Answer:
C) 2.0 kb
Explanation:
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.
The regression equation is:
y = 1.15 – 0.251 x₁ + 0.078 x₂ + 0.306 x₁x₂
<span>
<span>
<span>
Where,
x₁ = -0.2508
x₂ = 0.0777
x₁x₂ = 0.3058
Inserting the values,
<span>
<span>
</span></span>y = 1.15 – {0.251 × (-0.2508)} + {0.078 ×(0.0777)} + {0.306 × (0.3058)}
</span></span></span>
y = 1.15 - (- 0.0629508) + 0.0060606 + 0.935748
y = 1.15 + 0.0629508 + 0.0060606 + 0.935748
y = 1.3125862
Answer:
x₁ and x₂ are main effects (a type of tuna and type of packing liquid) that contribute significant information for the prediction of <span>y.</span>
Hi!
Could you please be more specific, so that I can answer your question properly? Layer of what?
