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

The correct answer is Option 2: A-C2'-endo; B-O4'-endo; C-C3'-endo; D-O4'-exo

Concept:

• The ribose sugar in nucleotides of nucleic acids adopts specific conformations, referred to as sugar puckers, to reduce steric hindrance and optimize interactions between neighboring atoms. These puckers are primarily described as "endo" or "exo," depending on whether a specific atom of the ribose ring is displaced above (endo) or below (exo) the plane of the ring.
• Key conformational states: The two most common sugar puckers in nucleic acids are:
  • C2'-endo: Common in B-DNA (B-form DNA), where the C2' carbon is positioned above the plane.
  • C3'-endo: Common in A-DNA (A-form DNA) and RNA, where the C3' carbon is positioned above the plane.
• In addition to C2'-endo and C3'-endo, other puckers such as O4'-exo, C2'-exo, and O4'-endo are observed but are less prevalent.

Explanation:

• A-C2'-endo: This is typical of B-form DNA, where the C2' atom is displaced above the plane of the sugar ring.

• B-O4'-endo: The O4' atom is displaced above the plane of the sugar ring in this conformation. In the O4'-endo conformation, the oxygen atom attached to the fourth carbon (O4') of the ribose ring is displaced out of the plane in the same direction as the C1'-C2' bond.
• C-C3'-endo: This is typical of A-form DNA and RNA, where the C3' atom is displaced above the plane of the sugar ring.

• D-O4'-exo: The O4' atom is displaced below the plane of the sugar ring in this conformation. In the O4'-exo conformation, the oxygen atom attached to the fourth carbon (O4') of the ribose ring is displaced out of the plane in the opposite direction to the C1'-C2' bond.

The correct answer is 1.65 × 10^-9 mol to 1.75 × 10^-9 mol

Explanation:

Step 1: Calculate the initial and final [H⁺] concentrations:
Using the formula [H⁺] = 10^-pH:

• Initial pH = 7.65 → Initial [H⁺] = 10^-7.65 = 2.24 × 10^-8 M
• Final pH = 6.87 → Final [H⁺] = 10^-6.87 = 1.35 × 10^-7 M

Step 2: Determine the change in [H⁺] concentration:

• Change in [H⁺] = Final [H⁺] - Initial [H⁺]
• = (1.35 × 10^-7) - (2.24 × 10^-8)
• = 1.13 × 10^-7 M

Step 3: Calculate the moles of H⁺ released:
The volume of the solution is 15 mL = 0.015 L.
Moles of H⁺ released = Change in [H⁺] × Volume
= (1.13 × 10^-7) × 0.015
= 1.695 × 10^-9 mol
Since each molecule of acetylcholine hydrolyzed releases one H⁺ ion, the moles of H⁺ released correspond directly to the moles of acetylcholine in the solution. Thus, the number of moles of acetylcholine is approximately 1.65 × 10^-9 mol to 1.75 × 10^-9 mol.

The correct answer is B and D

Explanation:

• Amino acids are synthesized or degraded in the body through specific metabolic pathways involving intermediates. These intermediates are part of larger metabolic processes, such as the citric acid cycle, glycolysis, or other biochemical pathways.
• Each amino acid is derived from or converted into specific metabolic intermediates. These intermediates serve as the link between amino acid metabolism and central metabolic pathways.

i) Citrate
ii) Oxaloacetate
iii) Succinyl CoA
iv) Pyruvate
v) α-ketoglutarate.

Fig:The metabolic relationship of amino acids to the citric acid (Krebs) cycle determines whether they are glucogenic or not.

• B. Alanine, valine, and leucine- Pyruvate: This is correct because alanine is derived from pyruvate (a key intermediate in glycolysis and the citric acid cycle), while valine and leucine are branched-chain amino acids that share a common metabolic pathway involving intermediates related to the citric acid cycle.
• D. Methionine, threonine, and lysine- This is correct, as these amino acids are linked to intermediates derived from the glycolytic pathway and other related processes.

Incorrect Options:

• A. Serine, Glycine, and Cysteine- This is incorrect because serine, glycine, and cysteine are derived from Pyruvate (iv).
• C. Glutamate, glutamine, and proline- This is incorrect because glutamate, glutamine, and proline are derived from α-ketoglutarate.
• E. Histidine- This is incorrect because histidine is derived from α-ketoglutarate.

The correct answer is It belongs to either alternate α/β or mixed α + β fold

Concept:

• Circular dichroism (CD) spectroscopy is a powerful technique used to study the secondary structure and folding of polypeptides and proteins. It involves analyzing the differential absorption of left-handed and right-handed circularly polarized light by chiral molecules.
• CD spectra in the far-UV region (190–260 nm) provide information about the secondary structure of the polypeptide chain, such as α-helices, β-sheets, and random coils.
• CD spectra in the near-UV region (250–350 nm) provide insights into the tertiary structure and aromatic side-chain environment.
• Different folds (α-helices, β-sheets, or mixed α + β structures) give distinct CD patterns, which help infer the polypeptide’s structural features.

Explanation:

It belongs to either alternate α/β or mixed α + β fold. This inference is based on the CD spectra data provided for both near-UV and far-UV regions.

• The far-UV CD spectrum indicates a mix of α-helices and β-sheets. This suggests the presence of both secondary structure elements, rather than exclusively one type.
• The near-UV CD spectrum shows features that correspond to tertiary interactions, which are characteristic of complex folds like alternate α/β or mixed α + β structures.
• Therefore, the polypeptide structure cannot be purely α-helical or β-sheet. The data aligns with either an alternate α/β fold (where α-helices and β-sheets alternate in the structure) or a mixed α + β fold (where both elements coexist without strict alternation).

The correct answer is Keq = C × e^1/T

Explanation:

• The relationship between the Gibbs free energy change (ΔG°) and the equilibrium constant (Keq) is given by the equation:
  ΔG° = -RT ln(Keq), where
  • R is the universal gas constant,
  • T is the absolute temperature
  • ln(Keq) is the natural logarithm of the equilibrium constant.
• ΔG° is temperature dependent, and this dependence can impact the equilibrium constant (Keq) of a reaction.
• For the given problem, ΔG° shows a specific temperature dependence which leads to the derived dependence of Keq on temperature.
• • From the equation ΔG° = -RT ln(Keq), we know that Keq is exponentially dependent on ΔG° (i.e., Keq = e^-ΔG°/RT).
  • In this case, ΔG° is expressed in a form that leads to Keq having a dependence on e^1/T. The constant C arises as a temperature-independent factor that encapsulates other parameters of the system.
  • Hence, the equilibrium constant (Keq) is expected to follow the relationship Keq = C × e^1/T.

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

\(V_0 =\frac {V_{max}[S]} {K_m + [S]}\)

• 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:

• \(V_0=\frac {100\times (10\times 10^3)}{(10\times 10^3)+(10\times 10^3)}\)
• 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. 

The correct answer is -(30 to 32) kJ/mol.

Explanation:

To determine the standard free energy (ΔG°) for the hydrolysis of ATP, we use the relationship between equilibrium constants and ΔG° for each reaction. The standard free energy change is calculated using the equation:

\(\Delta G^\circ = -RT \ln K_{eq}\)

Where:

• R is the gas constant (8.314 J/mol·K)
• T is the temperature in Kelvin (298 K for 25°C)
• K_eq is the equilibrium constant

For reaction [1]:

• Glucose 6-phosphate + H₂O → Glucose + Pi
• The equilibrium constant K_eq is given as 270.
• Using the equation ΔG° = -RT ln K_eq, we substitute: \( \Delta G^\circ_1 = -(8.314 \text{J/mol·K}) \times (298 \text{K}) \times \ln(270)\)

For reaction [2]:

• ATP + Glucose → ADP + Glucose 6-phosphate
• The equilibrium constant K_eq is given as 890.
• Using the same formula:
• \(\Delta G^\circ_2 = -(8.314 \, \text{J/mol·K}) \times (298 \, \text{K}) \times \ln(890) \)

Overall reaction and ATP hydrolysis:

ATP hydrolysis involves both reactions, and the net free energy change will give us the free energy for the hydrolysis of ATP.

• \( K_{\text{eq, coupled}} = K_{\text{eq, [1]}} \times K_{\text{eq, [2]}} \)
• \(K_{\text{eq, coupled}} = 270 \times 890\)
• \(K_{\text{eq, coupled}} = 240300 \)

By calculating the values, the standard free energy of ATP hydrolysis falls in the range of -30 to -32 kJ/mol.

• \( \Delta G^\circ_{\text{coupled}} = -(8.314 \text{J/mol} \ \text{K}) \times (298 \text{K}) \times \ln(240300)\)
• \(\ln(240300) \approx 12.391\)
• \( \Delta G^\circ_{\text{coupled}} = -(8.314 \text{J/mol} \text{K}) \times (298 \text{K}) \times 12.391 \)
• \(\Delta G^\circ_{\text{coupled}} \approx -30659.8 , \text{J/mol}\)

Therefore, the standard free energy change for the hydrolysis of ATP at 25°C is approximately  -30.66 kJ/mol

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:

\(\rm \mathrm{E}+\mathrm{S} \underset{k_{-1}}{\stackrel{k_1}{\rightleftharpoons}} \mathrm{ES} \stackrel{k_2}{\longrightarrow} \mathrm{E}+\mathrm{P}\)

• Here,\(k_1 =\) the rate constant for the formation of enzyme-substrate complex. 
• \(k_{-1} =\) rate constant of the reverse reaction.
• \(k_2=\) 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 \(V_{max}\) 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 \(k_m = \frac{k_{-1} +k_2}{k_1} \)
• A lower value of Km means that enzymes have a greater affinity for the substrate. 
• The Michaelis-Menten equation is given by: \(V =\frac{V_{max}[S]}{K_m + [S]}\)

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.

\(k_m = \frac{k_{-1} +k_2}{k_1} \\ k_m = \frac{4 \times 10^4 s^{-1} +8 \times 10^2 s^{-1}}{1 \times 10^8 M^{-1} s^{-1}} \\ k_m = \frac{ 40800 s^{-1}}{1 \times 10^8 M^{-1} s^{-1}}\\ k_m = 408 \times 10^{-6}M\\ k_m = 408 \mu M\)

• Now we will determine the \(K_s\)
• When the value of \(k_2\) is much smaller as compared to that of \(k_{-1}\), then Michaelis constant is expressed as follows:

\(k_m \approx \frac {k_{-1}}{k_1} = K_s\)

 \(\therefore K_s = \frac {4 \times 10^4 s^{-1}}{1 \times 10^8 M^{-1} s^{-1}}\\ K_s = 4 \times 10^{-4 }M \\ K_s = 400 \mu M\)

Hence, the correct answer is option 4.