Q.177 Zoology ⭐ 360° Option Elimination Active

NEET 2021 Zoology Question 177 Solution

Topic: Principles of Inheritance and Variation · Official NTA Booklet Code M1

Question

In a cross between a male and female, both heterozygous for sickle cell anaemia gene, what percentage of the progeny will be diseased?

(1)50%
(2)75%
(3)25%
(4)100%

✓

Verified Against NTA Final Answer Key

Option (3) : 25%

📚 Principles of Inheritance and Variation Easy

📋 Given Parameters & Problem Setup

Parental Genotypes Both male and female are heterozygous carriers: Hb^A Hb^S × Hb^A Hb^S

Disorder Type Autosomal Recessive (Sickle Cell Anaemia)

Question Objective Calculate the percentage of diseased progeny

⚡

The 30-Second Exam Hall Shortcut: ⚡ "Carrier × Carrier = 1 Normal : 2 Carriers : 1 Diseased (25% Diseased)"

⏱️ Save ~25 seconds

Sickle cell anaemia is an autosomal recessive disorder. Heterozygous parents ( $Hb^A Hb^S$ ) are clinically unaffected carriers. Only homozygous mutant progeny ( $Hb^S Hb^S$ ) manifest the diseased phenotype: exactly 1 out of 4 progeny (25%)!

1 Recognize disease pattern: Sickle cell anaemia is strictly autosomal recessive.

2 Set up cross:

Hb^A Hb^S \times Hb^A Hb^S

3 Determine progeny genotypes:

1\,Hb^A Hb^A

(Normal, 25%),

2\,Hb^A Hb^S

(Carriers, 50%),

1\,Hb^S Hb^S

(Diseased, 25%).

4 Diseased percentage = 25% -> Mark Option (3).

🛡️

Why You Can 100% Trust This Shortcut: NCERT Biology · Principles of Inheritance and Variation (Mendelian Disorders: Sickle-cell anaemia is an autosomal linked recessive trait that can be transmitted from parents to offspring when both partners are carrier/heterozygous (Hb^A Hb^S); 25% of offspring inherit the homozygous diseased genotype Hb^S Hb^S).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Genetics of Sickle Cell Anaemia

Sickle-cell anaemia is an autosomal recessive Mendelian disorder controlled by a single pair of alleles, $Hb^A$ (normal $\beta$ -globin allele) and $Hb^S$ (mutant $\beta$ -globin allele with a GAG $\to$ GUG transversion at codon 6, substituting glutamic acid with valine). - Normal homozygous ( $Hb^A Hb^A$ ): Completely unaffected (100% normal biconcave erythrocytes). - Heterozygous carriers ( $Hb^A Hb^S$ ): Phenotypically normal under routine atmospheric oxygen tensions; carry sickle-cell trait but do not suffer from anaemia. - Homozygous diseased ( $Hb^S Hb^S$ ): Express the full clinical syndrome of sickle cell anaemia with microvascular vaso-occlusion, acute sickle crises, and haemolytic anaemia.

Step 2

Punnett Square Derivation

Cross: Heterozygous male ( $Hb^A Hb^S$ ) $\times$ Heterozygous female ( $Hb^A Hb^S$ ): $\begin{array}{c|c|c} & Hb^A & Hb^S \\ \hline Hb^A & Hb^A Hb^A \, (25\%) & Hb^A Hb^S \, (25\%) \\ \hline Hb^S & Hb^A Hb^S \, (25\%) & Hb^S Hb^S \, (25\%) \\ \end{array}$ - Normal phenotype (disease-free): $Hb^A Hb^A$ (25%) + $Hb^A Hb^S$ (50%) = 75% - Diseased phenotype: $Hb^S Hb^S$ = $1/4 = 25\%$ .

P(\text{Diseased Progeny}) = P(Hb^S Hb^S) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 25\%

⭐

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 1

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. 50% represents the proportion of heterozygous carriers ( $Hb^A Hb^S$ ), who are clinically unaffected.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing carrier frequency (50%) with diseased frequency (25%).

2 2

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. 75% would be diseased if the condition were dominant, but sickle cell anaemia is autosomal recessive.

⚠️

Student Fallacy (Why candidates pick this trap): Treating the mutant allele as dominant, yielding 3:1 affected:unaffected.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Correct. Only the homozygous recessive genotype $Hb^S Hb^S$ manifests the diseased state ( $1/4 = 25\%$ ).

💡

Why this is airtight: None. Candidate accurately executes Mendelian recessive probability.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. 100% diseased progeny only occurs when both parents are homozygous affected ( $Hb^S Hb^S \times Hb^S Hb^S$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Assuming sickle cell trait always results in affected offspring.

⚠️

NTA Trap Alert: Carrier vs Diseased Phenotype Trap

Many candidates conflate "having the mutant allele" (75% carry at least one $Hb^S$ ) with "being diseased" (only the 25% homozygous $Hb^S Hb^S$ are diseased). Heterozygotes have sickle cell trait but are NOT diseased.

🛡️ How to avoid the -1 mark: Read the prompt carefully: "diseased" requires homozygous recessive (

Hb^S Hb^S

= 25%).

📖

Core NEET Syllabus Standard

NEET / NCERT Biology · Principles of Inheritance and Variation Standard Benchmark

Core Scientific Principle Tested:

"Sickle cell anaemia is an autosomal recessive disorder; a carrier × carrier cross produces 25% homozygous diseased progeny."

Syllabus Focus: Sickle-Cell Anaemia

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Zoology Question 177 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ "Carrier × Carrier = 1 Normal : 2 Carriers : 1 Diseased (25% Diseased)"

Step-by-Step Rigorous Workflow & Mathematical Proof

Genetics of Sickle Cell Anaemia

Punnett Square Derivation

360° Option Elimination Engine Key Feature

NTA Trap Alert: Carrier vs Diseased Phenotype Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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