Test 3 1 January Winter 2020, questions and answers PDF

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Biology 1A03 – Practice Test 3

PART A: MULTIPLE CHOICE QUESTIONS (worth 30 marks) Please answer Questions 1 to 30 on the OMR Scan Sheet in HB pencil, NOT pen. Select the option which best answers the question. There is no penalty for guessing. Each question is worth 1 mark. DR. KAJIURA’S MULTIPLE CHOICE QUESTIONS 1. Which of the following statements correctly describes the G0 phase? A. It is the phase that prepares the cell for the production of gametes. B. It is the phase that prepares the cell for mitosis and DNA replication. C. It describes a nondividing state. D. It is the phase during which cells display no evidence of passing through the S phase. E. Both C and D are correct 2.A biological male with a karyotype 47, XXY is expected to have any of the following characteristics, EXCEPT: A. reduced sperm count B. reduced testicular development C. development of breast enlargement D. very high sperm production E. tall stature with poor beard growth 3. What conditions must be met for the cell to pass the G1 checkpoint? A. All chromosomes must be attached to the spindle apparatus. B. Nutrient levels must be sufficient and the cell size must be adequate. C. Chromosomes must have replicated successfully and have undamaged DNA. D. B & C only E. A, B, and C 4. Which of the following experiments helped biologists to study the cell cycle regulation enzymes/proteins? A. Density centrifugation experiments conducted by Meselson & Stahl. B. Cell microinjection experiments similar to those conducted by Masui and Markert. C. Cell fusion experiments similar to those conducted by Johnson and Rao. D. Both A and C. E. Both B and C. 5. If a cell has 12 chromatids at Metaphase I Meiosis I, how many chromosomes would be present in each of the nuclei of Telophase II Meiosis II? A. 3 chromosomes would be present in each of the nuclei B. 4 chromosomes would be present in each of the nuclei C. 12 chromosomes would be present in each of the nuclei D. 6 chromosomes would be present in each of the nuclei E. 2 chromosomes would be present in each of the nuclei

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6. What is the difference between the leading strand and the lagging strand in DNA replication? A. The leading strand is synthesized continuously in a 5’ → 3’ direction, while the lagging strand is synthesized discontinuously in an overall direction of 3’ → 5’ with Okazaki fragment individually growing in the 5’ → 3’ direction. B. The leading strand is synthesized in a 3’→5’ direction in a discontinuous manner, while the lagging strand is synthesized in a 5’ → 3’ direction in a continuous manner. C. There are several different DNA polymerases involved in elongation of the leading strand and the lagging strand. D. The leading strand needs an RNA primer, whereas the lagging strand does not require an RNA primer. E. Both B & C 7. Once an animal cell finishes the process of mitosis, molecular division triggers must be turned off. Which of the following statements correctly describes what happens to mitosis-promoting factor during mitosis? A. The cyclin-dependent kinases are stimulated to start the process of cytokinesis. B. The MPF is totally degraded and both components of it are lost and not re-used. C. Cyclin is degraded, the concentration of cyclin-dependent kinase remains unchanged, but without cyclin, MPF is not formed. D. Cyclin-dependent kinase is broken down and degraded, and cyclin concentration remains constant, and MPF is synthesized in very large amounts. E. Both A & B 8. What is the aim of annotating a whole genome of an organism? A. The aim is to establish the probability of mutations in prokaryotes. B. The aim is to learn the number of nucleotides contained within the genome. C. The aim is to learn how gene products interact to produce phenotypes. D. The aim is to identify the genes and their positions within the genome. E. Both A & C 9. If you are studying a cell in G1 phase, what characteristics would you observe? A. You would observe actin-myosin rings causing the plasma membrane to begin pinching inwards to separate the cell into 2 other cells. B. You would observe the preparation of materials (protein machinery, primary growth phase) to get the cell ready for mitosis. C. You would observe the secretion of polysaccharides and cell plate formation. D. You would observe nuclear envelope breakdown and also spindle apparatus formation. E. Both A & C 10. Because dideoxy sequencing is based on the chain termination, why are normal deoxynucleotides also used in the reaction? A. to create DNA synthesis products long enough to allow running a gel B. to enhance the chain termination ability of the deoxynucleotides C. to produce a range of DNA synthesis products that terminate at every occurrence of a particular base D. to provide a strong substrate for the DNA polymerase E. Both B & D.

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11.How is a ddNTP that is used in Sanger sequencing different in comparison to the dNTP typically present in DNA? A. ddNTP’s have a 3’H rather than OH B. dNTP’s have a 3’H rather than OH C. dNTP’s have a 5’ OH rather than H D. ddNTP’s have a 5’OH rather than H E. Both A and C are 12. Which of the following statements correctly describes the characteristics of cancer cells? A. Cancers stop dividing after they have formed a completed single layer. B. Cancer cells do not display density dependent inhibition C. Cancer cells do not display anchorage dependence D. Cancer cells display low rates of cell division in a carefully regulated manner. E. Both B and C 13. Which of the following supports the legacy of Steve Jobs related to cancer research? A. The identification of the PIM 1 biomarker associated with his rare type of pancreatic cancer. B. The applied use of Afinitor, which is a drug used for treatment of pancreatic cancers. C. The continued investigation of the field of immunotherapy related to pancreatic cancers. D. Both A and C only. E. A, B, C are correct. 14. Which of the following statements is correct regarding the types of SNPs? A. Linked SNPs are located within the gene B. Linked SNPS are located outside of the gene C. Linked SNPs do not affect the way a protein functions D. Causative SNPs do not affect the way a protein functions E. Both B & C 15.Suppose that a you and your Biology 1A03 lab partner use similar methods to that of Johnson and Rao's cell-cell fusion experiments. If a Metaphase cell is fused to a G1 phase cell, what would you and your lab partner predict to happen to the chromosomes in the G1 phase cell? A. The chromosomes in the G1 phase cell would begin to immediately replicate its DNA B. The chromosomes in the G1 phase cell would immediately produce nonkinetochore regions at the centromere regions C. The chromosomes of the G1 phase cell would become thicker D. The chromosomes of the G1 phase cell would be broken down by enzymes E Both A and B 16. Which of the following statements correctly defines the term “TUFs”? A. Translation of Useful Function B. Transcripts of Unknown Function C. Thymine of Unique Function D. Transcripts of Unique Function E. Telomerase of Unique

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17. Which of the following can cause mutations in DNA? A. free radicals which are formed as by-products of aerobic respiration B. ultraviolet radiation from the sunlight C. X rays D. exposure to carcinogens E. All of the above 18. As discussed in Biology 1A03 lectures, which of the following terms correctly describes the field of study that sequences, interprets, and compares whole genomes? A. The field of study is called bioallometry B. The field of study is called systematics C. The field of study is called proteomics D. The field of study is called genomics 19. Which of the following statements correctly describes a cell during prometaphase? A. The chromosomes in the prometaphase cell are directly located on the metaphase plate. B. The chromosomes in the prometaphase cell are not aligned on the metaphase plate and they do not appear organized C. The chromosomes in the prometaphase cell have completed the process of disjunctional segregation. D. The chromosomes in the prometaphase cell begin to start the process of binary fission. E. Both A & C 20. Suppose that a medical physician suspects that a chromosomal defect is the cause of a patient’s condition. He requests a karyotype be made for the patient. What would be the usefulness of a karyotype? A. The karyotype would show the shape of the patient’s chromosomes B. The karyotype would indicate the total number of chromosomes present in the patient’s cells C. The karyotype would display the number of homologous pairs of chromosomes that the patient has D. A, B, and C E. Both A and B only 21. What is a frameshift mutation? A. A frameshift mutation is a silent mutation. B. A frameshift mutation occurs when there is the addition or deletion of a nucleotide that is not an exact multiple of 3 nucleotides, so that it shifts (changes) the reading frame of translation, which may lead to massive missense. C A frameshift mutation is a synonymous mutation. D frameshift mutation occurs when a nucleotide substitution does not change an amino acid of the translated protein and often leads to the premature termination of translation resulting in a truncated (short) protein. E. A frameshift mutation is a nonsense mutation

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22. Recall our Biology 1A03 discussion of the laser beam experiment. What is the error shown in the following figure?

A. The kinetochore microtubules are shortening. B. The chromosomes are attached to the kinetochore microtubules and the laser beam has cut the kinetochore microtubules. C. The chromosomes are attached to the nonkinetochore microtubules. D. The nonkinetochore microtubules are shortening. E. Both A and C. 23.Suppose that a medical doctor wants to examine the karyotype of a high-risk baby, who is still developing within the uterus of its mother. Which test should the medical doctor request in order to get the fastest karyotyping and biochemical tests results? A. Amniocentesis will provide the fastest results, since it can be conducted earlier than chorionic villus sampling. B. Chorionic villus sampling will provide the fastest results, since it can be carried out much earlier than amniocentesis. C. Northern blot testing will provide the fastest karyotyping results. D. Base titration will provide the fastest karyotyping results. E. Both A and D are correct. 24. A perspective of how the chimpanzee and human can share most of their nucleotide sequence yet display significant phenotypic differences is that many of the most important sequence differences alter or modify __________________. A. cultural factors B. regulatory sequences and proteins C. environmental factors D. structural genes E. both A and C DR. DA SILVA’S MULTIPLE CHOICE QUESTIONS 25. When considering the sex chromosomes of the human body A. a female will inherit two X chromosomes from her mother. B. recombination does not occur between the X and Y chromosomes in a male. C. a male will inherit a Y chromosome from his mother. D. the X and Y chromosomes have mostly similar genes.

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26. The X-linked inherited disorder, hemophilia, has been transmitted through many generations of the British Royal Family. Queen Victoria was a carrier of hemophilia with a genotype of Hh. Her husband, Prince Albert, was an unaffected male. What was the chance that daughters of this couple would be carriers of hemophilia? A. 0% B. 25% C. 50% D. 75% 27. Mendel’s second law of inheritance tells us that two genes sort independently during gamete formation. When is this least likely to occur? A. When the two genes are on the same arm of the same chromosome. B. When the two genes are on different chromosomes. C. When one gene is on an autosome and the other is on a sex chromosome. D. When the two genes are positioned far apart, on different arms of the same chromosome. 28. When studying a human population, you observe the following association between 2 markers and a disease phenotype. The gene responsible for the phenotype can be mapped, # of people Disease and marker A

5

Disease and marker B

40

Unaffected and marker A 40 Unaffected and marker B 5 A. equal in distance between markers A and B. B. on a different chromosome to markers A and B. C. closer to marker A. D. closer to marker B. 29. If you observe a recombinant phenotype in male offspring for X-linked traits, this is due to the formation of chiasma in which of the following? A. Meiosis in the father of the recombinant male offspring. B. Meiosis in the mother of the recombinant male offspring. C. Mitosis in the father of the recombinant male offspring. D. Mitosis in the mother of the recombinant male offspring. 30. Two phenotypically normal individuals have a child that is affected by an inherited disease trait. What can we conclude about the parents? A. They both likely carry the disease allele. B. They are not the parents of the child. C. They are affected as well. D. No conclusions can be drawn.

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PART B: WRITTEN ANSWER-STYLE QUESTIONS 31-34 (worth 10 marks) PLEASE COMPLETE THE FOLLOWING QUESTIONS IN BLUE OR BLACK PEN. ANSWERS WRITTEN IN PENCIL WILL NOT BE RE-GRADED. DR. KAJIURA’S WRITTEN ANSWER QUESTIONS (7 marks) 31. Describe what might happen to cells if they repeatedly divided without any gap phases? (1 mark + 1 mark + 1 mark = 3 marks).

ANY OF THE FOLLOWING IN ANY ORDER FOR A TOTAL MAXIMUM OF 3 MARKS If cells repeatedly divided without any gap phases, the cells would most likely get progressively smaller in size because they would not have sufficient levels of nutrients and materials (0.5 mark) The cells would have fewer organelles since the necessary “protein machinery” would not be fully developed (0.5 mark) The cells would not have sufficient amounts of replicated DNA for the daughter cells (0.5 mark) The cells would most likely not have the correct social signals present for adequate cell division (0.5 mark) The cells would most likely have missed situations where DNA damage had occurred (0.5 mark). The cells would most likely not function properly (0.5 mark).

32. Compare the specific functions of Meiosis I and Meiosis II in the overall generation of gametes (0.5 + 0.5 = 1 mark) and also state the specific events that are associated with these functions. (0.5 + 0.5 = 1 mark).

The function of Meiosis I is to provide opportunities for genetic variability to be introduced (0.5 mark). The function of Meiosis II is to create the haploid state (0.5 mark) after recombination and assortment created genetic variability. This is accomplished by events such as crossing over (0.5 mark) [also acceptable formation of chiasmata] which occurs during Prophase I of Meiosis I (0.5 mark) and via assortment (0.5 mark) during Anaphase II of Meiosis II (0.5 mark).

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33. Using the following karyotype shown below, a) state this specific aneuploid condition (0.5 mark), b) name the specific syndrome (0.5 mark), and c) describe any two of the associated characteristics of the patient with this specific karyotype. (0.5 + 0.5 = 1 mark).

a) The name of this specific aneuploid condition is Trisomy of Chromosome 13 [ also acceptable 47, 13+] (0.5 mark) b) The name of the specific syndrome is Patau Syndrome (0.5 mark) c) Any two of the following characteristics each worth 0.5 mark for a maximum of 1 mark: Severely mentally challenged (0.5 mark) Cleft palate (0.5 mark) Deaf (0.5 mark) Malformed organs (0.5 mark)

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DR. DA SILVA’S WRITTEN ANSWER QUESTIONS (3 marks) 34. The image below shows a pedigree with a mother, father and their two children. In this family, the father is affected with an X-linked recessive trait, colourblindness. The mother is an unaffected female. (Total = 3 marks)

A. What is the likelihood that this couple’s son will be born colourblind? Why? (0.5 + 0.5 +0.5 = 1.5 marks) -the son will NOT be born colourblind. (0.5 mark) This is because the father DOES NOT pass on (0.5 mark) the affected X chromosome to his son (0.5 mark) (Note: can also accept, because the father only passes on (0.5 mark) the Y chromosome to his son (0.5 mark) B. Will this couple’s daughter be colourblind? Why or why not? What is the possible outcome for the daughter? (0.5 + 0.5 +0.5 = 1.5 marks) -The daughter WILL NOT be colourblind (0.5 mark) because she only inherits ONE copy of the affected X chromosome from the father (0.5 mark) -the daughter can however be a carrier for colourblindness (0.5 mark)...