Molecular Basis of Inheritance, on the exam screen.

In the lac operon the structural gene z codes for beta-galactosidase, the enzyme that hydrolyses lactose into glucose and galactose. Hence (3) is correct. The y gene codes for permease, the a gene for transacetylase, and the separate regulatory i gene codes for the repressor protein.

A transcription unit is defined by a promoter, a structural gene and a terminator (Statement I); the promoter lies towards the 5'-end of the structural gene (Statement II), provides the binding site for RNA polymerase (Statement III), and by its position determines which strand is template and which is coding (Statement IV). The terminator lies towards the 3'-end of the coding strand and usually ends transcription (Statement V). All five statements are correct.

(Statement I)DNA fingerprinting proceeds by isolating DNA and digesting it with restriction enzymes,
(Statement V)separating the fragments by gel electrophoresis according to size,
(Statement III) blotting them onto a synthetic nylon/nitrocellulose membrane,
(Statement II)hybridising with a labelled variable number of tandem repeats (VNTR) probe,
and
(Statement IV) finally detecting the bound probe by autoradiography.

This gives the order I, V, III, II, IV.

Eight histone molecules assemble into a histone octamer (I correct), and histones are positively charged basic proteins rich in lysine and arginine (III correct), so Statement II is wrong. It is the negatively charged DNA (not positively charged) that wraps around the octamer to form a nucleosome, so Statement IV is wrong. Higher-order packaging needs additional non-histone chromosomal (NHC) proteins (V correct). Hence I, III and V only.

The regulator gene $i$ is transcribed continuously from its own promoter, independent of whether lactose is present, so the repressor protein is always available - that is what constitutive expression means. This is essential to the logic of the operon: the repressor must already exist in order to keep the structural genes switched off in the absence of lactose. NCERT Class 12, Molecular Basis of Inheritance (regulation of gene expression).

Eukaryotes use three nuclear RNA polymerases with a clean division of labour. RNA polymerase II transcribes the precursor of messenger RNA (mRNA), the heterogeneous nuclear RNA (hnRNA), which is then capped, tailed and spliced into mature mRNA. RNA polymerase I handles the ribosomal RNAs and RNA polymerase III the transfer RNAs (tRNAs), 5S rRNA and small nuclear RNAs. NCERT Class 12, Molecular Basis of Inheritance.

The length of the DNA can be calculated using the formula $L = n \times d$, where n is the number of base pairs and d is the distance between two consecutive base pairs. Substituting the values, $L = 6.6 \times 10^{9} \times 0.34 \times 10^{-9}$ m = 2.244 m $\approx$ 2.2 m. Chapter: Molecular Basis of Inheritance.

During transcription, DNA-dependent RNA polymerase binds the promoter and locally opens the DNA helix to expose the template strand. DNA helicase is primarily associated with replication, so the keyed answer is RNA polymerase. Chapter: Molecular Basis of Inheritance.

In double-stranded DNA, adenine pairs with thymine through two hydrogen bonds, while guanine pairs with cytosine through three. Complementary base pairing explains the hydrogen-bond count asked here.

Before peptide synthesis begins, each amino acid is activated and attached to its cognate transfer RNA (tRNA) by aminoacyl-tRNA synthetase. This aminoacylation or charging of tRNA is the first phase of translation. Chapter: Molecular Basis of Inheritance.