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Molecular Basis of Inheritance

 Questions - Answers

Very short answer type questions     

1. What is the function of histones in DNA packaging?

A: The positively charged histones form an octomer and are wrapped by negatively charged DNA thus forming the nucleosomes thus making the DNA neutral and stable. The nucleosomes appear on the chromatin like a string of beads.

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

A:

  
 

3. Who proved that DNA is genetic material? What is the organism they worked on?

A: Alfred Hershey and Martha Chase proved that DNA is the genetic material. They worked on the virus Bacteriophage.
 

4. What is the function of DNA polymerase?

A: DNA polymerase is an enzyme found in the nucleus which catalyses polymerisation of a large number of deoxyribonucleosides in a short time and with a high degree of accuracy.
 

5. What are the components of a nucleotide?

A: A nucleotide is composed of

A sugar (Ribose or Deoxyribose)

A Phosphate and 

A nitrogen base (which may be a Purine-Adenine and Guanine or a Pyrimidine- Thymine, Cytosine and Uracil)
 

6. Write the structure of chromatin.

A: Chromatin shows a “beads on string structure” and is composed of repeating units called nucleosomes and DNA. Each nucleosome is composed of an octomer of positively charged histones wrapped by negatively charged DNA. A typical nucleosome contains 200 bp of DNA helix.

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

A: As one of the parental strand is a lagging strand the synthesis of ENA is discontinuous. The short stretches of DNA are later joined by the enzyme ligase.
 

8. Given below is the sequence of coding strand of DNA in a transcription unit. 3' AAT GCA GCT ATT AGG 5'

Write the sequence of

     a. Its complementary strand 

     b. mRNA

A:    a) The complementary strand would be

             5' TTA CGT CGA TAA TCC 3'

        b) The mRNA would be

             5' UUA CGU CGA UAA UCC 3'

9. In a nucleus the number of ribonucleoside triphosphates is 10 times the number of deoxyribonucleoside triphosphates, but only deoxyribonucleotides are added during the DNA replication. Suggest a mechanism.

A: DNA replication occurs in the synthetic phase of inter phase before cell division during which the enzyme DNA polymerase is active. Hence the DNA polymerase catalyses polymerisation of deoxyribonucleotides during DNA replication.

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

A: Plant viruses – Tobacco Mosaic Virus

HIV – Human Immuno Deficient Virus and Wound tumour virus, Reo virus, Blue tongue virus (any three of the above examples)

11. Write the sequence of nucleotides after single base insertion and deletion in the given gene.

A: Gene. THE CAT ATE THE FAT RAT

a.     Insertion of a single base ‘B’ between THE and CAT.

       THE BCA TAT ETH EFA TRA T

b.     Deletion of a single base ‘C’ from CAT.

        THE ATA TET HEF ATR AT

12. Why was DNA chosen over RNA as genetic material in the majority of organisms?

A: DNA is structurally more stable and chemically less reactive and less mutagenic than RNA. Hence it is chosen to be the genetic material in majority of organisms.
 

13. What are the components of a transcription unit?

A: The components of a transcription unit are

Promoter

Structural gene

Terminator
 

14. What is the difference between exons and introns?

A: The coding or expressed sequences of DNA are called exons. They appear in processed RNA also.

The non-coding or unexpressed sequences of DNA are called introns. They do not appear in processed RNA.


15. What is meant by capping and tailing?

A: Addition of methyl guanosine tri phosphate at 5' end of hnRNA is called capping.

Addition of 200 to 300 adenylate residues to the 3' end of hnRNA is called tailing or polyadenylation.

16. What is meant by point mutation? Give an example.

A: The mutation that involves insertion or deletion of one or two bases on DNA is called a point mutation. e.g.: Sickle cell anaemia is caused by change in the amino acid glutamate to valine due to change in one nitrogen base in the GAA into GUA.
 

17. Define peptide bond. Why are proteins referred to as polypeptide chains?

A: The bond formed between NH2 group of one amino acid and COOH group of another amino acid liberating a water molecule is called peptide bond. 

Proteins contain long chains of amino acids associated with peptide bonds hence they are called polypeptide chains.
 

18. What is meant by charging of tRNA?

A: The process of activation of amino acid in presence of ATP and its linking with tRNA is called charging of tRNA or amino acylation of tRNA.
 

19. What is a regulator and a promoter?

A: Regulator is a part of transcription unit which acts as inhibitor and codes for repressor which prevents RNA polymerase from transcription. Promoter is located at the 5' end of the transcription unit. It is a DNA sequence that provides the binding site for RNA polymerase.

20. During DNA replication why is it that the entire molecule does not open in one go? Explain replication fork.

A: Due to very high energy requirement the DNA cannot be separated throughout the entire length in one stretch. Hence replication occurs within a small opening of the DNA called replication fork. The replication fork usually arises at a point called ‘ori’ (Origin of replication).
 

21. What is the function of the codon AUG?

A: The codon AUG has two functions:

It acts as initiating codon during translation (protein synthesis)

It codes for the amino acid methionine.
 

22. Define stop codon. Write the codons.

A: The codons which do not code for any amino acid are called stop codons or non-sense codons. The polypeptide chain comes to an end when this codons are present on the mRNA.

e.g.: UAA, UAG, UGA.

23. What is the difference between the template strand and a coding strand in a DNA molecule?

A: During transcription the strand of DNA that is coded and transcribed into mRNA is called template strand.

The coding strand is the strand of DNA that is not transcribed.
 

24. Write any two chemical differences between DNA and RNA.
A:

25. In a typical DNA molecule, the proportion of Thymine is 30% of the N bases. Find out the percentages of other nitrogen bases.
A: If Thymine is 30% Adenine is also 30% (Adenine and Thymine are complementary).
If Adenine and Thymine pair account for 60% then Guanine and Cytosine pair would account for 40% of total nitrogen bases.
Hence Guanine would be 20% and Cytosine 20% (Guanine and Cytosine are complementary).
... Adenine = 30%
Guanine = 20% and
Cytosine = 20%

 

26. The proportions of nucleotides in a given nucleic acid are: Adenine 18%, Guanine 30%, Cytosine 42% and Uracil 10%. Name the nucleic acid and mention the number of strands in it.
A: The nucleic acid is RNA. ( The given data does not show 1 : 1 ratio of purine and pyrimidine).
RNA contains one strand (single strand).

27. If the base sequence of a codon in mRNA is 5' AUG 3', what is the sequence of tRNA pairing with it?
A: UAC.

Short Answer Type Questions

1. Define transformation in Griffith’s experiment. Discuss how it helps in the identification of DNA as genetic material.
A: Transformation: It is the change in physical form of a living organism due to transfer of genetic material.
 * Griffith’s experiment helps in identification of DNA as genetic material.
 * Griffith conducted experiment by injecting mice with S strain (virulent) and R strain (non-virulent) Streptococcus pneumoniae bacteria and could identify the transforming principle.
 * When he injected living S strain bacteria into mice the mice died and living S strain cells were recovered from the body of the dead mice.
 * When he injected living R strain bacteria into the mice the mice lived and living R strain bacteria were recovered from the body of the mice.
* When he injected heat killed S strain bacteria into the mice the mice lived and no cells were recovered from the mice.
 * When he injected heat killed S strain along with living R strain bacteria the mice died and living S strain cells were recovered from dead mice.
 * Thus he concluded that the heat killed S strain bacteria somehow transformed R strain bacteria into S strain. This must be due to the transfer of genetic material.
 * These studies later helped to identify the transforming principle as DNA by Avery, MacLeod and McCarty
 * Hershey and Chase while working on bacteriophages gave proof that DNA is the genetic material.

 

2. Discuss the significance of heavy isotope of nitrogen in Meselson and Stahl’s experiment.
A: * Meselson and Stahl’s experiment on Escherichia coli with heavy isotope of nitrogen proved that DNA replicated semi conservatively.
 * They grew Escherichia coli on medium containing the heavy isotope of nitrogen ( 15NH4Cl) and allowed to replicate.
 * The 15N incorporated DNA could be distinguished from the normal DNA by centrifugation in Caesium chloride density gradient centrifugation.
 * Samples of DNA collected at regular time intervals produced light and hybrid DNA with definite proportions
 * Using heavy isotope of nitrogen it was proved that the DNA in chromosomes replicates semiconservatively.

 

3. Define a cistron. Differentiate between monocistronic and Polycistronic transcription unit with suitable examples.
A: Cistron: A cistron is a segment of DNA coding for a polypeptide.
Monocistronic unit: In a monocistronic unit structural genes have interrupted coding sequences called split genes. The coding or expressed sequences are called exons and the intervening non-coding sequences are called introns. The exons are seen in processed RNA while the introns do not appear on processed RNA.
e.g.: Eukaryotes

Polycistronic unit: A Polycistronic unit does not show exons and introns.
But the co-ordinated regulation of expression of several sets of genes is necessary for production of a character.
The Polycistronic unit contains regulatory sequences, operator and promoter sites to complete the process of transcription.
e.g.: Prokaryotes and Bacteria.

 

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

A: It is like a mother may have one, two, three or four children, but no child has more than one mother.
In the genetic code the code is unambiguous. That means a codon coding for one amino acid cannot code for another amino acid. Hence only one possible sequence of amino acids can be deduced from a given nucleotide sequence.
But the genetic code is degenerate. That means some amino acids are coded by more than one codon. Hence multiple nucleotide sequence can be deduced from a single amino acid sequence. 

 

5. 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.
A: No. The statement is not correct.
A single base mutation in a gene always results in loss or gain of function because it changes the reading frame.

e.g.: THE CAT ATE THE FAT RAT ( a sequence of nucleotides)
Insertion of a single base ‘B’ between THE and CAT.
THE BCA TAT ETH EFA TRA T
Deletion of a single base ‘C’ from CAT.
THE ATA TET HEF ATR AT .
In both the above examples a single base change leads to change in the whole sequence of nucleotides hence alters the phenotype.

 

6. How many types of RNA polymerases exist in cells? Write their names and functions.
A: Three types of RNA polymerases exist in cells. They are:
1. RNA polymerase I
2. RNA polymerase II
3. RNA polymerase III
Functions:
 RNA polymerase I transcribes rRNAs (28 S, 18 S and 5.8 S).
 RNA polymerase II transcribes for hnRNA (heterogenous RNA) which is the precursor of mRNA.
 RNA polymerase III transcribes for tRNA, 5srRNA and snRNAs.

7. What are the differences between DNA and RNA?
A:

8. Write the important features of Genetic code.
A: Genetic code is a combination of nitrogen bases on the nucleic acid molecule that directs the sequence of amino acids during the synthesis of proteins.
Important features of genetic code:
 The genetic code is a triplet code that means three nitrogen bases on the nucleic acid form a codon which codes for one amino acid.
 The genetic code is non-ambiguous or specific.
 The code is degenerate that means more than one codon may code for the same amino acid.
 The code is comma less. There are no punctuations or gaps in the nitrogen base sequence.
 The code is universal, which means that it is common in all organisms from bacteria to human beings (some exceptions are mitochondrial codons and some protozoans)
 AUG has dual functions. It codes for the amino acid Methionine and also acts as initiator codon.

 

Long Answer Type Questions (8 Marks)
1. Give the salient features of Double helix structure of DNA.

Ans:  Salient features of double helix structure of DNA:
Deoxy Ribo Nucleic Acid is a polymer of deoxyribonucleotides.
* The double helix model of DNA by Watson and Crick shows that the DNA is made up of two polynucleotide chains which run antiparallel to each other.
* One chain has the polarity 5' 3' and the other has 3' 5'.
* Each polynucleotide has a number of nucleotides. Each nucleotide has three components 
A nitrogen base which may be a Purine (Adenine, Guanine) or a Pyrimidine (Thymine, Cytosine).
A pentose sugar (deoxyribose C5H10O4).
A phosphate group.
* The nitrogen base is linked to the pentose sugar by a N-glycosidic linkage to form a nucleoside. (deoxyadenosine, deoxyguanosine, deoxycytidine or deoxythymidine)
* The nucleoside is linked to a phosphate group through a phospho ester bond to form a nucleotide.
* The nucleotides are linked to form a polynucleotide chain with one end having phosphate group at 5'-OH called 5' end and the other end has a free 3' OH group called 3' end.
* The sugar and phosphates from the back bone of the DNA polynucleotide.
* The nitrogen bases in two strands are paired through hydrogen bonds to form base pairs (bps)
* Adenine pairs with Thymine with 2 hydrogen bonds and Guanine pairs with Cytosine with three hydrogen bonds. Hence the purine pyrimidine ratio in DNA would be 1 : 1.
* The two strands show a uniform distance of 20A0.
* The two strands are coiled in right handed fashion. Each turn is 3.4 nm long and there are 10 base pairs with equal distances of 0.34 nm.
* Due to hydrogen bonds the DNA is more stable than RNA.

Posted Date : 16-09-2021

గమనిక : ప్రతిభ.ఈనాడు.నెట్‌లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.

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