Short & Long Question Answers with Solution: Molecular Basis of Inheritance | Biology for ACT PDF Download

Short Answer Type Questions

Q1: Explain about the dual polymerase present in E.coli.
Ans: The DNA polymerase found in E.coli is a type of DNA-dependent polymerase that plays crucial roles in DNA replication.
It has several functions, including:

• Facilitating the process of DNA replication.
• Engaging in polymerase activity from the 5′ to 3′ direction and also possessing 3′ to 5′ exonuclease activity.
• Serving as a proofreading mechanism by identifying incorrect nucleotides and replacing them with the correct ones.

Q2: What are the three types of RNA and what role do they play in protein synthesis?
Ans: There are three types of RNA:

• Messenger RNA (mRNA): This is a single-stranded RNA molecule responsible for carrying genetic information transcribed from DNA to the cell, where it is used in the process of protein synthesis.
• Transfer RNA (tRNA): tRNA is a molecule with an "anticodon loop" at one end, which reads the genetic code, and an amino acid acceptor end at the other, where it binds to a specific amino acid.
• Ribosomal RNA (rRNA): Ribosomes act as sites for protein synthesis and play a role in catalyzing the formation of peptide bonds during this process.

Q3: Why it is that transcription & translation could be coupled in prokaryotic cell but not in eukaryotic cell?
Ans: In prokaryotes, mRNA is synthesized without the need for additional processing, and both transcription and translation occur in the same cytosol. However, in eukaryotes, the primary transcript contains both exons and introns. This primary transcript undergoes a process called splicing, in which introns are removed, and exons are joined together in a specific order to form mature mRNA.

Q4: Sometimes, the young ones born have an extremely different set of eyes or limbs. Give a relevant explanation for the abnormality.
Ans: Various factors contribute to this abnormality, including maternal alcohol abuse during pregnancy, medication side effects or reactions affecting the womb, environmental factors like maternal exposure to chemicals and radiation, viral infections, and genetic factors, as well as miscoordination in the regulation of gene expression related to organ development.

Q5: Why can transcription and translation be coupled in prokaryotic cells but not eukaryotic cells?
Ans: In prokaryotes, mRNA is produced without the need for any processing to become active, and both transcription and translation take place in the same cytosol. On the other hand, in eukaryotes, the primary transcript includes both exons and introns and undergoes a process called splicing. During splicing, introns are excised, and exons are joined together in a precise sequence to form mature mRNA.

Q6: Mention any four important characteristics of genetic code.
Ans: The genetic codon possesses the following important characteristics:

• Each codon is composed of three bases, making it a triplet code.
• Each codon uniquely codes for a single amino acid, making it unambiguous.
• Some amino acids can be encoded by multiple codons, leading to degeneracy in the genetic code.
• Codons are read continuously in a specific direction without any punctuation.

Q7: Retroviruses do not follow central dogma.  Comment on this statement
Ans: Retroviruses deviate from the central dogma of molecular biology because they contain RNA as their genetic material, which is subsequently converted into DNA through the action of the enzyme reverse transcriptase.

Q8: Mention any four significant features of the genetic code.
Ans: A genetic codon possesses the following key traits:

• It consists of a triplet, which includes three bases.
• It uniquely represents one amino acid, ensuring no ambiguity.
• Certain amino acids are encoded by multiple codons, indicating degeneracy.
• Codons are interpreted continuously without interruptions.

Q9: Give two reasons why both the strands are not copied during transcription?
Ans: 

• If both strands code for RNA, two distinct RNA molecules and two distinct proteins are generated, which would make the genetic process more complex.
• Because the two RNA molecules produced would be complementary, they would combine to form double-stranded RNA (ds-RNA).

Q10: What do you mean by “Central Dogma of Molecular genetics?”
Ans: The central dogma of molecular genetics represents the transfer of genetic information as follows: from DNA to DNA through replication, from DNA to mRNA through transcription, and from mRNA to proteins through translation.
Replication (DNA → DNA) → Transcription (DNA → mRNA) → Translation (mRNA → proteins)

Long Answer Type Questions

Q1: Give a reason for the discontinuous synthesis of DNA on one of the parental strands?
Ans: DNA synthesis in biology naturally occurs in the 5' to 3' direction. In a double-stranded DNA molecule, the two strands run parallel and antiparallel to each other. When DNA is being synthesized, both strands serve as templates. However, only one strand (running in the 3' to 5' direction) can directly synthesize the complementary strand in the 5' to 3' direction. The other strand, which runs in the 5' to 3' direction, is synthesized in the opposite direction, resulting in the creation of small fragments of DNA called Okazaki fragments. This discontinuous synthesis on one of the parental strands is the reason behind the formation of Okazaki fragments during DNA replication.

Q2: What is transformation? Describe Grifith’s experiment to show transformation? What did he prove from his experiment?
Ans: Transformation refers to a change in the genetic makeup of an individual. Frederick Griffith conducted a series of experiments with Streptococcus pneumoniae bacteria, observing two distinct strains: one forming smooth colonies with capsules (S-type) and the other forming rough colonies without capsules (R-type).

• When live S-type cells were introduced into mice, they caused pneumonia, leading to the death of the mice. 
• When live R-type cells were introduced into mice, the disease did not manifest. 
• When heat-killed S-type cells were introduced into mice, the disease did not occur. 
• When heat-killed S-type cells were mixed with live R-cells and introduced into mice, the mice died.

Griffith concluded that the R-strain bacteria had undergone a transformation caused by heat-killed S-strain bacteria. This transformation was likely due to the transfer of genetic material between the two strains.

Q3: Two claimant fathers sued a woman who claimed to be the father of her only daughter. How could identifying the true biological father resolve this case?
Ans: The DNA fingerprinting technique could have been used to resolve this case and identify the true biological father.
Here's how the method works:

• First, the DNA of the two claimants who need to be tested is isolated.
• The isolated DNA is then digested with a restriction enzyme, and the resulting digest is separated by gel electrophoresis.
• Alkali treatment is used to denature the double-stranded DNA fragments, resulting in single-stranded DNA.
• The electrophoresed DNA is transferred onto a nitrocellulose filter paper and fixed in place.
• A known sequence of DNA, which has been labeled with the radioactive isotope 32p, is added to the nitrocellulose paper.
• Autoradiography is employed to capture an image of the nitrocellulose paper on X-ray film. The film is then examined for the presence of hybrid nucleic acid.
• The DNA fingerprints of the two claimants are compared with the DNA fingerprints of the woman and her daughter. Whoever has matching DNA fingerprints is declared to be the biological father of her daughter.

Q4: Enumerate the post-transcriptional modifications in a eukaryotic mRNA.
Ans: Transcription is the process of converting DNA into mRNA. After transcription, the mRNA undergoes several post-transcriptional modifications, which include:

• Capping at the 5'-end: A protective cap is added to the 5'-end of the mRNA molecule. This cap serves to shield the RNA from ribonucleases, enzymes that can degrade RNA.
• Poly-A tail at the 3'-end: A polyadenine (poly-A) tail is added to the 3'-end of the mRNA. This tail also plays a protective role by safeguarding the mRNA from enzymatic degradation.
• mRNA splicing: During mRNA splicing, introns (non-coding regions) are removed from the mRNA molecule, and exons (coding regions) are joined together. This process results in a continuous sequence that codes for a functional protein.

In summary, these post-transcriptional modifications help protect the mRNA molecule and ensure that it contains only the necessary coding information for protein synthesis.

Q5: What is an operon? Explain an inducible operon.
Ans: An operon is a functional unit of DNA containing a group of genes under the control of a single promoter.
It comprises the following elements:

• The DNA segment that serves as a template for mRNA synthesis.
• A regulator gene that encodes a repressor protein.
• An inducer molecule that prevents the repressor protein from binding to the operator.
• A promoter region where RNA polymerase attaches to initiate transcription.
• An operator sequence located adjacent to the promoter, which is the binding site for the repressor protein.

One well-known example of an operon is the lac operon found in E. coli, which operates as an inducible operon.

Q6: Define bacterial transformation? Who proved it experimentally & how?
Ans: Transformation is a process by which genetic information is transferred between organisms or from one organism to another. Frederick Griffith conducted experiments to demonstrate transformation using two strains of Diplococci, which showed the following steps:

• When live S-III bacterial strains were injected into mice, pneumonia developed, and the mice died.
• When R-II bacterial strains were injected into mice, they did not develop pneumonia, and the mice survived.
• When heat-killed S-III bacterial strains were injected into mice, no pneumonia symptoms developed, and the mice remained healthy.
• When a mixture of heat-killed S-III strains and live R-II strains was injected into mice, they developed pneumonia and died.
• Based on these results, Griffith concluded that the presence of heat-killed S-III bacteria must have transformed the living R-II type bacteria into S-III type, restoring their capacity for capsule formation. This phenomenon was termed "BACTERIAL TRANSFORMATION."

Summary of results:

• S strain injected into mice: Mice die.
• R strain injected into mice: Mice survive.
• S strain (heat-killed) injected into mice: Mice survive.
• Mixture of heat-killed S strain and live R strain injected into mice: Mice die.

Q7: Explain the process of DNA replication.
Ans: DNA replication is the process by which a cell makes an identical copy of its DNA molecule. It's a crucial process in cell division and the transmission of genetic information from one generation to the next.
Here's a simplified explanation of the steps involved in DNA replication:

• Initiation: DNA replication begins at specific sites called origins of replication. In the double-stranded DNA molecule, the two strands are separated, creating a "bubble" or replication fork. Enzymes called helicases unwind and separate the DNA strands.
• Primer Synthesis: To start DNA synthesis, a short RNA primer is synthesized on each strand by an enzyme called primase. This primer provides a 3'-OH group to which DNA polymerase can add nucleotides.
• DNA Polymerization: DNA polymerase, the main enzyme responsible for DNA replication, adds complementary nucleotides to each template strand. DNA polymerase can only add nucleotides in the 5' to 3' direction, so it proceeds in a continuous manner on one strand (leading strand) and in a discontinuous manner on the other strand (lagging strand).
• Leading Strand: On the leading strand, DNA polymerase continuously adds nucleotides in the 5' to 3' direction as the replication fork opens up. This is relatively straightforward and efficient.
• Lagging Strand: On the lagging strand, DNA polymerase adds short fragments of DNA called Okazaki fragments in the opposite direction of the replication fork's movement (3' to 5'). These fragments are then connected by another enzyme called DNA ligase to form a continuous strand.
• Proofreading and Repair: DNA polymerase has a proofreading function, which checks for errors in nucleotide incorporation. Incorrectly paired nucleotides are removed and replaced with the correct ones. This ensures high fidelity in DNA replication.
• Termination: DNA replication proceeds until it reaches the end of the DNA molecule or encounters obstacles. At the termination point, the newly synthesized DNA strands are complete, and the process concludes.
• Result: After DNA replication, two identical DNA molecules, each consisting of one original strand and one newly synthesized strand, are produced. These daughter DNA molecules are now ready for cell division or other cellular processes.

Overall, DNA replication is a highly accurate and complex process that ensures the faithful transmission of genetic information from one generation of cells to the next. It's a fundamental aspect of genetics and cell biology.

Q8: What do you mean by DNA replication’s semi-conservative nature? Who and how did they prove it?
Ans: The concept of semiconservative DNA replication proposes that during replication, the two DNA strands separate, and each serves as a template for creating a new complementary strand. This process ensures that after replication, each DNA molecule consists of one original (parental) strand and one newly synthesized strand, preserving half of the genetic information over a generation. To validate this idea, Matthew Meselson and Franklin Stahl conducted an experiment using Escherichia coli (E. coli).

Their experiment involved the following steps:

• Labeling DNA Strands: E. coli bacteria were cultivated in a medium containing 15NH4Cl, leading to the incorporation of the heavy nitrogen isotope 15N into the two strands of newly synthesized DNA. Consequently, both DNA strands in the E. coli cells became "heavy" due to the presence of 15N.
• Separation of Heavy DNA: Heavy DNA, marked with 15N, could be differentiated from regular DNA using centrifugation in a cesium chloride (CsCl) density gradient. This method allowed for the isolation and differentiation of DNA based on density.
• Transition to Normal Medium: After one generation of replication in the 15N medium, the cells were transferred to a medium containing the standard nitrogen isotope, 14NH4Cl (X14NH4Cl).
• Sampling Over Time: Meselson and Stahl collected samples of E. coli cells at various time intervals after transferring them to the regular 14N medium. They then extracted DNA from these samples.
• Centrifugation and Density Determination: The extracted DNA underwent centrifugation in a CsCl density gradient again. This centrifugation was used to determine the density of DNA molecules in each sample.

Results:

• Following the first replication generation in the 14N medium, the DNA from the E. coli cells exhibited an intermediate density. This indicated a mixture of 15N-labeled (heavy) and 14N-labeled (light) DNA strands.
• Subsequent replication generations in the 14N medium resulted in a gradual shift in DNA density toward the lighter 14N DNA. This confirmed that the initial heavy DNA strands were undergoing semiconservative replication, where each new DNA molecule consisted of one strand from the original (15N-labeled) DNA and one newly synthesized strand (labeled with 14N).
• This well-structured experiment provided strong evidence supporting the semiconservative nature of DNA replication, which involves the separation of DNA strands and the use of each as a template for generating complementary strands during replication.

Q9: The length of DNA in an eukaryotic cell is N 2.2 m How can such a huge DNA be packaged in a nucleus of micrometer in diameter.
Ans: In eukaryotic cells, DNA is organized by being wound around positively charged histone octamers, creating a structure known as a nucleosome. Typically, a nucleosome consists of 200 base pairs of the DNA helix. Nucleosomes repeat and assemble to form chromatin fibers. During the metaphase stage of cell division, these chromatin fibers condense further to become visible chromosomes.
To achieve higher levels of chromatin packaging, additional proteins known as non-histone chromosomal proteins are involved. Within the cell nucleus, specific regions of chromatin are loosely packed, appearing lighter when stained; these regions are referred to as euchromatin. In contrast, other regions are densely packed, staining darker, and are termed heterochromatin.

Q10: Two claimant fathers filed a case against a lady claiming to be the father of her only daughter. How could this case be settled identifying the real biological father?
Ans: To resolve the paternity dispute and determine the true biological father, DNA fingerprinting can be employed.
Here's how the technique works:

• First, DNA samples are collected from the two individuals in question, the potential fathers.
• The isolated DNA is then treated with a specific restriction enzyme and subjected to gel electrophoresis, separating the DNA fragments based on size.
• The double-stranded DNA fragments are denatured into single-stranded DNA using an alkali treatment.
• These single-stranded DNA fragments are transferred onto a nitrocellulose filter paper and fixed in place.
• A known DNA sequence called a probe-DNA is prepared and labeled with a radioactive isotope, typically 32P. This labeled probe-DNA is then applied to the nitrocellulose paper.
• Autoradiography is used to capture an image of the nitrocellulose paper on X-ray film. The film is subsequently analyzed to detect the presence of hybrid nucleic acid.
• Finally, the DNA fingerprints of the two potential fathers are compared with the DNA fingerprint of the lady and her daughter. The individual whose DNA fingerprint matches with that of the daughter is declared as her biological father.