CUET UG Biology Booster Unit 5 Test (M5) -

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CUET Booster Biology Unit 5 Test (M5)

⏱️ 30 Minutes

❓ 20 Questions

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

CUET Booster Biology Unit 5 Test (M5)

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QUESTION 1 OF 20

Arrange the following regulatory events in the correct chronological order as they occur in a eukaryotic cell leading to successful protein synthesis:
1. Regulation of splicing
2. Transport of mRNA to the cytoplasm
3. Formation of the primary transcript
4. Translational level regulation

Answer Review

1 Arrange the following regulatory events in the correct chronological order as they occur in a eukaryotic cell leading to successful protein synthesis:
1. Regulation of splicing
2. Transport of mRNA to the cytoplasm
3. Formation of the primary transcript
4. Translational level regulation

A 3 -> 1 -> 2 -> 4

B 1 -> 3 -> 2 -> 4

C 3 -> 2 -> 1 -> 4

D 1 -> 2 -> 3 -> 4

💡 Short Explanation

Transcription forms primary transcript (3). Splicing/processing occurs (1). Mature mRNA is exported (2). Translation occurs at ribosomes (4).

✅ Correct Answer Explanation

In eukaryotes, gene expression is multi-layered. First, DNA is transcribed into pre-mRNA (primary transcript). Next, the pre-mRNA undergoes processing (splicing to remove introns). The mature mRNA is then transported out of the nucleus into the cytoplasm, where translation occurs.

❓ Why Other Options Are Incorrect

Options B, C, and D propose incorrect sequences, such as splicing or translation occurring before the primary transcript is formed.

🎯 Strategy

Used

Contextual/Tonal Matching: Following the standard Central Dogma flow of eukaryotic gene expression.

Final Logic: Transcription > Splicing > Transport > Translation.

🧠 Memory Trick

"Transcript-Spliced-Transferred-Translated."

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

2 If a eukaryotic cell successfully transcribes and splices a gene but completely fails to transport the mature mRNA from the nucleus to the cytoplasm, which subsequent level of gene expression will be directly and immediately prevented?

A Formation of primary transcript

B Transcriptional level regulation

C Translational level

D Processing level regulation

💡 Short Explanation

Translation happens in the cytoplasm. mRNA in the nucleus cannot reach ribosomes.

✅ Correct Answer Explanation

Translation requires ribosomes, which are located in the cytoplasm. If mRNA cannot leave the nucleus, it cannot encounter the translation machinery, directly halting the translational level of gene expression.

❓ Why Other Options Are Incorrect

A, B, and D occur before or during the transport stage, so they are not prevented by the failure of transport.

🎯 Strategy

Used

Elimination: Identify the step that physically requires the mRNA to be in the cytoplasm.

Final Logic: mRNA in nucleus cannot be translated.

🧠 Memory Trick

"No exit = No translation."

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

3 The Jacob and Monod model describes a polycistronic structural gene regulated by a common promoter. Why is this specific arrangement metabolically advantageous for bacteria like E. coli?

A It allows genes needed together in the same metabolic pathway to be regulated simultaneously.

B It ensures that the repressor is never synthesized.

C It allows the bacteria to utilize glucose as an inducer.

D It prevents transcription entirely when lactose is present.

💡 Short Explanation

Polycistronic arrangement coordinates pathway expression. Saves energy by turning off unused genes.

✅ Correct Answer Explanation

The operon model is efficient because it groups genes with related functions under a single regulatory control system. This allows the bacterium to switch on or off an entire metabolic pathway in response to environmental conditions.

❓ Why Other Options Are Incorrect

B, C, and D are factually incorrect regarding the biology of the lac operon.

🎯 Strategy

Used

Contextual/Tonal Matching: Understanding the biological purpose of an operon.

Final Logic: Polycistronic = Coordinated regulation.

🧠 Memory Trick

"Grouping = Coordination."

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

4 Which of the following is NOT true regarding the regulatory and operator sequences in prokaryotic operons?

A The operator region is adjacent to the promoter elements in most operons.

B Accessibility of promoter regions is regulated by the interaction of proteins with operators.

C Every operon in a bacterium utilizes the exact same operator and repressor protein.

D The lac operator interacts specifically with the lac repressor only.

💡 Short Explanation

Each operon is unique. Specific repressor proteins bind to specific operators.

✅ Correct Answer Explanation

Operons are highly specific. The lac operon has a unique operator sequence that recognizes only the lac repressor protein. Different operons have different regulators.

❓ Why Other Options Are Incorrect

A, B, and D are standard accurate descriptions of prokaryotic operon architecture.

🎯 Strategy

Used

Extreme Word Filter: "Every operon... same" is a false generalization.

Final Logic: Specificity is key in genetic regulation.

🧠 Memory Trick

"One operon = One repressor."

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

5 Match the specific mutations in E. coli lac structural genes to their direct physiological consequences:

Column I	Column II
1. Defective z gene	i. Repressor binding
2. Defective y gene	ii. Permease (transport)
3. Defective a gene	iii. Transacetylase (acetylation)
4. Defective i gene	iv. β-galactosidase (hydrolysis)

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

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

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

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

💡 Short Explanation

z = beta-galactosidase (hydrolysis). y = permease (transport). a = transacetylase (acetylation). i = repressor (binding).

✅ Correct Answer Explanation

Each structural gene codes for a specific protein. z codes for the enzyme breaking down lactose, y for the pump bringing it in, and a for the acetyl-transferring enzyme. Defect in i leads to constant expression as the repressor cannot bind.

❓ Why Other Options Are Incorrect

They misalign the structural gene function with its physiological result.

🎯 Strategy

Used

Option Grouping: Pairing genes to their specific protein function.

Final Logic: Match enzyme function to the corresponding gene.

🧠 Memory Trick

"Z=Hydrolysis, Y=Transport, A=Acetyl, I=Repress."

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

6 In an E. coli culture grown without glucose, if a spontaneous mutation completely inactivates the y gene (permease), what will occur when lactose is suddenly added to the surrounding medium?

A The operon will be rapidly induced because lactose diffuses freely without permease.

B Beta-galactosidase will immediately hydrolyze the external lactose.

C Lactose cannot enter the cells efficiently, thereby failing to induce the operon.

D The z gene will mutate to take over the function of permease.

💡 Short Explanation

Permease is needed to bring lactose in. No induction without internal lactose.

✅ Correct Answer Explanation

The lac operon requires a small amount of lactose to enter the cell to act as an inducer. Permease (y gene product) is essential for this transport. Without it, lactose levels in the cell remain too low to inactivate the repressor.

❓ Why Other Options Are Incorrect

A → Lactose does not diffuse freely. B → Enzymes are intracellular. D → Genes do not swap functions.

🎯 Strategy

Used

Elimination: Based on the physiological necessity of permease.

Final Logic: No transport = No induction.

🧠 Memory Trick

"Permease = Gateway."

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

7 Consider the following statements regarding the repressor in the lac operon:
Statement I: The repressor protein is synthesized all-the-time (constitutively) from the i gene.
Statement II: In the presence of allolactose, the repressor is activated to bind to the operator.

A Both Statement I and II are correct

B Statement I is correct, but Statement II is incorrect

C Statement I is incorrect, but Statement II is correct

D Both Statement I and II are incorrect

💡 Short Explanation

Statement I is true (constitutive expression). Statement II is false (allolactose inactivates the repressor).

✅ Correct Answer Explanation

The 'i' gene is expressed constitutively. However, allolactose is an inducer that inactivates the repressor (it prevents the repressor from binding the operator), it does not activate it.

❓ Why Other Options Are Incorrect

Options A, C, and D are based on the false claim that the inducer activates the repressor.

🎯 Strategy

Used

Substitution: Replace "activated" with "inactivated" in Statement II.

Final Logic: Inducer = Inactivation of Repressor.

🧠 Memory Trick

"Inducer = Inactivator."

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

8 Which of the following statements does NOT correctly describe the inducer's role in the lac operon?

A Lactose is the substrate for beta-galactosidase and regulates the switching on and off of the operon.

B The inducer interacts with the repressor, inactivating it.

C The inducer directly binds to the RNA polymerase to initiate transcription.

D Inactivation of the repressor by the inducer allows RNA polymerase access to the promoter.

💡 Short Explanation

Inducer binds to repressor, not RNA polymerase.

✅ Correct Answer Explanation

The inducer binds to the repressor protein, causing a conformational change that prevents it from binding to the operator. It does not interact with RNA polymerase.

❓ Why Other Options Are Incorrect

A, B, and D are accurate descriptions of the inducer mechanism.

🎯 Strategy

Used

Elimination: Identify the false interaction mechanism (inducer-polymerase).

Final Logic: Inducer > Repressor, NOT Polymerase.

🧠 Memory Trick

"Inducer hits Repressor."

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

9 Given that the human genome has approximately 3 x 10^9 base pairs, why did the Human Genome Project necessitate the rapid development of Bioinformatics?

A Because 99.9% of the base pairs are polymorphic and require manual alignment.

B Because the enormous amount of data required high-speed computational devices for storage, retrieval, and analysis.

C Because Bioinformatics was required to genetically engineer bacterial vectors.

D Because the ethical, legal, and social issues could only be solved by computers.

💡 Short Explanation

3 billion base pairs is massive data. Bioinformatics is needed for processing big data.

✅ Correct Answer Explanation

The HGP generated a massive amount of genomic data (billions of base pairs). Manual analysis was impossible, making sophisticated computational tools (Bioinformatics) essential.

❓ Why Other Options Are Incorrect

A, C, and D attribute incorrect or secondary reasons for the rise of Bioinformatics.

🎯 Strategy

Used

Contextual/Tonal Matching: Recognizing the role of technology in large-scale science.

Final Logic: Large data = Computational requirement.

🧠 Memory Trick

"Big Data = Bioinformatics."

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

10 Match the non-human model organisms sequenced as a result of expanding HGP goals to their descriptions:

Column I	Column II
1. Caenorhabditis elegans	i. Nematode
2. Drosophila	ii. Plant
3. Arabidopsis	iii. Fruit fly
4. Yeast	iv. Single-celled eukaryote

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

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

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

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

💡 Short Explanation

C. elegans = Nematode (i). Drosophila = Fruit fly (iii). Arabidopsis = Plant (ii). Yeast = Single-celled eukaryote (iv).

✅ Correct Answer Explanation

The HGP included sequencing other model organisms to facilitate comparative genomics. The matches provided in option A are correct definitions for these organisms.

❓ Why Other Options Are Incorrect

They misidentify the types of organisms.

🎯 Strategy

Used

Option Grouping: Associating organism names with their scientific classifications.

Final Logic: Correct classification matching.

🧠 Memory Trick

"Fly = Drosophila, Nematode = C. elegans."

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

11 If a genomics researcher strictly wants to identify and sequence only the segments of DNA that are responsible for coding cellular proteins, which specific methodology from the HGP would be the most efficient primary approach?

A Single Nucleotide Polymorphism mapping

B Variable Number of Tandem Repeats

C Expressed Sequence Tags (ESTs)

D Sequence Annotation of the entire genome

💡 Short Explanation

ESTs focus on RNA-transcribed sequences. RNA corresponds to expressed genes/proteins.

✅ Correct Answer Explanation

ESTs are focused on identifying the genes that are expressed. Since only coding sequences (exons) are typically expressed, this method effectively isolates protein-coding DNA segments for sequencing.

❓ Why Other Options Are Incorrect

A, B, and D are methodologies meant for mapping polymorphisms or the entire genome, not specifically for targeting expressed coding sequences.

🎯 Strategy

Used

Substitution: "Expressed sequence" = "Protein-coding gene."

Final Logic: ESTs map expressed RNA sequences.

🧠 Memory Trick

"EST = Expressed."

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

12 Which of the following is NOT a characteristic of the Sequence Annotation approach used in the Human Genome Project?

A It is a blind approach sequencing the whole set of the genome.

B It exclusively targets sequences that are expressed as RNA.

C It involves converting total DNA into random fragments and cloning them.

D It involves assigning different regions in the sequence with functions after sequencing.

💡 Short Explanation

Sequence Annotation is "blind" (whole genome), not restricted to expressed genes.

✅ Correct Answer Explanation

Sequence annotation is a holistic approach to sequencing the entire genome. ESTs are the method that exclusively targets expressed sequences. Therefore, B is false.

❓ Why Other Options Are Incorrect

A, C, and D are valid characteristics of the Sequence Annotation method.

🎯 Strategy

Used

Elimination: Differentiate between "blind sequencing" and "targeted sequencing."

Final Logic: Annotation = Everything, not just expression.

🧠 Memory Trick

"Annotation = Whole."

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

13 Consider the following statements drawn from the salient features of the human genome:
1. Chromosome 1 has the most genes (2968).
2. The Y chromosome has the fewest genes (231).
3. The largest known human gene is dystrophin at 2.4 million bases. 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

All three statements are NCERT facts.

✅ Correct Answer Explanation

These represent the specific salient features listed in the NCERT textbook: Chromosome 1 has 2968 genes, Y has 231 genes, and dystrophin is the largest at 2.4 million bases.

❓ Why Other Options Are Incorrect

They exclude one or more of the correct factual statements provided.

🎯 Strategy

Used

Contextual/Tonal Matching: Verifying facts against the textbook chapter data.

Final Logic: All listed facts are correct.

🧠 Memory Trick

"1-max, Y-min, Dystrophin-giant."

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

14 Which of the following is NOT a true conclusion drawn about repetitive DNA sequences by the Human Genome Project?

A They are thought to have no direct coding functions.

B They make up a very small, insignificant portion of the human genome.

C They are stretched of DNA repeated sometimes hundred to thousand times.

D They shed light on chromosome structure, dynamics, and evolution.

💡 Short Explanation

Repeated DNA makes up a very large portion of the genome.

✅ Correct Answer Explanation

Repetitive DNA sequences are a major component of the human genome, not a small/insignificant one.

❓ Why Other Options Are Incorrect

A, C, and D are all accurate descriptions of repetitive DNA according to the HGP findings.

🎯 Strategy

Used

Extreme Word Filter: "Small, insignificant" contradicts genomic reality.

Final Logic: Repeats = Large Genomic Fraction.

🧠 Memory Trick

"Repeats = Vast."

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

15 In the preparation of DNA for fingerprinting, how is satellite DNA physically distinguished from the bulk genomic DNA?

A By utilizing radioactive EST probes

B Through separation as different peaks during density gradient centrifugation

C By sequencing the entire 3 billion base pairs

D Through visual inspection under an electron microscope

💡 Short Explanation

Bulk DNA = Major peak. Satellite = Minor peaks due to density differences.

✅ Correct Answer Explanation

During density gradient centrifugation, bulk DNA (the majority) forms a large peak. Satellite DNA, due to different base composition (A:T/G:C ratio), forms smaller, distinct peaks.

❓ Why Other Options Are Incorrect

A, C, and D are not the physical methods used to isolate/distinguish the satellite peak from the bulk genome.

🎯 Strategy

Used

Contextual/Tonal Matching: Recalling the methodology of DNA isolation/separation.

Final Logic: Density = Separation method.

🧠 Memory Trick

"Density = Satellite."

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

16 Arrange the following factors contributing to satellite DNA classification into a logical sequence based on structural analysis:
1. Analysis of the length of the segment;
2. Separation of satellite DNA from bulk DNA via centrifugation;
3. Determination of the number of repetitive units;
4. Classification into micro-satellites or mini-satellites

A 2 -> 1 -> 3 -> 4

B 1 -> 2 -> 4 -> 3

C 3 -> 1 -> 2 -> 4

D 2 -> 3 -> 1 -> 4

💡 Short Explanation

Isolate DNA (2). Measure segment length (1). Count repeats (3). Classify (4).

✅ Correct Answer Explanation

First, the satellite DNA is isolated (2). Then, the segment's length is determined (1), followed by identifying the number of repeats (3). Finally, based on these parameters, it is classified as mini- or micro-satellite (4).

❓ Why Other Options Are Incorrect

They misorder the sequence of steps for classification.

🎯 Strategy

Used

Contextual/Tonal Matching: Logical scientific process of identification and classification.

Final Logic: Isolate > Size/Length > Repeats > Categorize.

🧠 Memory Trick

"Isolate-Analyze-RepeatCount-Label."

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

📄 Passage

The VNTR belongs to a class of satellite DNA referred to as mini-satellite. A small DNA sequence is arranged tandemly in many copy numbers. The copy number varies from chromosome to chromosome in an individual. The numbers of repeat show very high degree of polymorphism. As a result the size of VNTR varies in size from 0.1 to 20 kb. Consequently, after hybridisation with VNTR probe, the autoradiogram gives many bands of differing sizes.

According to the passage, why does the autoradiogram yield multiple bands of differing sizes when a VNTR probe is hybridized to an individual's DNA?

A Because restriction endonucleases completely destroy the VNTR sequences.

B Because the copy number of the tandem repeats varies from chromosome to chromosome.

C Because the VNTR probe specifically binds to protein-coding exons only.

D Because the size of VNTRs is uniformly 20 kb across all chromosomes.

💡 Short Explanation

Copy number variation causes different fragment lengths.

✅ Correct Answer Explanation

The passage states the copy number varies from chromosome to chromosome, leading to VNTRs of different sizes. This results in varying band positions in the autoradiogram.

❓ Why Other Options Are Incorrect

A, C, and D contradict the passage's explanation of polymorphism.

🎯 Strategy

Used

Contextual/Tonal Matching: Direct retrieval from the passage.

Final Logic: Variable copy number = Variable fragment size.

🧠 Memory Trick

"Copy number change = Band size change."

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

📄 Passage

Based strictly on the provided passage, to which specific sub-category of satellite DNA does VNTR belong?

A Micro-satellite

B Bulk genomic DNA

C Mini-satellite

D Expressed Sequence Tag

💡 Short Explanation

Passage explicitly says: "VNTR belongs to a class of satellite DNA referred to as mini-satellite."

✅ Correct Answer Explanation

The text explicitly identifies VNTR as a type of mini-satellite.

❓ Why Other Options Are Incorrect

They are either other categories or irrelevant terms not mentioned in that context in the passage.

🎯 Strategy

Used

Contextual/Tonal Matching: Direct extraction of information from the passage text.

Final Logic: Passage definition = Mini-satellite.

🧠 Memory Trick

"VNTR = Mini-satellite."

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

19 In the methodology of DNA fingerprinting, what is the primary purpose of the autoradiography step?

A To digest the DNA into smaller fragments

B To separate DNA fragments by their size

C To transfer DNA to a nylon membrane

D To detect the DNA fragments that have hybridized with the labelled VNTR probe

💡 Short Explanation

Autoradiography is the visualization step. Detects radioactive hybridization.

✅ Correct Answer Explanation

The radiolabelled VNTR probe allows visualization of the fragments on X-ray film via autoradiography, which reveals the specific fingerprint pattern.

❓ Why Other Options Are Incorrect

A, B, and C refer to earlier stages in the fingerprinting protocol (digestion, electrophoresis, blotting).

🎯 Strategy

Used

Contextual/Tonal Matching: Mapping the step name to its specific function in the process.

Final Logic: Autoradiography = Detection.

🧠 Memory Trick

"Autoradiograph = Reveal."

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

20 How has the application of DNA fingerprinting in forensic science been revolutionized to require only DNA from a single cell?

A By switching from VNTR to EST probes

B By using the polymerase chain reaction (PCR) to increase sensitivity

C By completely sequencing the human genome at a crime scene

D By eliminating the need for gel electrophoresis

💡 Short Explanation

PCR amplifies DNA. Allows forensic analysis from tiny samples.

✅ Correct Answer Explanation

PCR is a powerful tool that replicates specific DNA sequences millions of times, allowing even a single cell's DNA to be amplified to a level detectable by current fingerprinting techniques.

❓ Why Other Options Are Incorrect

A, C, and D are either ineffective or impossible methods for forensic DNA amplification.

🎯 Strategy

Used

Contextual/Tonal Matching: Connecting scientific method (PCR) to its specific application (Sensitivity/Amplification).

Final Logic: PCR = Amplification/Sensitivity.

🧠 Memory Trick

"PCR = Amplify."

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

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