Question

If R = any purine, Y = any pyrimidine, and N = any nucleotide, what is the probability of finding the sequence 5′ GRANTY 3′ in any molecule of DNA? Assume nucleotides are in equal percentages.

a) 1/64
b) 1/128
c) 1/256
d) 1/1024
e) 1/4096

Answer 1

The probability of finding the sequence 5′ GRANTY 3′ in any molecule of DNA, with all nucleotides equally frequent, is calculated by multiplying the individual probabilities of each nucleotide's occurrence. Given that G is guanine, R is any purine (adenine or guanine), A is adenine, N is any nucleotide (adenine, cytosine, guanine, or thymine), T is thymine, and Y is any pyrimidine (cytosine or thymine), the probability is 1/256.

Step-by-step explanation:

The question is asking for the probability of finding the sequence 5′ GRANTY 3′ in any molecule of DNA, assuming that each type of nucleotide occurs at an equal frequency. To calculate this, we break down the sequence into its components: G is guanine, R is any purine (adenine or guanine), A is adenine, N is any nucleotide (adenine, cytosine, guanine, or thymine), T is thymine, and Y is any pyrimidine (cytosine or thymine).

Each nucleotide choice has an equal chance of 1/4, except for R and Y which have a 1/2 chance since they can be one of two nucleotides. Thus, the probability of the sequence is the product of the individual probabilities:

P(G) = 1/4
P(R) = 1/2
P(A) = 1/4
P(N) = 1
P(T) = 1/4
P(Y) = 1/2

Multiplying these together: (1/4) × (1/2) × (1/4) × 1 × (1/4) × (1/2) = 1/256

Therefore, the probability is 1/256, which corresponds to option c) 1/256.