BIO325 discussions 2006

BIO325 discussions 2006 - penetics (BIO S325) , - ' I . ,...

Info iconThis preview shows pages 1–23. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
Background image of page 19

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 20
Background image of page 21

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 22
Background image of page 23
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: penetics (BIO S325) , - ' I . , ‘ Summer 2006 Discussion 2 Indicate whether the following 'statements are true or false: (a) Mothers transmit X chromosomes to both their sons and daughters. I 'l (b) Fathers tranSmit X chromosomes to their sons, but not to their daughters. .m _ (c) For X—linked‘ dominant traits, affected daughters always have affected fathers. (d) For autosomal dominant traits, affected individuals have at least one affected parent. .I .‘ T i (e) If both parents are affected by an autosomal recessive trait, all children will be affected. T ' ' (f) For X-linked dominant traits, the number of affected females in a population' is approximately equal to the number of affected males. F' I | l (g) Autosomal traits generally affect equal numbers of males and females,- (h) For X-linked recessive traits, the number of carrier females in a population is approximately equal to the number of affected males. T (i) For X-linked‘ dominant traits, affected so_an always have affected mothers. 0) If a child and both parents are affectedby a genetic trait, the trait is dominant. A female fruit fly with 'a yellow body is di8covered in a wild-type culture. The female is crossed with a wild—type male. In the F1 generation, the males are yellow-bodied and the females are wild-type. When the F1 flies are crossed among themselves, the F2 produced are both yellow-bodied and wild-type, equally split among males and females. Explain the pattern of inheiitance of this trait and ‘ show the genotypes of the parents, the F], and the F2 progeny. y - \ Aw A we HF“ : zJfi Wt >< 75>? WSW » xYxV , , l a) ‘ * l A. , ‘ \Nl'x 4C, \fel 'X-YIKY Xv x‘fxv xyxy W .Y, twirl ."l ’ (d) How many double-stranded DNA molecules are there in a somatic cell just Alligators have 32 chromosomes in somatic cells. (a) How many homologous pairs of chromosomes does an alligator have? i r” ' (b)' How many chromosomes are there in a haploid set? " ‘ ‘ llp n (c) How many chromosomes are present in a somatic cell just after mitosis? ($52 31' ’\ before mitosis? .' M (a). What is the inheritance pattern displayed in the.following pedigree? ' ‘ - ao+o§cwa~\ “(5552 r6 (b) Write the genotype of all the individuals in this pedigree. Me I [1 lax -* 9 'Jv‘ i: ,A Lg {CA N The following pedigree depicts the inheritance of a rare hereditary disease affecting muscles: What is the most likely mode of inheritance of this disease? Write the genotypes of all the individuals in this pedigree. \( r ‘ mil 6 r“ r (Mg. Va [0 ‘- A .I (C) A/\ A diploid plant has a chromosome content of Zn j—L 4: Each diploid cell has one long pair and one short pair of,chromosomes. Diagrammed below are individaul cells of a dihybrid plant (A/a; B/b)._ In each case mention if the diagram represents a cell in mitosis, meiosis I, meiosis II,'or not possible. Each line represents a chromatidi I A\/' .b B YX/‘{ p“ L. B ' b a. . (b) b /B\ ‘A WA (flue b ‘ , /\ ms“ IL A\/ b\/ m I l A B (d) A #15 ‘mg‘\ 09“) I. (e) A <> A a/\ a\/ Consider the following cross: A/a;B/b;C/c;D/d;E/e x a/a;B/b;C/C;d/d;E/E If there is complete dominance for each of the genes, (a) What proportion of the offspring will show the dominant phenotype for each gene? // \, Z «VZMSMUV 23m: “no (b) What proportion of the offspring will be heterozygous at each locus? LVZMVDC/ZY‘QMVD” V32 3 8. I I Consider the following peldigree of a rare autosom‘al recessive disease. Assume all people‘marrying into the pedigree do not carry the abnormal allele. ax I» '/ - r - . ‘ fl / I [\\ I, f I (a) If individuals A and B have a child, what is th probability that the child will have the disease? l 4. l; , ,\. . 2 ‘ fl ‘- ' ' 2V , "/7, 40 , at (b) If individuals C and D have ild, what is the probability that the child will 0 . have the disease? \ l «. a V 337 \‘V V .i I ‘ 1%, i. (c) If the first child (3ny x D is normal, what is the p obability that their second child will have the disease? 7 . in ’3 (d) If the first child of C x D has the disease, what is the probability that their" second child will have the disease? . ll/ ' ' ‘u. Huntington's disease is "inherited as an'autosomal dominant trait. A woman whose great-grandfather had Huntington's disease marries and is considering having a child. Both her father and grandfather were killed in wars while in their twenties. If she consulted you, how would you advise her about the risks to her child? (Make sure you give her probability values fOr your predictions.) ll [J b) , . GENETICS '(B10 3325) I ' Summer 2006 Discussion 3 . . ~ . I ——————_—-y—————H—'+w———t—_y———— (i) Chromatids joined together at the centromere are' called: ' - I l ‘ ‘ ' I ' ‘ (a) Homologs. ‘ V , . (b) Alleles _ ‘ It I i ‘ (c) Non-sisterchromatids ‘ - I i ' (djfiSister chromatidls . , - I ii) Meiosis differs from mitosis with respect to the presence of ' i I l (a) Chromatids I , (b). Homologs - .. . , I ' ' ' (c). Bivalents ' '- '(d) Centromeres l V ' I I I I . ' l Ignoring the effects of Chromosome crossovers. how many genetically different gametes can an individual human produce? In humans, each diploidcell has 23 pairs of chromosomes. 2 2.3 ‘ _ l I I ' For each of the foll0wing sentences. fill in the blanks with the correct word selected from the list given below. Use each word only once. , ' I (a) In eucaryotic chromosomes, DNA is complexed with proteins to form Q W‘W’ \ . i I . (b) The paternal and lnaternal copies of human chromosome 1 are . . .- (c) When histone proteins are removed from chromosomes, a densely staining central core of nonhistoneprotein called a l 5c5§£nhi . is observed. . (d) X chromosome inactivatioh is the mechanism used in M— for dosage compensation of X-linked genes between XX females and XY males. . I ‘ ‘ I Drosophila. mammals. scaffold, homologous, non—homologous. karyotype, chromatin, centromere. I A dihybrid plant with the genotype Ala; B/b was testcrossedfand the progenyl with the following genotypes and ,4 J a ‘ ’ distribution were obtained. ' I Z > I | ' V ‘ w' , 3: \ ‘ Ala - B/b 122 :: ~ 7— ' Ll a I .. 5 , , ’l../Jl \r/ ' mfi. ' I A/a - b/b 118 I , 3 ' a/a - B/b 81 - ’ i a/a - b/b 79 I; I 3» U I . ‘ t-IISH III I i‘ll. I m if» i'r-I -~ \ ’I'Itl Ii»<‘i‘\' Are the genes A and B assorting' = e um n m ‘-i~ w i-r ism . l . i . ,r .i - W "434 I N...‘ '11s: 7‘7» ii I» l I.: l‘» ‘t‘ I.‘ m I 4 independently? Test this hypOIhCSIS With a It; a... W. I a] 1.5. .. .. .. .N, I“... ... .. . a chi-square [651. . i 5—. ll\ i~ I I it 4:- 1). r| J l‘li IJ lI-ii _ . ‘ .\' .LH _I.\"I 12w ‘u-I i. In; i~~-i. I‘~=~ 5. For each of the followmg definitions. I: 7" 4 1m ~ w in»: i-v i~ i . . I| .‘Iv‘ ‘3‘ 4st.» "‘I.‘ hm. hm _'..Is~ give the appropriate term: |i : A| “a l‘l1'.I| , 1 I _-~ r |_| IIJ JJiu m: ll 24-. last. _]Ii>; _ _ II '> l I g (a) The linkage arrangement in a dihybrid 'f “W I: I‘- if :i :- I in . . . la 4”“ 33'3" "'I- l~ '-‘- I In»: Liv ,i.| w ;i|.'i triii 4 which the dominant alleles of two genes i< WI «' «at? H " "r W m r w are on one chromosome and their corresponding recessive alleles are on the homologous chromosome. C W\ b3 (b) The linkage arrangement in a dihybrid in which. on each chromosome of a homologous pair. there is one dominant allele and a recessive allele of the two genes. TESL“ g Im lfa plant with t e genotype A/A - 3/8 is crossed to a plant with the genotype a/a - 12/1). and the Fl is testcrossed. what percent of the testcross progeny will be (1/0 - b/b if the two genes are (a) unlinked 2570 v (b) completely linked SO’IO /1 , I , (c) 10 map units apart i ‘ ‘ i @ la) 40% "8. ll. (d) 24 map units apart 3870 ' ' ‘ | 4 . . ‘ A plant with genotype 'Ab/aB is testcrossed to a plant with the genotype ab/ab. In 60% of the cells undergoing meiosis. " ho chiasmata is observed between the linked genes A and B; in 40% of the cells. there is one chiasma between the two geries. What proportion of the progeny from the cross will be ab/ab'? ‘ 10% (b) 20% (d) 5% l A fruit fly of genotype a+/a - b+/b (parent 1) is crossed to another fruit fly of genotype a/a ' b/b (patient 2). The progeny of this cross were ' ' Genotype Number of individuals .. ' ‘ ' ‘ ' (If/a - b+/b 38 j . a/a - M) 37 ' *a+/a - b/b 12 ' + b 1'3 7‘ «H b 32, a/[l ' b ‘ | .‘n . a . (a) What gametes were produced by parent 1 and in what proportions? 5%70 “- 1' ‘0‘ 372° “lb J ago" (b) [Do these proportions demonstrate independent assortment of the two_genes? No ' _ (c) What can be deduced from these proportions? “th 1F?o;\2'+\3 225' t (d) Show the arrangement of the genes on the chromosomes in parents 1' and 2 using the appropriate sy'mbols. 05*‘0+/a~b amok 009/000 . -. . In Drosop/ula, the eye color mutation scarlet (st) and the bristle mutation spineless (ss) are located on an autosome. [4 map units apart. What phenotypes, in what proportions, would you expect in the progeny of the following mating? st+ ss+/sr 55 females x 51 ss/st ss males ' H570 wuokfipfi, “V570 ‘ l 7° I .1 a0 l The A, B, and C loci are linked in the order written. There are 0 map units between A and B, and 20 map units between B and C. If a plant with the genotype A B C/a b c is testcrossed, what prpportion 9f the progeny plants will have the \ genotype 0 b c/a b c if there is no interference? f“ (:5 '1 ' \7 g v A I r i ~ -. (_ I , U . I a b d -' w~ * «is K Given the map 10 20 . and with no chromosome interference. how many phenotypically (I+'b+ (I individuals would you expect in a progeny of 1000 ofthe following cross: " I t + + + A» 1— ‘ /‘ a bd/{lb dxabd/aba'? 0‘ Jun?) ; . \o‘t 1,9/ . 0‘ \ - . - . . Three genes of corn, R. D. and Y, lie on chromosome 9. The map of the three genes, usmg map units is: R 10 D 20 Y \ :V’ In a testcross of a trihybrid plant (R D WM! )2) arising from true-breeding R D Y/R D Yand r (I y/r d y parents, how many R (I Y/— d - plants would be expected in a progeny of 1000 if there were no interference? If the coefficient of coincidence were 0.3. how would this change your answer? 5'1“ .77 l , (6H it.) ' ’ '~ 7 "x \ r7 '\ Ky Ml t t1», -. ~ i e a “'/' 7K ',\ r) ‘ ' fl A 7 '7 .C.‘ K (i ll a (r l’ I’ V l t *1} / I \J \ x \ if,“ 1 L7“ L9 7:“ 2’ x , i \ (fig/Ni 7 i, M lb {4 0‘ l i i All") 1/ to) // . GENETICS (BIO 5325) SUMMER 2006 I ‘ Discussion sheet 4 ' ' | ' I —_____—._..l—_—___—'——o——t————— (Q 4. Five independent flower—color mutants of Petunia were isolated. The wild-type flower color was red and the mutants were colorless. To determine if mutations that resulted in the colorless phenotype were present in the sam'e gene or in different genes in the different mutants, each mutant was crossed with the other mutants and the following progeny phenotype was obtained: ‘ ‘ ' ‘ ’ 7 , r . l l r r Progeny phenotype , I Yup-"’2 i I" l" 4‘- ' [fr (1 I Mutant l x Mutant 2 ColorleSs /- ' , " . Mutant l x Mutant 3 Red ‘ .‘( ( W 3' I ' j, ‘ Mutant l xMutant4 ‘ Red OUR ‘-} 7 .ng '51 7 .4 l I. | Mutant l x Mutant 5 ' ‘ ‘. Colorless ,4 g, ' Mutant2 xMutant 3 Red I I l I (J Mutant 2 x Mutant 4 r _‘ U" _ Red ‘ Mutant2 xMutantS ' ‘ ' Colorless f , LKI Mutant 3 x Mutant4 ' Colorless fl 1 V I ' ' Mutant 3 x Mutant 5 I Red J ‘ ° Mutant 4 x Mutant 5 ‘ Red (a) Determine the number of complementation groups from these crosses. . . w . , /_ (b) Identify the mutants in.each complementa’tion group. ’ In humans. the three alleles IA. 18, and i constitute a multiple allelic series that detérmines the ABO blood group system. In the following examples, state (yes or no) whether the child mentioned can be produced from the marriage shown for each . example. ' ' v. l A 9 (a) A child with blood group 0 from marriage of individuals — both parents with blood group A. W \{C ,\ (b) A child with blood group 0 from marriage of individuals -rone parent with blood group A and the other parent with blood 0‘ ' i \ B' Ir ' z ' l r (c) A child with blood group AB from marriage of individuals — one parent with blood group A and the other parent with blood group O. '. ' v - . » Na. (d) A child with blood group 0 from marriage of individuals.- one parent with blood group AB and the other parent with " blood group A. we _ \f‘ " tr? ' (e) A child with blood group A ’r'i‘om'3marriage of individuals — one parent with, blood group AB and the otherlparent with blood group B. V. r \F‘ Pt; 9 o . What is the distinction between incomplete dominance and codominance? , \ ‘ ltCWlO/"lf r \wlm pal. , "0'51 ‘ottrlt vl ~ ‘ a (of: I r Genes that interact in the determination of a particular phenotype give rise to modified dihybrid ratios. Write the a phenotypic ratios for a dihybrid cross with the following types of interactions: UI (a) Dominant epistasis IT 7) l ,‘v 3 l (b) Recessive epistasis 54.. V " Ll fl f 3'. (c) Duplicate dominant genes [ST \ [S- i (d) Duplicate recessive genes a ; i) q . (6) Dominant and recessive interaction ‘ ,, ‘ I ’6 r 7 Prototrophic strains of Neurospora can make compound 2 from the constituents of minimal medium. Two mutant strains lack the ability to make compound Z. Mutant c can make Z if supplied with X. but not if supplied with Y. while mutant strain (I can make Z from either X or Y. Construct the simplest linear pathway of the synthesis on from the precursors X and Y. and show where the pathway is blocked in mutants c and (I. 5“ 7’ ‘ l X \r. (a a ) t + 222 a . a + 4 A C o v \ y *>‘ z I . l 6. Twelve urid-ine-requiring Neurospora mutants ((1—1) were isolated after ultraviolet irradiation. Complementation testing was done with the following results. where 0 denotes no growth and + denotes growth of the heterokaryon on minimal medium: .3 . , ‘ r u l l * flflflflflllll r tflflflflflflflfl I . ‘ Testing for the accumulation of intermediates in the pathway gave the following results, where the intermediate compounds are numbered 1 through 4 ' Compound Mutants a b c d e f y g . h i j k l 1 o 0 + 0 0 + 0 ‘ 0 + ‘+ 0 0 ' I ' 2 O 0 + O 0 O Q 0 + O ' 0 O 3 0 O + O O 0 0 O + + 0 0 4 O 0 O O O O .0 0 + . O 0 O (a) How many different genes are represented among the twelve mutants? Grou'p the mutants according to allelism.. (b) What is the probable order of compounds in the biosynthetic pathway?\l *\ _, _ x 7 ,_» __'7¥_ \ .t I T) I! ' L—l —a 2+}; vér l “I >24» 4 (c) Which genes affect which steps 0fthe pathway? ' ‘ 7. ‘ Four steps in a biochemical pathway in Neurospora are shown below. Five single mutants were independently isolated and were shown to be blocked in the following steps. (Capital letters indicate compounds; numbers indicate mutants.) 1 1—) 5—» 02—5) D Assume that each ofthe enzymes involved is encoded by a single gene. and F is essential for growth. The mutants are plated on minimal medium plus one of the intermediate compounds. Fill in the following table. Indicate + for growth. Leave a blank if no growth occurs. Growth on minimal medium and A B C D E Mutant 1 + “h 2 + + 3 ’l’ 4 5 A— 4’ 8. The biochemical pathway for methionine and threonine synthesis in Neurospora is given below. (Note: cells require both methionine and threonine for growth. Also note that A and D can be made from compounds in the minimal medium.) lJ 7/ .l ’ r . L‘- ‘ i . ' A_.‘-. B .C Threonine ' ' . l . g , . 1' ' I 9,§ ' l I, .I ‘. ‘ F —I"- Q—fi—D— Methionine“ ,‘ I . . . D ——n- E ‘ . l Four mutant strains were isolated witli lfie following growth properties (+ ,indicates growth. - indicates no growth): O Growth on minimal medium and: Mutant strain' , B C Threonine E F G Methionine. ° 1 — — — , .. — +‘ ' + + ' 2 — + — — — — — . I 3 — — — — — — + 4 — — + — — — - Determine the step in the biochemical pathway that is affected by each mutant. (Place the number of each mutant | 1 (3) onto the pathway.) (b) What, if‘any. intermediate compounds will accumulate in each mutant? 1’ cog ) 2' ts, sac“ Lt-C_ (c) Two new mutant strains (5 and 6) were isolated. These mutants gave the following results complementation tests with mutants 1—4: (+ = complementation, 0 = no complementation) in Place mutants 5 and 6 onto the pathway. . Six point mutations are known to reside in three genes. Complete the following table Where "+" = complementation and "O" = no complementation. A ' GENETICS (BIO S325) SUMNIER 2006 Discussion sheet 5 l. A gal' mutant (a) Cannot grow without galactose (b) Is resistant to galactose (c Can utilize galactose as a carbon source Cannot utilize galactose as a carbon source (e) Can make its own galactose 2. A StrR mutant Requires streptomycin Can grow in the presence of streptomycin (0) Cannot grow in the presence of streptomycin (d) Makes its own streptomycin (e) Cannot make its own streptomycin 3. If a mez‘thr' Hfr strain is mated with an F‘ of genotype leu' thi‘, prototrophic recombinants can be . . [W detected by platlng the mlxture on ‘ ' (a) Leucine and methionine (b) Threonine and thiamine Leucine and thiamine (d) Methionine, threonine, leucine, and thiamine 5‘. , fl; Minimal medium V J .. __/ "NJ (I‘m , 4. Transformation is a mode of gene transfer in bacteria f® that can be utilized for determining the linkage of genes on the bacterial chromosome (b) in which DNA is transferred from one cell to another cell by a bacteriophage (c) in which the donor and recipient cells make physical contact before DNA transfer takes place (d) in which both the donor and recipient cells lyse 5. A cross in Salmonella typhimurium is performed using the generalized transducing phage P22. The cross involves an Arg‘ Leu' His' recipient strain and bacteriophage P22 that was grown on an Arg" Leu+ His+ strain. 1000 Arg“ transductants were selected and tested on several selective media with the following results: , Vii l"? Arg+ Leu’ His‘ 585 l Arg+ Leu' His+ 300 t 7 IF Arg+Leu+His+ 114 a i , n‘ <‘ ' ' ‘ Arg+ Leu+ His' 1 0 What conclusion can you draw about the linkage relationship among the three marker genes? ,’\ ‘fll‘i’w ' [\l‘ "x l l l J *0 Hm; Ari clog/do “rankegl 6. In an interrupted mating experiment in E. coli, three markers were selected for in exconjugants at the following time intervals: met+ - 10 minutes; leu” - 12 minutes; arg+ - 25 minutes. (a) Construct a linkage map showing the arrangement of these markers with respect to the origin of transfer and showing the direction of transfer. \P'fi W‘C'l ._ WM 9.1% If?) 5\ (b) What units can you use to show the distance between any two markers? VY‘.kJ', (c) In this experiment, what is the donor strain (choose from P, Hfr, or F)? \ V, r ‘/ I ‘23 it; Li‘fiw‘ ‘5' “A Tr V" V / ‘(f /¢:va.\cl“pan (‘f.\\x At time zero an Hfr strain was mixed with an F' strain, and at various times after mixing, samples were removed and agitated to separate conjugating cells. Samples are then plated on media that contain streptomycin plus four of the five nutritional requirements. Each of the selective media is lacking only one of the nutritional requirements. The cross may be written as Hfr (strxade+pr0+glu+tlti+arg+) x F‘(str’ade'pro‘glu'thi'arg') The parentheses indicate that the order of these genes is unknown + = prototrophy ' = auxotrophy r = resistant s = sensitive str = streptomycin pro = proline ade = adenine glu = glutamine thi = thiamin arg = arginine + = colonies obtained - = no colonies obtained SUPPLEMENT MISSING FROM THE SELECTIVE MEDIA Time of mating Interruption Proline Adenine Arginine Glutamine Thiamin 6 min - — — + - 11 min + - - + — 17 min + - - + + 23 min + - - + + 27 min + + - + + 32 min + + — + + 36 min + + + + + 40 min + + + + + Draw a map of this region of the chromosome indicating the relative position of all genes and the order in which they are transferred by the Hfr strain. For each gene, indicate its approximate location in minutes relative to the origin of Hfr transfer. it)" a 6 (p “ix/frw ,t (/w *‘w‘ "m 10. In the cross of E. coli strain A (F) x E. coli strain B (F') during conjugation, (a) What is the donor strain? firm/M A (b) What is the reci ient strain? p :1) V6\ ‘V-\ g (c) What is the exconjugant (choose from F+ or F)? l " Wild-type E. coli can grow on medium lacking both methionine and leucine. A mutant met‘ strain of E. coli, which also contains an F plasmid, cannot grow on medium lacking methionine. A leu' strain, which lacks an F plasmid, cannot grow on medium lacking leucine. After mixing the met' and leu' strains together for a few hours you transfer the culture to medium lacking both methionine and leucine and obtain colonies growing in the new medium. Which of the following event is most likely to have occurred in giving rise to these colonies? (a) The mutant genes in both strains have been restored to normal function by further mutations. A functional met gene has been transferred by conjugation from the leu' strain to the " met' strain. E A leu gene has been transferred by conjugation from the met' strain to the leu' strain. (d) A met gene has been transferred by conjugation from the met" strain to the leu' strain. f/y ’ {‘4 _ E ‘7.. _ . A bacterial chromosome has four markers, A, B, C, and D, evenly spaced throughout the circle. A generalized transducing phage will (a) Pick up all the markers in one particle (b) Transduce only one specific marker such as A @ Transduce any of the markers by different transduction events K d) Never transduce more than one marker (e) Always transduce at least two of the four markers l I! ‘. ‘ ’1‘”, qt \ kca’§d>:~,-. GENETICS (BIO S325) Summer 2006 Discussion _7 /,.. 3 ,r 3*ANouaC The following sequence of nucleotides in a DNA strand was used as a template to synthesize an mRNA that was then translated into protein: Q—PPAAOGGOPTTTTPGT l . .bn nrw”'6 . . . . Predict the carboxy-termlna angino acid and theamino-termlnal ammo acnd of the resulting polypeptide. Assume that the mRNA is translated starting with the first nucleotide and without the need for a start codon. ammo'awA'aw (arlwm‘ LC\U r Avg Transcription of gene A takes place from left to right. Transcription of gene B takes place from right to left. Which strand of DNA (top or bottom) as shown in the following figure serves as the template for transcription of each gene? 57 &W A—JTQP 6%? er 2272614", “'V‘V‘ The following DNA sequence contains the start site of transcription of a bacterial gene. The sequences shown in bold represent the —l 0 (TATATT) and —35 (TTGACA) regions of the promoter. 5 ' - CCGTGGACCTACGTACAATATAAGCTAGCCCTAGTCGATTGTCAACGGTACCGATCTAAT- 3 ’ 3 ’ -GGCACCTGGATGCATGTTATATTCGATCGGGATCAGCTAACAGTTGCCATGGCTAGATTA- 5 ’ (a) Will the start site (+1) for RNA synthesis be to the left, right, or between the two sequences shown in bold? \e C l' (b) Which of the two strands (upper or lower) will be the template strand? upper The following diagram shows a protein-coding gene and specifies the positions of its translational start and stop signals.r — (n A Q (A, L) A Q / CTA CAT 5 __________________ _u_____/ /_______ ____________ “3 3 __________________ _“_____ / /_______ ____________ _-5 GAT GTA Which ofthe following is correct? A) The upper strand is the template, with the carboxy terminus of the protein to the right The upper strand is the template, with the carboxy terminus of the protein to the left ) The lower strand is the template, with the carboxy terminus of the protein to the right D) The lower strand is the template, with the carboxy terminus of the protein to the left E) Either strand could be the template on the basis of this information l0. The following segment of DNA occurs within the protein-coding region ofa gene: 5'—GGAACTCTAGGGGCTG—3' 3'—CCTTGAGATCCCCGAC—5' Which one ofthe following is true: A) The upper strand must be the transcribed strand. B) The lower strand must be the transcribed strand. < C)\,_..» Either could be the transcribed strand. Neither strand can be transcribed because both contain stop codons. A tRNA with the anticodon 3'—ACC—5' would carry the amino a 'd A) phenylalanine B) tyrosine C) serine D) threonine @ tryptophan All of the following are general purpose translation components and could be used in the translation of any gene, except for one. Which one? A) tRNA B) rRNA mRNA D) methionine E) initiation factors y">"v‘\ " What would the minimum word (codon) size be to encode all 20 amino acids ifthe number of different bases in the mRNA were (a) two? (b) three? (c) five? Give your reasoning. Assume that all other features of the translation of mRNA into protein are the same as what we now know. b3 (,3 2 Consider the following piece of messenger RNA: 5 '-AUGGAGUCGUUAAUUAAACCGGUGCGGAUCGUAUUUAGUCCCCAC-3 ' (a) Draw both strands of the segment of DNA that this mRNA was transcribed from. (b) State which of the DNA strands the RNA synthesizing enzyme used as template for the transcription process. (c) Using the codon chart, give the amino acid sequence of the protein that would be produced by translation of the mRNA, assuming the ribosome moved along the mRNA from left to right. In the silkmoth, a single gene encodes the protein that is the major constituent of the cocoon. (a) The mRNA encoding this protein was isolated from silkmoth larvae and hybridized with silkmoth DNA. The following electron micrograph was obtained: DNA 23511074 / mRNA (1) Explain the loops in the DNA strand. Jaws gt \ as a My :7 ill 12. (2) Show on the diagram which side ofthe mRNA is its 5' and which is its 3' end. Explain. (b) A mutant silkmoth was isolated that cannot make the cocoon protein. The mutant makes mRNA encoding the protein at the same concentration as the wild type, but the mRNA molecules are slightly longer than usual. In addition, an electron micrograph of the mRNA-DNA hybridization with mRNA from the mutant organism shows the following: DNA \ //‘O:—;—\\‘k mRNA Suggest a possible explanation for the nature of the mutation. One early piece of evidence that the genetic code is non-overlapping was the observation that hemoglobin mutants that had amino acid substitutions always had a single amino acid substitutied. The first 8 amino acids in the normal human beta chain are: l 2 3 4 5 6 7 8 Val — His - Leu - Thr - Pro - Glu - Glu - Lys A mutant hemoglobin is discovered that has the substitution of isoleucine for threonine at position 4. A second mutant hemoglobin is discovered that has the substitution of arginine in place of the same threonine. Based on these mutants, deduce the nucleotide sequence of normal human beta chain mRNA starting with the first nucleotide base for position 4 threonine. Continue as far as you can and write an unambiguous sequence. Show our reasoning. TM «- 0 7,310 r9 A C U x/ A 0“ [R‘s /\ 06. V, M <2 A gene codes for a protein 8 amino acids long which may be abbreviated: NH2 -phe-tyr—phe-tyr-phe-tyr-phe-tyr-COOH UUL) — QAQ‘WU‘ UAU v UU} Di? Lilo) L)!“ L) Assume phe = UUU and tyr = UAU in the mRNA Using comparable notation, give the amino acid sequence of the protein coded by the following mutant derivatives of the gene: (a) A mutant with an acridine-induced addition causing the insertion of U between the 4th and 5th nucleotide in the mRNA. UUU/UUA/UUU—UU/‘K v’Uuo—UUA ~UUU_UUA .0 (b) A mutant with an acridine-induced deletion of the 15th nucleotide. wooamoou’oAU—uUWAUU,wry/3.0 (c) A double mutant containing both alterations, (a) and (b). above. IQ GENETICS (BIO 8325) SUMMER 2006 Discussion sheet 8 A strain of yeast translates mRNA into protein with a high level of inaccuracy. Individual molecules of a particular protein isolated from this yeast have the following variations in the first ll amino acids compared with the sequence of the same protein isolated from normal yeast cells. What is the most likely cause of this variation in protein sequence? Explain your answer. (Refer to the Codon table) normal sequence Met Thr Ala Ile Val Ser Asn Thr Gln Ile Lys variants Met Thr Ala Ala Val Ser Asn Thr Gln Ile Lys Met Thr Ala 61y Val Ser Asn Thr Gln Ile Lys Met Thr Ala Val Val Ser Asn Thr Gln Ile Lys Met Thr Ala Ile Val Ser Asn Thr Gln Ala Lys Met Thr Ala Ile Val Ser Asn Th: Gln Gly Lye Met Thr Ala Ile Val Ser Asn Thr Gln Val Lys (a) A mutation in the DNA coding for the protein ,5 W at) 5A g“ “‘9 We“ 4.: rw‘ é, (b) A mutation in the anticodon of the isoleucine tRNA (tRNAIle) c)\ A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different amino acids (d) A mutation in the isoleucyl-tRNA synthetase that decreases its ability to distinguish between different [RNA molecules » "i l m t u— r =~ “K (e) A mutation in a component of the ribosome that allows binding of incorrect [RNA molecules to the A—site Second letter u c A G uuu ucu UAU} UGU} u U uuclphe ucc Ser UAC Ty' UGC Cy‘ c UUA}Leu UCA UAA Stop UGA Stop A UUG UCG UAG Stop UGG Trp G cuu ccu CAU}His CGU u cuc ccc CAC CGC c .4 E C CUA Le" CCA Pm CAA}GI CGA “'9 A a: ; CUG CCG CAG “ CGG G a 3 AUU ACU AAU AGU u E E A Auclue ACC Th AAclA‘" AGclser c '3' “' AUA ACA ' AAA AGA} A -‘ AUG Met ACG AAG}Ly5 AGG “9 G GUU ch GAU}A GGu u G Guc Vl Gcc M GAc 5" GGc G c GUA a GCA a GAA}GI GGA y A GUG GCG GAG “ GGG G A poison added to an in vitro translation mixture containing mRNA molecules with the sequence 5’-AUGAAAAAAAAAAAAUAA-3’ has the following effect: the only product made is a Met—Lys dipeptide that remains attached to the ribosome. What is the most likely way in which the poison acts to inhibit protein synthesis? Explain your answer. (a) It inhibits binding of the small subunit of the ribosome to mRNA. (b) It inhibits peptidyl transferase activity. Q It inhibits movement of the small subunit relative to the large subunit. ( ) It inhibits release factor. (e) It mimics release factor. fl our [~5ch W; VCng m fly ,c..n( W, at ’4'“; Y . ,, A.» What structural characteristic of an operon ensures that the genes contained in the operon will be coordinately controlled? gramoekef 3 w(+'x1rk'{_C/V/ i‘nh‘vbl’i 53/ /a(e\ Ive/“fro r Consider an E. coli cell with the following mutations. What effect would each mutation have on the function of the lac operon (assuming no glucose is present)? (a) A mutant lac operator that cannot bind repressor (QWQ Vt. but )I’ C 3 9.1 . mgr ' (b) A mutant lac repressor that cannot bind the lac operator (mg kad-‘rwt, 0/4 ‘M <>k (c) A mutant lac repressor that cannot bind to allolactose rte/Pm (fch A“ 4' \Vf '1‘ M (d) A mutant [ac promoter that cannot bind CAP plus CAMP .\ Q 6 9‘5? 3 r: ms 217*?” (S 'I ah {1 l/\ L: Ed ext/«Cf f3 Positive and negative control are different ways of controlling the activity of a gene. What is the difference, and can one gene be subject to both modes of control? PafipH re 7 acAbcAro" Wan x , int-sol t \f‘c‘. In the table below, state whether or not the genetic regions of the lac operon listed are transcribed in wild- type cells of E. coli: Transcribed in Transcribed in presence of lactose absence of lactose (a) Z gene _L _ l f I (b) I gene L 44 (c) lac promoter i i (d) Y gene L __ Would the following genotypes for the E. coli lac operon be non-inducible (non-ind), inducible (ind), or constitutive (con) for B-galactosidase production? (a) 1-P+Q+z— 6“ Wm M (b) 1+P+oCz- mm \Mal (c) I~P+o+z-/I+P+o+z+ ‘mal (d) 1+P+oCz-/I+P+o+z+ tool (6) ISP+o+z+ new NA (01+P'0+Z+ non molt Describe what effect each of the following mutations would have on induction of B-galactosidase by lactose. (a) A mutation that inactivates the lac! gene (b) A mutation that prevents the product of the lac] gene from binding lactose (c) A mutation that prevents the product of the lac] gene from binding DNA (d) A mutation that inactivates theéa‘gene gum»? V053 f“ I“ (e) A mutation in the operator that prevents the lac repressor from binding l I—Potylx l 10. ll. 12. A number of mutations affect the expression of the lactose operon in E. coli. Complete the table given below. Use + to indicate that the enzyme is synthesized at greater than basal levels, and 0 to indicate that enzyme is not synthesized. Inducer (IPTG) absent Inducer (IPTG) present W FEM mm Per—mas; (a) 1+ 0+ 2+ W 2 Q i i (b) IS 0+ 2+ Y+ Q Q 9 0 (c) 1- 0+ 2+ Y+ i i i :1: (d) 1+ oC 2+ Y+ i i i t (0 IS oC 2+ Y- i Q i 0 (f) 1+ 0C 2+ Y'/I' 0+ 2- Y+ : Q i j: Predict whether B-galactosidase will be produced by the following E. coli strains under the conditions noted. The diploid genotypes represent F' (lac) strains. Use + to indicate that the enzyme is synthesized at greater than basal levels, and 0 to indicate that the enzyme is not synthesized. Lactose absent Lactose present (a) 1+ P+ 0+ 2+ 3 i- (b) 1- P+ 0+ 2+ (c) 1+ P+ 0C z+ (d) 1- P+ 0C z+ (e) 1- P+ 0C 2- (0 1+ P+ 0+ 2- g) 1+ P+ 0C 2- / 1+ P+ 0+ 2+ (h) 1+ P+ 0+ 2- / 1+ P+ 0C 2+ (i) 1+ P+ 0+ z-/1- P+ 0+ 2+ 0) 1+ P+ 0+ 2+ / 1- P+ 0+ 2- (k) 1+ P+ 0C z-/1- P+ 0+ 2+ (I) 1+ P+ 0+ z-/1- P+ 0C z+ (m) 1- P+ 0+ 2- / 1+ P+ 0C z+ (n) 1- P+ 0C z-/1- P+ 0+ 2+ Which of the following is not involved with the initiation of transcription of human genes? A) TATA-binding protein D) Activators B RNA polymerase E) Coactivators DNA polymerase A DNA fragment that comes from the promoter region of a light-inducible plant gene is spliced t0 the 5' end of the promoter of another plant gene. The artificially constructed hybrid gene exhibits light inducibility. When the fragment from the original light-inducible promoter is "flipped-over" in its new place, this artificially constructed hybrid gene is also li t inducible. The above fragment contains A) A repressor element @ An enhancer B) A TATA box ) A GAGA box C) A CCAAT box GENETICS (BIO S325) Summer 2006 Discussion sheet 9 Here are the recognition sites of four restriction enzymes: BamHl ECORI Hindfll :9ng l l J, l 5'-GGATCC-3' 5'-GAATI‘C-3' 5'-A.AGC'IT—3' 5'-AGATCT-3' 3'-CCTAGTG-5' 3’-CTI‘AAI\G-5' 3'-'ITCGA’.FA-5' 3'-TCTAG1A-5' Each enzyme cuts between the bases indicated by the arrows. Which pair of these enzymes will create compatible (complementary) sticky ends? What is the sequence of this compatible sticky end? 60m Hl amok bgli Restriction endonucleases BamHl, NheI, XbaI, and EcoRI make sequence-specific cuts in DNA at the following sequences: BamHl ‘ NheI XbaI EcoRI 5'-GGATCC-3' 5'-GCTAGC-3' 5'-TCTAGA—3' 5'-GAATTC-3' 3'-CCTAGG—5 ' 3'-CGATCG—5' 3'-AGATCT-5' 3'-CTTAAG-5 ' All these enzymes break a phosphodiester bond between the first and second nucleotide of the recognition sequence (counting from the 5' end), leaving a 5' P and a 3' OH. (a) Could ligase covalently join two ends produced by the following enzymes? If so, would ligation regenerate one of the above restriction sites. (1) Two BamHl ends? x E \(68 (2) Two XbaI ends? (3) A BamHI and an EcoRI end? N O (4) An XbaI and an NheI end? «(S (5) A BamHI and an NheI end? No (b) A 3-kb supercoiled circular plasmid contains the restriction sites shown below. N and X represent NheI and XbaI sites, respectively; A—F mark 6 points on the plasmid. A preparation of plasmid DNA was separately digested to completion with XbaI alone, NheI alone, and a mixture of the two enzymes. The products were separated by agarose gel electrophoresis. Note that supercoiled circular DNA migrates faster than a linear DNA molecule of the same size during gel electrophoresis. Keeping this in mind, draw the predicted appearance of the gel, assuming that you ran the following in the various lanes: Lane l—A mixture of known markers increasing in size from 1 kb to 6 kb, in steps of 1 kb Lane 2—Sample from the original plasmid DNA preparation Lane 3—Products of complete digestion with XbaI alone Lane 4—Products of complete digestion with NheI alone Lane 5—-Products of complete digestion with XbaI plus NheI \ Z #3“ i 7, -W Ho r / < V, l l q 1p 2 \N y f ’2 l— —‘ l K» " l L r l l S 3. The restriction enzyme ClaI recognizes the sequence AT J, CGAT. It cleaves the phosphodiester backbone between the T and the C bases at the arrow. ‘ 1;" The restriction enzyme Tan recognizes the sequence T J; CGA. It cleaves the phosphodiester backbone between the T and the C bases at the arrow. /\ 5 All? (3) Indicate the ends of the DNA molecule left after cleavage by ClaI. Show the sequence, polarity, and overhang, if any. (b) On average, how many sites would Tan recognize in a random sequence of 1100 base pairs? “$370 (“5m : “l (c) If a Tan-digested DNA end is ligated to a ClaI-digested DNA end, will ClaI cleave the newly ligated DNA? Will Tan cleave the newly ligated DNA? gt. a at. A [Cher qko ) \(fls / L, r. \ f a. 4. A small virus isolated from unicorns is a cir<‘:ular duplex DNA molecule, 10 kb in length. When digested with the restriction endonucleases Bam and Hit, each digest was electrophoresed separately, and the set of gels illustrated below was produced. Purified fragments obtained from Barn digestion, separated and eluted off the gels, were each digested with Hit and electrophoresed with the results shown. The letters identify specific fragments. The numbers refer to the size of the fragments in kb. Ix Barn A Barn 6 Qflm + + Bam Hit Hit Hit Ekb ' V\ B 4 C 4.5 4.5 ‘ I. D 3.5 3 _ " ‘ E 2 1 1 « \O 0.5 / , \ ’ l‘ n ' , a ' \F (k W\ \l) \A \ I I, ,n \ , ‘r‘ 1. f 2 Show the positions of the sites cleaved by Ham and Hit on the map of the virus. Barn 1 i‘l HA M ' x 5. Plasmid pBR322 is a circular DNA molecule, 4 kb in size; it was one of the first vectors used for cloning any DNA sequence (whether from human, mouse, Drosophila, Tetrahymena, or yeast) in E. coli. The restriction map of pBR322 is illustrated below. E I — i‘ - m n: . -.~ 1 833 to”? . = Lu¥€ on; $5. Inner'cxrcle "' for distances in kb (a) Give the number and approximate size of each fragment that results after complete digestion with each of the following restriction enzymes: BamHl, BglI, EcoRI, and PstI. (b) If pBR322 is digested with both EcoRI and PstI, how many fragments would result, and what are their approximate sizes? Plasmid pBR322 carries genes conferring resistance to the antibiotics ampicillin (AmpR) and tetracycline (TetR), as shown on the map (gene length is shown by the double-headed arrows). (c) Which restriction enzyme would you choose for inserting a foreign DNA fragment into this plasmid for cloning, and why? (d) Outline the steps that you would carry out to select an E. coli strain that carries the foreign DNA fragment in plasmid pBR322. ’ l ‘ i l . t. “ . l r . ~O '.. . _ _. Ii - 2 ~ ’ \\) YCWLA. I c ( \'-.’ , . ~-:, g, \l‘. ‘ , I ,K r,, .n A v-f g ‘ . 4 (AAA §\‘~41,,.(f (“f/a. L4. (JV-1' ivgy/ I‘ (’,e§’35y V7,;J .r/VH { (r ‘i- rT$,.vf ir’ L; "f’,‘.—( ‘3. a Le (7“ [or (x *.. \«L d L—\. linkro('! '4‘ 6. A plasmid, pUC18, contains the ampicillin—resistance gene, the origin of replication, and the B-gal gene, which codes for the B-galactosidase protein. This protein can break down the synthetic chemical X—gal, producing a blue product that stains the entire cell blue (but is harmless to the bacteria). At the beginning of the B-gal gene there are several unique restriction sites (some of them are shown in the diagram below). Petl Xbal You wish to clone a 1.0—kb XbaI fragment into the pUC18 plasmid, so you cut the plasmid with XbaI and, after removing the enzyme, mix the XbaI-cut plasmid with the 1.0-kb fragment, ligate, and transform competent bacteria. (a) On what medium would you grow your transformed bacteria? AW?! (5." l \~ )7‘ w (b) Do you expect the bacteria carrying plasmid pUC18 (without the insert) to be blue or white when grown in the presence of X-gal? Explain. i \l .1 ‘ (c) Do you expect the bacteria carrying the recombinant plasmid with the 1.0-kb XbaI fragment (chimeric DNA) to be blue or white when grown in the presence of X-gal? Explain. MFHL ((1) Draw a diagram of the recombinant plasmid. 7. A piece of DNA was sequenced by the Sanger dideoxy method, using radioactively labeled primers to allow visualization of the fragments produced. The autoradiogram of the electrophoretic results is shown below. Determine the sequence and polarity of the DNA (newly synthesized) strand. ddATP ddTl'P titiGTFJ ddCTF’ — - Direction of migration GENETICS (B10 5325) Summer 2006 I Discussion sheet 10 4 —__—____—~__.—’%___———_— 1. Show the steps (i.e., several replications) by which 5-bromouracil causes the GC base pair shown below to change to an AT base pair. Show the point at which the rare (tautomeric) form of 5-brom0uracil gets incorporated in the DNA. Show both products of all divisions in which 5-bromouracil is involved. Do not 6C T show chemical structures. Mk ’Hi_‘ V W 7 3) “‘fi: 7 _G— W wra— " __,. ’41“ S I; U l __C___ __<‘._J. 2. Sequential mutations affecting one codon caused the changes arg —> t/zr —) met —) leu —> gln What base pair changes produced these amino acid changes? 3. The rare enol form of thymine pairs with guanine. If a thymine enolization occurs during replication, what , ,7 _x\ 1 1* T ) _ r i would be the mutational event. I, g I, ~ _ T 23 C7 m.» m __\ T” g g 6 1:7?) CGto AT /B’) GC to TA /C) AT t0 TA< D) TA to CG E) CG to CC a \5 2; A E fat/arc”. / , N V CW Tlér4. AftEr mutagen treatment, a molecule of 2-aminopurine (an adenine analogue) incorporates into DNA. C, .43 (g7 During replicati .he 2-AP protonates. The mutatio al event caused by this will be fl? (H) “J'Lfdca/ U/ 7K (‘ A)«-/ AT to [email protected] to AT )2 AT to TA 65%3 to GC XE) GC to CG taxi/T fl i? b"/ ’ -’ .I / I I r l r ’ > ‘” I ~ 7 , w ,. Lg __) 5. /B;agileXsyndrome is caused by 1,77 {I I I" \ ‘1. (\I C Extension of trinucleotide repeats D) 5—Bromouracil r/ ' l (l B) Free radicals E) Depurination ; ll C) Microdeletions )7)“ C7 l \ l l \\ £3? 6. During mutagenic treatment with nitrous acid, an adenine deaminates to form hypoxanthine, which bonds ,4 7,3) x . Fl “' 1 like guanine. The mutational event ould be ”Z\\ hgp‘v’v S l HM", ‘ ‘ L’ l yATtoCG B) ATtoGC )ZXATtoTA D) GCtoAT/EéGCtoTA 341'. .w\ ‘L ,r ‘ ,/L ‘ a e I I n ’T “wuttwl r "l . wag? l l 1‘? Fl " T 7. Discuss three possible mechanisms at may lead to the conversion of a proto—oncogene to an oncogene resulting in a pathway to cancer. 8. Which ofthe following mutations will result in cancer? A homozygous recessive mutation in which there is a deletion in the coding region of a proto-oncogene, leaving it non-functional. _ @Homozygous recessive point mutations in a tumor-suppressor gene coding for a non~functiona1 protein. (c) A dominant mutation in a proto-oncogene in which the protein product is overexpressed. W dominant mutation in a tumor—suppresor gene in which the protein product is overexpressed. 9. Multiple myeloma is a cancer of antibody-producing cells. Usually, a single type of antibody rather than a spectrum of antibodies is produced by a given tumor. How would you explain this observation? 10. In a cytogenetic study of a new form of atypical lymphoma, a translocation between chromosomes 3p and 22q is commonly seen. What are two possible mechanisms by which this translocation could result in cancers? (\pcl ) ,‘ ‘ \ \3 C. ‘ \ ‘ \ll ’ | GT ' ._ y a \ (n / g " N’v'lm PM? W. \ \ A zlwt /:/ A g 'P I F" l~> 63 C. l T \‘t H H. T k (5/ I l -. n ‘ . .IIV WC N: “Fri CV MW l l l r ‘ E V l l NFC, ...
View Full Document

This note was uploaded on 12/17/2009 for the course CH 310 N taught by Professor Blocknack during the Spring '08 term at University of Texas.

Page1 / 23

BIO325 discussions 2006 - penetics (BIO S325) , - ' I . ,...

This preview shows document pages 1 - 23. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online