Skip to content

Immunology & Molecular Biology

Latest questions in this category.

25 questions

For pre-implantation genetic diagnosis (PGD) which of the following does not take place?

PGD requires IVF, biopsy of 1-2 trophectoderm cells or single blastomere, not all embryonic cells, preserving viability. FISH detects unbalanced translocations, PCR identifies sex-linked disease via sex determination or direct mutation testing. Testing every cell would destroy embryo, therefore not performed.

Ref: Strachan and Read, Human Molecular Genetics, 5th Edition, Chapter 20 Genetic Testing and Screening, Section Preimplantation Genetic Diagnosis, describes IVF, FISH for aneuploidy and limited biopsy not involving all embryonic cells, Garland Science publisher.

Which of the methods can be used for gene therapy for cancer? (A) Inhibition of fusion protein (B) Stimulation of immune

Cancer gene therapy targets oncogenic fusion proteins like BCR-ABL via antisense, enhances anti-tumor immunity via CAR-T and cytokines, and silences oncogenes via siRNA RNAi. Overexpressing angiogenic factors would promote vascularization and tumor growth, opposite of therapeutic strategy, thus excluded.

Ref: Weinberg, The Biology of Cancer, 2nd Edition, Chapter 16 Applying Knowledge in Clinic, Section Gene Therapy Approaches, covers inhibition of fusion proteins, immune stimulation and RNA interference versus angiogenesis inhibition for cancer therapy, Garland Science.

Given below are two statements : Statement I : In lyonization of X chromosome inactivation, all the genes of one X chrom

Lyonization inactivates most, not all, genes on one X; ~15% escape inactivation, retaining biallelic expression. In 46,XX testicular DSD, ~90% have SRY translocated to Xp during paternal meiosis, explaining male phenotype despite XX karyotype, making statement II true and I false.

Ref: Thompson and Thompson, Genetics in Medicine, 8th Edition, Chapter 6 X-Chromosome Inactivation and Chapter 8 Disorders of Sexual Development, discusses Lyonization escape genes and SRY translocation to X in 46,XX males, Elsevier Saunders publisher.

Which of the chemicals are used in the preparation of metaphase chromosomes for analysis of light microscopy? (A) Phytoh

Phytohemagglutinin stimulates T-lymphocyte mitosis in blood cultures. Colchicine arrests cells at metaphase by depolymerizing microtubules. Hypotonic saline swells cells for chromosome spreading. Quinacrine is fluorochrome for Q-banding, not metaphase preparation per se, hence A, C, D combination is used.

Ref: Gardner, Sutherland and Shaffer, Chromosome Abnormalities and Genetic Counseling, 4th Edition, Chapter Laboratory Methods, describes phytohemagglutinin mitogen, colchicine metaphase arrest and hypotonic treatment for metaphase chromosome preparation, Oxford University Press.

Which of the following statements are correct for sickle cell disease? (A) Hbs differs from normal HbA by a single amino

HbS results from β6 Glu→Val substitution causing deoxy-HbS polymerization, not membrane binding. Polymerized fibers distort RBCs into sickles, precipitating vaso-occlusive pain, splenic infarctions causing autosplenectomy with severe infection risk, and thrombotic crises, so A and D are true.

Ref: Lehninger, Nelson and Cox, Principles of Biochemistry, 8th Edition, Chapter 5 Protein Structure and Hemoglobin, explains sickle cell disease as single amino acid Glu to Val substitution causing deoxy HbS polymerization and vaso-occlusion, W.H. Freeman publisher.

Which of the following options are correct for immunology? (A) the immunoglobulin molecule is made up of six polypeptide

Immunoglobulin comprises four chains, not six. Kappa locus contains L, V, J, C regions rearranged somatically. Organ sharing correlates with MHC/HLA haplotypes, not complement. Both immunoglobulin and TCR generate diversity via V(D)J recombination, junctional flexibility and combinatorial association, making B and D accurate.

Ref: Janis Kuby, Immunology, 8th Edition, Chapter 5 Structure of Immunoglobulin and Chapter 7 TCR Diversity, describes four-chain antibody, kappa locus organization with four regions and V(D)J diversity shared with TCR, W.H. Freeman publisher.

Which of the following is INCORRECT for mitochondrial disorders?

Mitochondrial disorders include mtDNA-encoded maternal inheritance and nuclear-encoded mitochondrial proteins following Mendelian patterns, so not all show matrilinear transmission. Mitochondrial genome encodes

Ref: Emery and Rimoin, Principles and Practice of Medical Genetics, 6th Edition, Chapter on Mitochondrial Disorders, details maternal and nuclear inheritance, small coding capacity of mtDNA, Leigh disease and retinopathy diabetes phenotypes, Academic Press publication.

Which of the following statement is correct for Heterozygote advantage?

Balancing selection maintains HbS allele because heterozygotes resist Plasmodium falciparum malaria while homozygotes suffer sickle disease. This heterozygote advantage explains high allele frequency in malaria-endemic regions, distorting Hardy-Weinberg equilibrium, unrelated to founder effect.

Ref: Hartl and Clark, Principles of Population Genetics, 4th Edition, Chapter 6 Natural Selection, discusses heterozygote advantage and balancing selection maintaining sickle cell allele in malaria regions with Hardy-Weinberg distortion, Sinauer Associates publisher.

Meta genomics refers to

Metagenomics involves culture-independent sequencing of total DNA extracted from environmental or host-associated communities, reconstructing microbiome genomes, functions and diversity. Distinct from single-organism genomics of human or plant, it enables unculturable microbial characterization.

Ref: Prescott et al., Microbiology, 11th Edition, Chapter 20 Microbial Ecology and Metagenomics, defines metagenomics as culture-independent genomic study of microbiomes through direct sequencing of community DNA, published by McGraw-Hill Education.

Two plant genotypes A/a ; b/b ; C/c ; D/d ; E/e. crossing with A/a B/b ; C/c ; d/d ; E/e. Assume the genes assort indepe

For independent genes multiply individual recessive frequencies. Aa×Aa gives 1/4 aa, b/b×B/b gives 1/2 bb, C/c×C/c gives 1/4 cc, D/d×d/d gives 1/2 dd, E/e×E/e gives 1/4 ee. Product 1/256, requiring at least 256 progeny screened to recover quintuple homozygote.

Ref: Griffiths et al., Introduction to Genetic Analysis, 12th Edition, Chapter 2 Single-Gene Inheritance and Independent Assortment, discusses calculating progeny frequencies by multiplying individual Mendelian ratios, published by W.H. Freeman and Company.

Mendel in his experiment did reciprocal crossing between two phenotypes – plants of purple flower with plant of white co

Reciprocal crosses yielding identical F1 purple phenotypes exclude X-linkage and cytoplasmic inheritance, indicating autosomal trait. F2 3:1 segregation with purple excess demonstrates purple as dominant allele masking white recessive allele in heterozygotes, following Mendelian monohybrid ratio.

Ref: Klug, Cummings and Spencer, Concepts of Genetics, 12th Edition, Chapter 2 Mendel's Principles, explains reciprocal crosses, dominance and 3:1 F2 segregation distinguishing dominant purple from recessive white, published by Pearson Education.

In mitochondrial mutation associated with disease, the following two crossings were made (i) mutant ♀ × wildtype ♂ → pro

Maternal inheritance dictates all progeny receive mitochondria from oocyte. Mutant female crossed to wild-type male transmits diseased mitochondria to every child, regardless of sex. Wild-type female crossed to mutant male transmits normal mitochondria, yielding all phenotypically normal progeny.

Ref: Hartl and Jones, Essential Genetics: A Genomics Perspective, 6th Edition, Chapter 6 Extranuclear Inheritance, illustrates mitochondrial disease crosses showing mutant mother gives all mutant offspring while normal mother gives all normal, Jones and Bartlett publication.