C1. The start codon begins at the fifth nucleotide. The amino acid sequence would be Met Gly Asn Lys Pro Gly Gln
C2. When we say the genetic code is degenerate, it means that more than one codon can specify the same amino acid.
For example, GGG, GGC, GGA, and GGU all specify glycine.
In general, the genetic code is nearly universal, because it is used in the same way by viruses, prokaryotes,
fungi, plants, and animals. As discussed in Table 13.3, there are a few exceptions, which occur primarily in
protozoa and organellar genetic codes.
C4. A.This mutant tRNA would recognize glycine codons in the mRNA but would put in tryptophan amino acids
where glycine amino acids are supposed to be in the polypeptide chain.
B. This mutation tells us that the aminoacyl-tRNA synthetase is primarily recognizing other regions of the tRNA
molecule besides the anticodon region. In other words, tryptophanyl-tRNA synthetase (i.e., the aminoacyl-
tRNA synthetase that attaches tryptophan) primarily recognizes other regions of the tRNA
other than the anticodon region), such as the T- and D-loops. If aminoacyl-tRNA synthetases recognized only
the anticodon region, we would expect glycyl-tRNA synthetase to recognize this mutant tRNA and attach
glycine. That is not what happens.
C5. As shown in Figure 13.11, the energy comes from ATP. It is this energy conversion that explains the term
C6. A.The answer is three. There are six leucine codons: UUA, UUG, CUU, CUC, CUA, and CUG. The anticodon
AAU would recognize UUA and UUG. You would need two other tRNAs to
recognize the other four
leucine codons. These could be GAG and GAU or GAA and GAU.
B. The answer is one. There is only one codon, AUG, so you need only one tRNA with the anticodon UAC.
C. The answer is three. There are six serine codons: AGU, AGC, UCU, UCC, UCA, and UCG. You would need
only one tRNA to recognize AGU and AGC. This tRNA could have the anticodon UCG or UCA. You would
need two tRNAs to efficiently recognize the other four tRNAs. These could be AGG and AGU or AGA and
C7. There are four proline codons, four glycine codons, one methionine codon, and six serine codons. We apply the
product rule to solve this problem.
6 = 96
C9. The codon is 5
, which specifies proline.
C10.It can recognize 5
, and 5
. All of these specify glycine.
C11.An anticodon that was 3
would recognize the two codons. To recognize 5
, it would have to
be modified to 3
C12.All tRNA molecules have some basic features in common. They all have a cloverleaf structure with three stem-