60 free MCQs on Principles of Inheritance and Variation with worked answers and explanations. Mendel's laws, inheritance patterns, DNA structure, and molecular biology. Core of NEET biology.
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Punnett square for a monohybrid cross between two heterozygous (Tt) parents, showing the 3:1 phenotypic ratio predicted by Mendel's Law of Segregation.
Q1.
DNA stands for:
Answer: B. Deoxyribonucleic acid
Why: DNA = Deoxyribonucleic acid. It is the genetic material in all cellular organisms.
Q2.
The father of genetics is:
Answer: B. Gregor Mendel
Why: Gregor Mendel (1822-1884) discovered the basic laws of inheritance through pea plant experiments. He is the father of genetics.
Q3.
In DNA, adenine pairs with:
Answer: C. Thymine
Why: In DNA: A pairs with T (2 hydrogen bonds); G pairs with C (3 hydrogen bonds). In RNA: A pairs with U.
Q4.
The monomer of DNA is called:
Answer: B. Nucleotide
Why: A nucleotide is the monomer of DNA/RNA. It consists of a phosphate group + deoxyribose sugar + nitrogenous base.
Q5.
A cross between a homozygous dominant (AA) and homozygous recessive (aa) gives:
Answer: B. All Aa
Why: AA x aa → 100% Aa (all heterozygous). All offspring show the dominant phenotype.
Q6.
In a monohybrid cross between two Aa parents, ratio of phenotypes in F2:
Answer: A. 3:1 dominant:recessive
Why: Aa x Aa → 1 AA : 2 Aa : 1 aa. Phenotype: 3 dominant (AA + 2 Aa) : 1 recessive (aa). Classic 3:1 ratio.
Q7.
The central dogma of molecular biology is:
Answer: B. DNA → RNA → Protein
Why: Central dogma: DNA is transcribed to mRNA, which is translated to protein. Information flows from nucleic acid to protein.
Q8.
mRNA carries genetic information from:
Answer: B. DNA (nucleus) to ribosome
Why: mRNA (messenger RNA) carries the genetic code from DNA in the nucleus to ribosomes in the cytoplasm for protein synthesis.
Q9.
Haemophilia is an example of:
Answer: C. X-linked recessive trait
Why: Haemophilia is X-linked recessive. Males (XY) who inherit one affected X have the disease; females need two affected X chromosomes.
Q10.
The genetic makeup of an organism is its:
Answer: B. Genotype
Why: Genotype is the genetic composition of an organism (actual allele combination). Phenotype is the observable trait.
Q11.
If both parents are carriers (Aa) of a recessive disease, probability of affected child is:
Answer: B. 25%
Why: Aa x Aa → 25% aa. There is a 25% (1 in 4) chance of the child being homozygous recessive and showing the disease.
Q12.
RNA differs from DNA in having:
Answer: C. Uracil instead of thymine; ribose sugar
Why: RNA: ribose sugar (not deoxyribose); uracil (not thymine); usually single-stranded.
Q13.
Sickle cell anaemia is caused by:
Answer: B. Single base mutation in haemoglobin gene (point mutation)
Why: Sickle cell anaemia: single point mutation (GAG→GTG in DNA; glutamic acid→valine in protein) causes abnormal haemoglobin.
Q14.
Which blood type is the universal donor?
Answer: D. O
Why: Blood type O negative has no A or B antigens. It can be donated to anyone (universal donor). AB is the universal recipient.
Q15.
ABO blood groups are an example of:
Answer: C. Co-dominance (A and B) with recessive O
Why: Blood type A and B are codominant (both expressed in AB). O is recessive. Blood type is controlled by multiple alleles (IA, IB, i).
Q16.
The structure of DNA was described by:
Answer: B. Watson and Crick (1953)
Why: Watson and Crick proposed the double helix structure of DNA in 1953, using X-ray data from Rosalind Franklin.
Q17.
A mutation in a body cell (somatic mutation) can cause:
Answer: B. Cancer, but cannot be passed to offspring
Why: Somatic mutations affect only body cells: they can cause cancer but are not passed to offspring (only germline mutations are heritable).
Q18.
Chromosomes are made of:
Answer: C. DNA and proteins (mainly histones)
Why: Chromosomes are made of DNA tightly wound around histone proteins. This DNA-protein complex is called chromatin.
Q19.
Transcription is the process of making:
Answer: B. mRNA from a DNA template
Why: Transcription: RNA polymerase reads the DNA template strand and synthesizes a complementary mRNA strand.
Q20.
Codon is:
Answer: B. Three bases on mRNA that code for one amino acid
Why: A codon is a triplet of bases on mRNA that codes for a specific amino acid (or start/stop signal).
Q21.
In incomplete dominance, F1 of RR x WW red and white snapdragons is:
Answer: C. Pink (intermediate phenotype)
Why: Incomplete dominance: neither allele is completely dominant. RW heterozygotes show an intermediate phenotype (pink in snapdragons).
Q22.
The law of independent assortment states that:
Answer: B. Genes on different chromosomes sort independently into gametes
Why: Independent Assortment (Mendel): alleles of different genes located on different chromosomes segregate independently during gamete formation.
Q23.
Which scientist used X-ray crystallography data to help discover DNA structure?
Answer: B. Rosalind Franklin
Why: Rosalind Franklin produced X-ray diffraction images (Photo 51) of DNA that revealed its helical structure, used by Watson and Crick.
Q24.
Chargaff's rules state that in DNA:
Answer: B. A = T and G = C (complementary base pairing)
Why: Chargaff's rules: A = T and G = C (in quantity). This holds for double-stranded DNA of any species.
Q25.
Point mutation that changes one amino acid to another is called:
Answer: C. Missense mutation
Why: Missense mutation: one nucleotide change causes one different amino acid to be incorporated. May affect protein function (e.g., sickle cell anaemia).
Q26.
Okazaki fragments are produced during:
Answer: B. Lagging strand DNA synthesis
Why: Okazaki fragments: short DNA fragments on the lagging strand template. DNA polymerase can only synthesize in 5' to 3' direction, so the lagging strand is synthesized discontinuously.
Q27.
A test cross involves crossing an individual with unknown genotype with:
Answer: C. A known homozygous recessive (aabb...)
Why: Test cross: unknown genotype x homozygous recessive. If offspring ratio is 1:1, the unknown is heterozygous; if all dominant, it is homozygous dominant.
Q28.
Sex-limited traits appear:
Answer: C. In both sexes but only fully expressed in one sex
Why: Sex-limited traits are encoded by autosomal genes but expressed only in one sex (e.g., milk yield in cows, beard in men) due to hormonal differences.
Q29.
Genetic mapping uses recombination frequencies to:
Answer: B. Estimate relative distances between genes on a chromosome
Why: Genetic (linkage) maps plot relative positions of genes based on recombination frequency. 1 map unit (cM) = 1% recombination frequency.
Q30.
The Hershey-Chase experiment proved that:
Answer: B. DNA, not protein, is the genetic material
Why: Hershey and Chase (1952) used radioactive S (protein) and P (DNA) in bacteriophages. Only P (DNA) entered bacteria, proving DNA is the genetic material.
Q31.
Epigenetics refers to:
Answer: B. Heritable changes in gene expression without DNA sequence change (e.g., methylation, acetylation)
Why: Epigenetics: heritable changes in gene expression not involving DNA sequence changes. Includes DNA methylation, histone modification.
Q32.
Repetitive DNA sequences that can move within the genome are called:
Answer: B. Transposons (jumping genes)
Why: Transposons (transposable elements/jumping genes) discovered by Barbara McClintock can move from one location to another in the genome.
Q33.
In which condition does a female carrier transmit a disease to half of her sons?
Answer: C. X-linked recessive
Why: X-linked recessive: carrier mother (XAXa) transmits the affected X to half her sons (XaY). Sons have only one X, so they show the disease.
Q34.
RNA polymerase reads the template strand in which direction?
Answer: B. 3' to 5' (template read), synthesizing mRNA 5' to 3'
Why: RNA polymerase reads DNA template 3' to 5' and synthesizes RNA in 5' to 3' direction (complementary to template).
Q35.
Post-transcriptional modification of pre-mRNA includes:
Answer: B. Capping (5 prime), poly-A tail (3 prime), and splicing out introns
Why: Pre-mRNA processing: 5-prime methyl guanosine cap + poly-A tail (added at 3-prime end) + splicing out introns by spliceosomes.
Q36.
The mutation that causes cystic fibrosis is most commonly:
Answer: B. Deletion of 3 nucleotides in CFTR gene (phenylalanine deleted at position 508)
Why: Most common CF mutation: DeltaF508 -- deletion of 3 nucleotides removes phenylalanine-508 from CFTR protein, causing misfolding and degradation.
Q37.
Which enzyme unwinds the DNA double helix during replication?
Answer: C. Helicase
Why: Helicase breaks hydrogen bonds between base pairs, unwinding the double helix and creating a replication fork.
Q38.
Lac operon in E. coli is an example of:
Answer: B. Negative regulation (lac repressor) with positive control (CAP-cAMP)
Why: Lac operon: repressed by lac repressor (no lactose). When lactose present, repressor removed (negative control). Optimally expressed when glucose absent (positive CAP control).
Q39.
Klinefelter syndrome (47,XXY) results in:
Answer: B. Male with extra X: infertile, some female features
Why: Klinefelter syndrome: XXY males. Usually taller, infertile (azoospermia), may have gynecomastia. Caused by nondisjunction.
Q40.
Genetic drift has the greatest effect in:
Answer: B. Small isolated populations
Why: Genetic drift: random changes in allele frequencies. Has the greatest effect in small populations where chance can significantly shift allele frequencies.
Q41.
Imprinting in genetics refers to:
Answer: B. Parent-of-origin-specific gene expression (some genes expressed only from maternal or paternal allele)
Why: Genomic imprinting: epigenetic silencing of one allele based on parental origin. E.g., IGF2 expressed from paternal allele only; H19 from maternal.
Q42.
Prader-Willi syndrome is caused by:
Answer: B. Deletion of paternal chromosome 15q11-q13 (or maternal UPD15)
Why: Prader-Willi: loss of paternal 15q11-q13 (imprinted region). If maternal copies of same region are lost, different disease (Angelman syndrome) results.
Q43.
The locus control region (LCR) regulates:
Answer: B. Long-range transcriptional regulation of gene clusters (e.g., globin gene cluster)
Why: LCR: regulatory element that controls chromatin accessibility and transcription of entire gene clusters over long distances (10s of kb). e.g., beta-globin LCR.
Q44.
ENCODE project revealed that the human genome is:
Answer: C. Largely transcribed with most non-coding regions having biochemical function
Why: ENCODE: found ~80% of genome has biochemical activity (transcription, protein binding, chromatin modification). Most non-coding DNA has regulatory or structural roles.
Q45.
Alternative splicing allows one gene to produce:
Answer: B. Multiple different proteins from the same pre-mRNA by including/excluding different exons
Why: Alternative splicing: different combinations of exons are joined to create multiple mRNA isoforms from one gene. Greatly expands protein diversity (~95% of human genes undergo it).
Q46.
The SOS response in E. coli is triggered by:
Answer: B. DNA damage (RecA coprotease activity)
Why: SOS response: DNA damage causes RecA to become activated as a coprotease. It cleaves LexA repressor, inducing DNA repair genes including RecA, UvrA, and error-prone polymerases.
Q47.
Heterochromatin position effect (position effect variegation) occurs when:
Answer: B. A gene is moved near heterochromatin and silenced in some cells but not others (mosaic phenotype)
Why: Position effect variegation (PEV): gene relocated near heterochromatin by rearrangement. Heterochromatin spreading silences gene variably, producing mosaic phenotype (e.g., eye color mosaicism in Drosophila white gene).
Q48.
Uniparental disomy (UPD) means:
Answer: B. Both copies of a chromosome pair from the same parent
Why: UPD: both copies of a chromosome pair inherited from one parent. Can unmask recessive mutations or disrupt imprinted genes (e.g., maternal UPD15 causes Prader-Willi syndrome).
Q49.
The chi-square test in genetics is used to:
Answer: B. Test whether observed phenotype ratios fit expected Mendelian ratios
Why: Chi-square test: statistical test for goodness-of-fit. Compares observed vs. expected offspring ratios to test if deviations are due to chance or real genetic differences.
Q50.
Trinucleotide repeat expansion diseases include:
Answer: B. Huntington disease, myotonic dystrophy, fragile X syndrome (repeat tracts expand each generation)
Why: Trinucleotide repeat expansion diseases: unstable DNA repeats expand in meiosis. Show anticipation (worsening each generation). Examples: HD (CAG), FXS (CGG), DM1 (CTG).
Q51.
X-inactivation in female mammals is:
Answer: B. Random inactivation of one X in each somatic cell (forming Barr body), creating cellular mosaicism
Why: Lyon hypothesis: one X is randomly inactivated per female somatic cell, forming a Barr body (inactive X condensed). Each cell expresses either maternal or paternal X alleles.
Q52.
XIST RNA is involved in:
Answer: B. Coating the inactive X chromosome and recruiting silencing factors (X-inactivation)
Why: XIST (X-inactive specific transcript): long non-coding RNA transcribed from the X inactivation center. Coats the X chromosome in cis, recruiting polycomb and other silencing factors.
Q53.
Anticipation in genetic diseases is caused by:
Answer: B. Trinucleotide repeat expansion through generations causing more severe disease at earlier onset
Why: Anticipation: disease onset earlier and more severe in successive generations. Caused by unstable trinucleotide repeat expansions (longer repeats = more severe). E.g., Huntington disease, myotonic dystrophy.
Q54.
In population genetics, the Hardy-Weinberg equilibrium requires:
Answer: B. No selection, no mutation, no migration, random mating, large population size (all 5 conditions)
Why: Hardy-Weinberg equilibrium (p^2 + 2pq + q^2 = 1): allele frequencies stay constant if no selection, no mutation, no migration, random mating, and large population.
Q55.
Quantitative trait loci (QTL) analysis identifies:
Answer: B. Chromosomal regions containing multiple genes that together contribute to complex quantitative traits
Why: QTL analysis: statistical method to map chromosomal regions containing genes that contribute to polygenic/complex traits (height, yield, disease risk).
Q56.
The selfish DNA theory proposes:
Answer: B. Repetitive and transposable DNA elements persist because they replicate themselves, not because they benefit the host
Why: Selfish DNA: transposons and other repetitive elements persist because they efficiently replicate and spread within genomes, even if they confer no benefit (or are slightly harmful) to the host.
Q57.
Epistasis occurs when:
Answer: B. One gene masks or modifies the expression of another gene
Why: Epistasis: one gene (epistatic) suppresses or modifies the phenotype of another gene (hypostatic). Causes modified ratios (e.g., 9:3:4, 9:7, 12:3:1 instead of 9:3:3:1).
Q58.
The founder effect is:
Answer: A. Loss of alleles when a small group colonizes a new area (low genetic diversity from few founders)
Why: Founder effect: a new population established by a small number of individuals has reduced genetic diversity. Certain allele frequencies can be very different from the source population.
Q59.
CpG islands in the genome are:
Answer: B. Regions with high C-G dinucleotide frequency, usually unmethylated near gene promoters; hypermethylation silences genes in cancer
Why: CpG islands: regions rich in CpG dinucleotides at gene promoters. Normally unmethylated (active genes). Hypermethylation silences tumor suppressor genes in cancer.
Q60.
Two-hit hypothesis (Knudson) for tumor suppressors means:
Answer: B. Both alleles of a tumor suppressor gene must be inactivated for cancer to occur (first hit germline, second somatic in familial cases)
Why: Knudson two-hit model: cancer requires loss of both alleles of a tumor suppressor (e.g., Rb). In familial retinoblastoma, one mutation is inherited (first hit); one somatic (second hit).