Chemistry Reference
In-Depth Information
NH
2
c.
N
N
N
N
NH
2
HO
O
N
H
H
O
OH
O
N
O
O
P
O
O
_
N
NH
O
H
H
N
O
OH
N
NH
2
O
P
O
O
_
O
H
H
OH
OH
18.18
The structure is identical to that of the T-A base pair shown in Sec. 18.8 of the text,
except that the methyl group in the thymine unit is replaced by a hydrogen.
18.19
3' T-A-C-C-A-G-A-T-C 5'
18.20
3' U-A-C-C-A-G-A-U-C 5'
The only difference between this sequence and the answer to Problem 18.19 is that
each T is replaced by U.
18.21
Given
m
RNA sequence: 5' A-G-C-U-C-G-U-A-C 3'
DNA strand from which
m
RNA was transcribed:
Note that the DNA strand that was
not
transcribed (the last segment shown above) is
identical with the given
m
RNA segment, except that each U is replaced by T.
18.22
For each T, there must be an A. For each G, there must be a C. As a consequence
of this base pairing, the mole percentages of T in any sample of DNA will be equal to
the mole percentages of A, and the same will be true for G and C. It is
not
necessary, however, that there be any special relationship between the percentages
of the two pyrimidines (T and C) or of the two purines (A and G).
18.23
The code reads from the 5' to the 3' end of
m
RNA.
5' C-A-C 3'
m
RNA
3' G-T-G 5'
DNA-transcribed chain
5' C-A-C 3'
DNA complement
