Question

Suppose we have access to the four nucleotides which compose DNA: A, T, C and G. Placed into a sequence with one another, they form a code which can be read by the cell to provide a blueprint for the creation of proteins. When forming a sequence of three nucleotides, the sequences TAA, TAG, and TGA are codes to tell the cell to stop reading the DNA. Assume each sequence of nucleotides is equally likely to be formed.

A. How many possible ways are there to form a sequence of three nucleotides? - 64
B. Of the possible ways to form a sequence of three nucleotides, what is the probability to form a stop sequence? = 0.0469
C. What is the probability that a sequence that is not a stop sequence is formed? 64-3 = 61 61 = 0.983
D. For a three-nucleotide sequence, G is engineered to be in the third place. For the remaining two nucleotides in the sequence, C cannot be in the second place (for example CTG is allowed but not TCG). How many sequences can be formed?
E. Of the number of sequences found in Part D, what is the probability that a stop sequence is formed?

Answer 1

The correct answer is :

A. we know there are four nucleotides and three sequence makes a triplet code then total possible ways are:

4^3 = 4×4×4= 64 ways.

B. The stop codons by 64 amino acid or triplet codes: TAA, TAG, and TGA then the probability would be = 3/64 = 0.0469

C. the probability that a sequence that is not a stop sequence is formed = total minus probability of stop codons:

64-3 = 61

61/64 = 0.953

D. G at end place and c should not be in the middle then:

the codes will be =

ATG

AGG

AAG

TTG

TGG

TAG

CTG

CGG

CAG

GTG

GGG

GAG

Total 12 triplet code sequences.

E. Among all these 12 codons there is only one stop codon that is TAG so the probability of this stop codon is :

1/12 = 0.083