Bis 2a midterm key 3 - Britt BIS2A midterm 3 2017 key 1 Which of the following activities does the ribosome perform a Establishment of the reading frame

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Unformatted text preview: Britt BIS2A midterm 3 2017 key 1. Which of the following activities does the ribosome perform? a. Establishment of the reading frame and formation of peptide bonds$ b. Polymerization of dNTPs c. Addition of the correct amino acid to the 3’OH of tRNA d. Removal of incorrect amino acids e. c and d are correct. 2. A missense mutation is … a. When the two bases in the double helix don’t match b. When the sequence of a codon is changed, but its meaning is not. c. A mutation that changes the amino acid encoded$ d. A frameshift mutation e. A mutation to a stop codon 3. The male black widow spider is eaten (head first!) by the female during mating. This is a good illustration of: a. Muller’s ratchet b. “Replicators” promoting their own survival-­‐ not their host’s$ c. Gene silencing d. Inbreeding depression e. Sexual selection 4. In prokaryotes, the correct reading frame for translation is precisely determined by: a. The position of the 5’ end of the mRNA b. The position of the methionine codon preceded by the Shine Dalgarno sequence $ c. The position of the most 3’ methionine codon d. The position of the most 5’ methionine codon i e. The position of the stop codon 5. The peacock’s tail is an example of… a. Genetic drift b. Natural selection c. Sexual selection d. Artificial selection e. Inbreeding 6. A very short eukaryotic mRNA has the sequence… 5’ ACGCCGAUCAUGCCGGGCAUGCGAGUAGAAUAAAAACUG 3’ How many amino acids are encoded by this message? (a codon table is provided) a. 12 b. 6 c. 8 d. 7$ e. 5 7. DNA polymerase proofreads its work. Which of the following components of gene expression also proofread? a. RNA polymerase b. Repressor proteins c. Ribosomes d. tRNA synthetases$ e. Enhancers 8. In eukaryotes, transcription and processing of mRNAs occurs in the _____, while translation occurs in the _________. a. nucleus, golgi b. nucleus, cytoplasm$ c. cytoplasm, nucleus d. cytoplasm, golgi e. golgi, nucleus 9. The Lac operator a. is a DNA sequence that binds the activator 1 pt b. is a DNA sequence located between the RNA polymerase binding site and the first coding sequence 1 pt c. is a DNA sequence that binds the repressor d. a and b e. b and c$ 10. The mutant allele coding for orange coat color in cats is located on the X chromosome. Cats heterozygous for this allele (X Xo cats) have patches of orange mixed with patches of black. They’re called “tortoise shell” cats. An orange male is crossed to a tortoise shell female. What fraction of their kittens are predicted to be orange? a. 0% b. 25% c. 50%$ d. 75% e. It cannot be predicted from the data provided. 11. Enhancers are frequently found in eukaryotic genes, but not in prokaryotic genes. One possible explanation for this difference is: a. The prokaryotic genome is too compact to include any regulatory sequences b. The prokaryotic genome is too large for enhancers to function c. The density of genes (# genes per kilobase) is too high for enhancers to specifically affect a single operon$ d. An entire operon is too long to be regulated by an enhancer e. the bacterial chromosome is circular and therefore cannot bend 12. Bacteria tend to locate their genes in functional clusters (operons)-­‐ where the genes involved in a single pathway involving several enzymes will be co-­‐located. In eukaryotes, however, the genes involved a biosynthetic pathway are randomly distributed throughout the genome. Why does this discrepancy exist? a. The eukaryotic genome is older, and the genes have moved apart due to random mutations b. DNA is often inherited via vertical, as well as horizontal, transmission in eukaryotes c. DNA is often inherited via horizontal, as well as vertical, transmission in prokaryotes$ d. Meiotic recombination breaks up gene clusters in eukaryotes, but not prokaryotes e. Meiotic recombination breaks up gene clusters in prokaryotes, but not eukaryotes 13. A mutant is isolated with a nonsense mutation in the 5’ end of the gene encoding the trp repressor. Such a mutant would be expected to: a. Strongly and constitutively express the trp operon$ b. Strongly express the trp operon only in the presence of trp c. Strongly express the trp operon only in the presence of lactose d. Strongly express the trp operon only when trp is absent e. Strongly express the trp operon only the presence of high levels of cAMP. 14. A mutant is isolated with a mutation in the trp repressor protein that prevents binding of trp to the repressor. Such a mutant would be expected to: a. Strongly and constitutively express the trp biosynthesis genes$ b. Strongly express the trp biosynthetic genes only in the presence of trp c. Strongly express the trp biosynthetic genes only in the presence of lactose d. Strongly express the trp biosynthetic genes only when trp is absent e. Strongly express the trp biosynthetic genes only the presence of high concentration of cAMP. 15. Only 10% of the maize (corn) nuclear genome encodes for proteins. The majority of the rest of the genome is made up of: a. Randomly assembled strings of nucleotides b. Ribosomal RNA and tRNA genes c. Retroviral and transposable element sequences$ d. Noncoding bacterial DNAs e. “Backup” copies of the genes. 16. The plant Coleus sometimes turns its leaves red. A hormone (here called “signal”) is involved in the regulation of flowering and leaf color. A pathway (see Figure above) is hypothesized to explain its action. All members of the signal transduction pathway are shown. The T-­‐bar symbol depicted between M and K is consistent with the action of a protein that: a. degrades the hormone “signal” b. binds to an enhancer, activating transcription of its target protein c. degrades protein K$ d. binds a regulatory region near the promoters of its target genes, inhibiting RNA polymerase II binding e. Bind to protein K, allosterically activating it there were accidentally two answers to question 16-­‐ either answer got full credit 17. "Signal" is a small molecule. Predict the correct outcome when “signal” is applied to the plant, according to the pathway shown in the Figure above, both in wild-­‐type (wt) and in mutants that do not express protein “M”. a. wt: Flowers and green leaves, mutant: flowers and green leaves$ b. wt: no flowers, red leaves, mutant: flowers, red leaves c. wt: flowers and red leaves, mutant: flowers and red leaves d. wt: no flowers, green leaves, mutant: no flowers, green leaves e. wt: no flowers, red leaves, mutant: no flowers, red leaves 18. Generally speaking, the genes for regulatory proteins that directly detect environmental signals (for example, the Lac Repressor protein) are transcribed… a. All the time (constitutively)$ b. Only when the genes they regulate are transcribed c. Only when the genes they regulate are not transcribed d. Only when the signal they detect is present 19. The fastest way to negatively regulate an enzymatic activity would be: a. Mutate the gene b. Turn off transcription of its gene c. Prevent translation of its mRNA d. Prevent degradation of the protein e. Allosteric inhibition of the protein$ 20. The least costly (least ATPs!) way to negatively regulate an enzymatic activity would be: a. Mutate the gene b. Turn off transcription of its gene$ c. Prevent translation of its mRNA d. Prevent degradation of the protein e. Allosteric inhibition of the protein 21. Which of the following eukaryotic processes reduce the effect of the relentless accumulation of mutations? a. Crossing over$ b. Asexual reproduction c. Mutational correction d. Post-­‐mitotic meioses e. Telomerase 22. Cassava (the source of tapioca) is a tree that is easily propagated by cutting the branches into short rods, sticking these rods into the ground, and allowing them to grow into new trees. It can also be propagated through seeds, either by self-­‐pollination or by crossing to other plants. Which reproductive system would be expected to eliminate recessive lethal alleles from the breeding population most rapidly? a. Clonal reproduction-­‐ the stick method b. Self-­‐pollination$ c. Crossing to other plants that are closely related d. Crossing to distantly related cassava plants e. Plants can’t carry lethal alleles-­‐ they would be dead. 23. The eukaryotic RNA polymerase II, required for mRNA production, cannot recognize promoters on purified DNA without assistance. What additional proteins/events are required for RNA pol II to bind to the promoter? a. The strands of the DNA must be separated by DNA polymerase first b. A primer must be added at the promoter c. Transcription factor II D must be bound to the TATA box$ d. The two RNA polymerase must be tethered to each other through and intermediary protein e. RNA polymerase I must initiate polymerization. 24. A DNA template sequence reads: 5' ATCGATCG 3' The sequence of the mRNA strand made when RNA polymerase transcribes this template will read… a. 5' CGATCGAT 3' b. 5' CGAUCGAU 3'$ c. 5' GCAUGCAU 3' d. 5' AUCGAUCG 3' e. 5' TACGTACG 3' 25. “Classic” hemophilia is a blood clotting disorder resulting a recessive defect in an X-­‐linked gene that produces a protein called “Coagulation Factor VIII”. People that lack the wild-­‐type allele for this gene have a blood clotting disorder. The frequency of males born with this clotting disorder is about 1/5000 worldwide. A woman who does not have the disorder (= who has no problem with blood clotting), but whose father did have it, marries a man who does not have the disorder. They have a son. What are the odds that their son has the disorder? a. We don’t know the genotype of the mother, so we can’t say b. 1/5000 c. 1/10,000 d. 50%$ e. 100% ...
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