NEET Biology — Cell Biology & Genetics 50 Important MCQs for 2027 with Answers

Key Fact: Cell Biology and Genetics together form the backbone of NEET Biology. These two areas contribute 35–40% of NEET Biology marks — typically 50–60 questions out of 90 Biology questions come from Cell Biology, Cell Division, Genetics, and Molecular Biology. Mastering these chapters is non-negotiable for scoring 340+ in NEET 2027.

Chapter Coverage & Expected Questions

Section A: 20 Solved MCQs on Cell Biology

Q1. Which organelle is called the “suicidal bag” of the cell?

A. Mitochondria   B. Lysosome   C. Ribosome   D. Golgi apparatus

Answer: B — Lysosome
Explanation: Lysosomes contain hydrolytic enzymes (proteases, lipases, nucleases) that can digest the cell’s own organelles during autophagy, earning the name “suicidal bag.”

Q2. Which of the following has the largest cell in the human body?

A. Neuron   B. Ovum   C. RBC   D. Muscle cell

Answer: A — Neuron (by length)
Explanation: Neurons can extend up to 1 metre in length. The human ovum is the largest cell by volume, but neurons are the longest cells.

Q3. Fluid mosaic model of plasma membrane was proposed by:

A. Watson and Crick   B. Singer and Nicolson   C. Davson and Danielli   D. Robertson

Answer: B — Singer and Nicolson (1972)
Explanation: Singer and Nicolson proposed the fluid mosaic model in 1972 — phospholipid bilayer is fluid, with proteins (peripheral and integral) embedded like tiles in a mosaic.

Q4. Which of the following is NOT a membrane-bound organelle?

A. Mitochondria   B. Chloroplast   C. Ribosome   D. Nucleus

Answer: C — Ribosome
Explanation: Ribosomes are non-membranous organelles made of rRNA and proteins. They are 70S in prokaryotes and 80S in eukaryotes.

Q5. The powerhouse of the cell is:

A. Nucleus   B. Chloroplast   C. Mitochondria   D. Golgi body

Answer: C — Mitochondria
Explanation: Mitochondria produce ATP via oxidative phosphorylation in the inner membrane (cristae). They have their own DNA (mtDNA) and ribosomes (70S).

Q6. Cristae are found in:

A. Chloroplast   B. Mitochondria   C. Nucleus   D. Lysosome

Answer: B — Mitochondria
Explanation: Cristae are infoldings of the inner mitochondrial membrane. They increase surface area for the electron transport chain (ETC) and ATP synthesis.

Q7. Which organelle is responsible for synthesis and secretion of proteins?

A. Ribosome alone   B. Rough ER + Golgi + Secretory vesicles   C. Smooth ER   D. Lysosome

Answer: B — Rough ER + Golgi + Secretory vesicles
Explanation: Secretory proteins are synthesised on ribosomes of rough ER, modified in Golgi apparatus, packaged in secretory vesicles, and released by exocytosis.

Q8. The cell wall of plants is primarily made of:

A. Chitin   B. Cellulose   C. Peptidoglycan   D. Lignin

Answer: B — Cellulose
Explanation: Plant cell walls are primarily made of cellulose microfibrils. Fungi have chitin; bacteria have peptidoglycan (murein). Secondary cell walls may contain lignin.

Q9. Stroma is found in:

A. Mitochondria   B. Nucleus   C. Chloroplast   D. Golgi body

Answer: C — Chloroplast
Explanation: Stroma is the fluid matrix of chloroplasts where the Calvin cycle (dark reactions) takes place. Thylakoids are membrane discs stacked in grana within the stroma.

Q10. Which phase of mitosis involves separation of sister chromatids?

A. Prophase   B. Metaphase   C. Anaphase   D. Telophase

Answer: C — Anaphase
Explanation: In anaphase, centromeres split and sister chromatids are pulled to opposite poles by spindle fibres shortening. Cell elongates; kinetochore microtubules shorten.

Q11. Meiosis results in:

A. 4 diploid cells   B. 2 haploid cells   C. 4 haploid cells   D. 2 diploid cells

Answer: C — 4 haploid cells
Explanation: Meiosis involves two divisions: Meiosis I (reductional) and Meiosis II (equational). The result is 4 genetically unique haploid cells (n).

Q12. Crossing over occurs during:

A. Zygotene of Meiosis I   B. Pachytene of Meiosis I   C. Diplotene   D. Diakinesis

Answer: B — Pachytene of Meiosis I
Explanation: Crossing over (exchange of segments between non-sister chromatids of homologous chromosomes) occurs at pachytene stage during prophase I of meiosis.

Q13. Which structure is visible as “X”-shaped during metaphase?

A. Chromatid   B. Chromosome   C. Chromatin   D. Nucleosome

Answer: B — Chromosome
Explanation: During metaphase, chromosomes are maximally condensed and visible as X-shaped structures (two sister chromatids joined at the centromere) aligned at the metaphase plate.

Q14. Golgi apparatus is also known as:

A. Golgiosome   B. Dictyosome   C. Chondriome   D. Microsome

Answer: B — Dictyosome
Explanation: In plant cells, the Golgi apparatus is called dictyosome. It is the packaging and distribution centre of the cell — modifies, sorts and ships proteins and lipids.

Q15. Which organelle contains hydrolytic enzymes at low pH?

A. Peroxisome   B. Lysosome   C. Vacuole   D. Mitochondria

Answer: B — Lysosome
Explanation: Lysosomes maintain an acidic pH (~4.5–5) inside, optimal for their hydrolytic enzymes. They digest macromolecules, worn-out organelles, and foreign particles.

Q16. In plant cells, the large central vacuole occupies up to what percentage of cell volume?

A. 10–20%   B. 30–50%   C. 60–90%   D. 5–10%

Answer: C — 60–90%
Explanation: The central vacuole in mature plant cells can occupy 60–90% of the cell volume. It stores water, ions, pigments (anthocyanins), and waste products; maintains turgor pressure.

Q17. The nucleosome consists of:

A. DNA wrapped around RNA   B. DNA wrapped around histone proteins   C. RNA wrapped around histone proteins   D. DNA and non-histone proteins only

Answer: B — DNA wrapped around histone proteins
Explanation: A nucleosome = 146 bp of DNA wound around an octamer of histone proteins (2 each of H2A, H2B, H3, H4). H1 is the linker histone that seals the DNA on the nucleosome.

Q18. Cell plate formation during cytokinesis occurs in:

A. Animal cells only   B. Plant cells only   C. Both plant and animal cells   D. Neither

Answer: B — Plant cells only
Explanation: Plant cells form a cell plate (phragmoplast) from vesicles of the Golgi apparatus that fuse at the equatorial plane. Animal cells divide by cleavage furrow (constriction).

Q19. Balbiani rings are found in:

A. Polytene chromosomes   B. Lampbrush chromosomes   C. Metaphase chromosomes   D. Sex chromosomes

Answer: A — Polytene chromosomes
Explanation: Balbiani rings (puffs) are areas of active gene transcription on polytene chromosomes of Drosophila salivary glands — sites of active RNA synthesis.

Q20. Which type of cell division is involved in growth and repair?

A. Meiosis   B. Mitosis   C. Amitosis   D. Binary fission

Answer: B — Mitosis
Explanation: Mitosis produces two genetically identical diploid daughter cells — used for growth, repair, and replacement of cells in somatic tissues. Meiosis is for gamete formation.

Section B: 20 Solved MCQs on Genetics

Q21. Mendel’s law of independent assortment applies when genes are:

A. On the same chromosome   B. On different chromosomes   C. Linked genes   D. Sex-linked genes

Answer: B — On different (non-homologous) chromosomes
Explanation: Independent assortment occurs when genes are on separate chromosomes (or very far apart). Genes on the same chromosome tend to be inherited together (linkage).

Q22. In a monohybrid cross (Tt × Tt), the expected phenotypic ratio is:

A. 1:1   B. 3:1   C. 1:2:1   D. 2:1

Answer: B — 3:1 (dominant: recessive)
Explanation: Tt × Tt gives TT : Tt : tt = 1:2:1 (genotypic ratio). Phenotypic ratio = 3 dominant (TT + Tt) : 1 recessive (tt) = 3:1.

Q23. Haemophilia is an example of:

A. Autosomal dominant disorder   B. X-linked recessive disorder   C. Autosomal recessive disorder   D. Y-linked disorder

Answer: B — X-linked recessive disorder
Explanation: Haemophilia A (Factor VIII deficiency) and B (Factor IX deficiency) are X-linked recessive. Males (XY) are affected if they carry one mutant allele; females need two copies.

Q24. In ABO blood group system, which genotype represents blood group AB?

A. I^A I^B   B. I^A i   C. I^B I^B   D. ii

Answer: A — I^A I^B
Explanation: Blood group AB has genotype I^A I^B — both alleles are co-dominant and expressed. Blood group O is ii (recessive homozygous). A is I^A I^A or I^A i; B is I^B I^B or I^B i.

Q25. Down syndrome results from:

A. Monosomy of chromosome 21   B. Trisomy of chromosome 21   C. Deletion of chromosome 21   D. Translocation of chromosome 13

Answer: B — Trisomy of chromosome 21
Explanation: Down syndrome (Trisomy 21) occurs due to non-disjunction during meiosis, resulting in 47 chromosomes (2n+1 = 47). Characterised by intellectual disability, simian crease, and cardiac defects.

Q26. Turner’s syndrome (45, X0) is characterised by:

A. Male with extra X chromosome   B. Female with only one X chromosome   C. Male with XXY   D. Female with XXX

Answer: B — Female with only one X chromosome (45, X0)
Explanation: Turner’s syndrome females have 45 chromosomes (one X missing). Features: short stature, webbed neck, shield chest, primary amenorrhoea, and infertility.

Q27. Colour blindness is more common in males because:

A. Males have stronger vision   B. The gene is on Y chromosome   C. Males have only one X chromosome — one copy of the gene is sufficient for expression   D. Females compensate with a second X

Answer: C
Explanation: Colour blindness gene is on the X chromosome (X-linked recessive). Males (XY) express the trait with just one recessive allele. Females (XX) need two copies to express it.

Q28. Which of the following represents incomplete dominance?

A. ABO blood groups   B. Flower colour in Antirrhinum (Snapdragon)   C. Haemophilia   D. Sickle cell anemia

Answer: B — Antirrhinum (snapdragon) flower colour
Explanation: In snapdragon: RR = red, rr = white, Rr = PINK (intermediate phenotype). This is incomplete dominance — the heterozygote shows a blended phenotype, unlike co-dominance.

Q29. Pleiotropy refers to:

A. One gene affecting multiple traits   B. Multiple genes affecting one trait   C. One gene, one enzyme   D. Two genes on same chromosome

Answer: A — One gene affecting multiple traits
Explanation: Pleiotropy = single gene controls multiple phenotypic traits. Example: Phenylketonuria (PKU) gene causes intellectual disability, fair complexion, and musty odour. Sickle cell gene causes anaemia and sickling.

Q30. The probability of a carrier female (X^H X^h) and normal male (X^H Y) having a haemophilic son is:

A. 25%   B. 50% of sons (25% overall)   C. 100% of sons   D. 0%

Answer: B — 50% of sons (25% of all children)
Explanation: Cross X^H X^h × X^H Y gives: X^H X^H (normal female), X^H X^h (carrier female), X^H Y (normal male), X^h Y (haemophilic male). So 1/4 overall = 25% or 50% of sons are affected.

Q31. Polygenic inheritance involves:

A. Multiple alleles at one locus   B. Multiple genes each contributing to one trait   C. One gene with multiple effects   D. Gene linkage

Answer: B — Multiple genes contributing to one trait
Explanation: Polygenic inheritance: skin colour, height, intelligence are controlled by multiple genes (2+ loci), each contributing additively. Results in continuous variation (bell curve distribution).

Q32. Linkage was discovered by:

A. Gregor Mendel   B. Thomas Hunt Morgan   C. Hugo de Vries   D. Watson and Crick

Answer: B — Thomas Hunt Morgan
Explanation: Morgan discovered linkage while working with Drosophila melanogaster. He found that genes on the same chromosome tend to be inherited together (violating independent assortment).

Q33. In a test cross, the unknown genotype is crossed with:

A. A dominant homozygous individual   B. A recessive homozygous individual   C. A heterozygous individual   D. Itself (self-pollination)

Answer: B — A recessive homozygous individual (aa)
Explanation: Test cross = cross with homozygous recessive. If offspring show both dominant and recessive phenotypes (1:1), the unknown is heterozygous. If all dominant, the unknown is homozygous dominant.

Q34. Klinefelter’s syndrome has the karyotype:

A. 45, X0   B. 47, XXY   C. 47, XXX   D. 47, XYY

Answer: B — 47, XXY
Explanation: Klinefelter’s syndrome: males with extra X chromosome (XXY = 47 chromosomes). Features: gynaecomastia, small testes, infertility, tall stature, learning difficulties.

Q35. When a dihybrid cross (AaBb × AaBb) is performed, the expected phenotypic ratio is:

A. 9:3:3:1   B. 1:1:1:1   C. 3:1   D. 1:2:1

Answer: A — 9:3:3:1
Explanation: Dihybrid cross AaBb × AaBb gives 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb phenotypic ratio. This assumes genes are on different chromosomes (independent assortment) and complete dominance.

Q36. Epistasis is when:

A. Two genes are linked   B. One gene masks the expression of another non-allelic gene   C. Multiple alleles exist at one locus   D. Gene mutation occurs

Answer: B — One gene masks another non-allelic gene
Explanation: Epistasis = gene interaction where one gene (epistatic gene) suppresses the expression of another gene (hypostatic gene). Example: Labrador coat colour — 9:3:4 modified ratio due to recessive epistasis.

Q37. The law of segregation states that:

A. Genes assort independently during gamete formation   B. Alleles separate during gamete formation so each gamete gets one allele   C. Dominant allele always masks recessive   D. Genes are located on chromosomes

Answer: B
Explanation: Mendel’s First Law (Law of Segregation): The two alleles of a gene separate during gamete formation, so each gamete carries only ONE allele of each gene.

Q38. Multiple allelism is exhibited by:

A. Seed colour in pea   B. ABO blood group system   C. Haemophilia   D. Height in humans

Answer: B — ABO blood group system
Explanation: ABO blood groups show multiple allelism — three alleles (I^A, I^B, i) exist in the population at the same locus. Any individual can have only two of these three alleles.

Q39. Sex determination in grasshopper (Locusta) is of type:

A. XX-XY type   B. XX-X0 type   C. ZW-ZZ type   D. Haplo-diploidy

Answer: B — XX-X0 type
Explanation: In grasshoppers, females are XX and males are X0 (have only one sex chromosome). Drosophila and humans are XX-XY. Birds are ZZ (male)-ZW (female). Honeybees show haplo-diploidy.

Q40. Maternal effect genes are those whose:

A. Expression is influenced by the father’s genotype   B. Effect is determined by the mother’s genotype, not the offspring’s   C. Genes are on mitochondrial DNA   D. Expression is sex-limited

Answer: B
Explanation: Maternal effect genes: the phenotype of the offspring is determined by the mother’s genotype (e.g., coiling direction in snail Limnaea). The mother’s mRNA and proteins stored in the egg control early development.

Section C: 10 High-Difficulty MCQs on Molecular Biology

Q41. In the lac operon, the operator is bound by:

A. RNA polymerase   B. Repressor protein   C. Inducer   D. CAP protein

Answer: B — Repressor protein
Explanation: The repressor (encoded by the i gene) binds to the operator when lactose is absent, preventing RNA polymerase from transcribing the structural genes (lacZ, lacY, lacA).

Q42. Which restriction enzyme produces sticky ends?

A. EcoRI   B. SmaI   C. HaeIII   D. AluI

Answer: A — EcoRI
Explanation: EcoRI cuts at G↓AATTC, leaving 5′ AATT overhangs (sticky/cohesive ends). SmaI, HaeIII, and AluI are blunt-end cutters. Sticky ends allow directional cloning of DNA fragments.

Q43. PCR (Polymerase Chain Reaction) requires all EXCEPT:

A. Taq polymerase   B. Primers   C. dNTPs   D. Ribosomes

Answer: D — Ribosomes
Explanation: PCR requires: template DNA, primers (forward + reverse), Taq polymerase (heat-stable), dNTPs (dATP, dTTP, dGTP, dCTP), MgCl2, and buffer. Ribosomes are for translation — not DNA amplification.

Q44. In gel electrophoresis, DNA migrates towards:

A. Negative electrode (cathode)   B. Positive electrode (anode)   C. Does not migrate   D. Both poles equally

Answer: B — Positive electrode (anode)
Explanation: DNA is negatively charged (due to phosphate groups). In gel electrophoresis, DNA migrates towards the positive electrode (anode). Smaller fragments travel farther in the gel.

Q45. The Human Genome Project revealed that the human genome has approximately how many protein-coding genes?

A. 30,000–35,000   B. 20,000–25,000   C. 100,000   D. 3 billion

Answer: B — 20,000–25,000 genes
Explanation: HGP (completed 2003) revealed ~20,000–25,000 protein-coding genes in the 3 billion base pair human genome. Over 98% of the genome is non-coding. There are ~1.4 million SNPs identified.

Q46. The Meselson-Stahl experiment proved that DNA replication is:

A. Conservative   B. Dispersive   C. Semi-conservative   D. Non-conservative

Answer: C — Semi-conservative
Explanation: Using N-14/N-15 isotope labelling in E. coli, Meselson and Stahl (1958) proved semi-conservative replication — each daughter DNA has one original strand and one new strand.

Q47. Which enzyme removes RNA primers during DNA replication in prokaryotes?

A. DNA Polymerase III   B. DNA Polymerase I   C. Primase   D. Helicase

Answer: B — DNA Polymerase I
Explanation: DNA Pol I has 5’→3′ exonuclease activity that removes RNA primers and fills the gaps with DNA. DNA Pol III is the main replication enzyme; Primase synthesises the RNA primers.

Q48. The “wobble hypothesis” (Crick, 1966) relates to:

A. Base pairing between DNA strands   B. Flexibility in codon-anticodon pairing at the third position   C. Protein folding   D. Mutation in genetic code

Answer: B — Flexibility at the third codon position
Explanation: The wobble hypothesis explains why fewer than 61 tRNAs are needed despite 61 sense codons. The third base of the codon can pair with non-Watson-Crick bases in the anticodon (wobble position).

Q49. In gene expression, splicing of pre-mRNA is carried out by:

A. Ribosome   B. Spliceosome   C. RNA polymerase II   D. Proteasome

Answer: B — Spliceosome
Explanation: The spliceosome (composed of snRNPs — small nuclear ribonucleoproteins) removes introns and joins exons during pre-mRNA processing in eukaryotes. This occurs in the nucleus before mRNA export.

Q50. The termination codon NOT coding for any amino acid is:

A. AUG   B. UAG   C. GUG   D. AAA

Answer: B — UAG (also UAA and UGA are stop codons)
Explanation: Three stop (termination) codons: UAA (ochre), UAG (amber), UGA (opal/umber) — none code for amino acids. AUG codes for methionine (start codon). Release factors recognise stop codons to terminate translation.

Key Diagrams to Memorise for NEET

• Cell organelle functions: Know each organelle, its membrane type, and function — especially mitochondria (double membrane, cristae), chloroplast (double membrane, grana, stroma), and nucleus (nuclear pore complex)
• Mitosis phases: IPMAT — Interphase, Prophase (chromatin condenses), Metaphase (chromosomes at equatorial plate), Anaphase (sister chromatids separate), Telophase (nuclear envelope reforms)
• Meiosis phases: Prophase I sub-stages — Leptotene, Zygotene (synapsis), Pachytene (crossing over), Diplotene (chiasmata visible), Diakinesis (terminalization)
• DNA double helix: Watson-Crick model — B-DNA, right-handed helix, 3.4 nm per turn, 10 base pairs per turn, 0.34 nm per base pair, antiparallel strands, complementary base pairing (A-T, G-C)

Revision Tips for Cell Biology and Genetics

• Make a table of all chromosomal disorders (Down, Turner, Klinefelter, Patau, Edwards) with chromosome number, karyotype, and key features
• Practice Punnett squares daily — dihybrid, trihybrid, sex-linked crosses. Time yourself to solve each in under 2 minutes
• Molecular biology flowchart: DNA → transcription → pre-mRNA → splicing → mRNA → translation → protein. Know every enzyme at each step
• NCERT diagrams: The lac operon diagram, the nucleosome, the replication fork — draw these from memory at least 3 times
• PYQ focus: Genetics and Molecular Biology had 20+ questions in NEET 2023 and 2024 combined. Solve all PYQs topic-by-topic

Frequently Asked Questions

Practice Quiz — NEET Cell Biology & Genetics 2027

Test yourself with these 10 NEET-standard MCQs covering Cell Biology (Q1–5) and Genetics (Q6–10):

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