(6 pts.) 2 & 3 ( 10 pts.) ( 18 pts.) ( 12 pts.) 8 partsa&b (13 pts.) 8 part C. Total out of 100

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1 Biol 321 Spring 2013 Midterm Exam [90 points] NAME REMEMBER: carefully inspect problems and pay attention to detail! YOU ARE NOT ALLOWED TO USE CALCULATORS OR CELL PHONES DURING THIS EXAM or to send or receive signals of any kind READ EACH QUESTION CAREFULLY BEFORE ANSWERING. Problems 1 (pts.) Score ( 3 pts.) 2 & 3 ( 10 pts.) 4 (18 pts.) 5 ( 18 pts.) 6 (6 pts.) 7 ( 12 pts.) 8 partsa&b (13 pts.) 8 part C ( 10 pts.) Total out of (3 pts.) Marfan s syndrome is an autosomal dominant disorder of connective tissue that is manifested in the ocular, skeletal and cardiovascular systems. This syndrome shows great clinical variability but all individuals with the mutation show some symptoms. In addition, all individuals studied so far with Marfan syndrome have a mutation in the same gene. Choose the best restatement of this information. a. b. c. d. e. f. Marfan syndrome is pleiotropic and incompletely penetrant. Marfan syndrome is pleiotropic and variably expressed. Marfan syndrome is polymorphic and variably expressed. Marfan syndrome is pleiotropic, variably expressed and genetically heterogeneous. Marfan syndrome is genetically heterogeneous and variably expressed. None of these statements is corrrect. 1

2 2. (6 pts.) Examine the diagram to the right. Each line represents a single molecule of doublestranded DNA, which will segregate as indicated. The dotted line indicates where the cytoplasm will be divided during cytokinesis. Fill in the 2 blanks for parts a & b. a. This diagram is consistent with mitosis in a 1 n = 3 organism b. This diagram is consistent with Meiosis II in a 2 n = 6 organism c. If the genotype the organism in (b) is AaBbDD, what fraction of the all of the Meiosis II divisions occurring in the gonad of the organism would resemble the diagram? [Ignore complications from crossing over events & no explanation needed.] Circle the correct answer: (i) 1.0 (ii) ½ (iii) ¼ (iv) 1/8 (v) not possible to determine from info given 3. (4 pts.) Recall our discussion of the evolution of the Y chromosome. Circle True or False. 1 pt if no explanation required. 2 pts if explanation required. True Despite its name, the differential region of the Y chromosome contains genes that are also present on the X chromosome. False The present day Y chromosome is a decayed version of its former self because mutations in the pairing region cannot be purged by recombination with a homolog. [By definition, the pairing regions of the X and Y chromosomes carry the same genes and therefore synapsis and cross-over during prophase of the first meiotic division.] True The evolution of the mammalian Y reflects an evolutionary trade-off between erosion (decay) of gene function versus ensuring that the Sry gene and genes required for male fitness stayed together. One sentence explanation The loss of the ability of the evolving ancestral Y chromosome to cross-over with the ancestral X chromosome resulted in the loss of most of the genes once carried in the region BUT allowed the Sry gene to be inherited as a (non-recombinant) unit along with genes required for sperm formation (and not represented elsewhere in the genome). 2

3 4. (18 pts.) Revised from your assignment set. You have a true-breeding wild-type strain of fish. A colleague of yours gives you a mutant fish that is naked (no scales). You do the following crosses: Parental naked X wild-type scales (scales all over) Nn LL (truebreeding) nn ll F1 1/2 linear scales (one line of scales) nn Ll 1/2 naked Nn Ll F1 Cross A F2 F1 naked X F1 naked Nn Ll Nn Ll 12/16 naked N- -- (9/16 N- L- 3/16 N- ll) 3/16 linear nn L- 1/16 wild-type nn ll F1 Cross B F1 linear X F1 linear nn Ll nn Ll F2 3/4 linear nn L- 1/4wild-type nn ll F1 Cross C F1 linear X F1 naked nn Ll Nn Ll F2 ½ Nn naked ¾ L- linear ½ nn has scales ¼ ll wildtype (scattered) OVERALL ½ (4/8) Nn -- naked 3/8 nn L- (linear) 1/8 nn ll (scattered) A: Define allele symbols. Clearly indicate dominance and describe (in a couple of words) the trait that each gene controls. N gene controls the production/development of scales N= no scales n = scales L gene controls the scale pattern L = linear (one line of scales) l =wildtype (scattered) B What type of gene interaction is operating here? Use proper terminology (no explanation required) dominant epistasis the N- genotype is epistatic to the L gene C. WRITE OUT GENOTYPES FOR EACH CROSS. For the cross C predict the the F2 genotypes and phenotype (indicate ratios for the latter). 3

4 5. (18 pts.) Revised from the S12 midterm. PART A (5 pts.) Examine Nature News blurb on pg 2 of the Info/Data sheet. The researchers isolated 7 different true-breeding mutant strains showing increased alcohol tolerance compared to the wildtype strain. Examine this first series of crosses and clearly & explicitly state all conclusions below each set of observations For mutant lines #1-5, reciprocal crosses between true-breeding wild-type (intolerant) and mutant (tolerant) lines produced male and female progeny with the wildtype phenotype. The mutant alleles are recessive to wild-type for each of these strains The genes mutated in these strains are all autosomal Be careful with wording here: traits and alleles show dominance but genes do not. For mutant line #6, when wildtype females were crossed with mutant males, all progeny were wildtype (alcohol intolerant). When the reciprocal cross was performed, all males were mutant and all females wildtype. The gene mutated in this strain is located on the X chromosome and the mutant allele is recessive to the wild-type allele. For mutant line #7, reciprocal crosses between true-breeding wild-type and mutant lines produced male and female progeny that were alcohol tolerant. The mutant allele is dominant to the wild-type allele and the gene mutated in this strain is located on an autosome. PART B (9 pts.) A second series of crosses is performed (below). This time the mutant strains are crossed with each other, and the following results are obtained. + = wildtype - = mutant Mutant 1 Mutant 2 Mutant 3 Mutant 4 Mutant 5 Mutant 1 _ + + _ + Mutant 2 _ + + _ Mutant 3 _ + + Mutant 4 _ + Mutant 5 _ (2 pts.) What kind of experiment is this? Use correct terminology. No explanation needed. This is a complementation test. (2 pts.) Why weren t mutants 6 & 7 included in these crosses? No need to test 6 since the gene defective in this strain is X-linked and all other strains have defects in autosomal genes. Since the mutant allele in 7 is dominant, a complementation test is not valid. problem continues on the next page 4

5 (5 pts) What can you conclude from this second set of crosses? Be very explicit Mutants # 1-5 defined 3 different genes: Gene A: the mutations in strain 1 and strain 4 are allelic (they fail to complement each other but complement the rest of the strains) and define one gene Gene B: the mutations in strain 2 and strain 5 are allelic (they fail to complement each other but complement the rest of the strains) and define one gene Gene C: mutant strain 3 complements all other mutant strains. PART C. (4 pts.) F1 progeny from the cross of mutants 1X3 are testcrossed What fraction of the offspring will be wild-type with respect to alcohol tolerance. Assume independent assortment. Show genotypes and indicate the logic of your calculation Mutant 1 aabbcc X Mutant 3 AABBcc F1 AaBBCc AaBBCc X aabbcc ¼ AaBbCc wildtype ¼ AaBbcc mutant ¼ aabbcc mutant ¼ aabbcc mutant Overall ¼ wildtype ¾ mutant 6. (6 pts.) Rexamine Nature News blurb on pg 2 of the Info/Data sheet. Your mom s you and asks you to explain this news article to her. She is an astute individual and a supporter of government-sponsored research, but is a bit mystified about the relationship between fly and human alcohol consumption and wonders how the implied significance of the fly research can be justified. Briefly discuss the rationale for this study. In your response you should: a. (2 pts.) Briefly state what is meant in general by a model. NOTE this question is not asking what features make a good model organism. Two sentences. Need to state explicitly that what we learn from a specific model about a specific biological process can often be applied generally to other organisms. b. (4 pts.) Provide a justification for focusing on an animal model rather than directly studying humans. Your justification should have two parts: practical and theoretical --based on your understanding of the history of life on earth. 2-4 well-crafted sentences. Practical : the usual short generation, large numbers of progeny, small size, cheap to maintain, etc. Theoretical: because of our common evolutionary origin we share a large percentage of our genes with all organisms including those that seem pretty different from us prokaryotes, fungi, plants, invertebrates, etc. 5

6 7. ( 12 pts.) A genetic counselor puzzles over the following pedigree for a relatively rare disease. She has interviewed individuals 1 and 3 and is trying to determine the probability that they will have an affected child. Read the entire question before answering. More copies of this pedigree are on the data sheet a. (4 pts) What or modes mode(s) of inheritance are NOT consistent the pedigree data? Circle all correct answers. No explanations necessary i. autosomal recessive (recessive allele is very rare in the general population) ii. autosomal dominant complete penetrance iii. autosomal dominant incomplete penetrance iv. X-linked dominant incomplete penetrance (sex-limited) v. X-linked recessive (recessive allele is very rare in the general population) vi. Y linked (gene for trait is on the differential region of the Y chromosome) b. (4 pts.) Assume that the disease allele show 100% penetrance. Making the assumption that all individuals who married in are homozygous for the wild-type allele, the genetic counselor concludes that there is a zero probability that 1 x 3 will have an affected child. Carefully examine the pedigree. What mode of inheritance did the counselor use for her calculation? X-linked recessive: male #1 would be hemizygous for the normal allele & his mate (#3) would be homozygous for the normal allele since father is normal and mom assumed to be homozygous for the wildtype allele Defend your answer with by showing genotypes of relevant individuals on the pedigrees. No credit if genotypes are not clearly written out (label: part b). [More copies of this pedigree are on the data sheet. Indicate if pedigrees on the data sheet need to be graded.] c. (4 pts.) As it turns out, this couple has a child who is affected with the disease. The parents sue the genetic counselor and hire you as their expert witness. You testify that the counselor should have realized that there was an alternative way to interpret the pedigree. You also speculate on why she made the mistake that she did. What exactly do you say in court? Twothree cogent sentences. ( The judge has a short attention span ) Also, defend your answer by writing out genotypes on one of the pedigrees (label part C). The alternative interpretation is that this is an autosomal recessive disease state and the observation that all affected individuals are male is just due to skew by chance in a small data set. Both #1 and #3 would be heterozygous (both have an affected great grandparent). 6

7 8 (13 pts.) This problem has 3 parts. You do NOT need to get the answers to one part correct in order to answer the next part. Carefully review information on pages 2 & 3 of the DATA/INFO sheet and answer each question using only the information given. Part A. (10 pts.) Indicate True, False or N (not enough information provided to assess). Answer False if any part of the statement if false. If you choose N, indicate what additional information you would need. No credit if no explanation. 2 pts each. T F N The Tph2 R441H mutant allele represents a polymorphism in the general population. One sentence defense/explanation: This mutant allele is very rare in the general population. T F N Figure B suggests that the Tph2 gene is haplosufficient in some tissues but not in others. One- two sentence defense/explanation: In two of the three tissues, one copy of the wild-type allele produces the same amount of 5HT as 2 wild-type copies; in the third tissue the het genotype produces less than the homozygous wild-type genotype. T F N Tph2 (R441H) mutant genotype is likely to be epistatic to the L-AADC gene. One sentence defense/explanation: The product of the Tph2 enzyme is the substrate for the L- AADC gene. If the Tph2 is defective, it doesn t matter whether the L-AADC gene is mutant or wild-type. T F N In data shown in Figure 2, the GSK3 heterozygote has no phenotype in the wildtype background. One sentence defense/explanation: The phenotype of the GSK3 het looks the same in the Tph2 wildtype and mutant backgrounds. T F N The authors of this paper conclude that targeting GSK3 may afford therapeutic advantages for the management of certain 5-HT (serotonin) related psychiatric conditions. This management would likely involve increasing the level of GSK3 function. One sentence defense/explanation The management would like involve decreasing the level of GSK3 function since the level of this enzyme is elevated in individuals with the Tph2 mutation and see also part B. Part B (3 pts.) Short Answer What type of gene interaction is clearly indicted in the data here? Defend your choice by citing the relevant observations. 1-2 sentences. Suppression. Loss-of-function alleles of GSK3 suppress the phenotype produced by Tph 2 mutations since mice with mutations in both Tph2 and GSK3 show a more wild-type phenotype. 7

8 Part C. (10 pts.) Probability calculations. As indicated in Figure 2, animal of various genotypes were generated by crossing heterozygotes for both genes (+ = wild-type; no plus =loss-of-function). The Tph2 gene is on chromosome 12 and the GSK3 gene on chromosome 3. Homozygotes for the GSK3 mutation (g) are inviable. t + t g + g females X t + t g + g males (i). (3 pts.) Calculate the probability that a given offspring (alive) is heterozygous for both mutant alleles. Show your work and indicate the logic of your calculation. Look at each gene separately and then combine probabilities t + t X t + t ¼ t + t + ½ t + t ¼ t t g + g X g + g ¼ g + g + ½ g + g ¼ gg of the viable progeny, 2/3 will be het overall probability that a live offspring is t + t g + g = [1/2 X 2/3] = 1/3 (ii) (2 pts.) You are especially interested in testing the behavior of mice of genotype t t g + g You pick one mouse at random. What is the probability that your mouse does not have the desired genotype? probability that a live offspring is t t g + g = [1/4 X 2/3] = 1/6 = ~17% probability that my mouse does not have the desired genotype = 1-1/6 = 5/6 (iii) (3 pts.) From this cross you select 10 mice to do behavioral tests on. What is the probability that only one of the mice is of genotype t t g + g? Use your answer to (ii) here. If you select10 mice to work with, the probability that the first mouse is t t g + g and the rest are not is [1/6][5/6] 9. But in any group of 10 it doesn t matter whether it is the first or last or 6 th or whatever mouse as long as one is t t g + g and the rest are not. There are 10 different ways that you can pick one mouse of genotype t t g + g and nine not. Overall probability = 10 [1/6] 1 [5/6] 9 = ~32% (iv) (2 pts.) You select 20 mice to work with. What is the probability that at least one is of genotype t t g + g. Note, this is not a messy calculation. Use the 1- rule here. Probability that none of the 20 mice are t t g + g = [5/6] 20 Probability that at least one is t t g + g = all other possible outcomes other than none = 1 - [5/6] 20 = =

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