Human Genetics — NEET Zoology MCQs with answers

Human Genetics is a Class 12 Zoology chapter in the NEET (UG) syllabus. NEET720 has 178 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Genomic Imprinting · Uniparental Disomy · DNA Fingerprinting · Multifactorial Inheritance · Pharmacogenomics · Chimerism · Genetic Linkage in Humans · Incomplete Penetrance · Genetic Anticipation · Sex-Limited and Sex-Influenced Traits · Nondisjunction · Pedigree Analysis · Heritability · Autosomal Dominant Disorders · Genetic Counselling · Mitochondrial Inheritance · Autosomal Recessive Disorders · Phenylketonuria · X-Linked Recessive Disorders · Y Chromosome Genes · Prenatal Diagnosis · Sickle-Cell Anaemia · Founder Effect and Genetic Drift · Genetic Testing Ethics · Preimplantation Genetic Diagnosis · Fetal-to-Adult Haemoglobin Switch · Twin Studies · Blood Groups · Sex Determination · Human Genetic Disorders · Hardy-Weinberg Principle · Rh Incompatibility · Chromosomal Disorders

8 free Human Genetics practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Twin Studies

   For a particular disease, researchers find that when one monozygotic (MZ) twin is affected, the co-twin is also affected in about 70% of pairs (concordance). For dizygotic (DZ) twins, the co-twin is affected in only about 20% of pairs. Statement I: This large concordance gap between MZ and DZ twins indicates a substantial genetic contribution to the disease. Statement II: A concordance rate of 70% in MZ twins (rather than 100%) also indicates that non-genetic (environmental) factors contribute to whether the disease actually develops. Evaluate these statements.

   • A.Both Statement I and Statement II are correct
   • B.Statement I is correct, but Statement II is incorrect, because MZ twins sharing identical genomes must always show 100% concordance for any genetically influenced disease
   • C.Statement I is incorrect, but Statement II is correct, because twin studies cannot provide any evidence about genetic contribution at all
   • D.Both statements are incorrect, because concordance rates in twin studies are meaningless without knowing each twin's exact genotype at every gene
   Show answer and explanation

   Answer: A. Both Statement I and Statement II are correct

   A much higher concordance rate in MZ twins (who share essentially all their genes) than in DZ twins (who share about half, like ordinary siblings) points to substantial genetic influence, while MZ concordance below 100% shows environmental or chance factors also matter.

   Twin studies compare monozygotic twins, who arise from a single zygote and are genetically near-identical, with dizygotic twins, who arise from two separately fertilised eggs and share on average half their segregating genes, like ordinary siblings. A markedly higher concordance rate in MZ twins compared to DZ twins for a given condition is classic evidence that genetic factors substantially influence susceptibility to that condition, supporting Statement I. However, because MZ twins share essentially the same genome, if the disease were purely and completely genetically determined, concordance would approach 100%; an MZ concordance of only about 70% shows that environmental exposures, developmental variation, or chance events also play a meaningful role in whether the disease actually manifests, supporting Statement II as well. Both statements are therefore correct and complementary, not contradictory.

   Common mistake: Assuming that because MZ twins are genetically identical, they must always be identically affected by any genetically influenced disease.

   Key point: A wide MZ-DZ concordance gap indicates genetic influence; MZ concordance below 100% indicates environmental/chance contribution as well.

2. Question 2 · medium · Twin Studies

   Two newborn twins are of the same sex and look very similar, and doctors wish to determine whether they are monozygotic or dizygotic. As a preliminary (though not fully definitive) check, their ABO and Rh blood groups are tested and found to be different from each other (e.g. one is blood group A, Rh-positive, and the other is blood group B, Rh-negative). What can be concluded from this specific finding?

   • A.The twins must be monozygotic, because blood group differences only ever arise between twins that came from the very same zygote
   • B.The twins cannot be monozygotic, because monozygotic twins arise from a single fertilised egg and therefore must share an identical genotype at every locus, including blood group genes
   • C.No conclusion can be drawn about zygosity from blood group results, since ABO and Rh genes are not inherited in a Mendelian manner
   • D.The twins must be dizygotic and, moreover, must have been conceived from eggs released in two entirely separate menstrual cycles
   Show answer and explanation

   Answer: B. The twins cannot be monozygotic, because monozygotic twins arise from a single fertilised egg and therefore must share an identical genotype at every locus, including blood group genes

   Since monozygotic twins share an identical genotype (arising from one zygote), any genotypic difference such as differing ABO/Rh blood groups rules out monozygosity, confirming the twins must be dizygotic.

   Monozygotic twins result from the splitting of a single fertilised zygote, so they carry identical genetic material at every locus, including the ABO and Rh blood group loci; they must therefore always share the same blood group. Dizygotic twins arise from two separate zygotes formed from two different eggs fertilised by two different sperm during the same ovulatory cycle, and like ordinary siblings, they can differ at many loci, including blood group genes. Observing different blood groups between the twins is therefore direct evidence against monozygosity, confirming they must be dizygotic; it does not, however, imply separate menstrual cycles, since dizygotic twins from a single pregnancy still arise from the same cycle's double ovulation. Blood group loci are classic Mendelian-inherited markers well suited to this kind of comparison, so this distractor is also incorrect.

   Common mistake: Reversing the logical direction, treating a blood group difference as evidence for rather than against monozygosity.

   Key point: Differing blood groups between twins rule out monozygosity (since MZ twins are genetically identical) and confirm dizygosity.

3. Question 3 · hard · Human Genetic Disorders

   A student analyses a pedigree for a certain disorder and observes: an affected father transmits the disorder to ALL of his daughters and NONE of his sons. The student concludes: 'This must be X-linked recessive inheritance, because the trait is carried on the X chromosome, and X-linked recessive traits always show affected fathers passing the condition to all their daughters.' Identify the specific error in this reasoning.

   • A.There is no error; X-linked recessive inheritance does produce exactly this pattern of an affected father transmitting to all daughters and no sons
   • B.The error is that affected fathers can never transmit any X-linked trait, recessive or dominant, to their daughters under any circumstances
   • C.The pattern described (affected father, all daughters affected, no sons affected) is actually the signature of X-linked DOMINANT inheritance, not X-linked recessive; in X-linked recessive inheritance, an affected father's daughters are typically unaffected carriers, not affected individuals
   • D.The error is that the trait must actually be autosomal dominant, since any trait transmitted to all daughters must be autosomal by definition
   Show answer and explanation

   Answer: C. The pattern described (affected father, all daughters affected, no sons affected) is actually the signature of X-linked DOMINANT inheritance, not X-linked recessive; in X-linked recessive inheritance, an affected father's daughters are typically unaffected carriers, not affected individuals

   An affected father who passes the disorder to all his daughters (as affected individuals, not just carriers) and none of his sons demonstrates X-linked DOMINANT inheritance, since each daughter receives his single X chromosome (carrying the dominant allele) and each son receives his Y chromosome instead.

   A father transmits his single X chromosome to every daughter and his Y chromosome to every son. If the father is affected by an X-linked dominant disorder, every daughter necessarily inherits his disease-bearing X allele and, because the allele is dominant, every daughter is herself affected; every son instead inherits the unaffected Y chromosome (and the mother's X, assumed unaffected), so no son is affected. This produces exactly the described pattern. In X-linked recessive inheritance, an affected father's daughters would each inherit his recessive disease allele on one X but would typically also inherit a normal allele from an unaffected mother on their other X, making them unaffected carriers rather than affected individuals, which is a critical distinction the student's reasoning missed. The trait pattern strongly implicates X-linkage (given the total absence of transmission to sons and total transmission to daughters), so an autosomal explanation is not supported either.

   Common mistake: Assuming any X-linked pattern involving affected daughters of an affected father must be recessive, without checking whether the daughters are affected or merely carriers.

   Key point: An affected father transmitting a disorder to all daughters (as affected, not carriers) and no sons signals X-linked dominant, not recessive, inheritance.

4. Question 4 · medium · Rh Incompatibility

   Two Coombs tests are used in the clinical workup of Rh incompatibility: one detects antibodies already bound to a patient's own red blood cells; the other detects free anti-Rh antibodies circulating in a patient's serum, by mixing that serum with reference red cells and then adding an anti-human-globulin reagent. Which statement correctly distinguishes the direct Coombs test from the indirect Coombs test in this context?

   • A.The direct Coombs test would be used on a newborn's blood to detect maternal anti-Rh antibodies already coating the baby's red cells, while the indirect Coombs test would be used on the Rh-negative mother's serum to detect circulating anti-Rh antibodies before or during pregnancy
   • B.The direct Coombs test is performed on maternal serum, while the indirect Coombs test is performed directly on the newborn's red cells, the reverse of their actual clinical use
   • C.Both tests are functionally identical and interchangeable, detecting the same antibody-antigen interaction regardless of whether antibodies are already bound to cells or still free in serum
   • D.The direct Coombs test detects ABO antibodies exclusively, while the indirect Coombs test detects Rh antibodies exclusively, making them specific to different blood group systems
   Show answer and explanation

   Answer: A. The direct Coombs test would be used on a newborn's blood to detect maternal anti-Rh antibodies already coating the baby's red cells, while the indirect Coombs test would be used on the Rh-negative mother's serum to detect circulating anti-Rh antibodies before or during pregnancy

   The direct Coombs test detects antibodies already bound to red cells (useful on a newborn's blood to detect maternal anti-Rh antibody coating), while the indirect Coombs test detects free antibodies in serum (useful on a mother's blood to check for circulating anti-Rh antibodies).

   The direct antiglobulin (Coombs) test is applied directly to a patient's red blood cells to detect antibodies already bound to their surface; in the context of Rh incompatibility, it is used on a newborn's cord blood to confirm that maternal anti-Rh (anti-D) antibodies have crossed the placenta and coated the infant's Rh-positive red cells, which is diagnostic of haemolytic disease of the newborn. The indirect antiglobulin (Coombs) test is instead applied to a patient's serum, mixing it with reference red cells of known antigen type and then adding anti-human-globulin reagent to detect free, unbound antibodies circulating in that serum; it is used to screen an Rh-negative mother's blood for the presence of anti-Rh antibodies, whether from a prior sensitising event or developing during the current pregnancy. Confusing which sample (cells versus serum) each test examines, or treating the two tests as interchangeable or system-restricted, misrepresents their distinct clinical roles.

   Common mistake: Confusing which sample type (red cells versus serum) is examined by the direct versus indirect Coombs test.

   Key point: Direct Coombs test: detects antibody already on red cells (e.g., newborn's blood). Indirect Coombs test: detects free antibody in serum (e.g., mother's blood).

5. Question 5 · medium · Chromosomal Disorders

   Two males are both diagnosed with Klinefelter-spectrum features. Karyotyping of Male 1 shows that every cell examined has the constitution 47,XXY. Karyotyping of Male 2 shows a mixture of cell lines: some cells are 46,XY and others are 47,XXY (mosaic Klinefelter). Which statement about their expected phenotypes is most biologically accurate?

   • A.Both males are expected to show identical severity of features, since the mere presence of any 47,XXY cells guarantees an identical phenotype regardless of the proportion of normal cells present
   • B.Male 2 is expected to show a MORE severe phenotype than Male 1, since having two different cell lines is inherently more disruptive than a single uniform abnormal cell line
   • C.Male 2 (mosaic 46,XY/47,XXY) generally tends to show a milder and more variable phenotype than Male 1 (uniform 47,XXY), because the proportion of normal 46,XY cells in his tissues can partially offset the effects of the extra X chromosome
   • D.Male 2's mosaicism means he cannot have any Klinefelter-associated features at all, since some of his cells are chromosomally normal
   Show answer and explanation

   Answer: C. Male 2 (mosaic 46,XY/47,XXY) generally tends to show a milder and more variable phenotype than Male 1 (uniform 47,XXY), because the proportion of normal 46,XY cells in his tissues can partially offset the effects of the extra X chromosome

   Mosaic Klinefelter individuals (46,XY/47,XXY), who carry a mixture of normal and trisomic cell lines, generally show a milder and more variable phenotype than individuals with uniform 47,XXY karyotype in every cell, because the proportion of normal cells can partially offset the extra-X effects.

   Klinefelter syndrome's phenotypic severity is influenced by the proportion of cells carrying the extra X chromosome. In non-mosaic Klinefelter (47,XXY in essentially all cells), the extra X chromosome affects tissues uniformly, typically producing the full spectrum of features (tall stature, small testes, reduced fertility, gynaecomastia, learning difficulties in some cases). In mosaic Klinefelter (a mixture of 46,XY and 47,XXY cell lines, arising from post-zygotic non-disjunction), the presence of normal 46,XY cells in various tissues, especially the gonads, can partially compensate for the trisomic cells, often resulting in a milder, more variable phenotype, and sometimes even preserved fertility, depending on the mosaic proportion and its tissue distribution. This variability means mosaicism does not guarantee an identical phenotype to full trisomy (B is wrong), is not inherently worse than uniform trisomy (C is wrong), and does not eliminate all features (D overstates the protective effect).

   Common mistake: Assuming mosaicism either has no effect on severity or makes the condition worse, rather than generally moderating it.

   Key point: Mosaic 46,XY/47,XXY Klinefelter generally produces a milder, more variable phenotype than uniform 47,XXY, due to the moderating presence of normal cells.

6. Question 6 · medium · Chromosomal Disorders

   Klinefelter syndrome (47,XXY) and XYY syndrome (47,XYY) are both sex chromosome aneuploidies involving one extra sex chromosome in a phenotypic male. Which statement correctly distinguishes the two conditions?

   • A.Both conditions produce an identical clinical picture, since both involve exactly one extra sex chromosome regardless of whether it is an X or a Y
   • B.XYY syndrome, not Klinefelter syndrome, is the condition typically associated with small testes and reduced fertility, the reverse of the correct association
   • C.Klinefelter syndrome arises from an extra Y chromosome and XYY syndrome arises from an extra X chromosome, the reverse of their actual chromosomal constitutions
   • D.Klinefelter syndrome (extra X) is typically associated with small testes, reduced fertility and often gynaecomastia, whereas XYY syndrome (extra Y) is typically associated with tall stature but usually normal testicular function and fertility, without the small-testes/infertility profile seen in Klinefelter
   Show answer and explanation

   Answer: D. Klinefelter syndrome (extra X) is typically associated with small testes, reduced fertility and often gynaecomastia, whereas XYY syndrome (extra Y) is typically associated with tall stature but usually normal testicular function and fertility, without the small-testes/infertility profile seen in Klinefelter

   Klinefelter syndrome (47,XXY, an extra X) typically involves small testes, reduced fertility and often gynaecomastia, while XYY syndrome (47,XYY, an extra Y) typically involves tall stature with usually normal testicular function and fertility.

   Klinefelter syndrome results from an extra X chromosome in a male (47,XXY), and the additional X commonly disrupts normal testicular development, leading to small, firm testes, reduced testosterone, often reduced fertility (frequently azoospermia), and in some cases gynaecomastia and tall stature. XYY syndrome results instead from an extra Y chromosome (47,XYY), and affected males are often somewhat taller than average but typically have normally functioning testes and normal (or near-normal) fertility, without the small-testes profile characteristic of Klinefelter syndrome; many XYY individuals are never clinically diagnosed because features can be subtle. The identity of the extra chromosome (X versus Y) is therefore biologically consequential, not interchangeable, and the testicular/fertility profile is specifically associated with the extra-X condition (Klinefelter), not the extra-Y condition (XYY).

   Common mistake: Assuming any extra sex chromosome produces the same clinical picture regardless of whether it is an X or a Y.

   Key point: Klinefelter (extra X) impairs testicular function/fertility; XYY (extra Y) typically does not, despite both being 'one extra sex chromosome' conditions.

7. Question 7 · medium · Chromosomal Disorders

   A Barr body is the condensed, inactivated X chromosome visible in the interphase nuclei of somatic cells; the number of Barr bodies present equals (number of X chromosomes) minus 1. Using this rule, how many Barr bodies would be expected in a buccal smear cell from an individual with a 47,XXX karyotype (triple X syndrome), compared with a typical 46,XX female?

   • A.The 47,XXX individual would show 3 Barr bodies, while a typical 46,XX female shows 2 Barr bodies, since Barr body number should simply equal total X count
   • B.The 47,XXX individual would show 2 Barr bodies, while a typical 46,XX female shows 1 Barr body
   • C.The 47,XXX individual would show 0 Barr bodies, while a typical 46,XX female shows 1 Barr body, since more X chromosomes would mean more, not fewer, active copies
   • D.Both the 47,XXX individual and a typical 46,XX female would show exactly 1 Barr body, since only ever one X chromosome per cell can ever be inactivated regardless of total X count
   Show answer and explanation

   Answer: B. The 47,XXX individual would show 2 Barr bodies, while a typical 46,XX female shows 1 Barr body

   By the rule Barr bodies = (X count − 1), a 47,XXX individual (three X chromosomes) shows 2 Barr bodies, while a typical 46,XX female (two X chromosomes) shows 1 Barr body.

   X-inactivation (dosage compensation) ensures that, regardless of how many X chromosomes a cell carries, only one X chromosome remains transcriptionally active per cell; all additional X chromosomes are condensed into Barr bodies. For a 46,XX female with two X chromosomes, one remains active and the other is inactivated, giving 1 Barr body (2 − 1 = 1). For a 47,XXX individual with three X chromosomes, one remains active and the other two are inactivated, giving 2 Barr bodies (3 − 1 = 2). This rule, Barr body count = X chromosome count − 1, applies generally: a 47,XXY (Klinefelter) male would show 1 Barr body, while a typical 46,XY male shows 0. Using total X count directly (B, without subtracting one) or reversing/capping the relationship (C, D) misapplies this well-established cytogenetic counting rule.

   Common mistake: Forgetting to subtract one active X chromosome, using total X chromosome count directly as the Barr body number.

   Key point: Barr body number = (number of X chromosomes) − 1, since exactly one X remains active per cell regardless of total X count.

8. Question 8 · medium · Human Genetic Disorders

   A rare autosomal recessive disorder has a very low carrier frequency in the general (unrelated-mating) population. Genetic counsellors note that the risk of two first-cousin parents (who share a recent common ancestor) both being carriers of the same rare recessive allele, and therefore having an affected child, is meaningfully higher than for two unrelated parents drawn at random from the population. What is the biological basis for this elevated risk in consanguineous unions?

   • A.Consanguineous mating directly causes new mutations to arise in the recessive allele at a higher rate than in unrelated matings
   • B.There is no genuine biological basis for this pattern; any apparent association between consanguinity and recessive disorders is purely a statistical artefact of small family sizes
   • C.Consanguineous unions increase risk only for X-linked disorders, never for autosomal recessive disorders, since autosomal genes are inherited identically regardless of parental relatedness
   • D.First cousins share a proportion of their genes inherited from a common ancestor, so if that ancestor carried the rare recessive allele, both cousins have an increased chance of having inherited a copy of the very same allele, unlike two unrelated individuals drawn independently from the population
   Show answer and explanation

   Answer: D. First cousins share a proportion of their genes inherited from a common ancestor, so if that ancestor carried the rare recessive allele, both cousins have an increased chance of having inherited a copy of the very same allele, unlike two unrelated individuals drawn independently from the population

   Related individuals such as first cousins have an elevated chance of both carrying the identical rare recessive allele inherited from a shared ancestor, raising the probability that both parents are carriers and, consequently, that their child inherits two copies of the disease allele.

   In an unrelated mating, the two parents' carrier statuses for a rare recessive allele are essentially independent events, each governed by the (low) general population carrier frequency, so the chance both happen to be carriers of the exact same rare allele is very small. In a consanguineous mating, such as between first cousins, both individuals may have inherited a copy of the same specific allele (rare or common) from a shared grandparent or great-grandparent, because they share a fraction of their genome by descent from that common ancestor. This means that if the shared ancestor carried the rare recessive disease allele, both cousins have a correlated, elevated chance of also carrying it, raising the probability of an affected child well above what independent unrelated carrier frequencies alone would predict. This mechanism applies to autosomal recessive disorders in general (making D incorrect), does not involve any increased mutation rate (making B incorrect), and is a well-documented, biologically grounded phenomenon, not a statistical artefact (making C incorrect).

   Common mistake: Attributing consanguinity's effect to increased mutation rate rather than to shared inheritance of an existing allele.

   Key point: Consanguinity raises recessive disorder risk because related parents have a correlated (not independent) chance of carrying the same rare allele inherited from a shared ancestor.

Questions about Human Genetics for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (178 active practice questions in this chapter today).