Cell cycle.
- Which phase consumes the maximum time in the cell cycle?
A. G1 phase
B. S phase
C. G2 phase
D. M phase
Answer: A
Explanation: The G1 (Gap 1) phase is typically the longest phase in the cell cycle, involving cell growth and preparation for DNA synthesis. In mammalian cells, it lasts 8–10 hours. - Synapsis of homologous chromosomes occurs during:
A. Leptotene
B. Zygotene
C. Pachytene
D. Diplotene
Answer: B
Explanation: Synapsis (pairing of homologous chromosomes) begins in Zygotene of Prophase I in meiosis. - The “restriction point” where the cell commits to division is in:
A. G1 phase
B. S phase
C. G2 phase
D. M phase
Answer: A
Explanation: The G1 restriction point (in animal cells) is where cells decide to divide based on growth factors, nutrients, and DNA integrity. - Which cyclin-CDK complex triggers the G2/M transition?
A. Cyclin D-CDK4
B. Cyclin E-CDK2
C. Cyclin A-CDK2
D. Cyclin B-CDK1
Answer: D
Explanation: Cyclin B-CDK1 (MPF) promotes entry into mitosis by phosphorylating nuclear lamins, histones, and other proteins. - In mitosis, spindle fibers attach to chromosomes via:
A. Centrioles
B. Kinetochores
C. Centromeres
D. Telomeres
Answer: B
Explanation: Kinetochores (protein structures on centromeres) serve as attachment sites for spindle microtubules. - Crossing over occurs during:
A. Pachytene
B. Diplotene
C. Diakinesis
D. Metaphase I
Answer: A
Explanation: Pachytene (Prophase I) involves crossing over between non-sister chromatids of homologous chromosomes. - Cytokinesis in plant cells occurs via:
A. Cleavage furrow
B. Cell plate formation
C. Septum formation
D. Binary fission
Answer: B
Explanation: Plant cells form a cell plate (from Golgi vesicles) that develops into the cell wall during cytokinesis. - The phase where DNA replication occurs is:
A. G1
B. S
C. G2
D. M
Answer: B
Explanation: S phase (Synthesis) is dedicated to DNA replication, resulting in duplicate chromatids. - Which checkpoint verifies DNA integrity before mitosis?
A. G1/S
B. G2/M
C. Spindle assembly
D. Metaphase-anaphase
Answer: B
Explanation: The G2/M checkpoint ensures DNA is fully replicated and undamaged before mitosis. - Meiosis II resembles mitosis because:
A. DNA replicates before division
B. Homologous chromosomes pair
C. Sister chromatids separate
D. Crossing over occurs
Answer: C
Explanation: In Meiosis II, sister chromatids separate (as in mitosis), but without DNA replication beforehand. - The term “cell cycle” was coined by:
A. Walther Flemming
B. Howard & Pelc
C. Sutton & Boveri
D. Robert Hooke
Answer: B
Explanation: Howard & Pelc (1953) defined the G1, S, G2, and M phases after studying radioactive DNA in beans. - Inactive cells (e.g., neurons) reside in which phase?
A. G0
B. G1
C. G2
D. S
Answer: A
Explanation: G0 phase is a quiescent state where metabolically active cells (e.g., liver/heart cells) exit the cycle. - The stage where chromosomes align equatorially:
A. Prophase
B. Metaphase
C. Anaphase
D. Telophase
Answer: B
Explanation: In Metaphase, chromosomes align at the metaphase plate via spindle fibers. - Which enzyme prevents telomere shortening?
A. DNA polymerase
B. Telomerase
C. Ligase
D. Helicase
Answer: B
Explanation: Telomerase (an RNA-dependent polymerase) adds telomeric repeats to chromosome ends to prevent shortening. - Chiasmata are observed during:
A. Leptotene
B. Zygotene
C. Diplotene
D. Pachytene
Answer: C
Explanation: Diplotene features chiasmata—X-shaped sites where homologous chromosomes remain attached after crossing over. - Cytokinesis in animal cells involves:
A. Phragmoplast
B. Cell plate
C. Cleavage furrow
D. Karyokinesis
Answer: C
Explanation: Cleavage furrow (formed by actin-myosin contractile ring) pinches animal cells during cytokinesis. - The number of chromatids in a human cell during G2 phase is:
A. 23
B. 46
C. 92
D. 184
Answer: C
Explanation: After S phase, each of the 46 chromosomes duplicates into two chromatids → 92 chromatids in G2. - Which protein regulates the G1/S transition?
A. p53
B. Rb (Retinoblastoma protein)
C. Cyclin D
D. Caspase
Answer: B
Explanation: Rb protein inhibits the G1/S transition; when phosphorylated by Cyclin D-CDK4/6, it releases E2F for S-phase entry. - Spindle formation in animal cells is initiated by:
A. Centrosomes
B. Kinetochores
C. Centromeres
D. Nucleolus
Answer: A
Explanation: Centrosomes (containing centrioles) organize microtubules to form the mitotic spindle in animal cells. - Apoptosis is controlled by:
A. Cyclins
B. Caspases
C. Kinases
D. Phosphatases
Answer: B
Explanation: Caspases are proteases that execute apoptosis by cleaving cellular components.
21. Which is NOT a stage of mitosis?
A. Interphase
B. Prophase
C. Anaphase
D. Telophase
Answer: A
Explanation: Interphase is not part of mitosis. It occurs before mitosis and includes G1, S, and G2 phases. - Meiosis produces:
A. Two identical diploid cells
B. Four identical haploid cells
C. Four genetically distinct haploid cells
D. Two genetically distinct diploid cells
Answer: C
Explanation: Meiosis results in four genetically diverse haploid gametes due to crossing over and independent assortment. - DNA damage halts the cell cycle at:
A. G1/S checkpoint
B. G2/M checkpoint
C. Both A and B
D. Spindle checkpoint
Answer: C
Explanation: DNA damage activates checkpoints at both G1/S and G2/M to prevent progression until repaired. - The shortest phase of mitosis is:
A. Prophase
B. Metaphase
C. Anaphase
D. Telophase
Answer: C
Explanation: Anaphase is the shortest, lasting only a few minutes, during which sister chromatids are rapidly pulled apart. - Recombination nodules appear in:
A. Leptotene
B. Zygotene
C. Pachytene
D. Diplotene
Answer: C
Explanation: Recombination nodules, essential for crossing over, appear in pachytene. - In meiosis, disjunction of chromosomes occurs in:
A. Anaphase I
B. Anaphase II
C. Metaphase I
D. Telophase I
Answer: A
Explanation: In Anaphase I, homologous chromosomes separate (disjunction), while sister chromatids remain joined. - Which cyclin is degraded via the anaphase-promoting complex (APC/C)?
A. Cyclin A
B. Cyclin B
C. Cyclin D
D. Cyclin E
Answer: B
Explanation: Cyclin B is targeted by APC/C for degradation, facilitating the exit from mitosis. - The term “mitosis” was introduced by:
A. Walther Flemming
B. Strasburger
C. Boveri
D. Sutton
Answer: A
Explanation: Walther Flemming coined the term “mitosis” in the 19th century based on his observations. - During which stage does the nucleolus reappear?
A. Prophase
B. Metaphase
C. Anaphase
D. Telophase
Answer: D
Explanation: In Telophase, the nuclear envelope reforms and the nucleolus reappears. - Polyploidy arises due to failure in:
A. Karyokinesis
B. Cytokinesis
C. Synapsis
D. Crossing over
Answer: B
Explanation: Polyploidy results from failure in cytokinesis, leading to cells with multiple chromosome sets. - The “quiescent center” in plant roots is in:
A. G0 phase
B. G1 phase
C. S phase
D. G2 phase
Answer: A
Explanation: Cells in the quiescent center are metabolically active but non-dividing, residing in G0 phase. - Which phase is absent in meiosis?
A. Interkinesis
B. S phase
C. G2 phase
D. G1 phase
Answer: B
Explanation: There is no S phase between Meiosis I and II; DNA is not replicated again. - Lampbrush chromosomes are observed in:
A. Mitotic prophase
B. Meiotic prophase I
C. Oocytes of amphibians
D. Polytene chromosomes
Answer: C
Explanation: Lampbrush chromosomes are extended meiotic chromosomes found in oocytes of amphibians and birds. - The spindle assembly checkpoint ensures:
A. DNA replication is complete
B. Chromosomes are aligned at metaphase plate
C. Chromosomes condense
D. Centrosomes duplicate
Answer: B
Explanation: This checkpoint prevents progression to anaphase until all kinetochores are attached to spindle fibers. - Which structure disappears during prophase?
A. Chromatin
B. Nucleolus
C. Spindle fibers
D. Chromatids
Answer: B
Explanation: The nucleolus disappears in prophase, and chromatin condenses into chromosomes. - Kinetochore is located on:
A. Chromosome arms
B. Telomeres
C. Centromere
D. Nucleolus
Answer: C
Explanation: The kinetochore forms on the centromere and serves as an attachment site for spindle fibers. - Cytokinesis in animals is aided by:
A. Cell wall
B. Phragmoplast
C. Actin-myosin ring
D. Vesicle fusion
Answer: C
Explanation: An actin-myosin contractile ring constricts to form a cleavage furrow during cytokinesis in animals. - Centrosome duplicates during:
A. G0 phase
B. G1 phase
C. S phase
D. M phase
Answer: C
Explanation: The centrosome (containing centrioles) duplicates during S phase alongside DNA replication. - Which phase involves the separation of sister chromatids?
A. Anaphase
B. Telophase
C. Prophase
D. Metaphase
Answer: A
Explanation: During Anaphase, sister chromatids are pulled to opposite poles by spindle fibers. - Meiosis ensures:
A. Chromosome duplication
B. Diploid zygote formation
C. Genetic uniformity
D. Genetic variation
Answer: D
Explanation: Meiosis promotes genetic variation through crossing over and independent assortment. - DNA content doubles in:
A. G1 phase
B. G2 phase
C. M phase
D. S phase
Answer: D
Explanation: DNA replication occurs during S phase, doubling the DNA content. - Chiasmata function in:
A. Chromatid cohesion
B. Spindle formation
C. Genetic recombination
D. Nuclear envelope breakdown
Answer: C
Explanation: Chiasmata are points where homologous chromosomes exchange genetic material. - How many cell divisions occur in meiosis?
A. One
B. Two
C. Three
D. Four
Answer: B
Explanation: Meiosis involves two successive cell divisions: Meiosis I and Meiosis II. - Colchicine blocks cell division at:
A. Prophase
B. Metaphase
C. Anaphase
D. Telophase
Answer: B
Explanation: Colchicine inhibits microtubule formation, arresting cells in metaphase. - Spindle fibers are composed of:
A. Actin
B. Tubulin
C. Myosin
D. Lamin
Answer: B
Explanation: Spindle fibers are made of tubulin microtubules that segregate chromosomes. - Number of chromosomes in human gametes:
A. 23
B. 46
C. 44
D. 22
Answer: A
Explanation: Gametes (sperm/egg) are haploid, containing 23 chromosomes. - Nondisjunction leads to:
A. Genetic recombination
B. Aneuploidy
C. Polyploidy
D. Mutation
Answer: B
Explanation: Nondisjunction is failure of chromosomes to separate properly, resulting in aneuploid cells. - Which organelle plays a role in spindle formation?
A. Golgi body
B. Nucleus
C. Centrosome
D. Lysosome
Answer: C
Explanation: The centrosome organizes spindle fibers for chromosome segregation. - How many chromatids are there in a human cell after S phase?
A. 46
B. 92
C. 23
D. 69
Answer: B
Explanation: Post-S phase, each of 46 chromosomes has two chromatids = 92 chromatids total. - Programmed cell death is termed:
A. Necrosis
B. Apoptosis
C. Lysis
D. Phagocytosis
Answer: B
Explanation: Apoptosis is a regulated process to eliminate damaged or unnecessary cells.
Allele concept, multiple allele (ABO blood group); pseudo allele; isoallele; allelic interaction.
- ABO blood groups in humans are controlled by:
A. Multiple alleles
B. Lethal genes
C. Complementary genes
D. Linked genes
Answer: A
Explanation: ABO blood groups involve three alleles (I<sup>A</sup>, I<sup>B</sup>, i) of a single gene, making it a classic example of multiple allelism. - A person with blood group AB has the genotype:
A. I<sup>A</sup>I<sup>A</sup>
B. I<sup>B</sup>I<sup>B</sup>
C. I<sup>A</sup>I<sup>B</sup>
D. ii
Answer: C
Explanation: Codominance between I<sup>A</sup> and I<sup>B</sup> alleles results in AB phenotype. - Isoalleles are defined as:
A. Alleles expressing identical phenotypes
B. Mutated alleles with lethal effects
C. Alleles with identical function but different protein sequences
D. Alleles that mask pseudoalleles
Answer: C
Explanation: Isoalleles produce functionally identical proteins but differ in DNA sequence, detectable only through molecular analysis. - Which allelic interaction explains the AB blood group?
A. Incomplete dominance
B. Codominance
C. Epistasis
D. Complementary genes
Answer: B
Explanation: I<sup>A</sup> and I<sup>B</sup> are codominant, expressing both antigens (A and B) equally. - Pseudoalleles are:
A. Non-functional alleles
B. Closely linked genes mimicking alleles
C. Alleles with identical mutations
D. Alleles showing incomplete dominance
Answer: B
Explanation: Pseudoalleles are tightly linked genes (e.g., in Drosophila eye color) that behave like alleles due to rare recombination. - A cross between blood group A (I<sup>A</sup>i) and B (I<sup>B</sup>i) parents can produce offspring with blood group:
A. A only
B. AB only
C. O, A, B, AB
D. O and AB only
Answer: C
Explanation: Genotypes: I<sup>A</sup>i × I<sup>B</sup>i → 25% I<sup>A</sup>I<sup>B</sup> (AB), 25% I<sup>A</sup>i (A), 25% I<sup>B</sup>i (B), 25% ii (O). - The allele for sickle-cell anemia shows:
A. Codominance
B. Overdominance
C. Incomplete dominance
D. Lethal dominance
Answer: A
Explanation: Heterozygotes (Hb<sup>A</sup>Hb<sup>S</sup>) express both normal and sickle hemoglobin, demonstrating codominance. - Which is NOT a feature of multiple alleles?
A. Occupy same locus
B. Arise from repeated mutations
C. Control same trait
D. Present in different individuals
Answer: D
Explanation: Multiple alleles exist in a population, but an individual diploid organism can possess only two alleles at a locus. - In chickens, the gene for feather color exhibits incomplete dominance: BB (black), Bb (blue), bb (white). A cross between blue chickens gives:
A. All blue
B. 1 black : 2 blue : 1 white
C. 3 blue : 1 white
D. 1 black : 1 white
Answer: B
Explanation: Bb × Bb → 25% BB (black), 50% Bb (blue), 25% bb (white). - The term “allele” was coined by:
A. Gregor Mendel
B. William Bateson
C. Thomas Morgan
D. Carl Correns
Answer: B
Explanation: William Bateson introduced “allele” in 1905 to describe variant forms of a gene. - Which phenomenon is observed in the Drosophila Bar eye mutation?
A. Multiple alleles
B. Pseudoallelism
C. Isoallelism
D. Overdominance
Answer: B
Explanation: Bar eye results from pseudoalleles (duplicated genes B<sup>+</sup> and B) where unequal crossing over causes phenotypic variation. - A gene with over 100 alleles in humans is:
A. ABO blood group
B. HLA complex
C. Rh factor
D. Hemoglobin gene
Answer: B
Explanation: The HLA complex (MHC) has >100 alleles per locus, crucial for immune diversity. - Complementary genes require:
A. Two dominant alleles to express a trait
B. One dominant allele to suppress another
C. Interaction of non-allelic genes
D. Multiple alleles at a single locus
Answer: A
Explanation: Complementary genes (e.g., sweet pea flower color) need dominant alleles of two different genes for trait expression. - Bombay phenotype arises due to:
A. Mutation in ABO gene
B. Epistasis by hh genotype
C. Codominance failure
D. Pseudoallelism
Answer: B
Explanation: Recessive hh genotype (epistatic gene) prevents ABO antigen formation, resulting in O phenotype regardless of ABO genotype. - Alleles I<sup>A</sup> and I<sup>B</sup> produce:
A. Antigens
B. Antibodies
C. Enzymes
D. Hormones
Answer: C
Explanation: I<sup>A</sup> encodes glycosyltransferase A (adds N-acetylgalactosamine), I<sup>B</sup> adds galactose to H antigen. - In mice, agouti coat color (AA) is dominant to black (aa). Heterozygotes (Aa) show intermediate phenotype. This is:
A. Codominance
B. Incomplete dominance
C. Epistasis
D. Lethal alleles
Answer: B
Explanation: Incomplete dominance results in an intermediate phenotype (e.g., yellow-brown) in heterozygotes. - Which represents allelic exclusion?
A. Random X-chromosome inactivation
B. Expression of only one immunoglobulin allele
C. Genomic imprinting
D. Overdominance
Answer: B
Explanation: Allelic exclusion ensures B-cells express only one heavy/light chain allele, producing monospecific antibodies. - The Rh<sup>+</sup> allele is:
A. Dominant
B. Recessive
C. Codominant
D. Incompletely dominant
Answer: A
Explanation: Rh<sup>+</sup> is dominant (DD/Dd); Rh<sup>-</sup> is recessive (dd). - Lethal alleles were first discovered in:
A. Drosophila
B. Mice
C. Snapdragon
D. Neurospora
Answer: B
Explanation: Cuenot (1905) observed yellow coat lethality in mice (Agouti gene). - In ABO system, universal donor is O because:
A. It lacks A/B antigens
B. It produces anti-A and anti-B antibodies
C. Both A and B
D. It is codominant
Answer: C
Explanation: Group O RBCs lack A/B antigens and plasma has anti-A/B antibodies, minimizing transfusion reactions. - Pseudoalleles can be distinguished by:
A. Phenotypic ratios
B. Complementation test
C. High recombination frequency
D. Lethal effects
Answer: B
Explanation: Complementation test reveals if mutations are in the same gene (fail to complement) or different genes (complement). - Epistasis differs from dominance in that:
A. It involves inter-genic interactions
B. It alters Mendelian ratios
C. Both A and B
D. It is intra-allelic
Answer: C
Explanation: Epistasis is inter-genic (non-allelic genes interact), often modifying 9:3:3:1 ratios. - The ABO gene is located on chromosome:
A. 1
B. 9
C. 11
D. 19
Answer: B
Explanation: ABO gene locus is on chromosome 9q34.2. - Which shows overdominance?
A. Sickle-cell trait in malaria zones
B. AB blood group
C. Pink snapdragons
D. Bombay phenotype
Answer: A
Explanation: Overdominance occurs when heterozygotes (Hb<sup>A</sup>Hb<sup>S</sup>) have higher fitness than homozygotes in malaria-endemic regions. - Isoalleles are identified through:
A. Phenotypic analysis
B. Biochemical assays
C. DNA sequencing
D. Complementation tests
Answer: C
Explanation: DNA sequencing detects subtle nucleotide differences in isoalleles producing identical phenotypes. - A type O mother and type AB father can have children with blood group:
A. A or B only
B. AB only
C. O only
D. A, B, AB, or O
Answer: A
Explanation: Mother (ii) × Father (I<sup>A</sup>I<sup>B</sup>) → offspring I<sup>A</sup>i (A) or I<sup>B</sup>i (B). - The H antigen in ABO system is synthesized by:
A. FUT1 gene
B. ABO gene
C. Rh gene
D. HLA gene
Answer: A
Explanation: FUT1 (H gene) on chromosome 19 encodes fucosyltransferase to produce H antigen. - Allelic exclusion in immunology ensures:
A. Self-tolerance
B. Clonal selection
C. Antibody specificity
D. MHC diversity
Answer: C
Explanation: By expressing only one allele, B-cells produce monospecific antibodies against a single epitope. - Which is an example of recessive lethal alleles?
A. Huntington’s disease
B. Tay-Sachs disease
C. Sickle-cell anemia
D. Cystic fibrosis
Answer: B
Explanation: Tay-Sachs is caused by recessive lethal alleles (homozygotes die in early childhood). - In Drosophila, the white-eye pseudoalleles were studied by:
A. Morgan
B. Lewis
C. Bridges
D. Sturtevant
Answer: B
Explanation: Edward B. Lewis demonstrated pseudoallelism in the bithorax complex using recombination studies. - Which best describes codominance?
A. One allele masks another
B. Both alleles express partially
C. Both alleles express fully
D. Neither allele expresses
Answer: C
Explanation: In codominance, both alleles contribute equally and independently to the phenotype (e.g., AB blood group). - Which blood group lacks both anti-A and anti-B antibodies?
A. O
B. AB
C. A
D. B
Answer: B
Explanation: AB individuals have both A and B antigens on RBCs, so they do not produce anti-A or anti-B antibodies. - A person with genotype ii has which blood group?
A. A
B. B
C. AB
D. O
Answer: D
Explanation: Genotype ii lacks A and B antigens, resulting in O blood group. - In humans, how many ABO alleles can an individual carry?
A. 1
B. 2
C. 3
D. 4
Answer: B
Explanation: Humans are diploid; they can have any two of the three alleles (I<sup>A</sup>, I<sup>B</sup>, i). - What is the genotype of a universal plasma donor?
A. I<sup>A</sup>I<sup>A</sup>
B. I<sup>B</sup>I<sup>B</sup>
C. I<sup>A</sup>I<sup>B</sup>
D. ii
Answer: C
Explanation: AB blood group lacks anti-A and anti-B antibodies in plasma, making them universal plasma donors. - Bombay phenotype individuals cannot express ABO antigens because:
A. They lack ABO alleles
B. They have IA and IB alleles
C. They lack H antigen
D. They produce abnormal antibodies
Answer: C
Explanation: The absence of H antigen (due to hh genotype) prevents expression of A or B antigens. - Which condition arises from incomplete dominance in humans?
A. Blood group AB
B. Tay-Sachs disease
C. Sickle-cell anemia
D. Pink flower in snapdragon
Answer: B
Explanation: In Tay-Sachs carriers, heterozygotes show intermediate enzyme activity, illustrating incomplete dominance at the biochemical level. - What type of interaction occurs between I<sup>A</sup> and I<sup>B</sup> alleles?
A. Dominance
B. Incomplete dominance
C. Codominance
D. Overdominance
Answer: C
Explanation: I<sup>A</sup> and I<sup>B</sup> alleles are codominant and both antigens are expressed in AB blood group. - A person with genotype I<sup>A</sup>i and I<sup>B</sup>i may produce which blood groups in offspring?
A. A only
B. A and B only
C. A, B, AB, O
D. AB and O only
Answer: C
Explanation: Cross between I<sup>A</sup>i × I<sup>B</sup>i gives all four possible ABO blood types. - What type of dominance is shown when heterozygote phenotype is intermediate?
A. Codominance
B. Incomplete dominance
C. Complete dominance
D. Overdominance
Answer: B
Explanation: In incomplete dominance, heterozygote shows a phenotype between the two homozygotes. - Which is true for lethal alleles?
A. They always cause death in heterozygous state
B. Only dominant lethal alleles are known
C. Homozygous individuals may not survive
D. They always affect eye color
Answer: C
Explanation: Recessive lethal alleles cause death only in homozygous condition (e.g., yellow coat in mice). - Which genetic principle explains why blood group O cannot receive AB blood?
A. Epistasis
B. Incomplete dominance
C. Antigen-antibody reaction
D. Codominance
Answer: C
Explanation: Blood group O has anti-A and anti-B antibodies which attack AB donor antigens. - The term ‘multiple alleles’ means:
A. Genes located on different chromosomes
B. More than two alternative forms of a gene
C. Many phenotypes in one individual
D. Two genes influencing one trait
Answer: B
Explanation: Multiple alleles refer to more than two allelic forms of a gene in a population (e.g., ABO). - A cross between two heterozygous snapdragons produces what flower color ratio?
A. 1 red : 2 pink : 1 white
B. 3 red : 1 white
C. 1 red : 1 white
D. All pink
Answer: A
Explanation: Incomplete dominance produces a 1:2:1 phenotypic ratio (RR:red, Rr:pink, rr:white). - A pseudoallele behaves like:
A. An allele due to linkage
B. A mutant allele
C. A silent gene
D. A non-functional protein
Answer: A
Explanation: Pseudoalleles are closely linked genes that mimic the behavior of alleles. - Which test differentiates between pseudoalleles and real alleles?
A. Punnett square
B. Complementation test
C. PCR
D. Northern blot
Answer: B
Explanation: Complementation tests determine if mutations affect the same or different genes. - In epistasis, one gene:
A. Is incompletely dominant
B. Masks the expression of another
C. Is codominant
D. Replaces another gene
Answer: B
Explanation: Epistasis is when one gene interferes with or suppresses the expression of another gene. - Allelic exclusion ensures:
A. Immunoglobulin diversity
B. Expression of both alleles
C. Single specificity of B-cells
D. Somatic recombination
Answer: C
Explanation: Allelic exclusion ensures each B-cell produces only one type of antibody. - In Rh incompatibility, a problem arises when:
A. Both parents are Rh-
B. Mother is Rh- and fetus is Rh+
C. Father is Rh- and mother is Rh+
D. Both fetus and mother are Rh+
Answer: B
Explanation: Rh- mother with Rh+ fetus may develop antibodies that attack fetal RBCs in later pregnancies. - Which gene is necessary for ABO antigen expression?
A. ABO
B. Rh
C. FUT1
D. MHC
Answer: C
Explanation: FUT1 gene encodes the enzyme that forms the H antigen, the base for ABO antigen development.
Sex determination with special reference to Drosophila and Man.
- In humans, sex is determined by:
A. X chromosome in sperm
B. Y chromosome in sperm
C. X chromosome in egg
D. Cytoplasm of egg
Answer: B
Explanation: Human sex determination follows the XY system. A sperm carrying a Y chromosome determines male development by contributing the SRY gene. - Drosophila sex determination relies on:
A. Number of X chromosomes
B. Presence of Y chromosome
C. Temperature
D. Hormones
Answer: A
Explanation: Drosophila determines sex based on the X:A ratio (number of X chromosomes relative to autosomes), not the presence of a Y chromosome. - The master regulator of male development in humans is:
A. DAX1 gene
B. SOX9 gene
C. SRY gene
D. WNT4 gene
Answer: C
Explanation: The SRY gene on the Y chromosome triggers male development by initiating testis formation. - A Drosophila with genotype XXY will be:
A. Male
B. Female
C. Intersex
D. Sterile male
Answer: B
Explanation: In Drosophila, the presence of two X chromosomes (X:A ratio = 1.0) determines female development, regardless of the Y chromosome. - Dosage compensation in humans occurs via:
A. Hypertranscription of X in males
B. Inactivation of one X in females
C. Y chromosome gene amplification
D. Autosome silencing
Answer: B
Explanation: One X chromosome in females is inactivated (Barr body) to balance gene dosage with males. - Klinefelter syndrome results from:
A. 44 + XXY
B. 44 + XO
C. 44 + XXX
D. 44 + YY
Answer: A
Explanation: Klinefelter syndrome (47,XXY) is a male with an extra X chromosome, resulting in infertility and feminized traits. - The Drosophila gene Sex-lethal (Sxl) is activated in:
A. Males (XY)
B. Females (XX)
C. Both sexes
D. Only in mutants
Answer: B
Explanation: Sxl is activated in XX embryos, initiating the female developmental pathway through alternative RNA splicing. - Turner syndrome is characterized by:
A. 45,XO karyotype
B. Male phenotype with infertility
C. Trisomy of X
D. XXY genotype
Answer: A
Explanation: Turner syndrome (45,XO) results in a sterile female with physical abnormalities such as short stature and webbed neck. - In Drosophila, the Y chromosome is essential for:
A. Male development
B. Sperm motility
C. Dosage compensation
D. Oogenesis
Answer: B
Explanation: While not involved in sex determination, the Y chromosome in Drosophila carries fertility genes required for sperm production. - The SRY gene encodes a:
A. Hormone receptor
B. Transcription factor
C. RNA-splicing protein
D. Membrane protein
Answer: B
Explanation: SRY encodes a transcription factor that promotes SOX9 expression, leading to testis formation. - A Drosophila gynandromorph occurs due to:
A. Non-disjunction
B. Loss of X chromosome during mitosis
C. Mutation in Sxl
D. Temperature sensitivity
Answer: B
Explanation: Gynandromorphs form when one X chromosome is lost in an early mitotic division, producing XX (female) and XO (male) cell lineages. - Androgen Insensitivity Syndrome (AIS) results from:
A. SRY deletion
B. Defective androgen receptors
C. Estrogen deficiency
D. XXY karyotype
Answer: B
Explanation: In AIS, individuals are genetically male (XY) but develop female phenotypes due to nonfunctional androgen receptors. - The Drosophila gene transformer (tra) is spliced into active form only in:
A. Males
B. Females
C. Both sexes
D. Larvae
Answer: B
Explanation: Sxl protein promotes female-specific splicing of tra mRNA, producing Tra protein only in females. - In humans, the default sex development pathway is:
A. Male
B. Female
C. Bisexual
D. Environment-dependent
Answer: B
Explanation: In the absence of SRY, the bipotential gonad develops into ovaries, making the female pathway default. - Drosophila males compensate for single X by:
A. X-inactivation
B. Hypertranscription of X
C. Y chromosome gene expression
D. Autosome silencing
Answer: B
Explanation: Drosophila males double the transcriptional output of their single X chromosome to equalize gene expression with females. - Swyer syndrome (46,XY female) is caused by:
A. SRY mutation
B. Extra X chromosome
C. Aromatase deficiency
D. Androgen excess
Answer: A
Explanation: Mutation in the SRY gene prevents testis development, leading to a female phenotype in an XY individual. - The Drosophila gene doublesex (dsx):
A. Determines X:A ratio
B. Is sex-specifically spliced
C. Activates Sxl
D. Encodes a hormone
Answer: B
Explanation: dsx mRNA is spliced into male- or female-specific isoforms depending on the sex, influencing sexual differentiation. - Which karyotype is non-viable?
A. 45,XO
B. 47,XXY
C. 47,XYY
D. 45,YO
Answer: D
Explanation: 45,YO embryos lack an X chromosome, which is essential for development, and are not viable. - Drosophila females with genotype tra/tra develop as:
A. Normal females
B. Males
C. Intersex
D. Sterile females
Answer: B
Explanation: tra is essential for female development. XX tra/tra mutants follow the male pathway due to lack of functional Tra protein. - The Barr body in human cells is:
A. An activated X chromosome
B. An inactivated Y chromosome
C. An inactivated X chromosome
D. A degenerate autosome
Answer: C
Explanation: In female cells, one X chromosome condenses into a Barr body and is transcriptionally inactive. - Drosophila with genotype X:AA = 0.5 will be:
A. Male
B. Female
C. Metafemale
D. Intersex
Answer: A
Explanation: An X:A ratio of 0.5 (e.g., XY or XO with diploid autosomes) results in male development in Drosophila. - SRY gene is located on:
A. Short arm of Y chromosome
B. Long arm of X chromosome
C. Autosome 7
D. Pseudoautosomal region
Answer: A
Explanation: The SRY gene is located on the short arm of the Y chromosome (Yp11.3 region). - In Drosophila, the msl-2 gene is repressed in females by:
A. Sxl protein
B. Tra protein
C. Dsx protein
D. Y chromosome
Answer: A
Explanation: Sxl protein inhibits translation of msl-2 mRNA in females, preventing dosage compensation machinery from acting. - Turner syndrome patients lack:
A. Barr bodies
B. Ovarian follicles
C. Both A and B
D. Functional SRY
Answer: C
Explanation: Turner syndrome individuals (45,XO) lack a second X chromosome (no Barr body) and functional ovaries. - Drosophila males have:
A. One Barr body
B. Hyperactive X chromosome
C. Inactive Y chromosome
D. Sxl gene expression
Answer: B
Explanation: Male Drosophila upregulate transcription of their single X chromosome to match female gene dosage. - The testis-determining factor (TDF) is produced by:
A. Leydig cells
B. SRY gene
C. Pituitary gland
D. Wolffian ducts
Answer: B
Explanation: The SRY gene encodes TDF, a transcription factor that initiates testis formation from bipotential gonads. - Drosophila with X:A ratio = 1.5 are:
A. Males
B. Females
C. Metafemales
D. Intersex
Answer: C
Explanation: A ratio of 1.5 (e.g., XXX:AA) produces metafemales—abnormal females with low fertility. - In humans, the Wolffian ducts develop into:
A. Ovaries
B. Uterus
C. Epididymis
D. Clitoris
Answer: C
Explanation: In male development, testosterone stimulates Wolffian ducts to form epididymis, vas deferens, and seminal vesicles. - The Drosophila gene fruitless (fru) controls:
A. Ovary development
B. Male courtship behavior
C. X:A ratio sensing
D. Dosage compensation
Answer: B
Explanation: The fru gene is essential for male courtship behavior and is expressed in a male-specific manner. - SRY activates which downstream gene?
A. DAX1
B. WNT4
C. SOX9
D. FOXL2
Answer: C
Explanation: SRY induces expression of SOX9, which promotes testis development and suppresses ovarian pathways.
Gene as a structural & functional unit cistron concept; one gene – one polypeptide; sickle cell anemia; thalassemia.
- The term “cistron” was coined by:
A. T.H. Morgan
B. Seymour Benzer
C. Beadle & Tatum
D. Jacob & Monod
Answer: B
Explanation: Seymour Benzer introduced “cistron” (1957) using complementation tests in bacteriophage T4, defining the smallest functional genetic unit. - The “one gene–one polypeptide” hypothesis emerged from studies on:
A. Drosophila eye color
B. Neurospora crassa mutants
C. E. coli lac operon
D. Human hemoglobin
Answer: B
Explanation: Beadle & Tatum’s work on Neurospora (1941) showed each gene controls a specific enzyme (later refined to polypeptide). - Sickle cell anemia results from a mutation in:
A. α-globin gene
B. β-globin gene
C. Heme synthesis enzyme
D. Spectrin protein
Answer: B
Explanation: A point mutation (GAG→GTG) in the β-globin gene (chromosome 11) substitutes valine for glutamic acid at position 6. - The cistron concept is validated by:
A. Recombination frequency
B. Complementation test
C. DNA sequencing
D. Transcription assays
Answer: B
Explanation: The complementation test determines if two mutations are in the same cistron (fail to complement) or different cistrons (complement). - Which is an exception to “one gene–one polypeptide”?
A. Alternative splicing
B. Sickle cell anemia
C. Thalassemia
D. Cistronic genes
Answer: A
Explanation: Alternative splicing allows one gene to produce multiple polypeptides (e.g., DSCAM gene in Drosophila yields 38,016 isoforms). - In sickle cell anemia, hemoglobin forms polymers due to:
A. Hydrophilic substitution
B. Hydrophobic valine at position 6
C. Frameshift mutation
D. Deletion in β-globin
Answer: B
Explanation: Hydrophobic valine causes HbS molecules to polymerize into rigid fibers under low oxygen. - β-thalassemia major results from:
A. Mutations in one β-globin allele
B. Complete absence of β-globin chains
C. Reduced α-globin synthesis
D. HbC mutation
Answer: B
Explanation: Homozygous mutations in β-globin genes cause severe anemia due to absent/nonfunctional β-chains. - The functional unit of a gene that specifies a polypeptide is a:
A. Muton
B. Recon
C. Cistron
D. Operon
Answer: C
Explanation: A cistron is the smallest unit of DNA that codes for a complete polypeptide (equivalent to a gene). - Heterozygote advantage is seen in:
A. β-thalassemia
B. Sickle cell trait (HbAS)
C. Cystic fibrosis
D. Hemophilia
Answer: B
Explanation: HbAS heterozygotes are resistant to Plasmodium falciparum malaria, providing selective advantage. - α-thalassemia is caused by:
A. β-globin mutations
B. Deletions in α-globin genes
C. Iron deficiency
D. Chromosome 11 defects
Answer: B
Explanation: Deletions in one or more of the four α-globin genes (chromosome 16) reduce α-chain synthesis. - Benzer’s work on rII mutants of T4 phage demonstrated:
A. Gene recombination
B. Cistron as a functional unit
C. One gene–one enzyme
D. Colinearity of gene and protein
Answer: B
Explanation: Complementation tests with rII mutants defined the cistron as the smallest genetic unit. - In sickle cell anemia, red blood cells become:
A. Biconcave discs
B. Crescent-shaped
C. Spherocytes
D. Target cells
Answer: B
Explanation: HbS polymerization deforms RBCs into sickle/crescent shapes, causing vaso-occlusion. - The “Hb Bart’s” hydrops fetalis syndrome results from:
A. β-thalassemia major
B. Deletion of all four α-globin genes
C. HbS homozygosity
D. Iron overload
Answer: B
Explanation: Deletion of all four α-globin genes causes severe γ4 tetramers (Hb Bart’s), leading to fetal death. - Which concept is challenged by overlapping genes?
A. Cistron
B. One gene–one polypeptide
C. Central dogma
D. Genetic code universality
Answer: B
Explanation: Overlapping genes (e.g., in ΦX174 phage) share nucleotides, allowing one DNA sequence to code for multiple polypeptides. - Sickle cell anemia inheritance is:
A. Autosomal dominant
B. X-linked recessive
C. Autosomal recessive
D. Mitochondrial
Answer: C
Explanation: Requires homozygous recessive (HbSS) genotype for full disease expression. - In β-thalassemia, excess unpaired α-globin chains:
A. Form stable tetramers
B. Precipitate in RBCs
C. Enhance oxygen transport
D. Degrade heme
Answer: B
Explanation: Excess α-chains precipitate, causing hemolysis and ineffective erythropoiesis. - The minimal genetic element defined by a complementation test is a:
A. Gene
B. Cistron
C. Exon
D. Operon
Answer: B
Explanation: Cistron is defined by complementation: mutations in the same cistron fail to complement. - Treatment for β-thalassemia major involves:
A. Iron supplements
B. Hydroxyurea
C. Regular blood transfusions
D. Gene therapy
Answer: C
Explanation: Lifelong blood transfusions correct anemia but cause iron overload, requiring chelation. - The genetic defect in sickle cell anemia is a:
A. Silent mutation
B. Missense mutation
C. Nonsense mutation
D. Frameshift mutation
Answer: B
Explanation: Missense mutation changes codon 6 (GAG→GTG), substituting valine for glutamate. - α-thalassemia silent carrier has:
A. One α-globin gene deleted
B. Two α-globin genes deleted
C. Three α-globin genes deleted
D. HbH disease
Answer: A
Explanation: Silent carriers lose one α-globin gene (e.g., -α/αα), showing no symptoms. - Which supports “one gene–one polypeptide”?
A. RNA editing
B. Collagen (trimeric protein)
C. Polycistronic mRNA
D. Monomeric enzymes
Answer: D
Explanation: Monomeric enzymes (e.g., lysozyme) are encoded by single genes, aligning with the hypothesis. - Hydroxyurea treats sickle cell disease by:
A. Correcting the β-globin mutation
B. Increasing fetal hemoglobin (HbF)
C. Chelating iron
D. Preventing RBC sickling
Answer: B
Explanation: Hydroxyurea induces HbF (γ-globin), which dilutes HbS and inhibits polymerization. - In humans, the β-globin gene cluster is on chromosome:
A. 11
B. 16
C. 9
D. X
Answer: A
Explanation: β-globin genes (HBB) are located on chromosome 11p15.4. - Hemoglobin Lepore is a fusion product seen in:
A. α-thalassemia
B. β-thalassemia
C. Sickle cell anemia
D. HbC disease
Answer: B
Explanation: Unequal crossover between δ- and β-globin genes creates a δβ fusion protein, causing β-thalassemia. - The cistron concept applies to:
A. Prokaryotes only
B. Eukaryotes only
C. Both prokaryotes and eukaryotes
D. Viruses only
Answer: C
Explanation: Complementation tests work in all organisms to define functional gene units. - Target cells and microcytosis are hallmarks of:
A. Sickle cell anemia
B. Thalassemia
C. Hemophilia
D. Leukemia
Answer: B
Explanation: Thalassemia causes microcytic, hypochromic RBCs with target cells due to imbalanced globin chains. - Which gene therapy product is approved for β-thalassemia?
A. Zynteglo (betibeglogene autotemcel)
B. Luxturna
C. Spinraza
D. Kymriah
Answer: A
Explanation: Zynteglo uses autologous CD34+ cells transduced with functional β-globin gene. - The molecular basis of “one gene–one polypeptide” is:
A. Colinearity of DNA and protein
B. Genetic code degeneracy
C. Polyribosomes
D. RNA interference
Answer: A
Explanation: Colinearity (e.g., in E. coli tryptophan synthetase) shows nucleotide sequence directly corresponds to amino acid sequence. - In α-thalassemia, HbH disease results from deletion of:
A. One α-globin gene
B. Two α-globin genes
C. Three α-globin genes
D. Four α-globin genes
Answer: C
Explanation: Three-gene deletion (e.g., –/-α) causes β4 tetramers (HbH), leading to hemolytic anemia. - Which confirms sickle cell diagnosis?
A. Complete blood count (CBC)
B. Hemoglobin electrophoresis
C. Peripheral smear
D. Iron studies
Answer: B
Explanation: Hemoglobin electrophoresis shows HbS migration pattern (slower than HbA) for definitive diagnosis.
Genetics and molecular biology of replication, transcription and translation.
- Semi-conservative DNA replication was demonstrated by:
A. Watson & Crick
B. Meselson & Stahl
C. Hershey & Chase
D. Taylor et al.
Answer: B
Explanation: Meselson & Stahl (1958) used ⁿ¹⁵N labeling in E. coli to show each daughter DNA has one parental strand. - The enzyme that synthesizes RNA primers during DNA replication is:
A. DNA polymerase III
B. Primase
C. Helicase
D. Ligase
Answer: B
Explanation: Primase synthesizes short RNA primers for DNA polymerase to initiate replication. - In eukaryotes, transcription of mRNA is performed by:
A. RNA polymerase I
B. RNA polymerase II
C. RNA polymerase III
D. Primase
Answer: B
Explanation: RNA pol II transcribes protein-coding genes into pre-mRNA. - The Shine-Dalgarno sequence functions in:
A. DNA replication initiation
B. Translation initiation in prokaryotes
C. Transcription termination
D. mRNA splicing
Answer: B
Explanation: This ribosome-binding site (AGGAGG) aligns mRNA with the 16S rRNA in prokaryotes. - Okazaki fragments are synthesized during:
A. Leading strand replication
B. Lagging strand replication
C. Transcription
D. Translation
Answer: B
Explanation: Lagging strand synthesis produces short, discontinuous Okazaki fragments (1,000-2,000 nt in prokaryotes). - The genetic code is degenerate because:
A. Multiple codons specify one amino acid
B. Codons are non-overlapping
C. AUG is the start codon
D. Stop codons terminate translation
Answer: A
Explanation: Degeneracy allows multiple codons (e.g., UCU, UCC, UCA) to code for serine. - Telomerase is a:
A. DNA-dependent DNA polymerase
B. RNA-dependent DNA polymerase
C. RNA-dependent RNA polymerase
D. Reverse transcriptase
Answer: D
Explanation: Telomerase (a reverse transcriptase) uses its RNA template to add telomeric repeats (e.g., TTAGGG) to chromosome ends. - Rho (ρ)-dependent termination occurs in:
A. DNA replication
B. Prokaryotic transcription
C. Eukaryotic transcription
D. Translation
Answer: B
Explanation: ρ-factor binds RNA and terminates transcription in E. coli by unwinding RNA-DNA hybrids. - Aminoacyl-tRNA synthetases catalyze:
A. Peptide bond formation
B. Charging tRNA with amino acids
C. mRNA splicing
D. Transcription initiation
Answer: B
Explanation: These enzymes link amino acids to their cognate tRNAs (ATP-dependent). - The TATA box is a component of:
A. Prokaryotic promoters
B. Eukaryotic promoters
C. Ribosome binding sites
D. Transcriptional enhancers
Answer: B
Explanation: TATA box (consensus TATAAA) in eukaryotes binds TFIID to position RNA pol II. - Proofreading during DNA replication is performed by:
A. 3’→5′ exonuclease activity of DNA pol
B. Primase
C. DNA ligase
D. Topoisomerase
Answer: A
Explanation: DNA pol I/III remove mismatched nucleotides via 3’→5′ exonuclease activity. - The initiator tRNA in bacteria carries:
A. Methionine
B. Formylmethionine (fMet)
C. N-formyl serine
D. Alanine
Answer: B
Explanation: fMet-tRNAⁿ⁺⁾⁺ initiates translation in prokaryotes. - CRISPR-Cas9 is derived from a:
A. Prokaryotic immune system
B. Eukaryotic RNAi pathway
C. Viral replication mechanism
D. Transposon-based system
Answer: A
Explanation: CRISPR-Cas9 is adapted from bacterial/archaeal adaptive immunity against viruses. - Introns are removed by:
A. RNA polymerase
B. Ribosomes
C. Spliceosomes
D. Aminoacyl-tRNA synthetases
Answer: C
Explanation: Spliceosomes (snRNPs) catalyze intron excision and exon ligation. - The enzyme relieving DNA supercoiling during replication is:
A. Helicase
B. Ligase
C. Topoisomerase II (DNA gyrase)
D. Primase
Answer: C
Explanation: DNA gyrase (prokaryotes) and topoisomerase II (eukaryotes) introduce negative supercoils. - Polycistronic mRNA is found in:
A. Prokaryotes
B. Eukaryotes
C. Both
D. Viruses only
Answer: A
Explanation: Prokaryotes have polycistronic mRNA (e.g., lac operon), encoding multiple proteins. - The wobble hypothesis explains:
A. Degeneracy of genetic code
B. Transcription errors
C. DNA repair
D. Frameshift mutations
Answer: A
Explanation: Wobble base pairing (3rd codon position) allows one tRNA to recognize multiple codons. - Antibiotics like streptomycin inhibit:
A. DNA replication
B. Transcription
C. Translation initiation
D. Peptide bond formation
Answer: C
Explanation: Streptomycin binds 30S ribosomal subunit, disrupting initiation complex formation. - The Kozak sequence (GCCRCCAUGG) optimizes:
A. Prokaryotic translation initiation
B. Eukaryotic translation initiation
C. Transcription termination
D. Splicing
Answer: B
Explanation: Kozak sequence enhances translation initiation efficiency in eukaryotes. - Huntington’s disease results from:
A. Trinucleotide repeat expansion
B. Point mutation in globin gene
C. Chromosomal deletion
D. Telomerase dysfunction
Answer: A
Explanation: CAG repeat expansion in HTT gene (>40 repeats) causes Huntington’s disease. - DNA polymerase III holoenzyme includes:
A. Primase activity
B. 5’→3’ polymerase and 3’→5’ exonuclease
C. Ligase function
D. Helicase activity
Answer: B
Explanation: DNA polymerase III synthesizes DNA in the 5’→3’ direction and proofreads via 3’→5’ exonuclease activity. - The lac operon is regulated by:
A. cAMP-CRP complex
B. TATA-binding protein
C. Rho factor
D. Release factors
Answer: A
Explanation: cAMP-CRP activates lac operon transcription in the absence of glucose by binding the promoter region. - Release factors (RF1/RF2) in translation recognize:
A. Start codons
B. Stop codons
C. Shine-Dalgarno sequence
D. Promoters
Answer: B
Explanation: RF1 (recognizes UAA/UAG) and RF2 (UAA/UGA) bind to stop codons and facilitate release of the nascent polypeptide chain. - Fragile X syndrome is caused by:
A. Trinucleotide (CGG) repeat expansion
B. Chromosome 15 deletion
C. Trisomy 21
D. Point mutation in FMR1
Answer: A
Explanation: Fragile X results from >200 CGG repeats in the FMR1 gene promoter, leading to gene silencing and intellectual disability. - The Pribnow box in prokaryotes is analogous to:
A. TATA box
B. CAAT box
C. GC box
D. Kozak sequence
Answer: A
Explanation: The Pribnow box (TATAAT) is a promoter element in prokaryotes analogous to the TATA box in eukaryotes. - Amino acid activation requires:
A. GTP
B. ATP
C. CTP
D. UTP
Answer: B
Explanation: ATP is hydrolyzed during amino acid activation to form aminoacyl-AMP before transfer to tRNA. - Which antibiotic inhibits peptidyl transferase?
A. Tetracycline
B. Streptomycin
C. Chloramphenicol
D. Rifampicin
Answer: C
Explanation: Chloramphenicol inhibits the peptidyl transferase activity of the 50S ribosomal subunit in prokaryotes. - The central dogma describes:
A. DNA → RNA → Protein
B. RNA → DNA → Protein
C. Protein → RNA → DNA
D. DNA → Protein → RNA
Answer: A
Explanation: Crick’s central dogma (1958) defines the unidirectional information flow from DNA to RNA to protein. - RNA editing occurs in:
A. Trypanosomes (guide RNA)
B. E. coli
C. Saccharomyces
D. Drosophila
Answer: A
Explanation: Trypanosomes use guide RNAs (gRNAs) for RNA editing, inserting or deleting uridines in mitochondrial transcripts. - The nucleolus is the site of:
A. DNA replication
B. rRNA synthesis and ribosome assembly
C. mRNA splicing
D. Protein degradation
Answer: B
Explanation: The nucleolus is where RNA polymerase I transcribes rRNA, and ribosomal subunits are assembled.
Mutation-types, detection, molecular mechanism, chromosomal aberration.
- A point mutation that changes a codon to a stop codon is called:
A. Missense mutation
B. Nonsense mutation
C. Silent mutation
D. Frameshift mutation
Answer: B
Explanation: Nonsense mutations (e.g., CAG→UAG) create premature stop codons, truncating proteins. - The Ames test detects:
A. Carcinogens
B. Aneuploidy
C. Chromosomal deletions
D. Polyploidy
Answer: A
Explanation: This bacterial reverse mutation assay (using Salmonella His⁻ mutants) identifies mutagens that may be carcinogens. - Cri-du-chat syndrome results from:
A. 5p⁻ deletion
B. Trisomy 18
C. Philadelphia chromosome
D. Fragile X expansion
Answer: A
Explanation: Terminal deletion of 5p causes microcephaly and cat-like cry due to CTNND2 gene loss. - UV radiation primarily causes DNA damage by forming:
A. Thymine dimers
B. 8-oxoguanine
C. AP sites
D. Single-strand breaks
Answer: A
Explanation: Thymine dimers (cyclobutane pyrimidine dimers) distort the DNA helix, blocking replication. - Which mutation is NOT a frameshift?
A. Insertion of 1 bp
B. Deletion of 2 bp
C. Insertion of 3 bp
D. Deletion of 4 bp
Answer: C
Explanation: Insertion/deletion of 3 bp (multiples of 3) maintains the reading frame, causing in-frame mutation. - Philadelphia chromosome is a:
A. Reciprocal translocation (9;22)
B. Deletion in chromosome 22
C. Inversion in chromosome 9
D. Trisomy 21
Answer: A
Explanation: t(9;22)(q34;q11) fuses BCR and ABL genes, producing oncogenic tyrosine kinase in CML. - Mutagens like 5-bromouracil cause:
A. Transitions
B. Transversions
C. Deletions
D. Insertions
Answer: A
Explanation: Base analogs (5-BU) induce A-T → G-C or G-C → A-T transitions by tautomeric shifts. - Klinefelter syndrome (47,XXY) is an example of:
A. Autopolyploidy
B. Autosomal trisomy
C. Sex chromosome aneuploidy
D. Triploidy
Answer: C
Explanation: Extra X chromosome (47,XXY) causes sex chromosome aneuploidy, leading to male infertility. - Fluorescence In Situ Hybridization (FISH) detects:
A. Point mutations
B. Chromosomal aberrations
C. Protein misfolding
D. mRNA expression
Answer: B
Explanation: FISH uses fluorescent probes to visualize specific DNA sequences for deletions/translocations. - A silent mutation:
A. Alters amino acid sequence
B. Creates a stop codon
C. Changes codon without altering amino acid
D. Shifts reading frame
Answer: C
Explanation: Synonymous mutations exploit codon degeneracy (e.g., UCU→UCC, both code for serine). - Down syndrome is most commonly caused by:
A. Robertsonian translocation
B. 21q21q translocation
C. Mosaicism
D. Free trisomy 21
Answer: D
Explanation: 95% of cases involve free trisomy 21 due to maternal meiotic non-disjunction. - Alkylating agents (e.g., EMS) cause mutations by:
A. Cross-linking DNA strands
B. Adding methyl/ethyl groups to bases
C. Intercalating into DNA
D. Breaking phosphodiester bonds
Answer: B
Explanation: Alkylation (e.g., O⁶-methylguanine) leads to mispairing (e.g., with T instead of C). - Pericentric inversions involve:
A. Centromere excluded
B. Centromere included
C. Telomeric regions
D. p-arm only
Answer: B
Explanation: Pericentric inversions include the centromere, altering chromosome arm ratios. - Sickle cell anemia results from a:
A. Missense mutation (Glu→Val)
B. Nonsense mutation
C. Frameshift mutation
D. Splice-site mutation
Answer: A
Explanation: β-globin codon 6 mutation (GAG→GTG) substitutes valine for glutamic acid. - Mutations in tumor suppressor genes typically exhibit:
A. Dominant inheritance
B. Recessive loss-of-function
C. Gain-of-function
D. X-linked dominance
Answer: B
Explanation: Knudson’s two-hit hypothesis: Both alleles must be inactivated (e.g., RB1, TP53). - Aneuploidy arises from:
A. Non-disjunction
B. Inversion loops
C. Transposition
D. Base excision repair
Answer: A
Explanation: Meiotic/mitotic non-disjunction causes abnormal chromosome numbers (e.g., trisomy, monosomy). - Which mutation detection method uses PCR and gel electrophoresis?
A. Karyotyping
B. FISH
C. RFLP analysis
D. Comet assay
Answer: C
Explanation: RFLP detects mutations via altered restriction enzyme sites visualized on gels. - Huntington’s disease is caused by:
A. CAG trinucleotide expansion
B. CTG repeat expansion
C. FMR1 deletion
D. Chromosome 4 translocation
Answer: A
Explanation: >40 CAG repeats in HTT exon 1 lead to toxic polyglutamine aggregates. - Reciprocal translocations may cause:
A. Position effect variegation
B. Balanced gametes only
C. Reduced fertility due to unbalanced gametes
D. Triploidy
Answer: C
Explanation: Adjacent-1/2 segregation produces gametes with duplications/deletions, causing infertility. - Xeroderma pigmentosum results from defects in:
A. Mismatch repair
B. Nucleotide excision repair (NER)
C. Base excision repair
D. Double-strand break repair
Answer: B
Explanation: NER deficiency impairs thymine dimer removal, increasing UV-induced skin cancer risk. - A robertsonian translocation involves:
A. Acrocentric chromosomes
B. Telomeric fusions
C. X and Y chromosomes
D. Paracentric inversion
Answer: A
Explanation: Fusion of acrocentric chromosomes (13,14,15,21,22) at centromeres forms derivative chromosomes. - Triploidy (69,XXY) commonly results from:
A. Dispermy
B. Meiotic non-disjunction
C. Mitotic error
D. Robertsonian translocation
Answer: A
Explanation: Dispermy (two sperm fertilizing one egg) is the most frequent cause of triploidy. - Mutagens causing frameshifts include:
A. Ethyl methanesulfonate (EMS)
B. Proflavin
C. 5-Bromouracil
D. UV light
Answer: B
Explanation: Intercalating agents (proflavin, acridine orange) insert between bases, causing insertions/deletions. - In pedigree analysis, autosomal dominant disorders show:
A. Vertical transmission
B. Skips generations
C. Male-to-male transmission absent
D. Higher male prevalence
Answer: A
Explanation: Vertical transmission (affected parent → child) with 50% recurrence risk. - Isochromosomes arise from:
A. Horizontal centromere division
B. Robertsonian translocation
C. Inversion
D. Deletion
Answer: A
Explanation: Transverse centromere cleavage creates mirror-image chromosomes with identical arms (e.g., i(17q)). - The “comet assay” detects:
A. Chromosomal translocations
B. DNA strand breaks
C. Point mutations
D. Telomere length
Answer: B
Explanation: Single-cell gel electrophoresis visualizes DNA fragmentation (“comet tail”) from strand breaks. - Which is a chromosomal duplication syndrome?
A. Charcot-Marie-Tooth disease (CMT1A)
B. Cri-du-chat
C. Williams syndrome
D. Angelman syndrome
Answer: A
Explanation: CMT1A involves 1.5 Mb duplication of PMP22 on 17p11.2, causing peripheral neuropathy. - Mutagenesis via oxidative damage commonly produces:
A. 8-oxoguanine
B. Thymine dimers
C. AP sites
D. Methylated cytosine
Answer: A
Explanation: 8-oxoG mispairs with adenine, causing G→T transversions. - Uniparental disomy (UPD) causes:
A. Prader-Willi syndrome
B. Down syndrome
C. Turner syndrome
D. Klinefelter syndrome
Answer: A
Explanation: Maternal UPD15 silences paternally expressed genes (e.g., SNRPN), causing Prader-Willi. - Which technique detects microdeletions?
A. Karyotyping
B. FISH
C. Southern blot
D. All of the above
Answer: D
Explanation:
- Karyotyping: Detects >5 Mb deletions
- FISH: Detects 1–10 Mb deletions
- Southern blot: Detects smaller deletions if probe-targeted
Elementary idea of DNA finger printing, PCR, cloning, oncogene.
- DNA fingerprinting is based on the principle of:
A. Variable Number Tandem Repeats (VNTRs)
B. Single Nucleotide Polymorphisms (SNPs)
C. Copy Number Variations (CNVs)
D. DNA methylation patterns
Answer: A
Explanation: VNTRs (hypervariable regions) show high polymorphism between individuals, forming the basis of DNA fingerprinting. - The enzyme used in PCR to synthesize DNA is:
A. Taq DNA polymerase
B. Reverse transcriptase
C. DNA ligase
D. Restriction endonuclease
Answer: A
Explanation: Taq polymerase from Thermus aquaticus is thermostable and synthesizes DNA during PCR amplification. - The first cloned mammal, Dolly the sheep, was produced by:
A. Somatic Cell Nuclear Transfer (SCNT)
B. Embryo splitting
C. CRISPR-Cas9
D. mRNA vaccination
Answer: A
Explanation: SCNT involved transferring a nucleus from an adult mammary cell into an enucleated egg, cloned in 1996. - Oncogenes are derived from:
A. Proto-oncogenes
B. Tumor suppressor genes
C. Viral genes
D. Housekeeping genes
Answer: A
Explanation: Proto-oncogenes regulate normal cell growth; mutations convert them into oncogenes that cause cancer. - In DNA fingerprinting, restriction enzymes are used to:
A. Cut DNA into fragments
B. Amplify DNA
C. Join DNA fragments
D. Sequence DNA
Answer: A
Explanation: Restriction enzymes (e.g., HaeIII) cut DNA at specific sites, generating VNTR fragments for analysis. - The denaturation step in PCR involves:
A. Heating to 94–98°C to separate DNA strands
B. Cooling to 50–65°C for primer annealing
C. Heating to 72°C for DNA synthesis
D. Adding probes for detection
Answer: A
Explanation: Denaturation (94–98°C) breaks hydrogen bonds, separating double-stranded DNA into single strands. - The p53 gene is a:
A. Tumor suppressor gene
B. Proto-oncogene
C. Retroviral oncogene
D. DNA repair enzyme
Answer: A
Explanation: p53 (“guardian of the genome”) arrests the cell cycle for DNA repair or triggers apoptosis if damage is irreparable. - Plasmid vectors used in cloning typically contain:
A. Origin of replication (ori)
B. Ribosomal RNA genes
C. Telomeric sequences
D. Centromeric DNA
Answer: A
Explanation: ori allows autonomous replication of plasmids in host cells (e.g., E. coli). - Short Tandem Repeats (STRs) are preferred over VNTRs in forensic DNA fingerprinting because they:
A. Are smaller and more amenable to PCR
B. Show higher mutation rates
C. Require radioactive probes
D. Are found only in exons
Answer: A
Explanation: STRs (2–6 bp repeats) are shorter, allowing robust PCR amplification and automated analysis. - The ras oncogene is activated by:
A. Point mutations
B. Chromosomal translocations
C. Gene amplification
D. Viral integration
Answer: A
Explanation: Point mutations (e.g., codon 12 in KRAS) lock the protein in a GTP-bound “ON” state, promoting proliferation. - RT-PCR is used to amplify:
A. DNA
B. RNA
C. Proteins
D. Carbohydrates
Answer: B
Explanation: Reverse Transcriptase PCR converts RNA to cDNA before amplification, enabling gene expression studies. - The main application of therapeutic cloning is to produce:
A. Genetically identical animals
B. Embryonic stem cells
C. Recombinant proteins
D. Vaccines
Answer: B
Explanation: Therapeutic cloning generates patient-specific embryonic stem cells for regenerative medicine. - In DNA fingerprinting, Southern blotting is used to:
A. Transfer DNA fragments to a membrane
B. Amplify DNA
C. Sequence DNA
D. Visualize proteins
Answer: A
Explanation: Southern blotting transfers gel-separated DNA to a nitrocellulose membrane for hybridization with probes. - The Philadelphia chromosome results from:
A. t(9;22) translocation
B. 5p⁻ deletion
C. Trisomy 21
D. BRCA1 mutation
Answer: A
Explanation: t(9;22) fuses BCR and ABL genes, producing a hyperactive tyrosine kinase in chronic myeloid leukemia. - Primers in PCR are:
A. Short synthetic oligonucleotides
B. Restriction enzymes
C. DNA ligases
D. Thermostable polymerases
Answer: A
Explanation: Primers (18–22 bp) are designed to flank the target DNA and initiate synthesis. - The HER2/neu oncogene is overexpressed in:
A. Breast cancer
B. Lung cancer
C. Colon cancer
D. Leukemia
Answer: A
Explanation: HER2/neu amplification occurs in 20–30% of breast cancers and is targeted by trastuzumab (Herceptin). - A common vector for cloning large DNA fragments is:
A. Bacterial Artificial Chromosome (BAC)
B. Plasmid
C. Cosmid
D. Phagemid
Answer: A
Explanation: BACs can carry inserts up to 300 kb, used in genome sequencing projects. - The number of cycles required to amplify a single DNA molecule into 1,048,576 copies is:
A. 20
B. 25
C. 30
D. 35
Answer: A
Explanation: After n cycles, copies = 2ⁿ. 2²⁰ = 1,048,576. - Which virus carries the src oncogene?
A. Rous sarcoma virus (RSV)
B. Human papillomavirus (HPV)
C. Epstein-Barr virus (EBV)
D. Hepatitis B virus (HBV)
Answer: A
Explanation: RSV carries v-src, a viral oncogene derived from the cellular c-src proto-oncogene. - The function of DNA ligase in cloning is to:
A. Join DNA fragments
B. Cut DNA
C. Synthesize DNA
D. Degrade RNA
Answer: A
Explanation: DNA ligase seals nicks between adjacent nucleotides, joining insert and vector DNA. - STR analysis in DNA fingerprinting detects variations in:
A. Number of tandem repeats
B. Single base substitutions
C. Methylation patterns
D. Telomere length
Answer: A
Explanation: STR loci vary in repeat number (e.g., 5–20 repeats), creating unique profiles. - BRCA1 is a:
A. Tumor suppressor gene
B. Proto-oncogene
C. Retroviral gene
D. Housekeeping gene
Answer: A
Explanation: BRCA1 repairs DNA double-strand breaks; mutations increase breast/ovarian cancer risk. - The purpose of the annealing step in PCR is to:
A. Allow primers to bind to DNA
B. Separate DNA strands
C. Synthesize new DNA strands
D. Digest non-specific products
Answer: A
Explanation: Annealing (50–65°C) enables primers to hybridize with complementary template sequences. - Agrobacterium-mediated gene transfer is used in cloning for:
A. Plants
B. Mammals
C. Bacteria
D. Yeast
Answer: A
Explanation: Agrobacterium tumefaciens transfers T-DNA to plant cells, making it a key tool in plant biotechnology. - Real-time PCR (qPCR) differs from standard PCR by:
A. Quantifying DNA during amplification
B. Using RNA templates exclusively
C. Omitting primers
D. Requiring gel electrophoresis
Answer: A
Explanation: qPCR uses fluorescent dyes/probes to measure DNA accumulation in real-time, enabling quantification. - Knudson’s “two-hit hypothesis” explains the mechanism of:
A. Tumor suppressor genes
B. Proto-oncogenes
C. Viral oncogenes
D. DNA repair genes
Answer: A
Explanation: Knudson’s model requires both alleles of a tumor suppressor gene (e.g., RB1) to be inactivated. - In DNA fingerprinting, the probability of two individuals having identical profiles is:
A. Extremely low (1 in billions)
B. 50%
C. 10%
D. 1%
Answer: A
Explanation: High polymorphism in 13–20 STR loci makes matches exceedingly rare (e.g., 1 in 10¹⁵ for CODIS). - The MYC oncogene is activated by:
A. Translocation in Burkitt lymphoma
B. Point mutation in lung cancer
C. Deletion in neuroblastoma
D. Viral integration in cervical cancer
Answer: A
Explanation: t(8;14) places MYC under immunoglobulin enhancer control, causing overexpression in lymphoma. - Electroporation is a method used in cloning to:
A. Introduce DNA into host cells
B. Amplify DNA
C. Digest DNA
D. Sequence DNA
Answer: A
Explanation: Electroporation uses electrical pulses to create pores in cell membranes for DNA uptake. - The main ethical concern regarding reproductive cloning is:
A. Health risks to clones
B. Cost of the procedure
C. Difficulty of the technique
D. Low success rate
Answer: A
Explanation: Health risks include premature aging, organ defects, and high miscarriage rates, as seen in animal cloning. - Inverse PCR is used to amplify:
A. DNA flanking known sequences
B. RNA templates
C. Whole genomes
D. Mitochondrial DNA
Answer: A
Explanation: Inverse PCR amplifies unknown regions adjacent to known sequences by circularizing and cutting DNA with restriction enzymes. - The primary function of a selectable marker in plasmid vectors is to:
A. Identify transformed cells
B. Promote DNA replication
C. Enhance transcription
D. Facilitate DNA sequencing
Answer: A
Explanation: Antibiotic resistance genes (e.g., ampᵣ) allow growth of only transformed cells in selective media. - Which oncogene encodes a transcription factor activated in Burkitt lymphoma?
A. MYC
B. RAS
C. SRC
D. FOS
Answer: A
Explanation: t(8;14) translocation places MYC under immunoglobulin enhancer control, causing overexpression. - In multiplex PCR:
A. Multiple genes are amplified in one reaction
B. PCR products are cloned directly
C. RNA is reverse transcribed
D. DNA is sequenced in real-time
Answer: A
Explanation: Uses multiple primer pairs to amplify several targets simultaneously (e.g., pathogen detection). - The “VNTR” in DNA fingerprinting refers to:
A. Variable Number Tandem Repeats
B. Viral Nucleotide Transfer RNA
C. Verified Non-Transcribed Region
D. Vector-mediated Nucleotide Transfer
Answer: A
Explanation: VNTRs (10–100 bp repeats) were used in early DNA fingerprinting but largely replaced by STRs. - TP53 gene mutations are most commonly associated with:
A. Li-Fraumeni syndrome
B. Familial adenomatous polyposis
C. Neurofibromatosis type 1
D. Von Hippel-Lindau syndrome
Answer: A
Explanation: Germline TP53 mutations cause Li-Fraumeni syndrome, increasing risk of multiple cancers. - The main advantage of nanopore sequencing over PCR is:
A. Direct long-read sequencing without amplification
B. Higher error rate
C. Requirement for DNA amplification
D. Limited to short DNA fragments
Answer: A
Explanation: Nanopore tech (e.g., Oxford Nanopore) sequences DNA/RNA directly in real-time, avoiding PCR bias. - Induced pluripotent stem cells (iPSCs) are generated by:
A. Reprogramming somatic cells with transcription factors
B. Somatic cell nuclear transfer
C. Embryo splitting
D. CRISPR-Cas9 editing
Answer: A
Explanation: Yamanaka factors (OCT4, SOX2, KLF4, c-MYC) reprogram adult cells to pluripotent state. - The APC gene is a tumor suppressor associated with:
A. Colorectal cancer
B. Breast cancer
C. Melanoma
D. Glioblastoma
Answer: A
Explanation: APC mutations cause familial adenomatous polyposis (FAP), leading to colorectal cancer. - In DNA fingerprinting, the Combined DNA Index System (CODIS) utilizes:
A. 13 core STR loci
B. VNTR polymorphisms
C. Mitochondrial DNA SNPs
D. Whole-genome sequencing
Answer: A
Explanation: CODIS database uses 13 STR loci for forensic identification with high discrimination power. - Gateway cloning relies on:
A. Site-specific recombination (λ phage)
B. Restriction enzyme digestion
C. T4 DNA ligase
D. Homologous recombination
Answer: A
Explanation: Uses λ phage att sites and BP/LR clonase enzymes for rapid, high-efficiency cloning. - The MDM2 oncogene promotes cancer by:
A. Degrading p53 protein
B. Activating DNA repair
C. Inhibiting angiogenesis
D. Promoting apoptosis
Answer: A
Explanation: MDM2 ubiquitinates p53 for proteasomal degradation, inactivating tumor suppression. - Digital PCR (dPCR) differs from qPCR in its ability to:
A. Quantify absolute copy number without standards
B. Measure fluorescence in real-time
C. Amplify RNA targets
D. Detect large chromosomal deletions
Answer: A
Explanation: dPCR partitions samples into nanoreactors for endpoint quantification, enabling absolute copy number measurement. - The ethical concern specific to reproductive cloning is:
A. Identity and autonomy of the clone
B. Cost of the procedure
C. Efficiency of somatic cell transfer
D. Use of embryonic stem cells
Answer: A
Explanation: Primary issues include loss of genetic uniqueness and psychological impact on the cloned individual. - CRISPR-Cas9 gene editing was adapted from:
A. Bacterial adaptive immunity
B. Retroviral replication
C. Eukaryotic RNA interference
D. Archaeal DNA repair
Answer: A
Explanation: Based on CRISPR arrays in bacteria that store viral DNA memories for targeted cleavage.
