Exam 1. CSS/Hort All questions worth 2 points

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1 Exam 1. CSS/Hort All questions worth 2 points 1. A general definition of plants is they are eukaryotic, multi-cellular organisms and are usually photosynthetic. In this definition, eukaryotic means that plants a. contain chlorophyll b. have membrane-bound nuclei c. lack the power of locomotion d. have sporophytic and gametophytic generations 2. Genes, as defined as the basic units of inheritance, consist of a. DNA b. RNA c. Protein d. Starch 3. An allele can be best described as a. a homozygous genotype b. a heterozygous genotype c. one of several possible forms of a gene d. a phenotype of an organism 4. Genome size is always directly proportional to plant size and longevity, i.e. short annual grasses have smaller genomes then tall woody perennials. 5. If a plant genome is 4,000 Mb in size, this means that a cell in the gametophytic generation (n) will have a. 4,000 base pairs of DNA b. 4,000,000 base pairs of DNA c. 4,000,000,000 base pairs of DNA d. 8,000,000,000 base pairs of DNA 6. If a germplasm repository is responsible for maintaining and distributing highly heterozygous varieties of a self-incompatible species, such as the hazel nut, the varieties will most likely be maintained and distributed via a. Asexual propagation b. Sexual propagation c. Genetic engineering d. Marker assisted selection 1

2 7. Which of the following terms best describes the process of using a DNA test for the presence or absence of a specific allele in order to assist in the selection of plants with the target allele. a. Linkage mapping b. Traditional plant breeding c. Marker assisted selection d. Genetic engineering 8. A plant geneticist observes that in a very large (n = 1000) F2 population derived from the cross of two completely homozygous parents, two specific combinations of traits are always inherited together: all blue-seed progeny are susceptible to a fungal disease and all white-seed progeny are resistant to the disease. This is most likely a case of a. Pleiotropy b. Linkage (with 10% recombination between genes) c. Incomplete penetrance d. Epigenetics 9. A plant geneticist observes a qualitative frequency distribution for flower color (48 blue:50 white) in 100 testcross progeny. This result suggests that a. Flower color is controlled by many genes lower color is controlled by one gene c. There is excellent fit to a dihybrid ratio d. The blue and white flower color genes are linked 10. Genetic analysis is simpler in plant than in animals because plants have only one (the nuclear) genome. 11. A trait determined by a gene in the chloroplast is expected to show a 3:1 phenotypic ratio in the F2 progeny of two parents differing for the trait. 12. A phenotype could be due to allelic variation at a single locus in the nuclear genome but incomplete penetrance could make you reject the null hypothesis of a 1:1 ratio in doubled haploid progeny. 13. You would expect to see autosomal and sex-linked inheritance patterns in a dioecious plant with defined sex chromosomes. 2

3 14. The Vrs1 gene of barley is over 1000 nucleotides long. Therefore, many alleles are possible at this locus but any given (diploid) barley plant can have a maximum of two of these alleles. 15. A homozygous purple flowered clematis (PP) is crossed with a homozygous white flowered clematis (pp). The F1 plants all have purple flowers. If you produce 100 doubled haploid progeny from the purple F1 plant, the expected ratio is a. 50 purple: 50 white b. All purple c. All white d. 75 purple: 25 white 16. Using the parents and F1 described in question # 15, if you produce 100 F2 progeny by selfing the purple F1 plant, the expected ratio is a. 50 purple: 50 white b. All purple c. All white d. 75 purple: 25 white 17. In the cross described in question # 15, testing observed vs. predicted ratios for the doubled haploid and the F2 progeny will be based on the same number of degrees of freedom (1) but different expected ratios will be tested for each population type (1:1 for doubled haploid and 3:1 for F2). 18. If the F2 population described in question # 16 gave an observed ratio of 72: 28 and the calculated chi square value was 0.48, you would conclude with statistical certainty that a. Flower color is controlled by a single dominant gene lower color is controlled by 2 genes, each showing complete dominance c. Chi square tests are useful for F2 populations but not for doubled haploids d. Flower color is a trait showing incomplete penetrance 19. Chi square tests are only useful for phenotype data (e.g. two row vs. 6-row). They cannot be used to test goodness of fit of expected and observed ratios for DNA sequence data (e.g. single nucleotide polymorphisms in the Vrs1 gene). 3

4 20. Segregation can be observed in both monohybrid and dihybrid crosses but independent assortment can only be observed in dihybrid crosses. 21. The expected monohybrid genotypic ratio in the F2 is 1AA:2Aa:1aa. If you chose one of the F2 Aa individuals and selfed it, you would expect its F3 progeny to segregate in a genotypic ratio of 1AA:2Aa:1aa. 22. Hermaphroditic flowers are very rare in angiosperms. 23. Which of the following best describes a perfect flower? a. Functional male and female organs in the same flower b. Typical of dioecious plants c. Has petals and sepals d. A structural mechanism to ensure 100% out-crossing 24. Many plants with perfect flowers do not self-pollinate, or self-pollination is relatively infrequent compared to cross-pollination. This is because a. Perfect flowers do not have male organs b. Recombination between X and Y chromosomes c. There are often advantages to heterozygosity, including avoidance of inbreeding depression d. They are self-compatible 25. Self-incompatibility, as in the hazelnut, is best defined as a. Male sterility due to non-functional pollen ailure to set seed when self-pollinated with viable pollen c. Female sterility due to a lack of stigmas d. Pollen failure due to a mutant gene in the cytoplasm 26. T cytoplasm corn plants with a mutant T-urf13 allele in the mitochondrial genome are male sterile and susceptible to southern corn leaf blight. Billion dollar losses due to southern corn leaf blight were caused by the fact that so many corn hybrids were produced using the T cytoplasm. This case study shows that: a. Only genes in chromosomes in the nucleus can show pleiotropy b. No important characters are determined by genes in the mitochondria, c. Plant breeders and farmers need should ensure that completely genetically identical varieties are not planted over extensive acreages d. Turning monoecious maize plants into females by removing the tassel is a bad idea 4

5 27. Which of the following mechanisms will encourage cross pollination in hermaphroditic and/ or monoecious plants? a. Spatial distribution of sexes b. Differential maturation rates of pollen and stigma c. Population density d. All of the above 28. A key advantage of an X,Y sex chromosome system in which males are XY and females are XX is that it ensures there will always be more females than males. 29. In a species with defined X and Y chromosomes, where males are XY and there is no recombination between the X and Y chromosomes, all genes on the Y chromosome will show complete linkage with each other. 30. In plants, the vegetative meristem undergoes a transition to a floral meristem. Subsequently, there is floral organ initiation in this meristem. Which of the following statements best describes sex determination in monoecious and dioecious plants? a. In all species with unisexual flowers, the floral organ initiation step is blocked. loral development is blocked at different stages (e.g. floral organ initiation, pollen development, egg maturation) in different species. 31. Which of the following statements best describes a plant nuclear chromosome? here are two DNA molecules per chromosome, attached at the centromere b. Each chromosome consists of multiple DNA molecules c. Chromosomes have two centromeres and one telomere d. Each chromosome consist of a single DNA molecule complexed with histone proteins 32. Which of the following chromosome landmarks is the site of attachment of spindle fibers at mitosis and meiosis? a. Centromere b. Telomere c. Nucleolus organizer region d. Nucleolus 5

6 33. If you were looking for regions of the genome most likely to contain lots of genes, you would look in a. Heterochromatin b. Euchromatin c. The synaptonemal complex d. The nucleolus 34. Which stage in the mitotic cell cycle lasts the longest? a. Prophase b. Metaphase c. Anaphase d. Telophase e. Interphase 35. If a plant is 2n = 2x = 20, how many total chromatids will there be in a nucleus at the prophase stage of mitosis? a. 5 b. 10 c. 20 d At which stage of mitosis would you see sister chromatids migrating to opposite poles? a. Prophase b. Metaphase c. Anaphase d. Telophase 37. In mitosis, each 2n cell gives rise to two 2 cells, each of which is 2n whereas in meiosis each 2n cell gives rise to 4 cells, each of which is n. 38. Pairing of homologous chromosomes occurs at Zygonema of both mitosis and meiosis. 39. Crossing over between non-sister chromatids in each bivalent occurs at which stage of meiosis? a. Zygonema b. Pachynema c. Metaphase I d. Metaphase II e. Telophase II 6

7 40. The random alignment of non-homologous chromosomes that accounts of independent assortment occurs at which stage of meiosis? a. Zygonema b. Pachynema c. Metaphase I d. Metaphase II e. Telophase II 41. Which of the following is the physical basis of segregation? a. Pairing of homologous chromosomes b. Sister chromatid exchange c. Anaphase I d. Anaphase II 42. From birth until death, both mitosis and meiosis are occurring in all cells of the organism, all the time. 43. Alleles at loci on different chromosomes will always show independent assortment. 44. Crossing over always involves a significant loss of chromatin and these deletions are the principal source of genetic variation attributable to meiosis. 45. Mutation is the source of new alleles and recombination is the source of new combinations of alleles. 46. If a species is 2n = 2X = 8, the number of linkage groups will be: a. 2 b. 4 c. 8 d. 16 7

8 Assume you have two homozygous varieties in a diploid species (2n = 14). One variety is VVWWNNLL. The other variety is vvwwnnll. You have a goal of developing a homozygous variety that is VVwwnnLL. The V locus is in chromosome 2. The W, N, and L loci are in chromosome 7. W and N are far enough apart on chromosome 7 that they show independent assortment. N and L are close enough together that crossovers can occur between the two loci, but not often (usually 10 crossovers per 100 meioses). You cross the two homozygous varieties and derived doubled haploids from the F Your goal of developing a homozygous variety that is VVwwnnLL cannot be achieved because all doubled haploids will be heterozygous. 48. Alleles at the V locus will show independent assortment with alleles at the W, N, and L loci. 49. The W, N, and L loci are members of the same linkage group. 50. The reason that W and N show independent assortment is because many crossovers occur between the two loci. Therefore the frequencies of WWNN, WWnn, wwnn, and wwnn genotypes will each be

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