5/5 - (2 votes)
  1. Which of the following reactions involves dehydration?
    I. Peptide bond formation
    II. Phosphodiester formation
    Ill. Glycosidic bond formation
    IV. Hydrogen bond formation between nitrogen bases

Dehydration synthesis is a type of chemical reaction that involves the removal of a water molecule from two or more molecules to form a new molecule. Peptide bond formation, glycosidic bond formation, and phosphodiester bond formation are all examples of dehydration synthesis reactions123Hydrogen bond formation between nitrogen bases is not an example of dehydration synthesis because it does not involve the removal of a water molecule4.

  1. Which of the following is best suited method for production of virus free plants?
    (a) Embryo culture (b) Meristem culture
    (c) Anther culture (d) Ovule culture

Meristem culture is the best-suited method for production of virus-free plants.

Meristem culture is the best-suited method for production of virus-free plants. This is because the meristem (apical and axillary) remains infection-free even when the plant is infected with a virus, and it can be removed and grown in vitro to obtain virus-free plants1234Embryo culture, anther culture, and ovule culture are not the best-suited methods for production of virus-free plants1.

  1. All are plant-derived alkaloids, except
    (a) menthol
    (h) nicotine
    (c) quinine
    (d) codeine

Menthol is not an alkaloid, but a terpenoid. Nicotine, quinine, and codeine are all plant-derived alkaloids1Therefore, the answer to the question is (a) menthol1.

  1. Which of the following genes is responsible for resistance against chilling?
    (a) Glycerol-1-phosphate acyl transferase
    (b) Polygalactouranase
    (c) ACC deaminase
    (d) Cellulose

(a) Glycerol-1-phosphate acyl transferase is responsible for resistance against chilling. This gene helps in the production of lipids that maintain the fluidity of cell membranes at low temperatures, thereby providing cold resistance.

  1. Starch content of potatoes can be increased by using a bacterial gene, known as
    (a) sucrose phosphate synthase gene
    (b) ADP-glucose pyrophosphorylase gene
    (c) amylose synthase
    (d) starch dehydrogenase

The starch content of potatoes can be increased by using the ADP-glucose pyrophosphorylase gene (b). This gene plays a key role in starch biosynthesis.

The ADP-glucose pyrophosphorylase gene is involved in the first step of starch synthesis in plants. It encodes an enzyme that converts glucose-1-phosphate and ATP (adenosine triphosphate) into ADP-glucose and pyrophosphate. ADP-glucose is the activated precursor of glucose residues that are incorporated into starch. Therefore, by increasing the activity of this gene, more ADP-glucose can be produced, leading to an increase in starch synthesis.

  1. Formaldehyde is used in RNA gel electrophoresis as a
    (a) denaturant
    (b) preservative
    (c) buffer
    (d) chelator
  • Formaldehyde acts as a denaturant in RNA gel electrophoresis by disrupting the hydrogen bonds that hold the RNA molecule in its three-dimensional shape. This causes the RNA to unfold into a linear form, which allows it to be separated by size during electrophoresis.
  • The RNA molecules move through the gel at different rates based on their size, with smaller molecules moving faster. This allows for the separation and analysis of RNA fragments.
  1. A growing culture of bacterial cells was inhibited by the edition of 100 micrograms
    per mL streptomycin. This action of the antibiotic can be attributed to
    (a) protein synthesis inhibition
    (b) oxidative phosphorylation inhibition
    (c) cell wall synthesis inhibition
    (d) DNA synthesis inhibition

The action of the antibiotic streptomycin can be attributed to (a) protein synthesis inhibition.

  • Streptomycin works by binding to the bacterial ribosome, a cellular structure where protein synthesis occurs. Specifically, it binds to the 30S subunit of the ribosome, causing a misreading of the genetic code.
  • This results in the production of abnormal proteins, which can harm the bacteria and lead to its death. Therefore, streptomycin is said to inhibit protein synthesis.
  1. In phenol-chloroform method for DNA extraction, the DNA is separated in the
    (a) precipitate (b) aqueous layer
    (c) organic layer (d) interphase

In the phenol-chloroform method for DNA extraction, the DNA is separated in the (b) aqueous layer.

  • The phenol-chloroform method for DNA extraction works on the principle of phase separation. When the mixture is centrifuged, it separates into two layers: an organic layer (containing phenol and chloroform) and an aqueous layer (containing water and other water-soluble components).
  • DNA is a polar molecule due to its phosphate backbone, so it is more soluble in water than in the organic layer. Therefore, the DNA ends up in the aqueous layer.
  1. Agrobacterium tumefaciens can bring about horizontal transfer of genes into plant via
    (a) transposons
    (b) chromosomes
    (c) plasmids
    (d) cosmids

Agrobacterium tumefaciens can bring about horizontal transfer of genes into plants via (c) plasmids.

  • Agrobacterium tumefaciens, a bacterium, has a special type of plasmid known as the Ti (tumor-inducing) plasmid. This plasmid can integrate a segment of its DNA, known as T-DNA, into the DNA of a host plant cell. This process is a form of horizontal gene transfer because it involves the transfer of genetic material from one organism to another that is not its offspring.
  • The transferred genes can cause the plant cells to grow into a tumor-like structure called a crown gall. This mechanism is also used in genetic engineering to introduce new genes into plants.
  1. Phase-contrast microscopy is preferred over bright-field microscopy to study
    (a) stained samples (b) colourless samples
    (c) viruses (d) plant cells

Phase-contrast microscopy is preferred over bright-field microscopy to study (b) colourless samples. This is because phase-contrast microscopy enhances the contrast in unstained, transparent samples by exploiting differences in refractive index.

  • In phase-contrast microscopy, light passes through different parts of the sample, which have different thicknesses and densities. This causes the light waves to slow down to varying degrees, leading to a phase shift.
  • The phase-contrast microscope has a special device called a phase plate that further shifts the phase of some of the light (specifically, the light that didn’t pass through the sample). When this light is recombined with the light that passed through the sample, the phase differences cause interference.
  • This interference enhances the contrast in the image, making it possible to see details that are not visible with a bright-field microscope. This technique is especially useful for observing transparent and colorless samples, such as living cells, which cannot be stained without killing them.
  1. Bacteria depends on permeability barrier of
    (a) cell wall
    (b) exopolysaccharide
    (c) peptidoglycan
    (d) cell membrane
  • The cell membrane, also known as the plasma membrane, acts as a permeability barrier for bacteria because it controls the movement of substances in and out of the cell. It is selectively permeable, meaning it allows certain molecules to pass through while blocking others.
  • This is crucial for maintaining the internal environment of the cell and for processes such as nutrient uptake and waste removal. The cell membrane is composed of a lipid bilayer, which is impermeable to most water-soluble molecules, thus acting as an effective barrier.
  1. A mutation that can confer ampicillin resistance is likely to be in the gene coding for
    (a) transpeptidase
    (b) amylase
    (c) racemase
    (d) viability staining of f β-lactamase

Ampicillin is a type of antibiotic that works by inhibiting the enzyme transpeptidase, which bacteria use to build their cell walls. If a mutation occurs in the gene coding for transpeptidase, the structure of the enzyme could change in a way that prevents ampicillin from binding to it. This would make the bacteria resistant to ampicillin, as the antibiotic could no longer inhibit the enzyme and stop the bacteria from growing.

  • Amylase is an enzyme that breaks down starch into sugars. It’s not involved in the mechanism of action of ampicillin, so a mutation in the gene coding for amylase wouldn’t confer resistance to this antibiotic.
  • Racemase is an enzyme that converts L-amino acids to D-amino acids and vice versa. Again, this process is not related to the action of ampicillin, so a mutation in the racemase gene wouldn’t confer resistance.
  • Viability staining of β-lactamase is not a gene or an enzyme. It’s a method used in the lab to determine whether cells are alive or dead. A mutation in a gene related to this process wouldn’t confer resistance to ampicillin.
  1. Many proteins are found completely outside bilayer on either extracellular side or
    cytoplasmic side but are covalently linked to the membrane bilayer. This is achieved by
    (a) carbohydrate anchor
    (b) peptide anchor
    (c) lipid anchor
    (d) oligonucleotide anchor
  • Lipid anchors are a type of post-translational modification where a lipid molecule is covalently attached to a protein. This lipid can then insert into the lipid bilayer of the cell membrane, effectively anchoring the protein to the membrane.
  • There are several types of lipid anchors, including fatty acyl groups, isoprenyl groups, and glycosylphosphatidylinositol (GPI) anchors. The type of lipid and the amino acid it is attached to can vary, and this can affect the function and location of the protein.
  • For example, a GPI anchor is a glycolipid that can be attached to the C-terminus of a protein. This anchor allows the protein to be located on the outer leaflet of the membrane, where it can interact with the extracellular environment.
  • The process of adding a lipid anchor to a protein is complex and requires several enzymes. It can also be reversible, allowing the cell to control where a protein is located.
  1. Other than cytoplasma, intermediate filaments can be found inside which of the
    following organelles?
    (a) Endoplasmic reticulum
    (b) Nucleus
    (c) Golgi apparatus
    (d) Lysosomes
  • Intermediate filaments are a type of protein fiber that make up the cytoskeleton of a cell. They are found throughout the cell, including in the nucleus. In the nucleus, they form a mesh-like structure known as the nuclear lamina, which provides structural support to the nucleus. The nuclear lamina is located on the inner side of the nuclear envelope, which is the membrane that surrounds the nucleus. The intermediate filaments in the nuclear lamina are made up of proteins called lamins.
  1. Cyanobacteria carry out photosynthesis but some of them can also convert nitrogen gas
    into reduced forms of nitrogen in a process called
    (a) denitrification
    (b) nitrification
    (c) nitrogen fixation
    (d) ammonia assimilation
  • Cyanobacteria convert nitrogen gas (N2) into a usable form through a process called nitrogen fixation. This process is facilitated by an enzyme called nitrogenase. Nitrogenase breaks the triple bond in N2 and adds hydrogen to form ammonia (NH3). This ammonia can then be used by the cyanobacteria to make amino acids and other nitrogen-containing compounds. This process is crucial because it makes nitrogen available to other organisms in the ecosystem.
  1. Given four different factors (1) size, (2) color, (3) internal organization and (4) strength;
    the prokaryotic and eukaryotic cells can be distinguished by
    (a) I only
    (b) 3 and 4
    (c) I and 3
    (d) 1 and 4

Prokaryotic and eukaryotic cells can be distinguished by (c) 1 and 3.

  • 1. Size: Prokaryotic cells are generally smaller than eukaryotic cells.
  • 3. Internal organization: Prokaryotic cells lack a nucleus and other membrane-bound organelles, while eukaryotic cells have a nucleus and other organelles. Color and strength are not typically used to distinguish between these two types of cells.
  1. Short sequence of DNA used for identification of complementary sequence is called as
    (a) probe
    (b) marker
    (c) aptamer
    (d) expressed sequence tag
  • A DNA probe is a short, single-stranded sequence of DNA that is complementary to the sequence of interest. It is used to identify a specific sequence of DNA in a complex mixture. The process works as follows:
  • 1. The DNA probe is labeled with a radioactive or fluorescent marker so it can be detected.
  • 2. The mixture of DNA is heated to separate the double-stranded DNA into single strands.
  • 3. The probe is added to the mixture and it binds, or hybridizes, to its complementary sequence.
  • 4. The location of the probe (and therefore the sequence of interest) can be detected because of the radioactive or fluorescent marker.
  1. Resistance to methotrexate, a drug commonly used in cancer therapy, arises due to the
    process of
    (a) amplification in dihydrofolatereductase gene
    (b) deletion in the dihydrofolatereductase gene
    (c) mutation in the dihydrofolatereductase gene
    (d) transposition in the dihydrofolatereductase gene
  • Methotrexate works by inhibiting the enzyme dihydrofolate reductase (DHFR), which is necessary for DNA synthesis. When the DHFR gene is amplified, it means that multiple copies of the gene are produced. This results in the production of more DHFR enzyme than normal. With more enzyme available, methotrexate is less effective because it cannot inhibit all of the excess enzyme. This is how gene amplification can lead to resistance to methotrexate.
  • How Methotrexate Works: Methotrexate is a drug that inhibits the enzyme dihydrofolate reductase (DHFR). This enzyme is involved in the synthesis of DNA, so by inhibiting it, methotrexate prevents cells from dividing and growing. This is particularly useful in cancer treatment, where the goal is to stop the growth of cancer cells. However, if the cancer cells have amplified their DHFR gene and are producing excess DHFR enzyme, methotrexate may not be able to inhibit all of it, leading to drug resistance.
  1. The value of e”‘ . is given by
    (a) 0
    (b) 1
    (c) -in
    (d) e

The value of e^0 is given by the property of exponents that any number raised to the power of 0 is 1. So, e^0=1. The correct answer is (b) 1.

  1. Retrotransposons transpose by
    (a) cut-paste mechanism
    (b) copy-paste mechanism
    (c) gene amplification mechanism
    (d) gene deletion mechanism
  • Retrotransposons are a type of transposable element that can move within the genome. They do this through a “copy-paste” mechanism.
  • Here’s how it works:
  • 1. The Retrotransposon is first transcribed into RNA by the cell’s machinery.
  • 2. This RNA is then reverse transcribed back into DNA by an enzyme called reverse transcriptase, which is encoded by the Retrotransposon itself. This creates a DNA copy of the Retrotransposon.
  • 3. This DNA copy is then inserted back into the genome at a new location. So, the original Retrotransposon remains at its original location, and a new copy is inserted elsewhere in the genome. This is why it’s called a “copy-paste” mechanism.
  1. The smooth Endoplasmic reticulum is especially abundant in cells that synthesize
    extensive amounts of
    (a) nucleic acid
    (b) lipids
    (c) oligosaccharides
    (d) enzymes

139. Two Plasmids are of the same compatibility group, if
(a) they can coexist in the same bacterial cell
(b) they cannot coexist in the same bacterial cell
(c) they carry the same antibiotic gene
(d) they carry the same toxin gene

  • Plasmids are small, circular pieces of DNA that exist independently of the bacterial chromosome.
  • They can replicate independently and can be transferred from one bacterium to another. Plasmids that belong to the same compatibility group can coexist in the same bacterial cell because they use different mechanisms or times for replication, which prevents them from interfering with each other.
  • This allows multiple plasmids from the same compatibility group to be present in a single cell without competing for resources.
  1. Which one of the following techniques cannot be used to determine the molecular
    weight of a protein?
    (a) UV absorption
    (b) Viscosity
    (c) Light scattering
    (d) Osmotic pressure measurement
  • UV absorption is not a direct method for determining the molecular weight of a protein. It is used to measure the concentration of protein in a solution based on the protein’s ability to absorb ultraviolet light, particularly at a wavelength of 280 nm. This absorption is due to the presence of aromatic amino acids in the protein, such as tryptophan and tyrosine.
  • However, this method does not provide information about the size or weight of the protein molecule. On the other hand, methods like viscosity, light scattering, and osmotic pressure measurement can provide information about the size or weight of the protein molecule, as these properties are directly influenced by the molecular weight of the protein.
  • (b) Viscosity: The viscosity of a solution can be affected by the size of the molecules in it. Larger molecules, like proteins, can increase the viscosity of a solution. By measuring the change in viscosity when a protein is added to a solution, you can estimate the size of the protein molecules, and thus their molecular weight.
  • (c) Light scattering: When light is shone on a solution containing protein molecules, the light is scattered in different directions. The amount and pattern of this scattering can be used to estimate the size of the protein molecules, and thus their molecular weight.
  • (d) Osmotic pressure measurement: Osmotic pressure is the pressure that needs to be applied to a solution to prevent the inward flow of water across a semipermeable membrane. The osmotic pressure of a solution can be affected by the size and number of the molecules in it. By measuring the osmotic pressure of a solution containing protein molecules, you can estimate the number of protein molecules in the solution, and thus their molecular weight.
  1. Triple-helical protein collagen is rich in the amino acid
    (a) glycine
    (b) alanine
    (c) proline
    (d) valine

Collagen, a triple-helical protein, is rich in the amino acid (a) glycine

  • Collagen is rich in the amino acid glycine because of its structure. Collagen is a triple-helical protein, meaning it has three polypeptide chains twisted around each other. Glycine is the smallest amino acid, and it fits well into the tight spaces in the triple helix. This is why every third amino acid in collagen is glycine. Without glycine, the three strands wouldn’t be able to pack together as closely, and the collagen wouldn’t be as strong.
  1. A right-handed alpha helix has
    (a) 3·0 residues of amino acids per tum
    (b) 3·3 residues of amino acids per tum
    (c) 3·6 residues of amino acids per tum
    (d) 4·0 residues of amino acids per tum
  • The structure of an alpha helix is determined by the properties of the amino acids and the peptide bonds that connect them. In an alpha helix, the amino acids are arranged in a right-handed coiled structure, like a spiral staircase. Each turn of the helix involves 3.6 amino acid residues.
  • This is due to the specific angles at which the peptide bonds are formed and the hydrogen bonding that occurs between the amino acids, which stabilizes the helix structure. The hydrogen bonds are formed between the carbonyl oxygen of one amino acid and the amide hydrogen of the amino acid that is 3.6 residues ahead in the sequence. This pattern repeats along the length of the helix, creating the characteristic alpha helix structure.
  1. Urea acts as a strong denaturant of proteins as it
    (a) perturbs electrostatic interactions only
    (b) perturbs hydrophobic interactions only
    (c) perturbs hydrophobic interactions as well as binds to peptide groups
    (d) perturbs hydrophobic interactions as well as binds to nonpolar side chains
  • Urea denatures proteins by disrupting the non-covalent bonds that hold the protein in its specific shape. It does this in two ways:
  • 1. Urea perturbs hydrophobic interactions: Proteins fold into specific shapes to keep their hydrophobic (water-repelling) parts hidden inside, away from the water in the cell. Urea molecules, being polar, can interact with these hydrophobic parts, disrupting the protein’s shape.
  • 2. Urea binds to peptide groups: The peptide groups in a protein are the parts of the molecule that link one amino acid to the next. Urea can form hydrogen bonds with these groups, which can interfere with the hydrogen bonds that help hold the protein in its specific shape.
  • By disrupting these interactions, urea causes the protein to unfold or “denature,” losing its specific shape and, therefore, its function.
  1. The concentration of a bovine serum albumin solution determined using a
    UV spectrophotometer and the knowledge of its extinction coefficient was found to
    be 1·4 mg/ml. Given that the molecular weight of the protein is 70 kDa, its
    concentration in molar units will be
    (a) 20×10^6M
    (b) 50×10^6M
    (c) 20 mM
    (d) 50 mM
  • To convert the concentration from mg/mL to molar units (M), you need to know the molecular weight of the protein. The molecular weight of bovine serum albumin is given as 70 kDa, which is equivalent to 70,000 g/mol.
  • First, convert the concentration of the protein from mg/mL to g/L:
  • 1.4 mg/mL = 1.4 g/L
  • Then, divide this concentration by the molecular weight to get the molar concentration:
  • (1.4 g/L) / (70,000 g/mol) = 0.00002 M or 2.0 x 10-5 M

133. Which of the following methods is used for estimation of sugars?
(a) Lowry
(b) Orcinol
(c) DNSA
(d) Diphenylamine

  • The DNSA method works on the principle of reduction. Here’s a simplified step-by-step process:
  • 1. The sugar solution is mixed with DNSA reagent, which contains 3,5-Dinitrosalicylic Acid.
  • 2. This mixture is then heated. During heating, the sugar reduces the DNSA to 3-amino,5-nitrosalicylic acid.
  • 3. 3-amino,5-nitrosalicylic acid has a reddish-orange color. The intensity of this color is directly proportional to the concentration of sugar in the solution.
  • 4. The colored solution is then cooled and its absorbance is measured using a spectrophotometer at a specific wavelength (usually around 540 nm).
  • 5. The absorbance is compared to a standard curve (a graph of absorbance versus known concentrations of sugar) to estimate the sugar concentration in the original solution.
  1. Which of the following staining techniques is used to check viability of cells?
    (a) Gram’s staining
    (b) Giemsa staining
    (c) Trypan blue staining
    (d) Coomassie staining
  • Trypan blue staining works on the principle that live cells possess intact cell membranes that exclude certain dyes, such as trypan blue, whereas dead cells do not. In this procedure, a cell sample is mixed with trypan blue and then examined under a microscope.
  • The live cells will appear clear while the dead cells will be stained blue because they’ve absorbed the dye. This allows researchers to determine the proportion of living cells to dead cells in the sample
  • 1. Gram’s Staining: This is a differential staining technique used to classify bacteria into two major groups: Gram-positive and Gram-negative. It involves the application of a series of dyes that leaves a characteristic color in the two types of bacteria.
  • 2. Giemsa Staining: This is used primarily for the microscopic detection of malaria parasites. It can also be used to stain the chromosomes in a cell, making it useful in cytogenetics.
  • 3. Coomassie Staining: This is a method of staining proteins in gel electrophoresis. It uses Coomassie Brilliant Blue dye to color the proteins, making them visible.
  1. Sarcoma is cancer of
    (a) skin
    (b) bones
    (c) connective tissue/ organ
    (d) lung
  • (a) Skin: Skin cancers are typically classified as melanomas or non-melanomas. Melanoma is a cancer that begins in the melanocytes, the cells that produce the skin coloring or pigment known as melanin. Non-melanoma skin cancers include basal cell carcinoma and squamous cell carcinoma.
  • (b) Bones: Cancers that originate in the bone are typically referred to as bone cancers. Examples include osteosarcoma (originates in the bone cells), Ewing sarcoma (often found in the bone or soft tissue), and chondrosarcoma (starts in cartilage cells).
  • (d) Lung: Cancers that originate in the lung are referred to as lung cancers. There are two main types: Non-Small Cell Lung Cancer (NSCLC) and Small Cell Lung Cancer (SCLC). NSCLC is the most common type of lung cancer, making up about 85% of all cases.
  1. Which of the following hereditary blood disease?
    (a) Thalassemia
    (b) Pernicious anemia
    (c) Megaloblastic anemia
    (d) Galactosemia

(a) Thalassemia and (d) Galactosemia are hereditary blood diseases.

  • Thalassemia and Galactosemia are considered hereditary diseases because they are caused by mutations in genes that are passed down from parents to their children. In the case of Thalassemia, the disease is caused by mutations in the genes that code for hemoglobin, the protein in red blood cells that carries oxygen. These mutations can cause the body to produce fewer healthy red blood cells and less hemoglobin, leading to anemia.
  • Galactosemia, on the other hand, is caused by mutations in the genes that code for enzymes needed to break down galactose, a type of sugar. When these enzymes are missing or not working properly, galactose can build up in the blood, leading to various health problems.
  1. Which of the following groups of antibodies can cross placenta?
    (a) lgM
    (b) !gO
    (c) lgD
    (d) lgA
  • IgG antibodies are unique because they have the ability to cross the placenta. This is due to a special receptor on placental cells that binds to IgG antibodies and transports them from the mother’s blood into the fetal blood. This process provides the fetus with passive immunity, protecting it from infections during the early months of life until its own immune system matures.
  1. In sickle-cell anemia, which of the following hemoglobin subunits is affected?
    (a) Alpha subunit
    (b) Beta subunit
    (c) Zeta subunit
    (d) Gamma subunit
  • Sickle-cell anemia is caused by a mutation in the gene that instructs the body to make the beta-globin subunit of hemoglobin. This mutation results in the production of an abnormal version of beta-globin called hemoglobin S.
  • When oxygen levels are low, hemoglobin S molecules can stick together and form long, rigid rods within a red blood cell. This causes the cell to become rigid and take on a sickle shape, hence the name of the disease. These sickle-shaped cells can block blood flow, leading to the symptoms of sickle-cell anemia.
  1. Lysozyme breaks down
    (a) α (1-4) linkage between NAM and NAG
    (b) β ( 1-4) linkage between NAM and NAG
    (c) covalent crosslinks in the peptidoglycan
    (d) β
    (1-4) linkage between two glucose molecules
  • Lysozyme, an enzyme found in many animals, breaks down the (1-4) linkage between NAM and NAG as part of its antimicrobial activity. The peptidoglycan layer of bacterial cell walls is composed of alternating units of NAM and NAG. By breaking this linkage, lysozyme weakens the cell wall, which can lead to the death of the bacteria. This is one way our immune system helps protect us from bacterial infections.
  • the linkage between NAM (N-acetylmuramic acid) and NAG (N-acetylglucosamine) in the peptidoglycan layer of bacterial cell walls is a beta (β) (1-4) linkage, not an alpha (α) linkage. The orientation of the glycosidic bond in a beta linkage is such that it resists cleavage by human digestive enzymes. This is part of what makes bacterial cell walls tough and resilient.
  1. Enzymes alcohol dehydrogenase belongs to class
    (a) oxidoreductase
    (b) transferase
    (c) hydrolases
    (d) lyases
  • Alcohol dehydrogenase belongs to the class of enzymes known as oxidoreductases because it catalyzes a type of reaction called oxidation-reduction, or redox, reactions. In the case of alcohol dehydrogenase, it specifically catalyzes the oxidation of alcohols into aldehydes or ketones, with the reduction of NAD+ to NADH. This is why it’s classified as an oxidoreductase.
  • Why not others
  • Hydrolases are a class of enzymes that catalyze hydrolysis reactions, where a molecule is split into two parts through the addition of a water molecule. Alcohol dehydrogenase does not perform this type of reaction. Instead, it catalyzes the removal of a hydrogen from an alcohol, converting it into an aldehyde or ketone, and transferring the hydrogen to NAD+ to form NADH. This is an oxidation-reduction reaction, not a hydrolysis reaction, which is why alcohol dehydrogenase is classified as an oxidoreductase, not a hydrolase
  • Alcohol dehydrogenase is not classified as a transferase, lyase, or any other class of enzyme because it does not perform the specific types of reactions that define those classes.
    • Transferases are enzymes that catalyze the transfer of a functional group from one molecule to another. Alcohol dehydrogenase does not do this; it removes a hydrogen from an alcohol and transfers it to NAD+, which is an oxidation-reduction reaction.
    • Lyases are enzymes that catalyze the breaking of various chemical bonds by means other than hydrolysis and oxidation. Alcohol dehydrogenase does not do this either; it catalyzes an oxidation-reduction reaction, not a bond-breaking reaction.
  1. The main product of glycolysis in skeletal muscles under heavy exercise conditions is
    {a) lactate
    (b) pyruvate
    (c) a-ketoglutarate
    (d) succinate
  • During heavy exercise, the demand for energy in skeletal muscles is high. Glycolysis, the process of breaking down glucose to produce energy, is accelerated. However, the oxygen supply may not be sufficient for the complete breakdown of glucose in the mitochondria (aerobic respiration). In such conditions, the cells switch to anaerobic respiration, where the end product of glycolysis, pyruvate, is converted into lactate. This process is called lactic acid fermentation. The conversion of pyruvate to lactate allows the regeneration of NAD+, which is essential for glycolysis to continue at a fast rate, even in the absence of oxygen. So, under heavy exercise conditions, the main product of glycolysis in skeletal muscles is lactate.
  1. DNA fingerprinting has proved in forensic science. It involves the use of
    (a) mini-satellites
    (b) r-RNA
    (c) e-DNA
    {d) bacterial DNA
  1. In crosses involving different pairs of genes A & B, C & D and E & F, the following
    proportions of recombinants were obtained ,
    A & B =52%, C & D = 13·8% and E & F ~ 26-4%
    The pair of genes which are not linked is
    {a) A & B
    {b) C & D
    {c) E & F
    {d) All of the above
  • The pair of genes that are not linked is A & B.
  • In genetics, a recombination frequency of 50% or more indicates that the genes are not linked and are likely located on different chromosomes. Here, A & B have a recombination frequency of 52%, suggesting they are not linked.
  1. A protein oligomerizes to form a dimer in acidic pH and a tetramer at neutral pH. Which
    of the following techniques can be used to separate the two species?
    (a) SDs-PAGE
    {b) Ion-exchange chromatography
    (c) Hydrophobic interaction chromatography
    (d) Gel permeation chromatography
  • GPC can be used to separate the two species. This technique separates molecules based on their size, and since a dimer is smaller than a tetramer, they would elute at different times.
  • Gel permeation chromatography (GPC) is often used because it allows for the separation of polymers based on their size in solution. Other methods may not be as effective at separating polymers of different sizes, especially when the differences in size are small. GPC is also relatively simple and quick, making it a popular choice for many applications.
  1. A double-stranded DNA will be more stable in
    (a) pure water
    (b) 0·05 M NaCI
    (c) 1 M urea
    {d) 20% formamide
  • A double-stranded DNA will be more stable in (b) 0.05 M NaCl. This is because the sodium chloride helps to stabilize the negatively charged phosphate backbone of the DNA, reducing repulsion between the strands and thus increasing stability.
  1. Which of the following is a ketose sugar?
    (a) Glucose
    (b) Ribulose
    (c) Galactose
    (d) Xylose
  • In the case of ribulose, it is a pentose (5-carbon sugar) and the ketone group is on the second carbon atom. This makes ribulose a ketose sugar.
  • Glucose, galactose, and xylose are not considered ketose sugars because they are aldose sugars. Aldose sugars have an aldehyde group (-CHO) at the end of the molecule, rather than a ketone group (C=O) within the molecule. In glucose, galactose, and xylose, the carbonyl group (C=O) is at the end of the carbon chain, making them aldoses. In contrast, in ribulose, the carbonyl group is at the second position in the carbon chain, making it a ketose.
  • Aldose sugars include glucose, galactose, ribose, and mannose
  • Ketose sugars include fructose, ribulose, and psicose.
  1. Which of the following bacteria is not a plant pathogen?
    (a) Pseudomonas
    (b) Xanthomonas
    (c) Shigella
    (d) Erwinia
  • It is not a plant pathogen. It is a type of bacteria that causes disease in humans, specifically a form of dysentery called shigellosis.
  • The bacteria Pseudomonas, Xanthomonas, and Erwinia are all plant pathogens. They cause various diseases in plants. For example, Pseudomonas syringae causes bacterial speck in tomatoes, Xanthomonas campestris causes black rot in crucifers, and Erwinia amylovora causes fire blight in apple and pear trees. There are many other bacteria that can cause diseases in plants, including Agrobacterium, Ralstonia, and Burkholderia.
  1. Which of the following statements is not true for facilitated diffusion?
    (a) Takes place along a concentration gradient
    (b) Takes place against a concentration gradient
    (c) Can be bidirectional
    (d) Follows Michael is-Menten kinetics
  • Facilitated diffusion always occurs along a concentration gradient, from an area of higher concentration to an area of lower concentration, not against it.
  • Facilitated diffusion is influenced by several factors:
    • 1. Concentration Gradient: As mentioned earlier, facilitated diffusion occurs along a concentration gradient, from an area of higher concentration to an area of lower concentration.
    • 2. Presence of Carrier Proteins: Facilitated diffusion requires specific carrier proteins in the cell membrane to transport substances across.
    • 3. Temperature: The rate of facilitated diffusion can increase with temperature, as higher temperatures can increase the kinetic energy of the molecules, making them move faster.
    • 4. Size and Charge of Molecules: Larger or highly charged molecules may require specific carrier proteins to help them cross the cell membrane.
    • 5. Saturation: The rate of facilitated diffusion can reach a maximum when all the carrier proteins are occupied, a situation known as saturation.
  1. Diffusibility of ions and molecules through a biological membrane increases in the order
    (a) indole >Water> glycerol> glucose >Na +
    {b) water> indole> glycerol> glucose> Na +
    (c) indole >water> glucose >glycerol>Na +
    {d) Na + >glycerol> glucose> indole> water
  • The order of diffusibility is determined by the size and polarity of the molecules and ions.
  • Smaller and less polar substances diffuse more easily through biological membranes.
  • Water is a small molecule and can easily pass through the membrane.
  • Indole, glycerol, and glucose are larger molecules, but they are still smaller and less polar than sodium ions (Na+), which makes them diffuse more easily than Na+.
  • Sodium ions are larger and more polar, which makes it harder for them to pass through the lipid bilayer of the membrane. They usually require a specific transport mechanism, like ion channels or pumps, to cross the membrane.
  1. Refractive index is best described as the
    {a) extent by which a medium slows the velocity of light
    {b) specific point at which the rays focus
    (c) focusing of a cone of light on a slide
    (d) coefficient of angle of refraction

The refractive index is best described as the (a) extent by which a medium slows the velocity of light.

  1. Transmission electron microscopy is best for high magnification viewing of
    (a) internal structure of fixed cells
    (b) internal structure of live and motile cells
    (c) surface structure of fixed cells
    (d) surface structure of live and motile cells
  • A light microscope, specifically a phase-contrast microscope or a differential interference contrast (DIC) microscope, is typically used to observe live and motile cells. These types of microscopes allow for the visualization of live cells without the need for staining or other treatments that could harm the cells.
  1. Which of the following statements is true for circularly polarized light?
    (a) The magnitude of the electric vector is constant, but its direction varies
    {b) The magnitude of the electric vector varies, but its direction is constant
    (c) Both the magnitude and direction vary
    (d) All the above conditions are possible depending upon the medium of propagation
  • Circularly polarized light is a type of light in which the electric field vector (which represents the direction and magnitude of the electric field) rotates in a circular motion as the light wave propagates forward.
  • In circularly polarized light, the magnitude of the electric field vector remains constant. This means that the strength of the electric field does not change. However, the direction of the electric field vector changes continuously. It rotates in a circle, which is why we call it circularly polarized light.
  1. In an electromagnetic spectrum, if the wavelength decreases, then
    (a) energy increases, frequency increases and wave number increases
    (b) energy increases, frequency increases and wave number decreases
    (c) energy decreases, frequency increases and wave number decreases
    {d) energy decreases, frequency decreases and wave number increases
  • Because frequency, wavenumber, energy are inversely proportional to wave length
  1. In enzyme assays, initial rates are used to
    (a) increase the sensitivity of the assay
    (b) prevent the substrate inhibition
    (c) promote substrate inhibition
    (d) minimize the contribution of reverse reaction
  • Enzyme assays are laboratory procedures that measure the activity of enzymes, which are proteins that catalyze biochemical reactions. These assays can determine the rate at which an enzyme catalyzes a reaction under specific conditions, such as temperature and pH. This information can be used to understand the enzyme’s function, its role in a biological process, or how it might be affected by certain substances or conditions.
  • Enzyme assays measure the rate of enzyme-catalyzed reactions. By focusing on the initial rates.The initial rate of a reaction is the rate at which the reaction occurs immediately after it has started, we are primarily observing the forward reaction where the enzyme converts the substrate into the product. This is because, at the start of the reaction, the concentration of the substrate is high and the concentration of the product is low. As the reaction progresses, the product concentration increases and the substrate concentration decreases, which can lead to the reverse reaction (product converting back to substrate). By using initial rates, we minimize the impact of this reverse reaction on our measurements, providing a more accurate assessment of the enzyme’s activity.
  1. When the substrate concentration is much lower than Km in an enzyme assay, the rate
    (a) approaches vmax
    (b) shows zero-order kinetics
    (c) is proportional to substrate concentration
    (d) is constant
  • When the substrate concentration is much lower than Km (the Michaelis constant), the rate of the reaction is proportional to the substrate concentration. This is because, under these conditions, most of the enzyme molecules are not yet saturated with substrate. Therefore, as you add more substrate, more enzyme-substrate complexes can be formed, leading to an increase in the rate of the reaction. This is known as first-order kinetics, where the rate of the reaction is directly proportional to the concentration of one reactant.
  • (a) The rate approaches Vmax when the substrate concentration is much higher than Km. At this point, nearly all enzyme active sites are occupied by the substrate. This is the maximum rate of reaction, or Vmax.
  • (b) Zero-order kinetics occur when the enzyme is saturated with substrate. At this stage, adding more substrate doesn’t increase the rate of reaction because all the enzyme’s active sites are occupied. The reaction rate is constant and independent of substrate concentration.
  • (d) The rate is not constant when the substrate concentration is much lower than Km. It increases as more substrate is added because more enzyme-substrate complexes can be formed.

110. Superantigens can bind to
(a) MHC I
(b) MHC II
(c) TCR and MHC II together
(d) BCR and MHC I together

(c) TCR and MHC II together

  • Superantigens are a type of antigen that can bind to both the T-cell receptor (TCR) and the Major Histocompatibility Complex class II (MHC II) simultaneously. This is because they have the ability to cross-link these two molecules. This binding is unique and causes a massive activation of the immune system, as it activates a large number of T cells, much more than a normal antigen would. This can lead to an excessive immune response, which can be harmful to the body.
  • Superantigens are a class of antigens that cause non-specific activation of T-cells, leading to polyclonal T cell activation and massive cytokine release. They are able to bind to Major Histocompatibility Complex (MHC) class II and T-cell receptors simultaneously, which is different from conventional antigens. This binding leads to a strong immune response, which can result in symptoms of toxic shock. Some bacteria and viruses produce superantigens as a way to disrupt the immune system and increase their chances of survival.

109.Which class of B cell receptors is expressed in naive B cells?
(a) IgG, IgA
(b) IgE, lgG
(c) IgM
(d) lgM, IgD

  • Naive B cells, which are B cells that have not yet encountered an antigen, express both IgM and IgD receptors on their surface. This is because during B cell development, the B cell receptor (BCR) initially expresses IgM. Then, through a process called class switching, the B cell begins to also express IgD. Both IgM and IgD can bind to the same antigen but have different roles in the immune response. IgM is usually the first antibody produced in response to an antigen, while IgD function is less clear but is thought to play a role in B cell activation and differentiation.
  1. Maturation of B cells occurs in the following order
    (a) progenitor B cells, affmity maturation, lg gene rearrangement, class switching
    (b) Ig gene rearrangement, progenitor B cells, class switching, affmity maturation
    (c) progenitor B cells, Ig gene rearrangement, affinity maturation, class switching
    (d) class switching, progenitor B cells, Ig gene rearrangement, affmity maturation
  • 1. Progenitor B cells: These cells are produced in the bone marrow from hematopoietic stem cells. They are the earliest form of B cells and have not yet undergone the process of gene rearrangement to produce a unique B cell receptor (BCR).
  • 2. Ig gene rearrangement: This process involves the rearrangement of the immunoglobulin (Ig) genes that code for the BCR. This rearrangement is random, which allows for a large diversity of B cells, each capable of recognizing a different antigen. This process is crucial for the adaptive immune response, as it allows the immune system to recognize and respond to a wide variety of pathogens.
  • 3. Affinity maturation: Once a B cell has encountered its specific antigen and become activated, it undergoes a process called affinity maturation. This process occurs in the germinal centers of lymph nodes. During affinity maturation, the B cell undergoes rapid cell division and mutation of the Ig genes. This results in B cells with BCRs that have an even better fit to the antigen. The B cells with the highest affinity BCRs are selected for survival, while others undergo apoptosis (programmed cell death).
  • 4. Class switching: The final step in B cell maturation is class switching. Initially, B cells produce IgM antibodies. However, depending on the signals received from helper T cells, B cells can switch the class of antibody they produce to IgG, IgA, or IgE. This process allows the immune system to mount the most effective response against the pathogen.
  1. Color blindness in human being is an X-linked trait. A color-blind man has a 45-X
    daughter who is also color-blind. The nondisjunction that led to the 45,X child occur in
    which parents and in which meiotic division?
    (a) Father; First meiotic division
    (b) Both father and mother; First meiotic division
    (c) Mother; First meiotic division
    (d) XYY
  • Nondisjunction is an error that can occur during meiosis, the process of cell division that produces gametes (sperm or eggs). In a normal meiosis, chromosomes are supposed to separate so that each gamete ends up with one copy of each chromosome. However, during nondisjunction, the chromosomes do not separate properly. This can result in a gamete that has an extra chromosome (n+1) or is missing a chromosome (n-1).
  • In the case of the 45,X daughter, the nondisjunction occurred in the mother during the first meiotic division. This means that instead of separating, both X chromosomes went into one daughter cell, leaving the other daughter cell without an X chromosome. The egg that was fertilized to produce the daughter was the one without an X chromosome, resulting in a 45,X karyotype.
  1. A child with Edward’s syndrome (18 trisomy) having his mother where nondisjunction
    of chromosome 18 occurred in the division of the secondary oocytes. What is the chance
    that a mature egg arising from this cell division will receive two numbers of
    chromosome 18?
    (a) 1/4
    (b) 1/2
    (c) 1/8
    (d) 3/4
  • Nondisjunction is an error in cell division that results in daughter cells with the wrong number of chromosomes. In this case, nondisjunction occurred in the mother’s secondary oocyte, which means that the chromosomes did not separate properly during meiosis II.
  • In meiosis II, the sister chromatids of each chromosome are supposed to separate, with one going to each of the two resulting cells. If nondisjunction occurs, both sister chromatids of chromosome 18 go to the same cell. This means that one cell will have two copies of chromosome 18, and the other will have none.
  • Therefore, there are two possible outcomes for the mature egg: it could have an extra chromosome 18 (if it came from the cell that received both), or it could have the normal number of chromosomes (if it came from the cell that received none). Since these outcomes are equally likely, the chance that a mature egg will receive two copies of chromosome 18 is 1/2.
  1. Transition type of gene mutation is caused, when
    (a) GC is replaced by TA
    (b) CG is replaced by GC
    (c) AT is replaced by CG
    (d) AT is replaced by GC
  • Transition type of gene mutation is caused when (d) AT is replaced by GC.
  • In this case, a purine (adenine, A) is replaced by another purine (guanine, G) and a pyrimidine (thymine, T) is replaced by another pyrimidine (cytosine, C). This is known as a transition mutation.
  1. Which of the following would yield only one type of monomer after complete hydrolysis?
    (a) Glycogen
    (b) DNA
    (c) Lipoprotein
    (d) Triacylglycerol
  • Triacylglycerol, also known as a triglyceride, yields only one type of monomer after complete hydrolysis because it is made up of three fatty acid chains attached to a glycerol molecule. When hydrolyzed, it breaks down into its constituent monomers: glycerol and fatty acids. Even though there are three fatty acid chains, they are all considered the same type of monomer because they are all fatty acids
  1. Which of the following is not a general feature of a nucleotide?
    (a) A phosphate is attached to the 2′ hydroxyl of ribose.
    (b) A base ring nitrogen atom is attached to the 1′ carbon atom of ribose.
    (c) A hydroxyl group is attached to the 3′ carbon atom of ribose.
    (d) A hydrogen atom or hydroxyl group is attached to the 2′ carbon atom of ribose.

This is not a general feature of a nucleotide. In a nucleotide, the phosphate group is attached to the 5′ carbon atom of the sugar (ribose or deoxyribose), not the 2′ hydroxyl.

  1. What will be the probability of obtaining a plant with AaBBCc genotypes from trihybrid
    (AaBbCc) parents?
    (a) 4 out of 64
    (b) I out of 64
    (c) 8 out of 64
    (d) 0 out of 64

The probability of obtaining a plant with AaBBCc genotypes from trihybrid (AaBbCc) parents is 4 out of 64. This is because the probability of obtaining Aa is 1/2, BB is 1/4, and Cc is 1/2. When these probabilities are multiplied together (1/2 x 1/4 x 1/2), the result is 1/16, which is equivalent to 4 out of 64.

  1. Substrate combines more firmly with enzyme, when
    (a) Km is high
    (b) Ki is high
    (c) Ki is low
    (d) Km is low
  • Km and Ki are both constants related to enzyme kinetics.
  • Km, or the Michaelis constant, is the substrate concentration at which the reaction rate is half of its maximum. A lower Km value indicates a higher affinity between the enzyme and substrate, meaning the enzyme can bind to the substrate more easily.
  • On the other hand, Ki is the inhibition constant. A high Ki value means the inhibitor binds less tightly to the enzyme, so it’s less likely to interfere with the enzyme-substrate interaction. However, Ki doesn’t directly affect how firmly the substrate combines with the enzyme, it’s more about the inhibitor’s effect on this process.
  • So, while a high Ki value can indirectly favor substrate-enzyme binding by reducing inhibitor interference, it’s the low Km value that directly indicates a stronger, more efficient binding between the enzyme and substrate.
  1. In the fluid mosaic model for membrane structure
    (a) carbohydrate is on the outer membrane surface
    (b) lipids but not proteins can diffuse in the plane of the membrane
    (c) proteins occur only in the inner leaflet of the membrane
    (d) the polar ends of phospholipids face the interior of the membrane
  • (a) Carbohydrate is on the outer membrane surface – This statement is true. Carbohydrates are often attached to proteins or lipids on the outer surface of the cell membrane, forming glycoproteins and glycolipids, respectively.
  • (b) Lipids but not proteins can diffuse in the plane of the membrane – This statement is false. Both lipids and proteins can diffuse laterally within the plane of the membrane, although proteins move more slowly due to their larger size.
  • (c) Proteins occur only in the inner leaflet of the membrane – This statement is false. Proteins can be found in both the inner and outer leaflets of the membrane. Some proteins span the entire membrane (integral proteins), while others are associated with only one side (peripheral proteins).
  • (d) The polar ends of phospholipids face the interior of the membrane – This statement is false. The polar (hydrophilic) ends of phospholipids face the aqueous environment, both inside and outside the cell, while the nonpolar (hydrophobic) tails face the interior of the membrane.
  1. Which of the following statements is true for trypsin, chymotrypsin and elastase?
    (a) They have similar reaction kinetics.
    (b) They use ATP for catalysis.
    (c) They are serine proteases.
    (d) They have similar thermostability.

Trypsin, chymotrypsin, and elastase are all classified as serine proteases because they all have a serine (an amino acid) in their active site. This serine is crucial for the enzyme’s ability to catalyze the breakdown of proteins. The serine’s side chain contains a hydroxyl (-OH) group which is able to donate a hydrogen atom, making it reactive and able to participate in the catalysis of peptide bonds in proteins.

  • (a) They do not have similar reaction kinetics. Each of these enzymes has different substrate specificities and therefore, their reaction rates or kinetics can vary depending on the protein they are breaking down.
  • (b) They do not use ATP for catalysis. These enzymes use a mechanism known as the “serine protease mechanism” which does not require ATP. Instead, it involves a nucleophilic attack by the serine residue at the active site on the peptide bond of the protein substrate.
  • (d) They do not have similar thermostability. Thermostability refers to an enzyme’s ability to function at high temperatures. The thermostability of an enzyme is determined by its specific structure and amino acid composition, which can vary among different enzymes.
  1. In a sample from a population, there were 65 individuals with the ‘BB’ genotype, 30
    individuals with the ‘Bb’ genotype, and 15 individuals with the ‘bb’ genotype. The
    frequency of the ‘b’ allele was
    (a) 0·27
    (b) 0·59
    (c) 0·41
    (d) 0·73
  • To find the frequency of the ‘b’ allele, we first need to find the total number of alleles in the population.
  • Each individual has two alleles, so the total number of alleles is 2 times the total number of individuals.
  • The total number of individuals is 65 (BB) + 30 (Bb) + 15 (bb) = 110 individuals.
  • So, the total number of alleles is 2 * 110 = 220 alleles.
  • The number of ‘b’ alleles is the number from Bb individuals (1 allele per individual) plus twice the number from bb individuals (2 alleles per individual). So, the number of ‘b’ alleles is 30 (from Bb) + 2*15 (from bb) = 60 ‘b’ alleles.
  • The frequency of the ‘b’ allele is the number of ‘b’ alleles divided by the total number of alleles.
  • So, the frequency of the ‘b’ allele is 60 / 220 = 0.27.

OR

  1. In a cross between two black Labrador retrievers, the phenotypic ratio of the offspring
    is 9 black puppies to 3 chocolate puppies to 4 yellow puppies. This is an example of
    (a) partial recessiveness
    (b) incomplete penetrance
    (c) incomplete dominance
    (d) epistasis
  • Epistasis is a phenomenon that consists of the effect of one gene being dependent on the presence of one or more ‘modifier genes’, i.e., the genetic background. In this case, the color of the Labrador retriever’s coat is determined by two genes, not just one.
  • In Labrador retrievers, coat color is determined by two genes: the E gene and the B gene. The E gene determines whether the coat color will be black or brown, while the B gene determines whether the color will be expressed or not.
    • If a dog has at least one E allele (Ee or EE), the B gene will be expressed. If the dog has BB or Bb, it will be black; if it has bb, it will be brown (chocolate). However, if the dog has ee (regardless of the B gene), it will be yellow because the e allele inhibits the expression of the B gene.
    • This is an example of epistasis, where one gene (E) masks the effect of another gene (B).
    • So, in the given cross, the 9:3:4 ratio suggests that both parents are EeBb, leading to 9 black (E-B-, E-bb), 3 chocolate (eeBB, eeBb), and 4 yellow (eebb) puppies.
  1. Which of the following is the most likely outcome of a cross between two heterozygous
    tall plants?
    (a) 63 tall : 59 short
    (b) 76 tall : 23 short
    (c) 24 tall : 49 medium : 25 short
    (d) 53 tall : 14 7 short
  • The most likely outcome of a cross between two heterozygous tall plants (Tt x Tt) is a 3:1 ratio of tall to short plants.
  • None of the options provided exactly match this ratio, but option (b) 76 tall : 23 short is the closest approximation.
  1. What function might you postulate for a polypeptide having a Zn-finger motif?
    (a) Signal transduction
    (b) Transcription factor
    (c) Growth hormone receptor
    (d) Cytoskeletal component
  • Zn-finger motifs are structural motifs in proteins that are characterized by the coordination of one or more zinc ions in order to stabilize the fold. They are commonly found in transcription factors, which are proteins that control the rate of transcription of genetic information from DNA to messenger RNA.
  • The Zn-finger motif allows the protein to interact with the DNA molecule. The zinc ion stabilizes the structure of the motif, allowing it to insert into the major groove of the DNA helix and make sequence-specific contacts with the DNA. This interaction allows the transcription factor to regulate the transcription of specific genes.
  1. Which of the following methods is true regarding the extraction of membrane protein?
    (a) Integral membrane protein is removed by change in pH
    (b) Peripheral membrane protein is removed by urea
    (c) Integral membrane protein is extracted by salt
    (d) Amphitropic protein is removed by chelating agents
  • Peripheral membrane proteins are not as tightly bound to the membranes as integral membrane proteins. They are associated with the lipid bilayer mainly through interactions with other proteins.
  • Urea is a denaturing agent that can disrupt these protein-protein interactions, allowing the peripheral membrane proteins to be removed. It does this by breaking the hydrogen bonds that hold the secondary and tertiary structure of the protein together. This makes the protein unfold into a linear chain, which can then be easily separated from the membrane.
  • (a) Integral membrane proteins are embedded within the lipid bilayer of the cell membrane and cannot be removed simply by a change in pH. They typically require detergents or other strong solvents to extract. ‘
  • (c) Salt is not typically used to extract integral membrane proteins. Instead, salts are often used in the process of salting out, which is a method of precipitating proteins out of solution. However, this method is not specific to integral membrane proteins.
  • (d) Chelating agents are used to bind and remove metal ions. Amphitropic proteins can associate with the membrane in a reversible manner and their interaction with the membrane can be regulated by various factors, not just the presence of metal ions. Therefore, a chelating agent would not necessarily remove these proteins.
  1. In response to DNA damage, p53 can mediate
    (a) cell cycle arrest only
    (b) apoptosis only
    (c) either cell cycle arrest or apoptosis
    (d) cell division
  1. Which of the following is the Incorrect statement about mature human red blood cells?
    (a) They lack nuclei and membrane bound organelles.
    (b) Cytoplasm is rich in hemoglobin.
    (c) They have a biconcave shape.
    (d) The membrane is lipid monolayer.
  1. Dosage compensation in humans is brought about by
    (a) inactivity of one X-chromosome in females
    (b) hyperactivity of single X-chromosome in males
    (c) hypoactivity of both X-chromosomes in females
    (d) hyperactivity of autosomes in females
  1. C-value paradox tells us about
    {a) linearity of the relationship between genome size and complexity of organism
    (b) non-linearity of the relationship between genome size and complexity of organism
    (c) dosage compensation
    (d) number of chromosomes

The C-value paradox refers to the (b) non-linearity of the relationship between genome size and the complexity of an organism. It describes the lack of correlation between the amount of DNA (C-value) a species has and its biological complexity.

  • The C-value paradox is a term used in molecular biology to describe the puzzling observation that the amount of DNA (or the genome size, also known as the C-value) in an organism’s cells doesn’t always correlate with the complexity of the organism.
  • For example, you might expect that a complex organism like a human would have a larger genome than a simple organism like a bacterium. However, this is not always the case. Some relatively simple organisms have much larger genomes than humans. This is the paradox. The reason for this paradox is not fully understood, but it’s thought to be due to the presence of non-coding DNA, or “junk” DNA, that doesn’t code for proteins. Some organisms have a lot of this non-coding DNA, which increases their genome size without necessarily increasing their complexity.
  1. Polytene chromosomes are formed due to
    (a) repeated S-phase, but no M-phase
    (b} repeated karyokinesis, but no cytokinesis
    (c) repeated S-phase, but M-phase without anaphase
    (d) non-disjunction of chromosomes

Polytene chromosomes are formed due to (a) repeated S-phase, but no M-phase. This process is known as endoreduplication or endoreplication, where DNA replication occurs without cell division, leading to an increase in the DNA content of the cell and the formation of polytene chromosomes.

  • let’s take the example of the salivary gland cells in the fruit fly, Drosophila melanogaster. These cells undergo repeated rounds of DNA replication without cell division, a process known as endoreduplication.
  • In a normal cell cycle, the cell undergoes a series of phases: G1 (growth), S (DNA synthesis), G2 (growth and preparation for mitosis), and M (mitosis). However, in the salivary gland cells of Drosophila, after the S phase, the cell does not proceed to the M phase. Instead, it goes back to the G1 phase and then enters the S phase again. This cycle repeats several times, leading to multiple copies of the same chromosome lying side by side. These multiple copies of chromosomes are tightly bound together, forming a structure known as a polytene chromosome.
  • The polytene chromosomes are visible under a light microscope and appear as a series of parallel bands. Each band represents a specific region of the chromosome. This unique structure allows for high levels of gene expression, which is necessary for the function of the salivary gland cells in Drosophila.

88. Holocentric chromosomes are
(a) chromosomes with multiple centromeres
(b) supernumerary chromosomes
(c) short chromosome with many genes
(d) chromosomes with centromere at the centre

(a) Holocentric chromosomes are chromosomes with multiple centromeres.

  • Holocentric chromosomes have multiple centromeres because the centromeric function (the part of the chromosome that links sister chromatids) is spread over a large area or potentially the entire chromosome, rather than being localized to a specific region.
  • This means that during cell division, spindle fibers can attach along the length of the chromosome, allowing for proper segregation of the chromosomes into the daughter cells. This is different from most organisms, which have monocentric chromosomes where the centromere is localized to a specific region.
  • An example of organisms with holocentric chromosomes are nematodes, such as Caenorhabditis elegans. Certain species of insects and plants also have holocentric chromosomes.
  1. Which of the following modifications leads to protein degradation?
    (a) Methylation
    (b) Acetylation
    (c) Acylation
    (d) Ubiquitination

Ubiquitination is a process in which ubiquitin molecules are attached to lysine residues of a protein. The attachment of multiple ubiquitin molecules, termed polyubiquitination, marks proteins for degradation to the proteasome 1. Thus, ubiquitination leads to protein degradation.

  1. The single most abundant protein in animal tissues is
    (a) collagen
    (b) actin
    (c) fibronectin
    (d) RuBisCO

a. Collagen is the most abundant protein in animal tissues. It is a structural protein that makes up the structure or framework of cells and tissues.

  1. Which of the following incorrectly matches the organelle with its function?
    (a) Glyoxysome-seed growth
    (b) Mitochondrion-photophosphorylation
    (c) Peroxysome-ROS quenching
    (d) ER-glycosylation

(b) Mitochondrion-photophosphorylation is incorrect. Mitochondria are involved in oxidative phosphorylation, not photophosphorylation. Photophosphorylation is a process that occurs in chloroplasts during photosynthesis.

  • Mitochondria are often referred to as the “powerhouses” of the cell because they generate most of the cell’s supply of adenosine triphosphate (ATP), which is used as a source of chemical energy. They do this through a process called oxidative phosphorylation, which involves the transfer of electrons from electron donors to electron acceptors such as oxygen, in reactions that release energy.
  • On the other hand, chloroplasts are found in plant cells and some algae and are responsible for photosynthesis. During photosynthesis, light energy is converted into chemical energy, producing glucose and oxygen. Photophosphorylation is a specific part of photosynthesis where light energy is used to create a high-energy electron donor and an electron acceptor.
  • (a) Glyoxysomes are involved in seed growth. They are specialized peroxisomes found in plants (especially in the fat storage tissues of germinating seeds) and play a crucial role in converting fatty acids to sugar for the emerging seedling.
  • (c) Peroxisomes are involved in ROS (Reactive Oxygen Species) quenching. They contain enzymes that neutralize harmful substances like hydrogen peroxide, thus protecting the cell from oxidative damage.
  • (d) The Endoplasmic Reticulum (ER) is involved in glycosylation. This is a process where sugars are added to proteins or lipids, which is crucial for their stability and function. The ER is a major site of protein synthesis and processing in the cell.
  1. An enzyme that relieves torsional strain while double-stranded DNA is being
    unwound is
    (a) DNA ligase
    (b) DNA gyrase
    (c) DNA relaxase
    (d) DNA helicase

(b) DNA gyrase is the enzyme that relieves torsional strain while double-stranded DNA is being unwound1

  1. Infectious proteins are present in
    (a) satellite viruses
    (b) Gemini viruses
    (c) viroids
    (d) prions

(d) prions

  1. One map unit of one centimorgan corresponds to recombination frequency of
    (a) 1%
    (b) 10%
    (c) 100%
    (d) 0·1%

One map unit of one centimorgan corresponds to a recombination frequency of 1% 1.

  1. What special types of cells are produced during the gametophyte stage of a plant’s
    life cycle
    {a) Haploid gametes
    (b) Zygotes
    (c) Spores
    (d) Seed cells

The gametophyte is a stage in the life cycle that is found in all plants and certain species of algae 1. The primary job of gametophyte is the production of gametes, which are the haploid reproductive cells, such as sperm and eggs 1.

  1. The main function of centrosomes is
    (a) osmoregulation
    (b) secretion
    (c) protein synthesis
    {d) formation of spindle fibre

The main function of centrosomes is (d) formation of spindle fibers 1.

  1. A stone is initially at rest and then it is released from height has measured from earth
    surface, where gravitational potential energy is mgh, ( m ; mass of stone ). While falling
    {a) kinetic energy of the stone will remain same
    {b) potential energy of the stone will remain same
    {c) total energy will remain same
    (d) None of the above

Thus, the correct option is © total energy will remain same.

  1. A particle at rest is decaying to two particles and their momentums are P1 and P2. The
    relation between P1 and P 2 is
    (a) equal magnitudes of P1 and P2
    (b) p1 = p2
    (c) P1 = P2 /2
    (d) P2 = P1 /2

The correct answer is (a) equal magnitudes of P1 and P2, p1=p2

  • This is because the law of conservation of momentum states that the total momentum before decay (which is zero, as the particle is at rest) must be equal to the total momentum after decay.
  • Therefore, the magnitudes of P1 and P2 must be equal and opposite in direction to maintain a total momentum of zero.
  1. In prokaryotes, the lagging strand primers are removed by
    (a) DNA polymerase I
    (b) DNA polymerase llI
    (c) 3′ to 5′ exonuclease
    (d) DNA ligase

In prokaryotes, the lagging strand primers are removed by (a) DNA polymerase I 1.

  1. Which of the following cell types is unlikely to be infected by viruses?
    (a) Nerve cell
    (b) Red blood cell
    (c) Liver cell
    (d) White blood cell
  1. For radioactive decay, which of. the following statements is false?
    (a) Alpha decay is seen in heavier elements.
    (b) Beta particles are Jess massive than alpha particle.
    (c) Gamma rays are more energetic than alpha or beta particles.
    (d) A nucleus can emit an electron.
  1. In eukaryotes, euchromatin replicates predominantly during
    (a) early S-phase
    (b) mid S-phase
    (c) G2 -phase
    (d) late S-phase

Euchromatin replicates predominantly during (a) early S-phase.

  • Euchromatin is a form of chromatin that is loosely packed and accessible to the cellular machinery responsible for transcription and replication. During the S-phase of the cell cycle, DNA replication occurs. Euchromatin replicates early in the S-phase because it contains genes that are actively being transcribed into RNA. The cell needs to duplicate this DNA early to ensure that the new cells will have the necessary genetic information to function properly.
  • Mid S-phase: This is when replication of heterochromatin, the more condensed form of DNA, begins. Heterochromatin contains genes that are not actively transcribed, so it’s not as crucial to replicate it early.
  • G2-phase: This is a period of rapid cell growth and protein synthesis during which the cell prepares itself for mitosis. DNA replication does not occur during this phase
  • Late S-phase: By this time, most of the DNA replication has already occurred. The remaining replication is typically of the tightly packed heterochromatin.
  1. Which of the following metal constituents of menbrane associated protein is involved in respiratory electron transport?
    (a) Magnesium
    (b) Manganese
    (c) Iron
    (d) Potassium
  1. In the following reaction, the molecules labelled with numbers from 1 to 6 are
    a.-Ketoglutarate -(–r-1
    =——-+) Glutamate~ Glutamine
    3 4 5 t
    (a) Glutamate dehydrogenase, Glutamine Synthetase, NAD(P)H + H+, NAD(P) +, ATP
    and ADP respectively
    (b) Glutamine Synthetase, Glutamine dehydrogenase, NAD(P)H + H+, NAD(P)+, ATP
    and ADP respectively
    (c) Glutamine Synthetase, Glutamine dehydrogenase, ATP, ADP, NAD(P)H + H+ and
    NAD(P) + respectively
    (d) Glutamate dehydrogenase, Glutamine Synthetase ATP ADP
    NAD(P) + respectively ‘ ‘ • NAD(P)H + H+ and

The reaction given in the question is the reversible conversion of α-ketoglutarate to glutamate and glutamine. The enzymes and molecules involved in this reaction are:

  1. Beer’s law states that
    (a) absorbance is proportional to both the path length and concentration of the
    absorbing species
    (b) absorbance is proportional to the log of the concentration of the absorbing species
    (c) absorbance is equal to P0 I P
    (d) absorbance is inversely proportional to both the path length and concentration of
    the absorbing species

70.Bone marrow transplantation in immune compromised patients can potentially cause
(a) GVHD
(b) T cell leukemia
(c) delayed hypersensitivity
(d) inability to use live donor

69. lf you measure the ability of cytotoxic T cells from an HLA-B27 person to kill virus
X-infected target cells, which one of the following statements is correct?
(a) Any virus X-infected target cell will be killed
(b) Only virus X-infected cells of HLA-B27 type will be killed
(c) Any HLA-827 cell will be killed
(d) No HLA-B27 cell will be killed

68. Viral replication within cells is inhibited directly by
(a) IFN-alpha (b) TNF-alpha
(c) IL- 1 (d) lL-4

Therefore, the correct answer is (a) IFN-alpha.

69. ______express CD3 and CD4 molecules (a) Th1 and Th2 cells (b) Ts and Tc cells (c) Th 1 and Tc cells (d) All T cells

(a) Th1 and Th2 cells.

  • CD4 is primarily expressed on helper T cells (Th cells), while CD8 is primarily expressed on cytotoxic T cells (Tc cells).
  • However, all T cells, including Th cells, Tc cells, and regulatory T cells (Treg cells), express CD3. This is a key part of the T cell receptor complex.
  • So, to clarify, CD3 is expressed by all T cells, but CD4 is not.
  1. Interactions between ·-________________- are not restricted by MHC molecules.
    (a) Th cell and dendritic cell
    (b) Tc cell and target cell
    (c) NK cell and target cell
    (d) Th cell and B cell
  1. The Class I MHC processing pathway primarily
    (a) processes antigens that are present in the cytosol
    (b) generates peptides, complexes them with Class I MHC molecules for presentation
    to helper T cells
    (c) generates peptides, complexes them with Class I MHC molecules for presentation
    to NK cells
    (d) generates peptides, complexes them with Class I MHC molecules for presentation
    to Kupfer cells

Therefore, the correct answer is (a) processes antigens that are present in the cytosol.

  1. Cell with specific killing effects is
    (a) NK cell
    (b) Neutrophils
    (c) CTL
    (d) Macrophage
  1. Which of the following is characteristic of B but not T cells?
    (a) Class I MHC (b) CD3
    (c) Measles virus receptor (d) Surface immunoglobulin

  1. CD8 is a marker of
    (a) B cells (b) helper T cells
    (c) cytotoxic T cells (d) an activated macrophage

CD8 is a marker of cytotoxic T cells (CTLs) 12. CTLs are a type of T cell that recognize and kill cells that are infected with intracellular pathogens such as viruses and bacteria 13. CD8 is a surface protein that is expressed on the surface of mature CTLs and is involved in the recognition of antigens presented by Class I MHC molecules 12.

  1. During the cell interactions involved in generating a cytotoxic T cell response, the
    T helper cell receives the necessary signal 2 from an antigen-presenting cell through
    which of the following?
    (a) IL-2 with IL-2R (b) B7 with CD-28
    (c) TCR with MHC Class I (d) IgD with antigen

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