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

BBCCT-117: Gene Organization Replication and Repair

IGNOU Solved Assignment Solution for 2023

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 2023 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/2023

Course Code: BBCCT-117

Assignment Name: Gene Organisation, Replication and Repair

Year: 2023

Verification Status: Verified by Professor


Attempt all questions. The marks for each question are indicated against it.

Maximum marks: 50



Q1) (a) Draw the structure of nitrogenous bases.

Ans) Emil Fischer found out about nitrogenous bases in the year 1880. These are heterocyclic organic compounds that have a flat shape. Based on how they are made, nitrogenous bases are put into two groups: purines and pyrimidines. Purines are structures with two rings that are numbered in the opposite direction of a clock.


Adenine (A) and Guanine (G) are two important purines that you may have heard of before (G). In our cells, there are also several other purine derivatives that are made by nature. But Adenine and Guanine are the only ones known to be part of nucleic acids.

Q1) (b) Define primary and secondary structure of DNA.


Primary Structure of DNA

The main structure of DNA is a single strand of DNA with a sugar-phosphate backbone that is linked by strong phosphodiester bonds. This is called a polynucleotide arrangement.


Secondary Structure of DNA

Our cells have a double-stranded DNA, which is made up of two single strands that run in opposite directions. The bases on the two strands interact with each other. This is the second part of DNA's structure. With the help of hydrogen bonds, the bases can talk to each other. These hydrogen bonds are weak compared to the strong phosphodiester bonds that make up the backbone of nucleic acids. This makes it easy to break the bonds between the strands so that they can be pulled apart. A lot of biological processes, like replication, transcription, etc., depend on DNA strands being separated.


Q2) (a) Describe DNA denaturation and melting curve.

Ans) Denaturation used to always be done by heating a DNA solution, so a melting curve is a graph that shows how a property changes with temperature. Denaturation can also be studied by changing the concentration of a denaturant at a constant temperature. This is because reagents can break hydrogen bonds or weaken interactions between molecules that don't like water. The easiest way to tell if DNA has become denatured is to watch how well DNA in a solution absorbs UV light at 260 nm.


At 260 nm, the bases of nucleic acid absorb the lightest. With a value of 0.02 units per microgram of DNA/ml, absorption at 260 nm is proportional to the concentration. How much light nucleic acid can take in depends on how the molecules are put together. The less light is taken in, the more organised the structure is. Since free nucleotides absorb lighter than a single-stranded (ss) polymer of DNA or RNA, ss polymers absorb lighter than double-stranded (ds) DNA molecules.


Q2) (b) Write a short note on types of promoters in a prokaryotic gene.

Ans) A piece of DNA called the promoter is needed for transcription to start in the right way. It is upstream of the structural gene and in front of the initiation codon. It can be read by RNA polymerase. The promoters of different genes are only separated by one or two bases. Some of the promoters are:

  1. Pribnow Box or -10 Sequence: It has a 6 base pair sequence that is -10 bp upstream from the start site. This is called the Pribnow Box or the -10 sequence. The start site sequence is 5'ATAAT3, and it tells the RNA polymerase's sigma factor how to find the Pribnow box.

  2. The -35 Sequence: The -35 sequence is also made up of 6 bp. It is found -35 bp before the start site and is recognised by the RNA polymerase sigma factor. There are about 16–18 bps between the 10 and 35 sequences.

  3. UP Element: Some highly expressed genes may have a part of the gene that is full of ATs. This part is called the upstream promoter element or UP element. It is between 40 and 60 nucleotides upstream and binds to the alpha subunit of RNA polymerase.


Q3) (a) Differentiate between (i) positive and negative supercoiling, (ii) relaxed and supercoiled DNA.


  1. Positive and Negative Supercoiling: Positive or negative supercoiling is caused by twisting. Depending on how the DNA molecule has been twisted and how tightly it has wound up, it can show either positive or negative supercoiling.

  2. Relaxed and Supercoiled DNA: In general, two strands of closed circular DNA that are not twisting and are completely relaxed will cross over each other a certain number of times without using any energy. This is the number that links. In the circular planar state, the DNA that is at rest doesn't change, so it stays at rest. For many biological tasks, like replication, the DNA strands need to be split apart. As they unwind, the strands get twisted and misshapen in other places.


Q3) (b) Explain the functions of Topoisomerases.

Ans) Since the DNA is tightly bound and packed in the nucleus, it needs to be opened from time to time for different biological processes to happen. This includes things like copying, writing, recombining, etc. So, DNA topoisomerases are needed to help some parts of the DNA open so that other molecules can reach them while keeping the rest of the DNA tightly packed.


Biological Functions of Topoisomerase

  1. Making the DNA more accessible.

  2. When the DNA helix gets overwound or underwound during replication and transcription because of unwinding or coiling at the site of the process, the DNA supercoils need to be removed.

  3. During recombination, the strands must break before the chromosomes can move away from each other during cell division.

  4. Condensation of chromosomes at different stages of a cell, where the chromosomes must alternately be packed together and spread out.

  5. DNA must be untangled by the action of DNA topoisomerase.


Q4) (a) Differentiate between euchromatin and heterochromatin.

Ans) Euchromatin is less tightly wound and holds the genes that are active or could be active. In the interphase nucleus, it makes up all the genome except for the heterochromatin. It is a type of chromatin that is loosely packed, has a lot of genes, and is often in the process of being copied.

Heterochromatin is the name for the parts of the genome that stain dark and are made of a very dense, tightly packed form of DNA. They don't get transcribed because enzymes or transcription factors have a hard time getting to them. They also replicate slowly. Heterochromatin helps control how genes work and keeps chromosomes from getting damaged.


Q4) (b) Mention the characteristics of eukaryotic chromosomes.

Ans) Eukaryotic chromosomes can divide and copy themselves, and they are successfully passed on to their daughter cells to make sure that their offspring have a variety of genes and stay alive. During the interphase, when cells don't divide, the chromosomes look like thin, coiled, flexible threads. This is why they are called "chromatin threads" under low magnification. During the metaphase stage of mitosis and the prophase stage of meiosis, these chromatin threads pack together and separate into ribbon-shaped chromosomes. Along the length of these chromosomes, there is a clear area called the kinetochore or centromere.


Q4) (c) DNA replication is semi-conservative. Give experimental evidence.

Ans) The semi-conservative model shows how DNA replication happens. Matthew Meselson and Franklin Stahl showed this in 1958. They did their research with E. coli cells, which were incubated with naturally occurring nitrogen and heavy but not radioactive nitrogen isotopes (N15) to label DNA with different densities, since nitrogen is a part of bases. When E. Coli cells are incubated with 15NH4Cl for a long time, the DNA will be marked with N15 and have a high density because of this. When these E. coli cells were taken, broken up, and their DNA was separated using CsCl Density Gradient Centrifugation, a single heavy density band was seen. This showed that the DNA was labelled with N15.


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

Ans) An enzyme called DNA polymerase can make new strands of DNA. There are three different types of prokaryotic DNA polymerases. DNA polymerase I and II are mostly used to fix DNA, while DNA polymerase III oversees copying DNA.


In eukaryotes, DNA polymerases are found in more than one place. The main enzymes involved in nuclear DNA replication in eukaryotes are DNA polymerase alpha, delta, and epsilon. Polymerase beta helps fix and recombine DNA in the nucleus. Trans-lesion DNA synthesis involves the DNA polymerases zeta, eta, and kappa. DNA repair is done with the help of the DNA polymerases theta, lambda, and mu. DNA polymerase sigma is involved in nuclear DNA replication, DNA repair, and the sticking together of sister chromatids. The enzyme DNA polymerase gamma is found in mitochondria and helps copy mitochondrial DNA.


Q5) (b) Describe the enzymatic activity associated with prokaryotic DNA polymerases.

Ans) DNA polymerase-I adds nucleotides one at a time from the 5' end to the 3' end. DNAP-I has three functions: 5'3' exonuclease activity removes RNA primers at the end of each Okazaki fragment, 5'3' polymerase activity fills the space between Okazaki fragments, and 3'5' exonuclease activity works as a proof-reader. The RNA primers at the end of each Okazaki fragment are taken out by this activity. 5'3' polymerase activity takes out ribonucleotides from the primers one by one and puts in deoxyribonucleotides in their place. Nick (gap) is moved from one place to another during this process. This movement of nick is called nick translation. Once all of the ribonucleotides have been changed to deoxyribonucleotides, the nick is closed by an enzyme called DNA ligase. The activity of 5'3' exonuclease is what makes DNA polymerase-I stand out, since no other DNA polymerase has this activity.



Q1) Describe the Holliday model for homologous recombination.

Ans) Robin Holliday came up with a model for homologous recombination in 1964 to explain how fungi change their genes.

  1. Step I: When two homologous chromosomes with similar sequences are next to each other, they make a pair. They are made up of two strands of DNA.

  2. Step II: An endonuclease cuts the homologous strands of paired duplexes in the same places on both strands.

  3. Step III: The ends of the nicked strands separate from their matching strand, and each strand moves into the other duplex. This happens in a way that makes a heteroduplex region made up of one strand from each duplex parent.

  4. Step IV: The ligase seals the nick, making a stable joint molecule in which one strand from each parent duplex crosses over into the other parent duplex. The structure that looks like a "X" is called a Holliday intermediate or a Chi structure. Branch migration then makes the heteroduplex area bigger. The stable joint can move along the pair of duplexes, letting more of each invading strand in and making the heteroduplex region bigger.

  5. Step V: Resolution can happen in either of the two ways, but only one of them makes the flanking markers switch places after recombination. In horizontal mode of resolution, the nicks are made in the same DNA strands that were nicked in the parental duplexes. After joining the two ends, this makes two duplex molecules with a patch of heteroduplex but no recombination in the flanking regions.


Q2) (a) Differentiate between replicative and non-replicative transposition mechanisms.

Ans) In replicative transposition, the element is copied. One copy stays at the donor site, while the other copy moves to the target site and becomes a part of it. Retro-element transposition can make copies of itself because the RNAs from the elements are turned into cDNA so that they can be added to the target site.


Conservative transposition is also called the "cut-and-paste" mechanism because the element is taken out of one place in the genome and put into a different place in the genome. In general, the number of copies of the element does not go up when this mechanism is at work.


The only difference between these two types of transposition is when the donor cut reaction takes place. In both cases, the first step is a donor cut at the 3' end of the element, which gives the free ends of the element a chance to interact with the target site. In the conservative mechanism, the donor DNA is released by making an extra cut at the 5' end of the element. This makes a gap in the donor DNA.


Q2) (b) Write a note on uses of transposons in post genomic era.

Ans) Now that many microbial genomes have been sequenced, it is important to find out what each gene does so that the important biological information can be learned. New families of transposons have been made that can live on a wide range of hosts and have short recognition sequences. Mariner is a common animal transposon that is found in many types of insects. Mos1 and Himar1 are the mariner transposons that target AT di-nucleotide recognition sites that have been studied the most.


Structural Reorganization and Genetic Instability Induced by Retrotransposons

Because they repeat so often, retrotransposons make chromatin unstable and cause changes to the genome. Humans can get genetic diseases like haemophilia, thalassemia, and muscular dystrophy because of an insertion in a certain part of the genome and a change in how it is put together.mThese genetic rearrangements can be broadly classified as:

  1. Non-allelic homologous recombination

  2. Insertional mutagenesis

  3. 3’ and 5’ transduction

  4. Trans mediated mobilization of RNAs


Q3) (a) What is RNA editing?

Ans) According to the central dogma of molecular biology, the information in DNA is copied to RNA, and the sequence of proteins that are made is determined by the information in the RNA. In other words, RNA is used to turn the information in DNA into a protein. But it was only recently found that some RNA products that are used to make proteins do not have the same information as the DNA they come from. Biochemical pathways that can change the sequence of the final transcription product are found in mitochondria and chloroplasts. This is called "editing of RNA."


Q3) (b) Give an overview of mitochondrial inheritance.

Ans) Mitochondrial genes are inherited from the maternal parent (female) because the egg contains tens of thousands of mtDNA molecules. On the other hand, there aren't many mitochondria in sperm, or they break down in the egg. Mitochondria are passed down from mother to child, so they can be used to find a person's mother-line ancestry. Mitochondrial DNA is passed down in a very different way than nuclear DNA is. Nuclear DNA comes from a lot of different ancestors, while mitochondrial DNA comes from only one line of female ancestors.


Mutations in mitochondrial genes have been linked to several human diseases. Most of these diseases affect the parts of the electron transport chain. In Kearns-Sayre syndrome, the body doesn't get enough energy. This causes the muscles that move the eyelids and eyes to be weak, the retina to deteriorate, and heart disease to develop. Another disease is Leber Hereditary Optic Neuropathy, or LHON. People with this disease quickly lose their sight in both eyes because the optic nerve breaks down. This loss of sight happens when a person is a young adult.


Q4) (a) Illustrate reporter assay and electrophoretic mobility shift assay.

Ans) Most of the time, the reporter assay is used to study how genes are expressed at the transcriptional level. It can be used in many areas of cell and molecular biology where you want to measure or track how a cloned gene is being used. It involves co-transfection of cloned gene of interest in cell line, out of which one will express the desired protein that will bind to the regulatory sequence element of gene of interest tagged with reporter gene.


It is a way to find interactions between proteins and nucleic acids. It works on the idea that a protein-nucleic acid complex moves more slowly in an electrophoresis than a free nucleic acid. So, if the specific nucleic acid sequences for binding a certain protein are changed, the properties of the proteins that bind to the changed nucleic acid would change. So, the way bound, and free protein-nucleic acid complexes move in an electric field would be different.


Q4) (b) Describe different types of point mutations.

Ans) In a point mutation, a single base is changed by being deleted, added, or substituted with another base. This could happen owing to a defect in DNA replication or a chemical alteration of the DNA.

  1. Base substitutions, which have the subtypes of transition and transversion, entail the replacement of one base pair by another.

  2. One amino acid in the encoded protein is altered because of the missense mutation.

  3. A stop codon is created because of a nucleotide base change in a nonsense mutation. As a result, translation is prematurely stopped, leading to the production of a protein that is incomplete.

  4. The reading frame from the place of mutation to the carboxyl terminus of the protein is altered by addition or deletion mutations, which are single-base-pair additions or deletions of a single base pair or multiple base pairs of nucleotides.


Q5) (a) Explain the base excision repair mechanism with the help of a suitable diagram.

Ans) A group of enzymes called DNA glycosylases can find the damage caused by mutagenic agents. The enzyme finds a certain kind of damaged base and gets rid of it by breaking the N-glycosyl bond. This is a key part of base excision repair. For instance, deamination turns a cytosine base into uracil, which is a base that is usually only found in RNA. Because the change from cytosine to uracil hasn't been fixed, uracil pairs with adenine during DNA replication.


A mutation is caused by this process. A glycosylase from the base excision repair pathway looks for de-aminated cytosine and gets rid of it. This keeps mutations from happening. Once the base has been taken out, the "empty" piece of DNA backbone is also taken out. The gap is then filled and sealed by the enzyme DNA ligase.

Q5) (b) What is xeroderma pigmentosum? Describe the cause of the disease.

Ans) Xeroderma pigmentosum is an exceptional genetic disease of DNA repair and occurs worldwide in all races and ethnic groups. This is an autosomal recessive disease. Xeroderma pigmentosum was first described in 1874 by dermatologist Moriz Kaposi based on a series of four patients with thin, dry skin, showing wrinkling, checkered pigmentation, small dilatations of the vessels, skin contraction, and development of skin-based tumours.


Cause of Xeroderma Pigmentosum

Xeroderma pigmentosum is a rare genetic disease of DNA repair that can affect people of any race or ethnicity in any part of the world. This is a disease that is passed down from parent to child. The skin gets Xeroderma Pigmentosum when it is exposed to the ultra-violet radiation that comes from the sun. This UV light is the most likely to cause mutations. If the mutations aren't fixed, they can cause cells to grow in an abnormal way, which makes cancer more likely to happen.


UV light from the sun is what does most of the photochemical reactions. In this reaction, the pyrimidine bases of DNA absorb UV light, which makes two separate DNA photoproducts, cyclobutane pyrimidine dimers, and a lot of other photoproducts. When UV light hits dipyrimidine sites, it makes CC (cytosine-cytosine), CT (cytosine-thymine), TT (thymine-thymine), and TC (thymine-thymine) (thymine-cytosine).

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