Name KEY Note Total points added up to only 98 points so everyone received 2 free points to make total points 100. Biology 201 (Genetics) Exam #3 23 November 2004 Read the question carefully before answering. Think before you write. You will have up to 85 minutes hour to take this exam. After that, you MUST stop no matter where you are in the exam. If I can not read your handwriting, I will count the question wrong. Sign the honor pledge if applicable. Good luck! I pledge that I have neither given nor received unauthorized assistance during the completion of this work. Signature: Happy thanksgiving! 1
Multiple choice section: (24 points total 4 points per question) Choose the BEST ANSWER for each question. Write your answer in the blank provided to the left. IF you want to explain your answer, you can do so next to the question. E 1. Which of the following examples is most likely to be caused by a somatic mutation? a. A child with a defective UV repair system b. An apple tree in a very large orchard produces all of its apples two weeks earlier than the other trees c. One child in seven in a family has sickle cell anemia d. One child in seven in a family is albino e. A purple flower has a small patch of white tissue A 2. Expression of the meow gene in normal cats causes the cats to meow. When tuna is present, the cats meow, and when tuna is absent, the cats do NOT meow. Transcription of meow gene is regulated by the MeoR protein. Deletion of meor causes loss of meowing (Dr R-J likes those cats). MeoR is a DNA binding protein that binds to a site in the meow promoter called the MeoR box. Deletion of the MeoR box eliminates meowing. Choose the description below that best matched the regulation of meowing: a) Expression of meow is inducible by tuna and positively regulated by MeoR. b) Expression of meow is repressible by tuna and positively regulated by MeoR. c) Expression of meow is inducible by tuna and negatively regulated by MeoR. d) Expression of meow is repressible by tuna and negatively regulated by MeoR. e) Either A or D is possible. Based on concepts in Ch 15 #2, 3, 6, 12 A 3. A recent report in the journal nature indicated that there are only 25,000 genes in humans. However, it is likely that there are more than 25,000 different proteins made in humans. A possible explanation for this is: a. Alternative splicing of mrna b. Regulation of gene expression by steroid hormones c. Regulation of export of mrna from the cytoplasm to the nucleus d. Acetylating of histones to make DNA more accessible for transcription e. All of the above. Question mentioned directly in class 2
B or D 4. Expression of the gene erg1 is induced by the steroid hormone estrogen. You isolate a mutant mouse that no longer produces the protein Erg1. The mutation in this mutant mouse could be a. a. deletion of erg1 enhancer b. mutation of estrogen receptor gene c. deletion of the erg1 repressor d. A and B e. All of the above A 5. Insertion of a plasmid into a bacterium is done by using. a. transformation of bacterium with plasmid via heat shock b. injecting the plasmid into the chromosomal DNA of the bacterium c. injecting the bacterium into the plasmid d. Both A and B Based on expt done in lab e. Either B or C. A 6. A piece of plant DNA is cloned into the SalI site of the plasmid shown below. Bacteria transformed with the correct clone that contains the plant gene will be a. resistant to ampicillin only b. resistant to tetracycline only c. resistant to both antibiotics Based on Chapter 17 #12 and lab expt d. sensitive to both antibiotics e. sensitive to ampicillin only 3
Short answer questions: 10 pts. 1. Briefly explain the four levels of protein structure. In your answer refer to any protein that we have talked about in class as an example (ie CAP, CFTR, b-globin/hemoglobin) where applicable. This question was on the Exam 3 review sheet. Also, you had to think about this topic in Chapter 14 #22. Primary structure is the linear sequence of the amino acids in a protein (ie the linear amino acid sequence of the CAP protein). Secondary structure refers to the localized folding of the protein. In secondary structure, closely aligned amino acids in the polypeptide chain form structures such as alpha helices and beta sheets (ie the alpha helix in CAP that binds to DNA). Tertiary structure refers to the 3-dimensional form that the protein adopts after folding up. This is determined by various atoms from distant amino acids in the protein interacting. CAP folds up into a structure that allows camp to bind in a pocket. Quaternary structure applies to proteins that have multiple subunits and refers to the interactions and arrangements of the subunits, as in the CAP dimer. Note I did NOT deduct points if you chose not to use a particular protein in your explanation. 10 pts. 2. Compare the effects of base substitution/point mutations and frame shift mutation changing a nucleotide in the CFTR gene on the ability to cause the genetic disease cystic fibrosis? This question was partially addressed in Chapter 15 #14. Base substitutions/point mutations change just one nucleotide in the gene sequence, with three possible results. If the mutation is in a wobble position and changes the mutation to a wobble nucleotide, there will be no change in the amino acid in the CFTR protein (thus wildtype CFTR protein and no CF). If the mutation changes the codon so to encode a different amino acid and if that amino acid is NOT important for CFTR function, there will be a change in the amino acid but no change in activity of the CFTR protein (thus functional CFTR protein and no CF). If the mutation changes the codon so to encode a different amino acid and if that amino acid is important for function of the CFTR protein, there will be a change in the amino acid and a defective CFTR protein (thus mutant CFTR protein and CF will result). Frameshift mutations insert 1 or 2 nucleotides into the gene. This changes the reading frame of the entire gene and so all the amino acids encoded after the frameshift will be different. This results in a defective CFTR protein or a truncated CFTR protein if an early stop codon results from the frame shift (thus mutant CFTR protein and CF will result). 4
5 pts. 3. During mismatch repair using the Mut system: This question was addressed in Ch 15 Slide 25. a) why is it necessary to distinguish between the template strand and the newly made daughter strand? The Mut system repairs errors that remain after DNA replication is complete. Since the newly made daughter strand (and not the template) has the mutation, it is critical that the repair system remove the mismatched base from the new strand and not the old strand. If the old strand nucleotide was removed and repair was based on new nucleotide on the new strand, the repair system would actually create a permanent mutation. b) how is this distinguishing between the template strand and the newly made daughter strand accomplished? DNA is methylated at particular DNA sequences after DNA replication is complete. However, immediately after DNA replication, there is a time frame where the newly made strand has not yet been methylated. It is during this time that the Mut system can distinguish the old methylated strand from the new unmethylated strand of DNA. 6 pts. 4. LIST the most likely way that a cell would repair the following types of DNA damage? This question was addressed in Chapter 15 #17 and in slide 24. a) Removal of the NH3 group from the base cytosine in non-dividing cells. Base excision or nucleotide excision repair. b) A change in the DNA sequence caused by DNA polymerase (that was not corrected by the proofreading aspect of DNA polymerase Mismatch repair. c) A thymine-thymine dimmer in the DNA of an actively dividing cell Photoreactivation, nucleotide excision repair, base excision, SOS/post-replication repair. 5
14 pts. 5. Fill in the table below to indicate how lacz gene expression would be affected in each strain after growth in media listed. Note that none of the media listed above has glucose in it. Use the following symbols as needed: + = transcription occurs at totally induced levels - = transcription does not occur at a detectable level w = very weak levels of transcription (higher than but not as high as +) lacz transcription No lactose present Lactose present Wild type - + Lac operator mutant + + Lac promoter mutant - - LacI mutant + + CAP mutant - w Adenylate cyclate mutant * - w LacA mutant - + * hint adenylate cyclase makes camp Based on Ch 16 #4, 5, 7, and lab questions 6 pts. 6. Using two examples, explain how allostery is important in the regulation of gene expression. The concept of was mentioned numerous times in class in the context of transcriptional regulator proteins. Allostery refers to the conformational change in a protein which is brought about by binding of a small molecule to the protein. The conformational change alters the activity of the protein. For example: 1. LacI repressor protein binds to the lac operator and negatively regulates transcription. Binding of lactose to the LacI repressor alters the conformation of the LacI protein so LacI can no longer bind the lac operator, therby allowing transcription of the lac operon to occur. 2. Tryptophan binds to the TrpR repressor protein causing an allosteric change in the TrpR protein so that TrpR can now bind to the trp operator and negatively regulate transcription. In the absence of tryptophan, the TrpR protein is in an inactive conformation and can no longer bind the trp operator; thus transcription of the trp operon occurs. 3. Other examples include allosteric changes in CAP when camp binds to it and allosteric changes in steroid receptor proteins when steroids bind to them. In both cases, the allosteric change upon binding of the small molecule allows CAP or the steroid receptor to bind DNA and function as positive regulators of transcription. 6
10 pts. 7. Analysis of VNTR or STR loci has been used in the past to establish paternity in cases where the biological father of a child is in question. Below are the DNA fingerprints of 5 people in such as case: the child (C), the mother (M), and three potential fathers of the child (F1, F2, F3). In this experiment, the total DNA of each individual was cut with a restriction enzyme, ran on three gels, and Southern blotted. Each blot was probed with a radioactive probes to a different VNTR loci (A, B, and C) and the results are shown below. M C F1 F2 F3 M C F1 F2 F3 M C F1 F2 F3 VNTR A probe VNTR B probe VNTR C probe This problem was similar to Insights and Solutions Problems on page 454-455 and problem #22 on Page 398. These problems were specifically mentioned in class and also emailed to you as problems to attempt. a) Which male or males can be eliminated as the biological father of child C based on these data? F1: In gel A, examination of lane with child s VNTR A alleles shows that the allele that had to be contributed by father (bottom band) is NOT in F1 lane. Thus, F1 can not be the father. Same is true for VNTR B gel. F3: In gel B, examination of lane with child s VNTR alleles shows that the allele that had to be contributed by father (bottom band) is NOT in F3 lane. Thus, F3 can not be the father. b) Which male or males can theoretically be the biological father of child C based on these data? F2 c) Consider that the frequency of each VNTR allele shown above is 1/10 in the general population. What is the chance that the male or males from part B are NOT the biological father of the child C? 1/1000 7
6 pts. 8. What is a microarray? List two uses for microarray technology. This question was similar to Ch 19 #12. A microarray is a glass slide that contains an ordered arrangement (or array) of known DNA molecules. These DNA molecules can be (1) each gene from an organism or (2) alleles of a particular set of genes from an organism. DNA from a particular source can be isolated, fluorescently labeled, and hybridized to the array. If one of the spots on the array becomes fluorescent, it means that the DNA added to the array contains that allele or gene. Some uses of microarrays include: 1. Examining gene expression in an organism in response to two different conditions. 2. Examining DNA for particular alleles, such as disease alleles. 3. Comparing DNA from two different species. 4. I also accepted studying cancer since this was the example from the textbook. 7 pts. 9. The restriction enzyme SmaI recognizes the DNA sequence CCCGGG and cuts the sequence at that point. A portion of the sequence of the normal gene X is CCCGGG. A portion of the sequence of the mutant gene X allele has the sequence CCGGGG. Which allele will the restriction enzyme SmaI recognize and cut? NORMAL gene X The difference in restriction fragments at these alleles is an example of. RFLP (restriction fragment length polymorphism). 8