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Cell Division Jan 96,27 Use the following information to answer the next question In the life cycle of this jellyfish, the polyp stage is sessile. The medussa is a diploid, freeswimming animal. The planula is ciliated and motile. 1. Meiosis occurs in this life cycle when A. B. C. D. an egg and a sperm fuse to form a zygote a mature polyp produces a medusa a planula produces a young polyp a medusa produces gametes Jan 96,28 Use the following information to answer the next question 2. Which structures normally have a haploid number of chromosomes? A. B. C. D. Seed plant spores and animal zygotes Seed plant spores and animal gametes Seed plant zygotes and animal zygotes Seed plant gametes and animal zygotes Jan 96,4 Use the following information to answer the next question Some Events that Occur During Meiosis 1. 2. 3. 4. Gametes are produced Spindles form and homologous pairs of chromosomes separate Centromeres divide and chromatid pairs separate Chromosomes are replicated Numerical Response 4. Provide the correct sequence of these four events that occur during meiosis (Record your four-digit answer in the numerical-response section of the answer sheet.) Answer: ___________ Jan 96,29 Use the following information to answer the next question Mature red blood cells lack a nucleus. Approximately 2 million red blood cells are destroyed every second in the circulatory system. All blood cells differentiate from a common source – a population of pluripotent stem cells in the bone marrow. Pluripotent refers to the potential of these cells to form any type of blood cell. 3. Which statement best explains why the umber of red blood cells in the human body can be maintained? A. B. C. D. Mature red blood cells retain sufficient mRNA for replication and transcription Pluripotent stem cells undergo cell division continuously to produce new red blood cells Other types of mature blood cells undergo differentiation to form new red blood cells Mature red blood cells, before they are destroyed, undergo cytoplasmic cleavage repeatedly to produce more red blood cells Jun 96,24 Use the following information to answer the next three questions The Life Cycle of Baker’s Yeast The life cycle of baker’s yeast demonstrates a number of biological processes. Gene a and its allele a+ are present in this yeast population. Genotypes of cells are shown below each cell or group of cells. 4. Which row correctly identifies processes I, II, III and IV? Process Row A B C D I fusion fusion fission fission II mitosis meiosis meiosis mitosis III meiosis mitosis mitosis meiosis IV mitosis mitosis meiosis meiosis Jun 96,25 5. Processes II and IV illustrate budding. The information in the diagram indicates that budding is a type of A. B. C. D. asexual reproduction that produces haploid or diploid cells asexual reproduction that produces diploid cells sexual reproduction that produces haploid or diploid cells sexual reproduction that produces haploid cells Jun 96,26 6. Which genetic principle is best demonstrated by process III? A. B. C. D. Dominance Segregation Crossing-over Incomplete dominance Jun 97,25 Use the following information to answer the next two questions 7. What does process Y represent? A. B. C. D. Oogenesis Fertilization Gastrulation Spermatogenesis Jun 97,26 8. The cell processes diagrammed are more advantageous than budding because A. B. C. D. new genetic combinations are produced recessive gene combinations cannot occur less variation occurs increasing survival rates only successful genetic combinations are produced Jun 97,27 Use the following information to answer the next two questions. To prepare a karyotype, cells are broken open and the chromosomes are stained with a dye. A photograph of the chromosomes is taken, and the chromosomes are arranged in pairs. Below is an example of a human karyotype that provides information about an individual. 9. According to this karyotype, his individual has A. B. C. D. one less autosome than normal one more autosome than normal one less sex chromosome than normal one more sex chromosome than normal Jun 97,28 10. This unusual number of chromosomes is the result of A. B. C. D. synapsis during gametogenesis crossing over during gametogenesis nondisjunction during gametogenesis lack of cytokinesis during gametogenesis Jun 98,23 Use the following information to answer the next two questions Organism Horse Dog Cat Human Chromosome Number (2n) 66 78 38 46 11. Which of the following statements about cell types and chromosome numbers is correct? A. B. C. D. The ovum of a horse contains 66 chromosomes A somatic cell of a cat contains 76 chromosomes The spermatid of a dog contains 39 chromosomes A somatic cell of a human contains 23 chromosomes Jun 98,24 Use the additional information to answer the next question 12. Which process illustrates reduction division, as in spermatogenesis? A. B. C. D. Process 1 Process 2 Process 3 Process 4 Jun 98,4 Use the following information to answer the next question Some Events That Occur During Cell Division 1 2 3 4 5 6 Centromeres divide Cytokinesis occurs Identical cells are produced DNA is replicated Haploid cells are produced Spindle fibres form Numerical Response 5. Four events that occur in both human asexual and sexual cell reproductions are represented by numbers __________ (Record your four-digit answer in lowest-to-highest numerical order in the numerical-response section of the answer sheet.) Answer: _____ _____ _____ _____ Jun 98,25 13. In a human cell cycle, the event that occurs before mitosis begins and the event that occurs at or near the completion of mitosis are, respectively, A. B. C. D. crossing over and synapsis cytokinesis and crossing over replication of DNA and synapsis replication of DNA and cytokinesis Jun 98,26 Use the following information to answer the next question During his study of genetics, Gregory Mendel cross-pollinated many pea plants. He recorded the number and types of offspring produced and applied his knowledge of mathematics to create explanations for his observations. He hypothesized that factors are inherited separately and proposed the law of segregation. 14. The modern-day interpretation of Mendel’s law of segregation is that A. B. C. D. alleles are expressed independently during mitosis alleles are expressed independently during meiosis paired alleles separate during mitosis and are distributed into different gametes paired alleles separate during meiosis and are distributed into different gametes Jan 00,23 Use the following information to answer the next question 15. Which structures in the life cycle of the Ulva are haploid (monoploid)? A. B. C. D. Zoospores and the zygote The sporophyte and zygote Zoospores and the gametophytes The sporophyte and the gametophytes Jan 00,24 Use the following information to answer the next four questions Chromosome Content of Human Cells During a Series of Events 16. In humans, what process must have occurred to obtain the cells at U? A. B. C. D. Mitosis Meiosis Fertilization Differentiation Jan 00,25 17. In humans, what process occurs between U and V? A. B. C. D. Mitosis Meiosis Fertilization Differentiation Jan 00,26 18. In humans, what process must occur before cell V forms cells W and X? A. B. C. D. Mitosis Meiosis Recombination Nondisjunction Jan 00,27 19. In humans, cells Y and Z represent individual cells that A. B. C. D. are two eggs will no longer divide will become a 4n cell could develop into identical twins Jan 00,3 Use the following information to answer the next question Phases of Mitosis 1 2 3 4 Anaphase Metaphase Prophase Telophase Numerical Response 6. The phases of mitosis in the sequence in which they occur are _____, _____, _____, and _____. (Record your four-digit answer in the numerical-response section on the answer sheet.) Jun 00,30 Use the following information to answer the next three questions Investigators were interested in determining the role chromosomes play in the formation of the mitotic spindle. Using extracts of eggs from the African frog Xenopus laevis, they monitored spindle assembly in a test tube. The researchers replaced the chromosomes with beads, coated with random sequence of DNA. The beads served as substitute genetic material, but centrosomes (centrioles) were absent. As well, a part of the centromere was missing. Simplified Diagram of Normal Mitotic Cell -from Travis, 1996 20. Which of the structures numbered above was replaced by the beads in the experimental setup? A. B. C. D. 1 2 3 4 Jun 00,31 Use the following additional information to answer the next question The investigators observed that the genetic material on the beads condensed and microtubules began to form. Within 90 minutes, the microtubules formed a spindle-like structure that lined up the beads along the centre of the cell. -from Travis, 1996 21. Based on the results of this research, the structure or molecule that does not appear to be necessary for mitosis is A. B. C. D. DNA a spindle centrosomes microtubules Jun 00,32 Use the following additional information to answer the next question Other studies showed that the phase that involves pulling chromosomes to the two poles of mitotic cells can be delayed for up to 4.5 h by pulling a chromosome out of line form the centre of the cell. -from Travis, 1996 22. The phase that is delayed and the phase where the chromosomes line up at the equator are, respectively, A. B. C. D. telophase and anaphase metaphase and prophase interphase and telophase anaphase and metaphase Jan 01,23 Use the following information to answer the next three questions. Meiosis is a process that results in the reduction of the chromosomes number from diploid to haploid. Sometimes chromosomes fail to separate, which results in an abnormal number of sex chromosomes. 23. Process Z represents A. B. C. D. fertilization crossing-over nondisjunction spermatogenesis Jan 01,24 Use the following additional information to answer the next question Scientists studying Klinefelter and Turner syndromes wanted to determine which of several hypotheses about gender determination was most likely. These hypotheses were: presence of a Y chromosome causes maleness lack of a second X chromosome causes maleness the presence of two X chromosomes causes femaleness the Y chromosome is not involved in gender determination Evidence noted by the scientists included the following points. Individuals with Klinefelter syndrome (XXY) have genitalia and internal ducts that are usually male, but their testes are underdeveloped. Individuals with Turner syndrome (XO) have female external genitalia and internal ducts; however, the ovaries are underdeveloped. -from Cummings and Klug, 1997 24. This evidence best supports which of the scientists’ hypotheses about gender determination? A. B. C. D. The presence of a Y chromosome causes maleness The lack of a second X chromosome causes maleness The presence of two X chromosomes causes femaleness The Y chromosome is not involved in gender determination Jan 01,25 Use the following additional information to answer the next question Partial Human Karyotype 25. This partial human karyotype represents the last six chromosome pairs, in numerical order. The karyotype presented is that of a A. B. C. D. male with trisomy 21 female with trisomy 21 male with Turner syndrome female with Turner syndrome Jan 01,26 Use the following information to answer the next four questions Tay-Sachs disease is a hereditary disease that kills 1 in 360,000 individuals in the general population, but 1 in 4,800 among the Ashkenazi (Eastern European) Jews. The disease disrupts or halts proper formation of lysosomes and increases fat deposition around the nerve sheath. Individuals that are homozygous for the defective allele have Tay-Sachs disease and die at an early age. Studies suggest the heterozygous individuals have a higher survival rate against tuberculosis than the rest of the population. Biochemical tests can be done to determine if parents are carriers. -from Cummings 1994 26. What type of inheritance is demonstrated in Tay-Sachs disease? A. B. C. D. Autosomal recessive Autosomal dominant Sex-linked recessive Sex-linked dominant Jan 01,27 27. If tuberculosis regained its former role as one of the world’s deadliest diseases, then the frequency of the Tay-Sachs allele over time would A. B. C. D. decrease because of a decreased selective advantage increase because of an increased selective advantage decrease because of an increased selective advantage remain the same as a result of Hardy – Weinberg equilibrium Jan 01,4 Numerical Response 6. A young couple decided to have genetic screening done to determine if they were carriers of TaySachs disease. if both individuals were carriers, what percentage of their offspring would be predicted to have protection from tuberculosis but not have Tay-Sachs disease? Answer: __________% (Record your answer as a whole number percentage in the numerical-response section on the answer sheet.) Jan 01,28 28. Genetic screening results show that an individual is a carrier of Tay-Sachs if the individual’s DNA binds to A. B. C. D. none of the DNA probes two of the normal allele DNA probes two of the defective allele DNA probes one of the normal allele DNA probes and one of the defective allele DNA probes Jan 01,5 Use the following information to answer the next question Numerical Response 7. Identify the stages in the conifer life cycle, as numbered above, that correspond with the letters that represent these stages on the diagram. Stages: _____ Diagram: A _____ B _____ C _____ D (Record your four-digit answer in the numerical-response section on the answer sheet.) Jan 02,35 Use the following information to answer the next six questions Sickle cell anemia is an autosomal recessive genetic disorder. Because individuals affected by sickle cell anemia have defective hemoglobin proteins, their blood cannot transport oxygen properly. There appears to be a relationship between the incidence of malaria and sickle cell anemia. Individuals with sickle cell anemia and carriers of the sickle cell allele have some resistance to malaria. Malaria is caused by the parasite Plasmodium and is transmitted between humans by mosquitoes. 29. The probability of two carriers parents having a child with sickle cell anemia is A. B. C. D. 25 % 50 % 75 % 100 % Jan 02,36 30. If scientists are successful in significantly reducing or eliminating malaria, the best prediction for what will happen to the allele for sickle cell anemia in the population is that it will A. B. C. D. not be affected by the elimination of malaria increase as its selective advantage is increased be reduced as its selective advantage is decreased quickly disappear as its selective advantage is increased Jan 02,37 Use the following additional information to answer the next two questions 31. The row below that identifies process 1 and process 2 is Row A. B. C. D. Process 1 mitosis mitosis meiosis meiosis Process 2 meiosis mitosis mitosis meiosis Jan 02,38 32. The row below that identifies the chromosome number at the first stage and the chromosome number at the second stage is Row A. B. C. D. First stage diploid diploid haploid haploid Second stage haploid diploid diploid haploid Jan 02,39 Use the following additional information to answer the next two questions Insecticides have been used to control mosquito populations in order to prevent the spread of malaria, but mosquitoes in malaria-infested areas are developing resistance to these insecticides. In addition, the antimalarial drug chloroquine, once very effective in protecting individuals against Plasmodium, has become ineffective, which ahs resulted in a resurgence of malaria. Scientists have identified a gene, called cg2, in Plasmodium that allows the Plasmodium to mount resistance to chloroquine. This research could be used by scientists to develop new versions of chloroquine that will sidestep the parasite’s resistance and, therefore, efffectively protect people against malaria. - from Travis, 1997 33. Some investigators have suggested that some strains of Plasmodium have become chloroquineresistant because these strains have an increased ability to pump chloroquine from their bodies. Other investigators suggest that the resistance stems from changes in some strains of Plasmodium that prevent chloroquine from entering the parasites in the first place. These two suggestions can best be described as A. B. C. D. theories hypotheses conclusions observations Jan 02,40 34. A possible reason that the Plasmodium parasite may have resistance to chloroquine is that the cg2 gene codes for protein that seems to play a role in membrane transport of the drug. If this true, researchers may want to develop compounds that specifically block this resistance mechanism by A. B. C. D. preventing mutation of the cg2 gene stimulating translation of the cg2 gene preventing transcription of the cg2 gene stimulating DNA replication of the cg2 gene