NEET Biology Molecular Basis of Inheritance Important Questions Set 02

These Class 12 NEET Biology important questions on Molecular Basis of Inheritance give 21 questions with step-by-step answers, arranged as very short, short and long answer types, taken from a model test paper. They help NEET aspirants revise the definitions, structures, differences and key terms of the chapter.

Very Short Answer Type Questions

Question 1. Mention the number of base pairs, N-glycosidic bonds and phosphate radicles in lambda bacteriophage.
Answer: In λ-bacteriophage, No. of bp → 48502 No. of N-glycosidic bonds → 48502 × 2 = 97004 No. of phosphate radicles → 97004

Question 2. Name the protein required in packaging of chromatin at higher level.
Answer: Set of acidic proteins or non-histone chromosomal proteins (NHCs)

Question 3. Which nucleic acid is better for the transmission of genetic information?
Answer: RNA

Question 4. What is the average rate of polymerisation of DNA polymerase III?
Answer: 2000 base pairs per second

Question 5. Mention the transcribed products of RNA polymerase I.
Answer: RNA polymerase I – 5.8S, 28S, 18S rRNA

Question 6. How many codons specify valine and serine?
Answer: Valine – 4 codons; Serine – 6 codons

Question 7. Discuss the significance of heavy isotopes of nitrogen in Meselson and Stahl's experiment.
Answer: The heavy DNA molecule with incorporated 15N could be distinguished from the normal DNA by centrifugation in CsCl density gradient.

Question 8. Consider the DNA strand given below and write down the codons sequence of codons in mRNA strand. Template strand 3′ - A TG CATGCATTTC TCTCC - 5′ Coding strand 5′ - T ACGTACGTAAAGAGAGG - 3′
Answer: Codon sequence on mRNA is given below: 5′-UACGUACGUAAAGAGAGG-3′

Short Answer Type Questions

Question 9. P y O x A
(i) Identify the above given molecule.
(ii) What type of linkage are represented by 'x' and 'y'?
Answer: (i) AMP (Adenosine monophosphate)
(ii) x - N-glycosidic bond y - Phosphoester bond

Question 10. "A low level of expression of lac operon occurs all the time." Explain the logic behind this phenomenon.
Answer: The lactose is transported into the cells through the action of permease which is the product of y-gene. So, low level of expression of lac operon is always present otherwise lactose cannot enter the cell.

Question 11. Why capping and tailing is required for hnRNA?
Answer: Capping or addition of 7mG (methylated gnanosine) is required for binding of smaller subunit of ribosome (40S) and formation of mRNA-ribosome complex. Tailing for efficient-translation mechanism.

Question 12. What is the cause of discontinuous synthesis of DNA on one of the parental strands? What happens to these short stretches of synthesised DNA?
Answer: Both strands are antiparallel and DNA polymerase activity occurs only in 5′ → 3′ direction for new chain elongation. Replication is discontinuous over the strand template with 5′ → 3′ polarity over which only small stretches of DNA are open due to opposite running of template strand. These short stretches of synthesised DNA are Okazaki fragments that are joined by DNA ligase.

Question 13. Define various methods used in sequencing the human genome.

Answer:

Various methods used in sequencing the human genome are:

(i) Expressed Sequence Tags (ESTs) : It is a method to identify all the genes that are expressed as RNAs and sequence the same.

(ii) Sequence Annotation : The whole genome, including both coding and non-coding regions, is first sequenced. Later, functions are assigned to different regions.

Question 14.

Replication

What is represented in the above diagram? How it is different in some viruses?

Answer:

It is central dogma which states that the genetic information flows from DNA → RNA → Protein. The flow of information is in reverse direction, that is from RNA to DNA in some viruses (e.g., HIV)

Question 15. How deoxyribonucleoside triphosphates serve dual purposes in replication?

Answer:

In addition to acting as substrates, they provide energy for polymerisation reaction.

The two terminal phosphates in a deoxyribonucleoside triphosphates are high energy phosphates same as in case of ATP.

Question 16. Mention the types of structural genes with examples.
Answer: The structural gene in a transcriptional unit could be monocistronic (mostly in eukaryotes) or polycistronic (mostly in bacteria).

Question 17. What is a major difference between RNA polymerase and DNA polymerase?
Answer: The DNA polymerases on their own cannot initiate the process of replication while RNA polymerase can initiate the process of transcription.

Question 18. Why both the strands of DNA are not copied during transcription? Give one reason.
Answer: If both strands act as a template, they would code for RNA molecule with different sequences, and in turn, if they code for proteins, the sequence of amino acids in the proteins would be different. Hence, one segment of the DNA would be coding for two different proteins, and this would complicate the genetic information transfer machinery.

Question 19. Transcription and translation can be coupled in bacteria. Explain it.
Answer: (i) There is no separation of cytosol and nucleus in bacteria, therefore transcription and translation take place in the same compartment.
(ii) mRNA does not require any processing to become active.

Long Answer Type Questions

Question 20. Discuss the process of translation in detail.

Answer:

Translation

It is the mechanism by which the triplet base sequences of mRNA molecules are converted into a specific sequence of amino acids in a polypeptide chain. It occurs on ribosomes. The major steps are :

(a) Activation of amino acids. In the presence of enzyme aminoacyl - tRNA synthetase (E), specific amino acids (AA) bind with ATP.

AA₁ + ATP ⎯⎯⎯⎯→E1, Mg²⁺ AA₁ - AMP - E₁ complex + PPi
(b) Charging of tRNA. The AA₁-AMP-E₁ complex formed in first step reacts with a specific tRNA. Thus, amino acid is transferred to tRNA. As a result, the enzyme and AMP are liberated.

AA₁-AMP-E₁ complex + tRNA₁ → AA₁-tRNA₁ + AMP + E₁ (Charged tRNA)
(c) Formation of polypeptide chain. It is completed in three steps.

(i) Chain Initiation. It requires three initiation factors in prokaryotes.

Nine intiation factors are required in eukaryotes. The cellular factory responsible for synthesising proteins is the ribosome. The ribosome consists of structural RNA and about 80 different proteins. In its inactive state, it exists as two subunits, a large subunit and a small subunit.

(a) Binding of mRNA with smaller subunit of ribosomes (30S/40S).

(b) Binding of 30S/40S-mRNA complex with tRNA. Non-formylated methionine is attached with tRNA in eukaryotes and formylated methionine in prokaryotes. ⎯⎯⎯→ 40S/30S-mRNA complex + tRNAmet GTP 40S/30S-mRNA-tRNAmet complex.

(c) Attachment of larger subunit of ribosomes.

(ii) Chain elongation. Ribosomes have two sites for binding amino acyl tRNA (i) Amino acyl or A site (acceptor site) (ii) Peptidyl site or P site (donor site). A second charged tRNA molecule along with its appropriate amino acid approaches the ribosome at A site close to the P site. Its anticodon binds to complementary codon of mRNA chain. A peptide bond is formed between COOH group of first amino acid (methionine) and NH₂ group of second amino acid. The formation of peptide bond requires energy and is catalysed by enzyme peptidyl transferase (23S rRNA in bacteria and 28S rRNA in eukaryotes). The initiating formyl methionine or methionine tRNA can bind only with P site. All other newly coming aminoacyl tRNA bind to A site.Translocation is movement of ribosome on mRNA.

(iii) Chain termination. The termination of polypeptide is signalled by one of the three terminal codons in the mRNA. The three terminal codons are UAG, UAA and UGA. A GTP-dependent factor known as release factor is associated with termination codon. It is eRF₁ in eukaryotes and RF₁ and RF₂ in prokaryotes.

At the time of termination, the terminal codon immediately follows the last amino acid codon. After this the polypeptide chain, tRNA, mRNA are released and the subunits of ribosomes get dissociated. xxx yyy zzz mRNA xxx yyy zzz mRNA 5′ 3′ 5′ 3′

Initiating 1 1 2 amino acid P A P A Site Site Site Site (A) Binding of first charged tRNA at P site (B) Binding of next charged tRNA at A site

mRNA xxx yyy zzz 5′ 3′ mRNA xxx yyy zzz 5′ 3′

2 2 P A Site Site P A 1 Site Site 1 (C) Peptide bond formation by peptidyl (D) Translocation movement of ribosome on transferase mRNA, so peptidyl tRNA is at P site. A site is open for the next charged tRNA

Question 21.

Give the correct word or phenomenon in column B for statements mentioned in column A.

Answer:

a - Heterochromatin

b - Primer

c - Nucleosome

d - Topoisomerase

e - Exons