Download Use the genetic code table below when solving the following problems

MCMP 208 Biochemistry I
Practice Problems for Translation
If needed, use the genetic code table below when solving the following problems.
U
C
A
G
UUU Phe (F)
UUC "
U
UUA Leu (L)
UUG "
UCU Ser (S)
UCC "
UCA "
UCG "
UAU Tyr (Y)
UAC
UAA Stop
UAG Stop
UGU Cys (C)
UGC
UGA Stop
UGG Trp (W)
CUU Leu (L)
CUC "
C
CUA "
CUG "
CCU Pro (P)
CCC "
CCA "
CCG "
CAU His (H)
CAC "
CAA Gln (Q)
CAG "
CGU Arg (R)
CGC "
CGA "
CGG "
AUU Ile (I)
AUC "
A
AUA "
AUG Met (M)
ACU Thr (T)
ACC "
ACA "
ACG "
AAU Asn (N)
AAC "
AAA Lys (K)
AAG "
AGU Ser (S)
AGC "
AGA Arg (R)
AGG "
GUU Val (V)
GUC "
G
GUA "
GUG "
GCU Ala (A)
GCC "
GCA "
GCG "
GAU Asp (D)
GAC "
GAA Glu (E)
GAG "
GGU Gly (G)
GGC "
GGA "
GGG "
1. Given the following sequences of exonic coding DNA (from the middle of the ORF) predict the translated
sequence (using single letter codes for the amino acids). Assume the reading frame is aligned with the start
of each of the sequences provided.
A.
B.
C.
D.
gctgcggggc
gagagcacag
gccctccggc
aagaaccaca
cgccgccgag
tgcgcttcgc
agaagaacgt
aattcaccgc
cgagggcgag
ccgcaaaggc
gcacgaggtg
ccgcttcttc
2. The following sequences are unspliced coding strand of DNA from the middle of the ORF and each
contains an intron with the splice junctions shown ( = splice donor junction; ↑ = splice acceptor junction).
Part of the intron sequence is not shown for brevity. Predict the translated sequence. Assume the reading
frame is aligned with the start of the first exonic DNA sequences provided.
A. aagcagcccacctcg…(intron DNA)…ac↑tctgcagccactgcacc
B. gacttcacg…(intron DNA)…ac↑tctggggcttcgggaagcaggga
C. ttccagtgtcaagtctgctgctcg…(intron DNA)…ac↑ttgttgta
D. cacacg…(intron DNA)…ac↑agcgctgccatgagttcgtcacgttc
3. Given the protein sequence provided below, specify a single coding DNA sequence that would encode that
protein.
A.
B.
C.
D.
LIHQGMKCDT
CMMNVHKRCV
MNVPSLCGTD
HTERRGRIYI
4. The sequence below is from the coding strand of a single exon for a gene and is translated with its
translation frame starting with the first nucleotide shown. There are 120 nucleotides shown, 30 per line. The
first nucleotide on each line given a number which precedes the sequence on each line. Use this information
to answer parts A through E below which ask about the impact of a specific mutation in this DNA sequence
on the translated protein sequence.
301
331
361
391
tcctgccccg
tctgatgacc
aagatccaca
tgtgaccact
gtgcggacaa
cccggagcaa
cgtactccag
gtggatcgct
gggcccggcc
acacaagttt
tcctaccttc
gctgtatgga
A.
B.
C.
D.
How would the amino acid sequence be changed if nucleotide 390 is changed to an a?
How would the amino acid sequence be changed if nucleotide 343 is changed to an a?
How would the amino acid sequence be changed if nucleotide 361 is changed to a g?
How would the amino acid sequence be changed if the three nucleotides ccc were inserted immediately
after the c at 308?
E. How would the amino acid sequence be changed if the nucleotides a was inserted immediately after the
g at 392?
5. Given the genetic code table and the sequence of a peptide, calculate how many different sequences encode
that specific peptide (i.e., its degeneracy)
A.
B.
C.
D.
E.
F.
IKCS
QAHIDREVLI
VVVRDAKNLV
PMDPNGLSDP
YVKLKLIPDP
KSESKQKTKT
6. Diagrammed below for each problem are the three reading frames in prokaryotic mRNA, with the start
codons shown as a vertical line (“|”) and a stop codon shown as an asterisk (“*”). All other positions are
simply shown as a dash (“-”). Codons are grouped by 10 to enhance readability. For each problem, indicate
which frame has is the longest ORF and also the length of that ORF in codons (not counting the stop
codon).
A. Frame 1:
Frame 2:
Frame 3:
----*--|-- -------|-- -----*---- ---|------ --*--------|-*----- |--------- ---------| ---------- ------*|--*-------- --|------- ---*--|--- --------|- ------*---
B. Frame 1:
Frame 2:
Frame 3:
----|*---- ---------- ---------- ---------- |-------*-|-------- ---------- ----|-*--- ---------- ------------------- ----*----- |--------- ---------- -|*-------
C. Frame 1:
Frame 2:
Frame 3:
-*-----|-- ---------- ----|----- ---------- ------|------------ ------*--- ---------- -----*---- --|-----*---|------ ---------- ---------- --*----*-- ----*-----
D. Frame 1:
Frame 2:
Frame 3:
|-------*- ---------- ---------- ---*------ |------------------ ---------* ---|------ *--------- ---------*--|------ -----|---- ---------- --------|- -------*--
E. Frame 1:
Frame 2:
Frame 3:
----*----- --------*- -------|-- ---------- ---**-----------**- ---------- --------*- ---------| ---------*
--*------- ---------- -*-------- |------*-- --------*-
The answers are on the following pages.
Answers:
1.
A. AAGPPPSEGE
B. ESTVRFARKG
C. ALRQKNVHEV
D. KNHKFTARFF
2.
A. KQPTFCSHCT
B. DFIWGFGKQG
C. FQCQVCCFVV
D. HKRCHEFVTF
3. The answers are shown below in a format where alternative nucleotides that might be used are shown above
the main line for the answer. The initial nucleotides on the above lines are not shown if the alternative
codons start with one or two of the same letters as the codon on the last line. For each problem, any
sequence with the acceptable codons in the proper order is correct (i.e., they do not need to be on the same
line within the answer below for that problem).
tta
ttg
g
t
t
c
a
a
a
a
c
t
a
c
g
t
t
g
A. ctt att cac cag ggg atg aaa tgc gac acc
a
t
c
t
c
B. tgt atg atg aac gtg
t
t
a
a
t
g
g
C. atg aac gtc ccc
agg
aga
a
t
t
a
t
a
g
t
c
cac aag cgc tgc gtg
tct tta
tcg ttg
tca
a
t
g
tcc
t
a
c
t
g
t
g
a
t
agc ctc tgc ggc act gac
t
a
t
c
D. cac acg
agg agg
agg
aga aga
aga
g
g
g
a
a
a
a
a
g
t
t
t
t
c
t
gaa cgc cgc ggc cgc atc tat atc
4. For reference, below is the encoded amino acid sequence with relative numbering, starting at 1:
1
11
21
31
SCPGADKGPA
SDDPRSKHKF
KIHTYSSPTF
CDHCGSLLYG
A. F at relative position 30 would be changed to L (a conservative changes, both hydrophobic)
B. Although this would change the first nucleotide in the codon, it would still encode R at position 15.
C. K at relative position 21 would be changed to an E (a rather drastic substitution, an acid for a base).
D. This would result in a second P being inserted after the P that is at relative position 3.
E. This would convert the codon coding for C at relative position 31 into a stop codon, which would
produce a C-terminally truncated translation product.
5. Note: to do this, multiply the degeneracies of each amino acid in the sequence.
A. 3 x 2 x 2 x 6 = 72
B. 2 x 4 x 2 x 3 x 2 x 6 x 2 x 4 x 6 x 3 = 24 x 32 x 42 x 62 = 210 x 34 = 1024 x 81 = 82,944
C. 4 x 4 x 4 x 6 x 2 x 4 x 2 x 2 x 6 x 4 = 23 x 45 x 62 = 215 x 32 = 32,768 x 9 = 294,912
D. 4 x 1 x 2 x 4 x 2 x 4 x 6 x 6 x 2 x 4 = 11 x 23 x 44 x 62 = 213 x 32 = 8192 x 9 = 73,728
E. 2 x 4 x 2 x 6 x 2 x 6 x 3 x 4 x 2 x 4 = 24 x 31 x 43 x 62 = 212 x 33 = 4096 x 27 = 110,592
F. 2 x 6 x 2 x 6 x 2 x 2 x 2 x 4 x 2 x 4 = 26 x 42 x 62 = 212 x 32 = 4096 x 9 = 36,864
6. Note that in the answers below, the end position of the ORF that is specified is the location of the stop
codon, which does not encode and amino acid.
A. Frame 2. The ORF is from 11 to 47, so it is 36 amino acids long.
B. Frame 2. The ORF is from 2 to 27, so it is 26 amino acids long. The ORF in frame 3 is only 22 long
C. Frame 3. The ORF is from 4 to 33, so it is 29 amino acids long. There might be a longer ORF in frame
1, since there are only start codons, however, without a stop codon, we cannot be sure that frame 1 has
an ORF.
D. Frame 3. The ORF is from 4 to 48, so it is 44 amino acids long.
E. Frame 1. The ORF is from 28 to 44, so it is 16 amino acids long.