B250WT1F04 - M .CO 411 MED - PR E BIL 250/GU “Genetics”...

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Unformatted text preview: M .CO 411 MED - PR E BIL 250/GU “Genetics” 9/28/04 P D41 1.C OM Exam 1 OM - PR EME Corrected answer to each questions are in bold and underlined fond. - PR EME D41 1.C I: Multiple choice format (select the most appropriate choice to answer each question. Each question is worth 10 points) D41 1.C OM 1. Which of the following clusters of terms accurately describes DNA as it is generally viewed to exist in prokaryotes and eukaryotes? Double-stranded, parallel, (A+T)/C+G)=variable, (A+G)/(C+T)=1.0 Double-stranded, antiparallel, (A+T)/C+G)=variable, (A+G)/(C+T)=1.0 Single-stranded, antiparallel, (A+T)/C+G)=1.0, (A+G)/(C+T)=1.0 Double-stranded, parallel, (A+T)/C+G)=1.0, (A+G)/(C+T)=1.0 Double-stranded, antiparallel, (A+T)/C+G)=variable, (A+G)/(C+T)=variable - PR EME D41 1.C OM - PR EME A. B. C. D. E. - PR EME D41 1.C OM 2. Assume that the molar percentage of thymine in a double-stranded DNA is 20. What are the percentages of the four bases (G, C, T, A)? If the DNA is single-stranded, would you change your answer? G = 20%, C = 20%, A = 20%, T = 20%; Yes. G = 30%, C = 30%, A = 20%, T = 20%; Yes. G = 30%, C = 30%, A = 20%, T = 20%; No. G = 20%, C = 20%, A = 20%, T = 20%; No. OM - PR EME D41 1.C OM A. B. C. D. - PR EME D41 1.C 3. The base content of a sample of DNA is as follows: A = 31%, G = 31%, T = 19%, C = 19%. What conclusion can be drawn from this information? A. B. C. D. PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 11. COM - PR EME D4 11. COM - PR EME D41 1.C OM The sample DNA is single-stranded. The sample DNA is double-stranded. The sample DNA is short fragment. The sample DNA is GC rich. M .CO 411 MED - PR E OM D41 1.C OM DNA; 5’-end DNA; 3’-end RNA; 5’-end RNA; 3’-end - PR EME A. B. C. D. U - PR EME D41 1.C OM A - PR EME D41 1.C 4. Given the figure to the right, is it DNA or RNA? Is the circle closest to the 5' or 3' end? D41 1.C OM 5. Which of the following terms accurately describes the replication of DNA? conservative dispersive semi-conservative nonlinear nonreciprocal - PR EME D41 1.C OM - PR EME A. B. C. D. E. - PR EME D41 1.C OM 6. Which cluster of terms accurately reflects the nature of DNA replication in prokaryotes? fixed point of initiation, bidirectional, conservative fixed point of initiation, unidirectional, conservative random point of initiation, bidirectional, semiconservative fixed point of initiation, bidirectional, semiconservative random point of initiation, unidirectional, semiconservative OM - PR EME D41 1.C OM A. B. C. D. E. PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 - PR EME polymerase slippage trinucleotide repeats the 5' to 3' polarity restriction topoisomerases cutting the DNA in a random fashion sister-chromatid exchanges COM 11. - PR EME D4 11. COM A. B. C. D. E. D41 1.C 7. The discontinuous aspect of replication of DNA in vivo is caused by M .CO 411 MED - PR E OM 1.C D41 - PR EME OM 1.C D41 - PR EME OM 1.C D41 - PR EME OM 1.C D41 1.C OM - PR EME The above diagram of a generalized tetranucleotide will serve as a basis for the following three questions. - PR EME D41 8. Is this a DNA or an RNA molecule? State which _________ D41 1.C OM A. DNA B. RNA - PR EME 9. Which circle is near the 3' end of this tetranucleotide? - PR EME D41 1.C OM A. the upper circle B. the lower circle 11. COM - PR EME 5’-TGTC-3’ 5’-ACAG-3’ 3’-TGTC-5’ 3’-ACAG-5’ - PR EME D4 A. B. C. D. D41 1.C OM 10. Given that the DNA strand which served as a template for the synthesis of this tetranucleotide was composed of the bases 5'- A C A G - 3', fill in the parentheses (in the diagram) with the expected bases. PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 11. COM 11. That some organisms contain much larger amounts of DNA than apparently "needed" and that some relatively closely related organisms may have vastly different amounts of DNA is more typical in A. viruses than in bacteria. B. RNA viruses than in DNA viruses C. eukaryotes than in prokaryotes D. the family "alphoid" rather than the diphloid family E. prokaryotes than in eukaryotes M .CO 411 MED - PR E - PR EME D41 1.C OM 12. When considering the initiation of transcription one often finds consensus sequences located in the region of the DNA where RNA polymerase(s) bind. Which are these common consensus sequences? A. B. C. D. E. 1.C OM - PR EME D41 1.C OM CAAT, TATA GGTTC, TTAT TTTTAAAA, GGGGCCCC any trinucleotide repeat satellite DNAs 1.C OM - PR EME D41 13. In 1964, Nirenberg and Leder used the triplet binding assay to determine specific codon assignments. A complex of which of the following components was trapped in the nitrocellulose filter? ribosomes and DNA free tRNAs charged tRNA, mRNA triplet, and ribosome uncharged tRNAs and ribosomes sense and antisense strands of DNA - PR EME D41 1.C OM - PR EME D41 A. B. C. D. E. D41 1.C OM 14. Select three posttranscriptional modifications often seen in the maturation of mRNA in eukaryotes. A. B. C. D. E. - PR EME D41 1.C OM - PR EME 5'-capping, 3'-poly(A) tail addition, splicing 3'-capping, 5'-poly(A) tail addition, splicing removal of exons, insertion of introns, capping 5-poly(A) tail addition, insertion of introns, capping heteroduplex formation, base modification, capping 5/6 X 5/6 X 5/6 5/6 1/6 X 1/6 X 1/6 1/6 PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 11. COM A. B. C. D. - PR EME D4 11. COM - PR EME D41 1.C OM 15. “Breaking the Genetic Code” has been referred to as one of the most significant scientific achievements in modern times. Use of polynulceotide phosphorylase to produce synthetic “mRNAs” by Nirenberg, Matthaei, and Ochoa in a cell-free protein-synthesizing system was one of these ground breaking experiments. Suppose they used mixed copolymers, A and C in a ratio of 1A:5C. What is the probability for a CCC sequence occurring? M .CO 411 MED - PR E OM - PR EME D41 1.C OM 5’3’first third OM A. B. C. D. - PR EME D41 1.C 16. A base at the first position of an anticodon on the tRNA would base pair with a base at the ____ position of the mRNA. - PR EME - PR EME D41 1.C polar, nonpolar linear, circular alpha, omega long, short primary, secondary OM A. B. C. D. E. D41 1.C 17. Side groups of amino acids are typically grouped under which of the following? - PR EME D41 1.C OM 18. Studies of Neurospora led to the ________________ statement whereas studies of human hemoglobin led to the _______________ statement . one-gene:one-enzyme, one-gene:one-polypeptide semi-conservative replication, discontinuous synthesis transcription, translation autosomal dominant, autosomal recessive - PR EME D41 1.C OM A. B. C. D. D41 1.C OM 19. Which of the following does not happen during hnRNA processing? a 7-methylguanosine cap is added to the 5’-end of the RNA introns are spliced out a poly A tail is added exons are spliced together ribosomes bind and begin translation - PR EME D41 1.C OM - PR EME A. B. C. D. E. PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 RNA polymerase II transcription factors TATA box CAAT box sigma subunit - PR EME D4 11. COM A. B. C. D. E. 11. COM 20. Initiation of transcription in eukaryotes requires following factors, except ________. M .CO 411 MED - PR E OM - PR EME D41 1.C OM ambiguity degeneracy colinearity universality OM A. B. C. D. - PR EME D41 1.C 21. The codons AGA and AGG both encode the amino acid arginine. This illustrates their property of _________. - PR EME D41 1.C OM continuous 3’ to 5’ semi conservative discontinuous all of the above 1.C OM A. B. C. D. E. - PR EME D41 1.C 22. The discovery of Okazaki fragments suggests that DNA synthesis is sometimes __________. - PR EME D41 23. What activity of DNA polymerase I reported by Arthur Kornberg in 1957 is responsible of the removal of the RNA primer in E. Coli? 3’ to 5’ exonuclease primase 5’ to 3’ polymerase 3’ to 5’ polymerase 5’ to 3’ exonuclease D41 1.C OM - PR EME D41 1.C OM A. B. C. D. E. OM - PR EME 24. During DNA replication, which molecule serves to destabilize the DNA helix in order to open it up, creating a replication fork? SSBPs DNA polymerase, the holoenzyme DNA helicase DNA gyrase DNA ligase - PR EME D4 11. COM - PR EME D41 1.C A. B. C. D. E. 25. What is the role of aminoacyl tRNA synthase in protein formation? PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 11. COM A. It is the primary energy source causing binding of the amino acid to tRNA. B. It serves as a catalyst for the reaction binding a specific amino acid to a tRNA. C. It catalyzes the formation of specific peptide bonds. D. It serves as a ribosomal initiation factor. M .CO 411 MED - PR E D41 1.C OM Extra credit question (20 points) 1.C OM - PR EME D41 1.C OM - PR EME D41 1.C OM - PR EME D41 1.C OM - PR EME D41 1.C OM - PR EME Supply the correct response for the questions (26 –27) provided below. The direction of the RNA polymerase is given by the arrow. 1.C OM - PR EME D41 26. The letter B is nearest the 5' or 3' end of the molecule? State which_________. A. 5’-end B. 3’-end - PR EME D41 27. Which terminus (N or C) of the growing polypeptide chain is nearest the letter E? State which________. PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 11. COM - PR EME D4 11. COM - PR EME D41 1.C OM A. N- terminus C- terminus M .CO 411 MED - PR E OM 1.C D41 Short answer format (each question is worth 10 points): OM - PR EME II. D41 1.C OM - PR EME D41 1.C OM - PR EME D41 1.C 1. Genetics is a subdiscipline of biology concerned with the study of heredity and variation. The central question of genetics is the study of the chemical basis of genetic process, including identifying which molecule serves as the genetic material. Based on what you have learned from this course: List the major characteristics a molecule must possess to serve the role of the genetic material; name this molecule which serves as the genetic material in eukaryotes and prokaryotes; and explain what chemical and biological properties of this molecule allow it to serve as the genetic material. 1.C OM - PR EME Answer: Four major characteristics of a genetic material in a cell: OM - PR EME D41 Replication = duplication of genetic material. Expression = production of a phenotype. Storage = stable maintenance and passage of information. Variation = capable of alteration. - PR EME D41 1.C DNA is the genetic material in eukaryotes and prokaryotes. Not RNA! Deduct a point, if they include RNA here. PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 11. COM - PR EME D4 11. COM - PR EME D41 1.C OM - PR EME D41 1.C OM The chemical and biological properties of DNA: they should include the following information: The basic structure of DNA: each nucleotide has a specific nitrogenous base, known as A, T, G, and C. The sequence of these nucleotides, represented by the four bases provides the molecular basis for the storage of enormous information code. The double helix structure and the base pairing rule: A pairs with T, G pairs with C are the foundation of the copying mechanism, which allow the replication of the molecule to occur. The central dogma, DNA transcription to RNA, and then translate to protein to produce a phenotype demonstrates the expression capability of DNA. Lastly, DNA sequence can be altered by the change of DNA bases at the DNA and chromosomal level, which is the foundation of genetic variation. M .CO 411 MED - PR E OM 1.C D41 OM - PR EME 2. Does the design of the Hershey-Chase experiment distinguish between DNA and RNA as the molecule serving as the genetic material? Why or why not? - PR EME D41 1.C Answer: D41 1.C OM - PR EME D41 1.C OM The answer is “No”. This is because phosphorus is found in approximately equal amount in DNA and RNA. Therefore labeling with 32P would “tag” both RNA and DNA. So, their design would not be able to distinguish between DNA and RNA as the molecule serving as the genetic material. 1.C OM - PR EME 3. State differences known to exist between prokaryotic and eukaryotic DNA transcription: - PR EME D41 1.C OM - PR EME D41 Answer: In eukaryotes, transcription and translation occur in separate compartments. No separate compartments for this process in prokaryotic cells. In bacteria, mRNA is polycistronic; in eukaryotes, mRNA is usually monocistronic. Polycistronic: one mRNA codes for more than one polypeptide moncistronic: one mRNA codes for only one polypeptide OM - PR EME D41 1.C OM Three RNA polymerases in euk., One in prok. Binding of Basal Transcription Factors required for euk. RNA Pol II binding is the key enzyme responsible for mRNA transcription. Sigma subunit is the only binding factor required for prokaryotic RNA polymerase. PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 11. COM - PR EME D4 11. COM - PR EME D41 1.C Processing of mRNA in eukaryotes such as: 5’ 7-methylguanosine (7mG) cap added, 3’ Poly-A tail added, Splicing out of introns are required. But, not in prokayotic cells. M .CO 411 MED - PR E - PR EME D41 1.C OM - PR EME D41 1.C OM 4. A nucleosome consists of histone octomers and DNA. While many mutations have been found in many different genes, very few mutations have been discovered in histones after surveying across all eukaryotic orgainisms. Explain the reason. Would an anti-histone antibiotic drug be effective in combating human diseases caused by eukaryotic pathogens? Would the FDA approve it? Explain your answer. 1.C OM Answer: - PR EME D41 1.C OM - PR EME D41 1.C OM - PR EME D41 Histones represent one of the most conserved molecules in nature because they are involved in a fundamental and important function relating to chromosome structure. Mutations are probably lethal. Even if there was a mutation at this region, the organism with this mutaton might already been eliminated during evolution. Since all antibody-producing organisms have essentially the same histones, it would be difficult to find an organism to produce histone antibodies, for to do so would be self-destructive. - PR EME D41 1.C OM 5. To understand the function of tRNA in translation and the nature of genetic code, Crick postulated the wobble hypothesis in 1966. Describe the wobble hypothesis and explain how his hypothesis helped us to decipher the genetic code? D41 1.C OM Answer: PRE MED 411 .CO MPRE M ED4 11. COM - PR EME D4 11. COM - PR EME D4 11. COM - PR EME D41 1.C OM - PR EME Crick’s wobble hypothesis: The first two positions of the codon are precise in their complementary relationships with tRNA. However, the third position is less specific. In a sense that the pairing between the third position of the codon and the first position of the anticodon does not need to following Watson and Crick’s base pairing rule. His hypothesis pointed out the degeneracy nature of the genetic code, and explained why there are over 30 tRNA molecules in a cell, while there are only 20 amino acids available for the production of protein. ...
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This document was uploaded on 10/27/2011 for the course BIL 250 at University of Miami.

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