Genetics%20466%20Exam%201%20v2

Genetics%20466%20Exam%201%20v2 - Gen 466 exam 1, 2.9.2006...

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Unformatted text preview: Gen 466 exam 1, 2.9.2006 Your name _________________ Genetics 466 exam #1. 1. Choose the best matching phrase in the right column for each of the terms in the left column. Write the corresponding number in the parenthesis after the term. (7 pts) a. phenotype ( 5 ) b. genotype ( 7 ) 1. alternate forms of a gene 2. alleles of these can subtly alter phenotypes produced by the action of other genes c. gene ( 8 ) d. allele ( 1 ) e. null mutation ( 4 ) f neomorph ( 6 ) g. conditional allele ( 3 ) 3. has different phenotypic strengths in different environmental circumstances 4. complete loss of function 5. observable characteristic of an organism 6. acquisition of a new function 7. genetic make up of an organism 8. heritable entity that determines a characteristic 2. Choose the best matching phrase referring to the cell cycle stage in the left column. Write the corresponding number in the parenthesis after the term. (7 pts) a. interphase ( 2 ) 1. pairs of sister chromatids migrate to each pole without centromere division b. S-phase ( 8 ) c. Prophase I (meiosis) ( 3 ) d. Metaphase I (meiosis) ( 4 ) e. Anaphase I (meiosis) ( 1 ) f. Anaphase II (meiosis) ( 6 ) g. Telophase ( 7 ) 2. chromosomes are decondensed 3. synapsis and chiasmata formation 4. tetrads aligned at the spindle equator 5. phase between cytokinesis of meiosis I and prophase of meiosis II 6. sister chromatids migrate to each pole after division of centromeres 7. chromosomes reach the poles and decondense, nuclear membrane reforms. 8. DNA replication 1 Gen 466 exam 1, 2.9.2006 Your name _________________ 3. For each mode of inheritance specified below, choose the pedigree that best matches the mode of inheritance. Each mode of inheritance should be matched to a single, different pedigree. Place the pedigree number in the parenthesis to the right of the mode of inheritance. (10 pts) Assume the following: a) trait is rare, so that in all cases assume that trait is carried in only one of the two founding family members (except in the case of the autosomal recessive, where you if necessary you may assume that the founding members are both carriers). b) there are no sex chromosome abnormalities or other cases of non-disjunction. c) Each trait is determined by a single gene. d) penetrance is complete except in case (e) Possible modes of inheritance a) autosomal recessive ( 5 ) b) autosomal dominant ( 1 ) c) X-linked dominant ( 2 ) d) autosomal dominant, sex influenced (e.g. premature baldness in humans) ( 4 ) e) autosomal dominant, incomplete penetrance ( 3 ) 2) 1) 4) 3) 5) 2 Gen 466 exam 1, 2.9.2006 Your name _________________ Questions 4- 11: Multiple choice. Choose the answer the best matches each statement or question. Circle one and only one option. (3 pts ea.) 4. the cross between two different true breeding strains to determine whether two alleles are forms of the same or different genes is called: a) a test cross b) a monohybrid cross c) a complementation cross d) a dihybrid cross e) none of the above 5. the phenomenon where alleles in a gene affect more than one phenotype is called a) epistasis b) dominance c) pleiotropy d) expressivity e) multiple allelism 6. An individual heterozygous for a dominant allele that causes Huntington's disease, a neurological disorder. If the individual has 4 children, what is the probability that 2 of them will be affected? a) 1 / 4 b) 3 / 8 c) 1 / 2 d) 5 / 8 e) none of the above 7. Consider the following pedigree for a rare, recessive trait. The probability that the progeny from this consanguineous mating is affected is: a) 1 / 16 b) 1 / 64 c) 1 / 128 d) 1 / 256 e) 1 / 512 ? 3 Gen 466 exam 1, 2.9.2006 Your name _________________ 8. m is a hypermorphic allele. M gene function is dosage dependent, so that the genotype m / m results in a visible phenotype. Dosage sensitive phenotypes can be modified by the manipulation of gene dosage. In which of the following genotypes is this visible phenotype likely made more severe (enhanced, in genetic terminology): a) m1 / m+, Dl (m), where Dl (m) is a deletion encompassing gene m. b) m1 / m+, Dp (m), where Dp (m) is a duplication encompassing gene m c) m1 / m2, where m2 is a hypomorphic allele d) m1 / m3, where m3 is an antimorphic mutation e) m1 / m0, where m0 is an amorphic mutation 9. Recessive, X-linked alleles that severely affect the reproductive potential of individuals are transmitted through generations primarily in the following pattern: a) from females to females in the heterozygous form b) from affected males to grandsons, through an intermediate female generation c) from affected males to their sons d) from non-affected carrier males to their grandsons, through an intermediate female generation e) from non-affected carrier males directly to their sons 10. Consider the following human pedigree containing a Klinefelter syndrome male (XXY, indicated with the shaded square). A and B refer to codominant alleles of the Xlinked gene GP6D. The phenotypes of each individual are shown on the pedigree. The Klinefelter male is likely the result of non-disjunction in: a. the father, meiosis I AB B b. the father, meiosis II c. the mother, meiosis I d. the mother, meiosis II e. can not be determined: either the father or the mother B AB AB B 11. In the cross AaBbCcdd x aabbCcdd, the fraction of progeny that will exhibit the quadruple recessive phenotype is: a. 1/16 b. 1/32 c. 9/64 d. 1/256 e. 9/512 1 + 1 4 Gen 466 exam 1, 2.9.2006 Your name _________________ Questions 12-21: State whether each of the following statements is true or false. Circle the right answer (2 point ea.) 12. A wild-type allele can be dominant to some alleles of the same gene, but recessive to other alleles of the same gene. ( true -- false ) 13. Using cytological methods, a paracentric inversion is usually easier to detect than a pericentric inversion. ( true -- false ) (pericentric is easier) 30. The brown and orange patchy pattern of a calico cat is caused by expression of the brown and orange alleles and X-chromosome inactivation ( true -- false ) 28. Male honeybees, which are haploid, rely on mitosis to create their gametes. -- false ) ( true 16. Penetrance is an extreme form of expressivity, where some individuals with the mutant genotype do not show a phenotype at all. ( true -- false )) 17. A synchronous culture of cells in interphase can be used for karyotyping. false ) (chromosomes would be in decondensed state) ( true -- 18. In human males, primary spermatocytes are arrested for many years at prophase of meiosis I. ( true -- false ) only human females have meiotic arrest 19. Plants typically have two multicellular generations, which alternate between haploid and diploid DNA content. ( true -- false ) 20. In humans, recombination between genes in the pseudoautosomal region results in genetic interchange between regions of the X and Y chromosomes. ( true -- false ) 21. Female snakes can produce male offspring in the absence of mating by producing diploid eggs though a modified version of mitosis. ( true -- false ) (a true note for this question: snakes have a ZW sex determination system) (it must be a modified version of meiosis note that if it was a modified version of mitosis, a ZW female would not be able to produce a ZZ male) 5 Gen 466 exam 1, 2.9.2006 Your name _________________ 22) A true-breeding plant strains produces extra petals. A second true-breeding strain that has pink petals instead of the normal red color. The two strains are intercrossed and all the resulting F1 exhibits extra, red-colored petals. a) For each gene, which alleles and corresponding phenotypes are dominant and which are recessive? Label alleles affecting petal number as A and a, and alleles affecting color as B and b (use capital letters to refer to the dominant alleles). (3 pts) A: extra petal number (dominant) a: normal petal number (recessive) B: red color (dominant) b: pink color (recessive) b) You want to test the segregation of these two genes using a standard genetic tool: a test cross. Which strain would you use as the tester strain (other true-breeding mutant combinations are available if necessary). Specify its phenotype. (3 pts) cross the F1 to aabb (pink color, normal number of petals) 6 Gen 466 exam 1, 2.9.2006 Your name _________________ c) The test cross results in four genotypic classes: Phenotype observed observed number of progeny I) Dominant for both genes 18 II) Dominant for petal number, recessive for petal color 10 III) Recessive for petal number dominant for petal color 7 IV) Recessive for both genes 15 Use the Chi Square analysis and the table below to determine if these two genes follow the principle of independent assortment (5 pts) Degrees of Freedom 1 2 3 4 5 0.99 0.02 0.11 0.30 0.55 0.90 0.02 0.21 0.58 1.06 1.61 0.50 0.45 1.39 2.37 3.36 4.35 P values 0.10 2.71 4.61 6.25 7.78 9.24 0.05 3.84 5.99 7.81 9.49 11.07 0.01 6.64 9.21 11.35 13.28 15.09 0.001 10.83 13.82 16.27 18.47 20.52 null hypothesis: if independent hypothesis, then expect 1 : 1 : 1 : 1 ratio class # observed I 18 II 10 III 7 IV 15 total 50 c2 = [(18 12.5) / 12.5] + [(10 12.5) / 12.5] + [(7 12.5) / 12.5] + [(15 12.5) / 12.5] = 5.84 degrees of freedom = df = n 1 = 3 For df of 3, critical P0.05 value is 7.81 c2 = 5.84 < 7.81, so null hypothesis stands (can not be rejected) 2 2 2 2 # expected 12.5 12.5 12.5 12.5 50 7 Gen 466 exam 1, 2.9.2006 Your name _________________ 23) A breeder finds a rare light colored cow in his normally dark colored stock. She crosses the rare cow back to dark-colored bulls of his farm's stock. Some of the crosses result in all dark F1 progeny, other crosses yield F1 dark to light individuals in a 1:1 ratio, while yet other crosses result in an F1 dark to light ratio of 3:1. The breeder suspects that something unusual may be happening in her stock. Thus she orders a bull of the same dark-colored strain from across the country and performs a similar cross. The F1 progeny from the light colored cow and this new, dark bull are all dark-colored. Once they have matured, these F1 siblings are crossed amongst themselves. The progeny in these F2 crosses now all exhibit the same ratio: 15 darkcolored to 1 light-colored. a) How can the breeder explain the 15:1 ratio? (For all possible genotypes in this cross, specify the corresponding phenotype) (4 pts) A_ B_ 9 aa B_ 3 A_ bb 3 15 dark colored aa bb 1 light colored b) Diagram a genetic model that explains this type of genetic interaction. In your model, specify all relevant genes, pigments and precursors (4 pts) A and B have redundant functions gene A light -------------> color --------------> gene B dark color c) Explain the 3:1 ratios obtained in the crosses when she uses bulls from her own farm (3 pts) A_ B_ ( x aa bb 3:1 8 Gen 466 exam 1, 2.9.2006 Your name _________________ 24) Cells from a diploid plant X with 3 pairs of chromosomes (1 basic chromosomal set = 3 chromosomes) undergo meiosis. However, one of the two cytokinesis events that normally occur during meiosis is defective. a) In terms of the ploidy number n, what is the ploidy number of the gametes derived from aberrant meiosis? (reminder: ploidy number is different from the total number of chromosomes) (2 pts) 2n b) these gametes fertilize a gamete, derived from normal meiosis, from a nearby plant of the same species. How many basic chromosomal sets does the resulting plant have? (2 pts) 3n c) Repeated self fertilization of the plant resulting from (b) produce only aborted seeds. Propose an explanation for this apparent sterility. (2 pts) unbalanced gametes due to odd # of homolog chromosomes (odd ploidy #) d) You reproduce the hybrid plant asexually via cuttings and expose them to agents that inhibit cytokinesis. Some of the plants produce flowers that are fully fertile when they undergo self-fertilization. What event led to the at least parts of the plant becoming fertile? (2 pts) inhibition of cytokinesis during mitosis resulted in clones of cells with double the original ploidy number 3n ----> 6n e) The viable seeds resulting from (d) produce fertile plants. What is the ploidy number of these fertile plants? (2 pts) 6n 9 ...
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