NCERT Exemplar Class 12 Biology Solutions Chapter 6 Molecular Basis of Inheritance

Question:2

A nucleoside differs from a nucleotide. It lacks the:
a. base
b. sugar
c. phosphate group
d. hydroxyl group

Answer:

The answer is option (c) Phosphate group

Explanation: A nucleotide has three parts: a sugar, a nitrogenous base, and a phosphate group. A nucleoside has only the sugar and base, so it lacks the phosphate group.

Question:3

Both deoxyribose and ribose belong to a class of sugars called:
a. trioses
b. hexoses
c. pentoses
d. polysaccharides

Answer:

The answer is option (c) pentoses
Explanation: Ribose (in RNA) and deoxyribose (in DNA) are 5-carbon sugars, which are called pentoses. They form the sugar part of the nucleotide.

Question:4

The fact that a purine base always pairs through hydrogen bonds with a pyrimidine base in the DNA double helix leads to:
a. the antiparallel nature
b. the semiconservative nature
c. uniform width throughout DNA
d. uniform length in all DNA

Answer:

The answer is option (c) uniform width throughout DNA

Explanation: A purine always pairs with a pyrimidine (A with T, G with C), which keeps the width of the DNA double helix consistent throughout, helping it stay stable and properly structured.

Question:5

The net electric charge on DNA and histones is:
a. both positive
b. both negative
c. negative and positive, respectively
d. zero

Answer:

The answer is option (c) negative and positive respectively
Explanation: The negatively charged DNA making a structure called nucleosome wraps around the positively charged histone. This offers a reasonable explanation for the efficient packing of DNA in such a small space inside the nucleus.

Question:6

The promoter site and the terminator site for transcription are located at:
a. 3′ (downstream) end and 5′ (upstream) end, respectively of the transcription unit
b. 5′ (upstream) end and 3′ (downstream) end, respectively of the transcription unit
c. the 5′ (upstream) end
d. the 3′ (downstream) end

Answer:

The answer is option (b) 5' (upstream) end and 3' (downstream) end, respectively, of the transcription unit.

Explanation: In transcription, the promoter is found at the 5′ end of the gene and helps start the process. The terminator is at the 3′ end and signals where transcription should stop.

Question:7

Which of the following statements is the most appropriate for sickle cell anaemia?
a. It cannot be treated with iron supplements
b. It is a molecular disease
c. It confers resistance to acquiring malaria
d. All of the above

Answer:

The answer is option (d) All of the above
Explanation: Sickle cell anaemia is a genetic disorder. Iron supplements cannot be used to treat it. Altered shape of RBCs confers resistance to malaria in people suffering from sickle cell anaemia.

Question:8

Which of the following is true with respect to AUG?
a. It codes for methionine only
b. It is an initiation codon
c. It codes for methionine in both prokaryotes and eukaryotes
d. All of the above

Answer:

The answer is the option (d) All of the above

Question:9

The first genetic material could be:
a. protein
b. carbohydrates
c. DNA
d. RNA

Answer:

The answer is option (d) RNA
Explanation: Many conclusive proofs favour the fact that RNA was the first genetic material. But being a catalyst, it was reactive and unstable. This paved the way for DNA as the genetic material in the living world.

Question:10

With regard to mature mRNA in eukaryotes:
a. Exons and introns do not appear in the mature RNA
b. exons appear but introns do not appear in the mature RNA
c. introns appear but exons do not appear in the mature RNA
d. both exons and introns appear in the mature RNA

Answer:

The answer is option (b) Exons appear but introns do not appear in the mature RNA

Explanation: In eukaryotes, introns are removed during RNA processing, and only exons are joined to form the mature mRNA that is used in protein synthesis.

Question:11

The human chromosome with the highest and least number of genes in they are respectively:
a. Chromosome 21 and Y
b. Chromosome 1 and X
c. Chromosome 1 and Y
d. Chromosome X and Y

Answer:

The answer is the option (c) Chromosome 1 and Y
Explanation: Chromosome 1 has the most genes (about 2968), while the Y chromosome has the fewest (about 231).

Question:12

Who amongst the following scientists had no contribution to the development of the double-helix model for the structure of DNA?
a. Rosalind Franklin
b. Maurice Wilkins
c. Erwin Chargaff
d. Meselson and Stahl

Answer:

The answer is the option (d) Meselson and Stahl

Explanation: Meselson and Stahl worked on proving the semi-conservative nature of DNA replication, not the structure of DNA.

Question:13

DNA is a polymer of nucleotides which are linked to each other by 3'-5’ phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?
a. Replace purine with pyrimidines
b. Remove/Replace 3′ OH group in deoxyribose
c. Remove/Replace 2′ OH group with some other group in deoxy ribose
d. Both 'b' and 'c'

Answer:

The answer is option (b) Remove/replace 3' OH group in deoxyribose

Explanation: The 3′ OH group is necessary for forming phosphodiester bonds. Removing it stops the chain from elongating.

Question:14

Discontinuous synthesis of DNA occurs in one strand, because:
a. DNA molecule being synthesised is very long
b. DNA-dependent DNA polymerase catalyses polymerisation only
in one direction (5' → 3')
c. it is a more efficient process
d. DNA ligase joins the short stretches of DNA

Answer:

The answer is option (b) DNA dependent DNA polymerase catalyses polymerisation only in one direction (5' -> 3')

Explanation: Because DNA polymerase only works in one direction, the lagging strand is made in short fragments called Okazaki fragments.

Question:15

Which of the following steps in transcription is catalysed by RNA polymerase?
a. Initiation
b. Elongation
c. Termination
d. All of the above

Answer:

The answer is option (d) All of the above

Explanation: RNA polymerase performs all three steps of transcription which are starting, extending, and ending the RNA chain.

Question:16

Control of gene expression in prokaryotes takes place at the level of:
a. DNA-replication
b. Transcription
c. Translation
d. None of the above

Answer:

The answer is option (b) Transcription
Explanation: It is the first step of gene expression. A particular segment of DNA is copied into mRNA in this process. Thus, it controls gene expression.

Question:17

Which of the following statements is correct about the role of regulatory proteins in transcription in prokaryotes?
a. They only increase expression
b. They only decrease expression
c. They interact with RNA polymerase but do not affect the expression
d. They can act both as activators and as repressors

Answer:

The answer is option (d) They can act both as activators and as repressors.
Explanation: The ability of RNA to recognise initiation sites is affected by the regulatory proteins. Regulatory proteins have both negative (repressor) and positive (activator) role.

Question:18

Which was the last human chromosome to be completely sequenced:
a. Chromosome 1
b. Chromosome 11
c. Chromosome 21
d. Chromosome X

Answer:

The answer is option (a) Chromosome 1

Explanation: Chromosome 1 is the largest human chromosome and was the last to be fully sequenced due to its size and complexity.

Question:19

Which of the following are the functions of RNA?
a. It is a carrier of genetic information from DNA to ribosomes synthesising polypeptides.
b. It carries amino acids to ribosomes.
c. It is a constituent component of ribosomes.
d. All of the above.

Answer:

The answer is the option (d) All of the above

Explanation: RNA plays multiple roles mRNA carries genetic codes, tRNA brings amino acids, and rRNA forms the core of ribosomes.

Question:20

While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff's rule, it can be concluded that:
a. it is a double-stranded circular DNA
b. It is single-stranded DNA
c. It is a double-stranded linear DNA
d. No conclusion can be drawn

Answer:

The answer is option (b) It is single-stranded DNA
Explanation: DNA from any cell of all organisms should have a 1:1 ratio (base pair rule) of pyrimidine and purine bases according to Chargaff's rules. This implies that the amount of Guanine is equal to cytosine, and the amount of adenine is equal to Thymine. This pattern is common in both strands of the DNA. In this case, the percentage of adenine is not equal to that of Guanine, and the same holds true for cytosine and Thymine. Thus, it is single-stranded DNA.

Question:21

In some viruses, DNA is synthesised by using RNA as a template. Such a DNA is called:
a. A-DNA
b. B-DNA
c. cDNA
d. rDNA

Answer:

The answer is the option (c) cDNA
Explanation: Complementary DNA or cDNA is double-stranded DNA synthesised from a messenger RNA (m RNA) template in a reaction catalysed by the enzyme reverse transcriptase.

Question:22

If Meselson and Stahl's experiment is continued for four generations in bacteria, the ratio of N15/N15: N15/N14: N14/N14 containing DNA in the fourth generation would be:
a. 1:1:0
b. 1:4:0
c. 0:1:3
d. 0:1:7

Answer:

The answer is option (d), 0:1:7

Explanation: The ratio of 15_N /15_N remains 0 in the subsequent generation. The ratio of 15_N/14_N remains constant while that of 14_N /14_N increases.

Question:23

If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is:
5′ – A T G A A T G – 3′, the sequence of bases in its RNA transcript would be;
a. 5′ – A U G A A U G – 3′
b. 5′ – U A C U U A C – 3′
c. 5′ – C A U U C A U – 3′
d. 5′ – G U A A G U A – 3′

Answer:

The answer is the option (a) 5' – A U G A A U G -3'

Explanation: The RNA sequence is the same as the coding strand of DNA except thymine (T) is replaced by uracil (U).

Question:24

The RNA polymerase holoenzyme transcribes:
a. the promoter, structural gene and the terminator region
b. the promoter and the terminator region
c. the structural gene and the terminator region
d. the structural gene only

Answer:

The answer is option (c) the structural gene and the terminator regions

Explanation: RNA polymerase binds at the promoter but transcription starts after it, covering the structural gene and terminator region.

Question:25

If the base sequence of a codon in mRNA is 5′-AUG-3′, the sequence of tRNA pairing with it must be:
a. 5′ – UAC – 3′
b. 5′ – CAU – 3′
c. 5′ – AUG – 3′
d. 5′ – GUA – 3′

Answer:

The answer is option (b) 5'- CAU -3'

Explanation: The anticodon on tRNA pairs with the mRNA codon in an antiparallel manner, so AUG pairs with CAU.

Question:26

The amino acid attaches to the tRNA at its:
a. 5′ – end
b. 3′ – end
c. Anticodon site
d. DHU loop

Answer:

The answer is option (b) 3' – end

Explanation: The amino acid binds to the 3′ end of the tRNA at the CCA sequence, which helps in protein synthesis.

Question:27

To initiate translation, the mRNA first binds to:
a. The smaller ribosomal sub-unit,
b. The larger ribosomal sub-unit
c. The whole ribosome
d. No such specificity exists.

Answer:

The answer is the option (a) The smaller ribosomal subunit

Explanation: The smaller subunit of the ribosome recognizes and binds to the mRNA first during the initiation of translation.

Question:28

In E.coli, the lac operon gets switched on when:
a. lactose is present, and it binds to the repressor
b. repressor binds to operator
c. RNA polymerase binds to the operator
d. lactose is present, and it binds to RNA polymerase

Answer:

The answer is the option (a) Lactose is present, and it binds to the repressor.

Explanation: Lactose inactivates the repressor by binding to it, preventing it from blocking the operator, thus allowing transcription.

Question:2

Distinguish between heterochromatin and euchromatin. Which of the two is transcriptionally active?

Answer:

Heterochromatin
(i) They are darkly staining and are scattered or accumulated near the nuclear envelope.
(ii) These are transcriptionally less active or inactive.
Euchromatin
(I) Euchromatin is not readily stainable and is dispersed.
(II) These are transcriptionally active.

Question:3

The enzyme DNA polymerase in E. coli is a DNA dependent polymerase and also has the ability to proofread the DNA strand being synthesised. Explain. Discuss the dual polymerase.

Answer:

DNA polymerase uses a DNA template to catalyse the polymerisation of deoxynucleotides, and hence it is called DNA-dependent. When a new strand of DNA is being processed, this enzyme moves along to increase the speed of polymerisation. While doing so, it "proofreads" the strand being formed. Doing so, helps in speeding up the process. Hence, its nature can be said to be dual, i.e. of reading the template and then proofreading the new strand.

Question:4

What is the cause of discontinuous synthesis of DNA on one of the parental strands of DNA? What happens to these short stretches of synthesised DNA?

Answer:

DNA polymerase catalyses polymerisation in only one direction, i.e. 5' - 3'. Due to this, replication is continuous on one strand (3'-5’), while it is discontinuous on another strand (5' – 3') The fragment which is discontinuous is later joined by DNA ligase.

Question:5

Given below is the sequence of the coding strand of DNA in a transcription unit 3 'A A T G C A G C T A T T A G G – 5' write the sequence of a) its complementary strand b) the mRNA.

Answer:

(a) 5'-TTACGTCGATAACC-3'
(b) 5'-UUACGUCGAUAACC-3'

Question:6

What is DNA polymorphism? Why is it important to study?

Answer:

DNA polymorphism is the appearance of inheritable mutation in a population at high frequency. Since Inheritable mutations finally lead to evolution, hence study of DNA polymorphism is important from the evolutionary perspective.

Question:7

Based on your understanding of genetic code, explain the formation of any abnormal haemoglobin molecule. What are the known consequences of such a change?

Answer:

Normally human beings have following types of haemoglobin: Hb^A , Hb^A2 and Hb^F. Alteration in genes for the beta chain on haemoglobin results in the formation of HbS type of haemoglobin. This type of haemoglobin molecule is responsible for sickle cell anaemia.

Question:8

Sometimes cattle or even human beings give birth to their young ones that are having extremely different sets of organs like limbs/position of the eye(s) etc. Comment.

Answer:

The presence of any different sets of organs in an animal is due to a disturbance in coordinated regulation of expression of sets of genes.

Question:9

In a nucleus, the number of ribonucleoside triphosphates is 10 times the number of deoxy x10 ribonucleoside triphosphates, but only deoxyribonucleotides are added during the DNA replication. Suggest a mechanism.Answer:

DNA polymerase is highly specific to recognise only deoxyribonucleoside triphosphates. Therefore, it cannot hold RNA nucleotides.

Question:10

Name a few enzymes involved in DNA replication other than DNA polymerase and ligase. Name the key functions for each of them.

Answer:

Following are some other enzymes involved in DNA replication other than DNA polymerase and ligase and their key functions:

• Primase: It adds RNA primers to template strands.

• RNAse: Removes the RNA primer.

• Exonuclease: Initiates cleaving of nucleotides one at a time.

Question:11

Name any three viruses which have RNA as the genetic material.

Answer:

The following viruses have RNA as genetic material:

• Ebola virus

• Tobacco Mosaic Virus

• SARS

Question:2

Who revealed the biochemical nature of the transforming principle? How was it done?Answer:

Oswald Avery, Colin MacLeod and Maclyn McCarty (1933 - 1944) conducted experiments to reveal the biochemical nature of the transforming principle.
They purified biochemical (proteins, DNA and RNA) form the heat-killed S cells. They wanted to see which one of them was able to transform the R strain. So, they discovered that protein-digesting enzymes (proteases) and RNA-digesting enzymes (RNases) did not affect transformation, but DNase inhibits transformation. Thus, they concluded that it was the DNA which was able to transform the R strain.

Question:3

Discuss the significance of heavy isotope of nitrogen in the Meselson and Stahl's experiment.

Answer:

The heavy isotope of nitrogen was used in Meselson and Stahl's experiment for various reasons. The heavy DNA molecule (containing 15_N) could be easily distinguished from the normal DNA by centrifugation in a Cesium chloride (CsCI) density gradient. It could be easily separated from lighter nitrogen (14_N) on the basis of density. The use of lighter and heavier nitrogen made the task of identifying the transfer of DNAs through subsequent generations quite easier.

Question:4

Define a cistron. Giving examples differentiate between monocistronic and polycistronic transcription unit.

Answer:

A segment of DNA coding for a polypeptide is called cistron. A cistron is basically a gene. If a stretch of replicating DNA contains a single cistron (or gene), it is called monocistronic. e.g. eukaryotes. If a stretch of replicating DNA contains more than one cistron, it is called polycistronic, e.g. bacteria and prokaryotes.

Question:5

Give any six features of the human genome.

Answer:

The six features of the human genome are as follows:

• The human genome contains 3164.7 million nucleotides.

• The average gene in the human genome contains 3000 bases.

• The total number of genes is estimated to be 30,000.

• Almost all (about 99.9%) of nucleotides are the same in all human beings.

• Less than 2 per cent of the genome codes for the protein.

• Chromosome 1 has the most genes (2968), and chromosome Y has the least (231).

Question:6

During DNA replication, why is it that the entire molecule does not open in one go? Explain the replication fork. What are the two functions that the monomers (d NTPs) play?

Answer:

DNA replication is an energy-Intensive process which requires a very high amount of energy. So, the practical solution is to replicate a DNA segment by segment. Due to this, the entire DNA molecule does not open in one go.
Replication Fork: DNA replication happens in a small opening of DNA helix, called replication fork.
Monomers as nucleotide triphosphate (NTP) are molecules containing a nucleotide bound to three phosphates. The NTPs present in DNA are called dNTP. They are the basic building blocks of life. They play an important role in various metabolic functions.

Question:7

Retroviruses do not follow central Dogma. Comment.

Answer:

Francis and Crick proposed the central conviction in molecular biology. According to this, genetic information flows from DNA —> RNA —> Protein. In retroviruses, genetic information flows in the reverse direction, i.e. Protein to RNA to DNA. Hence, it is said that retroviruses do not follow central coniction.
The process followed by retroviruses is also called reverse transcription because of the opposite sequence of the process involved.

Question:8

In an experiment, DNA is treated with a compound which tends to place itself amongst the stacks of nitrogenous base pairs. As a result of this, the distance between two consecutive base increases from 0.34nm to 0.44 nm calculate the length of DNA double helix (which has 2×109 bp) in the presence of saturating amount of this compound.

Answer:

The length of the DNA double helix can be calculated by multiplying the distance between two consecutive base pairs with the total number of base pairs.
0.44 * 10^-9m * 2 * 10⁹ bp
= 0.88 m

Question:9

What would happen if histones were to be mutated and made rich in acidic amino acids such as aspartic acid and glutamic acid in place of basic amino acids such as lysine and arginine?

Answer:

Aspartic and glutamic acid are acidic amino acids, while lysine and arginine are basic amino acids. Lysine and arginine carry positive charge on their side chains which is not the case with aspartic acid and glutamic acid. DNA is negatively charged and hence is wrapped around the positively charged histone octamer. DNA will not be able to wrap around itself if acidic amino acids are present in histone because of mutation.
Thus, long strands of DNA will not be able to fit inside the small space in the nucleus. This will mean an end to the nuclear organisation, which is possible because of efficient packaging.

Question:10

Recall the experiments done by Frederick Griffith, Avery, MacLeod and McCarty, where DNA was speculated to be the genetic material. If RNA, instead of DNA was the genetic material, would the heat-killed strain of Pneumococcus have transformed the R-strain into virulent strain? Explain.

Answer:

RNA is less stable than DNA, and hence, DNA replaced RNA as the genetic material in the living world. If RNA was the genetic material in Griffith's experiment, it would have been destroyed by heat. Thus, the heat-killed strain of Pneumococcus could not have transformed the R-strain into a virulent strain.

Question:11

You are repeating the Hershey-Chase experiment and are provided with two isotopes: 32P and 15N (in place of 35S in the original experiment). How do you expect your results to be different?

Answer:

The facts that DNA contains phosphorus while protein contains sulphur supported the selection of phosphorus and sulphur. In this experiment, phosphorus was used as a marker for DNA and Sulphur was used as a marker for protein. By tracing the movement of sulphur and phosphorus, it was easier to trace the movement of DNA and protein through subsequent generations. But nitrogen is present in DNA as well as in protein. Hence, the use of 15_N will not help in finding whether the DNA or protein is the genetic material.

Question:12

There is only one possible sequence of amino acids when deduced from given nucleotides. But multiple nucleotides sequence can be deduced from a single amino acid sequence. Explain this phenomenon.

Answer:

There are 61 codons and 20 amino acids. Hargobind Khorana and Marshall Nierenberg worked on this principle. It was proposed that a codon for an amino acid is made up of 3 nucleotides. It was also seen that one codon codes for only one amino acid (unambiguous and specific).
Some amino acids are coded by more than one codon (degeneracy of codon). In simple terms, it can be said that there is only one possible sequence of amino acids when deduced from a given set of nucleotides. But multiple nucleotide sequences can be deduced from a single amino acid sequence.

Question:13

A single base mutation in a gene may not 'always' result in loss or gain of function. Do you think the statement is correct? Defend your answer.

Answer:

A single-base mutation in a gene may not always result in loss or gain of function. We know that a codon Is composed of three nucleotides. In simplified terms, a codon can be taken as a word, which is composed of 3 letters. For making any meaningful sentence, we need a complete word. Addition or deletion of a single letter may not result In a meaningful word. Hence, in most of the cases, there is a need for a mutation in three bases to affect loss or gain of function. This can be illustrated by the following example of a sentence being altered:
RAM HAS RED CAP
RAM HAS BRE DCA P
RAM HAS BIR EDC AP
RAM HAS BIG RED CAP
It is quite clear that a meaningful sentence is made only when at least three letters are inserted in this sequence.

Question:14

A low level of expression of Lac Operon occurs at all the time. Can you explain the logic behind this phenomenon?

Answer:

A very low level of expression of Lac Operon has to be present in the cell all the time; otherwise, lactose cannot enter the cells.

Question:15

How has the sequencing of the human genome opened new windows for the treatment of various genetic disorders? Discuss amongst your classmates.

Answer:

The sequencing of the human genome has opened new windows for the treatment of various genetic disorders. It is widely known that genetic disorders are caused by some alteration in genes. At present, we do not have exact information about the base pair sequence where this alteration takes place.
Hence, we are unable to devise any tool to prevent genetic disorders. By proper understanding of the particular sequence responsible for a particular genetic disorder, the scientist may be able to come up with some tools to prevent genetic disorders. A future may come when nobody is suffering from genetic disorders; especially those who create serious disability.

Question:16

The total number of genes in humans is far less (< 25,000) than the previous estimate (up to 1,40,000 gene). Comment.

Answer:

When scientists began estimating the number of human genes, they began with a very high figure, i.e. more than 100,000. At that time, the technology for studying human genes was not sophisticated or advanced enough, and the estimate was qualitative in nature as it was mainly based on assumptions. With the gradual progress of technology and knowledge about the human genes, the estimated number began to come down. The present knowledge tells us that the total number of genes in humans is between 20,000 to 25,000.

Question:17

Now, the sequencing of total genomes is getting less expensive day by day. Soon it may be affordable for a common man to get his genome sequenced. What is your opinion could be the advantage and disadvantage of this development?

Answer:

Affordable Genome Sequencing can help in settling disputes which may arise in case of parentage of a child. This can also help in disputes of property inheritance by finding the authentic beneficiary. The human genome can also help in preparing a database on people with a criminal record. It can help in identifying the chances of genetic disorders in a family.
The main disadvantage is that Genome sequencing can have serious issues of privacy. Some employers may misuse the data to blackmail their employees. Many private matters may leak into the public domain; creating embarrassment for the affected person.

Question:18

Would it be appropriate to use DNA probes such as VNTR in DNA fingerprinting of a bacteriophage?

Answer:

VNTR (Variable Number Tandem Repeat) is a location in the genome where a short nucleoside is organised as a simultaneous repeat. Analysis of VNTR is used for many purposes, including DNA fingerprinting. But bacteriophage does not have too many DNAs rather only a few strands of DNA are available in bacteriophage. This does not leave room for repeating sequences in DNA. Hence, VNTR cannot be used in DNA fingerprinting of a bacteriophage.

Question:19

During in vitro synthesis of DNA, a researcher used 2', 3' – dideoxy cytidine triphosphate as raw nucleotide in place of 2'-deoxycytidine. What would be the consequence?

Answer:

2', 3' - dideoxy cytidine triphosphate is a reverse transcriptase inhibitor. Reverse transcriptase Is a viral DNA polymerase which facilitates DNA replication In HIV and other retroviruses. The commercial name of ddC is Zalcitabine, and it is sold as a pharmaceutical product for the management of HIV. If 2', 3'- dideoxy cytidine triphosphate is used as a raw nucleotide in place of 2' - deoxycytidine, it will stop DNA replication. The researcher will not be able to proceed on his or her experiment because of the contrary effect of his or her chosen reagent.

Question:20

What background information did Watson and Crick have made available for developing a model of DNA? What was their contribution?

Answer:

Watson and Crick made the following background Information available for developing a model of DNA:
• Pairing between the two strands of polynucleotide chains.
• Base pairing of polynucleotide chains is complementary in nature.
• If the base sequence of one strand is known, then the base sequence of another strand can be predicted.
• If each strand from a DNA acts like a template, then both the daughter DNAs would be similar to the mother DNA.
Contributions of Watson and Crick:
• A simple model of DNA was available because of them.
• Genetic implications of DNA replication could be easily understood.
• The model brought a revolution in the understanding of biology at a molecular level.

Question:21

What are the functions of

(i) methylated guanosine cap,

(ii) poly-A "tail" in a mature on RNA?

Answer:

The function of Methylated Guanosine Cap: It regulates nuclear export of mRNA. It promotes translation. (Fully processed hnRNA Is called mRNA).
The function of Poly-A Tail: Protects RNA from degradation by exonucleases. Plays an important role in transcription termination.

Question:22

Do you think that the alternative splicing of exons may enable a structural gene to code for several isoproteins from the same gene? If yes, how? If not, why so?

Answer:

In humans, about 95% of multi-exonic genes are alternatively spliced. Alternative splicing helps in generating many proteins from one and the same gene. In this process, a particular exon may be excluded from or included in a specific RNA. Splicing, which results in a single gene coding for multiple proteins, is called alternative splicing.

Question:23

Comment on the utility of variability in the number of tandem repeats during DNA fingerprinting.

Answer:

Variability in the number of tandem repeats (VNTR) is highly useful in DNA fingerprinting. The DNA sample is subjected to gel electrophoresis or Southern blotting. After that, VNTR exhibits as a pattern of lines of different lengths. The variability in lengths of lines and their respective arrangement varies from one individual to another. This is more or less unique the way a person's fingerprint is. Thus. VNTR helps in establishing the exact identity of an individual through DNA fingerprinting.

Question:2

During evolution, why DNA was chosen over RNA as genetic material? Give reasons by first discussing the desired criteria in a molecule that can act as genetic material and in the light of biochemical differences between DNA and RNA.

Answer:

The desired criteria in a molecule that can act as genetic material are as follows:

• It should have the capability of replication.

• It should be chemically and structurally stable.

• It should be able to incorporate slow changes (mutation), which are required for evolution.

• It should be able to express itself in the form of Mendelian characters.

Biochemistry of DNA and RNA:

• Both DNA and RNA show complementarity of base pairs and hence are capable of replication.

• As shown by Griffith's experiment, DNA is more stable than RNA because it could survive even heat-killing during the experiment.

• 2'-011 group is present in RNA. This makes RNA labile and degradable, which is not the case with DNA.

• Both RNA and DNA can carry on mutations. But DNA being more stable is better suited for long term storage of mutations.

Hence, DNA was preferred as the genetic material during the course of evolution.

Question:3
Give an account of post-transcriptional modifications of a eukaryotic mRNA.

Answer:

The following events happen during the transformation of hnRNA (precursor of mature mRNA):

• Polymerase II facilitates transcription of hnRNA into mature mRNA,

• Primary transcripts contain both the introns and exons and these are non-functional. Splicing takes place which results in the removal of introns and joining of exons in define order.

• Capping and tailing happen in hnRNA. It acquires a cap of methyl guanosine and a tail of poly adenylate. Cap Is added at 5' end, and Poly-A tall is added at 3' end of hnRNA.

• Now, the hnRNA changes into mature mRNA.
  

Question:4

Discuss the process of translation in detail.

Answer:

The translation is the process of polymerisation of amino acid to form a polypeptide. Thus, the biological process through which protein is synthesised is called translation. Translation happens in the following main steps:
Initiation: The Ribosome assembles around the target mRNA, and we know that the ribosome is the site of protein synthesis. The first tRNA gets attached at the start codon. A codon is a triplet of amino acids.
Elongation: The tRNA transfers an amino add to the tRNA corresponding to the next codon. This phase involves the addition of subsequent amino acids to form a long chain. This step forms the bulk of protein synthesis.
Translocation: The ribosome then moves to the next mRNA codon and continues the process. This creates an amino acid chain.
Termination: When a stop codon is reached, the ribosome releases the polypeptide.

Question:5

Define an operon. Giving an example, explain an Inducible operon.

Answer:

A functioning unit of genomic DNA containing a cluster of genes under the control of a single promoter is called an operon. An operon Is generally transcribed into polycistronic mRNA. A polycistronic mRNA is a single mRNA which codes for more than one protein. An operon Is made up of 3 basic DNA components:
(a) Promoter: A nucleotide sequence that enables a gene to be transcribed is called a promoter. It is recognised by RNA polymerase, which then initiates transcription.
(b) Operator: A segment of DNA to which a repressor binds is called an operator.
(c) Structural genes: The genes that are co-regulated by the operon are called structural genes.
Inducible Operon: When the operon is regulated by an Inducer. It is called Inducible operon. An inducer can switch on or off the operon. Lac Operon is an example of the inducible operon. Lactose Is a substrate of enzyme beta-galactosidase and is the Inducer of the lac operon.
In the presence of an inducer, the repressor becomes inactive. This allows transcription in the operator region, which results in the release of mRNA. Subsequently, mRNA promotes translation, and protein synthesis Is accomplished.

Question:6

'There is a paternity dispute for a child'. Which technique can solve the problem? Discuss the principle involved.

Answer:

DNA fingerprinting can be used to resolve a dispute regarding paternity for a child. DNA fingerprinting is based on the following principle:
DNA fingerprinting: This involves identifying the difference in some specific regions of DNA. The sequence in such regions is called repetitive DNA. A small stretch of DNA is repeated much time in such sequences.
During density gradient centrifugation, these sequences are separated from bulk DNA as different peaks. The bulk DNA forms major peaks, and other small peaks are called satellite DNA. Satellite DNA can be classified into various types, depending on the base composition, length of segment and number of repetitive units. The base composition reveals whether the sequence is A: T rich or G: C rich. These sequences show a high degree of polymorphism and hence form the basis of DNA fingerprinting.
DNA from every tissue shows the same degree of polymorphism in the case of an individual. Hence, DNA from any tissue can be utilised to analyse the DNA fingerprinting of an individual. Moreover, polymorphism is inherited from parents to children. Hence. DNA fingerprinting can be utilised to determine the paternity of a child.

Question:7

Give an account of the methods used in sequencing the human genome.

Answer:

Two approaches were involved in sequencing the human genome.
Using Expressed Sequence Tags (ESTs): In this approach, all the genes that are expressed as RNA are identified and then sequenced.
Blind Approach: This approach involved sequencing the whole set of the genome and then assigning different regions in the sequence with functions. This is referred to as sequence annotation. This approach is comprised of the following steps:

• Total DNA from a cell is isolated and converted into random fragments of smaller sizes.

• These fragments are cloned in a suitable host by using specialised vectors. The cloning results in the amplification of each fragment, thus making it easier to sequence the fragment. Bacteria and yeast are the commonly used hosts for this purpose. The vectors were called BAC (bacterial artificial chromosomes) and YAC (yeast artificial chromosomes).

• Automated DNA sequencers were used to sequence the fragments. Then these sequences were arranged on the basis of some overlapping regions present in them.

• For generating overlapping fragments in these sequences, help of computer programmes was taken because it was not possible for humans to do so.

• Then, the sequences were annotated and assigned to each chromosome.

• Genetic physical mapping of the genome was done based on polymorphism in some segments of the DNA.
  

Question:8

List the various markers that are used in DNA fingerprinting.

Answer:

A DNA marker Is a gene sequence on a known chromosome which can be used to identify an individual or a species. A genetic marker or DNA marker can be a short sequence or a long sequence. Following are the commonly used markers for DNA fingerprinting.

• RFLP (Restriction fragment length polymorphism)

• SSLP (Simple sequence length polymorphism)

• AFLP (Amplified fragment length polymorphism)

• RAPD (Random amplification of polymorphic DNA)

• VNTR (Variable number tandem repeat)

• SSR Microsatellite polymorphism. (Simple sequence repeat)

• SNP (Single nucleotide polymorphism)

• STR (Short tandem repeat)

• SFP (Single feature polymorphism)

• DArT (Diversity Arrays Technology)

• RAD markers (Restriction site associated DNA markers)

Question:9

Replication was allowed to take place in the presence of radioactive deoxynucleotides precursors in E. coli that was a mutant for DNA ligase. Newly synthesised radioactive DNA was purified, and strands were separated by denaturation. These were centrifuged using density gradient centrifugation. Which of the following would be a correct result?

Answer:

Let us assume that heavier nitrogen was used in this experiment. This nitrogen molecule from parent's cell would be transmitted equally in the daughter cells. Each daughter cell will have half of the DNA with heavier nitrogen and another half with lighter nitrogen.
In the F₂ generation, 50% of daughter cells will have a combination of radioactive and non-radioactive DNAs. The rest 50% of daughter cells will have non-radioactive DNA.
This is the reason why the graph shows two peaks with each peak representing a particular form of nitrogen in DNA.