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Molecular Basis of Inheritance EXERCISE QUESTIONS
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Q. 1 Multiple-Choice Questions
1. Griffith worked on .............
a. Bacteriophage
b. Drosophila
c. Frog eggs
d. Streptococci
ANS: d. Streptococci
Frederick Griffith was a British bacteriologist who worked on the bacteria Streptococcus pneumoniae, which causes pneumonia. In his famous experiment, he discovered the phenomenon of bacterial transformation, where genetic material could be transferred from one strain of bacteria to another.
2. The molecular knives of DNA are …………..
a. Ligases
b. Polymerases
c. Endonucleases
d. Transcriptase
ANS: c. Endonucleases
Endonucleases are enzymes that cleave the phosphodiester bonds within a DNA molecule. They are often referred to as molecular knives because they can cut DNA at specific locations, allowing researchers to manipulate DNA sequences.
3. Translation occurs in the ...............
a. Nucleus
b. Cytoplasm
c. Nucleolus
d. Lysosomes
ANS: b. Cytoplasm
Translation is the process by which the information in messenger RNA (mRNA) is used to synthesize proteins. It occurs in the cytoplasm of the cell, specifically on the ribosomes.
4. The enzyme required for transcription is ..................
a. DNA polymerase
b. RNA polymerase
c. Restriction enzyme
d. RNAase
ANS: b. RNA polymerase
Transcription is the process of synthesizing RNA from a DNA template. RNA polymerase is the enzyme that catalyzes this reaction.
5. Transcription is the transfer of genetic information from ..............
a. DNA to RNA
b. tRNA to mRNA
c. DNA to mRNA
d. mRNA to tRNA
ANS: c. DNA to mRNA
Transcription is the process by which genetic information stored in DNA is transferred to RNA. Specifically, DNA is used as a template to synthesize mRNA
6. Which of the following is NOT part of protein synthesis?
a. Replication
b. Translation
c. Transcription
d. All of these
ANS; a. Replication
Protein synthesis is the process by which cells build proteins. It consists of two main stages: transcription and translation. Replication, on the other hand, is the process by which cells make an exact copy of their DNA prior to cell division.
7. In the RNA molecule, which nitrogen base is found in place of thymine?
a. Guanine
b. Cytosine
c. Thymine
d. Uracil
ANS: d. Uracil
RNA contains four nitrogenous bases: adenine (A), guanine (G), cytosine (C), and uracil (U). Uracil is used in RNA in place of thymine, which is used in DNA.
8. How many codons are needed to specify three amino acid?
a. 3
b. 6
c. 9
d. 12
ANS; c. 9
A codon is a sequence of three nucleotides in mRNA that specifies a particular amino acid. Therefore, three codons are needed to specify three amino acids.
9. Which out of the following is NOT an example of inducible operon?
a. Lactose operon
b. Histidine operon
c. Arabinose operon
d. Tryptophan operon
ANS: (b) Histidine operon.
The lactose operon, arabinose operon, and tryptophan operon are all examples of inducible operons, which means that the transcription of the operon is induced or increased in the presence of a specific molecule. The lactose operon is induced by lactose, the arabinose operon is induced by arabinose, and the tryptophan operon is repressed by tryptophan. However, the histidine operon is a repressible operon, which means that the transcription of the operon is normally on and is repressed in the presence of a specific molecule (in this case, histidine).
10. Place the following event of translation in the correct sequence
i. Binding of met-tRNA to the start codon.
ii. Covalent bonding between two amino acids.
iii. Binding of second tRNA.
iv. Joining of small and large ribosome subunits.
A. iii, iv, i, ii
B. i, iv, iii, ii
C. iv, iii, ii, i
D. ii, iii, iv, i
ANS: (b) i, iv, iii, ii.
The correct sequence of events in translation is:
i. Binding of met-tRNA to the start codon.
iv. Joining of small and large ribosome subunits.
iii. Binding of second tRNA.
ii. Covalent bonding between two amino acids.
So, the correct order is i, iv, iii, ii, which is option (b).
Q. 2 Very Short Answer Questions:
1. What is the function of an RNA primer during protein synthesis?
ANS; An RNA primer serves as a starting point for DNA synthesis during protein synthesis. It provides a free 3' OH group to which nucleotides can be added by DNA polymerase. The RNA primer is later removed and replaced with DNA by DNA polymerase.
2. Why the genetic code is considered as commaless?
ANS: The genetic code is considered commaless because there are no commas or gaps between codons. Each codon consists of three nucleotides, and these nucleotides are read continuously without any breaks. This allows the code to be read without ambiguity, and ensures that each codon is read as a complete unit.
3. What is genome?
ANS: A genome is the complete set of genetic material that an organism possesses. It includes all of an organism's genes, as well as non-coding DNA, regulatory sequences, and other elements that contribute to the organism's genetic makeup.
4. Which enzyme does remove supercoils from replicating DNA?
ANS: The enzyme that removes supercoils from replicating DNA is called topoisomerase. Topoisomerase works by breaking one or both strands of DNA, allowing the DNA to relax and remove the supercoils. Once the supercoils have been removed, the DNA strands are re-ligated and the enzyme is released.
5. Why are Okazaki fragments formed on lagging strand only?
ANS: Okazaki fragments are formed on the lagging strand of DNA because DNA polymerase can only add nucleotides in the 5' to 3' direction, and the lagging strand is oriented in the opposite direction to the replication fork. As a result, the lagging strand is synthesized in short fragments, which are later joined together by ligase to form a continuous strand.
6. When does DNA replication take place?
ANS; Okazaki fragments are formed on the lagging strand of DNA because DNA polymerase can only add nucleotides in the 5' to 3' direction, and the lagging strand is oriented in the opposite direction to the replication fork. As a result, the lagging strand is synthesized in short fragments, which are later joined together by ligase to form a continuous strand.
7. Define term- codon and codogen.
ANS: A codon is a sequence of three nucleotides that encodes a specific amino acid or a stop signal during protein synthesis. Codogen is a nucleotide sequence that can potentially code for an amino acid but is not actually used as a codon in the genetic code.
8. What is the degeneracy of genetic code?
ANS; The degeneracy of the genetic code refers to the fact that more than one codon can code for the same amino acid. For example, there are four different codons that code for the amino acid leucine. This redundancy in the code allows for some degree of error tolerance and helps to protect against mutations.
9. Which are the nucleosomal 'core' histones?
ANS: The degeneracy of the genetic code refers to the fact that more than one codon can code for the same amino acid. For example, there are four different codons that code for the amino acid leucine. This redundancy in the code allows for some degree of error tolerance and helps to protect against mutations.
Q. 3 Short Answer Questions:
1. Write a short note on DNA packaging in eukaryotic cell.
ANS: DNA packaging in eukaryotic cells is the process of compacting long strands of DNA into a smaller space to fit inside the nucleus. DNA is wrapped around histone proteins to form nucleosomes, which then further compact to form chromatin fibers. These fibers are then organized into loops and domains that form chromosomes. DNA packaging is essential for gene regulation and expression, as it allows specific regions of DNA to be accessible or inaccessible to transcription factors and other regulatory proteins.
2. Enlist the characteristics of genetic code.
ANS: The characteristics of the genetic code are:
- It is universal: The same codons code for the same amino acids in all organisms.
- It is degenerate: More than one codon can code for the same amino acid.
- It is non-overlapping: Each nucleotide is only part of one codon.
- It is commaless: There is no punctuation or space between codons.
- It is triplet: Each codon consists of three nucleotides.
- It is continuous: The codons are read in a continuous sequence along the mRNA molecule.
3. Write a note on applications of DNA fingerprinting.
ANS: DNA fingerprinting is a technique used to identify individuals based on unique features in their DNA. It has a variety of applications, including:
- Forensic science: DNA fingerprinting is commonly used in criminal investigations to match suspects to crime scenes or to identify victims.
- Paternity testing: DNA fingerprinting can determine whether two individuals are biologically related, such as in paternity testing or identifying missing persons.
- Medical diagnosis: DNA fingerprinting can help diagnose genetic disorders, such as cystic fibrosis or sickle cell anemia.
- Evolutionary studies: DNA fingerprinting can be used to study the evolutionary relationships between species or populations.
4. Explain the role of lactose in ‘Lac Operon’.
ANS: In Lac Operon, lactose plays a role in regulating the expression of the genes involved in lactose metabolism. When lactose is present, it binds to the repressor protein, causing a conformational change that makes the repressor unable to bind to the operator region of the operon. This allows RNA polymerase to transcribe the genes involved in lactose metabolism, leading to the production of enzymes that can break down lactose. In the absence of lactose, the repressor binds to the operator region, preventing RNA polymerase from transcribing the genes. This system allows the bacteria to use lactose only when it is present in the environment, conserving energy when lactose is absent.
Q. 4 Short Answer Questions:
1. Write a note on the Human genome project (HGP)
ANS:
- The HGP was a research project that aimed to identify and map all the genes in the human genome.
- The project was launched in 1990 and completed in 2003.
- The project provided a comprehensive understanding of the structure and function of human genes.
- The HGP has had a significant impact on the field of genetics and helped scientists understand the genetic basis of many diseases and conditions.
2. Describe the structure of ‘Operon’
ANS:
- An operon is a group of genes located in close proximity to each other on a chromosome.
- The operon typically includes a promoter, an operator, and one or more structural genes.
- The promoter is the site where RNA polymerase binds to initiate transcription.
- The operator is a regulatory sequence that controls the expression of the operon.
- The structural genes code for proteins that are involved in a common biochemical pathway or function.
- The structure of the operon allows for coordinated gene expression and efficient regulation of gene activity in response to environmental signals.
3. In the figure below A, B, and C are three types of ___________________
ANS:4. Identify the labeled structures on the following diagram of translation.
Part A is the ________________________.
Part B is the ________________________.
Part C is the ________________________.
ANS: A- Anticodon present on anticodon loop of tRNA
B- amino acid
C- large subunit of amino acid
5. Match the entries in column I with those of column II and choose the correct answer
ANS: A. Alkali treatment - ii. Split DNA fragments into single strands
Q. 5 Long Answer Questions:
- Initiation: The process of DNA replication begins when a protein called helicase unwinds the double helix of the DNA molecule, breaking the hydrogen bonds that hold the two strands together. This creates a replication fork, which is the point where the DNA strands separate and DNA synthesis begins.
- Elongation: Once the DNA strands have separated, a protein called DNA polymerase binds to each strand and begins synthesizing a new complementary strand by adding nucleotides one by one. The new strand is synthesized in the 5’ to 3’ direction, while the original strand is read in the 3’ to 5’ direction. This means that the two strands are synthesized in opposite directions, creating a leading and a lagging strand.
- Termination: The process of DNA replication ends when the DNA polymerase reaches the end of the template strand or encounters a termination sequence. At this point, the newly synthesized strand is checked for errors and the DNA strands are rewound back into a double helix structure.
- Initiation: The process of transcription begins when an enzyme called RNA polymerase binds to a specific region of DNA called the promoter. The promoter sequence signals the beginning of the gene to be transcribed.
- Elongation: Once RNA polymerase is bound to the promoter, it begins to move along the DNA strand in the 3’ to 5’ direction, reading the DNA template strand and synthesizing a complementary RNA strand in the 5’ to 3’ direction. The RNA polymerase reads the template strand and pairs complementary RNA nucleotides to the template strand.
- Termination: The process of transcription ends when the RNA polymerase reaches the end of the gene being transcribed. At this point, the RNA polymerase and the newly synthesized RNA molecule are released from the DNA template.
- Initiation: The process of translation begins when the small ribosomal subunit binds to the mRNA molecule at the start codon (AUG). Then, the large ribosomal subunit binds to the complex.
- Elongation: Once the ribosome is assembled, it begins to move along the mRNA molecule, reading the sequence of codons and matching each codon with the appropriate amino acid. tRNA (transfer RNA) carries the amino acid to the ribosome, and the ribosome catalyzes the formation of a peptide bond between adjacent amino acids. This process continues until the ribosome reaches a stop codon (UAA, UAG, or UGA).
- Termination: The process of translation ends when the ribosome reaches the stop codon, which signals the end of the protein-coding sequence. At this point, the ribosome releases the newly synthesized protein and dissociates into its component subunits.
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