Midterm 1 Answer Key WQ 2010

Midterm 1 Answer Key WQ 2010 - BlS101-01 ANSWER KEY Winter...

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Unformatted text preview: BlS101-01 ANSWER KEY Winter 2010/MlDTERM 1 Professor 8. O’Neill Student ID #: MIDTERM 1 EXAM January 25, 2010 There are 12 questions and a total of 11 pages to this examination. Write your name at the top of each page. Please show your work for partial credit! HONOR CODE: My signature below affirms that I wrote this exam in the spirit of the honor system of UCD. I neither received nor furnished any “help” during the exam, nor did I use any unauthorized references. Signature Student Authorization: l authorize the University to distribute my grades electronically, posted to SmartSite, using my student ID and to return publicly this graded exam (e.g., handed out after class, or after that left in a bin for me to pick up. Lp+$ Bl8101-01 ANSWER KEY I Winter 2010/MIDTERM 1 Professor 8. O’Neill 1. (8 points) The cells shown below were all from the same individual. Indicate in the table below: 1. the type of cell division (mitosis vs. meiosis). 2. which cell divisions stage is depicted in each figure. b. a. Type of Cell Division Cell Division Stage What is the diploid number of the organism? BlS101—01 ANSWER KEY Winter 2010/MIDTERM 1 Professor 8. O’Neill 2. (10 points) In a human diploid cell there are 46 chromosomes and at Gl approximately 3.5 nanograms of DNA per nucleus. For each of the following stages of the cell cycle identify (1) the number of chromosomes, (2) the number of chromatids, and (3) the amount of DNA per cell. ~ Go“ Cr: we CT a.) End ofS phase: ', i” e 1. 46 chromosomes 0 N g H “‘ ) . . _ ' & 2. 92chromatids ,4 all arAI$rN~etS 0» \ooNu S Quin/T— 3. 7 nanograms (ng) LN” at b.) End of telophase l: t 1. 23 chromosomes 2. 46 chromatids 3. 3.5 nanograms (ng) c.) End of telophase ll: 1. 23 chromosomes 2. 23 chromatids 3. 1.75 nanograms (ng) 3. (8 points) Complete the table below, indicating the sex of the individuals with each of the genotypes. m Drosophila BlS101-01 ANSWER KEY Winter 2010/MlDTERM 1 Professor S. O’Neill 4. (10 points total) Fill in the “blank” answers (2 points each answer):- Please write your answer in the box. a.) is a DNA sequence variation and can be a source of phenotypic differences. A 7/ Single nucleotide polymorphism b.) The region at the end of a chromosome is called the and helps to protect the chromosome end from deteriorating. Telomere x ’L c.) Spindle fibers attach to the during mitosis and meiosis. Centromere A. v) d.) Rapid growth, metabolic activity and replication of the centrioles occur in of interphase. G 47/ l e.) In meiosis, pairing of homologous chromosomes starts in this phase: and becomes visible in this phase: Zygotene l Pomi’ Pachytene lgow'l/ BlS101-01 ANSWER KEY ~—— 5’ Winter 2010/MIDTERM 1 ( ’ 1U Professor 8. O'Neill V‘e 09 I 0 L 5. (8 points) Mendelian genetics: Assume independent assortment; A pure breeding yellow (Y) wrinkled (R) pea plant is crossed to a pure breeding green (y) smooth (r) plant. The F1 generation is then selfed. When the F2 generation is test crossed, what fraction of the F2 green wrinkled individuals would yield half green-smooth, and half green-wrinkled F3 progeny? Place your final answer in the box below. Show your work. P YYKR X 15an 5902MB Cw 1F“ {2v X far 2 9+5 ‘Cov owed" +6T' (L iv CWSS ‘. l \\> ‘Oljz‘r $3 l/L‘yflkr ‘; /L ‘3an FINAL ANSWER: (an new wflinklwb( u; . 63 321.19%) (3%?” 3M0“) ipf‘flO/L Z :iw'CtAL .5,” W 94 art. ONL‘O Z. 0w+ 0‘9 1:7, ojflflfl wminklwfi wk“ “Jive ‘H’C‘S (L‘L‘HOJ Answea KEV * Q 5 BlS101—O1 Name: vs I cm i Winter 2010/MIDTERM 1 Last, First Professor 8. O’Neill 5. (8 points) Mendelian genetics: Assume independent assortment. A pure breeding yellow (Y) wrinkled (R) pea plant is crossed to a pure breeding green (y) smooth (r) plant. The F1 generation is then selfed. When the F2 generation is test crossed, what fraction of the F2 green wrinkled individuals would yield half green-smooth, and half green—wrinkled F3 progeny? Place your final answer in the box below. Show your work. P Y Y X 33rv~ use 60“ “St \L 0L ‘1 X ‘I pu‘M'hcf-(‘i E R V X Z V“ gqirtav’fi, IMP/flooz ’ i ltd. FlNALANSWER: Tlfvmilfl wyarre‘r 0” 3”“ W’LN“ ~—%—~ J; if” 12;» 51 5mm” 9% qneeN WfliN‘C'Ue) _ SNOW {gr __I__ AL.“ : M“ v 2 8 4;. ML gfitfi wILlukluL ‘3 (j/Lr_(_t313 RR ‘ g 5 T§+fip BlSlO1—01 ANSWER KEY Winter 2010/MIDTERM 1 Professor 8. O’Neill 6. (6 points) Genes a, b, and c assort independently and are recessive to their respective alleles A, B, and C. Two triply heterozygous (AaBch) individuals are crossed. What is the probability that a given offspring will be phenotypically ABC, that is exhibits all three dominant traits? Show your work. AmBLCC X 'Aa BLCC What is the probability that a given offspring will be genotypically - homozygous dominant for all three traits? A a POW 3 17 L WM r a “a ; ’31- “ Lor swam. if“) L . (ls/h Gd) x94) 3 l C; 3n BlS101—01 ANSWER KEY Winter 2010/MlDTERM 1 Professor 8. O’Neill 7. (10 points) The pedigree below is for a rare genetic disease in humans. Assume that all people marrying into the family do not carry the abnormal allele. A. Fill in the pedigree, using genotypes for as many individuals as possible. What is the most liker mode of transmission of this disease? (5 points) X-linked dominant plus completed pedigree symbols. Xx X‘Y XAX )0] Xy XAX VAX Xx X‘)’. yy X‘X’X‘ X" K X7 Xfl)’ X7 YA} x7\x"x What is the biggest clue about determining the mode of inheritance? (2 points) All the daughters of an affected father are affected while none of the sons is affected. B. is there a less likely possibility that it could have a different mode of transmission? If so, what is it, and explain your answer below? (3 points) It is possible, but less likely, that it could be autosomal dominant inheritance, and that just by chance or coincidence all daughters and no sons in the 2"d and 4th generations got the disease allele from their father. BlS101-01 Vlfinter 2010/MIDTERM 1 Professor 8. O'Neill ANSWER KEY 8. (10 points) In mice, dwarfism is caused by an X-linked recessive allele, and tan coat is caused by an autosomal recessive allele (coat color is normally brown). If a dwarf, brown female from a pure line is crossed with a normal, tan male, what will be the phenotypic ratios in the F1 and F2? Be sure to include sex of the progeny in your phenotype description. Show yourwork. Use these symbols: XD = normal B = brown Xd=dwarf b=tan P: dwarf, brown female x normal, tan male X4 X XDY Poid's'f \/1 WW I: 5%“me cg +i B\,< Vi A brown) O/V'FI‘ Bio x” x A x B l) x W 3/ B "‘ L " D J rowu wovuoL ‘le°XA< H ”' BXX ‘0 ' 9+ V“! X’Xd +a“) movML $4,, 3 ‘. v.1 Bu. '7: B" brow»!J dwaaoé; 2+, \ /‘l ‘31:: bedxa 41K, AW¢4‘( 945/ 3 ‘ ~ \/.i YD), < 8 «3’7" 8—- Xpy byoup’flOlMCL &, l [’1 —)’~: hn’ywvmoL k, ’15:. B"X‘1Y brougawad' 02H JR; fingdwwtl 6+, BlS101-01 ANSWER KEY Winter 2010/MIDTERM 1 Professor S. O’Neill ' 9. (10 points) Angelina and Brad are contemplating having children‘- As illustrated by the pedigree below, both Angelina and Brad have relatives with galactosemia, a rare autosomal recessive disease. Unless the pedigree suggesfiotherwise, assume all people marrying into the family do not carry the abnormal allele. Show your work. Use the symbols: G = normal 9 = abnormal A. (2 points) What is the probability that Angelina is a carrier for galactosemia? mm B. (2 points) What is the probability that Brad is a carrier for galactosemia? G) G) 3 Brad '5 .WVS+ 60% if? E: %( Aug: {—0 preswce 0C W37qu re assure 3 ‘ .» “(RV/m? '2 ,2— C. (2 points) What is the probability that Brad and Angelina s first child will beaglrl with galactosemia? FCAHTJW (7‘3) ; l T /\« {7(Bfac} Qffl : 3—— g. \z s a ’ - 3 ‘ Mama 33) r 2% Pflg) : BlS101-O1 ANSWER KEY Winter 2010/MIDTERM 1 Professor S. O’Neill 10. (6 points) In peas, the allele for full pod (C) is dominant to the allele for constricted pod (c) and the allele for round seed (W) is dominant to the one for wrinkled seed (w). The two loci assort independently. A cross is made between pea plans of unknown genotypes and unknown phenotypes. The offspring have the following ratios: 3/8 full, round 3/8 full, wrinkled 1/8 constricted, round 1/8 constricted, wrinkled Determine the genotype and phenotype of each parent. Show your work. The dominant character full pod is seen in 3/4 of the progeny and it’s recessive counterpart in the remaining 1/4 of the progeny. This suggests the possibility of mating between heterozygotes. The dominant character round is seen in half of the progeny and the recessive wrinkled in the other half. This suggests mating between a heterozygote and a homozygous recessive. The most likely mating is between full, round and a full, wrinkled parental plants. The parental genotypes are Cch and chw. 11. (4 points) in the Petunia plant, purple flowers are dominant to white flowers. When a purple—flowered plant is self—pollinated, its progeny are 28 purple—flowered plants are 10 white-flowered plants. What proportion of the purple-flowered progeny plants will breed true if they are self—fertilized? Show your work. This example indicated that there is approximately a 3:1 ratio between dominant (purple) and recessive (white) phenotypes among progeny of two dominant parental plants. The parental plant is heterozygous and the expected proportion of homozygous (true breeding) purple progeny is 1/3, since 2/3 of the purple plants are heterozygous and 1/3 are homozygous. 1/3 BlSlO1-O1 ANSWER KEY Winter 2010/MIDTERM 1 Professor 3. O’Neill 12. (10 points total) Define the following vocabulary terms (2 points each answer): a. Mendel’s first law: Two members of a gene pair (alleles) segregate from each other in meiosis; each gamete has an equal probability of obtaining either member of the gene pair. b. Reciprocal cross: A mating in which the traits of the male and female parents are switched in the next cross. Specifically, this term refers to a pair of crosses of the type geneotype A (female) x genotype B (male) and B (female) x A (male). c. X chromosome inactivation: The process by which the genes of an X chromosome in a mammal can be completely repressed as part of the dosage-compensation mechanism. d. Hemizygosity: The kind of inheritance seen when a gene is present in only one copy in a diploid organism; for example, an X-linked gene in a male mammal. e. Bivalents: Two homologous, synapsed chromosomes paired at meiosis during the first meiotic division. 11 ...
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This note was uploaded on 03/29/2010 for the course BIO SCI 101 taught by Professor Kimsey during the Spring '10 term at UC Davis.

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Midterm 1 Answer Key WQ 2010 - BlS101-01 ANSWER KEY Winter...

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