Principles of Inheritance and Variation Previous Year Questions — NEET Biology
7 past-year Principles of Inheritance and Variation questions from NEET (Biology).
Q113 (2024)
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?
- Haemophilia
- Thalassemia
- Sickle-cell anaemia
- Phenylketonuria
### Related Formula
Sickle Cell Anaemia: Single point mutation at $6^{\text{th}}$ codon of $\beta$-globin gene ($GAG \rightarrow GUG$), 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 $\rightarrow$ Val substitution at 6th position of $\beta$-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 $\beta$-globin = Sickle-cell anaemia.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Principles of Inheritance and Variation
Q114 (2024)
Match List I with List II :
<div style="overflow-x: auto; margin: 1rem 0;"><table style="width: 100%; border-collapse: collapse; text-align: left; font-size: 14px;"><thead><tr><th style="border: 1px solid #888; padding: 8px;"></th><th style="border: 1px solid #888; padding: 8px;">List-I</th><th style="border: 1px solid #888; padding: 8px;"></th><th style="border: 1px solid #888; padding: 8px;">List-II</th></tr></thead><tbody><tr><td style="border: 1px solid #888; padding: 8px;">A.</td><td style="border: 1px solid #888; padding: 8px;">Incomplete dominance</td><td style="border: 1px solid #888; padding: 8px;">I.</td><td style="border: 1px solid #888; padding: 8px;">Human skin colour</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">B.</td><td style="border: 1px solid #888; padding: 8px;">Co-dominance</td><td style="border: 1px solid #888; padding: 8px;">II.</td><td style="border: 1px solid #888; padding: 8px;">Inheritance of flower colour in Antirrhinum sp.</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">C.</td><td style="border: 1px solid #888; padding: 8px;">Pleiotropy</td><td style="border: 1px solid #888; padding: 8px;">III.</td><td style="border: 1px solid #888; padding: 8px;">Phenylketonuria disease in humans</td></tr><tr><td style="border: 1px solid #888; padding: 8px;">D.</td><td style="border: 1px solid #888; padding: 8px;">Polygenic inheritance</td><td style="border: 1px solid #888; padding: 8px;">IV.</td><td style="border: 1px solid #888; padding: 8px;">ABO blood groups</td></tr></tbody></table></div>
Choose the correct answer from the options given below :
- A-II, B-IV, C-III, D-I
- A-I, B-III, C-II, D-IV
- A-II, B-I, C-III, D-IV
- A-I, B-IV, C-III, D-II
### Related Formula
Incomplete dominance: *Antirrhinum* flower colour
Codominance: ABO Blood Group ($I^A I^B$)
Pleiotropy: Single gene $\rightarrow$ Multiple phenotypic effects (PKU)
Polygenic: Multiple genes $\rightarrow$ Single trait (Skin colour)
### Core Logic
- Incomplete dominance: Flower colour in *Antirrhinum sp.* (Snapdragon) $\rightarrow$ II
- Co-dominance: ABO blood group system $\rightarrow$ IV
- Pleiotropy: Phenylketonuria in humans $\rightarrow$ III
- Polygenic inheritance: Human skin colour $\rightarrow$ 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).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Principles of Inheritance and Variation
Q118 (2024)
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 :
- B, C, D and E only
- A, B, C and D only
- A, B, D and E only
- A, B, C and E only
### Related Formula
Honeybee Sex Determination: Haplodiploid system.
Female ($2n = 32$) $\xrightarrow{\text{Meiosis}}$ Eggs ($n = 16$)
Male ($n = 16$) $\xrightarrow{\text{Mitosis}}$ Sperms ($n = 16$)
### Core Logic
- A: Fertilized egg ($2n$) $\rightarrow$ Female (True).
- B: Unfertilized egg ($n$) $\rightarrow$ Male by parthenogenesis (True).
- C: Male ($n=16$) has half the chromosome count of female ($2n=32$) (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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Principles of Inheritance and Variation
Q137 (2024)
The sixth mutant codon of beta globin gene causing polymerization of Haemoglobin and change in RBC shape is:
- CAG
- GUG
- AUG
- GAG
### Related Formula
$\text{Normal codon (GAG)} \xrightarrow{\text{Point mutation (A } \rightarrow \text{ U)}} \text{Mutant codon (GUG)} \rightarrow \text{Valine (Val)}$
### Core Logic
Sickle-cell anaemia is caused by a point mutation in the beta-globin gene. The normal 6th codon GAG is mutated to GUG in mRNA. This leads to the substitution of Glutamic acid (Glu) by Valine (Val) at the 6th position of the beta-globin polypeptide chain, causing polymerisation under low oxygen tension and distorting RBCs into a sickle shape.
### Step 1: Identification of Mutant Codon
The normal codon is GAG and the mutant codon is GUG.
### Pattern Recognition
Sees: "sixth mutant codon beta globin gene".
Shortcut: Normal = GAG (Glutamic acid), Mutant = GUG (Valine).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Principles of Inheritance and Variation
Q144 (2024)
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?
- 50%
- 0%
- 75%
- 25%
### Related Formula
$$\text{Parent Genotypes: } I^A i \times I^B i$$
$$\text{Offspring Genotypes: } I^A I^B \text{ (AB)}, \quad I^A i \text{ (A)}, \quad I^B i \text{ (B)}, \quad ii \text{ (O)}$$
### Core Logic
Mother's genotype (heterozygous A) = $I^A i$
Father's genotype (heterozygous B) = $I^B i$
Crossing these parents yields four possible genotypes in equal proportions (1 : 1 : 1 : 1):
1. $I^A I^B$ (Blood group AB)
2. $I^A i$ (Blood group A)
3. $I^B i$ (Blood group B)
4. $i i$ (Blood group O)
### Step 1: Calculation of Probability
$$\text{Probability of 'O' blood group } (i i) = \frac{1}{4} = 25\%$$
### Pattern Recognition
Sees: "heterozygous A and B parents blood group O probability".
Shortcut: Cross $I^A i \times I^B i$ gives 1/4 = 25% for each blood group (A, B, AB, O).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Biology: Principles of Inheritance and Variation
