Test-2-practice-ques_44452

Test-2-practice-ques_44452 - Practice questions...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Practice questions representative of those that will be included in exam 2. 1. What is the name of the enzyme that, during DNA replication, extends the lagging strand from the 3’ end of each RNA primer? A. DNA polymerase I. B. DNA polymerase III. C. Primase. D. Replicase. E. Topoisomerase. 2. What would you expect to happen during DNA replication in a cell lacking topoisomerase? A. DNA ahead of the replication fork would become increasingly tightly coiled. B. Okazaki fragments would not be joined. C. DNA polymerase I could not displace the RNA primer at the 5’ end of Okazaki fragments. D. Only lagging strand DNA synthesis would halt. E. All of the above. 3. Which answer combines the correct responses to the questions below? (i) What is the name of the structure shown below? (ii) What is its function in transcription? (iii) What is the probable base composition of the region marked by the “A” arrow? (iv) What is the probable base composition of the region marked by the “B” arrow? A B 5’ 3’ *A. (i) Hairpin loop; (ii) Termination of transcription in prokaryotes; (iii) G/C rich; (iv) string of U’s. B. (i) Hairpin loop; (ii) Termination of transcription in prokaryotes; (iii) string of U’s; (iv) G/C rich. C. (i) tRNA; (ii) Termination of transcription in prokaryotes; (iii) G/C rich; (iv) string of G’s. D. (i) Hairpin loop; (ii) Termination of transcription in prokaryotes; (iii) A/U rich; (iv) string of U’s. E. None of the above. 4. In this picture of DNA, which number identifies the leading strand and how do you know? A. 1 because the 3’ end is extended in the same direction as the replication fork B. 2 because the 5’ end is extended in the same direction as the replication fork C. 1 because the 5’ end is extended in the same direction as the replication fork D. 2 because the 3’ end is extended in the same direction as the replication fork E. Both strands are leading strands because they are copied from complementary template strands 5. This parental helix of DNA is replicated in a test tube containing all the proteins necessary for DNA replication, GTP (guanosine triphosphate) and radioactive CTP (cytosine triphosphate  ­ 32P): 5’-GGGGGGGGGGG-3’ 3’-CCCCCCCCCCC-5’ What pattern of radioactivity would you expect among the new helices? A. Only 1 of the new helices would be radioactive it is impossible to predict which one B. Only 1 of the new helices would be radioactive it is possible to predict which one C. Both of the new helices will be radioactive D. There is insufficient information given to predict the likely pattern of radioactivity E. Both of the new helices will be radioactive, but only one at a time 6. The functional equivalent of the TATA binding protein in prokaryotes is: A. Sigma factor. B. Holoenzyme. C. Rho factor. D. TFIID. E. Hairpin loops. 7. You delete a gene coding one of the proteins required for splicing (removing) introns from pre ­mRNA. This deletion mutation stops growth of a yeast cell. Why? A. Failure to remove introns imposes a high energy cost because proteins will be longer. B. Introns are important in directing the export of mRNA to the cytoplasm where it is translated. C. Failure to remove introns will lead to non ­functional proteins because their primary sequence will be derived from introns as well as exons. D. Intron removal is essential for addition of the 5’ cap and 3’ polyA ­tail to mRNA. E. All of the above. 8. What experimental observation tested the possibility that the genetic code was overlapping? A. A mutation changing a single nucleotide changed one amino acid in the coded protein. B. Different mutations had similar phenotypic effects. C. A mutation changing a single nucleotide changed one nucleotide in the transcribed messenger RNA. D. A mutation changing a single nucleotide changed several adjacent amino acids in the coded protein. E. Determining whether or not the genetic code was overlapping had to wait for the entire code to be determined (that is, the full mapping between codons and amino acids). 9. A newly discovered set of genes – the hou operon – are transcribed from the same promoter. They allow the bacteria to use a rare sugar, houstose. This sugar doesn’t give much energy when it is broken down, but, when nothing else is available, it is better than nothing. In what combination of environmental conditions would you expect bacteria to express the hou operon? (Assume that conserving energy used in expressing a gene is important – if a bacteria does not need to express a gene, it won’t.) A. In the absence of houstose and the absence of any other more effective nutrient source. B. In the absence of houstose and the presence of any other more effective nutrient source. C. In the presence of houstose and the presence of any other more effective nutrient source. D. If houstose does not produce much energy when it is used hou operon should always be expressed. E. In the presence of houstose and the absence of any other more effective nutrient source. 10. Does the lac repressor (LacI) act as a cis or a trans regulator? A. Trans. B. It is both a cis and trans regulator. C. It is neither a cis or trans regulator. D. Cis. E. Depends on the presence of another copy of the lac operon (as in a partial diploid). 11. How does tryptophan inhibit transcription of the trp operon through attenuation? A. Pausing of the ribosome at two consecutive trp codons in the leader region causes mRNA to fold in a way that terminates transcription by RNA polymerase B. The ribosome NOT pausing at two consecutive trp codons in the leader region causes mRNA to fold in a way that terminates transcription by RNA polymerase C. trp operon mRNA always folds in a way that cause transcription to terminate prior to transcribing the structural genes D. The TrpR repressor binds to an operator region preventing RNA polymerase binding if trp is present E. The LacI repressor can bind to the trp operator and block the initiation of transcription 12. A mutant strain of E. coli has the following genotype at its lac operon (assume all other genes are functional): I ­ P+ O+ Z+ Y+ A+ (I – repressor; P – promoter; O – operator; Z,Y,A ­ structural genes). Which of the following best matches the environments in which the operon be expressed at a high level? A. Glucose is absent, regardless of the presence of lactose. B. Lactose is present, regardless of the presence of glucose. C. Glucose is present, regardless of the presence of lactose. D. Lactose is present and glucose is absent. E. It won’t be expressed in any environment. 13. A second copy of the lac operon is introduced into the above strain (this second copy is on an F’lac plasmid). This second copy has the genotype I+ P+ O+ Z ­ Y ­ A ­ (the original chromosomal copy is still I ­ P+ O+ Z+ Y+ A+). In which of the following environments will functional copies of the LacZ, LacY and LacA proteins be expressed? A. Glucose is absent, regardless of the presence of lactose. B. Lactose is present, regardless of the presence of glucose. C. Glucose is present, regardless of the presence of lactose. D. Lactose is present and glucose is absent. E. It won’t be expressed in any environment. Answers: 1. B: polymerase III is the main polymerase, acting to extend DNA strands from RNA primers. 2. A: topoisomerase ‘relaxes’ tightly coiled DNA. 3. A: picture is of an intrinsic termination region. 4. A: DNA synthesis occurs from 5’ to 3’ – the leading strand follows the replication fork in this direction. 5. B: the new helix using the poly ­G strand as a template will be radioactive because it directs the incorporation of the radioactive ‘C’ nucleotides. 6. A: the TATA binding protein directs the eukaryotic RNA polymerase to the promoter. In prokaryotes this function is performed by a sigma factor. 7. C: if introns are translated the primary, and thus secondary and tertiary, protein structure will be altered. This is likely to reduce functionality. 8. A: the test used to determine if the genetic code was overlapping was to alter a single nucleotide and examine how many amino acids were changed. If only one amino acid was changed, the code was non ­overlapping. 9. E: a low efficiency resource (sugar) will be used when it is present and no better resource is also available. If a better resource was available, that would be used instead. 10. A: as a diffusible protein, the LacI repressor will act as a trans regulator. 11. B: the ribosome does not pause when tryptophan is present. D is true, but this mechanism is not called attenuation. 12. A: the lacI ­ mutation means that the LacI repressor is not produced so cannot bind to the operator and inhibit transcription – in this case the presence of lactose is irrelevant (normally lactose binds to LacI and removes it from the operator). The glucose mediated cAMP ­CAP positive activator is still functional, so the operon will only be fully expressed when glucose is low. 13. D: this second copy of the lac operon will provide functional LacI in the cell. In this case regulation will be restored to the normal wild type. ...
View Full Document

This note was uploaded on 08/16/2011 for the course BIOL 1361-1362 taught by Professor Any during the Spring '08 term at University of Houston.

Ask a homework question - tutors are online