i. This type of DNA damage will result in (circle all correct answers) b. a transversion mutation

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1 Biol 321 Winter 2010 Quiz 5 40 points NAME ANSWERS IN RED COMMENTS IN BLUE 1. (2 pts.) Recall the article entitled: Human Genetic Variation By each statement circle True/False/Not addressed in the paper. Answer false if any part of the statement is false. False This article states that copy number variations (CNVs) have turned out to be very unusual in humans. True People with high starch diets such as in Japan have extra copies of a gene encoding a starch-digesting enzyme as compared with members of hunter-gatherer societies. 2. (4 pts.) Circle True or False. If there are two statements, the first statement is true and you are to decide whether the second statement is True or False. No credit if no explanation. True Both spontaneous and induced mutations are random events. For this reason, most nonneutral mutations in the proto-oncogene class of genes will not result in cancer-inducing (driver) alleles. One sentence explanation: Most mutations will not be driver mutations since they will likely cause a loss-of-function rather than a gain-of-function. REMEMBER: when you take a sledgehammer to a gene, most of the time you will break it completely. Only gf mutations convert proto-oncogenes to oncogenes. False The tumor suppressor class of genes (often found mutated in cancer cells) are named for their loss-of-function phenotype. One sentence explanation: The name describes the normal wild-type function of the gene product: to reduce the rate of cell division or slow the movement of the cell through the cell cycle. 3. (3 pts) Reactive oxygen species such as hydrogen peroxide and hydroxyl radicals are produced as by-products of normal aerobic metabolism. Such species can cause damage to the bases of DNA. Consider an oxidative lesion to guanine (G ox ). G ox can pair with adenine or cytosine. Based on this information, what kind of mutation(s) can result from the oxidation of a guanine either before or after it is incorporated into a DNA polymer? NOTE: to get the correct answer you should draw out the pairing events. I noticed many student answered this problem with out sketching out any rounds of replication or pairing between G ox and A. i. This type of DNA damage will result in (circle all correct answers) b. a transversion mutation ii. This type of DNA damage will result in the following point mutations (circle all correct answers) a. GC TA A guanine already incorporated into a DNA strand is damaged and pairs with A in the next round of replication the A then pairs with T b. AT CG A free guanine (or in the dgtp form) is damaged. It is incorporated into a new DNA strand opposite an A. The G* then pairs with C in the next round of rep 1

2 4. (17 pts.) Huntington disease (HD) is caused by a variable expressed but fully penetrant autosomal dominant mutation that causes late onset (post-reproductive) neurodegeneration. The mutations that cause HD involve an expansion of a triplet repeat located in the coding region of the gene (HTT --see below). Normal alleles of this gene have CAG repeats. Alleles with >40 repeats confer a clear HD phenotype. Interestingly, as the number of repeated triplets increases, the age of onset in the patient decreases. Furthermore, because the unstable trinucleotide repeat can lengthen when passed from parent to child, the age of onset can decrease from one generation to the next. As discussed in lecture, by convention, a gene s sequence is always presented 5 to 3 as the mrna sequence would read Homo sapiens huntingtin (HTT), mrna (in cdna language) >gi ref NM_ Homo sapiens huntingtin (HTT), mrna GCTGCCGGGACGGGTCCAAGATGGACGGCCGCTCAGGTTCTGCTTTTACCTGCGGCCCAGAGCCCCATTC ATTGCCCCGGTGCTGAGCGGCGCCGCGAGTCGGCCCGAGGCCTCCGGGGACTGCCGTGCCGGGCGGGAGA CCGCCATGGCGACCCTGGAAAAGC TGATGAAGGCCTTCGAGTCCCTCAAGTCCTTCCAGCAGCAGCAGCA GCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAGCAACAGCCGCCACCGCCGCCG CCGCCGCCGCCGCCTCCTCAGCTTCCTCAGCCGCCGCCGCAGGCACAGCCGCTGCTGCCTCAG CCGCAGC CGCCCCCGCCGCCGCCCCCGCCGCCACCCGGCCCGGCTGTGGCTGAGGAGCCGCTGCACCGACCAAAGAA AGAACTTTCAGCTACCAAGAAAGACCGTGTGAATCATTGTCTGACAATATGTGAAAACATAGTGGCACAG TCTGTCAGAAATTCTCCAGAATTTCAGAAACTTCTGGGCATCGCTATGGAACTTTTTCTGCTGTGCAGTG ATGACGCAGAGTCAGATGTCAGGATGGTGGCTGACGAATGCCTCAACAAAGTTATCAAAGCTTTGATGGA a. (2 pts.) The 5 end of the huntintin mrna is shown above. Note that by convention all database sequence files are in DNA language. Circle the translation start codon. See ANSWER POSTED FOR take home quiz b. (4 pts. all or nothing since one correct and one incorrect primer will not give you a partial PCR product ) You are planning to set up PCR to amplify the shaded region between the s. The PCR product should include all of (but not extend beyond) the designated region. List the first five bases of each primer. Your primer sequences must read in the 5 to 3 direction. Primer A: 5 TGATG 3 Primer B: 5 CTGAG 3 c. (1pt.) DNA polymerase is a special needs enzyme. What property common to all DNA polymerases is critical to the amplification specificity of a PCR? One complete sentence. DNA polymerases require a correctly based-pair 3 end in the form of a primer base-paired to the template in order to initiate DNA synthesis d. (2 pts.) Will an expansion or contraction of the number of CAG repeats result in a frameshift at the protein level: NO One sentence defense of your answer (no credit if no explanation): An mrna is translated 3 bases at a time. The addition or deletion of 3 contiguous bases will add or delete an amino acid but will not shift the reading frame. 2

3 e. (2 pts.) On the sequence below, label the strand that acts as the template during transcription template strand in red f. (6 pts.) DRAW a clearly labelled diagram to show how an increase of one repeat unit can occur as a spontaneous mistake during DNA replication: Be sure to label the 5 and 3 ends of all DNA strands. Be sure to label the parental and daughter DNA strands. Only show relevant parts of sequence Hint: the initial mistake can occur during the replication of either parental strand so follow the replication of just one of the parentals PARENTAL DNA CTCAAGTCCTTCCAGCAGCAGCAG GAGTTCAGGAAGGTCGTCGTCGTC Lower strand is the template for RNA polymerase during transcription Upper strand is the mrna like strand see previous page See drawings on pg 64 of 2/10/10 lecture To get an increase in one repeat unit, the daughter and parental (template) strands must slip relative to each other during DNA replication generating an unpaired loop equivalent to one repeat unit in the daughter strand. In the next round of replication the strand with the extra repeat unit will serve as a template for DNA polymerase and the resulting product with have an extra repeat in both strands. 3

4 5. (14 pts.) See pedigree and PCR data on the extra sheet. Both families shown have mutations in the GATA gene. The affected individuals have similar symptoms including heart defects, immune deficiencies, deafness and renal malformations..the data for genotyped individuals is shown directly below their pedigree symbol. N1 and N2 represent normal control individuals. A codon table is on the extra sheet. a. (1 pt.) Examine panel a. At the protein level, what type of mutation is seen in this family? One word. No explanation. Frameshift (deletion of 49 bp) b. (1 pt.) Answer the same question for family 12/99. Nonsense: CGA (ARG) UGA (STOP) Circle True/False/Not enough info to decide. 1 pt if no explanation is required. 2 pts if an explanation is required. For the latter, no credit given if there is no explanation COMMENTS: To assess these statements you must look at the inheritance patterns (pedigrees) AND the PCR/agarose gel data. One tells you about phenotype and the other directly assesses genotype. TRUE The mutation in Family 12/99 could have been originally generated by environmental exposure to a base analog mutagen such as 2 aminopurine. Base analogs typically cause transition mutations. TRUE In family 12/99, the mutant allele is codominant at the molecular level of assessment but dominant (completely or incompletely) at the organismal level of assessment. All affected individuals are heterozygous so the mutant allele must be dominant to the wildtype allele. FALSE The GATA gene is likely to be X-linked. One sentence defense of your answer: Affected males are heterozygous so the GATA gene must be autosomal. If X linked, a male would be hemizygous for either the mutant or the wildtype allele. FALSE In both families the mutant alleles are likely to result in a gain-of-function. Frameshift and nonsense mutations typically result in a loss of gene function. FALSE The mutant GATA allele is recessive in Family 26/99 and dominant in Family 12/99 One sentence defense of your answer: All affected individuals in both families are heterozygous for the mutant allele. NOTE in family 26/99 the affected son is carrying a mutation not present in either parent. TRUE Based on the data presented here, GATA mutations are completely penetrant. All heterozygotes for the mutant allele express a mutant phenotype. FALSE In family 12/99, the mutant phenotype results from an extra copy of the chromosome that the GATA gene is located on. 4

5 TRUE Setting niceties aside, if individual II #1 from family 26/99 mates with I #2 from family 12/99, the probability of a normal offspring is 1/4. Both individuals are het for the wildtype allele so there is a 25% probability that they would have an offspring homozygou WT. FALSE If the parents in either family have additional children, there is a 50:50 probability of an affected offspring. One sentence defense of your answer: The statement is true for family 12/99 since the parents are g + g + father X G g + mother (g+= wildtype allele and G = mutant allele) = ½ chance of Gg + offspring. The situation is different for family 26/99 since both parents are homozygous for the g + allele. Their affected son must be the product of a new mutation that occurred in the germline of one of his parents. It is very unlikely that they will have another affected child. [BUT if the mutation occurred early in the development of the germline such as during fetal development of one of the parents, it is possible that the gonad is a mosaic of cells that are g + g + and Gg + and that additional affected children would be produced.] 5