In a population of a grasshopper species, the chromosome number of some members is 23$23$ and some other members possess 24$24$ chromosomes. The 23$23$ and 24$24$ chromosome-bearing members in this species are

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

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$I^{A}I^{B}$) (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

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$n=16$), having half the chromosomes of females (2n=32$2n=32$).
• 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

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

Core Logic

The genotypes of the parents are:

• Mother: Heterozygous 'A' blood group arrow IAi$\rightarrow I^{A}i$
• Father: Heterozygous 'B' blood group arrow IBi$\rightarrow I^{B}i$

When calculating the genetic cross via Punnett square:

• Gametes from mother: IA$I^{A}$ and i$i$
• Gametes from father: IB$I^{B}$ and i$i$

The possible offspring genotypes are:

• IAIB$I^{A}I^{B}$ (Blood group AB)
• IAi$I^{A}i$ (Blood group A)
• IBi$I^{B}i$ (Blood group B)
• ii$ii$ (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$1/4$ or 25%$25\%$.

Pattern Recognition

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

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation