লগ ইননিবন্ধন
পরীক্ষা আর্কাইভ

৪৯তম বিসিএস ⎯ প্রাণ রসায়ন [৬০১]

পরীক্ষা৪৯তম বিসিএস ⎯ প্রাণ রসায়ন [৬০১]তারিখতারিখ অনির্ধারিতসময়30 minutes
মোট প্রশ্ন৫০
সিলেবাস
Exam 9 Metabolism — Enzymes; Carbohydrate and Protein metabolism [Source: Class–6 and relevant books]
প্রশ্ন দেখুনঅনুশীলন
ঘনত্ব
উত্তর
উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

৪৯তম বিসিএস ⎯ প্রাণ রসায়ন [৬০১]

৪৯তম বিসিএস ⎯ প্রাণ রসায়ন [৬০১] · তারিখ অনির্ধারিত · ৫০ প্রশ্ন

১.
Which statement best describes a fundamental characteristic of enzymes?
  1. ক
    They are consumed during the reaction they catalyze.
  2. খ
    They alter the final equilibrium constant (Keq​) of a reaction.
  3. গ
    They are highly specific in their choice of substrate.
  4. ঘ
    They are typically small, simple inorganic molecules.
ব্যাখ্যা

Answer: C
Explanation: Enzymes exhibit remarkable substrate specificity. This is due to the unique three-dimensional structure of their active site, which is precisely shaped to bind to a specific substrate, much like a lock and key. Enzymes are catalysts and are not consumed; they do not change the reaction's equilibrium, only the rate at which it is reached; and they are large.

২.
An enzyme that catalyzes the transfer of electrons from a donor (reductant) to an acceptor (oxidant) belongs to which class?
  1. ক
    Transferases
  2. খ
    Hydrolases
  3. গ
    Oxidoreductases
  4. ঘ
    Lyases
ব্যাখ্যা

Explanation: Oxidoreductases (EC 1) are the class of enzymes that catalyze oxidation-reduction (redox) reactions. A common example is lactate dehydrogenase, which catalyzes the interconversion of pyruvate and lactate with the concomitant interconversion of NADH and NADH H+

৩.
The enzyme pepsin uses water to break peptide bonds in dietary proteins. To which major enzyme class does it belong?
  1. ক
    Lyases
  2. খ
    Ligases
  3. গ
    Transferases
  4. ঘ
    Hydrolases
ব্যাখ্যা

Explanation: Hydrolases (EC 3) are enzymes that catalyze the cleavage of a chemical bond through the addition of a water molecule (a process called hydrolysis). Proteases like pepsin and chymotrypsin are classic examples of hydrolases.

৪.
DNA ligase joins two separate DNA fragments, a process that requires energy from ATP. This enzyme is classified as a: 
  1. ক
    Isomerase
  2. খ
    Hydrolase
  3. গ
    Ligase
  4. ঘ
    Lyase
ব্যাখ্যা

Answer: C
Explanation: Ligases (EC 6) catalyze the joining (ligation) of two molecules, a reaction that is coupled with the hydrolysis of a high-energy phosphate bond in ATP or another nucleotide. This energy input is required to form the new chemical bond, such as the phosphodiester bond formed by DNA ligase.

৫.
The "induced-fit" model of enzyme-substrate interaction, proposed by Daniel Koshland, suggests that the:
  1. ক
    Active site is a rigid, pre-shaped pocket for the substrate.
  2. খ
    Substrate binds to a random location on the enzyme surface.
  3. গ
    Binding of the substrate causes a conformational change in the active site for an optimal fit.
  4. ঘ
    Enzyme and substrate are perfect complements before any binding occurs.
ব্যাখ্যা

Answer: C 
Explanation: The induced-fit model refines the older "lock-and-key" theory. It posits that the active site is flexible. The initial binding of the substrate induces a subtle change in the enzyme's three-dimensional structure, optimizing the alignment of catalytic groups and strengthening the binding interaction, which facilitates catalysis.

৬.
 How is one International Unit (IU) of enzyme activity defined?
  1. ক
    The amount of enzyme that catalyzes the conversion of 1 mole of substrate per second.
  2. খ
    The amount of enzyme that catalyzes the conversion of 1 micromole (μmol) of substrate per minute.
  3. গ
    The concentration of enzyme needed to achieve half of the maximum velocity.
  4. ঘ
    The amount of enzyme that produces 1 milligram of product per second.
ব্যাখ্যা

Explanation: The International Unit (IU) is a standard measure of enzyme activity. It is precisely defined as the amount of enzyme that catalyzes the formation of one micromole of product per minute under specified assay conditions (e.g., pH, temperature, substrate concentration).

৭.
The term kcat​, or the turnover number, represents the:
  1. ক
    Affinity of the enzyme for its substrate.
  2. খ
    Number of substrate molecules converted to product per unit time by a single enzyme molecule at saturation.
  3. গ
    Concentration of substrate required for maximal enzyme activity.
  4. ঘ
    Overall free energy change of the reaction.
ব্যাখ্যা

Answer: B
Explanation: The turnover number (kcat​) is a measure of an enzyme's maximum catalytic efficiency. It tells us how many reactions a single active site can catalyze per unit of time when it is fully saturated with substrate. It is calculated as kcat​=Vmax​/[Et​], where [Et​] is the total enzyme concentration.

৮.
Extreme changes in pH away from the optimum can cause an irreversible loss of enzyme activity primarily by:
  1. ক
    Breaking the peptide bonds that form the enzyme's primary structure.
  2. খ
    Altering the ionization state of key amino acid residues and disrupting the enzyme's tertiary structure.
  3. গ
    Increasing the kinetic energy of the substrate molecules.
  4. ঘ
    Causing the enzyme to aggregate through hydrophobic interactions.
ব্যাখ্যা

Explanation: An enzyme's activity is highly dependent on pH. Deviations from the optimum pH alter the protonation state of ionizable amino acid side chains (like Asp, Glu, His, Lys). This can disrupt the precise ionic and hydrogen bonds maintaining the enzyme's crucial 3D structure (leading to denaturation) and can also affect the charge of residues directly involved in substrate binding and catalysis at the active site.

৯.
 As temperature increases from low to optimal, the rate of an enzyme-catalyzed reaction increases because:
  1. ক
     The substrate becomes more soluble.
  2. খ
    The enzyme's primary structure is altered.
  3. গ
    The equilibrium of the reaction shifts towards the products.
  4. ঘ
    The kinetic energy of both enzyme and substrate molecules increases, leading to more frequent and energetic collisions.
ব্যাখ্যা

Explanation: According to collision theory, increasing the temperature increases the kinetic energy of molecules. This results in more frequent collisions between the enzyme and substrate and ensures that more of these collisions have sufficient energy to overcome the activation energy barrier. Beyond the optimal temperature, the enzyme rapidly denatures, causing a sharp drop in activity.

১০.
At a substrate concentration [S] that is not limiting, the rate of an enzyme-catalyzed reaction is:
  1. ক
    Inversely proportional to the enzyme concentration.
  2. খ
    Directly proportional to the enzyme concentration.
  3. গ
    Independent of the enzyme concentration.
  4. ঘ
    Proportional to the square of the enzyme concentration.
ব্যাখ্যা

Answer: B Explanation: When the substrate is in excess (saturating), the reaction rate is limited only by the amount of enzyme available to process it. Therefore, if you double the concentration of the enzyme, you double the number of active sites available, and the overall reaction rate (V0​) will also double. This demonstrates a direct, linear relationship.

১১.
  1. ক
    Initial reaction velocity and temperature.
  2. খ
    Enzyme concentration and maximum velocity.
  3. গ
    Initial reaction velocity and substrate concentration.
  4. ঘ
    Substrate concentration and pH.
ব্যাখ্যা

Answer: C
Explanation: The Michaelis-Menten equation is a cornerstone of enzyme kinetics. It mathematically models how the initial rate of reaction (V0​) changes in response to varying concentrations of the substrate ([S]), for a fixed amount of enzyme.

১২.
The Michaelis constant, Km​, is operationally defined as:
  1. ক
    The substrate concentration at which the reaction velocity is equal to Vmax​.
  2. খ
    The maximum velocity divided by the total enzyme concentration.
  3. গ
    The substrate concentration at which the reaction velocity is half of Vmax​.
  4. ঘ
    The rate constant for the dissociation of the enzyme-substrate complex.
ব্যাখ্যা

Answer: C
Explanation: By definition, Km​ is the substrate concentration required for the enzyme to achieve half of its maximum velocity (Vmax​/2). It is a characteristic property of an enzyme for a particular substrate and is an indicator of the affinity between them.

১৩.
The maximum velocity, Vmax​, of an enzyme-catalyzed reaction is:
  1. ক
    Directly proportional to the Km​ value.
  2. খ
    Achieved when the substrate concentration is equal to Km​.
  3. গ
    Independent of the enzyme concentration.
  4. ঘ
    Directly proportional to the total enzyme concentration [Et​].
ব্যাখ্যা

Answer: D
Explanation: Vmax​ is the theoretical maximum rate when all enzyme active sites are saturated with substrate. This rate is limited by the speed of catalysis. Therefore, the more enzyme molecules you have, the higher the maximum possible rate. This relationship is expressed as Vmax = kcat​[Et​].

১৪.
Which parameter is considered the best measure of an enzyme's overall catalytic efficiency and substrate specificity?
  1. ক
    Vmax​ only
  2. খ
    Km​ only C
  3. গ
    The specificity constant, kcat​/Km
  4. ঘ
    The ratio Km​/Vmax​
ব্যাখ্যা

Answer: C Explanation: The specificity constant (kcat​/Km​) combines the enzyme's catalytic rate (kcat​) and its affinity for the substrate (inversely related to Km​). It describes how efficiently an enzyme converts substrate into product at low substrate concentrations, making it the most comprehensive measure of catalytic performance.

১৫.
 In a Lineweaver-Burk plot, which is a double reciprocal plot of enzyme kinetics, the y-intercept represents:
  1. ক
    Km​
  2. খ
    Vmax​
  3. গ
    1/Vmax
  4. ঘ
    −1/Km​
ব্যাখ্যা

১৬.
 In competitive inhibition, the inhibitor molecule:
  1. ক
    Binds to a site on the enzyme other than the active site.
  2. খ
    Binds irreversibly to the active site, forming a covalent bond.
  3. গ
    Structurally resembles the substrate and binds reversibly to the active site.
  4. ঘ
    Binds only to the enzyme-substrate (ES) complex.
ব্যাখ্যা

Answer: C
Explanation: A competitive inhibitor "competes" with the substrate for the same active site because it typically has a chemical structure similar to the substrate. This binding is reversible, and its effect can be overcome by increasing the substrate concentration, which will outcompete the inhibitor for access to the active site.

১৭.
How does a pure non-competitive inhibitor affect the kinetic parameters of an enzyme?
  1. ক
    It increases Km​ and does not change Vmax​.
  2. খ
    It decreases Vmax​ and does not change Km​.
  3. গ
    It decreases both Km​ and Vmax​.
  4. ঘ
    It does not change Km​ and does not change Vmax​.
ব্যাখ্যা

Answer: B
Explanation: A non-competitive inhibitor binds to an allosteric (non-active) site on the enzyme, regardless of whether the substrate is bound. This binding alters the enzyme's conformation, reducing its catalytic efficiency without affecting its ability to bind the substrate. As a result, Vmax​ is decreased, but the affinity for the substrate, and thus Km​, remains unchanged.

১৮.
An uncompetitive inhibitor is characterized by its ability to:
  1. ক
    Bind to either the free enzyme or the enzyme-substrate complex.
  2. খ
    Bind only to the free enzyme at the active site.
  3. গ
    Bind only to the enzyme-substrate (ES) complex.
  4. ঘ
    Increase Vmax​ while decreasing Km​.
ব্যাখ্যা

Answer: C
Explanation: Uncompetitive inhibition is a unique case where the inhibitor can bind only to the pre-formed enzyme-substrate (ES) complex, forming an inactive ESI complex. This action effectively removes the ES complex from the reaction, leading to a decrease in both Vmax​ and Km​.

১৯.
Penicillin works by forming a stable, covalent bond with the enzyme glycopeptide transpeptidase, which is essential for bacterial cell wall synthesis. Penicillin is therefore an example of a:
  1. ক
    Allosteric activator
  2. খ
    Competitive inhibitor
  3. গ
    Uncompetitive inhibitor
  4. ঘ
    Irreversible inhibitor
ব্যাখ্যা

Answer: D
Explanation: Irreversible inhibitors, also known as inactivators or suicide substrates, bind very tightly to an enzyme, often by forming a covalent bond with a key amino acid residue in the active site. This permanently inactivates the enzyme molecule.

২০.
The net ATP gain from one molecule of glucose in glycolysis under anaerobic conditions is:
  1. ক
    1 ATP
  2. খ
    2 ATP
  3. গ
    3 ATP
  4. ঘ
    4 ATP
ব্যাখ্যা

Answer: B) 2 ATP
Explanation: Glycolysis produces 4 ATP, but 2 ATP are consumed during the preparatory phase, resulting in a net gain of 2 ATP.

২১.
The enzyme responsible for the irreversible phosphorylation of glucose to glucose-6-phosphate in the first step of glycolysis is:
  1. ক
    Hexokinase
  2. খ
    Phosphofructokinase-1
  3. গ
    Glucokinase
  4. ঘ
    Pyruvate kinase
ব্যাখ্যা

Answer: A) Hexokinase
Explanation: Hexokinase catalyzes the first step of glycolysis, converting glucose to glucose-6-phosphate using 1 ATP.

২২.
Phosphofructokinase-1 (PFK-1) is allosterically activated by:
  1. ক
    ATP
  2. খ
    Citrate
  3. গ
    AMP
  4. ঘ
    NADH
ব্যাখ্যা

Answer: C) AMP
Explanation: AMP signals low energy status and activates PFK-1, promoting glycolysis. ATP and citrate inhibit it.

২৩.
The end product of glycolysis under aerobic conditions is:
  1. ক
    Lactate
  2. খ
     Acetyl-CoA
  3. গ
    Pyruvate
  4. ঘ
    Ethanol
ব্যাখ্যা

Answer: C) Pyruvate
Explanation: In aerobic conditions, glucose is broken down into two molecules of pyruvate, which then enter the TCA cycle.

২৪.
Which step of glycolysis produces NADH?
  1. ক
    Glucose → Glucose-6-phosphate
  2. খ
    Fructose-1,6-bisphosphate → Glyceraldehyde-3-phosphate
  3. গ
    Glyceraldehyde-3-phosphate → 1,3-bisphosphoglycerate
  4. ঘ
    Phosphoenolpyruvate → Pyruvate
ব্যাখ্যা

Answer: C) Glyceraldehyde-3-phosphate → 1,3-bisphosphoglycerate
Explanation: This reaction, catalyzed by G3P dehydrogenase, produces NADH.

২৫.
Pyruvate kinase catalyzes which type of reaction in glycolysis?
  1. ক
    Oxidation-reduction
  2. খ
    Substrate-level phosphorylation
  3. গ
    Hydrolysis
  4. ঘ
    Decarboxylation
ব্যাখ্যা

Answer: B) Substrate-level phosphorylation
Explanation: Pyruvate kinase converts phosphoenolpyruvate (PEP) to pyruvate, producing ATP directly.

২৬.
Which molecule is a common inhibitor of both PFK-1 and pyruvate kinase?
  1. ক
    AMP
  2. খ
    Citrate
  3. গ
    ATP
  4. ঘ
    NAD+ 
ব্যাখ্যা

Answer: C) ATP
Explanation: ATP acts as a negative feedback inhibitor, indicating high energy availability, thus slowing glycolysis.

২৭.
The total NADH yield per glucose molecule from glycolysis is:
  1. ক
    1
  2. খ
    2
  3. গ
    3
  4. ঘ
    4
ব্যাখ্যা

Answer: B) 2
Explanation: Each glyceraldehyde-3-phosphate produces 1 NADH, and since 2 G3P are formed per glucose, the total is 2 NADH.

২৮.
The first step of the TCA cycle involves the formation of:
  1. ক
    Oxaloacetate
  2. খ
    Citrate
  3. গ
    Succinyl-CoA
  4. ঘ
    Fumarate
ব্যাখ্যা

Answer: B) Citrate
Explanation: Acetyl-CoA combines with oxaloacetate to form citrate, catalyzed by citrate synthase.

২৯.
How many molecules of CO2 are released per turn of the TCA cycle?
  1. ক
    1
  2. খ
    2
  3. গ
    3
  4. ঘ
    4
ব্যাখ্যা

Answer: B) 2
Explanation: Each acetyl-CoA entering the cycle results in the release of two CO2 molecules.

৩০.
The enzyme isocitrate dehydrogenase produces:
  1. ক
    NADPH
  2. খ
    NADH and CO2
  3. গ
    ATP
  4. ঘ
    FADH2
ব্যাখ্যা

Answer: B) NADH and CO2
Explanation: This enzyme catalyzes oxidative decarboxylation of isocitrate to α-ketoglutarate, releasing NADH and CO2.

৩১.
The TCA cycle generates how many ATP equivalents per acetyl-CoA?
  1. ক
    6
  2. খ
    8
  3. গ
    10
  4. ঘ
    12
ব্যাখ্যা

Answer: C) 10
Explanation:

3 NADH = 7.5 ATP
1 FADH2 = 1.5 ATP
1 GTP = 1 ATP
Total = 10 ATP

৩২.
Which step of the TCA cycle produces FADH2?
  1. ক
    Citrate → Isocitrate
  2. খ
    Succinate → Fumarate
  3. গ
    Malate → Oxaloacetate
  4. ঘ
    α-Ketoglutarate → Succinyl-CoA 
ব্যাখ্যা

answer: B) Succinate → Fumarate
Explanation: Succinate dehydrogenase catalyzes this reaction, reducing FAD to FADH2.

৩৩.
The TCA cycle is inhibited by:
  1. ক
    NAD+
  2. খ
    ADP
  3. গ
    NADH
  4. ঘ
    Pyruvate
ব্যাখ্যা

Answer: C) NADH
Explanation: High levels of NADH signal sufficient energy, inhibiting key enzymes like isocitrate dehydrogenase.

৩৪.
Which enzyme links glycolysis and the TCA cycle by converting pyruvate to acetyl-CoA?
  1. ক
     Pyruvate kinase
  2. খ
    Pyruvate carboxylase
  3. গ
    Pyruvate dehydrogenase
  4. ঘ
    Citrate synthase
ব্যাখ্যা

Answer: C) Pyruvate dehydrogenase
Explanation: The pyruvate dehydrogenase complex (PDC) catalyzes this key irreversible step.

৩৫.
he primary purpose of the pentose phosphate pathway is to produce:
  1. ক
    ATP and FADH2
  2. খ
    NADH and Ribose-5-phosphate
  3. গ
    NADPH and Ribose-5-phosphate
  4. ঘ
    ATP and NADPH
ব্যাখ্যা

Answer: C) NADPH and Ribose-5-phosphate
Explanation: NADPH is used for biosynthesis, and Ribose-5-phosphate is essential for nucleotide synthesis.

৩৬.
Which enzyme catalyzes the rate-limiting step of the oxidative phase of PPP?
  1. ক
    Glucose-6-phosphatase
  2. খ
    Glucose-6-phosphate dehydrogenase
  3. গ
    Transketolase
  4. ঘ
    Transaldolase
ব্যাখ্যা

Answer: B) Glucose-6-phosphate dehydrogenase
Explanation: This enzyme converts glucose-6-phosphate to 6-phosphoglucono-δ-lactone, producing NADPH.

৩৭.
Deficiency of Glucose-6-phosphate dehydrogenase (G6PD) leads to:
  1. ক
    Lactic acidosis
  2. খ
    Hemolytic anemia
  3. গ
    Diabetes mellitus
  4. ঘ
    Scurvy
ব্যাখ্যা

Answer: B) Hemolytic anemia
Explanation: G6PD deficiency reduces NADPH, impairing glutathione regeneration, leading to oxidative damage to RBCs.

৩৮.
The non-oxidative phase of PPP mainly involves:
  1. ক
    ATP production
  2. খ
    Interconversion of sugars
  3. গ
    NADPH production
  4. ঘ
    CO2 release
ব্যাখ্যা

Answer: B) Interconversion of sugars
Explanation: This phase generates various sugars like ribose-5-phosphate and fructose-6-phosphate.

৩৯.
Gluconeogenesis occurs primarily in the:
  1. ক
    Brain and muscle
  2. খ
    Liver and kidney
  3. গ
    Pancreas
  4. ঘ
    Adipose tissue
ব্যাখ্যা

Answer: B) Liver and kidney
Explanation: These organs maintain blood glucose levels during fasting by synthesizing glucose.

৪০.
Which enzyme bypasses pyruvate kinase in gluconeogenesis?
  1. ক
    Glucose-6-phosphatase
  2. খ
    Pyruvate carboxylase and PEP carboxykinase
  3. গ
    Fructose-1,6-bisphosphatase
  4. ঘ
    Hexokinase
ব্যাখ্যা

Answer: B) Pyruvate carboxylase and PEP carboxykinase
Explanation: Pyruvate → Oxaloacetate (via pyruvate carboxylase) → PEP (via PEP carboxykinase).

৪১.
The first committed step of gluconeogenesis is catalyzed by:
  1. ক
    PEP carboxykinase
  2. খ
    Glucose-6-phosphatase
  3. গ
    Pyruvate carboxylase
  4. ঘ
    Fructose-1,6-bisphosphatase
ব্যাখ্যা

Answer: C) Pyruvate carboxylase
Explanation: This enzyme converts pyruvate to oxaloacetate using ATP and biotin.

৪২.
Which of the following is not a substrate for gluconeogenesis?
  1. ক
    Glycerol
  2. খ
    Alanine
  3. গ
    Lactate
  4. ঘ
    Acetyl-CoA 
ব্যাখ্যা

Answer: D) Acetyl-CoA
Explanation: Acetyl-CoA cannot be converted back to glucose because the pyruvate dehydrogenase reaction is irreversible.

৪৩.
The regulation of gluconeogenesis is strongly influenced by the energy status of the cell. High ATP levels will:
  1. ক
    Activate glycolysis
  2. খ
    Inhibit gluconeogenesis
  3. গ
    Activate gluconeogenesis
  4. ঘ
    Activate both glycolysis and gluconeogenesis
ব্যাখ্যা

Answer: C) Activate gluconeogenesis
Explanation: High ATP indicates sufficient energy, shifting metabolism toward glucose production rather than breakdown.

৪৪.
The enzyme fructose-1, 6-bisphosphatase is inhibited by:
  1. ক
    ATP
  2. খ
    AMP
  3. গ
    Citrate
  4. ঘ
    Acetyl-CoA
ব্যাখ্যা

Answer: B) AMP
Explanation: AMP signals low energy status and inhibits gluconeogenesis, favoring glycolysis instead.

৪৫.
 The central molecule that accepts amino groups in transamination reactions is:
  1. ক
    Glutamate
  2. খ
    Alpha-ketoglutarate
  3. গ
    Oxaloacetate
  4. ঘ
    Pyruvate
ব্যাখ্যা

Explanation: (b) Alpha-ketoglutarate is the primary amino group acceptor in most transamination reactions, forming glutamate. Glutamate (a) is often the donor in subsequent reactions. Oxaloacetate (c) and pyruvate (d) can also act as acceptors in specific transaminase reactions, but alpha-ketoglutarate is the most universal.

৪৬.
The reaction catalyzed by glutamate dehydrogenase is significant because it:
  1. ক
    Irreversibly fixes nitrogen into an organic molecule.
  2. খ
    Provides a direct link between amino acid and carbohydrate metabolism.
  3. গ
    Uses NAD+ to oxidatively deaminate glutamate, releasing free NH₄⁺.
  4. ঘ
    Is the only way to synthesize the amino acid glutamate.
ব্যাখ্যা

Explanation: (c) Glutamate dehydrogenase catalyzes the oxidative deamination of glutamate, using NAD⁺ (or NADP⁺) as a cofactor, which results in the production of alpha-ketoglutarate and free ammonium ion (NH₄⁺). This is a key step in channeling amino nitrogen into the urea cycle.

৪৭.
An amino acid that is solely ketogenic is:
  1. ক
    Leucine
  2. খ
    Lysine
  3. গ
    Phenylalanine
  4. ঘ
    Isoleucine
ব্যাখ্যা

Explanation: (b) Lysine and leucine are the only two amino acids that are exclusively ketogenic, meaning their carbon skeletons are degraded only to acetyl-CoA or acetoacetate, not to glucose precursors. Phenylalanine (c) and isoleucine (d) are both glucogenic and ketogenic.

৪৮.
The urea cycle occurs primarily in the:
  1. ক
    Kidneys and cytosol
  2. খ
    Liver and mitochondria
  3. গ
    Muscle and cytosol
  4. ঘ
    Brain and mitochondria
ব্যাখ্যা

Explanation: (b) The urea cycle is a hepatic (liver) process. Its first two steps occur in the mitochondrial matrix, and the subsequent three steps occur in the cytosol.

৪৯.
The purpose of transamination is to:
  1. ক
    Remove the amino group as ammonia.
  2. খ
    Transfer an amino group from one amino acid to a keto acid, forming a new pair.
  3. গ
    Synthesize non-essential amino acids from ammonia.
  4. ঘ
    Decarboxylate amino acids to form neurotransmitters.
ব্যাখ্যা

Explanation: (b) Transamination is a reversible reaction that involves the transfer of an α-amino group from an amino acid to an α-ketoacid, resulting in a new amino acid and a new ketoacid. This is the primary mechanism for redistributing nitrogen.

৫০.
The molecule carbamoyl phosphate, a substrate for the urea cycle, is synthesized in the:
  1. ক
    Cytosol from citrulline and ATP
  2. খ
    Mitochondria from aspartate and ATP
  3. গ
    Cytosol from glutamine and HCO3-
  4. ঘ
     Mitochondria from CO2, NH4+, and ATP
ব্যাখ্যা

Explanation: Carbamoyl phosphate is synthesized in the mitochondrial matrix by the enzyme carbamoyl phosphate synthetase I (CPS I), which uses CO2, NH4+ (ammonium ion), and 2 ATP molecules.