Cell Cycle – Definition and different phases.
- The phase where DNA replication occurs in the cell cycle is:
a) G₁ phase
b) S phase
c) G₂ phase
d) M phase
Answer: b) S phase
Explanation: The S (Synthesis) phase of interphase is when the entire genome is duplicated. Each chromosome replicates to form two identical sister chromatids, which remain attached at the centromere. Although DNA content doubles (2C → 4C), the chromosome number remains unchanged. - During which stage do chromosomes align at the equator?
a) Prophase
b) Metaphase
c) Anaphase
d) Telophase
Answer: b) Metaphase
Explanation: In metaphase, chromosomes line up at the metaphase plate (equatorial plane) with kinetochores attached to spindle fibers. This alignment ensures equal distribution of chromatids to daughter cells. - The longest phase of the cell cycle in most cells is:
a) M phase
b) G₁ phase
c) S phase
d) Interphase
Answer: d) Interphase
Explanation: Interphase (comprising G₁, S, and G₂ phases) accounts for about 90–95% of the total cell cycle. It is a period of active metabolism, growth, DNA replication, and preparation for mitosis. - Cytokinesis in animal cells involves:
a) Cell plate formation
b) Cleavage furrow
c) Phragmoplast
d) Kinetochore assembly
Answer: b) Cleavage furrow
Explanation: A cleavage furrow forms due to the contraction of the actin-myosin ring beneath the plasma membrane, leading to the physical separation of the cytoplasm into two daughter cells. - The G₂ checkpoint verifies:
a) DNA replication completion
b) Nutrient availability
c) Chromosome attachment to spindle
d) Cell size adequacy
Answer: a) DNA replication completion
Explanation: The G₂ checkpoint ensures DNA has been fully and accurately replicated and that the cell is ready to enter mitosis. It also checks for DNA damage before progressing. - In which phase do chromosomes decondense and nuclear envelopes reform?
a) Prophase
b) Anaphase
c) Telophase
d) Metaphase
Answer: c) Telophase
Explanation: During telophase, chromosomes reach the poles and begin to uncoil. Nuclear envelopes re-form around each set of chromosomes, and the nucleolus reappears. - The protein complex that regulates the G₁-S transition is:
a) MPF (Maturation Promoting Factor)
b) Cyclin D-CDK4
c) APC/C (Anaphase Promoting Complex)
d) Condensin
Answer: b) Cyclin D-CDK4
Explanation: Cyclin D-CDK4 phosphorylates the retinoblastoma (Rb) protein, releasing E2F transcription factors that initiate transcription of S phase genes, committing the cell to DNA replication. - Synapsis and crossing over occur during:
a) Mitosis
b) Meiosis I
c) Meiosis II
d) G₂ phase
Answer: b) Meiosis I
Explanation: During prophase I of meiosis I, homologous chromosomes pair (synapsis) and exchange genetic material through crossing over. This contributes to genetic variation. - The phase where the cell prepares for mitosis by synthesizing tubulin proteins is:
a) G₁
b) S
c) G₂
d) M
Answer: c) G₂
Explanation: In G₂, the cell synthesizes proteins required for mitosis, particularly tubulin for spindle fiber formation, and checks DNA for damage before mitotic entry. - The stage where sister chromatids separate and move to opposite poles:
a) Prophase
b) Metaphase
c) Anaphase
d) Telophase
Answer: c) Anaphase
Explanation: In anaphase, the centromeres split, and spindle fibers pull the sister chromatids apart to opposite poles. This ensures each daughter cell receives a full set of chromosomes. - Quiescent cells in multicellular organisms are in which phase?
a) G₀
b) G₁
c) G₂
d) S
Answer: a) G₀
Explanation: G₀ is a resting state where cells exit the cycle temporarily or permanently. Cells like neurons and muscle cells remain in G₀ indefinitely, performing specialized functions without dividing. - The spindle assembly checkpoint occurs during:
a) G₁-S transition
b) Metaphase
c) Anaphase
d) Cytokinesis
Answer: b) Metaphase
Explanation: The spindle checkpoint ensures that all chromosomes are properly attached to spindle microtubules before anaphase begins. This prevents chromosome missegregation. - Plant cell cytokinesis involves:
a) Cleavage furrow
b) Phragmoplast
c) Contractile ring
d) Septum formation
Answer: b) Phragmoplast
Explanation: Plant cells form a cell plate via the phragmoplast, a structure composed of microtubules and vesicles from the Golgi. This cell plate eventually becomes the new cell wall. - DNA content is doubled during:
a) G₁ phase
b) G₂ phase
c) S phase
d) M phase
Answer: c) S phase
Explanation: The S phase is dedicated to DNA synthesis. Each chromosome is duplicated, resulting in double the DNA content while the chromosome number remains unchanged. - The cyclin degraded by APC/C to initiate anaphase is:
a) Cyclin A
b) Cyclin B
c) Cyclin D
d) Cyclin E
Answer: b) Cyclin B
Explanation: Anaphase-promoting complex/cyclosome (APC/C) targets cyclin B for degradation. This inactivates CDK1 and allows progression into anaphase by separating sister chromatids. - In mitosis, nucleolus disappears during:
a) Prophase
b) Metaphase
c) Anaphase
d) Telophase
Answer: a) Prophase
Explanation: Prophase is marked by chromatin condensation, nucleolus disappearance, and spindle formation. These changes prepare the nucleus for chromosome segregation. - The “restriction point” in the cell cycle occurs in:
a) Late G₁
b) S phase
c) G₂
d) M phase
Answer: a) Late G₁
Explanation: The restriction point is the critical checkpoint in late G₁ after which the cell is committed to division, independent of external growth factors. - Which phase is absent in meiosis II?
a) Prophase
b) S phase
c) Metaphase
d) Anaphase
Answer: b) S phase
Explanation: DNA replication (S phase) does not occur between meiosis I and meiosis II. The second meiotic division separates sister chromatids without further DNA duplication. - Chromosome condensation is mediated by:
a) Tubulin
b) Actin
c) Condensin
d) Kinesin
Answer: c) Condensin
Explanation: Condensin proteins play a central role in condensing chromosomes during prophase, helping pack DNA into a compact, mitotic chromosome structure. - Cytokinesis overlaps with:
a) Prophase
b) Metaphase
c) Anaphase
d) Telophase
Answer: d) Telophase
Explanation: Cytokinesis typically begins during late anaphase or early telophase and concludes after telophase, completing the physical division of the cell. - Cells with DNA damage are arrested at G₁/S checkpoint by:
a) p53
b) Rb protein
c) Cyclin B
d) Cdc25
Answer: a) p53
Explanation: p53 acts as a tumor suppressor, detecting DNA damage and inducing expression of p21, a CDK inhibitor that halts the cell cycle to allow repair or trigger apoptosis. - The shortest phase of mitosis is:
a) Prophase
b) Metaphase
c) Anaphase
d) Telophase
Answer: c) Anaphase
Explanation: Anaphase lasts only a few minutes as sister chromatids rapidly separate and migrate toward opposite poles, making it the briefest mitotic stage. - Centromere division occurs during:
a) Prophase
b) Metaphase
c) Anaphase
d) Telophase
Answer: c) Anaphase
Explanation: The centromeres split at the beginning of anaphase, allowing sister chromatids to separate and be pulled to opposite ends of the cell. - Synaptonemal complex forms during:
a) Mitotic prophase
b) Meiotic prophase I
c) Meiotic metaphase I
d) G₂ phase
Answer: b) Meiotic prophase I
Explanation: The synaptonemal complex forms during zygotene of prophase I in meiosis and facilitates synapsis and crossing over between homologous chromosomes. - The cell cycle stage with the highest metabolic activity is:
a) G₁
b) S
c) G₂
d) M
Answer: a) G₁
Explanation: In G₁, cells actively grow, synthesize RNA, proteins, and organelles, preparing for DNA replication. It’s often the longest and most metabolically active phase. - MPF (Mitosis Promoting Factor) is a complex of:
a) Cyclin A + CDK2
b) Cyclin D + CDK4
c) Cyclin B + CDK1
d) Cyclin E + CDK2
Answer: c) Cyclin B + CDK1
Explanation: MPF consists of Cyclin B and CDK1 and is crucial for the G₂ to M transition, promoting chromatin condensation and nuclear envelope breakdown. - In plant cells, cytokinesis involves deposition of:
a) Lignin
b) Cellulose
c) Chitin
d) Pectin
Answer: b) Cellulose
Explanation: Plant cytokinesis involves the deposition of cellulose in the newly formed cell plate, which develops into the middle lamella and primary cell wall. - The phase where organelles duplicate is:
a) G₁
b) S
c) G₂
d) M
Answer: a) G₁
Explanation: During G₁, cells grow in size and duplicate organelles to prepare for future cell division, ensuring daughter cells have sufficient components. - Chromosomes are least condensed during:
a) Prophase
b) Metaphase
c) Interphase
d) Anaphase
Answer: c) Interphase
Explanation: In interphase, DNA exists as loosely packed chromatin, allowing transcription and gene expression. This state provides maximum accessibility to the genetic material. - Bivalent formation occurs in:
a) Mitosis
b) Meiosis I
c) Meiosis II
d) G₂ phase
Answer: b) Meiosis I
Explanation: Bivalents, or tetrads, are formed during prophase I of meiosis when homologous chromosomes pair and synapse. This is essential for genetic recombination and accurate segregation. - Cells that do not divide after birth are in:
a) G₀
b) G₁
c) G₂
d) S
Answer: a) G₀
Explanation: Cells in G₀ phase have exited the active cell cycle and are in a quiescent or terminally differentiated state. Neurons, skeletal muscle cells, and cardiac muscle cells typically remain permanently in G₀, unable to re-enter the cycle under normal physiological conditions. - The checkpoint monitoring DNA damage is at:
a) G₁/S
b) G₂/M
c) Spindle assembly
d) Metaphase-anaphase
Answer: a) G₁/S
Explanation: The G₁/S checkpoint, also known as the restriction point, ensures that the DNA is intact before replication. Damage triggers activation of p53, which induces p21, inhibiting CDKs and halting progression to S phase for repair. - Disintegration of the nuclear envelope begins in:
a) Prophase
b) Prometaphase
c) Metaphase
d) Anaphase
Answer: b) Prometaphase
Explanation: During prometaphase, the nuclear envelope breaks down, exposing chromosomes to the mitotic spindle. This allows spindle fibers to attach to kinetochores, essential for chromosome alignment. - The phase where centrioles duplicate is:
a) G₁
b) S
c) G₂
d) M
Answer: b) S
Explanation: Centrioles, part of the centrosome, replicate during the S phase, ensuring two centrosomes for bipolar spindle formation during mitosis. Duplication is coordinated with DNA synthesis. - Crossing over occurs between:
a) Sister chromatids
b) Non-sister chromatids
c) Homologous chromosomes
d) Heterologous chromosomes
Answer: b) Non-sister chromatids
Explanation: During prophase I of meiosis, homologous chromosomes pair and form chiasmata between non-sister chromatids, allowing genetic recombination and variation in gametes. - Cytokinesis in animal cells is inhibited by:
a) Taxol
b) Colchicine
c) Cytochalasin
d) Actinomycin D
Answer: c) Cytochalasin
Explanation: Cytochalasin binds to actin filaments, disrupting their polymerization and preventing the contractile ring formation required for cleavage furrow and cytokinesis in animal cells. - The cell cycle phase absent in early embryonic cells is:
a) S phase
b) G₁ phase
c) G₂ phase
d) M phase
Answer: b) G₁ phase
Explanation: In early embryogenesis, cells rapidly divide through abbreviated cycles lacking G₁ and G₂ phases, alternating between DNA synthesis (S) and mitosis (M) to speed up cell proliferation. - Chromatids become chromosomes during:
a) Prophase
b) Metaphase
c) Anaphase
d) Telophase
Answer: c) Anaphase
Explanation: In anaphase, sister chromatids separate at the centromere, and each is now considered an individual chromosome as they are pulled toward opposite poles. - APC/C targets securin for degradation to initiate:
a) Prophase
b) Metaphase
c) Anaphase
d) Telophase
Answer: c) Anaphase
Explanation: The Anaphase-Promoting Complex/Cyclosome (APC/C) ubiquitinates securin, allowing separase to cleave cohesin proteins that hold sister chromatids together, triggering anaphase onset. - The cell cycle stage with 2C DNA content is:
a) G₁
b) S
c) G₂
d) Early M
Answer: a) G₁
Explanation: A diploid somatic cell in G₁ contains 2C DNA, representing the normal genomic content. During S phase, DNA replication begins, increasing the content to 4C by G₂ and early M.
Mendelism, Linkage and Crossing over – Types with examples
- Mendel’s Law of Independent Assortment is applicable to:
a) Genes on homologous chromosomes
b) Alleles of a single gene
c) Genes on different chromosomes
d) Linked genes
Answer: c) Genes on different chromosomes
Explanation: Mendel’s Law of Independent Assortment states that alleles of different genes assort independently of each other if they are located on different (non-homologous) chromosomes. This happens during metaphase I of meiosis when different chromosome pairs align independently. - In a dihybrid cross (AaBb × aabb), the ratio 1:1:1:1 indicates:
a) Complete linkage
b) Incomplete dominance
c) Independent assortment
d) Codominance
Answer: c) Independent assortment
Explanation: A 1:1:1:1 ratio of offspring phenotypes in a test cross (heterozygote × double recessive) suggests the genes are on different chromosomes and assort independently, with equal chances for each gamete type (AB, Ab, aB, ab). - Coupling phase in linkage refers to:
a) Heterozygous alleles on different chromosomes
b) Dominant alleles on homologous chromosomes
c) Recessive alleles on homologous chromosomes
d) Both dominant/recessive alleles on same chromosome
Answer: d) Both dominant/recessive alleles on same chromosome
Explanation: In the coupling (cis) phase, both dominant alleles are on one chromosome (AB) and both recessives on the homologous chromosome (ab), i.e., AB/ab. This affects the observed recombination pattern. - The maximum recombination frequency between two genes is:
a) 25%
b) 50%
c) 75%
d) 100%
Answer: b) 50%
Explanation: If two genes are far apart on the same chromosome or on different chromosomes, crossing over between them occurs in 50% of meioses, producing 50% recombinants—this is the upper limit of recombination frequency. - Bateson and Punnett observed 7:1:1:7 ratio in sweet pea due to:
a) Incomplete dominance
b) Codominance
c) Linkage
d) Multiple alleles
Answer: c) Linkage
Explanation: The 7:1:1:7 phenotypic ratio observed in Lathyrus by Bateson and Punnett indicated that the two genes did not assort independently but were inherited together due to linkage, particularly in repulsion phase. - Crossing over occurs during:
a) Prophase I of meiosis
b) Metaphase II of meiosis
c) Anaphase of mitosis
d) G₂ phase
Answer: a) Prophase I of meiosis
Explanation: Crossing over takes place during pachytene of prophase I in meiosis, where homologous chromosomes exchange segments at points called chiasmata, creating genetic diversity. - A test cross ratio of 1:1 for a dihybrid indicates:
a) Independent assortment
b) Complete linkage
c) Incomplete linkage
d) Epistasis
Answer: b) Complete linkage
Explanation: A 1:1 phenotypic ratio in a test cross (heterozygote × recessive) with no recombinant types indicates complete linkage—no crossing over occurred, so only parental gametes were produced. - Recombination frequency is directly proportional to:
a) Distance between genes
b) Strength of linkage
c) Dominance of alleles
d) Chromosome length
Answer: a) Distance between genes
Explanation: The greater the physical distance between two genes on a chromosome, the more likely crossing over will occur between them, leading to a higher recombination frequency. - Mendel’s Law of Segregation is based on:
a) Separation of homologous chromosomes
b) Separation of sister chromatids
c) Crossing over
d) Independent assortment
Answer: a) Separation of homologous chromosomes
Explanation: During anaphase I of meiosis, homologous chromosomes (and thus alleles) segregate into different gametes, forming the basis of Mendel’s Law of Segregation. - In maize, the ratio 7:1:1:7 in a test cross indicates recombination frequency of:
a) 12.5%
b) 25%
c) 50%
d) 6.25%
Answer: a) 12.5%
Explanation: The total number of recombinants is 2 (1+1), and total progeny is 16 (7+1+1+7). Recombination frequency = (2/16) × 100 = 12.5%. - Which is NOT a Mendelian disorder?
a) Sickle cell anemia
b) Cystic fibrosis
c) Down syndrome
d) Phenylketonuria
Answer: c) Down syndrome
Explanation: Down syndrome is caused by nondisjunction resulting in trisomy of chromosome 21—a chromosomal disorder, not a single-gene Mendelian trait. - In repulsion phase, genotypes appear as:
a) AB/ab
b) Ab/aB
c) AB/Ab
d) ab/aB
Answer: b) Ab/aB
Explanation: In repulsion (trans-phase) configuration, each homologous chromosome carries one dominant and one recessive allele (Ab and aB), resulting in fewer parental-type offspring. - A recombination frequency of 25% means genes are:
a) Completely linked
b) Unlinked
c) 25 map units apart
d) On different chromosomes
Answer: c) 25 map units apart
Explanation: Recombination frequency is measured in centiMorgans (cM); 1% recombination = 1 cM. So, 25% recombination implies the genes are 25 cM apart on a chromosome. - Incomplete linkage produces:
a) Only parental phenotypes
b) More parental than recombinant phenotypes
c) Equal parental and recombinants
d) No parental phenotypes
Answer: b) More parental than recombinant phenotypes
Explanation: In incomplete linkage, crossing over occurs occasionally between genes, resulting in fewer recombinants than parental types. - Which factor reduces crossing over?
a) Higher temperature
b) X-ray exposure
c) Age of organism
d) Heterochromatin regions
Answer: d) Heterochromatin regions
Explanation: Heterochromatin is tightly packed, transcriptionally inactive DNA that inhibits crossing over and recombination due to reduced accessibility. - A dihybrid cross gives 13:3 ratio. This is due to:
a) Linkage
b) Complementary genes
c) Inhibitory genes
d) Codominance
Answer: c) Inhibitory genes
Explanation: The 13:3 phenotypic ratio arises in dominant epistasis when one gene product suppresses the expression of another gene, such as in certain grain color patterns in maize. - The statistical test to determine linkage is:
a) t-test
b) Chi-square test
c) ANOVA
d) z-test
Answer: b) Chi-square test
Explanation: The chi-square (χ²) test assesses whether observed genetic ratios significantly deviate from expected Mendelian ratios, helping to identify linkage. - ABO blood group inheritance illustrates:
a) Pleiotropy
b) Multiple alleles
c) Codominance
d) Both b and c
Answer: d) Both b and c
Explanation: The ABO system involves three alleles (Iᴬ, Iᴮ, i). Iᴬ and Iᴮ are codominant, while both are dominant over i, demonstrating both codominance and multiple alleles. - Morgan’s experiment with white-eyed Drosophila showed:
a) Autosomal inheritance
b) Sex-linked inheritance
c) Cytoplasmic inheritance
d) Codominance
Answer: b) Sex-linked inheritance
Explanation: Morgan’s studies with Drosophila revealed that the white-eye mutation is X-linked, showing how traits can be inherited through sex chromosomes. - Map distance between genes A and B is 20 cM. The recombination frequency is:
a) 10%
b) 20%
c) 40%
d) 80%
Answer: b) 20%
Explanation: A genetic distance of 20 centiMorgans corresponds directly to a 20% recombination frequency between the two genes. - In a test cross, parental types exceed recombinants. This indicates:
a) Complete linkage
b) Incomplete linkage
c) Independent assortment
d) Mutation
Answer: b) Incomplete linkage
Explanation: In incomplete linkage, the genes are located on the same chromosome but far enough apart to allow occasional crossing over. This produces more parental combinations and fewer recombinant types—i.e., recombination frequency is less than 50% but greater than 0%. - Crossing over results in:
a) Gene duplication
b) Chromosomal aberrations
c) Genetic recombination
d) Non-disjunction
Answer: c) Genetic recombination
Explanation: During meiosis, homologous chromosomes exchange segments through crossing over, resulting in the formation of recombinant chromosomes with new combinations of alleles. This increases genetic variation. - A couple has normal vision. Their colorblind son indicates the mother is:
a) Homozygous dominant
b) Homozygous recessive
c) Heterozygous
d) Affected
Answer: c) Heterozygous
Explanation: Colorblindness is an X-linked recessive trait. A son inherits his X chromosome from his mother. Since the son is colorblind (XᶜY), the mother must carry one colorblind allele (XᴮXᶜ), making her a carrier (heterozygous). - Chromosomal theory of inheritance was proposed by:
a) Mendel
b) Sutton and Boveri
c) Morgan
d) Bateson
Answer: b) Sutton and Boveri
Explanation: In 1902, Sutton and Boveri proposed that genes are located on chromosomes and follow Mendel’s laws during meiosis, unifying genetics with cytology. - In sweet pea, purple flower (P) is dominant over red (p), and long pollen (L) dominant over round (l). A cross PpLl × ppll gives 1:1 ratio. This suggests:
a) Independent assortment
b) Linkage
c) Incomplete dominance
d) Epistasis
Answer: b) Linkage
Explanation: A 1:1 phenotypic ratio in a test cross suggests that the two genes are linked and inherited together, not assorting independently. This is complete linkage between P and L loci. - Three-point test cross is used for:
a) Detecting mutation
b) Gene sequencing
c) Ordering genes on chromosome
d) Measuring dominance
Answer: c) Ordering genes on chromosome
Explanation: A three-point test cross involves analyzing the frequency of double and single crossovers to determine the linear sequence and distance between three genes on the same chromosome. - If recombination frequency between A-B is 10%, B-C is 20%, and A-C is 30%, the gene order is:
a) A-B-C
b) A-C-B
c) B-A-C
d) C-A-B
Answer: a) A-B-C
Explanation: Since recombination frequencies are additive when genes are in linear order, the smallest distances are between A and B (10%) and B and C (20%), making the gene order A-B-C. - A test cross ratio of 1:1:1:1 for two genes indicates:
a) Complete linkage
b) Independent assortment
c) Incomplete linkage
d) Sex-linkage
Answer: b) Independent assortment
Explanation: Equal proportions of all four genotypes (two parental and two recombinant) indicate the genes are on different chromosomes or far apart on the same chromosome, showing independent assortment. - Which is a completely sex-linked trait?
a) Haemophilia
b) Sickle cell anemia
c) Cystic fibrosis
d) Albinism
Answer: a) Haemophilia
Explanation: Haemophilia is an X-linked recessive disorder. Males (XY) are more often affected because they inherit only one X chromosome from the mother; if it’s defective, no second X is present to compensate. - In Drosophila, recombination frequency between genes yellow body (y) and white eye (w) is 1%. The distance is:
a) 1 cM
b) 10 cM
c) 100 cM
d) 0.1 cM
Answer: a) 1 cM
Explanation: The distance between two genes on a chromosome is measured in centiMorgans (cM), where 1% recombination frequency corresponds to 1 cM. - A dihybrid cross produces 9:7 ratio. This is due to:
a) Complementary genes
b) Duplicate genes
c) Dominant epistasis
d) Recessive epistasis
Answer: a) Complementary genes
Explanation: In complementary gene interaction, two dominant non-allelic genes (e.g., A and B) are both required for the expression of a particular phenotype. Their absence leads to the alternate phenotype, producing a 9:7 ratio. - Which law is violated by linkage?
a) Law of Segregation
b) Law of Dominance
c) Law of Independent Assortment
d) Law of Purity of Gametes
Answer: c) Law of Independent Assortment
Explanation: Linked genes are located close together on the same chromosome and tend to be inherited together, violating Mendel’s Law of Independent Assortment. - In a three-point cross, the parental genotype is ABC/abc. The double crossover genotype is:
a) AbC/aBc
b) Abc/abC
c) aBC/Abc
d) ABc/abC
Answer: a) AbC/aBc
Explanation: In double crossovers, the middle gene changes position while the end genes remain the same. Thus, recombination occurs only in the middle gene (B), converting ABC → AbC and abc → aBc. - Coefficient of coincidence = 0.8. Interference is:
a) 20%
b) 80%
c) 0.8%
d) 0.2%
Answer: a) 20%
Explanation: Interference = 1 – Coefficient of coincidence = 1 – 0.8 = 0.2 → 20%. It indicates the percentage of expected double crossovers that were not observed due to interference. - A woman with normal vision (father colorblind) marries a colorblind man. The probability of a colorblind son is:
a) 25%
b) 50%
c) 75%
d) 100%
Answer: b) 50%
Explanation: The woman must be a carrier (XᴮXᶜ) since her father was colorblind. Her husband is colorblind (XᶜY). Half of their sons (XᶜY) will be colorblind and half (XᴮY) will be normal. So, 50% probability for a colorblind son. - In maize, gene for coloured (C) is dominant over colourless (c), and full (F) dominant over shrunken (f). A cross CcFf × ccff gives 4:4:1:1 ratio. Recombination frequency is:
a) 10%
b) 20%
c) 40%
d) 50%
Answer: b) 20%
Explanation: Total progeny = 4+4+1+1 = 10; Recombinants = 1+1 = 2. Recombination frequency = (2/10) × 100 = 20%. - Crossing over frequency is highest at:
a) Telomeres
b) Centromeres
c) Gene-rich regions
d) Heterochromatin
Answer: c) Gene-rich regions
Explanation: Crossing over occurs more frequently in euchromatin, which is gene-rich and loosely packed, unlike heterochromatin near centromeres which is tightly packed and recombinationally inactive. - Mendel chose pea plants because they:
a) Have short life cycle
b) Show incomplete dominance
c) Are cross-pollinated
d) Have linked genes
Answer: a) Have short life cycle
Explanation: Pea plants grow quickly, are easy to cultivate, have distinct contrasting traits, and allow controlled pollination, making them ideal for genetic studies. - If genes A and B are 40 cM apart, the probability of no crossover is:
a) 60%
b) 40%
c) 20%
d) 10%
Answer: a) 60%
Explanation: If the recombination frequency is 40%, the remaining 60% represent parental (non-recombinant) combinations, meaning no crossover occurred between genes A and B. - A cross between AABB × aabb produces F₁ AaBb. F₁ test cross ratio 1:1 indicates:
a) Complete linkage
b) Independent assortment
c) Codominance
d) Both a and b
Answer: a) Complete linkage
Explanation: The F₁ heterozygote (AaBb) crossed with aabb gives only two types of offspring (AaBb and aabb) in 1:1 ratio without recombinants. This indicates no crossing over — i.e., complete linkage between A and B. - In Drosophila, white eye (w) and miniature wing (m) are 8 cM apart. A cross wm/wm × WM/Y produces F₁ females. If F₁ females are test crossed, recombinant males are:
a) 4%
b) 8%
c) 16%
d) 92%
Answer: b) 8%
Explanation: The map distance between the genes is 8 cM, which equals an 8% recombination frequency. In a test cross (F₁ wm/WM × wm/Y), only recombinant gametes from F₁ females produce recombinant males. Therefore, the expected proportion of recombinant males is 8%. - Which ratio in F₂ suggests recessive epistasis?
a) 9:3:3:1
b) 9:7
c) 12:3:1
d) 9:3:4
Answer: d) 9:3:4
Explanation: A 9:3:4 ratio in F₂ indicates recessive epistasis, where the presence of two recessive alleles at one locus masks the expression of alleles at another locus. Example: coat color in mice where “cc” (recessive) masks expression of pigment genes. - Two genes show 35% recombination. The distance in map units is:
a) 35
b) 65
c) 3.5
d) 0.35
Answer: a) 35
Explanation: Recombination frequency is directly proportional to genetic distance. 1% recombination = 1 map unit (centimorgan). So 35% recombination = 35 map units. - The ratio 15:1 in F₂ generation indicates:
a) Complementary genes
b) Duplicate dominant genes
c) Dominant epistasis
d) Recessive epistasis
Answer: b) Duplicate dominant genes
Explanation: A 15:1 ratio arises when either of two dominant alleles is sufficient to express the phenotype. Only double homozygous recessive (aabb) fails to express the trait. Seen in traits controlled by duplicate genes like awnless condition in wheat. - In humans, mitochondrial genes exhibit:
a) X-linked inheritance
b) Autosomal inheritance
c) Maternal inheritance
d) Codominance
Answer: c) Maternal inheritance
Explanation: Mitochondrial DNA is transmitted through the maternal line only, as sperm contributes negligible cytoplasm. All offspring inherit mitochondrial genes from the mother regardless of sex. - A test cross gives 45% parental and 55% recombinant progeny. This suggests:
a) Complete linkage
b) Independent assortment
c) Incomplete linkage
d) Experimental error
Answer: d) Experimental error
Explanation: Recombinant progeny cannot exceed 50% in a two-gene test cross. If recombinants exceed 50%, this suggests data error or misinterpretation. Normally, unlinked genes yield 50% recombinants, and linked genes yield less. - Mendel’s monohybrid cross ratio is:
a) 1:1
b) 3:1
c) 9:3:3:1
d) 1:2:1
Answer: b) 3:1
Explanation: In F₂ generation of a monohybrid cross (e.g., Aa × Aa), Mendel observed 3 dominant : 1 recessive phenotype ratio. The genotypic ratio is 1:2:1, but phenotypically it’s 3:1 when dominant allele masks the recessive. - Linkage maps are constructed using:
a) Physical distance
b) Recombination frequency
c) Gene expression
d) Mutation rate
Answer: b) Recombination frequency
Explanation: Genetic (linkage) maps are based on observed recombination frequencies between genes during meiosis. Genes farther apart show higher recombination rates. Unlike physical maps, they do not reflect base-pair distances. - A cross between AaBb and aabb produces progeny in ratio 45:5:5:45. The configuration of genes in AaBb parent is:
a) Coupling
b) Repulsion
c) Independent
d) Unlinked
Answer: a) Coupling
Explanation: The parental types (AB and ab) appear at high frequency (45%), and recombinants (Ab and aB) at low frequency (5%). This suggests coupling phase (cis configuration), where dominant alleles are on one chromosome (AB/ab). - The phenomenon where linked genes inherit together is called:
a) Crossing over
b) Recombination
c) Linkage
d) Segregation
Answer: c) Linkage
Explanation: Linkage refers to the tendency of genes located close together on the same chromosome to be inherited together during meiosis. This reduces the likelihood of recombination between them and alters Mendelian ratios.
Chromosome aberration, Aneuploidy, Euploidy;
- Cri-du-chat syndrome results from:
a) Trisomy 13
b) Deletion in chromosome 5p
c) Reciprocal translocation
d) Inversion in chromosome 9
Answer: b) Deletion in chromosome 5p
Explanation: Cri-du-chat syndrome is caused by a terminal deletion of the short arm (p arm) of chromosome 5. Affected infants have a high-pitched cry (resembling a cat), microcephaly, intellectual disability, and facial dysmorphism. - Klinefelter syndrome has the karyotype:
a) 45,X
b) 47,XXY
c) 47,XYY
d) 47,XXX
Answer: b) 47,XXY
Explanation: Klinefelter syndrome is a condition in males where they have an extra X chromosome. Common features include tall stature, small testes, infertility, and sometimes gynecomastia. It is the most common cause of male hypogonadism. - Autopolyploidy arises from:
a) Hybridization between species
b) Doubling of chromosomes within a species
c) Deletion of chromosome segments
d) Translocation between non-homologous chromosomes
Answer: b) Doubling of chromosomes within a species
Explanation: Autopolyploidy occurs when a species’ chromosome set is duplicated without hybridization. The resulting organism has multiple chromosome sets from the same species (e.g., AAA). It often occurs naturally or can be induced. - Philadelphia chromosome is formed by:
a) t(9;22) translocation
b) t(8;14) translocation
c) 5p deletion
d) Isochromosome 17q
Answer: a) t(9;22) translocation
Explanation: The Philadelphia chromosome is formed by a reciprocal translocation between chromosomes 9 and 22. This translocation produces the BCR-ABL fusion gene, a hallmark of chronic myeloid leukemia (CML). - Down syndrome is caused by:
a) Monosomy 21
b) Trisomy 21
c) Robertsonian translocation
d) Both b and c
Answer: d) Both b and c
Explanation: Down syndrome is most commonly due to trisomy 21 (three copies of chromosome 21). In some cases (~5%), it is caused by Robertsonian translocation involving chromosome 21 and another acrocentric chromosome, such as 14. - Aneuploidy involves:
a) Loss or gain of whole chromosome sets
b) Loss or gain of single chromosomes
c) Structural changes in chromosomes
d) Point mutations
Answer: b) Loss or gain of single chromosomes
Explanation: Aneuploidy refers to an abnormal number of chromosomes, such as 2n+1 (trisomy) or 2n–1 (monosomy). Unlike polyploidy, which involves entire sets, aneuploidy affects individual chromosomes and can lead to syndromes like Down or Turner. - Turner syndrome patients have:
a) 47 chromosomes
b) 45 chromosomes
c) Triploid genome
d) Tetraploid genome
Answer: b) 45 chromosomes
Explanation: Turner syndrome occurs in females with a single X chromosome (45,X). It leads to short stature, webbed neck, lack of sexual development, and infertility. It is the only monosomy that is compatible with life in humans. - Allopolyploidy is exemplified by:
a) Triticum aestivum (wheat)
b) Oryza sativa (rice)
c) Solanum tuberosum (potato)
d) Musa acuminata (banana)
Answer: a) Triticum aestivum (wheat)
Explanation: Bread wheat is an allohexaploid (AABBDD), created by hybridization and chromosome doubling involving three different species. Allopolyploidy combines chromosome sets from distinct species, enabling fertility. - A pericentric inversion includes:
a) Only one chromosome arm
b) The centromere
c) Telomeres
d) The p-arm only
Answer: b) The centromere
Explanation: In pericentric inversion, the inverted segment includes the centromere. It differs from paracentric inversions, which occur only on one arm and exclude the centromere. Pericentric inversions can cause recombination errors. - Edward syndrome is due to:
a) Trisomy 18
b) Trisomy 13
c) Monosomy X
d) Trisomy 21
Answer: a) Trisomy 18
Explanation: Edward syndrome results from trisomy 18. Features include clenched fists, overlapping fingers, rocker-bottom feet, and heart defects. It is a severe disorder with high infant mortality. - Inversion heterozygotes produce abnormal gametes due to:
a) Crossing over within inverted region
b) Non-disjunction
c) Deletion
d) Gene duplication
Answer: a) Crossing over within inverted region
Explanation: During meiosis, crossing over within an inversion loop may result in abnormal chromosomes—dicentric or acentric (paracentric), or with duplications and deletions (pericentric), leading to inviable gametes. - A diploid organism with 2n=24 becomes tetraploid. Its chromosome number is:
a) 24
b) 36
c) 48
d) 72
Answer: c) 48
Explanation: Tetraploidy means four sets of chromosomes (4n). For a diploid with 2n=24, tetraploidy gives 4n=48 chromosomes. - Patau syndrome is characterized by:
a) Rocker-bottom feet
b) Microcephaly and cleft palate
c) Cat-like cry
d) Long arms and learning disability
Answer: b) Microcephaly and cleft palate
Explanation: Patau syndrome (trisomy 13) presents with severe congenital anomalies, including microcephaly, cleft lip and palate, polydactyly, and heart defects. It is associated with poor prognosis and high early mortality. - Balanced translocation carriers are phenotypically normal because:
a) They have extra chromosomes
b) No genetic material is lost
c) Inactivation of abnormal chromosome
d) Compensation by miRNAs
Answer: b) No genetic material is lost
Explanation: In balanced translocations, chromosomal segments are exchanged without gain or loss of genetic content. Though phenotypically normal, carriers may have reproductive issues due to unbalanced gamete formation. - Aneuploidy in plants is tolerated better than in animals due to:
a) Absence of circulatory system
b) Flexible development
c) Photosynthetic ability
d) Lack of immune response
Answer: b) Flexible development
Explanation: Plants exhibit developmental plasticity, allowing them to tolerate genetic imbalances like aneuploidy better than animals. They can often complete development even with abnormal chromosome numbers. - Terminal deletion of chromosome 4 causes:
a) Williams syndrome
b) Wolf-Hirschhorn syndrome
c) Angelman syndrome
d) Prader-Willi syndrome
Answer: b) Wolf-Hirschhorn syndrome
Explanation: Deletion of the distal part of chromosome 4’s short arm (4p-) leads to Wolf-Hirschhorn syndrome. Features include intellectual disability, growth delay, seizures, and facial features described as a “Greek warrior helmet”. - Uniparental disomy is associated with:
a) Down syndrome
b) Angelman syndrome
c) Turner syndrome
d) Klinefelter syndrome
Answer: b) Angelman syndrome
Explanation: Angelman syndrome may result from maternal deletion of 15q11–q13 or paternal uniparental disomy, meaning both copies of chromosome 15 are inherited from the father, leading to absence of maternal gene expression. - A woman with 45,XX,t(14;21) karyotype is phenotypically normal. Risk of Down syndrome in her child is:
a) 0%
b) 10–15%
c) 50%
d) 100%
Answer: b) 10–15%
Explanation: A woman with Robertsonian translocation involving chromosome 21 can produce unbalanced gametes, leading to Down syndrome in about 10–15% of her offspring due to inheritance of extra chromosome 21 material. - Isochromosomes form when:
a) Centromere divides horizontally
b) Two breaks occur in one arm
c) Centromere is lost
d) Telomeres fuse
Answer: a) Centromere divides horizontally
Explanation: Isochromosomes result when the centromere splits horizontally instead of vertically, producing chromosomes with identical arms (e.g., i(17q)). Common in cancers and some syndromes. - Which is NOT a cause of aneuploidy?
a) Non-disjunction
b) Robertsonian translocation
c) Inversion
d) Anaphase lag
Answer: c) Inversion
Explanation: Inversion involves rearrangement within a chromosome, not gain or loss. Aneuploidy results from errors like non-disjunction, translocations, or anaphase lag during cell division. - A ring chromosome forms due to:
a) Deletion of both telomeres
b) Duplication of centromere
c) Pericentric inversion
d) Isochromosome formation
Answer: a) Deletion of both telomeres
Explanation: Loss of telomeric regions causes chromosome ends to fuse, forming a ring. Ring chromosomes are unstable and may cause developmental disorders. - Triploid watermelons are seedless because:
a) They lack flowers
b) Meiosis produces unbalanced gametes
c) Pollen is sterile
d) Ovaries degenerate
Answer: b) Meiosis produces unbalanced gametes
Explanation: Triploids (3n) have uneven chromosome sets, preventing normal pairing during meiosis. This leads to sterility, resulting in seedless fruit. - Reciprocal translocation heterozygotes form:
a) Univalent
b) Bivalent
c) Trivalent
d) Quadrivalent
Answer: d) Quadrivalent
Explanation: In meiosis, individuals heterozygous for reciprocal translocations form a quadrivalent structure to allow synapsis between translocated segments. - Aneuploidy screening in pregnancy uses:
a) Amniocentesis
b) ELISA
c) Western blot
d) PCR
Answer: a) Amniocentesis
Explanation: Amniocentesis collects amniotic fluid to analyze fetal karyotype for chromosomal abnormalities like trisomy 21. It’s done in the second trimester. - Trisomy 16 in humans is:
a) Viable
b) Common in spontaneous abortions
c) Associated with longevity
d) Caused by translocation
Answer: b) Common in spontaneous abortions
Explanation: Trisomy 16 is incompatible with life and is the most frequent autosomal trisomy in miscarriages. It leads to early embryonic death. - Allopolyploids can be fertile because:
a) Chromosomes pair with identical homologs
b) They self-pollinate
c) Mitosis replaces meiosis
d) They avoid odd ploidy
Answer: a) Chromosomes pair with identical homologs
Explanation: Fertility in allopolyploids occurs when homologous chromosomes from the same ancestral genome can pair properly during meiosis. - A 47,XYY male is produced by:
a) Maternal non-disjunction
b) Paternal non-disjunction in meiosis II
c) Translocation
d) Deletion
Answer: b) Paternal non-disjunction in meiosis II
Explanation: Failure of Y chromatids to separate during paternal meiosis II results in a sperm with two Y chromosomes, leading to a 47,XYY zygote after fertilization. - Terminal deficiency in chromosome 22 causes:
a) DiGeorge syndrome
b) Cri-du-chat syndrome
c) Williams syndrome
d) Turner syndrome
Answer: a) DiGeorge syndrome
Explanation: Deletion at 22q11.2 leads to DiGeorge syndrome, characterized by thymic aplasia, cleft palate, cardiac defects, and hypocalcemia due to parathyroid agenesis. - Polyploidy in plants is induced by:
a) Colchicine
b) Ethidium bromide
c) Actinomycin D
d) Streptomycin
Answer: a) Colchicine
Explanation: Colchicine disrupts spindle formation during mitosis, preventing chromosome separation and causing chromosome doubling—producing polyploid cells. - Monosomy of autosomes is lethal in humans due to:
a) Haploinsufficiency
b) Genomic imprinting
c) Uniparental disomy
d) Dominant mutations
Answer: a) Haploinsufficiency
Explanation: Monosomy causes loss of gene copies. Many essential genes require two functional copies; a single copy is insufficient for survival (haploinsufficiency). - A Robertsonian translocation involves:
a) Exchange between acrocentric chromosomes
b) Fusion of telomeres
c) Deletion of p-arms
d) Inversion in q-arms
Answer: a) Exchange between acrocentric chromosomes
Explanation: Robertsonian translocation is a type of chromosomal rearrangement that typically occurs between two acrocentric chromosomes (e.g., chromosomes 13, 14, 15, 21, 22). The long arms (q-arms) of these chromosomes fuse at the centromere, and the short arms (p-arms) are often lost. Despite this loss, the person is usually phenotypically normal because the p-arms contain redundant rRNA genes. - A karyotype with 69 chromosomes indicates:
a) Triploidy
b) Trisomy
c) Tetraploidy
d) Monosomy
Answer: a) Triploidy
Explanation: Triploidy is a form of polyploidy where there are three sets of chromosomes (3n). In humans, the diploid number is 46 (2n), so triploidy equals 69 chromosomes. This condition often results from fertilization by two sperm (dispermy) or failure of meiotic division. - Paracentric inversion crossover produces:
a) Duplications
b) Dicentric and acentric fragments
c) Isochromosomes
d) Rings
Answer: b) Dicentric and acentric fragments
Explanation: In paracentric inversions (not involving the centromere), crossover within the inverted segment results in one chromosome with two centromeres (dicentric) and another without any centromere (acentric). These abnormal chromosomes lead to genetic instability. - The mutation causing Burkitt lymphoma is:
a) t(8;14)
b) t(9;22)
c) del(5p)
d) inv(16)
Answer: a) t(8;14)
Explanation: The hallmark of Burkitt lymphoma is the translocation t(8;14)(q24;q32), which places the MYC proto-oncogene under the control of the immunoglobulin heavy chain promoter, leading to uncontrolled MYC expression and cell proliferation. - Karyotype 47,XXX results from:
a) Paternal meiosis I error
b) Maternal meiosis I error
c) Paternal meiosis II error
d) Either maternal or paternal non-disjunction
Answer: d) Either maternal or paternal non-disjunction
Explanation: The extra X chromosome in Triple X syndrome can arise due to nondisjunction in either maternal or paternal meiosis, although it is more commonly maternal (especially during meiosis I). The individual is phenotypically female and usually asymptomatic. - Autotetraploids are represented as:
a) AABB
b) AA
c) AAAA
d) ABCD
Answer: c) AAAA
Explanation: Autotetraploids contain four identical sets of chromosomes from the same species (e.g., AAAA). This is different from allopolyploids which arise from the fusion of genomes from different species (e.g., AABB). - “Jacobsen syndrome” is due to:
a) 11q deletion
b) 11p deletion
c) 17p deletion
d) 22q deletion
Answer: a) 11q deletion
Explanation: Jacobsen syndrome, also known as 11q deletion disorder, results from partial deletion of the long arm of chromosome 11. It is associated with intellectual disabilities, distinctive facial features, and bleeding disorders due to Paris-Trousseau syndrome. - Aneuploidy is commonly caused by:
a) Deletion
b) Non-disjunction
c) Inversion
d) Duplication
Answer: b) Non-disjunction
Explanation: Aneuploidy refers to the presence of an abnormal number of chromosomes and is primarily caused by nondisjunction during meiosis or mitosis. This leads to gametes with missing or extra chromosomes (e.g., trisomy 21 or monosomy X). - Triticale is an example of:
a) Autopolyploidy
b) Allopolyploidy
c) Aneuploidy
d) Trisomy
Answer: b) Allopolyploidy
Explanation: Triticale is a man-made hybrid between wheat (Triticum) and rye (Secale), combining genomes from both parents. It is an allopolyploid because it has complete sets of chromosomes from two different species. - In a reciprocal translocation heterozygote, adjacent-1 segregation produces:
a) Balanced gametes
b) Duplications
c) Gametes with duplications/deletions
d) Haploid gametes
Answer: c) Gametes with duplications/deletions
Explanation: In adjacent-1 segregation, homologous centromeres segregate to opposite poles, but the resulting gametes carry parts of both translocated and normal chromosomes. This leads to unbalanced gametes with both duplications and deletions. - Genomic imprinting explains phenotypic differences in:
a) Down syndrome
b) Angelman vs. Prader-Willi syndromes
c) Turner syndrome
d) Klinefelter syndrome
Answer: b) Angelman vs. Prader-Willi syndromes
Explanation: Both syndromes involve deletions on chromosome 15q11-13, but differ based on parental origin. Maternal deletion causes Angelman syndrome (happy demeanor, seizures), while paternal deletion leads to Prader-Willi syndrome (obesity, hypotonia). - Polyploidy is rare in animals due to:
a) Sex determination mechanisms
b) Lack of self-fertilization
c) Complex development
d) All of the above
Answer: d) All of the above
Explanation: Polyploidy interferes with gene dosage balance, sex determination (especially in XY systems), and the intricacy of embryonic development, making it more compatible with plants than animals. - A pericentric inversion in chromosome 9 is associated with:
a) Increased risk of leukemia
b) Infertility
c) No phenotypic effect
d) Alzheimer’s disease
Answer: c) No phenotypic effect
Explanation: inv(9)(p12q13) is a common and usually benign variant found in many normal individuals. It is considered a balanced polymorphism with no clinical symptoms unless it disrupts a gene at the breakpoints. - If a diploid gamete fuses with a haploid gamete, the zygote is:
a) Diploid
b) Triploid
c) Tetraploid
d) Aneuploid
Answer: b) Triploid
Explanation: A diploid (2n) gamete fertilizing a haploid (n) gamete results in a triploid zygote (3n), which is usually non-viable in humans and leads to spontaneous abortion or developmental defects. - “Double monosomy” refers to:
a) 2n-1-1
b) 2n+1+1
c) 2n-2
d) 2n+2
Answer: a) 2n-1-1
Explanation: Double monosomy means the loss of one copy each of two different chromosomes, resulting in a total chromosome count of 44 in humans (2n=46 minus two different chromosomes). - Terminal deletion of 7q causes:
a) Williams syndrome
b) Cri-du-chat syndrome
c) Jacobsen syndrome
d) Smith-Magenis syndrome
Answer: d) Smith-Magenis syndrome
Explanation: Smith-Magenis syndrome is caused by deletion of 17p11.2, not 7q. However, among the options, it is the only one associated with a deletion; 7q deletions are not linked to Smith-Magenis. Be cautious—this could be a distractor or miskeyed. - Trisomy rescue can lead to:
a) Uniparental disomy
b) Polyploidy
c) Triploidy
d) Tetraploidy
Answer: a) Uniparental disomy
Explanation: In trisomy rescue, a zygote with three copies of a chromosome loses one to become disomic. If both remaining chromosomes come from the same parent, uniparental disomy (UPD) results, which can affect imprinting disorders. - A 45,XY,rob(13;14) karyotype indicates:
a) Trisomy 13
b) Robertsonian translocation
c) Monosomy 14
d) Inversion
Answer: b) Robertsonian translocation
Explanation: This karyotype indicates a balanced Robertsonian translocation between chromosomes 13 and 14, with one fewer total chromosome (45), but no loss of genetic material from the q-arms. - Nullisomy in wheat (2n=42) has chromosomes:
a) 40
b) 41
c) 43
d) 44
Answer: a) 40
Explanation: Nullisomy is the complete loss of a homologous chromosome pair. In hexaploid wheat (2n = 6x = 42), nullisomy would result in 42 – 2 = 40 chromosomes. - “Pallister-Killian syndrome” is caused by:
a) Trisomy 12p
b) Tetrasomy 12p
c) Deletion 12p
d) Inversion 12p
Answer: b) Tetrasomy 12p
Explanation: Pallister-Killian syndrome is due to mosaic tetrasomy 12p, often involving an isochromosome i(12p). It causes intellectual disability, hypotonia, and distinctive facial features.
Gene Mutation – Definition, types and importance,
- Which mutation type involves a single nucleotide change without altering the amino acid sequence?
(a) Missense
(b) Nonsense
(c) Silent
(d) Frameshift
Answer: (c) Silent
Explanation: Silent mutations involve base substitutions that do not change the amino acid coded due to the redundancy in the genetic code (degeneracy), typically occurring at the third position of a codon. - A point mutation converting a codon to a stop codon is called:
(a) Silent mutation
(b) Nonsense mutation
(c) Missense mutation
(d) Deletion
Answer: (b) Nonsense mutation
Explanation: Nonsense mutations change a codon that encodes an amino acid into a stop codon, resulting in premature termination of translation and often leading to a nonfunctional truncated protein. - Frameshift mutations result from:
(a) Substitution of one base
(b) Insertion or deletion of bases not divisible by 3
(c) Duplication of a codon
(d) Inversion of a segment
Answer: (b) Insertion or deletion of bases not divisible by 3
Explanation: Such mutations shift the reading frame of mRNA, altering all downstream amino acids, and typically yield nonfunctional or truncated proteins. - Sickle-cell anemia is caused by which type of mutation?
(a) Nonsense
(b) Missense
(c) Silent
(d) Frameshift
Answer: (b) Missense
Explanation: A single base change (A→T) in the HBB gene changes codon GAG (glutamic acid) to GTG (valine), altering hemoglobin’s structure and causing red blood cell sickling. - Which mutation is LEAST likely to affect protein function?
(a) Silent mutation
(b) Frameshift mutation
(c) Nonsense mutation
(d) Splice-site mutation
Answer: (a) Silent mutation
Explanation: Silent mutations do not alter the amino acid sequence and thus typically have minimal to no impact on protein function. - In a gene, if 3 nucleotides are deleted, the mutation is:
(a) Frameshift
(b) In-frame deletion
(c) Nonsense
(d) Insertion
Answer: (b) In-frame deletion
Explanation: Deleting three nucleotides removes one codon but maintains the reading frame, potentially leading to a shorter yet still partially functional protein. - Mutagens like UV light primarily cause:
(a) Chromosomal translocations
(b) Pyrimidine dimers
(c) Aneuploidy
(d) Trisomy
Answer: (b) Pyrimidine dimers
Explanation: UV light induces covalent bonding between adjacent thymine or cytosine bases, forming dimers that distort DNA and impede replication. - Which enzyme repairs thymine dimers via nucleotide excision repair?
(a) DNA glycosylase
(b) DNA ligase
(c) Exinuclease
(d) Photolyase
Answer: (c) Exinuclease
Explanation: In humans, nucleotide excision repair removes bulky lesions like thymine dimers through endonuclease (exinuclease) activity. Photolyase is used in other organisms, not humans. - A mutation in the promoter region of a gene most likely affects:
(a) Protein folding
(b) Transcription efficiency
(c) mRNA splicing
(d) tRNA binding
Answer: (b) Transcription efficiency
Explanation: The promoter contains binding sites for transcription factors and RNA polymerase. Mutations here affect the rate of transcription initiation. - Which mutation type contributes to evolutionary diversity?
(a) Lethal mutation
(b) Somatic mutation
(c) Neutral mutation
(d) Conditional mutation
Answer: (c) Neutral mutation
Explanation: Neutral mutations do not affect fitness and accumulate over time, providing raw material for evolution through genetic drift. - Cri-du-chat syndrome results from which mutation?
(a) Inversion
(b) Translocation
(c) Duplication
(d) Deletion
Answer: (d) Deletion
Explanation: Caused by deletion on the short arm of chromosome 5 (5p15.2), leading to microcephaly, developmental delay, and cat-like cry. - The Ames test detects:
(a) Carcinogenicity
(b) Teratogenicity
(c) Mutagenicity
(d) Pathogenicity
Answer: (c) Mutagenicity
Explanation: The Ames test uses Salmonella bacteria with a known mutation; reversion to normal growth in the presence of a chemical suggests mutagenic potential. - Which is a transition mutation?
(a) AT → GC
(b) AT → TA
(c) AT → CG
(d) GC → TA
Answer: (a) AT → GC
Explanation: A→G or T→C are transitions, meaning a purine is replaced with another purine or a pyrimidine with another pyrimidine. - BRCA1 gene mutations increase risk for:
(a) Lung cancer
(b) Breast cancer
(c) Colon cancer
(d) Leukemia
Answer: (b) Breast cancer
Explanation: BRCA1 is involved in DNA repair. Mutations disrupt repair pathways, increasing susceptibility to breast and ovarian cancers. - Huntington’s disease involves which mutation?
(a) Point mutation
(b) Frameshift mutation
(c) Dynamic mutation
(d) Silent mutation
Answer: (c) Dynamic mutation
Explanation: Huntington’s is caused by CAG trinucleotide repeat expansion (>36 repeats) in the HTT gene, worsening with each generation (anticipation). - Which mutagen causes alkylation of DNA bases?
(a) UV light
(b) Acridine orange
(c) Nitrous acid
(d) EMS (Ethyl methanesulfonate)
Answer: (d) EMS (Ethyl methanesulfonate)
Explanation: EMS alkylates guanine bases, causing mispairing during replication (G pairs with T instead of C). - Somatic mutations are crucial for:
(a) Cancer development
(b) Inherited disorders
(c) Speciation
(d) Antibiotic resistance
Answer: (a) Cancer development
Explanation: Somatic mutations in genes like TP53 or RAS can lead to uncontrolled cell growth and tumor formation. - Ionizing radiation (e.g., X-rays) primarily causes:
(a) Thymine dimers
(b) Double-strand breaks
(c) Depurination
(d) Cross-linking
Answer: (b) Double-strand breaks
Explanation: Ionizing radiation produces reactive radicals that break both strands of the DNA, which can lead to deletions, rearrangements, or cell death. - A missense mutation in the CFTR gene causes:
(a) Sickle-cell anemia
(b) Alzheimer’s
(c) Hemophilia
(d) Cystic fibrosis
Answer: (d) Cystic fibrosis
Explanation: The common ΔF508 mutation removes phenylalanine from CFTR protein, causing misfolding and impaired chloride transport. - Spontaneous deamination converts cytosine to:
(a) Thymine
(b) Adenine
(c) Uracil
(d) Guanine
Answer: (c) Uracil
Explanation: Deamination replaces the amino group in cytosine with a carbonyl, converting it to uracil—a base not typically found in DNA. - Knockout mice study:
(a) Loss-of-function mutations
(b) Gain-of-function mutations
(c) Neutral mutations
(d) Regulatory mutations
Answer: (a) Loss-of-function mutations
Explanation: Knockout models help determine the function of a gene by observing what happens when it is completely inactivated. - Which mutagen is used in crop improvement?
(a) 5-Bromouracil
(b) Ethyl methanesulfonate (EMS)
(c) Aflatoxin B1
(d) Mustard gas
Answer: (b) Ethyl methanesulfonate (EMS)
Explanation: EMS induces random point mutations in plant genomes, useful for creating new traits like disease resistance. - The Philadelphia chromosome involves:
(a) Deletion of chromosome 22
(b) Inversion of chromosome 9
(c) Translocation t(9;22)
(d) Trisomy 21
Answer: (c) Translocation t(9;22)
Explanation: The BCR-ABL fusion from t(9;22) translocation results in a constitutively active tyrosine kinase, driving CML. - Xeroderma pigmentosum results from defective:
(a) Mismatch repair
(b) Base excision repair
(c) Nucleotide excision repair (NER)
(d) Direct repair
Answer: (c) Nucleotide excision repair (NER)
Explanation: XP patients lack the ability to remove UV-induced thymine dimers, increasing skin cancer risk. - “Anticipation” in pedigrees suggests:
(a) Genomic imprinting
(b) Trinucleotide repeat expansion
(c) Somatic mosaicism
(d) Epigenetic silencing
Answer: (b) Trinucleotide repeat expansion
Explanation: Disorders like Huntington’s and Fragile X show earlier and more severe symptoms in successive generations due to expanding repeats. - A silent mutation is exemplified by:
(a) UUU → UUC (Phe → Leu)
(b) UGG → UAG (Trp → Stop)
(c) AAA → AAG (Lys → Lys)
(d) CAU → CAC (His → His)
Answer: (c) AAA → AAG (Lys → Lys)
Explanation: Both codons code for lysine, representing a classic example of a silent (synonymous) mutation. - Ethidium bromide mutates DNA by:
(a) Base alkylation
(b) Intercalation
(c) Deamination
(d) Cross-linking
Answer: (b) Intercalation
Explanation: Ethidium bromide slips between stacked base pairs, distorting DNA structure and causing frameshift mutations during replication. - Germline mutations are important because they:
(a) Cause aging
(b) Trigger autoimmune diseases
(c) Are heritable
(d) Repair chromosomal breaks
Answer: (c) Are heritable
Explanation: Germline mutations are passed from parents to offspring and contribute to inherited genetic traits and diseases. - Frameshift mutations arise from inserting:
(a) 1 nucleotide
(b) 3 nucleotides
(c) 6 nucleotides
(d) 9 nucleotides
Answer: (a) 1 nucleotide
Explanation: Insertions or deletions not divisible by three disrupt the triplet codon reading frame, altering the downstream protein sequence. - Which is a point mutation subtype?
(a) Deletion
(b) Inversion
(c) Transition
(d) Duplication
Answer: (c) Transition
Explanation: A transition is a specific type of point mutation involving a purine-purine (A↔G) or pyrimidine-pyrimidine (C↔T) substitution. - Tautomeric shifts in DNA bases cause:
(a) Induced mutations
(b) Spontaneous mutations
(c) Only transversions
(d) Chromosomal breaks
Answer: (b) Spontaneous mutations
Explanation: Tautomeric shifts are rare, spontaneous changes in base structure (e.g., keto ↔ enol in thymine, amino ↔ imino in cytosine), causing mispairing during DNA replication. For example, an enol form of thymine can pair with guanine instead of adenine, leading to base substitution mutations without exposure to external mutagens. - A nonsense mutation is exemplified by:
(a) UAC → UAG (Tyr → Stop)
(b) UUU → UUC (Phe → Phe)
(c) GAA → GAG (Glu → Glu)
(d) UGG → UAG (Trp → Stop)
Answer: (d) UGG → UAG (Trp → Stop)
Explanation: Nonsense mutations convert a sense codon into a stop codon, leading to premature termination of translation. UGG (tryptophan) is converted to UAG (a stop codon), which stops protein synthesis early. Option (a) is misleading as UAC codes for tyrosine, and UAG is a stop, but this mutation doesn’t occur naturally from UAC. - Aflatoxin B1 primarily targets which organ for mutagenesis?
(a) Liver
(b) Lung
(c) Kidney
(d) Skin
Answer: (a) Liver
Explanation: Aflatoxin B1, a potent mycotoxin produced by Aspergillus flavus, is metabolized in the liver into a reactive epoxide that binds to guanine in DNA, leading to G→T transversions. It often mutates the TP53 tumor suppressor gene, contributing to hepatocellular carcinoma. - DNA polymerase proofreading prevents:
(a) Transcription errors
(b) Translation errors
(c) Replication errors
(d) Post-translational errors
Answer: (c) Replication errors
Explanation: DNA polymerases have a 3’→5’ exonuclease activity, enabling proofreading during DNA replication. This function removes misincorporated nucleotides immediately after they are added, ensuring high fidelity in DNA synthesis. - Mutational “hot spots” occur frequently in:
(a) Telomeres
(b) Repetitive sequences
(c) Centromeres
(d) Introns
Answer: (b) Repetitive sequences
Explanation: Repetitive sequences, such as microsatellites, are prone to replication slippage and strand misalignment, making them hot spots for mutations. These errors can lead to insertions or deletions, as seen in diseases like Huntington’s and Fragile X syndrome. - Constitutive expression of the lac operon results from mutations in:
(a) Operator
(b) Structural gene
(c) Repressor gene
(d) Promoter
Answer: (d) Promoter
Explanation: A promoter mutation that enhances RNA polymerase binding can lead to constitutive (continuous) transcription, even in the absence of inducers. Such “up” promoter mutations override regulatory controls. - Mutations are agriculturally important for developing:
(a) Disease-resistant crops
(b) Hybrid seeds
(c) Polyploid fruits
(d) Genetically identical clones
Answer: (a) Disease-resistant crops
Explanation: Induced mutagenesis (e.g., using EMS, X-rays, gamma rays) creates novel alleles that confer resistance to pathogens. Many commercial crops (e.g., disease-resistant wheat, rice) have been developed using this approach. - The unit for absorbed dose of ionizing radiation is:
(a) Sievert
(b) Becquerel
(c) Gray
(d) Curie
Answer: (c) Gray
Explanation: Gray (Gy) is the SI unit for absorbed radiation dose, where 1 Gy = 1 joule of energy absorbed per kilogram of tissue. Sievert (Sv) accounts for biological effect; Becquerel and Curie measure radioactivity, not absorbed dose. - Site-directed mutagenesis uses:
(a) CRISPR only
(b) PCR-based methods
(c) X-ray exposure
(d) Chemical mutagens
Answer: (b) PCR-based methods
Explanation: Site-directed mutagenesis involves designing primers with specific mutations to introduce base changes during PCR. This technique is used to study gene function or modify protein-coding regions precisely. - Hydrogen peroxide (H₂O₂) causes mutations via:
(a) Alkylation
(b) Intercalation
(c) Deamination
(d) Oxidative damage
Answer: (d) Oxidative damage
Explanation: H₂O₂ produces reactive oxygen species (ROS) like hydroxyl radicals that oxidize guanine to 8-oxoguanine, which pairs with adenine instead of cytosine, causing G→T transversions and DNA damage. - Hereditary nonpolyposis colorectal cancer (HNPCC) involves defective:
(a) Nucleotide excision repair
(b) Base excision repair
(c) Mismatch repair
(d) Double-strand break repair
Answer: (c) Mismatch repair
Explanation: HNPCC, or Lynch syndrome, is caused by mutations in mismatch repair (MMR) genes such as MLH1, MSH2. These mutations lead to microsatellite instability and accumulation of replication errors in oncogenes and tumor suppressor genes. - A transversion mutation is:
(a) Purine → Pyrimidine substitution
(b) Pyrimidine → Pyrimidine substitution
(c) Insertion of 3 nucleotides
(d) Deletion of a codon
Answer: (a) Purine → Pyrimidine substitution
Explanation: Transversions occur when a purine (A/G) is replaced by a pyrimidine (C/T) or vice versa. These mutations are more structurally disruptive than transitions (purine↔purine or pyrimidine↔pyrimidine). - The SOS response in E. coli is:
(a) Error-free repair
(b) Base excision repair
(c) Photoreactivation
(d) Error-prone repair
Answer: (d) Error-prone repair
Explanation: The SOS response is triggered by DNA damage and involves LexA repression removal, activating error-prone polymerases (e.g., Pol IV, Pol V), which can bypass DNA lesions but often insert incorrect bases. - Hermann Muller induced mutations in Drosophila using:
(a) UV radiation
(b) X-rays
(c) Mustard gas
(d) Ethanol
Answer: (b) X-rays
Explanation: Hermann Muller demonstrated that X-rays can induce mutations in fruit flies, proving the mutagenic potential of radiation and earning the Nobel Prize in 1946. - Benzo[a]pyrene in tobacco smoke causes mutations by:
(a) Deaminating cytosine
(b) Creating thymine dimers
(c) Forming DNA adducts
(d) Cross-linking strands
Answer: (c) Forming DNA adducts
Explanation: Benzo[a]pyrene is metabolized into a diol epoxide that binds guanine bases in DNA, forming bulky adducts that distort the helix and result in G→T transversions, commonly observed in smoking-related cancers. - A null mutation results in:
(a) Complete loss of protein function
(b) Reduced gene expression
(c) Enhanced protein activity
(d) Altered protein localization
Answer: (a) Complete loss of protein function
Explanation: Null mutations, such as nonsense or frameshift mutations, produce nonfunctional proteins or truncated polypeptides, eliminating gene function entirely (e.g., gene knockouts). - 5-Bromouracil acts as a mutagen by functioning as a:
(a) Intercalating agent
(b) Alkylating agent
(c) Base analog
(d) Deaminating agent
Answer: (c) Base analog
Explanation: 5-Bromouracil resembles thymine and can incorporate into DNA. When tautomerized, it pairs with guanine instead of adenine, causing transition mutations (A→G or T→C). - Suppressor mutations restore function by:
(a) Increasing mutation rate
(b) Compensating for the original mutation
(c) Silencing the gene
(d) Activating repair pathways
Answer: (b) Compensating for the original mutation
Explanation: Suppressor mutations (in the same or different gene) restore normal function by offsetting the effects of a previous mutation. For instance, a tRNA suppressor can insert an amino acid at a premature stop codon. - AP endonuclease repairs DNA damage caused by:
(a) UV radiation
(b) Double-strand breaks
(c) Depurination
(d) Mismatched bases
Answer: (c) Depurination
Explanation: AP endonucleases act at apurinic/apyrimidinic (AP) sites—locations where bases have been lost, often through depurination. These enzymes cleave the DNA backbone to allow repair synthesis. - Depurination of DNA is caused by:
(a) Heat/acidic conditions
(b) UV light
(c) Ionizing radiation
(d) Base analogs
Answer: (a) Heat/acidic conditions
Explanation: Depurination is a spontaneous reaction where purine bases (adenine or guanine) are lost from the DNA backbone, often under acidic or high-temperature conditions. This creates AP sites, leading to mutation if unrepaired.
Gene regulation (Operation concept), genetic code, protein synthesis,
- In the lac operon, what binds to the operator to prevent transcription?
(a) Allolactose
(b) cAMP
(c) Repressor protein
(d) RNA polymerase
Explanation: The lac operon repressor protein, encoded by the lacI gene, binds to the operator region in the absence of lactose, blocking RNA polymerase from initiating transcription. Allolactose (an inducer) binds the repressor, preventing it from attaching to the operator. - The trp operon is repressed when:
(a) Tryptophan levels are low
(b) Tryptophan acts as a co-repressor
(c) Attenuation occurs
(d) cAMP levels are high
Explanation: When tryptophan is abundant, it binds to the trp repressor protein, allowing it to bind the operator and block transcription—this is classic negative feedback control. - CAP-cAMP complex enhances lac operon transcription by binding to:
(a) The operator
(b) The promoter
(c) Structural genes
(d) The repressor
Explanation: When glucose is low, cAMP binds to CAP. The CAP-cAMP complex then binds upstream of the promoter to enhance RNA polymerase binding, increasing transcription of the lac operon. - Attenuation in the trp operon involves:
(a) Repressor binding
(b) Premature transcription termination
(c) mRNA degradation
(d) Protein inactivation
Explanation: Attenuation involves the formation of a terminator hairpin in the mRNA leader sequence when tryptophan is abundant. This causes RNA polymerase to terminate transcription early. - A constitutive mutant of the lac operon could result from a mutation in:
(a) lacZ gene
(b) lacI gene
(c) lacY gene
(d) CAP binding site
Explanation: A mutation in the lacI gene prevents the repressor protein from being made or functioning properly, leading to continuous (constitutive) transcription regardless of lactose presence. - Which operon is inducible?
(a) Trp operon
(b) Lac operon
(c) Arg operon
(d) His operon
Explanation: The lac operon is inducible—it’s off by default and turns on in the presence of lactose. In contrast, the trp operon is repressible and is turned off when tryptophan is present. - In the absence of glucose, cAMP activates:
(a) Repressor protein
(b) CAP protein
(c) Beta-galactosidase
(d) Permease
Explanation: Low glucose leads to high cAMP levels. cAMP binds to CAP, forming the CAP-cAMP complex that activates transcription of the lac operon by facilitating RNA polymerase binding. - The ara operon is regulated by:
(a) Only repression
(b) Both activation and repression
(c) Attenuation
(d) Riboswitches
Explanation: The AraC protein regulates the ara operon. In the absence of arabinose, it acts as a repressor; in the presence, it activates transcription—a dual regulatory function. - Catabolite repression prioritizes:
(a) Glucose over lactose
(b) Lactose over glucose
(c) Tryptophan over glucose
(d) Arabinose over lactose
Explanation: Bacteria preferentially metabolize glucose first. When glucose is available, cAMP levels drop, inactivating CAP and repressing transcription of alternative sugar operons (like lac). - A mutation preventing repressor binding in the trp operon would cause:
(a) No transcription
(b) Constitutive transcription
(c) Attenuation only
(d) Enhanced repression
Explanation: If the repressor cannot bind the operator, the operon remains active regardless of tryptophan levels, leading to constitutive (continuous) transcription. - The genetic code is degenerate because:
(a) Codons overlap
(b) Multiple codons code for one amino acid
(c) One codon codes for multiple amino acids
(d) Stop codons are absent
Explanation: Degeneracy refers to redundancy—most amino acids are coded by more than one codon (e.g., leucine has six codons), providing error tolerance in mutations. - Wobble hypothesis explains:
(a) Universality of genetic code
(b) Flexible pairing at the third base of codon
(c) Non-overlapping codons
(d) Start codon function
Explanation: The third base of the codon can pair loosely with tRNA anticodon (e.g., G-U pairing), allowing fewer tRNAs to recognize multiple codons for the same amino acid. - Which is a stop codon?
(a) AUG
(b) UGA
(c) GUG
(d) UUG
Explanation: UGA is one of three stop codons (UAA, UAG, UGA) that signal termination of translation. AUG is a start codon. - Mitochondrial genetic code differs in that AUA codes for:
(a) Isoleucine
(b) Methionine
(c) Stop
(d) Tyrosine
Explanation: In the standard genetic code, AUA codes for isoleucine. In mitochondria, it codes for methionine, reflecting evolutionary divergence. - The start codon AUG codes for:
(a) Formyl methionine (prokaryotes)
(b) Methionine (eukaryotes)
(c) Both (a) and (b)
(d) Methionine in all organisms
Explanation: AUG universally codes for methionine. In prokaryotes, it is modified to N-formyl methionine (fMet), initiating translation. - Shine-Dalgarno sequence is essential for:
(a) Eukaryotic transcription
(b) Prokaryotic translation initiation
(c) Splicing
(d) tRNA charging
Explanation: This purine-rich sequence upstream of the start codon in prokaryotic mRNA pairs with 16S rRNA in the ribosome to initiate translation. - Peptidyl transferase activity is a function of:
(a) mRNA
(b) Small ribosomal subunit
(c) 23S rRNA in large subunit
(d) tRNA
Explanation: The 23S rRNA in the large ribosomal subunit acts as a ribozyme, catalyzing peptide bond formation during translation. - Aminoacyl-tRNA synthetase ensures:
(a) Transcription fidelity
(b) Correct amino acid-tRNA pairing
(c) mRNA splicing
(d) Ribosome assembly
Explanation: This enzyme attaches the correct amino acid to its matching tRNA, a critical step in ensuring translational accuracy. - Kozak sequence enhances:
(a) Prokaryotic transcription
(b) Eukaryotic translation initiation
(c) tRNA charging
(d) Splicing efficiency
Explanation: The Kozak sequence flanks the start codon in eukaryotic mRNAs and helps ribosomes identify the correct AUG for translation initiation. - Release factors recognize:
(a) Start codons
(b) Stop codons
(c) Shine-Dalgarno sequence
(d) Promoters
Explanation: Release factors bind to stop codons in the A site of the ribosome, triggering hydrolysis of the polypeptide from tRNA and ending translation. - Antibiotic inhibiting the 30S ribosomal subunit is:
(a) Chloramphenicol
(b) Erythromycin
(c) Tetracycline
(d) Cycloheximide
Explanation: Tetracycline binds to the 30S subunit of prokaryotic ribosomes, blocking aminoacyl-tRNA entry into the A site. Cycloheximide targets eukaryotic ribosomes. - snRNPs are primarily involved in:
(a) Transcription
(b) Splicing
(c) Translation
(d) DNA replication
Explanation: snRNPs (small nuclear ribonucleoproteins) form the core of the spliceosome, removing introns from pre-mRNA in eukaryotic cells. - An example of post-translational modification is:
(a) 5′ capping
(b) Glycosylation
(c) Polyadenylation
(d) Splicing
Explanation: Glycosylation involves adding sugar groups to proteins, altering their stability, folding, and localization—occurs after translation. - Chargaff’s rule states:
(a) A = T and G = C
(b) A = G and T = C
(c) A + C = T + G
(d) A + T = G + C
Explanation: In double-stranded DNA, the number of adenines equals thymines, and guanines equal cytosines—evidence of base pairing. - Rho-dependent transcription termination requires:
(a) Sigma factor
(b) Rho helicase
(c) Release factors
(d) snRNPs
Explanation: Rho is a helicase that binds to the rut site on RNA and moves along the transcript, causing RNA polymerase to disassociate at termination sites. - An amino acid is activated by attaching to:
(a) mRNA
(b) tRNA
(c) rRNA
(d) DNA
Explanation: Aminoacyl-tRNA synthetases link amino acids to their corresponding tRNAs, forming aminoacyl-tRNA (charged tRNA) for use in translation. - The eukaryotic transcription factor that binds the TATA box is:
(a) TFIIA
(b) TFIID
(c) TFIIH
(d) Sigma factor
Explanation: TFIID, specifically the TBP (TATA-binding protein) subunit, recognizes and binds the TATA box to initiate pre-initiation complex assembly. - Formylmethionine (fMet) is used in translation initiation by:
(a) Prokaryotes
(b) Eukaryotes
(c) Mitochondria
(d) Chloroplasts
Explanation: fMet is the first amino acid incorporated during translation in prokaryotes and organelles like mitochondria and chloroplasts. - Isoaccepting tRNAs:
(a) Carry the same amino acid
(b) Have different anticodons
(c) Both (a) and (b)
(d) Recognize stop codons
Explanation: Isoacceptor tRNAs recognize different codons but attach the same amino acid, contributing to the degeneracy of the genetic code. - An example of a ribozyme is:
(a) DNA polymerase
(b) RNase P
(c) Aminoacyl-tRNA synthetase
(d) Helicase
Explanation: RNase P is an RNA molecule with catalytic activity involved in processing tRNA precursors—one of the first discovered ribozymes. - Intron removal occurs via:
(a) 5′ capping
(b) Polyadenylation
(c) Transesterification reactions
(d) Charging
Answer: (c) Transesterification reactions
Explanation: In eukaryotes, introns are removed through a precise splicing process involving two sequential transesterification reactions, catalyzed by the spliceosome complex. The first reaction occurs at the 5′ splice site, where the 2′-OH of an internal adenosine within the branch point attacks the 5′ end of the intron. The second reaction links the two exons and releases the intron in the form of a lariat structure. - Polycistronic mRNA is commonly found in:
(a) Prokaryotes
(b) Eukaryotes
(c) Fungi
(d) Plants
Answer: (a) Prokaryotes
Explanation: In prokaryotes, polycistronic mRNAs are produced from operons and encode multiple proteins from a single mRNA molecule. For example, the lac operon produces a single transcript for lacZ, lacY, and lacA genes, allowing coordinated regulation. In contrast, eukaryotic mRNA is typically monocistronic. - An antibiotic targeting peptidyl transferase activity is:
(a) Streptomycin
(b) Chloramphenicol
(c) Tetracycline
(d) Rifampicin
Answer: (b) Chloramphenicol
Explanation: Chloramphenicol inhibits the peptidyl transferase activity of the 50S ribosomal subunit in prokaryotes. This enzyme catalyzes peptide bond formation between amino acids during translation elongation. By blocking this, protein synthesis is halted. - The 5′ cap of eukaryotic mRNA consists of:
(a) 7-methyladenosine
(b) 7-methylguanosine
(c) Poly-A tail
(d) TATA box
Answer: (b) 7-methylguanosine
Explanation: The 5′ cap is a 7-methylguanosine (m⁷G) residue linked via a 5′-5′ triphosphate bond to the first nucleotide of the pre-mRNA. This modification protects mRNA from degradation, facilitates nuclear export, and promotes ribosome binding during translation initiation. - Selenocysteine is incorporated during translation using:
(a) AUG codon
(b) UGA codon in a specific context
(c) Stop codon readthrough
(d) Frameshifting
Answer: (b) UGA codon in a specific context
Explanation: Selenocysteine, known as the 21st amino acid, is inserted at UGA codons (normally stop codons) in the presence of a SECIS (Selenocysteine Insertion Sequence) in the mRNA. Specialized elongation factors and tRNAs recognize this context-specific recoding. - The holoenzyme for prokaryotic transcription is:
(a) Core RNA polymerase
(b) RNA polymerase + Sigma factor
(c) TFIID complex
(d) Ribosome
Answer: (b) RNA polymerase + Sigma factor
Explanation: The RNA polymerase holoenzyme in prokaryotes comprises the core enzyme (α₂ββ’ω) and a σ (sigma) factor, which is essential for promoter recognition and transcription initiation. The core alone cannot initiate transcription accurately. - The ribosomal E-site binds:
(a) Aminoacyl-tRNA
(b) Peptidyl-tRNA
(c) Deacylated tRNA
(d) mRNA
Answer: (c) Deacylated tRNA
Explanation: The E-site (exit site) of the ribosome temporarily holds the uncharged (deacylated) tRNA after its amino acid has been transferred to the growing polypeptide chain, prior to its release from the ribosome. - The consensus sequence for the splice donor site is:
(a) AGGU
(b) GU
(c) AG
(d) CCA
Answer: (b) GU
Explanation: The 5′ end (donor site) of eukaryotic introns typically begins with GU, while the 3′ end (acceptor site) ends with AG. These conserved sequences are critical for spliceosome recognition and correct intron removal. - The Signal Recognition Particle (SRP) binds to:
(a) Cytosolic proteins
(b) Nascent polypeptide with signal sequence
(c) tRNA
(d) DNA promoter
Answer: (b) Nascent polypeptide with signal sequence
Explanation: The SRP is a ribonucleoprotein that recognizes and binds to the signal sequence emerging from ribosomes synthesizing secretory proteins. It pauses translation and directs the complex to the endoplasmic reticulum membrane for co-translational translocation. - A nonsense mutation converts:
(a) Amino acid to another amino acid
(b) Codon to a synonymous codon
(c) Amino acid codon to a stop codon
(d) Stop codon to an amino acid codon
Answer: (c) Amino acid codon to a stop codon
Explanation: Nonsense mutations change a sense codon (coding for an amino acid) into a premature stop codon (e.g., UAG, UGA, UAA). This leads to truncated proteins, often nonfunctional. Example: CAG → UAG can disrupt β-globin in beta-thalassemia. - tmRNA rescues stalled ribosomes by:
(a) Adding a poly-A tail to mRNA
(b) Tagging the incomplete protein for degradation
(c) Repairing damaged DNA
(d) Charging tRNAs
Answer: (b) Tagging the incomplete protein for degradation
Explanation: tmRNA functions as both tRNA and mRNA in bacteria. When ribosomes stall on defective mRNAs, tmRNA enters the A-site and allows translation to continue, appending a peptide degradation tag to the incomplete protein, marking it for proteolysis. - The editing function of DNA polymerase I is due to its:
(a) 5’→3′ polymerase activity
(b) 3’→5′ exonuclease activity
(c) 5’→3′ exonuclease activity
(d) Helicase domain
Answer: (b) 3’→5′ exonuclease activity
Explanation: DNA polymerase I proofreads DNA through its 3’→5′ exonuclease activity, removing incorrectly incorporated nucleotides during DNA synthesis, thus enhancing fidelity. Its 5’→3′ exonuclease activity removes RNA primers. - The amino acid encoded by only one codon is:
(a) Serine (6 codons)
(b) Arginine (6 codons)
(c) Methionine (AUG)
(d) Leucine (6 codons)
Answer: (c) Methionine (AUG)
Explanation: Methionine is encoded solely by AUG, which also serves as the start codon. Another example is Tryptophan, encoded only by UGG. All other amino acids have multiple codons, showing redundancy of the genetic code. - IRES (Internal Ribosome Entry Site) enables:
(a) Intron splicing
(b) Cap-independent translation initiation
(c) Transcription activation
(d) Polyadenylation
Answer: (b) Cap-independent translation initiation
Explanation: IRES elements are RNA sequences that allow ribosomes to bind internally, bypassing the need for the 5′ cap. Found in viral RNAs (e.g., poliovirus) and some eukaryotic stress-response genes, they ensure translation under adverse conditions. - Attenuation in the trp operon requires:
(a) Low tryptophan for termination
(b) Ribosome stalling at trp codons in the leader sequence
(c) Repressor binding
(d) CAP activation
Answer: (b) Ribosome stalling at trp codons in the leader sequence
Explanation: In the trp operon, attenuation is a transcriptional control mechanism. When tryptophan is scarce, the ribosome stalls at Trp codons in the leader region, allowing formation of an anti-terminator structure that permits transcription continuation. - Ubiquitin targets proteins for:
(a) Glycosylation
(b) Proteasomal degradation
(c) Nuclear export
(d) Phosphorylation
Answer: (b) Proteasomal degradation
Explanation: Ubiquitination is a post-translational modification where ubiquitin molecules are covalently attached to lysine residues of proteins, tagging them for degradation by the 26S proteasome, a major regulatory system in eukaryotic cells. - Proofreading by aminoacyl-tRNA synthetase occurs at:
(a) The catalytic site
(b) A separate editing site
(c) Ribosomal A-site
(d) mRNA codon
Answer: (b) A separate editing site
Explanation: Many aminoacyl-tRNA synthetases have a proofreading (editing) domain that removes incorrectly attached amino acids, ensuring the correct amino acid is charged to its corresponding tRNA. This helps maintain translation fidelity. - The Ribosome Recycling Factor (RRF) dissociates:
(a) mRNA from tRNA
(b) The post-termination ribosomal complex
(c) DNA-RNA hybrids
(d) Spliceosomes
Answer: (b) The post-termination ribosomal complex
Explanation: After translation termination, RRF along with EF-G and GTP promotes disassembly of the 70S ribosome into its 50S and 30S subunits, releasing tRNA and mRNA to allow reuse of ribosomal components. - CRISPR-Cas9 originated from:
(a) Eukaryotic RNAi
(b) Prokaryotic adaptive immunity
(c) Viral replication machinery
(d) Transposon jumping
Answer: (b) Prokaryotic adaptive immunity
Explanation: The CRISPR-Cas system is a bacterial and archaeal defense mechanism against foreign genetic elements. It uses CRISPR-derived RNAs (crRNAs) to guide Cas9 endonuclease to specific DNA sequences for cleavage. - Molecular chaperones assist in:
(a) Transcription initiation
(b) Protein folding
(c) tRNA charging
(d) DNA replication
Answer: (b) Protein folding
Explanation: Chaperones, such as Hsp70, GroEL-GroES, bind to nascent or misfolded proteins, helping them achieve proper tertiary structure. They prevent aggregation and assist in refolding or degradation under stress conditions.
Brief Knowledge about Transposon.
- Who discovered transposons in maize?
(a) James Watson
(b) Barbara McClintock
(c) Gregor Mendel
(d) Thomas Hunt Morgan
Explanation: Barbara McClintock discovered transposable elements while studying color variegation in maize kernels. She identified the Ac-Ds system, revealing genes could change positions within the genome. She received the Nobel Prize in Physiology or Medicine in 1983 for this work. - Transposons that move via a “cut-and-paste” mechanism are:
(a) Retrotransposons
(b) Class II DNA transposons
(c) SINEs
(d) LINEs
Explanation: Class II DNA transposons move without an RNA intermediate, relying on a transposase enzyme to excise themselves from one site and integrate into another. - Which is a retrotransposon?
(a) Tn5
(b) Alu element
(c) Ac element
(d) P element
Explanation: Alu elements are short interspersed nuclear elements (SINEs) that replicate via reverse transcription. They are abundant in the primate genome and classified as non-autonomous retrotransposons. - The inverted terminal repeats (ITRs) in DNA transposons are essential for:
(a) Translation
(b) Transposase recognition
(c) RNA splicing
(d) Protein folding
Explanation: ITRs are sequences at both ends of DNA transposons that are recognized by transposase, guiding precise excision and insertion. - In maize, the autonomous element in the Ac-Ds system is:
(a) Ds (Dissociation)
(b) Ac (Activator)
(c) Spm (Suppressor-mutator)
(d) Mu (Mutator)
Explanation: The Ac element encodes functional transposase, making it autonomous, whereas Ds lacks this gene and requires Ac to move. - P elements in Drosophila are widely used for:
(a) DNA replication studies
(b) Germline transformation
(c) Antibiotic resistance
(d) RNA interference
Explanation: Engineered P elements are used as vectors to insert foreign DNA into the germline of Drosophila for genetic studies. - Composite transposons in bacteria (e.g., Tn10) carry:
(a) Only transposase genes
(b) Transposase + antibiotic resistance genes
(c) Reverse transcriptase
(d) Ribosomal RNA
Explanation: Composite transposons consist of two IS elements flanking additional genes like antibiotic resistance, enabling co-movement. - LINEs (Long Interspersed Nuclear Elements) encode:
(a) Transposase
(b) Reverse transcriptase and endonuclease
(c) DNA polymerase
(d) RNA polymerase
Explanation: LINEs are autonomous retrotransposons. In humans, LINE-1 encodes both reverse transcriptase and endonuclease to mediate their own integration. - Transposons contribute to genome evolution by:
(a) Preventing mutations
(b) Facilitating exon shuffling
(c) Inhibiting recombination
(d) Stabilizing telomeres
Explanation: Transposons can induce recombination between non-homologous regions, contributing to novel exon arrangements and gene evolution. - Hybrid Dysgenesis in Drosophila results from:
(a) Ac-Ds transposition
(b) P element mobilization
(c) Alu insertion
(d) Tn5 excision
Explanation: When wild-type males with active P elements are crossed with lab females lacking them, uncontrolled transposition causes sterility and mutations. - The Mutator (Mu) system in maize is a:
(a) DNA transposon
(b) Retrotransposon
(c) Group II intron
(d) Plasmid
Explanation: The Mu system consists of highly active DNA transposons that induce high mutation rates; MuDR is the autonomous master element. - Transposon tagging identifies genes by:
(a) Inhibiting transcription
(b) Insertional mutagenesis
(c) DNA methylation
(d) RNA splicing
Explanation: When transposons insert into genes and cause mutations, the tagged locus can be isolated and sequenced to identify gene function. - Which enzyme is encoded by retrotransposons?
(a) DNA polymerase
(b) Reverse transcriptase
(c) RNA polymerase
(d) DNA ligase
Explanation: Reverse transcriptase synthesizes DNA from RNA, a key step in retrotransposon life cycles (e.g., LINE-1). - Alu elements belong to which category?
(a) DNA transposons
(b) SINEs
(c) LINEs
(d) LTR retrotransposons
Explanation: Alu is a SINE, short and non-autonomous, relying on LINE-1 enzymes for retrotransposition. They are primate-specific. - The IS (Insertion Sequence) element in bacteria lacks:
(a) Terminal repeats
(b) Antibiotic resistance genes
(c) Transposase gene
(d) Inverted repeats
Explanation: IS elements only carry genes for their own transposition, unlike composite transposons which have extra genes. - Transposons can cause mutations by:
(a) Correcting DNA errors
(b) Inserting into functional genes
(c) Stabilizing chromosomes
(d) Enhancing DNA repair
Explanation: Transposon insertion disrupts coding or regulatory sequences, leading to loss or gain of function mutations. - The mariner transposon is notable for:
(a) Host-specificity to plants
(b) Horizontal gene transfer across species
(c) Encoding reverse transcriptase
(d) RNA-mediated silencing
Explanation: Mariner transposons are DNA-based elements found in diverse animal species, suggesting they spread via horizontal transfer. - Retrotransposons replicate via:
(a) Direct DNA replication
(b) “Copy-and-paste” mechanism
(c) “Cut-and-paste” mechanism
(d) Conservative transposition
Explanation: Retrotransposons make RNA copies, which are reverse-transcribed to DNA and integrated elsewhere, increasing copy number. - Which transposon is used in genome editing (e.g., Sleeping Beauty)?
(a) Ac
(b) Alu
(c) Tc1/mariner family
(d) P element
Explanation: Sleeping Beauty is a synthetic Tc1/mariner transposon used in vertebrate gene transfer, including human gene therapy trials. - Transposons influence epigenetics by:
(a) Degrading histones
(b) Triggering DNA methylation
(c) Activating telomerase
(d) Preventing recombination
Explanation: Cells methylate transposons to prevent their movement, which can also spread into neighboring genes, silencing them epigenetically.
Oncogene,
- Oncogenes are derived from:
(a) Tumor suppressor genes
(b) Proto-oncogenes
(c) DNA repair genes
(d) Apoptotic genes
Explanation: Proto-oncogenes are normal genes that regulate cell growth and differentiation. When mutated, overexpressed, or misregulated, they become oncogenes that drive uncontrolled proliferation, contributing to cancer. - Which mechanism activates proto-oncogenes to oncogenes?
(a) DNA methylation
(b) Chromosomal translocation
(c) Telomere shortening
(d) Histone deacetylation
Explanation: Chromosomal translocations can bring proto-oncogenes under strong promoters or fuse them with other genes, producing oncogenic fusion proteins. A classic example is the BCR-ABL fusion from t(9;22) in CML. - The RAS oncogene is most commonly activated by:
(a) Point mutations
(b) Gene amplification
(c) Deletion
(d) Viral insertion
Explanation: RAS activation typically involves point mutations at codons 12, 13, or 61, locking it in the GTP-bound active state, leading to continuous cell signaling for proliferation. - HER2/neu (ERBB2) oncogene amplification is associated with:
(a) Lung cancer
(b) Breast cancer
(c) Colon cancer
(d) Leukemia
Explanation: HER2/neu gene amplification leads to overexpression of the HER2 receptor tyrosine kinase in a subset of breast cancers, often resulting in aggressive tumor behavior. - The Philadelphia chromosome results from:
(a) t(8;14)
(b) t(9;22)
(c) t(15;17)
(d) t(11;14)
Explanation: The t(9;22)(q34;q11) translocation forms the Philadelphia chromosome, fusing BCR with ABL to create a constitutively active tyrosine kinase that drives CML. - Which viral oncoprotein inactivates p53?
(a) HPV E6
(b) HPV E7
(c) SV40 T-antigen
(d) HTLV Tax
Explanation: SV40 large T-antigen binds and inactivates p53 and RB tumor suppressors, disrupting cell cycle checkpoints and promoting transformation. - MYC oncogene dysregulation often occurs via:
(a) Point mutations
(b) Translocation (e.g., Burkitt lymphoma)
(c) Deletion
(d) Viral integration
Explanation: In Burkitt lymphoma, t(8;14) places MYC under the control of the immunoglobulin heavy chain enhancer, leading to uncontrolled MYC expression. - Oncogene addiction refers to:
(a) Tumor reliance on angiogenesis
(b) Cancer cell dependence on a single oncogene
(c) Immune evasion by tumors
(d) Metastasis initiation
Explanation: Certain cancers are highly dependent on a single oncogene for survival and proliferation. Targeted inhibition (e.g., imatinib for BCR-ABL) can induce tumor regression. - The SRC oncogene encodes a:
(a) G-protein
(b) Non-receptor tyrosine kinase
(c) Growth factor receptor
(d) Transcription factor
Explanation: SRC is the first identified oncogene and encodes a cytoplasmic tyrosine kinase that promotes signal transduction leading to proliferation and survival. - BRAF V600E mutation is common in:
(a) Glioblastoma
(b) Melanoma
(c) Pancreatic cancer
(d) Prostate cancer
Explanation: The BRAF V600E mutation results in constitutive activation of the MAPK/ERK pathway and is a common driver in melanomas. It is targetable by drugs like vemurafenib. - Tumor suppressor genes differ from oncogenes in that they:
(a) Promote cell division
(b) Require two-hit inactivation
(c) Are viral in origin
(d) Amplify during cancer
Explanation: Tumor suppressors like RB or TP53 require both alleles to be inactivated (Knudson’s two-hit hypothesis), while a single activating mutation is often sufficient for oncogenes. - TP53 is a tumor suppressor, but mutant p53 acts as:
(a) Cell cycle inhibitor
(b) Dominant-negative oncoprotein
(c) DNA repair enzyme
(d) Angiogenesis promoter
Explanation: Mutant p53 can interfere with wild-type p53 activity and acquire gain-of-function mutations that promote metastasis, resistance, and poor prognosis. - BCR-ABL is targeted by:
(a) Cetuximab
(b) Imatinib
(c) Trastuzumab
(d) Rituximab
Explanation: Imatinib specifically inhibits the BCR-ABL tyrosine kinase by blocking ATP binding, effectively treating CML and revolutionizing targeted cancer therapy. - Which oncogene is a growth factor receptor?
(a) RAS
(b) EGFR
(c) MYC
(d) JUN
Explanation: EGFR (Epidermal Growth Factor Receptor) is a receptor tyrosine kinase; overexpression or mutation leads to activation of downstream pathways promoting cancer. - MYC oncogene’s primary function is:
(a) DNA repair
(b) Transcriptional activation of cell cycle genes
(c) Apoptosis induction
(d) Angiogenesis inhibition
Explanation: MYC is a transcription factor that upregulates genes essential for cell cycle progression, growth, metabolism, and ribosome biogenesis. - RAS mutations are prevalent in:
(a) Breast cancer
(b) Pancreatic cancer (90%)
(c) Leukemia
(d) Thyroid cancer
Explanation: KRAS mutations are found in over 90% of pancreatic ductal adenocarcinomas, making it one of the most RAS-driven cancers. - APC gene mutations cause:
(a) Melanoma
(b) Familial adenomatous polyposis (FAP)
(c) Li-Fraumeni syndrome
(d) Neurofibromatosis
Explanation: APC is a tumor suppressor that regulates β-catenin. Its loss leads to Wnt pathway activation and hundreds of polyps in FAP patients. - MDM2 amplification in cancers:
(a) Activates p53
(b) Degrades p53
(c) Repairs DNA
(d) Promotes apoptosis
Explanation: MDM2 is an E3 ubiquitin ligase that binds and degrades p53. Overexpression of MDM2 mimics TP53 loss-of-function. - JAK2 V617F mutation is associated with:
(a) Lung adenocarcinoma
(b) Myeloproliferative neoplasms (e.g., polycythemia vera)
(c) Glioblastoma
(d) Ovarian cancer
Explanation: JAK2 V617F is a gain-of-function mutation that leads to cytokine-independent proliferation in blood-forming cells, seen in polycythemia vera and related diseases. - CDK4 amplification drives cancer by:
(a) Inhibiting cyclins
(b) Phosphorylating RB to release E2F
(c) Activating p53
(d) Degrading MYC
Explanation: CDK4, when overexpressed or amplified, partners with cyclin D to phosphorylate RB, thereby releasing E2F and allowing G1/S cell cycle progression. - PIK3CA mutations activate:
(a) MAPK pathway
(b) PI3K-AKT pathway
(c) JAK-STAT pathway
(d) TGF-β pathway
Explanation: PIK3CA encodes the catalytic subunit of PI3K. Mutations lead to increased PIP3 levels, activating AKT and promoting survival, proliferation, and metabolism. - NOTCH1 acts as an oncogene in:
(a) Colon cancer
(b) T-cell acute lymphoblastic leukemia (T-ALL)
(c) Breast cancer
(d) Glioblastoma
Explanation: Gain-of-function mutations in NOTCH1 in T-ALL promote T-cell proliferation and block differentiation, acting as an oncogenic driver. - ALK translocations targetable by crizotinib occur in:
(a) Non-small cell lung cancer (NSCLC)
(b) Chronic lymphocytic leukemia
(c) Osteosarcoma
(d) Bladder cancer
Explanation: EML4-ALK fusion is present in a subset of NSCLC patients. Crizotinib is a selective ALK inhibitor used in targeted therapy. - HPV E6 oncoprotein promotes cancer by:
(a) Activating RAS
(b) Degrading p53 via ubiquitin ligase
(c) Inhibiting RB
(d) Amplifying MYC
Explanation: HPV E6 binds p53 and recruits E6AP, a ubiquitin ligase, leading to proteasomal degradation of p53 and loss of cell cycle control. - MET oncogene amplification causes resistance to:
(a) Imatinib
(b) EGFR inhibitors (e.g., gefitinib)
(c) Trastuzumab
(d) Vemurafenib
Explanation: MET amplification activates bypass signaling through ERBB3 and PI3K, undermining EGFR inhibitor therapy in NSCLC. - BCL2 is an oncogene that:
(a) Promotes DNA repair
(b) Inhibits apoptosis
(c) Activates cell cycle
(d) Degrades oncoproteins
Explanation: BCL2 prevents mitochondrial outer membrane permeabilization, blocking cytochrome c release and apoptosis. Overexpression is common in follicular lymphoma. - RET mutations are linked to:
(a) Ovarian cancer
(b) Medullary thyroid carcinoma
(c) Prostate cancer
(d) Liver cancer
Explanation: Activating mutations in RET lead to constitutive kinase signaling, causing MEN2 syndromes, including medullary thyroid cancer. - HRAS mutations are characteristic of:
(a) Chronic myeloid leukemia
(b) Bladder cancer
(c) Retinoblastoma
(d) Wilms tumor
Explanation: HRAS mutations are frequent in low-grade urothelial carcinomas. These mutations activate downstream signaling pathways promoting cell growth. - VHL gene loss leads to oncogenesis via:
(a) RAS activation
(b) HIF-1α stabilization
(c) MYC amplification
(d) EGFR overexpression
Explanation: VHL normally degrades HIF-1α under normoxia. Its loss results in HIF accumulation, driving angiogenesis and cell survival, especially in renal cancer. - IDH1 mutations in gliomas cause:
(a) Enhanced DNA repair
(b) Production of oncometabolite D-2-HG
(c) Telomerase activation
(d) RB phosphorylation
Explanation: Mutant IDH1 converts α-ketoglutarate to D-2-hydroxyglutarate, an oncometabolite that inhibits DNA demethylation and blocks cell differentiation.
Gene cloning,
- Gene cloning primarily involves:
(a) Mutating specific genes
(b) Producing identical copies of a gene/DNA sequence
(c) Sequencing entire genomes
(d) Synthesizing artificial genes
Explanation: Gene cloning refers to the process of creating multiple, identical copies of a specific gene or DNA segment. This is commonly achieved by inserting the target DNA into a cloning vector (like a plasmid), which is then introduced into a host organism (e.g., E. coli), where it replicates. - Which enzyme is essential for cutting DNA at specific sites during cloning?
(a) DNA ligase
(b) Restriction endonuclease
(c) DNA polymerase
(d) Reverse transcriptase
Explanation: Restriction endonucleases recognize specific palindromic sequences in DNA and cut at or near these sites. For example, EcoRI cuts at GAATTC. These enzymes are vital for preparing DNA fragments for cloning. - Plasmids are preferred as cloning vectors because they:
(a) Integrate into host chromosomes
(b) Self-replicate and carry selectable markers
(c) Can hold DNA inserts >100 kb
(d) Are viral in origin
Explanation: Plasmids are small, circular DNA molecules that replicate independently of the host genome. They contain an origin of replication and selectable markers (e.g., antibiotic resistance genes), which are crucial for selecting transformed cells. - The function of DNA ligase in cloning is to:
(a) Cut DNA
(b) Join DNA fragments
(c) Denature DNA
(d) Proofread sequences
Explanation: DNA ligase forms phosphodiester bonds between the 3’-OH and 5’-phosphate ends of DNA, sealing gaps and joining the vector and insert into a continuous molecule. - A “sticky end” produced by EcoRI has:
(a) Blunt terminus
(b) 5′-overhang (AATT-)
(c) 3′-overhang (-AATT)
(d) No overhang
Explanation: EcoRI cuts DNA asymmetrically between G and A in the GAATTC site, creating short single-stranded 5’ overhangs (AATT), which are known as sticky ends and facilitate ligation. - Blue-white screening identifies recombinants based on:
(a) Antibiotic resistance
(b) Disruption of β-galactosidase gene (lacZ)
(c) Fluorescence
(d) Plasmid size
Explanation: In blue-white screening, the lacZ gene encodes β-galactosidase. Insertional inactivation by foreign DNA leads to white colonies (recombinants), while blue colonies are non-recombinant. - Which vector is used for cloning large DNA inserts (100–300 kb)?
(a) Plasmid
(b) Lambda phage
(c) BAC (Bacterial Artificial Chromosome)
(d) Cosmid
Explanation: BACs, derived from F plasmids, are designed to carry large DNA fragments and are used to construct genomic libraries, especially in genome projects. - The purpose of a polylinker in a vector is to:
(a) Enhance replication
(b) Provide multiple restriction sites for cloning
(c) Promote transcription
(d) Stabilize the insert
Explanation: A polylinker or multiple cloning site (MCS) is a short region containing several unique restriction sites, allowing flexibility in choosing enzymes for insertion of foreign DNA. - Reverse transcriptase is used to clone:
(a) Genomic DNA
(b) cDNA from mRNA
(c) Mitochondrial DNA
(d) Viral DNA
Explanation: Reverse transcriptase synthesizes complementary DNA (cDNA) from an mRNA template, which is useful for cloning expressed genes without introns. - Electroporation is a method for:
(a) DNA amplification
(b) Introducing DNA into host cells
(c) DNA sequencing
(d) DNA labeling
Explanation: Electroporation uses high-voltage electric pulses to create temporary pores in cell membranes, through which DNA can enter the host cell. - The first recombinant DNA molecule was constructed using:
(a) Human insulin and plasmid
(b) SV40 virus and lambda phage
(c) pBR322 and EcoRI
(d) Bacterial plasmid and Salmonella gene
Answer: (d) Bacterial plasmid and Salmonella gene
Explanation: In 1973, Stanley Cohen and Herbert Boyer constructed the first recombinant DNA molecule by inserting a gene from Salmonella into a plasmid (pSC101) from E. coli. This marked the beginning of modern genetic engineering. - Shuttle vectors can replicate in:
(a) Only bacteria
(b) Multiple hosts (e.g., bacteria and yeast)
(c) Only eukaryotes
(d) Viruses only
Answer: (b) Multiple hosts (e.g., bacteria and yeast)
Explanation: Shuttle vectors contain origins of replication and selectable markers for more than one type of organism, such as E. coli and yeast. They are especially useful in eukaryotic gene expression systems. - Alkaline phosphatase treatment prevents:
(a) DNA cutting
(b) Vector self-ligation
(c) Insert degradation
(d) Host cell transformation
Answer: (b) Vector self-ligation
Explanation: Alkaline phosphatase removes 5′ phosphate groups from the cut ends of plasmid vectors. This prevents the vector from closing back on itself without incorporating the foreign DNA insert. - A genomic library contains:
(a) Only expressed genes
(b) All DNA fragments of an organism
(c) Mutated genes
(d) Viral sequences
Answer: (b) All DNA fragments of an organism
Explanation: A genomic library is a collection of DNA fragments that represent the entire genome of an organism, including both coding and non-coding regions, introns, and regulatory elements. - Ti plasmid is used for cloning genes in:
(a) Bacteria
(b) Plants
(c) Mammals
(d) Fungi
Answer: (b) Plants
Explanation: The Ti (tumor-inducing) plasmid from Agrobacterium tumefaciens naturally transfers DNA to plant cells. Scientists modify it to deliver genes without causing tumors, making it a tool in plant biotechnology. - The selectable marker in pBR322 is:
(a) LacZ
(b) Ampicillin and tetracycline resistance
(c) GFP
(d) T7 promoter
Answer: (b) Ampicillin and tetracycline resistance
Explanation: pBR322 contains two antibiotic resistance genes: ampᵣ (ampicillin) and tetᵣ (tetracycline), which allow selection of transformed bacterial cells and identification of recombinant colonies. - Cosmid vectors combine features of:
(a) Plasmid and BAC
(b) Plasmid and phage lambda
(c) YAC and plasmid
(d) BAC and phage
Answer: (b) Plasmid and phage lambda
Explanation: Cosmids are plasmids containing cos sites from lambda phage. They replicate like plasmids but are packaged into phage particles, enabling the cloning of larger DNA fragments (~40–45 kb). - Southern blotting detects:
(a) Proteins
(b) Specific DNA sequences
(c) RNA
(d) Carbohydrates
Answer: (b) Specific DNA sequences
Explanation: Southern blotting involves transferring DNA from a gel onto a membrane and probing it with a labeled complementary DNA probe to detect specific sequences. - Biolistic gene transfer employs:
(a) Viral vectors
(b) Gene gun
(c) Electroporation
(d) Heat shock
Answer: (b) Gene gun
Explanation: The biolistic method (gene gun) fires DNA-coated gold or tungsten particles into plant or animal tissues. It is especially useful for species that are not easily transformed by Agrobacterium. - Expression vectors require:
(a) Restriction sites only
(b) Promoters and ribosome-binding sites
(c) Telomeres
(d) Introns
Answer: (b) Promoters and ribosome-binding sites
Explanation: Expression vectors must have promoters for transcription initiation and ribosome binding sites for translation. This ensures the cloned gene is expressed as a protein in the host organism. - cDNA libraries lack:
(a) Exons
(b) Introns and regulatory sequences
(c) Coding regions
(d) Poly-A tails
Answer: (b) Introns and regulatory sequences
Explanation: cDNA is synthesized from mature mRNA and therefore includes only the exons (coding regions). Introns and non-coding regulatory DNA are absent, making cDNA libraries useful for expressing eukaryotic genes in prokaryotes. - The Cre-lox system is used for:
(a) DNA amplification
(b) Site-specific recombination
(c) DNA sequencing
(d) Protein purification
Answer: (b) Site-specific recombination
Explanation: The Cre-lox system, derived from bacteriophage P1, uses Cre recombinase to catalyze recombination between loxP sites, allowing controlled gene deletion, inversion, or integration. - Golden Rice was developed by cloning genes for:
(a) Pest resistance
(b) β-carotene biosynthesis
(c) Vitamin C synthesis
(d) Nitrogen fixation
Answer: (b) β-carotene biosynthesis
Explanation: Golden Rice contains genes from daffodil and Pantoea bacteria that enable the biosynthesis of β-carotene, a vitamin A precursor, in rice endosperm, addressing vitamin A deficiency. - Colony hybridization identifies clones using:
(a) Antibodies
(b) DNA probes
(c) Enzymes
(d) Fluorescent tags
Answer: (b) DNA probes
Explanation: Colony hybridization uses labeled DNA probes to hybridize with complementary sequences in bacterial colonies. Colonies containing the target gene are identified via autoradiography or fluorescence. - YACs (Yeast Artificial Chromosomes) mimic:
(a) Bacterial chromosomes
(b) Eukaryotic chromosomes
(c) Viral genomes
(d) Mitochondrial DNA
Answer: (b) Eukaryotic chromosomes
Explanation: YACs contain yeast centromeres, telomeres, and autonomous replication sequences (ARS), allowing them to replicate and segregate like natural eukaryotic chromosomes. - Recombinant human insulin is produced in:
(a) Saccharomyces cerevisiae
(b) Escherichia coli
(c) Pichia pastoris
(d) Mammalian cells
Answer: (b) Escherichia coli
Explanation: The synthetic A and B chain genes of human insulin are expressed separately in E. coli, purified, and chemically joined to form functional insulin. It was the first recombinant therapeutic protein approved for human use. - Phagemids are vectors with:
(a) Only plasmid ori
(b) Plasmid and phage origins
(c) Only phage ori
(d) BAC features
Answer: (b) Plasmid and phage origins
Explanation: Phagemids (e.g., pBluescript) combine plasmid replication and phage packaging elements. They can be packaged into phage particles when helper phages are present, allowing efficient single-stranded DNA production. - The main goal of the Human Genome Project was:
(a) Gene therapy
(b) Sequencing the entire human genome
(c) Cloning humans
(d) Creating GMOs
Answer: (b) Sequencing the entire human genome
Explanation: The Human Genome Project (1990–2003) aimed to determine the complete sequence of human DNA (~3 billion base pairs) and identify all human genes. It laid the foundation for genomics and personalized medicine. - PCR is NOT typically used for:
(a) Amplifying DNA inserts
(b) Site-directed mutagenesis
(c) Ligating DNA fragments
(d) Cloning gene fragments
Answer: (c) Ligating DNA fragments
Explanation: PCR (Polymerase Chain Reaction) is used to amplify DNA and introduce mutations but does not ligate DNA. Ligation is carried out by DNA ligase enzymes to form recombinant molecules. - The primary application of gene cloning is:
(a) Forensic analysis
(b) Evolutionary studies
(c) Recombinant protein production
(d) Fossil dating
Answer: (c) Recombinant protein production
Explanation: Gene cloning allows the production of recombinant proteins like insulin, human growth hormone, and interferons in host organisms (e.g., E. coli, yeast), revolutionizing medicine and biotechnology.
Gene transfer
- Bacterial conjugation involves gene transfer via:
(a) Naked DNA uptake
(b) Direct cell-to-cell contact
(c) Viral vectors
(d) Electroporation
Explanation: Conjugation is a horizontal gene transfer mechanism requiring physical contact between donor and recipient cells, typically via a sex pilus. The F⁺ (fertility factor-containing) cell forms a pilus to connect with an F⁻ cell and transfer plasmid or chromosomal DNA. - Griffith’s experiment demonstrated:
(a) Transduction
(b) Bacterial transformation
(c) Conjugation
(d) Transfection
Explanation: Griffith’s 1928 experiment showed that dead virulent S. pneumoniae could transform non-virulent live bacteria into virulent forms. This demonstrated transformation, a process where bacteria uptake and incorporate foreign DNA from the environment. - Which virus is commonly used for gene therapy?
(a) Influenza virus
(b) Adenovirus
(c) HIV
(d) Herpes simplex virus
Explanation: Adenoviruses are widely used in gene therapy because they can efficiently infect dividing and non-dividing cells, deliver genes episomally (without integrating into host DNA), and express transgenes transiently. - T-DNA transfer in Agrobacterium tumefaciens is mediated by:
(a) Conjugative pilus
(b) Type IV secretion system
(c) Phage tail
(d) Membrane fusion
Explanation: Agrobacterium uses a Type IV secretion system (VirB/D4 complex) to deliver T-DNA from the Ti plasmid into plant host cells, where it integrates into the nuclear genome, causing crown gall tumors. - Transduction is gene transfer via:
(a) Free DNA
(b) Bacteriophages
(c) Cell fusion
(d) Nanotubes
Explanation: In transduction, bacteriophages (viruses that infect bacteria) transfer bacterial DNA between cells. It occurs in two forms—generalized transduction (random DNA) and specialized transduction (specific DNA near prophage). - Biolistic gene transfer uses:
(a) Electric pulses
(b) DNA-coated microprojectiles
(c) Lipid vesicles
(d) Viral envelopes
Explanation: Also called the gene gun method, this technique uses gold or tungsten particles coated with DNA, which are shot into cells using high-velocity propulsion. It is used in plant and animal cells, especially for chloroplast or embryo transformation. - The F factor in E. coli enables:
(a) Transformation
(b) Conjugation
(c) Transduction
(d) Transposition
Explanation: The F factor (fertility plasmid) carries genes necessary for pilus formation and enables DNA transfer by conjugation. Cells with the F plasmid are called F⁺, and recipients are F⁻. - Electroporation facilitates gene transfer by:
(a) Chemical permeabilization
(b) Temporary membrane pores via electric pulses
(c) Viral fusion
(d) Mechanical injection
Explanation: Electroporation uses short electric pulses to create transient pores in the plasma membrane, allowing DNA or other molecules to enter the cell. It is used in both prokaryotic and eukaryotic systems. - Retroviral vectors integrate into the host genome using:
(a) DNA ligase
(b) Integrase
(c) Transposase
(d) Recombinase
Explanation: Retroviruses reverse-transcribe their RNA into DNA and use a viral enzyme called integrase to insert the DNA into the host genome, enabling stable gene expression in target cells. - Lipofection uses:
(a) Viral capsids
(b) Cationic lipid-DNA complexes
(c) Gold nanoparticles
(d) Protein carriers
Explanation: In lipofection, cationic liposomes encapsulate DNA or RNA and fuse with the cell membrane, delivering the genetic material into the cell. It is a non-viral and efficient transfection method. - Hfr (High-frequency recombination) strains transfer:
(a) Only F plasmid
(b) Chromosomal DNA
(c) Phage DNA
(d) Transposons
Explanation: In Hfr strains, the F factor is integrated into the bacterial chromosome, so during conjugation, the chromosomal genes adjacent to the F factor are transferred to the recipient cell at high frequency. - Competent cells in transformation are:
(a) Resistant to antibiotics
(b) Able to uptake extracellular DNA
(c) Virally infected
(d) Conjugative donors
Explanation: Competent cells are in a physiological state that allows them to take up foreign DNA from their environment. Competency can be natural (e.g., Bacillus) or artificially induced (e.g., CaCl₂ treatment in E. coli). - Which is a horizontal gene transfer mechanism?
(a) Mitosis
(b) Transduction
(c) Meiosis
(d) Replication
Explanation: Horizontal gene transfer involves the movement of genetic material between organisms, not from parent to offspring. Transduction is an example of horizontal transfer, as it involves gene transfer via viruses. - Microinjection is commonly used for:
(a) Bacterial transformation
(b) Animal cell transgenesis
(c) Plant protoplasts
(d) Fungal spores
Explanation: Microinjection directly introduces DNA into animal cells, often into zygotes or oocytes using a fine glass needle. It is a precise but labor-intensive method used for creating transgenic animals. - In A. tumefaciens, the vir genes are activated by:
(a) Light
(b) Plant phenolic compounds (e.g., acetosyringone)
(c) Nitrogen limitation
(d) pH change
Explanation: When Agrobacterium detects plant wound compounds such as acetosyringone, it activates vir genes on the Ti plasmid, initiating T-DNA processing and transfer into the plant genome. - Phage λ-mediated specialized transduction transfers:
(a) Any bacterial gene
(b) Specific genes near attB
(c) Plasmid DNA
(d) Ribosomal RNA
Explanation: In specialized transduction, only bacterial genes near the phage integration site (attB) are transferred (e.g., gal or bio operons in λ phage), due to imprecise excision of the prophage. - Sonoporation uses:
(a) Magnetic fields
(b) Ultrasound waves
(c) Lasers
(d) Heat shock
Explanation: Sonoporation uses ultrasound waves to create temporary pores in the cell membrane, allowing DNA or drugs to enter. It is often enhanced by microbubbles to increase permeability. - Gene doping in athletes refers to illicit use of:
(a) Antibiotics
(b) Gene therapy vectors (e.g., EPO gene)
(c) Hormone blockers
(d) CRISPR-Cas9
Explanation: Gene doping involves introducing genetic material to enhance physical performance, such as using EPO (erythropoietin) genes to increase red blood cells and oxygen-carrying capacity illegally. - The Ti plasmid is disarmed for plant genetic engineering by deleting:
(a) vir genes
(b) Oncogenes (e.g., iaaM, ipt)
(c) Opine synthesis genes
(d) Origin of replication
Explanation: In genetic engineering, oncogenes (e.g., causing tumor growth) on the Ti plasmid are deleted, rendering it non-pathogenic, while retaining T-DNA border sequences and vir genes for DNA transfer. - Magnetofection enhances gene transfer using:
(a) Electric fields
(b) Magnetic nanoparticles
(c) Light pulses
(d) Centrifugation
Explanation: Magnetofection uses DNA bound to magnetic nanoparticles, which are guided by external magnetic fields to target cells, enhancing transfection efficiency. - Conjugative transposons transfer via:
(a) Transformation
(b) Conjugation-like mechanism
(c) Transduction
(d) Electroporation
Explanation: Conjugative transposons can excise from one genome, form a circular intermediate, and transfer via conjugation-like mechanisms, independent of plasmids or phages. - The first gene therapy treated:
(a) Cystic fibrosis
(b) Severe Combined Immunodeficiency (SCID)
(c) Duchenne muscular dystrophy
(d) Sickle-cell anemia
Explanation: In 1990, the first successful human gene therapy treated a girl with ADA-SCID, using retroviral vectors to introduce the functional ADA gene into her white blood cells. - Cre-lox system requires:
(a) Integrase
(b) Site-specific recombinase
(c) Transposase
(d) DNA ligase
Explanation: The Cre-lox system uses Cre recombinase to recognize and recombine at loxP sites, allowing for precise insertion, deletion, or inversion of DNA, often in transgenic mouse models. - Gene gun is used for:
(a) Bacterial conjugation
(b) Plant and animal cell transformation
(c) Viral transduction
(d) DNA sequencing
Explanation: The gene gun delivers DNA-coated particles into plant and animal cells, especially hard-to-transform tissues like chloroplasts or embryonic plant cells. - In situ hybridization detects:
(a) Proteins
(b) Specific DNA/RNA in tissues
(c) Carbohydrates
(d) Lipids
Explanation: In situ hybridization (ISH) uses labeled nucleic acid probes to detect specific DNA or RNA sequences in fixed tissues, retaining spatial information. - Calcium phosphate transfection is used for:
(a) In vivo gene therapy
(b) In vitro mammalian cell transfection
(c) Bacterial transformation
(d) Plant protoplasts
Explanation: This classical method forms DNA-calcium phosphate precipitates that enter mammalian cells by endocytosis and is often used in laboratory cell culture systems. - Dendrimers in gene transfer act as:
(a) Viral envelopes
(b) Synthetic polymer carriers
(c) Electroporation enhancers
(d) Nucleases
Explanation: Dendrimers are branched, tree-like polymers that compact DNA into nanoparticles, facilitating cellular uptake and endosomal escape during gene delivery. - The Flp-FRT system is analogous to:
(a) CRISPR-Cas9
(b) Cre-lox
(c) TALENs
(d) Zinc-finger nucleases
Explanation: Like Cre-lox, the Flp-FRT system uses Flp recombinase to mediate site-specific recombination at FRT sites, allowing for controlled gene editing. - Viroids transfer via:
(a) Conjugation
(b) Mechanical damage (e.g., tools)
(c) Transduction
(d) Electroporation
Explanation: Viroids are naked infectious RNAs that spread mechanically, such as via contaminated agricultural tools, or through sap and insect vectors in plants. - Germline gene transfer refers to modification of:
(a) Somatic cells
(b) Gametes or zygotes
(c) Stem cells
(d) Cancer cells
Explanation: Germline gene transfer modifies the heritable genome (e.g., in sperm, eggs, or embryos). Changes can be passed to future generations, raising ethical and safety concerns.
PCR.
- The inventor of PCR awarded the Nobel Prize in Chemistry is:
(a) James Watson
(b) Kary Mullis
(c) Francis Collins
(d) Craig Venter
Explanation: Kary Mullis developed the polymerase chain reaction (PCR) technique in 1983, which revolutionized molecular biology by enabling rapid amplification of DNA. He was awarded the Nobel Prize in Chemistry in 1993 for this contribution. - The denaturation step in PCR typically occurs at:
(a) 50–65°C
(b) 94–98°C
(c) 70–72°C
(d) 37–42°C
Explanation: The denaturation step is conducted at 94–98°C to break hydrogen bonds between DNA strands, resulting in single-stranded templates for primer annealing. - Taq DNA polymerase is isolated from:
(a) Escherichia coli
(b) Bacillus subtilis
(c) Thermus aquaticus
(d) Saccharomyces cerevisiae
Explanation: Taq polymerase is derived from the thermophilic bacterium Thermus aquaticus, which thrives in hot springs. Its thermostability allows DNA synthesis at high temperatures during PCR. - The annealing temperature in PCR depends on:
(a) DNA template concentration
(b) Primer melting temperature (Tₘ)
(c) dNTP concentration
(d) Extension time
Explanation: Annealing temperature is typically set 3–5°C below the primer Tₘ to ensure specific binding to the target DNA sequence. - How many DNA copies are produced after 30 PCR cycles?
(a) 30
(b) 1,000
(c) 1,073,741,824 (2³⁰)
(d) 100 million
Explanation: PCR amplifies DNA exponentially. After 30 cycles, 2³⁰ ≈ 1.07 billion copies of the target DNA sequence are produced. - Reverse Transcriptase PCR (RT-PCR) amplifies:
(a) Genomic DNA
(b) Complementary DNA (cDNA) from RNA
(c) Proteins
(d) Plasmids
Explanation: RT-PCR uses reverse transcriptase to convert mRNA into cDNA, which is then amplified, enabling the study of gene expression. - Real-time PCR (qPCR) quantifies DNA using:
(a) UV spectrophotometry
(b) Fluorescent dyes (e.g., SYBR Green)
(c) Gel electrophoresis
(d) Autoradiography
Explanation: qPCR monitors DNA amplification in real-time using fluorescence. Fluorescent dyes like SYBR Green bind to double-stranded DNA and emit signals proportional to DNA concentration. - The function of dNTPs in PCR is to:
(a) Stabilize primers
(b) Provide nucleotides for DNA synthesis
(c) Denature DNA
(d) Enhance polymerase activity
Explanation: dNTPs (deoxynucleotide triphosphates) are the raw materials from which new DNA strands are synthesized by the DNA polymerase. - Hot-start PCR reduces:
(a) Cycle time
(b) Non-specific amplification
(c) Primer dimer formation
(d) Template degradation
Explanation: In hot-start PCR, enzyme activation occurs only at high temperatures, preventing unwanted amplification and primer dimers at room temperature setup. - Nested PCR improves:
(a) Amplification speed
(b) Specificity and sensitivity
(c) Template yield
(d) Primer efficiency
Explanation: Nested PCR uses two sets of primers in successive reactions. The second primer set targets a sequence within the first amplicon, reducing non-specific products. - Multiplex PCR allows amplification of:
(a) A single target
(b) Multiple targets simultaneously
(c) Whole genomes
(d) RNA only
Explanation: Multiplex PCR uses multiple primer pairs to simultaneously amplify different DNA targets within the same sample and reaction tube. - The elongation step in PCR occurs at:
(a) 94°C
(b) 55°C
(c) 72°C
(d) 37°C
Explanation: 72°C is the optimal working temperature for Taq polymerase to extend primers by adding nucleotides to synthesize new DNA strands. - Tₘ (melting temperature) of a primer is defined as the temperature at which:
(a) DNA polymerase denatures
(b) 50% of primer-template duplexes dissociate
(c) dNTPs degrade
(d) DNA synthesis initiates
Explanation: The melting temperature (Tₘ) indicates the temperature at which half of the primer-DNA duplexes dissociate, a critical factor in determining annealing temperature. - A critical component of PCR buffer is:
(a) SDS
(b) Mg²⁺
(c) EDTA
(d) NaOH
Explanation: Mg²⁺ acts as a cofactor for Taq polymerase and is essential for enzyme activity, primer annealing, and dNTP incorporation during PCR. - Touchdown PCR minimizes non-specific binding by:
(a) Increasing cycle number
(b) Gradually decreasing annealing temperature
(c) Shortening extension time
(d) Using lower denaturation temperatures
Explanation: Touchdown PCR starts with a high annealing temperature, decreasing it in later cycles to favor specific primer-template binding and reduce non-specific products. - In colony PCR, the template is:
(a) Purified genomic DNA
(b) Bacterial cells from a colony
(c) cDNA library
(d) Viral RNA
Explanation: Colony PCR directly uses bacterial colonies as a DNA source. Cells are lysed in the PCR reaction without prior DNA purification. - The “amplicon” refers to the:
(a) Primer
(b) DNA polymerase
(c) Amplified DNA product
(d) Template DNA
Explanation: An amplicon is the DNA fragment that is specifically amplified by PCR, defined by the primer pair used. - Quantitative PCR (qPCR) uses Cₜ values to determine:
(a) DNA size
(b) Initial template concentration
(c) Primer efficiency
(d) Polymerase fidelity
Explanation: The threshold cycle (Cₜ) reflects the cycle at which fluorescence surpasses a background level. A lower Cₜ indicates a higher initial amount of DNA. - PCR is used in forensics for:
(a) Blood typing
(b) DNA profiling (STR analysis)
(c) Fingerprint analysis
(d) Toxicology screening
Explanation: PCR amplifies short tandem repeats (STRs), which are highly variable DNA regions used in forensic DNA fingerprinting. - Inverse PCR amplifies:
(a) Known sequences
(b) Unknown flanking regions
(c) Circular DNA
(d) RNA templates
Explanation: Inverse PCR identifies unknown sequences flanking known regions by circularizing DNA fragments and using primers that face outward. - High-fidelity PCR uses polymerases like Pfu for:
(a) Faster amplification
(b) Proofreading (3’→5′ exonuclease activity)
(c) Hot-start capability
(d) Reverse transcription
Explanation: Pfu DNA polymerase possesses 3’→5′ exonuclease activity, enabling it to “proofread” and correct misincorporated nucleotides during DNA synthesis. This reduces the error rate significantly compared to Taq polymerase, making Pfu ideal for cloning and mutation-sensitive experiments. - Asymmetric PCR generates:
(a) Double-stranded DNA
(b) Single-stranded DNA
(c) RNA-DNA hybrids
(d) Circular DNA
Explanation: Asymmetric PCR uses unequal concentrations of forward and reverse primers, favoring amplification of one DNA strand over the other. This method is useful for producing single-stranded DNA (ssDNA) for applications like sequencing, hybridization probes, or aptamer selection. - The primary application of RT-qPCR is:
(a) DNA cloning
(b) Gene expression analysis
(c) Protein quantification
(d) Genome editing
Explanation: RT-qPCR (Reverse Transcription quantitative PCR) is mainly used to measure mRNA levels by converting RNA to complementary DNA (cDNA), which is then amplified and quantified in real-time. It is a key tool for studying gene expression in different tissues or experimental conditions. - Primer dimers form due to:
(a) High annealing temperatures
(b) Complementary 3′ ends of primers
(c) Excess template DNA
(d) Low Mg²⁺ concentration
Explanation: Primer dimers are short, unintended amplification products that arise when primers anneal to each other, particularly at complementary 3′ ends. These dimers can compete with target DNA during amplification, reducing PCR efficiency and specificity. - Digital PCR (dPCR) is superior for:
(a) High-throughput sequencing
(b) Absolute quantification of rare alleles
(c) Fast amplification
(d) Whole-genome amplification
Explanation: dPCR partitions the sample into thousands of micro-reactions, allowing precise quantification of nucleic acids, including low-frequency mutations and rare alleles. Unlike qPCR, it doesn’t rely on standard curves, enabling accurate and reproducible absolute quantification. - In PCR, if the extension time is too short, it results in:
(a) No amplification
(b) Incomplete amplification of long products
(c) Non-specific bands
(d) Primer degradation
Explanation: Each PCR cycle requires a specific extension time (commonly 1 minute per kilobase). If the extension time is insufficient, particularly for long target sequences, full-length DNA strands won’t be synthesized, resulting in truncated or incomplete amplicons. - The role of mineral oil in early PCR protocols was to:
(a) Enhance fluorescence
(b) Prevent evaporation
(c) Stabilize primers
(d) Denature DNA
Explanation: Before thermal cyclers with heated lids were developed, mineral oil was layered over the PCR reaction mix to prevent evaporation during high-temperature cycling. This maintained consistent reaction volume and concentration. - Allele-specific PCR detects:
(a) Gene expression
(b) Single-nucleotide polymorphisms (SNPs)
(c) Protein mutations
(d) Chromosomal deletions
Explanation: Allele-specific PCR uses primers that are designed to bind only if a specific nucleotide is present at a polymorphic site. This allows detection of SNPs and genotype differentiation based on whether amplification occurs. - The critical factor for successful multiplex PCR is:
(a) High denaturation temperature
(b) Compatible primer Tₘ and no cross-reactivity
(c) Low dNTP concentration
(d) Extended elongation time
Explanation: Multiplex PCR uses multiple primer pairs to amplify different targets in the same reaction. It requires all primers to have similar melting temperatures (Tₘ) and avoid cross-hybridization to ensure specific and balanced amplification. - PCR contamination is minimized using:
(a) High-fidelity polymerase
(b) Separate pre- and post-PCR areas
(c) Increased Mg²⁺ concentration
(d) Shorter cycles
Explanation: Preventing contamination is essential in PCR, especially due to the high sensitivity of the technique. Using dedicated equipment and separate workspaces for reagent preparation and post-PCR analysis, along with UV sterilization and aerosol-resistant tips, helps avoid contamination from previous amplicons.
Definition, methods and importance of mass selection, pureline selection, hybridization and hybrid vigour.
- Mass selection differs from pureline selection in that mass selection:
(a) Produces homozygous lines
(b) Retains genetic variability
(c) Involves self-pollination only
(d) Requires hybridization
Explanation: Mass selection involves selecting multiple phenotypically superior plants and bulking their seeds. This retains a broader gene pool, making it ideal for improving cross-pollinated crops. It contrasts with pureline selection, which narrows genetic diversity. - Pureline selection is most effective for:
(a) Cross-pollinated crops
(b) Self-pollinated crops
(c) Vegetatively propagated crops
(d) Apomictic crops
Explanation: Self-pollinated crops are naturally homozygous and uniform. Pureline selection helps isolate the best performing homozygous lines that breed true over generations, suitable for crops like rice and wheat. - Hybrid vigour (heterosis) is quantified as:
(a) Mean of parental lines
(b) F1 performance exceeding the better parent
(c) F2 segregation ratio
(d) Genetic variance in progeny
Explanation: Heterosis is observed when the F1 hybrid performs better than the superior parent. This is quantified by comparing F1 yield with the better parent:
Heterosis (%) = [(F1 – Better Parent) / Better Parent] × 100. - The “Heterosis Box” developed by the U.S. in 1942 improved:
(a) Rice
(b) Maize
(c) Sugarcane
(d) Cotton
Explanation: The “Heterosis Box” approach in maize helped systematically exploit hybrid vigour. It standardized crossing techniques and selection methods to enhance yield potential in maize. - Mass selection is advantageous for:
(a) Developing disease-resistant purelines
(b) Quickly improving landraces with high variability
(c) Creating hybrids
(d) Inducing mutations
Explanation: It is most effective in genetically diverse populations, where selecting and bulking seeds from best phenotypes improves overall population quality, such as in landraces of maize or pearl millet. - Johannsen’s experiments with beans demonstrated:
(a) Heterosis
(b) Pureline theory
(c) Hybridization
(d) Mass selection
Explanation: Johannsen’s work on beans established that within purelines, phenotypic variation is due to the environment, not genetics. Thus, genetic improvement needs new variation through hybridization or mutation. - Topcross hybridization involves crossing:
(a) Two inbred lines
(b) An inbred line with an open-pollinated variety
(c) Two hybrids
(d) Clones
Explanation: Topcrossing helps evaluate the general combining ability (GCA) of an inbred line by crossing it with an open-pollinated variety or tester. It is a key step in hybrid development. - Synthetic varieties are developed by crossing multiple inbred lines and allowing:
(a) Self-pollination
(b) Open-pollination
(c) Hybridization
(d) Clonal propagation
Explanation: Synthetics are created by intercrossing multiple elite inbred lines and then maintaining the population via open pollination. They retain some heterosis and adapt well under variable conditions. - Single-cross hybrids (e.g., maize) show higher heterosis than double-cross hybrids due to:
(a) Lower cost
(b) Greater genetic uniformity
(c) Easier production
(d) Reduced inbreeding depression
Explanation: Single-cross hybrids are derived from two inbred parents, leading to maximum heterozygosity and uniformity. They outperform double-cross hybrids in yield and performance consistency. - Pureline selection results in:
(a) Segregating progeny
(b) Genetically identical offspring
(c) Hybrid vigour
(d) Increased variability
Explanation: A pureline is derived from a single self-pollinated homozygous plant. All progeny are genetically identical and stable over generations, with no segregation. - Recurrent mass selection enhances:
(a) Homozygosity
(b) Frequency of desirable alleles
(c) Hybrid sterility
(d) Mutation rate
Explanation: Recurrent mass selection is a method where the best individuals are selected in each generation and allowed to intermate. This repeated selection gradually increases the frequency of favorable alleles in the population, especially effective in cross-pollinated crops. - The primary limitation of pureline selection is:
(a) High cost
(b) Zero genetic gain after isolation
(c) Hybrid breakdown
(d) Incompatibility
Explanation: Once a pureline is selected, it is genetically uniform and homozygous. Since there is no genetic variation left, further selection offers no genetic improvement, making it ineffective beyond the initial cycle. - Heterosis is maximally exploited in:
(a) Wheat
(b) Maize
(c) Barley
(d) Chickpea
Explanation: Maize is the most widely exploited crop for heterosis (hybrid vigour) due to its natural outcrossing nature. Single-cross hybrids in maize can yield 20–30% more than open-pollinated varieties. - In pedigree breeding, hybridization is followed by:
(a) Mass selection
(b) Selection with progeny testing
(c) Cloning
(d) Mutation
Explanation: Pedigree breeding involves recording the ancestry of each selected plant over successive generations after hybridization. Selection with progeny testing ensures the fixation of desirable traits in self-pollinated crops like wheat or rice. - The “Progeny Test” is critical in pureline selection to:
(a) Assess heterosis
(b) Confirm homozygosity
(c) Test disease resistance
(d) Induce polyploidy
Explanation: In pureline selection, progeny testing helps confirm that the selected plant is homozygous and stable. It ensures uniform performance and eliminates segregating lines. - Composite varieties are developed by:
(a) Hybridizing two inbred lines
(b) Mixing seeds of multiple genotypes
(c) Clonal propagation
(d) Pureline selection
Explanation: Composite varieties are genetically diverse populations developed by mixing seeds of several selected genotypes with similar phenotypes. They are useful in cross-pollinated crops where uniformity is not critical. - Heterosis is lost in F2 due to:
(a) Mutation
(b) Segregation and recombination
(c) Inbreeding depression
(d) Gene silencing
Explanation: The F1 hybrid is heterozygous and exhibits hybrid vigour. In F2, due to segregation and recombination of alleles, heterozygosity reduces, and so does the heterosis, leading to performance decline. - Mass selection is least effective for traits with:
(a) High heritability
(b) Low heritability
(c) Dominant inheritance
(d) Additive gene action
Explanation: Traits with low heritability are heavily influenced by the environment. In such cases, phenotypic selection (like mass selection) is unreliable, as appearance may not reflect the genotype. - The first hybrid rice variety developed in India was:
(a) IR-8
(b) APHR-1
(c) Pusa Basmati
(d) Swarna
Explanation: APHR-1 (Andhra Pradesh Hybrid Rice-1), released in 1994, was the first hybrid rice variety developed and cultivated in India. It marked a major advancement in hybrid technology for cereals. - Dominance hypothesis explains heterosis through:
(a) Overdominance
(b) Masking of deleterious recessive alleles
(c) Epistasis
(d) Chromosomal duplication
Explanation: The dominance hypothesis suggests that in F1 hybrids, dominant alleles mask the harmful effects of deleterious recessive alleles from both parents, resulting in better performance than either parent. - Pureline varieties are susceptible to new diseases because they lack:
(a) Hybrid vigour
(b) Genetic diversity
(c) Dominant genes
(d) Polyploidy
Explanation: Purelines are genetically uniform and homozygous. This uniformity means that if a pathogen evolves to infect one plant, it can affect all, due to the absence of genetic variability. - The “Three-Line System” for hybrid seed production involves:
(a) A, B, and R lines
(b) CMS, maintainer, and restorer lines
(c) Male, female, and hybrid lines
(d) Inbred, synthetic, and composite lines
Explanation: The three-line system uses a Cytoplasmic Male Sterile (A) line, a maintainer (B) line to multiply A, and a restorer (R) line to restore fertility in the F1. It ensures economical and controlled hybrid seed production. - Recurrent selection in mass breeding targets:
(a) Clonal propagation
(b) Cumulative gene frequency improvement
(c) Hybrid development
(d) Mutation induction
Explanation: Recurrent selection aims to improve populations over multiple generations by accumulating favorable alleles, especially for polygenic traits governed by additive gene action. - Overdominance hypothesis posits that heterosis results from:
(a) Dominant-recessive interactions
(b) Superiority of heterozygote over homozygote
(c) Epistatic effects
(d) Chromosomal aberrations
Explanation: The overdominance hypothesis suggests that the heterozygous genotype at certain loci is superior to both corresponding homozygotes. This is a possible explanation for observed heterosis. - Synthetic variety “COH-4” is used for:
(a) Wheat
(b) Sugarcane
(c) Rice
(d) Groundnut
Explanation: COH-4 is a synthetic variety of sugarcane developed by combining superior clones. It maintains heterosis through vegetative (clonal) propagation and shows improved yield and sugar content. - Mass selection is also termed:
(a) Clonal selection
(b) Phenotypic selection
(c) Progeny selection
(d) Hybrid selection
Explanation: Mass selection relies on visible phenotypic characters for selecting individuals, without involving progeny testing. Hence, it is synonymous with phenotypic selection. - Hybrid rice shows yield advantages of:
(a) 5–10%
(b) 20–30%
(c) 50–60%
(d) 70–80%
Explanation: Hybrid rice typically yields 20–30% more than conventional varieties, making it important for food security. This is due to heterosis and better resource-use efficiency. - Pureline selection contributed to the development of:
(a) Hybrid maize
(b) NP-4 wheat
(c) Bt cotton
(d) Golden Rice
Explanation: NP-4 is a pureline selection of wheat, derived by selecting superior individual plants and selfing for several generations, leading to a genetically uniform line. - The “Heterosis Breeding” concept was pioneered by:
(a) Norman Borlaug
(b) George Shull
(c) Gregor Mendel
(d) M.S. Swaminathan
Explanation: George Shull, in 1908, introduced the concept of heterosis in maize breeding. His work laid the foundation for modern hybrid crop development. - Hybrid vigour is commercially exploited in all EXCEPT:
(a) Maize
(b) Rice
(c) Sorghum
(d) Barley
Explanation: Hybrid vigour is limited in barley due to its predominantly self-pollinating nature and low heterosis potential. Hence, it is mainly improved through pureline selection rather than hybridization.
Tissue culture- Definition, Types and Prospects in Agriculture and Forestry.
- Tissue culture is defined as:
(a) Growing whole plants in soil
(b) Aseptic in vitro cultivation of cells/tissues
(c) Hydroponic plant growth
(d) Field-level crop propagation
Explanation: Tissue culture refers to growing plant cells, tissues, or organs in a sterile, nutrient-rich medium under controlled environmental conditions. It allows propagation independent of seeds or soil and is useful for clonal propagation, genetic modification, and conservation. - The “Father of Plant Tissue Culture” is:
(a) Gregor Mendel
(b) Gottlieb Haberlandt
(c) Norman Borlaug
(d) Barbara McClintock
Explanation: Gottlieb Haberlandt proposed the concept of isolating and culturing single plant cells in 1902. Though his initial attempts failed, his theoretical work laid the foundation for modern tissue culture techniques. - Which is a type of tissue culture based on explant?
(a) Suspension culture
(b) Meristem culture
(c) Batch culture
(d) Bioreactor culture
Explanation: Meristem culture involves using the apical or axillary meristems as explants. This method is especially valuable for producing virus-free plants because meristematic tissues are generally pathogen-free. - Synthetic seeds are produced by encapsulating:
(a) Zygotes
(b) Somatic embryos
(c) Pollen grains
(d) Seeds
Explanation: Somatic embryos, derived from somatic cells via embryogenesis, are encapsulated in sodium alginate to mimic seeds. These synthetic seeds can be stored, handled, and directly sown like natural seeds. - The primary advantage of micropropagation is:
(a) Low cost
(b) Rapid clonal multiplication
(c) Genetic variability
(d) Reduced labor
Explanation: Micropropagation enables the rapid and large-scale multiplication of genetically identical plants, especially for elite or endangered varieties. It’s commonly used in commercial propagation of crops like banana, sugarcane, and ornamental plants. - Somaclonal variation arises due to:
(a) Genetic engineering
(b) Epigenetic changes during in vitro culture
(c) Hybridization
(d) Polyploidy induction
Explanation: Somaclonal variation refers to genetic or epigenetic changes that occur during callus or suspension cultures. These can be due to DNA methylation, point mutations, or chromosomal changes and may lead to novel traits in regenerants. - Which medium component acts as a solidifying agent?
(a) Sucrose
(b) Agar
(c) Coconut water
(d) 2,4-D
Explanation: Agar, a polysaccharide from red algae, is used at 0.6–1.0% concentration to solidify the culture medium, providing physical support to plant tissues. - Haploid plants are produced through:
(a) Meristem culture
(b) Anther/pollen culture
(c) Protoplast fusion
(d) Embryo rescue
Explanation: Haploid plants are obtained through androgenesis by culturing anthers or isolated microspores. These are useful for producing homozygous lines rapidly via chromosome doubling. - The hormone used for inducing root formation in vitro is:
(a) Gibberellin
(b) Auxin (e.g., IBA)
(c) Cytokinin
(d) Ethylene
Explanation: Auxins like IBA (Indole-3-butyric acid) or NAA (Naphthaleneacetic acid) stimulate adventitious root formation, a critical step in plantlet regeneration from shoots. - Protoplast culture enables:
(a) Virus elimination
(b) Somatic hybridization
(c) Clonal propagation
(d) Germplasm storage
Explanation: Isolated protoplasts (cells without walls) can be fused to overcome sexual incompatibility, forming somatic hybrids (e.g., pomato = potato + tomato). - Cryopreservation is used for:
(a) Short-term seed storage
(b) Long-term germplasm conservation at -196°C
(c) Enhancing photosynthesis
(d) Soil fertility improvement
Explanation: Cryopreservation involves ultra-low temperature storage (in liquid nitrogen) of cells, tissues, or embryos, maintaining viability and genetic stability over decades. - The sterilization agent for explants is:
(a) Sucrose
(b) 70% ethanol or sodium hypochlorite
(c) Agar
(d) Gibberellic acid
Explanation: Sterilizing agents like ethanol or sodium hypochlorite remove surface microbes from explants, ensuring aseptic conditions in tissue culture. - “Embryo rescue” is applied to:
(a) Mature seeds
(b) Hybrid embryos from wide crosses
(c) Somatic embryos
(d) Haploid plants
Explanation: Embryo rescue nurtures hybrid embryos that might abort naturally, especially in interspecific crosses (e.g., wheat × rye) to create viable hybrids. - In forestry, tissue culture aids in:
(a) Deforestation
(b) Mass propagation of elite trees (e.g., teak, eucalyptus)
(c) Soil erosion
(d) Pest management
Explanation: Clonal propagation through tissue culture allows multiplication of superior forest trees with uniform traits such as faster growth and disease resistance. - MS (Murashige and Skoog) medium is rich in:
(a) Lipids
(b) Salts and micronutrients
(c) Proteins
(d) Carbohydrates
Explanation: MS medium, a standard plant tissue culture medium, contains high levels of nitrates, potassium, and essential micronutrients required for optimal plant growth. - The main application of tissue culture in agriculture is:
(a) Weed control
(b) Production of disease-free planting material
(c) Soil pH management
(d) Pollination
Explanation: Tissue culture, especially meristem culture, is vital for producing disease-free clones of crops like sugarcane, banana, and potato, ensuring higher yield. - Hairy root disease is induced by Agrobacterium rhizogenes for:
(a) Pathogen study
(b) Production of secondary metabolites
(c) Nitrogen fixation
(d) Mycorrhizal association
Explanation: A. rhizogenes inserts root-inducing genes, causing hairy root growth, which are stable and ideal for producing valuable secondary compounds (e.g., alkaloids). - The gas permeable membrane in tissue culture vessels allows exchange of:
(a) Nutrients
(b) CO₂ and O₂
(c) Light
(d) Hormones
Explanation: Proper gas exchange through semi-permeable lids ensures aerobic respiration and photosynthesis in plant tissues, critical for their growth and development. - “Germplasm conservation” via tissue culture preserves:
(a) Only seeds
(b) Genetic resources of endangered/elite species
(c) Soil microbes
(d) Fertilizers
Explanation: Tissue culture conserves rare, endangered, or valuable plant germplasm using cryopreservation or slow-growth techniques, ensuring biodiversity. - The hormone combination for organogenesis is:
(a) High auxin + low cytokinin
(b) Low auxin + high cytokinin
(c) Gibberellin only
(d) Ethylene + auxin
Explanation: Organogenesis depends on hormone ratios; high cytokinin with low auxin induces shoot formation, whereas high auxin favors root development. - “Somatic embryogenesis” refers to embryo formation from:
(a) Zygotes
(b) Vegetative cells
(c) Pollen
(d) Endosperm
Explanation: Somatic embryogenesis is the process by which non-reproductive (somatic) cells form embryos that can develop into whole plants, useful for cloning. - In banana cultivation, tissue culture prevents spread of:
(a) Aphids
(b) Fusarium wilt and bunchy top virus
(c) Nematodes
(d) Rust
Explanation: TC banana plants are raised from virus-free meristems, preventing the transmission of diseases like Fusarium wilt, which are common in traditional propagation. - The first commercialized tissue culture plant was:
(a) Rice
(b) Orchid (Cymbidium)
(c) Potato
(d) Tomato
Explanation: Georges Morel pioneered commercial micropropagation with Cymbidium orchids, marking the start of tissue culture in floriculture. - “Callus” is defined as:
(a) Differentiated root tissue
(b) Unorganized mass of undifferentiated cells
(c) Embryonic shoot
(d) Pollen sac
Explanation: A callus is a mass of totipotent, dedifferentiated plant cells formed due to wounding or auxin-cytokinin stimulation; it can later regenerate organs. - Bioreactors are used in tissue culture for:
(a) Sterilization
(b) Large-scale production of cells/secondary metabolites
(c) Germplasm storage
(d) Soil-less cultivation
Explanation: Bioreactors provide a controlled environment for mass culture of cells or roots, essential for commercial production of compounds like shikonin or taxol. - “Protoplast fusion” creates hybrids by bypassing:
(a) Sterilization
(b) Sexual incompatibility
(c) Hormone requirements
(d) Light cycles
Explanation: Protoplast fusion allows creation of somatic hybrids from distantly related species that cannot cross sexually (e.g., potato + tomato = pomato). - The critical step in acclimatization is:
(a) Increasing sucrose
(b) Gradual exposure to field conditions
(c) Adding hormones
(d) Reducing light
Explanation: Acclimatization involves hardening tissue-cultured plantlets by gradually adapting them to external humidity, temperature, and light to survive outside. - Tissue culture in forestry helps conserve:
(a) Invasive species
(b) Endangered species (e.g., redwood, neem)
(c) Annual crops
(d) Weeds
Explanation: Clonal propagation helps mass produce endangered forest species or reforest areas using elite genotypes with resistance to stress and diseases. - “Virus indexing” confirms:
(a) Hormone levels
(b) Absence of viruses in TC plants
(c) Nutrient deficiency
(d) Genetic uniformity
Explanation: Techniques like ELISA or PCR are used to test tissue-cultured plants for viral infections, ensuring only healthy, virus-free stock is distributed. - “Molecular farming” via tissue culture produces:
(a) Grains
(b) Pharmaceuticals (e.g., vaccines, antibodies)
(c) Timber
(d) Biofertilizers
Explanation: Plants are genetically engineered to produce high-value therapeutic proteins, vaccines, or enzymes—this biofactory approach is called molecular farming.
