Restriction enzyme HinfI cleaves a five nucleotide DNA sequence GA(A/T)TC. The ambiguity in the central position - (A/T) - means
that either A or T can occur in the cleavage site. Assuming that each of four nucleotides is equally likely to occur at any position on a DNA molecule, an average HinfI cleavage fragment is about A) 0.5 kb B) 1.0 kb C) 2.0 kb D) 4.0 kb E) 8.0 kb
It is given that out of the 4 nucleotides A, T, C & G each one has equal probability to occur at any position on the DNA molecule which simply means that the probability of occurrence of any nucleotide at a position is 1/4.
Also, it is given that probability of occurrence of either A or T at 3rd position is equal which means that the probability at that particular position will be 2/4 = 1/2.
Now, GA(A/T)TC is the DNA sequence where Restriction enzyme HinfI cleaves so the total probability of an average HinfI cleavage fragment will be = 1/4 x 1/4 x 1/2 x 1/4 x 1/4 = 0.00195 = 0.2 i.e. 2 kb.
As you can see in the table provided below when you do a punnet's square for the given genotypes, you get the possible genotypes of their offspring. Marked in blue are the genotypes homozygous for gene T. There are 4 out of 16 of them or expressed as a decimal 0,25.