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UCI / BIO SCI / BIO 97 / the following statements refer to events occurring during mitosis

the following statements refer to events occurring during mitosis

the following statements refer to events occurring during mitosis

Description

School: University of California - Irvine
Department: BIO SCI
Course: Genetics
Professor: R. warrior
Term: Fall 2015
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Cost: 50
Description: 2013 B1 Name __KEY_____________________________________ If you were approved to take the exam with a different section than you are enrolled in, or are a UCI extension student, check here: BIO SCI 97, GENETICS, FALL 2013 FINAL EXAMINATION DO NOT OPEN EXAM UNTIL TOLD TO DO SO PHONES OFF, BAGS UP FRONT BEFORE BEGINNING: 1
Uploaded: 06/30/2017
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2013 B1 Name __KEY_____________________________________ If you were approved to take the exam with a different section than  you are enrolled in, or are a UCI extension student, check here:  BIO SCI 97, GENETICS, FALL 2013 FINAL EXAMINATION DO NOT OPEN EXAM UNTIL TOLD TO DO SO PHONES OFF, BAGS UP FRONT BEFORE BEGINNING: 1. Write your name in the upper right corner of this page (and check the box if you’re an extension  student) 2. Write your name and ID number on your Scantron card, and bubble in the code box for your ID. Also,  if you’re an extension student, put an “X” in front of your written ID number, e.g X0698337 3. Bubble in your exam version number (see top right of this page, B1=A, B2=B, B3=C) on the Scantron  4. Read the rules below and sign the pledge. Academic dishonesty (cheating) will not be tolerated in any form.  The following are the EXAM RULES: 1) You must sit in your assigned seat with your ID on the desk and face up (unless given permission  or moved by a professor) 2) The ONLY items allowed on your desk in addition to the exam are: student ID (removed from  any covers/cases), pencil, and an eraser. THIS IS A CLOSED BOOK TEST, NO NOTES ARE  ALLOWED. 3) You may not leave your seat (unless directed by a proctor) except to turn in your exam. 4) Your exam must be completed by you alone—keep your eyes on your own paper. No discussion  with other students. Shield your own answers from other students. 5) Other rules may be announced during the exam and must be followed. 6) You must sign this page to indicate that you understand these rules. IF YOU DO NOT SIGN  BELOW, WE WILL NOT GRADE YOUR EXAM. FAILURE TO FOLLOW ANY OF THESE RULES MAY RESULT IN AN “F” FOR THE EXAM OR CLASS.  I UNDERSTAND THESE RULES AND AGREE TO FOLLOW THEM:  ________________________________________________ (sign your name here) NOTES for all questions: • Affected = has the disease. Unaffected = does not have the disease (but may be a carrier if the  disease is recessive). Normal = does not have the disease and is not a carrier. • For all questions, assume any trait or disease mentioned has 100% penetrance, unless it is  stated otherwise. For inheritance questions, assume no new mutations, unless it is stated  otherwise. • All diseases mentioned are genetic diseases, unless it is stated otherwise. • Pedigrees will only show you if someone is affected (black symbol) or unaffected (white symbol).  If someone is a heterozygous carrier of a recessive disease, they will be depicted as unaffected.B1-Page 2 MULTIPLE CHOICE – ANSWER ON YOUR SCANTRON CARD Each question is worth one point. 1 Consider the following 4 Statements about the events of meiosis and mitosis and then pick the  one Answer that most accurately and most completely describes the these events. Statements: 1. Disjunction of sister chromatids occurs in mitosis 2. Disjunction of sister chromatids occurs in meiosis 3. Disjunction of homologous chromosomes occurs in mitosis 4. Disjunction of homologous chromosomes occurs in meiosis Answer: A. 1 and 3 are TRUE B. 2 and 4 are TRUE C. 1, 2 and 4 are TRUE D. 1, 2, and 3 are TRUE E. 1, 2, 3 and 4 are TRUE 2 Which of the following statements (A-D) is TRUE? A. In any given population, most people will express the dominant phenotype B. A dominant allele will eventually become fixed in the population C. An individual must be homozygous in order to express a dominant phenotype  D. An individual must be homozygous in order to express a recessive phenotype  E. More than one of the above statements (A-D) is true. 3 In humans, both freckling (that is, having freckles) and dimpling (that is, having dimples) are  autosomal dominant traits. If a woman with freckles and dimples (Felicia) mates with a non dimpled, non-freckled man (David), what can we say with certainty about their kids? A. All of their kids will be both freckled and dimpled B. We would expect some of their kids to be dimpled and some not to be C. None of their kids will be both freckled and dimpled D. Whether or not the kids are freckled or dimpled will depend upon the sex of the kids E. None of the above (A-D) can be said with certainty 4 Which of the following statements are FALSE? A. The forced sterilization of individuals judged to be genetically “inferior” has never been  practiced in the United States B. Pre-implantation genetic diagnosis is a technique that can be used by couples to choose  the sex of their child-to-be. C. By saving the lives of patients with recessive genetic diseases who would have otherwise  died, modern medical science is significantly reducing the fitness of the human gene pool D. More than one of the above statements are false E. None of the above the statements are false 5 Which of the following statements about forensic DNA typing is FALSE? A. DNA typing is generally very reliable, provided enough sample material is available B. The preferred method involves the analysis of four independent loci that are  polymorphic due to short tandem repeats C. Over a dozen people have been freed from death row after DNA evidence proved they  were innocent D. Mixed samples (more than one person’s DNA in the sample) are frequent and can be hard  to interpret E. None of the above statements are false


If a woman with freckles and dimples (Felicia) mates with a non dimpled, non-freckled man (David), what can we say with certainty about their kids?



Don't forget about the age old question of uci scantron

B1-Page 3 6 The following statements refer to genetically-modified (GM) organisms. Which of the following  statements (A-D) are TRUE? A. GM organisms are banned in the U.S. B. GM animals are banned in the U.S., but not GM crops C. GM insects have never been intentionally released into the wild D. GM crops have been proven to be less nutritious than organic crops, and have been shown  to increase cancer risk in people who eat them. E. None of the above statements (A-D) is true. 7 Next 2 questions: Imagine a world where there are three human sexes, males, females, and  bivalves. Bivalves have two sex chromosomes: X and Z. Bivalves can mate with either males or  females, but conceptions with the YZ genotype are early embryonic lethal. The Z chromosome  has no genes in common with either X or Y, but (just as Y does) it pairs with X during meiosis. Which couple could have kids of any of the three sexes? A. Male-Bivalve B. Female-Bivalve C. Male-Female D. A and B E. Neither A, B or C 8 A male and a bivalve mate, and a kid is born with the karyotype 47, XYZ. What could have  caused this? A. Non-disjunction during meiosis I or meiosis II in the male or the bivalve B. Non-disjunction during meiosis I in the male or the bivalve C. Non-disjunction during meiosis II in the male or the bivalve D. This could only result from a non-disjunction event in the male E. This could only result from a non-disjunction event in the bivalve 9 (i) Can Turner syndrome (45, X) be the result of a non-disjunction event occurring during  gametogenesis in a man? (ii) Can Klinefelter syndrome (47, XXY) be the result of a non-disjunction event occurring during  gametogenesis in a woman? A. (i) = yes; (ii) = yes B. (i) = yes; (ii) = no C. (i) = no; (ii) = yes D. (i) = no; (ii) = no 10 Non-disjunction of chromosome 21 occurs during meiosis I in a human female. After meiosis II  occurs normally, what are the total numbers of chromosomes in the four resulting gametes? A. 24, 24, 22, 22 B. 24, 22, 23, 23 C. 47, 45, 23, 23 D. 46, 46, 0, 0 E. 47, 47, 0, 0


What could have caused this?



If you want to learn more check out what does 97 mean

B1-Page 4 11 6. A deamination occurs on the cytosine residue in the following DNA sequence. This cytosine  residue happens to be methylated on the 5-position of the aromatic ring.  5'-GAATCG-3'.  (Note: the top strand is shown; this is the strand where the deamination occurs.) If the mutation  is not repaired, and a round of DNA replication occurs, then the sequence of the newly replicated complementary strand (i.e., the bottom strand) will be: A. 5'-CAATTC-3' B. 5'-CGATTC-3' C. 5'-GAATCG-3' D. 5'-GAATTC-3' E. None of the above 12 (i) A missense mutation results in the change of one amino acid for another. (ii) A silent mutation is likely to be more damaging to a protein’s function than a frameshift. A. (i) = TRUE; (ii) = TRUE B. (i) = TRUE; (ii) = FALSE C. (i) = FALSE; (ii) = TRUE D. (i) = FALSE; (ii) = FALSE 13 (i) On average, each new kid born carries one new mutant allele that was not present in the  somatic cells of either parent. (ii) Heterozygote superiority is a force that removes disease-causing alleles from the gene pool A. (i) = TRUE; (ii) = TRUE B. (i) = TRUE; (ii) = FALSE C. (i) = FALSE; (ii) = TRUE D. (i) = FALSE; (ii) = FALSE 14 Which of the following statements (A-C) is FALSE? A. Multifactorial traits are determined by multiple genes and the environment B. Variance in a trait between identical twins must be due to genetic variance C. Variance in a trait between fraternal twins cannot be due to environmental variance D. None of the above statements (A-C) is false. E. More than one of the above statements (A-C) is false. 15 Next two questions: Consider two small populations of lizards on two islands in the south seas.  In each population there are 9 lizards. The lizards are colored green (genotypes GG or Gg at the  skin color locus) or white (genotype gg). • On Fire Island, all the lizards are green (3 are GG and 6 are Gg). • On Skull Island, there 8 green lizards (4 are GG and 4 are Gg) and 1 white lizard. As it turns out, the allele frequencies on both islands are the same.  What is the allele frequency, p, of the G allele? What is the allele frequency, q, of the g allele? A. p = 1/3, q = 2/3 B. p = ¾, q = ¼ C. p = 2/3, q = 1/3 D. p = ¼, q = ¾ E. p = 8/9, q = 1/9 16 Does the population of lizards on Fire Island seem to be at Hardy-Weinberg equilibrium? Does the population of lizards on Skull Island seem to be at Hardy-Weinberg equilibrium? (Ignore the small size of these populations when answering this question) A. Fire = yes Skull = yes B. Fire = yes Skull = NO C. Fire = NO Skull = yes D. Fire = NO Skull = NO E. The term “Hardy-Weinberg” is not used when talking about populations or equilibria.


After meiosis II occurs normally, what are the total numbers of chromosomes in the four resulting gametes?



If you want to learn more check out How many centimorgans separate two independently-assorting loci?

B1-Page 5 17 Which of the following do NOT contribute to keeping low-fitness alleles in our gene pool? A. Late age of disease onset B. Efficient selection against rare recessive disease-causing alleles  C. New mutations D. All of the above (A-C) contribute to keeping low-fitness alleles in our gene pool E. None of the above (A-C) contribute to keeping low-fitness alleles in our gene pool 18 Which of the following statements is TRUE? A. Most cancers occur in people who have inherited a cancer predisposition disease B. Somatic mutations hardly ever occur C. Most somatic mutations don’t matter to the organism D. More than one of the above statements is true E. None of the above statements are true 19 24. Fill in the blanks. A gene that encodes a protein required for the repair of certain types of DNA  damage is likely to be a _______. A gene that encodes a protein required to relay a signal telling a  non-dividing cell to divide is likely to be a _______. A. oncogene; tumor-suppressor gene B. tumor-suppressor gene; oncogene C. oncogene; oncogene D. tumor-suppressor gene; tumor-suppressor gene E. malignant, hemizygous 20 Which of the following statements is FALSE:? A. Cancer is a genetic disease of somatic cells B. Somatic mutations are involved in many major diseases in the U.S., including cancer,  heart disease, diabetes and schizophrenia C. Mutations in several distinct genes are required for a normal cell to become cancerous D. Oncogenes are usually dominant E. None of the above statements is false 21 Which of the following is true about someone with an inherited predisposition to cancer such a  familial adenomatous polyposis or BRCA-associated familial breast cancer? A. Every cell of their body contains a defective, loss-of-function allele of a tumor  suppressor gene B. Every cell of their body contains a gain-of-function allele of an oncogene C. Most cells in their body contain multiple cancer-causing mutations D. More than one of the above (A-C) is true E. None of the above (A-C) is true 22 A boy is born with an autosomal recessive genetic disease that is determined by a single-gene  defect. Both his parents are unaffected. What can we say about the parents genotypes with  respect to the relevant locus?: A. One parent is homozygous B. Both parents are homozygous C. Only one of the two parents is a carrier D. Mom is a heterozygote, and dad is hemizygous wild-type E. Both his parents must be heterozygotes 23 A man and a woman, both unaffected with Spinkle’s disease, mate and have an affected daughter.  What mechanism(s) of inheritance can we NOT rule out for Spinkle’s disease? (DO NOT assume the disease is rare) A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. More than one of the above cannot be ruled out

Don't forget about the age old question of two pure-breeding mutant plants produce white flowers. when they are crossed, all of the progeny have wild-type purple flowers. what does this genetic complementation tell you?

B1-Page 6 24 This is a two-part true false question. I. Recombination mistakes are repaired by nucleotide excision repair II. Recombination mistakes usually result in point mutations A. I = TRUE; II = TRUE B. I = TRUE; II = FALSE C. I = FALSE; II = TRUE D. I = FALSE; II = FALSE 25 Sickle cell anemia is an autosomal recessive disease with an incidence of 1/400 among African Americans. What is the approximate frequency of heterozygote carriers for this disease within this  population? Assume Hardy-Weinberg conditions apply. A. 1/400 B. 1/200 C. 1/20 D. 1/10 E. 1/4 26 Sally’s maternal grandmother (i.e. her mother’s mother) is affected by a rare x-linked recessive  condition, but no one else in the family is. What are the chances that Sally is a carrier of this  condition? A. 100% B. 67% (2/3) C. 50% (1/2) D. 25% (1/4) E. 0% 27 Next two questions: 1 2 3 4 5 6 7 89 10 11 12 The pedigree shows the inheritance of a genetic condition in four generations of a family, which  includes a first cousin mating. Based on the pedigree, the mode of inheritance of this condition could be… A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. A or B 28 Assume the condition in the above pedigree is very rare in the general population. Then the  mode of inheritance of this condition is most likely… A. Autosomal recessive B. X-linked recessive C. Autosomal dominant D. X-linked dominant E. C or D

Don't forget about the age old question of oct 23, 2015 uploaded

B1-Page 7 Next 4 questions: You want to create a restriction map of a plasmid that is 8 kb long, using enzymes A  and B. A colleague has helpfully performed digests and run a gel, shown below. Lane 2 contains plasmid  cut with enzyme A, lane 3 plasmid cut with enzyme B, and lane 4 plasmid cut with a mix of enzymes A  and B. Your colleague imaged the gel in a way that the intensity of band fluorescence does not reflect the  quantity of DNA in each band. Help fill in the restriction map shown below, in which 2 sites have already  been filled in for you. Two consecutive sites are 1kb apart, and no site can be cut by both A and B. Hint: start by figuring out how many sites each enzyme cuts.  B  + B A A   s :::r423 e  keeernnnaaaaLMLL6kb5kb 4kb 3kb 2kb 1kb Site 1: A Site 2: neither B Site 8 nor A cut here Site 3 Site 7 Site 4 Site 6 Site 5 29 What enzyme cleaves at site 3? A: A cleaves B: B cleaves C: Neither A nor B cleave D: There is not enough information to answer the question 30 What enzyme cleaves at site 4? A: A cleaves B: B cleaves C: Neither A nor B cleave D: There is not enough information to answer the question 31 What enzyme cleaves at site 5? A: A cleaves B: B cleaves C: Neither A nor B cleave D: There is not enough information to answer the question 32 4. What enzyme cleaves at site 6? A: A cleaves B: B cleaves C: Neither A nor B cleave D: There is not enough information to answer the question

We also discuss several other topics like the ancient greeks saw the human form even in their temple styles.

ANSWER: A sites are at 1 and 7, B sites are at 4 and 8 END OF 1 POINT QUESTIONSB1-Page 8 2 POINT QUESTIONS – EACH QUESTION HAS ONLY ONE RIGHT ANSWER FILL IN THAT ANSWER FOR BOTH NUMBERS ON YOUR SCANTRON 33&34 The DNA of three people (X, Y and Z) is digested with the same restriction enzyme and probed  with the same molecular probe using the Southern blotting procedure. The pattern shown  above is obtained. What can be said about the locus that the probe detects? A. X = heterozygote, Y = homozygote, Z = heterozygote  B. X = heterozygote, Y = homozygote, Z = homozygote C. X = homozygote, Y = heterozygote, Z = homozygote D. X = homozygote, Y = heterozygote, Z = heterozygote E. X = heterozygote, Y = heterozygote, Z = homozygote 35&36 In Drosophila, the mutations eyeless (ey) and eyegone (eg) are two different genes that map to the second  chromosome and both result in eyes that are smaller than wildtype (+). If a homozygous eyeless mutant fly  is mated to a homozygous eyegone mutant, which of the following offspring would you expect? a. all ey+ eg+ heterozygotes b. 1/2 eyeless and 1/2 eyegone flies c. all homozygous + (wild type) flies d. 3/4 eyeless to 1/4 eyegone ratio e. 1/2 eyeless and eyegone to 1/2 normal 37&38 A dihybrid plant was crossed; the F2 generation consisted of: 860 tall plants with purple  flowers; 285 dwarf, purple plants; 340 tall, pink plants; and 115 dwarf, pink plants. The data  remind you of a 9:3:3:1 ratio, but you’re not sure whether the progeny numbers are consistent with your  null hypothesis that this is simple mendelian inheritance. After referring to the Chi-square table you  should:  A Chi-Square Table Probability Degrees of Freedom 0.9 0.5 0.1 0.05 0.01 1 0.02 0.46 2.71 3.84 6.64 2 0.21 1.39 4.61 5.99 9.21 3 0.58 2.37 6.25 7.82 11.35 4 1.06 3.36 7.78 9.49 13.28 5 1.61 4.35 9.24 11.07 15.09 a. reject your hypothesis;  b. accept your hypothesis c. perform a backcross;  d. fudge your data;  e. use a lower critical chi-square value

B1-Page 9 39&40 The following fictitious enzymes cut at positions shown by triangles.  5’ 3’  … GGGCCC … 5’ 3’  … TTTAAA … 3’ 5’ 3’ 5’ … CCCGGG … … AAATTT … Enzyme 1 Enzyme 2 TTGCAA  … GGTACC … 5’ 3’ 5’ 3’ 3’ 5’ 3’ 5’ … AACGTT … … CCATGG … Enzyme 3 Enzyme 4 You are given a DNA fragment that has been cut with enzyme 1 at one end, and with enzyme 2  at the other end. You would like to ligate this fragment into a vector whose map is given below;  the position at which enzymes cut the vector are shown on the map. Amp Enzyme 2 resistance n tioain licrigpoeREnzyme 1 Enzyme 2 Enzyme 3 Enzyme 4 With which enzymes will you digest the vector? A. Enzymes 1 and 3 B. Enzymes 2 and 3 C. Enzymes 1 and 4 D. Enzymes 1 and 2 E. Enzymes 3 and 4 41&42 Skin pattern in a new frog species is controlled by three alleles of the same gene. Two alleles,  Green Spots and White Stripes, are codominant and are both recessive to Black. Which of the  choices below includes ALL the phenotypes expected to exist in the frog population in the wild ? A. White Stripes, Green Spots, White Stripes and Green spots B. Green Stripes, White Spots and uniform Black C. Green spots, White Stripes, Green Spots and White Stripes, Uniform black D. Uniform Black, Black spots, White Stripes, Green Stripes and White spots E. White Stripes, Green Spots, Black Stripes, Black Spots, Green spots & white stripes

END OF EXAM – HAPPY HOLIDAYS !
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