1
$\begingroup$

In my Biology class we were asked this question:

Simple DNA Strand

This DNA strand consists of eight pairs of nitrogenous bases. How many different sequences of eight bases can you make? Explain how you found your answer.

I guessed either 28 or 8!. Apparently, the answer is 8! ÷ 24. I asked my teacher, but she did not know the answer. Does anyone know why this would be?

Improve this question
$\endgroup$
2
  • 2
    $\begingroup$ The answer your teacher gave you might be the answer to the question of how many sequences of 8 bases can be formed using only the bases shown in the diagram, each one can be used once. The factorial comes from the fact that once you pick a base there are n-1 options left and so on. 2^4 is 2*2*2*2 which accounts for there being four duplicate bases so that count only unique sequences. $\endgroup$
    – Cell
    Mar 15, 2019 at 0:47
  • 2
    $\begingroup$ Don't guess! You need to learn to approach this sort of simple statistics logically as laid out by @Remi.b. The relevence of this sort of problem to biology is more in relation to the frequency of restriction sites, which are smaller and a better place to start. You can find lots of practice questions and an explanation of how to answer them on a self-teaching resource I put up for students at my own university. Give it a try. $\endgroup$
    – David
    Mar 15, 2019 at 18:32

1 Answer 1

Reset to default
2
$\begingroup$

At each base, you can have 4 different bases (A,T,C or G). Therefore for the first base there are 4 possibilities, namely

  • A
  • T
  • C
  • G

For the first two base pairs there are $4^2 = 16$ possible combinations

  • AA
  • AT
  • AC
  • AG
  • TA
  • TT
  • TC
  • TG
  • CA
  • CT
  • CC
  • CG
  • GA
  • GT
  • GC
  • GG

For the first three bases, there are $4^3$ possible combinations. For 8 base pairs, there are $4^8 = 65536$ possible combinations. $2^8$, $8!$ and $8! + 2^4$ are all wrong.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ I think the problem asks combinations with bases in the given sequence. There is 2 of each of A, T, G, and C. So the combinations come out to 8!/(2*2*2*2) which is the answer their teacher provided. $\endgroup$
    – Roni Saiba
    Oct 28, 2020 at 6:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.