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EXAM 2 1. Chronic myeloid leukemia is often caused by which kind of chromosomal abnormality? a. Reciprocal translocation b. Duplication c. Inversion d. Deletion e. Robertsonian translocation 2. This protein inhibits an important tumor suppressor protein a. MPF b. E2F c. P53 d. APC/C e. Mdm2 3. Why are the pseudoautosomal regions of the sex chromosome important? a. They contain most of the genes on X and Y b. They allow for crossing over and pairing of X and Y in meiosis c. They determine the sex of the organism d. They are essential for X-inactivation e. They are the sites where spindle fibers bind X and Y 4. In an individual who is heterozygous for a sizable inversion, the affected chromosome will form a _______ while interacting in Prophase I. a. Loop b. Bivalent c. Quadrivalent d. Hairpin e. Barr body 5. In an individual who is heterozygous for a sizable reciprocal translocation, the affected chromosomes will form a ______ while interacting in Prophase I. a. Loop b. Bivalent c. Quadrivalent d. Hairpin e. Barr body 6. P53 is one of our most important tumor repressors. When DNA damage is sensed, p53 may do all of the following except… a. Induce apoptosis b. Lengthen telomerase c. Arrest the cell cycle d. Inhibit angiogenesis e. Inhibit metastasis 7. What would be the result of a cell acquiring a mutation that causes a gain of function of E2F ( it can no longer be inhabited)? a. The cell would lose its ability to stop the cell cycle to repair DNA
b. The cell would divide before all kinetochores attach spindle fibers c. The cell would begin DNA replication even if no more cells are needed d. The cell would get stuck at metaphase e. The cell would lose the ability to undergo apoptosis 8. What would be the result of a cell acquiring a mutation that causes production of non- functional APC/C protein? a. The cell would lose its ability to repair DNA b. The chromosome would divide before all spindle fibers attach, causing aneuploidy c. Regulation of DNA replication would be lost d. The cell would get stuck at metaphase e. The cell would lose the ability to undergo apoptosis 9. What would be the result of a cell acquiring a mutation that causes production of non- functional Securin protein? a. Regulation of DNA replication would be lost b. Telomeres would be constantly lengthened c. The cell would lose the ability to undergo apoptosis d. The cell would lose its ability to stop the cell cycle to repair DNA e. The chromosome would divide before all spindle fibers attach, causing aneuploidy 10. Why is aneuploidy of the Y chromosome less deleterious than aneuploidy of an autosome? a. The cell can produce extra gene expression from the Y chromosome b. The cell has a mechanism for silencing all but one Y chromosome c. The pseudoautosomal regions pick up the slack d. There are few genes on the Y chromosome e. A new Y chromosome can be created from X 11. Why is aneuploidy of the X chromosome less deleterious than aneuploidy of an autosome? a. There are few genes on the X chromosome b. The genes on the X chromosome are only relevant to fertility c. The cell can produce extra gene expression from a long X chromosome d. The cell has a mechanism for silencing all but one X chromosome e. The X chromosome is not required for life 12. In order to promote cancer, tumor suppressor genes must have a ________, and proto- oncogenes must have a _______. a. Duplication; delection b. Loss of function; gain of function c. Translocation; inversion d. Gain of function; loss of function e. Loss of function; loss of function 13. Once a gene becomes duplicated, the second copy is less likely to … a. Maintain its original function b. Be duplicated again c. Be deleted d. Gain a new function e. Be inverted 14. Ectopic recombination between direct repeats on the same chromosome would result in …
a. Inversion b. Deletion c. Duplication d. Reciprocal translocation e. Robertsonian translocation 15. Ectopic recombination between direct repeats on two homologous chromosomes would result in … a. Inversion b. Deletion c. Duplication d. Reciprocal translocation e. Robertsonian translocation 16. Ectopic recombination between repeats on two-homologous chromosomes would result in a. Inversion b. Deletion c. Duplication d. Reciprocal translocation e. Roberstsonian translocation 17. What is the chromosomal location circled below? *** p = short arm, q = long arm *** a. 7-2-31 b. 31p7 c. 7q31 d. 31q7 e. 7p31 18. What is the chromosomal location circles below?
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a. 13q4 b. 4q13 c. 4-1-13 d. 13p4 e. 4p13 19. You created a monoploid plant. What do you need to do to make it produce seeds? a. Nothing. Let it grow b. Treat it with auxin c. Breed it with a 2n plant d. Treat it with colchicine e. Cold-shock it 20. You examine the chromosomal content of four sperm cells that arose from the same spermatocyte. What genetic anomaly occurred to give rise to the sperm? n = normal haploid number of chromosomes (2 in this case) a. Chromosomal fusion b. Nondisjunction in meiosis I c. Nondisjunction in meiosis II d. Chromosomal deletion e. Nondisjunction in mitosis 21. You examine the chromosomal content of four sperm cells that arose from the same spermatocyte. What genetic anomaly occurred to give rise to the sperm? n = normal haploid number of chromosomes (2 in this case) a. Chromosomal fusion b. Nondisjunction in meiosis I c. Nondisjunction in meiosis II d. Chromosomal deletion e. Nondisjunction in mitosis 22. If a tumor develops but does not affect the organism’s health, it is best described as … a. Metastatic b. In remission c. Malignant d. Senescent
e. Benign 23. If a tumor develops and spreads to other areas of the body, it is best described as … a. Malignant b. Senescent c. Benign d. Metastatic e. In remission 24. If a tumor develops and affects the organism’s health, but it has not spread to other tissues, it is best described as …. a. Malignant b. Senescent c. Benign d. Metastatic e. In remission 25. A crossover falling within a pericentric inversion loop would produce … a. Acentric and dicentric chromosomes b. Viable gametes c. A Robertsonian translocation d. Deletions and duplications e. A reciprocal translocation 26. A crossover falling within a paracentric inversion loop would produce … a. Acentric and dicentric chromosomes b. Viable gametes c. A Robertsonian translocation d. Deletions and duplications only e. Polyploidy 27. Whole acrocentric chimp Chromosomes 2A and 2B bear remarkable similarity to the q and p arms of Human Chromosome 2, respectively. A(n) _____ event likely occured in the human lineage, after the point when human ancestors diverged from chimp ancestors, in order to create modern Human Chromosome 2. a. Reciprocal translocation b. Duplication c. Inversion d. Nonreciprocal transversion e. Roberstsonian translocation 28. In Georgii Karpechenko’s only stroke of good luck, this genetic event led to the ability of Brassicoraphanus to produce seeds a. Nondisjunctions b. Endoreduplication c. Reciprocal translocation d. Roberstsonian translocation e. Monoploidy 29. Why wouldn’t Georgii Karpecheko’s plant brassicorphanus produce seeds? a. Its three homologs can’t be divided evenly in meiosis
b. The biochemistry of cabbages and radishes are incompatible c. Brassicorphanus is aneuploid d. Radishes and cabbages reproduce through budding e. Cabbage and radish chromosomes can't pair with each other in meiosis 30. What might result if M checkpoint did not function properly? a. Chromosomal fusion b. Polyploidy c. Too many cells of that kind d. Division with damaged DNA e. Nondisjunction 31. Cancer is sometimes treated with radiation therapy, but we know radiation can cause cancer. Why doesn’t the radiation harm the patient? a. The radiation is aimed away from the healthy cells b. The DNA of healthy cells is protected by histones c. Healthy cells take DNA damage, but they are able to repair it d. The radiation therapy creates less harmful benign tumor cells that out-compete the cancer e. Patients take iodine pills which make them impervious to radiation damage 32. Familial retinoblastoma is often caused by an inherited mutation in the RB gene. As the cellular level, this mutation appears to be ______, while at the organismal level, it appears to be ________ a. Recessive; dominant b. Malignant; benign c. Oncogenic; tumor suppressive d. Dominant; recessive e. Rare; common 33. Chemotherapy drugs cause hair loss and nausea because hair root cells and epithelial cells of the digestive tract … a. Accumulate many mutations b. Divide rapidly c. Are extremely sensitive to toxins d. Are unable to control the cell cycle e. Have low expression of tumor suppressors 34. Loss of function mutations are usually … a. Harmless b. Dominant c. Malignant d. Recessive e. Homozygous 35. Gain of function mutations are usually … a. Dominant b. Harmless c. Cancerous d. Recessive e. Homozygous
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36. You wish to create a monoploid plant from a diploid plant. What is the first step required to generate a haploid plant tumor that can later be grown up into an adult plant in culture? a. Colchicine treatment b. Auxin treatment c. Cold shock d. Self-fertilization e. Mutagenication 37. If a cell at the restriction point does not receive a signal to divide, it will enter _____ and possibly never return a. Meiosis b. Apoptosis c. G2 d. Mitosis e. G0 38. One function of p53 is to respond to DNA damage. When activated, it will … a. Halt the cell cycle b. Inhibit angiogenesis c. Inhibit metastasis d. Induce apoptosis e. All of the above 39. If a cell has telomeres that have become too short, at which cell cycle checkpoint would the cell cycle likely be arrested? a. G1 b. G2 c. G0 d. M e. None of the above 40. If a cell telomeres that have become too short, it will usually enter a state called _____ where it continues living but stops dividing, usually for the rest of its life a. Apoptosis b. Mitopause c. G 0 d. G 3 e. Senescence 41. Look at the karyotype above. This person has …. a. Kilefelter’s Syndrome b. Down Syndrome c. Chronic Myeloid Leukemia d. Turner Syndrome e. No genetic disorder
42. What part of the genome was used to show that Genghis Khan is the most successfully breeding human in all of history? a. Y chromosome b. Mitochondrial DNA c. Non-coding autosomal DNA d. X-chromosome e. Coding autosomal DNA 43. Two organisms heterozygous for gene A are bred together. Two alleles of gene A show complete dominance . What would be the ratio of phenotypes in the offspring? a. 1:1 b. 1:2:1 c. 3:1 d. 9:3:3:1 e. All offspring are the same 44. Two organisms heterozygous for gene A are bred together. Two alleles of gene A show incomplete dominance . What would be the ratio of phenotypes in the offspring? a. 1:1 b. 1:2:1 c. 3:1 d. 9:3:3:1 e. All offspring are the same 45. You breed two pink snapdragons together. What phenotypic ratio would you expect in the offspring? a. 1 red: 1 white b. 1 red: 2 pink: 1 white c. 3 red: 1 white d. 3 pink: 1 white e. All pink 46. Which of the following is an accurate paraphrasing of Mendel’s Law of Segregation? a. Genes exist as pairs, and in gamete formation, each gamete gets one member of each pair b. In meiosis I, non-homologous chromosomes will line up at segregates positions c. During gamete formation, the way ne pair of alleles assort into gametes does not affect any other pair of alleles d. Dominant and recessive alleles are kept segregated from one another in mitosis e. In meiosis I, the two members of pair of genes will line up in metaphase at independent positions 47. Which of the following is an accurate paraphrasing of Mendel’s Law of Independent Assortment? a. Genes exist as pairs and in gamete formation, each gamete gets one member of each pair b. During gamete formation, the two members of a pair of alleles can assort into gametes in any manner c. During gamete formation, the way on pair of alleles assort into gametes does not affect any other pair of alleles
d. The dominant alleles in a pair will always assort together e. During gamete formation, the two members if a pair of genes will line up in metaphase at independent positions 48. A testcross is performed by breeding an individual of dominant phenotypes/unknown genotype with a … a. Sibling b. Parent c. Heterozygote d. Homozygous dominant e. Homozygous recessive 49. A testcross is performed on a plant derived from a round, yellow pea. The offspring from this testcross are 50% round green, 50% round yellow. What is the genotype of the parent plant? W = round, w = wrinkled, G = yellow, g = green a. WWGG b. WwGg c. WWGg d. WwGG e. wwGg 50. A testcross is performed on a plant derived from a round, yellow pea. The offspring from this testcross are 50% wrinkled yellow, 50% round yellow. What is the genotype of the parent plant? W = round, w = wrinkled, G = yellow, g = green a. WWGG b. WwGg c. WWGg d. WwGG e. wwGg 51. Genetically speaking, an organism that is true-breeding is … a. Hybrid b. Homozygous c. Heterozygous d. Hemizygous e. Dominant 52. Mendel’s wrinkly peas were created when a(n) ______ inserted into the gene SBEI, causing the pea to be unable to create amylopectin from starch a. Virus b. Transposable element c. Intron d. Centromere 53. Gregor Mendel chose to study traits in his pea plants that are diametrical and mutually exclusive. A trait with these properties usually … a. Is controlled by multiple genes b. Results from aneuploidy c. Exhibits incomplete dominance d. Is controlled by a single gene
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e. Is X-linked For the following genes/proteins, state whether they are tumor suppressors (Fill A) or proto- oncogene (Fill B) 1. APC/C ------------------ Oncogene 2. Cyclins and Cdks ------ Oncogene 3. Mdm2 ------------------ Oncogene 4. E2F ----------------------- Oncogene 5. Bax ----------------------- Tumor Supressor 6. Growth factor receptors Oncogene 7. Securin ------------------- Tumor Suppressor 8. Separase ----------------- Oncogene 9. Telomerase -------------- Oncogene 10. Bcl2 ---------------------- Oncogene 11. BRCA1------------------- Oncogene 12. RB ------------------------ Tumor Supressor 13. BRCA2 ------------------ Tumor Suppresor Shown below are four sets of chromosomes pairing with their homologs in prophase I/metaphase O. Match them with the kind of chromosomal aberration each one displays. 35. Inversion 36. Reciprocal translocation 37. Roberstsonian translocation 38 . deletion
Around genes A-D are three homologous repetitive regions (1-3) with various orientations Ectopic recombination occurs between regions 1 and 3. a. What type of repeats are 1 and 3, with respect to their orientation? Direct repeats b. Draw the resulting molecule(s) after recombination has occurred (drawing the intermediate is not required). Include the genes as well as the repetitive regions, with orientation arrows. If more than one molecule would result, draw both. c. What is the name for this type of chromosomal alteration? Deletion Around genes A-D are three homologous repetitive regions (1-3) with various orientations Ectopic recombination occurs between regions 1 and 3. d. What type of repeats are 1 and 3, with respect to their orientation? Inverse / opposites e. Draw the resulting molecule(s) after recombination has occurred (drawing the intermediate is not required). Include the genes as well as the repetitive regions, with orientation arrows. If more than one molecule would result, draw both. --- A --- ← ( ) --- C --- ←(2) --- B --- → ( ) --- D --- f. What is the name for this type of chromosomal alteration? Inversion Describe the molecular events regulating M checkpoint. Mention the functions of securin, separase, and APC/C, as well as the overall goal of M checkpoint. g. Normally, APC/C is being inhibited until it receives a signal to go to securin and inhibit it. When secring releases separase, separase will then go to the chromosome to A B C D 1/3
then cut the cohesions, by cutting the cohesions, the chromosomes are freely able to divide and go into anaphase. The goal is to ensure each kinetochore gets a spindle fiber → no nondisjunction 1. Taxol was one of the first chemotherapy drugs. What does this drug do on a molecular level, AND why does it cause nausea and hair loss? Blocks the cell cycle in its G1 or M phases by stablizing the microtubule cystoskeleton against depolymerization Cell cycle phases G0 and senescence both share the property that the cell has left the cell cycle proper and stopped dividing. For each of these two phases, answer the following. At what cell cycle checkpoint would a cell usually enter the phase, AND what is a common reason for the cell to enter that phase and leave the cell cycle… G0 Senescence Enter @ At G1 After G2 checkpoint Why Enter? No growth signal Cells in senescence will survive but not divide Seedless fruits and hybrid organisms (like the mule) are sterile for different reasons. Explain why each of these TWO conditions results in sterility, at the chromosomal level. a. A hybrid organism (like a mule) is sterile because when P1 created an offspring they have a different number of chromosomes (however, they are similar enough to create a viable offspring) so when F1 wants to produce offspring they cannot. Because their odd matching chromosomes cannot cross over in meiosis to create germ cells. While with a seedless fruit they are not diploid so when meiosis tries to occur chromosomes are unable to make a regular haploid cell. There is an odd number of homologs and so they cannot divide evenly. 2. Explain in detail why watermelons from the store are seedless. How and why, precisely, do they fail to make seeds? a. Watermelon are seedless because they are polyploid which allows them to grow large and vigorous; they can not make seeds because their chromosomes cannot divide evenly during anaphase since they are triploidy 3. Except for trisomy 21, all aneuploidies for autosomes are generally fatal, but aneuploidy of the sex chromosomes is generally survivable. i) why is aneuploidy of the X chromosome less disruptive? AND ii) why is aneuploidy of the Y chromosome less disruptive? a. Aneuploidy of X chromosomes less disruptive because X can inactivate extraX’s to make more Barr Body’s (turns off expression) b. Y chromosome aneuploidy is less disruptive because Y is largely heterochromatic (except pseudoautosomal regions)
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4. Name two genetic modifications that allowed the icefish to inhabit waters below zero degrees Celsius. a. No hemoglobin (RBC’s) b. Natural antifreeze The “end replication problem” states that our chromosomes will shrink by losing a bit of DNA off the ends of the chromosomes every time they replicate. Explain why this shrinkage occurs AND explain how the cell counteracts this shrinkage. a. After DNA replication occurs, the DNA shortens a little bit because at the end there is a primer in place, however, DNA polymerase (I or III) may not be able to code it because of the lack of 3’ OH. This is where telomerase comes into play and places an non-coding strand of nucleic acids to help lengthen the strand and prevent important DNA sequences from being deleted. In class, we highlighted two points during meiosis where genetic diversity is increased among the pool of gametes created. What are these two points and how does each increase genetic diversity? 1) Crossing over in the pachytene phase of prophase because of the mixing of genes 2) Lining up of the homologous chromosomes during metaphase Histones are proteins that help condense out DNA a) Histone tails are sometimes chemically modified by addition of certain functional groups. What are the two most common modifications and what are their consequences? i) Modification #1: Acetylation → away (1) DNA can fall off easily (2) Over express!!! ii) Modification #2: Methylation (1) DNA is held on too much (2) Under express!!! b) Which amino acid below is likely found in higher amounts in histone proteins and WHY? i) Lysine - it is “+” positively charged and DNA is “-” negatively charged. We want DNA to “stick” to histones and +/- charges attract