The Molecular Basis Of Inheritance

Step 1: परिचय.
उत्परिवर्तन (Mutation) DNA अनुक्रम में होने वाले परिवर्तन को कहते हैं। यह परिवर्तन प्रोटीन संश्लेषण को प्रभावित कर सकता है क्योंकि प्रोटीन का निर्माण mRNA के कोडन से होता है।

Step 2: प्रभाव.
- यदि उत्परिवर्तन Silent mutation है, तो प्रोटीन पर कोई प्रभाव नहीं पड़ता।
- यदि उत्परिवर्तन Missense mutation है, तो प्रोटीन में गलत अमीनो अम्ल जुड़ जाता है।
- Nonsense mutation की स्थिति में समय से पहले stop codon आ जाता है और प्रोटीन अधूरा बन जाता है।
- Frameshift mutation (Deletion/Insertion) से पूरा reading frame बदल जाता है और प्रोटीन पूर्णतः विकृत हो जाता है।

Step 3: निष्कर्ष.
उत्परिवर्तन प्रोटीन की संरचना और कार्य दोनों को प्रभावित कर सकता है और कई बार यह आनुवंशिक रोगों का कारण भी बनता है।

Step 1: Gene expression control levels.
जीन अभिव्यक्ति (Gene Expression) को कई स्तरों पर नियंत्रित किया जा सकता है — मुख्य रूप से transcription और translation स्तर पर।

Step 2: विकल्पों का विश्लेषण.
(A) प्रतिलेखन → Transcription control सबसे महत्वपूर्ण है।
(B) अनुवादन → Translation control भी होता है।
(C) उत्पन्न प्रतिक्रिया → ये gene expression control का standard level नहीं है।
(D) (a) और (b) दोनों → सही, क्योंकि transcription और translation दोनों स्तरों पर regulation होता है।

Step 3: निष्कर्ष.
सही उत्तर है (D) (a) और (b) दोनों।

Step 1: tRNA structure.
tRNA (transfer RNA) की संरचना cloverleaf model (तिपर्णी जैसी) होती है। इसमें anticodon loop और amino acid binding site पाई जाती है।

Step 2: अन्य विकल्पों का विश्लेषण.
(A) mRNA → Linear structure होती है।
(B) rRNA → Ribosome का हिस्सा है, लेकिन तिपर्णी जैसी संरचना नहीं।
(C) hnRNA → Heterogeneous nuclear RNA precursor होता है।

Step 3: निष्कर्ष.
सही उत्तर है (D) tRNA।

Step 1: Nature of DNA replication.
DNA replication is semi-conservative — each new DNA has one old and one new strand. DNA polymerase is the key enzyme.
Step 2: Leading and lagging strands.
Replication is continuous on the leading strand but discontinuous on the lagging strand (Okazaki fragments).
Step 3: Analyzing options.
(A) Correct — DNA replication is semi-conservative.
(B) Correct — DNA polymerase is the main enzyme.
(C) Correct — Errors during replication can cause mutations.
(D) Incorrect — Only leading strand is continuous; lagging is discontinuous.
Step 4: Conclusion.
Thus, the incorrect statement is Replication on both strands of DNA is continuous.

Step 1: Functions of RNA polymerases in eukaryotes.
- RNA polymerase I: Transcribes rRNA (except 5S rRNA).
- RNA polymerase II: Transcribes mRNA, snRNA, some miRNAs.
- RNA polymerase III: Transcribes tRNA and 5S rRNA.
Step 2: Analyzing options.
(A) RNA polymerase: General term, not specific.
(B) RNA polymerase I: Correct — Transcribes most ribosomal RNA.
(C) RNA polymerase II: Incorrect — Responsible for mRNA synthesis.
(D) RNA polymerase III: Incorrect — Transcribes tRNA and 5S rRNA.
Step 3: Conclusion.
Hence, RNA polymerase I is responsible for transcription of rRNA.

Step 1: Nature of genetic code.
The genetic code has redundancy — multiple codons can code for the same amino acid. This is called degeneracy of the genetic code.
Step 2: Example.
Leucine is coded by six different codons.
Step 3: Analyzing options.
(A) Unambiguous: Incorrect — Each codon specifies only one amino acid.
(B) Degenerate/redundant: Correct — More than one codon codes for same amino acid.
(C) Universal: Incorrect — The code is nearly universal across organisms.
(D) Specific: Incorrect — Misleading, as redundancy is the point here.
Step 4: Conclusion.
Hence, such a code is called degenerate/redundant.

Step 1: Genes in lac operon.
The lac operon in E. coli has three structural genes:
- lac Z → β-galactosidase
- lac Y → Permease
- lac A → Transacetylase
Step 2: Function of lac Y.
The lac Y gene codes for permease, which facilitates lactose entry into the bacterial cell.
Step 3: Conclusion.
Thus, the y gene in lac operon codes for permease.

Step 1: Main aim.
The Human Genome Project (HGP) was an international research program to map and sequence the entire human genome.
Step 2: Specific goals.
1. Identify all genes present in human DNA (around 20,000–25,000).
2. Determine the sequence of 3 billion DNA base pairs.
3. Store genetic information in databases.
4. Improve tools for data analysis.
5. Transfer related technologies to the private sector.
6. Address ethical, legal, and social issues (ELSI).
Step 3: Conclusion.
HGP created a foundation for modern genomics and personalized medicine.

Chemical Composition of DNA.
DNA (Deoxyribonucleic acid) is the molecule that carries the genetic information in living organisms. Its chemical composition includes:
1. Nucleotides: The building blocks of DNA, consisting of:
- Sugar: Deoxyribose in DNA.
- Phosphate group: Forms the backbone of the DNA structure.
- Nitrogenous bases: Adenine (A), Thymine (T), Cytosine (C), and Guanine (G).
2. Double Helix Structure: DNA forms a double helix with two complementary strands held together by hydrogen bonds between the bases. Adenine pairs with thymine, and cytosine pairs with guanine.
3. Chromatin: DNA in the nucleus is wrapped around histone proteins, forming structures called nucleosomes, which make up chromatin.
Conclusion.
DNA is made up of nucleotides with a specific structure that encodes genetic information.

Step 1: Stability of DNA compared to RNA.
DNA is more stable than RNA due to the following reasons:
1. Double-stranded structure: DNA exists as a double-stranded helix, which provides greater stability. The double-strand structure helps in protecting the genetic information against damage.
2. Deoxyribose sugar: DNA contains deoxyribose, which lacks a hydroxyl group at the 2' position (compared to RNA, which contains ribose with a hydroxyl group at the 2' position). The absence of this hydroxyl group makes DNA less reactive and more stable.
3. Longer lifespan: DNA is designed to store genetic information for long periods, whereas RNA is more transient and involved in protein synthesis.
Step 2: The roles of the three parts of the transcription unit.
The transcription unit consists of three main parts:
1. Promoter: The promoter is a region of DNA that initiates transcription. It is recognized by RNA polymerase, which binds to it and begins the transcription process.
2. Coding Region: The coding region is the part of the DNA that contains the gene to be transcribed into RNA. This is the portion that is ultimately used to synthesize proteins.
3. Terminator: The terminator is a sequence that signals the end of transcription. When RNA polymerase reaches this sequence, it stops synthesizing RNA and releases the RNA transcript.
Step 3: Conclusion.
DNA is more stable than RNA due to its structural properties, while the three parts of the transcription unit (promoter, coding region, and terminator) play crucial roles in the process of gene transcription.

Step 1: Experiment details.
Hershey and Chase (1952) performed experiments using bacteriophage (virus) and bacteria. They labeled viral DNA with radioactive phosphorus ( ) and viral protein coat with radioactive sulfur ( ).
Step 2: Observation.
After infection, only the -labeled DNA entered the bacterial cells, while the protein coat remained outside.
Step 3: Conclusion.
This proved that DNA is the genetic material in viruses, not proteins.

Step 1: Recall stop codons.
There are three termination (stop) codons in the genetic code: UAG, UAA, and UGA.
Step 2: Function of AUG.
AUG is not a termination codon. It is the initiation/start codon, which codes for Methionine and signals the beginning of protein synthesis.
Step 3: Conclusion.
Thus, the codon which is not a termination codon is AUG.

Step 1: Concept of charging of t-RNA.
Charging of t-RNA refers to the process where an amino acid is chemically linked to its corresponding t-RNA by an enzyme called aminoacyl-tRNA synthetase. This prepares the t-RNA for protein synthesis.
Step 2: Analysis of options.
(1) Correct — This is the definition of charging of t-RNA.
(2) Incorrect, this occurs during translation, not charging.
(3) Incorrect, translation is the entire process of protein synthesis.
(4) Incorrect, charging is not modification but amino acid attachment.
Step 3: Conclusion.
Therefore, the correct answer is (1) Linking of amino acid with cognate t-RNA.

Step 1: Discovery of DNA structure.
The double helical structure of DNA was proposed by Watson and Crick in 1953.
Step 2: X-ray diffraction.
The crucial experimental X-ray diffraction data came from Rosalind Franklin and Maurice Wilkins, which revealed DNA’s helical nature.
Step 3: Conclusion.
Hence, the X-ray diffraction data of DNA was provided by Maurice Wilkins and Rosalind Franklin.

Step 1: Understanding the lac operon.
The lac operon is a genetic system found in bacteria, such as E. coli, which controls the metabolism of lactose. It consists of three structural genes: z, y, and a. The z-gene specifically codes for the enzyme -galactosidase, which breaks down lactose into glucose and galactose.
Step 2: Analyze the options.
- z-gene: This gene codes for -galactosidase, the enzyme responsible for lactose metabolism. - a-gene: This gene codes for an enzyme called transacetylase, not -galactosidase. - y-gene: This gene codes for permease, which facilitates the transport of lactose into the cell but not the breakdown of lactose. - i-gene: This gene codes for the repressor protein that regulates the lac operon but does not directly code for -galactosidase.
Step 3: Conclusion.
The correct gene responsible for coding -galactosidase in the lac operon is the z-gene.
Final Answer:

Step 1: Understanding the terms.
The terms "muton," "cistron," and "recon" were introduced by J. Lederberg. These terms are used in the field of genetics, particularly in the context of gene structure and function.
Step 2: Analyze the options.
- J. Lederberg: He was the scientist who first proposed these terms to describe the functional units of genes and genetic material. - W. Ingram: Known for his work on the molecular genetics of hemoglobin but did not coin these terms. - Bateson: A prominent biologist known for his work on inheritance but not the coining of these terms. - S. Benzer: Known for his work in genetic recombination and the molecular genetics of bacteriophages but not for the terms muton, cistron, and recon.
Step 3: Conclusion.
The correct answer is J. Lederberg, who introduced the terms "muton," "cistron," and "recon."
Final Answer:

Step 1: Understand codons.
An initial codon (start codon) is a specific sequence in mRNA that signals the start of translation. The most common start codon is AUG, which codes for methionine.
Step 2: Review the options.
- UGA is a stop codon, not a start codon. - GUA is a codon for valine but is not the start codon. - AUG is the correct start codon, coding for methionine. - "All of these" is incorrect because only AUG is the start codon.
Step 3: Conclude.
The correct initial codon is AUG, making option (C) the correct answer.
Final Answer:

Step 1: Understand the process.
Transcription is the process where RNA is synthesized from a DNA template. This is the first step in protein synthesis.
Step 2: Review the options.
- Translation refers to the process by which mRNA is translated into a protein at the ribosome, not the synthesis of RNA itself.
- Transcription is the correct answer as it directly involves the synthesis of RNA from a DNA template.
- Replication refers to the process of copying DNA, not RNA synthesis.
- Transduction is a process of gene transfer in bacteria, unrelated to RNA formation.
Step 3: Conclude.
Transcription is the process in which RNA is formed from DNA, making option (B) the correct answer.
Final Answer:

Step 1: Recall nucleotide bonding in DNA.
The nucleotides in DNA are attached through covalent bonds, forming the backbone of the DNA strand.
Step 2: Review the options.
- Hydrogen bonds form between complementary base pairs (A-T and C-G), but the nucleotides are connected by covalent bonds.
- Covalent bonds form the sugar-phosphate backbone, connecting the nucleotides in a DNA strand.
- Electrovalent bonds are not involved in nucleotide bonding in DNA.
- Option (D) "All of these" is incorrect since only covalent bonds are involved in nucleotide bonding.
Step 3: Conclude.
The nucleotides in DNA are connected by covalent bonds, making option (B) the correct answer.
Final Answer:

Step 1: Understand RNA types.
There are several types of RNA involved in protein synthesis, each with distinct roles.
Step 2: Review the options.
- m-RNA (messenger RNA) is soluble and carries genetic information from DNA to the ribosome for protein synthesis.
- r-RNA (ribosomal RNA) is part of the ribosome structure but is not considered soluble.
- c-RNA is not a recognized standard RNA type.
- t-RNA (transfer RNA) is involved in protein synthesis but is not classified as soluble RNA.
Step 3: Conclude.
m-RNA is the soluble form of RNA, making option (A) the correct answer.
Final Answer: