Molecular Biology And Genetics — Part 2 MCQs with Explanations
Practice high-yield Biotechnology Engineering MCQs on Molecular Biology And Genetics — Part 2 with worked academic solutions for PPSC, FPSC, NIH, NTS.
Questions
40
Difficulty Level
Exam-Level
Target Exams
PPSC, FPSC, NIH...
Estimated Time
48 mins
Exam-Level PPSC FPSC NIH NTS
Verified for Commission Syllabus • Oct 14, 2026
40
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Question 1 of 40
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The RNA primer synthesized during the replication process in bacteria is removed by
Correct Answer: Option C — DNA polymerase I
Detailed Academic Explanation:
DNA polymerase I is the correct answer to “The RNA primer synthesized during the replication process in bacteria is removed by”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 2 of 40
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A disease is inherited by a child with a probability of $${1 \over 4}$$. In a family with two children, the probability that exactly one sibling is affected by this disease is
Correct Answer: Option B — $${3 \over 8}$$
Detailed Academic Explanation:
$${3 \over 8}$$ is the correct answer to “A disease is inherited by a child with a probability of $${1 \over 4}$$. In a family with two children, the probability that exactly one sibling is affected by this disease is”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 3 of 40
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The term 'protein splicing' refers to
Correct Answer: Option C — Removal of intervening protein sequences from the translated protein
Detailed Academic Explanation:
Removal of intervening protein sequences from the translated protein is the correct answer to “The term 'protein splicing' refers to”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 4 of 40
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Match the terms in Group I with the ploidy in Group II . Group I Group II P. Disome 1. 2n + 1 Q. Monosome 2. 2n - 1 R. Nullisome 3. n - 1 S. Trisome 4. n + 1
Correct Answer: Option A — P-4, Q-2, R-3, S-1
Detailed Academic Explanation:
P-4, Q-2, R-3, S-1 is the correct answer to “Match the terms in Group I with the ploidy in Group II . Group I Group II P. Disome 1. 2n + 1 Q. Monosome 2. 2n - 1 R. Nullisome 3. n - 1 S. Trisome 4. n + 1”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 5 of 40
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A dioecious plant has XX sexual genotype for female and XY for male. After double fertilization, what would be the genotype of the embryos and endosperms?
Correct Answer: Option D — 50% ovules will have XXX endosperm and XX embryo, while the other 50% will have XXY endosperm and XY embryo
Detailed Academic Explanation:
50% ovules will have XXX endosperm and XX embryo, while the other 50% will have XXY endosperm and XY embryo is the correct answer to “A dioecious plant has XX sexual genotype for female and XY for male. After double fertilization, what would be the genotype of the embryos and endosperms”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 6 of 40
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An in vitro translation system can synthesize peptides in all three reading frames of the RNA template. When 5' - UCUCUCUC-(UC) n -UCUCUCUC - 3' was used as the template in this in vitro translation system, the synthesized peptides contained 50% each of serine and leucine. When 5' - CCUCCUCCU-(CCU) n -CCUCCU - 3' was used as the template, the synthesized peptides contained 33.3% each of serine, leucine, and proline. Deduce the codon for proline.
Correct Answer: Option C — CCU
Detailed Academic Explanation:
CCU is the correct answer to “An in vitro translation system can synthesize peptides in all three reading frames of the RNA template. When 5' - UCUCUCUC-(UC) n -UCUCUCUC - 3' was used as the template in this in vitro translation system, the synthesized peptides contained 50% each of serine and leucine. When 5' - CCUCCUCCU-(CCU) n -CCUCCU - 3' was used as the template, the synthesized peptides contained 33.3% each of serine, leucine, and proline. Deduce the codon for proline.”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 7 of 40
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During protein synthesis, tRNAs are not involved in
Correct Answer: Option D — Termination
Detailed Academic Explanation:
Termination is the correct answer to “During protein synthesis, tRNAs are not involved in”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 8 of 40
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If the nucleotide composition (%) of a viral genome is A = 10, U = 20, C = 40 and G = 30, which one of the following is this genome?
Correct Answer: Option B — Single stranded RNA
Detailed Academic Explanation:
Single stranded RNA is the correct answer to “If the nucleotide composition (%) of a viral genome is A = 10, U = 20, C = 40 and G = 30, which one of the following is this genome”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 9 of 40
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Which set of the following events occurs during the elongation step of translation? P. Attachment of mRNA with the smaller subunit of ribosome Q. Loading of correct aminoacyl-tRNA into the A site. R. Formation of a peptide bond between the amino acyl-tRNA in the A site and the peptide chain that is attached to the peptidyl-tRNA in the P site. S. Dissociation of the ribosomal subunits. T. Translocation of peptidyl-tRNA from the A site to the P site of the ribosome.
Correct Answer: Option C — Q, R and T
Detailed Academic Explanation:
Q, R and T is the correct answer to “Which set of the following events occurs during the elongation step of translation? P. Attachment of mRNA with the smaller subunit of ribosome Q. Loading of correct aminoacyl-tRNA into the A site. R. Formation of a peptide bond between the amino acyl-tRNA in the A site and the peptide chain that is attached to the peptidyl-tRNA in the P site. S. Dissociation of the ribosomal subunits. T. Translocation of peptidyl-tRNA from the A site to the P site of the ribosome.”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 10 of 40
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In a genetic cross between the genotypes WWXX and wwxx, the following phenotypic distributions were observed among the F 2 progeny: WX, 562; wx, 158; Wx, 38; and wX, 42. Likewise, a cross between XXYY and xxyy yielded the following results: XY, 675; xy, 175; Xy, 72; and xY, 78. Similarly, a cross between WWYY and wwyy yielded: WY, 292; wy, 88; Wy, 12; and wY, 8. In all the genotypes, capital letters denote the dominant allele. Assume that the F 1 progeny were self-fertilized in all three crosses. Also, double cross-over does not occur in this species. Which of the following is correct?
Correct Answer: Option C — Relative position: Y-W-X Distances: Y-W = 5 map units, W-X = 11 map units
Detailed Academic Explanation:
Relative position: Y-W-X Distances: Y-W = 5 map units, W-X = 11 map units is the correct answer to “In a genetic cross between the genotypes WWXX and wwxx, the following phenotypic distributions were observed among the F 2 progeny: WX, 562; wx, 158; Wx, 38; and wX, 42. Likewise, a cross between XXYY and xxyy yielded the following results: XY, 675; xy, 175; Xy, 72; and xY, 78. Similarly, a cross between WWYY and wwyy yielded: WY, 292; wy, 88; Wy, 12; and wY, 8. In all the genotypes, capital letters denote the dominant allele. Assume that the F 1 progeny were self-fertilized in all three crosses. Also, double cross-over does not occur in this species. Which of the following is correct”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 11 of 40
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The major groove of DNA is lined by
Correct Answer: Option C — N7 of purine and C4 of pyrimidine
Detailed Academic Explanation:
N7 of purine and C4 of pyrimidine is the correct answer to “The major groove of DNA is lined by”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 12 of 40
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An example for template independent DNA polymerase is
Correct Answer: Option C — Terminal deoxynucleotidyl transferase
Detailed Academic Explanation:
Terminal deoxynucleotidyl transferase is the correct answer to “An example for template independent DNA polymerase is”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 13 of 40
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Synteny refers to
Correct Answer: Option D — Local conservation of gene order
Detailed Academic Explanation:
Local conservation of gene order is the correct answer to “Synteny refers to”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 14 of 40
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Which of the following processes can increase genetic diversity of bacteria in nature? P. Conjugation Q. Transformation R. Transduction S. Transfection
Correct Answer: Option C — P, Q and R only
Detailed Academic Explanation:
P, Q and R only is the correct answer to “Which of the following processes can increase genetic diversity of bacteria in nature? P. Conjugation Q. Transformation R. Transduction S. Transfection”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 15 of 40
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"Determine the correctness or otherwise of the following Assertion (A) and the Reason (R) . Assertion: N-methyl-N'-nitro-N-nitrosoguanidine (NTG) is an effective chemical mutagen. Reason: Mutations induced by NTG mainly are the GC → AT transitions."
Correct Answer: Option A — Both (A) and (R) are true and (R) is the correct reason for (A)
Detailed Academic Explanation:
Both (A) and (R) are true and (R) is the correct reason for (A) is the correct answer to “"Determine the correctness or otherwise of the following Assertion (A) and the Reason (R) . Assertion: N-methyl-N'-nitro-N-nitrosoguanidine (NTG) is an effective chemical mutagen. Reason: Mutations induced by NTG mainly are the GC → AT transitions."”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 16 of 40
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If a segment of a sense strand of DNA is 5'-ATGGACCAGA-3', then the resulting RNA sequence after transcription is
Correct Answer: Option D — 5'-AUGGACCAGA-3'
Detailed Academic Explanation:
5'-AUGGACCAGA-3' is the correct answer to “If a segment of a sense strand of DNA is 5'-ATGGACCAGA-3', then the resulting RNA sequence after transcription is”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 17 of 40
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Which of the following statements with respect to the orientation of the nitrogenous bases to the pentose sugars, and the puckering of the sugar, is correct?
Correct Answer: Option B — Anti, and 2'-endo in B form DNA
Detailed Academic Explanation:
Anti, and 2'-endo in B form DNA is the correct answer to “Which of the following statements with respect to the orientation of the nitrogenous bases to the pentose sugars, and the puckering of the sugar, is correct”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 18 of 40
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Of the two diploid species, species I has 36 chromosomes and species II has 28 chromosomes. How many chromosomes would be found in an allotriploid individual?
Correct Answer: Option B — 46 or 50
Detailed Academic Explanation:
46 or 50 is the correct answer to “Of the two diploid species, species I has 36 chromosomes and species II has 28 chromosomes. How many chromosomes would be found in an allotriploid individual”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 19 of 40
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Ames test is used to determine
Correct Answer: Option C — Both mutagenicity and carcinogenicity of a chemical
Detailed Academic Explanation:
Both mutagenicity and carcinogenicity of a chemical is the correct answer to “Ames test is used to determine”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 20 of 40
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A mutation in a gene that codes for a polypeptide results in a variant polypeptide that lacks the last three amino acids. What type of mutation is this?
Correct Answer: Option B — Nonsense mutation
Detailed Academic Explanation:
Nonsense mutation is the correct answer to “A mutation in a gene that codes for a polypeptide results in a variant polypeptide that lacks the last three amino acids. What type of mutation is this”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 21 of 40
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Match the entries in Group I with the entries in Group II . Group I Group II P. RNAse P 1. Polyadenylation Q. RNase H 2. Splicing R. snRNAs 3. Ribozymes S. CstF 4. DNA-RNA hybrids
Correct Answer: Option A — P-3, Q-4, R-2, S-1
Detailed Academic Explanation:
P-3, Q-4, R-2, S-1 is the correct answer to “Match the entries in Group I with the entries in Group II . Group I Group II P. RNAse P 1. Polyadenylation Q. RNase H 2. Splicing R. snRNAs 3. Ribozymes S. CstF 4. DNA-RNA hybrids”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 22 of 40
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Topological winding number of a 2.0 kb covalently closed circular DNA was found to be 191 with a writhing number of -4. Hence, its linking number and the number of base pair per turn when the molecule is laid flat on the surface is . . . . . . . . and . . . . . . . ., respectively.
Correct Answer: Option A — 187, 10.69
Detailed Academic Explanation:
187, 10.69 is the correct answer to “Topological winding number of a 2.0 kb covalently closed circular DNA was found to be 191 with a writhing number of -4. Hence, its linking number and the number of base pair per turn when the molecule is laid flat on the surface is . . . . . . . . and . . . . . . . ., respectively.”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 23 of 40
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PKU is one of the best-known hereditary disorders in amino acid metabolism. The defect is attributed to a lesion in one of the following enzymatic activities:
Correct Answer: Option B — Phenylalanine hydroxylase
Detailed Academic Explanation:
Phenylalanine hydroxylase is the correct answer to “PKU is one of the best-known hereditary disorders in amino acid metabolism. The defect is attributed to a lesion in one of the following enzymatic activities:”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 24 of 40
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A truncated polypeptide is synthesized due to a nonsense mutation. Where would you introduce another mutation to obtain a full-length polypeptide?
Correct Answer: Option B — Transfer RNA gene
Detailed Academic Explanation:
Transfer RNA gene is the correct answer to “A truncated polypeptide is synthesized due to a nonsense mutation. Where would you introduce another mutation to obtain a full-length polypeptide”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 25 of 40
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Diploid Drosophila has eight chromosomes. Which one of the following terms should not be used to describe Drosophila with sixteen numbers of chromosomes?
Correct Answer: Option B — Aneuploid
Detailed Academic Explanation:
Aneuploid is the correct answer to “Diploid Drosophila has eight chromosomes. Which one of the following terms should not be used to describe Drosophila with sixteen numbers of chromosomes”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 26 of 40
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How many 3-tuples are possible for the following amino acid sequence?
Correct Answer: Option C — 11
Detailed Academic Explanation:
11 is the correct answer to “How many 3-tuples are possible for the following amino acid sequence”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 27 of 40
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Molecular chaperones are class of proteins that facilitate
Correct Answer: Option A — The proper folding of newly synthesized proteins
Detailed Academic Explanation:
The proper folding of newly synthesized proteins is the correct answer to “Molecular chaperones are class of proteins that facilitate”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 28 of 40
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Which one of the following is not a therapeutic agent based on nucleic acid for the treatment of genetic disorders?
Correct Answer: Option D — Avidin
Detailed Academic Explanation:
Avidin is the correct answer to “Which one of the following is not a therapeutic agent based on nucleic acid for the treatment of genetic disorders”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 29 of 40
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The enzyme that transcribes the eukaryotic genes encoding precursor ribosomal RNAs (pre-rRNAs) of 28S, 18S and 5.8S rRNAs is
Correct Answer: Option A — RNA polymerase I
Detailed Academic Explanation:
RNA polymerase I is the correct answer to “The enzyme that transcribes the eukaryotic genes encoding precursor ribosomal RNAs (pre-rRNAs) of 28S, 18S and 5.8S rRNAs is”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 30 of 40
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Caspases are involved in the process of
Correct Answer: Option D — Apoptosis
Detailed Academic Explanation:
Apoptosis is the correct answer to “Caspases are involved in the process of”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 31 of 40
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Which of the following statements are true with respect to a special complex called 'dicer'? P. It consists of deoxyribonuclease and DNA fragments. Q. It consists of ribonuclease and RNA fragments. R. It is involved in gene silencing. S. It triggers apoptosis.
Correct Answer: Option B — Q, R
Detailed Academic Explanation:
Q, R is the correct answer to “Which of the following statements are true with respect to a special complex called 'dicer'? P. It consists of deoxyribonuclease and DNA fragments. Q. It consists of ribonuclease and RNA fragments. R. It is involved in gene silencing. S. It triggers apoptosis.”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 32 of 40
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During transcription,
Correct Answer: Option C — Both DNA gyrase and DNA topoisomearse I introduce negative supercoils
Detailed Academic Explanation:
Both DNA gyrase and DNA topoisomearse I introduce negative supercoils is the correct answer to “During transcription,”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 33 of 40
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What will be the binding status of regulatory proteins in lac operon when concentrations of both lactose and glucose are very low in the culture medium?
Correct Answer: Option D — Both the repressor and cAMP-CAP complex remain bound to their respective binding sites
Detailed Academic Explanation:
Both the repressor and cAMP-CAP complex remain bound to their respective binding sites is the correct answer to “What will be the binding status of regulatory proteins in lac operon when concentrations of both lactose and glucose are very low in the culture medium”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 34 of 40
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The 5' ends of the mature forms of the prokaryotic mRNAs and tRNAs are
Correct Answer: Option A — A triphosphate group in mRNAs and a monophosphate group in tRNAs
Detailed Academic Explanation:
A triphosphate group in mRNAs and a monophosphate group in tRNAs is the correct answer to “The 5' ends of the mature forms of the prokaryotic mRNAs and tRNAs are”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 35 of 40
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Nuclease-hypersensitive sites in the chromosomes are sites that appear to be
Correct Answer: Option D — Nucleosome free
Detailed Academic Explanation:
Nucleosome free is the correct answer to “Nuclease-hypersensitive sites in the chromosomes are sites that appear to be”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 36 of 40
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How many different protein sequences of 100 residues can be generated using 20 standard amino acids?
Correct Answer: Option C — 20100
Detailed Academic Explanation:
20100 is the correct answer to “How many different protein sequences of 100 residues can be generated using 20 standard amino acids”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 37 of 40
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The G + C content of bacteriophage M13 double stranded DNA is 68%. What would you expect the G + C content of its mRNA?
Correct Answer: Option A — About 68%
Detailed Academic Explanation:
About 68% is the correct answer to “The G + C content of bacteriophage M13 double stranded DNA is 68%. What would you expect the G + C content of its mRNA”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 38 of 40
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Efficient expression of a heterologous protein product is influenced by
Correct Answer: Option D — All of the above
Detailed Academic Explanation:
All of the above is the correct answer to “Efficient expression of a heterologous protein product is influenced by”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 39 of 40
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The 4-amino or 4-keto group of pyrimidine bases is located in the
Correct Answer: Option A — Major groove of the double stranded DNA
Detailed Academic Explanation:
Major groove of the double stranded DNA is the correct answer to “The 4-amino or 4-keto group of pyrimidine bases is located in the”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
Question 40 of 40
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Which one of the following modifications is common to both protein and DNA?
Correct Answer: Option C — Methylation
Detailed Academic Explanation:
Methylation is the correct answer to “Which one of the following modifications is common to both protein and DNA”. This question is standard in competitive examination syllabi for Molecular Biology And Genetics.
💡Key Conceptual Anchor:Molecular Biology And Genetics
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Analytical Tools In Biotechnology — Part 1
Master the next curriculum module with 40 exam-standard MCQs.
💡Examiner's Advisory for Biotechnology Engineering
In competitive commission exams (FPSC, PPSC, KPPSC, SPSC, CSS, PMS), questions are frequently phrased with distractor options designed to catch rote learners. Whenever you answer an MCQ, read the Detailed Academic Explanation and Conceptual Anchor to understand why the remaining options are incorrect.