General Genetics Practice Exam 1 PDF

BIOL 23373 General Genetics Practice Exam 1 1. Histones have a_________charge that helps their interaction with DNA, that has a __________charge provided by the __________. A. positive, negative, phosphates B. positive, negative, bases C. negative,...

BIOL 23373 General Genetics Practice Exam 1 1. Histones have a_________charge that helps their interaction with DNA, that has a __________charge provided by the __________. A. positive, negative, phosphates B. positive, negative, bases C. negative, positive, phosphates D. negative, positive, bases E. negative, positive, ribose 2. Each molecule of tRNA is configured into a number of stems and loops. The anticodon loop matches up with a codon of mRNA. At least how many of the bases on this loop have to base pair exactly with the bases of the codon, and why? A. one, because of the wobble position B. two, because of the wobble position C. three, because it pairs with a codon D. one or two, depending on the codon E. none of the above 3. What do DNA polymerases need to successfully synthesize DNA? A. primer B. template DNA C. deoxynucleoside triphosphates (dNTPs) D. all of the above E. only A and C 4. In the bacterium E. coli, the genetic material is composed of A. circular, double-stranded DNA. B. linear, double-stranded DNA. C. circular double-stranded DNA and histone proteins. D. circular, double-stranded RNA. E. linear, double-stranded RNA. 5. Why was the DNase treatment used by Avery, MacLeod, and McCarty an important step to elucidate the molecular nature of the “transforming principle”? A. This allowed them to isolate pure DNA samples. B. This allowed them to isolate pure protein samples. C. This allowed them to demonstrate that removing the DNA prevents transformation. D. This allowed them to demonstrate that mixing rough cells with DNA prevents transformation. E. This allowed them to demonstrate that the acting factor was not RNA. 6. DNA isolated from a fungus has an adenine content of 25%. What would be the % (G+C) within the fungus DNA? A. 0%  B. 12.5% C. 25% D. 50% E. 75% 7. Introns A. are spliced out by snRNPs. B. form a lariat structure during splicing C. have specific nucleotide sequences at their 5’- and 3’-ends D. are removed from the primary mRNA before translation E. all of the above 8. What was the critical observation in Griffith’s 1928 experiments that demonstrated a genetic transformation of non-virulent into virulent bacteria (Streptococcus pneumoniae)? A. Mice injected with a mix of dead S (virulent) bacteria and living R (non-virulent) bacteria died. B. Virulent strains of bacteria could be isolated from dead mice that had been injected with a mix of dead S (virulent) bacteria and living R (non-virulent) bacteria. C. Uptake of a bacteriophage transformed non-virulent into virulent bacteria. D. DNA released from virulent bacteria was taken up by non-virulent bacteria. E. Virulent bacteria were transformed into non-virulent bacteria when they lost their polysaccharide capsule. 9. The compaction leading to a metaphase chromosome involves which of the following A. the formation of nucleosomes B. the formation of a 30-nm fiber C. anchoring and further compaction of the radial loops D. B and C E. A, B, and C 10. In the promoter region of a gene one would find A. cis-acting sequences B. consensus sequences for RNA polymerase binding C. consensus sequences for transcription termination D. A and B are correct E. A and B and C are correct 11. During DNA replication, once the replication bubble is formed, which of the following protein(s) are recruited to maintain the DNA template “open”? A. single-stranded binding proteins B. DNA primase C. Topoisomerases D. DNA helicases E. DNA polymerases 12. Eukaryotic chromosomes contain two general domains that relate to the degree of condensation. These two regions are… A. void of introns. B. called nucleoplasm and heteroplasm. C. called heterochromatin and euchromatin. D. separated by large stretches of repetitive DNA. E. both void of typical protein-coding sequences of DNA. 13. Which of the following is not true of tRNA? A. encoded by tRNA genes B. translated into protein in cytoplasm C. has secondary structure D. has anticodon loop E. has amino acid accepting site 14. When examining the genetic code, it is apparent that A. there can be more than one amino acid for a particular codon. B. there are multiple start codons C. there can be more than one codon for a particular amino acid. D. the code is ambiguous in that the same codon can code for two or more amino acids. E. there are 44 non-coding triplets because there are only 20 amino acids. 15. The ___________ is the cis-acting DNA sequence that serves as the binding site for RNA polymerase. A. promoter B. terminator C. enhancer D. regulator E. Shine-Dalgarno 16. What is the major difference between the lagging and leading strands? A. On the leading strand, DNA synthesis occurs from 5' to 3', while DNA synthesis occurs from 3' to 5' on the lagging strand. B. DNA polymerase is able to continuously add new nucleotides on the leading strand while it must keep 'starting over' on the lagging strand. C. The lagging strand requires only a single primer while the leading strand requires many. D. Helicase opens the leading strand at a faster rate than the lagging strand. E. The leading strand can not complete synthesis at telomeres. 17. Which of the following experimental observations gave Watson and Crick essential clues that helped them come up with the double helix model of DNA? A. Avery, MacLeod and McCarty’s experiment showing that DNA was the transforming principle. B. Chargaff’s finding that DNA contains the same numbers of purines and pyrimidines. C. Rosalind Franklin’s X-ray diffraction data that revealed the constant diameter and helical nature of DNA.  D. A and B are correct. E. B and C are correct. 18. During DNA replication, the enzymes below function in the following order: A. DNA ligase, DNA polymerase, primase, DNA helicase B. DNA helicase, primase, DNA polymerase, DNA ligase C. DNA helicase, DNA polymerase, primase, DNA ligase D. primase, DNA polymerase, DNA helicase, DNA ligase E. DNA ligase, primase, DNA polymerase, DNA helicase 19. Okasaki fragments are a consequence of A. the inability of the DNA polymerase to initiate a new DNA strand. B. the inability of the DNA polymerase to use RNA primers. C. random strand breakages resulting from supercoiling. D. mutations in the gene for DNA ligase. E. the inability of the DNA polymerase to polymerize in the 3' to 5' direction. 20. The bacterial sigma factor allows A. RNA polymerase to elongate the transcript. B. RNA polymerase to terminate transcription. C. RNA polymerase to identify and tightly bind promoter elements. D. DNA polymerase III to bind at the ori and initiate DNA synthesis. E. RNA polymerase to identify and tightly bind the TATA box sequence. 21. Which of these is NOT a feature of RNA polymerases? A. Transcribe DNA into RNA B. Synthesize DNA during the process of replication C. Do not require a primer D. They synthesize in the 5’ to 3’ direction E. They use ribonucleotides as substrate 22. The figure indicates a molecule of ________; the arrow indicates the location of______ A. RNA, 3’ hydroxyl B. DNA, 3’ hydroxyl C. RNA, 5’ phosphate D. DNA, 5’ phosphate E. RNA, 5’ oxygen 23. The complementary sequence of 5'-AATTCGCTTA-3' is: A. 5'-TAAGCGAATT-3' B. 5'-AATTCGCTTA-3' C. 3'-AATTCGCTTA-5' D. 5'-TAACGCTTAA-3' E. 3'-TAAGCGAATT-5' 24. When considering the initiation of transcription in prokaryotes, one finds consensus sequences located in the ________ region that are important for the binding of RNA polymerase and an associated ________ to the promoter. A. -30 ; TBP B. -10 and -35; sigma factor C. Shine-Dalgarno; sigma factor D. -35; RNA pol II E. -30; sigma factor 25. Telomerase A. synthesizes telomeric repeats at the end of chromosomes B. it is made of RNA and protein C. it uses RNA that is part of the telomerase as template for DNA synthesis D. binds to the 3’ overhang region of the telomere E. all of the above 26. The figure below depicts a replication fork. Find the correct designations of the components marked by letters. A. A: leading strand; B: Okazaki fragment (DNA); C: primer (RNA); D: lagging strand; E: template strand B. A: template strand; B: primer (RNA); C: Okazaki fragment (DNA); D: lagging strand; E: leading strand C. A: template strand; B: Okazaki fragment (RNA); C: primer (DNA); D: lagging strand; E: leading strand D. A: template strand; B: Okazaki fragment (DNA); C: primer (RNA); D: lagging strand; E: leading strand E. A: template strand; B: Okazaki fragment (RNA); C: primer (RNA); D: discontinuous strand; E: lagging strand 27. Splicing of transcripts normally occurs: A. only in the cytoplasm of prokaryotes B. only in the cytoplasm of eukaryotes C. in both prokaryotes and eukaryotes D. only in the nucleus of eukaryotes E. in mitochondria, as they have their own DNA 28. Occasionally, a mutation may occur in the telomerase enzyme in a cell that renders it nonfunctional. 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. Chromosomes will not undergo recombination. 29. What is the function of general transcription factors in eukaryotes? A. They recognize the splice donor and acceptor sites. B. They recognize the -10 and -35 sequences. C. They serve as sequences to which RNA polymerase binds. D. They recognize the TATA box and recruit RNA polymerase II. E. They initiate binding to the Shine-Dalgarno sequence. 30. The spliceosome: A. removes introns from primary mRNA B. recognizes specific nucleotides within the introns C. forms a lariat in the process of splicing D. A and B are correct E. A, B, and C are correct 31. Based on their proposed structure of DNA published in the Nature 1953 paper, Watson and Crick suggested: A. a mechanism that explained how mRNA is translated. B. a copying mechanism for the genetic material. C. a mechanism for how RNA polymerases can transcribe DNA. D. a repair mechanism for DNA nicks. E. a mechanism for genetic recombination. 32. The following coding section of a eukaryotic mRNA transcript specifies ______ in the polypeptide chain. Assume the sequence is in reading frame. 5’-ACCGACAUCGUCGUAGCAACUUUUUGGGGACCC-3’ A. 9 amino acids B. 11 amino acids C. 5 amino acids D. 6 amino acids E. 10 amino acids 33. Given the following gene sequence, which of the following would be the correct sequence of the corresponding mRNA transcript? Coding strand 5’-ATCGTCTTAGCCGT-3’ Template strand 3’-TAGCAGAATCGGCA-5’ A. 5’-ATCGTCTTAGCCGT-3’  B. 5’-ACGGCUAAGACGAU-3’ C. 5’-UGCCGAUUGUGCUA-3’ D. 5’-AUCGUCUUAGCCGU-3’ E. 3’-UAGCAGAAUCGGCA-5’ 34. The human genome contains approximately 20,000 protein-coding genes, yet has the capacity to produce several hundred thousand gene products. What can account for the vast difference in gene number and product number? A. A large number of human genes produce more than one protein by alternative splicing. B. There are more introns than exons. C. There are more exons than introns. D. Degeneracy in the third codon position. E. Every gene can be read in both directions, and each gene can have inversions and translocations. 35. Which of the following is not a function of the 5’ cap on eukaryotic mRNA? A. It protects the mRNA from RNase degradation. B. It is bound by proteins that export the mRNA out of the nucleus. C. Facilitates 5’ intron excision. D. It is required to orient the mRNA in the ribosome for translation. E. All of the above are true functions of the 5’ cap.