If only two nucleotides A and G occur in a 4-base pair DNA sequence, how
many such sequences are possible?
4
8
16
32
Number of Sequences
For a 4-base pair DNA sequence using only adenine (A) and guanine (G) nucleotides, exactly 16 such sequences are possible. This result comes from the combinatorial principle where each of the 4 positions can independently be either A or G, yielding 24 = 16 combinations.
Option Analysis
| Option | Calculation | Explanation | Correct? |
|---|---|---|---|
| 4 | 2+2 or permutations without repetition | Assumes only 2 distinct sequences per strand or unique nucleotides without replacement, ignoring repetition allowed in DNA. | No |
| 8 | 23 | Matches 3 base pairs (not 4), undercounting one position. | No |
| 16 | 24 | Correct: 2 choices per position × 4 positions, listing all like AAAA, AAAG, up to GGGG. | Yes |
| 32 | 25 or double-strand consideration | Fits 5 base pairs or mistakenly doubles for complementary strands (but question specifies sequence, typically single-strand count). | No |
Detailed Combinatorics Explanation
DNA sequences treat nucleotides as an ordered arrangement with replacement, so permutations with repetition apply: for n positions and k bases, total is kn. Here, k=2 (A, G) and n=4, so 24 = 16.
All 16 sequences are:
AAAA, AAAG, AAGA, AAGG, AGAA, AGAG, AGGA, AGGG, GAAA, GAAG, GAGA, GAGG, GGAA, GGAG, GGGA, GGGG.
This mirrors genetic code math where even standard 4 bases yield 43 = 64 codons.
CSIR NET Life Sciences Relevance
CSIR NET genetics questions often test such calculations for short DNA sequences, emphasizing molecular biology basics like nucleotide variability. Understanding 24 = 16 prepares students for related topics such as base pairing (A-T, G-C normally, but here purine-only), sequencing limits, and combinatorial libraries in biotech. Practice listing sequences reinforces exam speed for multiple-choice questions.
