Essential Cell Biology 5th edition

A:20 Answers
cells, and reactions that produce and further hydrolyze PP i
are therefore virtually irreversible (see Figure 3−41).
ANSWER 7–16
A. A titin molecule is made of 25,000 (3,000,000/120) amino
acids. It therefore takes about 3.5 hours
[(25,000/2 ) × (1/60) × (1/60)] to synthesize a single
molecule of titin in muscle cells.
B. Because of its large size, the probability of making a titin
molecule without any mistakes is only 0.08
[= (1 – 10 –4 ) 25,000 ]; that is, only 8 in 100 titin molecules
synthesized are free of mistakes. In contrast, over 97%
of newly synthesized proteins of average size are made
correctly.
C. The error rate limits the sizes of proteins that can be
synthesized accurately. If a eukaryotic ribosomal protein
were synthesized as a single molecule, a large portion
(87%) of this hypothetical giant ribosomal protein
would be expected to contain at least one mistake.
It is therefore more advantageous to make ribosomal
proteins individually, because in this way only a small
proportion of each type of protein will be defective, and
these few bad molecules can be individually eliminated
by proteolysis to ensure that there are no defects in the
ribosome as a whole.
D. To calculate the time it takes to transcribe a titin mRNA,
you would need to know the size of its gene, which is
likely to contain many introns. Transcription of the exons
alone (25,000 × 3 = 75,000 nucleotides) requires about
42 minutes [(75,000/30) × (1/60)]. Because introns can
be quite large, the time required to transcribe the entire
gene is likely to be considerably longer.
ANSWER 7–17 Mutations of the type described in (B) and
(D) are often the most harmful. In both cases, the reading
frame would be changed, and because this frameshift
occurs near the beginning or in the middle of the coding
sequence, much of the protein will contain a nonsensical
and/or truncated sequence of amino acids. In contrast,
a reading-frame shift that occurs toward the end of the
coding sequence, as described in (A), will result in a largely
correct protein that may be functional. Deletion of three
consecutive nucleotides, as described in (C), leads to the
deletion of an amino acid but does not alter the reading
frame. The deleted amino acid may or may not be important
for the folding or activity of the protein; in many cases,
such mutations are silent—that is, they have no or only
minor consequences for the organism. Substitution of
one nucleotide for another, as in (E), is often completely
harmless. In some cases, it will not change the amino acid
sequence of the protein; in other cases, it will change a
single amino acid; at worst, it may create a new stop codon,
giving rise to a truncated protein.
ANSWER 7–18 The RNA transcripts that are growing from
the DNA template like bristles on a bottlebrush tend to be
shorter at the left-hand side of each gene and longer on
the right-hand side. Because RNA polymerase synthesizes
in the 5ʹ-to-3′ direction it must move along the DNA
template strand in the 3ʹ-to-5ʹ direction (see Figure 7−7).
The longest RNAs, therefore, should appear at the 5ʹ end of
the template strand—when transcription is nearly complete.
Hence the 3′ end of the template strand is toward the left of
the image (Figure A7−18). The RNA transcripts, meanwhile,
are synthesized in the 5ʹ-to-3ʹ direction. Thus, the 5ʹ end of
each transcript can be found at the end of each bristle (see
Figure A7−18); the 3ʹ end of each transcript can be found
within the RNA polymerase molecules that dot the spine of
the DNA template molecule.
Chapter 8
ANSWER 8–1
A. Transcription of the tryptophan operon would no longer
be regulated by the absence or presence of tryptophan;
the enzymes would be permanently turned on in
scenarios (1) and (2) and permanently shut off in scenario
(3).
B. In scenarios (1) and (2), the normal tryptophan
repressor molecules would restore the regulation of
the tryptophan biosynthesis enzymes. In contrast,
expression of the normal protein would have no effect
in scenario (3), because the tryptophan operator would
remain occupied by the mutant protein, even in the
presence of tryptophan.
ANSWER 8–2 Contacts can form between the protein
and the edges of the base pairs that are exposed in the
major groove of the DNA (Figure A8–2). These sequencespecific
contacts can include hydrogen bonds with the
highlighted oxygen, nitrogen, and hydrogen atoms, as
well as hydrophobic interactions with the methyl group on
thymine (yellow). Note that the arrangement of hydrogenbond
donors (blue) and hydrogen-bond acceptors (red) of
a T-A pair is different from that of a C-G pair. Similarly, the
arrangements of hydrogen-bond donors and hydrogenbond
acceptors of A-T and G-C pairs are different from one
another and from the two pairs shown in the figure. These
differences allow recognition of specific DNA sequences
via the major groove. In addition to the contacts shown in
the figure, electrostatic attractions between the positively
charged amino acid side chains of the protein and the
negatively charged phosphate groups in the DNA backbone
usually stabilize DNA–protein interactions. Finally, some
DNA-binding proteins also contact bases from the minor
3′ end of template DNA strand 5′ ends of RNA transcripts 5′ end of template DNA strand
Figure A7−18
1 μm