Biochemistry MCQ Quiz - Objective Question with Answer for Biochemistry - Download Free PDF

The correct answer is 1) A (iii), B (ii), C (i), D (i)

Concept:

• DNA and RNA can adopt various structural conformations based on their sequence, environment, and functional requirements. These conformations are classified into A-form, B-form, and Z-form.
• Each form has unique structural features such as handedness, base pair orientation, base pair positioning, and the number of base pairs per helical turn.
• The A-form is typical for RNA double helices and dehydrated DNA, the B-form is the most common conformation for DNA under physiological conditions, and the Z-form is a left-handed helical structure observed in specific DNA sequences.

Explanation:

A (iii): Left-handed → Z form

• The Z-form DNA is a left-handed helix, in contrast to the typical right-handed helix observed in A and B forms.
• It is observed in sequences with alternating purines and pyrimidines (e.g., CGCGCG) and under high salt concentrations.

B (ii): Number of base pairs per turn is 10 → B form

• The B-form DNA is the most common form of DNA in physiological conditions.
• It has 10 base pairs per helical turn and is right-handed.

C (i): The base pairs are off-centered → A form

• The A-form helix is more compact than the B-form, and its base pairs are tilted and displaced from the central axis, giving an off-centered appearance.
• This form is typical for RNA double helices and DNA-RNA hybrids.

D (i): RNA double helix → A form

• RNA typically adopts the A-form helix due to the presence of the 2'-hydroxyl group on the ribose sugar, which favors this structure.
• The A-form is more compact and has a deep, narrow major groove compared to the B-form.

The correct answer is (21 to 22) and (-7.5 to -7.7) kJ/mol

Concept:

• To determine the equilibrium constant (K'ₑ_q) and Gibbs free energy change (ΔG°') for a reaction, we use principles from thermodynamics and chemical equilibrium.
• The reaction in question is Glucose 1-phosphate ⇋ Glucose 6-phosphate
• The equilibrium constant (K'_eq) is calculated using the ratio of the concentrations of products to reactants at equilibrium.

ΔG°' (standard Gibbs free energy change) is calculated using the equation:
ΔG°' = -RT ln(K'ₑ_q), where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (25°C = 298 K).

Explanation:

Equilibrium Constant Calculation (K'_eq):

• At equilibrium, the concentrations are given:
  • Glucose 1-phosphate = 4.5 × 10⁻³ M
  • Glucose 6-phosphate = 9.5 × 10⁻² M
• Equilibrium constant formula: K'_eq = [Glucose 6-phosphate] / [Glucose 1-phosphate]
• Substitute the values: K'_eq = (9.5 × 10⁻²) / (4.5 × 10⁻³) = 21.11 (approximately 21 to 22).

Gibbs Free Energy Change Calculation (ΔG°'):

• Formula: ΔG°' = -RT ln(K'_eq), where:
  • R = 8.314 J/mol·K
  • T = 298 K
  • K'_eq = 21.11
• First, calculate ln(K'_eq): ln(21.11) ≈ 3.05.
• Substitute into the equation: ΔG°' = - (8.314 × 298 × 3.05) J/mol = -7,540 J/mol.
• Convert to kJ/mol: ΔG°' ≈ -7.5 to -7.7 kJ/mol.

The correct answer is Cys mutated to Ile causes steric hindrance in the core of the protein, while mutation to Ala does not.

Concept:

• Proteins are composed of amino acids, and their structure and function are highly dependent on the precise arrangement of these amino acids in three-dimensional space.
• A mutation in an amino acid can disrupt the protein's stability, solubility, or functionality due to changes in size, charge, or hydrophobicity.
• Cysteine (Cys) has unique properties due to its sulfur-containing side chain. It can participate in disulfide bonds or stabilize proteins in a non-surface exposed region.
• Isoleucine (Ile) is a bulky, hydrophobic amino acid, while alanine (Ala) is smaller and less bulky compared to Ile.

Explanation:

• When Cysteine (Cys) is mutated to Isoleucine (Ile), the bulky side chain of Ile introduces steric hindrance in the core of the protein. This disrupts the tightly packed arrangement of amino acids within the protein, leading to protein instability and aggregation, which causes precipitation.
• In contrast, when Cys is mutated to Alanine (Ala), the smaller side chain of Ala does not introduce significant steric hindrance. This allows the protein to maintain its structural integrity, solubility, and functionality similar to its native form.
• The non-surface exposed region where Cys is located is likely crucial for the protein's stability, and introducing a bulky side chain like Ile interferes with this stability.
• This steric hindrance is a key factor that explains why the mutation to Ile leads to precipitation, while the mutation to Ala does not.

Other Options:

• "Cys mutated to Ile alters the net charge of the protein, while mutation to Ala does not.": This is incorrect because neither Ile nor Ala alters the net charge of the protein significantly. Both residues are nonpolar and neutral, so the observed effects are not due to changes in charge.
• "Cys mutated to Ala alters the net charge of the protein, while mutation to Ile does not.": This is incorrect for the same reason as Option 1. Ala, like Ile, is nonpolar and neutral, so the net charge remains unaffected.
• "Cys mutated to Ala causes steric hindrance in the core of the protein, while mutation to Ile does not.": This is incorrect because Ala is smaller and does not introduce steric hindrance. Steric hindrance is caused by the bulkier Ile side chain, not Ala.

The correct answer is 1.65 × 10^-3 M to 1.75 × 10^-3 M

Concept:

• Enzymes follow Michaelis-Menten kinetics, which describe the relationship between the reaction rate (v), the substrate concentration ([S]), the Michaelis constant (Km), and the maximum reaction rate (Vmax).
• The equation is given as: v = (Vmax × [S]) / (Km + [S])
• Vmax is the maximum rate achieved by the enzyme, and it can be calculated as: Vmax = k_cat × [E], where [E] is the enzyme concentration.
• The Michaelis constant (Km) is the substrate concentration at which the reaction rate is half of Vmax.
• To find the substrate concentration at which the reaction is one-quarter of the maximum rate, we solve for [S] when v = 0.25 × Vmax using the Michaelis-Menten equation.

Explanation:

• The enzyme's k_cat is 30 s^-1, and Km is 0.005 M.
• At one-quarter of the maximum rate, v = 0.25 × Vmax.

Substitute v = 0.25 × Vmax into the Michaelis-Menten equation: 

• 0.25 × Vmax = (Vmax × [S]) / (Km + [S])

Simplify the equation by canceling out Vmax: 0.25 = [S] / (Km + [S])

Rearrange to solve for [S]: 

[S] = 0.25 × (Km + [S])

Expand and rearrange:

• 0.25Km + 0.25[S] = [S]
• 0.25Km = [S] - 0.25[S]
• 0.25Km = 0.75[S]
• [S] = 0.25Km / 0.75
• [S] = Km / 3

Substitute Km = 0.005 M:
[S] = 0.005 / 3 = 0.00167 M, approximately in the range of 1.65 × 10^-3 M to 1.75 × 10^-3 M.

The correct answer is Peptide 1

Concept:

• Beta-sheets are a type of secondary structure in proteins formed by hydrogen bonding between peptide strands. The strands can be oriented in parallel or antiparallel fashion.
• The length of a peptide within a beta-sheet is determined by the number of residues involved in the hydrogen bonding, as well as the connecting sequences between strands.
• In the given scenario, three strands of a beta-sheet are shown with specific orientations (parallel or antiparallel), and the smallest length of connecting sequences is assumed for determining the peptide lengths.

Explanation:

• The correct answer is Peptide 1 .
  • Peptide 1 is the shortest because it follows an antiparallel Beta sheet with minimal loop length.
  • Peptides 3 and 4 are of equal length as they are positioned similarly in the beta-sheet and have identical residues, along with similar connectivity and orientation.
  • Peptide 2 is the longest because it indicates a parallel beta sheet with the longest loop length.

Concept:

• Enzyme inhibitor is any substance that prevents the enzyme-substrate reaction.
• In reversible inhibition, an inhibitor called reversible inhibitor binds noncovalently to the enzyme and dissociates rapidly from the enzyme.
• The effect of a reversible inhibitor can be reversed after the dissociation of inhibitor from the enzyme.
• There are three types of reversible inhibition- competitive, uncompetitive and mixed (noncompetitive inhibition).

Explanation:

• Uncompetitive inhibition is a type of inhibition in which an inhibitor binds to an enzyme-substrate complex (ES complex). It does not bind to free enzymes.
• Uncompetitive inhibitors decrease both Km and Vmax. This is due to the binding of the inhibitor to the ES complex.
• ES complex does not break into products and free enzymes.
• In uncompetitive inhibitors, apparent Vmax and Km both decrease.
• In mixed inhibition, the inhibitor binds to the enzyme at a site other than the active site but it binds to either free enzyme or enzyme-substrate complex.
• A special case of mixed inhibition is non-competitive inhibition. In this substrate and inhibitor bind at different sites on the enzyme and the binding of the inhibitor does not affect the binding of substrate.
• So, in noncompetitive inhibition, Vmax decreases, and Km remains unchanged.

Additional information:

• Km is also known as Michaelis Menten Constant. It is defined as the substrate concentration at which the reaction rate reaches half of its maximum value.
• Vmax or maximum velocity is the rate of reaction at which enzyme is saturated with substrate.

So, the correct answer is option 1. 

Concept:

Michaelis Menten Equation

• V₀ = measured initial velocity of an enzymatic reaction,
• V_max = reaction's maximum velocity
• K_m = Michaelis-Menten constant

Explanation:

Given - 

• Km = 10 mM 
• Vmax = 100 μmol/min
• S = 10 mM

Applying Michaelis Menten Equation:

• V₀ = 50 μmol/min

Therefore, the correct answer is option A (50 μmol/min).

Concept:

• Electrons are transferred from NADH/FADH₂ to oxygen through a series of electron carriers present on the inner mitochondrial membrane.
• The process of electron transport begins when the hydride ion is removed from NADH and is converted into a proton or two electrons.
• The electron transport chain consists of four major respiratory enzyme complexes in the inner mitochondrial membrane.

Important Points

• Iron-sulfur (Fe-S) clusters are a prosthetic group that consists of inorganic sulfide-linked non-heme iron. They are an important part of metalloproteins involved in the electron transport chain.
• They are best known for participation in oxidation-reduction reactions in photosynthetic electron transport in thylakoid membranes and respiratory electron transport in the inner mitochondrial membrane.

NADH-Coenzyme Q reductase or NADH dehydrogenase -

• It is the complex I that consists of 46 subunits and FMN (flavin mononucleotide) and Fe-S as the prosthetic group.
• It transports electrons from NADH to coenzyme Q.
• During the transport of each pair of electrons from NADH to coenzyme Q, complex I pumps four protons across the inner mitochondrial membrane.  

Succinate-Coenzyme Q reductase (succinate dehydrogenase) -

• It is known as complex II and consists of 4 subunits, FAD (flavin adenine dinucleotide) and Fe-S as the prosthetic group.
• Succinate dehydrogenase converts succinate to fumarate during Kreb’s cycle.
• The two electrons released are first transferred to FAD, then to the Fe-S cluster, and finally to coenzyme Q.

Coenzyme Q-cytochrome c reductase or cytochrome bc₁ complex -

• It is known as complex III and consists of 11 subunits and heme and Fe-S as the prosthetic groups.
• In this complex, the electrons released from coenzyme Q follow two paths.
• In one path, electrons move through via Rieske iron-sulfur clusters and cytochrome c₁, directly to cytochrome c.
• In other pathways, electrons move through b-type cytochromes and reduce oxidized coenzyme Q.

Cytochrome c oxidase -

• It is known as complex IV and consists of 13 subunits and heme and Cu⁺ as the prosthetic group.
• It catalyzes the transfer of electrons from the reduced form of cytochromes c to molecular oxygen.

So, the correct answer is option 3. 

The correct answer is Pyruvate kinase : Fructose-1,6-bisphosphate

Explanation:

• Pyruvate kinase is a key enzyme in glycolysis that catalyzes the conversion of phosphoenolpyruvate (PEP) to pyruvate, generating ATP.
• It is allosterically activated by Fructose-1,6-bisphosphate. This is an example of feedforward regulation, where an earlier product in the glycolytic pathway (Fructose-1,6-bisphosphate) activates an enzyme downstream (pyruvate kinase) to ensure the continuation of glycolysis.

Other Options:

• Phosphofructokinase (PFK) : Citrate – Citrate is actually an allosteric inhibitor of phosphofructokinase, not an activator. PFK is allosterically activated by AMP and inhibited by ATP and citrate, regulating glycolysis in response to energy needs.
• Pyruvate dehydrogenase : NADH – NADH is an allosteric inhibitor of pyruvate dehydrogenase, not an activator. High levels of NADH indicate a high energy state, which inhibits pyruvate dehydrogenase to prevent further conversion of pyruvate into acetyl-CoA in the citric acid cycle.
• Pyruvate carboxylase : ADP – ADP is not an activator of pyruvate carboxylase. Acetyl-CoA is the allosteric activator of pyruvate carboxylase, which converts pyruvate into oxaloacetate in gluconeogenesis.

Additional Information

Concept:

• Proteins are polymers of amino acids. There are 22 naturally occurring amino acids.
• Each amino acid consists of an alpha-carbon surrounded by four groups- amino group, carboxyl group, hydrogen, and a variable group (R). Amino acids are classified based on the type of R group present.

Explanation:

• A strong acid dissociates completely whereas a weak acid does not dissociate completely, that is, has a lower percentage of molecules in a dissociated state.
• K_eq=[H⁺] [A^-)/[HA]=K_a
• The equilibrium constant for the above reaction is expressed as an acid ionization constant or acid dissociation constant or acidity constant, represented by K_a. So, it is a measure of the strength of an acid in a solution.
• Stronger acids will have a larger value of K_a. Weak acids do not dissociate completely. Hence, a measure of k_a for a weak acid is given by its pK_a, which is equivalent to the negative logarithm of K_a.
• pKa is the number that defines the acidity of a particular molecule. The lower the pKa, the stronger the acid and it will donate more protons.
• Isoelectric point (pI) is a pH at which a molecule does not carry any charge, that is, the net charge is zero. pI is the mean of pKa values.
• Lysine has three ionizable groups, α -COOH, α- amino, and ε-amino group.
• There are two basic groups (amino group) with pKa-9.06 and 10.54 respectively.
• So, I will be equal to the mean to pKa, 9.06+10.54/2=9.8
• So, the pI of lysine will be 9.8.

So, the correct answer is option 4. 

Concept:

• Alpha helix is a rigid rod like structure that forms when a polypeptide chain twists into helical conformation.
• Helical structure can rotate right handed clockwise or left handed counterclockwise concerning its axis. All alpha helices found in proteins are right handed.

Important Points

• In alpha helix there are 3.6 amino acid residues per turn of the helix and the pitch length of one complete turn is 0.54 nm.
• A single turn of a helix involves 13 atoms from O to the H bond of the hydrogen bonded loop. That is why the alpha helix is known as 3.6₁₃ helix.
• The alpha helix is stabilized by intrachain hydrogen bonds between NH and CO groups of the main chain.
• The CO group of each amino acid forms a hydrogen bond with the NH group of the amino acid that is situated four residues ahead in the sequence.
• In an alpha-helical polypeptide, the carbonyl oxygen (CO) of the nth amino acid forms a hydrogen bond with the amide hydrogen (NH) of the (n + 4)th amino acid. This specific pattern of hydrogen bonding stabilizes the helical structure.
• The number of hydrogen bonds that can form in an α-helix of n amino acids can be approximated by the formula: n - 4.
• So, the alpha helix of 15 residues will have 11 hydrogen bonds.
• All the hydrogen bonds lie parallel to the helix axis and point in the same direction.
• The side chains of amino acids extend outwards from the helix.

Additional Information

• Beta pleated sheets form when two or more polypeptide chain segments line up side by side. Each segment is referred to as a beta strand.
• Instead of being copied, each beta strand is fully extended.
• The distance between adjacent amino acids along a beta strand is approximately 3.5 Å, in contrast with 1.5 Å along an alpha helix.
• Beta pleated sheets are stabilized by interchain hydrogen bonds that form between the polypeptide backbone N H and carbonyl groups of adjacent strands.

So, the correct answer is option 2. 

Concept:-

• One of the most effective strategies that have survived through evolution for controlling flux across biochemical pathways is feedback allosteric inhibition of metabolic enzymes.
• Allosteric enzymes usually work on the first step of the pathway.
• Enzyme regulation known as allostery occurs when binding at one location affects binding at succeeding sites.
• Efficiency results from a precise, immediate, and direct effect that enables dynamic management of the flux through biochemical channels.
• Feedback inhibition is also unaffected by complex signal transduction cascades, translation, or transcription.​

Explanation:

Option 1:- K inhibits F → G and L inhibits F → H; D → E is inhibited at equal amounts of K and L

• In concerted feedback inhibition pathways, end products inhibit their own synthesis by blocking their respective enzymes, and more than one end product or all end products must be present in excess to repress the first enzyme, according to the representation of some feedback reactions that are prevalent in metabolic pathways.
• As above told, allosteric enzymes work at the first step of the pathway. therefore, the first step of each step is inhibited which is k will inhibit  F → G by feedback mechanism and so do L will inhibit F → H and D → E is inhibited by both K and L.
• Hence, this option is correct.

Option 2:- D → E is inhibited at equal amounts of K and L; K inhibits F → H and L inhibits F → G

• The first part of this option is correct in that D → E is inhibited at equal amounts of K and L but for K,  F → H is another step of the bifurcated metabolic pathway that is not related to it and the same goes for L.
• Hence, this option is incorrect. 

Option 3:- D → E is inhibited at equal amount of G and H; K inhibits  F → H and L inhibits F → G

• G and H are not the end products of the pathways, but the first step of the bifurcating pathways.
• Therefore, D → E cannot be inhibited at equal amounts of G and H.
• ​Hence, this option is incorrect.

Option 4:- K inhibits F → H and L inhibits F → G.

• For K,  F → H is another step of the bifurcated metabolic pathway that is not related to it and the same goes for L.
• ​Hence, this option is incorrect.

Concept:

• All alpha amino acids except glycine are chiral molecules. A chiral amino acid exists in two configurations that are non-superimposable mirror images of each other. These two are known as enantiomers.
• An enantiomer is identified by its absolute configuration. Different systems have been developed to specify the absolute configuration of a chiral molecule.

Important Points

• The DL system refers to the absolute configuration of the alpha-carbon of an amino acid to that of the 3-carbon aldose sugar glyceraldehyde.
• Glyceraldehyde can have two absolute configurations, that D or L.
• When the hydroxyl group attached to the chiral carbon is on the left in a Fischer projection, the configuration is L and when the hydroxyl group is on right, the configuration is D.
• DL system refers to the absolute configuration of the four substituents bonded to the chiral carbon.
• All amino aids which are ribosomally incorporated into proteins exhibit L-configuration. So, all are L-alpha amino acids. The basis for preference for L-amino acids is not known.
• D-form of amino acids is not found in ribosomally synthesized proteins, although they exist in some peptide antibiotics and tetrapeptide chains of the peptidoglycan cell walls. Peptides with D-amino acids exist in archaea where they are made by the presence of racemases. Racemases convert L-amino acids to D-amino acids.

Additional Information

• For compounds with more than one chiral center, the most useful system to describe absolute configuration is the RS system.
• Using the RS system, one can define the configuration of a chiral compound in the absence of a reference compound.
• The configuration is described based on an atomic number of four different substituents bonded to an asymmetrical carbon.

So, the correct answer is option 3.

Concept:

• Michaelis -Menten describes the relationship between substrate concentration and reaction velocity. 
• Following reactions shows the reactions:

• Here, the rate constant for the formation of enzyme-substrate complex. 
•  rate constant of the reverse reaction.
•  rate constant of conversion of ES complex to enzyme and product. 
• According to Michaelis-Menten, the rate of an enzyme-catalyzed reaction is measured at varying substrate concentrations then the rate of the reaction is dependent on the concentration of the substrate. 
• Initial velocity increases linearly when the concentration of substrate is low at the start of the reaction. When the reaction reaches its maximum velocity  then further increases in the rate of reaction are not observed.
• The concentration of substrate at which the reaction reaches half its maximum velocity is called the Michaelis constant (K_m). 
• The km is given by 
• A lower value of Km means that enzymes have a greater affinity for the substrate. 
• The Michaelis-Menten equation is given by: 

Explanation:

Given - 

k1 = 1 × 108 M-1 s-1

k-1 = 4 × 104 s-1

k2 = 8 × 102 s-1

• To determine the rate constant we will use the following formula and substitute values.

• Now we will determine the 
• When the value of  is much smaller as compared to that of , then Michaelis constant is expressed as follows:

Hence, the correct answer is option 4.