NEET · Biology —

Principles of Inheritance and Variation appeared 7 times across 1 year — 7.8% of Biology. This question is from Sex Determination in Honey Bee.

Year 2024 Total
Questions 7 7

In which animal do haploid cells divide mitotically to produce gametes?

Solution & Explanation

Core Logic

The male honeybee is haploid and females are diploid. The gamete formation in female honey bee is by meiosis, whereas male honeybees form gametes by mitosis. Thus, a haploid cell undergoes mitosis to produce gametes in male honeybees.

Pattern Recognition

Haplodiploid sex-determination system: Males (drones) develop parthenogenetically and are haploid. Therefore, they must produce sperm via mitosis to maintain the haploid state.

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation

Reference Study Guides

More Principles of Inheritance and Variation Previous-Year Questions

Q113 neet_2026_03_may_morning Genetic Disorders
Which one of the following disorders is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule?
  • A. Haemophilia
  • B. Thalassemia
  • C. Sickle-cell anaemia
  • D. Phenylketonuria

Solution

Core Logic

Sickle-cell anaemia is an autosome-linked recessive trait. At the molecular level, it is caused by a point mutation leading to a single amino acid substitution: Glutamic acid (Glu) is replaced by Valine (Val) at the sixth position of the beta-globin chain of the haemoglobin molecule. This occurs due to a single base substitution at the sixth codon of the beta-globin gene from GAG to GUG.

Step 1: Final Conclusion

This specific substitution is the classic hallmark of Sickle-cell anaemia.

Pattern Recognition

GAG to GUG -> Glu to Val -> position 6 of beta-globin = Sickle-cell anaemia. Universally tested standard fact.

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation Class 12 Biology: Molecular Basis of Inheritance

Q114 neet_2026_03_may_morning Genetic Principles
Match List I with List II :
List-IList-II
A. Incomplete dominanceI. Human skin colour
B. Co-dominanceII. Inheritance of flower colour in Antirrhinum sp.
C. PleiotropyIII. Phenylketonuria disease in humans
D. Polygenic inheritanceIV. ABO blood groups
Choose the correct answer from the options given below :
  • A. A-II, B-IV, C-III, D-I
  • B. A-I, B-III, C-II, D-IV
  • C. A-II, B-I, C-III, D-IV
  • D. A-I, B-IV, C-III, D-II

Solution

Core Logic
  • Incomplete dominance (A) occurs when a heterozygote has an intermediate phenotype. Example: Flower colour in snapdragon (Antirrhinum sp.) (A-II).
  • Co-dominance (B) occurs when both alleles are fully expressed. Example: ABO blood groups, specifically individuals with AB blood group (IAIB) (B-IV).
  • Pleiotropy (C) is when a single gene influences multiple phenotypic traits. Example: Phenylketonuria (PKU) in humans leads to mental retardation and reduced hair/skin pigmentation (C-III).
  • Polygenic inheritance (D) is when a single trait is controlled by three or more genes. Example: Human skin colour is controlled by three pairs of non-allelic genes (D-I).
Step 1: Final Conclusion

The correct matches are A-II, B-IV, C-III, D-I.

Pattern Recognition

Match the most classic examples: Co-dominance = ABO blood group. Polygenic = Skin colour. Pleiotropy = PKU.

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation

Q118 neet_2026_03_may_morning Sex Determination
Which of the following statements are true with reference to the sex-determination in honeybees? A. An offspring formed from the union of a sperm and an egg, develops as a female (queen or worker). B. An unfertilized egg develops as a male by parthenogenesis. C. A male has half the number of chromosomes than that of a female. D. Males produce sperms by meiosis. E. Honeybees have a haplodiploid sex-determination system. Choose the correct answer from the options given below :
  • A. B, C, D and E only
  • B. A, B, C and D only
  • C. A, B, D and E only
  • D. A, B, C and E only

Solution

Core Logic

Honeybees exhibit haplodiploid sex determination.

  • Statement A is true: Fertilized eggs (diploid) develop into females (queens or workers).
  • Statement B is true: Unfertilized eggs develop into males (drones) via parthenogenesis.
  • Statement C is true: Males are haploid (n=16), having half the chromosomes of females (2n=32).
  • Statement D is false: Because males are already haploid, they cannot undergo meiosis to form gametes; they produce sperms via mitosis.
  • Statement E is true: This entire mechanism is known as the haplodiploid sex-determination system.
Step 1: Final Conclusion

Statements A, B, C, and E are correct, while D is incorrect.

Pattern Recognition

Haploid males cannot undergo meiosis (which halves ploidy). Therefore, D is false. Eliminating options with D leaves only option 4.

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation

Q137 neet_2026_03_may_morning Genetic Disorders
The sixth mutant codon of beta globin gene causing polymerization of Haemoglobin and change in RBC shape is ____.
  • A. CAG
  • B. GUG
  • C. AUG
  • D. GAG

Solution

Core Logic

Sickle cell anaemia is caused by a point mutation at the sixth position of the beta globin chain. The normal codon is GAG, which codes for Glutamic acid. The mutation changes this normal codon to GUG on the mRNA, which codes for Valine. This mutant valine causes hemoglobin molecules to polymerize under low oxygen tension, changing the RBC shape to sickle-like.

Step 1: Final Conclusion

The mutant codon is GUG.

Pattern Recognition

GAG = Normal (Glutamic Acid). GUG = Mutant (Valine). The question explicitly asks for the 'mutant codon'.

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation Class 12 Biology: Molecular Basis of Inheritance

Q144 neet_2026_03_may_morning Multiple Alleles and Blood Groups
What is the probability of having children with 'O' blood group, where both mother and father are heterozygous for 'A' and 'B' blood group, respectively?
  • A. 50%
  • B. 0%
  • C. 75%
  • D. 25%

Solution

Core Logic

The genotypes of the parents are:

  • Mother: Heterozygous 'A' blood group arrow IAi
  • Father: Heterozygous 'B' blood group arrow IBi
  • When calculating the genetic cross via Punnett square:

  • Gametes from mother: IA and i
  • Gametes from father: IB and i
  • The possible offspring genotypes are:

  • IAIB (Blood group AB)
  • IAi (Blood group A)
  • IBi (Blood group B)
  • ii (Blood group O)
Step 1: Final Conclusion

Out of 4 possible combinations, only 1 results in blood group 'O'. Therefore, the probability is 1/4 or 25%.

Pattern Recognition

Heterozygous A (AO) × Heterozygous B (BO) yields all four blood types in equal 1:1:1:1 ratio (25% each).

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation

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