Match List-I with List-II: Choose the correct answer from the options given below:List-I (Genes) List-II (Proteins – codes for lac operon) (A) ‘i’ (I) permease (B) ‘a’ (II) β-galactosidase (C) ‘y’ (III) transacetylase (D) ‘z’ (IV) repressor
(A) - (IV), (B) - (III), (C) - (I), (D) - (II)
The lac operon is a classic example of gene regulation found in *E. coli* and some other bacteria. It is responsible for the metabolism of lactose. The operon consists of regulatory genes and structural genes.
Let's look at the key genes involved in the lac operon and what proteins they code for:
Based on the functions described above, we can match the genes in List-I with the proteins they code for in List-II:
List-I (Genes)
List-II (Proteins – codes for lac operon)
Let's make the correct pairings:
The correct matching is therefore: (A) - (IV), (B) - (III), (C) - (I), (D) - (II).
| Gene (List-I) | Protein (List-II) |
|---|---|
| (A) ‘i’ | (IV) repressor |
| (B) ‘a’ | (III) transacetylase |
| (C) ‘y’ | (I) permease |
| (D) ‘z’ | (II) $\beta$-galactosidase |
Remembering which gene codes for which protein is key to understanding the lac operon mechanism. The structural genes (z, y, a) are transcribed together into a single messenger RNA (mRNA) molecule, which is then translated into the individual proteins. The 'i' gene is transcribed separately.
| Lac Operon Gene | Protein Coded | Function |
|---|---|---|
| i | Repressor | Binds to operator, inhibits transcription when lactose is absent. |
| z | $\beta$-galactosidase | Breaks down lactose into glucose and galactose. |
| y | Permease | Transports lactose into the cell. |
| a | Transacetylase | Function in lactose metabolism unclear, possibly detoxification. |
The lac operon is a classic example of an inducible operon. This means that its transcription is usually turned off but can be turned on in the presence of an inducer, which is lactose (or more precisely, allolactose, an isomer of lactose). Here's a simplified view of the regulation:
Transcription of the lac operon is also influenced by glucose levels through a mechanism involving the catabolite activator protein (CAP) and cyclic AMP (cAMP). This ensures that the cell primarily uses glucose, its preferred energy source, when available, before switching to lactose.
Arrange the following steps of DNA fingerprinting in proper sequence:
(A) Hybridisation using labelled VNTR probe
(B) Separation of DNA fragments by electrophoresis
(C) Digestion of DNA by restriction endonucleases
(D) Blotting of separated DNA fragments to nylon
(E) Isolation of DNA
Choose the correct answer from the options given below:
Nucleosome is:
“Transforming Principle” was given by:
Select the incorrect statement:
Amino acid is attached to which site of tRNA?
Given below is the DNA coding sequence. Its complementary strand would read as:
5′ – GTATTACG – 3′
The size of VNTR varies from:
In which phase of the cell cycle does replication of DNA take place?
In Hershey and Chase experiment, some viruses grew on medium that contained:
Who performed experiments on Vicia faba to prove that DNA replicates semi-conservatively?
What will be the chromosome number in the gamete of fruit fly if its meiocyte has 8 chromosomes?
Match List-I with List-II:
| List-I (Organism) | List-II (Sex Chromosomes) |
|---|---|
| (A) Male grasshopper | (I) XY |
| (B) Male Drosophila | (II) XX |
| (C) Female bird | (III) XX |
| (D) Female grasshopper | (IV) XO |
Choose the correct answer from the options given below:
Match List-I with List-II:
| List-I | List-II |
|---|---|
| (A) Bacteriophage lambda | (I) 231 gene |
| (B) Y-chromosome of human | (II) 48502 bp |
| (C) Haploid content of human DNA | (III) 3.3 × 109 bp |
| (D) Escherichia coli DNA | (IV) 4.6 × 106 bp |
Choose the correct answer from the options given below:
Central dogma in molecular biology states that genetic information flows from:
Read the following and select the set of correct statements. (A) Euchromatin is transcriptionally inactive (B) Heterochromatin is more densely packed (C) Heterochromatin is loosely packed (D) Euchromatin is transcriptionally active (E) Euchromatin stains lighter