DNA MUTATION, REPAIR, AND TRANSPOSITION

Transcription

1 Hyde Chapter 18 Solutions 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 ( NH 2 ) 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

2 136 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 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 F 1 hybrid offspring are sterile. On the other hand, the reciprocal cross between an M male and a P female yields normal F 1 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 F 1 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 F 1 hybrid, causing it to become sterile.

3 Hyde Chapter 18 Solutions 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 G C A T I II III 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.

4 138 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 False. The his culture used in the Ames test is that of the bacterium Salmonella typhimurium. 2. False. The revertant colonies were formed on agar lacking histidine. (The bacterial cultures are plated on minimal media.) 3. True. 4. False. Tube III approximates what would happen in the human body. 5. False. Compound A is not mutagenic nor are its metabolites. (The number of colonies is roughly the same in all three tubes.) 6. False. Compound B is not mutagenic but its metabolites are. (Tube III has nine times more colonies than Tube II.) 7. True. 8. 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 and 5 -PBCG-3 3 -BPGC-5 The second and third rounds of DNA replication would yield the following sequences:

5 Hyde Chapter 18 Solutions ATCG-3 3 -BPGC-5 Round II 5 -ATCG-3 5 -GCCG-3 3 -BPGC-5 Round III 5 -ATCG-3 5 -ATCG-3 5 -GCCG-3 3 -CGGC-5 5 -GCCG-3 3 -BPGC-5 5 -PBCG-3 Round II 5 -PBCG-3 3 -CGGC-5 5 -ATCG-3 Round III 5 -PBCG-3 3 -CGGC-5 5 -GCCG-3 3 -CGGC-5 5 -ATCG-3 5 -ATCG 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 = 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 =

6 140 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 F 1 offspring of 1/2 X*X:1/2 XY, or 1 female:1 male. b. The F 1 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 F 2 will be 2: 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) = ( ) + ( ) = = b. The probability that gene T does not undergo a mutation in one generation is 1 ( ) = Similarly, the probability that gene Z does not undergo a mutation in one generation is 1 ( ) = 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) = = =

7 Hyde Chapter 18 Solutions 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 Mutant Codon (Encoded Amino Acid) Stop Codon CGA CAA (glutamine) UAA CGG UGG (tryptophan) UAG UGG (tryptophan) UGA CAG (glutamine) UAG CGA (arginine) UGA 52. a. 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 Wild type = Codon = 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

8 142 Hyde Chapter 18 Solutions which single mutation (addition or deletion) can explain these amino acid differences. Codon Wild type = Codon = Mutant 2 = 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 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 Codon = 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.

9 Hyde Chapter 18 Solutions 143 Codon Wild type = Codon = Mutant 3 = 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 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 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 Codon = AUG UAC AUY ACG UGG GAU GAA CCU GUA AAA UAR U A G GA

10 144 Hyde Chapter 18 Solutions CHAPTER INTEGRATION PROBLEM a. The 30-bp DNA fragment shows the sequence of codons of the fb + gene. The schematic of the fb + gene is shown here (with distances in base pairs): UTR E1 I1 E2 I2 E3 UTR The entire coding sequence of the fb + gene is 375 (exon 1) (exon 2) (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 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 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.

11 Hyde Chapter 18 Solutions 145 e. i. 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-to- G 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.

12 146 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 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 Therefore, exon 1 will begin at nucleotide +154 and end 375 bases later at Similar reasoning yields the following table: Region Nucleotides (Coding Strand) 5' Untranslated region (5'-UTR) +1 to +153 Exon to +528 Intron to +639 Exon to Intron to Exon to ' untranslated region (3'-UTR) to Note that the translation initiation codon spans nucleotides +154 to +156, while the termination codon is located at nucleotides to Now back to nucleotide It is located = 52 bases downstream of the translation initiation codon. This is equivalent to 52/3 = codons. This simply

13 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 to 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 falls = 1201 bases downstream of the intron 1 exon 2 junction. This is equivalent to 1201/3 = codons. This simply means that there are 400 full codons between that nucleotide and the start of exon 2, and so nucleotide is the 5' end of the very next codon. Therefore, nucleotides to do make up a single codon, and thus the TTG to CTG mutation has no effect on the Fb + protein. viii. Nucleotide 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'-CAA- 3' 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 = 121 bases downstream of the intron 1 exon 2 junction. This is equivalent to 121/3 = 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

14 148 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 to 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 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 is located = 366 bases into exon 2. This corresponds to the 3' base of the 122nd codon in exon 2. Therefore, nucleotide 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 = 1801 bases or 1801/3 = codons between nucleotide and the beginning of exon 2. Therefore, nucleotides to 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 is located at a distance of 15 nucleotides from the end of exon 3. Therefore, nucleotides 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 = 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 = 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 Therefore, nucleotide +692 is the middle T of the codon 5'-TTA-3' (coding strand of DNA).

15 Hyde Chapter 18 Solutions 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! i. 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.