Sankalp NEET Full Test-5 Question-130 Solution

Question: 130: Match List-I with List-II :

List-IList-II
(a) Bacteriophage \phi \times 174(i) 48502 base pairs
(b) Bacteriophage lambda(ii) 5386 nucleotides
(c) Escherichia coli(iii) 3.3 \times 10^{9} base pairs
(d) Haploid content of human DNA(iv) 4.6 \times 10^{6} base pairs

Choose the correct answer from the options given below:

(1) (a) – (i), (b) – (ii), (c) – (iii), (d) – (iv)

(2) (a) – (ii), (b) – (iv), (c) – (i), (d) – (iii)

(3) (a) – (ii), (b) – (i), (c) – (iv), (d) – (iii)

(4) (a) – (i), (b) – (ii), (c) – (iv), (d) – (iii)

Answer: Option (3)

Explanation:

Bacteriophage \phi \times 174 is a small virus with a single-stranded DNA genome.

Its genome size is 5386 nucleotides, so (a) correctly matches with (ii).

Bacteriophage lambda is a double-stranded DNA virus.

The length of its DNA is 48502 base pairs, hence (b) matches with (i).

Escherichia coli is a prokaryotic organism.

The size of its genome is approximately 4.6 \times 10^{6} base pairs, so (c) matches with (iv).

The haploid content of human DNA represents the DNA present in one set of chromosomes.

It contains approximately 3.3 \times 10^{9} base pairs, therefore (d) matches with (iii).

Thus, the correct matching is given in option (3).

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