Explanation:
Problem Set 4 Answers
1a. The template DNA strand, from which the mRNA is synthesized, is 5’ CAAACTACCCTGGGTTGCCAT 3’
(RNA synthesis proceeds in a 5’ à 3’ direction, so the template strand and the mRNA will be complementary to each other)
b. The coding DNA strand, which is complementary to the template strand, is 5’ ATGGCAACCCAGGGTAGTTTG 3’
c. The sequence of the mRNA is 5’ AUGGCAACCCAGGGUAGUUUG 3’
(the sequence of the mRNA is complementary to the template strand and identical to the coding strand with U substituted for T)
d. The third codon is 5’ ACC 3’. Therefore, the corresponding anti-codon is 5’ GGU 3’
2. Below is a table for the genetic code:
T
C
A
G
T
TTT Phe (F)
TTC "
TTA Leu (L)
TTG "
TCT Ser (S)
TCC "
TCA "
TCG "
TAT Tyr (Y)
TAC "
TAA Stop
TAG Stop
TGT Cys (C)
TGC "
TGA Stop
TGG Trp (W)
C
CTT Leu (L)
CTC "
CTA "
CTG "
CCT Pro (P)
CCC "
CCA "
CCG "
CAT His (H)
CAC "
CAA Gln (Q)
CAG "
CGT Arg (R)
CGC "
CGA "
CGG "
A
ATT Ile (I)
ATC "
ATA "
ATG Met (M)
ACT Thr (T)
ACC "
ACA "
ACG "
AAT Asn (N)
AAC "
AAA Lys (K)
AAG "
AGT Ser (S)
AGC "
AGA Arg (R)
AGG "
G
GTT Val (V)
GTC "
GTA "
GTG "
GCT Ala (A)
GCC "
GCA "
GCG "
GAT Asp (D)
GAC "
GAA Glu (E)
GAG "
GGT Gly (G)
GGC "
GGA "
GGG "
a. The following codons can be mutated by one base to produce an amber codon:
CAG Gln
AAG Lys
GAG Glu
TCG Ser
TTG Leu
TGG Trp
TAA Stop
TAT Tyr
TAC Tyr
