All questions of Replication of DNA for Grade 9 Exam

RNA polymerase moves along the template strand, synthesising an mRNA molecule. In prokaryotes RNA polymerase is a holoenzyme consisting of a number of subunits, including a sigma factor (transcription factor) that recognises the promoter. In eukaryotes there are three RNA polymerases: I, II and III. The process includes a proofreading mechanism.

Splicing is the process of removing introns and joining exons in a defined order in a transcription unit. This process takes place in the nucleus of eukaryotic cells and is mediated by a complex called the spliceosome.

Explanation:

• Transcription is the process of synthesizing RNA from a DNA template. During transcription, the entire DNA sequence is transcribed into RNA, including both the introns and exons.

• Introns are non-coding sequences within a gene, while exons are the coding sequences that specify the amino acid sequence of a protein. Introns need to be removed from the RNA sequence before translation can occur.

• Splicing is the process of removing introns and joining exons in a defined order, resulting in a mature messenger RNA (mRNA) molecule that can be translated into a protein.

• The splicing process is mediated by the spliceosome, a complex that consists of small nuclear RNAs (snRNAs) and proteins. The snRNAs base-pair with the intron sequences, while the proteins catalyze the chemical reactions that remove the introns and join the exons.

• Splicing is essential for the proper expression of genes, as it allows for the creation of multiple protein variants from a single gene by alternative splicing. Mutations that affect splicing can lead to a variety of diseases, including cancer and genetic disorders.

In conclusion, splicing is the process of removing introns and joining exons in a defined order in a transcription unit, and it is essential for the proper expression of genes.

Nucleobases, also known as nitrogenous bases  are nitrogen-containing biological compounds that form nucleosides, which in turn are components of nucleotides, with all of these monomersconstituting the basic building blocks of nucleic acids. The ability of nucleobases to form base pairs and to stack one upon another leads directly to long-chain helical structures such as ribonucleic acid (RNA) and deoxyribonucleic acid (DNA)....

TRNA with an attached amino acid is said to be "charged". The enzyme that attaches the amino acid to the 3'-OH is called an aminoacyl tRNA synthetase (aaRS). There is a specific tRNA for each amino acid, 20 in all. Similarly, there is a specific aaRS for each tRNA.

The site at which RNA polymerase binds during transcription is called the promoter. It is a specific region of DNA that is recognized by RNA polymerase, which then initiates the process of transcription. The promoter is located upstream of the transcriptional start site and contains specific DNA sequences that are recognized by RNA polymerase and other transcription factors.

The promoter plays a crucial role in regulating gene expression, as different promoters can activate or repress transcription depending on the cellular context. The strength of the promoter can also affect the rate of transcription, with stronger promoters resulting in higher levels of mRNA production.

There are different types of promoters, including constitutive promoters, which are active in all cells, and inducible promoters, which are activated in response to specific signals or conditions. The sequence and structure of the promoter can vary depending on the gene being transcribed and the organism in which it is expressed.

In summary, the promoter is a critical element in the process of transcription, serving as the site at which RNA polymerase binds and initiating the production of mRNA from DNA. It plays a crucial role in regulating gene expression and can vary in strength and specificity depending on the cellular context and the gene being transcribed.

The genetic code is degenerate: Some amino acids are encoded by more than one codon, inasmuch as there are 64 possible base triplets and only 20 amino acids. In fact, 61 of the 64 possible triplets specify particular amino acids and 3 triplets (called stop codons) designate the termination of translation. Thus, for most amino acids, there is more than one code word.

The discovery that DNA has genetic properties was first revealed by Oswald Avery and his team of scientists in 1944. Avery was a molecular biologist who worked at the Rockefeller Institute for Medical Research in New York City.

Experiment
Avery and his team conducted a series of experiments to determine whether DNA was the genetic material responsible for the transformation of bacteria. They used two strains of Streptococcus pneumoniae, one that was virulent (able to cause disease) and one that was non-virulent. The virulent strain had a capsule made of a complex sugar that protected it from the immune system, while the non-virulent strain lacked this capsule and was easily destroyed by the immune system.

Results
Avery and his team extracted various biochemical components from the virulent strain of bacteria, including proteins, lipids, RNA, and DNA. They then mixed each of these components with the non-virulent strain to see if they could induce transformation. Only the DNA extract was able to transform the non-virulent strain into a virulent one, proving that DNA was the genetic material responsible for the transformation.

Conclusion
Avery's discovery was groundbreaking because it showed that DNA, which was previously thought to be a simple molecule with no biological significance, was in fact the carrier of genetic information. This discovery paved the way for the development of the field of molecular biology and our current understanding of genetics.

Meselson and Stahl prove experimentally semi conservation of DNA replication not say but structure of DNA.

A typical nucleosome contains 200 bp of DNA double helix wrapped(2 turns) around a core of histone octamer having two copies of each of four types of histone proteins ..H2A,H2B,H3,&H4.....H1 histone molecule lies outside the nucleosome core & seals the two turns of DNA by binding at the point where DNA enters and leaves the core..

In 1928 Griffith experiment on streptococcus pneumonia and mice

The Hershey and Chase Experiment

The Hershey and Chase experiment, conducted in 1952 by Alfred Hershey and Martha Chase, provided important evidence supporting the concept that DNA, rather than protein, is the genetic material. This experiment was crucial in the development of the field of molecular biology and helped solidify the understanding of DNA as the molecule responsible for transmitting hereditary information.

Experimental Setup

In the Hershey and Chase experiment, the researchers used bacteriophages, which are viruses that infect bacteria. Bacteriophages consist of a protein coat, called the capsid, that encapsulates their genetic material, either DNA or RNA. The researchers wanted to determine whether the genetic material of the bacteriophage was DNA or protein.

To accomplish this, Hershey and Chase used two separate batches of bacteriophage T2. In one batch, they labeled the DNA of the bacteriophage with radioactive phosphorus-32 (32P), while in the other batch, they labeled the protein coat with radioactive sulfur-35 (35S).

Infection and Blending

The labeled bacteriophages were then used to infect separate cultures of E. coli bacteria. The infection process allowed the bacteriophage to inject its genetic material into the bacterial cell. To separate the bacteriophages from the bacterial cells, Hershey and Chase used a blender to shear off the protein coats that remained on the outside of the bacteria.

Centrifugation

After blending, the mixture was subjected to centrifugation, a process that separates particles based on their density. The heavy bacterial cells settled at the bottom of the centrifuge tube, forming a pellet, while the lighter bacteriophages remained in the liquid portion, called the supernatant.

Results

The researchers observed that the pellet in the 32P-labeled experiment was radioactive, indicating that the bacteriophage's genetic material had entered the bacterial cells. Conversely, the supernatant in the 32P-labeled experiment was not radioactive, suggesting that the protein coats remained outside the bacterial cells.

In the 35S-labeled experiment, the opposite pattern was observed. The pellet was not radioactive, indicating that the protein coats did not enter the bacterial cells, while the supernatant was radioactive, suggesting that the labeled protein coats had remained outside the bacterial cells.

Conclusion

Based on these results, Hershey and Chase concluded that the genetic material of the bacteriophage is DNA, not protein. The radioactive 32P-labeled DNA had entered the bacterial cells and was responsible for transmitting the genetic information, while the 35S-labeled protein coats remained outside the bacterial cells.

This experiment provided strong evidence supporting the role of DNA as the genetic material and laid the foundation for future research and discoveries in molecular biology.

Four different types of nitrogenous bases are found in DNA: two purines adenine (A) and guanine (G) & two pyrimidines cytosine (C) and thymine (T). In RNA, the thymine is replaced by uracil (U).

For gene transfer into the host cell without using vector microparticles made of tungsten and gold coated with foregin DNA are bombarded into target cells at a very high velocity. This method is called biolistics or gene gun which is suitable for plants.So the correct option is 'gold or tungsten'.

Option A is correct becose, the nucleic acid (DNA or RNA) is consists of polymers of nucleo tides..

The correct answer is b.

Control of gene expression takes place at the level of transcription, which is the process of synthesizing RNA from DNA template. Transcription is a highly regulated process that involves the binding of transcription factors and other regulatory proteins to specific DNA sequences, known as cis-acting elements, located in the promoter and enhancer regions of genes. The control of gene expression at the level of transcription involves various mechanisms that can enhance or repress transcription, such as:

1. Gene regulatory proteins: These proteins bind to specific DNA sequences and either activate or repress transcription. Examples of gene regulatory proteins include transcription factors, coactivators, and corepressors.

2. Epigenetic modifications: These modifications can alter the accessibility of DNA to transcriptional machinery, either by loosening or tightening the chromatin structure. Examples of epigenetic modifications include DNA methylation, histone modifications, and chromatin remodeling.

3. RNA processing: Alternative splicing, polyadenylation, and RNA editing can generate multiple mRNA isoforms from a single gene, which may have different stability, localization, and translation efficiency.

4. Regulatory RNAs: Non-coding RNAs, such as microRNAs and long non-coding RNAs, can regulate gene expression by binding to mRNAs and either repressing their translation or promoting their degradation.

In summary, the control of gene expression at the level of transcription is a complex process that involves the interplay of various regulatory mechanisms. Understanding these mechanisms is essential for developing treatments for diseases that result from dysregulated gene expression.

Restriction endonuclease is a type of enzyme that can cleave molecules of DNA at a particular site called restriction site having polindromic sequence. These enzymes are produced by many bacteria and protect the cell by cleaving and destroying the DNA of invading viruses. Now a days, restriction enzymes are widely used in the techniques of genetic engineering.Therefore, the correct answer is option 'A'.

Overview of Restriction Endonucleases
Restriction endonucleases, also known as restriction enzymes, are crucial in molecular biology for manipulating DNA. Let's evaluate the statements provided regarding these enzymes.
Statement Analysis
- i. Hind II was the first restriction endonuclease discovered and cuts DNA at a specific sequence of six base pairs.
- This statement is incorrect. Hind II was not the first restriction enzyme discovered; that title goes to Hind II's predecessor, Hind I. Additionally, while Hind II does cut DNA at a specific site, it does not necessarily cut at a sequence of six base pairs.
- ii. The convention for naming these enzymes is the first 2 letters of the name comes from the genus and the second one letter comes from the species of the prokaryotic cell from which they were isolated.
- This statement is correct. For example, EcoRI is derived from Escherichia coli (genus: Eco, species: R).
- iii. Exonucleases and endonucleases are both types of restriction enzymes that function by cutting DNA at specific sites.
- This statement is incorrect. Exonucleases remove nucleotides from the ends of DNA, while endonucleases cut within the DNA strand. Only endonucleases are classified as restriction enzymes.
- iv. Each restriction endonuclease can only recognize and cut palindromic sequences in the DNA.
- This statement is correct. Most restriction enzymes recognize specific palindromic sequences in the DNA, which are sequences that read the same forwards and backwards.
Correct Statements
Based on the analysis, the correct statements are ii and iv. Thus, the correct answer is option 'c' (i and iv).

Functions of RNA:
RNA, or ribonucleic acid, plays a crucial role in various cellular processes. It is involved in genetic information transfer, protein synthesis, and is a constituent component of ribosomes.

1. RNA as a carrier of genetic information from DNA to ribosomes synthesizing polypeptides:
One of the primary functions of RNA is to act as an intermediary between DNA and protein synthesis. This process involves three types of RNA: messenger RNA (mRNA), transfer RNA (tRNA), and ribosomal RNA (rRNA).

- mRNA: mRNA carries the genetic information encoded in DNA to the ribosomes, where it serves as a template for protein synthesis. It carries the instructions for the order of amino acids in a polypeptide chain.

- tRNA: tRNA plays a crucial role in protein synthesis by transporting amino acids to the ribosomes. It binds to specific amino acids and carries them to the ribosome, where they are added to the growing polypeptide chain according to the instructions provided by mRNA.

- rRNA: rRNA is a major component of ribosomes, which are the cellular organelles responsible for protein synthesis. rRNA combines with proteins to form ribosomes, providing the structural framework for protein synthesis.

2. RNA carrying amino acids to ribosomes:
As mentioned above, tRNA molecules are responsible for carrying amino acids to the ribosomes during protein synthesis. Each tRNA molecule has a specific anticodon sequence that matches with the codon on mRNA, ensuring the correct amino acid is added to the growing polypeptide chain.

3. RNA as a constituent component of ribosomes:
rRNA, along with proteins, forms the structure of ribosomes. Ribosomes consist of a large subunit and a small subunit, both of which contain rRNA molecules. These ribosomal subunits work together to catalyze the process of protein synthesis.

Conclusion:
In summary, RNA serves multiple functions in the cell, including carrying genetic information from DNA to ribosomes, transporting amino acids to ribosomes, and being a constituent component of ribosomes. These functions are essential for the synthesis of proteins, which are vital for various cellular processes and functions.

Beadle and Tatum proposed one gene one enzyme hypothesis and gene is present on DNA..

Understanding the Assertion and Reason
The assertion and reason provided in the question relate to the use of alternative selectable markers in recombinant DNA technology.
Assertion (A): The use of alternative selectable markers in recombinant DNA technology simplifies the identification of recombinant colonies.
- Selectable markers are crucial in recombinant DNA technology.
- They help distinguish between colonies that contain recombinant DNA and those that do not.
- By using alternative markers, researchers can streamline the identification process.
Reason (R): These markers allow for the differentiation of recombinants from non-recombinants based solely on color change in the presence of a chromogenic substrate.
- Chromogenic substrates react with specific markers to produce a color change.
- This color change serves as a visual indicator of successful recombination.
- It provides a straightforward method for identifying recombinant colonies.
Why Option A is Correct
- Both the assertion and reason are true.
- The assertion accurately reflects that alternative selectable markers simplify the identification process.
- The reason correctly explains how these markers function, specifically highlighting the role of color change in differentiating recombinants from non-recombinants.
Conclusion
- Therefore, option "A" is correct, as both statements are true, and the reason provided is indeed the correct explanation for the assertion.
- This understanding enhances the efficiency of selecting recombinant DNA in laboratory settings, showcasing the importance of these techniques in biotechnology.

Background of Hershey and Chase Experiment
The Hershey-Chase experiment, conducted in 1952, was pivotal in demonstrating that DNA is the genetic material of viruses. This experiment primarily involved the use of T2 bacteriophage, a virus that infects bacteria.
Medium Used for Cultivation
In their experiment, Alfred Hershey and Martha Chase used a specific medium for cultivating the viruses:

• Radioactive Phosphorus (P): They labeled the DNA of the T2 bacteriophage with radioactive phosphorus-32 (32P). This was crucial because DNA contains phosphorus in its backbone, while proteins do not.
• Radioactive Sulfur (S): To label the proteins, they used radioactive sulfur-35 (35S) since proteins contain sulfur in amino acids, while DNA does not.

Significance of Using Radioactive Phosphorus
The use of radioactive phosphorus allowed Hershey and Chase to track the DNA's incorporation into the infected bacteria, establishing that it was indeed DNA that carried the genetic instructions necessary for the production of new viruses.

• DNA vs. Protein: By using these isotopes, they could definitively show that only the labeled DNA entered the bacterial cells, while the protein coat remained outside.
• Conclusion: This led to the conclusion that DNA, not protein, is the genetic material, which was a groundbreaking discovery in molecular biology.

Conclusion
Thus, the correct answer to the question about the medium used in their experiment is indeed option 'C', a medium containing radioactive phosphorus (P), which was essential for identifying the genetic material of the virus.

Gel electrophoresis is used to separate macromolecules like DNA, RNA and proteins. DNA fragments are separated according to their size. Proteins can be separated according to their size and their charge (different proteins have different charges).

Understanding Selectable Markers in E. coli
Selectable markers are essential tools in molecular biology, particularly for the transformation and selection of genetically modified organisms. In the case of E. coli, certain markers are commonly used, while others may not be effective.
Common Selectable Markers
- Ampicillin Resistance Gene: This gene allows E. coli to survive in the presence of ampicillin, making it a widely used selectable marker for plasmids.
- Chloramphenicol Resistance Gene: Similar to the ampicillin resistance gene, this marker enables E. coli to grow in the presence of chloramphenicol, facilitating the selection of transformed cells.
- Tetracycline Resistance Gene: This gene provides E. coli with resistance to tetracycline, allowing for the selection of successfully transformed bacteria.
Why Penicillin Resistance is NOT a Useful Marker
- Drug Resistance E.coli Gene Against Penicillin: While penicillin resistance may seem like a viable option, it is not considered a useful selectable marker in E. coli for several reasons:
- Ineffectiveness in E. coli: E. coli is inherently resistant to penicillin due to its cell wall structure. Therefore, using a penicillin resistance gene does not provide a selection advantage, as E. coli can grow in penicillin-containing media without any genetic modification.
- Lack of Selection Pressure: Since E. coli can already survive in penicillin, it fails to distinguish transformed from non-transformed cells, rendering it ineffective as a selectable marker.
In conclusion, when selecting markers for E. coli, it is crucial to choose those that provide a clear selection advantage, such as ampicillin, chloramphenicol, or tetracycline resistance genes, while avoiding options like penicillin resistance that do not serve this purpose.

Incorrectly Matched Pair: Anticodons - RNA
The correct answer to the question about incorrectly matched pairs is option 'D', which states "Anticodons - RNA". Let's delve into why this is not accurate.
Understanding Anticodons
- Anticodons are indeed sequences of three nucleotides that are complementary to codons on mRNA.
- However, they are specifically found on tRNA (transfer RNA), not just any RNA.
Role of Anticodons
- Each anticodon pairs with a corresponding codon on the mRNA during the process of translation.
- This pairing ensures that the correct amino acid is brought to the growing polypeptide chain.
Other Options Explained
- Option A: Initiation codons - AUG, GUG
- Correct: AUG is the primary start codon, and GUG can also function as a start codon in some contexts.
- Option B: Stop codons - UAA, UAG, UGA
- Correct: These are the three stop codons that signal the termination of protein synthesis.
- Option C: Methionine - AUG
- Correct: Methionine is encoded by the start codon AUG, which is the first amino acid in protein synthesis.
Conclusion
In summary, while anticodons are related to RNA, they specifically refer to the tRNA molecules that carry them. Therefore, option 'D' is incorrectly matched, as it lacks the specificity of tRNA. Understanding these distinctions is crucial for grasping the nuances of genetic translation in molecular biology.

Explanation:

Recognition sequences are specific DNA sequences that are recognized and cleaved by Type II restriction enzymes. These enzymes typically recognize palindromic sequences, which read the same forward and backward on both strands of DNA. The recognition sequence is usually 4-8 base pairs in length.

Let's analyze each option to determine if it can be a recognition sequence for a Type II restriction enzyme:

a) GAATTC: This sequence is the recognition sequence for the restriction enzyme EcoRI. It is a palindromic sequence, reading the same forward and backward on both strands: 5'-GAATTC-3' on one strand and 3'-CTTAAG-5' on the complementary strand.

b) AGCT: This sequence is not palindromic and therefore cannot be a recognition sequence for a Type II restriction enzyme. It reads 5'-AGCT-3' on one strand and 3'-TCGA-5' on the complementary strand.

c) GCGGCCGC: This sequence is the recognition sequence for the restriction enzyme NotI. It is a palindromic sequence: 5'-GCGGCCGC-3' on one strand and 3'-CGCCGGCG-5' on the complementary strand.

d) ATGCCT: This sequence is not palindromic and therefore cannot be a recognition sequence for a Type II restriction enzyme. It reads 5'-ATGCCT-3' on one strand and 3'-TACGGA-5' on the complementary strand.

Therefore, option D (ATGCCT) cannot be a recognition sequence for a Type II restriction enzyme because it is not palindromic. The correct answer is D.