Answer:
Explanation:
Given
There are three stop codon
TAA, TAG, TGA.
Out of 64 available codons, 3 are stop codon thus the remaining are non-stop codon.
So the probability of choosing a non stop codon is
Let us suppose the number of trials are "x" then the probability of choosing a non stop codon is
Probability of choosing a stop codon is equal to
probability of choosing a non stop codon
Answer:
a. =
b. i. 0.75
ii. 0.000061
iii. 0.012
iv. 0.17
c. 0.67
Explanation:
a. The expansion of the binomial (p + q)7 would be such that:
=
b. Both couples are heterozygous:
Aa x Aa
AA Aa Aa aa
Since A is dominant over a,
probability of having mole (aa) = 1/4
probability of not having moles = 3/4
<em>Therefore, the probability of the first child not having moles </em>= 3/4 or 0.75
ii. Let the probability of not having mole = p and the probability of having mole = q. From the binomial expansion:
=
<em>Probability that all of the children will have moles </em>=
since p = 3/4 and q = 1/4
=
= 0.000061
iii. <em>Probability that the first two children will have no moles and the last five will have moles</em> =
=
= 0.012
iv. <em>Probability that 4 will have no moles and 3 will have moles out of the 7 children</em> =
=
= 0.17
c. <em>Probability that the child born without moles is a carrier of the a-allele = probability of heterozygou</em>s.
From the cross in (b), the genotypes of those born without moles are AA and 2Aa. Therefore, the probability of not having moles and be Aa is:
= 2/3 or 0.67
Answer:
c) methyltransferase. DNAse.
Explanation:
The way DNA is folded can affect gene expression, or which genes are turned on. When the chromosomes are tightly condensed, the DNA is wrapped very tightly which makes it difficult for transcription factors to bind to the DNA.
