Q.152 Zoology ⭐ 360° Option Elimination Active

NEET 2025 Zoology Question 152 Solution

Topic: Principles of Inheritance and Variation · Official NTA Booklet Code 45

Question

With the help of given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in F3 generation.

✓

Verified Against NTA Final Answer Key

Option (1) : 1/4

📚 Principles of Inheritance and Variation Medium

📋 Given Parameters & Problem Setup

Mother in F2 Carrier female (XᴬXᵃ, circle with central dot)

Father in F2 Affected male (XᵃY, filled black square)

Target Condition in F3 Child having no disease AND being a carrier (mutation in one allele)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Pedigree Punnett Cross Fingerprint

⏱️ Save ~40 seconds

Carrier mother (XᴬXᵃ) × Affected father (XᵃY) -> Zygotes: 1/4 XᴬXᵃ (carrier female, healthy), 1/4 XᵃXᵃ (affected female), 1/4 XᴬY (healthy male, non-carrier), 1/4 XᵃY (affected male)! Probability of healthy carrier child = 1/4!

1 Deduce the mode of inheritance from the pedigree:

2 - F0 affected female (

X^a X^a

) passes disease allele to all sons and makes all daughters carriers (

X^A X^a

3 - Carrier female (

X^A X^a

) in F1 mated to normal male (

X^A Y

) produces an affected son (

X^a Y

) in F2.

4 - Thus, the trait is classical X-linked recessive (e.g. Haemophilia or Colour blindness).

5 Identify parental genotypes mating to yield F3:

6 - Mother (F2): Carrier female =

X^A X^a

7 - Father (F2): Affected male =

X^a Y

8 Perform the genetic cross:

X^A X^a \times X^a Y

9 - Gametes: Mother produces

X^A

(1/2) and

X^a

(1/2); Father produces

X^a

(1/2) and

Y

(1/2).

10 - Offspring distribution (total 4 equally probable classes):

X^A X^a

: Carrier female, no disease (1/4).

X^a X^a

: Affected female, has disease (1/4).

X^A Y

: Normal male, no disease, NOT a carrier (1/4).

X^a Y

: Affected male, has disease (1/4).

15 Who satisfies "child having no disease and being a carrier"? ONLY

X^A X^a

! Probability = 1/4 (Option 1)!

16 Select Option (1) 1/4!

🛡️

Why You Can 100% Trust This Shortcut: Mendelian vs Chromosomal Genetic Disorders (NCERT Biology · Principles of Inheritance and Variation: Aneuploidy, Down, Turner, Klinefelter).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Deciphering the Pedigree and Mode of Inheritance

Analyzing the transmission pattern across generations:

• In F0, an affected female (black circle, $X^a X^a$ ) mates with an unaffected male (empty square, $X^A Y$ ). All daughters inherit one paternal $X^A$ and one maternal $X^a$ , making them obligate carrier females (circles with central dot, $X^A X^a$ ) in F1.
• In F1, a carrier female ( $X^A X^a$ ) mates with an unaffected male ( $X^A Y$ ). They produce an affected son ( $X^a Y$ ) in F2. This confirms X-linked recessive inheritance beyond doubt.
• The consanguineous marriage in F2 involves a carrier female ( $X^A X^a$ ) and an affected male ( $X^a Y$ ).

Step 2

Constructing the F2 × F2 Punnett Square for F3 Offspring

Determining all possible genotypic and phenotypic outcomes:

• Maternal gametes: $X^A$ ( $50\%$ ) and $X^a$ ( $50\%$ ).
• Paternal gametes: $X^a$ ( $50\%$ ) and $Y$ ( $50\%$ ).
• Cross outcomes ( $4$ equally likely quadrants, each $p = 0.25$ ):
• 1. $X^A X^a$ : Phenotypically normal female, carries one mutated allele (Heterozygous carrier female).
• 2. $X^a X^a$ : Phenotypically affected female (Homozygous mutant).
• 3. $X^A Y$ : Phenotypically normal male (Hemizygous wild-type, carries zero mutant alleles, cannot be a carrier).
• 4. $X^a Y$ : Phenotypically affected male (Hemizygous mutant).

Step 3

Evaluating the Probability Condition

The question specifically requires the birth of a child (of either sex) having: (i) NO disease, AND (ii) being a carrier (bearing the disease mutation in one allele).

P(\text{Carrier child}) = P(X^A X^a) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

Step 4

Conclusion

Out of all 4 possible zygotic combinations, exactly 1 corresponds to an unaffected carrier child ( $X^A X^a$ ). The probability is 1/4.

\text{Probability} = \frac{1}{4} \text{ (Option 1)}

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. Only the XᴬXᵃ genotype (1 out of 4 total offspring) is both disease-free and a carrier of one mutated allele.

💡

Why this is airtight: None; flawless pedigree probability calculation.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. 1/2 is the probability of a daughter being a carrier, but the question asks for the probability of a "child" (which includes sons and daughters).

⚠️

Student Fallacy (Why candidates pick this trap): Conditioning on female sex only instead of all children.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Unnecessary multiplication by extra generation factors.

⚠️

Student Fallacy (Why candidates pick this trap): Over-multiplying parental probabilities.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Carrier daughters are readily produced with 25% overall probability.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming affected fathers cannot produce carrier daughters.

⚠️

NTA Trap Alert: Child vs Daughter Probability Trap

A classic genetics trap: If asked "probability of a DAUGHTER being a carrier", the answer is 1/2 (since among daughters, 1 is XᴬXᵃ and 1 is XᵃXᵃ). But when asked "probability of a CHILD being a carrier", all 4 children are considered, so probability is 1/4!

🛡️ How to avoid the -1 mark: Check the wording: "child" = divide by 4; "daughter" = divide by 2.

📖

Core NEET Syllabus Standard

NEET Biology Core Syllabus · Principles of Inheritance and Variation Standard Benchmark

Core Scientific Principle Tested:

"In an X-linked cross of XᴬXᵃ × XᵃY, the probability of an unaffected carrier child (XᴬXᵃ) is 1/4."

Syllabus Focus: Mendelian Disorders · Pedigree Analysis (Sex-Linked Recessive)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

1 in 4 is the Carrier Door: XᴬXᵃ = 1/4 total progeny!

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NEET 2025 Zoology Question 152 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Pedigree Punnett Cross Fingerprint

Step-by-Step Rigorous Workflow & Mathematical Proof

Deciphering the Pedigree and Mode of Inheritance

Constructing the F2 × F2 Punnett Square for F3 Offspring

Evaluating the Probability Condition

Conclusion

360° Option Elimination Engine Key Feature

NTA Trap Alert: Child vs Daughter Probability Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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