The sixth mutant codon of beta globin gene causing polymerization of Haemoglobin and change in RBC shape is:

Related Formula

Sickle Cell Anaemia: Single point mutation at 6th$6^{\text{th}}$ codon of β$\beta$-globin gene (GAG arrow GUG$GAG \rightarrow GUG$), replacing Glutamic acid (Glu) with Valine (Val).

Core Logic

Sickle-cell anaemia is an autosomal recessive genetic defect caused by point mutation resulting in Glu arrow$\rightarrow$ Val substitution at 6th position of β$\beta$-globin polypeptide chain, leading to RBC polymerisation under low oxygen tension.

Step 1: Conclusion

The disorder is Sickle-cell anaemia. Option (3) is correct.

Pattern Recognition

Glu to Val at 6th position of β$\beta$-globin = Sickle-cell anaemia.

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation

Related Formula

Incomplete dominance: Antirrhinum flower colour Codominance: ABO Blood Group (I^A I^B$I^A I^B$) Pleiotropy: Single gene arrow$\rightarrow$ Multiple phenotypic effects (PKU) Polygenic: Multiple genes arrow$\rightarrow$ Single trait (Skin colour)

Core Logic

• Incomplete dominance: Flower colour in Antirrhinum sp. (Snapdragon) arrow$\rightarrow$ II
• Co-dominance: ABO blood group system arrow$\rightarrow$ IV
• Pleiotropy: Phenylketonuria in humans arrow$\rightarrow$ III
• Polygenic inheritance: Human skin colour arrow$\rightarrow$ I

Step 1: Match Verification

The correct combination is A-II, B-IV, C-III, D-I. Correct option is (1).

Pattern Recognition

Incomplete dominance = Antirrhinum (A-II). Polygenic = Human skin colour (D-I). Matches option (1).

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation

Related Formula

Honeybee Sex Determination: Haplodiploid system. Female (2n = 32$2n = 32$) Meiosis$\xrightarrow{\text{Meiosis}}$ Eggs (n = 16$n = 16$) Male (n = 16$n = 16$) Mitosis$\xrightarrow{\text{Mitosis}}$ Sperms (n = 16$n = 16$)

Core Logic

• A: Fertilized egg (2n$2n$) arrow$\rightarrow$ Female (True).
• B: Unfertilized egg (n$n$) arrow$\rightarrow$ Male by parthenogenesis (True).
• C: Male (n=16$n=16$) has half the chromosome count of female (2n=32$2n=32$) (True).
• D: Males are haploid, so they produce sperms by MITOSIS, not meiosis (False).
• E: It is a haplodiploid system (True).

Step 1: Conclusion

Correct statements are A, B, C, and E. Option (4) is correct.

Pattern Recognition

Haploid male honeybees produce sperm by MITOSIS (D is False). Eliminating D immediately isolates A, B, C, E.

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation

Related Formula

Core Logic

Mother's genotype (heterozygous A) = I^A i$I^A i$ Father's genotype (heterozygous B) = I^B i$I^B i$

Crossing these parents yields four possible genotypes in equal proportions (1 : 1 : 1 : 1):

• I^A I^B$I^A I^B$ (Blood group AB)
• I^A i$I^A i$ (Blood group A)
• I^B i$I^B i$ (Blood group B)
• i i$i i$ (Blood group O)

Step 1: Calculation of Probability

Pattern Recognition

Sees: "heterozygous A and B parents blood group O probability". Shortcut: Cross I^A i × I^B i$I^A i \times I^B i$ gives 1/4 = 25% for each blood group (A, B, AB, O).

Chapter Mix

Class 12 Biology: Principles of Inheritance and Variation