Exam%20I%202008%20Key004

Exam%20I%202008%20Key004 - PRINCIPLES OF GENETICS I...

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Unformatted text preview: PRINCIPLES OF GENETICS I (BSC1222) Spring Semester, 2008 c// EXAMINATION I Name University ID Lab Section Instructions: 1. Put your name on each page of the exam. 2. Read each question carefully. 3. Place all answers in the space provided. 4. You have 75 minutes for this exam. GOOD LUCK! Total Pts. M ulti le Choice: Name N CIRCLE ANSWERS ON LETTERS TO THE LEFT OF EACH QUESTION! Each question has 1 correct answer and is worth 5 points. AB@DE ABCD A A man and a women, both with long index fingers, have four daughters. Three of the daughters have long index fingers and one has short index fingers. The allele for short index fingers has to be: A. On the X’chromosome. B. Dominant. C. Recessive. D. Sex linked. E. On an autosome. What is the probability that a cross between aaBbCC x AABbcc will produce a AaBch offspring: A. 1/16. B. 1/8. C. 1/4. D. 1/2. E. 1. How many unique gametes could be produced through independant assortment by an individual with the genotype AaBbCCDdEE: A. 32 B. 4 C. 16 D. 8. E. 64. If there are 20 chromatids in a cell at metaphase, how many chromosomes are there in each daughter cell: A. 10. B. 20 C. 30 D. 40. E. 80. In horses, the color black is due to a dominant allele (B) and chest- nut color is due to the recessive allele (b). The trotting gait is due to a dominant allele (T) and the pacing gait to the recessive (I). What will be the gametes produced by a homozygous black pacer? A. All Bt. B. All tt. C. 50% BB and 50% tt. D. All BB. Name A B C D E 6. Suppose we cross the black pacer from question 5 above with a homozygous chestnut trotter. What will be the gametes produced by the F1 generation (the genes are not linked)? A. All Bt. B. All bT C. 50% Bb and 50% Tt. D. All Tt. E. 25% BT, 25% Bt, 25% bT and 25% bt A B C @ E 7. A geneticist determined the pedigree shown in the diagram with filled symbols showing the affected individuals. This disease state appears to be caused by a gene, and Susan’s geno- type is while jenny’s genotype is —— /Susan Jenny/ A. Dominant; aa; Aa B. Recessive; AA; aa C. Dominant; AA; aa D. Recessive; aa, Aa E. Semidominant; AA; AA. A (B) C D E 8. Red—green color blindness is X-linked in humans. If a male is red- green color blind, and both parents have normal color vision, which of the male's grandparents is most likely to be red-green color blind Maternal grandmother. Maternal grandfather. Paternal grandmother Paternal grandfather. Either grandfather is equally likely. F1905”? ABCDE 9. 10. 11. 12. Name Whenever either or both allele A or allele B is present, red color is produced. A cross of AaBb x AaBb can be expected to yield how many red offspring out of 16? Assume the genes are not linked. A. 1 B. 4 C. 12 D. 15 E. 16 Two heterozygous F1 hummingbirds with long, orange beaks are mated. Among the F2 offspring the phenotypic ratio for beak length is 3 long: 1 short and for beak color is 1 red: 2 orange: 1 yellow. Consequently: A. The beak color gene has incomplete dominance. B. The beak color gene is epistatic. C. The yellow beak gene is recessive. D. The long beak allele is co-dominant. E. beak color is polygenic. A tan colored male cat and a black female have 5 kittens, 3 of whom are female. All the kittens are tabbies (black and tan patches). This is an example of : A. Incomplete dominance. B. Codominance. C. Epistasis. D. Multiple alleles. E. X chromosome inactivation. Using the information in question 12, what percent of kittens would have tan fur if a tabby cat is crossed with a black cat? A. 0. B. 25 C. 75. D. 100. Name M Short answer questions. 1) True or false (2 points each) In cytoplasmic inheritance maternal phenotypes are expressed in both males and females Cytoplasmic inheritance violates Mendel’s principles i Extensive phenotypic variation is observed in cytoplasmic inheritance i Cytoplasmic inheritance involves pleiotropy i: True breeding individuals are heterozygotes F A cross between true breeding individuals will yield progeny in a 3:1 ratio E The allele for wrinkled seed shape is considered recessive because individuals with the allele have lower fitness than individuals with the dominant allele The alleles found in haploid individuals cannot be dominant or recessive T XXY Drosophila are male 3: Compared to asexual reproduction, sex increases mutations F Compared to asexual reproduction, sex produces new variants “V- Anaphase 1 of meiosis leads to seperation of genetically identical chromosomes 1: 2) Fill in the gaps (2 points per gap) Interactions among the human ABO blood group alleles involve nd a (plc rom omman incomp ete ominance c\mplete domi 'flnariQQ/epistasis/continuous variation) Multifactorial traits are influenced by and Mum" R (pick fronrfiiiiltiple ge lpleiotropy/dominance/Wnticipation) Name In species in which males are the heterogametic sex, they cannot be homozygous or heterozygous for sex—linked genes. Instead they are UVH‘L r q- a A human karyotype showing two X chromosomes and one Y chromosome is associated with (A i mg TU“ syndrome and the sex of the person is {:1 (ax/(L 3) A diploid body cell from a rat has a total of 42 chromosomes (5 points each). What is the total number of DNA molecules in the nucleus of a rat cell in 02? What is the total number of chromosomes in a polar body cell from a rat? 4) Describe what a complementation test is used for (10 pints) H“ \‘S To Jae/71th (ALE/UL»! Twit AC WT mot/mm; Owe carzfifiL 5amm_hmus<akkfit mxax JaQwatbch%Mauua 5) Outline Nancy Wexler’s aim and methodology Aim (3 points) 13 {gig/174+? (t1 36mg it}! Quit/ell up WW‘TV‘t‘jTUWII a ye/z/leflis— rem” Wflnjrumj/ WemTwh/ We Methodology (7 points) (mart/mated _m [flea/WM: :A lot/7e 97mch ’ w V‘s/Vie Ewe/low th‘ l/l‘Ciid/lCQ HMTW77Td/l T‘an [fill/qu IMLLF 257 TM Wit KIWI: olwrm W; //M4/ fiféc'A/c/éiflm Liam“ mg, was rvliw‘vy WWW v54 $275k my amt/W, Wmflpw‘ey/ /tv%' t 6) The two basic coat colors in Labrador Retrievers are black (BB or Rb) and chocolate (bb). However, for the black or chocolate to be expressed requires the presence of a dominant allele at the pigment transporter (E) locus. If the genotype at this gene is ee, the dog will be yellow regardless of its genotype at the B locus. This is an example of 6 Ce 57) V”? epistasis (2 points) What is/are the potential genotype(s) of a male if all we know is he has a black coat (3 points)? ’7 x I E I)! Q” W Wit/Te {Mala M am?) What is/are the potential genotype(s) of a black male if you know his father has a yellow coat (3 points)? .1 fl;ee. Name 7). Distinguish between negative eugenics and positive eugenics . Negative eugenics (3 points each) . $5 _ r" m‘ f, iuwvefvv’xj (Willi/l; m o/Urw/wwpw/ {’7 Odom/Tl my, S'T-er. 1 mm W7 Positive eugenics (3 points each) ‘ V ‘ ’ " any/LU 7” f? a? Mijw) fyamdémon/m #1 j-chauay @442 «fix/0336 Galton believed social worth reflected genetic worth. We live in a more meritocratic (fairer) society than existed in Galtons day. How and why would this likely influence the impact of genes on sporting or academic success (10 points) WW1" “‘«O/V’VvWfl/‘VV/S’oecc c;ij OV‘J cl /'U’VTVWN71€? Mat/(fr m r") M N my») flvT-MT—‘V‘j ~ /m m Qjc’V/IRVIOM l wu — C W"! Q/“(TOG/m’l'lé ) WV’U’d/l/WT \ y’a—S ~— "dp ‘JV’f—L/M/HTTQ; flacbe 5 “dqcfiuwfi WVU MM ‘ . fl . f. x. 44.25.. mug/x; . C eye/J can fit L c/UrH/Ve Wfldk Mvm or) fm‘mm 77,331: M w” T.“ “we; c ' Mow C’lfingv @1756” (net ’ ‘ N C e .Two pea pla ts With purple owers are crossed. Am g the offspring, 63 have pu le flowers, and 17 have white flowers. With a chi-square test, compare the observed numbers with a 3:1 ratio and determine the probability that the difference between observed and expected could be the result of chance (probability table is at the back of this exam). Provide a null hypothesis, show workings and state conclusion (20 points). 16/1/er A e’X/éém/ flVUIjJ Ly/‘RCZDI/fi .10101 d Wm (/{o / "T‘ W K L’( G d 4 (dd/"fax MMN 5Ww'fl :1: d (“/11 LJ‘jLJHj swag/00%)! CV‘})C“ l Que .4u1 ' 7- um/o’) MWU [W4 / o Life/VJ 4’» fl‘rf/ flfl" CA won/VIC? . Name You will probably want to use Punnet squares to answer questions 10 and 11 9). In chickens, "barring" is a pattern of feather coloring that was useful in sexing young chicks. The locus for barring is on the sex chromosome, Z. The allele B produces the barred phenotype and b does not. B is dominant to b. For the mating ZBW x 2be determine the progeny genotypic and phenotypic ratios. Indicate the sex of each genotypic class (10 points): 10) In mice, there are several alleles at the C locus that control intensity of pigmentation in the hair. The designations for alleles and phenotype are: C Full color 8°" Chinchilla ce Extreme dilution h " ilayan (pigment restricted to extremities) or maximum 14;""‘*ino listed in order of dominance with C most dominant. What is the progeny ratio for each of the following matings (Answers should be presented as, for example, 3 full color: 1 chinchilla) (5 points each): ...
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Exam%20I%202008%20Key004 - PRINCIPLES OF GENETICS I...

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