Principles of Inheritance and Variation — NEET Zoology MCQs with answers

Principles of Inheritance and Variation is a Class 12 Zoology chapter in the NEET (UG) syllabus. NEET720 has 133 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

Sex Determination in Insects · Human Sex Determination · Non-disjunction · Sex Determination in Animals · Sex-linked Inheritance · Human Linkage and Crossing Over · Genetic Counselling · Blood Group Genetics · Dosage Compensation · Pedigree Analysis · Mendelian Disorders in Man · Multiple Alleles · Population Genetics of Human Disorders · Sex Determination in Birds · Sex Determination in Humans · Sex Determination Mechanisms · Chromosomal Disorders · ABO Blood Groups · Rh Blood Group · Linkage and Recombination

8 free Principles of Inheritance and Variation 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 · Sex Determination in Insects

   In the honeybee, a drone develops from an unfertilised egg and is haploid, while queens and workers develop from fertilised eggs and are diploid. Which single statement correctly describes spermatogenesis in a drone and its genetic consequence?

   • A.Because the drone is already haploid, its spermatogonia cannot undergo a reductional division; sperm are produced by a modified meiosis that behaves like mitosis, so every sperm a drone makes is genetically identical to every other.
   • B.The drone doubles its chromosome number in the testis and then undergoes a normal two-division meiosis, so its sperm show the usual recombinational variety.
   • C.The drone produces both haploid and diploid sperm in equal numbers, which is why some daughters are queens and others are workers.
   • D.The drone's sperm are produced by normal meiosis I and II and therefore differ from one another by independent assortment of its maternal and paternal chromosomes.
   Show answer and explanation

   Answer: A. Because the drone is already haploid, its spermatogonia cannot undergo a reductional division; sperm are produced by a modified meiosis that behaves like mitosis, so every sperm a drone makes is genetically identical to every other.

   A haploid drone has no homologous pairs, so the first meiotic (reductional) division is abortive and all of its sperm are clones of its single genome.

   Haplodiploid sex determination means the drone's entire body, including its germ line, carries a single chromosome set inherited from the queen. Meiosis I normally separates homologues, but a haploid cell has no homologues to separate, so in drone spermatogenesis the first division is abortive (an unequal, non-reductional division) and only an equational division follows. Every functional spermatozoon therefore carries exactly the same haploid genome. The genetic consequence is important: a drone transmits 100 per cent of its genes, unaltered, to every daughter it fathers. Option B invents a doubling step, option C wrongly makes the queen-worker difference genetic when it is caused by royal-jelly feeding of the larva, and option D forgets that independent assortment requires two parental chromosome sets.

   Common mistake: Assuming meiosis proceeds normally in a haploid drone and produces recombinant sperm.

   Key point: A haploid male has no homologues, so all his sperm are genetically identical.

2. Question 2 · medium · Sex Determination in Insects

   In grasshoppers and some bugs the male has one unpaired sex chromosome and the female has two (XO/XX), whereas in other bugs the male has an X and a morphologically different Y (XY/XX). Comparing these two male-heterogametic arrangements, which statement is correct?

   • A.In the XO system females produce two kinds of eggs, so the mother decides the sex of the offspring.
   • B.The XO system is female-heterogametic and therefore comparable to the ZW system of birds.
   • C.Both are male-heterogametic: the XO male makes X-bearing and nullo-X sperm in equal numbers, while the XY male makes X-bearing and Y-bearing sperm, and in both cases a 1:1 sex ratio is expected.
   • D.In the XO system the male is sterile because an unpaired chromosome cannot pass through meiosis at all.
   Show answer and explanation

   Answer: C. Both are male-heterogametic: the XO male makes X-bearing and nullo-X sperm in equal numbers, while the XY male makes X-bearing and Y-bearing sperm, and in both cases a 1:1 sex ratio is expected.

   XO and XY insects are both male-heterogametic; the XO male's two sperm classes are X-bearing and X-lacking, again giving a 1:1 sex ratio.

   In the XO (Protenor) type the female is XX and the male is XO. At meiosis the male's single X behaves as a univalent and passes undivided into one secondary spermatocyte, so half the sperm carry an X and half carry no sex chromosome at all. Fertilisation by an X-bearing sperm gives XX (female) and by a nullo-X sperm gives XO (male), a 1:1 ratio. In the XY (Lygaeus) type the male's X pairs with a Y, giving X- and Y-bearing sperm in equal numbers and the same 1:1 expectation. The essential shared feature is male heterogamety; the difference is only whether the X's partner is a Y or is absent. The univalent X is not a barrier to meiosis, so the XO male is fully fertile.

   Common mistake: Treating the absence of a Y chromosome as evidence of female heterogamety.

   Key point: XO and XY are both male-heterogametic; only the presence of a Y differs.

3. Question 3 · medium · Sex Determination in Insects

   In Drosophila, sex depends on the ratio of X chromosomes to autosomal sets (the X:A ratio): a ratio of 1.0 gives a normal female and 0.5 a normal male. A rare fly is recovered that carries two X chromosomes together with three complete autosomal sets. Its expected sexual phenotype is:

   • A.A normal fertile female, because two X chromosomes are present.
   • B.A normal fertile male, because the extra autosomal set acts like a Y chromosome.
   • C.An intersex, because the X:A ratio is 2/3 = 0.67, which lies between the male value of 0.5 and the female value of 1.0.
   • D.A metafemale (superfemale), because adding chromosomes always exaggerates the female phenotype.
   Show answer and explanation

   Answer: C. An intersex, because the X:A ratio is 2/3 = 0.67, which lies between the male value of 0.5 and the female value of 1.0.

   X:A = 2/3 = 0.67 falls between 0.5 and 1.0, producing an intersex fly.

   In Drosophila the Y chromosome carries fertility factors but does not determine sex; sex is set by the balance between female-determining genes on the X and male-determining genes on the autosomes. Compute the ratio: two X chromosomes divided by three autosomal sets gives 0.67. Values of 1.0 or more give female (1.5 gives a sterile metafemale), 0.5 gives male, values below 0.5 give a sterile metamale, and intermediate values such as 0.67 give an intersex with a mixture of male and female characters. This contrasts sharply with humans, where an XXY individual is male because the Y carries SRY regardless of the number of X chromosomes.

   Common mistake: Counting only X chromosomes and forgetting the autosomal denominator.

   Key point: In Drosophila sex is decided by X:A ratio, not by the presence of a Y.

4. Question 4 · easy · Human Sex Determination

   A normal human secondary oocyte and a normal human spermatozoon are examined for their chromosome content. The correct description is:

   • A.The oocyte is 22 + X or 22 + Y, and the sperm is always 22 + X.
   • B.The oocyte is always 22 + X, while the sperm is either 22 + X or 22 + Y.
   • C.Both the oocyte and the sperm are 23 + X, since every gamete carries 23 autosomes plus a sex chromosome.
   • D.Both are 44 + XX or 44 + XY, since gametes retain the somatic chromosome number until fertilisation.
   Show answer and explanation

   Answer: B. The oocyte is always 22 + X, while the sperm is either 22 + X or 22 + Y.

   Human females are homogametic (all eggs 22 + X); males are heterogametic (22 + X or 22 + Y sperm).

   The human diploid number is 46, made of 22 pairs of autosomes plus one pair of sex chromosomes. Meiosis reduces this to 23 per gamete, that is 22 autosomes plus one sex chromosome. A female is XX, so every egg receives one X and is described as 22 + X; the female is homogametic. A male is XY, so half his sperm carry an X and half a Y, giving 22 + X and 22 + Y classes; the male is heterogametic. This is the chromosomal reason why the sperm, not the egg, determines the sex of the child and why a 1:1 sex ratio is expected at conception.

   Common mistake: Writing 23 autosomes plus a sex chromosome instead of 22 plus one.

   Key point: 22 + X eggs; 22 + X or 22 + Y sperm.

5. Question 5 · hard · Non-disjunction

   A boy is found to have a 47,XYY karyotype. Reasoning only from which gametes could combine to produce this constitution, the error that produced him must have been:

   • A.Non-disjunction of the sister chromatids of the Y chromosome during the second meiotic division in the father, producing a YY sperm that fertilised a normal X egg.
   • B.Non-disjunction at the first meiotic division in the father, producing an XY sperm that fertilised a normal X egg.
   • C.Non-disjunction at the first meiotic division in the mother, producing an XX egg.
   • D.Failure of the second polar body to be extruded, so the zygote retained an extra maternal chromosome set.
   Show answer and explanation

   Answer: A. Non-disjunction of the sister chromatids of the Y chromosome during the second meiotic division in the father, producing a YY sperm that fertilised a normal X egg.

   Two Y chromosomes can only come from the father, and two identical Ys in one sperm require sister-chromatid non-disjunction at meiosis II.

   Start from the karyotype: 47,XYY needs one X and two Y chromosomes. The mother is XX and cannot contribute a Y, so both Ys must be paternal and the X must be maternal. Now ask which paternal error yields a YY sperm. At meiosis I the homologues X and Y separate; failure there gives an XY secondary spermatocyte and hence XY sperm, which with an X egg would give 47,XXY (Klinefelter), not XYY. At meiosis II the sister chromatids of an already-separated Y fail to disjoin, giving a YY sperm; fertilising a normal 22 + X egg this produces 47,XYY. Hence the error is necessarily paternal and necessarily at meiosis II. This reasoning also explains why XYY, unlike trisomy 21, shows no maternal-age effect.

   Common mistake: Assuming all aneuploidies arise from maternal meiosis I non-disjunction.

   Key point: Two identical sex chromosomes in one gamete point to a meiosis II (sister chromatid) error.

6. Question 6 · easy · Sex Determination in Animals

   In birds and in the silkworm Bombyx mori, the male is ZZ and the female is ZW. In such species, the sex of the offspring is decided by:

   • A.The sperm, because the male carries two different sex chromosomes.
   • B.Neither gamete, because sex in birds and silkworms is fixed by incubation temperature.
   • C.The egg, because the female is heterogametic and produces Z-bearing and W-bearing eggs in equal numbers.
   • D.Both gametes equally, because Z and W chromosomes are contributed by both parents.
   Show answer and explanation

   Answer: C. The egg, because the female is heterogametic and produces Z-bearing and W-bearing eggs in equal numbers.

   In ZZ/ZW systems the female is heterogametic, so the egg decides the offspring's sex.

   The Z and W letters are used precisely to signal that heterogamety lies with the female, in contrast to the XX/XY convention where the male is heterogametic. A ZZ male produces only Z-bearing sperm. A ZW female produces Z-bearing and W-bearing eggs in equal numbers; a Z egg gives ZZ (male) and a W egg gives ZW (female), so a 1:1 ratio is still expected. This is why, in poultry, the mother rather than the father determines the chick's sex. Temperature-dependent sex determination is a separate, non-chromosomal system found in several reptiles and does not apply to birds or silkworm.

   Common mistake: Assuming the male always determines the sex of the offspring.

   Key point: ZW notation signals female heterogamety; the egg determines sex.

7. Question 7 · medium · Human Sex Determination

   A tall adolescent with typical female external appearance is investigated for absence of menstruation. Testes are found in the abdomen, blood testosterone is in the normal male range, and the karyotype is 46,XY. The defect lies in the X-linked gene encoding the androgen receptor. The best explanation of this combination is:

   • A.The SRY gene functioned normally so testes formed and secreted testosterone, but the target tissues cannot respond to androgen, so the external genitalia followed the default female pathway despite the male karyotype.
   • B.The Y chromosome must be absent from the gonadal tissue, making this a case of 45,X mosaicism.
   • C.Testosterone was never produced, which is why female external features developed.
   • D.The individual has two SRY genes, and the extra copy reversed the direction of sexual differentiation.
   Show answer and explanation

   Answer: A. The SRY gene functioned normally so testes formed and secreted testosterone, but the target tissues cannot respond to androgen, so the external genitalia followed the default female pathway despite the male karyotype.

   Testis determination by SRY is intact, but a non-functional androgen receptor blocks the hormone's action, so external development defaults to female.

   Human sexual development happens in two steps. First, SRY on the Y chromosome directs the indifferent gonad to become a testis; this step is independent of hormones. Second, the testis secretes testosterone (masculinising the external genitalia and Wolffian ducts) and anti-Mullerian hormone (regressing the Mullerian ducts). In complete androgen insensitivity the androgen receptor gene on the X chromosome is defective, so step one succeeds but the androgen arm of step two fails: external genitalia develop along the default female route. Anti-Mullerian hormone still acts, which is why the uterus and upper vagina are absent and menstruation never occurs. Circulating testosterone is normal or high, which is exactly why the hormone-deficiency explanation in option C is excluded by the data given.

   Common mistake: Equating a female phenotype with absence of a Y chromosome or of testosterone.

   Key point: A normal male karyotype and normal testosterone can still give a female phenotype if the receptor fails.

8. Question 8 · medium · Sex-linked Inheritance

   A man with haemophilia A marries a woman who is not a carrier. They have several daughters, all with normal clotting. One of these daughters marries a man with normal clotting. For any single child born to this daughter, the probability of being haemophilic is:

   • A.Zero, because the disorder was not expressed in the daughter and has therefore been lost from the family.
   • B.One quarter, because the daughter is an obligate carrier and the child must be both a son and a recipient of her affected X.
   • C.One half, because the daughter is a carrier and transmits the allele to half her children.
   • D.One eighth, because the daughter has only a one-half chance of being a carrier in the first place.
   Show answer and explanation

   Answer: B. One quarter, because the daughter is an obligate carrier and the child must be both a son and a recipient of her affected X.

   An affected father's every daughter is an obligate carrier, so her child's risk is 1/2 (male) x 1/2 (affected X) = 1/4.

   An affected man is X(h)Y. He passes his Y to every son and his single X(h) to every daughter, so every daughter is X(H)X(h), an obligate carrier, with probability one, not one half. Her cross with a normal man X(H)Y gives four equally likely classes: X(H)X(H) normal daughter, X(H)X(h) carrier daughter, X(H)Y normal son, X(h)Y haemophilic son. One class in four is affected, so the risk for any single child is 1/4. This pattern, in which the trait skips the daughters and reappears in the grandsons, is the reason X-linked recessive disorders are described as passing from a grandfather to his grandsons through unaffected daughters.

   Common mistake: Assigning a one-half carrier probability to the daughter of an affected father.

   Key point: Every daughter of an affected man is a carrier with certainty.

Questions about Principles of Inheritance and Variation for NEET

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