NEET Biology Molecular Basis of Inheritance MCQs Set 10

Answer: (d) DNA

Explanation:
1. Most living organisms use DNA as genetic material.
2. RNA is the genetic material only in some viruses.

Teacher's Note:
1. DNA is stable and so suits long-term storage of information.
2. RNA is more reactive.

Answer: (a) 5' end

Explanation:
1. The 5' end carries a free phosphate on the 5' carbon of the terminal sugar.
2. The 3' end carries a free OH group.

Teacher's Note:
1. A chain is written 5' to 3'.
2. New nucleotides are added at the 3' end.

Answer: (a) sugars and phosphates

Explanation:
1. Sugar and phosphate groups joined by 3'-5' phosphodiester bonds form the backbone.
2. The bases project from it.

Teacher's Note:
1. The backbone is the same in every DNA.
2. Only the base sequence differs.

Answer: (c) Purine-pyrimidine hydrogen bonding

Explanation:
1. A purine always pairs with a pyrimidine.
2. The pair then has the same width all along the helix.

Teacher's Note:
1. Two purines would be too wide and two pyrimidines too narrow.
2. Hence A pairs with T and G with C.

Answer: (c) reverse transcription

Explanation:
1. Reverse transcriptase makes DNA from an RNA template, as in retroviruses.
2. This is a reversal of normal transcription.

Teacher's Note:
1. Temin and Baltimore discovered it.
2. HIV uses this process.

Answer: (b) cistron

Explanation:
1. A cistron is a DNA segment that codes for one polypeptide.
2. A codon is a triplet of bases on mRNA.

Teacher's Note:
1. Cistron means functional gene unit.
2. An operon is a group of genes under one control.

Answer: (d) elongation

Explanation:
1. RNA polymerase catalyses elongation.
2. It needs the sigma factor for initiation and the rho factor for termination.

Teacher's Note:
1. Initiation and termination need extra factors.
2. This is true in bacteria.

Answer: (c) George Gamow

Explanation:
1. George Gamow reasoned that 4 bases must give 64 combinations to code 20 amino acids.
2. So the code must be a triplet.

Teacher's Note:
1. A triplet code gives 64 codons.
2. Khorana later helped to decipher them.

Answer: (b) Digestibility

Explanation:
1. The genetic material must be able to replicate, be stable, allow slow mutation and express itself as phenotype.
2. Digestibility is not a requirement.

Teacher's Note:
1. The four qualities are replication, stability, mutability and expressibility.
2. RNA is less stable than DNA.

Answer: (b) CsCl density gradient centrifugation

Explanation:
1. They used caesium chloride density gradient centrifugation with 15N and 14N labelled DNA.
2. The bands showed the heavy, hybrid and light DNA.

Teacher's Note:
1. The first generation showed a single hybrid band.
2. This ruled out conservative replication.

Answer: (d) polycistronic and non-split

Explanation:
1. Prokaryotic genes are usually polycistronic and have no introns.
2. Several cistrons form one transcription unit.

Teacher's Note:
1. The lac operon is polycistronic.
2. Eukaryotic genes are the opposite.

Answer: (c) sickle cell anaemia

Explanation:
1. Sickle cell anaemia results from a single base change, GAG to GUG, in the beta globin gene.
2. Glutamic acid becomes valine.

Teacher's Note:
1. This is a base substitution.
2. Down syndrome is a chromosomal disorder.

Answer: (a) an inverted L

Explanation:
1. In two dimensions tRNA looks like a clover leaf.
2. In three dimensions it folds into an inverted L.

Teacher's Note:
1. Do not mix up the 2-D and 3-D shapes.
2. The anticodon loop and the amino acid end lie at opposite ends.

Answer: (c) untranslated regions (UTRs)

Explanation:
1. UTRs lie at the 5' end (before the start codon) and the 3' end (after the stop codon).
2. They help in efficient translation.

Teacher's Note:
1. Do not confuse UTRs with introns.
2. Introns are removed by splicing.

Answer: (c) repressor

Explanation:
1. The regulatory gene i codes for the repressor protein.
2. It is expressed all the time.

Teacher's Note:
1. The i gene is not part of the operon proper.
2. It has its own promoter.

Answer: (b) inducer

Explanation:
1. Lactose (allolactose) binds the repressor and makes it inactive.
2. Transcription then starts.

Teacher's Note:
1. An inducer switches on an inducible operon.
2. Lactose is the substrate and the inducer.

Answer: (c) negative regulation

Explanation:
1. The repressor switches the operon off, which is negative regulation.
2. An activator would be positive regulation.

Teacher's Note:
1. Negative regulation prevents transcription.
2. The inducer removes the block.

Answer: (c) inducer

Explanation:
1. The inducer binds the repressor and changes its shape.
2. It can no longer bind the operator.

Teacher's Note:
1. The inducer is lactose or allolactose.
2. Then RNA polymerase moves on.

Answer: (c) satellite DNA

Explanation:
1. Repetitive DNA separates as small peaks from the bulk DNA.
2. It is called satellite DNA.

Teacher's Note:
1. Satellite DNA does not code for proteins.
2. It shows high variability.

Answer: (c) Alec Jeffreys

Explanation:
1. Alec Jeffreys developed DNA fingerprinting.
2. He used VNTR probes.

Teacher's Note:
1. Fingerprinting is used in forensics and paternity tests.
2. It needs very little DNA when PCR is used.

Answer: (b) mini-satellite

Explanation:
1. VNTR (variable number tandem repeats) are mini-satellites.
2. The repeats are about 10 to 100 bases long.

Teacher's Note:
1. The copy number of VNTR varies between individuals.
2. This gives a unique pattern.

Answer: (d) polymerase chain reaction

Explanation:
1. PCR amplifies very small samples of DNA.
2. The fingerprint can then be made from a tiny sample.

Teacher's Note:
1. PCR was developed by Kary Mullis.
2. It uses Taq polymerase.

Answer: (b) amino acids

Explanation:
1. Amino acids join by peptide bonds on the ribosome.
2. This forms a polypeptide.

Teacher's Note:
1. Transcription is the synthesis of RNA.
2. Translation is the synthesis of polypeptides.

Answer: (a) catalyst

Explanation:
1. The 23S rRNA acts as a ribozyme in bacteria.
2. It catalyses peptide bond formation.

Teacher's Note:
1. This suggests RNA was the first catalyst.
2. It supports the RNA world idea.

Answer: (a) start codon; stop codon

Explanation:
1. A translational unit lies between the start codon (AUG) and the stop codon.
2. The UTRs lie outside them.

Teacher's Note:
1. A transcription unit is flanked by promoter and terminator.
2. Do not mix up the two units.