CUET UG Biology Booster Unit 5 Test (D5) -

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CUET Booster Biology Unit 5 Test (D5)

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET Booster Biology Unit 5 Test (D5)

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QUESTION 1 OF 20

Consider a hypothetical mutation in a eukaryotic cell that selectively prevents the splicing machinery from recognizing introns in the primary transcript (hnRNA). Analytically, what is the most direct downstream consequence on gene expression?

Answer Review

1 Consider a hypothetical mutation in a eukaryotic cell that selectively prevents the splicing machinery from recognizing introns in the primary transcript (hnRNA). Analytically, what is the most direct downstream consequence on gene expression?

A The RNA polymerase II will fail to bind to the promoter.

B The mRNA transported to the cytoplasm will contain non-coding sequences, leading to an incorrect or non-functional polypeptide during translation.

C Transport of the mRNA from the cytoplasm to the nucleus will be halted.

D The primary transcript will immediately undergo reverse transcription back into DNA.

💡 Short Explanation

Splicing removes introns (non-coding). Failure to splice results in introns remaining in mature mRNA. These sequences disrupt the reading frame during translation.

✅ Correct Answer Explanation

Eukaryotic pre-mRNA contains both exons (coding) and introns (non-coding). The splicing machinery is required to remove introns. If splicing fails, the introns remain in the mRNA, which is then translated into a nonsense or non-functional polypeptide because introns usually contain stop codons or shift the reading frame.

❓ Why Other Options Are Incorrect

Option A → Relates to transcription initiation, not processing.
Option C → Transport direction is nucleus to cytoplasm, not reverse.
Option D → Reverse transcription is not part of the standard processing pathway.

🎯 Strategy

Used

Contextual/Tonal Matching: Following the flow of information from gene to protein.

Final Logic: Introns = Non-coding; if kept, protein = non-functional.

🧠 Memory Trick

\\\"Intron in = Nonsense protein.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: RNA Splicing – .

2 Which of the following scenarios does NOT correctly pair a level of eukaryotic gene regulation with its biological location or process?

A Transcriptional level: Formation of primary transcript in the nucleus.

B Processing level: Removal of introns from hnRNA in the nucleus.

C Transport level: Movement of mature mRNA from the nucleus to the cytoplasm.

D Translational level: Splicing of exons by ribosomes in the cytoplasm.

💡 Short Explanation

Splicing is a processing step, not a translational step. Ribosomes perform translation, not splicing.

✅ Correct Answer Explanation

Splicing (the removal of introns and joining of exons) is a processing event occurring in the nucleus before transport to the cytoplasm. Translation (protein synthesis) is the process ribosomes perform in the cytoplasm.

❓ Why Other Options Are Incorrect

Options A, B, and C are correct definitions and locations of their respective regulatory levels.

🎯 Strategy

Used

Elimination: Identify the statement that misattributes a biological process (splicing) to the wrong location/machinery (ribosomes/translation).

Final Logic: Ribosomes = Protein synthesis; Spliceosome = Splicing.

🧠 Memory Trick

\\\"Translation \\\\neq Splicing.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Regulation of Gene Expression – .

3 In the Jacob and Monod model, the constitutive (all-the-time) expression of the i gene is vital. Analytically, if the i gene were heavily mutated such that its product could no longer physically interact with the operator region, what would be the metabolic state of the E. coli cell?

A The operon would be permanently shut off, even in the presence of lactose.

B The structural genes would be transcribed continuously regardless of the presence or absence of lactose.

C The cell would use glucose to induce the operon.

D RNA polymerase would immediately degrade the z gene.

💡 Short Explanation

\\\'i\\\' gene makes repressor. Mutated repressor cannot bind operator. Operator is \\\"open\\\" = constitutive expression.

✅ Correct Answer Explanation

The repressor protein (product of \\\'i\\\' gene) acts as a negative regulator by binding the operator. If it cannot interact with the operator, the \\\"brake\\\" is removed, and RNA polymerase can transcribe the structural genes constitutively.

❓ Why Other Options Are Incorrect

Option A → Describes a super-repressor or promoter mutation.
Option C → Glucose causes repression, not induction.
Option D → Polymerase does not degrade genes.

🎯 Strategy

Used

Substitution: Replace \\\"repressor function\\\" with \\\"loss of binding\\\" to predict \\\"constitutive transcription.\\\"

Final Logic: Repressor loss = Permanent \\\"ON\\\" switch.

🧠 Memory Trick

\\\"No Repressor = Always ON.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Lac Operon – .

4 Match the functional sequences in the lac operon to their dynamic analytical interactions:

Column I	Column II
1. Promoter	i. Repressor binding site
2. Operator	ii. RNA polymerase binding site
3. i gene	iii. Repressor synthesis
4. Inducer	iv. Repressor inactivation

A 1-ii, 2-i, 3-iii, 4-iv

B 1-i, 2-ii, 3-iii, 4-iv

C 1-ii, 2-i, 3-iv, 4-iii

D 1-iii, 2-i, 3-ii, 4-iv

💡 Short Explanation

Promoter = RNA Polymerase binding (ii). Operator = Repressor binding (i). i gene = Repressor synthesis (iii). Inducer = Repressor inactivation (iv).

✅ Correct Answer Explanation

The lac operon functions through precise molecular interactions: Promoter binds polymerase, Operator binds the inhibitory repressor, \\\'i\\\' gene creates that repressor, and the Inducer interacts with the repressor to turn the system on.

❓ Why Other Options Are Incorrect

They misalign the roles of the regulatory sequences with their molecular partners.

🎯 Strategy

Used

Option Grouping: Mapping regulatory elements to their binding targets/functions.

Final Logic: Promoter-Polymerase; Operator-Repressor.

🧠 Memory Trick

\\\"Promoter for Polymerase.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Lac Operon – .

5 Arrange the following events in the correct analytical sequence describing the induction and action of the structural genes in a lac operon upon the addition of lactose:
1. RNA polymerase gains access to the promoter and initiates transcription of polycistronic mRNA.
2. Lactose enters the cell via a low baseline level of permease.
3. Beta-galactosidase begins hydrolyzing lactose into galactose and glucose.
4. Lactose interacts with the constitutively produced repressor, inactivating it.

A 2 -> 4 -> 1 -> 3

B 4 -> 2 -> 1 -> 3

C 2 -> 1 -> 4 -> 3

D 1 -> 2 -> 4 -> 3

💡 Short Explanation

Lactose entry (2). Repressor inactivation (4). Transcription begins (1). Metabolism occurs (3).

✅ Correct Answer Explanation

The sequence is: Low baseline permeability allows lactose in (2) > Lactose/Allolactose binds/inactivates the repressor (4) > Operator is cleared, RNA polymerase transcribes (1) > Enzymes are produced and perform hydrolysis (3).

❓ Why Other Options Are Incorrect

Other sequences misplace the order of interaction (e.g., transcription before inactivation).

🎯 Strategy

Used

Contextual/Tonal Matching: Logical progression of metabolic signaling.

Final Logic: Entry > Inactivation > Transcription > Hydrolysis.

🧠 Memory Trick

\\\"Enter-Inactivate-Transcribe-Hydrolyze.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Lac Operon – .

6 Which of the following analytical statements is NOT an accurate deduction regarding the production of enzymes by the lac operon structural genes?

A The synthesis of beta-galactosidase is essentially regulated by its own substrate.

B The z, y, and a genes are transcribed as a single polycistronic mRNA.

C The lac operon relies on a completely different promoter for each of its three structural genes to allow for independent expression levels.

D A very low level of expression of the lac operon must be present in the cell at all times to allow initial entry of lactose.

💡 Short Explanation

The lac operon is a polycistronic unit. It shares one common promoter for all three genes.

✅ Correct Answer Explanation

The lac operon is defined by its polycistronic nature, meaning it uses one promoter for all structural genes, ensuring they are all expressed together, not independently.

❓ Why Other Options Are Incorrect

Options A, B, and D are accurate analytical deductions regarding lac operon function.

🎯 Strategy

Used

Extreme Word Filter: \\\"Completely different promoter\\\" contradicts the fundamental definition of an \\\"operon.\\\"

Final Logic: Operon = Single Promoter.

🧠 Memory Trick

\\\"One Operon = One Promoter.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Lac Operon – .

7 Consider the following statements regarding the negative regulation of the lac operon:
1. Negative regulation occurs because the repressor protein actively turns off the operon by blocking RNA polymerase.
2. The repressor binds specifically to the operator sequence, not the promoter.
3. The lac operon is exclusively under negative control and possesses no mechanisms for positive regulation. Which of the statements given above are correct analytically?

A 1 and 2 only

B 2 and 3 only

C 1 and 3 only

D 1, 2, and 3

💡 Short Explanation

1 is true (repressor blocks). 2 is true (binds operator). 3 is false (there is positive regulation by CAP-cAMP).

✅ Correct Answer Explanation

Statement 1 and 2 correctly describe negative control. Statement 3 is false because bacteria utilize positive regulation (via CAP-cAMP binding) to increase efficiency when glucose is low.

❓ Why Other Options Are Incorrect

They include Statement 3, which is biologically incorrect.

🎯 Strategy

Used

Substitution: Recognize that the lac operon involves both negative (repressor) and positive (CAP) regulation.

Final Logic: Repressor = Negative; CAP = Positive.

🧠 Memory Trick

\\\"Lac = Negative AND Positive control.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Lac Operon Regulation – .

8 If a synthetic analogue of lactose is introduced to an E. coli culture that successfully binds to and inactivates the repressor but CANNOT be hydrolyzed by beta-galactosidase, what will be the long-term transcriptional status of the operon?

A The operon will be rapidly shut down as the analogue cannot provide energy.

B The operon will remain continually induced because the analogue is never cleared by hydrolysis

C The operon will be permanently repressed because the analogue cannot be metabolized into glucose and galactose.

D The operon will show only a brief induction before the repressor regains its activity despite the analogue remaining bound

💡 Short Explanation

o Inducer (analogue) keeps repressor inactive.

o Since it\\\'s not hydrolyzed, it persists.

✅ Correct Answer Explanation

The inducer must be present to keep the operon \\\"on.\\\" If an analogue cannot be hydrolyzed, it remains in the cell indefinitely, continuously binding to the repressor and preventing it from stopping transcription.

❓ Why Other Options Are Incorrect

o Option A → Misinterprets the trigger (binding, not energy).

o Options C and D → Are biological impossibilities.

🎯 Strategy

Strategy Used:

o Contextual/Tonal Matching: Analyzing the feedback loop of operon induction.

Final Logic: Inducer persists = Repressor stays inactive = Transcription stays ON.

🧠 Memory Trick

Analogue stays = Operon stays ON

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Lac Operon Induction

9 Which of the following is NOT a logical or technological consequence derived from the challenge of sequencing the 3 billion base pair human genome?

A It required high-speed computational devices due to the sheer volume of data (Bioinformatics).

B It necessitated the development of automated DNA sequencers based on Frederick Sanger\\\'s principles.

C It required the isolation and random fragmentation of DNA because sequencing a 3 billion bp intact polymer was technically impossible.

D It proved that exactly 50% of the 3 billion base pairs code for functional proteins.

💡 Short Explanation

Less than 2% of the genome codes for proteins. Statement D is false.

✅ Correct Answer Explanation

The HGP revealed that a very small fraction (less than 2%) of the genome codes for proteins. Claiming \\\"exactly 50%\\\" is incorrect.

❓ Why Other Options Are Incorrect

A, B, and C are logically correct technological and methodological consequences of the HGP.

🎯 Strategy

Used

Extreme Word Filter: \\\"Exactly 50%\\\" is factually incorrect.

Final Logic: Coding DNA = < 2%.

🧠 Memory Trick

\\\"Coding = Small (<2%).\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: HGP Findings – .

10 Analytically, what was a surprising outcome regarding the HGP goal to identify all human genes, given the initial scientific expectations prior to the project\\\'s completion?

A Scientists discovered that the human genome contained nearly 1,40,000 genes, far exceeding estimates.

B The total number of identified genes was roughly 30,000, which was much lower than the previous estimates of 80,000 to 1,40,000.

C It was found that 99% of the genome coded directly for proteins.

D Scientists identified that chromosome Y had the highest concentration of functional genes.

💡 Short Explanation

Initial estimates were high (80-140k). Final result was low (~30k).

✅ Correct Answer Explanation

Prior to the HGP, it was widely hypothesized that the human genome would contain many more genes to account for human complexity. The discovery that there were only about 20,000–30,000 genes was a major and surprising finding.

❓ Why Other Options Are Incorrect

A, C, and D are factually false.

🎯 Strategy

Used

Contextual/Tonal Matching: Recalling a major HGP surprise finding.

Final Logic: Surprise = Low gene count.

🧠 Memory Trick

\\\"HGP = 30k genes (lower than expected).\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: HGP Findings

11 Consider the following statements differentiating HGP methodologies:
1. Expressed Sequence Tags (ESTs) focus selectively on identifying genes that are transcribed into RNA.
2. Sequence Annotation is a blind approach that involves sequencing all coding and non-coding parts of the genome before assigning functions.
3. Sequence Annotation requires the fragmentation and amplification of DNA using vectors like BAC and YAC.
Which of the statements given above are correct?

A 1 and 2 only

B 2 and 3 only

C 1 and 3 only

D 1, 2, and 3

💡 Short Explanation

ESTs target expressed genes (RNA). Annotation is \\\"blind\\\"/whole-genome. Annotation uses cloning vectors (BAC/YAC).

✅ Correct Answer Explanation

All three statements correctly define the dual approaches used in HGP. ESTs target RNA-transcribed sequences, while Sequence Annotation covers the entire genome blindly using cloning vectors for sequencing.

❓ Why Other Options Are Incorrect

They are incomplete as all three statements provided are factually correct descriptions of HGP methodologies.

🎯 Strategy

Used

Contextual/Tonal Matching: Identifying accurate descriptions of genomic mapping techniques.

Final Logic: All statements are standard HGP facts.

🧠 Memory Trick

\\\"EST=RNA; Annotation=Everything/Vectors.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: HGP Methodology – .

12 Match the specific cloning tools to their functional roles in the Sequence Annotation methodology of the HGP:

Column I	Column II
1. Bacteria and Yeast	i. Vectors used to clone and amplify DNA pieces
2. BAC and YAC	ii. Used to align overlapping sequences generated during sequencing
3. Automated DNA sequencers	iii. Hosts used for the cloning of fragmented DNA
4. Specialised computer-based programs	iv. Devices working on Sanger\\\'s principle to sequence DNA

A 1-iii, 2-i, 3-iv, 4-ii

B 1-i, 2-iii, 3-ii, 4-iv

C 1-iii, 2-ii, 3-iv, 4-i

D 1-iv, 2-i, 3-iii, 4-ii

💡 Short Explanation

Hosts = Bacteria/Yeast (iii). Vectors = BAC/YAC (i). Sequencers = Sanger\\\'s (iv). Computer = Alignment (ii).

✅ Correct Answer Explanation

This aligns the laboratory tools (hosts, vectors, sequencers) and bioinformatic tools with their specific biological or analytical functions in the sequencing pipeline.

❓ Why Other Options Are Incorrect

They incorrectly pair the tools with their functions (e.g., swapping host/vector).

🎯 Strategy

Used

Option Grouping: Aligning technical terms with their roles.

Final Logic: Host-Host; Vector-Vector; Machine-Sequencing; Software-Alignment.

🧠 Memory Trick

\\\"Host=Bacteria, Vector=BAC/YAC.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: HGP Methodology – .

13 If the human genome contains 3164.7 million base pairs, the average gene is 3000 bases, and there are roughly 30,000 genes, what does this mathematically imply regarding the overall composition of the human genome?

A The vast majority of the 3164.7 million base pairs do not consist of protein-coding gene sequences.

B Over 90% of the genome consists of active structural genes.

C The dystrophin gene at 2.4 million bases accounts for 50% of the genome.

D Chromosome 1 contains exactly 30,000 genes.

💡 Short Explanation

Calculation: 30,000 \\\\times 3,000 = 90 million bp (coding). Total is ~3,164 million bp. Coding fraction is very small.

✅ Correct Answer Explanation

Multiplying 30,000 genes by an average of 3,000 bases per gene gives ~90 million base pairs, which is less than 3% of the total genome size of 3,164.7 million bp. Thus, most of the genome is non-coding.

❓ Why Other Options Are Incorrect

B, C, and D contradict the mathematical reality of genomic coding fraction.

🎯 Strategy

Used

Dimensional/Unit Analysis: Comparing the calculated coding sequence to the total genomic size.

Final Logic: 90 < 3164; therefore, most is non-coding.

🧠 Memory Trick

\\\"Coding is <3%, most is non-coding.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Salient Features – .

14 Which of the following is NOT an analytical deduction regarding repetitive sequences based on HGP findings?

A Despite having no direct coding functions, they are retained in the genome because they shed light on chromosome dynamics.

B They are represented by stretches of DNA that can be repeated hundreds to thousands of times.

C They are the primary reason why less than 2 percent of the genome codes for proteins.

D They consist entirely of Single Nucleotide Polymorphisms (SNPs) which code for functional enzymes.

💡 Short Explanation

Repetitive DNA \\\\neq SNP. Repetitive DNA is generally non-coding.

✅ Correct Answer Explanation

Repetitive sequences are large segments of non-coding DNA, not SNPs (which are single-base variations). They certainly do not code for functional enzymes.

❓ Why Other Options Are Incorrect

A, B, and C are accurate deductions regarding repetitive DNA.

🎯 Strategy

Used

Elimination: Identifying the false statement regarding the definition of repetitive DNA.

Final Logic: Repetitive DNA = Non-coding, NOT SNP-based.

🧠 Memory Trick

\\\"Repeats \\\\neq SNP.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Salient Features – .

15 In the context of genetic polymorphism, how does a Single Nucleotide Polymorphism (SNP) fundamentally differ analytically from the polymorphism used in DNA fingerprinting (VNTRs)?

A SNPs involve a single-base DNA difference, whereas VNTRs involve variations in the copy number of a tandemly repeated DNA sequence.

B SNPs are only found in RNA, while VNTRs are found in DNA.

C SNPs code for proteins, whereas VNTRs are exclusively located in the promoter region.

D VNTRs are used to trace human history, while SNPs are exclusively used for paternity testing.

💡 Short Explanation

SNP = Single nucleotide change. VNTR = Repeat number change.

✅ Correct Answer Explanation

This defines the two primary forms of human genomic polymorphism accurately.

❓ Why Other Options Are Incorrect

B, C, and D contain incorrect biological premises.

🎯 Strategy

Used

Contextual/Tonal Matching: Identifying the core definition of genomic polymorphism types.

Final Logic: SNP = Base; VNTR = Repeat count.

🧠 Memory Trick

\\\"SNP = Single; VNTR = Repeat.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Genetic Polymorphism – .

16 Arrange the following evolutionary steps to demonstrate how a silent genetic variation analytically transitions into a useful polymorphic marker for a population:
1. The mutation is transmitted to offspring without impairing reproductive ability.
2. An individual develops a mutation in a non-coding DNA sequence within a germ cell.
3. The allelic variant establishes a frequency greater than 0.01 in the human population.
4. The mutation spreads extensively through sexual reproduction across generations.

A 2 -> 1 -> 4 -> 3

B 2 -> 4 -> 1 -> 3

C 1 -> 2 -> 3 -> 4

D 4 -> 2 -> 1 -> 3

💡 Short Explanation

Mutation occurs (2). Inheritance (1). Spreads (4). Reaches polymorphic threshold (3).

✅ Correct Answer Explanation

Genetic variation starts with a germ-cell mutation (2). If non-lethal (1), it spreads via sexual reproduction (4). Once it becomes prevalent (>0.01), it is officially a polymorphism (3).

❓ Why Other Options Are Incorrect

They misorder the progression from initial mutation to population-wide marker status.

🎯 Strategy

Used

Contextual/Tonal Matching: Logical evolutionary process.

Final Logic: Mutation > Survival > Spread > Frequency.

🧠 Memory Trick

\\\"Mutate-Survive-Spread-Frequency.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Genetic Polymorphism – .

📄 Passage

It differs from individual to individual in a population except in the case of monozygotic (identical) twins. The sensitivity of the technique has been increased by use of polymerase chain reaction (PCR). Consequently, DNA from a single cell is enough to perform DNA fingerprinting analysis. In addition to application in forensic science, it has much wider application, such as in determining population and genetic diversities.

Analytically connecting the passage to the principles of VNTRs, why does the DNA fingerprinting pattern differ from individual to individual (excluding identical twins)?

A Because every individual has completely unique restriction endonucleases that cut DNA randomly.

B Because the copy number of tandem repeats at specific chromosomes inherited from parents is highly polymorphic and unique to an individual\\\'s lineage.

C Because the PCR technique introduces random mutations during amplification.

D Because monozygotic twins do not possess satellite DNA.

💡 Short Explanation

VNTR = Variable Number of Tandem Repeats. Inheritance + Mutation = Uniqueness.

✅ Correct Answer Explanation

Polymorphism in VNTRs arises because the repeat number varies, creating unique fragment patterns that are inherited.

❓ Why Other Options Are Incorrect

A, C, and D are incorrect biological claims.

🎯 Strategy

Used

Contextual/Tonal Matching: Defining polymorphism source (VNTRs).

Final Logic: VNTR = Inheritance of specific repeat counts.

🧠 Memory Trick

\\\"VNTR = Lineage-specific repeats.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: DNA Fingerprinting – .

📄 Passage

Which analytical combination of factors dictates whether a specific repetitive sequence is classified as a micro-satellite or mini-satellite?

A Base composition (A:T rich or G:C rich), length of the segment, and number of repetitive units.

B The presence of exons, introns, and promoter regions.

C The ability to code for beta-galactosidase and permease.

D Only the overall size of the bulk genomic DNA peak.

💡 Short Explanation

Satellite DNA classification depends on repeat size and composition.

✅ Correct Answer Explanation

Satellite DNA classification as mini- or micro-satellite is based on the length of the repetitive unit and the number of repetitions.

❓ Why Other Options Are Incorrect

B, C, and D are irrelevant to the structural classification of satellite DNA.

🎯 Strategy

Used

Contextual/Tonal Matching: Recall structural biology of satellite DNA.

Final Logic: Classification = Length/Composition.

🧠 Memory Trick

\\\"Satellite = Length + Composition.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: Satellite DNA – .

📄 Passage

According to the passage\\\'s analytical premise, why is DNA fingerprinting fundamentally ineffective in distinguishing between monozygotic (identical) twins in a forensic setting?

A Because monozygotic twins share identical germ-line mutations and therefore possess the exact same copy number of VNTRs across their chromosomes.

B Because PCR amplification fails on DNA derived from identical twins.

C Because identical twins do not possess any repetitive DNA sequences.

D Because the DNA from monozygotic twins degrades instantly upon isolation.

💡 Short Explanation

Twins share the same DNA/genotype. Fingerprinting relies on DNA genotype.

✅ Correct Answer Explanation

Monozygotic twins develop from a single fertilized egg, resulting in identical genetic material, including the number of VNTRs, so their DNA fingerprints are identical.

❓ Why Other Options Are Incorrect

B, C, and D are false scientific statements.

🎯 Strategy

Used

Contextual/Tonal Matching: Recognizing the implication of \\\"identical\\\" in genetics.

Final Logic: Same DNA = Same Fingerprint.

🧠 Memory Trick

\\\"Identical Twins = Identical VNTRs.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: DNA Fingerprinting – .

📄 Passage

Based on the passage, the introduction of the Polymerase Chain Reaction (PCR) fundamentally improved the DNA fingerprinting technique by addressing which critical technical limitation?

A The inability to transfer DNA from gels to nylon membranes.

B The requirement for massive amounts of starting biological material, increasing sensitivity so a single cell is now sufficient.

C The inability of VNTR probes to become radiolabelled.

D The cross-reactivity of human DNA with bacterial vectors like BACs.

💡 Short Explanation

PCR = Amplification. Solves the \\\"low sample\\\" issue.

✅ Correct Answer Explanation

Before PCR, you needed a large amount of sample to detect DNA markers. PCR allows researchers to replicate tiny amounts (like from a single cell) until they are detectable.

❓ Why Other Options Are Incorrect

A, C, and D are not technical issues solved by PCR.

🎯 Strategy

Used

Contextual/Tonal Matching: Identifying the main utility of PCR in diagnostics.

Final Logic: PCR = Amplification = High Sensitivity.

🧠 Memory Trick

\\\"PCR = Amplify = Sensitive.\\\"

📚 Class XII – Biology – Chapter: Molecular Basis of Inheritance – Topic: DNA Fingerprinting/PCR

CUET Booster Biology Unit 5 Test (D5)

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