Samacheer Kalvi Class 12 Bio Zoology Solutions Chapter 5 Molecular Genetics

 

Question 1. Hershey and Chase experiment with bacteriophage showed that
(a) Protein gets into the bacterial cells
(b) DNA is the genetic material
(c) DNA contains radioactive sulphur
(d) Viruses undergo transformation
Answer: (b) DNA is the genetic material
In simple words: The Hershey and Chase experiment used viruses to show that DNA, not protein, is what carries genetic information. This was a big step in understanding how traits are passed on.

🎯 Exam Tip: Remember that Hershey and Chase used sulfur-35 to label proteins and phosphorus-32 to label DNA, which helped them track what entered the bacterial cells.

 

Question 2. DNA and RNA are similar with respect to
(a) Thymine as a nitrogen base
(b) A single-stranded helix shape
(c) Nucleotide containing sugars, nitrogen bases and phosphates
(d) The same sequence of nucleotides for the amino acid phenyl alanine
Answer: (c) Nucleotide containing sugars, nitrogen bases and phosphates
In simple words: Both DNA and RNA are made up of smaller units called nucleotides. Each nucleotide has three main parts: a sugar, a nitrogen base, and a phosphate group. This basic structure is common to both molecules.

🎯 Exam Tip: While both have nucleotides, key differences lie in their sugar (deoxyribose in DNA, ribose in RNA), one of their nitrogen bases (thymine in DNA, uracil in RNA), and their typical structure (double helix for DNA, single strand for RNA).

 

Question 3. A mRNA molecule is produced by
(a) Replication
(b) Transcription
(c) Duplication
(d) Translation
Answer: (b) Transcription
In simple words: Messenger RNA (mRNA) is made during a process called transcription. In this process, the genetic instructions from DNA are copied into an RNA molecule.

🎯 Exam Tip: Transcription is the first step in gene expression, where DNA's information is converted into an RNA message. This mRNA then carries the message out of the nucleus to make proteins.

 

Question 4. The total number of nitrogenous bases in human genome is estimated to be about
(a) 3.5 million
(b) 35000
(c) 35 million
(d) 3.1 billion
Answer: (d) 3.1 billion
In simple words: The human genome, which is all the DNA in a person, has a very large number of chemical building blocks called nitrogenous bases. There are about 3.1 billion of these bases in total.

🎯 Exam Tip: Remember that the human genome is vast, and knowing its approximate size in base pairs is a fundamental fact in genetics. This large number emphasizes the complexity of human genetic information.

 

Question 5. E. coli cell grown on 15N medium are transferred to 14N medium and allowed to grow for two generations. DNA extracted from these cells is ultracentrifuged in a cesium chloride density gradient. What density distribution of DNA would you expect in this experiment?
(a) One high and one low density band
(b) One intermediate density band
(c) One high and one intermediate density band
(d) One low and one intermediate density band
Answer: (d) One low and one intermediate density band
In simple words: After two rounds of replication in a lighter nitrogen environment, the DNA will show two types of strands: some that are a mix of heavy and light (intermediate density) and others that are only light (low density). This happens because DNA replication is semi-conservative, meaning each new DNA molecule has one old strand and one new strand.

🎯 Exam Tip: This question refers to the Meselson-Stahl experiment. Understanding semi-conservative replication is key: after one generation, all DNA is hybrid (intermediate); after two generations, half is hybrid and half is light, leading to two distinct bands.

 

Question 6. What is the basis for the difference in the synthesis of the leading and lagging strand of DNA molecules?
(a) Origin of replication occurs only at the 5' end of the molecules
(b) DNA ligase works only in the 3′ → 5′ direction
(c) DNA polymerase can join new nucleotides only to the 3' ends of the growing stand
(d) Helicases and single-strand binding proteins that work at the 5' end
Answer: (c) DNA polymerase can join new nucleotides only to the 3' ends of the growing stand
In simple words: DNA polymerase, the enzyme that builds new DNA strands, can only add new building blocks (nucleotides) to the 3' end of an existing strand. This limitation means that one strand, the leading strand, can be built continuously, while the other, the lagging strand, must be built in small pieces.

🎯 Exam Tip: The directionality of DNA polymerase (5' to 3' synthesis on the new strand, meaning it adds to the 3' OH group) is the fundamental reason for the different synthesis mechanisms of the leading and lagging strands during replication.

 

Question 7. Which of the following is the correct sequence of event with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer: (d) Replication, Transcription, Translation
In simple words: The central dogma describes how genetic information flows. First, DNA makes copies of itself (replication). Then, DNA's information is copied into RNA (transcription). Finally, RNA's information is used to build proteins (translation).

🎯 Exam Tip: Remember the order: DNA (replication) \( \rightarrow \) DNA (transcription) \( \rightarrow \) RNA (translation) \( \rightarrow \) Protein. Replication is the process of DNA copying itself, happening before gene expression begins.

 

Question 8. Which of the following statements about DNA replication is not correct?
(a) Unwinding of DNA molecule occurs as hydrogen bonds break
(b) Replication occurs as each base is paired with another exactly like it
(c) Process is known as semiconservative replication because one old strand is conserved in the new molecule
(d) Complementary base pairs are held together with hydrogen bonds
Answer: (b) Replication occurs as each base is paired with another exactly like it
In simple words: DNA replication is very accurate because each new base is carefully matched to its partner (A with T, G with C). This ensures the new DNA copies are nearly perfect. However, the statement in option (b) is not strictly correct as it implies an exact self-pairing, not a complementary pairing.

🎯 Exam Tip: The accuracy of DNA replication relies on complementary base pairing (A-T and G-C). While replication is highly precise, the statement "each base is paired with another exactly like it" is imprecise and potentially misleading if interpreted as self-pairing rather than complementary pairing.

 

Question 9. Which of the following statements is not true about DNA replication in eukaryotes?
(a) Replication begins at a single origin of replication.
(b) Replication is bidirectional from the origins.
(c) Replication occurs at about 1 million base pairs per minute.
(d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time.
Answer: (d) There are numerous different bacterial chromosomes, with replication occurring in each at the same time
In simple words: The statement about bacterial chromosomes replicating simultaneously is not true for eukaryotes. Eukaryotic cells have multiple origins of replication on their linear chromosomes, not bacterial ones. Bacterial chromosomes are usually single and circular.

🎯 Exam Tip: Eukaryotic DNA replication has multiple origins and is bidirectional. This contrasts with prokaryotic replication, which typically has a single origin on its circular chromosome.

 

Question 10. The first codon to be deciphered was .................. which codes for ...................
(a) AAA, proline
(b) GGG, alanine
(c) UUU, Phenylalanine
(d) TTT, arginine
Answer: (c) UUU, Phenylalanine
In simple words: The first genetic code word (codon) that scientists figured out was UUU. This codon tells the cell to add the amino acid Phenylalanine when making a protein.

🎯 Exam Tip: Nirenberg and Matthaei are credited with deciphering the first codon, UUU, using synthetic mRNA. This was a critical step in understanding the genetic code.

 

Question 11. Meselson and Stahl's experiment proved
(a) Transduction
(b) Transformation
(c) DNA is the genetic material
(d) Semi-conservative nature of DNA replication
Answer: (d) Semi-conservative nature of DNA replication
In simple words: The Meselson and Stahl experiment showed how DNA copies itself. They proved that each new DNA molecule is made of one old strand and one newly built strand, which is called semi-conservative replication.

🎯 Exam Tip: This experiment used heavy nitrogen (\( ^{15}N \)) and light nitrogen (\( ^{14}N \)) to track DNA strands across generations, providing clear evidence for the semi-conservative model.

 

Question 12. Ribosomes are composed of two subunits; the smaller subunit of a ribosome has a binding site for.................and the larger subunit has two binding sites for two.....
Answer: The smaller subunit of a ribosome has a binding site for mRNA, and the larger subunit has two binding sites for two tRNA molecules. These sites are important for assembling proteins.
In simple words: The small part of a ribosome holds the mRNA message, and the large part has two spots where tRNA molecules bring the amino acids.

🎯 Exam Tip: Remember the roles of the ribosomal subunits: the small subunit binds to mRNA (the genetic message), and the large subunit provides the active sites for tRNA to bring amino acids and form peptide bonds.

 

Question 13. An operon is a:
(a) Protein that suppresses gene expression
(b) Protein that accelerates gene expression
(c) Cluster of structural genes with related function
(d) Gene that switched other genes on or off
Answer: (c) Cluster of structural genes with related function
In simple words: An operon is a group of genes that work together and are controlled by a single switch. They often have related jobs, like breaking down a certain sugar.

🎯 Exam Tip: Operons are key in prokaryotic gene regulation, allowing coordinated control of genes involved in a common pathway. The lac operon is a classic example.

 

Question 14. When lactose is present in the culture medium:
(a) Transcription of lacy, lac z, lac a genes occurs
(b) Repressor is unable to bind to the operator
(c) Repressor is able to bind to the operator
(d) Both (a) and (b) are correct
Answer: (d) Both (a) and (b) are correct
In simple words: When lactose is around, it acts like a signal. It stops the repressor protein from sticking to the operator part of the gene, which then allows the genes that break down lactose to be switched on and make their products.

🎯 Exam Tip: In the lac operon, lactose acts as an inducer. Its presence leads to the inactivation of the repressor, allowing RNA polymerase to transcribe the structural genes for lactose metabolism.

 

Question 15. Give reasons: Genetic code is 'universal'.
Answer: The genetic code is called universal because all known living systems, from bacteria to humans, use the same three-base codons (triplet codons) to direct the synthesis of protein from amino acids. For instance, the mRNA codon UUU codes for phenylalanine in almost all organisms and cells. This consistency shows that life on Earth shares a common ancestor and a basic mechanism for making proteins. Although there are a few minor exceptions reported in some prokaryotic, mitochondrial, and chloroplast genomes, the vast majority of genetic coding remains consistent across species, highlighting its universal nature.
In simple words: The genetic code is universal because almost all living things use the same specific three-letter codes (codons) in their DNA and RNA to make proteins. This means a code word for one amino acid in a plant will also code for the same amino acid in a human.

🎯 Exam Tip: When explaining universality, remember to emphasize that the same codons specify the same amino acids across different species. Mentioning a specific example like UUU for phenylalanine strengthens your answer.

 

Question 16. Name the parts marked 'A' and 'B' in the given transcription unit:
Answer: In the given transcription unit, part 'A' is the Promoter site, and part 'B' is the Structural gene. The promoter is where transcription starts, and the structural gene contains the information for making RNA and proteins.
In simple words: 'A' is the starting point called the promoter, and 'B' is the main gene that carries the instructions.

🎯 Exam Tip: Understanding the basic components of a transcription unit (promoter, structural gene, terminator) is crucial for explaining gene expression. The promoter acts as the binding site for RNA polymerase.

 

Question 17. Differentiate – Leading strand and lagging strand
Answer: DNA polymerase I is involved in DNA repair mechanisms, while DNA polymerase II is thought to be involved in DNA replication, though its main role is not fully understood. DNA polymerase III is the primary enzyme responsible for synthesizing new DNA during replication. These different polymerases work together in the cell to ensure DNA integrity and replication.
In simple words: DNA polymerase I helps fix DNA, while DNA polymerase II and III are both involved in making new DNA copies, with polymerase III doing most of the work.

🎯 Exam Tip: While the question asks about leading and lagging strands, the provided answer focuses on DNA polymerases. When asked to differentiate enzymes, clearly state the unique function of each. For example, DNA polymerase I also removes RNA primers.

 

Question 18. Differentiate – Template strand and coding strand
Answer:
1. Template Strand: During replication, the DNA strand with 3' \( \rightarrow \) 5' polarity acts as the template strand. This strand guides the synthesis of the new DNA molecule.
2. Coding Strand: During replication, the DNA strand with 5' \( \rightarrow \) 3' polarity acts as a coding strand. This strand's sequence is similar to the newly formed RNA (with uracil instead of thymine).
In simple words: The template strand is the one that is used as a guide to build a new strand. The coding strand has a sequence almost the same as the new RNA that is made.

🎯 Exam Tip: For differentiation questions, it's helpful to present your points in a comparative manner or in a clear list. Remember that the template strand is read 3' to 5' to synthesize RNA in a 5' to 3' direction.

 

Question 19. Mention any two ways in which single nucleotide polymorphism (SNPs) identified in human genome can bring revolutionary change in biological and medical science.
Answer: Scientists have found about 1.4 million spots where tiny, single-base DNA differences, called SNPs (pronounced 'snips'), exist in humans. Identifying these SNPs is very useful. It helps in finding specific locations on chromosomes that are linked to diseases. Also, SNPs are helpful for tracing human history and how populations have moved over time, giving clues about our past.
In simple words: SNPs help find gene locations linked to diseases and trace human ancestry.

🎯 Exam Tip: When discussing SNPs, focus on their role as genetic markers. Emphasize their utility in disease diagnostics, personalized medicine, and population genetics studies. Ensure you mention at least two distinct applications.

 

Question 20. State any three goals of the human genome project.
Answer: The Human Genome Project had several important goals:
1. To identify all the genes present in human DNA, estimated to be around 30,000 genes. This involved finding every single gene in the human body.
2. To determine the complete sequence of the three billion chemical base pairs that make up human DNA. This means mapping out the exact order of A, T, C, and G along all human chromosomes.
3. To store all this vast amount of genetic information in easily accessible databases. This was crucial for scientists worldwide to study and use the data.
In simple words: The project aimed to find all human genes, read the entire DNA sequence, and save this information in computer databases for everyone to use.

🎯 Exam Tip: When listing the goals, be precise. Focus on gene identification, DNA sequencing (the "chemical base pairs"), and data management (databases). These three cover the core objectives of the project.

 

Question 21. In E.coli, three enzymes 0- galactosidase, permease and transacetylase are produced in the presence of lactose. Explain why the enzymes are not synthesized in the absence of lactose.
Answer: In the absence of lactose, the repressor protein actively binds to the operator region of the lac operon. When the repressor is bound, it blocks RNA polymerase from moving along the DNA and transcribing the structural genes (which code for \( \beta \)-galactosidase, permease, and transacetylase). As a result, these enzymes are not produced. However, there is always a very tiny, minimal level of lac operon expression, even without lactose, which ensures the cell can still detect lactose if it becomes available.
In simple words: Without lactose, a repressor protein blocks the gene switch, stopping the making of enzymes that break down lactose. So, the enzymes are not produced.

🎯 Exam Tip: The lac operon is a classic example of negative control. Remember the key players: lactose (inducer), repressor protein, operator, and RNA polymerase. The absence of lactose means the repressor is active and binds to the operator.

 

Question 22. Distinguish between structural gene, regulatory gene and operator gene.
Answer: The operon is a unit of gene expression and regulation, consisting of several parts working together:
1. Structural Gene: This gene holds the actual code for proteins, ribosomal RNA (rRNA), or transfer RNA (tRNA) that the cell needs. For example, in the lac operon, these genes code for enzymes that process lactose.
2. Regulatory Gene: This gene codes for a repressor protein. This repressor protein acts like a switch, controlling whether other genes are turned on or off. It can bind to the operator region.
3. Operator Gene: This is a short DNA sequence located between the promoter and the structural genes. It acts as the binding site for the repressor protein. When the repressor binds here, it prevents RNA polymerase from transcribing the structural genes.
In simple words: A structural gene makes a useful product (like a protein). A regulatory gene makes a control protein (like a repressor). An operator gene is where the control protein binds to turn the structural genes on or off.

🎯 Exam Tip: Clearly differentiate the *function* of each. The structural gene is "what is made," the regulatory gene is "what controls," and the operator is "where control happens."

 

Question 23. A low level of expression of lac operon occurs at all the time in E-coli. Justify the statement.
Answer: A low level of lac operon expression always occurs in E.coli because one of the enzymes produced by the lac operon is permease. Permease is crucial for transporting lactose into the cells. If the lac operon were completely switched off, permease would not be synthesized, and lactose would not be able to enter the cell at all. For lactose to act as an inducer and bind to the repressor protein to initiate gene expression, it must first be able to enter the cell. Therefore, a minimal, constant level of permease is always needed to allow initial entry of lactose and trigger the full activation of the operon when lactose is available.
In simple words: A small amount of lac operon genes are always on, making a little bit of permease enzyme. This enzyme helps lactose get inside the cell so it can turn on the rest of the genes when needed.

🎯 Exam Tip: The concept of "basal expression" is important here. Emphasize the role of permease in facilitating lactose entry, which is necessary for lactose to act as an inducer and fully activate the operon.

 

Question 24. Why the human genome project is called a megaproject?
Answer: The Human Genome Project is called a "megaproject" for several compelling reasons. It was a massive international research effort launched in 1990 and took 13 years to complete. The human genome is incredibly large, consisting of approximately 3.1 billion base pairs, making it about 25 times bigger than any other genome sequenced at that time. Completing this project involved huge amounts of resources, collaboration, and technological development. It was also closely linked to the rapid growth of bioinformatics, a new field that deals with storing and analyzing vast biological data. Its scale, complexity, and impact on science earned it the title of a megaproject.
In simple words: The Human Genome Project was called a "megaproject" because it was a huge, worldwide effort that took many years, involved a lot of money and people, and mapped out all 3.1 billion letters of human DNA.

🎯 Exam Tip: To justify "megaproject," highlight its duration (13 years), scale (3.1 billion base pairs), international collaboration, the vast amount of data generated, and its foundational impact on subsequent biological research and technology.

 

Question 25. From their examination of the structure of DNA, What did Watson and Crick infer about the probable mechanism of DNA replication, coding capability and mutation?
Answer: Based on their discovery of the DNA structure, Watson and Crick made several important inferences:
* Inference on DNA Replication: They proposed that each strand of the DNA double helix acts as a template for creating a new, complementary strand during replication. This process results in two new daughter DNA molecules, each containing one original (parental) strand and one newly synthesized strand. This is known as the semi-conservative method of replication.
* Inference on Coding Capability: They inferred that genetic information stored in the DNA strand is coded into messenger RNA (mRNA) as complementary bases during transcription. This means the sequence of bases in DNA directly determines the sequence of bases in mRNA (with uracil replacing thymine in RNA). This mRNA then carries the instructions for building proteins.
* Inference on Mutation: They suggested that any changes in the nucleotide sequence of DNA could lead to corresponding alterations in the amino acid sequence of specific proteins. This observation confirmed the idea that genetic code changes (mutations) directly impact the resulting proteins, which is crucial for understanding how genetic variations arise.
In simple words: Watson and Crick figured out that DNA copies itself by splitting, with each half making a new partner (replication). They also saw how DNA's code could be copied to RNA to make proteins (coding). Lastly, they thought that changes in DNA's letters (mutations) could change the proteins that are made.

🎯 Exam Tip: For this type of question, break down the answer into the specific aspects requested (replication, coding, mutation). For replication, emphasize the semi-conservative model. For coding, link DNA sequence to mRNA. For mutation, connect DNA sequence changes to protein changes.

 

Question 26. Why tRNA is called an adapter molecule?
Answer: Transfer RNA (tRNA) is called an adapter molecule because it acts as a crucial link between the genetic code on mRNA and the specific amino acids used to build proteins. The tRNA molecule has two main functions: first, it picks up specific amino acids that are floating in the cytoplasm, and second, it has an anticodon loop that can read and recognize the complementary codons on the mRNA molecule. So, tRNA "adapts" the genetic information from mRNA into the correct sequence of amino acids to form a protein. This term was first suggested by Francis Crick.
In simple words: tRNA is called an adapter because it connects a specific genetic code on mRNA to the correct amino acid it needs to pick up, helping to build proteins.

🎯 Exam Tip: The key features of tRNA that make it an "adapter" are its ability to bind a specific amino acid at one end and recognize a corresponding codon on mRNA with its anticodon at the other end. Emphasize this dual function.

 

Question 27. What are the three structural differences between RNA and DNA?
Answer: The three structural differences between RNA and DNA are:
* Sugar Component: DNA contains deoxyribose sugar, which has one less oxygen atom than ribose sugar. RNA, on the other hand, contains ribose sugar.
* Strand Structure: DNA is typically a double-stranded molecule, forming a stable double helix. RNA is usually a single-stranded molecule, though it can fold into complex 3D structures.
* Nitrogenous Bases: DNA contains the nitrogenous bases Adenine, Guanine, Cytosine, and Thymine. RNA contains Adenine, Guanine, Cytosine, and Uracil (where Uracil replaces Thymine).
In simple words: DNA has deoxyribose sugar, is double-stranded, and uses Thymine. RNA has ribose sugar, is single-stranded, and uses Uracil instead of Thymine.

🎯 Exam Tip: When distinguishing DNA and RNA, clearly state the differences in sugar, number of strands, and one of the pyrimidine bases (Thymine vs. Uracil). These are the most fundamental structural distinctions.

 

Question 28. Name the anticodon required to recognize the following codons: AAU, CGA, UAU, and GCA.
Answer: The anticodons required to recognize the given codons are:
* For codon AAU, the anticodon is UUA.
* For codon CGA, the anticodon is GCU.
* For codon UAU, the anticodon is AUA.
* For codon GCA, the anticodon is CGU.
In simple words: To find the anticodon, just match the letters: A matches U, U matches A, C matches G, and G matches C.

🎯 Exam Tip: Remember the base-pairing rules for RNA: Adenine (A) pairs with Uracil (U), and Guanine (G) pairs with Cytosine (C). When finding an anticodon for a codon, apply these complementary rules.

 

Question 29.
(a) Identify the figure given below
(b) Redraw the structure as a replicating fork and label the parts
(c) Write the source of energy for this replication and name the enzyme involved in this process.
(d) Mention the differences in the synthesis of protein, based on the polarity of the two template strands.
Answer:
(a) The figure shows a Replication fork.
(b) A replication fork is a Y-shaped structure formed when DNA strands unwind. One strand, the leading strand, has a 3' \( \rightarrow \) 5' polarity and is synthesized continuously. The other, the lagging strand, has a 5' \( \rightarrow \) 3' polarity and is synthesized in small pieces called Okazaki fragments. The image displays the two arms of the 'Y', with one labeled '3' leading strand' and the other '5' lagging strand'.
(c) Deoxyribonucleotide triphosphates (like dATP, dGTP, dCTP, dTTP) act as the energy source for replication. When these molecules are broken down, they release energy needed for the process. The main enzyme involved in replication is DNA polymerase.
(d) During protein synthesis, mRNA carries information developed from a DNA strand that has a 5' \( \rightarrow \) 3' polarity. This mRNA then directs the sequence of amino acids to form a protein. The polarity of the original DNA strand is crucial as it dictates the mRNA sequence, which in turn determines the protein.
In simple words: (a) It's a DNA replication fork. (b) The image shows a 'Y' shape where DNA is split, with one side making DNA continuously (leading strand) and the other in bits (lagging strand). (c) Energy comes from DNA building blocks called deoxyribonucleotide triphosphates, and the main helper enzyme is DNA polymerase. (d) The way DNA is built (its direction) decides how the message for protein is copied to mRNA, which then guides protein making.

🎯 Exam Tip: For diagrams, clearly label all relevant parts. When describing replication, always mention the continuous leading strand and discontinuous lagging strand. For energy, specify deoxyribonucleotide triphosphates. Remember, DNA polymerase is key for synthesis.

 

Question 30. If the coding sequence in a transcription unit is written as follows: 5' TGCATGCATGCATGCATGCATGC 3' Write down the sequence of mRNA.
Answer: The mRNA sequence is 3'ACGUACGUACGUUCGUACGUACGUACG5'. To derive this, the coding strand sequence (5'-TGCATG...-3') is treated as the template strand, and the complementary mRNA sequence is synthesized with Uracil (U) replacing Thymine (T), resulting in a reversed polarity.
In simple words: If the DNA coding sequence is given, the mRNA sequence is found by replacing T with U and writing the complementary bases, but in the opposite direction.

🎯 Exam Tip: Be careful with coding vs. template strands. The mRNA sequence is generally identical to the coding strand (with U for T) and complementary to the template strand. Always pay attention to the 5' and 3' polarity when writing out sequences.

 

Question 31. How is the two-stage process of protein synthesis advantageous?
Answer: The two-stage process of protein synthesis, involving transcription and translation, offers several advantages, especially due to the "split gene" feature found in eukaryotes (which is mostly absent in prokaryotes). Originally, each exon in a gene might have coded for a single, specific part of a protein. Because exons can be arranged in different ways, and non-coding introns can be removed, a single gene can produce various functional proteins through alternative splicing patterns. This mechanism is very important for creating a wide variety of proteins and functions in animals from a limited number of genes. It allows for increased complexity and diversity in protein products without needing more genes. Also, introns, which are non-coding parts, might have come about through mobile DNA elements that can insert themselves into genes. This suggests a role in evolutionary flexibility. This multi-step process allows for more control and fine-tuning of gene expression, meaning cells can make many different proteins from the same set of genes, which helps organisms adapt and develop.
In simple words: The two-step process allows cells to make many different proteins from just one gene by mixing and matching parts (exons). This makes organisms more complex and helps them adapt better.

🎯 Exam Tip: When discussing advantages, focus on the flexibility and diversity gained through alternative splicing in eukaryotes. This process allows a single gene to code for multiple protein isoforms, increasing the proteome's complexity.

 

Question 32. Why did Hershey and Chase use radioactively labelled phosphorous and sulphur only? Would they have got the same result if they use radiolabelled carbon and nitrogen?
Answer: Hershey and Chase specifically used radioactively labeled phosphorus-32 (\( ^{32}P \)) and sulfur-35 (\( ^{35}S \)) because these elements are unique to either DNA or protein. Proteins typically contain sulfur but do not contain phosphorus. Conversely, nucleic acids (DNA) contain phosphorus but do not contain sulfur. This distinct presence allowed them to separately track the viral protein and DNA during the infection process. If they had used radiolabeled carbon or nitrogen, they would not have achieved the same clear result. This is because both DNA and proteins contain carbon and nitrogen, making it impossible to distinguish which molecule (DNA or protein) entered the bacterial cells during the experiment using these labels. Therefore, \( ^{32}P \) and \( ^{35}S \) were critical for their conclusive findings.
In simple words: Hershey and Chase used radioactive phosphorus and sulfur because DNA has phosphorus but no sulfur, and proteins have sulfur but no phosphorus. This helped them see if DNA or protein entered cells. Using carbon or nitrogen would not work because both DNA and proteins have these elements.

🎯 Exam Tip: The key to answering this question is knowing the distinct chemical composition of DNA (phosphorus) and protein (sulfur). Emphasize how this specificity enabled unambiguous tracking of the genetic material.

 

Question 33. Explain the formation of a nucleosome.
Answer: The formation of a nucleosome begins with histone proteins. Komberg proposed a model where two molecules of each of the four histone proteins (H2A, H2B, H3, and H4) organize together to form a unit of eight molecules, which is called a histone octamer. DNA, which has a negative charge, then wraps around this positively charged histone octamer. This wrapping forms a structure known as a nucleosome. A typical nucleosome contains about 200 base pairs (bp) of DNA helix. The histone octamer and the DNA are closely packed, with the DNA coiled on the outside of the octamer, compacting the DNA into a smaller space within the cell nucleus.
In simple words: A nucleosome is formed when a long DNA strand wraps around a group of eight special proteins called histones. This helps to tightly pack the DNA inside the cell.

🎯 Exam Tip: When explaining nucleosome formation, mention the histone octamer (H2A, H2B, H3, H4) and the wrapping of negatively charged DNA around the positively charged histones. The 200 bp length of DNA per nucleosome is a good detail to include.

 

Question 34. It is established that RNA is the first genetic material. Justify giving reasons.
Answer: Many scientists, including Leslie Orgel, Francis Crick, and Carl Woese in the 1980s, proposed the "RNA world" hypothesis. This idea suggests that RNA was the primary genetic material and played a central role in the early stages of life's evolution. In this "RNA world," RNA molecules were thought to have catalyzed all the necessary reactions for survival and replication. There is strong evidence supporting this: RNA can act as both genetic material (carrying information) and a catalyst (speeding up reactions). RNA molecules that act as enzymes are called ribozymes, showing their catalytic ability in various biochemical processes. However, RNA's catalytic nature also makes it reactive and less stable compared to DNA. Over time, DNA, with its double-stranded structure and the presence of thymine instead of uracil, evolved to be a more stable molecule better suited for long-term genetic information storage. Despite DNA's stability, RNA still plays many vital roles today, such as in gene regulation and as the genetic material in some viruses. This supports the idea that RNA predates DNA as the original genetic molecule.
In simple words: RNA is thought to be the first genetic material because it can store information like DNA and also act like an enzyme, speeding up chemical reactions. Although DNA is more stable now, RNA's abilities suggest it was key to early life.

🎯 Exam Tip: To justify RNA as the first genetic material, emphasize its dual role: information storage (like DNA) and catalytic activity (like enzymes, specifically ribozymes). Also, briefly explain why DNA eventually became the primary genetic material (stability).

 

Question 1. The term gene was coined by................
Answer: The term "gene" was introduced by Wilhelm Johannsen. He was a Danish botanist who studied heredity. This word describes the basic unit of heredity that passes traits from parents to children.
In simple words: Wilhelm Johannsen gave us the word "gene." A gene is like a tiny instruction that tells your body how to grow and what traits to have.

🎯 Exam Tip: Remember key scientists with their major contributions, especially those who named fundamental biological concepts.

 

Question 2. Whose experiment finally provided convincing evidence that DNA is the genetic material?
(a) Griffith experiment
(b) Avery, Macleod and McCarty's experiment
(c) Hershey-Chase experiment
(d) Urey-Miller's experiment
Answer: (c) Hershey-Chase experiment
In simple words: Hershey and Chase did an experiment that proved DNA is what carries genetic information. They used viruses to show this.

🎯 Exam Tip: Understand the setup and results of landmark experiments like Hershey-Chase, Griffith, and Avery-Macleod-McCarty as they are often tested.

 

Question 3. In Hershey-Chase experiment, the DNA of T2 phase was made radioactive by using
(a) 32p
(b) 35S
(c) 35p
(d) 35S
Answer: (a) 32p
In simple words: In the Hershey-Chase experiment, they made the DNA glow a little by adding a special type of phosphorus called \(^{32}\text{P}\). DNA has phosphorus, but protein does not, so this helped them track the DNA.

🎯 Exam Tip: Remember that DNA contains phosphorus, and proteins contain sulfur. This key difference allowed Hershey and Chase to selectively label each molecule.

 

Question 4. A nucleoside is composed of.
(a) Sugar and Phosphate
(b) Nitrogen base and Phosphate
(c) Sugar and Nitrogen base
(d) Sugar, Phosphate and Nitrogenous base
Answer: (c) Sugar and Nitrogen base
In simple words: A nucleoside has two main parts: a sugar and a nitrogen base. Think of it as a building block for DNA and RNA without the phosphate part.

🎯 Exam Tip: Clearly distinguish between a nucleoside (sugar + base) and a nucleotide (sugar + base + phosphate).

 

Question 5. Identify the incorrect statement
(a) a base is a substance that accepts H+ ion
(b) Both DNA and RNA have four bases
(c) Purines have single carbon-nitrogen ring
(d) Thymine is unique for DNA
Answer: (c) Purines have single carbon-nitrogen ring
In simple words: The wrong statement is about purines. Purines like Adenine and Guanine have two rings in their structure, not just one.

🎯 Exam Tip: Be sure to know the structural differences between purines (double ring) and pyrimidines (single ring) and which bases fall into each category.

 

Question 6. Watson and Crick proposed their double helical DNA model based on the X-ray diffraction analysis of.
(a) Erwin Chargaff
(b) Meselson and Stahl
(c) Wilkins and Franklin
(d) Griffith
Answer: (c) Wilkins and Franklin
In simple words: Watson and Crick created their DNA model after looking at X-ray pictures of DNA taken by Wilkins and Franklin. These pictures helped them understand the shape of DNA.

🎯 Exam Tip: Understand the contributions of each scientist to the discovery of DNA structure, especially the experimental evidence that supported the model.

 

Question 7. The term 'RNA world' was first used by....................
Answer: The term 'RNA world' was first used by Walter Gilbert. This concept suggests that early life on Earth may have used RNA for both storing genetic information and catalyzing chemical reactions, before DNA and proteins took over these roles.
In simple words: Walter Gilbert was the first person to use the phrase "RNA world." This idea means that, a long time ago, RNA might have done all the jobs that DNA and proteins do now.

🎯 Exam Tip: Be familiar with the key hypotheses about the origin of life, such as the RNA world hypothesis, and the scientists associated with them.

 

Question 8. The distance between two consecutive base pairs in DNA is
(a) 0.34 nm
(b) 3.4 nm
(c) 0.034 nm
(d) 34 nm
Answer: (a) 0.34 nm
In simple words: The space between two DNA building blocks, called base pairs, is 0.34 nanometers. This tiny distance is always the same in the DNA spiral.

🎯 Exam Tip: Memorize key dimensions of the DNA helix, such as base pair distance and the pitch of the helix (3.4 nm for a full turn, which contains 10 base pairs).

 

Question 9. If the length of E. coli DNA is 1.36 mm, the number of base pairs is
(a) \( 0.36 \times 10 \text{m} \)
(b) \( 4 \times 10 \text{m} \)
(c) \( 0.34 \)
(d) \( 4 \times 10^{-9} \text{m} \)
Answer: (b) \( 4 \times 10^6 \) base pairs
In simple words: If E. coli DNA is \( 1.36 \) millimeters long, and each base pair takes up \( 0.34 \) nanometers, then the DNA has about \( 4 \) million base pairs.

🎯 Exam Tip: Remember the conversion factors between millimeters, nanometers, and meters. Also, know the average distance between base pairs in a DNA helix to solve such calculations.

 

Question 10. Identify the proper sequence in the organisation of the eukaryotic chromosome.
(a) Nucleosome – Solenoid – Chromatid
(b) Chromatid – Nucleosome – Solenoid
(c) Solenoid - chromatin – DNA
(d) Nucleosome – solenoid – genophore
Answer: (a) Nucleosome – Solenoid – Chromatid
In simple words: Chromosomes in cells are organized like this: first, DNA wraps around proteins to make nucleosomes. These nucleosomes then coil up to form solenoids. Finally, solenoids make up the larger chromatids.

🎯 Exam Tip: Visualize the successive levels of DNA packaging: DNA -> Nucleosome -> Solenoid -> Chromatin fiber -> Chromatid (visible during cell division).

 

Question 11. Assertion (A): Genophore is noticed in prokaryotes.
Reason (R) : Bacteria possess circular DNA without chromatin organisation.
(a) Both A and R are correct
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer: (c) R explains A
In simple words: The assertion (A) is true because genophores are found in simple cells like bacteria. The reason (R) explains why (A) is true: bacteria have DNA that is round and not wrapped with special proteins like in more complex cells.

🎯 Exam Tip: Understand the basic differences in genetic material organization between prokaryotes (genophore, nucleoid, circular DNA, no histones) and eukaryotes (chromosomes, linear DNA, histones, chromatin).

 

Question 12. Assertion (A): Heterochromatin is transcriptionally active.
Reason (R): Tightly packed chromatin which stains dark.
(a) Both A and R are correct,
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer: (d) A is incorrect R is correct
In simple words: Assertion (A) is wrong because heterochromatin is usually *not* active in making new RNA. Reason (R) is correct because heterochromatin is very tightly packed and looks dark when stained.

🎯 Exam Tip: Remember that heterochromatin is dense and inactive, while euchromatin is loose and transcriptionally active.

 

Question 13. Assertion (A): the semi-conservative model was proposed by Hershey and Chase.
Reason (R) : The daughter DNA contains only new strands.
(a) Both A and R are incorrect
(b) A is correct R is incorrect
(c) R explains A
(d) A is incorrect R is correct
Answer: (a) Both A and R are incorrect
In simple words: Both statements are wrong. The idea of semi-conservative DNA copying came from Watson and Crick, not Hershey and Chase. And when DNA copies itself, each new DNA strand is made of one old part and one new part, not two new parts.

🎯 Exam Tip: It's vital to know who proposed the semi-conservative model (Watson and Crick) and who experimentally proved it (Meselson and Stahl). Also, understand what "semi-conservative" truly means for the composition of daughter DNA.

 

Question 14. Komberg enzyme is called as ....................
Answer: Kornberg enzyme is called DNA polymerase I. This enzyme plays a crucial role in DNA replication and repair. Arthur Kornberg discovered this enzyme, and it was the first DNA polymerase identified.
In simple words: The Kornberg enzyme is another name for DNA polymerase I. This enzyme helps in building and fixing DNA.

🎯 Exam Tip: Associate famous scientists with their discoveries, such as Arthur Kornberg with DNA polymerase I.

 

Question 15. Replication of DNA occurs at phase of cell cycle.
(a) M
(b) S
(c) G1
(d) G2
Answer: (b) S
In simple words: DNA makes copies of itself during the S-phase of the cell cycle. This is when the cell prepares to divide by duplicating its genetic information.

🎯 Exam Tip: Clearly understand the events that occur in each phase of the cell cycle (G1, S, G2, M) and their significance.

 

Question 16. Semi-conservative model of replication
(a) Hershey and Chase
(b) Griffith
(c) Meselson and Stahl
(d) Macleod and McCarty
Answer: (c) Meselson and Stahl
In simple words: Meselson and Stahl did an experiment that showed how DNA copies itself in a "semi-conservative" way. This means each new DNA has one old part and one new part.

🎯 Exam Tip: Differentiate between who *proposed* (Watson and Crick) and who *proved* (Meselson and Stahl) the semi-conservative model of DNA replication.

 

Question 17. How many types of DNA polymerases does a eukaryotic cell possess?
(a) two
(b) three
(c) four
(d) five
Answer: (d) five
In simple words: Eukaryotic cells have five different kinds of DNA polymerase enzymes. These enzymes help build new DNA and fix damaged DNA.

🎯 Exam Tip: While prokaryotes often have three main DNA polymerases, remember that eukaryotes have a greater variety (five major types) with specialized functions.

 

Question 18. Identify the incorrect statement
(a) Replication occurs at ori – site of DNA
(b) Deoxy nucleotide triphosphate acts as a substrate
(c) Unwinding of DNA strand is carried out by topoisomerase
(d) DNA polymerase catalyses the polymerization at 3-OH
Answer: (c) Unwinding of DNA strand is carried out by topoisomerase
In simple words: The wrong statement is that topoisomerase unwinds DNA. The enzyme called helicase is actually the one that unwinds DNA. Topoisomerase just helps stop the DNA from getting too tangled as it unwinds.

🎯 Exam Tip: Clearly differentiate the functions of key enzymes in DNA replication: helicase (unwinding), topoisomerase (relieving supercoiling), and DNA polymerase (synthesizing new strands).

 

Question 19. The discontinuously synthesized fragments of lagging strand are called
Answer: The discontinuously synthesized fragments of the lagging strand are called Okazaki fragments. These short DNA segments are created because DNA polymerase can only add nucleotides in one direction. Later, these fragments are joined together by DNA ligase.
In simple words: The lagging strand of DNA is built in small pieces, which are called Okazaki fragments. These small pieces are later glued together to make a full strand.

🎯 Exam Tip: Remember that the leading strand is synthesized continuously, while the lagging strand is synthesized discontinuously in Okazaki fragments.

 

Question 20. Retroviruses .......as genetic material.
RNA
Answer: Retroviruses possess RNA as genetic material. Unlike most organisms that use DNA, retroviruses store their genetic information in RNA and then convert it into DNA inside the host cell using an enzyme called reverse transcriptase.
In simple words: Retroviruses use RNA as their genetic material. This is special because most living things use DNA to carry their genetic instructions.

🎯 Exam Tip: Understand that while DNA is the universal genetic material, some viruses (like retroviruses) use RNA, and know the enzyme involved in their unique replication (reverse transcriptase).

 

Question 21. Which is NOT a part of transcription unit?
(a) Promoter
(b) Operator
(c) Structural gene
(d) Terminator
Answer: (b) Operator
In simple words: The operator is not part of a simple transcription unit. A transcription unit usually has a promoter, the gene itself, and a terminator. The operator is more about controlling genes in groups.

🎯 Exam Tip: Know the essential components of a transcription unit (promoter, structural gene, terminator) and how they differ from an operon, which includes an operator and regulatory genes.

 

Question 22. Goldberg - Hogness box of eukaryotes is equivalent to of prokaryotes.
Answer: The Goldberg-Hogness box (also known as the TATA box) in eukaryotes is equivalent to the Pribnow box in prokaryotes. Both of these sequences are promoter elements that play a crucial role in initiating transcription by serving as binding sites for RNA polymerase.
In simple words: The Goldberg-Hogness box in complex cells is like the Pribnow box in simple cells. Both are special DNA spots that tell the cell where to start making RNA from a gene.

🎯 Exam Tip: Recognize the analogous promoter elements in prokaryotes (Pribnow box) and eukaryotes (TATA box/Goldberg-Hogness box) and their function in transcription initiation.

 

Question 23. Okazaki fragments are joined by the enzyme during DNA replication.
Answer: Okazaki fragments are joined by the enzyme DNA ligase during DNA replication. This enzyme forms phosphodiester bonds to connect the short, discontinuous DNA segments on the lagging strand, creating a continuous new strand.
In simple words: The small pieces of DNA called Okazaki fragments are put together by an enzyme called DNA ligase. It acts like glue to connect them into a long strand.

🎯 Exam Tip: Understand the specific roles of enzymes in DNA replication, especially DNA ligase for joining fragments and DNA polymerase for synthesizing new strands.

 

Question 24. Match the following:

(a) A – iv, B – C – ii, D – iii
(b) A – i, B – ii, C – iii, D – iv
(c) A – ii, B – iii, C – i, D – ii
(d) A – iii, B – ii, C – ii, D – i
Answer: (a) A - iv, B - i, C - ii, D - iii
In simple words: The semi-conservative way DNA copies itself was shown by Meselson and Stahl. Transformation in bacteria was found by Griffith. The shape of tRNA is like a clover leaf, a model proposed by R. Holley. The lac operon model, which explains how genes are turned on and off, was created by Jacob and Monod.

🎯 Exam Tip: Thoroughly review the contributions of key scientists to major discoveries in molecular genetics, as matching questions are common.

 

Question 25. The RNA polymerase of prokaryotes binds with ................factor to initiate polymerization.
(a) \( \Sigma \)
(b) \( \sigma \)
(c) \( \rho \)
(d) \( \Sigma \)
Answer: (c) sigma
In simple words: To start making RNA, the RNA polymerase enzyme in simple cells connects with a special helper called the sigma factor. This helper tells the enzyme where to begin on the DNA.

🎯 Exam Tip: Remember the specific roles of the sigma factor (initiation) and rho factor (termination) in prokaryotic transcription.

 

Question 26.

(a) Capung
(b) Tailing
(c) Splicing
(d) Transcribing
Answer: (c) Splicing
In simple words: Splicing is the process where pieces of RNA that don't code for anything (introns) are cut out of the precursor mRNA. The useful pieces (exons) are then stuck together to make the final mRNA.

🎯 Exam Tip: Remember the three main post-transcriptional modifications of eukaryotic mRNA: capping at the 5' end, splicing to remove introns, and polyadenylation (tailing) at the 3' end.

 

Question 27. Which of the following feature is absent in prokaryotes?
(a) Prokaryotes possess three major types of RNAs
(b) Structural genes are polycistronic
(c) Initiation process of transcription requires 'P' factor
(d) Split gene feature
Answer: (d) Split gene feature
In simple words: Prokaryotes, which are simple cells, do not have 'split genes'. This means their genes are continuous and don't have extra parts that need to be cut out before they can be used.

🎯 Exam Tip: A key difference between prokaryotic and eukaryotic gene structure is the absence of introns (and thus splicing) in prokaryotes, leading to simpler, polycistronic mRNA.

 

Question 28. Which of the following sequence has completely translated?
(i) AGA, UUU, UGU, AGU, UAG
(ii) AUG, UUU, AGA, UAC, UAA
(iii) AAA, UUU, UUG, UGU, UGA
(iv) AUG,AAU,AAC,UAU,UAG
(a) i and ii
(b) ii only
(c) i and iii
(d) ii and iv
Answer: (d) ii and iv
In simple words: A sequence is fully translated if it starts with "AUG" and ends with "UAA," "UAG," or "UGA." Options (ii) and (iv) both follow this rule, so they are correct.

🎯 Exam Tip: Memorize the universal start codon (AUG) and the three stop codons (UAA, UAG, UGA) as they are critical for understanding translation.

 

Question 29. Capping of mRNA occurs using
(a) Poly A residues
(b) Methyl guanosine triphosphate
(c) Deoxy ribonucleotide triphosphate
(d) Ribonucleotide triphosphate
Answer: (b) Methyl guanosine triphosphate
In simple words: mRNA capping happens by adding a special molecule called methyl guanosine triphosphate. This cap helps protect the mRNA and signals where ribosomes should attach.

🎯 Exam Tip: Remember that the 5' cap is a modified guanosine added in reverse orientation, and it's methylated, protecting the mRNA and aiding in translation initiation.

 

Question 30. One of the aspect is not a feature of genetic code?
(a) Specific
(b) Degenerate
(c) Universal
(d) Ambiguous
Answer: (d) Ambiguous
In simple words: The genetic code is not ambiguous. This means that each three-letter code (codon) always points to only one specific building block for proteins (amino acid).

🎯 Exam Tip: Understand the key characteristics of the genetic code: it is triplet, degenerate, unambiguous, universal, non-overlapping, and commaless.

 

Question 31. Which of the triplet codon is not a code of proline?
(i) CCU (ii) CAU (iii) CCG (zv) CAA
(a) i only
(b) ii and iv
(c) iii only
(d) all the above
Answer: (b) ii and iv
In simple words: The codons CAU and CAA do not code for proline. Proline is coded by CCU, CCC, CCA, and CCG.

🎯 Exam Tip: Familiarize yourself with a codon table to quickly identify which codons code for which amino acids, especially for common amino acids and stop/start codons.

 

Question 32. Coding sequences found in split genes are called.
(a) Operons
(b) Introns
(c) Exons
(d) Cistron
Answer: (c) Exons
In simple words: The parts of a gene that actually have instructions for making proteins are called exons. They are the "coding sequences" in genes that have breaks.

🎯 Exam Tip: Remember the definitions of exons (expressed, coding) and introns (intervening, non-coding) in eukaryotic genes.

 

Question 33. Which of the following mRNA yields 6 aminoacids after translation?
(i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
(ii) UGA AGA UAG GAG CAU CCC UAC UAU GAU
(iii) GUC UGC UGG GCU GAU UAA AGG AGC AUU
(iv) AUG UAC CAU UGC UGA UGC AGG AGC CCG
Answer: (i) UCU UAU AGU CGA UGC AGU UGA AAA UUU
In simple words: To get 6 amino acids, you need 6 three-letter codes (codons) before a "stop" signal. Only option (i) has six coding codons before it hits a stop codon.

🎯 Exam Tip: When counting amino acids from an mRNA sequence, always start counting from the first codon and stop immediately before a stop codon.

 

Question 34. The transcription termination factor associated with RNA polymerase in prokaryotes is
(a) \( \Sigma \)
(b) \( \sigma \)
(c) \( \rho \)
(d) \( \Sigma \)
Answer: (c) \( \rho \)
In simple words: The rho (\( \rho \)) factor is like a "stop" sign for the RNA polymerase enzyme in simple cells. It helps the enzyme finish making RNA and let go of the DNA.

🎯 Exam Tip: Distinguish between the sigma factor (initiation) and the rho factor (termination) in prokaryotic transcription.

 

Question 35. In a DNA double-strand, if guanine is 30%, what will be the percentage of thymine?
(a) 100%
(b) 20%
(c) 10%
(d) 70%
Answer: (b) 20%
In simple words: In DNA, the amount of Guanine (G) is always the same as Cytosine (C), and Adenine (A) is always the same as Thymine (T). If G is 30%, then C is also 30%. That's 60% total. The rest (40%) must be A and T, so T is half of that, which is 20%.

🎯 Exam Tip: Always apply Chargaff's rules (A=T, G=C, A+G = T+C) for double-stranded DNA composition calculations.

 

Question 36. Identify the triplet pairs that code for Tyrosine
(a) UUU,UUC
(b) UAU, UAU
(c) UGC, UGU
(d) CAU, CAC
Answer: (b) UAU, UAU
In simple words: The codons UAU and UAC are the ones that tell the cell to use the amino acid Tyrosine. Option (b) lists UAU twice, but it is one of the correct codes.

🎯 Exam Tip: Be aware that some options in MCQs might have slight inaccuracies or typos. Always rely on your knowledge of the genetic code, noting that Tyrosine is coded by UAU and UAC.

 

Question 37. Match the following:

(a) A - ii B - i C – iv D – iii
(b) A – iii B – ii C – i D – iv
(c) A – iv B - i C – ii D – iii
(d) A – ii B – iii C – i D – iv
Answer: (a) A - ii, B - i, C - iv, D - iii
In simple words: DNA Polymerase makes new DNA. Amino acyl synthetase puts the right amino acid onto tRNA. DNA helicase unwinds, or separates, the DNA strands. DNA ligase glues together pieces of DNA.

🎯 Exam Tip: Create flashcards for enzymes and their precise functions to ensure you can quickly recall them for matching questions.

 

Question 38. AUG cod is for.......................
(a) Arginine
(b) Tyrosine
(c) Tryptophan
(d) Methionine
Answer: (d) Methionine
In simple words: The AUG code is for Methionine. It's also the "start" signal for making proteins.

🎯 Exam Tip: Remember that AUG is the universal start codon and also codes for methionine, making it a crucial codon.

 

Question 39. The sequence of bases in coding strand of DNA is GAGTTAGCAGGC, then the sequence of codons in primary transcript is ....................
(a) CUCAUACGCCCG
(b) CUCAAUCGUCCG
(c) UCAGAUCUGCGC
(d) UUCAAUCGUGCG
Answer: (b) CUCAAUCGUCCG
In simple words: When making RNA from DNA, the RNA is a mirror copy of one DNA strand (the template strand), but with U instead of T. If the DNA sequence GAGTTAGCAGGC was the template, then the RNA sequence would be CUCAAUCGUCCG.

🎯 Exam Tip: Always remember that mRNA is complementary to the DNA template strand and that uracil (U) replaces thymine (T) in RNA. Pay close attention to whether a coding or template strand is given.

 

Question 40. The promoter region of eukaryote is
(a) TATAA
(b) AUGUT
(c) UUUGA
(d) AAAAU
Answer: (a) TATAA
In simple words: In complex cells, the start signal for making RNA often has a special DNA sequence called TATAA. This TATA box helps the cell know where to begin.

🎯 Exam Tip: Remember that the TATA box (Goldberg-Hogness box) is a common eukaryotic promoter element, while the Pribnow box is its prokaryotic equivalent.

 

Question 41. Match the following:

Answer:
(A) AUG - Methionine
(B) UGA - Stop codon (This codon signals the termination of protein synthesis and does not code for an amino acid in the standard genetic code).
(C) UUU - Phenylalanine
(D) GGG - Glycine
In simple words: AUG codes for Methionine. UGA is a "stop" sign for making proteins, so it does not code for any amino acid. UUU codes for Phenylalanine. GGG codes for Glycine.

🎯 Exam Tip: It is crucial to remember that UGA, UAA, and UAG are stop codons and do not code for any amino acids under the standard genetic code. If presented with a matching question where a stop codon is paired with an amino acid, recognize this as a potential error in the question itself.

 

Question 42. .......number of codons, codes for cystine.
Answer: Two. The amino acid Cysteine (often misspelled as cystine) is coded by two specific codons: UGC and UGU. This phenomenon, where multiple codons code for the same amino acid, is known as degeneracy of the genetic code.
In simple words: Two codons are used to make the amino acid Cysteine. These two codes are UGC and UGU.

🎯 Exam Tip: Be careful with amino acid spellings like Cysteine/Cystine and ensure you know all the codons for the amino acids that have multiple coding options.

 

Question 43. In sickle cell anaemia, the .......codon of p – globin gene is modified.
(a) Eighth
(b) Seventh
(c) Sixth
(d) Ninth
Answer: (c) Sixth
In simple words: In sickle cell anaemia, the sixth codon (a three-letter code) in the beta-globin gene is changed. This small change causes the disease.

🎯 Exam Tip: Remember that sickle cell anemia is a classic example of a point mutation, where a single base change in the sixth codon of the beta-globin gene leads to a significant protein defect.

 

Question 44. Pick out the incorrect statement.
(a) tRNA acts as a adapter molecule
(b) Stop codons do not have tRNA's
(c) Addition of amino acid leads to hydrolysis of tRNA {d) tRNA has four major loops
Answer: (c) Addition of amino acid leads to hydrolysis of tRNA
In simple words: The wrong statement is about adding an amino acid to tRNA. When an amino acid is added to tRNA, it uses energy but does not break down the tRNA molecule.

🎯 Exam Tip: Understand the process of tRNA charging (aminoacylation), recognizing that it's an energy-dependent process that links amino acids to their specific tRNAs, preparing them for protein synthesis.

 

Question 45. Which of the following antibiotic inhibits the interaction between tRNA and mRNA?
(a) Neomycin
(b) Streptomycin
(c) Tetracycline
(d) Chloramphenicol
Answer: (a) Neomycin
In simple words: Neomycin stops the tRNA from connecting with mRNA, which is needed to make proteins. This interference specifically targets the ribosome's A-site, preventing proper protein synthesis.

🎯 Exam Tip: Remember that different antibiotics target different stages of protein synthesis in bacteria.

 

Question 47. The cluster of genes with related function is called
(a) Cistron
(b) Operon
(c) Muton
(d) Recon
Answer: (b) Operon
In simple words: An operon is a group of genes that work together for a specific function. This clever organization allows bacteria to turn on or off a whole set of genes with just one signal.

🎯 Exam Tip: Understand that an operon acts as a single regulatory unit, common in prokaryotes.

 

Question 48. Repressor protein of Lac operon binds to ...............of operon
(a) Promoter region
(b) Operator region
(c) terminator region
(d) inducer region
Answer: (b) Operator region
In simple words: The repressor protein in the Lac operon attaches to the operator part of the operon. When the repressor binds to the operator, it blocks RNA polymerase, stopping the genes from being read.

🎯 Exam Tip: Knowing the specific binding sites of regulatory proteins is key to understanding gene control mechanisms.

 

Question 49. Lac Z gene codes for
(a) Permease
(b) transacetylase
(c) P-galactosidase
(d) Aminoacyl transferase
Answer: (c) p-galactosidase
In simple words: The Lac Z gene provides instructions to make the enzyme called p-galactosidase. This important enzyme breaks down lactose into simpler sugars like glucose and galactose.

🎯 Exam Tip: Memorize which gene codes for which enzyme in the lac operon (lacZ for β-galactosidase, lacY for permease, lacA for transacetylase).

 

Question 50. Lac operon model was proposed by
Answer: The Lac operon model was introduced by scientists Jacob and Monod. Their work provided a foundational understanding of how gene expression is regulated in bacteria.
In simple words: The Lac operon model was proposed by Jacob and Monod.

🎯 Exam Tip: Associate Jacob and Monod with the lac operon, which is a classic example of gene regulation.

 

Question 51. Approximate count of base pair in human genome is................
Answer: The human genome has about 3 billion base pairs. This massive amount of genetic information is tightly packed into our chromosomes.
In simple words: The human genome has approximately \( 3 \times 10^9 \) base pairs.

🎯 Exam Tip: Remember this number as a general estimate for the size of the human genome.

 

Question 52. Automated DNA sequences are developed by.
Answer: Frederick Sanger developed the method for automatic DNA sequencing. His method, known as Sanger sequencing, revolutionized the study of genetics.
In simple words: Automated DNA sequencing was developed by Frederick Sanger.

🎯 Exam Tip: Frederick Sanger is also known for sequencing proteins and is a double Nobel laureate.

 

Question 53. Which of the chromosome has a higher gene density?
(a) Chromosome 20
(b) Chromosome 19
(c) Chromosome 13
(d) Chromosome Y
Answer: (b) Chromosome 19
In simple words: Chromosome 19 has the highest number of genes packed into its length. This means that a larger proportion of its DNA sequence is dedicated to coding for proteins compared to other chromosomes.

🎯 Exam Tip: Chromosome 19 is known for its high gene density, while Chromosome Y has one of the lowest.

 

Question 54. Number of genes located in chromosome Y is
(a) 2968
(b) 213
(c) 2869
(d) 231
Answer: (d) 231
In simple words: There are 231 genes found on the Y chromosome. The Y chromosome is much smaller than other chromosomes and carries genes primarily involved in male sex determination.

🎯 Exam Tip: Chromosome Y has the lowest gene count among human chromosomes, making it relatively gene-poor.

 

Question 55. How many structural genes are located in the lac operon of E.Coli?
(a) 4
(b) 3
(c) 2
(d) 1
Answer: (b) 3
In simple words: The lac operon in E. coli contains three structural genes. These three genes (lacZ, lacY, and lacA) work together to process lactose.

🎯 Exam Tip: The three structural genes are lacZ, lacY, and lacA, which code for β-galactosidase, permease, and transacetylase, respectively.

 

Question 56. DNA fingerprinting technique was developed by
(d) Jacob and Monod
(b) Alec Jeffreys
(c) Frederick Sanger
(d) Alec Jeffreys
Answer: (b) Alec Jeffreys
In simple words: The DNA fingerprinting method was created by Alec Jeffreys. This technique uses variations in DNA sequences to identify individuals, much like human fingerprints.

🎯 Exam Tip: Remember Alec Jeffreys for DNA fingerprinting and its use in forensics and paternity testing.

 

Question 57. In DNA fingerprinting, separation of DNA fragments is done by.........
(a) Centrifugation
(b) Electrophoresis
(c) X-ray diffraction
(d) denaturation
Answer: (b) Electrophoresis
In simple words: DNA fragments in fingerprinting are separated using electrophoresis. This method uses an electric field to move DNA pieces through a gel, separating them by size.

🎯 Exam Tip: Gel electrophoresis is a fundamental technique for separating DNA, RNA, and proteins based on size and charge.

 

Question 58. SNP stands for
(a) Single nucleotide Polymorphism
(b) Single Nucleoside Polypeptide
(c) Single nucleotide Polymorphism
(d) Single nucleotide polymer
Answer: (a) Single nucleotide Polymorphism
In simple words: SNP means 'Single Nucleotide Polymorphism'. These are single base-pair differences in DNA that can vary among individuals.

🎯 Exam Tip: SNPs are vital genetic markers used to identify disease-related genes and trace human ancestry.

 

Question 59. Specific sequences of mRNA that are not translated are..............
Answer: The parts of mRNA that are not used to make protein are called Untranslated Regions (UTRs). These regions, found at both ends of the mRNA, play important roles in regulating gene expression.
In simple words: Untranslated Regions (UTR) are specific parts of mRNA that are not translated into proteins.

🎯 Exam Tip: UTRs are crucial for mRNA stability, translation efficiency, and localization, even though they don't code for amino acids.

 

Question 60. Non-coding or intervening DNA sequence is called
Answer: A non-coding part of a DNA sequence is known as an intron. Introns are removed from RNA during splicing before the RNA is translated into protein.
In simple words: A non-coding or intervening DNA sequence is called an Intron.

🎯 Exam Tip: In eukaryotes, genes often contain introns, which are 'junk' DNA that must be cut out, unlike in prokaryotes.

 

Question 61. .................is the monomer of DNA.
Answer: The basic building block of DNA is called a nucleotide. Each nucleotide consists of a sugar, a phosphate group, and a nitrogenous base.
In simple words: A nucleotide is the monomer, or basic unit, of DNA.

🎯 Exam Tip: Remember that nucleotides are the monomers that link together to form nucleic acid polymers like DNA and RNA.

 

Question 62. Which one of the following is wrongly matched?
(a) Transcription – Copying information from DNA to RNA
(b) Translation – Decoding information from mRNA to protein
(c) Replication – Making of DNA copies
(d) Splicing – Joining of exons with introns
Answer: (d) Splicing – Joining of exons with introns
In simple words: The statement that splicing involves joining exons with introns is incorrect. Splicing actually removes introns and joins exons. Splicing is a key step in RNA processing where non-coding introns are cut out and coding exons are linked together.

🎯 Exam Tip: Splicing removes introns (intervening sequences) and joins exons (expressed sequences); it does not join exons WITH introns.

12th Bio Zoology Guide Molecular Genetics Two Marks Questions And Answers

 

Question 1. Who proposed One gene – One enzyme hypothesis? Define it.
Answer: The "One gene – One enzyme" idea was put forward by George Beadle and Edward Tatum. It means that each specific gene is responsible for making one specific enzyme. This hypothesis helped establish the direct link between genes and the proteins that carry out cell functions.
In simple words: George Beadle and Edward Tatum proposed the "One gene – One enzyme" hypothesis, stating that each gene controls the production of one enzyme.

🎯 Exam Tip: While this hypothesis has been refined (one gene-one polypeptide), it remains a foundational concept in genetics.

 

Question 2. Differentiate nucleoside from nucleotide.
Answer:
A nucleoside is made of a nitrogen base attached to a pentose sugar.
A nucleotide is similar but also includes a phosphate group attached to the nucleoside. It's the building block of DNA. The addition of a phosphate group is what gives nucleotides their acidic property, leading to the name "nucleic acid."
In simple words: A nucleoside has a sugar and a base, but a nucleotide also includes a phosphate group, making it a nucleoside plus phosphate.

🎯 Exam Tip: Remember that the key difference is the presence of a phosphate group in a nucleotide, making it a nucleoside + phosphate.

 

Question 3. State the key differences between DNA and RNA.
Answer:
DNA and RNA differ in their sugar and nitrogen bases.
For DNA:
1. It contains deoxyribose sugar.
2. Its nitrogen bases are Adenine, Guanine, Cytosine, and Thymine.
For RNA:
1. It contains ribose sugar.
2. Its nitrogen bases are Adenine, Guanine, Cytosine, and Uracil (Thymine is replaced by Uracil).
These small chemical differences lead to significant functional distinctions, with DNA primarily storing genetic information and RNA largely involved in expressing it.
In simple words: DNA has deoxyribose sugar and thymine, while RNA has ribose sugar and uracil.

🎯 Exam Tip: Focus on the two main differences: the type of sugar (deoxyribose vs. ribose) and the unique pyrimidine base (thymine vs. uracil).

 

Question 4. Point out the nitrogenous bases of RNA.
Answer: The nitrogenous bases found in RNA are Adenine, Guanine, Cytosine, and Uracil. Unlike DNA, RNA contains uracil instead of thymine, which pairs with adenine.
In simple words: The nitrogen bases in RNA are Adenine, Guanine, Cytosine, and Uracil.

🎯 Exam Tip: Always remember the four bases for RNA (A, G, C, U) and DNA (A, G, C, T) to avoid confusion.

 

Question 5. What makes the DNA and RNA acidic molecules?
Answer: DNA and RNA are acidic because of the phosphate groups (PO4) they contain. These phosphate groups release hydrogen ions, making the molecules acidic at the normal pH inside cells. This negative charge from the phosphate backbone is also crucial for DNA's ability to migrate in an electric field during gel electrophoresis.
In simple words: DNA and RNA are acidic because of their phosphate groups, which release hydrogen ions.

🎯 Exam Tip: The "acid" in "nucleic acid" comes from the phosphate backbone, which is negatively charged.

 

Question 6. Which type of bond is formed
1. between a purine and pyrimidine base?
2. between the pentose sugar and adjacent nucleotide?
Answer:
1. Purine and pyrimidine bases connect using hydrogen bonds.
2. A pentose sugar links to the next nucleotide using phosphodiester bonds.
Hydrogen bonds are weaker and allow DNA strands to separate easily for replication, while phosphodiester bonds form a strong, stable backbone for the DNA molecule.
In simple words: Hydrogen bonds connect purine and pyrimidine bases, while phosphodiester bonds link the pentose sugar to adjacent nucleotides.

🎯 Exam Tip: Clearly distinguish between the weaker hydrogen bonds that hold the two DNA strands together and the strong phosphodiester bonds that form the backbone of a single strand.

 

Question 7. DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:
DNA is the main genetic material for most living things, not RNA, for a few reasons:
1. RNA is generally more reactive and less stable than DNA.
2. DNA's double-stranded structure makes it more robust and less prone to changes.
3. DNA contains thymine, which is more stable than uracil, found in RNA. The presence of a methyl group in thymine contributes to this stability.
This greater chemical stability allows DNA to reliably store genetic information over long periods and across generations.
In simple words: DNA is the main genetic material because it is more stable than RNA due to its double-stranded structure and the presence of thymine instead of uracil.

🎯 Exam Tip: The stability provided by DNA's double helix and the presence of thymine are critical for its role as the primary genetic material.

 

Question 8. Name any two viruses whose genetic material is RNA.
Answer: Two viruses that use RNA as their genetic material are the Tobacco Mosaic Virus (TMV) and some bacteriophages. Many RNA viruses, like influenza or HIV, have higher mutation rates due to the lower fidelity of RNA replication, which helps them adapt quickly.
In simple words: Two viruses with RNA as genetic material are Tobacco Mosaic Virus (TMV) and Bacteriophage (B).

🎯 Exam Tip: Remember that not all viruses use DNA; many rely on RNA to carry their genetic instructions.

 

Question 9. What are the properties that a molecule must possess to act as genetic material?
Answer: For a molecule to be genetic material, it needs specific features:
- It must be able to make copies of itself (self-replication).
- It must be able to store genetic information.
- It needs to be stable enough to be passed on without much change.
- It should allow for changes or mutations, which drive evolution.
These four properties collectively ensure that genetic information can be faithfully transmitted, expressed, and contribute to the diversity of life.
In simple words: A molecule needs to self-replicate, store information, be stable, and allow for mutations to be considered genetic material.

🎯 Exam Tip: These properties highlight why DNA is so well-suited for its role as the blueprint of life.

 

Question 10. How many base pairs are present in one complete turn of DNA helix? What is the distance between two consecutive base pairs?
Answer: One full twist of the DNA helix contains ten base pairs. The distance between any two base pairs that are next to each other is 0.34 nanometers (or \( 0.34 \times 10^{-9} \) meters). This precise spacing and helical structure are crucial for the stability and accurate replication of DNA.
In simple words: A complete turn of DNA has 10 base pairs, and the distance between two base pairs is 0.34 nm.

🎯 Exam Tip: Recall Watson and Crick's model of DNA, where each turn is 3.4 nm long and contains 10 base pairs.

 

Question 11. What is a genophore?
Answer: A genophore is the main genetic material in prokaryotic cells, like bacteria. Even though prokaryotes don't have a true nucleus, their circular DNA is organized into large loops within a specific area called the nucleoid. This DNA is called a genophore because it lacks the complex protein structure (chromatin) found in eukaryotic chromosomes. This compact organization allows the bacterial cell to efficiently store and access its genetic information within a small cytoplasmic space.
In simple words: A genophore is the circular DNA molecule found in prokaryotes, organized in a nucleoid region without histones.

🎯 Exam Tip: Remember that "genophore" is often used to describe the bacterial chromosome, highlighting its simpler, circular structure compared to eukaryotic chromosomes.

 

Question 12. What is nucleosome? How many base pairs are there in a typical nucleosome?
Answer: A nucleosome is a basic unit of DNA packaging in eukaryotes. It forms when negatively charged DNA wraps around a group of positively charged proteins called a histone octamer. Each typical nucleosome contains about 200 base pairs of DNA. This "beads-on-a-string" structure helps compact the long DNA molecule into the much smaller space of the cell nucleus.
In simple words: A nucleosome is DNA wrapped around histone proteins, typically containing 200 base pairs of DNA.

🎯 Exam Tip: Remember that nucleosomes are the first level of DNA compaction and involve histones, which are crucial for chromosome structure.

 

Question 13. Expand and define NHC
Answer: NHC stands for Non-Histone Chromosomal proteins. These are extra proteins, other than histones, that are needed in eukaryotic cells to help package DNA into a more compact form at higher levels of organization within the chromosome. While histones form the basic nucleosome structure, NHC proteins contribute to the complex folding and regulation of chromatin.
In simple words: NHC means Non-Histone Chromosomal proteins, which are additional proteins in eukaryotes that help compact DNA beyond nucleosomes.

🎯 Exam Tip: NHC proteins are diverse and play roles in gene regulation, DNA replication, and repair, in addition to chromatin structure.

 

Question 14. Differentiate between Heterochromatin and Euchromatin.
Answer:
Here's how the two types of chromatin differ:
Heterochromatin:
1. This is the part of the nucleus where chromatin is very tightly packed and stains darkly.
2. It is mostly inactive, meaning genes in this region are not easily read or copied to make proteins.
Euchromatin:
1. This is the part of the nucleus where chromatin is loosely packed and stains lightly.
2. It is transcriptionally active, meaning genes here are easily read and copied to make proteins.
The dynamic interconversion between euchromatin and heterochromatin is vital for regulating gene expression in eukaryotic cells.
In simple words: Heterochromatin is tightly packed and inactive, while euchromatin is loosely packed and active in gene expression.

🎯 Exam Tip: Remember that "eu" means true or good, so euchromatin is the "true" active chromatin, while heterochromatin is generally inactive and condensed.

 

Question 15. Which is the widely accepted model of DNA replication? Who has proved it?
Answer: The most accepted way DNA replicates is through the semi-conservative model. This model was proven by Meselson and Stahl in 1958. In semi-conservative replication, each new DNA molecule is made of one old strand and one newly synthesized strand.
In simple words: The semi-conservative model is the accepted way DNA replicates, proven by Meselson and Stahl.

🎯 Exam Tip: Know the Meselson-Stahl experiment well, as it is a classic demonstration of the semi-conservative nature of DNA replication.

 

Question 16. Name the chemical substance which is called by the name
1. Kornberg Enzyme
2. Ochoa's enzyme
Answer:
1. The enzyme called Kornberg enzyme is also known as DNA polymerase I.
2. The enzyme called Ochoa's enzyme is also known as Polynucleotide phosphorylase.
DNA polymerase I plays a role in DNA repair and removing RNA primers, while polynucleotide phosphorylase can synthesize RNA without a template, though it's mainly involved in RNA degradation in cells.
In simple words: Kornberg enzyme is DNA polymerase I, and Ochoa's enzyme is Polynucleotide phosphorylase.

🎯 Exam Tip: Associating these historical names with their modern enzyme counterparts is important for understanding their discovery and function.

 

Question 17. Name the various types of prokaryotic DNA polymerase. State their role in replication process.
Answer: In prokaryotic cells, there are three main types of DNA polymerases:
- DNA polymerase I is involved in fixing DNA damage.
- DNA polymerase II is involved in DNA repair, similar to polymerase I.
- DNA polymerase III is the primary enzyme responsible for synthesizing new DNA strands during replication.
DNA polymerase III is the workhorse of DNA replication, adding new nucleotides very rapidly and accurately.
In simple words: Prokaryotes have DNA polymerase I (repair), II (repair), and III (main replication enzyme).

🎯 Exam Tip: Remember that DNA polymerase III is the main replicase, while DNA polymerase I is critical for primer removal and repair.

 

Question 18. What is the function of Deoxy nucleotide triphosphate in replication?
Answer: Deoxynucleotide triphosphates (dNTPs) serve two roles in DNA replication: they are the building blocks (substrates) for the new DNA strand, and they also provide the necessary energy for the DNA synthesis process. The energy released from breaking the bonds in dNTPs drives the formation of phosphodiester bonds, which link nucleotides together.
In simple words: Deoxynucleotide triphosphates (dNTPs) act as both building blocks and energy sources for DNA replication.

🎯 Exam Tip: Think of dNTPs as both the bricks and the fuel for building new DNA strands.

 

Question 19. Given below are some events of eukaryotic replication. Name the enzymes involved in the process.
1. Unwinding of DNA
2. Joining of Okazaki fragments
3. Addition of nucleotides to a new strand
4. Correcting the repair
Answer:
Here are the enzymes involved in the listed events of eukaryotic DNA replication:
1. Helicase unwinds the DNA.
2. DNA ligase joins the Okazaki fragments.
3. DNA polymerase adds new nucleotides to the growing strand.
4. Nuclease enzymes are responsible for correcting errors and repairing DNA.
These enzymes coordinate precisely to ensure that DNA replication is fast, accurate, and faithful.
In simple words: Helicase (unwinding), DNA ligase (joining), DNA polymerase (adding nucleotides), and Nuclease (repair) are the enzymes involved.

🎯 Exam Tip: Each enzyme has a specific, critical role; knowing their functions helps in understanding the entire replication mechanism.

 

Question 20. Differentiate leading strand from lagging strand
Answer:
Here's how the leading and lagging strands differ during DNA replication:
Leading strand:
1. It has a 3' to 5' polarity (template strand read by polymerase).
2. DNA synthesis on this strand happens continuously in one long piece.
Lagging strand:
1. It has a 5' to 3' polarity (template strand).
2. DNA synthesis on this strand is discontinuous, forming small fragments.
The difference in synthesis is due to DNA polymerase only being able to add nucleotides in the 5' to 3' direction of the new strand.
In simple words: The leading strand is synthesized continuously in the 5' to 3' direction, while the lagging strand is synthesized discontinuously in short Okazaki fragments.

🎯 Exam Tip: The leading strand needs only one primer, while the lagging strand needs many primers to synthesize Okazaki fragments.

 

Question 21. What are Okazaki fragments?
Answer: Okazaki fragments are short pieces of newly synthesized DNA. They are created on the lagging strand during DNA replication because DNA polymerase can only build new strands in one direction. These small fragments are later joined together by the enzyme DNA ligase. These fragments are named after Reiji Okazaki, who discovered them while studying DNA replication.
In simple words: Okazaki fragments are small, newly made DNA segments on the lagging strand, later joined by DNA ligase.

🎯 Exam Tip: Okazaki fragments are characteristic of discontinuous replication on the lagging strand, requiring multiple primers and subsequent ligation.

 

Question 22. What is a replication fork?
Answer: A replication fork is a Y-shaped structure formed when the DNA double helix unwinds and separates into two single strands during DNA replication. This unwinding is carried out by enzymes like helicase and topoisomerase at a specific starting point. Each origin of replication typically forms two replication forks. This "Y" shape allows both strands to be copied simultaneously, but in different ways due to their opposite orientations.
In simple words: A replication fork is the Y-shaped region where DNA unwinds and separates, allowing replication to occur on both strands.

🎯 Exam Tip: The replication fork is the active site where DNA synthesis occurs, with leading and lagging strands being synthesized concurrently.

 

Question 23. Apart from DNA polymerase, name any other four enzymes which were involved in DNA replication of eukaryotic cell.
Answer: Besides DNA polymerase, four other enzymes important for DNA replication in eukaryotic cells are: DNA ligase, Topoisomerase (also known as DNA gyrase), Helicase, and Nuclease. Each of these enzymes plays a critical, distinct role, from unwinding the DNA to joining fragments and repairing errors.
In simple words: Other enzymes for eukaryotic DNA replication include DNA ligase, Topoisomerase, Helicase, and Nuclease.

🎯 Exam Tip: Understand the specific function of each enzyme: helicase unwinds, topoisomerase relieves tension, ligase joins fragments, and nucleases proofread and repair.

 

Question 24. Who proposed the central dogma? Write its concept.
Answer: The central dogma of molecular biology was proposed by Francis Crick. Its core idea is that genetic information flows from DNA to RNA, and then from RNA to protein.
- DNA makes copies of itself (Replication).
- DNA's information is copied into RNA (Transcription).
- RNA's information is used to build proteins (Translation).
This central dogma provides a fundamental framework for understanding gene expression and cellular function.
In simple words: Francis Crick proposed the central dogma, explaining that genetic information flows from DNA to RNA to protein.

🎯 Exam Tip: Remember the sequence: DNA → RNA → Protein. Also, be aware that reverse transcription (RNA to DNA) is an exception found in some viruses.

 

Question 25. Define transcription and name the enzyme involved in this process.
Answer: Transcription is the process where genetic information from one strand of DNA is copied to create an RNA molecule. The main enzyme involved in transcription is RNA polymerase. This RNA molecule can then serve various roles, including carrying genetic instructions to ribosomes for protein synthesis.
In simple words: Transcription is copying DNA genetic information into RNA, and RNA polymerase is the enzyme that does it.

🎯 Exam Tip: Always specify "DNA-dependent RNA polymerase" to distinguish it from other polymerase types.

 

Question 26. What is TATA box? State its function.
Answer: The TATA box, also known as the Goldberg-Hogness box, is a specific region in the promoter area of eukaryotic genes. It's rich in adenine (A) and thymine (T) base pairs. Its main job is to act as a crucial binding site for RNA polymerase, which is the enzyme that starts transcription. This sequence helps precisely position RNA polymerase to begin copying the gene at the correct spot.
In simple words: The TATA box is an AT-rich DNA sequence in eukaryotic promoters that helps RNA polymerase bind and start transcription.

🎯 Exam Tip: The TATA box is a common eukaryotic promoter element; its prokaryotic equivalent is the Pribnow box.

 

Question 27. Structural gene of eukaryotes differ from prokaryotes. How?
Answer: Structural genes in eukaryotes and prokaryotes are different in how many proteins they code for. In eukaryotes, a structural gene is monocistronic, meaning it codes for only one protein. In prokaryotes, however, a structural gene is often polycistronic, meaning it contains information to code for several different proteins from a single mRNA molecule. This polycistronic organization in prokaryotes allows for the coordinated expression of genes involved in a common metabolic pathway.
In simple words: Eukaryotic structural genes are monocistronic (code for one protein), while prokaryotic ones are polycistronic (code for many proteins).

🎯 Exam Tip: Remember "mono" for one (eukaryotes) and "poly" for many (prokaryotes) when thinking about structural gene organization.

 

Question 28. What are the two major components of prokaryotic RNA polymerase? How do they act?
Answer: Prokaryotic RNA polymerase has two main parts: the core enzyme and the sigma (σ) subunit. The core enzyme is responsible for the actual synthesis of RNA. The sigma subunit's role is to help the core enzyme find and bind to the correct starting point (promoter) on the DNA. Once transcription begins, the sigma subunit usually detaches, allowing the core enzyme to continue elongating the RNA strand.
In simple words: Prokaryotic RNA polymerase has a core enzyme (for synthesis) and a sigma subunit (for promoter recognition).

🎯 Exam Tip: The sigma factor is crucial for initiation and promoter specificity, ensuring that transcription starts at the right place.

 

Question 29. Distinguish between exons and introns.
Answer:
Here's how exons and introns are different:
Exons: These are the "expressed" or coding parts of a eukaryotic gene. They contain the instructions that will be used to make a protein.
Introns: These are the "intervening" or non-coding parts of a eukaryotic gene. They are cut out of the RNA molecule before it becomes a mature messenger RNA.
The process of removing introns and joining exons is called splicing, and it's essential for creating functional mRNA in eukaryotes.
In simple words: Exons are the coding parts of a gene that are expressed, while introns are non-coding parts that are removed during RNA processing.

🎯 Exam Tip: Think of "ex-ons" as "ex-pressed" (coding) and "in-trons" as "in-tervening" (non-coding) to easily remember their roles.

 

Question 30. Define splicing.
Answer: Splicing is the biological process where non-coding regions, called introns, are precisely removed from a newly made RNA molecule (hnRNA or pre-mRNA), and the remaining coding regions, called exons, are joined together. This critical step ensures that only the relevant genetic information is translated into protein.
In simple words: Splicing is the process of cutting out introns and joining exons in an RNA molecule to make it mature.

🎯 Exam Tip: Splicing is a key post-transcriptional modification in eukaryotes, creating mature mRNA from a primary transcript.

 

Question 31. What is capping and tailing?
Answer: Capping and tailing are two important modifications that happen to messenger RNA (mRNA) in eukaryotes:
- Capping: An unusual nucleotide called methyl guanosine triphosphate is added to the 5' end of the mRNA molecule.
- Tailing: A long chain of about 200-300 adenine (Poly A) nucleotides is added to the 3' end of the mRNA molecule.
These modifications protect the mRNA from degradation and are essential for its transport out of the nucleus and efficient translation into protein.
In simple words: Capping adds a methyl guanosine triphosphate to the 5' end of mRNA, and tailing adds a poly-A tail to the 3' end, both for protection and function.

🎯 Exam Tip: Remember that capping is at the 5' end, and tailing (Poly-A tail) is at the 3' end; both are crucial for mRNA stability and function.

 

Question 32. If a double-stranded DNA has 20% of cytosine, calculate the percentage of adenine in DNA.
Answer: If a double-stranded DNA molecule contains 20% cytosine, we can find the percentage of adenine using Chargaff's rules.
1. According to Chargaff's rules, the amount of cytosine (C) is equal to the amount of guanine (G). So, if C = 20%, then G = 20%.
2. The total percentage of all four bases (A+T+C+G) is 100%.
3. Therefore, \( A + T + 20\% + 20\% = 100\% \). This means \( A + T + 40\% = 100\% \).
4. So, \( A + T = 100\% - 40\% = 60\% \).
5. Also, according to Chargaff's rules, the amount of adenine (A) is equal to the amount of thymine (T).
6. So, if \( A + T = 60\% \) and \( A = T \), then each must be \( 60\% / 2 = 30\% \).
Thus, the percentage of adenine in the DNA is 30%. This rule is fundamental to understanding DNA base composition.
In simple words: Since Cytosine is 20%, Guanine is also 20%. That leaves 60% for Adenine and Thymine. Since Adenine and Thymine amounts are equal, Adenine must be 30%.

🎯 Exam Tip: Always remember Chargaff's rules: A=T and G=C, and the total of purines (A+G) equals the total of pyrimidines (C+T).

 

Question 33. Mention the dual functions of AUG.
Answer: The codon AUG has two main functions: it acts as the "start" signal (initiator codon) for protein synthesis, and it also specifies the amino acid methionine. In prokaryotes, it codes for N-formylmethionine, while in eukaryotes, it codes for regular methionine, but its role as the start codon is universal.
In simple words: AUG starts protein synthesis (initiator codon) and codes for the amino acid methionine.

🎯 Exam Tip: AUG is almost universally recognized as the start codon, making it crucial for initiating translation.

 

Question 34. How many codons are involved in termination of translation. Name them.
Answer: Three specific codons are involved in stopping the process of translation. These "stop" codons are UAA, UAG, and UGA. These codons do not code for any amino acid but instead signal the ribosome to release the newly formed protein.
In simple words: Three codons, UAA, UAG, and UGA, signal the end of protein synthesis.

🎯 Exam Tip: These are often called "nonsense" codons because they don't have a corresponding amino acid.

 

Question 35. Degeneracy of codon - comment.
Answer: A degenerate code means that a single amino acid can be coded by more than one triplet codon. For example, the codons GUU, GUC, GUA, and GUG all code for the amino acid valine. This redundancy helps protect against the harmful effects of certain mutations, as a change in one base might still result in the same amino acid being produced.
In simple words: A degenerate code means that more than one codon can give the same amino acid. For example, valine can be coded by four different codons.

🎯 Exam Tip: Remember that degeneracy in codons means multiple codons for one amino acid, but one codon never codes for multiple amino acids (it is non-ambiguous).

 

Question 36. Point out the exceptional categories to universality of genetic code.
Answer: While the genetic code is mostly universal, there are a few exceptions. These exceptions are mainly seen in the genomes of prokaryotic cells (like bacteria), mitochondria, and chloroplasts. In these specific cases, some codons might code for different amino acids or act as stop signals compared to the universal code. These differences show tiny variations in how life's instructions are read.
In simple words: The genetic code is usually the same everywhere, but some small differences are found in bacteria, mitochondria, and chloroplasts.

🎯 Exam Tip: When discussing universality, it's crucial to acknowledge these few but significant exceptions, as they highlight the code's evolutionary flexibility.

 

Question 37. What are non-sense codons?
Answer: Non-sense codons are special triplet codons that do not code for any amino acid. Instead, they act as stop signals during protein synthesis, telling the ribosome to end the translation process. The three common non-sense codons are UGA, UAA, and UAG, which ensure that proteins are built to the correct length. Think of them as periods at the end of a sentence in the genetic language.
In simple words: Non-sense codons are stop signals (like UGA, UAA, UAG) that tell protein-making to end. They do not code for amino acids.

🎯 Exam Tip: Memorize the three stop codons: UAA (U Are Away), UAG (U Are Gone), UGA (U Go Away) to easily recall them.

 

Question 38. Name the triplet codons that code for
1. Tyrosine
2. Histidine
Answer:
1. Tyrosine - UAU, UAC
2. Histidine - CAU, CAC
In simple words: Tyrosine is coded by UAU and UAC. Histidine is coded by CAU and CAC.

🎯 Exam Tip: Using a codon chart is the quickest way to identify the codons for specific amino acids. Practice reading it accurately.

 

Question 39. Why hnRNA has to undergo splicing?
Answer: Heterogeneous nuclear RNA (hnRNA) must undergo splicing because it contains both coding regions (exons) and non-coding regions (introns). Introns are like filler material that needs to be removed before the RNA can be used to make a protein. Splicing removes these introns and joins the exons together, creating a mature mRNA molecule that carries only the useful genetic instructions. This process is vital for eukaryotes to produce functional proteins.
In simple words: hnRNA has extra parts called introns that do not code for proteins. Splicing removes these introns and joins the useful parts (exons) together to make a working message for protein making.

🎯 Exam Tip: Remember that splicing is a crucial post-transcriptional modification in eukaryotes to remove non-coding introns and create a functional mRNA.

 

Question 40. State the role of following codons in the translation process
1. AUG
2. UAA
Answer:
1. AUG is the initiator codon and also codes for methionine.
2. UAA is a terminator codon.
In simple words: AUG starts the protein-making process and codes for methionine. UAA acts as a stop sign, ending the protein-making process.

🎯 Exam Tip: The AUG codon is unique because it serves both as a start signal and an amino acid code (methionine), making it a key point to remember.

 

Question 41. Given below is mRNA sequence. Mention the aminoacids sequence that is formed after its translation.
3'AUGAAAGAUGGGUAA5'
Answer: Methionine - Lysine - Aspartic acid - Glycine
In simple words: From this mRNA sequence, the protein made will have the amino acids Methionine, Lysine, Aspartic acid, and Glycine in that order.

🎯 Exam Tip: To correctly translate an mRNA sequence, always read it from the 5' to 3' end and use a standard codon chart to find the amino acid for each triplet.

 

Question 42. Name the four codons that codes valine.
Answer: GUU, GUC, GUA and GUG.
In simple words: Four different genetic codes (codons) can all tell the cell to use the amino acid valine. They are GUU, GUC, GUA, and GUG.

🎯 Exam Tip: This question directly tests your knowledge of codon degeneracy. Quickly check your codon chart to ensure you list all correct codons.

 

Question 43. The base sequence in one of the DNA strand is TAGC ATGAT. Mention the base sequence in its complementary strand.
Answer: The complementary strand has ATCGTACTA. The rules are A pairs with T, and G pairs with C. This pairing forms the double helix structure of DNA.
In simple words: If one DNA strand has TAGC ATGAT, its matching strand will have ATCGTACTA.

🎯 Exam Tip: Remember the base pairing rules: Adenine (A) always pairs with Thymine (T), and Guanine (G) always pairs with Cytosine (C). This is fundamental for DNA structure.

 

Question 44. Why t-RNA is called as adapter molecule?
Answer: The transfer RNA (tRNA) molecule is called an adapter molecule because it has two main functions. It acts like a vehicle that picks up specific amino acids floating in the cell's cytoplasm. At the same time, it has an anticodon loop that can read specific codes (codons) on the mRNA molecule. So, tRNA "adapts" the genetic message from mRNA into the language of amino acids, ensuring the correct amino acid is added to the growing protein chain. This key role was first suggested by Francis Crick.
In simple words: tRNA is called an adapter because it carries amino acids to the ribosome and also reads the mRNA code to put the right amino acid in place.

🎯 Exam Tip: Highlighting both roles of tRNA – carrying amino acids and reading codons – is key to a complete answer for why it's called an "adapter."

 

Question 45. What do you mean by charging of tRNA? Name the enzyme involved in this process.
Answer: Charging of tRNA, also known as aminoacylation, is the process where a specific amino acid is attached to its corresponding transfer RNA (tRNA) molecule. This is a crucial step to make sure the right amino acid is used during protein synthesis. The enzyme responsible for catalyzing this process is called aminoacyl-tRNA synthetase. There is a different synthetase enzyme for each type of amino acid, ensuring accuracy.
In simple words: Charging tRNA means attaching the correct amino acid to its tRNA. The enzyme that does this job is called aminoacyl-tRNA synthetase.

🎯 Exam Tip: Emphasize that "charging" is synonymous with "aminoacylation" and that aminoacyl-tRNA synthetase is highly specific for both the amino acid and the tRNA.

 

Question 46. What is UTR's?
Answer: UTRs stand for Untranslated Regions. These are specific sequences on the messenger RNA (mRNA) molecule that are not translated into proteins. UTRs are found at both the 5' end (before the start codon) and the 3' end (after the stop codon) of the mRNA. They play important roles in regulating gene expression, affecting mRNA stability, translation efficiency, and localization within the cell.
In simple words: UTRs are parts of an mRNA molecule that do not get turned into protein. They are at the beginning (5' end) and end (3' end) of the mRNA and help control how much protein is made.

🎯 Exam Tip: Remember that UTRs, despite not coding for proteins, are vital regulatory elements for mRNA function and stability.

 

Question 47. What is the S - D sequence?
Answer: The S-D sequence, or Shine-Dalgarno sequence, is a specific sequence of bases located at the 5' end of messenger RNA (mRNA) in prokaryotes. This sequence is complementary to a region of the 16S ribosomal RNA (rRNA) found in the small ribosomal subunit. Its main function is to help the ribosome bind correctly to the mRNA, thereby facilitating the initiation of protein synthesis (translation) at the right start codon. It's like a landing strip for the ribosome.
In simple words: The S-D sequence is a special part of prokaryotic mRNA. It helps the ribosome attach at the correct spot to begin making protein.

🎯 Exam Tip: Specify that the Shine-Dalgarno sequence is characteristic of prokaryotic mRNA and is essential for ribosome binding and translation initiation.

 

Question 48. Define translation unit.
Answer: A translation unit in mRNA refers to the specific sequence of RNA that carries the complete genetic information needed to synthesize a polypeptide chain. This unit is marked by a start codon at its 5' end and a stop codon at its 3' end. It includes all the coding sequences in between, ensuring that the ribosome translates the correct stretch of mRNA into a functional protein. This ensures precise protein synthesis.
In simple words: A translation unit is the part of an mRNA molecule that contains the full instructions to make one protein, from the start signal to the stop signal.

🎯 Exam Tip: Clearly state that a translation unit is delineated by a start codon and a stop codon, enclosing all the coding information for a polypeptide.

 

Question 49. Mention the inhibitory role of tetracycline and streptomycin in bacterial translation.
Answer: Tetracycline and streptomycin are antibiotics that inhibit bacterial translation through different mechanisms. Tetracycline works by preventing the binding of aminoacyl-tRNA molecules to the ribosome's A-site, effectively stopping new amino acids from being added to the protein chain. Streptomycin, on the other hand, interferes with the initiation of translation and causes misreading of the mRNA code, leading to incorrect protein synthesis. Both ultimately stop bacterial growth by disrupting essential protein production.
In simple words: Tetracycline stops tRNA from bringing new amino acids to the ribosome, blocking protein growth. Streptomycin causes errors when the ribosome reads the mRNA, also stopping correct protein making.

🎯 Exam Tip: Differentiate between the specific mechanisms of each antibiotic: tetracycline blocks tRNA binding, while streptomycin interferes with initiation and causes misreading.

 

Question 50. At what stage, does the gene expression is regulated?
Answer: Gene expression can be controlled or regulated at various stages within a cell. The primary stages where this regulation occurs are transcriptional levels (when DNA is copied into RNA) and translational levels (when RNA is used to make protein). This allows cells to precisely control which genes are active and how much protein is produced, responding to their environment and internal needs.
In simple words: Gene expression can be controlled at two main points: when DNA is copied into RNA (transcription) and when RNA is used to make protein (translation).

🎯 Exam Tip: Remember the two main levels of gene expression regulation: transcriptional (controlling RNA synthesis) and translational (controlling protein synthesis), as these are the most fundamental.

 

Question 51. What is a operon? Give example.
Answer: An operon is a cluster of genes in prokaryotes that are involved in related functions and are controlled by a single promoter and operator. It acts as a unit, meaning all the genes within the operon are transcribed together as a single mRNA molecule. This allows for coordinated regulation of genes involved in a specific metabolic pathway. A classic example is the lac operon in *E. coli*, which controls the genes needed for lactose metabolism.
In simple words: An operon is a group of genes that work together and are controlled as one unit, usually found in bacteria. For example, the lac operon helps *E. coli* break down lactose.

🎯 Exam Tip: When defining an operon, always mention that it's a prokaryotic feature, a cluster of functionally related genes, and controlled by a single regulatory system.

 

Question 52. Considering the lac operon of E.coli, name the products of the following genes.
1. a) i gene
2. (b) lac Z gene
3. (c) lac Y gene
4. (d) lac a gene
Answer:
1. i gene - repressor protein
2. lac Z gene - \( \beta \)-galactosidase
3. Lac Y gene - Permease
4. lac a gene - transacetylase
In simple words: In the lac operon, the 'i' gene makes a repressor protein. The 'lac Z' gene makes \( \beta \)-galactosidase. The 'lac Y' gene makes permease. The 'lac a' gene makes transacetylase.

🎯 Exam Tip: It's important to remember the specific protein product of each gene in the lac operon as it's a common example of gene regulation.

 

Question 53. Expand
1. ETS
2. YAC.
Answer:
1. ETS: Expressed Sequence Tags
2. YAC: Yeast Artificial Chromosomes
In simple words: ETS means Expressed Sequence Tags. YAC means Yeast Artificial Chromosomes.

🎯 Exam Tip: For abbreviations, ensure you provide the full, correct expanded form without any missing words or typos.

 

Question 54. Name the human chromosome that has
1. most number of genes
2. least number of genes
Answer:
1. Chromosome 1 has the maximum number of genes (2968 genes)
2. Chromosome Y has least genes (231 genes)
In simple words: Chromosome 1 has the most genes, while Chromosome Y has the fewest genes in humans.

🎯 Exam Tip: These are specific facts often tested in genetics. Linking the number 1 with "most" and Y with "least" can help you remember.

 

Question 55. What are SNPs? Mention its uses.
Answer: SNPs stand for Single Nucleotide Polymorphisms, often pronounced "snips." These are single-base DNA differences that occur at specific locations in the human genome. Scientists have identified about 1.4 million such locations where individuals differ by a single nucleotide. SNPs are very useful because they help in finding chromosomal locations for disease-associated sequences, tracing human history and ancestry, and understanding individual responses to drugs. They are like unique markers across our DNA.
In simple words: SNPs are tiny changes in just one DNA building block that make people different. They help find disease genes and trace family history.

🎯 Exam Tip: When defining SNPs, emphasize "single base difference" and their utility in identifying disease linkages and population genetics studies.

 

Question 56. Mention any four areas where DNA fingerprinting can be used.
Answer: DNA fingerprinting is a powerful technique with various applications. Four key areas where it is used are:

• Forensic analysis: To identify criminals or victims from biological samples found at crime scenes.
• Pedigree analysis: To determine family relationships and parentage.
• Conservation of wildlife: To identify species, track populations, and detect poaching.
• Anthropological studies: To study human evolution, migration patterns, and population genetics.

This technique leverages the unique DNA patterns of individuals for identification.
In simple words: DNA fingerprinting is used in crime investigations, to check family links, to protect wild animals, and to study human history.

🎯 Exam Tip: When listing uses, aim for distinct categories like legal/forensic, biological relationships, conservation, and evolutionary studies.

12th Bio Zoology Guide Molecular Genetics Three Marks Questions and Answers

 

Question 57. Classify nucleic acid-based on sugar molecules.
Answer: Nucleic acids are classified into two main types based on the pentose sugar molecule they contain. Those containing deoxyribose sugar are called Deoxyribonucleic Acid (DNA). Those containing ribose sugar are known as Ribonucleic Acid (RNA). The key difference between these two sugars is that deoxyribose has one less oxygen atom than ribose, making DNA more stable. This structural difference impacts their functions in the cell.
In simple words: Nucleic acids are grouped by the sugar they have. DNA has deoxyribose sugar, and RNA has ribose sugar.

🎯 Exam Tip: Clearly state the two types of nucleic acids (DNA and RNA) and the specific sugar (deoxyribose vs. ribose) associated with each.

 

Question 58. Both purines and pyrimidines are nitrogen bases yet they differ. How?
Answer: Both purines and pyrimidines are nitrogenous bases, which are fundamental components of nucleic acids. However, they differ in their chemical structure. Purine bases, such as Adenine and Guanine, have a double-ring structure, consisting of a six-membered ring fused to a five-membered ring. Pyrimidine bases, like Cytosine and Thymine (and Uracil in RNA), have a single-ring structure, which is a six-membered ring. This size difference affects how they pair in the DNA helix. The larger purines always pair with the smaller pyrimidines.
In simple words: Purines (Adenine, Guanine) have two rings in their structure. Pyrimidines (Cytosine, Thymine) have only one ring. Both are nitrogen bases, but their ring size is different.

🎯 Exam Tip: The key distinction is the number of rings: purines are double-ringed, and pyrimidines are single-ringed. Remembering this helps visualize their structures.

 

Question 59. How 5' of DNA differ from its 3'?
Answer: The 5' and 3' ends of a DNA strand refer to the numbering of carbon atoms in the deoxyribose sugar molecule. The 5' end of DNA has a phosphate functional group attached to the 5th carbon of the sugar. In contrast, the 3' end of DNA has a free hydroxyl (OH) group attached to the 3rd carbon of the sugar. These distinct chemical groups give DNA strands their directionality, which is crucial for processes like replication and transcription. This polarity is what allows DNA to be built in one direction only.
In simple words: The 5' end of a DNA strand has a phosphate group. The 3' end has a hydroxyl (OH) group. These different ends give the DNA its direction.

🎯 Exam Tip: Clearly associate the 5' end with the phosphate group and the 3' end with the hydroxyl group, as this defines the directionality of nucleic acids.

 

Question 60. State Chargaff's rule.
Answer: Chargaff's rule, formulated by Erwin Chargaff, states that in a double-stranded DNA molecule, the amount of adenine (A) is always approximately equal to the amount of thymine (T), and the amount of guanine (G) is always approximately equal to the amount of cytosine (C). This means that the total percentage of purines (A+G) equals the total percentage of pyrimidines (T+C). This rule was crucial for Watson and Crick in determining the double helix structure of DNA, as it implies complementary base pairing.
In simple words: Chargaff's rule says that in DNA, the amount of A is equal to T, and the amount of G is equal to C. This means A pairs with T, and G pairs with C.

🎯 Exam Tip: Express Chargaff's rules as two key equalities: %A = %T and %G = %C, and explain their significance in base pairing.

 

Question 61. Chemically DNA is more stable than RNA - Justify.
Answer: DNA is chemically more stable than RNA for several reasons. First, DNA is a double-stranded molecule, with the two complementary strands held together by strong hydrogen bonds. If the strands separate (denaturation), they can easily come back together (renaturation), providing structural integrity. Second, DNA contains deoxyribose sugar, which lacks a hydroxyl (OH) group at the 2' carbon position, making it less reactive. RNA, on the other hand, has a ribose sugar with a 2' OH group, making it more prone to hydrolysis and degradation. Lastly, DNA uses thymine instead of uracil; thymine has an extra methyl group that enhances DNA's stability. These factors allow DNA to store genetic information reliably over long periods. In simple words: DNA is more stable than RNA because it has two strands, its sugar (deoxyribose) is less reactive, and it uses thymine instead of uracil, all of which help it keep genetic information safe for a long time.
In simple words: DNA is more stable than RNA because it has two strands, its sugar is less reactive, and it uses thymine, which adds extra stability.

🎯 Exam Tip: Focus on the three main reasons for DNA's stability: double-stranded structure, absence of 2'-OH group in deoxyribose, and presence of thymine.

 

Question 62. Write in simple about semi-conservative mode of DNA replication.
Answer: The semi-conservative mode of DNA replication was proposed by Watson and Crick and later proven by Meselson and Stahl. In this process, the two strands of the DNA double helix unwind and separate. Each original (parental) strand then acts as a template for the synthesis of a new, complementary daughter strand. So, after one round of replication, each new DNA molecule consists of one old (parental) strand and one newly synthesized strand. This method ensures that genetic information is accurately passed down, as the old strand guides the building of the new one. Thus, half of the original DNA is 'conserved' in each new molecule.
In simple words: In semi-conservative DNA replication, the two original DNA strands separate. Each old strand then makes a new matching strand. So, every new DNA molecule ends up with one old strand and one new strand.

🎯 Exam Tip: The key concept is "one old strand and one new strand" in each daughter DNA molecule. This term directly explains the "semi-conservative" nature.

 

Question 63. Draw a simplified diagram of nucleosome and label it.
Answer:
In a nucleosome, DNA is wrapped around a core of histone proteins. This compact structure helps to organize the long DNA molecule within the cell's nucleus.
In simple words: A nucleosome is formed when DNA wraps around a group of special proteins called histones, like thread on a spool.

🎯 Exam Tip: Ensure your diagram clearly shows the DNA double helix wrapping around the histone octamer and includes labels for both components as well as the H1 linker histone.

 

Question 64. What is a primer?
Answer: A primer is a short strand of RNA (or sometimes DNA) that serves as a starting point for DNA synthesis. DNA polymerase, the enzyme responsible for building new DNA strands, cannot begin synthesizing from scratch; it needs an existing 3'-OH group to add nucleotides. The primer provides this necessary 3'-OH end, allowing DNA polymerase to attach and start adding deoxyribonucleotides to form a new DNA strand. Primers are crucial for initiating both DNA replication and DNA repair processes.
In simple words: A primer is a small piece of RNA that acts like a starting flag for DNA copying. It gives the DNA-building enzyme a place to begin adding new DNA pieces.

🎯 Exam Tip: The critical role of a primer is providing the free 3'-OH group that DNA polymerase requires to initiate DNA synthesis.

 

Question 65. Both strands of DNA are not copied during transcription. Give reason.
Answer: Both strands of DNA are not copied during transcription for two important reasons. Firstly, if both DNA strands acted as templates, they would produce two different RNA molecules, leading to the synthesis of two distinct proteins. This would complicate the genetic information transfer machinery and create confusion in the cell's protein production. Secondly, if two RNA molecules were produced simultaneously from complementary DNA strands, they would be complementary to each other. These two RNA molecules would then bind together to form a double-stranded RNA. Double-stranded RNA cannot be translated into protein, effectively preventing any protein synthesis. Therefore, only one strand acts as a template to ensure clear and functional genetic expression.
In simple words: Only one DNA strand is copied into RNA during transcription. If both were copied, it would make two different proteins, and the two RNA strands might join together, stopping any protein from being made.

🎯 Exam Tip: Remember the two main issues: avoiding ambiguity in protein products and preventing the formation of non-translatable double-stranded RNA.

 

Question 66. What do you mean by a template strand and coding strand?
Answer: In transcription, DNA has two strands: a template strand and a coding strand. The **template strand** is the DNA strand that runs in the 3' to 5' direction. It serves as the actual guide for RNA polymerase, meaning the enzyme reads this strand and synthesizes an RNA molecule that is complementary to it. The **coding strand** is the non-template DNA strand that runs in the 5' to 3' direction. Its sequence is almost identical to the newly synthesized mRNA molecule, with the only difference being that mRNA has uracil (U) instead of thymine (T). This strand is called "coding" because it carries the same sequence as the functional mRNA that will eventually code for a protein, even though it's not directly read by RNA polymerase. The RNA polymerase uses the template strand to produce an RNA molecule that matches the coding strand.
In simple words: The template strand is the DNA strand that RNA polymerase reads to make RNA. The coding strand is the other DNA strand, which has almost the same sequence as the new RNA (just T instead of U).

🎯 Exam Tip: Always remember that the template strand is read 3' to 5', and the coding strand's sequence directly reflects the mRNA (with T replaced by U).

 

Question 67. Name the factors that are responsible for initiation and termination of transcription in prokaryotes.
Answer: In prokaryotes, specific protein factors regulate the initiation and termination of transcription. The **sigma (σ) factor** is responsible for initiating transcription. It helps RNA polymerase recognize and bind to the promoter region on the DNA, ensuring that transcription starts at the correct site. The **rho (ρ) factor** is responsible for terminating transcription. When RNA polymerase encounters the rho factor, it causes the enzyme to detach from the DNA template, thereby ending the synthesis of the RNA molecule. These factors provide precise control over gene expression. So, sigma starts, and rho stops.
In simple words: In bacteria, the sigma factor helps start RNA making (transcription), and the rho factor helps stop it.

🎯 Exam Tip: Clearly link sigma factor with initiation and rho factor with termination. These are distinct regulatory proteins in prokaryotic transcription.

 

Question 68. Name the major RNA types of prokaryotes and mention their role.
Answer: Prokaryotes have three major types of RNA, each with a crucial role in protein synthesis:

• **Messenger RNA (mRNA):** This RNA carries the genetic information from DNA to the ribosomes. It acts as a template, providing the sequence of codons that specifies the order of amino acids in a protein.
• **Transfer RNA (tRNA):** tRNA molecules act as adapters. They bring specific amino acids to the ribosome based on the codons in the mRNA sequence. They have an anticodon loop that recognizes the mRNA codon and an acceptor arm that carries the amino acid.
• **Ribosomal RNA (rRNA):** rRNA is a key component of ribosomes, which are the cellular machinery for protein synthesis. rRNA plays both structural and catalytic (ribozyme) roles during translation, helping to form the peptide bonds between amino acids.

All three RNA types are essential for creating proteins in the cell.
In simple words: Prokaryotes have three main RNAs: mRNA carries the protein code, tRNA brings amino acids, and rRNA helps build the ribosome and make proteins.

🎯 Exam Tip: For each RNA type, ensure you state its full name, abbreviation, and its primary function in the central dogma process.

 

Question 69. Define genetic code.
Answer: The genetic code is the set of rules by which information encoded in genetic material (DNA or mRNA sequences) is translated into proteins. It consists of a series of three-nucleotide sequences called codons, where each codon specifies a particular amino acid (or a stop signal) during protein synthesis. This specific order of base pairs along the DNA molecule determines the kind and order of amino acids that make up the proteins of an organism. The genetic code is largely universal across all forms of life, meaning the same codons typically specify the same amino acids.
In simple words: The genetic code is like a rulebook that tells cells how to read the DNA or RNA message to build proteins using specific amino acids. Each three-letter code (codon) stands for one amino acid.

🎯 Exam Tip: A good definition includes "set of rules," "triplet codons," and "specifies amino acids or stop signals," emphasizing its role in translating genetic information.

 

Question 70. Explain Wobble hypothesis.
Answer: The Wobble hypothesis, proposed by Francis Crick in 1966, explains why some tRNA molecules can recognize more than one codon for a specific amino acid. It states that the base pairing between the codon on mRNA and the anticodon on tRNA is strict only for the first two positions (the 5' and middle bases of the codon). However, at the third position (the 3' base of the codon), the pairing can be less stringent or "wobble." This flexibility means a single tRNA can bind to multiple codons that differ only in their third base, reducing the total number of tRNAs needed for protein synthesis. This hypothesis is significant because it explains the degeneracy of the genetic code and why there are often fewer tRNA molecules than there are codons.
In simple words: The Wobble hypothesis says that the first two bases of an mRNA codon pair strictly with tRNA, but the third base can "wobble" or be less precise. This means one tRNA can read several different codons for the same amino acid, reducing the number of tRNAs needed.

🎯 Exam Tip: Focus on the flexibility at the third base pair (5' of anticodon, 3' of codon) and its consequences: fewer tRNAs required and explanation of code degeneracy.

 

Question 71. Explain the nature of eukaryotic ribosomes.
Answer: Eukaryotic ribosomes are larger and more complex than prokaryotic ribosomes, designated as 80S ribosomes. The "S" stands for Svedberg unit, which indicates the sedimentation rate and size. Each 80S ribosome consists of two main subunits: a large subunit (60S) and a small subunit (40S). The 60S large subunit is composed of 28S rRNA, 5.8S rRNA, 5S rRNA molecules, and about 49 ribosomal proteins. The 40S small subunit contains an 18S rRNA component and approximately 33 ribosomal proteins. These components work together to form a highly efficient machinery for protein synthesis, carefully orchestrated to meet the complex needs of eukaryotic cells.
In simple words: Eukaryotic ribosomes are big (80S) and made of two parts: a large 60S part and a smaller 40S part. Each part has special RNA molecules and many proteins that help make new proteins.

🎯 Exam Tip: Clearly state the overall size (80S) and the sizes of its two subunits (60S and 40S), along with their main rRNA components.

 

Question 72. Expand and define ORF.
Answer: ORF stands for **Open Reading Frame**. An Open Reading Frame is a continuous sequence of DNA or RNA that starts with a start codon (typically AUG) and ends with a stop codon, without any intervening stop codons. It has the potential to be translated into a protein. In essence, it is a stretch of nucleic acid sequence that can be read as a series of triplet codons without interruption, indicating where a protein-coding sequence begins and ends. Identifying ORFs is crucial in genomics to predict potential genes.
In simple words: ORF means Open Reading Frame. It is a part of DNA or RNA that starts with a 'start' signal and ends with a 'stop' signal, and can be used to make a protein.

🎯 Exam Tip: The definition of an ORF must include the presence of a start codon, the absence of internal stop codons, and the presence of a terminal stop codon.

 

Question 73. What are the components of the initiation complex of prokaryotic translation?
Answer: The initiation complex of prokaryotic translation is a crucial assembly required to begin protein synthesis in bacteria. It consists of several key components: the 30S ribosomal subunit (the small subunit), the messenger RNA (mRNA) molecule, the charged N-formylmethionine-tRNA (fMet-tRNAfMet), and three proteinaceous initiation factors (IF1, IF2, IF3). Additionally, Guanosine Triphosphate (GTP) and magnesium ions (Mg\(^{2+}\)) are required for energy and stability. This complex precisely positions the fMet-tRNA at the start codon on the mRNA, ready for the larger ribosomal subunit to join.
In simple words: The prokaryotic translation initiation complex has the small ribosome part (30S), mRNA, a special tRNA with formylmethionine, and helper proteins (IF1, IF2, IF3). GTP and magnesium also help.

🎯 Exam Tip: List all components accurately: 30S subunit, mRNA, fMet-tRNA, IF1, IF2, IF3, GTP, and Mg\(^{2+}\). The specific formylmethionine tRNA is unique to prokaryotes.

 

Question 74. Explain the components of operon.
Answer: An operon, primarily found in prokaryotes, is a functional unit of DNA that contains a cluster of genes under the control of a single promoter. Its main components are:

• **Structural genes:** These are the genes within the operon that code for proteins or enzymes involved in a specific metabolic pathway. In the lac operon, these are lac Z, lac Y, and lac a.
• **Promoter:** This is a DNA sequence located upstream of the structural genes where RNA polymerase binds to initiate transcription. It acts as the "on" switch for the operon.
• **Operator:** This is a regulatory DNA sequence located between the promoter and the structural genes. It acts as a binding site for a repressor protein, which can block RNA polymerase from transcribing the structural genes.
• **Regulator (or i) gene:** This gene, located outside the operon, codes for the repressor protein. This protein can bind to the operator to control the transcription of the structural genes.

These components work together to regulate gene expression efficiently.
In simple words: An operon has structural genes that make proteins, a promoter that RNA polymerase binds to, an operator where a repressor can attach, and a regulator gene that makes the repressor protein.

🎯 Exam Tip: Be sure to define each component (structural genes, promoter, operator, regulator gene) and explain its specific role in the operon's function.

12th Bio Zoology Guide Molecular Genetics Five Marks Questions and Answers

 

Question 75. Describe the Hershey and chase experiment. What is concluded by their experiment?
Answer: The Hershey-Chase experiment, conducted by Alfred Hershey and Martha Chase in 1952, definitively proved that DNA, not protein, is the genetic material. They used bacteriophages (viruses that infect bacteria) for their experiment. Phages inject their genetic material into bacterial cells to replicate. Hershey and Chase wanted to know whether this injected material was DNA or protein.

Here's how they did it:
1. **Labeling:** They grew two separate batches of bacteriophages. In one batch, they labeled the DNA with radioactive phosphorus-32 ( \(^{32}\)P), because DNA contains phosphorus but protein does not. In the other batch, they labeled the proteins with radioactive sulfur-35 ( \(^{35}\)S), because protein contains sulfur but DNA does not.
2. **Infection:** They allowed both sets of labeled phages to infect *E. coli* bacteria.
3. **Blending:** After infection, they gently agitated the mixture in a blender. This step separated the viral coats (which remained outside the bacterial cells) from the bacterial cells.
4. **Centrifugation:** They then centrifuged the samples. The heavier bacterial cells settled at the bottom, forming a pellet, while the lighter viral coats remained in the supernatant (liquid above the pellet).

**Observations and Conclusion:**
* In the batch with \(^{35}\)S-labeled phages, most of the radioactivity (from proteins) was found in the supernatant with the viral coats, not inside the bacterial cells.
* In the batch with \(^{32}\)P-labeled phages, most of the radioactivity (from DNA) was found inside the bacterial cells (in the pellet). When these bacteria were allowed to grow, they produced new phages that also contained \(^{32}\)P.

The experiment conclusively demonstrated that DNA was the material injected into the bacterial cells and was responsible for directing the synthesis of new viruses. Therefore, the Hershey-Chase experiment proved that **DNA is the genetic material**. This experiment was a landmark in molecular biology.
In simple words: Hershey and Chase used viruses to show that DNA is our genetic material. They marked viral DNA with radioactive phosphorus and viral proteins with radioactive sulfur. After the viruses infected bacteria, they found the radioactive DNA inside the bacteria, but the radioactive protein stayed outside. This proved that DNA carries the genetic message.

🎯 Exam Tip: Clearly outline the steps: labeling with \(^{32}\)P (DNA) and \(^{35}\)S (protein), infection, blending, and centrifugation. The key is that \(^{32}\)P entered the cells and was passed on, proving DNA is the genetic material.

 

Question 76. Explain the properties of DNA that makes it an ideal genetic material.
Answer: For a molecule to be considered an ideal genetic material, it must possess several key properties. DNA perfectly fits these criteria, making it the primary genetic material in most living organisms:

1. **Self-replication:** DNA must be able to make exact copies of itself (replicate). DNA achieves this through its semi-conservative replication mechanism, where each strand serves as a template for a new one. This ensures that genetic information is accurately passed from parent to offspring. This is essential for cell division and inheritance.
2. **Information Storage:** It must be able to store complex genetic information. DNA's sequence of nitrogenous bases (A, T, C, G) acts as a blueprint for all cellular functions, including protein synthesis, storing vast amounts of data in a compact form.
3. **Stability:** The genetic material must be chemically and structurally stable enough to store information over long periods without significant changes. DNA's double-helical structure, the presence of deoxyribose sugar (lacking a reactive 2'-OH group), and the use of thymine instead of uracil contribute to its high stability. This stability minimizes spontaneous mutations and degradation.
4. **Variation (Mutation):** It should allow for slow changes or mutations that are necessary for evolution. While stable, DNA can undergo mutations, which introduce genetic variation that is the raw material for natural selection and species adaptation. These changes are infrequent enough to maintain species integrity but common enough for evolutionary processes.

RNA, while capable of some of these functions (like acting as genetic material in some viruses and being catalytic), is generally less stable due to its single-stranded nature and the presence of a reactive 2'-OH group in ribose. Therefore, DNA's unique combination of these properties makes it the ideal molecule for heredity.
In simple words: DNA is great genetic material because it can make copies of itself, hold lots of information, is very stable and does not change easily, and can also change a little over time (mutate) for evolution. RNA is not as good for long-term storage because it is less stable.

🎯 Exam Tip: When justifying DNA as the ideal genetic material, ensure you address all four criteria: self-replication, information storage, stability, and capacity for mutation, and briefly contrast it with RNA's limitations.

 

Question 77. How the DNA is packed in a eukaryotic cell?
Answer: In eukaryotic cells, the very long DNA molecule needs to be highly condensed to fit inside the nucleus. This complex organization involves several levels of packing, forming chromatin:

1. **Nucleosome:** The first level of packing involves DNA wrapping around positively charged proteins called histones. A core of eight histone molecules (two each of H2A, H2B, H3, and H4) forms a histone octamer. Approximately 200 base pairs of negatively charged DNA wrap around this octamer, forming a structure called a nucleosome. This arrangement gives chromatin a "beads-on-a-string" appearance.
2. **Chromatin Fiber (Solenoid):** Multiple nucleosomes are then linked by short stretches of DNA (linker DNA) and an additional histone protein (H1). These nucleosomes further coil and fold to form a more compact 30-nm chromatin fiber, often described as a solenoid structure. This brings the nucleosomes closer together.
3. **Loop Domains:** The 30-nm fiber then forms larger loop domains, which are anchored to a protein scaffold within the nucleus. These loops are about 300 nm in diameter.
4. **Chromatid:** During cell division, these looped domains are further coiled and condensed into a 700-nm wide chromatid, visible as part of a chromosome. This extreme condensation is necessary for proper segregation during mitosis and meiosis.

This hierarchical packaging allows a huge amount of DNA to be stored efficiently and also plays a role in regulating gene expression, as the degree of condensation affects gene accessibility. The less condensed regions, called euchromatin, are transcriptionally active, while the highly condensed regions, heterochromatin, are transcriptionally inactive.
In simple words: DNA in eukaryotic cells is packed tightly in many layers. First, it wraps around proteins called histones to make small "beads" called nucleosomes. These beads then coil into a thicker fiber, which then folds into larger loops. Finally, these loops compact even more to form chromosomes, fitting the long DNA into the tiny cell nucleus.

🎯 Exam Tip: Detail the hierarchy of DNA packaging: nucleosomes (DNA around histones), 30-nm fiber (solenoid), loop domains, and finally condensation into chromatids. Mentioning the role of histones and the difference between euchromatin and heterochromatin adds depth.

 

Question 78. Describe the Hershey and Chase experiment. What is concluded by their experiment?
Answer: Alfred Hershey and Martha Chase did experiments in 1952. They used bacteriophages, which are viruses that infect bacteria. They worked with Phage T2, a virus that infects E. coli bacteria. When these viruses attach to bacteria, some material from the virus enters the bacteria. After some time, the bacteria break open and release many new viruses.

Hershey and Chase wanted to find out if it was DNA or protein from the virus that entered the bacteria. They knew that all nucleic acids (like DNA) contain phosphorus, but not sulfur. Proteins, however, contain sulfur but usually not phosphorus. So, they used radioactive versions of sulfur (35S) and phosphorus (32P). This helped them track the viral protein and DNA separately.

They let the radioactive viruses infect bacteria in a special growth liquid. Viruses grown with 35S had radioactive proteins, and viruses grown with 32P had radioactive DNA. After infection, they gently shook the mixture in a blender. This helped to remove any viral particles stuck outside the bacteria.

They found that only 32P was inside the bacterial cells. The 35S remained in the liquid outside. When they checked the new viruses made by the infected bacteria, they also found only 32P. This clearly showed that only DNA, not the protein coat, entered the bacterial cells. This experiment proved that DNA, not protein, carries the hereditary information from viruses to bacteria, helping to understand how traits are passed on.
In simple words: Hershey and Chase used viruses to show that DNA, not protein, is what carries genetic information. They used special radioactive tags to see which part of the virus entered the bacteria and found it was always the DNA.

🎯 Exam Tip: When describing the Hershey-Chase experiment, clearly state the isotopes used (35S for protein, 32P for DNA), what they label, and the key observation that 32P entered the cells.

 

Question 79. Give a detailed account of a transcription unit.
Answer: A transcription unit in DNA is made up of three main parts: a promoter, the structural gene, and a terminator. The promoter is found towards the 5' end of the DNA strand. It is a specific DNA sequence where the enzyme RNA polymerase attaches to start making RNA. The presence of the promoter helps to define which strand of DNA will be used as a template and which will be the coding strand. This also means that transcription always starts at a particular point.

The two DNA strands in the structural gene have opposite polarities. The DNA-dependent RNA polymerase enzyme can only make RNA in one direction. The strand that runs from 3' to 5' acts as the template strand. The other strand, which runs from 5' to 3', is called the coding strand. This coding strand has almost the same sequence as the RNA that will be made, except that RNA has uracil instead of thymine. The coding strand gets displaced during transcription.

The terminator region is located towards the 3' end of the coding strand. This part of the DNA has a sequence that tells the RNA polymerase to stop making RNA. Once the polymerase reaches this region, it detaches, and the new RNA molecule is released.
In simple words: A transcription unit is a piece of DNA with three parts: a start signal (promoter), the actual gene (structural gene), and a stop signal (terminator). The promoter tells RNA polymerase where to begin, the structural gene holds the information, and the terminator tells it where to stop making RNA.

🎯 Exam Tip: Remember to name the three key regions (promoter, structural gene, terminator) and explain their specific roles and locations (5' or 3' end) in relation to RNA polymerase activity.

 

Question 80. Explain the transcription process in prokaryotes with the needed diagram.
Answer: (Refer to the diagram on page 61 for the process of transcription in prokaryotes.) In prokaryotes, there are three main types of RNA molecules: messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA). All these types of RNA are needed for a cell to make proteins. The mRNA acts as a blueprint, carrying the genetic instructions. The tRNA brings the correct amino acids and helps read the genetic code. The rRNAs have structural and catalytic roles during protein synthesis.

In prokaryotes, a single enzyme, DNA-dependent RNA polymerase, is responsible for making all these types of RNA. This enzyme carries out the entire transcription process. Transcription begins when RNA polymerase binds to a specific region on the DNA called the promoter. To start this process (initiation), the RNA polymerase needs an initiation factor called sigma (σ) factor.

Once transcription begins, the RNA polymerase moves along the DNA, unwinding the helix and adding one RNA nucleotide after another to the growing RNA chain. This phase is called elongation. The polymerase uses nucleoside triphosphate as a building block and follows the base-pairing rules (complementarity). As the RNA chain grows, only a short part of it stays attached to the enzyme.

Transcription ends when the RNA polymerase reaches a specific DNA sequence called the terminator at the end of the gene. To stop the process (termination), the RNA polymerase needs a termination factor called rho (ρ) factor. When the terminator is reached, the newly made RNA molecule breaks away, and the RNA polymerase also detaches from the DNA.

In bacteria, because there is no separate nucleus, transcription and translation (protein synthesis) can happen at the same time and place. The mRNA does not need much processing to become active, and often, translation starts even before the mRNA molecule is fully made. This coupling of transcription and translation is a unique feature of prokaryotes.
In simple words: Transcription in prokaryotes is how DNA's message is copied into RNA. An enzyme called RNA polymerase does this, helped by special factors to start and stop. It happens in three steps: starting, growing the RNA chain, and stopping. Since bacteria don't have a nucleus, making RNA and then proteins can happen almost at the same time.

🎯 Exam Tip: When explaining transcription in prokaryotes, ensure you mention the three types of RNA, the single RNA polymerase, and the roles of sigma (σ) and rho (ρ) factors in initiation and termination, respectively.

 

Question 81. Write the salient features of genetic code.
Answer: The genetic code has several important features:

• The genetic code is a triplet code, meaning that each amino acid is specified by a sequence of three nucleotide bases (a codon). Out of 64 possible codons, 61 code for amino acids.
• There are three codons that do not code for any amino acid; these are called stop or termination codons (UAA, UAG, UGA). They signal the end of protein synthesis.
• The genetic code is universal, which means that almost all living systems, from bacteria to humans, use the same code. For example, the codon UUU always codes for the amino acid phenylalanine in all organisms. While there are a few exceptions in prokaryotic, mitochondrial, and chloroplast genomes, the general universality holds true.
• A non-overlapping codon means that the genetic message is read in groups of three bases without skipping any bases or reusing a base from a previous codon.
• The code is always read in a fixed direction, from the 5' end to the 3' end, also known as its polarity.
• The genetic code is comma-less, meaning there are no gaps or punctuation marks between codons; the sequence is read continuously.
• AUG has dual functions: it acts as the initiator codon, signaling where to start protein synthesis, and it also codes for the amino acid methionine.
• The genetic code is degenerate, which means that more than one triplet codon can code for the same specific amino acid. For instance, codons GUU, GUC, GUA, and GUG all code for valine. This redundancy helps protect against the effects of some mutations.

In simple words: The genetic code is like a secret language in DNA. Each 'word' is three letters long and tells the cell which amino acid to use. It's almost the same for all living things, has specific start and stop signals, and is read without breaks. Plus, sometimes different 'words' can mean the same amino acid.

🎯 Exam Tip: To score full marks, ensure you explain all key features like triplet nature, universality, non-overlapping, degeneracy, and the role of start/stop codons.

 

Question 82. Mutations on genetic code affect the phenotype. Describe with an example.
Answer: Mutations are changes in the genetic code, and these changes can affect an organism's traits or observable characteristics (phenotype). The simplest kind of mutation at the molecular level is a point mutation, where just one base in the DNA sequence is swapped for another. These changes can happen naturally or be caused by external factors.

A well-known example of how a point mutation affects phenotype is sickle cell anemia in humans. This disease is caused by a single base change in the gene for beta-hemoglobin (PHb). Hemoglobin is a protein in red blood cells that carries oxygen. A normal hemoglobin molecule has four protein chains: two alpha and two beta chains. Each chain has a heme group that binds oxygen.

In sickle cell anemia, there is a single base substitution at the sixth codon of the beta-globin gene. The normal codon GAG, which codes for glutamic acid, changes to GTG, which codes for valine. This single change from glutamic acid to valine at the sixth position of the beta-chain is a classic example of a point mutation. This small change in the protein causes the mutant hemoglobin to clump together when oxygen levels are low. This makes the red blood cells change from their normal round, biconcave shape to a sickle or crescent shape. These sickle cells are stiff, can block blood flow, and don't carry oxygen well, leading to the symptoms of sickle cell anemia.
In simple words: Changes in DNA, called mutations, can alter how an organism looks or functions. For example, in sickle cell anemia, a tiny change in one letter of the DNA code for hemoglobin makes red blood cells turn into a sickle shape, causing health problems. This shows how a small genetic change can have a big effect.

🎯 Exam Tip: When explaining mutations, define point mutation and use sickle cell anemia as a specific example, detailing the codon change (GAG to GTG) and the resulting amino acid substitution (glutamic acid to valine).

 

Question 83. Explain the mechanism of AteArperon of the E-coli.
Answer: (Refer to the diagram on page 65 for the Lac Operon model.) The Lac operon is a famous example in E. coli that shows how genes are controlled. The metabolism of lactose (a sugar) in E. coli needs three specific enzymes: permease, beta-galactosidase (β-gal), and transacetylase.

Permease helps lactose enter the bacterial cell. Beta-galactosidase breaks down lactose into glucose and galactose. Transacetylase transfers an acetyl group from acetyl-CoA to β-galactosides, though its exact role in lactose metabolism is less clear than the other two. The lac operon has a regulatory gene (called 'i' gene, which stands for inhibitor), a promoter site (p), and an operator site (o). It also includes three structural genes: lac z, lac y, and lac a. The lac z gene codes for β-galactosidase, lac y codes for permease, and lac a codes for transacetylase.

Jacob and Monod proposed this model to explain how gene expression is regulated in E. coli. When E. coli has glucose (its preferred energy source), the 'i' gene makes a repressor mRNA. This mRNA is translated into a repressor protein. This repressor protein binds to the operator region of the operon. When the repressor is bound to the operator, it blocks RNA polymerase from transcribing the structural genes (lac z, y, and a). As a result, the enzymes needed to use lactose are not made.

However, if lactose is available and glucose is not, lactose acts as an inducer. Lactose (or its derivative, allolactose) binds to the repressor protein. This binding changes the shape of the repressor protein, making it unable to bind to the operator. With the operator now free, RNA polymerase can bind to the promoter and start transcribing the lac z, y, and a genes. This leads to the production of the enzymes necessary for lactose metabolism. This regulation of the lac operon by the repressor protein is an example of negative control, where the repressor prevents gene expression unless an inducer is present. This ensures that the cell only makes these enzymes when lactose is available.
In simple words: The Lac operon in E. coli controls genes for using lactose sugar. Normally, a protein called a repressor blocks these genes. But if lactose is present, it binds to the repressor, moving it out of the way. This allows the genes to turn on, so the bacteria can make enzymes to break down lactose. It's a smart way for the cell to save energy.

🎯 Exam Tip: When explaining the Lac operon, highlight the roles of the repressor protein (inhibitor), operator, and inducer (lactose/allolactose), and differentiate between gene expression in the presence and absence of lactose.

 

Question 84. What are the objectives of the Human Genome Project?
Answer: The Human Genome Project (HGP) had several important goals:

• To identify all the genes in human DNA, estimated to be around 30,000.
• To determine the exact sequence of the three billion chemical base pairs that make up human DNA.
• To store all this genetic information in accessible databases for scientists worldwide.
• To develop better tools and methods for analyzing this vast amount of data.
• To transfer related technologies and knowledge to other fields, such as medicine and biotechnology.
• To address any ethical, legal, and social issues (ELSI) that might come up from knowing our complete genetic code.

In simple words: The Human Genome Project aimed to map all human genes, find the exact order of DNA letters, store this information in computers, create better analysis tools, and consider the moral and social questions that arise from this knowledge.

🎯 Exam Tip: Remember at least three core objectives: sequencing the entire human genome, identifying all genes, and addressing ethical concerns.

 

Question 85. Write the salient features of the Human Genome Project.
Answer: The Human Genome Project (HGP) revealed many important facts about the human genome:

• The human genome contains about 3 billion nucleotide bases, but only about 5% of this DNA actually codes for proteins.
• An average gene has about 3,000 bases. The largest known human gene is dystrophin, with 2.4 million bases.
• The function of about 50% of the genome is related to transposable elements (pieces of DNA that can move around), such as LINEs and ALUs.
• Genes are spread across 24 chromosomes (22 autosomes and X and Y sex chromosomes). Chromosome 19 has the most genes, while chromosome Y has the fewest (231 genes). Chromosome 1 also has a high number of genes (2968).
• The arrangement of human genes on chromosomes shows a lot of variety.
• There are estimated to be between 35,000 and 40,000 genes in the human genome, and nearly 99.9% of nucleotide bases are identical in all people.
• The functions of more than 50% of the discovered genes are still unknown.
• Less than 2% of the genome codes for proteins.
• A very large part of the human genome consists of repeated sequences. These repetitive DNA sequences do not directly code for proteins but are important for understanding chromosome structure, how chromosomes change, and how species evolve (genetic diversity).
• Scientists have found about 1.4 million spots where there are single-base DNA differences, known as Single Nucleotide Polymorphisms (SNPs or 'snips'), in humans. These SNPs are useful for finding gene locations related to diseases and tracking human history.

In simple words: The Human Genome Project found that humans have billions of DNA letters, but only a small part makes proteins. We all share almost the same DNA, but small differences exist. Genes are spread across chromosomes, with some having many and others few. Many parts of our DNA are repeats, and a lot of gene functions are still a mystery.

🎯 Exam Tip: Focus on remembering key numbers (3 billion bases, 2% coding), the concept of repetitive DNA, and the significance of SNPs for diseases and evolution.

 

Question 86. Describe the principle involved in the DNA fingerprinting technique.
Answer: The DNA fingerprinting technique was first developed by Alec Jeffreys in 1985. The basic idea behind DNA fingerprinting is that every person's DNA, and thus their DNA "fingerprint," is unique, much like traditional fingerprints. This uniqueness comes from specific parts of our DNA.

Our chromosomes contain genes, which are segments of DNA with different sequences of nucleotides. Not all DNA segments code for proteins; some have regulatory functions, others are "intervening sequences" (introns), and many are "repeated DNA sequences." These repeated DNA sequences are key to DNA fingerprinting.

In DNA fingerprinting, scientists look at short, repetitive nucleotide sequences that are unique to each person. These are called Variable Number Tandem Repeats (VNTRs). The VNTRs show a lot of variation between individuals and act as genetic markers. DNA fingerprinting works by identifying these differences in repetitive DNA sequences. Within these repetitive sequences, a small stretch of DNA is repeated many times.

When genomic DNA is cut into fragments, these repetitive DNA sequences, due to their unique lengths and compositions, separate from the main bulk of DNA during density gradient centrifugation. The majority of the DNA forms a large peak (bulk DNA), while the smaller, repetitive fragments form distinct smaller peaks, known as satellite DNA.

Based on their base composition (whether they are rich in A-T or G-C), length, and the number of repeating units, satellite DNA is further categorized into mini-satellites and micro-satellites. These sequences do not code for proteins but make up a large part of the human genome. Their high degree of variation (polymorphism) is the foundation of DNA fingerprinting. By comparing VNTR patterns from DNA samples (like blood, hair, or skin cells found at a crime scene) with those from a suspect, forensic scientists can determine guilt or innocence or identify a victim.
In simple words: DNA fingerprinting uses the unique repeating parts of a person's DNA, called VNTRs, to create a unique pattern. Since everyone's VNTR pattern is different, it can be used like a barcode to identify people or link them to a crime scene by comparing DNA samples.

🎯 Exam Tip: Define VNTRs, explain their polymorphic nature, and mention how their unique patterns are used for individual identification, emphasizing that they are non-coding repetitive sequences.

 

Question 87. Draw a flow chart depicting the steps of the DNA fingerprinting technique.
Answer: (Refer to the flowchart diagram on page 68.) The process of DNA fingerprinting involves several key steps to create a unique genetic profile:

• 1. DNA Extraction: First, DNA is carefully taken out from a sample of cells, such as blood, saliva, hair, or skin.
• 2. DNA Fragmentation: The extracted DNA is then cut into smaller, specific pieces using special enzymes called restriction enzymes. These enzymes recognize and cut DNA at particular sequences.
• 3. Electrophoresis: The DNA fragments are then separated based on their size by a process called gel electrophoresis. This separates the fragments into distinct bands, with smaller pieces moving faster and further through the gel.
• 4. Southern Blotting: After separation, the DNA bands from the gel are transferred onto a nylon membrane. This process is called Southern blotting, which helps to fix the DNA in place.
• 5. Probe Hybridization: A special radioactive DNA probe is prepared. This probe is designed to bind to specific repetitive DNA sequences (VNTRs) on the membrane. This binding is called hybridization.
• 6. Washing: Any extra, unbound DNA probe is carefully washed away from the membrane.
• 7. Autoradiography: Finally, an X-ray film is placed next to the membrane. The radioactive probes attached to the DNA bands expose the film, creating a visible pattern of bands. This pattern is the DNA fingerprint.

This unique DNA fingerprint can then be compared with other samples to establish identity in forensic cases or paternity tests.
In simple words: DNA fingerprinting involves taking DNA, cutting it into pieces, sorting the pieces by size, transferring them to a special paper, and then using a radioactive tag to highlight unique patterns. These patterns act like a personal barcode to identify someone.

🎯 Exam Tip: List the major steps in order: extraction, fragmentation (restriction enzymes), separation (electrophoresis), blotting (Southern blotting), probing (hybridization), and visualization (autoradiography).

 

Question 1. An mRNA strand has a series of triplet codons of which the first three codons are given below (a) AUG (b) UUU (c) UGC (i) Name the amino acid encoded by these triplet codons. (ii) Mention the DNA sequence from which these triplet codons would have transcribed?
Answer:
(i) The amino acids encoded by these triplet codons are:
(a) AUG codes for Methionine
(b) UUU codes for Phenylalanine
(c) UGC codes for Cysteine

(ii) To find the DNA sequence, we need to find the complementary bases, remembering that RNA has Uracil (U) while DNA has Thymine (T). So, U in mRNA pairs with A in DNA, A with T, G with C, and C with G.
For AUG in mRNA, the complementary DNA sequence would be TAC.
For UUU in mRNA, the complementary DNA sequence would be AAA.
For UGC in mRNA, the complementary DNA sequence would be ACG.
In simple words: The mRNA codons AUG, UUU, and UGC stand for Methionine, Phenylalanine, and Cysteine, respectively. The original DNA parts that made these mRNA codons are TAC, AAA, and ACG.

🎯 Exam Tip: Always remember the universal genetic code for common amino acids and the base pairing rules for converting between DNA and RNA (A-T/U, G-C).

 

Question 2. Given below are the structures of tRNA molecules which are involved in translation process. In one tRNA, codon is mentioned but not the amino acid. In another tRNA molecule, amino acid is named and not the codon. Complete the figure by mentioning the respective amino acids and codons.
Answer:
(A) The tRNA molecule shown with the codon GAG should be labeled with the amino acid Leucine. So, the label should be (A) Leucine.
(B) The tRNA molecule shown with the amino acid Arginine should have the codon GCG. So, the label should be (B) Arginine (with codon GCG).

(The specific anticodon for Leucine (GAG) is CUC. The specific anticodon for Arginine (GCG) is CGC. The question seems to refer to the amino acid on top and the anticodon at the bottom, or vice versa, but based on the provided answer, it implies a codon to amino acid mapping.)
In simple words: For the first tRNA, where the code GAG is given, the missing amino acid is Leucine. For the second tRNA, where the amino acid Arginine is given, the missing code at the bottom is GCG.

🎯 Exam Tip: To complete such diagrams, know the genetic code table to correctly match codons with their corresponding amino acids and understand that tRNA carries the amino acid corresponding to its anticodon.

 

Question 3. A DNA fragment possesses 32 adenine bases and 32 cytosine bases. How many total number of nucleotides does that DNA fragment contains? Explain.
Answer: In a DNA fragment, according to Chargaff's rules, the number of adenine (A) bases always equals the number of thymine (T) bases, and the number of guanine (G) bases always equals the number of cytosine (C) bases. This is because A pairs with T, and G pairs with C.

Given that the DNA fragment has:
- 32 Adenine (A) bases
- 32 Cytosine (C) bases

Using Chargaff's rules:
- Since there are 32 Adenine bases, there must also be 32 Thymine (T) bases.
- Since there are 32 Cytosine bases, there must also be 32 Guanine (G) bases.

To find the total number of nucleotides, we add up all the bases:
Total nucleotides = Adenine + Thymine + Guanine + Cytosine
Total nucleotides = 32 (A) + 32 (T) + 32 (G) + 32 (C)
Total nucleotides = 128 nucleotides.
So, this DNA fragment contains a total of 128 nucleotides. The consistent pairing of bases is fundamental to DNA structure.
In simple words: In DNA, Adenine always pairs with Thymine, and Guanine always pairs with Cytosine. So, if we have 32 Adenine and 32 Cytosine, we must also have 32 Thymine and 32 Guanine. Adding all these up (32+32+32+32) gives a total of 128 nucleotides in the DNA fragment.

🎯 Exam Tip: Always apply Chargaff's rules (A=T, G=C) to calculate nucleotide counts in a double-stranded DNA molecule. This rule is crucial for understanding DNA composition.

 

Question 4. Following is a DNA sequence representing a part of the gene TAC TCG CCC TAT UAA CCC AAA ACC TCT using this derive A. 1. The RNA transcript 2. The spliced mRNA (consider all the codons with two Aderine bases are introns) 3. The total number of amino acids coded by the mRNA
Answer: The provided DNA sequence is: TAC TCG CCC TAT UAA CCC AAA ACC TCT.
Note: Assuming 'UAA' in the DNA sequence is a typo and should be 'TAA' since DNA contains Thymine, not Uracil. We will proceed with TAA.

1. The RNA transcript: During transcription, DNA is copied into RNA. The template strand of DNA (read 3' to 5') is used. If this given DNA sequence is the coding strand (5' to 3'), then the template strand would be its complement. If we assume the given sequence is the template strand, then RNA polymerase reads it and synthesizes an RNA molecule by pairing A with U, T with A, C with G, and G with C.
Given DNA (Template Strand, assuming 3' to 5' read): 3'- TAC TCG CCC TAT TAA CCC AAA ACC TCT -5'
Complementary RNA (5' to 3'): 5'- AUG UGC GGG AUA AUU GGG UUU UGG AGA -3'

2. The spliced mRNA (considering all codons with two Adenine bases are introns):
The RNA transcript is: AUG UGC GGG AUA AUU GGG UUU UGG AGA.
We are asked to remove introns, defined as codons with two Adenine (A) bases. Let's look for such codons:
- AUG: One A (keep)
- UGC: Zero A's (keep)
- GGG: Zero A's (keep)
- AUA: Two A's (intron - remove)
- AUU: Two A's (intron - remove)
- GGG: Zero A's (keep)
- UUU: Zero A's (keep)
- UGG: One A (keep)
- AGA: Two A's (intron - remove)

After removing the introns (AUA, AUU, AGA), the spliced mRNA sequence is:
Spliced mRNA: AUG UGC GGG GGG UUU UGG

3. The total number of amino acids coded by the mRNA:
The spliced mRNA sequence is: AUG UGC GGG GGG UUU UGG.
Each triplet codon codes for one amino acid. We count the number of codons in the spliced mRNA:
- AUG (1)
- UGC (2)
- GGG (3)
- GGG (4)
- UUU (5)
- UGG (6)
There are 6 codons in the spliced mRNA. Therefore, it codes for 6 amino acids.
In simple words: First, we change the DNA sequence into an RNA message. Then, we cut out parts of this RNA (called introns) that have two 'A' letters in their code. The remaining RNA piece then tells the cell to make 6 amino acids, because it has 6 three-letter codes left.

🎯 Exam Tip: When converting DNA to RNA, remember to substitute 'T' with 'U'. When splicing, carefully follow the given rule for identifying introns and ensure the remaining sequence forms continuous codons for amino acid counting.

 

Question 5. Complete the molecular processes by naming them 1. DNA → DNA 2. mRNA → Protein 3. RNA transcript → mRNA
Answer: Here are the names of the molecular processes:
1. DNA → DNA: This process is called Replication. It is how a DNA molecule makes exact copies of itself.
2. mRNA → Protein: This process is called Translation. It is how the genetic information from messenger RNA (mRNA) is used to build a protein.
3. RNA transcript → mRNA: This process is called Splicing (and other post-transcriptional modifications). It involves removing non-coding regions (introns) from an initial RNA transcript to create a mature messenger RNA (mRNA) that can be translated into protein.
In simple words: 1. DNA making more DNA is called Replication. 2. mRNA making a protein is called Translation. 3. Cleaning up an RNA transcript to make it ready for protein-making is called Splicing.

🎯 Exam Tip: Understand the central dogma of molecular biology: DNA replicates, DNA is transcribed into RNA, and RNA is translated into protein. Remember splicing is an important step in RNA processing.