top of page
BBCCT-117: Gene Organization Replication and Repair

BBCCT-117: Gene Organization Replication and Repair

IGNOU Solved Assignment Solution for 2022

If you are looking for BBCCT-117 IGNOU Solved Assignment solution for the subject Gene Organization Replication and Repair, you have come to the right place. BBCCT-117 solution on this page applies to 2022 session students studying in BSCBCH courses of IGNOU.

Looking to download all solved assignment PDFs for your course together?

BBCCT-117 Solved Assignment Solution by Gyaniversity

Assignment Solution

Assignment Code: BBCCT-117/TMA/2022

Course Code: BBCCT-117

Assignment Name: Gene Organisation, Replication and Repair

Year: 2022

Verification Status: Verified by Professor


Maximum marks:100

Note: Attempt all questions. The marks for each question are indicated against it. For any question worth 2 marks, the word limit is 50 words, for 5 marks question it is 100 words; and for 10 marks it is 250-300 words.


PART-(A) Maximum marks: 50


Q1 (a) Define the terms genetic material, gene and DNA, DNA melting and Tm. 5

Ans) Definitions:


Genetic material: Genetic material is the hereditary substance in the cell. It carries all information specific to an organism. It is known as DNA (deoxyribonucleic acid) or RNA (ribonucleic acid).

Gene and DNA: Gene and DNA are two terminologies used primarily in the field of genetics. In general, a gene is a short section of DNA and DNA- Deoxyribonucleic acid is a molecule which carries the genetic instructions or the hereditary materials.

DNA melting: Dissociation of the double stranded DNA helix into single coils is referred to as DNA melting.

Tm: The melting point for a double-stranded nucleic acid. Technically, this is defined as the temperature at which 50% of the strands are in double-stranded form and 50% are single-stranded, i.e. midway in the melting curve.


(b) What are the different types of bonds present in a double stranded DNA molecule? Explain. 5

Ans) The DNA double helix polymer of nucleic acid held together by nucleotides which base pair together. In B-DNA, the most common double helical structure found in nature, the double helix is right-handed with about 10–10.5 base pairs per turn. The backbone of the DNA is made up of phosphodiester bonds and the two strands are connected to each other by hydrogen bonds. There are two hydrogen bonds in A-T base pairing and three hydrogen bonds in G-C base pairing.


Q2 (a) Explain Cot curves. 2

Ans) The Cot curve is defined as the extent of DNA renaturation plotted versus log Cot. Cot data has provided significant insight into the structure and evolution of eukaryotic genomes, including estimation of genome size, estimation of the proportion of single-copy(sequences present in only one locatable region in the DNA molecule) and repetitive sequences, estimation of genome kinetic complexity, and interspecific comparison.


(b) In what way does the DNA topology influence the expression of genes? 5

Ans) Euchromatin, heterochromatin, histone and non-histone proteins, changes in accessibility, and dynamic looping domains all influence gene expression. As a macromolecule, chromatin has the unique property of creating numerous connections that are required for the smooth operation of many biological processes critical to a cell's survival and proliferation.


(c) Write a short note on DNA viruses. 3

Ans) The major genetic material of these viruses is DNA, and they replicate utilising DNA polymerases that are DNA dependent. Although most are double stranded, some are single stranded. These polymerases are frequently found within the viral genome in viruses with large genomes; however, small genome viruses must rely on host polymerases because they do not encode the polymerase. The genomes of DNA viruses can be circular (SV40) or linear (Adenoviruses).


Q3 (a) Draw and label structure and parts of chromosome. 5

Ans) Detailed graphical representation of different parts of chromosome

(b) What are chromosome bands, and how are they produced? 5

Ans) Any of the transverse bands that emerge on a chromosome after staining are referred to as chromosomal bands. Each type of chromosome has a distinct banding pattern that allows for identification. After staining with a dye, chromosomal banding is defined as alternating light and dark patches along the length of a chromosome. The section of a chromosome that is readily recognisable from its adjacent segments by appearing darker or lighter using one or more banding techniques is referred to as a band.


Q4 (a) Differentiate between conservative and semiconservative replication. 5

Ans) The differences between conservative and semiconservative replication are:

(b) Explain the formation of phosphodiester bond. 5

Ans) While DNA synthesis occurs from 5' to 3', the template strand is duplicated from 3' to 5'. Watson Crick base pairing rule between incoming nucleotide and template nucleotide selects a new nucleotide from a pool of nucleotides.

If A is present at the template nucleotide, T will be chosen as the incoming nucleotide, and vice versa; similarly, if G is present at the template nucleotide, C will be chosen as the incoming complementary nucleotide, and vice versa. The free 3' hydroxyl group of the final nucleotide at the 3' end of the expanding chain attacks the alpha phosphorus of the incoming nucleotide during the polymerization event, forming a phosphodiester link and releasing PPi. This is an endergonic, thermodynamically unfavourable process.


Phosphohydrolase (phosphatase) will hydrolyse this PPi, resulting in Pi. This mechanism releases energy, allowing the DNA synthesis reaction to be thermodynamically possible. As a result, the entering monomeric nucleotide contributes to the energy necessary for the insertion of a new nucleotide. These reactions are known as tail growth reactions, and they can be proof read, which means that if the final nucleotide contributed is incorrect, it may be changed with a correct nucleotide.


Q5 (a) Differentiate between prokaryotic and eukaryotic DNA polymerases. 5

Ans) The differences between prokaryotic and eukaryotic DNA polymerases are:



  1. Prokaryotic DNA: Prokaryotic DNA is found in the cytoplasm of prokaryotic cells as well as circular plasmids.

  2. Eukaryotic DNA: Eukaryotic DNA is found in the nucleus of the cell, inside the chloroplast and mitochondria.


Organelle DNA

Prokaryotic DNA: Prokaryotic DNA is not found inside organelles.

Eukaryotic DNA: Some of the eukaryotic DNA is found inside chloroplast and mitochondria as well.



  1. Prokaryotic DNA: The size of the DNA is less than 0.1 pg in prokaryotes.

  2. Eukaryotic DNA: The size of the DNA is high in eukaryotes, usually more than 1 pg.


GC/AT Content

  1. Prokaryotic DNA: GC content is more than the AT content.

  2. Eukaryotic DNA: AT content is more than 4xGC content.


Number of Copies

  1. Prokaryotic DNA: Prokaryotic DNA consists of one copy of the genome.

  2. Eukaryotic DNA: Eukaryotic DNA consists of more than one copies of the genome.


Number of Genes

  1. Prokaryotic DNA: Prokaryotic DNA contains a small number of genes.

  2. Eukaryotic DNA: Eukaryotic DNA contains a large number of genes.

(b) Name any two inhibitors that bind to DNA covalently. 2

Ans) The two inhibitors that bind to DNA covalently are:

Alkylating agents such as Nitrogen mustards, Alkyl sulfonates and Nitrosoureas

Mitomycin C


(c) List any three role of Topoisomerases in replication. 3

Ans) The three role of Topoisomerases in replication are:


Topoisomerase I (topoisomerase I) and recombination are important functions of type I topoisomerases (topoisomerase III).

Type II topoisomerase is involved in chromosomal condensation (topoisomerase II) and daughter chromosome segregation during cell division (topoisomerase IV).

In neurons, the human Type II protein performs a more specific role in reducing recombination and aiding transcription.



PART-(B) Maximum marks: 50


Q1 (a) Describe different types of recombination. 5

Ans) The following types of recombination processes occur in nature:

  1. Homologous (generic) recombination occurs when DNA molecules with comparable sequences come together. During meiosis, live cells undergo widespread recombination.

  2. Non-homologous (illegitimate) recombination: This type of recombination happens when DNA molecules are not homologous. There is often some resemblance between the sequences, but not to the amount that there is in homologous recombination.

  3. Site-specific recombination is a sort of recombination that occurs between specific, very short, and specific sequences that are frequently comparable.

  4. Mitotic recombination is a mechanism that occurs during interphase and is comparable to meiotic recombination. When exposed to radiation, this sort of recombination becomes more common, resulting in tumours.


(b) Differentiate between replicative and non-replicative transposition mechanisms. 5

Ans) The differences between replicative and non-replicative transposition mechanisms are:


  1. Transposons that replicate, making a new copy and leaving the old copy behind, are known as replicative transposons, whereas those that migrate from one location to another without leaving a gap are known as non-replicative transposons.

  2. The replicative transposition method is also known as copy-paste, whereas the non-replicative transposition technique is known as cut-past transposition.

  3. The donor and receptor DNA sequences produce a characteristic intermediate "theta" configuration in the replicative transposition mechanism, which is frequently referred to as a "Shapiro intermediate." Non-replicative transposition, on the other hand, begins with the action of a transposase protein. The transposon's transposase protein aids in the transposon's migration within a genome.


Q2 (a) Describe the importance of transposable elements. 3

Ans) Transposons are natural tools for genetic engineering that have evolved over time. The majority of the constructs are produced from bacteriophages or insertion sequences incorporating composite element transposase. Transposons can also be utilised as mutagens, causing deletions and inversions in DNA. For molecular genetics, gene and genome mapping, and in vivo cloning, transposable elements are a useful tool.


(b) What is extra nuclear inheritance? 3

Ans) Extracellular inheritance is a type of non-Mendelian inheritance in which a trait is passed down from one generation to the next via organellar genetic material rather than nuclear material. Extranuclear inheritance has resulted in the manifestation of characteristics in several eukaryotes. The mitochondria, for example, contain genetic material that is not included within the nucleus' chromosomes.


(c) Discuss genomic imprinting. 4

Ans) Genomic imprinting is an epigenetic process that determines whether genes are expressed or not based on whether they were acquired from the mother or father. Partially imprinted genes are also possible. Instead of total expression and repression of one parent's allele, partial imprinting occurs when alleles from both parents are expressed variably. In fungi, plants, and animals, different types of genomic imprinting have been discovered.


Q3 (a) What is RNA editing? 5

Ans) RNA editing (also known as RNA modification) is a biological mechanism by which some cells can make discrete changes to nucleotide sequences within an RNA molecule after it has been created by RNA polymerase. It is one of the most evolutionarily conserved features of RNAs, appearing in all living creatures. The insertion, deletion, and base substitution of nucleotides inside the RNA molecule are all examples of RNA editing. Splicing, 5'-capping, and 3'-polyadenylation are examples of typical kinds of RNA processing that aren't commonly considered editing. It has been associated to human disorders and can influence the activity, location, and stability of RNAs.


(b) Discuss methods of mutagenesis. 5

Ans) The methods of mutagenesis are:


Based on the structure, function, catalytic mechanism, and catalytic residues of enzymes, site-directed mutagenesis is a useful technique for modifying genes and studying the structural and functional features of a protein. It is a method for producing a specific mutation in a known sequence in vitro. When site-directed mutagenesis is performed via PCR, the primers are intended to include the desired mutation, which could be a base substitution, addition, or deletion. The mutation is absorbed into the amplicon during PCR and replaces the original sequence.

The Gene Knockout method causes the deletion of a gene, resulting in the loss of function of that gene, by manipulating the organism's DNA in a laboratory and then breeding to establish a population of homozygotes for the modified gene. It's a common method for researching the function of a gene by deleting it from a healthy animal. Most of the time, a specific gene loss in a specific tissue is of relevance.


Q4 (a) Illustrate methods of Ames test and its application. 5

Ans) The Ames test is a method of determining a chemical compound's capacity to cause DNA mutations. It was created in the early 1970s by Bruce Ames. It is based on the hypothesis that any material that is mutagenic for the bacteria used in the histidine test could also be a carcinogen or cancer-causing agent.

Figure: Ames test is used to identify mutagenic, potentially carcinogenic chemicals. A bacteria strain Salmonella histidine auxotroph is exposed to a potential mutagen/carcinogen. The number of reversion mutants capable of growing in the absence of histidine is counted and compared with the number of natural reversion mutants that arise in the absence of the potential mutagen/carcinogen.


The test animal in the Ames test is Salmonella typhimurium, which has a faulty (mutant) gene involved in histidine synthesis, preventing it from synthesising the amino acid histidine (His) in non-selective culture medium. Many non-mutagenic/non-carcinogenic compounds are occasionally converted into mutagens/carcinogens by our bodies' metabolic processes. As a result, the Ames test includes a variety of liver enzymes as well as a potential mutagen.


A wide variety of compounds used in industry and agriculture produce Ames test positive results. Ethylene dibromide (EDB), for example, is added to leaded gasoline solely to evaporate lead deposits in the engine and expel them through the exhaust. In this test, Ziram, a fungicide, and medications like isoniazid (used to prevent tuberculosis), were found to be mutagenic.


(b) How does damage of Genomic and mitochondrial DNA occur? 5

Ans) mtDNA, like other DNA, can be damaged not only by radiation and genotoxic substances, but also by reactive oxygen species (ROS), which are created often in mitochondria. Because of mistakes in DNA replication and the lack of typical histone proteins in mitochondria, mtDNA damage might get even worse. Several endogenous variables produced by metabolism, as well as external factors such as radiation and environmental stresses, have been identified as potential inducers of DNA alterations. Radiation, genotoxic compounds, and reactive oxygen species can all harm mtDNA.


Q5 (a) How do DNA lesions cause genomic instability? 5

Ans) The accumulation of DNA lesions and damage response occurs when DNA damage repair is lost. The most dangerous type of DNA lesion is the DNA double-strand break (DSB). DSB can induce genomic instability and a variety of other genetic disorders if it is not repaired. The most dangerous type of DNA lesion is the DNA double-strand break (DSB). DSB can induce genomic instability and a variety of other genetic disorders if it is not repaired. The accumulation of DNA lesions and damage response occurs when DNA damage repair is lost. Several proteins, including as cohesin and BRCA, have a role in DNA damage and repair.


(b) What is Cockayne syndrome? Describe the cause of the disease. 5

Ans) Cockayne syndrome is a rare disorder characterised by an abnormally small head size (microcephaly), failure to gain weight and grow at the expected rate (failure to thrive), and delayed development. In the United States and Europe, Cockayne syndrome is predicted to affect 2 to 3 per million new-borns.


Mutations in the ERCC6 gene (also known as CSB) or the ERCC8 gene cause Cockayne syndrome (also known as CSA). These genes provide instructions for creating proteins that are involved in repairing DNA damage caused by UV rays from the sun, toxic chemicals, radiation, and unstable molecules known as free radicals.

100% Verified solved assignments from ₹ 40  written in our own words so that you get the best marks!
Learn More

Don't have time to write your assignment neatly? Get it written by experts and get free home delivery

Learn More

Get Guidebooks and Help books to pass your exams easily. Get home delivery or download instantly!

Learn More

Download IGNOU's official study material combined into a single PDF file absolutely free!

Learn More

Download latest Assignment Question Papers for free in PDF format at the click of a button!

Learn More

Download Previous year Question Papers for reference and Exam Preparation for free!

Learn More

Download Premium PDF

Assignment Question Papers

Which Year / Session to Write?

Get Handwritten Assignments

bottom of page