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Exam1Spring2000

Exam1Spring2000 - Name EXAM#1 Principles of Genetics(Biol...

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Unformatted text preview: Name: EXAM #1 Principles of Genetics (Biol 2153), Larkin Spring 2000 Remember to fill in your name and ID number on the answer sheet! Choose the best answer for each question by filling in the appropriate space on your answer sheet. There are 33 questions. i. The process by which a DNA sequence is copied to produce an mRNA sequence is known as: a) replication. b} translation. c) transcription. d) mutation. e) cytokinesis. 2. Hugo de Vries is known for: a) discovering the role of chromosomes in heredidty. b) discovering sex linkage. c) using fruitflies to study genetics. d) rediscovering Mendel’s principles of heredity. e) none of the above. 3. If a spindle microtubule fails to attach to the kinetochore, the result could be: a) failure of cytokinesis. b) nondisjunction. c) complementation. d) the cell becomes haploid. e) replication. 4. The stage of meiosis shown to the right is: I __,_ a) anaphase I. x b) anaphase H. c) metaphase 11. r d) prophase I. J//// e} Cannot be meiosis—must be mitosis. "QL/ 5. The ends of a chromosome are known as the: a) telomeres. b) centrioles. c) centromeres. d) kinetochors 6) none of the above. 6. When a wound heals, cell division occurs. This type of division is called: at) mitosis. b) meiosis. c) nondisjunction. d) independent assortment. 7'. Which stage of meiosis in an organism has a value of c equal to half that of a diploid cell in G2 of mitosis? a) Prophasel b) Prophase II c) Garnetes. d) None of the above. Name: 8. Of the following stages of meiosis, which is the earliest stage where the chromosome number (n) has been reduced from 2n to in? a) ProphaseI b) Prophase II c) Metaphase H d) Telophase ll 6) Gametes Questions 9 through 12 refer to the following cross involving the independently asserting genes A, B, C, D, E, and F. Assume that the capitalized allele is fully dominant, and that these genes Show no genetic interactions (epistasis, etc.) Aa BB cc Dd Ee Ff x aa BB Cc DD Ee Ff 9. How many different types of gametes can the parent on the left produce? a) 2 b) 8 c) 15 d) 16 e) 32 10. How many different genotypes can be produced from this cross? a) 8 b) 16 c) 64 (:1) T2 e} 144 ll. How many different phenotypes can be produced from this cross? a) 4 b) 8 c) 16 d) 64 e) 72 12. What proportion of AA BB CC DD EE. FF individuals will be produced from this cross? a) 182 b) 1fl6 0) US d) li’4 e) 0 13. In many molecular biology labs, genetic markers are used that are detected by bands on electrophoresis gels. Each allele results in a band of DNA or protein of different size from other alleles. Hornozygotes have only a single band. In hetei‘ogrgotes a specific size band can be detectedfor each difiérenr allele present. BAND SIZE IS THE PHENOTYPE! An example gel is shown to the right for two alleles, at! and a‘. These genetic markers show: ~- __M._F__._. a) codominance. b) incomplete dorninance. c) epistasis. d) quantitative inheritance. "" "" Name: 14. When alleles of one gene mask or hide the ability to detect alleles of a different gene, this is called: a) codominance. b) incomplete dominance. c) epistasis. d) quantitative inheritance. e) none of the above. 15. You discover two varieties of a lizard,~blue and white. You cross a blue lizard with a white lizard, and the Fl lizards are all blue. When an F1 lizard is crossed with a white lizard, 8 distinct phenotypic classes of colored lizards are detected, each with a frequency of US. How many genes are likely to be involved? a) 1 b) 2 c) 3 d) 4 e) 5 16. What is the most probable genotype for the white lizard? a) Homozygous dominant for all genes. b) Homozygous recessive for all genes. c) Heterozygous for all genes. d) Can’t distinguish between a), b), and c) from information given. Questions 17, 18 and 19 refer to a species of plant with three independently asserting genes, T, G and S. T=short, t=tall, Tt=intermediate. zyellow, g=green, G dominant. Szrough seed, s=srnooth seed, S dominant. 1?. From the cross Tt Gg 85 x Tt Gg ss, what is the probability of getting an intermediate, yellow, smooth plant? a) 0 b) l! 16 c) 3! 16 d) NS e) 3i’8 18. A short, yellow, smooth plant was crossed with an intermediate, yellow, rough plant, and the following offspring are obtained: short, yellow, rough 294 short, green, rough 106 intermed, yellow, rough 305 intermed, green, rough fl 800 total What are the genotypes of the two parents? . a) TT YY as x Tt Yy 35 b) TT Yy ss x Tt YY SS c) TT Yy ss x Tt Yy SS d) TT Yy ss x tt Yy SS e) TT YY ss x ttyy ss 19. The alleles T and I show: a) codominance. b) incomplete dominance. c) epistasis. d) quantitative inheritance. e) none of the above. Name: Questions 20, 21, and 22 refer to the following pedigree. Filled circles=affected individuals. T GTE; ? ‘51??? E I 3. “EL “3’ 20. Does this pedigree indicate that the affected phenotype is caused by a dominant or a recessive allele? a) Dominant. b) Recessivc. c) Can’t tell. 21. What is the probability that a child of 11—2 and II-3 (A?) will be affected? a) U4 (0.25) b) 223 (0.66) 0) U3 (0.33) Cl) 18 (0.125) 6) U9 (0.11) 2’2. What is the probability that a child of 11-4 and 11-5 (B?) will be a carrier? Notice til-tar H-5 has children from a previous marriage to 11-6. 21) 1X4 (0.25) b) 2K3 (0.66) C) l/3 (0.33) (1) NB (0.125) e) U9 (0.11) 23. The pedigree to the right shows inheritance of a dominant disease. How can we explain the fact that H4 is not affected? L a) Incomplete dominance H b) Codominance 7- c) Incomplete penetrance d) Variable expressivin Name: 24. A strain of mice is found that lacks a tail. When tailless mice are crossed with each other, the result is always 213 tailaless mice and U3 wild-type mice. How can this result be explained? a) Two genes that both must be mutant to see tailuless phenotype. b) Single gene with codominant alleles. c) Incomplete penetrancc of tail-less trait. d) Tail—less phenotype caused by dominant allele lethal as homozygotc 6) None of the above. 25. What would you expect in the F1 if you cross a tail—less mouse with a normal mouse? a) All tail—less. b) lf2 tail-less, U2 normal. c) 23 tail—less, 1K3 normal. d) All normal. e) No offspring (all die). 26. Four different recessive mutations (a, b, c and d) are found that result in both big ears and political talent. To determine whether these mutations represent mutations in one gene or in different genes, you perform all possible crosses between big-carted politicians, with the following results: aa X bb:all normal an x cc:all big-carted politicians aa x ddzall big-earred politicians bl) x cc=all normal bb x dd=all normal cc x dd:all big—earred politicians a) a, c, and d are. alleles of same gene, b is allele of different gene. b) a and b are alleles of one gene, c and cl are alleles of a different gene. c) All are alleles of same gene. d) All are alleles of different genes. 27. A fifth big-cared politician mutation (E) is identified. This mutation is dominant to normal, and when crossed with aa, bb, cc, or dd gives big-cared politicians in the F1. What can we conclude about which mutations E is allelic to? a} E is allelic to a, b, c, and d. b) E is not allelic to any of the other mutations. c) E is allelic to a, c, and d, but not b. d) Because E is dominant, we can conclude nothing. 28. A series of alleles in a coat color gene (c) results in the following phenotypes in homozygous and heteroz 'gous combinations: 01" ch =brown ea c3 =tan cc cc =white c‘1 c11 =brown ch cC =white c c”: =white What is the order of alleles from most dominant to most recessive? a) C2, > cb> cc b) cb> c”) cc c) c°> Cb> C3 d) cc > c‘"‘> cb e") ca> c°> cb Name: 29. For a mother with type A blood and a father with type AB blood, which of the following genotypes is n_0t possible. a) A b) B c) AB d) 0 6) all are possible. 30. From a cross of two brown mice, the following results were obtained: brown coat 910 black coat 295 white coat fl 1600 Use a x2 test to determine if the results are consistent with the hypothesis that these three phenotypes are caused by a single gene with incomplete dominance. The results are. a) Not different from expected, x2: 0.5, hypothesis probably true. b) Not different from expected, x2: 42.7,hypothesis probably true. 0) Significantly different from expected, x2: 15.1,hypothesls probably false. d) Significantly different from expected, x2: 42.7,hypothesis probably false. e) Significantly different from expected, x2: 42.7,hypothesis probably true. 31. Hemophilia is an X-linked recessive disease. The father of a normal (unaffected) man had hemophilia. What is the chance that this man will pass a hemophilia allele on to his own son (grandson of the affected man)? a) 0 b) US c) U4 d) U2 e) 1 32. In Drosophila (fruitfly) the X chromosome in females is: a) transcribed twice as much as the male X. b) transcribed half as much as the male X. c) not transcribed; it is inactivated. 33. In Drosophila, a cross involving the sexwlinked w gene was performed. A wild-type female was crossed with a white—eyed male. The results were as follows: females: U2 white-eyed, U2 wild-type males: U2 white-eyed, U2 wild-type What were the genotypes in the cross? a) n'l if“ X WY b) w+w 1: WM." 0) WW x wY d) ww x wlY e) none of the above. TABLE 3.2 Table of Chi-Square 0(2) 5% Critical Values" Degrees of Freedom 5% Critical Value 1 3.841 2 5.991 3 7.815 -'-1 9.488 5 11.0?0 6 12.592 7 14.067 8 15.507 9 16.919 10 18.307 15 24.996 20 31.410 25 37.652 30 43.773 E‘Selected entries from R. A. Fisher and Yates, 1943, Statistical Tables for Biological, Agricufrum.’ and Medica! Research. Oliver and Boyd, London. ...
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Exam1Spring2000 - Name EXAM#1 Principles of Genetics(Biol...

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