IIT JAM 2024 Biotechnology (BT) Question Paper (Available)- Download Solution pdf with Answer Key

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IIT JAM 2024 Biotechnology (BT) Question Paper with Answer Key pdf is available for download. IIT JAM 2024 BT exam was conducted by IIT Madras in shift 2 on February 11, 2024. In terms of difficulty level, IIT JAM 2024 Biotechnology (BT) paper was of moderate level. IIT JAM 2024 question paper for BT comprised a total of 60 questions.

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IIT JAM 2024 Biotechnology (BT) Question Paper with Solution PDFs

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IIT JAM 2024 Biotechnology Questions with Solutions

Question 1:

Which one of the following is a simple tissue system in plants?

Epidermis
Parenchyma
Phloem
Xylem

Correct Answer: (1) Epidermis

View Solution

Explanation:

Simple tissues are composed of one type of cell. The epidermis is a simple tissue that forms the outer protective covering of plants. It helps prevent water loss and provides protection against mechanical injury. Complex tissues, like xylem and phloem, are composed of more than one type of cell.

Step-by-step breakdown:

Identify the definition of simple tissue: A simple tissue is composed of a single cell type.
Consider each option:
- Epidermis: Forms the outer layer of the plant.
- Parenchyma: Can have different cell types and is not always a simple tissue.
- Phloem: A complex tissue with multiple cell types.
- Xylem: A complex tissue with multiple cell types.
Conclude: Epidermis fits the definition of a simple tissue since it’s made of only one cell type, making it the correct choice.

Question 2:

In DNA replication, the Okazaki fragments are joined by:

DNA helicase
DNA ligase
DNA polymerase
DNA primase

Correct Answer: (2) DNA ligase

View Solution

Explanation:

Okazaki fragments are short DNA segments synthesized on the lagging strand during DNA replication. DNA ligase plays a crucial role by sealing the nicks or gaps between these fragments to form a continuous DNA strand. Other enzymes like helicase and primase have distinct roles; helicase unwinds the DNA, and primase synthesizes RNA primers.

Step-by-step breakdown:

Understand the process of DNA replication: DNA replication is a complex process involving many enzymes. Okazaki fragments are produced on the lagging strand.
Identify the function of the listed enzymes:
- DNA helicase: Unwinds the DNA double helix.
- DNA ligase: Joins DNA fragments.
- DNA polymerase: Synthesizes new DNA strands.
- DNA primase: Synthesizes RNA primers.
Connect the function of each to Okazaki fragments: DNA ligase is responsible for joining the Okazaki fragments.
Conclude: The function of DNA ligase is specific to sealing nicks between DNA fragments, making it the correct answer.

Question 3:

The most abundant type of RNA in a metabolically active mammalian cell is:

mRNA
rRNA
snoRNA
tRNA

Correct Answer: (2) rRNA

View Solution

Explanation:

Ribosomal RNA (rRNA) constitutes about 80% of the total RNA in a cell. It's a key component of ribosomes, where protein synthesis occurs. Messenger RNA (mRNA) and transfer RNA (tRNA) make up smaller proportions and serve as intermediaries in protein synthesis.

Step-by-step breakdown:

Understand different types of RNA: There are mainly three types of RNA: mRNA, rRNA, and tRNA, each with specific roles.
Identify the function of the listed RNA types:
- mRNA: Carries genetic code for protein synthesis.
- rRNA: Is a key component of ribosomes.
- snoRNA: Guides chemical modifications of other RNAs.
- tRNA: Transfers amino acids to ribosomes during protein synthesis.
Compare the abundance of different RNA types: rRNA constitutes the largest percentage in a cell.
Conclude: rRNA is the most abundant type of RNA in a cell.

Question 4:

Which organelle in a eukaryotic cell is the site of the electron transport chain?

Endoplasmic reticulum
Golgi apparatus
Mitochondrion
Peroxisome

Correct Answer: (3) Mitochondrion

View Solution

Explanation:

The electron transport chain is a series of protein complexes embedded in the inner mitochondrial membrane. These complexes transfer electrons and pump protons to create an electrochemical gradient, driving ATP synthesis. Other organelles, like the Golgi apparatus or peroxisomes, are not involved in this process.

Step-by-step breakdown:

Identify the electron transport chain (ETC): ETC is a process that generates ATP, and it occurs in a specific organelle.
Understand the functions of each given organelle:
- Endoplasmic reticulum: Involved in protein and lipid synthesis.
- Golgi apparatus: Processes and packages proteins.
- Mitochondrion: Site of cellular respiration, including the ETC.
- Peroxisome: Involved in fatty acid and hydrogen peroxide metabolism.
Connect the location of ETC to one of the organelles: The electron transport chain is localized in the inner mitochondrial membrane.
Conclude: Mitochondrion is the correct answer because it contains the ETC.

Question 5:

RNA is a polymer of:

Glycosides
Ribonucleosides
Ribonucleotides
Riboses

Correct Answer: (3) Ribonucleotides

View Solution

Explanation:

RNA is composed of ribonucleotides, each containing a ribose sugar, a phosphate group, and a nitrogenous base (adenine, uracil, cytosine, or guanine). This structure forms the backbone of RNA, enabling its various functions in cells, including protein synthesis.

Step-by-step breakdown:

Understand the basic structure of nucleic acids: Nucleic acids like RNA are polymers of repeating units called nucleotides.
Identify the building blocks of RNA:
- Glycosides: Molecules composed of a sugar and another non-sugar group.
- Ribonucleosides: Contain a ribose sugar and a nitrogenous base.
- Ribonucleotides: Contain a ribose sugar, a nitrogenous base, and a phosphate group.
- Riboses: Five-carbon sugars.
Connect the building blocks to RNA: RNA is made of ribonucleotides, which has the ribose sugar, nitrogenous base, and phosphate group.
Conclude: Ribonucleotides are the correct answer since they are the monomers of RNA.

Question 6:

Which one of the following is present in a bacterial cell?

28S rRNA
70S ribosome
Chitinous cell wall
Histones

Correct Answer: (2) 70S ribosome

View Solution

Explanation:

Bacteria possess 70S ribosomes, which are smaller than the 80S ribosomes found in eukaryotes. They lack chitinous cell walls, which are characteristic of fungi, and histones, which are associated with eukaryotic chromatin.

Step-by-step breakdown:

Understand the characteristics of bacteria: Bacterial cells have unique structural features compared to eukaryotic cells.
Consider each option:
- 28S rRNA: A component of the 80S ribosome found in eukaryotes.
- 70S ribosome: A ribosome type found in bacteria.
- Chitinous cell wall: A cell wall component found in fungi, not bacteria.
- Histones: Proteins associated with eukaryotic DNA.
Connect the features with the appropriate cell type: Bacterial cells have 70S ribosomes.
Conclude: 70S ribosome is the correct answer because bacteria are characterized by having 70S ribosome.

Question 7:

Which color of light excites a natural GFP to emit green fluorescence?

Blue
Green
Infrared
Red

Correct Answer: (1) Blue

View Solution

Explanation:

Green fluorescent protein (GFP) absorbs blue light and emits green fluorescence due to its specific chromophore structure. The absorption of higher-energy light (blue) is necessary to excite the electrons within the GFP molecule so that they can release energy as green light.

Step-by-step breakdown:

Understand the principle of fluorescence: Fluorescence involves absorption of light at one wavelength, followed by emission at another wavelength.
Know the excitation properties of GFP: GFP absorbs blue light.
Identify the relationship between absorption and emission wavelengths: The emission wavelength is usually longer than the absorption wavelength (blue to green).
Conclude: Blue light excites GFP to emit green fluorescence.

Question 8:

Which one of the following hormones promotes fruit ripening?

Abscisic acid
Auxin
Ethylene
Gibberellin

Correct Answer: (3) Ethylene

View Solution

Explanation:

Ethylene is a plant hormone that promotes fruit ripening by triggering enzymes involved in cell wall degradation, starch-to-sugar conversion, and pigment synthesis. Other hormones like abscisic acid, auxin, and gibberellin have other functions.

Step-by-step breakdown:

Understand the role of plant hormones: Different plant hormones have specific functions in plant development.
Identify the functions of the given hormones:
- Abscisic acid: Involved in stress response and dormancy.
- Auxin: Involved in cell elongation and growth.
- Ethylene: Promotes fruit ripening.
- Gibberellin: Involved in stem elongation and seed germination.
Connect hormones with fruit ripening: Ethylene is known for its role in fruit ripening.
Conclude: Ethylene is the hormone that promotes fruit ripening.

Question 9:

Which one of the following has a catalytic RNA?

Ribonuclease H
Ribozyme
RNA polymerase I
RNA polymerase II

Correct Answer: (2) Ribozyme

View Solution

Explanation:

Ribozymes are RNA molecules with catalytic activity, capable of cleaving RNA and participating in RNA splicing. They are distinct from proteins with enzymatic activity. Other enzymes listed are proteinaceous.

Step-by-step breakdown:

Understand catalytic RNA: Catalytic RNA refers to RNA molecules that possess enzymatic activity.
Identify the nature of each option:
- Ribonuclease H: A protein enzyme that degrades RNA in DNA/RNA hybrids.
- Ribozyme: An RNA molecule that acts as an enzyme.
- RNA polymerase I: A protein enzyme involved in transcription.
- RNA polymerase II: A protein enzyme involved in transcription.
Connect catalytic function to one of the options: Ribozyme has catalytic activity because it's an RNA enzyme.
Conclude: Ribozyme is the correct answer because it's an RNA molecule with a catalytic activity.

Question 10:

The number of significant figures in a reported measurement of 0.00361 is:

Correct Answer: (2) 4

View Solution

Explanation:

Significant figures are all the digits that are known accurately, plus one uncertain digit. In 0.00361, the first three zeros are placeholders and do not count as significant figures.

Step-by-step breakdown:

Understand rules for significant figures: Leading zeros in a decimal are not significant, while non-zero numbers are.
Apply the rules to the given number:
- 0.00: These leading zeros are placeholders and not significant.
- 3: A non-zero number and is significant.
- 6: A non-zero number and is significant.
- 1: A non-zero number and is significant.
Count the significant figures: The total number of significant digits is 3.
Conclude: The number of significant figures is 3.

Question 11:

Match the terminology in Group I with the stimulus in Group II that generates growth response of plants.

Group I: P. Gravitropism Q. Phototropism R. Thigmotropism S. Chemotropism

Group II: 1. Light 2. Touch 3. Chemical 4. Gravity

P – 3, Q – 4, R – 2, S – 1
P – 2, Q – 1, R – 3, S – 4
P – 4, Q – 1, R – 2, S – 3
P – 4, Q – 2, R – 1, S – 3

Correct Answer: (3) P – 4, Q – 1, R – 2, S – 3

View Solution

Explanation:

Gravitropism refers to growth towards or away from gravity. Phototropism involves a response to light. Thigmotropism is a response to touch, seen in climbing plants. Chemotropism is the response to chemical stimuli, such as pollen tube growth.

Step-by-step breakdown:

Understand the definition of each term in Group I:
- Gravitropism: Directional growth response to gravity.
- Phototropism: Directional growth response to light.
- Thigmotropism: Directional growth response to touch.
- Chemotropism: Directional growth response to chemical stimuli.
Match the terms in Group I with the stimuli in Group II:
- Gravitropism (P) matches with Gravity (4).
- Phototropism (Q) matches with Light (1).
- Thigmotropism (R) matches with Touch (2).
- Chemotropism (S) matches with Chemical (3).
Conclude: The correct matching is P-4, Q-1, R-2, S-3.

Question 12:

The correct hierarchy of taxa in the Linnaean classification of eukaryotes is:

Kingdom, class, phylum, order, family, genus
Kingdom, order, class, phylum, family, genus
Kingdom, phylum, order, family, class, genus
Kingdom, phylum, class, order, family, genus

Correct Answer: (4) Kingdom, phylum, class, order, family, genus

View Solution

Explanation:

The Linnaean system classifies living organisms into hierarchical ranks: Kingdom > Phylum > Class > Order > Family > Genus > Species. This taxonomy reflects increasing specificity from Kingdom to Genus.

Step-by-step breakdown:

Understand the hierarchy of Linnaean classification: It's a system that organizes living organisms.
Recall the correct order of taxa: The correct order from broader to narrower is: Kingdom, Phylum, Class, Order, Family, Genus, Species.
Compare options with the correct order: Option 4 has the correct sequence.
Conclude: The correct order is Kingdom, phylum, class, order, family, and genus.

Question 13:

Which one of the following statements about polyploidy is correct?

Autopolyploids are derived from a single species.
Autopolyploids are derived from two different species.
Allopolyploids are derived from a single species.
Allopolyploids are not fertile when mated with each other.

Correct Answer: (1) Autopolyploids are derived from a single species.

View Solution

Explanation:

Polyploidy involves an organism having multiple sets of chromosomes. Autopolyploids originate from chromosome duplication within a single species. Allopolyploids arise from hybridization between two species.

Step-by-step breakdown:

Understand the definition of polyploidy: Polyploidy is having multiple sets of chromosomes.
Distinguish between autopolyploids and allopolyploids:
- Autopolyploids: Derived from chromosome duplication within the same species.
- Allopolyploids: Derived from hybridization between two species.
Evaluate each option: Option 1 correctly states the origin of autopolyploids.
Conclude: The correct statement is that autopolyploids are derived from a single species.

Question 14:

Which one of the following hormones is a tyrosine derivative?

Epinephrine
Estradiol
Progesterone
Testosterone

Correct Answer: (1) Epinephrine

View Solution

Explanation:

Epinephrine, also known as adrenaline, is synthesized from the amino acid tyrosine. Other options like estradiol, progesterone, and testosterone are steroid hormones derived from cholesterol.

Step-by-step breakdown:

Understand the classification of hormones: Hormones are classified based on their synthesis pathways.
Know the precursor of each hormone:
- Epinephrine: Synthesized from tyrosine.
- Estradiol: Derived from cholesterol.
- Progesterone: Derived from cholesterol.
- Testosterone: Derived from cholesterol.
Connect each hormone with its precursor: Epinephrine is synthesized from tyrosine.
Conclude: Epinephrine is the only hormone derived from tyrosine.

Question 15:

Which one of the following immunoglobulins crosses the human placenta?

Correct Answer: (3) IgG

View Solution

Explanation:

IgG is the only immunoglobulin that can cross the placenta, providing passive immunity to the fetus. Other antibodies like IgA and IgM do not cross the placenta.

Step-by-step breakdown:

Understand the role of immunoglobulins: Immunoglobulins are antibodies involved in immune responses.
Know the characteristics of the listed immunoglobulins:
- IgA: Found in mucous secretions and doesn't cross the placenta.
- IgE: Involved in allergic reactions and does not cross the placenta.
- IgG: Can cross the placenta and provide passive immunity.
- IgM: Involved in primary immune responses and does not cross the placenta.
Connect each immunoglobulin with placental crossing: IgG can cross the placenta.
Conclude: IgG is the only immunoglobulin that crosses the placenta.

Question 16:

Determine the correctness or otherwise of the following Assertion (A) and the Reason (R).

Assertion (A): The resolving power of a transmission electron microscope is higher than that of the light microscope.

Reason (R): The wavelength of electrons is shorter than that of visible light.

Both (A) and (R) are true and (R) is the correct reason for (A).
Both (A) and (R) are true but (R) is not the correct reason for (A).
Both (A) and (R) are false.
(A) is true but (R) is false.

Correct Answer: (1) Both (A) and (R) are true and (R) is the correct reason for (A).

View Solution

Explanation:

The resolving power of a microscope depends on the wavelength of the source used. Electron microscopes use electrons with wavelengths much shorter than visible light, allowing them to achieve higher resolution.

Step-by-step breakdown:

Understand the concept of resolving power: It is the ability of a microscope to distinguish between two adjacent points.
Know the principle of resolving power: Resolving power depends on the wavelength of the source; shorter wavelengths give better resolution.
Evaluate the given assertion and reason:
- Assertion (A): Correct, electron microscopes have higher resolving power.
- Reason (R): Correct, electrons have shorter wavelengths than visible light.
Connect the Assertion and Reason: The shorter wavelength of electrons is the reason for the higher resolution in the electron microscope.
Conclude: Both A and R are true, and R explains A, making option 1 correct.

Question 17:

Match the morphology in Group I with the corresponding microorganism in Group II.

Group I: P. Coccus Q. Rod R. Comma S. Spiral

Group II: 1. Treponema 2. Bacillus 3. Neisseria 4. Vibrio

P – 3, Q – 2, R – 4, S – 1
P – 4, Q – 1, R – 3, S – 2
P – 2, Q – 4, R – 1, S – 3
P – 1, Q – 2, R – 3, S – 4

Correct Answer: (1) P – 3, Q – 2, R – 4, S – 1

View Solution

Explanation:

Coccus refers to spherical bacteria, e.g., Neisseria. Rod morphology is seen in Bacillus. Comma-shaped bacteria are represented by Vibrio. Spiral morphology is characteristic of Treponema.

Step-by-step breakdown:

Understand different bacterial morphologies: Bacterial shapes are used to classify them.
Match each bacterial morphology in Group I to examples in Group II:
- Coccus (P): Spherical bacteria, e.g., Neisseria (3).
- Rod (Q): Bacillus (2).
- Comma (R): Vibrio (4).
- Spiral (S): Treponema (1).
Conclude: P matches to 3, Q to 2, R to 4, and S to 1. Therefore option 1 is correct.

Question 18:

Which one of the following genetic crosses and their results indicates cytoplasmic inheritance?

Wild-type male × mutant female → 100% progeny are mutant
Wild-type male × mutant female → 25% progeny are wild-type
Mutant male × wild-type female → 50% progeny are mutant
Mutant male × wild-type female → 75% progeny are wild-type

Correct Answer: (1) Wild-type male × mutant female → 100% progeny are mutant

View Solution

Explanation:

Cytoplasmic inheritance involves genes in organelles like mitochondria, which are inherited maternally. All offspring inherit the mother's cytoplasmic traits.

Step-by-step breakdown:

Understand cytoplasmic inheritance: It’s the inheritance of genes in cytoplasmic organelles from mother.
Identify the key characteristics of cytoplasmic inheritance:
- Traits are inherited from the maternal parent.
- All offspring of a mutant female will show the mutant phenotype.
Evaluate the given options: Option 1 represents maternal inheritance, in which offspring inherit traits from the mother.
Conclude: Option 1 shows a scenario consistent with cytoplasmic inheritance, where all offspring show the maternal trait.

Question 19:

Which of the following is NOT a characteristic morphological feature of apoptotic cells?

Disassembly of nuclear envelope
DNA fragmentation
Increased cell size
Membrane blebbing

Correct Answer: (3) Increased cell size

View Solution

Explanation:

During apoptosis, cells undergo shrinkage, not enlargement. Other features like DNA fragmentation and membrane blebbing are typical of apoptosis.

Step-by-step breakdown:

Understand the characteristics of apoptosis: Apoptosis is programmed cell death, having distinct features.
Identify the morphological features of apoptosis:
- Disassembly of nuclear envelope: Happens during apoptosis.
- DNA fragmentation: Happens during apoptosis.
- Increased cell size: Cells shrink during apoptosis, so this is not a characteristic feature.
- Membrane blebbing: Happens during apoptosis.
Find out the option that is not characteristic of apoptosis: Increased cell size is the opposite of what happens.
Conclude: Increased cell size is the feature that is NOT associated with apoptotic cells.

Question 20:

Competition between two populations in an ecosystem is:

Beneficial (+) to both the populations
Deleterious (-) to both the populations
Beneficial (+) to one population, but deleterious (-) to the other population
Beneficial (+) to one population, but no effect (0) on the other population

Correct Answer: (2) Deleterious (-) to both the populations

View Solution

Explanation:

Competition occurs when two populations vie for the same limited resource, negatively impacting both populations as resources become scarce.

Step-by-step breakdown:

Understand ecological competition: Competition happens when two or more species need the same limited resources.
Identify the impacts of competition: Competition has negative effects on both populations, due to limited resources.
Evaluate the given options: Option 2 reflects that competition is a negative relationship for both populations.
Conclude: Competition is deleterious to both populations, making option 2 correct.

Question 21:

Adenine constitutes 0.16 mole fraction in a given single-stranded DNA. What is the mole fraction of uracil in the resultant RNA, if this entire DNA fragment is transcribed?

0.16
0.32
0.34
0.68

Correct Answer: (1) 0.16

View Solution

Explanation:

In transcription, adenine in the DNA template pairs with uracil in RNA. Since adenine's mole fraction in DNA is 0.16, uracil's mole fraction in RNA will also be 0.16.

Step-by-step breakdown:

Understand the base pairing rules in transcription: Adenine in DNA pairs with uracil in RNA.
Recognize the relationship between the mole fraction of pairing bases: The mole fraction of adenine in DNA will be the same as the mole fraction of uracil in RNA transcribed from it.
Conclude: Because the mole fraction of adenine is 0.16 in DNA, the mole fraction of uracil in RNA will be 0.16.

Question 22:

Which one of the following is NOT used as a component in subunit vaccines?

Capsular polysaccharide
Inactivated exotoxin
Inactivated virus
Viral glycoprotein

Correct Answer: (3) Inactivated virus

View Solution

Explanation:

Subunit vaccines include only specific parts of the pathogen, such as capsular polysaccharides, inactivated exotoxins, or viral glycoproteins, and not the entire inactivated virus. Using only specific parts of the pathogen minimizes risks by excluding infectious components.

Step-by-step breakdown:

Understand subunit vaccines: These contain only specific components of the pathogen, not whole pathogens.
Identify the components of subunit vaccines:
- Capsular polysaccharide: Specific part of bacteria.
- Inactivated exotoxin: Modified toxin from bacteria.
- Inactivated virus: Whole virus, thus not a component of subunit vaccine.
- Viral glycoprotein: Specific proteins from a virus.
Evaluate each option for being part of subunit vaccine: Inactivated virus is a complete component and not a part of subunit vaccine.
Conclude: An inactivated virus is not a component of a subunit vaccine.

Question 23:

Metabolic acidosis is associated with decreased plasma level of:

Bicarbonate
Lactate
Oxygen
Urea

Correct Answer: (1) Bicarbonate

View Solution

Explanation:

Metabolic acidosis is caused by the accumulation of acids or loss of bicarbonate, leading to reduced plasma bicarbonate levels and a drop in blood pH. Other options may be elevated.

Step-by-step breakdown:

Understand metabolic acidosis: It’s a condition caused by an increase in acid or loss of base in the body.
Identify the changes during metabolic acidosis: A drop in blood bicarbonate is characteristic of metabolic acidosis.
Conclude: Metabolic acidosis is associated with a reduced plasma level of bicarbonate.

Question 24:

Genes in two species that are derived from the same ancestral gene in their most recent common ancestor are called:

Analogs
Heterologs
Orthologs
Paralogs

Correct Answer: (3) Orthologs

View Solution

Explanation:

Orthologs are genes in different species that evolved from a common ancestral gene. They retain the same function across species, unlike paralogs, which arise by duplication within the same genome. Analogs and heterologs are different classifications for genes.

Step-by-step breakdown:

Understand the relationship between genes: Genes can have different relationships based on evolutionary origins.
Define each type of gene:
- Analogs: Have similar functions but different evolutionary origins.
- Heterologs: Genes transferred between species.
- Orthologs: Genes from different species derived from the same ancestral gene.
- Paralogs: Genes that arise by duplication within the same genome.
Connect the given description with gene classification: The question is describing orthologs, which are from the same ancestral gene.
Conclude: Genes derived from the same ancestral gene in different species are called orthologs.

Question 25:

An object is placed 15 cm in front of a convex mirror, which has a radius of curvature 30 cm. Which one of the following is true of the image formed?

Real and inverted
Real and upright
Virtual and inverted
Virtual and upright

Correct Answer: (4) Virtual and upright

View Solution

Explanation:

Convex mirrors always produce virtual, upright, and diminished images regardless of the object's position. Using the mirror equation confirms this result. The image is not real or inverted.

Step-by-step breakdown:

Understand the properties of a convex mirror: Convex mirrors are diverging mirrors.
Recall the image properties of convex mirrors: They always produce virtual, upright, and diminished images.
Conclude: The image formed will be virtual and upright.

Question 26:

If a variable z shows a standard normal distribution, then the percent probability that 0 < z² < 1 is:

Correct Answer: (2) 68

View Solution

Explanation:

The standard normal distribution has 68% of its probability within one standard deviation of the mean. Since z² is bounded between 0 and 1, the percentage probability is 68%. The given range signifies the region within 1 standard deviation from the mean.

Step-by-step breakdown:

Understand standard normal distribution: It is a bell-shaped distribution of data.
Recall the empirical rule: 68% of data is within 1 standard deviation from the mean.
Relate the z value to the empirical rule: z² being bounded between 0 and 1 implies z is between -1 and 1.
Conclude: The probability is within 1 standard deviation of the mean and thus is 68%.

Question 27:

In the chick embryo, the ectoderm generates:

Alveolar cells
Germ cells
Neurons
Red blood cells

Correct Answer: (3) Neurons

View Solution

Explanation:

The ectoderm layer gives rise to the nervous system, including neurons, and also forms the epidermis of the skin. Other options are from mesoderm or endoderm.

Step-by-step breakdown:

Understand the germ layers of an embryo: Germ layers are the foundation for all the different tissues of the body.
Identify the derivatives of the ectoderm: The ectoderm produces tissues like the nervous system and skin.
Connect the given options to the germ layers: Neurons are derived from ectoderm.
Conclude: Ectoderm generates neurons.

Question 28:

The boiling points of Iodomethane, Dibromomethane, Bromomethane, Chloromethane follow the order:

Bromomethane > Dibromomethane > Iodomethane > Chloromethane
Bromomethane > Iodomethane > Chloromethane > Dibromomethane
Dibromomethane > Iodomethane > Bromomethane > Chloromethane
Iodomethane > Bromomethane > Chloromethane > Dibromomethane

Correct Answer: (3) Dibromomethane > Iodomethane > Bromomethane > Chloromethane

View Solution

Explanation:

Boiling points depend on molecular weight and intermolecular forces. Dibromomethane has the highest boiling point due to its greater molecular weight and polarity compared to the others. Molecular weight increases down the periodic table, and hence halogens increase boiling point with increase in size. More polar bonds also cause a higher boiling point due to stronger intermolecular forces.

Step-by-step breakdown:

Understand factors affecting boiling points: These include molecular weight, intermolecular forces, polarity.
Compare the given molecules:
- Dibromomethane: Higher molecular weight and polar, so highest boiling point.
- Iodomethane: Higher molecular weight, so higher boiling point than Bromomethane.
- Bromomethane: Intermediate molecular weight, intermediate boiling point.
- Chloromethane: Lowest molecular weight, so lowest boiling point.
Arrange the options by boiling points: Dibromomethane has the highest boiling point followed by Iodomethane, Bromomethane, and then Chloromethane.
Conclude: The correct order is Dibromomethane > Iodomethane > Bromomethane > Chloromethane.

Question 29:

Chromosome duplication during the cell cycle occurs in:

G1 phase
G2 phase
M phase
S phase

Correct Answer: (4) S phase

View Solution

Explanation:

DNA replication occurs in the S (synthesis) phase of the cell cycle, ensuring that each daughter cell receives an identical set of chromosomes. Other phases are for different functions like cell growth, checking for errors, or mitosis.

Step-by-step breakdown:

Understand the cell cycle: Different phases of cell cycle have specific functions.
Know the function of different cell cycle phases:
- G1 phase: Cell growth.
- G2 phase: Preparation for mitosis.
- M phase: Mitosis (cell division).
- S phase: DNA replication or chromosome duplication.
Connect chromosomal duplication with specific cell cycle phase: DNA replication occurs in the S phase.
Conclude: S phase is where chromosome duplication happens.

Question 30:

Ionic character of the covalent bonds in the compounds Cl₂, HCl, NaCl, NaF follows the order:

Cl₂ > NaCl > HCl > NaF
HCl > Cl₂ > NaF > NaCl
HCl > NaCl > NaF > Cl₂
NaF > NaCl > HCl > Cl₂

Correct Answer: (4) NaF > NaCl > HCl > Cl₂

View Solution

Explanation:

Ionic character depends on the electronegativity difference between bonded atoms. NaF has the highest ionic character due to the large difference between sodium and fluorine. Electronegativity difference increases with difference in group number.

Step-by-step breakdown:

Understand electronegativity: It’s the ability of an atom to attract electrons towards itself.
Relate electronegativity difference to ionic character: The greater the electronegativity difference, the more ionic the bond.
Compare the electronegativity differences in the compounds:
- NaF: Largest electronegativity difference due to Na and F.
- NaCl: Intermediate electronegativity difference due to Na and Cl.
- HCl: Smaller electronegativity difference due to H and Cl.
- Cl₂: No electronegativity difference since the atoms are the same, so not ionic.
Arrange them according to increasing ionic character: The order of ionic character is NaF > NaCl > HCl > Cl₂
Conclude: NaF has the highest ionic character, while Cl₂ has the lowest.

Question 31:

Which of the following is/are lateral meristems?

Cork cambium
Procambium
Protoderm
Vascular cambium

Correct Answer: (1) and (4) Cork cambium and Vascular cambium

View Solution

Explanation:

Lateral meristems, such as cork cambium and vascular cambium, are responsible for secondary growth in plants, increasing the girth. Procambium and protoderm are primary meristems involved in elongation.

Step-by-step breakdown:

Understand types of meristems: Meristems are plant tissues where cell division occurs.
Differentiate between lateral and primary meristems:
- Lateral meristems: For secondary growth (girth).
- Primary meristems: For primary growth (elongation).
Identify given options:
- Cork cambium: Lateral meristem.
- Procambium: Primary meristem.
- Protoderm: Primary meristem.
- Vascular cambium: Lateral meristem.
Conclude: The correct options that are lateral meristems are cork cambium and vascular cambium.

Question 32:

Which of the following statement(s) about Golden Rice is/are correct?

Consumption of it increases vitamin A levels
Consumption of it increases vitamin D levels
Consumption of it increases vitamin K levels
It is a transgenic crop containing β-carotene

Correct Answer: (1) and (4) Consumption of it increases vitamin A levels and It is a transgenic crop containing β-carotene

View Solution

Explanation:

Golden Rice is genetically engineered to contain β-carotene, a precursor to vitamin A, addressing vitamin A deficiency. It does not affect vitamin D or K levels.

Step-by-step breakdown:

Understand the nature of golden rice: It is a genetically modified rice variety.
Know the key characteristics of Golden Rice:
- It contains β-carotene.
- It increases vitamin A levels.
- It does not affect vitamin D or K levels.
Evaluate given statements: Statement 1 and 4 correctly describes the characteristics of Golden Rice.
Conclude: The correct statements about Golden Rice are that its consumption increases vitamin A levels and it is a transgenic crop containing β-carotene.

Question 33:

Which of the following statement(s) about eukaryotic DNA topoisomerase is/are correct?

Topoisomerase I creates transient single-strand breaks
Topoisomerase I creates transient double-strand breaks
Topoisomerase II creates transient single-strand breaks
Topoisomerase II creates transient double-strand breaks

Correct Answer: (1) and (4) Topoisomerase I creates transient single-strand breaks and Topoisomerase II creates transient double-strand breaks

View Solution

Explanation:

Topoisomerase I relaxes supercoils by creating single-strand breaks, while Topoisomerase II creates double-strand breaks to untangle DNA during replication. Each has a distinct mechanism of action.

Step-by-step breakdown:

Understand DNA topoisomerases: They’re enzymes that regulate DNA topology.
Identify the activities of topoisomerase I and II:
- Topoisomerase I: Creates transient single-strand breaks.
- Topoisomerase II: Creates transient double-strand breaks.
Evaluate given statements: Statement 1 and 4 accurately describes the function of the topoisomerases.
Conclude: The correct statements are that Topoisomerase I creates transient single-strand breaks and Topoisomerase II creates transient double-strand breaks.

Question 34:

Which of the following method(s) is/are used to estimate protein concentration?

Anthrone
Biuret
Bradford
Lowry

Correct Answer: (2), (3) and (4) Biuret, Bradford, and Lowry

View Solution

Explanation:

The Biuret, Bradford, and Lowry methods involve colorimetric techniques for estimating protein concentrations. Anthrone is used for carbohydrate estimation. Each of these uses different chemical properties of proteins.

Step-by-step breakdown:

Understand methods for protein quantification: These methods quantify the concentration of protein in a sample.
Know the principle of each method:
- Anthrone: Used for carbohydrate quantification.
- Biuret: Measures peptide bonds.
- Bradford: Measures protein content using Coomassie dye.
- Lowry: Measures tyrosine/tryptophan residues.
Identify which options are for protein estimation: Biuret, Bradford, and Lowry are for protein quantification.
Conclude: Biuret, Bradford and Lowry methods are used to estimate protein concentration.

Question 35:

Which of the following is/are example(s) of a lotic ecosystem?

Lake
Pond
River
Stream

Correct Answer: (3) and (4) River and Stream

View Solution

Explanation:

Lotic ecosystems involve flowing water, such as rivers and streams. Lakes and ponds are lentic (standing water) ecosystems.

Step-by-step breakdown:

Understand lotic and lentic ecosystems: Lotic refers to flowing water and lentic refers to still water.
Categorize the given options:
- Lake: Lentic (still water).
- Pond: Lentic (still water).
- River: Lotic (flowing water).
- Stream: Lotic (flowing water).
Connect the given options with lotic or lentic: Rivers and streams are examples of lotic systems.
Conclude: Rivers and streams are examples of lotic ecosystems.

Question 36:

Which of the following statement(s) about the effect of genetic drift is/are correct?

It can cause changes in the frequency of alleles at random
It is a mechanism of evolution
It can lead to loss of genetic variation within small populations
It is significant in large populations

Correct Answer: (1), (2) and (3) It can cause changes in the frequency of alleles at random, It is a mechanism of evolution, and It can lead to loss of genetic variation within small populations

View Solution

Explanation:

Genetic drift is a random process that can cause unpredictable changes in allele frequencies, particularly in small populations. It is one of the mechanisms of evolution and can lead to a loss of genetic variation over time. Its effects are not significant in large populations due to the stabilizing impact of larger gene pools.

Step-by-step breakdown:

Understand genetic drift: It’s a random change in allele frequencies in populations.
Identify the features of genetic drift:
- Random change in allele frequencies.
- Mechanism of evolution.
- Can cause loss of genetic variation in small populations.
- Not significant in large populations.
Evaluate the given statements: Statements 1, 2 and 3 are the true characteristics of genetic drift.
Conclude: Genetic drift can cause random changes in allele frequency, is a mechanism of evolution, and can lead to a loss of genetic variation in small populations.

Question 37:

Which of the following technique(s) can be used to determine the three-dimensional structure of an organic compound?

Mass spectrometry
NMR spectroscopy
UV-visible spectroscopy
X-ray crystallography

Correct Answer: (2) and (4) NMR spectroscopy and X-ray crystallography

View Solution

Explanation:

NMR spectroscopy and X-ray crystallography are widely used for determining the three-dimensional structure of organic compounds. NMR spectroscopy provides information about molecular structure through interactions of nuclear spins, while X-ray crystallography directly visualizes the molecular arrangement using crystal diffraction data. Mass spectrometry is used to determine molecular weight not structure.

Step-by-step breakdown:

Understand structure determination methods: Various techniques provide information about molecular structures.
Know the purpose of the given techniques:
- Mass spectrometry: Determines molecular weight.
- NMR spectroscopy: Determines 3D molecular structure using nuclear spin interactions.
- UV-visible spectroscopy: Measures electronic transitions.
- X-ray crystallography: Directly visualizes molecular structure using crystal diffraction.
Identify methods that determine 3D structure: NMR and X-ray crystallography are used for determining 3D structure.
Conclude: NMR and X-ray crystallography can determine 3D structure of an organic compound.

Question 38:

Which of the following entity(ies) is/are found inside the intact nucleus of eukaryotic cells?

Centrosome
Lysosome
Nucleolus
Nucleosome

Correct Answer: (3) and (4) Nucleolus and Nucleosome

View Solution

Explanation:

Within the nucleus of eukaryotic cells, the nucleolus is responsible for ribosomal RNA synthesis and assembly, while nucleosomes are the fundamental units of chromatin, consisting of DNA wrapped around histone proteins. Centrosomes and lysosomes are located outside the nucleus. Understanding their location is key to understanding cellular function.

Step-by-step breakdown:

Understand cell structures: Organelles are structures inside eukaryotic cells with specific functions.
Identify the location of the given organelles/entities:
- Centrosome: Found outside the nucleus.
- Lysosome: Found outside the nucleus.
- Nucleolus: Found inside the nucleus.
- Nucleosome: Found inside the nucleus.
Connect the structure with its location within the cell: Nucleolus and nucleosomes are located in nucleus.
Conclude: Nucleolus and nucleosomes are found inside the nucleus.

Question 39:

Which of the following is/are trace element(s)?

Correct Answer: (1) and (4) Mn and Zn

View Solution

Explanation:

Trace elements such as manganese (Mn) and zinc (Zn) are required in small quantities for enzymatic and metabolic activities. Phosphorus (P) and sulfur (S) are macronutrients needed in larger amounts for cellular functions. Trace elements are needed in lower amounts.

Step-by-step breakdown:

Understand the role of minerals: Minerals play various roles in biological systems.
Know the classification of minerals:
- Trace elements: Needed in very small amounts.
- Macronutrients: Needed in larger amounts.
Identify the given minerals:
- Mn: Trace element.
- P: Macronutrient.
- S: Macronutrient.
- Zn: Trace element.
Conclude: Mn and Zn are trace elements.

Question 40:

Which of the following is/are true about Retrovirus?

It contains double-stranded RNA genome
It can cause cancer
It contains reverse transcriptase
It contains double-stranded DNA genome

Correct Answer: (2) and (3) It can cause cancer and It contains reverse transcriptase

View Solution

Explanation:

Retroviruses, such as HIV, possess reverse transcriptase, which converts RNA into DNA during infection. They can integrate into the host genome, sometimes leading to oncogenesis. They do not have a double-stranded RNA or DNA genome; their genome is single-stranded RNA.

Step-by-step breakdown:

Understand the nature of retroviruses: Retroviruses have unique features and replication strategies.
Know key characteristics of retroviruses:
- Contain reverse transcriptase.
- Have a single stranded RNA genome.
- Can cause cancer.
Identify correct properties of retroviruses: Option 2 and 3 accurately describe retroviruses.
Conclude: Retroviruses can cause cancer and contain reverse transcriptase.

Question 41:

A wooden plant accumulates 10 mg/kg of ¹⁴C during its life span. A fossil of this plant was discovered and contains 2.5 mg/kg of ¹⁴C. The age of this fossil at the time of discovery is (use 5730 years as the half-life of ¹⁴C):

Correct Answer: 11460 to 11460 years

View Solution

Explanation:

Using the half-life formula:

t = (ln(N₀/N) / ln(2)) * T_1/2

where N₀ = 10, N = 2.5, and T_1/2 = 5730 years. Substituting the values:

t = (ln(10/2.5) / ln(2)) * 5730 = 11460 years.

Step-by-step breakdown:

Understand radiocarbon dating: It’s a method to determine the age of organic material.
Identify the initial amount (N₀) and current amount (N) of ¹⁴C: N₀ = 10 mg/kg and N = 2.5 mg/kg.
Recall the half-life (T_1/2) of ¹⁴C: T_1/2 = 5730 years.
Use the half-life formula: t = (ln(N₀/N) / ln(2)) * T_1/2.
Plug values into the formula and calculate: t = (ln(10/2.5) / ln(2)) * 5730 = 11460 years.
Conclude: The age of the fossil is 11460 years.

Question 42:

A cylinder contains 50 L of an ideal gas at a pressure of 50 atm. Assuming that the temperature remains unchanged, the volume of the gas at 1 atm is:

Correct Answer: 2500 L

View Solution

Explanation:

Using Boyle's law: P₁V₁ = P₂V₂

Substitute P₁ = 50 atm, V₁ = 50 L, and P₂ = 1 atm:

V₂ = (P₁V₁) / P₂ = (50 * 50) / 1 = 2500 L.

Step-by-step breakdown:

Understand Boyle's Law: It states that at constant temperature, pressure and volume are inversely related for a fixed amount of gas.
Recall Boyle's Law formula: P₁V₁ = P₂V₂.
Identify the given parameters: P₁ = 50 atm, V₁ = 50 L, P₂ = 1 atm.
Rearrange the formula to solve for V₂: V₂ = (P₁V₁) / P₂.
Plug the values into the formula and calculate: V₂ = (50 * 50) / 1 = 2500 L.
Conclude: The volume of the gas at 1 atm is 2500 L.

Question 43:

One molecule of the protein myoglobin contains one atom of iron. A myoglobin sample was found to contain 0.34% iron. The molecular weight of myoglobin is (use atomic mass of iron = 55.9 g/mol):

Correct Answer: 16440 to 16445 g/mol

View Solution

Explanation:

The molecular weight of myoglobin can be calculated as:

Molecular weight = (Mass of Fe * 100) / % of Fe = (55.9 * 100) / 0.34 = 16441.18 g/mol.

Step-by-step breakdown:

Understand the relationship between molecular weight and percentage composition: The percentage composition is helpful to estimate molecular weight.
Recall the formula for molecular weight based on composition: Molecular weight = (Mass of Fe * 100) / % of Fe.
Identify the given values: Atomic mass of iron = 55.9 g/mol and the percentage of iron = 0.34%.
Plug in the values into the formula and calculate: Molecular weight = (55.9 * 100) / 0.34 = 16441.18 g/mol.
Conclude: The molecular weight of myoglobin is approximately 16441 g/mol.

Question 44:

The wavelength of visible light for the green color is 600 nm. The energy of photons of this color is (Planck's constant h = 6.63 × 10^-34 J.s, 1 eV = 1.6 × 10^-19 J, speed of light c = 3 × 10⁸ m/s):

Correct Answer: 2.0 to 2.2 eV

View Solution

Explanation:

Using the energy formula: E = h * c / λ

Substitute h = 6.63 × 10^-34, c = 3 × 10⁸, and λ = 600 nm = 600 × 10^-9 m:

E = (6.63 × 10^-34 * 3 × 10⁸) / (600 × 10^-9) = 3.315 × 10^-19 J.

Convert to electronvolts: E = (3.315 × 10^-19) / (1.6 × 10^-19) = 2.07 eV.

Step-by-step breakdown:

Understand the relationship between energy, wavelength, and frequency: The energy of a photon is related to its wavelength and frequency.
Recall the formula for energy: E = h * c / λ where h= Planck’s constant, c= speed of light and λ= wavelength.
Convert the wavelength from nm to m: 600 nm = 600 * 10^-9 m.
Substitute the values and calculate: E = (6.63 × 10^-34 * 3 × 10⁸) / (600 × 10^-9) = 3.315 × 10^-19 J.
Convert joules to electronvolts: E = (3.315 × 10^-19) / (1.6 × 10^-19) = 2.07 eV.
Conclude: The energy of the photons is approximately 2.07 eV.

Question 45:

A ball dropped from a bridge hits the surface of the water in 3 s. The height of the bridge, ignoring air resistance, is (use g = 9.8 m/s²):

Correct Answer: 44.1 to 44.1 m

View Solution

Explanation:

Using the equation of motion: s = (1/2) * g * t²

Substitute g = 9.8 m/s² and t = 3 s:

s = (1/2) * 9.8 * (3)² = 44.1 m.

Step-by-step breakdown:

Understand the equation for free fall: The motion under gravity can be calculated using the equation.
Recall the equation of motion for distance during free fall: s = (1/2) * g * t².
Identify the given values: g = 9.8 m/s² and t = 3 s.
Substitute the given values into the formula and calculate: s = (1/2) * 9.8 * (3)² = 44.1 m.
Conclude: The height of the bridge is 44.1 m.

Question 46:

For a given square, if the area of its incircle is 100 cm², then the area of its circumcircle is:

Correct Answer: 200 cm²

View Solution

Explanation:

The radius of the incircle is given by: r = √(Area of incircle / π) = √(100/π)

The radius of the circumcircle is √2 times the radius of the incircle: R = √2 * r = √2 * √(100/π)

The area of the circumcircle is: Area = πR² = π * (√2 * √(100/π))² = 200 cm²

Step-by-step breakdown:

Understand the concept of incircle and circumcircle: Incircle is inside and circumcircle outside of the square.
Find the radius of the incircle: r = √(Area of incircle / π) = √(100/π).
Know the relationship between radii of incircle and circumcircle: Radius of the circumcircle = √2 times the radius of incircle. So, R = √2*r = √2 * √(100/π)
Calculate area of the circumcircle: Area = πR² = π * (√2 * √(100/π))² = 200 cm².
Conclude: The area of the circumcircle is 200 cm².

Question 47:

The number of peaks in the ¹H NMR spectrum of methoxymethane (CH₃OCH₃) is:

Correct Answer: 1

View Solution

Explanation:

Methoxymethane (dimethyl ether) has two equivalent methyl groups (CH₃) attached to the same oxygen atom. Since all hydrogen atoms in the molecule are chemically equivalent, the ¹H NMR spectrum shows only one peak.

Step-by-step breakdown:

Understand NMR spectroscopy: It’s a technique used to determine the structure of organic compounds based on their nuclei.
Analyze the structure of methoxymethane: It has two identical methyl groups.
Identify equivalent protons: The protons of equivalent methyl groups appear as one signal in the ¹H NMR spectrum.
Conclude: There will be 1 peak in the ¹H NMR spectrum of methoxymethane.

Question 48:

The amount of agarose required to prepare 250 mL of 0.8% agarose gel is:

Correct Answer: 2 g

View Solution

Explanation:

The weight of agarose required is calculated as: Weight = Concentration × Volume.

Substitute 0.8% (0.8 g/100 mL) and 250 mL:

Weight = 0.8 * 250 / 100 = 2 g.

Step-by-step breakdown:

Understand percentage solution calculations: Percentage is expressed as grams per 100mL.
Recognize the relationship between concentration, volume, and amount: The amount (weight) is the product of concentration and volume.
Convert the percentage into g/mL: 0.8 % = 0.8g/100ml.
Plug values into formula and calculate: Weight = 0.8 * 250 / 100 = 2 g.
Conclude: 2 g of agarose is needed for the gel preparation.

Question 49:

Three genes x, y, and z are located on a chromosome in a linear order. If the recombination frequencies between x and y is 0.15, and between y and z is 0.10, the expected frequency of double crossovers is:

Correct Answer: 0.015

View Solution

Explanation:

The frequency of double crossovers is the product of individual recombination frequencies: Double crossover frequency = 0.15 * 0.10 = 0.015.

Step-by-step breakdown:

Understand recombination frequency: It's a measure of genetic distance between genes on a chromosome.
Recall the relationship between single and double crossovers: Double crossover frequency = product of the single crossover frequencies.
Identify recombination frequencies: The recombination frequencies between x and y is 0.15, and y and z is 0.10.
Calculate double crossover frequency: Double crossover frequency = 0.15 * 0.10 = 0.015.
Conclude: The expected frequency of double crossovers is 0.015.

Question 50:

A bacterial cell suspension contains 2 × 10⁵ cells/mL. The volume of this suspension required to obtain 1.4 × 10⁶ cells is:

Correct Answer: 7 mL

View Solution

Explanation:

The required volume is calculated using the formula: Volume = Total cells required / Cell concentration per mL

Substitute 1.4 × 10⁶ cells and 2 × 10⁵ cells/mL: Volume = 1.4 × 10⁶ / 2 × 10⁵ = 7 mL.

Step-by-step breakdown:

Understand the relationship between cell concentration and volume: Cell concentration, total number of cells, and volume are related.
Recall the formula for volume calculation: Volume = Total cells required / Cell concentration per mL.
Identify given values: Total cells = 1.4 × 10⁶, and cell concentration per mL = 2 × 10⁵ cells/mL.
Plug the values into formula and calculate: Volume = 1.4 × 10⁶ / 2 × 10⁵ = 7 mL.
Conclude: The required volume of the suspension is 7 mL.

Question 51:

The data provided in the table were obtained from the following reaction, carried out at 273 K:

A + B → C

Initial concentration of [A]	Initial concentration of [B]	Initial rate of formation of [C]
0.2 mol/L	0.2 mol/L	0.3 mol/L/s
0.4 mol/L	0.2 mol/L	0.6 mol/L/s
0.4 mol/L	0.4 mol/L	2.4 mol/L/s

The order of the reaction with respect to [A] is:

Correct Answer: 1

View Solution

Explanation:

To determine the order with respect to A, we compare the initial rates for the same concentration of [B]. The rate doubles as [A] doubles, indicating the order is 1 with respect to A.

Step-by-step breakdown:

Understand reaction order: The reaction order indicates how the rate of a reaction depends on the concentration of reactants.
Compare experimental data: Compare runs where [B] is constant and [A] changes.
Evaluate the effect of changing [A] on rate: As [A] doubles from 0.2 M to 0.4 M (while [B] remains at 0.2 M), the rate of formation of C also doubles from 0.3 mol/L/s to 0.6 mol/L/s.
Conclude: Since the rate doubles when the concentration of A doubles, the reaction is first order with respect to [A].

Question 52:

Ammonia is synthesized in the Haber process in the following reaction:

N₂(g) + 3H₂(g) → 2NH₃(g)

The temperature above which the reaction becomes spontaneous is:

Correct Answer: 463.0 K

View Solution

Explanation:

The temperature for spontaneity can be found using the Gibbs free energy equation: ΔG = ΔH – TΔS

For spontaneity, ΔG = 0. Substituting the given values ΔH = -92.2 kJ/mol and ΔS = -199 J/K:

0 = -92.2 * 10³ - T * (-199)

Solving for T: T = (92.2 * 10³) / 199 = 463.0 K.

Step-by-step breakdown:

Understand spontaneity and Gibbs free energy: A spontaneous reaction has a negative ΔG (change in Gibbs free energy) value.
Recall the Gibbs free energy formula: ΔG = ΔH – TΔS.
Set ΔG = 0 for the point at which the reaction becomes spontaneous: 0 = ΔH – TΔS.
Rearrange to solve for T: T = ΔH/ΔS.
Substitute the given values: T = (-92.2 * 10³) / (-199) = 463.0 K.
Conclude: The temperature at which the reaction becomes spontaneous is 463.0 K.

Question 53:

In the given molecule, the number of chiral centers is:

chiral centers

Correct Answer: 5

View Solution

Explanation:

A chiral center is a carbon atom bonded to four different groups. The molecule shown contains 5 such centers.

Look for carbon atoms bonded to four different groups to identify chiral centers.

Question 54:

Two resistors of 2 Ω and 4 Ω are combined in parallel. If this combination is connected to a battery of 16 V, the maximum current that can be drawn from the battery is:

Correct Answer: 12 A

View Solution

Explanation:

The total resistance R_total of the parallel combination is: 1/R_total = 1/2 + 1/4 = 3/4, so R_total = 4/3 Ω. Using Ohm's law: I = V/R_total = 16 / (4/3) = 12 A.

Step-by-step breakdown:

Understand parallel resistance: When resistors are connected in parallel, the reciprocal of total resistance is the sum of reciprocals of individual resistances.
Calculate the reciprocal of the total resistance: 1/R_total = 1/2 + 1/4 = 3/4.
Determine the total resistance: R_total = 4/3 Ω.
Recall Ohm's Law: I = V/R, where I is current, V is voltage, and R is resistance.
Substitute given values into Ohm's law: I = 16 / (4/3) = 12 A.
Conclude: The maximum current that can be drawn is 12 A.

Question 55:

A box of mass 20 kg is pulled at constant speed across a floor by a rope. The rope makes an angle of 45° with the horizontal. Assuming that friction is negligible, the work done in pulling the box by a distance of 20 m is:

Correct Answer: 2740 J

View Solution

Explanation:

The work done W is given by: W = F * d * cos(θ), where F is the force, d is the distance, and θ is the angle. The force applied is F = mg, so: W = 20 * 9.8 * 20 * cos(45°) = 2740 J.

Step-by-step breakdown:

Understand the concept of work: Work is done when a force causes a displacement.
Recall the work equation for constant force: W = F * d * cos(θ).
Identify the given values: Mass (m) = 20 kg, distance (d) = 20 m, acceleration due to gravity (g) = 9.8 m/s², and angle (θ) = 45°.
Find the applied force: Since friction is negligible and the box is moving with constant speed, the applied force F equals the gravitational force F = mg. F = 20 * 9.8 = 196 N.
Plug in the values to calculate work done: W = 196 * 20 * cos(45°) = 2740 J.
Conclude: The work done is approximately 2740 J.

Question 56:

Consider an enzyme that follows simple Michaelis-Menten kinetics, and has a K_M of 5 μM. The initial velocity of the reaction will be 10 percent of the maximum velocity at a substrate concentration of:

Correct Answer: 0.54 μM

View Solution

Explanation:

For Michaelis-Menten kinetics, the initial velocity (v₀) at 10% of Vmax is: v₀ = 0.1 * Vmax.

The formula for the reaction velocity is: v₀ = Vmax * [S] / (K_M + [S])

Setting v₀ = 0.1 * Vmax and solving for [S]: 0.1 = [S] / (K_M + [S]).

Substitute K_M = 5: 0.1 = [S] / (5 + [S]), so 0.1(5 + [S]) = [S] or 0.5 + 0.1[S] = [S], finally [S] = 0.5/0.9 = 0.54 μM.

Step-by-step breakdown:

Understand Michaelis-Menten kinetics: Describes the relationship between enzyme activity and substrate concentration.
Identify the given information: K_M = 5 μM, v₀ = 0.1 * Vmax.
Recall the Michaelis-Menten equation: v₀ = Vmax * [S] / (K_M + [S]).
Substitute v₀ = 0.1 * Vmax into equation: 0.1 * Vmax = Vmax * [S] / (K_M + [S]).
Simplify the equation and solve for [S]: 0.1 = [S] / (K_M + [S]), so 0.1(K_M + [S]) = [S]
Substitute K_M = 5 and solve for [S]: 0.1(5 + [S]) = [S], so 0.5 + 0.1[S] = [S], finally [S] = 0.5/0.9 = 0.54 μM.
Conclude: The substrate concentration is approximately 0.54 μM.

Question 57:

The value of: lim (x² - 9) / (x² - 4x + 3) as x approaches 3

Correct Answer: 3

View Solution

Explanation:

Factorize both the numerator and denominator: x² - 9 = (x - 3)(x + 3) and x² - 4x + 3 = (x - 3)(x - 1).

So, the expression becomes: (x - 3)(x + 3) / (x - 3)(x - 1)

Cancel out the (x - 3) terms: (x + 3) / (x - 1)

Substitute x = 3: (3 + 3) / (3 - 1) = 6 / 2 = 3.

Step-by-step breakdown:

Understand limits: Limits determine the behavior of a function near a certain point.
Check direct substitution: Substituting x = 3 leads to a 0/0 form.
Factorize numerator and denominator: x² - 9 = (x - 3)(x + 3) and x² - 4x + 3 = (x - 3)(x - 1)
Rewrite the expression and cancel the common factor: (x - 3)(x + 3) / (x - 3)(x - 1) = (x + 3) / (x - 1).
Substitute x = 3 into simplified expression: (3 + 3) / (3 - 1) = 6 / 2 = 3.
Conclude: The limit is equal to 3.

Question 58:

A population of 1000 plants are in Hardy-Weinberg equilibrium. Two alleles R and r determine a particular trait in this population. If the number of plants with RR genotype is 640, Rr genotype is 320, and rr genotype is 40, the frequency of the r allele (in percentage) in this population is:

Correct Answer: 20

View Solution

Explanation:

The frequencies of the genotypes are: Frequency of RR = 640/1000 = 0.64; Frequency of Rr = 320/1000 = 0.32; Frequency of rr = 40/1000 = 0.04.

The frequency of the r allele (q) can be calculated using the formula: q² = frequency of rr = 0.04.

Thus q = √0.04 = 0.2 or 20%.

Step-by-step breakdown:

Understand Hardy-Weinberg equilibrium: Describes the genetic variation in a non-evolving population.
Calculate the frequencies of each genotype:
- Frequency of RR = Number of RR / total number of population = 640/1000 = 0.64.
- Frequency of Rr = Number of Rr / total number of population = 320/1000 = 0.32.
- Frequency of rr = Number of rr / total number of population = 40/1000 = 0.04.
Relate the frequency of the homozygous recessive genotype to the allele frequency: The frequency of rr = q².
Solve for q which is frequency of the r allele: q = √(frequency of rr) = √0.04 = 0.2.
Convert to percentage: 0.2 * 100 = 20%.
Conclude: The frequency of the r allele is 20%.

Question 59:

If a fair coin is tossed two times, the probability that the first or the second toss will be heads is:

Correct Answer: 0.75

View Solution

Explanation:

The probability of getting heads on either toss is: P(first head) = 1/2, P(second head) = 1/2.

Using the formula for the probability of either event occurring: P(first or second head) = P(first head) + P(second head) – P(both heads).

P(both heads) = 1/4, so P = 1/2 + 1/2 – 1/4 = 0.75.

Step-by-step breakdown:

Understand probability for multiple events: Probability can be calculated for either/or conditions.
Identify the probability of each independent event: The probability of getting a head on one toss is 1/2.
Calculate the probability of two independent events both occuring: The probability of getting two heads is (1/2)*(1/2) = 1/4.
Use the formula for the probability of either event occurring: P(first or second head) = P(first head) + P(second head) – P(both heads)
Substitute calculated probability values into formula: 1/2 + 1/2 – 1/4 = 0.75.
Conclude: The probability that the first or the second toss will be heads is 0.75.

Question 60:

The number of bands that will be visible in the gel when exposed to UV light after complete digestion of a circular plasmid with EcoRI and XhoI is:

the number of bands

Correct Answer: 3

View Solution

Explanation:

EcoRI cuts the plasmid into two fragments (2 kb and 3 kb), and XhoI cuts it into 2 fragments (2 kb and 5 kb). The restriction map shows two EcoRI and two XhoI sites, leading to a total of three distinct bands when the plasmid is completely digested.

Multiple restriction enzymes lead to multiple bands in gel electrophoresis, corresponding to the sizes of the fragments.

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IIT JAM 2024 Biotechnology (BT) Question Paper with Solution PDFs

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1.The divergence of a 3-dimensional vector \(\frac{𝑟̂}{𝑟^3}\) (𝑟̂ is the unit radial vector) is:

4.In the x-y plane, a vector is given by \(\overrightarrow𝐹 (𝑥, 𝑦) = \frac{−𝑦𝑥̂ + 𝑥𝑦̂}{𝑥^ 2 + 𝑦^2} . \)The magnitude of the flux of \(\overrightarrow∇ ×\overrightarrow 𝐹\) , through a circular loop of radius 2, centered at the origin, is:

6.In the Taylor expansion of function, 𝐹(𝑥)=𝑒𝑥 sin 𝑥, around 𝑥 = 0, the coefficient of 𝑥5 is ______. (Rounded off to three decimal places)

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1.
The divergence of a 3-dimensional vector \(\frac{𝑟̂}{𝑟^3}\) (𝑟̂ is the unit radial vector) is:

4.
In the x-y plane, a vector is given by
\(\overrightarrow𝐹 (𝑥, 𝑦) = \frac{−𝑦𝑥̂ + 𝑥𝑦̂}{𝑥^ 2 + 𝑦^2} . \)
The magnitude of the flux of \(\overrightarrow∇ ×\overrightarrow 𝐹\) , through a circular loop of radius 2, centered at the origin, is:

6.
In the Taylor expansion of function, 𝐹(𝑥)=𝑒^𝑥 sin 𝑥, around 𝑥 = 0, the coefficient of 𝑥⁵ is ______. (Rounded off to three decimal places)