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BIOL . 303 EXAM II 3/22/07 -----------------------------------------------------------------------------------------------------------------This exam consists of 40 multiple choice questions worth 2.5 points each. On the separate testing form, please fill-in the single best choice for each question. Be sure to fill-out your student number and name on the answer sheet. Good luck! -----------------------------------------------------------------------------------------------------------------1. It is usually easier to identify spontaneous mutations in bacteria than most eukaryotes because 1. mutations are visible as color changes within a bacterial colony. 2. mutations can be induced via exposure to the bacteriophage T1. 3. most bacteria are haploid and so mutations are phenotypically expressed readily in bacterial cells. 4. mutations occur at a much higher frequency in bacteria. 2. In conjugation, genetic information is transferred 1. from one cell into the culture medium, where it is taken up by another cell. 2. with the help of a viral go-between. 3. in a bidirectional fashion between two cells. 4. from one bacterium to another. 3. Hfr bacterial cells exhibit the highest levels of gene transfer because 1. they contain two distinct F factors. 2. they induce the production of significantly more sex pili than normal F+ cells. 3. rather than existing as an isolated plasmid, the F factor is incorporated within the main chromosome where it can readily transfer genes. 4. they do not require conjugation for gene transfer. 4. The name for the general category of DNA molecules into which double-stranded circular extrachromosomal DNA elements such as F factors would fall is 1. capsid 2. plasmid 3. plaque 4. partial diploid 5. One bacterial Hfr strain transfers genes in the order A--B--C--D--> while a second Hfr strain transfers genes in the order B--C--D--A-->. The most likely explanation for this is that 1. one strain actually carries an F' element and is a merozygote. 2. the F factor integrated at the same site but in opposite orientations in the two strains. 3. the F factor integrated at different sites and in opposite orientations in the two strains. 4. the F factor integrated at different sites but in the same orientation in the two strains. 6. In humans, the genetic basis for determining the sex “male” is accomplished by the presence of 1. the Y chromosome 2. one X chromosome 3. a balance between the number of X chromosomes and the number of haploid sets of autosomes 4. multiple alleles scattered throughout the autosomes 7. Klinefelter syndrome in humans, which leads to underdeveloped testes and sterility, is caused by which chromosomal condition? 1. 2. 3. 4. 47, XXY 47, 21+ 45, X 47, XYY 8. In birds, sex is determined by a ZW chromosome scheme, which is much like the typical XY scheme seen in humans and many other organisms. However, in birds the system is reversed. Males are ZZ (similar to XX in humans) and females are ZW (similar to XY in humans). A lethal recessive allele that causes death of the embryo occurs on the Z chromosome in pigeons. What would be the sex ratio in the offspring of a cross between a male heterozygous for the lethal allele and a normal female? 1. 1:1 male to female 2. 3:1 male to female 3. 1:2 male to female 4. 2:1 male to female 9. Hemophilia is caused by a sex-linked recessive gene. Assume that a man with hemophilia marries a normal woman whose father had hemophilia. What is the probability that they will have a daughter with hemophilia? (Note: in this problem you must include the probability of having a daughter in your computation of the final probability.) 1. 1/16 2. 1/8 3. 1/4 4. 1/2 10. What is the expected number of Barr bodies in an individual with the karyotype XXYYY? 1. 0 2. 1 3. 2 4. 3 11. When an organism gains or loses one or more chromosomes but not a complete haploid set, the condition is known as 1. polyploidy 2. aneuploidy 3. triploidy 4. trisomy 12. The condition known as cri-du-chat syndrome in humans has a genetic constitution designated as. 1. 45, X 2. 46,5p3. triploidy 4. trisomy 13. A genomic condition that may be responsible for some forms of fragile-X syndrome, as well as Huntington disease, involves 1. F plasmids inserted into the FMR-1 gene. 2. expansion of trinucleotide repeats. 3. multiple inversions in the X chromosome. 4. single translocations in the X chromosome. 14. Trisomy 21, or Down syndrome, occurs when there is a normal diploid chromosomal complement of 46 chromosomes plus one (extra) chromosome #21. Such individuals therefore have 47 chromosomes. While there is impaired fertility of both sexes, females are more likely to be fertile than males. Assume that children are born to a female with Down syndrome and a normal 46-chromosome male. What proportion of the offspring would be expected to have Down syndrome? 1. One half of the offspring will be expected to have Down syndrome. 2. None of the offspring would be expected to have Down syndrome. 3. All the children would be expected to have Down syndrome. 4. Two-thirds of the offspring would be expected to have Down syndrome. 15. While the most frequent forms of Down syndrome are caused by a random nondisjunction of chromosome #21, Down syndrome occasionally runs in families with apparently normal individuals having children with Down syndrome. The cause of this form of familial Down syndrome is 1. 2. 3. 4. 16. an inversion involving chromosome #21 too many X chromosomes a translocation between chromosome #21 and another chromosome a maternal age effect Nondisjunction 1. can lead to aneuploidy. 2. does not alter a karyotype. 3. turns a diploid genome into a haploid genome. 4. is responsible for familial Down syndrome. 17. The observation that a fly heterozygous for the double Bar allele and the wildtype allele (genotype = BD/B+) has a different phenotype than a fly that is homozygous for the Bar eye allele (genotype = B/B) serves as an example of the phenomenon known as 1. chaos. 2. inversions. 3. incomplete penetrance. 4. position effect. 18. The genetic material of most living things is 1. deoxyribonucleic acid. 2. ribonucleic acid. 3. RNA. 4. polysaccharide. 19. In the classic experiment conducted by Hershey and Chase, why were the bacteria radioactive after they had been infected with viruses which had been grown on medium containing 32P? 1. The bacteria had incorporated radioactive proteins into their cell membranes. 2. The radioactive viruses (coats plus DNA) were injected into the bacteria. 3. The bacteria contained radioactive viral DNA 4. None of the above. 20. The basic structure of a nucleotide includes the following components: 1. amino acids. 2. base, sugar, phosphate. 3. phosphorous and sulfate. 4. all of the above. 21. If 15% of the nitrogenous bases in a sample of DNA from a particular organism is thymine, what percentage should be cytosine? 1. 15% 2. 30% 3. 70% 4. 35% 22. Which of the following clusters of terms accurately describes DNA as it is generally viewed to exist in prokaryotes and eukaryotes? 1. Double-stranded, parallel, (A+T)/(C+G)= variable, (A+G)/(C+T)=1.0 2. Single-stranded, antiparallel, (A+T)/(C+G)=1.0, (A+G)/C+T)=1.0 3. Double-stranded, antiparallel, (A+T)/(C+G)=variable, (A+G)/(C+T)=1.0 4. Double-stranded, antiparallel, (A+T)/C+G)=1.0, (A+G)/(C+T)=variable Refer to the following figures for question 23 a d. 23. b. c. e. Which diagram shows a nucleotide with a purine base? 1. Figure a 2. Figure b 3. Figure c 4. Figure d 24. The two strands of a DNA helix are held together (to each other) by 1. covalent bonds. 2. hydrogen bonds. 3. phosphodiester bonds. 4. glycosidic bonds. For questions 25-26 refer to the replication fork depicted below: 5’ 3’ B A 3’ 5’ B B C 5’ 3’ 25. Item A is called 1. 2. 3. 4. 26. the leading strand. primase. single strand binding protein. an Okazaki fragment. B is most likely: 1. an RNA primer. 2. a DNA primer. 3. DNA polymerase III. 4. an exonuclease. 27. During replication of DNA, strand elongation proceeds 1. in a 5' to 3' direction on the leading strand, but in a 3' to 5' direction on the lagging strand. 2. in a 3' to 5' direction on the leading strand, but in a 5' to 3' direction on the lagging strand. 3. in a 5' to 3' direction on both the leading and lagging strands. 4. in a 3' to 5' direction on both the leading and lagging strands. 28. The enzyme that appears to be inappropriately expressed in cancer cells and helps to maintain the very ends of eukaryotic chromosomes is called 1. topoisomerase. 2. ligase. 3. telomerase. 4. terminase. 29. In addition to highly repetitive and unique DNA sequences, a third category of DNA sequences exists. What is it called and what types of elements are involved? 1. 2. 3. 4. composite DNA, telomeres and heterochromatin dominant DNA, euchromatin and heterochromatin multiple gene family DNA, hemoglobin and 5.0S RNA middle repetitive DNA, SINEs, LINEs, and VNTRs 30. Chromatin of eukaryotes is organized into repeating interactions with protein octomers called nucleosomes. Nucleosomes are composed of which class of molecules? 1. 2. 3. 4. 31. histones glycoproteins lipids nonhistone chromosomal proteins In eukaryotic chromatin, the next highest level of organization above "beads-ona-string" is 1. the 30 nm fiber, or “solenoid”. 2. chromatin loops. 3. the nuclear matrix. 4. the metaphase chromosome. 32. Most DNA occurring in nature 1. 2. 3. 4. is relaxed. is supercoiled. is “naked” and is not complexed with proteins has ends that are free to rotate. 33. Mutations which arise in nature from no particular artificial agent are called 1. induced mutations. 2. spontaneous mutations. 3. chromosomal aberrations. 4. cosmic mutations. 34. Two terms used to describe categories of mutational nucleotide substitutions in DNA are called 1. base analogues and frameshift. 2. error prone and spontaneous. 3. transversions and transitions. 4. euchromatic and heterochromatic. 35. A tautomeric shift 1. is typically induced by high-energy radiation. 2. occurs only in prokaryotes. 3. can lead to a transversion mutation. 4. can produce a transition mutation. 36. A class of mutations that results in multiple contiguous amino acid changes in proteins is likely to be which of the following? 1. base analogue. 2. transversion. 3. transition. 4. frameshift. 37. Ultraviolet light causes pyrimidine dimers to form in DNA. Some individuals are genetically incapable of repairing some dimers at normal rates. Such individuals are likely to suffer from 1. xeroderma pigmentosum. 2. phenylketonuria. 3. muscular dystrophy. 4. Huntington disease. 38. In mammals, DNA double-strand breaks may be repaired by 1. mismatch repair. 2. base excision repair. 3. nucleotide excision repair. 4. nonhomologous end-joining 39. In the first step of base excision repair 1. 2. 3. 4. 40. a phosphodiester bond is broken two nicks are made surrounding the DNA damage a break in DNA is sealed by DNA ligase a base is removed by the action of a DNA glycosylase The simplest type of transposable elements in bacteria are known as 1. transposons. 2. insertion sequences. 3. Ds elements. 4. Ty elements. That’s all! ITEM NUMBER 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 CORRECT ANSWER 3 4 3 2 4 1 1 4 3 2 2 2 2 1 3 1 4 1 3 2 4 3 4 2 4 1 3 3 4 1 1 2 2 3 4 4 1 4 4 2