Download DNA MUTATION, REPAIR, AND TRANSPOSITION

135
Hyde Chapter 18 – Solutions
18
DNA MUTATION, REPAIR,
AND TRANSPOSITION
CHAPTER SUMMARY QUESTIONS
2. Silent mutations are those that do not produce a detectable phenotypic change. They
can occur in the region between genes, in introns, in the 5' and 3' untranslated
regions of a gene, and in the wobble position of a codon.
4. Both types of mutation restore the wild-type function of a gene without restoring its
original sequence. An intragenic suppression occurs at a different site in the same
gene than the first mutation, while an intergenic suppression occurs in a different
gene, called a suppressor gene (which is typically a tRNA gene).
6. The mutation rate is the number of mutations that arise per gene for a specific length
of time in a population. The mutation frequency for a given gene is the number of
mutant alleles per the total number of copies of the gene within a population.
Therefore, the mutation rate represents newly generated mutations, whereas the
mutation frequency represents the prevalence of a specific mutation, both old and
newly generated.
8. Both chemicals are base modifiers. Ethylethane sulfonate is an alkylating agent that
can add an ethyl group to any of the four bases. It can promote transitions and
transversions. Nitrous acid replaces amino groups (–NH2) on nucleotides with keto
groups (=O). It causes transitions by converting cytosine into uracil, which acts like
thymine, and adenine into hypoxanthine, which acts like guanine.
10. Ultraviolet light induces cross-linking, or dimerization, between adjacent
pyrimidines in DNA. The principal products of UV irradiation are thymine–thymine
dimers, although cytosine–cytosine and cytosine–thymine dimers are occasionally
produced. The dimers distort the DNA and result in the failure of the pyrimidines to
base-pair with their complementary purines. This type of DNA damage can be
repaired by three different mechanisms: (1) photoreactivation, which involves
photolyase-mediated reversal of the thymine dimer without removing the bases,
(2) nucleotide excision repair, which involves removal and replacement of a short
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Hyde Chapter 18—Solutions
stretch of DNA that includes the pyrimidine dimer, (3) postreplicative repair, which
involves a specific group of DNA polymerases that are capable of replicating
damaged DNA.
12. The SOS response is a complex, inducible repair system in E. coli that is used to
repair DNA when excessive damage has occurred. The RecA protein interacts with
single-stranded DNA, which stimulates a protease activity that is normally silent in
the RecA protein. The RecA* protease activity cleaves the LexA protein. LexA is a
transcriptional repressor of about 18 genes, many of which are involved in DNA
repair and the inhibition of cell division. Therefore, the inactivation of the LexA
repressor by RecA* protease allows the transcription of these genes.
14. IS elements, or insertion sequences, are relatively small transposable elements in
bacteria. Unlike other transposons, they carry no bacterial genes. They are
sometimes referred to as “selfish DNA” because they may exist without a noticeable
benefit to their host cell.
16. Transposons may induce deletions and inversions through pairing and recombination
between copies of the transposable elements. If the copies of a transposable element
are present in a directly repeated orientation, pairing followed by a single crossover
event between the two copies can result in the deletion of one copy and the region
between transposons. If the copies of the transposable element are present in an
inverted orientation, pairing followed by a single crossover event between the two
copies can result in an inversion of the region between the transposable elements.
Refer to figure 18.41.
18. Hybrid dysgenesis is the production of sterile Drosophila offspring due to
transposable element activity in germ cells. When a male of a P stock is crossed to a
female of an M stock, the resulting F1 hybrid offspring are sterile. On the other hand,
the reciprocal cross between an M male and a P female yields normal F1 offspring.
Analysis of the stocks revealed that P flies contain one or more copies of a
transposon, the P element, while the M flies either lacked or contained shorter
nonfunctional versions of the P element. The sterility in the F1 hybrids resulted from
transposition of the P element within the genomes of germ cells, causing
chromosomal abnormalities, insertions into essential genes, and an overall disruption
of gamete formation. Therefore, the cross of a P male with an M female results in
sterile hybrid progeny. Hybrid dysgenesis does not occur if the cross is between a
female fly from a P stock and a male fly from an M stock. The P element encodes a
transposase enzyme as well as a repressor of transposase. The egg cells from a
P female Drosophila contain enough of this repressor protein in the cytoplasm to
prevent the transposition of the P element. Therefore, a zygote produced from an
M male and a P female will have the repressor and so the offspring will be normal.
On the other hand, a zygote produced from a P male and an M female (who lacks a
full-length P element) would not have the repressor. The P elements passed on
through the sperm would transpose in the germ cells of the F1 hybrid, causing it to
become sterile.
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Hyde Chapter 18 – Solutions
20. Bacterial and eukaryotic transposable elements typically have inverted repeats at
their ends and at least one gene in the middle. The transposons are also similar in that
they generate direct repeats during their insertion at target sites.
EXERCISES AND PROBLEMS
22. UV light causes pyrimidine dimers, the most common of which are thymine–
thymine dimers. Therefore the DNA molecule with the highest percentage of
thymine will be expected to be the most sensitive to UV light damage, while that
with the smallest thymine percentage should be the least sensitive. (Note: The
thymines have to be adjacent to each other on the same DNA strand.) The known %
GC content of each molecule allows us to calculate the percentage of each of the four
bases.
% Content
Molecule
I
II
III
G
C
A
T
25
20
15
25
20
15
25
30
35
25
30
35
Therefore, DNA molecule I is the least sensitive, while molecule III is the most
sensitive.
24. Frameshift mutations are caused by insertions or deletions of bases (that are not
multiples of 3). These will shift the reading frame for all codons downstream from
the mutation. Single base-substitutions, on the other hand, only affect a single codon.
Therefore, frameshift mutations will have a more deleterious effect than basesubstitution mutations because they affect a larger proportion of amino acids in the
polypeptide.
26. A genetic system that correctly repaired every damaged base would not be advantageous to
an organism. New mutations are necessary to increase genetic diversity within a species.
While it is important that most damaged bases and mutations be repaired, a low mutation
rate is desirable because of the advantages of genetic diversity.
28. IS elements have terminal inverted repeats. The pair in (a) is a direct repeat. The pair
in (b) is inverted; however, it is not complementary. The pair in (c) is
complementary but not inverted. The pair in (d) is complementary and inverted, and
could therefore be found at the ends of an IS element.
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Hyde Chapter 18—Solutions
30. The histidine auxotroph probably contains a deletion. If a few bases are missing,
nothing is available to cause transitions or transversions. It is highly unlikely that the
correct number of missing bases could be spontaneously and correctly inserted.
32. (a) Glu Asp, and (d) Ala Gly could result from a single transversion event.
34. This event would not be considered a mutation because it did not involve a
“permanent” change in the DNA.
36. 1.
2.
3.
4.
5.
6.
7.
8.
False. The his– culture used in the Ames test is that of the bacterium Salmonella
typhimurium.
False. The revertant colonies were formed on agar lacking histidine. (The
bacterial cultures are plated on minimal media.)
True.
False. Tube III approximates what would happen in the human body.
False. Compound A is not mutagenic nor are its metabolites. (The number of
colonies is roughly the same in all three tubes.)
False. Compound B is not mutagenic but its metabolites are. (Tube III has nine times
more colonies than Tube II.)
True.
False. Compound C could be carcinogenic. (Mutagenic agents are potentially,
but not necessarily, carcinogenic.)
38. The two mutagens are base analogs: 2-aminopurine substitutes for adenine during
DNA replication, but may base-pair with cytosine, while 5-bromouracil substitutes
for thymine, but may base-pair with guanine. Therefore, after one round of DNA
replication, the two DNA molecules would be
5’-ATCG-3’
3’-BPGC-5’
and
5’-PBCG-3’
3’-TAGC-5’
The second and third rounds of DNA replication would yield the following
sequences:
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Hyde Chapter 18 – Solutions
5’-ATCG-3’
3’-TAGC-5’
5’-ATCG-3’
3’-TAGC-5’
5’-ATCG-3’
3’-TAGC-5’
5’-ATCG-3’
3’-BPGC-5’
Round II
Round III
5’-GCCG-3’
3’-CGGC-5’
5’-GCCG-3’
3’-BPGC-5’
5’-GCCG-3’
3’-BPGC-5’
5’-PBCG-3’
3’-CGGC-5’
5’-PBCG-3’
3’-CGGC-5’
5’-GCCG-3’
3’-CGGC-5’
5’-PBCG-3’
3’-TAGC-5’
Round II
Round III
5’-ATCG-3’
3’-TAGC-5’
5’-ATCG-3’
3’-TAGC-5’
5’-ATCG-3’
3’-TAGC-5’
40. a.
The mutation rate in this population is the number of new mutations per gamete.
The number of new mutations is 21 – 9 = 12. Two gametes are required to
produce an individual, and therefore the mutation rate is 12/(420,316 2) =
12/840,632 = 1.43 10–5.
The mutation frequency is the number of mutant alleles divided by the total
number of gene copies in this population. (There is no need to differentiate
between new and old mutations.) Therefore, the mutation frequency is
21/840,632 = 2.5 10–5.
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Hyde Chapter 18—Solutions
b. When calculating the mutation rate and frequency we have assumed that the
mutant gene is completely penetrant and that it does not lead to the death of the
individual in utero. These assumptions may not be true, and therefore the
estimate would be inaccurate.
42. a.
Let X* = mutant X chromosome in irradiated male fly. The parental
cross is thus: X*Y XX. This will produce an F1 offspring of 1/2 X*X:1/2 XY,
or 1 female:1 male.
b. The F1 cross (X*X XY) produces the following progeny: 1/4 X*X:1/4
XX:1/4 X*Y:1/4 XY. The X*Y males will die in utero because they are
hemizygous for an X-linked recessive lethal mutation. Therefore, the female-tomale ratio in the F2 will be 2:1.
44. The preferential binding of the Tn10 transposase to hemi-methylated DNA is similar
to the binding of the MutS and MutL proteins to hemi-methylated sequences during
mismatch repair. This mechanism ensures that the Tn10 element has just been
replicated, and the replication machinery is nearby and available to assist in the
transposition of the Tn10 element.
46. The wild-type DNA sequence of this mRNA is (with the coding strand on top):
5’-CCGAC ATG TGG ACA AGT GAA CCG TCA GCA TAA GCACG-3’
3’-GGCTG TAC ACC TGT TCA CTT GGC AGT CGT ATT CGTGC-5’
Proflavin is an intercalating agent that causes insertion or deletion mutations in
DNA. It typically causes frameshift mutations. The fact that there was a silent
mutation with only a single base difference between the mutant and wild-type DNA
sequences suggests that the insertion or deletion affected the stop codon TAA. An
insertion of a G at the second or third position of this stop codon would produce
TGA or TAG stop codons, respectively. Insertion of an A at the second or third
position of the stop codon would produce TAA stop codons. Insertion of a T at the
second or third position of this stop codon would produce TAA or TAG stop codons,
respectively. A deletion could also explain the proflavin-induced mutation. The base
immediately following the TAA codon is a G. Thus, a deletion of either A would
produce a TAG stop codon.
48. a.
The probability that either gene will undergo a mutation in a single generation is
given by the “or” (sum) rule of probability: P(mutation in T) + P(mutation in Z)
= (7 10–5) + (3 10–6) = 0.000073 = 7.3 10–5.
b. The probability that gene T does not undergo a mutation in one generation is 1 –
(7 10–5) = 0.99993. Similarly, the probability that gene Z does not undergo a
mutation in one generation is 1 – (3 10–6) = 0.999997. Therefore, the
probability that neither gene will undergo a mutation in a single generation is
given by the “and” (product) rule of probability: P(no mutation in T) P(no
mutation in Z) = 0.99993 0.999997 = 0.999927 = 9.99927 10–1.
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Hyde Chapter 18 – Solutions
50. There are six mRNA codons specifying arginine: 5'-CGU-3', 5'-CGC-3', 5'-CGA-3',
5'-CGG-3', 5'-AGA-3', and 5'-AGG-3'. The wild-type arginine codon could not be 5'AGA-3' or 5'-AGG-3' because the 5' A base would have to undergo a transversion in
order for the arginine codon to be converted to a stop codon in two steps. Similarly,
the wild-type arginine codon could not be 5'-CGU-3' or 5'-CGC-3' because the 3'
pyrimidine base would have to undergo a transversion before the arginine codon can
be converted to a stop codon. Therefore, the only possibilities for the arginine codon
are 5'-CGA-3' and 5'-CGG-3'. The missing amino acids and stop codons are given in
the following table:
Arginine Codon
CGA
CGG
52. a.
Mutant Codon (Encoded Amino Acid)
CAA (glutamine)
UGG (tryptophan)
UGG (tryptophan)
CAG (glutamine)
CGA (arginine)
Stop Codon
UAA
UAG
UGA
UAG
UGA
We can begin by writing the sequence of the wild-type mRNA, using the
symbols N = any nucleotide, R = any purine (A or G), and Y = any pyrimidine
(C or U).
Codon
1
Wild type =
Codon =
2
3
4
5
6
7
8
9
10
Met Tyr Ile Thr Trp Asp Glu Pro Val Lys Stop
AUG UAY AUY ACN UGG GAY GAR CCN GUN AAR UAR
AUA
UGA
We can now examine each mutant and methodically try to determine the identity
of the N, Y, and R bases.
Mutant 1: The only difference between this mutant and the wild-type sequence
is the presence of an isoleucine instead of a lysine at position number 10. This is
the result of a missense mutation that transformed an AA(A/G) triplet into
AU(A/C/U). The only possibility for the single base difference is an AAA AUA point mutation. This can be caused by an AT-to-TA transversion in the
DNA. Therefore, the wild-type lysine codon (10) is AAA.
Mutant 2: The amino acid sequence shows three missense mutations (codons
4–6), which are immediately followed by a premature termination signal. This
suggests the presence of a frameshift mutation in the DNA near codons 3 and 4.
A comparison of the mutant and wild-type mRNA sequences should reveal
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Hyde Chapter 18—Solutions
which single mutation (addition or deletion) can explain these amino acid
differences.
Codon
1
2
3
4
5
6
7
8
9
10
Wild type =
Codon =
Met Tyr Ile Thr Trp Asp Glu Pro Val Lys Stop
AUG UAY AUY ACN UGG GAY GAR CCN GUN AAA UAR
AUA
UGA
Mutant 2 =
Codon =
Met Tyr Ile His Val Gly Stop
AUG UAY AUY CAY GUN GGN UAR
AUA
UGA
Let’s closely examine the wild-type sequence of codons 4–6 = ACN UGG
GAY. If the A in codon 4 is deleted, the resulting codons 4 and 5 would be CNU
GGG. CNU can be rewritten as CAY, if N = A, and so it will specify histidine
(amino acid 4 in mutant 2). However, the codon GGG specifies glycine and not
valine (amino acid 5 in mutant 2). Therefore, the deletion of the A cannot cause
mutant 2. A similar reasoning demonstrates that the deletions of the second or
third bases of the wild-type codon 4 (ACN) cannot account for the amino acid
differences between mutant 2 and the wild-type protein. Therefore, mutant 2
must have been the result of an insertion. The mRNA sequence of mutant 2
shows the first base in its codon 4 is a C. This suggests that the insertion of a C
between the wild-type codons 3 and 4 could account for mutant 2. Inserting a C
in that position yields the following mutant mRNA sequence:
Codon
1
Codon =
2
3
4
5
6
7
8
9
10
AUG UAY AUY CAC NUG GGA YGA RCC NGU NAA AUA
A
G
This sequence can indeed explain the origin of mutant 2. The new codon 4,
CAC, specifies histidine. Codon 5, NUG, would encode valine, if N = G. GGA
(codon 6) specifies glycine. Codon 7, YGA, can be a termination signal, if Y =
U. Therefore, mutant 2 resulted from the insertion of a C between codons 3 and
4 in the wild-type DNA.
Mutant 3: The amino acid sequence shows three missense mutations
(codons 6–8), which are immediately followed by a premature termination
signal. This suggests the presence of a frameshift mutation in the DNA near
codons 5 and 6. A comparison of the mutant and wild-type mRNA sequences
should reveal which single mutation (addition or deletion) can explain these
amino acid differences.
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Hyde Chapter 18 – Solutions
Codon
1
2
3
4
5
6
7
8
9
10
Wild type =
Codon =
Met Tyr Ile Thr Trp Asp Glu Pro Val Lys Stop
AUG UAY AUY ACG UGG GAU GAR CCN GUN AAA UAR
AUA
UGA
Mutant 3 =
Codon =
Met Tyr Ile Thr Trp Met Asn Leu Stop
AUG UAY AUY ACN UGG AUG AAY CUN UAR
AUA
UUR UGA
(Note: The identity of the underlined bases in the wild-type sequence was
obtained from the sequences of mutants 1 and 2.)
Let’s closely examine the wild-type sequence of codons 6–8 = GAU GAR
CCN. Any single base insertion in codon 6, or between codons 5 and 6, would
generate an immediate stop codon (UGA) at position 7. Since mutant 3 is longer
than six amino acids, it could not have resulted from a base insertion. Therefore,
mutant 3 must have been produced by a deletion in the wild-type DNA
sequence. Indeed, the deletion of any of the two G’s in codon 5 or the G in
codon 6 would account for the mRNA sequence of mutant 3. If such a deletion
occurs, the mRNA and amino acid sequences of mutant 3 would be
Codon
1
2
3
4
5
6
7
8
9
10
AUG UAY AUY ACG UGG AUG ARC CNG UNA AAU
A
Met Tyr Ile Thr Trp Met Asn Leu Stop
The new codon 6, AUG, specifies methionine. Codon 7, ARC, would encode
asparagine, if R = A. CNG (codon 8) would specify leucine, if N = U. Codon 9,
UNA, can be a termination signal, if N = A or G. Therefore, mutant 3 resulted
from the deletion of one of the three Gs in codons 5 and 6 in the wild-type DNA.
b. Based on the sequences of mutants 1–3, the wild-type mRNA sequence is
Codon
1
Codon =
2
3
4
5
6
7
8
9
10
AUG UAC AUY ACG UGG GAU GAA CCU GUA AAA UAR
U
A
G
GA
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Hyde Chapter 18—Solutions
CHAPTER INTEGRATION PROBLEM
a.
+
The 30-bp DNA fragment shows the sequence of codons 136–145 of the fb gene.
+
The schematic of the fb gene is shown here (with distances in base pairs):
5
153
375
UTR E1
111
1971
290
672
179
I1
E2
I2
E3
UTR
3
+
The entire coding sequence of the fb gene is 375 (exon 1) + 1971 (exon 2) + 672
(exon 3) = 3018 bp. Because each 3 bp constitute a codon, this coding sequence
corresponds to 3018/3 = 1006 codons (including the start and stop codons). Codons
136–145 must be located in exon 2, and so the 30-bp sequence should not contain a
stop codon in its coding or nontemplate strand. Let’s start by examining the top
strand. This strand contains termination codons in both directions: a TGA (codon 5,
going from left to right), and a TAA (codon 3, going from right to left). Therefore,
the top strand cannot be the coding strand, and so it must be the template strand.
Examination of the bottom strand reveals a TAG stop codon (number 2) from right
to left but none from left to right. Therefore, the bottom strand is the coding strand
and its polarity is 5' 3' from left to right. The polarity of the double-stranded DNA
sequence is shown here:
3 -GAC CTT CGA GTT TGA AAA AAA AAT CTA ATG-5
5 -CTG GAA GCT CAA ACT TTT TTT TTA GAT TAC-3
The top strand is the template strand and is transcribed from left to right.
b. Remember that the mRNA sequence is identical to the DNA coding strand with U’s
replacing T’s. Therefore, the mRNA and polypeptide sequences of codons 136–145 of the
+
fb gene are
5 -CUG GAA GCU CAA ACU UUU UUU UUA GAU UAC-3
Leu Glu Ala Gln Thr Phe Phe Leu Asp Tyr
c.
There are 30 bases in the coding sequence. Each of these can undergo three different basesubstitution mutations. For example, the first C (DNA coding strand) can be substituted for a
T (transition) or either an A or a G (transversions). Therefore, 30 3 = 90 different single
base substitutions are possible.
d. The DNA sequence contains nine consecutive AT base pairs. This repeated sequence is
prone to addition and deletion mutations because of potential misalignment of the template
and daughter strands during DNA replication (refer to figure 18.16). In addition, the run of
T’s on the bottom strand makes the DNA prone to the formation of thymine dimers.
Hyde Chapter 18 – Solutions
e.
i.
145
Codon 145 is 5'-TAC-3'. The deletion of the C makes the new codon 5'-TAN-3', where
N is the base immediately downstream of the deleted C. N can be an A, C, G, or T. For
5'-TAN-3' to be a termination codon, N would have to be either an A or a G (5'-TAA-3'
and 5'-TAG-3' are two of the three stop codons). Because the four bases are found in
equal frequency, the probability that N is A or G is 50%. Therefore, the probability that
the deletion of C will generate a translation termination signal in codon 145 is 1/2.
ii. The insertion of a G–C base pair immediately following codon 145 creates a new
codon 146 that has the sequence 5'-GNN-3'. None of the three stop codons has a G as its
5' base. Therefore, the probability that the new codon 146 is a termination triplet is 0.
Let’s now consider the new codon 147. We don’t know anything about its sequence.
However, we do know that the insertion of the G–C base pair produced a random codon
sequence downstream. We also know that there are 64 possible codons, three of which
signal the end of protein synthesis. Therefore, the probability that codon 147 is a
termination signal is 3/64, or approximately 4.7%.
f.
i.
The DNA sequence in question is that of codon 136:
3 -GAC-5
5 -CTG-3
Tautomerization to the imino form of the A will make it pair like a G, causing an A-toG transition at that spot. Consequently, there will be a T-to-C transition in the
nontemplate or coding strand. The mutant DNA sequence would be
3 -GGC-5
5 -CCG-3
The transcribed mRNA codon would be 5'-CCG-3', which encodes the amino acid
proline. Therefore, the mutation is a missense mutation at the level of the protein.
ii. The DNA sequence in question is that of codon 138:
3 -CGA-5
5 -GCT-3
Tautomerization to the enol form of the T will make it pair like a C, causing a T-to-C
transition at that spot. Consequently, there will be an A-to-G transition in the template
or transcribed strand. The mutant DNA sequence would be
3 -CGG-5
5 -GCC-3
The transcribed mRNA codon would be 5'-GCC-3', which also encodes the amino acid
alanine. Therefore, the mutation is a silent mutation at the level of the protein.
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Hyde Chapter 18—Solutions
iii. The DNA sequence in question is that of codon 139:
3 -GTT-5
5 -CAA-3
Deamination of the C will convert it to a U, thereby causing a C-to-T transition at that
spot. Consequently, there will be an G-to-A transition in the template or transcribed
strand. The mutant DNA sequence would be
3 -ATT-5
5 -TAA-3
The transcribed mRNA codon would be 5'-UAA-3', which is a termination signal.
Therefore, the mutation is a nonsense mutation that would cause a premature end to
protein synthesis.
+
iv. Nucleotide number –173 is located in the promoter region of the fb gene, upstream of
the transcribed sequence. It is likely not within any of the known conserved elements of
promoters. Therefore, the A-to-G transition will have no effect on the mRNA or its
encoded protein.
v.
The triplet TGA encodes a stop codon. However, it will cause premature termination
only if it is inserted in the correct reading frame (that is, in between two wild-type
codons). Therefore, to find the effect of this insertion, we have to determine the exact
location of nucleotide +208.
First, it is important to remember that the first transcribed base is given the
designation +1. With this in mind, let us determine the exact locations of all the regions
of the fb+ gene transcript. The 5'-UTR (5' untranslated region) is 153 bases, and so it
will stretch from nucleotides +1 to +153. Therefore, exon 1 will begin at nucleotide
+154 and end 375 bases later at +528. Similar reasoning yields the following table:
Region
5' Untranslated region (5'-UTR)
Exon 1
Intron 1
Exon 2
Intron 2
Exon 3
3' untranslated region (3'-UTR)
Nucleotides (Coding Strand)
+1 to +153
+154 to +528
+529 to +639
+640 to +2610
+2611 to +2900
+2901 to +3572
+3573 to +3751
Note that the translation initiation codon spans nucleotides +154 to +156, while the
termination codon is located at nucleotides +3570 to +3572.
Now back to nucleotide +208. It is located 208 – 156 = 52 bases downstream of the
translation initiation codon. This is equivalent to 52/3 = 17.33 codons. This simply
Hyde Chapter 18 – Solutions
147
means that there are 17 full codons between that nucleotide and the translational start
site, and so nucleotide +208 is the 5' end of the very next codon (number 19, including
the start codon). Therefore, insertion of the TGA triplet after this nucleotide will not
cause premature termination of protein synthesis. However, this insertion will alter the
meaning of the next two codons, after which the correct reading frame is maintained. It
does not matter what the identities of bases +208 to +210 are. The insertion of TGA
between bases 208 and 209 will only influence this codon and the next one. Moreover,
this insertion cannot generate a premature termination signal. It is hard to determine the
effect of this mutation on the wild-type protein. The mutation adds one amino acid,
while potentially modifying another. If these changes are in the active site of the
protein, its function may be significantly affected.
vi. Nucleotides +529 to +538 constitute the first 10 bases of intron 1. Their deletion will
not affect the primary mRNA transcript. However, because the deleted region contains
conserved sequences required for intron removal, intron 1 cannot be excised from the
primary transcript. This will most likely produce a nonfunctional Fb+ protein.
vii. Nucleotides +1840 to +1842 fall in exon 2. The TTG to CTG conversion is a transition
at the level of the DNA. If the three nucleotides constitute a single codon, the mutation
would be silent at the level of the protein since both TTG and CTG encode leucine.
However, the three nucleotides may span two different codons, and so the change may
affect the protein. Let’s determine the exact location of these nucleotides.
Base +1840 falls 1840 – 639 = 1201 bases downstream of the intron 1–exon 2
junction. This is equivalent to 1201/3 = 400.33 codons. This simply means that there are
400 full codons between that nucleotide and the start of exon 2, and so nucleotide +1840
is the 5' end of the very next codon. Therefore, nucleotides +1840 to +1842 do make up
a single codon, and thus the TTG to CTG mutation has no effect on the Fb+ protein.
viii. Nucleotide +3570 is the 5' base of the stop codon, so it is not a coincidence that it is a T.
(Remember the three stop codons all start with a T: 5'-TAA-3', 5'-TAG-3', 5'-TGA-3'.)
Therefore, a T-to-C transition in the DNA coding strand will convert the stop codon
into a sense codon, which can only encode one of two amino acids: glutamine (5'-CAA3' and 5'-CAG-3') or arginine (5'-CGA-3'). Regardless, this mutation will cause run-on
protein synthesis. The result is a longer and most likely nonfunctional Fb+ protein.
g.
Mutant 1. The deletion of nucleotides –35 to –17 removed the TATA (Hogness) box which
is usually centered at –25. This box is important for promoter recognition by RNA
polymerase and its associated transcription factors. Therefore, the fb+ gene in this mutant
could not be transcribed resulting in the absence of a sense of humor.
Mutant 2. The deletion contains nine base pairs. Nucleotide +760 falls 760 – 639 =
121 bases downstream of the intron 1–exon 2 junction. This is equivalent to 121/3 =
40.33 codons. This simply means that there are 40 full codons between that nucleotide and
the start of exon 2, and so nucleotide +760 is the 5' end of the very next codon. Therefore,
nucleotides +760 to +768 account for three consecutive codons, and so the 9-bp deletion
does not cause a frameshift mutation. The fact that the sense of humor of this mutant is OK
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Hyde Chapter 18—Solutions
suggests that the missing amino acids are not critical for the function of the Fb+ protein.
They may be in a region of the protein that is distant from the active site.
Mutant 3. Nucleotides +1005 to +1007 fall in exon 2. The CTC-to-CTA conversion is a
transversion at the level of the DNA. If the three nucleotides constitute a single codon, the
mutation at nucleotide +1007 would be silent at the level of the protein since both CTC and
CTA encode leucine. However, there is no sense of humor in this mutant, so the mutation
must have had an effect at the level of the protein. Nucleotide +1005 is located 1005 – 639 =
366 bases into exon 2. This corresponds to the 3' base of the 122nd codon in exon 2.
Therefore, nucleotide +1007 occupies the middle position of the 123rd codon. The wild-type
codon must be 5'-TCN-3', which encodes serine. The mutant codon has the sequence
5'-TAN-3'. If the N is an A or G, this would create a premature termination codon. This in
turn would produce a much shorter and nonfunctional Fb protein and therefore no sense of
humor in the mutant.
Mutant 4. There are 2440 – 639 = 1801 bases or 1801/3 = 600.33 codons between nucleotide
+2440 and the beginning of exon 2. Therefore, nucleotides +2440 to +2442 constitute codon
number 601 of exon 2. The AAG-to-GAG conversion is a transition at the level of the DNA,
and a missense (lysine to glutamic acid) at the level of the protein. This particular amino
acid must be critical to the function of the wild-type Fb protein. Changing the amino acid
from a lysine (positively charged) to a glutamic acid (negatively charged) must have
disrupted the function of the protein. This produced a mutant with no sense of humor.
Mutant 5. Nucleotide +3558 is located at a distance of 15 nucleotides from the end of exon
3. Therefore, nucleotides +3558 to +360 make up a single codon. The CGA-to-TGA
conversion is a nonsense mutation, changing an arginine codon into a stop signal. While this
would cause a premature termination of translation, the Fb protein would only be shortened
by four amino acids. The wild-type protein has a primary structure of 1005 amino acids, and
so the mutant protein would be 1001 amino acids long. This mutant protein is still capable of
folding into its active 3-D structure and still able to perform its function. The missing amino
acids must not have a noticeable effect on the protein’s activity, and thus the mutant
individual still possesses a sense of humor.
h. The wild-type nucleotide +692 is located 692 – 639 = 53 bases into exon 2. But what is
it? Well, the exact nature of this base can be deduced by analysis of the codon structure
of the fb+ gene.
Exon 1 contains 375 bp/3 bp = 125 codons, and so exon 2 begins with codon 126.
But exon 2 starts at nucleotide +640; therefore, codon 126 has nucleotide +640 at its
5' end. Nucleotide +692 is 53 bases downstream of the start of exon 2, or 53/3 = 17.66
codons. This simply means that nucleotide +692 is the second base of the 18th codon in
exon 2. This particular codon is actually the 143rd codon in the fb+ gene (125 from exon
1 plus 18 from exon 2). The 30-bp DNA sequence provided in this problem spans
codons 136–145. Therefore, nucleotide +692 is the middle T of the codon 5'-TTA-3'
(coding strand of DNA).
Hyde Chapter 18 – Solutions
i.
149
Transversion of this T produces an A or a G, resulting in either a 5'-TAA-3' or a
5'-TGA-3' codon. Both of these are nonsense codons which cause the synthesis of the
Fb+ protein to terminate prematurely. The mutant protein would only contain 142 out of
the wild-type 1005 amino acids and would surely be nonfunctional. Therefore,
unfortunately, Zorkan Tarazan lacks a sense of humor!
Zorkan and Zelda are homozygous for the transversion mutant and wild-type alleles,
respectively. Their child Zultan will inherit each of these two alleles and will therefore
be heterozygous. He in turn will have a 50:50 chance of transmitting the mutant allele to
his son. Therefore, the probability that Zultan’s son will inherit the transversion mutant
allele from his grandfather Zorkan is 1/2.
For Zoltan’s son to be homozygous for the mutant transversion allele, Zoltan’s wife
would have to be a “carrier” for this allele (P = 1/12,500), and she would also have to
transmit it to her son (P = 1/2). The probability that Zoltan will transmit the mutant allele
to his son is 1/2. Therefore, the probability that Zoltan’s son will be homozygous for the
transversion mutant allele is 1/2 1/12,500 1/2 = 1/50,000.