Name_________________________________________ November 9, 2004 BIO184 Genetics Exam 2 This examination is worth 100 points. There are questions on pages. Read through the questions completely before answering the questions. If you are unclear about a question, please ask me. Good luck Part A: Multiple Choice: Answer with the best choice. Make sure that you clearly circle the correct answer and erase any erroneous marks. (4 pts each) 1. AG = 1 CT for double-stranded helical DNA. a) True b) False Answer: A 2.The DNA of a eukaryotic chromosome is one long double helix. a) True b) False Answer: A 3. If DNA polymerase could catalyze sugar phosphate bonds in both directions equally well what would no longer be seen during synthesis? a) replication bubble *b) Okazaki fragments c) RNA primer d) telomerase e) both a and b 4.If phages are labeled with radioactive sulfur and allowed to infect bacterial cells, the phage progeny resulting from lysis are expected * a) to be non-radioactive. b) to have radioactive DNA. c) to have radioactive proteins. d) to have radioactive DNA and proteins. e) to have radioactive carbohydrates. 5) Which of the following is not true of DNA? a) It is acidic b) It contains deoxyribose c) It is found in cell nuclei d) It contains phosphate e) it contains proteins 6) a) b) c) d) e) The group of enzymes able to relax supercoils in DNA is called: primases helicases topoisomerases telomeres ligases 7) Ocassionally, a loss of function mutation may occur in the telomerase enzyme in a cell. What is likely to be the result of this mutation on the DNA in the cell over the course of several rounds of mitosis? a) Chromosome length will gradually increase b) Chromosome length will gradually decrease c) Chromosome length will stay constant d) Chromosomes will fail to dissociate after replication e) None of the above 8) In general, which of the following usually has a greater chance of lethality than the others? a) inversions b) duplications c) deletions d) translocations e) none of the above 9) Inversions may be hard to detect because they: a) never visibly change chromosome banding patterns b) increase recombination in heterozygotes c) do not usually cause an abnormal phenotype d) normally are removed immediately in natural populations 10) Down syndrome can result from: a) three copies of chromosome 21 b) a translocation of a part of chromosome 21 c) a reciprocal translocation between any two autosomes d) a and b e) a, b and c 11) Which of the following is not an aneuploidy? a) monosomy b) tetraploid c) trisomy d) tetrasomy 12) Hybrids in which the chromosome sets come from two distinct, though related, species are known as: a) autopolyploids b) allopolyploids c) aneuploids d) bivalents 13) Nondisjunction can occur at either the first or second division of meiosis, XXY individuals could arise from nondisjunction at the _____ meiotic division in the _____. a) first, mother b) second, mother c) first, father d) second, father e) more than one of the above could give rise to XXY individuals 14) Histones and DNA have a strong attraction for each other because: a) DNA is positively charged and histones are negatively charged b) both DNA and proteins are hydrophobic c) DNA is negatively charged and histones are positively charged d) like substances share common charges e) none of the above 15) In mammalian cells, replication proceeds: a) unidirectionally b) bidirectionally c) unidirectionally from many origins d) bidirectionally from only one origin 16) How many replication forks depart from an origin of replication? a) one b) two c) three d) four e) none of the above 17) From DNA research, sites of transcription and therefore most of the genes along the length of the chromosome appear to be found in: a) heterochromatin b) euchromatin c) constitutive heterochromatin d) centromeric regions 18) When a chromosomal rearrangement such as an inversion occurs placing a known gene into a heterochromatic region, the gene’s expression: a) will turn on b) may be amplified c) may cease d) none of the above 19) In which of the following cases will a Barr body be seen? a) XX b) XY c) XO d) XXY e) both a and d 20) Telomerase is not needed in bacterial DNA replication because bacteria A) have a polymerase that can synthesize in the 3’-to-5’ direction B) have plasmids that allow transitive polymerization of the telomeres C) are able to tolerate the loss of chromosomal ends *D) have circular chromosomes E) do not replicate chromosomal DNA PART B - Short answer. Please write clearly. 20) Match the region or item on the DNA replication diagram with the correct term from the list below. (7 pts) d____ Leading strand c____ Lagging strand a____ Origin e____ RNA Primer f____ Okazaki fragment g____ Helicase b____ Polymerase 22) Griffith found that smooth (S) forms of S. pneumoniae have a polysaccharide capsule and rough (R) forms do not. Only S forms cause infection. Briefly describe how Griffith demonstrated transformation using live R form and heat killed S form bacteria. (6 pts) Answer: S heat killed does not yield infection; R live does not yield infection; S heat killed plus R live = infection. R has been transformed by S. 21) When Meselson and Stahl performed the experiment which showed that replication is a semiconservative process, they utilized E. coli, and various isotopes of nitrogen (15N and 14N). Explain briefly what their results would have been if DNA replicated conservatively (be sure to describe the first two generations).(7 pts) Answer: Following centrifugation, the first generation of replication would yield 2 bands- 15N and 14N (no hybrid). The second generation would again result in the same pattern with no hybrid pattern ever revealed.