Central dogma in molecular biology states that genetic information flows from:
(b) DNA → mRNA → Protein
The central dogma of molecular biology is a fundamental principle that describes the flow of genetic information within a biological system. It explains how the instructions encoded in DNA are converted into functional products, primarily proteins.
The central dogma outlines the typical sequence of events in gene expression:
Therefore, the main flow of genetic information described by the central dogma is DNA → mRNA → Protein.
Let's evaluate each provided option based on the principles of the central dogma:
Based on the analysis, the sequence that accurately represents the primary flow of genetic information according to the central dogma is DNA → mRNA → Protein.
| Process | Information Flow | Description |
|---|---|---|
| Replication | DNA → DNA | Copying DNA |
| Transcription | DNA → mRNA | Synthesizing mRNA from DNA template |
| Translation | mRNA → Protein | Synthesizing protein from mRNA template |
| Concept | Explanation | Key Molecule(s) |
|---|---|---|
| Central Dogma | Describes the flow of genetic information | DNA, RNA, Protein |
| Replication | DNA copying itself | DNA polymerase |
| Transcription | DNA to RNA | RNA polymerase |
| Translation | RNA to Protein | Ribosomes, tRNA |
While the central dogma states DNA → mRNA → Protein as the main flow, there are some exceptions or additions:
However, even with these additions, the transfer of information from protein back to nucleic acid (DNA or RNA) is not known to occur naturally.
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:
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
A double stranded DNA fragment has 500 Adenine bases. If total number of base pairs in this fragment is 2500, then what will be the number of guanine bases?