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Genome 361, Spring 2013 Exam 2 Name Student # Your TA Section Instructions
- There are 12 pages in total. If you do not have all 12 pages, let us know.
- You have 80 minutes to complete the exam.
- Questions 1 - 8 are multiple-choice questions. Please answer them on the bubble sheet.
- Answers to questions 9 - 13 must be written on the exam. Answers written on this cover sheet, the tear-off page, or scratch paper will not be graded.
- The last page is a tear-off sheet.
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Please write your name on the front page.
You may tear off the last page to use it as scratch paper.
Sections: AA Wed 9 :30 – 10:20am Tanya AB Wed 10:30 – 11:20am Tanya AC Wed 11:3 0 – 1 2:20pm Jenny AD Wed 12:30 – 1:20pm Jenny AE Wed 1:30 – 2 :20pm Sergey AF Wed 2:30 – 3:20pm Sergey AG Wed 4:30 – 5:20pm Frances AH Wed 5:30 – 6:20pm Frances
Genome 361 Spring 2013, Exam 2 2 Multiple choice questions (6 pts each; choose only one answer)
- Mutations in the tumor suppressor gene BRCA1 can lead to breast and ovarian cancer. A point mutation in the 5’ untranslated region (5’ UTR) of BRCA1 leads to less efficient translation of the BRCA1 protein. As a result, cells go into uncontrolled proliferation and form tumors. This mutation is most likely: A. A gain-of-function mutation B. A loss-of function mutation C. A nonsense mutation D. A missense mutation E. Both B and C Note that the point mutation occurs in the 5’ untranslated region, not in the protein-coding region, so it won’t be a nonsense or missense mutation. Because the mutation results in less efficient translation of the protein, such that less functional protein will be made, it is a loss-of- function mutation. Remember that loss- or gain-of function mutation is defined by the comparison with the normal properties of the gene, not of the organism or the resulting phenotype. For example, though the cell with the mutated BRCA1 gene goes into uncontrolled proliferation, it is not a gain-of- function in respect to BRCA1. It is a loss-of-function mutation because less functional BRCA protein is made.
- In class we discussed the fact that lactase persistence (or lactose tolerance) is common in populations that have a history of animal domestication. You look at the sequence of the lactase gene in several homozygous individuals. They all produce lactase protein with the same amino acid sequence: Individual Ethnicity Lactase sequence Lactase protein I Northern European A promoter mutation that leads to lactase persistence Fully functional protein; production persists into adulthood. II East African A promoter mutation at a different site from individual 1, which also leads to lactase persistence Fully functional protein; production persists into adulthood. III East Asian No mutation in the promoter or coding region Fully functional protein; production does not persist into adulthood. IV East Asian No mutation in the promoter; a silent mutation in the coding region Fully functional protein; production does not persist into adulthood.
Genome 361 Spring 2013, Exam 2 4 Answer: C Individual 3’s allele produces a single band after digestion by BamHI. The question states that the allele from individual 4 has a new BamHI site in its lactase sequence. So the BamHI digest should cut individual 4’s allele into two smaller fragments, which will poduce two bands on the gel. Individuals 3 and 4 are both homozygous, and their child will receive one allele from each. Therefore the child should have one band from individual 3’s allele and two bands at lower molecular weight from individual 4’s allele.
- The figure to the right shows yeast’s adenine biosynthesis pathway. Recessive lof mutants requiring adenine to grow have been isolated. ade1 mutants are red due to accumulation of the intermediate X, which UNK1 converts to a red pigment. You plate a bunch of cells from an ade1 culture and find a white colony among the red ade1 mutants on a complete plate. This white colony does not grow on a plate without adenine. Which of the following statements about this white colony could be true? A. It has a reversion mutation from ade1 back to ADE1. B. It has acquired a lof mutation in ADE. C. It has acquired a lof mutation in UNK. D. Both B or C are possible. E. A, B or C are all possible.
Genome 361 Spring 2013, Exam 2 5 If the colony had a reversion back to ADE1, it would be able to grow on a plate without adenine. Since the white colony cannot grow on a plate without adenine, it cannot be a reversion back to ADE1. However, a loss-of-function mutation in either ADE3 or UNK1 will turn the colony white, while still leaving it unable to grow without adenine.
- In rabbits, gene R encodes the enzyme tyrosinase, which is involved in the first step of pigment formation. Rabbits with the functional wild-type allele R can express pigment, while rabbits homozygous for the non- functional allele r are pure white. Himalayan rabbits have a temperature sensitive allele rh, such that only their ears and cooler extremities can express pigment. R is dominant over r h , but r h is dominant over r. Gene B controls whether the pigment produced is black or brown. B? rabbits are black, and bb rabbits are brown. You cross a uniformly black rabbit with a white rabbit of genotype bbrr, and you find that 50% of the F1 offspring are uniformly black and 50% of them are black only at their extremities (Himalayan black). You allow the Himalayan black rabbits to cross among each other. What is the expected phenotypic ratio in the next generation? A. 9 black : 3 Himalayan black : 3 Himalayan brown : 1 white B. 9 Himalayan black : 3 Himalayan brown : 4 white C. 3 Himalayan black : 1 white D. 9 black: 3 brown : 4 white E. 9 Himalayan black : 7 Himalayan brown The parental cross is B?R? (uniformly black) x bbrr (white). Since the white parent can only contribute a gamete with genotype br, the uniformly black F1 must have genotype BbRr (B and R from the black parent). The Himalayan black F1 genotype must be Bbrhr (br from the white parent; Brh^ from the black parent). We can now infer that the parental black parent’s genotype is BBRr h . We allow the F1 himalayan black rabbits (Bbr h r) to cross among each other. The resulting F will have the classic 9:3:3:1 ratio of genotype classes: 9 B_r h _ (Himalayan black) 3 bbr h _ (Himalayan brown) 3 B_rr (white) 1 bbrr (white) The resulting phenotypic ratio is 9 Himalayan black : 3 Himalayan brown : 4 white. Himalayan rabbit
Genome 361 Spring 2013, Exam 2 7
- Reginald Punnett, of Punnett square fame, also reported one of the first examples of genetic interaction. Comb morphology in chickens is determined by two loci. A purebred chicken with a rose comb (RRee) was crossed with a rooster with a pea comb (rrEE), giving all F1 offspring with walnut combs (RrEe). When the F1s were crossed together, the progeny segregated in the ratio 9 walnut : 3 rose : 3 pea : 1 single comb. Dr. Imsland is trying to identify the genes for comb morphology. She crosses an F2 rooster with a walnut comb to a chicken with a single comb. She gets chicks in the ratio 1 walnut comb : 1 rose comb. What was the rooster’s genotype? A. RREE B. RREe C. RrEE D. RrEe E. rree Based on the given parental genotypes and the resulting F1 ratio, we can determine the following information: Parental: RRee (rose) x rrEE (pea) F1: RrEe (walnut) x RrEe (walnut) F2: 9 R_E_ (walnut) 3 R_ee (rose) 3 rrE_ (pea) 1 rree (single) The F2 cross performed is therefore R?E? (walnut) x rree (single), resulting in F3 ratio of 1 R?E? (walnut) to 1 R?ee (rose). Since the F2 single comb parent can only contribute re gametes, the walnut F3’s genotype must be RrEe, and the rose F3 must be Rree. We can deduce from the F phenotype ratio that half of the gametes the walnut F2 rooster contributed had the genotype RE, and half had the genotype Re. So the F2 rooster’s genotype must have been RREe in order to produce the 1 walnut : 1 rose ratio in the next generation.
Genome 361 Spring 2013, Exam 2 8
- (12 pts) Two cat breeds, the Highland fold and the Scottish fold, have folded ear morphology. In both breeds, the trait is dominant over normal ear shape. A cat breeder has one Highland fold, one Scottish fold, and a few normal-eared cats. He is not sure whether his Scottish fold is true- breeding or not. Trivia for cat-lovers: Scottish folds and Highland folds are both supposed to have originated from a farm cat named Snooks. Scottish folds are short-haired, and Highland folds are long- haired. (i) Propose a cross to test whether his Scottish fold is homozygous or heterozygous. You can use any cats this breeder has (Scottish fold, Highland fold, or normal cats). State the expected phenotypic ratio for each conclusion. (5 pts) Breed 1 Breed 2 Proposed cross: Scottish fold x Normal cat Expected phenotypic ratio if homozygous (write none if you don’t expect a particular phenotypic class to appear): None Normal-eared : All Folded ear Expected phenotypic ratio if heterozygous (write none if you don’t expect a particular phenotypic class to appear): __ 1 ___ Normal-eared : ___ 1 __ Folded ear Only a testcross will work here. Crossing the Scottish fold with Highland fold will only give you folded-ear kittens regardless of whether the Scottish fold is homozygous or heterozygous. Question 9 continues. (ii) The cat breeder establishes that both the Scottish fold and the Highland fold are homozygous. He is curious about whether the mutations that lead to folded ear morphology in these two breeds are in the same gene or different genes. He mates the Scottish fold to the Highland fold, and all of the F1 kittens have folded ears. Highland fold Scottish fold
Genome 361 Spring 2013, Exam 2 10 If F1 is mated to normal-eared cats, then the cross can be written as AaBb x aabb, which will result in the following genotypic ratio: 1 AaBb (folded) 1 aaBb (folded) 1 Aabb (folded) 1 aabb (normal) The resulting F2 phenotypic ratio is: 1 normal-eared F2 : 3 folded ear F
- (7 pts) Papaya ringspot virus (PRV) causes one of the most destructive diseases affecting papaya crops. Infected plants die early as seedlings. You want to clone a gene that confers resistance, so that you can develop a commercial papaya strain that is PRV resistant. Assume that papayas are diploid. Suppose that you have identified a strain that is resistant due to a dominant mutation. Outline the steps you would perform to clone the gene conferring resistance. Step 1. Make a genomic library by extracting the genomic DNA from wild-type papaya resistant papaya (circle one) We will then clone the genomic fragments into a vector that can be propagated in papaya and E. coli. Next we transform the ligated plasmids into E. coli , grow the bacteria, and extract the library from E. coli. Step 2. Transform the genomic library into wild-type papaya resistant papaya (circle one) Step 3. Screen / Select (circle one) by looking for plants that survive after PRV infection
- (9 pts) Shown below is Gene B’s sequence (shaded) with its restriction enzyme cut sites. You obtain a plasmid (shown below) that let you perform cloning more efficiently than using a lacZ vector. You want to clone Gene B’s sequence into this plasmid, then transform the ligated plasmid into E. coli to make many copies. The ccdb 1 gene encodes a protein toxic to E. coli.
Genome 361 Spring 2013, Exam 2 11 (i) In what selective medium would you grow the transformed bacteria? What strain of bacteria would you transform the plasmid into? (2 pts) Ampicillin. Use ampicillin-sensitive bacteria. (ii) How does this plasmid help you distinguish bacteria that have taken up a plasmid with an insert from those that have taken up a plasmid without an insert? (4 pts) Bacteria that takes up a plasmid without an insert will produce the toxic ccdb1 protein and will not survive; bacteria that takes up a plasmid with an insert that destroys the ccdb1 gene will not produce the toxic protein and will survive. (iii) Which restriction enzyme(s) should you use for cloning Gene B’s sequence into the plasmid, such that only plasmids with inserts will be recovered? (3 pts) HindIII only EcoRI and XbaI cut within Gene B, so they can be eliminated. PvuII cuts the ORI, so it can be eliminated. XbaI can be eliminated again since it cuts the Ampicllin resistance gene. NotI is not present in the given restriction map of the gene, so it can be ruled out. BamHI does not cut within ccdb1 sequence, so we can’t be certain that colonies that survive on Ampicillin plate have an insert.
- (12 pts) One of your classmates is interested in mutations in the gene Sonic Hedgehog (SHH). SHH is involved in many developmental processes, including limb growth. A single base mutation in the regulatory region of SHH causes a dominant form of polydactyly. SHH is normally expressed in limbs during early development. This mutation causes expression to stay on later in development than usual, resulting in additional fingers. (i) Is this allele a gain-of-function or loss-of-function mutation? Briefly describe why. (3 pts) Gain-of-function. SHH is expressed later in development when it normally should not be. A loss- of-function allele would have no SHH expression.
Genome 361 Spring 2013, Exam 2 13
- (12 pts) You are studying alcohol metabolism using fruit flies as a model system. Mutants with different ethanol sensitivity can be isolated using a device nicknamed the “inebriometer.” Flies are put into the top and stand on a series of baffles. Enough ethanol vapor is pumped into the device to cause the flies to pass out and fall out the bottom. Wild-type flies can stay awake for 20 minutes of ethanol exposure. You want to collect mutants that are sensitive to ethanol. You mutagenize a population of flies, make them homozygous, and test the whole population in the inebriometer. You collect 10 flies that pass out after only 10 minutes of ethanol exposure. You first perform a complementation test to determine how many genes are represented in your mutant collection. ‘+’ indicates wild-type progeny; ‘-’ indicates mutant progeny; and question marks represent crosses that haven’t been performed yet.
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Genome 361 Spring 2013, Exam 2 14 m1 m2 m3 m4 m5 m6 m 7 m8 m9 m10 WT m1 - + + + +? + + + - + m2 -? + +? + - + + + m3 - + + + + - + + + m4 - + + -?? + + m5 - + + + + + + m6 - + + + - + m7 - + - + + m8 - + + + m9 - + + m10 - + WT + (i) How many complementation groups are there, and which mutants belong to each one? (4 pts) There is/are ___ 4 _ complementation group(s). Write the mutants that are in the same complementation group on the same line. You may not need all the rows provided: 1, 6, 10 2, 3, 8 4, 7, 9 5 (ii) You cross m4 to m8. The progeny resulting from this cross (circle one): (2 pts)
- all pass out after 10 minutes of ethanol exposure
- all stay awake for 20 minutes of ethanol exposure
- half pass out after 10 minutes, half stay awake for 20 minutes of ethanol exposure m4 and m8 belong to different complementation groups, so they have mutations in different genes. Their progeny will have the wild-type phenotype, which can tolerate up to 20 minutes of ethanol exposure.
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- Genome 361 Spring 2013, Exam
- Questions 1- 8 __________ /
- Question 9 __________ /
- Question 10 __________ /
- Question 11 __________ /
- Question 12 __________ /
- Question 13 __________ /
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