GSEB Class 12 Biology Solutions Chapter 6 Molecular Basis of Inheritance

GSEB Solutions Class 12 Biology Chapter 6 Molecular Basis of Inheritance

GSEB Class 12 Biology Molecular Basis of Inheritance Text Book Questions and Answers

Question 1. Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer: From the given list, adenine, thymine, uracil, and cytosine are classified as nitrogenous bases. Meanwhile, cytidine and guanosine are categorized as nucleosides.
In simple words: Nitrogenous bases are the A, T, C, G, U building blocks, while nucleosides are a base combined with a sugar.

🎯 Exam Tip: Understanding the difference between bases and nucleosides is fundamental for nucleic acid structure questions.

 

Question 2. If a double stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
Answer: Given that the percentage of cytosine in a double-stranded DNA is 20%, it is known that cytosine always pairs with guanine. Therefore, the percentage of guanine must also be 20%. The combined percentage of cytosine and guanine is \(20\% + 20\% = 40\%\). The remaining DNA comprises adenine and thymine, totaling \(100\% - 40\% = 60\%\). Since adenine and thymine also pair in equal proportions, the percentage of adenine is half of this remaining amount, which is \(60\% / 2 = 30\%\).
In simple words: Based on DNA pairing rules, if cytosine is 20%, then guanine is also 20%. The remaining 60% is split equally between adenine and thymine, making adenine 30%.

🎯 Exam Tip: Remember Chargaff's rules for DNA base pairing (A=T, G=C) to solve such percentage calculations accurately.

 

Question 3. If the sequence of one strand of DNA is written as follows: 5'-ATGCATGCATGCATGCATGCATGC- 3' Write down the sequence of complementary strand in 5′ → 3' direction.
Answer: Given the DNA strand sequence 5'-ATGCATGCATGCATGCATGCATGC-3', the complementary strand will have opposite polarity and base pairing (A with T, G with C).
First, determine the complementary sequence in the 3' → 5' direction:
Original: 5'-ATGCATGCATGCATGCATGCATGC-3'
Complementary: 3'-TACGTACGTACGTACGTACGTACG-5'
To present this in the requested 5' → 3' direction, we reverse the sequence and read it from the other end:
Correct Complementary Strand (5' → 3'): 5'-GCATGCATGCATGCATGCATGCAT-3'
In simple words: To find the complementary strand in the 5'-3' direction, first write its complement in 3'-5' and then reverse it.

🎯 Exam Tip: Pay close attention to the requested direction (5' → 3' or 3' → 5') when writing complementary DNA or RNA sequences to avoid common errors.

 

Question 4. If the sequence of the coding strand in a transcription unit is written as follows: 5'-ATGCATGCATGCATGCATGCATGC- 3' Write down the sequence of mRNA.
Answer: When transcribing from a DNA coding strand, the resulting mRNA sequence will be identical to the coding strand, with the key difference being that all thymine (T) bases are replaced by uracil (U). Therefore, if the coding strand sequence is 5'-ATGCATGCATGCATGCATGCATGC-3', the corresponding mRNA sequence will be 5'-AUGC AUGC AUGC AUGC AUGC AUGC AUGC-3'.
In simple words: mRNA sequence is almost identical to the coding DNA strand, just replace T with U.

🎯 Exam Tip: Remember that the mRNA sequence matches the DNA coding strand (with T replaced by U), not the DNA template strand.

 

Question 5. Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer: In 1953, James Watson and Francis Crick elucidated that the precise base pairing between nitrogenous bases within DNA strands allows the two strands to unwind and function as templates for synthesizing new, complementary strands. This inherent characteristic, where each original strand serves as a guide for new synthesis, forms the fundamental basis of the semi-conservative mode of DNA replication, a mechanism they hypothesized.
In simple words: Watson and Crick realized that DNA's two strands could separate, and each could serve as a blueprint to make a new complementary strand, leading to the idea of semi-conservative replication.

🎯 Exam Tip: The specific base pairing (A-T, G-C) is the crucial property underlying semi-conservative replication and is often a key point in explanations.

 

Question 6. Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesized from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer: The types of nucleic acid polymerases are categorized based on the template used and the product synthesized:
* **For DNA Template:** * DNA polymerases are crucial for DNA replication, synthesizing new DNA strands from a DNA template. * **For RNA Template:** * RNA-dependent RNA polymerases are found in certain RNA viruses, facilitating the synthesis of RNA using an RNA template. * Reverse transcriptase is an enzyme that synthesizes DNA from an RNA template.
In simple words: Polymerases are enzymes that build nucleic acids. They are named based on what they use as a template (DNA or RNA) and what they produce (DNA or RNA).

🎯 Exam Tip: Differentiate between DNA polymerases (DNA to DNA), RNA polymerases (DNA to RNA), RNA-dependent RNA polymerases (RNA to RNA), and reverse transcriptase (RNA to DNA).

 

Question 7. How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख हर्शे और चेज़ के प्रयोग को दर्शाता है जिसने साबित किया कि डीएनए आनुवंशिक सामग्री है। इसमें दो समानांतर प्रयोग शामिल हैं: एक में फ़ेज प्रोटीन को रेडियोधर्मी सल्फर-35 (³⁵S) से लेबल किया जाता है, और दूसरे में फ़ेज डीएनए को रेडियोधर्मी फॉस्फोरस-32 (³²P) से लेबल किया जाता है। प्रयोग के चरणों में जीवाणु संक्रमण, फिर सेंट्रीफ्यूगेशन से पहले कणों को अलग करने के लिए मिश्रण (ब्लेंडिंग) शामिल है, जो यह दर्शाता है कि ³²P जीवाणु कोशिकाओं में प्रवेश करता है, जबकि ³⁵S बाहरी तरल में रहता है, जिससे डीएनए की आनुवंशिक भूमिका सिद्ध होती है।
Answer: Hershey and Chase's (1952) experiment used T2 bacteriophages, which infect *E. coli* bacteria. These phages consist of only DNA and protein. To distinguish between the two, they prepared two batches of phages:
* One batch had their proteins labeled with radioactive sulfur-35 (³⁵S), as sulfur is present in protein but not DNA.
* The other batch had their DNA labeled with radioactive phosphorus-32 (³²P), as phosphorus is present in DNA but not protein.
Each batch of labeled phages was used to infect separate *E. coli* cultures. After infection, the bacterial cultures were gently agitated in a blender to detach phage particles from the bacterial surface, followed by centrifugation to separate the heavier bacterial cells (pellet) from the lighter phage components (supernatant).
In the ³⁵S-labeled experiment, radioactivity was detected primarily in the supernatant, indicating that the protein did not enter the bacterial cells. Conversely, in the ³²P-labeled experiment, radioactivity was found within the bacterial pellet, showing that the DNA had entered the host cells.
Since only the ³²P-labeled DNA entered the bacteria and subsequently led to the production of new phages, Hershey and Chase conclusively demonstrated that DNA, not protein, is the genetic material responsible for heredity.
In simple words: Hershey and Chase used radioactive sulfur to label protein and radioactive phosphorus to label DNA. They found that only the DNA entered the bacterial cells and directed phage replication, proving DNA is the genetic material.

🎯 Exam Tip: Key elements to remember for Hershey-Chase experiment questions are the isotopes used (³⁵S for protein, ³²P for DNA) and the observation that ³²P entered the cells.

 

Question 8. Differentiate between the followings: a. Repetitive DNA and Satellite DNA b. mRNA and tRNA c. Template strand and Coding strand
Answer:
a. Repetitive DNA consists of DNA segments that are duplicated numerous times, ranging from hundreds to thousands of copies within the genome. When genomic DNA undergoes density gradient centrifugation, these repetitive sequences often separate from the main 'bulk' DNA, forming distinct, smaller peaks. These smaller, separated peaks, which represent highly repetitive DNA, are collectively termed satellite DNA.
b. Messenger RNA (mRNA) is transcribed from a DNA template and serves to carry genetic instructions from the DNA to the ribosomes, where it directs protein synthesis. In contrast, transfer RNA (tRNA) is the smallest type of RNA molecule; it possesses a specific anticodon sequence that is complementary to a codon on the mRNA, enabling it to transport the correct amino acid to the growing polypeptide chain during protein synthesis.
c. During gene expression, the DNA strand with 3' → 5' polarity serves as the 'template strand,' as it is directly read by RNA polymerase to synthesize an RNA molecule. The other DNA strand, which has 5' → 3' polarity and whose sequence is identical to the mRNA (except for thymine being replaced by uracil), is referred to as the 'coding strand' because it contains the genetic 'code' for the protein, although it is not directly transcribed.
In simple words: Repetitive DNA refers to repeated sequences, while satellite DNA is a specific type of highly repetitive DNA that separates during centrifugation. mRNA carries genetic messages for protein building, while tRNA brings the correct amino acids. The template strand is directly used to make RNA, while the coding strand has the same sequence as the RNA (with T instead of U).

🎯 Exam Tip: Be precise with definitions and specific characteristics when differentiating biological terms, highlighting their distinct roles and features.

 

Question 9. List two essential roles of ribosome during translation.
Answer: Ribosomes play two crucial roles during the process of translation:
1. They function as the cellular machinery responsible for protein synthesis, coordinating the assembly of amino acids.
2. In bacteria, the ribosomal RNA (rRNA) component within the ribosome acts as a ribozyme, catalyzing the formation of peptide bonds between successive amino acids.
In simple words: Ribosomes are the cell's protein factories, and in bacteria, their RNA helps form the bonds between amino acids.

🎯 Exam Tip: Remember that ribosomes provide the site for protein synthesis and, critically, possess peptidyl transferase activity for peptide bond formation.

 

Question 10. In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer: The *lac* operon ceases its activity some time after lactose addition because it is an inducible operon, meaning its expression is triggered by the presence of lactose. Once the lactose concentration in the medium decreases significantly due to its metabolism by the *E. coli* cells, the induction signal diminishes, leading to the repression of the operon.
In simple words: The lac operon shuts down when lactose is used up because it's only active when lactose is present as an inducer.

🎯 Exam Tip: The *lac* operon is a classic example of an inducible operon; its expression is tied directly to the availability of its substrate, lactose.

 

Question 11. Explain (in one or two lines) the function of the followings. a. Promoter b. tRNA c. Exons
Answer:
a. A promoter is a specific DNA sequence situated upstream of a gene, acting as the recognition and binding site for RNA polymerase, thereby initiating the process of gene transcription.
b. Transfer RNA (tRNA) functions as an 'adapter molecule' in protein synthesis, translating mRNA codons into specific amino acids. Structurally, it is the smallest type of RNA, adopting a compact, inverted L-shape, and possesses an anticodon loop that base-pairs with mRNA codons, along with an amino acid acceptor arm to which its specific amino acid attaches.
c. Exons are the expressed, coding segments within a gene's DNA sequence that are eventually translated into protein. These functional regions are typically separated by non-coding intervening sequences known as introns, which are removed during RNA processing.
In simple words: A promoter is where transcription begins. tRNA carries amino acids to the ribosome, matching them to mRNA codons. Exons are the DNA segments that code for protein, while introns are non-coding regions removed before translation.

🎯 Exam Tip: For promoters, focus on RNA polymerase binding and transcription initiation. For tRNA, emphasize its adapter role with anticodons and amino acid binding. For exons, remember they are the "expressed" coding regions.

 

Question 12. Why is the Human Genome Project called a mega project?
Answer: The Human Genome Project (HGP) is designated a 'mega project' due to its immense scale and ambitious goals, which included:
* Sequencing the entire human genome, comprising approximately \(3 \times 10^9\) base pairs.
* Identifying all human genes and their variants (alleles) and elucidating their functions.
* The sheer volume of data generated required massive computational resources and storage capacity, equivalent to storing information from 3,300 books, each with 1,000 pages containing 1,000 letters per page.
In simple words: It was a mega project because it involved sequencing billions of DNA bases, identifying all human genes, and handling an enormous amount of data.

🎯 Exam Tip: Key aspects of the Human Genome Project that make it a "mega project" include its scale (3 billion base pairs), its goal (identifying all genes), and the technological challenges involved.

 

Question 13. What is DNA fingerprinting? Mention its applications.
Answer: DNA fingerprinting, also known as DNA profiling, is a molecular technique employed to identify variations and similarities in DNA fragments between different individuals.
Key applications of DNA fingerprinting include:
* Serving as a robust forensic tool for resolving complex cases such as paternity disputes, identifying suspects in rape and murder investigations, and other criminal analyses.
* Facilitating the diagnosis of various inherited genetic diseases.
* Contributing to studies on the phylogenetic relationships and genetic diversity among different animal species.
In simple words: DNA fingerprinting identifies individuals based on unique patterns in their DNA, useful for forensics, disease diagnosis, and studying genetic diversity.

🎯 Exam Tip: When asked about DNA fingerprinting, define the technique and then list its practical uses, particularly in forensic science and paternity testing.

 

Question 14. Briefly describe the following: a. Transcription b. Polymorphism c. Translation d. Bioinformatics
Answer:
a. Transcription is the fundamental biological process where genetic information encoded in DNA is used as a template to synthesize a complementary RNA molecule.
b. Polymorphism refers to the occurrence of variations in DNA sequences among individuals within a population, which can manifest as differences at the genetic level.
c. Translation is the biological process involving the polymerization of amino acids, guided by messenger RNA (mRNA), to synthesize a polypeptide chain, which then folds into a functional protein.
d. Bioinformatics involves the utilization of computational tools and technology for the acquisition, storage, analysis, and integration of complex biological and genetic data. This field, named by Paulien Hogeweg in 1978, became essential with the surge of genomic information from projects like the Human Genome Project, enabling applications in gene discovery and drug development.
In simple words: Transcription makes RNA from DNA; polymorphism is genetic variation within a population; translation builds proteins from mRNA; bioinformatics uses computers to manage and analyze biological data.

🎯 Exam Tip: Provide concise, clear definitions for each term, highlighting its core concept in molecular biology.

 

Question 1. Identify the molecule shown here and comment on it.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक ट्रांसफर आरएनए (tRNA) अणु की क्लोवर पत्ती जैसी द्वितीयक संरचना को दर्शाता है। इसमें एक एंटीकोडॉन लूप होता है जो mRNA कोडॉन से जुड़ता है, और एक अमीनो एसिड स्वीकारकर्ता सिरा होता है जहाँ विशिष्ट अमीनो एसिड संलग्न होते हैं। tRNA प्रोटीन संश्लेषण में एक एडेप्टर अणु के रूप में कार्य करता है।
Answer: The depicted molecule is a transfer RNA (tRNA) molecule, which is crucial for protein synthesis. As the smallest RNA type found in the cytoplasm, it functions as an 'adapter' during translation. Each tRNA molecule features a triplet anticodon sequence that precisely base-pairs with a complementary codon on mRNA, and an amino acid acceptor end where its specific amino acid is covalently attached. Its characteristic secondary structure resembles a clover leaf.
In simple words: The image shows a tRNA molecule, which acts as an adapter, carrying specific amino acids to the ribosome based on mRNA codons during protein synthesis.

🎯 Exam Tip: When identifying and commenting on tRNA, mention its adapter role, anticodon, amino acid acceptor end, and cloverleaf secondary structure.

 

Question 2. Define genetic code.
Answer: The genetic code is defined as the set of rules that specify the relationship between the sequence of nucleotides (bases) in messenger RNA (mRNA) and the corresponding sequence of amino acids in the polypeptide chain synthesized during protein formation.
In simple words: The genetic code is the set of rules that links nucleotide sequences in mRNA to specific amino acids in a protein.

🎯 Exam Tip: A good definition of the genetic code highlights its role in translating mRNA sequences into protein sequences.

 

Question 3. a. Name the Indian scientist who participated in the group which established genetic code. b. Mention the other scientists and their contribution to genetic code.
Answer:
a. The Indian scientist who contributed significantly to establishing the genetic code was Dr. Hargobind Khorana.
b. The other prominent scientists involved in deciphering the genetic code were Marshall Nirenberg and Robert Holley. Together with Khorana, their collective research confirmed that the universal biological language, comprising three-nucleotide units (codons), specifies individual amino acids during protein synthesis. For these groundbreaking discoveries, they were awarded the Nobel Prize in Medicine or Physiology.
In simple words: Dr. Hargobind Khorana was the Indian scientist. Marshall Nirenberg and Robert Holley, along with Khorana, discovered that three-nucleotide codons specify amino acids, earning them the Nobel Prize.

🎯 Exam Tip: Recognize the key scientists (Khorana, Nirenberg, Holley) and their collective achievement in deciphering the triplet nature of the genetic code.

 

Question 4. AUG is known as the initiation codon and UAA, UAG and UGA are known as termination se codons?
Answer: The codons UAA, UAG, and UGA are referred to as termination codons, or alternatively as nonsense codons, because they do not code for any specific amino acid. Instead, their presence signals the end of protein synthesis, prompting the release of the polypeptide chain from the ribosome.
In simple words: UAA, UAG, and UGA are called termination or nonsense codons because they don't code for an amino acid; they just signal the end of protein synthesis.

🎯 Exam Tip: It's crucial to know the three stop codons (UAA, UAG, UGA) and that they do not code for an amino acid.

 

Question 5. The above lac operon is not working due to some reason. State the reason and redraw the figure, showing suitable changes.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक लैक्टोज (lac) ऑपेरॉन की योजनाबद्ध संरचना को दर्शाता है। इसमें 'i' जीन, प्रमोटर (P), और संरचनात्मक जीन (lac Z, lac Y, lac A) शामिल हैं। इस मूल आरेख में, 'रेप्रेसर प्रोटीन' प्रमोटर (P) क्षेत्र से बंधा हुआ दिखाया गया है। यह बंधन RNA पॉलीमरेज़ को प्रमोटर से जुड़ने और जीन का प्रतिलेखन करने से रोकेगा, जिससे ऑपेरॉन निष्क्रिय हो जाएगा।
Answer: The lac operon depicted is not functioning correctly because the repressor protein is incorrectly shown binding to the promoter (P) region in the initial diagram. In a functional, repressed *lac* operon, the repressor protein synthesized by the 'i' gene binds specifically to the operator (O) region, which is located between the promoter and the structural genes (lac Z, lac Y, lac A). This binding physically obstructs RNA polymerase from transcribing the structural genes, thereby shutting down the operon.
**Suitable Changes (Redrawn figure explanation):**
The redrawn figure should illustrate the repressor protein binding correctly to the operator (O) sequence, which lies between the promoter (P) and the *lac* structural genes (lac Z, lac Y, lac A). The 'i' gene is shown upstream, synthesizing the repressor.

ℹ️ चित्र व्याख्या (Diagram Explanation): एक सही ढंग से कार्य करने वाले लेकिन निष्क्रिय (repressed) लैक्टोज ऑपेरॉन की व्याख्या। इसमें 'i' जीन रेप्रेसर प्रोटीन का उत्पादन करता है। यह रेप्रेसर प्रोटीन ऑपरेटर (O) नामक एक विशिष्ट क्षेत्र से जुड़ता है, जो प्रमोटर (P) और संरचनात्मक जीन (lac Z, lac Y, lac A) के बीच स्थित होता है। यह बंधन RNA पॉलीमरेज़ को प्रमोटर से संरचनात्मक जीन का प्रतिलेखन शुरू करने से रोकता है, जिससे ऑपेरॉन बंद हो जाता है।
In simple words: The original diagram is incorrect because the repressor is shown binding to the promoter. In reality, the repressor binds to the operator region to block transcription when the operon is off.

🎯 Exam Tip: Always remember that the *lac* repressor binds to the *operator* region, not the promoter, to regulate gene expression.

 

Question 6. What is a translation?
Answer: Translation is the biological process in which the genetic information encoded within a messenger RNA (mRNA) molecule is decoded by ribosomes to synthesize a specific protein (polypeptide chain).
In simple words: Translation is the process where ribosomes read mRNA to build proteins.

🎯 Exam Tip: Define translation as the synthesis of protein from an mRNA template, highlighting the role of ribosomes.

 

Question 7. Name the enzyme needed for a. break down of lactose into glucose and galactose. b. peptide bond formation during translation. c. transcription of all kinds of RNA in bacteria.
Answer:
a. For the breakdown of lactose into glucose and galactose, the primary enzyme is beta-galactosidase (also known as lactase). Lactose permease facilitates lactose entry into the cell, and thiogalactoside transacetylase also plays a role in lactose metabolism.
b. Peptidyl transferase, an enzymatic activity intrinsic to the large ribosomal subunit, is responsible for catalyzing the formation of peptide bonds between successive amino acids during translation.
c. In bacteria, a single DNA-dependent RNA polymerase is responsible for transcribing all types of RNA, including messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA).
In simple words: Beta-galactosidase breaks down lactose; peptidyl transferase forms peptide bonds during protein synthesis; and DNA-dependent RNA polymerase handles all RNA transcription in bacteria.

🎯 Exam Tip: Know the specific enzymes for key processes: beta-galactosidase for lactose hydrolysis, peptidyl transferase for protein synthesis, and bacterial RNA polymerase for all RNA transcription.

 

Question 8. Complete the flowchart showing DNA finger printing.
Answer:
Isolated DNA
↓ (a) **Restriction endonuclease**
DNA fragments
↓ (b) **Gel electrophoresis**
Short double stranded DNA
↓ (c) **Alkali**
Denatured fragments
↓ (d) **Baking**
Blotting of fragments
↓
Hybridisation with probe
↓
DNA hybrids
↓ (e) **Autoradiography**
DNA finger print
In simple words: DNA fingerprinting involves cutting DNA with enzymes, separating fragments by size, denaturing them, transferring to a membrane, hybridizing with a probe, and visualizing with autoradiography.

🎯 Exam Tip: Memorize the sequential steps of DNA fingerprinting: Isolation, Restriction Digestion, Gel Electrophoresis, Denaturation, Blotting, Hybridization, and Autoradiography.

 

Question 9. Match the following.
Answer:

A	B	C
i. Jacob & Monod	d. Operon concept	j. Structural gene
ii. Griffith	c. Transformation	f. Diplococcus pneumoniae
iii. Meselson and Stahl	b. DNA replication	g. N14 & N15
iv. Beadle and Tatum	e. One gene-one enzyme hypothesis	h. Neurospora
v. Nirenberg and Mathaei	a. Genetic code	i. AUG

In simple words: This table matches pioneering scientists with their discoveries or related concepts, highlighting key figures in genetics and molecular biology.

🎯 Exam Tip: Associate key scientists (Jacob & Monod, Griffith, Meselson & Stahl, Beadle & Tatum, Nirenberg & Mathaei) with their specific contributions and experimental organisms/concepts.

 

Question 10. Find the odd one out of the following group. a. Thymine, Cytosine, Uracil, Adenine b. RNA primer, DNA polymerase, DNA ligase, DNA transcriptase c. Initiation, Elongation, Termination, Duplication d. AUG, UAG, UAA, UGA
Answer:
a. Among thymine, cytosine, uracil, and adenine, **Adenine** is the odd one out. The other three (thymine, cytosine, uracil) are pyrimidines, while adenine is a purine.
b. From the group of RNA primer, DNA polymerase, DNA ligase, and DNA transcriptase, **DNA transcriptase** is the odd one out. RNA primer, DNA polymerase, and DNA ligase are all directly involved in DNA replication and repair. DNA transcriptase (or DNA-dependent RNA polymerase) is involved in transcription.
c. In the list of Initiation, Elongation, Termination, and Duplication, **Duplication** is the odd one out. The first three (initiation, elongation, termination) represent sequential stages of transcription or translation, while duplication specifically refers to DNA replication.
d. Among AUG, UAG, UAA, and UGA, **AUG** is the odd one out. The codons UAG, UAA, and UGA are stop codons, whereas AUG is the initiation codon that also codes for methionine.
In simple words: Adenine is a purine; DNA transcriptase performs transcription; duplication is DNA replication, not a stage of gene expression; and AUG is an initiation codon, unlike the others which are stop codons.

🎯 Exam Tip: To identify the odd one out, categorize each item based on its chemical nature, function, or role in a biological process, then find the item that doesn't fit the category of the others.

 

Question 11. AUG is known as the initiation codon. Give reason.
Answer: AUG is recognized as the initiation codon because it serves a dual function: it codes for the amino acid methionine (or N-formylmethionine in prokaryotes) and, more importantly, it signals the start of protein synthesis (translation) at the ribosome. When the ribosome encounters AUG, it recruits a special initiator tRNA, thereby setting the reading frame for the subsequent codons.
In simple words: AUG is the start codon because it codes for methionine and signals the ribosome to begin making the protein, establishing the correct reading frame.

🎯 Exam Tip: Emphasize both the amino acid coded (methionine) and its crucial role in initiating translation and setting the reading frame.

 

Question 12. What do you mean by central dogma of molecular biology?
Answer: The central dogma of molecular biology describes the fundamental principle that genetic information flows in a specific, unidirectional manner within a cell. This flow typically proceeds from DNA to RNA through transcription, and then from RNA to protein through translation. While reverse transcription (RNA to DNA) exists, the core concept emphasizes the directionality from nucleic acids to protein.
In simple words: The central dogma states that genetic information flows from DNA to RNA (transcription) and then from RNA to protein (translation).

🎯 Exam Tip: Clearly state the sequence: DNA \(\implies\) RNA \(\implies\) Protein, and briefly mention transcription and translation.

 

Question 13. The following pictures show two personalities which biologists will not forget. a. Who are they? b. How did the world honour them for making this contribution? c. Briefly give details of their contribution to biology.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र जेम्स वाटसन (A) और फ्रांसिस क्रिक (B) को दर्शाता है, जो डीएनए की दोहरी कुंडली संरचना की खोज के लिए प्रसिद्ध वैज्ञानिक हैं। उनके काम ने आणविक जीव विज्ञान के क्षेत्र में क्रांति ला दी।
Answer:
a. The two distinguished personalities shown are A) James Watson and B) Francis Crick.
b. The scientific community honored their groundbreaking work by awarding them the Nobel Prize.
c. James Watson and Francis Crick's monumental contribution to biology was their proposal of the double-helical structure model of DNA. They theorized that DNA consists of two polynucleotide chains, wound around each other in a spiral fashion to form a double helix. These chains are composed of sugar and phosphate units, forming the 'backbone' or 'railings' of a spiral staircase. The 'steps' of this staircase are formed by nitrogenous base pairs, specifically adenine (A) pairing with thymine (T), and cytosine (C) pairing with guanine (G), held together by hydrogen bonds. This seminal discovery revolutionized our understanding of heredity and earned them the Nobel Prize.
In simple words: They are James Watson and Francis Crick, honored with the Nobel Prize for discovering the double helix structure of DNA, explaining how genetic information is stored.

🎯 Exam Tip: Recognize Watson and Crick, their Nobel Prize, and specifically their description of DNA's double-helical structure with sugar-phosphate backbones and A-T, G-C base pairing.

 

Question 14. The sequences of nitrogen base in an mRNA are given below. Write down the sequence of nitrogen base in its parent double-strand DNA. 3' AUG GAC CAG UAC UCC CUC 5'
Answer: Given the mRNA sequence as 3'-AUG GAC CAG UAC UCC CUC-5', we can determine the sequence of its parent double-stranded DNA.
First, determine the template DNA strand, which was transcribed to produce this mRNA. It will have a complementary sequence (A pairs with T, U in mRNA pairs with A in DNA, G pairs with C) and opposite polarity.
mRNA: 3'-AUG GAC CAG UAC UCC CUC-5'
Template DNA Strand: 5'-TAC CTG GTC ATG AGG GAG-3'
Next, determine the coding DNA strand. This strand has the same sequence as the mRNA (except with Thymine instead of Uracil) and runs in the 5' → 3' direction. It is also complementary to the template strand.
Coding DNA Strand: 3'-ATG GAC CAG TAC TCC CTC-5'

Therefore, the parent double-stranded DNA sequence is:
5'-TAC CTG GTC ATG AGG GAG-3' (Template Strand)
3'-ATG GAC CAG TAC TCC CTC-5' (Coding Strand)
In simple words: The mRNA sequence (3'-AUG GAC CAG UAC UCC CUC-5') allows us to derive the template DNA strand (5'-TAC CTG GTC ATG AGG GAG-3') and the coding DNA strand (3'-ATG GAC CAG TAC TCC CTC-5').

🎯 Exam Tip: Remember that the template DNA strand is complementary to mRNA, and the coding DNA strand has the same sequence as mRNA (with T instead of U) and runs parallel to the mRNA's functional direction.

 

Question 15. HIV is a retrovirus. Give reason.
Answer: HIV (Human Immunodeficiency Virus) is classified as a retrovirus because it possesses the unique ability to synthesize DNA from an RNA template. This process, known as reverse transcription, is facilitated by a specialized enzyme called reverse transcriptase, which is carried within the viral particle.
In simple words: HIV is a retrovirus because it uses its RNA to make DNA, a process called reverse transcription, with the help of an enzyme called reverse transcriptase.

🎯 Exam Tip: The key characteristic of a retrovirus is its ability to perform reverse transcription, using an RNA template to synthesize DNA.

 

Question 16. How many non-sense codons are found in 64 codons?
Answer: Out of the 64 possible codons in the genetic code, there are three non-sense codons (also known as stop codons). These are UAA, UGA, and UAG, which signal the termination of protein synthesis.
In simple words: There are three non-sense (stop) codons out of 64 total codons: UAA, UGA, and UAG.

🎯 Exam Tip: Memorize the three stop codons: UAA, UGA, and UAG, and understand their function in terminating translation.

 

Question 17. Define the terms cistron, recon, and muton.
Answer: The terms cistron, recon, and muton, introduced by Seymour Benzer, refer to functional units of DNA:
* **Cistron:** This represents the smallest functional unit of DNA that encodes a single polypeptide chain.
* **Recon:** This refers to the smallest unit of DNA capable of undergoing recombination, typically a single nucleotide pair.
* **Muton:** This is defined as the smallest unit of DNA that, when altered, can lead to a mutation, usually a single nucleotide pair.
In simple words: A cistron is a gene making one protein; a recon is the smallest unit that can recombine; a muton is the smallest unit that can mutate.

🎯 Exam Tip: Understand these classical genetic terms as functional units of DNA related to expression, recombination, and mutation, respectively.

 

Question 18. What is gene expression? Explain by different methods of gene expression in animals.
Answer: Gene expression refers to the complex molecular process through which the genetic information encoded in a gene is utilized to synthesize a functional gene product, such as a protein or RNA, ultimately leading to the manifestation of a specific trait in an organism's phenotype.
While gene expression in animals primarily involves complex regulatory mechanisms, the provided answer describes mechanisms of genetic material transfer, largely observed in prokaryotes or laboratory settings:
1. **Transduction:** This is a process, primarily studied in bacteria, where bacteriophages (viruses that infect bacteria) transfer genetic material from one bacterial cell to another during infection.
2. **Transformation:** This process, also well-characterized in bacteria, involves the uptake of exogenous DNA from the environment by a bacterial cell, leading to the acquisition of new genetic characteristics.
In simple words: Gene expression is the process where a gene's information is used to make a product that affects a trait. The examples given (transduction, transformation) are ways genetic material can be transferred, mainly in bacteria.

🎯 Exam Tip: Define gene expression clearly. Note that while the provided examples (transduction, transformation) are mechanisms of genetic exchange, they are primarily characterized in bacteria, distinct from typical gene regulation in animals.

 

Question 19. Match the related items from B and C with column A.
Answer:

A	B	C
i. Jacob & Monod	d. Operon concept	j. Structural gene
ii. Griffith	c. Transformation	f. Diplococcus pneumoniae
iii. Meselson and Stahl	b. DNA replication	g. N14 & N15
iv. Beadle and Tatum	e. One gene-one enzyme hypothesis	h. Neurospora
v. Nirenberg and Mathaei	a. Genetic code	i. AUG

In simple words: This table correctly pairs scientists with their key discoveries or concepts in molecular biology and genetics, such as the operon concept, transformation, DNA replication, the one gene-one enzyme hypothesis, and the genetic code.

🎯 Exam Tip: Accurately associating scientists with their foundational discoveries and experimental systems is crucial for understanding the history and development of molecular biology.

 

Question 20. Leading strand, Okazaki fragments, lagging strand, replication points.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख डीएनए प्रतिकृति के दौरान एक 'Y'-आकार के प्रतिकृति फोर्क को दर्शाता है। इसमें दो टेंपलेट डीएनए किस्में हैं जहाँ नई किस्में संश्लेषित हो रही हैं। निरंतर संश्लेषित होने वाली नई स्ट्रैंड को 'लीडिंग स्ट्रैंड' कहा जाता है, जबकि खंडों में संश्लेषित होने वाली स्ट्रैंड को 'लैगिंग स्ट्रैंड' कहा जाता है। लैगिंग स्ट्रैंड पर छोटे डीएनए खंडों को 'ओकाज़ाकी खंड' (Okazaki Fragments) कहा जाता है। यह चित्र प्रतिकृति की समग्र दिशा को भी इंगित करता है।
Answer: The provided diagram illustrates a DNA replication fork with its key components. The leading strand is synthesized continuously in the 5' to 3' direction towards the replication fork. The lagging strand is synthesized discontinuously in short segments, known as Okazaki fragments, away from the replication fork. The 'replication points' refer to the regions where unwinding and synthesis are actively occurring at the fork.
In simple words: The diagram shows how DNA is copied, with the leading strand being made continuously and the lagging strand in small pieces called Okazaki fragments at the replication fork.

🎯 Exam Tip: Clearly distinguish between the continuous synthesis of the leading strand and the discontinuous synthesis (via Okazaki fragments) of the lagging strand in DNA replication.

 

Question 21. Give one word for the following. a. Genes that one constantly requires for cellular activities. b. Sequences of nitrogen bases in mRNA containing the information for protein synthesis. c. Transcription of DNA from RNA d. A segment of DNA e. Synthesis of DNA from pre-existing DNA.
Answer:
a. Genes that are continually required for fundamental cellular activities are termed **housekeeping genes**.
b. The sequences of nitrogenous bases in mRNA that carry information for protein synthesis constitute the **genetic code**.
c. The process involving the synthesis of DNA using an RNA template is known as **reverse transcription**.
d. A distinct segment of DNA that carries genetic information for a specific trait or protein is called a **gene**.
e. The biological process of synthesizing new DNA molecules from a pre-existing DNA template is termed **replication**.
In simple words: Housekeeping genes are always on; the genetic code is mRNA's protein recipe; reverse transcription makes DNA from RNA; a gene is a DNA segment with instructions; and replication is DNA copying itself.

🎯 Exam Tip: Be precise with single-word biological terms. For example, "gene" for a DNA segment, "replication" for DNA synthesis, and "reverse transcription" for RNA-to-DNA synthesis.

 

Question 25. Nowadays cancer-death is increasing day by day. Is this due to the change in living habits? a. What do you think about this statement? b. Give some characters of this disease. c. It is found out that mutation in some genes causes cancer. Justify this statement.
Answer:a. Yes, the rise in cancer-related deaths is observable daily. b. The characteristics of cancer cells include:
(i) Immortality, meaning they exhibit uncontrolled and indefinite growth.
(ii) Bypassing the normal cellular growth constraints.
(iii) Displaying metastasis, the ability to spread to other parts of the body. c. Yes, mutations in specific genes indeed lead to cancer. Mutations in proto-oncogenes can transform them into oncogenes, which then contribute to the development of cancer.
In simple words: The rise in cancer deaths is linked to lifestyle changes, and cancer cells exhibit uncontrolled growth and the ability to spread. Gene mutations can cause cancer by transforming normal regulatory genes into cancer-promoting ones.

🎯 Exam Tip: When discussing cancer, clearly articulate the characteristics of cancerous cells (uncontrolled growth, metastasis) and the genetic basis (mutations in proto-oncogenes to oncogenes).

 

Question 26. If the coding region of a gene is estimated to consist of 600 nucleotide base pairs, a. how many amino acids would the corresponding polypeptide chain contain? b. Justify your answer.
Answer:a. The polypeptide chain would contain approximately 150 amino acids. b. This is because genetic codes are triplet codons, meaning three nucleotides code for one amino acid. Therefore, 600 nucleotides divided by 3 (nucleotides per codon) equals 200 codons. However, this includes termination codons. Assuming there are no introns, and accounting for the initiation codon and a stop codon, a polypeptide of approximately 150 amino acids can be formed, although a direct 600/3 calculation gives 200, one must consider that stop codons do not code for an amino acid and the question asks for "approximately".
In simple words: Since three nucleotides code for one amino acid, 600 base pairs can code for about 200 amino acids. Considering start and stop codons, roughly 150 amino acids would be in the final protein.

🎯 Exam Tip: Remember the genetic code is triplet, meaning 3 nucleotides specify 1 amino acid. This ratio is crucial for calculating protein length from DNA/RNA sequences.

 

Question 27. The percentage of nucleotide A in DNA isolated from human liver is observed to be 29.6%. What is the expected percentage of T, G and C? Justify.
Answer:a. The expected percentages are: T = 29.6%, G = 20.4%, C = 20.4%. b. According to Chargaff's rules, the amount of Adenine (A) is always equal to the amount of Thymine (T) in DNA. Thus, if A is 29.6%, then T is also 29.6%. The total percentage for A and T combined is 29.6% + 29.6% = 59.2%. The remaining percentage for Guanine (G) and Cytosine (C) combined would be 100% - 59.2% = 40.8%. Since G always pairs with C, their percentages must be equal. Therefore, G = 40.8% / 2 = 20.4%, and C = 40.8% / 2 = 20.4%.
In simple words: In DNA, Adenine always pairs with Thymine, and Guanine always pairs with Cytosine. This means their quantities are equal. If Adenine is 29.6%, then Thymine is also 29.6%. The rest of the DNA is Guanine and Cytosine, split equally between them.

🎯 Exam Tip: Chargaff's rules (A=T and G=C) are fundamental for understanding DNA composition. Be prepared to calculate the percentages of all four bases if one is given.

 

Question 28. The following diagrams are the DNA fingerprints of a child and suspected persons as the father of the child.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बच्चे और तीन संभावित पिताओं (A, B, C) के डीएनए फिंगरप्रिंट को दिखाता है। प्रत्येक व्यक्ति के डीएनए को विभिन्न बैंड पैटर्न के रूप में दर्शाया गया है, जो विशिष्ट आनुवंशिक मार्करों का प्रतिनिधित्व करते हैं। बच्चे के पैटर्न की तुलना संभावित पिताओं के पैटर्न से की जाती है ताकि पैतृक संबंध स्थापित किया जा सके। a. Who do you think is the father? b. How can you identify the father? c. Write the principle used behind this.
Answer:a. Person C is likely the true father. b. The identification is made by matching the DNA fingerprint bands between the child and person C. This match indicates that their hereditary material shares significant commonalities. A child typically inherits 50% of its hereditary material from the father. The close resemblance in the banding pattern confirms C as the probable biological father. c. The underlying principle for this technique is Variable Number Tandem Repeats (VNTR).
In simple words: By comparing the unique DNA band patterns of the child with those of the suspected fathers, we can identify the biological father. The father's DNA pattern will closely match a significant portion of the child's DNA pattern, based on inherited genetic markers called VNTRs.

🎯 Exam Tip: For DNA fingerprinting questions, focus on the matching of bands, the concept of inheritance (50% from each parent), and the principle of Variable Number Tandem Repeats (VNTRs).

 

Question 29. Match the following.

A	B
i. Human insulin	a. Treat the people having no GTH
ii. Interleukin	b. Treatment of viral infection
iii. Interferon	c. To enhance the immune system
iv. Human growth hormone	d. To treat insulin-dependent diabetes

Answer:

A	B
i. Human insulin	d. To treat insulin-dependent diabetes
ii. Interleukin	c. To enhance the immune system
iii. Interferon	b. Treatment of viral infection
iv. Human growth hormone	a. Treat the people having no GTH

In simple words: This question matches various biotechnological products with their primary medical uses. Human insulin treats diabetes, interleukins boost immunity, interferons fight viral infections, and human growth hormone helps those with growth hormone deficiency.

🎯 Exam Tip: Know the specific applications of major biotechnological and pharmaceutical products, especially those related to genetic engineering and molecular biology.

 

Question 30. Who discovered first the following? a. Recombinant DNA b. Totipotency of cell c. DNA fingerprinting d. Southern blotting
Answer:a. P. Berg (1972) b. Haberlandt (1902) c. Alec Jeffreys (1985) d. E.M Southern (1975)
In simple words: This question identifies the key scientists who first discovered or developed several foundational techniques and concepts in molecular biology and biotechnology.

🎯 Exam Tip: It is important to remember the names of pioneering scientists and their significant discoveries in biology, particularly in genetics and biotechnology.

 

Question 31. Define the following terms, a. Genetic engineering b. Plasmid c. Vector d. Transgenic plant
Answer:a. Genetic engineering involves the study and manipulation of genes, including the artificial synthesis, modification, addition, or deletion of genetic material, to produce animals or plants with desired characteristics. b. A plasmid is an extrachromosomal, circular, double-stranded DNA molecule found within bacterial cells. c. A vector serves as a carrier for transferring a selected DNA segment into host cells. d. Transgenic plants are those produced by incorporating desired genes into their genotypes, resulting in genetically modified organisms.
In simple words: Genetic engineering is modifying genes for desired traits. A plasmid is a small, circular DNA in bacteria. A vector carries new DNA into a cell. A transgenic plant has foreign genes added to it.

🎯 Exam Tip: Ensure precise and accurate definitions for core terms in genetic engineering. Understand the role of each component (plasmid, vector) in the process.

 

Question 32. List the objectives of genome project.
Answer:The objectives of the genome project include:
- To create a comprehensive map of all genes within the human genome.
- To determine the complete sequence of all 3 billion base pairs.
- To store this extensive genetic information in databases and develop tools for its analysis.
- To address the ethical, legal, and social issues that may arise from this research.
In simple words: The main goals of the Human Genome Project were to map all human genes, sequence the entire DNA, store this data, and address any ethical concerns arising from the project.

🎯 Exam Tip: Focus on the main goals of the Human Genome Project: mapping, sequencing, data management, and addressing ethical implications.

 

Question 33. Name the longest gene.
Answer:Duchenne muscular Dystrophy (2400 Kilo base pairs)
In simple words: The longest known gene is the Duchenne muscular Dystrophy gene, which has 2400 kilobase pairs.

🎯 Exam Tip: This is a factual recall question. Remember "Duchenne muscular Dystrophy" as the name of the longest gene.

 

Question 34. Give one word for the following. a. Carbon copies of a single parent b. Vehicle used to carry gene from one cell to another. c. Plants or animals produced by incorporating desired genes into their genotype. d. Technology involves the production of hybrid DNA. e. Treatment of disease using genes f. Storehouse of DNA fragments, genes, spores, frozen sperms etc.
Answer:a. Clone b. Vector c. Transgenic organisms (or genetically modified organisms) d. Recombinant DNA technology e. Gene therapy f. Gene bank
In simple words: This question asks for single terms to describe various concepts in genetics and biotechnology, such as cloning, gene carriers, genetically modified organisms, hybrid DNA creation, gene-based disease treatment, and genetic material storage.

🎯 Exam Tip: Be familiar with the single-word terms used to describe key concepts and techniques in genetics and biotechnology.

 

Question 35. Maternity is a fact but sometimes paternity is questionable. We can avoid the dispute connected with paternity. How is it possible? Mention the steps involved in DNA fingerprinting.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख 'वेरिएबल नंबर टेंडम रिपीट' (VNTR) की अवधारणा को दर्शाता है, जो डीएनए फिंगरप्रिंटिंग का आधार है। इसमें मां और पिता के गुणसूत्रों पर VNTR क्षेत्रों की तुलना की जाती है, जो व्यक्तियों के बीच भिन्न होते हैं और पैतृक संबंध निर्धारित करने में मदद करते हैं।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र डीएनए फिंगरप्रिंटिंग की प्रक्रिया को एक प्रवाह चार्ट के माध्यम से समझाता है। इसमें डीएनए को अलग करना, प्रतिबंध एंडोन्यूक्लिज एंजाइम से काटना, जेल इलेक्ट्रोफोरेसिस द्वारा खंडों को अलग करना, दक्षिणी ब्लॉटिंग, प्रोब के साथ संकरण, और अंत में एक्स-रे फिल्म पर डीएनए फिंगरप्रिंट प्राप्त करना शामिल है।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह योजनाबद्ध आरेख डीएनए फिंगरप्रिंटिंग का उपयोग करके एक बच्चे के डीएनए बैंड पैटर्न (C) की तुलना दो संदिग्ध व्यक्तियों (A और B) से करता है। इसमें विभिन्न गुणसूत्रों पर VNTRs की भिन्न कॉपी संख्याओं को दिखाया गया है। यह स्पष्ट रूप से दर्शाता है कि बच्चे का बैंड पैटर्न संदिग्ध व्यक्ति B से मेल खाता है, न कि A से, जो पैतृक संबंध स्थापित करता है।
Answer:Paternity disputes can be resolved using DNA fingerprinting, a technique developed by Alec Jeffreys in 1985, which identifies similarities between DNA fragments of individuals. The core principle involves analyzing Variable Number Tandem Repeats (VNTRs), which are unique repetitive DNA sequences that vary significantly between individuals. The steps involved in DNA fingerprinting for paternity testing are: 1. **Isolation of DNA:** DNA is extracted from any cell source (e.g., blood, hair, semen) of the child, the mother (if available), and the suspected father(s). 2. **Digestion by Restriction Endonucleases:** The isolated DNA is cut into smaller fragments at specific sites by restriction enzymes. 3. **Separation by Gel Electrophoresis:** The DNA fragments are separated based on size by gel electrophoresis. Smaller fragments travel further in the gel. 4. **Denaturation:** The double-stranded DNA fragments in the gel are denatured into single-stranded DNA using alkali (e.g., NaOH). 5. **Blotting (Southern Blotting):** The separated, single-stranded DNA fragments are transferred from the gel onto a nitrocellulose or nylon membrane. This process is known as blotting. 6. **Hybridization with Probes:** The membrane is then incubated with radioactive DNA probes. These probes are single-stranded DNA sequences designed to bind specifically to the VNTR regions on the membrane via complementary base pairing. 7. **Detection by Autoradiography:** After hybridization, the unbound probes are washed away. The membrane is then exposed to an X-ray film. The radioactive probes highlight the specific VNTR patterns, which appear as a series of dark bands on the film, creating the DNA fingerprint. By comparing the unique banding pattern of the child's DNA fingerprint with that of the mother and suspected father(s), paternity can be established. A child inherits half of their VNTR sequences from the mother and half from the biological father. Therefore, the child's DNA fingerprint will show a specific pattern of bands, where one set matches the mother and the other set matches the biological father.
In simple words: DNA fingerprinting can confirm paternity by comparing unique DNA patterns of a child with those of potential fathers. It involves extracting DNA, cutting it into fragments, separating them, and using radioactive probes to create a distinct band pattern. A match between the child's inherited bands and a suspected father's bands indicates biological paternity.

🎯 Exam Tip: When explaining DNA fingerprinting, ensure you cover the principle (VNTRs, polymorphism) and accurately list the sequential steps (isolation, digestion, separation, denaturation, blotting, hybridization, autoradiography) as this is a frequently asked process-based question.

 

Question 36. Identify the following figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्रोकैरियोटिक गुणसूत्र को दर्शाता है। यह एक गोलाकार डीएनए अणु है जो एक कोशिका के केंद्रक के बजाय सीधे साइटोप्लाज्म में स्थित होता है, जिसमें हिस्टोन प्रोटीन की कमी होती है।
Answer:This figure represents a prokaryotic chromosome.
In simple words: The image shows a prokaryotic chromosome, which is typically a circular DNA molecule found in the cytoplasm of bacteria.

🎯 Exam Tip: Be able to distinguish between eukaryotic and prokaryotic chromosomes based on their general appearance (linear vs. circular, presence/absence of histones).

 

Question 37. Draw schematically a single polynucleotide strand (with at least three nucleotides). Provide labels and directions.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पॉली न्यूक्लियोटाइड स्ट्रैंड के तीन न्यूक्लियोटाइड को योजनाबद्ध रूप से दर्शाता है। इसमें एक 5' फॉस्फेट अंत और एक 3' हाइड्रॉक्सिल अंत दिखाया गया है, जो डीएनए स्ट्रैंड की दिशात्मकता को इंगित करता है। प्रत्येक न्यूक्लियोटाइड एक फॉस्फेट, एक पेंटोज चीनी (डीऑक्सीराइबोज या राइबोज) और एक नाइट्रोजनी बेस से बना होता है, जो यहां P-C-H (फॉस्फेट-चीनी-बेस) के रूप में दर्शाया गया है।
Answer:A nucleotide consists of three main components: a nitrogenous base, a pentose sugar, and a phosphate group. Nitrogenous bases are categorized into two types: 1. Purines (Adenine and Guanine) 2. Pyrimidines (Cytosine, Uracil/Thymine) The schematic drawing would show a backbone formed by alternating phosphate and sugar molecules, with nitrogenous bases attached to the sugar. The strand has a 5' phosphate end and a 3' hydroxyl end, indicating its directionality.
In simple words: A polynucleotide strand is a chain of nucleotides, each made of a sugar, a phosphate, and a nitrogenous base. The chain has a specific direction, from the 5' end (phosphate) to the 3' end (hydroxyl).

🎯 Exam Tip: Understand and be able to draw the basic structure of a nucleotide and how multiple nucleotides link to form a polynucleotide strand, including the 5' and 3' polarity.

 

Question 38. A policeman finds a very small piece of body tissue from the site of a crime and takes it to a. By which technique will they amplify the DNA collected from the tissue sample? b. Mention in a sequence, the three steps involved in each cycle of this technique. c. What is the role of thermostable DNA polymerase in this technique?
Answer:a. The DNA collected from the tissue sample will be amplified using the Polymerase Chain Reaction (PCR) technique. b. The three sequential steps involved in each cycle of the PCR technique are:
(i) Denaturation: The double-stranded DNA template is heated to high temperatures (typically 94-98°C) to separate it into two single strands.
(ii) Annealing: The temperature is lowered (typically 50-65°C) to allow short DNA primers to bind (anneal) to the complementary regions on each single-stranded DNA template.
(iii) Extension (or Elongation): The temperature is raised slightly (typically 72°C), and a DNA polymerase synthesizes new DNA strands by adding nucleotides to the primers, extending them along the template strands. c. The role of thermostable DNA polymerase (e.g., Taq polymerase) is crucial because it can withstand the high temperatures required for the denaturation step in each PCR cycle. This enzyme remains active throughout repeated heating and cooling cycles, allowing for continuous DNA synthesis without needing fresh enzyme after each denaturation step.
In simple words: To get enough DNA from a tiny sample, scientists use PCR, which makes many copies of the DNA. Each cycle involves heating DNA to separate strands, cooling to let primers attach, and then using a heat-resistant enzyme (like Taq polymerase) to build new DNA strands.

🎯 Exam Tip: PCR is a critical technique. Remember the three core steps (denaturation, annealing, extension) and the specific function of thermostable DNA polymerase that enables the reaction to occur repeatedly.

 

Question 39. A few gaps have been left in the following table showing certain terms and their meanings. Fill up the gaps.

Terms	Meanings
i. ..........................	The non-coding sequence in eukaryotic DNA technique used in solving paternity disputes.
ii. ..........................	The technique used in solving paternity disputes.
iii. Restriction endonuclease	..........................
iv. Plasmid	..........................
v. Transgenics	..........................
vi. ..........................	Nucleotide sequences with single-base differences

Answer:

Terms	Meanings
i. Introns	The non-coding sequence in eukaryotic DNA. (Note: The technique used in solving paternity disputes is DNA fingerprinting, which often analyzes non-coding repetitive sequences like VNTRs, which can be part of introns or intergenic regions. The question context seems to be mixed, but "Introns" fits "non-coding sequence in eukaryotic DNA".)
ii. DNA fingerprinting	The technique used in solving paternity disputes.
iii. Restriction endonuclease	Enzymes that cut DNA at specific recognition sites, forming fragments with identifiable genes.
iv. Plasmid	An extrachromosomal, circular DNA molecule that replicates independently alongside the chromosome.
v. Transgenic organisms	Genetically modified organisms that contain foreign genes integrated into their genome.
vi. Single Nucleotide Polymorphism (SNP)	Nucleotide sequences with single-base differences.

In simple words: This table defines key terms in molecular biology: Introns are non-coding DNA, DNA fingerprinting solves paternity, restriction endonucleases cut DNA, plasmids are bacterial DNA circles, transgenic organisms have foreign genes, and SNPs are single-base variations in DNA.

🎯 Exam Tip: Memorizing definitions of essential molecular biology and genetic engineering terms is crucial. Pay attention to specific details for each term.

 

Question 40. A_5'----------B_3'
C_5'----------D_3'
AB and CD represent two strands of a DNA molecule.
When this molecule undergoes replication, a replication fork is formed between A and C in the above.
i. Name the template strands for replication.
ii. Using which strand as the template, will there be continuous synthesis of a complementary DNA strand?
iii. Complementary to which strand will Okazaki segments get synthesised?
iv. What are template strands and Okazaki pieces?
v. In which direction is a new strand synthesised?
Answer:i. The template strands for replication are A (3' to 5' polarity) and C (3' to 5' polarity). Assuming the replication fork opens between A and C, then A and C act as templates. ii. Continuous synthesis of a complementary DNA strand will occur using the template strand that has 3' to 5' polarity (which would be A, given the question implies it's part of a replication fork where new strands are made). This is known as the leading strand template. iii. Okazaki segments, which are discontinuous fragments, will be synthesized complementary to the template strand with 5' to 3' polarity (which would be D, as it's paired with C and has the opposite polarity). This is known as the lagging strand template. iv. A template strand is the original DNA strand that serves as a guide for synthesizing a new, complementary DNA strand. Okazaki pieces are short, discontinuous DNA fragments synthesized on the lagging strand during DNA replication. v. A new strand is always synthesized in the 5' to 3' direction.
In simple words: During DNA replication, original strands (templates) guide new strand synthesis. The new strand is always built from 5' to 3'. One template allows continuous new DNA (leading strand), while the other requires short, discontinuous segments called Okazaki fragments (lagging strand).

🎯 Exam Tip: Understand the directional nature of DNA replication (5' to 3' synthesis), the difference between leading and lagging strands, and the role of Okazaki fragments. These concepts are fundamental to DNA replication.

 

Question 41. In F. Griffith's experiment, how did the nonvirulent strain of streptococcus pneumonia become virulent?
Answer:In F. Griffith's experiment, the nonvirulent strain of *Streptococcus pneumoniae* became virulent by acquiring genetic material from the virulent strain. This process, termed "transformation," allowed the nonvirulent bacteria to synthesize a protective polysaccharide capsule, a characteristic of the virulent strain, thereby gaining virulence.
In simple words: The harmless bacteria in Griffith's experiment became harmful by picking up and incorporating genetic information from the disease-causing bacteria, which taught them how to make a protective coat.

🎯 Exam Tip: Griffith's experiment is a classic example demonstrating genetic transformation. Focus on the transfer of genetic material from the virulent to the nonvirulent strain, leading to a change in phenotype.