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Biochemistry

Description: GATE: Biotechnology: Biochemistry
Number of Questions: 20
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Tags: Biochemistry
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The number of base pairs per turn of a helix of A DNA is __________.

  1. 10

  2. 12

  3. 11

  4. 13

  5. 11.5


Correct Option: C
Explanation:

A DNA has 11 base pairs per turn of its helix.

Which one of the following is an inhibitor of oxidative phosphorylation?

  1. Fluroacetate

  2. Arsenite

  3. Malonate

  4. Piercidin A

  5. Fluoride


Correct Option: D
Explanation:

Piercidin A inhibits NADH-CoQ reductase (Complex I) to block electron flow from the latter to CoQ. So oxidations of substrates such as pyruvate, isocitrate, α-ketogluterate, malate and 3-hudroxybutyrate, whose electrons enter the respiratory chain through complex I are prevented.

In salvage pathway of purine biosynthesis, the 7th nitrogen molecule comes from which of the following?

  1. CO2

  2. Aspartate

  3. Glycine

  4. Glutamine

  5. Formyltetrahydrofolate


Correct Option: C
Explanation:

4th and 5th carbon and 7th nitrogen come from glycine.

Epimerism is an isomerism that occurs depending on whcih of the following?

  1. Presence of asymmetric carbon atom.

  2. Spatial orientation of groups of single carbon atom.

  3. Orientation of H-OH group in C1 and C4.

  4. Rotation of plane of polarized light.

  5. Spatial arrangement of groups in asymmetric carbon atoms.


Correct Option: B
Explanation:

Difference in spatial orientation of groups on a single asymmetric carbon of a molecule having more than one asymmetric carbon. For example, glucose and galactose are 4 epimers of each other differing with respect to groups on C4.

The rate limiting step of glycolysis is controlled by

  1. Pfk - I

  2. Pfk - II

  3. Glyceraldehyde-3-phosphate dehydrogenase

  4. Hexokinase

  5. Aldolase


Correct Option: A
Explanation:

Physiological concentration of ATP in cell inhibits Pfk-I allosterically and reduces its substrate affinity for fructose 6 phosphate. A 20% fall in ATP concentration rises the the rate of glycolysis 10-fold and Pfk activity to 50%. So low ATP/AMP ratio regulates the Pfk-I activity and thus, regulates glycolysis. Thus, Pfk-I is the enzyme for rate limiting step of glycolysis.

During protein synthesis in eukaryotes, when does the tRNA-met bind to 5’ mRNA?

  1. Before binding of the small ribosomal subunit.

  2. After binding of the small ribosomal subunit.

  3. Before binding of the large ribosomal subunit.

  4. After binding of the large ribosomal subunit.

  5. Before binding of both the ribosomal subunits.


Correct Option: A
Explanation:

The initiator tRNA (tRNAimet) binds to the 5’ end of mRNA before binding of the small (40S) ribosomal subunit. The large ribosomal subunit in eukaryotes is 60S subunit, which binds to 5’mRNA after initiator tRNA, the 40S subunit and initiation factors bind to mRNA.

In complex III, electrons are transferred _________.

  1. from NADH to CoQ

  2. from QH2 to Cytochrome C

  3. from FAD to CoQ

  4. from cytochrome C to O2

  5. from FMN to CoQ


Correct Option: B
Explanation:

In complex III, electrons are transferred from ubiquinol (QH2) to cytochrome c by the enzyme cytochrome c reductase.

Meselson and Stahl proved that DNA replication is semiconservative. E.coli bacteria was grown in N15-NH4Cl medium, then washed and incubated with N14 compounds for 3 generations. Then CsCl density gradient centrifugation of isolated DNA showed semiconservative DNA replication. What is the pattern of band formation, if formed in the following way: N14, N14/15, N15?

  1. Thin band at N14, thick band at N15

  2. Thin bands at N15 and N14/15

  3. Thick band at N14 and thin band at N15

  4. Thick band at N14/15 and thin band at N15

  5. Thick band at N14


Correct Option: C
Explanation:

Since, the Bacteria has been transferred from N15 to N14 medium, the source of nitrogen for synthesis of DNA will be non radioactive. Now, the bacteria was grown in N15 medium after which it was grown for several generations in N14 medium. So, after centrifugation thick band will be formed in N14 zone and a thin band will be formed in the N15 zone as initially the DNA contained N15 nitrogen.

Which of the following are yielded in a citric acid cycle?

  1. 3 NADH and 2 CO2

  2. 6 NADH and 2 CO2

  3. 6 NADH and 4 CO2

  4. 2 NADH and 4 CO2

  5. 2 NADH and 2 CO2


Correct Option: A
Explanation:

NADH is yielded in three steps- (a) oxidation of isocitrate to oxalosuccinate (b) oxidation of α-ketogluterate to succinyl CoA and (c) oxidation of malate to oxaloacetate. Thus, 3 NADH molecules are formed and CO2 is produced during conversion of oxalosuccinate to α-ketogluterate and the latter to succinyl CoA. So in 1 cycle 3 NADH and 2 CO2 are formed.

Which of the following is mismatched?

  1. Retinol - Synthesis of rhodopsin

  2. Riboflavin - Synthesis of neurotransmitter

  3. Cobalamin - Production of blood cells

  4. Niacin - Energy transfer reactions

  5. Biotin - Metabolism of carbohydrates, proteins and fats


Correct Option: B
Explanation:

Riboflavin or vitamin B2 plays an important role in the production of energy in electron transport chain, the citric acid cycle and also in β-oxidation of fatty acid and it has no role in synthesis of neurotransmitter.

Deletion of the leader sequence of trp operon of E.coli causes _________.

There is a leader sequence in gene controlling synthesis of tryptophan.

  1. increased transcription of trp operon

  2. decreased transcription of trp operon

  3. decreased transcription of trp operon in presence of tryptophan

  4. no effect on transcription

  5. transcription to occur constitutively


Correct Option: A
Explanation:

Attenuator produces terminator loop in presence of high tryptophan and anti terminator loop in presence of low tryptophan. The attenuator sequence is present within the leader sequence of the operon. In absence of leader sequence, there will be no attenuation. So, no attenuator loop will be formed and transcription of operon will increase.

The final product of β-oxidation of a 16-carbon fatty acid is ________.

The metabolism of fatty acids occur by α and β-oxidation.

  1. 2 molecules of acetyl CoA

  2. 1 acetyl CoA and 14-carbon fatty acyl CoA

  3. Two acyl CoA

  4. 1 acetyl CoA

  5. 1 acyl CoA


Correct Option: A
Explanation:

In each β-oxidation cycle, 2 carbons are lessoned from a fatty acyl CoA for the production of 1 molecule of acetyl CoA. In this way a 16-carbon fatty acid is oxidized by reduction of 2 carbon atoms at each cycle and finally the last fatty acyl CoA is none other than an acetyl CoA.

Match the following: Pathways and rate limiting enzymes P. Glycogenesis a. Carbamoyl phosphate synthetase Q. Urea synthesis b. α-ketogluterate dehydrogenase R. TCA cycle c. Acetyl CoA carboxylase S. Purine Biosynthesis d. Glycogen synthase T. Fatty acid synthesis e. Glutamyl amido transferase

  1. P - a, Q - b, R - c, S - d, T - e

  2. P - d, Q - a, R - c, S - d, T - e

  3. P - d, Q - a, R - b, S - e, T - c

  4. P - e, Q - a, R - c, S - b, T - d

  5. P - a, Q - c, R - b, S - d, T - e


Correct Option: C
Explanation:

Rate limiting enzyme for glycogenesis is glycogen synthase,urea is carbamoyl phosphate synthetase, TCA cycle is α-ketogluterate dehydrogenase, purine biosynthesis is glutamyl amido transferase and fatty acid synthesis is Acetyl CoA carboxylase.

It is the principal buffer of __________.

Bicarbonate buffer is a critical buffer of our body

  1. extracellular fluid (ECF)

  2. intracellular fluid

  3. blood plasma

  4. the RBC

  5. tissue fluid


Correct Option: C
Explanation:

Bicarbonate is the principal buffer of blood plasma. It consists of bicarbonate (HCO-3) and carbonic acid (H2CO3) as the base and acid members respectively. The bicarbonate buffer neutralizes stronger dietary and metabolic acids changing them to corresponding conjugate bases.

The active form of this enzyme is ______.

Glycogen synthase enzyme catalyses the rate limiting step of glycogenesis.

  1. glycogen synthase a

  2. glycogen synthase b

  3. glycogen synthase a and b are active

  4. glycogen synthase a and b are inactive

  5. none of the above


Correct Option: A
Explanation:

Glycogen synthase a is the active form of enzyme glycogen synthase.

The pH of bicarbonate buffer is _______.

Bicarbonate buffer is a critical buffer of our body

  1. 3.8

  2. 6.1

  3. 7.4

  4. 6.8

  5. 7.0


Correct Option: B
Explanation:

The pH of buffer is 6.1 in plasma as it exists there in equilibrium with CO2.

Active glycogen synthase is converted to inactive form by phosphorylation of ____________.

Glycogen synthase enzyme catalyses the rate limiting step of glycogenesis.

  1. proline residues

  2. tyrosine residues

  3. serine residues

  4. threonine residues

  5. alanine residues


Correct Option: C
Explanation:

Active glycogen synthase a is converted to inactive form glycogen synthase b by phosphorylation of its seven serine residues with ATP by different cAMP dependent protein kinase enzymes.

The number of ATP molecules produced from a 16 carbon fatty acid is _______________.

The metabolism of fatty acids occur by α and β-oxidation.

  1. 129

  2. 131

  3. 106

  4. 108

  5. 130


Correct Option: C
Explanation:

The ATP yield for every oxidation cycle is theoretically at maximum yield 17, as NADH produces 3 ATP, FADH2 produces 2 and a full rotation of the Citric Acid Cycle produces 12. In practice it is closer to 14 ATP for a full oxidation cycle as in practice the theoretical yield isn't attained, it's generally closer to 2.5 ATP per NADH molecule produced, 1.5 for each FADH2 Molecule produced and this equals to 10 per cycle of the TCA (according to the P/O ratio)Source ATP Total1 FADH2x 1.5 ATP = 1.5 ATP (Theoretically 2 ATP)1 NADHx 2.5 ATP = 2.5 ATP (Theoretically 3 ATP)1 acetyl CoAx 10 ATP = 10 ATP (Theoretically 12 ATP)TOTAL = 14 ATPFor a 16-carbon fatty acid, the number of ATPs generated:Source ATP Total7 FADH2 x 1.5 ATP = 10.5 ATP7 NADH x 2.5 ATP = 17.5 ATP8 acetyl CoA x 10 ATP = 80 ATP. Activation = -2 ATPNET = 106 ATPIn the first step of β-oxidation, the acyl CoA produced from a 16-C saturated fatty acid is transferred to mitochondria by carnitine carriers at the cost of 2 ATPs. So the net energy yield is 106 ATPs at the end of one complete β-oxidation of a 16-C fatty acid.

Match the following: Column A Column B A. Cycloheximide 1. Inhibits bacterial protein synthesis by blocking the A-site on the ribosome.

B. Puromycin 2. Inhibits bacterial protein synthesis by blocking peptidyl transfer.

C. Chloramphenicol 3. Inhibits protein synthesis by resembling an amino acid.

D. Tetracyclin 4. Misreads the genetic code and inhibits initiation of protein synthesis.

  1. A - 4, B - 3, C - 2, D - 1

  2. A - 3, B - 4, C - 2, D - 1

  3. A - 1, B - 2, C - 3,D - 4

  4. A - 4, B - 4, C - 1, D - 2

  5. A - 2, B - 3, C - 2, D - 4


Correct Option: A
Explanation:

Cycloheximide inhibits initiation of protein synthesis, puromycin inhibits protein synthesis by resembling an amino acid attached to terminal adenosine of t-RNA, chloramphenicol is an antibiotic that inhibits bacterial protein synthesis by blocking the translation elongation and tetracyclin is also an antibiotic that inhibits bacterial protein synthesis by blocking the a-site on ribosome.

Within the leader sequence, lies the __________.

There is a leader sequence in gene controlling synthesis of tryptophan.

  1. attenuator sequence

  2. promoter sequence

  3. operator sequence

  4. regulator gene

  5. structural genes


Correct Option: A
Explanation:

The attenuator sequence is present within leader sequence.

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