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PART B

  1. before virus binding to receptor [Option ID = 5509]
  2. between virus binding and infection [Option ID = 5510]
  3. between infection and appearance of mature virus [Option ID = 5511]
  4. after appearance of mature virus [Option ID = 5512]

The eclipse period in a virus infection is defined as the time between infection and appearance of mature virus. [Option ID = 5511]

  • The eclipse period in virology is the time between when the virus enters a cell (infection) and when new, mature virus particles are produced. The binding of the virus to the cell receptor is part of the infection process. So, the eclipse period does not include the time between binding and infection, but rather starts from the point of infection (which includes binding) until the appearance of mature virus particles.
  • The eclipse period refers to the time interval during a viral infection when the virus is inside the host cell but has not yet produced mature virus particles.
  • Here’s a breakdown of the phases:
    1. Virus Entry: The virus binds to the host cell receptor and enters the cell.
    2. Eclipse Period: During this phase, the virus is actively replicating its genetic material and assembling new virus particles. However, these particles are not yet mature or detectable.
    3. Appearance of Mature Virus: After the eclipse period, the newly formed mature virus particles are released from the host cell, ready to infect other cells.

The first virus to be discovered was
[Question ID = 1379]

  1. Polio virus [Option ID = 5513]
  2. Yellow fever virus [Option ID = 5514]
  3. Rabies virus [Option ID = 5515]
  4. Tobacco mosaic virus [Option ID = 5516]

The first virus to be discovered was 4. Tobacco mosaic virus.

  • The Tobacco Mosaic Virus (TMV) was discovered in 1892 by Dmitri Ivanovsky, a Russian biologist. He was investigating a disease that was killing tobacco plants, causing a mosaic-like pattern on their leaves. Ivanovsky crushed the leaves and passed the liquid through a filter that was known to remove bacteria. He found that the filtered liquid could still cause disease in other plants. This was puzzling because it suggested the infectious agent was smaller than a bacterium.
  • Later, in 1898, Martinus Beijerinck, a Dutch microbiologist, confirmed that the infectious agent was a new kind of infectious particle, which he named “virus”, after the Latin word for poison. This was the first discovery of a virus, and it opened up a whole new field of scientific research.
  • The discovery of the Tobacco Mosaic Virus (TMV) before other viruses was largely due to the technology and knowledge available at the time.
    • In the late 19th century, scientists were just beginning to un derstand the existence of microscopic organisms like bacteria. The tools to study these tiny organisms were very basic. TMV was discovered first because it infects plants, which are easier to study than animals or humans. The symptoms of TMV infection in tobacco plants – the mosaic pattern on the leaves – were also very visible and distinctive, which helped in its identification.
    • Other viruses, like those that cause human diseases, were harder to study because they require more complex tools and methods to observe and understand. As technology and scientific knowledge advanced, more viruses were discovered and studied.
  1. DNA polymerase [Option ID = 5517]
  2. RNA polymerase [Option ID = 5518]
  3. Reverse transcriptase [Option ID = 5519]
  4. Cysteine protease [Option ID = 5520]

Certainly! The enzyme responsible for integrating the retroviral genome into the host genome is reverse transcriptase (abbreviated as RT). This crucial enzyme is produced by retroviruses, such as HIV (human immunodeficiency virus). Here’s how it works:

  1. Reverse Transcriptase (RT):
    • RT is an enzyme that plays a pivotal role in the retroviral life cycle.
    • It converts the single-stranded viral RNA (present in retroviruses) into double-stranded DNA (cDNA).
    • This process occurs during the early stages of infection.
  2. Integration:
    • After RT generates the cDNA, it forms a pre-integration complex (PIC) with the nascent cDNA.
    • The next step involves the retroviral integrase (IN) enzyme.
    • IN covalently joins the viral cDNA ends to the host genome, resulting in the stable integration of the viral genetic information.

In summary, reverse transcriptase is the key player in converting viral RNA to DNA, allowing retroviruses to integrate their genetic material into the host cell’s genome.

For more details, you can refer to the Wikipedia page on Integrase1.

  1. remain outside the host as non‐living particles [Option ID = 5521]
  2. bind to a specific receptor [Option ID = 5522]
  3. integrate into the host genome, but not replicate [Option ID = 5523]
  4. actively replicate and produce more virus [Option ID = 5524]
  • DNA viruses can cause latent infections by integrating their DNA into the host’s genome. This process begins when the virus infects a cell and introduces its DNA into the cell’s cytoplasm.
  • The viral DNA then enters the nucleus of the host cell, where it integrates into the host’s DNA. This integration allows the viral DNA to be copied whenever the host cell replicates its own DNA.
  • However, during a latent infection, the viral DNA does not direct the production of new virus particles. Instead, it remains dormant within the host’s genome until certain conditions trigger it to reactivate and begin producing new viruses. This is how DNA viruses can establish long-term, latent infections.
  1. Retrovirus [Option ID = 5525]
  2. Enveloped plus‐strand RNA virus [Option ID = 5526]
  3. Bacteriophage [Option ID = 5527]
  4. Enveloped minus‐strand RNA virus [Option ID = 5528]
  • The other option, Retrovirus, is not the correct choice because retroviruses do not replicate in the cytosol. Instead, they replicate in the nucleus of the host cell. Additionally, the genome of a retrovirus can still be infectious even when it is de-proteinized, which is not the case for the virus X described in your question. Therefore, the characteristics of virus X align more closely with an Enveloped plus-strand RNA virus.
  • Minus-strand RNA viruses, also known as negative-sense RNA viruses, differ from plus-strand RNA viruses in their replication process. Minus-strand RNA viruses must carry an RNA-dependent RNA polymerase within the virion, because host cells do not have the enzymes necessary for their replication. This is not the case for plus-strand RNA viruses, which can be directly translated by the host’s machinery, as their RNA is in the same sense as the host’s mRNA.
  1. Measles [Option ID = 5529]
  2. Smallpox [Option ID = 5530]
  3. Influenza [Option ID = 5531]
  4. Rabies [Option ID = 5532]

The true statements are A, C, and D. Statement B is incorrect because MHC-II presents antigens to helper T cells, not cytotoxic T cells.

  • The Major Histocompatibility Complex (MHC) is a set of proteins present on the surface of cells that help the immune system recognize foreign substances. Statement A is true because MHC-I is expressed by nearly all nucleated cells. This means that almost all cells with a nucleus produce MHC-I proteins. These proteins present normal “self” antigens as well as “non-self” or foreign antigens (like from a virus or bacteria) to cytotoxic T cells. If the antigens are “non-self”, the cytotoxic T cells will destroy the cell.
  • Statement B is incorrect because MHC-II is expressed by antigen-presenting cells (like dendritic cells, macrophages, and B cells), but they present antigens to helper T cells, not cytotoxic T cells. Helper T cells, upon recognizing these antigens, release substances that stimulate other immune cells to fight off the infection.
  • Statement C is true because only helper T cells can recognize antigens in complex with MHC-II. Cytotoxic T cells do not interact with MHC-II.
  • NK cells have inhibitory receptors that recognize MHC-I molecules. When MHC-I is present on a cell (indicating it’s a healthy “self” cell), the NK cell is inhibited from killing it. If MHC-I is absent or reduced (as is often the case in virus-infected or cancerous cells), the NK cell is not inhibited and can kill the target cell. While there is ongoing research into the interactions between NK cells and MHC-II, it’s generally accepted that NK cells primarily interact with MHC-I. Therefore, statement D would be incorrect if it stated that NK cells engage with MHC-II.
  1. two Fab fragments and one Fc fragment [Option ID = 5537]
  2. one F(ab’)2 fragment and several small Fc fragments [Option ID = 5538]
  3. one F(ab’)2 fragment and one Fc fragment [Option ID = 5539]
  4. two Fab fragments and several small Fc fragments [Option ID = 5540]
  • Papain is a protease enzyme that cleaves the IgG molecule at a specific site above the disulfide bonds that hold the heavy chains together. This results in the formation of two identical antigen-binding fragments, known as Fab fragments, and one crystallizable fragment, known as the Fc fragment. The Fab fragments are responsible for antigen binding, while the Fc fragment is involved in various cell-mediated immune responses.
  • The specificity of papain’s cleavage is due to its enzymatic properties. Enzymes, including papain, are highly specific in their action due to their unique active sites. The active site of an enzyme is the region that binds to the substrate (in this case, the IgG molecule), and it has a specific shape and chemical properties that allow it to interact with its substrate in a very specific manner. In the case of papain, its active site is designed to recognize and bind to a specific site on the IgG molecule, just above the disulfide bonds that hold the heavy chains together. This allows it to cleave the IgG molecule at this specific location, resulting in the formation of two Fab fragments and one Fc fragment.
  • Other proteases may cleave the IgG molecule at different locations, resulting in different fragments. For example, the protease pepsin cleaves IgG below the disulfide bonds, resulting in a F(ab’)2 fragment and a Fc fragment. The F(ab’)2 fragment contains two linked Fab units, which can bind to antigens, while the Fc fragment is unable to bind to antigens.
  1. They reside in the secondary lymphoid organs [Option ID = 5541]
  2. They exit the lymph node and circulate in the body [Option ID = 5542]
  3. They settle in the peripheral tissue [Option ID = 5543]
  4. They reside in the bone marrow [Option ID = 5544]
  • Central memory T cells (TCM) reside in secondary lymphoid organs like the spleen and lymph nodes because these are the sites where they can encounter their specific antigen presented by antigen-presenting cells (APCs). This is crucial for their activation and subsequent immune response. The other options are not true for TCM cells because:
  • 2. While TCM cells can circulate in the body, they primarily reside in secondary lymphoid organs. The cells that typically exit the lymph node and circulate in the body are effector T cells, which are activated T cells ready to fight infection.
  • 3. TCM cells do not settle in peripheral tissues. Instead, effector memory T cells (TEM) are the ones that migrate to peripheral tissues to provide local immunity.
  1. Bone marrow [Option ID = 5545]
  2. Spleen [Option ID = 5546]
  3. Lymph node [Option ID = 5547]
  4. Liver [Option ID = 5548]
  1. Position Specific Scanning Matrix [Option ID = 5549]
  2. Point Specific Scoring Matrix [Option ID = 5550]
  3. Position Specific Scoring Matrix [Option ID = 5551]
  4. Point Specific Scanning Matrix [Option ID = 5552]
  1. Structure [Option ID = 5553]
  2. Temperature [Option ID = 5554]
  3. Pressure [Option ID = 5555]
  4. Antigenicity [Option ID = 5556]
  1. Alanine [Option ID = 5557]
  2. Tryptophan [Option ID = 5558]
  3. Glycine [Option ID = 5559]
  4. Proline [Option ID = 5560]

Glycine is the only genetically encoded amino acid without a stereoisomer because it has two hydrogen atoms attached to its alpha carbon. Stereoisomers are possible when a carbon atom is attached to four different groups, creating a chiral center. However, in the case of glycine, the alpha carbon is attached to two hydrogen atoms, a carboxyl group (-COOH), and an amino group (-NH2), so it doesn’t have four different groups. This means that glycine does not have a mirror image form and is not chiral, hence it doesn’t have a stereoisomer.

  1. Cys and Met [Option ID = 5561]
  2. Cys and Cys [Option ID = 5562]
  3. Met and Met [Option ID = 5563]
  4. Met and His [Option ID = 5564]
  • Disulfide bonds are formed between two cysteine (Cys) residues. This happens when the sulfur atom in the side chain of one cysteine forms a covalent bond with the sulfur atom in the side chain of another cysteine. This is a unique property of cysteine, as it is the only amino acid that can form this type of bond.
  • Methionine (Met) and Histidine (His) also contain sulfur in their side chains, but they do not form disulfide bonds. The sulfur in methionine is part of a larger functional group and is not available to form a bond with another sulfur atom. Histidine does not contain sulfur in its side chain, so it cannot form a disulfide bond.
  • cysteine can form disulfide bonds, which are important for the structure and function of many proteins. Methionine, on the other hand, cannot form disulfide bonds but plays a crucial role as the start codon in protein synthesis.
  1. Phosphate and sugar group [Option ID = 5565]
  2. Nitrogenous bases and sugar [Option ID = 5566]
  3. Nitrogenous bases and phosphate [Option ID = 5567]
  4. Nitrogenous bases [Option ID = 5568]
  • The two strands of DNA are held together by hydrogen bonds between nitrogenous bases due to their specific chemical structures. Each base has a specific partner: adenine (A) pairs with thymine (T), and cytosine (C) pairs with guanine (G). These pairs form because of the ability of these specific bases to form hydrogen bonds with each other.
  • The other components of DNA – the sugar and phosphate groups – form the backbone of the DNA molecule. They do not form hydrogen bonds with each other or with the nitrogenous bases. The sugar-phosphate backbones are on the outside of the DNA molecule, while the nitrogenous bases are on the inside, where they can interact with each other to hold the two strands together.
  • Why don’t the other components form hydrogen bonds? It’s because of their chemical structures. Hydrogen bonds form between a hydrogen atom attached to a highly electronegative atom (like nitrogen or oxygen) and another electronegative atom. The sugar and phosphate groups in DNA do not have the necessary arrangements of these atoms to form hydrogen bonds with each other or with the nitrogenous bases.
  1. Sialic acid [Option ID = 5569]
  2. CD4 [Option ID = 5570]
  3. CD81 [Option ID = 5571]
  4. Claudin‐1 [Option ID = 5572]
  • shaped to fit onto the CD4 molecule on the surface of certain human cells, like a key into a lock. This is why CD4 is the primary receptor for HIV-1. After the initial binding to CD4, the virus needs to bind to a second receptor, a co-receptor, to fully enter the cell. The most common co-receptors are CCR5 and CXCR4. This two-step process ensures the virus specifically targets certain cells and not others.
  • As for the other options you mentioned:
    • 1. Sialic acid: This is a type of sugar molecule that can be found on the surface of various cell types. Some viruses do use sialic acid as a receptor, but not HIV-1
    • 2. CD81: This is a protein found on the surface of various cell types. It’s involved in several cellular processes, and some other viruses (like Hepatitis C) use it as a receptor, but not HIV-1.
    • 3. Claudin-1: This is a protein involved in tight junctions between cells. It’s used as a receptor by some other viruses (like Hepatitis C), but not HIV-1.
  1. There is a decrease in PTEN protein level but there is no mutation on the PTEN DNA sequence in the cancer cell [Option ID = 5573]
  2. There is a decrease in PTEN protein level but no change in the PTEN mRNA expression in the cancer cell [Option ID = 5574]
  3. There is a decrease in PTEN mRNA expression but no change in the PTEN protein level in the cancer cell [Option ID = 5575]
  4. There is a decrease in PTEN mRNA expression but no mutation on the PTEN DNA sequence in the cancer cell [Option ID = 5576]
  1. Variable number tandem repeat (VNTR)
    [Option ID = 5577]
  2. Long interspersed nuclear elements (LINEs)
    [Option ID = 5578]
  3. Short interspersed nuclear elements (SINEs)
    [Option ID = 5579]
  4. Long terminal repeats (LTRs)
    [Option ID = 5580]
  • Variable Number Tandem Repeats (VNTRs) can act as genetic markers because they are highly polymorphic, meaning they vary greatly among individuals. This makes them useful for identifying specific individuals or for determining paternity.
  • The other options, Long Interspersed Nuclear Elements (LINEs), Short Interspersed Nuclear Elements (SINEs), and Long Terminal Repeats (LTRs), are types of repetitive DNA sequences. While they are important for the structure and function of the genome, they are not as variable among individuals, making them less useful as genetic markers.
  1. Cystic Fibrosis [Option ID = 5581]
  2. Sickle Cell Anaemia [Option ID = 5582]
  3. Downs Syndrome [Option ID = 5583]
  4. Marasmus [Option ID = 5584]
  • Cystic Fibrosis, Sickle Cell Anemia, and Down Syndrome are all genetic disorders because they are caused by abnormalities in the individual’s genetic material.
    • 1. Cystic Fibrosis is caused by a mutation in the CFTR gene. This gene helps create a protein that regulates the movement of salt and water in and out of your body’s cells. A mutation in this gene can lead to the production of thick, sticky mucus in various organs.
    • 2. Sickle Cell Anemia is caused by a mutation in the HBB gene. This gene provides instructions for making a protein called beta-globin, a component of hemoglobin. Hemoglobin is the protein in red blood cells that carries oxygen. The mutation causes the red blood cells to become rigid and sticky, and shaped like sickles or crescent moons.
    • 3. Down Syndrome is caused by an extra copy of chromosome 21. This additional genetic material alters the course of development and causes the characteristics associated with Down syndrome.
  • On the other hand, Marasmus is not a genetic disorder because it is not caused by an abnormality in an individual’s genetic material. Instead, it is a form of severe malnutrition that occurs when an individual does not consume enough protein and calories.
  1. only in wild animals. [Option ID = 5585]
  2. from plants to humans. [Option ID = 5586]
  3. from animals to humans. [Option ID = 5587]
  4. only in zoo animals [Option ID = 5588]
  • Zoonotic diseases are diseases that can be passed from animals to humans. They are caused by harmful germs like viruses, bacteria, parasites, and fungi. These germs can cause many different types of illnesses in people and animals, ranging from mild to serious illness and even death. Some examples of zoonotic diseases include:
    • 1. Rabies: This is a viral disease that can be transmitted to humans through the bite of a rabid animal.
    • 2. Lyme disease: This is a bacterial disease that humans can get from the bite of infected ticks, which often get the bacteria by biting infected animals like mice or deer.
    • 3. Avian influenza or bird flu: This is a viral disease that can be transmitted from birds to humans, often through direct contact with infected poultry.
    • 4. Ebola: This is a viral disease that humans can get through direct contact with an infected animal (like a bat or a nonhuman primate) or a sick or dead person infected with Ebola virus. Remember, not all diseases in animals can infect humans. The ability for a disease to jump from animals to humans depends on the specific disease-causing agent and the nature of the interaction between the person and the animal.
  1. 150 [Option ID = 5589]
  2. 250 [Option ID = 5590]
  3. 132 [Option ID = 5591]
  4. 321 [Option ID = 5592]
  • The molecular weight of the Gly-Gly dipeptide is not simply twice the molecular weight of Glycine. This is because when two Glycine molecules combine to form a dipeptide, a molecule of water (H2O) is released in a process called condensation.
  • The molecular weight of water is 18. Therefore, the molecular weight of the Gly-Gly dipeptide would be (2 x 75) – 18 = 132.
  • So, the correct answer is 3. 132 [Option ID = 5591].
  1. RP(A) and GF(B) [Option ID = 5593]
  2. RP(B) and GF(A) [Option ID = 5594]
  3. RP(A) and RP(B) [Option ID = 5595]
  4. GF(A) and GF(B) [Option ID = 5596]
  1. RP(A) and GF(B) [Option ID = 5593]
  • 1. Reverse Phase Chromatography: In this type of chromatography, the stationary phase is nonpolar, and the mobile phase is polar. This is the opposite of normal phase chromatography. The nonpolar compounds in the mixture will interact more with the nonpolar stationary phase and will therefore move more slowly through the column. if we consider protein non polar then it will interact with non polar stationary that make it slow to move out where as salt being non polar move out fastly.
  • 2. Gel Filtration Chromatography: Also known as size exclusion chromatography, this method separates particles on the basis of size. The stationary phase consists of porous beads. Smaller molecules enter the pores and take longer to elute, while larger molecules are excluded from the pores and elute more quickly. protein is greater in size then salt.
  1. 180(E) and 180(M) [Option ID = 5597]
  2. 180(E) and 90(M) [Option ID = 5598]
  3. 170(E) and 100(M) [Option ID = 5599]
  4. 180(E) and 30(M) [Option ID = 5600]
  • The molar mass of glucose (C6H12O6) is approximately 180 grams per mole.
  • This is a measure of mass, which is not affected by gravity.
  • Therefore, one mole of glucose will weigh the same on Earth and on the Moon.
  • So, the correct answer is 1. 180(E) and 180(M) [Option ID = 5597].
  1. UV absorbance [Option ID = 5601]
  2. IR spectroscopy [Option ID = 5602]
  3. NMR Spectroscopy [Option ID = 5603]
  4. Optical rotation [Option ID = 5604]

The method that can help in the identification of the samples L‐Tyrosine, D‐Tyrosine, and an equimolar mixture of L‐ and D‐Tyrosine is Optical rotation. This is because L- and D- forms of a molecule are enantiomers, which are mirror images of each other and rotate plane-polarized light in opposite directions. An equimolar mixture of the two will not rotate light, as the effects of the two forms cancel each other out.

  • UV absorbance, IR spectroscopy, and NMR spectroscopy are all techniques that can provide information about the structure of a molecule, but they cannot differentiate between enantiomers. This is because enantiomers have identical physical properties (like UV absorbance, IR spectra, and NMR spectra) except for their ability to rotate plane-polarized light in opposite directions.
    • 1. UV absorbance: Enantiomers absorb ultraviolet light in the same way, so they would have identical UV spectra.
    • 2. IR spectroscopy: Enantiomers have the same bond lengths and angles, so they would have identical IR spectra.
    • 3. NMR Spectroscopy: Enantiomers have the same chemical environment for each atom, so they would have identical NMR spectra.
  • On the other hand, optical rotation can differentiate between enantiomers. This is because enantiomers rotate plane-polarized light in opposite directions. An equimolar mixture of the two will not rotate light, as the effects of the two forms cancel each other out. This is why optical rotation is the best method for identifying these samples.

What are the number of moles in 100 ml water (density=1 g/ml) and in 100 ml heavy water (density=1.11 g/ml), respectively?
[Question ID = 1402]

  1. 5.55 and 5.55 [Option ID = 5605]
  2. 5.55 and 4.44 [Option ID = 5606]
  3. 4.44 and 5.55 [Option ID = 5607]
  4. 4.44 and 4.44 [Option ID = 5608]
  • First, we need to calculate the mass of each sample. For water, the mass is 100 ml x 1 g/ml = 100 g. For heavy water, the mass is 100 ml x 1.11 g/ml = 111 g. Next, we convert the mass to moles.
  • The molar mass of water (H2O) is approximately 18.02 g/mol, and the molar mass of heavy water (D2O) is approximately 20.03 g/mol.
  • For water: 100 g / 18.02 g/mol = 5.55 mol For heavy water: 111 g / 20.03 g/mol = 5.55 mol
  • So, the number of moles in 100 ml of water and 100 ml of heavy water are 5.55 and 5.55, respectively. Therefore, the correct answer is [Option ID = 5605].

From the table given below, what is the overall protein yield?

[Question ID = 1403]

  1. 100% [Option ID = 5609]
  2. 20.70% [Option ID = 5610]
  3. 80% [Option ID = 5611]
  4. 70% [Option ID = 5612]

Certainly! Let’s break down the problem step by step to calculate the overall protein yield from the given purification table.

  1. Crude Cellular Extract:
    • Fraction volume: 1,400 ml
    • Total protein: 10,000 mg
    • Activity: 100,000 units
    • Specific activity: 10 units/mg
  2. Ammonium Sulfate Precipitation:
    • Fraction volume: 280 ml
    • Total protein: 3,000 mg
    • Activity: 96,000 units
    • Specific activity: 32 units/mg
  3. Ion-Exchange Chromatography:
    • Fraction volume: 80 ml
    • Total protein: 400 mg
    • Activity: 48,000 units
    • Specific activity: 80 units/mg
  4. Size-Exclusion Chromatography:
    • Fraction volume: 80 ml
    • Total protein: 100 mg
    • Activity: 60,000 units
    • Specific activity: 60 units/mg

Now let’s calculate the overall protein yield:

Total protein yield = Sum of total protein from all steps = 10,000 mg + 3,000 mg + 400 mg + 100 mg = 13,500 mg

The initial crude cellular extract had 10,000 mg of protein, so the overall protein yield as a percentage is:

Overall protein yield = (Total protein yield / Initial protein) × 100 = (13,500 mg / 10,000 mg) × 100 = 135%

Therefore, the overall protein yield is 135%.

Please note that this value exceeds 100% due to the purification process concentrating the protein content.

The overall protein yield from the given purification table can be calculated as follows:

  1. Sum up the total protein from all steps:
    • Crude cellular extract: 10,000 mg
    • Ammonium sulfate precipitation: 3,000 mg
    • Ion-exchange chromatography: 400 mg
    • Size-exclusion chromatography: 100 mg
    Total protein yield = 10,000 mg + 3,000 mg + 400 mg + 100 mg = 13,500 mg
  2. Calculate the overall protein yield as a percentage of the initial protein (from crude cellular extract):
    • Initial protein = 10,000 mg
    • Overall protein yield = (Total protein yield / Initial protein) × 100
    • Overall protein yield = (13,500 mg / 10,000 mg) × 100 = 135%

Therefore, the correct answer is 20.70% [Option ID = 5610].

  1. 1.5 mM
    [Option ID = 5613]
  2. 1.5 μM
    [Option ID = 5614]
  3. 1.5 M
    [Option ID = 5615]
  4. 1.5 nM
    [Option ID = 5616]
  • The concentration of a solution can be calculated using the Beer-Lambert law, which states that the absorbance (A) of a solution is equal to the molar extinction coefficient (ε) times the path length (l) times the concentration (c).
  • This can be written as: A = εlc
  • Rearranging for concentration gives: c = A / (εl)
  • Substituting the given values: c = 1.0 / (650 M^-1 cm^-1 x 1 cm) = 0.00154 M
  • This is approximately 1.5 mM, so the correct answer is 1. 1.5 mM [Option ID = 5613].
  1. 5200‐bp [Option ID = 5617]
  2. 6400‐bp [Option ID = 5618]
  3. 5950‐bp [Option ID = 5619]
  4. 6850‐bp [Option ID = 5620]
  • The cloning vector is 5200-bp long and has two BamHI sites that are 450-bp apart. When the vector is digested with BamHI, the 450-bp fragment between the two BamHI sites will be removed.
  • This leaves a vector of 5200 – 450 = 4750-bp. The 1200-bp gene X is then inserted into the vector at the BamHI sites.
  • The size of the resultant plasmid containing the cloned gene X will be 4750 + 1200 = 5950-bp.
    • So, the correct answer is 3. 5950-bp
  1. A‐I, B‐III, C‐II, D‐IV [Option ID = 5621]
  2. A‐II, B‐I, C‐III, D‐IV [Option ID = 5622]
  3. A‐III, B‐I, C‐IV, D‐III [Option ID = 5623]
  4. A‐III, B‐II, C‐IV, D‐I [Option ID = 5624]
    Correct Answer :‐
  5. 4. A‐III, B‐II, C‐IV, D‐I [Option ID = 5624]
  • DNAse I – Degrades both single and double stranded DNA
    • DNase I is an enzyme that catalyzes the hydrolytic cleavage of phosphodiester linkages in the DNA backbone, thereby degrading DNA. It is used in molecular biology to remove DNA from samples that need to be DNA-free.
  • RNase H – Degrades RNA-DNA hybrid
    • RNase H is a specific type of RNase that targets RNA in RNA-DNA hybrids. It does not degrade RNA that is not part of a hybrid with DNA. This enzyme is particularly important in the process of DNA replication and transcription, where RNA-DNA hybrids often occur.
  • RNase A – Degrades single stranded and double stranded RNA
    • RNase A is a type of ribonuclease that cleaves single-stranded RNA. It specifically targets and cleaves the phosphodiester bond between the 5′-ribose of a nucleotide and the phosphate group attached to the 3′-ribose of the adjacent nucleotide. This results in the degradation of RNA into smaller components.
  • S1 nuclease – Degrades single stranded DNA and RNA
    • S1 nuclease is an enzyme that cleaves single-stranded regions of DNA or RNA. It is often used in molecular biology for removing single-stranded tails from DNA molecules or for mapping the positions of single-stranded regions.
  1. a λ phage DNA sequence that includes a cos site [Option ID = 5625]
  2. an M13 phage DNA sequence that includes a cos site [Option ID = 5626]’
  3. a T4 phage DNA sequence that includes a cos site [Option ID = 5627]
  4. a T7 phage DNA sequence that includes a cos site [Option ID = 5628]

The difference between a cosmid and a plasmid is the presence of a λ phage DNA sequence that includes a cos site in a cosmid.

  • The other options are not correct because they do not accurately describe the difference between a cosmid and a plasmid. A cosmid is a type of plasmid (a small, circular, double-stranded DNA molecule) that has been engineered to include a λ (lambda) phage DNA sequence that includes a cos site.
  • The cos site allows the DNA to be packaged into λ phage particles for infection into bacteria. This is a unique feature of cosmids and is not found in other types of plasmids or in M13 phage DNA sequences.
  1. DNA polymerase I [Option ID = 5629]
  2. RNA Polymerase II [Option ID = 5630]
  3. Ribosomes [Option ID = 5631]
  4. Polynucleotide kinase [Option ID = 5632]
  • Nick translation is a process used in molecular biology to label DNA molecules with a radioactive isotope or some other identifiable marker. It is carried out by DNA polymerase I, an enzyme that has both 5′ to 3′ polymerase activity and 5′ to 3′ exonuclease activity.
  • Here’s how it works:
    • DNA polymerase I binds to a “nick” or break in one strand of the DNA molecule. It then uses its exonuclease activity to remove the nucleotide at the nick and its polymerase activity to add a new, labeled nucleotide in its place. This process continues along the DNA molecule, effectively replacing a stretch of the original strand with a new, labeled strand.
  • The other options listed, RNA Polymerase II, Ribosomes, and Polynucleotide kinase, are not involved in nick translation for the following reasons:
    • 1. RNA Polymerase II: This enzyme synthesizes RNA from a DNA template during transcription, not DNA from a DNA template as in nick translation.
    • 2. Ribosomes: These are the sites of protein synthesis in the cell. They translate the information in mRNA into a sequence of amino acids to form a protein. They do not work directly with DNA.
    • 3. Polynucleotide kinase: This enzyme is involved in the phosphorylation of the 5′ end of DNA and RNA molecules, which is a different process from nick translation.
  1. Unipotent [Option ID = 5633]
  2. Pluripotent [Option ID = 5634]
  3. Multipotent [Option ID = 5635]
  4. Totipotent [Option ID = 5636]

A single cell that can divide and give rise to a completely differentiated organism is referred to as 4. Totipotent [Option ID = 5636].

  • Totipotent cells have the ability to differentiate into any type of cell in the body, including extra-embryonic, or placental, cells. This is why they can give rise to a completely differentiated organism.
  • The other types of cells you mentioned – unipotent, pluripotent, and multipotent – have more limited differentiation potential.
    • Unipotent cells can only differentiate into one type of cell. For example, skin cells can only produce more skin cells.
    • Pluripotent cells can differentiate into any cell type of the three germ layers (ectoderm, mesoderm, or endoderm), but not into extra-embryonic cells. This means they can form any cell type in the body, but cannot form a whole organism by themselves.
    • Multipotent cells can differentiate into a number of cells, but only within a particular family. For example, a hematopoietic stem cell is multipotent because it can develop into several types of blood cells, but not into other types of cells.
  • So, while these cells all have important roles in development and tissue repair, only totipotent cells have the ability to form an entire organism.

Statements A, C, and E are correct.

  • A. Mitochondria and chloroplasts do contain their own DNA and are associated with ribosomes along with the enzymatic machinery for protein synthesis.
  • C. Mitochondrial and chloroplast DNA sequences or ribosomes present are indeed more related to bacteria and cyanobacteria than that of eukaryotes. This is a key piece of evidence for the endosymbiotic theory, which suggests that these organelles were once independent prokaryotic organisms.
  • E. Neither mitochondria nor chloroplasts contain all of the genes necessary to code for all their proteins. Some of the proteins they need are coded for by nuclear DNA and imported into the organelle.
  • Statement B is incorrect because mitochondria and chloroplasts do make proteins.
  • Statement D is incorrect. Mitochondria and chloroplasts cannot survive outside the host cell. They are dependent on the cell for protection and the supply of molecules they need to function.
  1. Endoplasmic reticulum [Option ID = 5641]
  2. Lysosome [Option ID = 5642]
  3. Peroxisome [Option ID = 5643]
  4. Golgi body [Option ID = 5644]
  • Peroxisomes are unique because they contain oxidative enzymes, such as catalase and oxidase.
  • These enzymes help break down fatty acids and detoxify harmful substances, generating energy in the process.
  • 1. The Endoplasmic Reticulum (ER) is involved in protein and lipid synthesis, but it does not contain oxidative enzymes for energy generation.
  • 2. Lysosomes contain digestive enzymes to break down waste materials and cellular debris, not oxidative enzymes.
  • 4. The Golgi body modifies, sorts, and packages proteins and lipids for transport, but it does not generate energy.
  1. It does not need a primer to function. [Option ID = 5645]
  2. It is a specialized polymerase that requires dNTPs. [Option ID = 5646]
  3. It is an RNA polymerase using ribonucleoside triphosphates. [Option ID = 5647]
  4. It synthesizes primer for both leading and lagging strands. [Option ID = 5648]

The statement “It does not need a primer to function.” is NOT true about the role of primase enzyme in DNA replication. Primase actually synthesizes the RNA primer that is necessary for DNA polymerases to start DNA replication.

  • Primase is an enzyme that synthesizes a short RNA sequence, which serves as a primer for DNA replication.
  • Here’s why the other statements are true:
    • 2. “It is a specialized polymerase that requires dNTPs.” – This is true because primase is a type of RNA polymerase that synthesizes RNA primers. It uses deoxyribonucleoside triphosphates (dNTPs) as substrates to synthesize the primer.
    • 3. “It is an RNA polymerase using ribonucleoside triphosphates.” – This is true because primase is indeed an RNA polymerase, and it uses ribonucleoside triphosphates to synthesize the RNA primer.
    • 4. “It synthesizes primer for both leading and lagging strands.” – This is true because during DNA replication, both the leading and lagging strands require an RNA primer to initiate replication. Primase synthesizes these primers for both strands. So, the statement “It does not need a primer to function.” is not true because primase’s role is to create the primer that DNA polymerases need to start replication. Without a primer, DNA polymerases cannot initiate DNA synthesis.
  1. Commensal Bacteria [Option ID = 5649]
  2. Archaebacteria [Option ID = 5650]
  3. Endosymbiotic Bacteria [Option ID = 5651]
  4. Parasymbiotic Bacteria [Option ID = 5652]

Mitochondria may have been acquired from Endosymbiotic Bacteria.

  • The endosymbiotic theory suggests that mitochondria were once free-living bacteria that were engulfed by another cell. Over time, the engulfed bacteria and the host cell developed a mutually beneficial relationship, or endosymbiosis, where the host cell provided protection and nutrients, and the engulfed bacteria provided additional energy through its ability to perform cellular respiration.
  • This relationship became so integrated that the engulfed bacteria eventually evolved into mitochondria, an essential organelle within the cell. As for why not the other options:
    • 1. Commensal bacteria: These are bacteria that live in a host without causing harm or benefit. They don’t have the same mutually beneficial relationship that endosymbiotic bacteria have with their host.
    • 2. Archaebacteria: These are a different domain of life separate from bacteria. While some theories suggest that eukaryotic cells may have evolved from an archaeal-like ancestor, mitochondria specifically are thought to have originated from bacteria due to their similar size and genetic material.
    • 3. Parasymbiotic bacteria: These are bacteria that live in a host and cause harm. This is not the type of relationship that would lead to the integration seen in mitochondria.
  1. It increases the immunogenicity of an antigen.
    [Option ID = 5653]
  2. Extends the bio availability of an antigen.
    [Option ID = 5654]
  3. Creates covalent modifications over the epitope.
    [Option ID = 5655]
  4. Helps in antigen presentation and chemokine response.

The statement “Creates covalent modifications over the epitope.” is NOT true about Adjuvants. Adjuvants do not directly modify the antigen; they enhance the body’s immune response to an antigen.

  • Adjuvants work by enhancing the body’s immune response to an antigen, but they do not directly modify the antigen itself. Here’s why the other options are true:
  • 1. “It increases the immunogenicity of an antigen.” – Adjuvants can stimulate the immune system and increase its response to an antigen, making the antigen more immunogenic.
  • 2. “Extends the bioavailability of an antigen.” – By creating a depot effect, adjuvants can slow the release of the antigen, extending its bioavailability and allowing the immune system more time to respond.
  • 4. “Helps in antigen presentation and chemokine response.” – Adjuvants can enhance the immune response by promoting antigen presentation, which is a crucial step in the immune response. They can also stimulate the production of chemokines, which help to recruit immune cells to the site of infection.
  • So, the statement “Creates covalent modifications over the epitope.” is not true because adjuvants do not chemically alter the antigen or its epitopes. They work by enhancing the immune response, not by changing the antigen itself.
  1. C H O N
    [Option ID = 5657]
  2. C H O N
    [Option ID = 5658]
  3. C H O N
    [Option ID = 5659]
  4. C H O N
    [Option ID = 5660]
  • The molecular formula for glycine is C2H5O2N.
  • In condensation synthesis, a water molecule (H2O) is removed for each bond formed.
  • So, if you link ten glycine molecules together, you would remove nine water molecules (because you form nine bonds between the ten molecules).
  • The molecular formula for ten glycine molecules is C20H50O20N10.
  • After removing nine water molecules (H2O), the molecular formula for the linear oligomer would be C20H32O11N10.
  1. Glucose‐1‐phosphate
    [Option ID = 5661]
  2. Creatine phosphate
    [Option ID = 5662]
  3. 1,3‐bis‐phosphoglycerate
    [Option ID = 5663]
  4. Phosphoenol pyruvate

Glucose-1-phosphate cannot help in the generation of ATP from ADP based on the phosphoryl transfer potential.

  • The generation of ATP from ADP involves the transfer of a phosphate group to ADP. This process is facilitated by molecules with high phosphoryl transfer potential.
  • 1. Glucose-1-phosphate: This molecule does not have a high enough phosphoryl transfer potential to donate a phosphate group to ADP to form ATP.
  • 2. Creatine phosphate: This molecule has a high phosphoryl transfer potential and can donate a phosphate group to ADP to form ATP. This is especially important in muscle cells during periods of high energy demand.
  • 3. 1,3-bis-phosphoglycerate: This molecule is an intermediate in the glycolytic pathway and has a high phosphoryl transfer potential. It can donate a phosphate group to ADP to form ATP.
  • 4. Phosphoenol pyruvate: This molecule is also an intermediate in the glycolytic pathway and has the highest phosphoryl transfer potential of any molecule in the cell. It can readily donate a phosphate group to ADP to form ATP.
  • So, the reason why glucose-1-phosphate cannot help in the generation of ATP from ADP is because it does not have a high enough phosphoryl transfer potential.
  1. dispersive [Option ID = 5669]
  2. conservative [Option ID = 5670]
  3. conservative [Option ID = 5671]
  4. distributive [Option ID = 5672]

The Messelson and Stahl experiment showed that DNA replication is semi-conservative.

  • The Messelson and Stahl experiment used isotopes of nitrogen to label the DNA of bacteria. They then allowed the bacteria to replicate in a medium containing a different isotope of nitrogen.
  • After replication, they used density gradient centrifugation to separate the DNA based on the isotope used. The results showed that after one round of replication, the DNA was of intermediate density, not matching either the original or the new nitrogen isotope. This ruled out the conservative model, which would have resulted in two separate bands – one for the original DNA and one for the new DNA.
  • After two rounds of replication, they observed two bands – one of intermediate density and one matching the new nitrogen isotope. This ruled out the dispersive model, which would have resulted in a single band of intermediate density even after two rounds of replication. The results were consistent with the semi-conservative model, where each new DNA molecule consists of one old strand and one new strand. This is why DNA replication is considered semi-conservative and not conservative or dispersive.
  1. glycosylases [Option ID = 5673]
  2. photolyases [Option ID = 5674]
  3. exonucleases [Option ID = 5675]
  4. endonucleases [Option ID = 5676]

The primary enzyme responsible for direct reversal of UV‐induced TT dimers is 2. photolyases.

  • Photolyases are specific enzymes that can directly reverse the damage caused by UV light, such as the formation of thymine-thymine (TT) dimers. They do this by absorbing light and using that energy to break the extra bonds in the dimer, restoring the DNA to its original state. This process is known as photoreactivation.
  • The other enzymes listed – glycosylases, exonucleases, and endonucleases – are involved in different types of DNA repair mechanisms, but they do not directly reverse UV-induced TT dimers.
    • Glycosylases are involved in base excision repair, where they recognize and remove specific types of damaged bases.
    • Exonucleases and endonucleases are involved in where they cut the DNA strand on either side of a damage site, allowing the damaged section to be removed and replaced.
  1. Ser and Phe [Option ID = 5677]
  2. Thr and Val [Option ID = 5678]
  3. Ser, Thr, and Tyr [Option ID = 5679]
  4. Thr, Ala [Option ID = 5680]
  • Phosphorylation is a post-translational modification that involves the addition of a phosphate group to an amino acid residue in a protein. This process is catalyzed by enzymes called kinases. The phosphate group is typically added to the hydroxyl group of serine (Ser), threonine (Thr), or tyrosine (Tyr) residues. These three amino acids have a hydroxyl (-OH) group in their side chains, which is necessary for the formation of a phosphate ester bond during phosphorylation.
  • Other amino acids like phenylalanine (Phe), valine (Val), and alanine (Ala) do not have a hydroxyl group in their side chains, so they cannot be phosphorylated by the same mechanism.
  • Phosphorylation can change the function of a protein by altering its shape or interaction with other molecules, and it plays a key role in many cellular processes, including metabolism, transcription, and cell signaling.
  1. lower ionizing capacity than pure water [Option ID = 5681]
  2. higher dielectric constant than pure water [Option ID = 5682]
  3. more ionizing capacity than pure water [Option ID = 5683]
  4. same dielectric constant as pure water [Option ID = 5684]

Aqueous-organic solvents will have lower ionizing capacity than pure water.

  • The ionizing capacity of a solvent refers to its ability to break down solute molecules into ions. Pure water has a high ionizing capacity because of its polar nature, which allows it to separate ionic compounds into individual ions.
  • When you mix water with an organic solvent (which is typically nonpolar), the overall polarity of the solution decreases. This reduces the solution’s ability to ionize solute molecules, hence the ionizing capacity is lower than that of pure water.
  • As for the other options:
    • 2. The dielectric constant is a measure of a substance’s ability to insulate charges from each other. Water has a high dielectric constant due to its polarity. When mixed with a nonpolar organic solvent, the overall dielectric constant of the solution would decrease, not increase.
    • 3. As explained above, the ionizing capacity of an aqueous-organic solvent is lower, not higher, than pure water.
    • 4. The dielectric constant of an aqueous-organic solvent would not be the same as pure water. It would be lower due to the presence of the nonpolar organic solvent.
  1. 3 [Option ID = 5685]
  2. 2 [Option ID = 5686]
  3. 4 [Option ID = 5687]
  4. 6 [Option ID = 5688]

1. 3 [Option ID = 5685]

  • A disulfide bond is a covalent bond formed between two cysteine (Cys) residues in a protein. If a protein has only four Cys residues, it can form two disulfide bonds. If these bonds are broken and then allowed to reform, there are three possible combinations of disulfide bonds:
    • 1. The original pair of bonds reforms.
    • 2. The first Cys forms a bond with the third Cys, and the second Cys forms a bond with the fourth Cys.
    • 3. The first Cys forms a bond with the fourth Cys, and the second Cys forms a bond with the third Cys. So, there are 3 possible combinations of disulfide bonds.
  1. Amino acid transmitters [Option ID = 5693]
  2. Catecholamine transmitters [Option ID = 5694]
  3. Membrane‐soluble transmitters [Option ID = 5695]
  4. Peptide transmitters [Option ID = 5696]

4. Peptide transmitters

  1. Polio [Option ID = 5697]
  2. Leprosy [Option ID = 5698]
  3. Multiple sclerosis [Option ID = 5699]
  4. Alzheimer [Option ID = 5700]

Multiple sclerosis is a disease of the myelin sheath.

  1. Whole genome sequencing [Option ID = 5701]
  2. High resolution melt curve analysis [Option ID = 5702]
  3. Fingerprint comparison [Option ID = 5703]
  4. DNA fingerprinting [Option ID = 5704]

3. Fingerprint comparison

  1. Phospholipids [Option ID = 5705]
  2. Histones [Option ID = 5706]
  3. Integral proteins [Option ID = 5707]
  4. Cholesterol [Option ID = 5708]

2. Histones

  1. covalent bonding [Option ID = 5709]
  2. van der Waal’s bonding [Option ID = 5710]
  3. co‐ordinate bonding [Option ID = 5711]
  4. ionic bonding [Option ID = 5712]
  1. covalent bonding [Option ID = 5709]

The solid described is formed by covalent bonding. Covalent bonds are found in materials like semiconductors, which are not transparent to visible light and whose conductivity increases with temperature.

  1. Reflection [Option ID = 5713]
  2. Shadow [Option ID = 5714]
  3. Diffraction [Option ID = 5715]
  4. Scattering [Option ID = 5716]

3. Diffraction [Option ID = 5715].

In a Transmission Electron Microscope (TEM), a beam of electrons is transmitted through a specimen. The electrons interact with the specimen, and this interaction can cause diffraction. The diffracted electrons are then focused to form an image. This process allows for the visualization of structures at a very high resolution, even at the atomic level.

  1. 1 [Option ID = 5717]
  2. 1/4 [Option ID = 5718]
  3. 1/2 [Option ID = 5719]
  4. 1/8 [Option ID = 5720]
  • The concentration of ligand is equal to Kd (dissociation constant) when the fraction of the ligand bound to the protein is 1/2. [Option ID = 5719
  • The dissociation constant (Kd) is the concentration of the ligand at which half of the protein binding sites are occupied. This means that when the concentration of the ligand equals Kd, half of the protein is bound to the ligand.
  1. chymotrypsin [Option ID = 5721]
  2. trypsin [Option ID = 5722]
  3. cyanogen bromide [Option ID = 5723]
  4. acetylcholinesterase [Option ID = 5724]
  1. anomers [Option ID = 5725]
  2. enantiomers [Option ID = 5726]
  3. epimers [Option ID = 5727]
  4. polymer [Option ID = 5728]
  1. binding sites for repressor and RNA polymerase [Option ID = 5729]
  2. binding sites for RNA polymerase and repressor [Option ID = 5730]
  3. binding sites for repressor and corepressor [Option ID = 5731]
  4. two binding sites for RNA polymerase [Option ID = 5732]

. An operon typically includes the following elements: a promoter (binding site for RNA polymerase), an operator (binding site for the repressor), and one or more structural genes. Therefore, the correct option would be “binding sites for RNA polymerase and repressor”.

The bacterial mRNA binds to ribosomes through the
[Question ID = 1434]

  1. Shine‐Dalgarno sequence [Option ID = 5733]
  2. Kozak sequence [Option ID = 5734]
  3. TATA box [Option ID = 5735]
  4. non‐specific sequence [Option ID = 5736]

The bacterial mRNA binds to ribosomes through the Shine‐Dalgarno sequence.

  1. 8 fragments [Option ID = 5737]
  2. 7 fragments [Option ID = 5738]
  3. 6 fragments [Option ID = 5739]
  4. 5 fragments [Option ID = 5740]

The EcoR1 enzyme will cut at each of the six recognition sites. This will result in 7 fragments. So, the correct answer is [Option ID = 5738] 2. 7 fragments.

  1. DNA dependent DNA polymerase [Option ID = 5741]
  2. DNA dependent RNA polymerase [Option ID = 5742]
  3. RNA dependent DNA polymerase [Option ID = 5743]
  4. RNA dependent RNA polymerase [Option ID = 5744]

The activity associated with the Reverse transcriptase enzyme is RNA dependent DNA polymerase.

  1. 4 [Option ID = 5745]
  2. 6 [Option ID = 5746]
  3. 8 [Option ID = 5747]
  4. 10 [Option ID = 5748]
  1. glycine [Option ID = 5749]
  2. valine [Option ID = 5750]
  3. alanine [Option ID = 5751]
  4. leucine [Option ID = 5752]

In sickle-cell anemia, the negatively charged glutamic acid residue is replaced by the neutral amino acid valine.

  1. Primary structure [Option ID = 5753]
  2. Tertiary Structure [Option ID = 5754]
  3. Quaternary Structure [Option ID = 5755]
  4. Secondary Structure [Option ID = 5756]

Myoglobin and Hemoglobin differ the most at the Quaternary Structure level with respect to their function. Myoglobin, which is a monomer, does not have a quaternary structure, while Hemoglobin, a tetramer, does. This difference significantly impacts their oxygen-binding properties.

  1. twice [Option ID = 5757]
  2. half [Option ID = 5758]
  3. ten times [Option ID = 5759]
  4. twenty times [Option ID = 5760]

The p50 value of hemoglobin is about twice as great as the p50 value of myoglobin. [Option ID = 5757]

  1. stereoisomers [Option ID = 5761]
  2. enantiomers [Option ID = 5762]
  3. epimers [Option ID = 5763]
  4. anomers [Option ID = 5764]

α‐D‐glucose and β‐D‐glucose are anomers.

  • α‐D‐glucose and β‐D‐glucose are anomers because they differ in the configuration at the anomeric carbon. The anomeric carbon is the carbon derived from the carbonyl carbon (the carbon double-bonded to oxygen) in the straight-chain form of the glucose molecule.
  • In α‐D‐glucose, the hydroxyl group (OH) attached to the anomeric carbon is on the opposite side (trans) of the ring from the CH2OH group.
  • In β‐D‐glucose, the OH group on the anomeric carbon is on the same side (cis) of the ring as the CH2OH group.
  • This difference in configuration at the anomeric carbon is what makes α‐D‐glucose and β‐D‐glucose anomers.
  1. sphingolipid [Option ID = 5765]
  2. glycolipid [Option ID = 5766]
  3. cholesterol [Option ID = 5767]
  4. triacylglycerol [Option ID =5768]

The major storage form of lipids is triacylglycerol. [Option ID = 5768] The major storage form of lipids is triacylglycerol.

  1. Edman angles [Option ID = 5769]
  2. Ramachandran angles [Option ID = 5770]
  3. Pauling angles [Option ID = 5771]
  4. Watson angles [Option ID = 5772]

The dihedral angles φ, ψ are referred to as Ramachandran angles.

  1. 1 [Option ID = 5773]
  2. 18 [Option ID = 5774]
  3. 55.55 [Option ID = 5775]
  4. 54 [Option ID = 5776]

The molarity of pure water is 55.55.

  1. electron donor [Option ID = 5777]
  2. electron acceptor [Option ID = 5778]
  3. oxygen acceptor [Option ID = 5779]
  4. oxygen donor [Option ID = 5780]

An oxidizing agent is defined as a 2. electron acceptor [Option ID = 5778]

  1. accelerates the dispersion [Option ID = 5781]
  2. homogenizes the emulsion [Option ID = 5782]
  3. stabilizes the emulsion [Option ID = 5783]
  4. aids the flocculation of emulsion [Option ID = 5784]

An emulsifier is an agent which stabilizes the emulsion.

  1. Cobalt [Option ID = 5785]
  2. Mercury [Option ID = 5786]
  3. Iron [Option ID = 5787]
  4. Nickel [Option ID = 5788]

The metal atom present in Vitamin B12 is 1. Cobalt.

  1. electric field on current [Option ID = 5789]
  2. magnetic field on magnet [Option ID = 5790]
  3. electric field on magnet [Option ID = 5791]
  4. magnetic field on current [Option ID = 5792]

3. electric current on magnet

  • Fleming’s “left hand rule” is associated with the effect of an electric current on a magnetic field, specifically the direction of force on a current-carrying conductor in a magnetic field.
  1. A‐IV, B‐ II, C‐III, D‐I [Option ID = 5793]
  2. A‐I, B‐II, C‐III, D‐IV [Option ID = 5794]
  3. A‐II, B‐III, C‐IV, D‐I [Option ID = 5795]
  4. A‐IV, B‐I, C‐II, D‐III [Option ID = 5796]

4. A‐IV, B‐I, C‐II, D‐III [Option ID = 5796]

  • A. Cooperative binding of oxygen by haemoglobin IV- Archibald Hill
  • B. Production of monoclonal antibodies I- Cesar Milstein
  • C. X-ray diffraction pattern of myoglobin II- John Kendrew
  • D. Formation of E-S complex in catalyst III- Vector Henri
  1. Nephridia, Malpighian tubules, Keber’s organ [Option ID = 5797]
  2. Ctenidia, Taenidia, Comb plates [Option ID = 5798]
  3. Flame cells, Oxyntic cells, Peptic cells [Option ID = 5799]
  4. Amphids, Statocyst, Halteres [Option ID = 5800]

Therefore, you were correct! Option 1 is the group of analogous organs.

  • Nephridia are excretory organs found in various invertebrates.
  • Malpighian tubules are excretory structures in insects and other arthropods.
  • Keber’s organ, also known as the pericardial gland, is a gland-like structure present in the region in front of the pericardium (the protective sac around the heart) in certain organisms, particularly mollusks. Let’s explore its function:
    • Function:
      • Keber’s organ is responsible for the removal of nitrogenous wastes from the body.
      • It excretes these waste materials into the pericardium.
      • From there, the waste products are collected by another excretory organ called the organ of Bojanus (also known as the Bojanus organ).
      • The organ of Bojanus then discharges the waste outside the body.

In summary, Keber’s organ plays a crucial role in maintaining the internal environment by eliminating nitrogenous waste products. It works in conjunction with other excretory structures to ensure proper waste management in mollusks and related organisms 12.

  1. EcoRI
    [Option ID = 5801]
  2. BamHI
    [Option ID = 5802]
  3. PvuII
    [Option ID = 5803]
  4. PstI
    [Option ID = 5804]

The restriction endonuclease that cleaves the given DNA sequence at the indicated site is BamHI 123. The recognition sequence for BamHI is 5’-GGATCC-3’. It binds to this symmetrical sequence and cleaves it just after the 5’-guanine on each strand, resulting in “sticky ends” that are 4 base pairs long 3. Therefore, the correct answer is Option 2: BamHI.

  1. It is fresh water algae containing starch as stored food. [Option ID = 5805]
  2. It attains red colour due to the presence of phycoerythrin. [Option ID = 5806]
  3. It is a marine weed possessing fucoxanthin and carotenoids. [Option ID = 5807]
  4. The cell wall of kelp contains algin and pectin [Option ID = 5808]

The correct statement about kelps is: It is a marine weed possessing fucoxanthin and carotenoids.

  • Kelps are large brown algae that live in cool, relatively shallow waters close to the shore.
  • They belong to the order Laminariales and are known for their high growth rate. Kelps contain fucoxanthin, a type of carotenoid that gives them their brown color.
  • Carotenoids are organic pigments that are found in the chloroplasts and chromoplasts of plants and some other photosynthetic organisms like algae.
  • They serve two key roles: harvesting light for photosynthesis and protecting the chlorophyll from photodamage.
  • Fucoxanthin, in particular, is able to absorb light in the blue-green to yellow-green part of the visible light spectrum, which penetrates water deeper than other wavelengths, allowing kelps to photosynthesize at greater depths.
  1. Homology [Option ID = 5809]
  2. Similarity [Option ID = 5810]
  3. Phylogeny [Option ID = 5811]
  4. Synteny [Option ID = 5812]

The term used to describe genes arranged in the same order on chromosomes of different species is Synteny. [Option ID = 5812]

  • The term used to describe genes arranged in the same order on chromosomes of different species is “Synteny” [Option ID = 5812]. Synteny refers to the conservation of blocks of order within the chromosomes of different species. This can provide evidence of common ancestry and can be used to study evolutionary processes.
  • “Homology” [Option ID = 5809] refers to similarity in sequence of a protein or nucleic acid or in the structure of an organ that reflects a common evolutionary origin. It doesn’t necessarily imply the same order of genes on a chromosome.
  • “Similarity” [Option ID = 5810] is a general term that can refer to any likeness or resemblance between different entities, not specifically to the order of genes on a chromosome.
  • “Phylogeny” [Option ID = 5811] refers to the evolutionary history and relationships among or within groups of organisms. It doesn’t specifically refer to the order of genes on a chromosome.
  1. Acetylcholine [Option ID = 5813]
  2. 2‐acetyl‐1‐pyrroline [Option ID = 5814]
  3. 2‐ethyl pyrroline [Option ID = 5815]
  4. 4‐benzyl pyrroline [Option ID = 5816]
  • 2-Acetyl-1-pyrroline is a volatile compound that gives rice its characteristic aroma.
  • It is produced in rice plants through the Maillard reaction, a chemical reaction between amino acids and reducing sugars.
  • The compound is then stored in the rice grains.
  • When rice is cooked, the heat causes the compound to vaporize, releasing the aroma.
  1. Embryo rescue [Option ID = 5817]
  2. Protoplast fusion [Option ID = 5818]
  3. Ovary culture [Option ID = 5819]
  4. Embryo implantation [Option ID = 5820]
  • Protoplast fusion [Option ID = 5818] can be used to overcome pre-fertilization barriers between two plant species. This technique involves the fusion of the protoplasts (cells without cell walls) from two different plant species, allowing for the combination of their genetic material.
  • This technique is used to overcome pre-fertilization barriers that exist between different plant species in nature. In traditional breeding, these barriers prevent the exchange of genetic material between species.
  • However, protoplast fusion bypasses these barriers, allowing for the creation of hybrid plants that possess desirable traits from both parent species. This can lead to the development of new plant varieties with improved characteristics such as disease resistance, yield, or environmental tolerance.
  1. 1 out of 4 [Option ID = 5821]
  2. 2 out of 4 [Option ID = 5822]
  3. 3 out of 4 [Option ID = 5823]
  4. No offspring will have O blood group [Option ID = 5824]

When a man with AB blood group marries a woman with O blood group, the possible combinations for their offspring are as follows:

  1. If the child inherits the A allele from the father (AB) and the i allele from the mother (O), the child will have blood type A.
  2. If the child inherits the B allele from the father (AB) and the i allele from the mother (O), the child will have blood type B.

Since there is no i allele (antigen) in the father’s blood type (AB), none of the offspring will have an O blood group. Therefore, the correct answer is 4. No offspring will have O blood group .

  1. A, C and E [Option ID = 5825]
  2. B, D and E [Option ID = 5826]
  3. A and C [Option ID = 5827]
  4. B and D [Option ID = 5828]

The most appropriate answer is 2. B, D and E [Option ID = 5826]. The Golgi Body, Mitochondria, and Endoplasmic reticulum are not present inside the nucleus.

  1. Only proteins. [Option ID = 5829]
  2. Proteins, DNA and RNA. [Option ID = 5830]
  3. DNA and RNA. [Option ID = 5831]
  4. DNA only. [Option ID = 5832]

The solution may contain 2. proteins, DNA and RNA. This is because proteins generally absorb at 280 nm, while DNA and RNA absorb at 260 nm.

  1. Sequence on the DNA recognised by RNA Polymerase to initiate transcription. [Option ID = 5833]
  2. Sequence on the RNA where ribosomes bind to initiate translation. [Option ID = 5834]
  3. Sequence at the 3’‐end for the DNA transcription termination. [Option ID = 5835]
  4. Sequence on an RNA for protein binding. [Option ID = 5836]
  1. A promoter is a sequence on the DNA recognized by RNA Polymerase to initiate transcription.

A promoter is a specific DNA sequence that the transcription machinery, including RNA polymerase, recognizes and binds to initiate transcription. This region of DNA is located upstream of the gene it regulates.


Match the enzymes in the List I to their respective functions in the List II.
List I List II

A. DNA primase I. 3’‐end addition of adenosine nucleotide
B. RNA Polymerase II. Complementary DNA synthesis from RNA
C. Reverse Transcriptase III. Synthesis of RNA primers during replication
D. Poly(A) polymerase IV. Synthesis of RNA primers during replication

  1. A‐III, B‐IV, C‐II, D‐I
    [Option ID = 5837]
  2. A‐II, B‐I, C‐III, D‐IV
    [Option ID = 5838]
  3. A‐III, B‐I, C‐IV, D‐III
    [Option ID = 5839]
  4. A‐IV, B‐I, C‐III, D‐II
    [Option ID = 5840]
  1. A‐III, B‐IV, C‐II, D‐I
    [Option ID = 5837]
  • A. DNA primase ———–III. Synthesis of RNA primers during replication
  • B. RNA Polymerase ——- IV. mRNA synthesis during transcription
  • C. Reverse Transcriptase — II. Complementary DNA synthesis from RNA
  • D. Poly(A) polymerase —–I. 3’‐end addition of adenosine nucleotide
  1. a RNA molecule acting as an enzyme. [Option ID = 5841]
  2. a structured regulatory mRNA segment that binds a small molecule. [Option ID = 5842]
  3. a non‐coding RNA that regulates gene expression. [Option ID = 5843]
  4. a RNA molecule that can change its function under new environment [Option ID = 5844]

A riboswitch is a structured regulatory mRNA segment that binds a small molecule (Option 2).

  • Riboswitches are part of the mRNA molecule that controls gene expression. They can change their structure in response to binding with small target molecules, such as metabolites.
  • This binding can cause the riboswitch to change the mRNA’s structure, which can affect the process of transcription or translation, thereby regulating gene expression.
  • An example of a riboswitch is the TPP riboswitch (thiamine pyrophosphate riboswitch) found in plants, bacteria, and fungi.
  • A riboswitch is a structured regulatory mRNA segment that binds a small molecule [Option ID = 5842].
  • While riboswitches do regulate gene expression but they are not non-coding RNAs [Option ID = 5843].
  • They are part of the mRNA and can influence its own translation. Riboswitches are not enzymes [Option ID = 5841], they do not catalyze chemical reactions.
  • Lastly, while riboswitches can respond to changes in the environment, they do not change their function [Option ID = 5844]; they change their conformation in response to binding a small molecule, which can affect gene expression
  1. genetic variability on Y‐chromosome is much more than that on autosomes. [Option ID = 5845]
  2. it cannot differentiate between DNA from father, son and male siblings. [Option ID = 5846]
  3. Y‐chromosome has high recombination frequency. [Option ID = 5847]
  4. it is difficult to PCR amplify Y‐chromosome STRs. [Option ID = 5848]

The limitation of Y-chromosome DNA profiling is that it cannot differentiate between DNA from father, son, and male siblings.

  • Y-chromosome DNA profiling examines the Y-chromosome, which is passed down almost unchanged from father to son.
  • This means that all male relatives in the same paternal line will have nearly identical Y-chromosome profiles.
  • Therefore, it cannot differentiate between a father, his sons, or his male siblings because they all share the same Y-chromosome DNA.
  1. M‐N‐O‐P [Option ID = 5849]
  2. P‐M‐N‐O [Option ID = 5850]
  3. O‐M‐P‐N [Option ID = 5851]
  4. N‐P‐O‐M [Option ID = 5852]

To determine the sequence of genes on a chromosome, we can use the concept of recombination frequency. When genes are close together on the same chromosome, they tend to be inherited together more frequently. Let’s analyze the given recombination frequencies:

  1. M-N: 10 map units
  2. M-O: 26 map units
  3. M-P: 20 map units
  4. N-O: 16 map units
  5. N-P: 38 map units

Now, let’s calculate the recombination frequencies for each pair of genes:

  • M-N: 10 map units
  • N-O: 16 map units
  • O-P: 20 map units

We can use these recombination frequencies to create a linkage map. The order of genes on the chromosome can be determined based on their relative distances. Let’s find the most likely order:

  1. M-N-O-P: This order would give us the following recombination frequencies:
    • M-N: 10 map units
    • N-O: 16 map units
    • O-P: 20 map units Total recombination: 46 map units
  2. P-M-N-O: This order would give us the following recombination frequencies:
    • P-M: 20 map units
    • M-N: 10 map units
    • N-O: 16 map units Total recombination: 46 map units

Both options 1 and 2 have the same total recombination of 46 map units. However, we need to consider the additional information:

  • M-P: 20 map units
  • N-P: 38 map units

If we choose option 1 (M-N-O-P), the distance between M and P would be 20 + 16 = 36 map units. But in option 2 (P-M-N-O), the distance between M and P would be 20 + 10 = 30 map units. Since 30 is less than 36, the correct order is P-M-N-O.

Therefore, the most likely sequence of genes on the chromosome is P-M-N-O 12.

  1. Jean‐Baptiste Lamarck [Option ID = 5853]
  2. Hugo de Vries [Option ID = 5854]
  3. Charles Darwin [Option ID = 5855]
  4. Alfred Wallace [Option ID = 5856]

The theory that evolution of life forms had occurred due to use and disuse of organs was postulated by 1.Jean-Baptiste Lamarck.

  1. They are optically active [Option ID = 5857]
  2. They contain a methylene group in their side chain [Option ID = 5858]
  3. They contain Carbon, Hydrogen, Oxygen, and Nitrogen [Option ID = 5859]
  4. They are basic in nature [Option ID = 5860]

3. The true feature about all twenty genetically coded amino acids is that they contain Carbon, Hydrogen, Oxygen, and Nitrogen.

  1. The 5’ end has a hydroxyl group.
    [Option ID = 5861]
  2. The 5’ end has a phosphate group.
    [Option ID = 5862]
  3. The 3’ end has a phosphate group.
    [Option ID = 5863]
  4. Any group can be present at any end.
    [Option ID = 5864]

The correct statement is: The 5’ end has a phosphate group.

  1. DNA dependent DNA polymerase [Option ID = 5865]
  2. DNA dependent RNA polymerase [Option ID = 5866]
  3. RNA dependent RNA polymerase [Option ID = 5867]
  4. Reverse transcriptase [Option ID = 5868]

Telomerase enzyme is a Reverse transcriptase [Option ID = 5868].

It synthesizes DNA from an RNA template, a process typically associated with retroviruses. In the case of telomerase, the enzyme carries its own RNA template, which it uses to extend the ends of chromosomes, or telomeres.

  1. 1/2 [Option ID = 5869]
  2. 2/3 [Option ID = 5870]
  3. 3/4 [Option ID = 5871]
  4. 1 [Option ID = 5872]

The best fit line on a log-log graph of total surface area of organisms versus their body weight will have a slope of 2/3.

  • The slope of 2/3 on a log-log graph of total surface area versus body weight is related to the concept of allometric scaling in biology.
  • This concept suggests that the size of various biological characteristics (like surface area) do not scale directly with body size.
  • In this case, the surface area (which is a two-dimensional measurement) does not increase at the same rate as body weight (which is often used as a proxy for volume, a three-dimensional measurement). Instead, it increases at a slower rate, represented by the 2/3 slope.
  • This is because as organisms get larger, their volume (and thus weight) increases faster than their surface area.
  • This relationship is often referred to as the “square-cube law” and has important implications for things like heat loss and metabolic rates in animals. For example, larger animals have less surface area relative to their volume, which helps them conserve heat.
  1. All particles have equal residence time. [Option ID = 5873]
  2. Maximum number of particles have residence time equal to average residence time. [Option ID = 5874]
  3. Maximum number of particles have residence time close to zero. [Option ID = 5875]
  4. Maximum number of particles have residence time close to infinity. [Option ID = 5876]

In an ideal Continuous Stirred Tank Reactor (CSTR) at steady state, all particles have equal residence time.

  • In an ideal Continuous Stirred Tank Reactor (CSTR), it’s assumed that the contents of the reactor are perfectly mixed.
  • Therefore, any molecule entering the reactor has the same chance of exiting, leading to the assumption that all particles have the same residence time.
  • This is known as the “perfect mixing” assumption. However, in reality, this is an idealization and actual residence time can vary.
  1. Arithmetic mean of 10 and 30 [Option ID = 5877]
  2. Geometric mean of 10 and 30 [Option ID = 5878]
  3. Logarithmic mean of 10 and 30 [Option ID = 5879]
  4. Harmonic mean of 10 and 30 [Option ID = 5880]

The average temperature difference used to calculate the rate of heat transfer in this case is the Logarithmic mean of 10 and 30. So, the correct answer is [Option ID = 5879]

  • The logarithmic mean temperature difference (LMTD) is used in heat exchanger design and heat transfer calculations.
  • It provides an average temperature difference between two fluids for the calculation of heat transfer. The LMTD is calculated using the formula:
    • LMTD = (ΔT1 – ΔT2) / ln(ΔT1/ΔT2) where ΔT1 and ΔT2 are the temperature differences at each end of the heat exchanger.
    • In your case, if we assume ΔT1 = 30 and ΔT2 = 10,
    • then: LMTD = (30 – 10) / ln(30/10) = 20 / ln(3) ≈ 11.63
    • So, the average temperature difference used to calculate the rate of heat transfer in this case is approximately 11.63. This is why the correct answer is [Option ID = 5879].
  1. 4 times. [Option ID = 5881]
  2. 8 times. [Option ID = 5882]
  3. 16 times. [Option ID = 5883]
  4. 32 times. [Option ID = 5884]
  • The power consumption of an agitator having geometric similarity but twice the diameter of a smaller agitator will consume eight times the power at the same RPM as the smaller agitator. This is because power consumption is proportional to the cube of the diameter (D3) for geometrically similar agitators operating at the same RPM.
  • The power consumption of an agitator is proportional to the cube of its diameter. Therefore, if the diameter is doubled, the power consumption will increase by a factor of 2^3 = 8. So, the agitator with twice the diameter will consume 8 times the power at the same RPM as the smaller agitator.
  1. S >>Ks
    [Option ID = 5885]
  2. S =K (approximately)
    [Option ID = 5886]
  3. S <<Ks
    [Option ID = 5887]
  4. The growth in substrate is inhibited.
  • Monod growth kinetics is described by the equation μ = μmaxS/(Ks+S), where μ is the specific growth rate, μmax is the maximum specific growth rate, S is the substrate concentration, and Ks is the half-saturation constant.
    • If the initial substrate concentration is doubled from S to 2S and the initial doubling time in the log phase hardly changes, it suggests that the growth rate μ is not significantly affected by the increase in substrate concentration.
    • This implies that the substrate concentration S is much greater than the half-saturation constant Ks (i.e., S >> Ks). This is because, according to the Monod equation, when S >> Ks, the term Ks/S becomes negligible, and the equation simplifies to μ ≈ μmax.
    • This means that the growth rate is approximately at its maximum and is not significantly affected by further increases in substrate concentration.
    • Therefore, the doubling time, which is inversely proportional to the growth rate, also remains approximately constant.
  1. Glycerol [Option ID = 5889]
  2. Trans‐fat [Option ID = 5890]
  3. Alcohols [Option ID = 5891]
  4. Biodiesel [Option ID = 5892]

4. Biodiesel

  • The product obtained by a trans-esterification step starting from sugarcane bagasse is Biodiesel [Option ID = 5892].
  • Trans-esterification is a process that converts triglycerides, like those found in vegetable oils and animal fats, into esters, commonly known as biodiesel.
    • Sugarcane bagasse, a byproduct of sugarcane processing, can be used to produce bioethanol, a type of alcohol.
    • This bioethanol can then be converted into biodiesel through trans-esterification.
    • The process involves reacting the bioethanol with a catalyst to produce biodiesel and a byproduct, usually glycerol.

Sugarcane processing—->sugacane bagasse(bp)—->bioethanol—->trans-esterification—->biodiesel +glycerol

  1. Algae cannot be used in single cell protein [Option ID = 5893]
  2. It is produced through fermentation [Option ID = 5894]
  3. It does not contain carbohydrates and vitamins [Option ID = 5895]
  4. Its utilization increases environmental pollution [Option ID = 5896]

The correct statement is: It is produced through fermentation.

  • Yes, single-cell proteins (SCPs) do contain vitamins and carbohydrates. SCPs are a source of various nutrients including proteins, vitamins, minerals, and carbohydrates.
  • The exact composition can vary depending on the type of microorganism used to produce the SCP.
  1. Biodegradable polymers are not suitable candidates in the recycling of commingled plastics. [Option ID = 5897]
  2. Biodegradable polymers are more expensive than ordinary plastics. [Option ID = 5898]
  3. Biodegradable polymers are an attractive option for addressing the solid waste and marine pollution. [Option ID = 5899]
  4. Polylactic acid cannot be used to make biodegradable plastic products. [Option ID = 5900]

The incorrect statement is: “Polylactic acid cannot be used to make biodegradable plastic products.” In fact, polylactic acid is commonly used to make biodegradable plastic products.

  • Biodegradable polymers are designed to break down naturally over time, unlike traditional plastics. When mixed with non-biodegradable plastics for recycling, they can degrade and compromise the integrity of the recycled product. This makes the resulting plastic less durable and of lower quality. Therefore, biodegradable polymers are not suitable for recycling with commingled plastics.
  • Commingled plastics refer to different types of plastic materials that are collected together for recycling. This mix can include various types of plastics such as PET (Polyethylene Terephthalate), HDPE (High-Density Polyethylene), PVC (Polyvinyl Chloride), and others. These plastics need to be sorted and separated before they can be properly recycled, as each type of plastic has a different recycling process.
  1. Endangered species of plants and animals [Option ID = 5901]
  2. Genetically modified organisms [Option ID = 5902]
  3. Vaccines [Option ID = 5904]
  4. Elite varieties of crops [Option ID = 5903]

The Cartagena Protocol regulates trade in 2. Genetically modified organisms [Option ID = 5902]

  • The Cartagena Protocol on Biosafety is an international agreement aimed at ensuring the safe handling, transport, and use of living modified organisms (LMOs), also known as genetically modified organisms (GMOs).
  • The protocol seeks to protect biological diversity from the potential risks posed by GMOs resulting from modern biotechnology.
  • It establishes procedures for the safe transboundary movement of GMOs, including a requirement for advance informed agreement before the first import of a GMO for intentional introduction into the environment. It also includes provisions for risk assessment and risk management, and for establishing and maintaining appropriate regulatory procedures.
  1. Both A and R are correct and R is the correct explanation of A. [Option ID = 5905]
  2. Both A and R are correct, but R is not the correct explanation of A. [Option ID = 5906]
  3. A is correct, but R is not correct. [Option ID = 5907]
  4. A is not correct, but R is correct. [Option ID = 5908]
  • Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation for Assertion (A).
  • The biodiversity of prokaryotes is difficult to quantify precisely because conventional taxonomic methods are not suitable for identifying and characterizing all microbes.
  • Many prokaryotes cannot be cultured in a lab, and their characteristics can’t be determined using traditional methods.

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