Molecular structure of genes and chromosomes; Mutations and mutagenesis; Nucleic acid
replication, transcription, translation and their regulatory mechanisms in prokaryotes and
eukaryotes; Mendelian inheritance; Gene interaction; Complementation; Linkage, genetics
(plasmids, transformation, transduction, conjugation); Horizontal gene transfer and Transposable
elements; RNA interference; DNA damage and repair; Chromosomal variation; Molecular basis
of genetic diseases.
1. which is the type of chromatin that is less condensed and transcriptionally active.
a)Nucleosome b)Heterochromatin c)Euchromatin d)Histone
Correct answer:c
Reason:
EUCHROMATIN-Less condensed transcriptionally active, gene-rich regions.
HETEROCHROMATIN-Highly condensed, transcriptionally inactive, found at centromeres &
telomeres
2.______________ are the fundamental unit of chromatin, ~146 bp DNA wrapped around
histone octamer.
a)chromomere b)telomere c)euchromatin d)Nucleosome
Correct answer:d
Reason:
A nucleosome comprises approximately 146 base pairs (bp) of DNA wrapped around a histone
octamer (comprising two copies of H2A, H2B, H3, and H4). This structure compacts DNA and
regulates gene expression.
3. Which intercalating agents cause frameshift mutation on nucleotides that disrupt the structure
of the DNA?
i)acridine orange ii)EtBr iii)DMSO iv)EMS
a)iv only b)i only c)i&ii d)iii&iv
Correct answer:c
Reason:
Acridine orange (i) and Ethidium Bromide (EtBr) (ii) are intercalating agents that insert
themselves between stacked DNA bases. This leads to frameshift mutations by distorting the
helical structure of DNA, often causing insertions or deletions during replication.
4.____________ is the analog of thymine that can mispair with Guanine instead of Adenine.
a)5-bromo uracil b)7-bromo uracil c)4-bromo uracil d)bromo uracil
Correct answer:a
Reason:5-BU is an analog of thymine (T) but can mispair with guanine (G) instead of adenine
(A) due to its ability to shift between two tautomeric forms.
5. EMS alkylates guanine, forming _____________, which mispairs with thymine (T) instead of
cytosine (C).
a) O4-ethylguanine b)O6-ethylguanine c)O3-ethylguanine d)O-ethylguanine
Correct answer :b
Reason:
EMS alkylates guanine, forming O6-ethylguanine. During replication, O6-ethylguanine mispairs
with T instead of C. In the next round of replication, T pairs with A, converting the original G-C
pair into an A-T pair (G → A transition mutation).
6. Which point mutation induces stop codon and causes mutation?
a)missense mutation b)silent mutation c)nonsense mutation
d)frameshift mutation
Correct answer:c
Reason:
Nonsense mutations introduce a premature stop codon (UAA, UAG, or UGA) in the coding
sequence, leading to early termination of translation.This results in a truncated, nonfunctional
protein, often causing severe functional loss.
7. Identify the correct statement
i)Histone acetylation deactivates the transcription
ii)when lactose is present, it induces transcription by inactivating the repressor.
iii)Histone Deacetylation (HDACs) represses transcription.
iv)when tryptophan is present, it represses its synthesis (negative feedback).
a)ii only b)all the above c)ii,iii,iv d)iv only
Correct answer:c
Reason:
(ii) Correct: When lactose is present, it binds to and inactivates the lac repressor, allowing
transcription of the lac operon.
(iii) Correct: Histone deacetylation (HDACs) represses transcription by making chromatin more
compact and less accessible to transcription factors.
(iv) Correct: Tryptophan represses its own synthesis via negative feedback by activating the trp
repressor, blocking transcription.
(i) Incorrect: Histone acetylation activates transcription by loosening chromatin, making DNA
accessible for transcription factors.
8.In prokaryotic transcription, which subunit of RNA polymerase is responsible for recognizing
the promoter?
A) α subunit
B) β subunit
C) β’ subunit
D) σ (sigma) factor
Correct answer:d
Reason:
The sigma (σ) factor of prokaryotic RNA polymerase is responsible for recognizing and binding
to promoter sequences, initiating transcription.
Once transcription starts, sigma is released, and the core enzyme continues elongation.
9. Which of the following is NOT a feature of eukaryotic transcription?
A) Presence of multiple RNA polymerases
B) Coupled transcription and translation
C) Splicing of pre-mRNA
D) Presence of a TATA box in promoters
Correct answer:b
Reason:In eukaryotes, transcription occurs in the nucleus, while translation occurs in the
cytoplasm, making coupling impossible.In prokaryotes, transcription and translation occur
simultaneously in the cytoplasm.
10. Which of the following is an example of a transcriptional activator in eukaryotes?
A) Histone deacetylase (HDAC)
B) RNA polymerase II
C) TATA-binding protein (TBP)
D) CREB (cAMP response element-binding protein)
Correct answer:d
reason:CREB (cAMP response element-binding protein) is a transcriptional activator that binds
to cAMP response elements (CREs) in promoters and enhances gene expression.
11.In a three-point test cross, the observed double crossovers (DCO) were 8, but the expected
DCO (calculated from crossover frequencies) was 10.Find the coefficient of coincidence (CoC)
and interference (I).
a)0.8,0.2 b)0.8,0.5 c)0.8,0.6 d)0.2,0.8
Correct answer:a
Reason:
CoC= Expected DCO
Observed DCO
=10/8=0.8
Interference (I)=1−CoC=1−0.8=0.2
12. In a dihybrid test cross between two linked genes A and B, the following progeny numbers
were obtained:Find the map distance between genes A and B.
Phenotype Number of Offspring
Parental (AB, ab) 820
Recombinant (Ab, aB) 180
a)20 cM b)18 cM c)27 cM d)16cM
Correct answer:b
Reason:
Total offspring = 820+180=1000
Recombinant offspring = 180
Map Distance=(180/1000)×100=18 cM
13. choose the correct answer:
1. autosomal dominant -i)Hemophilia
2 .down syndrome-ii)Monosomy X
3. turner syndrome-iii)Huntington’s disease
4. x-linked diseases-iv)Trisomy 21
a)iii, iv, ii ,i
b)iii ,i, iv ,ii
c)ii, iv ,i, iii
d)iv, i , ii , iii
Correct answer:a
Reason:
Huntington’s disease is inherited in an autosomal dominant manner, meaning a single mutated
allele causes the disorder. Down syndrome is caused by an extra copy of chromosome 21
(Trisomy 21). Turner syndrome results from a missing X chromosome (Monosomy X, 45,X)
.Hemophilia is an X-linked recessive disorder affecting blood clotting.
14. What is the mode of replication in prokaryotes
a)semi-conservative , bidirectional b)semi-discontinuous , unidirectional
c)semi-conservative .unidirectional d)conservative ,bidirectional
Correct answer: a
Reason:Semi-conservative: Each new DNA molecule consists of one original strand and one
newly synthesized strand.
Bidirectional: Replication proceeds in both directions from the origin of replication, forming two
replication forks.
15. identify the correct match
i)Pol α: Initiator polymerase
ii)Pol δ: Synthesizes lagging strand
iii)Pol ε: Synthesizes leading strand
iv)RNA Pol III: tRNA & 5S rRNA
a)ii only b)i, ii,iii c)all the above d)iv only
Correct answer:c
Reason:(i) Pol α: Initiator polymerase → It initiates DNA synthesis by adding RNA primers.
(ii) Pol δ: Synthesizes the lagging strand → Extends Okazaki fragments on the lagging strand.
(iii) Pol ε: Synthesizes the leading strand → Primarily responsible for leading strand synthesis.
(iv) RNA Pol III: Synthesizes tRNA & 5S rRNA → Involved in transcribing tRNA and 5S rRNA
in eukaryotes.
16. In eukaryotes, the first amino acid formed during the translation process is _________.
a)Glycine b)Methionine c)lysine d)proline
Correct answer:b
Reason: In eukaryotic translation, the AUG start codon codes for methionine (Met), which is the
first amino acid incorporated into the growing polypeptide chain.In prokaryotes, the first amino
acid is N-formylmethionine (fMet) instead of regular methionine.
17. Which enzyme is primarily responsible for DNA replication in prokaryotes?
a) DNA ligase
b) DNA helicase
c) DNA-dependent DNA polymerase
d) RNA polymerase
Answer: c) DNA-dependent DNA polymerase
Reason: In prokaryotes, DNA-dependent DNA polymerase catalyzes the synthesis of new DNA
strands by adding nucleotides complementary to the template strand during replication.
18. In eukaryotic transcription, which RNA polymerase is responsible for transcribing tRNA
genes?
a) RNA polymerase I
b) RNA polymerase II
c) RNA polymerase III
d) RNA polymerase IV
Correct answer: c) RNA polymerase III
Reason: RNA polymerase III transcribes genes encoding tRNA, 5S rRNA, and other small
RNAs in eukaryotic cells.
19. During the initiation of transcription in prokaryotes, the sigma (σ) factor associates with
which enzyme to facilitate promoter recognition?
a) DNA ligase
b) RNA polymerase
c) DNA helicase
d) Topoisomerase
Correct answer: b
Reason: In prokaryotes, the sigma (σ) factor binds to RNA polymerase, forming a holoenzyme
that recognizes and binds to specific promoter sequences, initiating transcription.
20.Which of the following statements is true regarding the regulation of gene expression in
eukaryotes?
a) All genes are organized into operons.
b) Transcription and translation occur simultaneously in the cytoplasm.
c) Gene expression is primarily regulated at the level of transcription initiation.
d) mRNA processing does not involve splicing.
Correct answer: c
Reason: In eukaryotes, gene expression is mainly regulated during transcription initiation,
involving various transcription factors and regulatory elements. Additionally, mRNA processing
in eukaryotes includes splicing, where introns are removed, and exons are joined to form mature
mRNA.
