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PART B

(c) II and III

  • II. ATP formation during photosynthesis is indeed termed as photophosphorylation. During the light-dependent reactions of photosynthesis, light energy is used to create a high-energy carrier molecule (ATP) in a process called photophosphorylation.
  • III. The reduction of NADP+ to NADPH occurs during the Calvin cycle. The Calvin cycle is the light-independent phase of photosynthesis where carbon dioxide is fixed and reduced to produce glucose. During this process, NADP+ is reduced to NADPH, which provides the reducing power to convert carbon dioxide into glucose.

The correct answer is (c) II, III and IV.

Auxins control apical dominance, gibberellins promote shoot elongation, and abscisic acid enables seeds to withstand desiccation. Cytokinins, however, do not suppress the synthesis of chlorophyll; they actually promote cell division and growth.

(d) Streptomycin

  • specific mechanisms of how tetracycline and streptomycin inhibit protein synthesis:
  • 1. Tetracycline: It binds to the 30S subunit of the bacterial ribosome and prevents the attachment of the aminoacyl-tRNA to the A site of the ribosome. This blocks the addition of new amino acids to the growing peptide chain, effectively stopping protein synthesis.
  • 2. Streptomycin: It also binds to the 30S subunit of the bacterial ribosome, but it causes a misreading of the mRNA template. This leads to the incorporation of incorrect amino acids into the growing peptide chain and premature termination of protein synthesis.

(a) evolution of different species from a common ancestor

Adaptive radiation specifically refers to the process where organisms diversify rapidly from an ancestral species into a multitude of new forms, particularly when a change in the environment makes new resources available, creates new challenges, or opens new environmental niches.

(c) Di Georges syndrome

DiGeorge Syndrome, also known as 22q11.2 deletion syndrome, is a disorder caused by the deletion of a small segment of chromosome 22. This deletion results in the poor development of several body systems. The thymus gland, which is responsible for producing T cells for the immune system, is often affected. If the thymus gland is underdeveloped or absent, it can lead to a weakened immune system and increased susceptibility to infections, which is a common symptom of DiGeorge Syndrome.

The correct answer is (d) dd and DD.

In this case, the shell-coiling direction in snails is determined by the mother’s genotype. The F1 snail has a dextral coiling, which means it must have received the dextral (D) allele from its mother. However, when this F1 snail self-fertilizes, it produces only sinistral progeny, indicating that it does not carry the dextral (D) allele itself, but rather two sinistral (d) alleles. Therefore, the F1 snail’s genotype must be dd. The mother of the F1 snail must have been DD because she passed on the dextral (D) allele to the F1 snail.

(C) Presence of conjugated plasmids in tetracycline resistance cells

This is due to a process called horizontal gene transfer, specifically conjugation, where one bacterium transfers a copy of a plasmid (a small, circular piece of DNA) that carries the tetracycline resistance gene to another bacterium. The recipient bacterium can then express this gene and become resistant to tetracycline. This is a major factor in the spread of antibiotic resistance among bacteria.

(d) The addition of an inorganic phosphate to ADP by pyruvate kinase.

(d) 5′-AGC-3′

(d) Oh

(c) CXCR4.

  • HIV enters a cell through a complex process. First, the virus binds to the CD4 receptor on the surface of the cell. This binding changes the shape of the virus, allowing it to also bind to a coreceptor, which is usually either CCR5 or CXCR4. The binding to the coreceptor triggers the fusion of the virus with the cell, allowing the viral RNA to enter the cell and begin the process of infection.
  • The reason HIV specifically uses the CD4 receptor and either the CCR5 or CXCR4 coreceptors is due to the specific structure and function of these proteins. The HIV virus has evolved to recognize and bind to these specific receptors. Other receptors on the cell surface do not have the correct shape or chemical properties to allow the virus to bind and enter the cell. This specificity is also why HIV primarily infects CD4+ T cells, which have these receptors.

(c) T cell will not receive co-stimulatory signals.

(d) Bone marrow transplantation from a healthy donor with CCR 5 mutation.

The bone marrow transplantation works by replacing the patient’s own immune cells with those from a donor who has a genetic mutation in a protein called CCR5. HIV uses this protein to enter the immune cells. If the protein is mutated, HIV cannot enter the cells, thus preventing the virus from spreading. After the transplantation, the patient’s immune system is essentially rebuilt with cells that are resistant to HIV, which can lead to the virus being cleared from the body.

  • The turnover rate of an enzyme is the number of substrate molecules an enzyme can convert into product per second. In this case, the enzyme is carbonic anhydrase, the substrate is H2CO3, and the concentration of the enzyme is 10-6M.
  • Given that the enzyme is converting 0.6M of H2CO3 per second, the turnover rate is calculated by dividing the rate of conversion by the concentration of the enzyme.
  • Turnover rate = (0.6 M/sec) / (10-6 M) = 6 x 105 per sec
  • So, the correct answer is (b) 6 x 105 per sec.

(a) Arg+ His+ Trp

  • The bacterium can grow on agar supplemented with arginine (Arg), tryptophan (Trp) and histidine (His), which means it can utilize all three amino acids. However, it fails to grow on agar containing only Arg and Trp or only His and Trp. This indicates that the bacterium cannot synthesize Trp on its own and is dependent on the external supply.
  • On the other hand, it can grow if only Arg and His are present, indicating that it can synthesize these two amino acids on its own. Therefore, the genotype of the bacterium with respect to these three amino acids is Arg+ His+ Trp.

(b) II

Cooperativity in protein function is typically represented by a sigmoidal (S-shaped) curve.

(d) Multiple steps allow the same response in all cells from the same ligand.

This statement is incorrect because the response to a signal can vary depending on the type of cell and its specific collection of signal receptor proteins.

(d) Both clockwise or anticlockwise direction depending on the orientation of IS elements

  • The F plasmid integrates into the E. coli chromosome through a process called recombination. This process is facilitated by Insertion Sequences (IS elements) present on both the F plasmid and the E. coli chromosome.
  • These IS elements contain transposase genes, which code for enzymes that cut and rejoin DNA strands. The direction of integration (clockwise or anticlockwise) depends on the orientation of these IS elements. If the IS elements on the F plasmid and the E. coli chromosome align in the same direction, the integration will occur in a clockwise direction. If they align in opposite directions, the integration will occur in an anticlockwise direction.

The molecular target of Colicin E3, a ribonuclease produced by Col plasmid, is (b) 16S rRNA.

Colicin E3, a ribonuclease, targets 16S rRNA by binding to the ribosome, where the 16S rRNA is located. Once bound, it cleaves specific sites on the 16S rRNA, which disrupts protein synthesis and leads to cell death. This is how Colicin E3 exerts its antibacterial effect.

(a) Genes expressed during stationary phase

  • Sigma factors are proteins needed for the initiation of transcription in bacteria. They are part of the RNA polymerase holoenzyme that binds to specific promoter sequences on the DNA.
  • Sigma factor σ38, also known as RpoS, is known to recognize consensus sequences found in the promoters of genes that are expressed during the stationary phase of bacterial growth. This is a phase when the growth rate of the bacteria slows down due to nutrient limitation or environmental stress.
  • The σ38 factor helps the bacteria to adapt to these stressful conditions by initiating the transcription of genes that help the bacteria survive in the stationary phase. This includes genes involved in stress responses, nutrient acquisition, and other survival strategies.
  • So, when E. coli enters the stationary phase, the σ38 sigma factor becomes more active and binds to the RNA polymerase, guiding it to the promoters of the stationary phase genes and initiating their transcription.
  • Sigma factors are highly specific and each one recognizes different consensus sequences in the promoter regions of genes. This specificity allows the bacteria to precisely control which genes are transcribed under different conditions.
  • Sigma factor σ70 is the primary sigma factor in E. coli and is responsible for the transcription of most genes during the exponential growth phase. However, when the bacteria enter the stationary phase or face certain stress conditions, alternative sigma factors like σ38 (RpoS) become active.
  • Sigma factor σ38 does not initiate transcription of genes involved in heat shock response, flagellar synthesis, or response to misfolded proteins in the periplasm because these genes have different consensus sequences in their promoters that are recognized by other specific sigma factors. For example, heat shock genes are recognized by σ32, and flagellar synthesis genes are recognized by σ28.

(c) have different functions under similar mechanism of regulation.

  • Alternative splicing allows a single gene to code for multiple proteins. This happens when different combinations of exons (coding regions of the gene) are included in the final mRNA transcript during the process of RNA splicing.
  • Even though the proteins produced through alternative splicing come from the same gene, they can have different functions. This is because the different combinations of exons can result in proteins with different structures and, therefore, different functions.
  • The regulation of alternative splicing is typically similar for the different proteins produced from the same gene. This is because the same cellular machinery and regulatory elements are involved in the splicing process for that gene. So, while the proteins may have different functions, the mechanism of regulation (how the splicing process is controlled) is similar.

(c) carbamoyl phosphate synthetase

  • Carbamoyl phosphate synthetase is not involved in the normal pathway for the breakdown of amino acids in the liver because it is primarily involved in the urea cycle and the synthesis of pyrimidines, not in the breakdown of amino acids.
  • The other enzymes listed, aminotransferases, glutamate dehydrogenase, and alpha-ketoglutarate dehydrogenase, are all involved in the process of amino acid catabolism.
  • Aminotransferases transfer an amino group from an amino acid to a keto acid. Glutamate dehydrogenase converts glutamate into alpha-ketoglutarate, which can then enter the citric acid cycle. Alpha-ketoglutarate dehydrogenase is an enzyme in the citric acid cycle that converts alpha-ketoglutarate into succinyl-CoA.

(c) uncoupling of the electron transport chain from ATP synthase

  • Thermogenin, also known as uncoupling protein 1 (UCP1), works by creating a pathway for protons (H+) to cross the inner mitochondrial membrane without going through ATP synthase. Normally, the electron transport chain creates a proton gradient across the inner mitochondrial membrane, and these protons drive ATP synthase to produce ATP.
  • However, when thermogenin is active, it allows protons to bypass ATP synthase and return to the mitochondrial matrix directly. This process releases the energy that would have been used for ATP synthesis as heat instead. This is known as uncoupling because it separates the electron transport chain from ATP synthesis.

(c) binding of the CAP-cAMP complex to the activator site and facilitating interaction of RNA-polymerase to the lac promoter.

  • The Catabolite Activator Protein (CAP) is a regulatory protein in bacteria that assists in the transcription of genes in response to the presence or absence of certain nutrients. When glucose levels are low, cyclic AMP (cAMP) levels in the cell increase.
  • The cAMP binds to CAP, forming the CAP-cAMP complex. This complex then binds to the activator site on the DNA, near the lac promoter. The binding of the CAP-cAMP complex to the activator site changes the DNA structure in such a way that it facilitates the binding of RNA polymerase to the lac promoter. This, in turn, initiates the transcription of the lac operon, which contains genes necessary for the metabolism of lactose.

Low activity of the acyltransferase enzyme would result in the accumulation of glycerol-3-phosphate (c). This is because acyltransferase is responsible for adding fatty acids to glycerol-3-phosphate in the synthesis of fats and phospholipids. If this enzyme is not active, the fatty acids cannot be added, and glycerol-3-phosphate would accumulate.

The other intermediates listed – diacylglycerol phosphate (a), triglyceride (b), and phosphatidyl serine (d) – are further along in the synthesis pathway of fats and phospholipids. They are formed after the action of acyltransferase on glycerol-3-phosphate. If acyltransferase activity is low, the reaction it catalyzes would not occur efficiently, and the intermediates that are formed after this step would not accumulate. Instead, the substrate for the enzyme, glycerol-3-phosphate, would accumulate because it is not being converted into the next intermediate in the pathway.

The “UAU” codon will encode for the amino acid Tyrosine. This is determined by the fact that the first synthetic mRNA sequence, which only contains “AUA” and “UAU” codons, produces a polypeptide with Isoleucine and Tyrosine. Since “AUA” is known to code for Isoleucine, “UAU” must code for Tyrosine.

(d) IV

(d) assist error-prone DNA polymerase to synthesize across DNA lesions

  • RecA protein does not assist error-prone DNA polymerase to synthesize across DNA lesions because that’s not its function. RecA is primarily involved in the process of DNA repair and recombination. It binds to single-stranded DNA and facilitates the search for homologous sequences in double-stranded DNA, which is a crucial step in recombinational DNA repair. The process of synthesizing across DNA lesions, also known as translesion synthesis, is typically performed by specialized DNA polymerases, not by RecA.
  • (a) RecA stimulates the autocleavage of LexA, a repressor protein. This autocleavage leads to the derepression of the SOS response genes, which are involved in DNA repair.
  • (b) RecA plays a crucial role in the recombinational repair of double-strand breaks. It binds to single-stranded DNA and facilitates the search for homologous sequences in double-stranded DNA, which is a crucial step in recombinational DNA repair.
  • (c) RecA also protects the free single-stranded 5′ end generated by RecBCD from exonucleolytic degradation. This protection is important for the proper functioning of the recombinational repair process.

(b) in the presence of low level of tryptophan, expression would increase

  • The Shine-Dalgarno sequence is a ribosomal binding site in bacterial and archaeal messenger RNA, generally located around 8 bases upstream of the start codon AUG. The sequence helps recruit the ribosome to the mRNA to initiate protein synthesis by aligning the ribosome with the start codon.
  • In the case of the trp operon, the Shine-Dalgarno sequence is part of the leader sequence of the mRNA. If a mutation inactivates the Shine-Dalgarno sequence, the ribosome cannot bind as efficiently, and protein synthesis is reduced.
  • However, when tryptophan levels are low, the trp operon is normally upregulated to produce more tryptophan. If the Shine-Dalgarno sequence is inactivated, this upregulation still occurs, but it is less efficient. So, the expression of the trp operon would increase, but not as much as it would without the mutation.

The methyl donor in the reaction leading from dUMP to dTMP is (a) N5-N10-Methylenetetrahydrofolate.

  • The reaction leading from dUMP (deoxyuridine monophosphate) to dTMP (deoxythymidine monophosphate) is catalyzed by the enzyme thymidylate synthase. In this reaction, N5-N10-Methylenetetrahydrofolate donates a methyl group to dUMP, converting it to dTMP.
  • The process can be summarized as follows:
  • dUMP + N5-N10-Methylenetetrahydrofolate → dTMP + dihydrofolate
  • This reaction is crucial for DNA synthesis as dTMP is one of the building blocks of DNA.

(c) Magnesium and Iron

  • 1. Magnesium is essential because it forms the central atom of the chlorophyll molecule. The chlorophyll molecule has a structure known as a porphyrin ring, which is a large ring of carbon, nitrogen, and hydrogen atoms. In the center of this ring, a magnesium ion (Mg2+) is held.
  • 2. Iron plays a crucial role in the synthesis of chlorophyll as it is involved in the production of a precursor molecule called protoporphyrin IX. This molecule is later converted into chlorophyll with the help of enzymes.

(d) Preculture, Cocultivation, Callus formation, Shoot development, Plantlets

(c) Vir D2 and Vir E2

  • VirD2 and VirE2 are proteins produced by the Agrobacterium that facilitate the transfer of T-DNA into the plant cell. Here’s how:
  • 1. VirD2 binds to the ends of the T-DNA, protecting it from degradation and guiding it to the plant cell.
  • 2. VirE2, on the other hand, coats the T-DNA, forming a complex that can be transported across the bacterial and plant cell membranes.
  • The individuals are heterozygous at all three loci, meaning they have one dominant allele (A, B, C) and one recessive allele (a, b, c) at each locus.
  • The genotype of each individual is AaBbCc. When these individuals are crossed, the probability of getting a dominant phenotype (A-, B-, C-) at each locus is 3/4 (1/2 for AA, 1/4 for Aa, and 1/4 for aA).
  • Since the loci are independent, the probability of getting the dominant phenotype at all three loci is (3/4) x (3/4) x (3/4) = 27/64. So, the proportion of progeny showing all three wild type characters is 27/64.
  • Therefore, the correct answer is (d) 27/64.

(b) co-dominant

(a) Tetraploid

I understand your confusion, but you are mixing up the terms monoploid and haploid. The monoploid number (x) is the number of chromosomes in a single set, not in a gamete. The haploid number (n) is the number of chromosomes in a gamete, which is half of the total number in a somatic cell12

For example, in humans, the monoploid number is 23, because we have 23 different types of chromosomes. The haploid number is also 23, because our gametes have one copy of each chromosome. The total number of chromosomes in a somatic cell is 46, which is 2n or 2x1

In a tetraploid plant, the monoploid number is still the same as the haploid number, but the total number of chromosomes in a somatic cell is 4n or 4x. So, if the monoploid number is 12, then the haploid number is also 12, and the total number is 482

In an octaploid plant, the monoploid number is still the same as the haploid number, but the total number of chromosomes in a somatic cell is 8n or 8x. So, if the monoploid number is 12, then the haploid number is also 12, and the total number is 962

The question says that the monoploid number is ¼ of the total number in gametes. This means that x = ¼n. If we multiply both sides by 4, we get 4x = n. This is only true for a tetraploid plant, because 4x is the total number of chromosomes in a somatic cell, and n is the total number in a gamete2

(a) Neurospora crassa

  • The “one gene-one enzyme” hypothesis was proposed by George Beadle and Edward Tatum through their work with Neurospora crassa, a type of bread mold. They used this organism because it was easy to grow and had a simple life cycle.
  • They exposed the mold to X-rays to cause mutations and then observed the effects on the mold’s ability to grow in minimal medium (a medium that only contains the bare essentials for growth). They found that some mutants could not grow unless the medium was supplemented with a specific substance. For example, one mutant could not grow unless the medium contained the amino acid arginine. They concluded that the mutation had damaged a gene that coded for an enzyme involved in the synthesis of arginine.
  • This led to the “one gene-one enzyme” hypothesis, which states that each gene in an organism’s DNA specifies the structure of a single enzyme. This hypothesis has been modified to “one gene-one polypeptide” as we now know that not all proteins are enzymes and that some proteins are made up of more than one polypeptide chain.

(b) test cross

(b) reptiles

  • Reptiles, on the other hand, hatch from their eggs looking like smaller versions of their adult forms. They grow and mature, but they do not undergo a dramatic physical transformation as insects do.
  • Therefore, reptiles do not have larval stages that are very different from their adult stage.

(c) Batesian mimicry

  • Batesian mimicry occurs through the process of natural selection. Here’s a simplified step-by-step explanation:
    • 1. In a population of harmless butterflies, there might be a small number of individuals that, due to genetic variation, look a little bit like a toxic species.
    • 2. Predators, such as birds, who have had bad experiences with the toxic species, avoid butterflies that look like the toxic species.
    • 3. Therefore, the harmless butterflies that look like the toxic species are less likely to be eaten by predators.
    • 4. These surviving butterflies pass on their genes (including those that make them look like the toxic species) to the next generation.
    • 5. Over many generations, the number of harmless butterflies that look like the toxic species increases in the population. This is Batesian mimicry.
  • (a) Cryptic coloration: This is a defensive mechanism in animals where they blend in with their environment to hide from predators. An example would be a snowshoe hare turning white in winter to blend in with the snow.
  • (b) Aposematic coloration: This is when animals have bright, noticeable colors or patterns to warn predators that they are toxic, venomous, or otherwise harmful. An example would be the bright colors of poison dart frogs.
  • (d) Müllerian mimicry: This is when two or more harmful species evolve to look similar, reinforcing the avoidance behavior in predators. An example would be different species of stinging bees and wasps, which all have similar yellow and black striping.

(d) The organ’s cells have receptors for thyroxine but not for insulin

  • Hormones like thyroxine and insulin can only elicit a response in cells that have specific receptors for them.
  • These receptors are proteins located either on the cell surface (for water-soluble hormones like insulin) or inside the cell (for lipid-soluble hormones like thyroxine). When a hormone binds to its specific receptor, it triggers a series of events inside the cell that lead to the cell’s response.
  • If a cell doesn’t have a receptor for a particular hormone, that hormone can’t affect the cell. So, in question, the organ’s cells have receptors for thyroxine, allowing it to affect those cells, but they don’t have receptors for insulin, so insulin has no effect.

(c) 30

  • In angiosperms, the endosperm is formed as a result of a process called double fertilization. Here’s how it works:
  • 1. One sperm cell from the pollen grain fertilizes the egg cell to form the zygote, which is diploid (2n). In your case, this would have 20 chromosomes.
  • 2. The other sperm cell fertilizes the two polar nuclei in the central cell of the ovule to form the endosperm, which is triploid (3n). So, if the diploid number is 20, the triploid number would be 30. That’s why the endosperm cells would have 30 chromosomes.

(b) II and III

  • T-cell receptors (TCRs) and B-cell receptors (BCRs) are formed by a process called somatic recombination. This is a unique type of genetic recombination that occurs in the immune system to create diversity in TCR and BCR.
  • In T-cells, the TCR gene segments (V, D, and J segments) undergo somatic recombination during T-cell development in the thymus. This process is random and results in a large variety of different TCRs, each capable of recognizing a different antigen.
  • Similarly, in B-cells, the BCR (or immunoglobulin) gene segments (V, D, and J for the heavy chain, and V and J for the light chain) undergo somatic recombination during B-cell development in the bone marrow. This results in a large variety of different BCRs, each capable of recognizing a different antigen.
  • This process is crucial for the adaptive immune response, allowing the immune system to recognize and respond to a wide range of pathogens.

(c) multiple copies, single pair

  • Lampbrush chromosomes are found in the oocytes (immature eggs) of amphibians during a prolonged phase of meiosis. They are characterized by their “brush-like” appearance due to extended loops of chromatin that are actively transcribing RNA. These chromosomes are present in multiple copies because each oocyte contains a full set of chromosomes, and there are many oocytes in an ovary.
  • On the other hand, polytene chromosomes are found in the salivary glands of Drosophila (fruit flies) and other Diptera. They result from repeated rounds of DNA replication without cell division (endoreduplication), leading to a single, giant chromosome that contains many identical copies (up to 1000) of the same DNA sequence aligned side-by-side. These chromosomes are “banded” due to the alignment of the replicated DNA, and they are also transcriptionally active.
  • Lampbrush chromosomes are typically found in the oocytes of amphibians and other vertebrates because these cells require a high level of transcriptional activity to prepare for the rapid cell divisions that occur after fertilization. The extended loops of chromatin in lampbrush chromosomes allow for this increased transcription.
  • Polytene chromosomes, on the other hand, are found in certain tissues of Drosophila and other Diptera where there is a need for high levels of gene expression. The salivary glands of these insects, for example, produce large amounts of saliva, which requires high levels of protein synthesis. The multiple copies of genes in polytene chromosomes allow for this high level of protein synthesis.

(c) Individuals start as male and later become female

  • This is a biology question about protandry. Protandry is a condition in which an organism starts its life as a male and then changes into a female. It is a type of sequential hermaphroditism, which means that an organism can change its sex during its lifetime12
  • The correct answer is © Individuals start as male and later become female. This is the definition of protandry. Some examples of animals that exhibit protandry are clownfish, wrasses, and slipper limpets123
  • The other options are incorrect because:
  • (a) Individuals start as female and later become male. This is the definition of protogyny, which is the opposite of protandry. Some examples of animals that exhibit protogyny are angelfish, parrotfish, and gobies124
  • (b) It produces eggs at one time and sperm at a different time, rather than both together. This is the definition of dichogamy, which is a type of synchronous hermaphroditism. It means that an organism can produce both male and female gametes, but not at the same time. Some examples of plants that exhibit dichogamy are corn, sunflower, and date palm.
  • (d) They exhibit hermaphroditism in early life but not in adults. This is the definition of protogynous hermaphroditism, which is a type of partial hermaphroditism. It means that an organism is hermaphroditic in its larval or juvenile stage, but becomes either male or female in its adult stage. Some examples of animals that exhibit protogynous hermaphroditism are sea urchins, sea stars, and some snails.

The wrongly matched pair is (b) Stratified cuboidal – oesophagus epithelium. The oesophagus is lined with stratified squamous epithelium, not stratified cuboidal epithelium.

  • (c) Columnar epithelium – peritoneum of body cavity: The peritoneum, which lines the body cavity, is indeed lined with simple columnar epithelium. This type of epithelium is also found in the digestive tract and other areas where absorption and secretion are primary functions.
  • (d) Squamous epithelium – skin of frog: The skin of a frog is lined with squamous epithelium. Squamous cells are flat and scale-like, and in the case of the frog, they help to protect the body and allow for gas exchange.
  • Stratified cuboidal epithelium is typically found lining certain glands and ducts, such as sweat glands, mammary glands, and salivary glands. It’s also found in parts of the male and female reproductive systems. This type of epithelium provides protection and has the capacity to stretch to accommodate the passage of bodily secretions. Thank you for discussing epithelial tissue with me.

(c) II III IV I

  • The correct matches are:
    • A. Neuron – ii. Nissl bodies
    • B. Bone matrix – iii. Ossein
    • C. RBC of man – iv.non nucleated
  • Nissl bodies are found in neurons and are involved in protein synthesis. Ossein is the organic part of the bone matrix. Non nucleated cells are found in red blood cells (RBCs) of man.

(a) stimulates the synthesis and secretion of androgens

  • GnRH, or Gonadotropin-releasing hormone, is produced by the hypothalamus in the brain. It acts on the pituitary gland, stimulating it to produce and release two hormones: luteinizing hormone (LH) and follicle-stimulating hormone (FSH).
  • In males, LH stimulates the Leydig cells in the testes to produce androgens, such as testosterone. FSH, on the other hand, acts on the Sertoli cells in the testes to support sperm production, which is also influenced by the presence of androgens.
  • In females, LH and FSH stimulate the ovaries to produce estrogen and progesterone, which are crucial for the menstrual cycle and pregnancy. So, while GnRH itself doesn’t directly stimulate the synthesis and secretion of androgens, it triggers a chain of hormonal events that lead to this outcome.

(c) Epinephrine

  • Epinephrine, also known as adrenaline, is released by the adrenal glands during times of stress, such as fear or excitement. This hormone triggers several responses in the body to prepare it for a ‘fight or flight’ situation. One of these responses is the increase of blood glucose levels.
  • Epinephrine stimulates the liver and muscle cells to break down glycogen, a stored form of glucose, into glucose that is then released into the bloodstream. This process is known as glycogenolysis. The increase in blood glucose provides the body with the quick energy it needs to respond to the stressful situation.

(b) There will be no detachment of endocytic vesicles from the plasma membrane

  • Dynamin is a GTPase protein that plays a crucial role in the process of endocytosis. After the formation of the endocytic vesicle, dynamin assembles around the neck of the vesicle. When GTP (guanosine triphosphate) binds to dynamin, it causes a conformational change in the protein, leading to the constriction of the vesicle neck. This constriction, followed by the hydrolysis of GTP to GDP (guanosine diphosphate), results in the pinching off or detachment of the vesicle from the plasma membrane. If dynamin is inactive, this process cannot occur, and the vesicle remains attached to the plasma membrane.
  • Endocytosis is a cellular process in which substances are brought into the cell. The process can be broken down into several steps:
    • 1. Initiation: The process begins when the substance to be ingested binds to specific receptors on the cell’s surface. This binding triggers the cell membrane to invaginate or fold inward, forming a pocket around the substance.
    • 2. Vesicle Formation: The invagination continues until the pocket pinches off to form a vesicle inside the cell. This vesicle is known as an endocytic vesicle. The formation of the vesicle is facilitated by a protein called clathrin, which forms a coat around the vesicle, giving it structure and stability.
    • 3. Scission: This is where dynamin comes into play. Dynamin wraps around the neck of the budding vesicle and, upon hydrolysis of GTP, constricts, leading to the detachment or “scission” of the vesicle from the plasma membrane.
    • 4. Uncoating: After the vesicle is released into the cytoplasm, the clathrin coat is removed by the action of uncoating proteins. This leaves a naked vesicle that can fuse with other organelles in the cell.
    • 5. Fusion and Release: The vesicle can then fuse with a lysosome, where the ingested substance can be broken down, or it can fuse with the plasma membrane to release its contents outside the cell (in a process called exocytosis).

(c) 200 seconds

So, it would take 200 seconds for the aqueous polymer solution to flow from one point to another in the viscometer. 0.9/1.8=100/t2. t2=200s

The dihedral angle \varphi represents rotation around the N-Cα bond. So, the correct answer is (a) N-Cα bond.

  • The dihedral angles in a polypeptide chain are specific to certain bonds because of the way the molecule is structured.
  • The φ (phi) angle is associated with the rotation around the N-Cα bond, ψ (psi) is associated with the rotation around the Cα-C’ bond, and ω (omega) is associated with the rotation around the C’-N bond. Other bonds in the molecule, like the C-H bond, do not have associated dihedral angles because they do not significantly affect the overall conformation of the polypeptide chain.

The peptide Gly-Ala-Val-Ser-Gly-Ala would show absorption maximum at (a) 220 nm. This is because peptides and proteins typically absorb UV light at this wavelength due to the presence of peptide bonds.

  • The angular velocity (ω) in radians per second can be calculated from the rotational speed in revolutions per minute (rpm) using the conversion factor 2π radians/revolution and 1 minute/60 seconds.
  • So, ω = 42,000 rpm x (2π rad/rev) x (1 min/60 sec) = 4,400 rad/sec.
  • Therefore, the correct answer is (b) 4400.

The correct answer is (b) IV only, as only C6H6 (benzene) lacks sp hybridized atoms.

  1. CH3CN:
    • The carbon atom in CH3CN forms a triple bond with nitrogen.
    • This triple bond indicates that the carbon atom is sp hybridized.
    • So, CH3CN contains sp hybridized atoms.
  2. C2H2:
    • C2H2 (acetylene) consists of two carbon atoms connected by a triple bond.
    • Both carbon atoms are sp hybridized.
    • So, C2H2 contains sp hybridized atoms.
  3. +NO2 (Nitronium ion):
    • The nitrogen atom in NO2 has a positive charge.
    • Nitrogen in NO2 has a triple bond with oxygen.
    • The nitrogen atom is sp hybridized.
    • However, the positive charge does not affect its hybridization.
    • So, +NO2 contains sp hybridized atoms.
  4. C6H6 (Benzene):
    • Benzene (C6H6) has a hexagonal ring structure.
    • Each carbon atom in benzene is connected to two other carbon atoms and one hydrogen atom.
    • The carbon atoms in benzene are sp2 hybridized.
    • Benzene does not contain sp hybridized atoms.

(d) Benzoic acid, Benzoyl chloride, benzaldehyde

  • Compound A is Benzoic Acid (C7H6O2), Compound B is Benzoyl Chloride (C7H5ClO), and Compound C is Benzaldehyde (C7H6O).
  • Here’s the reaction sequence:
    • 1. Toluene (C7H8) is oxidized by KMnO4 to form Benzoic Acid (A).
    • 2. Benzoic Acid is then treated with thionyl chloride (SOCl2) to form Benzoyl Chloride (B), which is an acid chloride and can turn blue litmus paper pink due to its acidic nature.
    • 3. Benzoyl Chloride is then reduced by hydrogen in the presence of Pd/BaSO4 to form Benzaldehyde (C).

(b) Ultra-violet spectroscopy

  • Ultra-violet (UV) spectroscopy generally cannot differentiate isomers of hydroxy benzoic acid because it primarily provides information about the presence of conjugated systems (alternating single and double bonds) in a molecule. UV spectroscopy is sensitive to changes in the electronic structure of a molecule, but it doesn’t provide detailed information about the specific arrangement of atoms within the molecule.
  • In contrast, techniques like nuclear magnetic resonance (NMR) and infra-red (IR) spectroscopy can provide more detailed information about the structure of a molecule, including the positions of specific atoms and functional groups, which can help differentiate isomers. Mass spectroscopy can provide information about the molecular weight and fragmentation pattern of a molecule, which can also help differentiate isomers.

(b) It has two stereogenic centers and three stereoisomers, one of them optically inactive

  • Tartaric acid has two stereogenic centers, which means it has two carbon atoms where the arrangement of the attached groups can vary.
  • This allows for the possibility of different stereoisomers, which are molecules with the same molecular formula and sequence of bonded atoms, but a different orientation of their atoms in space.
  • For tartaric acid, there are three stereoisomers: two are “enantiomers” (mirror images of each other) and are optically active, meaning they rotate plane-polarized light. The third stereoisomer is a “meso compound”, which is optically inactive because it has an internal plane of symmetry. This means that the molecule is its own mirror image, so it does not rotate plane-polarized light.
  • So, the correct statement is: Tartaric acid has two stereogenic centers and three stereoisomers, one of them optically inactive.

Therefore, the correct answer is (d) 16 times.

  • Since every 10°C rise in temperature doubles the rate of the reaction, a 40°C rise (which is four times 10°C) would double the rate four times.
  • Doubling something four times means multiplying it by 2 four times, or 24. So, the rate of the reaction would increase by 16 times.

(b) Molecularity of the reaction changes with temperature or concentration is false.

Molecularity is a theoretical concept and is defined as the number of molecules coming together to react in a single step, which does not change with temperature or concentration.

(a) Polymerization

  • Polymerization is a process where small molecules, called monomers, combine to form a large molecule, called a polymer. This process is an example of an entropy decrease because it involves the organization of many small, disordered particles (the monomers) into a single, more ordered structure (the polymer).
  • In other words, the randomness or disorder (entropy) decreases because the system becomes more ordere

(b) 2 ethanol and 2 CO2 molecules

  • The process of ethanol fermentation involves the breakdown of glucose (C6H12O6) into ethanol (C2H5OH) and carbon dioxide (CO2).
  • The overall chemical reaction is: C6H12O6 → 2 C2H5OH + 2 CO2
  • This process occurs in two main steps:
    • 1. Glycolysis: One molecule of glucose is broken down into two molecules of pyruvate. This process also produces 2 ATP (energy molecules) and 2 NADH (electron carrier molecules).
    • 2. Fermentation: The two pyruvate molecules are then converted into two molecules of ethanol and two molecules of carbon dioxide. This step also regenerates the NAD+ needed for glycolysis to continue.
    • So, from one glucose molecule, we get 2 ethanol and 2 CO2 molecules.

(d) (I) dehydration (II) hydration reactions

  • The conversion of citrate to cis-aconitate is a dehydration reaction because it involves the removal of a water molecule. In this step, an enzyme called aconitase catalyzes the removal of a water molecule from citrate, forming cis-aconitate.
  • On the other hand, the conversion of fumarate to malate is a hydration reaction because it involves the addition of a water molecule. In this step, an enzyme called fumarase catalyzes the addition of a water molecule to fumarate, forming malate.
  • So, in the context of the citric acid cycle, these reactions are classified based on whether they involve the addition or removal of water.

(a) Use acylation and then carry out reduction of the carbonyl group.

  • In the Friedel-Crafts alkylation reaction, it is generally observed that rearrangement (shift of hydride or alkyl group) can occur as a side reaction. To prepare alkyl-substituted aromatic compounds using the Friedel-Crafts alkylation reaction, it is better to avoid the rearrangement.
  • This method involves first acylating the aromatic ring (using an acyl chloride or anhydride) and then reducing the carbonyl group to an alkyl group. By bypassing direct alkylation, this approach minimizes the chances of rearrangement while achieving the desired substitution.

(d) The alkyl carbon is nucleophilic. In a Grignard reagent (R-MgX), the carbon atom is nucleophilic because it carries a partial negative charge due to the polar covalent bond with the magnesium atom.

  1. Starting Compound:
    • We begin with an alcohol (ROH), where R represents an alkane structure (a hydrocarbon chain).
  2. Reagent:
    • The reagent used in this transformation is phosphorus tribromide (PBr3).
  3. Reaction:
    • PBr3 reacts with the alcohol (ROH) to replace the hydroxyl group (OH) with a bromine atom (Br).
    • The reaction proceeds as follows:ROH + PBr3 → RBr + HBr
      • The hydroxyl group (OH) is replaced by a bromine atom (Br) to form an alkyl bromide (RBr).
  4. Products:
    • The products formed are labeled as (I) and (II):
      • (I): Alkyl bromide (RBr) resulting from the substitution of the hydroxyl group.
      • (II): Hydrogen bromide (HBr) is also produced as a byproduct.
  5. Summary:
    • The overall transformation involves the conversion of an alcohol into an alkyl bromide.
    • Both (I) and (II) are the products of this reaction.

Remember that PBr3 is commonly used for converting alcohols to alkyl halides (such as bromides) through nucleophilic substitution reactions. In this case, it results in the formation of both (I) and (II) as the products.

  1. Starting Compound:
    • We have a chlorinated alkene as the starting compound.
  2. Reaction with OE^- (Sodium Ethoxide):
    • OE^- (sodium ethoxide) is a strong base.
    • It will remove the acidic hydrogen from the alkene, leading to the formation of an epoxide as an intermediate.
  3. Epoxide Intermediate:
    • The epoxide intermediate has a strained three-membered ring containing oxygen.
    • It is highly reactive and can be attacked by nucleophiles.
  4. Potential Products:
    • Let’s analyze the potential products labeled I to V:
      • Product I:
        • Direct elimination of the epoxide.
        • Not likely due to high ring strain.
      • Product II:
        • Nucleophilic attack at the less substituted carbon of the epoxide.
        • This is a plausible product.
      • Product III:
        • Anti addition across the alkene.
        • Not favored here.
      • Product IV:
        • Syn addition across the alkene.
        • Not favored here.
      • Product V:
        • Nucleophilic attack at the more substituted carbon of the epoxide.
        • Possible but less likely due to steric hindrance.
  5. Conclusion:
    • Based on the analysis, options I and II are most likely.
    • Therefore, the correct answer is (b) I and II only.
  1. Assign Priorities:
    • CO2H (carboxylic acid) has the highest priority (1).
    • HO (hydroxy group) has the second-highest priority (2).
    • Cl (chlorine) has the third-highest priority (3).
    • C2H5 (ethyl group) has the fourth-highest priority (4).
  2. Visualize the Compound:
    • Imagine looking at the compound from above, with the lowest-priority substituent (H) pointing away from you (dashed line).
  3. Determine Configuration:
    • Trace a curved arrow from the highest-priority (1) substituent (CO2H) to the lowest-priority (4) substituent (H).
    • If the arrow goes counterclockwise (left when leaving the 12 o’clock position), the configuration is S (from Latin “Sinister,” meaning left).
    • If the arrow goes clockwise (right when leaving the 12 o’clock position), the configuration is R (from Latin “Rectus,” meaning right).
  4. Result:
    • The arrow points counterclockwise, so the configuration at the stereocenter is S.
    • Therefore, the absolute configuration of the given compound is (2S, 3R).

The correct answer remains (2S, 3R).

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