Match List-I with List-II: Choose the correct answer from the options given below: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
(b) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
This question asks us to match different biological entities, specifically their genetic material, with properties like DNA length in base pairs (bp) or gene count. We need to carefully analyze the items in List-I and match them with the characteristics provided in List-II based on the options.
List-I contains the biological entities:
List-II contains the properties:
We are given that option (b) is the correct answer, which provides the following matches:
Let's examine each match according to this option:
| List-I Item | List-II Match | Explanation based on Option (b) |
|---|---|---|
| (A) Bacteriophage lambda | (II) ${48502}$ bp | According to option (b), Bacteriophage lambda DNA has a size of approximately ${48502}$ base pairs. This is a typical size for the genome of this virus. |
| (B) Y-chromosome of human | (I) ${231}$ gene | Option (b) matches the human Y-chromosome with ${231}$ genes. The human Y-chromosome is known to have a relatively small number of genes compared to other human chromosomes. |
| (C) Haploid content of human DNA | (IV) ${4.6 \times 10^6}$ bp | As per option (b), the haploid content of human DNA is ${4.6 \times 10^6}$ base pairs. This value represents the total length of DNA in a single set of human chromosomes. |
| (D) Escherichia coli DNA | (III) ${3.3 \times 10^9}$ bp | Option (b) matches Escherichia coli (E. coli) DNA with ${3.3 \times 10^9}$ base pairs. This value represents the size of the circular chromosome found in E. coli bacteria. |
Following the matches provided in option (b), we link each item from List-I to its corresponding property in List-II.
Therefore, based on the selected option:
| Entity | Type | Characteristic (from List-II) | Match in Option (b) |
|---|---|---|---|
| Bacteriophage lambda | Virus | ${48502}$ bp | (A) - (II) |
| Human Y-chromosome | Eukaryotic Chromosome | ${231}$ gene | (B) - (I) |
| Haploid human DNA | Eukaryotic Genome (Haploid) | ${4.6 \times 10^6}$ bp | (C) - (IV) |
| Escherichia coli DNA | Prokaryotic Genome | ${3.3 \times 10^9}$ bp | (D) - (III) |
Understanding the relative sizes of genomes and the number of genes in different organisms is fundamental in molecular biology and genetics. Here's some additional context:
These examples illustrate the vast differences in genetic content across different life forms and genetic elements.
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:
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
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?