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

The correct answer is 100

Given:

• Enzyme concentration  ([E_t]) = 15nM
• Substrate concentration ([S]) =  10µM
• Reaction velocity (V₀) =  5 µMs−1
• Michaelis constant (Km) = 5µM

The Michaelis-Menten equation is \( V_0 = \frac{V_{\text{max}} [S]}{K_{\text{M}} + [S]} \)

Rearranging the equation to solve for V_max: \(V_{\text{max}} = V_{0} \times \frac{K_{m} + [S]}{[S]} \)

• Vmax = 5 x 10^-6 Ms^-1 x 1.5
• Vmax = 7.5 x 10-6 Ms-1

The relationship between Vmax, Kcat and enzyme concentration is

V_max =K_cat [E]_t

• K_cat = V_max/[E_t]
• K_cat = 7.5 x 10-6 Ms-1 / 15 x 10^-9 M
• Kcat = 500 s^-1

Kinetic energy = K_cat/K_m

• K.E = 500 s-1/ 5 x 10^-6
• K.E = 100 x 10⁶ M^-1s^-1

The correct answer is Option 1 and Option 3

Explanation:

The standard flow of electrons from NADH to O₂ involves four main complexes and two mobile carriers:

• NADH dehydrogenase (Complex I): NADH donates two electrons to Complex I. This complex pumps protons from the mitochondrial matrix to the intermembrane space.
• Ubiquinone (Q or Coenzyme Q): Ubiquinone is a lipid-soluble mobile carrier that accepts electrons from Complex I (and also from Complex II, succinate dehydrogenase. Reduced ubiquinone (ubiquinol, QH₂) then moves within the membrane to Complex III.
• Cytochrome bc1 (Complex III): Ubiquinol donates its electrons to Complex III. This complex also pumps protons.
• Cytochrome c: Cytochrome c is a water-soluble mobile carrier that accepts electrons from Complex III and transports them to Complex IV.
• Cytochrome c oxidase (Complex IV): Cytochrome c donates its electrons to Complex IV. This complex is the terminal electron acceptor, reducing O₂ to water and pumping protons.
• O₂: Molecular oxygen acts as the final electron acceptor.

Alternative Pathways in Plants:

Plants possess additional components in their ETC that allow for more flexibility and regulation, particularly under stress conditions:

• Alternative NADH dehydrogenases: Plants have several alternative NAD(P)H dehydrogenases that can oxidize NADH (or NADPH) and pass electrons directly to ubiquinone, bypassing Complex I. These are often non-proton pumping, meaning they contribute less to ATP synthesis but allow for continued electron flow when Complex I might be inhibited or when less ATP is needed.
• Alternative Oxidase (AOX): This enzyme is unique to plants (and some fungi/protists). It can directly accept electrons from ubiquinol (reduced ubiquinone) and transfer them to O₂, bypassing both Complex III and Complex IV. AOX activity is non-proton pumping and generates heat rather than ATP. It's important for thermogenesis, especially in thermogenic plants, and for balancing redox states under various stresses.

Fig: The alternative oxidase shown as part of the complete electron transport chain. UQ is ubiquinol/ubiquinone, C is cytochrome c and AOX is the alternative oxidase. (Source)

The correct answer is 4.8

Concept:

• For a weak acid (HA) that partially dissociates in water, the dissociation can be represented as \(HA \rightleftharpoons H^+ + A^-\). The acid dissociation constant (K_a) is the equilibrium constant for this reaction, indicating the strength of the acid.
• A higher Ka value signifies a stronger acid (more dissociation) \(Ka = \frac{[H^+][A^-]}{[HA]}\)
• pKa Value: The pKa is a more convenient and commonly used scale to express the strength of an acid, especially for weak acids. It is defined as the negative base-10 logarithm of the Ka value pKa = log10(K_a)
• A lower pKa value corresponds to a stronger acid, and a higher pKavalue corresponds to a weaker acid. This inverse relationship makes pKa easier to compare than Ka values, which can span many orders of magnitude.
• Acetic acid (CH₃COOH) is a common example of a weak organic acid found in vinegar. Its dissociation constant is relatively small, hence it is a weak acid.

Calculation:

pKa = log10(Ka)

pKa = log 10 (1.74 x 10^-5)

pKa = 4.759

pKa = 4.8

The correct answer is Chondroitin-4-sulfate and Heparin

Explanation:

A polysaccharide is a large carbohydrate molecule made up of many monosaccharide units joined together by glycosidic bonds.

• A homopolysaccharide is composed of only one type of monosaccharide unit.
• A heteropolysaccharide is composed of two or more different types of monosaccharide units (or their derivatives).

Chondroitin-4-sulfate: Chondroitin-4-sulfate is a heteropolysaccharide. It is a glycosaminoglycan (GAG) composed of repeating disaccharide units of N-acetylgalactosamine (GalNAc) and glucuronic acid (GlcA), with a sulfate group typically on the 4-position of the N-acetylgalactosamine. 

Chitin: Chitin is a homopolysaccharide. It is composed solely of repeating units of N-acetylglucosamine (GlcNAc), which is a derivative of glucose. 

Cellulose: Cellulose is a homopolysaccharide. It is composed entirely of repeating units of glucose, linked by β(1→4) glycosidic bonds.

Heparin: Heparin is a heteropolysaccharide. It is a glycosaminoglycan (GAG) composed of repeating disaccharide units that typically consist of iduronic acid (IdoA) or glucuronic acid (GlcA), and N-acetylglucosamine (GlcNAc) or glucosamine (GlcN), with various sulfate groups. 

The correct answer is PLASNGK and GSQTKRL

Explanation:

The absorption of ultraviolet (UV) light by peptides and proteins above 250 nm wavelength is primarily due to the presence of aromatic amino acids:

• Tryptophan (W): Has the strongest absorption, with a maximum absorbance around 280 nm.
• Tyrosine (Y): Absorbs UV light with a maximum around 275-280 nm, though less strongly than Tryptophan.
• Phenylalanine (F): Has the weakest absorption among the aromatic amino acids, with a maximum around 257 nm.

Other standard amino acids (like Alanine, Glycine, Leucine, Proline, Serine, etc.) do not absorb significantly above 250 nm. The peptide bonds themselves absorb strongly below 200-220 nm, but not in the region above 250 nm.

• MQRTVWG: This peptide contains Tryptophan (W). Tryptophan strongly absorbs UV light around 280 nm, which is above 250 nm. Therefore, this peptide will absorb UV light above 250 nm.
• YDEIGVL: This peptide contains Tyrosine (Y). Tyrosine absorbs UV light around 275-280 nm, which is above 250 nm. Therefore, this peptide will absorb UV light above 250 nm.
• PLASNGK: This peptide does not contain Tryptophan (W), Tyrosine (Y), or Phenylalanine (F). Therefore, it will not absorb significant UV light above 250 nm.
• GSQTKRL: This peptide does not contain Tryptophan (W), Tyrosine (Y), or Phenylalanine (F). Therefore, it will not absorb significant UV light above 250 nm.

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. 

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. 

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:-

• 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.

The correct answer is 4 atoms; C1, N, Cα, C for ϕ;    N, Cα, C, N+1 for ψ 

Concept:

• Peptide bond is amide linkage that is formed when unshared electron pair of α-amino acid of one amino acid attacks the carboxyl carbon of another amino acid.
• This is a nucleophilic acyl substitution reaction, where one molecule of water is released as the by-product. 
• Peptide bond have 40% partial double bond characters because of which the 6-atom molecules are in rigid planar configuration and rotation of peptide bond is restricted. 
• Angle of rotation of peptide bond is given by ω and it usually have a value of 180°(trans) and occasionally its value is \(\psi\) = 0°.

​

• However rotation is permitted along the\(\ N - C\alpha\) and \(C\alpha - C \) bonds.
• The rotation along \(\ N - C\alpha \). is called ϕ while the rotation along the \(C\alpha - C \) is called \(\psi\).
• \(\psi\) and ϕ  can have value ranging from +180° to -180°, but most of the value are prohibited due to steric hinderence between the atoms in the backbone of polypeptide and side chains of amino acids.
• G. N. Ramachandran, was the first to determine the permitted values and this permitted values is indicated in the  ϕ- \(\psi\) plane and it is called Ramachandran plot.

Explanation:

• ϕ (phi): It is the torsion angle between the C⁻¹, N, Cα, and C atoms, defining the rotation around the N-Cα bond. It is defined by four atoms:
  • The C of the previous amino acid (C⁻¹) in the chain.
  • The N (amide nitrogen) of the current amino acid.
  • The Cα (alpha carbon) of the current amino acid.
  • The C (carbonyl carbon) of the current amino acid.
• ψ (psi): It is the torsion angle between the N, Cα, C, and N⁺¹ atoms, defining the rotation around the Cα-C bond. It is defined by four atoms:
  • The N (amide nitrogen) of the current amino acid.
  • The Cα (alpha carbon) of the current amino acid.
  • The C (carbonyl carbon) of the current amino acid.
  • The N of the next amino acid (N⁺¹).
• These four atoms are necessary for each angle as they define the rotations around the backbone of the peptide chain.