Biochemistry MCQs: AIIMS 2001-2005

AIIMS MAY 2001
Q-1. At physiological pH, the most stable amino acid is:
a) Histidine
b) Lysine
c) Arginine
d) Leucine

Answer: Histidine
Explanation:
Histidine plays unique roles in enzymatic catalysis.
The pK of its imidazole proton permits it to function at neutral pH or physiological pH as either a base or an acid catalyst.

Q-2. Amino acids secreted in cystinuria are all except
a) Ornithine
b) Histidine
c) Arginine
d) Lysine

Answer: Histidine
Explanation:
Cystinuria is an autosomal-recessive defect in re-absorptive transport of cystine and the dibasic amino acids ornithine, arginine, and lysine from the luminal fluid of the renal proximal tubule and small intestine.
Calculi are frequently multiple and bilateral, and they often form stag-horns.
Urinalysis:
Cystine is one of the sulfur-containing amino acids; therefore, the urine may have the characteristic odor of rotten eggs.
Urinalysis may show typical hexagonal or benzene crystals, which are essentially pathognomonic of cystinuria.
Sodium cyanide–nitroprusside test:
This is a rapid, simple, and qualitative determination of cystine concentrations.
Cyanide converts cystine to cysteine. Nitroprusside then binds, causing a purple hue.
Proton nuclear magnetic resonance spectroscopy of urine:
It is a very powerful technique that allows multi-component analysis useful in both diagnosis and follow-up.

Q-3. Amber codon refers to
a) Mutant codon
b) Stop codon
c) Initiating codon
d) Codon for more than one amino acids

Answer: Stop codon
Explanation:
Translation starts with a chain initiation codon or start codon.
The most common start codon is AUG, which is read as methionine or, in bacteria, as formyl-methionine.
Stop codons are also called “termination” or “nonsense” codons.
The three stop codons have been given names:
UAG: Amber
UGA: Opal
UAA: Ochre

Q-4. In chromatography mass movements of the substances in due to:
a) Diffusion
b) Electrophoresis
c) Osmosis
d) Paper chromatography

Answer: Diffusion
Explanation:
Chromatography is a method by which a mixture is separated by distributing its components between two phases. The stationary phase remains fixed in place while the mobile phase carries the components of the mixture through the medium being used.
Diffusion is the process whereby solute is transferred in a fluid from a point of high concentration to a point of lower concentration. Diffusion is a concentration driven mass transfer process.
Solute diffusivity affects the quality of a chromatographic separation in two ways. If the diffusivity of the solute is high in the two phases of a chromatographic system, solute exchange is rapid.

Q-5. Vitamin K is needed for which of these post translational modification processes:
a) Methylation
b) Carboxylation
c) Hydroxylation
d) Transketolation

Answer: Carboxylation
Explanation:
Gamma-glutamyl carboxylase is an enzyme that catalyzes the posttranslational modification of vitamin K-dependent proteins.
Many of these vitamin K-dependent proteins are involved in coagulation so the function of the encoded enzyme is essential for hemostasis.
The posttranslational modifications of proteins that depend upon vitamin C as a cofactor include proline and lysine hydroxylations and carboxy terminal amidation. The hydroxylating enzymes are identified as prolyl hydroxylase and lysyl hydroxylase.
The donor of the amide for C-terminal amidation is glycine.
The most important hydroxylated proteins are the collagens.
Several peptide hormones such as oxytocin and vasopressin have C-terminal amidation.

Q- 6. Which end product of citric acid cycle is used in detoxification of ammonia in brain?
a) Citrate
b) Succinate
c) Alfa-keto glutarate
d) Oxalo-acetate

Answer: Alfa-keto glutarate
Explanation:
Glutamine is an amide of Glutamic acid which is formed from alfa-keto glutarate.
Glutamine is the sole form of transport of ammonia in brain.
Glutamine provides non-toxic storage and transport form of ammonia.
The formation of glutamine occurs primarily in the muscles, liver but also important in the nervous system, where it is the major mechanism for the removal of ammonia in the brain.

Q-7. Dietary triglycerides are transported by
a) Chylomicrons
b) VLDL
c) LDL
d) HDL

Answer: Chylomicrons
Explanation:
Dietary triglycerides are transported as part of lipoprotein particles called chylomicrons.
Triglycerides synthesized in the liver are transported as part of lipoprotein particles called very low density lipoproteins or VLDLs.

Q-8. Thiamine is not used in which of the following reactions
a) Glucose to pentose
b) Lactate to Pyruvate
c) Oxidative decarboxylation of Alfa-keto amino acids
d) Alfa-keto glutarate to succinyl Co-A

Answer: Lactate to Pyruvate
Explanation:
Thiamine pyrophosphate, or thiamine di-phosphate, is a thiamine (vitamin B1) derivative which is produced by the enzyme thiamine pyrophosphatase.
Thiamine pyrophosphate is a cofactor catalyzes several biochemical reactions.
TPP works as a coenzyme in many enzymatic reactions, such as:
Pyruvate dehydrogenase complex
Pyruvate decarboxylase in ethanol fermentation
Alpha-keto-glutarate dehydrogenase complex
Branched-chain amino acid dehydrogenase complex
2-hydroxyphytanoyl-CoA lyase
Trans-ketolase
Lactate dehydrogenase: It catalyzes the inter-conversion of pyruvate and lactate with concomitant inter-conversion of NADH and NAD+. Co-enzyme is niacin.

Q-9. In which type of chromatography, the proteins are bound to another substance?
a) Gel Chromatography
b) Paper Chromatography
c) Hydrophobic Chromatography
d) Affinity Chromatography

Answer: Affinity Chromatography
Explanation:
Affinity Chromatography is a separation technique based upon molecular conformation, which frequently utilizes application specific resins.
These resins have ligands attached to their surfaces which are specific for the compounds to be separated.
Many membrane proteins are glyco-proteins and can be purified by lectin affinity chromatography.
Detergent-solubilized proteins can be allowed to bind to a chromatography resin that has been modified to have a covalently attached lectin.
Proteins that do not bind to the lectin are washed away and then specifically bound glyco-proteins can be eluted by adding a high concentration of a sugar that competes with the bound glyco-proteins at the lectin binding site.

AIIMS NOV 2001
Q-1. Dietary fibres are degraded by colonic bacteria to form which of the following?
a) Butyrate
b) Glycerol
c) Sucrose
d) Free Radicals

Answer: Butyrate
Explanation:
Some non-absorbed carbohydrates are fermented to short-chain fatty acids (chiefly acetic, propionic and n-butyric), and carbon dioxide, hydrogen and methane. Almost all of these short-chain fatty acids will be absorbed from the colon.
The production of short-chain fatty acids has several possible actions on the gut mucosa.
All of the short-chain fatty acids are readily absorbed by the colonic mucosa, but only acetic acid reaches the systemic circulation in appreciable amounts.
Butyric acid appears to be used as a fuel by the colonic mucosa as the preferred energy source for colonic cells.

Q-2. A child presents with hepatomegaly and bi-lateral lenticular opacities. Deficiency of which of the following enzymes will not cause such features?
a) Galacto-kinase
b) Galactose-1- phosphate uridyl transferase
c) UDP Galactose 4 Epimerase
d) Lactase

Answer: Lactase
Explanation:
Child is suffering from Galactosemia.
Lactase deficiency doesn’t cause Galactosemia.
Defect in the following enzymes can cause Galactosemia:
Galactose-1- phosphate uridyl transferase- Classical type
Galacto-kinase- Minor type
UDP Galactose 4 Epimerase-Rarely seen

Q-3. What is the possible cause of hyper-uricemia and gout in a patient who has a glucose-6-phosphatase deficiency?
a) More formation of pentose
b) Increased accumulation of sorbitol
c) Increased synthesis of glycerol
d) Impaired degradation of free radicals

Answer: More formation of pentose
Explanation:
Cause of hyper-uricemia and gout in glucose-6-phosphatase deficiency:
Increased production of uric acid:
G-6-phosphatase deficiency-> Increased G-6-phosphate-> Increased pentose formation (HMP shunt) -> Increased purine synthesis-> Increased uric acid formation
Decreased renal excretion of uric acid:
G-6-phosphatase deficiency-> Decreased conversion of G-6-phosphate to Glucose-> Hypoglycemia-> Increased catecholamines-> Increased glycogenolysis in muscles-> Increased lactic acidosis-> It competes with urate for excretion-> Decreased uric acid excretion

Q-4. Cancer cells derive nutrition from
a) Glycolysis
b) Oxidative phosphorylation
c) Increase in mitochondria
d) From a fast food joint

Answer: Glycolysis
Explanation:
Most cancer cells exhibit increased glycolysis and use this metabolic pathway for generation of ATP as a main source of their energy supply.
The paradox is that cancer cells rely on glycolysis even if oxygen is available. This phenomenon is called aerobic glycolysis or the Warburg effect.
Warburg effect is considered as one of the most fundamental metabolic alterations during malignant transformation.

Q-5. Northern blot test is used for
a) DNA analysis
b) RNA analysis
c) Analysis of proteins
d) Enzyme analysis

Answer: RNA analysis
Explanation:
Northern blot: RNA analysis
Southern blot: DNA analysis
Western blot: Analysis of proteins

AIIMS MAY 2002
Q-1. Which of these amino acids does not have anomeric carbon atom?
a) Valine
b) Alanine
c) Tyrosine
d) Glycine

Answer: Glycine
Explanation:
With the sole exception of glycine, the alfa-carbon of amino acids is chiral. Glycine lacks anomeric carbon.
Although some protein amino acids are dextrorotatory and some levorotatory, all share the absolute configuration of L-glyceraldehyde and thus are L-α-amino acids.
Several free L-α amio-acid fulfill important roles in metabolic processes. Examples include ornithine, citrulline, and arginine-succinate that participate in urea synthesis; tyrosine in formation of thyroid hormones; and glutamate in neurotransmitter biosynthesis.
D-Amino acids that occur naturally include free D-serine and D-Aspartate in brain tissue, D-alanine and D-glutamate in the cell walls of gram-positive bacteria, and D-amino acids in certain peptides and antibiotics produced by bacteria, fungi, reptiles and other non-mammalian species.

Q-2. In a mutation if valine is replaced by which of the following would not result in any change in function of protein (hemoglobin)
a) Proline
b) Glycine
c) Aspartic acid
d) Leucine

Answer: Aspartic acid
Explanation:
The codon for valine at position 67 of the beta chain of hemoglobin is not identical in all persons who possess a normal functional beta chain of hemoglobin.
Hemoglobin Milwaukee has at position 67 a Glutamic acid: hemoglobin Bristol contains aspartic acid at position 67. Hemoglobin Sydney contains alanine at position 67.

Q-3. A mutation in the codon, which causes a change in the amino acid being coded, is referred to as
a) Missense mutation
b) Recombination
c) Somatic mutation
d) Mitogenesis

Answer: Missense mutation
Explanation:
Point mutation:
A missense mutation is a point mutation in which a single nucleotide change results in a codon those codes for a different amino acid.
Missense mutations can render the resulting protein nonfunctional, and such mutations are responsible for diseases such as Epidermolysis bullosa, sickle-cell disease etc.
A nonsense mutation converts an amino acid codon into a termination codon. This causes the protein to be shortened because of the stop codon interrupting its normal code.
Silent mutations: Code for the same amino acid. A silent mutation has no effect on the functioning of the protein.

Q-4. In a patient with starvation for 72 hours which of the following would be seen?
a) Increased ketosis due to breakdown of fats
b) Increased gluconeogenesis by muscle protein breakdown
c) Increased Glycogenolysis
d) Increased gluconeogenesis

Answer: Increased ketosis due to breakdown of fats
Explanation:
During starvation, the metabolic changes providing energy can be divided into three stages:
First stage:
It lasts first 2 to 3 days.
Glycogenolysis (Liver glycogen is the first stored food that is mobilized for energy production.)
Gluconeogenesis- From alanine, glycerol, lactate
Second stage:
It lasts for over 2 weeks.
Fat is mobilized with ketosis and ketonuria.
Third stage:
Tissue protein is degraded for energy production.

Q-5. Which of the following membranes would be having highest protein content per gram tissue?
a) Inner mitochondrial membrane
b) Outer mitochondrial membrane
c) Plasma membrane
d) Myelin sheath

Answer: Inner mitochondrial membrane
Explanation:
Ratio of protein to lipid in different membranes:
Inner mitochondrial membrane- 3.2
Sarcoplasmic reticulum- 2.0
Outer mitochondrial membrane-1.1
Myelin sheath-0.23

Q-6. The presence of lone pair on the oxygen in water molecule results in
a) Makes water non-polar solvent
b) Electro positive charge on water molecule
c) Electro negative charge on water molecule
d) Covalent bond in ice

Answer: Electro negative charge on water molecule
Explanation:
In water, each hydrogen nucleus is covalently bound to the central oxygen atom by a pair of electrons that are shared between them. In H2O, only two of the six outer-shell electrons of oxygen are used for this purpose, leaving four electrons which are organized into two non-bonding pairs.
Because the two non-bonding pairs remain closer to the oxygen atom, these exert a stronger repulsion against the two covalent bonding pairs, effectively pushing the two hydrogen atoms closer together. The result is a distorted tetrahedral arrangement.
The H2O molecule is electrically neutral, but the positive and negative charges are not distributed uniformly.

Q-7. Defect in folding proteins would result in a clinical disease in which of the following
a) Myopia
b) Hypothyroidism
c) Migraine
d) Kuru

Answer: Kuru
Explanation:
The term “prions” refers to abnormal, pathogenic agents that are transmissible and are able to induce abnormal folding of specific normal cellular proteins called prion proteins that are found most abundantly in the brain.
Prion diseases are usually rapidly progressive and always fatal.
Human Prion Diseases:
Creutzfeldt – Jakob disease (CJD)
Variant Creutzfeldt – Jakob disease (vCJD)
Gerstmann-Sträussler-Scheinker Syndrome
Fatal Familial Insomnia
Kuru
Animal Prion Diseases:
Bovine Spongiform Encephalopathy (BSE)
Chronic Wasting Disease (CWD)
Scrapie
Transmissible mink encephalopathy
Feline spongiform encephalopathy
Ungulate spongiform encephalopathy

Q-8. When deoxy-hemoglobin gets converted into oxy-hemoglobin, the changes seen would include
a) Binding of O2 causes release of H+
b) Increased binding of 2, 3 DPG
c) pH of blood has no effect on the binding of O2
d) One mole of deoxy-hemoglobin binds two mole of 2, 3 DPG

Answer: Binding of O2 causes release of H+
Explanation:
Deoxy-hemoglobin binds one proton for every two O2 molecules released, contributing significantly to the buffering capacity of blood.
The somewhat lower pH of peripheral tissues, aided by carbamation, stabilizes the T state and thus enhances the delivery of O2 binds to deoxy-hemoglobin; protons are released and combine with bicarbonate to form carbonic acid.
Dehydration of H2CO3 catalyzed by carbonic anhydrase forms CO2+ which is exhaled. Binding of oxygen thus drives the exhalation of CO2.
This reciprocal coupling of protons and O2 binding is termed the Bohr Effect. The Bohr Effect is dependent upon cooperative interactions between the Hemes of the hemoglobin tetramer.
Myoglobin, a monomer, exhibits no Bohr Effect.

Q-9. Which of the following dose not cross cell membrane easily?
a) Glucose
b) Glucose 6 phosphate
c) Nitric oxide
d) Carbon monoxide

Answer: Glucose 6 phosphate
Explanation:
When digestion of a meal is complete, insulin levels fall, and enzyme systems in the liver cells begin to remove glucose molecules from strands of glycogen in the form of G6P. This process is termed glycogenolysis.
The G6P remains within the liver cell unless the phosphate is cleaved by glucose-6-phosphatase.
This de-phosphorylation reaction produces free glucose and free PO4 anions.
The free glucose molecules can be transported out of the liver cells into the blood to maintain an adequate supply of glucose to the brain and other organs of the body.

Q-10. Viscosity in synovial fluid depends upon
a) N-acetyl galactosamine
b) N-acetyl glucosamine
c) Hyaluronic acid
d) Glucuronic acid

Answer: Hyaluronic acid
Explanation:
Normal synovial fluid contains hyaluronan (hyaluronic acid), a polymer of disaccharides composed of D-glucuronic acid and D-N-acetyl glucosamine joined by glycosidic bonds.
Hyaluronan is synthesized by the synovial membrane and secreted into the joint cavity to increase the viscosity and elasticity of articular cartilages and to lubricate the surfaces between synovium and cartilage.
Hyaluronic acid is especially high in concentration in embryonic tissue and is thought to play an important role in permitting cell migration during morphogenesis and wound repair.
The high concentration of hyaluronic acid and chondroitin sulfates present in cartilage contribute to its compressibility.

Q-11. All of the following are extracellular proteins except
a) Laminin
b) Integrin
c) Collagen
d) Fibronectin

Answer: Integrin
Explanation:
Integrins are the part of the cell membrane that mediates cellular attachment to extracellular matrix.
Extra-cellular matrix:
Fibrous structural proteins-Collagen and Elastin
Adhesive glyco-proteins- Fibronectin and Laminin
Ptroteo-glycans and hyaluronan

Q-12. In a DNA, the coding region reads 5-CGT-3 in coding strand. This would code in the RNA as
a) 5-CGU-3
b) 5-GCA-3
c) 5-ACG-3
d) 5-UGC-3

Answer: 5-CGU-3
Explanation:
The strand that is transcribed or copied into an RNA molecule is referred to as the template strand of the DNA.
The other DNA strand, non template strand is frequently referred to as the coding strand of that gene.
It is called this because with the exception of T for U changes, it corresponds exactly to the sequence of the RNA primary transcript.

Q-13. RBC membrane is maintained by
a) Spectrin
b) Laminin
c) Collagen
d) Elastin

Answer: Spectrin
Explanation:
The shape, flexibility and the strength of RBC membrane is maintained by:
Spectrin (Most important)
Ankyrin
Protein 4.1
The deficiency of any one of the membrane skeletal proteins can cause hereditary spherocytosis.

Q-14. In synthesis of fatty acids, energy is supplied by
a) NAD
b) FAD
c) GTP
d) NADPH

Answer: NADPH
Explanation:
In humans, fatty acid synthesis occurs primarily in the liver and lactating mammary glands and to lesser extent, in adipose tissue.
The process incorporates carbons from acetyl Co-A into the growing fatty acid chain, using ATP and NADPH.
The HMP is the major supplier of NADPH for fatty acid synthesis.

AIIMS NOV 2002
Q-1. In hemoglobin the innate affinity of heme for carbon monoxide is diminished by the presence of:
a) His F9
b) His E7
c) Gly B6
d) Thr C4

Answer: His E7
Explanation:
The heme of Myoglobin lies in a crevice between helices E and F oriented with its polar propionate groups facing the surface of the globin. The remainder resides in the non-polar interior.
The fifth co-ordination position of the iron is linked to ring nitrogen of a proximal histidine, His F8.
The distal histidine, His E7, lies on the side of the heme ring opposite to His F8.
The innate affinity of heme for carbon monoxide is diminished by the presence of His E7. His E7 creates a hindering environment for CO by disturbing the orientation of atoms in the heme molecule.

Q-2. An alfa helix of a protein is most likely to be disrupted if a missense mutation introduces the following amino acid within the alpha helical structure?
a) Alanine
b) Aspartic acid
c) Tyrosine
d) Glycine

Answer: Glycine and Aspartic acid
Explanation:
Alfa helix of a protein can be disrupted due to the introduction of the following amino acids:
Proline: The imino group of proline is not geometrically compatible with the right handed spiral of the alpha-helix
Proline can only be stably accommodated within the first turn of an alfa helix. When present elsewhere, proline disrupts the conformation of the helix, producing a bend. Because of its small size, glycine also often induces bends in alfa helix.
Charged amino acids: Glutamate, aspartate, lysine, arginine and histidine
Amino acids with bulky side chains: tryptophan
Amino acids that branch at the beta-carbon: Valine and iso-Leucine.

Q-3. One of the following enzymes is not a protein:
a) DNAase
b) Abzyme
c) Eco RI
d) Ribozyme

Answer: Ribozyme
Explanation:
An Abzyme is a monoclonal antibody with catalytic activity.
Ribozymes are RNA molecules with catalytic activity. These generally involve trans-esterification reactions, and most are concerned with RNA metabolism (Splicing and endo-ribonuclease)
Ribozyme also performs the peptidyl transferase activity.

Q-4. During gluconeogenesis reducing equivalents from mitochondria to the cytosol are transported by:
a) Malate
b) Aspartate
c) Glutamate
d) Oxalo-acetate

Answer: Malate
Explanation:
Oxalo-acetate formed in mitochondria must enter the cytosol where the other enzymes of gluconeogenesis are located.
OAA is unable to directly cross the inner mitochondrial membrane; it must first be reduced to Malate by mitochondrial Malate dehydrogenase.
OAA + (NADH + H+) -> Malate + NAD+
Malate can be transported from mitochondria to the cytosol, where it is re-oxidized to OAA by cytosolic Malate dehydrogenase.
Malate + NAD±> OAA + (NADH + H+)

Q-5. The following separation technique depends on the molecular size of the protein:
a) Chromatography on a carboxy-methyl (CM) cellulose column
b) Iso-electric focusing
c) Get filtration chromatography
d) Chromatography on a di-ethyl-amino-ethyl (DEAE) cellulose column

Answer: Get filtration chromatography
Explanation:
Get filtration chromatography OR Size-exclusion chromatography (SEC):
Gel filtration (GF) chromatography separates proteins solely on the basis of molecular size.
Separation is achieved using a porous matrix to which the molecules, for steric reasons, have different degrees of access–i.e., smaller molecules have greater access and larger molecules are excluded from the matrix. Hence, proteins are eluted from the GF column in decreasing order of size.

Q-6. At the physiological pH, the DNA molecules are:
a) Positively charged
b) Negatively charged
c) Neutral
d) Amphipathic

Answer: Negatively charged
Explanation:
For all forms of DNA the global structure has an overall negative charge and the overall electrostatic potential is negative.
This is largely due to the phosphate groups of DNA which bear an overall negative charge due to the negatively charged oxygen atoms they have.
However, DNA has both positive and negative charges covering its outer structure. The charge dispersion over the global structure of DNA varies between A-DNA, B-DNA, and Z-DNA.
B-DNA has mostly negative electrostatic potential spread throughout its global exterior and concentrated mostly in major and minor grooves.

Q-7. The following hormone does not have any intracellular receptor:
a) Vitamin D3
b) Cortisone
c) Adrenaline
d) Thyroxin

Answer: Adrenaline
Explanation:
Intracellular receptor:
Cytoplasmic receptors:
Glucocorticoids
Mineralocorticoids
Estrogen
Progestin
Vitamin D
Nuclear receptors:
Thyroid hormone

Q-8. Accumulation of sphingomyelin in phagocytic cells is feature of:
a) Gaucher Disease
b) Niemann Pick Disease
c) Down Syndrome
d) Tay Sachs Disease

Answer: Niemann Pick Disease
Explanation:
Niemann-Pick disease (NPD) is a lipid storage disorder that results from the deficiency of a lysosomal enzyme, acid Sphingomyelinase.
The diagnosis of Niemann-Pick disease (NPD) is confirmed with measurement of enzyme activity in peripheral white blood cells or in cultured fibroblasts.

AIIMS MAY 2003
Q-1. The activity of following enzyme is affected by biotin deficiency
a) Trans-ketolase
b) Oxidase
c) Carboxylase
d) Dehydrogenase

Answer: Carboxylase
Explanation:
Biotin is a water-soluble B-vitamin (vitamin B7). Biotin is important in fatty acid synthesis, branched-chain amino acid catabolism, and gluconeogenesis.
Biotin deficiency is rare because, in general, intestinal bacteria produce biotin in excess of the body’s daily requirements.
Biotin is a cofactor responsible for carbon dioxide transfer in several carboxylase enzymes:
a) Acetyl-Co-A carboxylase
b) Propionyl-CoA carboxylase
c) Pyruvate carboxylase

Q-2. Acetyl Co-A acts as a substrate for all the enzymes except:
a) HMG-Co A synthetase
b) Malic enzyme
c) Malonyl Co A synthetase
d) Fatty acid synthetase

Answer: Malic enzyme
Explanation:
Coenzyme-A is a coenzyme, notable for its role in the synthesis and oxidation of fatty acids, and the oxidation of pyruvate in the citric acid cycle.
Acetyl Co-A acts as a substrate for the enzymes:
HMG-Co A synthetase
Malonyl Co A synthetase
Fatty acid synthetase

Q-3. At the physiological pH the DNA molecules are:
a) Positively charged
b) Negatively charged
c) Neutral
d) Amphipathic

Answer: Negatively charged
Explanation:
For all forms of DNA the global structure has an overall negative charge and the overall electrostatic potential is negative.
This is largely due to the phosphate groups of DNA which bear an overall negative charge due to the negatively charged oxygen atoms they have.
However, DNA has both positive and negative charges covering its outer structure. The charge dispersion over the global structure of DNA varies between A-DNA, B-DNA, and Z-DNA.
B-DNA has mostly negative electrostatic potential spread throughout its global exterior and concentrated mostly in major and minor grooves.

Q-4. Cholesterol present in LDL:
a) Represents primarily cholesterol that is being removed from peripheral cells.
b) Binds to a receptor and diffuses across the cell membrane.
c) On accumulation in the cell inhibits replenishment of LDL receptors.
d) When enters a cell, suppresses activity of acyl-CoA: cholesterol acyl-transferase (ACAT).

Answer: On accumulation in the cell inhibits replenishment of LDL receptors.
Explanation:
The mechanism of suppression of cholesterol biosynthesis by LDL-bound cholesterol involves specific LDL receptors that project from the surface of human cells.
The first step of the regulatory mechanism involves the binding of the lipoprotein LDL to these LDL receptors, thereby extracting the LDL particles from the blood.
Accumulation of intracellular cholesterol eventually inhibits the replenishment of LDL receptors on the cell surface, a phenomenon called down regulation, thereby blocking further uptake and accumulation of cholesterol.

Q-5. A newborn infant refuses breast milk since the 2nd day of birth, vomits on force-feeding but accepts glucose-water, develops diarrhea on the third day, by 5th day she is jaundiced with liver enlargement and eyes show signs of cataract. Urinary reducing sugar was positive but blood glucose estimated by glucose oxidation method was found low. The most likely cause is deficiency of:
a) Galactose-1-phosphate uridyl transferase
b) Beta galactosidase
c) Glucose-6-phosphatase
d) Galactokinase

Answer: Galactose-1-phosphate uridyl transferase
Explanation:
Galactose 1 phosphate uridyl-transferase is missing in individuals with classical Galactosemia.
Galactose 1 phosphate and therefore Galactose accumulate in cells.
Accumulated Galactose is shunted into side-pathways such as that of Galactitol production.
This reaction is catalyzed by aldose reductase, same enzyme that converts glucose to sorbitol.
More benign form of Galactosemia is caused by a deficiency of galactokinase.
Classical Galactosemia:
Uridyl-transferase deficiency
Autosomal recessive disorder
It causes Galactosemia and galactosuria, vomiting, diarrhoea and jaundice.
Accumulation of Galactose 1 phosphate and Galactitol in nerve, lens, liver and kidney tissue causes liver damage, severe mental retardation and cataracts.
Antenatal diagnosis is possible by chorionic villus sampling.
Therapy: Rapid removal Galactose i.e. lactose from the diet.

Q-6. Replication and transcription are similar processes in mechanistic terms because both
a) Use RNA primers for initiation
b) Use deoxy-ribo-nucleotides as precursors
c) Are semi conserved events
d) Involve phosphodiester bond formation with elongation occurring in the 5′-> 3′ direction

Answer: Involve phosphodiester bond formation with elongation occurring in the 5′-> 3′ direction
Explanation:
Replication and transcription: Similarities
The general steps of initiation, elongation and termination with 5′-> 3′ polarity with formation of phosphodiester bond
Large multi-component initiation complex
Adherence of Watson Crick base pairing rule

AIIMS NOV 2003
Q-1. A buffer that is most effective at a pH of about 4.5 is
a) Acetate buffer
b) Bicarbonate buffer
c) Phosphate buffer
d) Tris buffer

Answer: Acetate buffer
Explanation:
The solution resists changes in pH most effectively at pH values close to the pKa.
A solution of a weak acid and its conjugate base, buffers most effectively in the pH range pKa ± 1.0 pH unit.
Acid and pK:
Acetic: pK = 4.76
Carbonic: pK1=6.37 and pK2=10.25
Phosphoric: pK1= 2.15, pK2=6.82 and pK3=12.38

Q-2. NADPH is produced by:
a) Glycolysis
b) Citric acid cycle
c) HMP Shunt
d) Glycogenesis

Answer: HMP Shunt
Explanation:
HMP Shunt provides a major portion of the body’s NADPH which functions as a biochemical reductant.
Uses of NADPH:
Reductive biosynthesis- Synthesis of steroids and fatty acids
Reduction of hydrogen per-oxide
Cytochrome P450 mono-oxygenase system
Phagocytosis by WBCs
Synthesis of nitric oxide

Q-3. Mitochondria are involved in all of the following except:
a) ATP production
b) Apoptosis
c) Tri-carboxylic acid cycle
d) Fatty acid biosynthesis

Answer: Fatty acid biosynthesis
Explanation:
The main pathway for de novo synthesis of fatty acid occurs in the cytosol.
The biochemical processes taking place in mitochondria:
TCA or Kreb’s cycle and fatty acid oxidation for ATP production
Formation of acetyl Co-A
Part of urea cycle
Part of gluconeogenesis

Q-4. The conjugation of bilirubin to glucuronic acid in the liver:
a) Converts a hydrophilic compound to a hydrophobic one
b) Converts a hydrophobic compound to a hydrophilic one
c) Enables the bilirubin molecule to cross the cell membrane
d) Is increased during neonatal jaundice

Answer: Converts a hydrophobic compound to a hydrophilic one
Explanation:
Bilirubin is a tetra-pyrrole created by the normal breakdown of heme.
Most bilirubin is produced during the breakdown of hemoglobin and other hemo-proteins.
Because bilirubin is highly insoluble in water, it must be converted into a soluble conjugate before elimination from the body.
In the liver, uridine di-phosphate (UDP)-glucuronyl transferase converts bilirubin to a mixture of mono-glucuronides and di-glucuronides, referred to as conjugated bilirubin, which is then secreted into the bile by an ATP-dependent transporter.
The kidneys do not filter un-conjugated bilirubin because of its avid binding to albumin. For this reason, the presence of bilirubin in the urine indicates the presence of conjugated hyper-bilirubinemia.

Q-5. The 3 dimensional shape of a protein is maintained mainly by:
a) Strong covalent interactions
b) Interactions with other proteins
c) Multiple weak interactions
d) Interactions with prosthetic groups

Answer: Multiple weak interactions
Explanation:
Higher orders of protein structure are stabilized primarily and often exclusively by non-covalent interactions.
Principal among these are hydrophobic interactions and other significant contributors include hydrogen bonds and salt bridge.
Some proteins contain covalent disulfide bonds. Inter-polypeptide disulfide bonds stabilize the quaternary structure of certain oligomeric proteins.

AIIMS MAY 2004
Q-1. What one of the following is not true for an alfa helix?
a) It is one of the most important secondary structure.
b) It has a net dipole moment
c) All hydrogen bonds are aligned in the same direction.
d) Long stretches of left handed ∝-helices occur in proteins.

Answer: Long stretches of left handed ∝-helices occur in proteins.
Explanation:
Proteins contain only L-amino acids, for which a right-handed ∝ helix is by far the more stable and only right handed ∝ helices are present in proteins.

Q-2. Misexpression of which of the following homeobox genes alter the position of the forelimbs during development:
a) HOXA7
b) HOXB8
c) HOXC9
d) HOXDI0

Answer: HOXB8
Explanation:
Homeobox genes are a large family of similar genes that direct the formation of many body structures during early embryonic development.
Positioning of the limbs along the cranio-caudal axis in the flank regions of the embryo is regulated by the HOX genes expressed along this axis.
The cranial limit of expression of HOXB8 is at the cranial border of forelimb and misexpression of this gene alters the position of these limbs. (Ref: Langman’s Medical Embryology)

Q-3. All of the following statements about LDL are true except
a) It delivers cholesterol to cells
b) It contains only one apo-protein
c) It is a marker for cardiovascular disease
d) It contains Apo-B4

Answer: It contains Apo-B4
Explanation:
LDL particles contain much less triacylglycerol than their VLDL predecessors, and have high concentration of cholesterol and cholesteryl ester. LDL is a marker for cardiovascular disease.
The primary function of LDL particles is to provide cholesterol to the peripheral tissues or return it to the liver.
Apo-lipoprotein B-100 is associated with LDL.

Q-4. The most important source of reducing equivalents for fatty acid synthesis in the liver is
a) Glycolysis
b) TCA cycle
c) Uronic acid pathway
d) HMP pathway

Answer: HMP pathway
Explanation:
HMP Shunt provides a major portion of the body’s NADPH which functions as a biochemical reductant.
Uses of NADPH:
Reductive biosynthesis- Synthesis of steroids and fatty acids
Reduction of hydrogen per-oxide
Cytochrome P450 mono-oxygenase system
Phagocytosis by WBCs
Synthesis of nitric oxide

Q-5. Insulin increases the activities of all of the following enzymes, except:
a) Glucokinase
b) Pyruvate carboxylase
c) Glycogen synthase
d) Acetyl-CoA carboxylase

Answer: Pyruvate carboxylase
Explanation:
Insulin increases the activities of all of the following enzymes:
Glucokinase
Phospho-fructokinase
Pyruvate Kinase
Protein phosphatase- I
Insulin increases the activity of glycogen synthase by inhibiting cAMP.
Insulin-> Increased activity of Protein phosphatase- I-> De-phosphorylation of Acetyl-CoA carboxylase-> Acetyl-CoA carboxylase activated

Q-6. Which of the following combinations of biologically active molecules does vitamin A consists of?
a) Retinol, retinal and retinoic acid
b) Retinol, retinal and tetra-hydro-folate
c) Retinal, conjugase and retinoic acid
d) PABA, retinol and retinal-aldehyde

Answer: Retinol, retinal and retinoic acid
Explanation:
Biologically active forms of Vitamin A:
Retinol
Retinal
Retinoic acid
Beta carotene
The retinoids are related to retinol (Vitamin A) are essential for vision, reproduction, growth and maintenance of epithelial tissues.
Retinoic acid derived from oxidation of dietary retinol, mediates most of the action of the retinoids except for vision, which depends on retinal (Retinal-aldehyde), the aldehyde derivative of retinol.
Plant foods contain beta carotene which can be oxidatively cleaved in the intestine to yield two molecules of retinal.

Q-7. The primary structure of a protein refers to:
a) Linear structure and order of the amino acids present
b) Regular conformational forms of a protein
c) Complete three-dimensional structure of the polypeptide units of a given protein
d) Subunit structure of the protein

Answer: Linear structure and order of the amino acids present
Explanation:
Primary structure: The sequence of the amino acids in polypeptide chain
Secondary structure: The folding of short (3 to 30 residues), contiguous segments of polypeptide into geometrically ordered units
Tertiary structure: The assembly of secondary structural units into larger functional units such as the mature polypeptide and its components domains;
Quaternary structure: The number and types of polypeptide units of oligomeric proteins and their spatial arrangement

Q-8. Excessive ultraviolet (UV) radiation is harmful to life. The damage caused to the biological systems by ultraviolet radiation is by:
a) Inhibition of DNA synthesis
b) Formation of thymidine dimers
c) Ionization
d) DNA fragmentation

Answer: Formation of thymidine dimers
Explanation:
Exposure of cell to ultra-violet light can result in the covalent joining of two adjacent pyrimidines usually thymines producing a dimmer.
These thymine dimmers prevent DNA polymerase from the replicating the DNA strand beyond the site of dimmer formation.
First a UV specific endo-nuclease also called UV specific excinuclease recognizes the dimmer and cleaves the damage strand. Damaged oglio-nucleotide is released.

Q-9. Biological role of metallothioneins is to sequester harmful metal ions. These bind:
a) Cd++, Cu++ & Zn++
b) AI+++, Hg++ & NH4+
c) Pt++, As+++ & PO4–
d) Fe+++, Na+ & K+

Answer: Cd++, Cu++ & Zn++
Explanation:
Metallothioneins are the group of the small proteins found in the cytosol of the cells, particularly of liver, kidney and intestine.
They have a high content of cysteine and can bind copper, zinc, cadmium and mercury.
The SH groups of cysteine are involved in binding of metals.

AIIMS NOV 2004
Q-1. Which one of the following amino acids is most likely to be found in the trans-membrane region of a protein?
a) Lysine
b) Arginine
c) Leucine
d) Aspartate

Answer: Leucine
Explanation:
Proteins tend to fold with the R- groups of amino acids with hydrophobic side chains in the interior.
Amino acids with charged or polar amino acids side chains (arginine, glutamine, and serine) generally are present on the surface.
Amino Acids with Aliphatic or Non-Polar Side Chains are Glycine, Alanine, Valine, Leucine and Iso-Leucine.

Q-2. Molecular weight of a protein can be determined by
a) Native Poly Acryl-amide Gel Electrophoresis (PAGE)
b) Sodium Dodecyl Sulfate PAGE
c) Iso-electric focusing
d) Ion Exchange Chromatography

Answer: Sodium Dodecyl Sulfate PAGE
Explanation:
The most widely used method for determining the purity of a protein is SDS – PAGE in the presence of the anionic detergent sodium dodecyl sulfate (SDS). Electrophoresis separates charged bio-molecules based on the rates at which they migrate in an applied electrical field.
Since the charge-to-mass ratio of each SDS polypeptide complex is approximately equal, the physical resistance each peptide encounters as it moves through the acryl-amide matrix determines the rate of migration.
Since large complexes encounter greater resistance, polypeptides separate based on their relative molecular mass (M), Individual polypeptides trapped in the acryl-amide gel are visualized by staining with dyes such as Coomassie blue.

AIIMS MAY 2005
Q-1. Membrane fluidity is increased by
a) Stearic acid
b) Palmitic acid
c) Cholesterol
d) Linoleic acid

Answer: Linoleic acid
Explanation:
Lipids with shorter chains are less stiff and less viscous because they are more susceptible to changes in kinetic energy due to their smaller molecular size and they have less surface area to undergo stabilizing van der Waals interactions with neighboring hydrophobic chains.
Lipid chains with double bonds are more fluid than lipids that are saturated with hydrogen and thus have only single bonds. On the molecular level, unsaturated double bonds make it harder for the lipids to pack together by putting kinks into the otherwise straightened hydrocarbon chain.
Examples of unsaturated fats are palmitoleic acid, oleic acid, linoleic acid, and arachidonic acid.
Cholesterol acts as a bidirectional regulator of membrane fluidity because at high temperatures, it stabilizes the membrane and raises its melting point, whereas at low temperatures it intercalates between the phospholipids and prevents them from clustering together and stiffening.

Q-2. The molecule, which is the initiator of cataract formation in the eye lens and whose 1-phosphate derivative is responsible for liver failure is:
a) Sorbitol
b) Mannitol
c) Inositol
d) Galactitol

Answer: Galactitol
Explanation:
Galactose 1 phosphate uridyl-transferase is missing in individuals with classical Galactosemia.
Galactose 1 phosphate and therefore Galactose accumulate in cells.
Accumulated Galactose is shunted into side-pathways such as that of Galactitol production.
This reaction is catalyzed by aldose reductase, same enzyme that converts glucose to sorbitol.
More benign form of Galactosemia is caused by a deficiency of galactokinase.
Classical Galactosemia:
Uridyl-transferase deficiency
Autosomal recessive disorder
It causes Galactosemia and galactosuria, vomiting, diarrhoea and jaundice.
Accumulation of Galactose 1 phosphate and Galactitol in nerve, lens, liver and kidney tissue causes liver damage, severe mental retardation and cataracts.
Antenatal diagnosis is possible by chorionic villus sampling.
Therapy: Rapid removal Galactose i.e. lactose from the diet.

Q-3. What is most accurate effect of smoking cessation?
a) Shift of oxy-hemoglobin curve to the right
b) Increased ciliary function
c) Decreased mucous production
d) Decreased incidence of post operative pneumonia

Answer: Shift of oxy-hemoglobin curve to the right
Explanation:
Smoking causes an increase in carboxy-hemoglobin levels, resulting in a leftward shift.
There are two mechanisms responsible for the leftward shift of oxy-hemoglobin dissociation curve when carbon monoxide is present in the blood.
Carbon monoxide has a direct effect on oxy-hemoglobin, causing a leftward shift of the oxygen dissociation curve, and carbon monoxide also reduces the formation of 2, 3-DPG by inhibiting glycolysis in the erythrocyte.

AIIMS NOV 2005
Q-1. At physiological pH, the carboxy- terminal of a peptide is
a) Positively charged
b) Negatively charged
c) Neutral
d) Infinitely charged

Answer: Negatively charged
Explanation:
At physiological pH, all amino acids have both a negatively charged group, carboxyl ion and a positively charged group ammonium ion.

Q-2. Apart from occurring in nucleic acids, Pyrimidines are also found in:
a) Theophylline
b) Theobromine
c) Flavin mononucleotide
d) Thiamine

Answer: Thiamine
Explanation:
The pyrimidine ring system has wide occurrence in nature as substituted and ring fused compounds and derivatives, including the nucleotides, thiamine (vitamin B-1) and alloxan.
It is also found in many synthetic compounds such as barbiturates and the HIV drug, zidovudine.

Q-3. By which of the following anticoagulants used in estimating blood glucose, glycolysis is prevented:
a) EDTA
b) Heparin
c) Sodium fluoride
d) Sodium citrate

Answer: Sodium fluoride
Explanation:
Fluoride inhibits glycolysis by inhibiting enolase; this will help in correct determination of the glucose level as glycolysis will not be able to affect the level of glucose.

Q-4. Which of the following is a component of the visual pigment Rhodopsin?
a) b-carotene
b) Retinal
c) Retinol
d) Retinoic acid

Answer: Retinal
Explanation:
The retinoids are related to retinol (Vitamin A) are essential for vision, reproduction, growth and maintenance of epithelial tissues.
Retinoic acid derived from oxidation of dietary retinol, mediates most of the action of the retinoids except for vision, which depends on retinal (Retinal-aldehyde), the aldehyde derivative of retinol.

Q-5. Vitamin A is stored mainly as retinol esters in:
a) Kidney
b) Muscle
c) Liver
d) Retina

Answer: Liver
Explanation:
Retinol esters present in the diet are hydrolyzed in the intestinal mucosa, releasing retinol and free fatty acids.
Retinol is re-esterified to long chain fatty acids in the intestinal mucosa and secreted as a component of chylomicrons into the lymphatics system.
Retinol esters contained in chylomicrons are taken up by and stored in the liver.

Q-6. Xeroderma pigmentosa is caused due to a group of closely related abnormalities in:
a) Mismatch repair
b) Base excision repair
c) Nucleotide excision repair
d) SOS repair

Answer: Nucleotide excision repair
Explanation:
Xeroderma pigmentosum is a rare disorder transmitted in an autosomal recessive manner.
The basic defect in xeroderma pigmentosum is in nucleotide excision repair (NER), leading to deficient repair of DNA damaged by UV radiation.
The most common form of this disease is caused by the absence of the UV-specific excinuclease.

Q-7. Substitution of which one of the following amino acids in place of alanine would increase the absorbance of protein at 280 nm:
a) Leucine
b) Arginine
c) Tryptophan
d) Proline

Answer: Tryptophan
Explanation:
Amino acids do not absorb visible light and thus are colorless. However, tyrosine, phenylalanine, and especially tryptophan absorb high-wavelength (250-290 nm) ultraviolet light. Because it absorbs ultraviolet light about ten times more efficiently than either phenylalanine or tyrosine, tryptophan makes the major contribution to the ability of most proteins of absorb light in the region of 280 nm.

Q-8. Secretory proteins are synthesized in:
a) Cytoplasm
b) Endoplasmic reticulum
c) First in cytoplasm and then in Endoplasmic Reticulum
d) First in Endoplasmic Reticulum and then in cytoplasm

Answer: Endoplasmic reticulum
Explanation:
Rough Endoplasmic Reticulum -> Golgi body->Secretory vesicles

Q-9. To synthesize insulin on a large scale basis, the most suitable starting material obtained from the beta cells of the pancreas is:
a) Genomic DNA
b) Total cellular RNA
c) c-DNA of insulin
d) m-RNA of insulin

Answer: m-RNA of insulin
Explanation:
Insulin is synthesized on a large scale basis from complementary DNA (c-DNA), but this is not the starting material because these c-DNAs are derived from mRNAs with the help of Reverse transcriptase and DNA polymerase 1.
The c-DNA produced by this method is inserted into an appropriate cloning vector which produces a large number of c-DNA clones.

Q-10. If cellular proteins do not fold into a specific conformation, their functions are affected. Certain disorders arise, if specific proteins are mis-folded. Which of the following disorders arises due to conformational isomerization?
a) Fatal familial insomnia
b) Hepatitis delta
c) Pernicious anemia
d) Lesch-Nyhan syndrome

Answer: Familial fatal insomnia
Explanation:
The term “prions” refers to abnormal, pathogenic agents that are transmissible and are able to induce abnormal folding of specific normal cellular proteins called prion proteins that are found most abundantly in the brain.
Prion diseases are usually rapidly progressive and always fatal.
Human Prion Diseases:
Creutzfeldt – Jakob disease (CJD)
Variant Creutzfeldt – Jakob disease (vCJD)
Gerstmann-Sträussler-Scheinker Syndrome
Fatal Familial Insomnia
Kuru
Animal Prion Diseases:
Bovine Spongiform Encephalopathy (BSE)
Chronic Wasting Disease (CWD)
Scrapie
Transmissible mink encephalopathy
Feline spongiform encephalopathy
Ungulate spongiform encephalopathy