon (C’), respectively.
  • These rotations are crucial because they determine the conformation of the protein backbone. The ω (omega) angle, which describes the rotation around the peptide bond, is typically 180° and doesn’t vary much because of the double bond character of the peptide bond. Therefore, the φ and ψ angles are the most important for determining the conformation of the protein backbone.
  • The pH scale is logarithmic, meaning each unit represents a tenfold difference in acidity.
  • So, a solution with a pH of 2 is ten times more acidic than a solution with a pH of 3, and a hundred times (10 x 10) more acidic than a solution with a pH of 4. Therefore, a pH of 2 is 100-fold more acidic than a pH of 4
  • Sodium dodecyl sulphate (SDS) is a detergent that can bind to proteins and give them a uniform negative charge. The reason it binds to hydrophobic side chains is because SDS itself is amphipathic, meaning it has both hydrophilic (water-loving) and hydrophobic (water-fearing) parts.
  • The hydrophobic part of SDS interacts with the hydrophobic side chains of proteins, effectively “unfolding” them into a linear shape. This allows the proteins to be separated based on size during polyacrylamide gel electrophoresis.

SDS-PAGE, or Sodium Dodecyl Sulphate PolyAcrylamide Gel Electrophoresis, is a technique used to separate proteins based on their molecular weight. Here’s a step-by-step explanation:

1. Proteins are mixed with SDS, which denatures the proteins, meaning it breaks down their 3D structure into a linear form. As mentioned earlier, SDS binds to the hydrophobic parts of the protein, causing it to unfold.

2. The SDS also imparts a negative charge to the proteins. This is important because in electrophoresis, molecules are separated based on their charge. Since all proteins are given the same negative charge, they will move towards the positive end of the gel.

3. The gel itself acts like a sieve. Smaller proteins can move through the gel more easily and therefore travel further, while larger proteins move more slowly and don’t travel as far.

4. After the electrophoresis is complete, the proteins are stained so they can be seen. The result is a series of bands on the gel, each representing a different protein. The position of the band on the gel gives an indication of the protein’s size: the further the band is from the starting point, the smaller the protein.

  • In proton NMR (Nuclear Magnetic Resonance) spectroscopy, the number of resonance peaks corresponds to the number of different types of hydrogen atoms in a molecule.
  • A “type” of hydrogen atom is defined by its unique chemical environment, which is determined by the atoms and groups of atoms it is connected to. In the molecule C2H2Cl2Br2, there are two types of hydrogen atoms: those attached to a carbon with two chlorines, and those attached to a carbon with two bromines.
  • When placed in a magnetic field, these two types of hydrogen atoms will resonate at different frequencies due to their different chemical environments, resulting in two distinct peaks on the NMR spectrum.
  • Toll-like receptors (TLRs) recognize pathogen-associated molecular patterns (PAMPs) by binding to specific molecular structures that are commonly found on pathogens but not on host cells.
  • These structures can include bacterial cell wall components, viral RNA, or fungal cell wall components, among others.
  • When a TLR binds to a PAMP, it triggers an immune response in the host organism. This response can include the production of cytokines, which are signaling molecules that help to coordinate the immune response.
  • The most likely reason for a positive tuberculin skin test (TST), but a normal chest X-ray and negative Quantiferon assay, could be a latent tuberculosis infection (LTBI). In LTBI, the person has been infected with the tuberculosis bacteria, but it is not active in their body and they are not contagious. This can cause a positive reaction to the TST, but no symptoms or signs of active disease.
  • Another possibility is that the person was vaccinated with the Bacillus Calmette-Guérin (BCG) vaccine, which can also cause a positive TST result. This vaccine is often given in countries with high rates of TB, like India.
  • The tuberculin skin test (TST) works by injecting a small amount of tuberculin purified protein derivative (PPD) into the skin. If a person has been exposed to the tuberculosis bacteria, their immune system will react to the PPD by causing a bump or swelling at the injection site. This is why a person with a latent tuberculosis infection (LTBI) or who has been vaccinated with the Bacillus Calmette-Guérin (BCG) vaccine can test positive on the TST
    • In the case of LTBI, the immune system has contained the bacteria, preventing it from spreading and causing disease, but it still reacts to the PPD.
    • In the case of the BCG vaccine, the vaccine contains a strain of bacteria related to TB, so the immune system may react to the PPD because it recognizes it as similar to the bacteria used in the vaccine.
  • However, a chest X-ray and the Quantiferon assay can distinguish between LTBI and active TB disease. The chest X-ray can show if the bacteria have caused damage to the lungs, which would indicate active disease. The Quantiferon assay measures the immune system’s reaction to TB proteins and can help confirm whether a person has been infected with TB bacteria.
  • Athymic mice, also known as “nude” mice, have a genetic mutation that results in the absence of a thymus gland.
  • The thymus gland is where T lymphocytes, or T cells, mature. Without a thymus, these mice cannot produce mature T cells, which are crucial for the immune response. Therefore, athymic mice show a lack of mature T lymphocytes.
  • T cell anergy is a state of unresponsiveness by the immune system’s T cells to antigens.
  • It occurs when a T cell recognizes an antigen (via the T cell receptor) but doesn’t receive a second, costimulatory signal. This second signal is usually provided by an antigen-presenting cell and is necessary for full activation of the T cell. Without this second signal, the T cell becomes anergic, meaning it won’t respond to that antigen in the future. This mechanism helps prevent an overactive immune response and autoimmunity.
  • The rate of passive diffusion across a biological membrane depends on several factors, including the size of the molecule, its solubility in lipids, and its charge.
  • Urea is a small, uncharged molecule that is soluble in lipids, so it can pass through the membrane relatively easily.
  • Cl ions are also small, but they are charged, so they have a harder time passing through the lipid bilayer of the membrane. However, they can still pass through via protein channels.
  • Water, while small and uncharged, is not lipid-soluble, so it primarily moves across membranes through special protein channels called aquaporins.
  • Glucose is a large, polar molecule, so it has the hardest time diffusing passively across the membrane. It typically requires a transport protein to help it cross.
  • So, the order from easiest to hardest to passively diffuse across a biological membrane would be: Urea, Cl ions, Water, Glucose.
  • The protein has to cross three membranes to reach the thylakoid lumen from the cytosol.
  • Here’s how:
  • 1. The protein first crosses the outer membrane of the chloroplast. This is a porous membrane that allows the passage of small molecules and proteins.
  • 2. Next, the protein crosses the inner membrane of the chloroplast. This membrane is selective and only allows specific molecules to pass through
  • 3. Finally, the protein crosses the thylakoid membrane to reach the lumen. This is where the light-dependent reactions of photosynthesis occur.
  • The surface area that the lipids would cover once spread across the surface of water would be twice the surface area of the red blood cells.
  • This is because a biological membrane is a lipid bilayer, meaning it has two layers of lipids. Therefore, if the total surface area of the red blood cells is 36 μm2, the lipids would cover a surface area of 2 x 36 μm2 = 72 μm2 when spread out.
  • radioactive phosphorus (P-32) is readily incorporated into newly synthesized DNA molecules due to its similarity to phosphate, which is a building block of DNA.
  • Similarly, radioactive sulfur (S-35) can be used to label proteins because it is found in certain amino acids, the building blocks of proteins.

All of the organelles listed – lysosomes, Endoplasmic reticulum, peroxisomes, and nuclei – can be isolated in their intact form from the cell using techniques such as Centrifugation. Therefore, none of the options provided are correct.

Some organelles present isolation challenges due to their structure or function, it’s not accurate to say it’s impossible for all of them. Scientists have developed various techniques and approaches to isolate organelles with increasing levels of intactness and functionality.

  • Nitroglycerine is converted into nitric oxide in the body.
  • Nitric oxide is a potent vasodilator, meaning it relaxes and widens the blood vessels. This increases blood flow and reduces the heart’s workload, thereby relieving angina pain.
  • Nitroglycerin is metabolized in the body to produce nitric oxide. Nitric oxide then diffuses into the smooth muscle cells that line the blood vessels. Inside these cells, nitric oxide triggers a series of reactions that result in the relaxation and dilation of the blood vessels.
  • Hensen’s node in a chick embryo is equivalent to the dorsal lip of the blastopore in a frog embryo because both structures serve as organizers in early embryonic development
  • They are the regions where gastrulation begins. Gastrulation is the process by which the blastula, a single-layered ball of cells, reorganizes itself into a multilayered structure called the gastrula. This process forms the three primary germ layers of the embryo: the ectoderm, mesoderm, and endoderm, which later develop into all the organs and tissues of the organism.
  • Fish have a two-chambered heart that consists of one atrium and one ventricle.
  • The heart pumps deoxygenated blood from the body to the gills, where it gets oxygenated. The oxygenated blood then flows directly from the gills to the rest of the body. So, in fish, the heart only pumps deoxygenated blood.
  • Haemocyanin contains copper because it is a type of respiratory pigment used by some invertebrates, such as mollusks and arthropods, for oxygen transport.
  • Unlike hemoglobin, which contains iron, hemocyanin binds oxygen between two copper atoms. This gives the blood a blue color when it is oxygenated.
  • It had feathers and wings like birds, but it also had features typically associated with reptiles, such as teeth and a long, bony tail.
  • This combination of features suggests that it could be a transitional species, representing a stage in the evolution from reptiles to birds.
  • During the light-dependent reactions of photosynthesis, water molecules are split in a process called photolysis. This process releases electrons, hydrogen ions (H+), and oxygen.
  • The H+ ions are then captured by a coenzyme called NADP+ (Nicotinamide adenine dinucleotide phosphate) to form NADPH. This happens through an enzyme-mediated reaction where NADP+ combines with an H+ ion and two high-energy electrons. The NADPH is then used in the light-independent reactions (Calvin cycle) to help convert carbon dioxide into glucose.
  • The accumulation of food in assimilatory cells can decrease the rate of photosynthesis because these cells become filled with the products of photosynthesis, such as glucose.
  • This accumulation can inhibit the process of photosynthesis by feedback inhibition, where the end product of a process inhibits the process itself. Essentially, when the plant has produced enough food, it slows down the process of photosynthesis.
  • In the life cycle of Funaria, a type of moss, the sporophyte is the stage that reproduces asexually.
  • It does this by producing spores. These spores are produced in the sporangium, a specialized structure at the top of the sporophyte. When the spores are mature, they are released and can grow into a new organism, or gametophyte, which is the sexually reproducing stage.
  • This cycle of alternating between sexual and asexual reproduction is known as alternation of generations
  • Grafting is a horticultural technique where tissues of plants are joined so as to continue their growth together.
  • The success of this method largely depends on the existence of vascular cambium tissues in both plants. The cambium is a type of plant tissue located between the wood and the bark that can regenerate and grow. Monocots, such as grasses and lilies, lack this cambium layer, which makes grafting difficult because there’s no tissue that can regenerate and fuse the grafted parts together.
  • Puromycin inhibits protein synthesis by mimicking an aminoacyl-tRNA, which is a molecule that brings new amino acids to the growing peptide chain during translation.
  • When puromycin is incorporated into the peptide chain, it causes premature termination of the chain, thus inhibiting the addition of further amino acids. This is why it’s said that puromycin inhibits the addition of amino acids to the peptide chain.
  • The DNA in chromatin is organized into a repeating unit structure called a nucleosome. Each nucleosome consists of a segment of DNA wound around eight histone proteins, forming a “core particle”.
  • This is why we see DNA fragments of a size that are multiples of a smallest unit size when chromatin is digested with enzymes. The regular periodicity of these core particles along the DNA strand gives chromatin its characteristic ‘beads-on-a-string’ structure. This organization allows the DNA to be tightly packed while still allowing access for DNA replication and transcription.
  • The major groove of B-form DNA is wider and deeper than the minor groove.
  • This allows more space for proteins to interact with the functional groups of the DNA bases.
  • The edges of the bases that face the major groove contain unique arrangements of atoms that can form hydrogen bonds and other interactions with proteins.
  • This is why proteins typically recognize and bind to specific sequences of DNA bases via the major groove.
  • In the major groove of B-form DNA, the edges of the bases are exposed and contain unique arrangements of atoms.
  • For example, a protein might have a positively charged amino acid that can form an ionic bond with a negatively charged phosphate group in the DNA. Or it might have a hydrophobic amino acid that can interact with the hydrophobic parts of the DNA bases.
  • Each strand of DNA has a sequence of nucleotides that perfectly complements the sequence on the other strain.
  • It can happen quite quickly because of the precise matching between the sequences on the two strands.

Therefore, the correct answer is (c) 27/64.

  • The individuals are heterozygous at all 3 loci, which means they have one wild type allele and one recessive allele for each gene. The genotype can be represented as AaBbCc.
  • When these individuals are crossed, the probability of getting a wild type phenotype (dominant) for each gene is 3/4 (1/2 x 1/2 = 1/4 for homozygous dominant AA, and 1/2 x 1/2 = 1/4 for heterozygous Aa, so 1/4 + 1/4 = 1/2).
  • Since the genes are independent, the probability of getting a wild type phenotype for all three genes is the product of the individual probabilities. So, the proportion of progeny showing all three wild type characters is (3/4) x (3/4) x (3/4) = 27/64.
  • Raoult’s law becomes Henry’s law under the circumstance (b) When KH ≫ p10.
  • This is because Henry’s law is applicable for dilute solutions where the solute concentration is very low, and thus the partial pressure of the solute (p10) is much less than the Henry’s law constant (KH).
  • The ∆H for a reaction does not depend on (b) various intermediate reactions.
  • This is due to Hess’s Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction and is independent of the pathway of the reaction.
  • When an anion acts as a nucleophile, it donates an electron pair to an electrophile, which is often a positively charged atom in another molecule.
  • This process results in the replacement or “substitution” of the electrophile with the nucleophile, hence it’s called a substitution reaction. This is a fundamental concept in organic chemistry.

(a) (CH3)3C-N3.

  • The given reaction is a nucleophilic substitution reaction (SN2).
  • The azide ion (N3) from sodium azide (NaN3) will act as a nucleophile and replace the chloride ion (Cl) in the given compound. The product of this reaction will be (a) (CH3)3C-N3.

Non-radiative transitions are processes in which an excited electron returns to a lower energy state (ground state or a lower energy excited state) without emitting a photon.

  • Azulene is a unique molecule that consists of a seven-membered ring fused to a five-membered ring. The seven-membered ring in azulene is considered aromatic because it follows Hückel’s rule.
  • According to this rule, a ring is aromatic if it contains a planar, cyclic arrangement of p-orbitals with (4n+2) π electrons, where n is a whole number. In the case of the seven-membered ring in azulene, it has 6 π electrons (n=1), making it aromatic.
  • The compound C6H4Br2 with a 1:1 ratio of peaks in 1H NMR suggests that it has a symmetrical structure.
  • This means that the two bromine atoms are likely in para position on the benzene ring, because this is the only position that would give two types of hydrogen atoms in equal amounts. Therefore, the compound is 1,4-dibromobenzene (para-dibromobenzene).

this image is just for hint

  • Picric acid, also known as 2,4,6-trinitrophenol, has three nitro groups (-NO2) and one hydroxyl group (-OH) attached to a benzene ring.
  • The -OH group is the acidic functional group in picric acid. This is because the oxygen atom in the -OH group is more electronegative than the hydrogen atom, creating a polar bond. This allows the hydrogen atom to be easily donated as a proton (H+), which is the characteristic of an acid according to the Brønsted-Lowry definition of acids and bases.
  • The frequency at which ^13C-NMR will be recorded is 75 MHz. This is because the gyromagnetic ratio of ^13C is approximately 1/4 that of ^1H.
  • Therefore, at the same magnetic field strength, the ^13C-NMR frequency will be 1/4 of the ^1H-NMR frequency. So, if the ^1H-NMR is recorded at 300 MHz, the ^13C-NMR will be recorded at 300 MHz / 4 = 75 MHz. So, the correct answer is (b) 75 MHz.

This is incorrect because according to the kinetic theory of gases, collisions between gas molecules are perfectly elastic, meaning there is no loss of kinetic energy in the collisions.

  • The critical temperature of a gas is not the lowest temperature at which liquefaction occurs because it’s not about when a gas can first become a liquid. Instead, it’s about the highest temperature at which a gas can be converted into a liquid by increasing pressure. Above this temperature, no matter how much pressure you apply, the gas will not condense into a liquid.
  • This is why the critical temperature is defined as the highest temperature at which liquefaction of the gas first occurs.
  • The alpha helix is a common secondary structure of proteins and is stabilized by hydrogen bonds. In an alpha helix, every backbone N-H group donates a hydrogen bond to the backbone C=O group of the amino acid four residues earlier.
  • This pattern of bonding pulls the polypeptide chain into a helical structure that resembles a coiled spring or helix. The hydrogen bonds are responsible for maintaining the shape of the alpha helix.
  • The secondary structure of a protein refers to the local folding of the protein’s polypeptide chain into structures such as alpha-helices and beta-sheets. This folding is primarily driven by hydrogen bonding between the backbone atoms of amino acid residues that are near each other in the sequence. For example, in an alpha-helix, every fourth amino acid is involved in a hydrogen bond. This is why the amino acid residues involved in secondary structure are generally near each other in sequence.
  • In contrast, the tertiary structure of a protein, which is its overall three-dimensional shape, can be influenced by interactions between amino acid residues that are far apart in the sequence but brought close together by the folding of the protein. These interactions can include hydrogen bonding, ionic bonding, and disulfide bridges, among others.
  • First, we need to calculate the moles of oxalic acid.
  • We can do this by dividing the mass of the oxalic acid by its molar mass: moles of oxalic acid=3.38g/90.035g/mol
  • Next, we need to calculate the moles of KOH needed to neutralize the oxalic acid.
  • Since oxalic acid is a diprotic acid, it will react with 2 moles of KOH for every mole of oxalic acid.
  • So, we multiply the moles of oxalic acid by 2: moles of KOH=2×moles of oxalic acid
  • Then, we can calculate the volume of 0.8 M KOH solution needed by dividing the moles of KOH by the molarity of the KOH solution: volume of KOH solution=moles of KOH/0.8M
  • Finally, we convert the volume from liters to milliliters by multiplying by 1000: volume of KOH solution in mL=volume of KOH solution in L×1000
  • Let’s calculate these steps.
  • moles of oxalic acid=3.38g/90.035g/mol=0.0375mol
  • moles of KOH=2×0.0375mol=0.075mol
  • volume of KOH solution in L=0.075mol/0.8M=0.09375L
  • volume of KOH solution in mL=0.09375L×1000=93.75mL
  • So, the answer is not in the options given. The correct answer should be approximately 93.75 mL

(a) an aldol condensation

  • The reaction between dihydroxyacetone phosphate and glyceraldehyde 3-phosphate to form fructose 1,6-bisphosphate is an aldol condensation because it involves the combination of two carbonyl compounds (in this case, a ketone and an aldehyde) to form a larger molecule with a new carbon-carbon bond.
  • In the first step of an aldol condensation, an enolate ion is formed from one of the carbonyl compounds. This enolate ion then attacks the carbonyl carbon of the other compound, forming a new carbon-carbon bond and an alcohol group. This is the “aldol” part of the reaction.
  • In the second step, the alcohol group is removed as a water molecule, leaving a double bond. This is the “condensation” part of the reaction.
  • In the case of the reaction you’re asking about, the enolate ion is formed from dihydroxyacetone phosphate, and this attacks the carbonyl carbon of glyceraldehyde 3-phosphate. The resulting molecule then loses a water molecule to form fructose 1,6-bisphosphate.
  • The reason is that nitrogen-15 (15N) is a heavier isotope of nitrogen than the more common nitrogen-14 (14N). DNA molecules that incorporate 15N instead of 14N will be heavier and thus have a higher density.
  • This was key to Meselson and Stahl’s experiment, as they used density gradient centrifugation to separate DNA molecules based on their densities, allowing them to track the incorporation of 15N into DNA over time.
  • In genetics, if two genes are linked, they are located close together on the same chromosome and tend to be inherited together.
  • The frequency of recombination (or crossover) between linked genes is less than 50%. If the recombination frequency is 50% or more, the genes are likely not linked and are independently assorted.
  • From the given data, the recombination frequencies are: A & B = 52% C & D = 13.8% E & F = 26.4%
  • Only the pair A & B has a recombination frequency greater than 50%, suggesting that these genes are not linked. Therefore, the correct answer is (a) A & B.
  • Glycogen phosphorylase is the enzyme responsible for breaking down glycogen into glucose-1-phosphate, which can then be converted into glucose-6-phosphate and enter the glycolysis pathway to provide energy.
  • If this enzyme is overactive or unregulated, it could lead to an excessive breakdown of glycogen, resulting in a large amount of glycogen in the liver (as seen in the biopsy) and low blood glucose levels (as the glucose is being rapidly used up). This could also explain the patient’s lethargy, as she may not have enough glucose for energy.
  • (d) Pyruvate + HCO3 + ATP → Oxaloacetate + ADP + Pi + H+ is an anaplerotic reaction for the citric acid cycle. It replenishes the cycle by producing oxaloacetate, a key intermediate in the cycle, from pyruvate.
  • The equilibrium constant (Keq) for the reaction catalyzed by malate dehydrogenase is very small (5.9 x 10-6).
  • This indicates that the reaction strongly favors the reactant side (malate) under standard conditions. However, in the context of the citric acid cycle, the reaction proceeds towards the product side (oxaloacetate) because the product is continuously removed to be used in the next step of the cycle. This removal of product shifts the equilibrium towards the product side according to Le Chatelier’s principle.
  • The glyoxylate cycle is a series of reactions that allows organisms to convert fats (in the form of acetyl-CoA) into sugars (glucose). This is essentially a bypass of the decarboxylation steps of the citric acid cycle, which would otherwise result in the loss of carbon as CO2. In the glyoxylate cycle, isocitrate is converted into glyoxylate and succinate instead of being decarboxylated to alpha-ketoglutarate. The glyoxylate and acetyl-CoA are then combined to form malate, which can be converted into glucose.
  • Animals lack the necessary enzymes (isocitrate lyase and malate synthase) to perform the glyoxylate cycle. Therefore, they cannot convert fats into glucose via this pathway, which is what is referred to as gluconeogenesis from fatty acids. However, animals can perform gluconeogenesis from other substrates like amino acids and lactate.
  • The effective half-life of a substance in the body is determined by both its radioactive decay and its biological elimination.
  • The effective half-life (Teff) can be calculated using the formula: 1/Teff = 1/Tbio + 1/Trad where Tbio is the biological half-life and Trad is the radioactive half-life.
  • Substituting the given values: 1/Teff = 1/2 days + 1/8 days = 0.5 + 0.125 = 0.625 days-1
  • So, Teff = 1/0.625 = 1.6 days
  • This is the time it takes for half of the substance to be gone. To find the time at which 3/4 of the substance is gone, we need to find the time for two half-lives (since half of half is a quarter, and we start with a whole, so we have a quarter left). So, the time is 2 x 1.6 days = 3.2 days. The closest answer is (b) 3.2 days.
  • The molecular weight of sucrose (C12H22O11) is 342.30 g/mol. The molecular weights of glucose (C6H12O6) and fructose (C6H12O6) are both 180.16 g/mol.
  • The sum of the molecular weights of glucose and fructose is 360.32 g/mol.
  • The difference between the molecular weight of sucrose and the sum of the molecular weights of glucose and fructose is 342.30 g/mol – 360.32 g/mol = -18.02 g/mol. However, since we are looking for the absolute difference, the answer is 18.02 g/mol. So, the correct answer is (d) 18.
  • Sucrose and monosodium glutamate (MSG) are both polar substances, which means they have regions of positive and negative charge. Vinegar is also a polar substance, and “like dissolves like” in chemistry, meaning polar substances tend to dissolve in other polar substances.
  • Oil, on the other hand, is nonpolar, so the polar sucrose and MSG will not dissolve well in it. Therefore, most of the sucrose and MSG will be located in the vinegar following phase separation.

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