test2 - BIOL 230 Genetics Test 2 name Team name When you...

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Unformatted text preview: BIOL 230 Genetics Test 2 name: \ Team name: When you finish, please sign the pledge (but only if it is trust): . “On my honor, I pledge that l have neither given nor received any aid on this exam, and I will not/did not discuss it, directly or indirectly, With any other student until after 1:00 PM today. l understand that to do so would constitute a violation of the Xavier University policy on academic honesty and could result in a 0 on this test." Basic Questions (45/60 goints). 1 -— 5: 2 pts each. 1. in a Mendelian cross, which generation is the “hybrid” generation? A. P @Fi c. Q D. F2 E. z 2. Imagine a cross between an organism with the dominant phenotype and another organism with the recessive phenotype. if the cross produces offspring in a phenotypic ratio of 1 dominant to 1 recessive, what does this indicate about the parents’ genotypes? A. Two genes involved; one parent was homozygous dominant and the other parent was homozygous recessive. One gene involved; one parent was heterozygous and the other parent was homozygous recessive. C. Two genes involved; both parents were heterozygous. :5 a D. One gene involved, both parents were homozygous. a g a a a f" ,9 .3. "Linked alleles tend to stay together in meiosis." To what does this short sentence refer? A. If a heterozygote has the genotype AaBb, then after meiosis the majority of haploids will have the same AaBb genotype. B. During meiosis, the chromatids are linked together by mitochondria. Chiasmata are the result of telomere shortening during replication. if two genes are close neighbors on the chromosome, when a heterozygote of genotype AB/ab goes through meiosis, most of the gametes will be AB or ab and fewer will be Ab or aB. D. Recessive alleles are always on the same chromosome and dominant alleles are always on the other chromosome. 80 the only way to have a double heterozygote is AB/ab, but not Ab/aB. E. When a heterozygote AB/ab goes through meiosis, and the products are haploids of genotypes AB, ab, Ab, aB in equal proportions, this tells us that the two genes exhibit independent assortment and are close neighbors on the same chromosome. 9. A genetically important event called crossing-over occurs during _. {A} Prophase l C. Telophase ll B. Anaphasel D. Telophase l E. Metaphase ll 5. Genes on different chromosomes undergo independent assortment because nonhomologous chromosomes align randomly on the metaphase plate 9% . A. Mitosis @7. Meiosis | C. Meiosis ll f1 4 6. A certain strain of haploid fungus is transgenic for a firefly gene that is not normail foundin this fungus. The ‘ transgenic strain is crossed to a normal strain. A cloned firefly gene is used in a Southern blot analysis of individual haploid progeny from thiscross. Whatgpercentage of offspring will show a band on the Southern blot result? (3 pts) A. 0% B. 25% LCJSO% D. 75% E. 100% 7. In the movie Shrek, ogres and villagers don’t get along. imagine that relationships improve in the land of Far, Far Away over many decades such that ogres and villagers often fall in love and get married. Deduce the genotypes of the following very prolific couples: Give the genotypes for the parents in each cross; choose your own letters. (8 pts) green X green i”: in ' r» m *( In not green X green ‘i a“ green X green (,1 v. R ,_ 4.: f): green, cranky X green, cranky i; (a) 3.; g,» a I? w Jig 16 cranky 5 nice 46 cranky 14 nice 8. A certain plant comes in several varieties. They may be tall or short. green or yellow, and have large or small fruits. Tall, green and large are dominant. A homozygous short, green plant with large fruit is crossed with a homozygous tall, yellow plant with small fruit. Predict the genotype and phenotype of the offspring. (2 pts) /‘ ('0 r \. 3" ‘ l “9 )4 i*‘f‘CDY‘/ l “f /\ (/1, l—«' 01 a n a r r i U /,. 1 /\ \Fj 524, la" l by, l4 ' Next, the offspring are crossed to short, yellow plants with small fruit. What proportion of these offspring will be tall and yellow with large fruit? (4 pts) bl ’ l /‘ , ,_ 5? ,. i‘ M “y fl» <5. H” “U r 00. DO(( A ‘» ‘x{ We?» L ” ' t / x I L r /.> r . (1‘ j /\/1 ,Zr\ ~ / l'at 3““ 7 f, o ‘7: / I! _‘ \ o a” f/ 30 [b0 or ( :o‘< \\W 7 / / -; w r i l/U \ r, » ‘7, " : ‘ , 5 (~56? érC l 9. The following pedigree is for type I osteogenesis imperfecta, ainherited connective tissue disorder characterized by collagen defects. Affected individuals have brittle bones that break easily, and they may develop deafness caused by abnormalities in the bones in the middle ear. In addition, the whites of the eyes often appear bluish. (8 pts) -/» n a. What is the mode of inheritance for this trait? A] w a A. dominant and recessive I B. autosomal and X-linked C. autosomal and recessive 1 2 @autosomal and dominant E. X-linked and recessive F. X-linked and dominant " H ‘ 132a a i4 a. 1; b. ,3 :3 57 o 1 2 3 4 5 6 G. Y linked and recessive H. Y linked and dominant b. Determine the genotype for each person. Write the genotypes on the pedigree. c. If person “1 the person marries. what is the probability that a child of this marriage would have osteogenesis imperfecta?? g/f Ur: r') j, (.5 IT‘ 23"” {C(‘Mrv {.‘Clrvr 7 i" 3’ >11} J, "7’"! r \\,/ 0 9 L‘ 9‘ A 0 (a ‘ _‘ n lac M i i’ v I A C 7, 10. Wild-type fungus Neurospora crassa produces its own thiamin, methionine and arginine. An arginine-requiring strain of Neurospora crassa was crossed with a methionine—and—thiamin~requiring strain and the following offspring were obtained: mum number of ' fl ‘ 'z 1 offsrin left t +5 10 (\l f7 -‘/’\_ t e» — -+ + 3 33 L16 Map these genes. D (6 a £1 ‘ ’1. '. 7.3 _/O / ‘ / f, . ,x \/ 3].; t {H g, 1 / I , 3 3 11. Analyze and draw appropriate conclusions from this set of crosses: a. True-breeding dwarf snapdragon plants with white flowers were crossed to true-breeding tall snapdragon plants with red flowers. The F1 plants were all dwarf with pink flowers. (2 pts) to: w C , r 71. 2...: t .— My L- " “If: '- m A v r , "a grad 23;, , .. , Lt‘w‘i' at» H V]. a,“ ' finm'f” 5 pa F4 1.2 145 PM Challen in uestions 15/60 oints. 11 b. (Continued from 11 above. Analyze and draw appropriate conclusions from this set of crosses:) Here are the results of a cross between an F1 plant and a tall, white-flowered plant: (3 pts) - , L” -= a g0 \ {L at; phenotype _ _, of offspring number r;:~m__g;' {3,4 5, r” pink dwarf 13 :1 iii-“3&- a‘i—n gr. rt; pink tall 38 V white dwarf 37 -< 13;: v. 7 / '1? a) .i -'-'\ V 1) , white tall 12 total 100 in” “CH: 5"" , - . fir“ 5/3.; {V1.1}. 47;»: ’ 'v/ 12. Two Drosophila mutants with crooked wings were obtained in a mutant hunt. The researchers developed true- breeding populations for each mutant phenotype. Then they performed the following crosses. a. Draw conclusions from the results of Cross l and II. (2 pts) n"? K A fir i.’ J Cross l: crooked-winged 1 X wildlil’t‘ype -) all the offspring had crooked wings. ,. J Cross ll: crooked-winged 2 X wild type —) all the offspring had crooked wings. b. Next they crossed the mutants to each other: crooked-wing 1 X crooked-wing 2. While they wait for the larvae to develop into winged adults, you can help by making predictions about the potential offspring phenotypes. You can express your answer using letters and arrows, paragraphs are not necessary. Predict the F1 and F2 phenotypes (and ratios where appropriate) if crooked-wing 1 and crooked-wing 2 have mutations in different, unlinked genes: W (2915) v a. =77: <37: " {151;} ijr“_‘ W :a,/- -\.. J r " 7": I») . It} '.' W/ F; ' ’— ‘ i Chg; t 1: ME], '61. Predict the F1 and F2 phenotypes (and ratios where appropriate) if crooked—wing 1 and crooked-wing 2 have mutations in the same gene: _;_i (2 pts) WW . f i 3f: ‘ '4': 0—3) viii ,_ JJ 13. The mglglngs. - n who has Huntington disease, a rare autosomal d ‘ rait that causes neuromuscular degeneration later in life, typically after 40 years. His daughter, ll—1, is now 45 years old and is also affected. The two grandchildren in generation III are in their early 20's. Before they marry and start families, they would like to know the odds of inheriting the H allele of Huntington disease. Near the Huntington locus there is a molecular marker A. A is a simple sequence repeat polymorphism that produces bands of varying sizes when analyzed with molecular techniques. This loo 5 is linked to Huntington with a recombination frequency of 5%. The A genotype of each family member is indicated in the gel lane directly below the pedigree. What is the probability that Ill—1 will develop Huntington disease as she gets older? :1 (3 a (2 pts) _/ a "i’ if” N“, w ‘7 _ fin-j Q \I. Q It; * "Lit"; “‘ rr 3 — J ‘ pr} 7’ / L a 0 y 1 we _,,_-z._ to, ._ / “cg ‘ i! g 1 gm v j g ., r _ oz, it ' ,, , 5 -, ,i la - w '0 What is the probability that lll-2kwill'deve’lop[Huntington disease as he gets older? (2 pts) ‘ t 7 a" ‘: V, i’if 5 5(4) / . Consider the “molecular analysis” shown in the figure. Check off the items you would need to do that analysis (the one in the figure) with PCR. (2 pts) _\_{ gel electrophoresis _ pGEM __ human DNA polymerase __ lysozyme Taq DNA polymerase __ snacks for the lab techs ' “‘7- T CZligase thermal cycler :__ ampicillin __ 37° incubator __ X—gal a, ’I/ __ E. coli host cells __ probe for the Huntington disease _ DNA samples from random humans __ probe for A locus _‘f'DNA samples from the family members in the _§ primers for both ends of the A locus pedigree __ primers for the ends of the Huntington gene _ DNA samples from rats /_. restriction enzyme / Remember to sign the pledge if true. 5 ...
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