Exam 4 - 1 BIOL2323 Exam 4{Spring 3311 How many...

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Unformatted text preview: 1. BIOL2323 Exam 4 {Spring 3311} How many tpmtein-enceding} genes de yen have [approximate number per haploid genome}? A. E . e. E. Which type of loop is depicted here“? xi? 2“ L" v" 3. 5. The 44361110 25.13%} 2.500.0(1'3 4.4104} 1 .flflflflflfl inversion loop Duplication loop Deletlen loop Fruit iDep Traneleeatlen lepp ie lhe repeaiing stmewrel unit of chromatin, cansifiting el‘ approximately 200 base pairs of DNA and emaciated hietone proteins. A. 5*» (53/ D. E. nucleolus. trenecfipteeeme nucleeeeme enhanoeeeeme epiloeeeeme Reveree traneeriptaee..._.. e. D. E. transcribee RNA inte DNA. transcribee en FINA sequence into an amine- eeid sequenna. IS used in DNA sequencing by the eheln termination method. is a protein eneeeed by bacteria] viruses. [phages]. trenecribee DNA into RNA. How much protein-coding DNA {DRFe}, intron DNA. and repetitive DNA [metly tranepeeene] approximately does the human gennme contain fin this order)? A. B C. a r . L/ ‘13.; What is the approximate degree d! mmpaetien that is needed [:1 lil an average DNA molecule into the cell nude-us? A. B. D. E. 5%; 15%: 33% 50%: 25%; 15%. 25%: 25%; 50% net: 8%; 2% 1.5%.. 25%; 50% 10-fold BID-Fold MEI-fete 1D.DDD—tofd Tfls—fotd BIOLEEEE Exam 4 {Spring 2011‘} T. What does this lith microscopic image shew? A rnetaphase chromosome 3-. Bill-hm chromatin litter {solenoid} - . Iantpbrush chromosome é petytene chromosome . Elar body it. You have been assigned the task to identify and characterize genes that are specificalhr expressed in the human heart. A. search of the scouence databases retreer that one or the genes you identified shows a surprisingly high sequence similarity {sea identity} to a gene in Wise called heartless {lotion has been shew to toe essential for development ofthe fly heart]. The gene turns out to be the only one in the human genome that. shows this similarity. What some he the correct title of your publication? -- A. Human paralog {Dress-tha heartless gene shows shame expression in heart tissue «'3. Human analog [transmits beanie—as gene shows strong eapmsaien in heart tissue fig: Human men of Dresophifa heardessgene shows strong expression in heart tissue ~21 Drosophiie heartless gene shows strong expression In heart tissue a due: only the term hernotog ureuid be appropriate 1i Eu karyotic soils are state to precisaty regulate Ieyeis of trenscriptton of specific. protein—encoding ' genes through... I IL A. complex enhancereiernents that can associate with multiple activator and represscr U proteins. Et. covalent chemical modifications of nucleesernat histories. mg. production of different types of RNA polymerase. it! All efthe above I—- I—nml H G' “A L4 10. Why would U‘le transmission of Down syndrome resulting from a Hobertsonian translomljun L‘rH: I unaffected by the age of the mother“? or Down syndrome resulting from a Rebertsonian trarrslncetinn does not result from meiotic non-disjunction, which is affected by age. Robertsonisn translotations occur only in young mothers. This statement is incorrect; the age of the mother does affect the transmission of Down syndrome resetting frern Robertsenian translocations. ‘Ex‘ Robertsonian translocatiens cannot. load to Down Syndrome. Mothers with a Robertsoan kansfl-oeatinn can‘t have children_ _ EL stotssss Exam 4 [Spring 201 it Which of the following is a property of facullatiuo hoteroch romatin that it does not share with constitutive hetea'ochromatin? A. It can be converted into such rornotin. B. It is lows! in the centromere and telon'Iere regions. '3. It contains no or onfy very few genes. It consists r'nosttjtI of repeat DNA. It is a highly condensed form of chromatin in which the DNA is tightly packaged. 111. Which of the following statements is Ute most accurate? $53. 1.4. '15. .l 'x_ nigtp [It My genotype consists of about assoc to 50.00:] alleles. My genome consists of about 25.000 genes. My genome consists of about 250,000 alleles. My genotype consists of about 250.000 genes. My proteome has the same size as my genome. J's is} c. o. E. The genetic control of eye development in insects and vertebrates i5 A. acct-t;:t|-.=.~t-.=:!gltI different [organs pet-iced independently by convergent evolution}. B. controlled by the homologous eyeIess snct Per-E genes but Far-o cannot functionally- replace etrefess in flies. (Q3. dependent on cyciess and Pox-E“ but these are non-homologous genes. __ D. controlled by the same master regulatory genes toyetcssiF'sX-fi}. E. none ot'the strove. Which of the foEImcing mechanisms isn_ot used by most eukaryotes to control gene expression? ,—~ [5,} flontret of Lransonption initiation. “a. Deletion of genes that are not needeo‘. C. Translational control. " o. Protein activity control. E. None ofthe Elbows. What is the specific term that dcschb-es the toss of a single chromosome (Zn — 1}? A. Nulfisorny E. Trist’trttt,r @ i-t-lc oosomy Zr. Dipioicly E Mo noaloictr Pcsgcistronic mRNAs ....... .. P are frequent in bofi1 prokeqrotes and en karyotcs and represent ththtAs to which several ribosomes are bound at the some time. product-z different proteins by alternative splicing. oncooo more than one protein and- sre typical for eukaryotcs. are formed by poettranscriptional fusion of mRNfi-s derived from different genes. on code more than one protein one are wpioat for pnoltaryotes. BIDLEBEB Exam 4 {Spn'ng 21111] ft, Marry eukaryotic genes heye complex tempoiel and spatial expression patterns. These patterns are specified by ....... .. If (5.“ multiple independent enhancers. use different types of RNA polymerfle. Pol l. Pol II. and Pol III. I'M—f I3. different n' factors. that associate Iwith the RNA polymerase. D. cell and devetopmental stage-specific combinations of transcription factors. E. both A and D are correct. 13. Arrange the following in order of decreasing compactness: A. nucleosorrte ibeads-on-e-string form of chromatin] if B. radial-loop scaffold I3. BU-nm chromatin fiber D. DNA dou title helix a B. A. C. D ~. 3. c, s. A. o C {Ape B. E. D. A in. -' a. c. s. o Ir" rel . D: A. C. B U -' ts. Ii."'i'hich of the following ififll true of retroviruses? They can cause cancer or other serious disorders such as AIDS. They have a DNA genome. They can inserl iheir genetic information into host cell DNA in form of e primrirus. They are mobile genetic elements. All of the soc-ye. "LK !T 2|}. I.Iirhich of the following isn_ot a property of the histories? A e'.Ii:i[1.itii:ina.ri|}r highlyI conserved . g; highly acidic 'f . small [molecular mass 11.0% to 23.11%] El. pact-cage DNA in the cell nucleus E. $113 in 5 different types [H1 HEA. HEB. H3 H4} 22L. '..‘I.-'l'-ich of the folksy-ring is at least one explanation for large differences in genome size that are obsorycci eycn behyocn species of similar complexity? @ Differences in gene number r." "‘i . Differences in gene size '3 Differences in the number and size of gene families D Differences in the amount of non-profein-encoding DNA I: None of the sleeve is correct. because species of similar complexity do indeed always heye s '4'er similar genome sin-e. BIDLEEEE Exam J-l- [Spring 2011] 321 Which ottl'se following is a major cause for spontaneous abortions {miscarriages}? .-' 'l-_ 23. 25. so —'B. E. autoeomai trisomiee chromosomal inversions balanced translocations hemeen non-homologous chromosomes trisomiee ot the sex: chromosomes none ofthe above When E_ coir bacteria are suspended in a culture medium that contains both lactose and glucose. the giucose I.iriil he metabolired first one the tactose IrlIill only be used after the stores oi gluoose have been depleted. "l'll'l'rgiI does the bacterial cell not flatly express (transmitter) the genes ofthe iac opens in the presence ofglucose. even though the inducer of the operon {lactose} is present? A. C. D. E Thc lac repressor binds to the operator and blocks transcription oi the operon. Full induction of the opera-n requires activation at the Catch-elite Activator Protein {CAP} by cAMP. Irrhioh is only present at a low level as tong as gluoooe is present. Glucose binds to and inactivates CAP. which is required for induction of the ope-ran. Gluoose binds to and activates the loo represses: A and D is oorreot. In yeast chromosomes. oentrorneres... It. [fit 3.: o. E. 'i-. help distinguish one chromosome horn HHBthEi'. are about 125 no short and closerI related in sequence. are me ny thousands of base pairs long and are defined by o—setell'rte DNA. become shorter with every replication cycle. are the origins of replication. I."lu'hich ofthe following sex chromosome aneuotoiolies is not usually seen in human live births? A. 6': o. I:. X0 XXV YU KKK None of the above The histone code hypothesis poelulales the-L. A. E. C. : (ET-I Riser“ the rump-er anti type of histories within the nucleosorr'e core oontrois gene expression. the specific amino acid sequences of the N-lerminai tails ofthe histories oontrol transcriptjonal activation and :epreseion as wail as lathe: chromatin activities. specific posttrsnsletional modifications oftho nucleosomai histories control transcriptional activation and reptessiort es I.Ilrell as other chromatin ECU'I'iTiES. :he universai genetic omic is rec-Laced by a histone code when it corncs to gene regutetion. the specific sequence terrier) of the 5 histories {H1, HEA. HEB, H3, H4} bound to the promoter region controls transcriptional activaliort and repression 23. El]. 32. Bio L2323 Exam 4 {Spring 2cm} Mudoorsomal chromatin win: a diameter of t! I'IITI superceils into e 30 nm super-home {solenoid}. The hiatone that appears to be responsible for this compaction is: .a-‘-- tet H‘l H2a H3 Hri H213 a mans triaorrry for the X chromosome toads to Down Syndrome. KN}. Gag 5' Klinefeiter Syndrome. H‘- “.I' Turner Syndrome. Iii—I3 phenotypicelly normal females. supermodels. ‘mfipaa I'M. The tee operon of E. ootiwes the first major paradigm of gene regulation. Iil'li'hich of the following lessons learned from this ptoltaryotic system does ootopon to gene regulation in eukaryotes’r‘ The initiation of transcription is nonsense try transcription factors. Genes are ciusterecl in nogolallzurjyI units called operons. C. Transcription factors can activate or repress gene exprwsion. D. The activity ol' transcription factors can be regulated by small molecules. or: a. r—-E. None of the she-ire. Which oi the following chromosomes cannot be discerned as individual chromosomes by light microscopy? A. poly-tene chromosomes (B; larnotzrush chromosomes Tu} (normal) interphsse chromosomes D. {non‘nali rnetaphaso chromosomes . none or the eta-owe Steroid hormones iilte estracliol set by.. .. '-.xnf_; sin-cling to nuclear receptors. “'3. ainding to membrane receptors C. controlling the expression :1! specific: genes. Entercaiating between :he lenses in the DNA double heiix. — E. .5. and C Which of the following is ri_ot true of on he ricer DNA elemena? A. They can be located more than ‘IIEICIU lap away [rem lI'IeI gene may regulate. PE; They retain function if meir nutsootido sequenoe is inverted. They bind RNA polymerase II. . They usueily tune a particular combination oftreneoription footers. E. They earn he Located downstream of the transcription unit they control. BI'U'L2323 Exam 4 (Sp-ring 2011] '33. Gene expression can be controlled tr; small signaling molecules in prckartrotes and eoitartrcltes. ' 1t'll'l'tat is the common mechanism election of many cl u‘lraSE ligands? They... A. chemicain modifi- Erenscripticnel control regions of genes. i” "a tiirccutr modify- N—tarminal tails oF nuciccccmci histones. - k6 induce alioeteric changes in transcription tensors. D. alter electric potential across the cell membrane. E. do hope ofthe ahotrc. ZR. Which of the {elbowing is a major cause of gene duplications? C§ UV radiation _.-" . Unequal crossing over in“ C Chromosome breaks caused by ionizing radiation a“ D. Traosoocon insertion in the vican of a gene. E. Boflt B and C 35. Transcontionat tic-regulators {oo-acthators anti eo-reptoseoreji ........ ._ A. control altcmaiiue spficsing. B. are recruited o3..- transcription factors and denature the DNA Io facilitate transcription 0/ c. recruit transcription remote to tho one. which than directly activate RNA polymerase II. F El; are recruited by hanscnpiton lecture and are often members of protein complexes that chemically modiltr the nudeoeomal histcnes. E. are transcription factors that direcfiy hind to DNA and activate RNA synthesis. \_.- ...
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