NEET Biology Molecular Basis of Inheritance MCQs Set 08

Answer: (b) Chromosome 1 has the fewest genes (231)

Explanation:
1. Only about 2 percent of the human genome codes for proteins, repeats form a large part and over 50 percent of genes have unknown function.
2. Chromosome 1 has the most genes (2968) and the Y chromosome has the fewest (231), so (b) is wrong.

Teacher's Note:
1. Chromosome 1: 2968 genes; Y: 231 genes.

Answer: (a) All the time

Explanation:
1. The i gene is expressed constitutively.
2. The repressor is therefore made all the time at a low level.

Teacher's Note:
1. Constitutive means always expressed.

Answer: (b) A-ii, B-i, C-iv, D-iii

Explanation:
1. The phi X174 genome has 5386 nucleotides and lambda phage DNA has 48502 bp.
2. E. coli has 4.6 x 10⁶ bp and a human diploid somatic cell has 6.6 x 10⁹ bp.

Teacher's Note:
1. The haploid human genome is 3.3 x 10⁹ bp.

Answer: (c) Cytosine

Explanation:
1. Adenine and guanine are purines.
2. Cytosine is a pyrimidine found in both nucleic acids, while thymine occurs only in DNA and uracil only in RNA.

Teacher's Note:
1. DNA: A, G, C, T; RNA: A, G, C, U.

Answer: (d) N-glycosidic bond

Explanation:
1. A nitrogenous base is linked to the 1' carbon of the sugar by an N-glycosidic bond.
2. Phosphoester and phosphodiester bonds involve the phosphate group.

Teacher's Note:
1. Nucleoside = base + sugar.

Answer: (a) Phosphoester bond

Explanation:
1. The phosphate is attached to the 5'-OH of the sugar through an ester linkage, a phosphoester bond.
2. The phosphodiester bond joins two nucleotides.

Teacher's Note:
1. Nucleotide = nucleoside + phosphate.

Answer: (d) Sugars and phosphates

Explanation:
1. Phosphodiester bonds link the 3' carbon of one sugar to the 5' carbon of the next through a phosphate.
2. So alternating sugars and phosphates form the backbone and the bases project inwards.

Teacher's Note:
1. Bases are not part of the backbone.

Answer: (d) 2' position of ribose sugar

Explanation:
1. RNA has ribose, which has a hydroxyl group at the 2' position.
2. DNA's deoxyribose lacks this 2'-OH.

Teacher's Note:
1. The 2'-OH makes RNA more reactive and labile.

Answer: (b) X-ray crystallography data of Wilkins and Franklin

Explanation:
1. Watson and Crick used the X-ray diffraction pictures of DNA produced by Maurice Wilkins and Rosalind Franklin.
2. Chargaff provided the base-pairing rules, not X-ray data.

Teacher's Note:
1. Chargaff: A = T and G = C.

Answer: (b) The bases in the two strands are paired through phosphodiester bonds

Explanation:
1. The two strands run in opposite directions, A pairs with T by two hydrogen bonds and the pitch is 3.4 nm.
2. The bases are paired by hydrogen bonds, not phosphodiester bonds, so (b) is wrong.

Teacher's Note:
1. G pairs with C by three hydrogen bonds.

Answer: (d) The plane of one base pair stacks over the other

Explanation:
1. Hydrophobic stacking of base pairs one over another in the double helix adds to stability.
2. The other options are covalent bonds within one strand or pairing rules, which do not specifically explain helix stability.

Teacher's Note:
1. Stacking interactions add to hydrogen bonds.

Answer: (c) Pairing between one purine and one pyrimidine

Explanation:
1. Every base pair has one purine (two rings) and one pyrimidine (one ring).
2. Each pair therefore has the same width, which keeps the distance between the strands constant.

Teacher's Note:
1. Purine + purine would be too wide, pyrimidine + pyrimidine too narrow.

Answer: (b) 2.2 metre

Explanation:
1. One base pair spans 0.34 nm, so a haploid genome of 3.3 x 10⁹ bp is 1.1 m long.
2. A diploid somatic cell has twice this, 2.2 m.

Teacher's Note:
1. Length = bp x 0.34 x 10^-9 m.

Answer: (c) Phosphoric acid

Explanation:
1. Each phosphate group in the backbone carries a negative charge at cellular pH.
2. So DNA is a polyanion due to its phosphates.

Teacher's Note:
1. Negative DNA binds positive histones.

Answer: (c) 200 bp

Explanation:
1. A nucleosome has about 200 bp of DNA wrapped around the histone octamer.
2. The core itself holds about 146 bp.

Teacher's Note:
1. Nucleosome = beads on a string.

Answer: (b) NHC proteins

Explanation:
1. Packaging of chromatin into higher orders requires non-histone chromosomal (NHC) proteins.
2. Histones form only the nucleosome core.

Teacher's Note:
1. Euchromatin is loosely packed; heterochromatin is dense.

Answer: (a) Supernatant

Explanation:
1. 35S labels the protein coat of the phage.
2. After blending, the protein coats remain outside the bacteria and stay in the supernatant.

Teacher's Note:
1. 32P (DNA) is found in the pellet.

Answer: (c) It should be non-mutable

Explanation:
1. Genetic material must replicate, be stable and express itself as characters.
2. It must also be able to mutate slowly, which is essential for evolution, so insisting on it being non-mutable is not a criterion.

Teacher's Note:
1. Slow mutation gives variation.

Answer: (b) 2'-OH group on the sugar

Explanation:
1. The 2'-OH group of ribose makes RNA reactive and chemically less stable than DNA.
2. So RNA is easily degraded.

Teacher's Note:
1. DNA is more stable and so is the genetic material.

Answer: (c) It can directly code for the synthesis of proteins

Explanation:
1. RNA (mRNA) can directly code for protein synthesis.
2. DNA acts as the template for RNA, so RNA is the medium for expression.

Teacher's Note:
1. DNA is the better stable store; RNA is better for expression.

Answer: (b) Presence of highly reactive RNA

Explanation:
1. RNA is labile and reactive and has no proof reading.
2. This leads to fast mutation and a short life span of viruses with RNA genomes.

Teacher's Note:
1. Single-strandedness alone is not the reason.

Answer: (d) Both RNA used to act as genetic material and RNA acting as a catalyst

Explanation:
1. RNA can both carry genetic information and act as a catalyst.
2. Together these suggest that early life processes were RNA based (RNA world).

Teacher's Note:
1. Ribozymes support the RNA world hypothesis.

Answer: (a) 1 : 7

Explanation:
1. 80 minutes is 4 generations, giving 16 DNA molecules.
2. Two of them are hybrid (one old 14N strand) and 14 are fully heavy, so the ratio is 2 : 14 = 1 : 7.

Teacher's Note:
1. Two old strands always end up in two hybrid molecules.

Answer: (d) Taylor using tritiated thymidine

Explanation:
1. Taylor used radioactive thymidine in Vicia faba root tips.
2. This proved semiconservative replication in eukaryotic chromosomes; Meselson and Stahl used 15N in E. coli.

Teacher's Note:
1. Taylor: Vicia faba, tritiated thymidine.