8. LOCATE YOUR GSI. Turn in your SCANTRON and EXAM to your GSI. YOU MUST TURN IN BOTH or else you will get a ZERO.

Transcription

1 BIOLOGY 1A MIDTERM # 2 November 2, 2015 VA NAME SECTION # DISCUSSION GSI 1. Sit every other seat and sit by section number. Place all books and paper on the floor. Turn off all phones, pagers, etc. and place them in your backpack. They cannot be visible. No calculator is permitted. Scantron Instructions 2. Use a #2 pencil. ERASE ALL MISTAKES COMPLETELY AND CLEARLY. 3. Write in in your name (Last, First). For TEST write in the name of your GSI. For period put the day and time of your discussion section. Write in YOUR SID in the TOP 8 boxes of the ID field. Make sure you also BUBBLE in your SID. The top 8 boxes of the ID field are for your SID. You can bubble in 00 for the bottom two or leave it blank. EXAM Instructions: 4. Print your name on THIS COVER SHEET. (otherwise, you will get a ZERO). 5. Leave your exam FACE UP. When told to begin, check your exam to see that there are 10 numbered pages, 51 multiple-choice questions. The exam is worth 100 pts. Each question is worth 2 points unless otherwise indicated. You are NOT PENALIZED for guessing! 6. Read all questions & choices carefully before bubbling in your response. 7. Do not talk during the exam. The exam is closed book. You cannot use a calculator. If you have a question, raise your hand; a GSI will help you. They will not give you the answer nor can they explain scientific terms (e.g. binding affinity, etc.). 8. LOCATE YOUR GSI. Turn in your SCANTRON and EXAM to your GSI. YOU MUST TURN IN BOTH or else you will get a ZERO. 9. WHEN TOLD TO STOP- YOU MUST REALLY STOP, even if you are not finished! Bubble in guesses BEFORE THIS TIME. If you continue to write after time has been called you will risk getting a There is always only one best answer.

2 1. What is the number of homologous pairs of chromosomes and the number of double-strand DNA molecules in a female human cell that has just completed Meiosis 1? A. 23 homologous pairs and 23 double-strand DNA molecules. B. 23 homologous pairs and 46 double-strand DNA molecules. C. No homologous pairs and 23 double-strand DNA molecules. D. No homologous pairs and 46 double-strand DNA molecules. E. No homologous pairs and 92 double-strand DNA molecules. 2. Duchenne Muscular Dystrophy (DMD) is an X-linked genetic condition that causes progressive loss of muscle function and weakness. It is caused by a recessive allele. Two of Jane s four brothers show symptoms of DMD. Jane, her four sisters, her father, and her mother, show no symptoms of DMD. What is most likely to be true? A. Jane s father is heterozygous for the mutant DMD allele. Jane s mother is homozygous wild type. B. Jane s mother is heterozygous for the mutant DMD allele. Jane s father is hemizygous wild type. C. Both Jane s mother and father are heterozygous for the mutant DMD allele. D. Neither Jane s mother or father is heterozygous for the mutant DMD allele. Both are homozygous wild type. 3. Jane (see question 3) and Tom s first child is a boy that has DMD. There is no occurrence of Duchenne Muscular Dystrophy (DMD) in Tom s family history. Jane and Tom have four more children two boys and two girls. What is the probability that both boys will have DMA? What is the probability that both girls will have DMA? A. P = 0 for the two boys, P = 0 for the two girls. B. P = 0.25 for the two boys, P = 0 for the two girls. C. P = 0.5 for the two boys, P = 0.5 for the two girls. D. P = 0.5 for the two boys, P = 0.5 for the two girls. E. P = 1.0 for the two boys, P = 0.5 for the two girls. 4. What two unique events occur during Meiosis 1 but not during Mitosis. A. recombination and assortment of homologous chromosomes into different cells. B. recombination and assortment of chromatids into different cells. C. DNA replication and assortment of homologous chromosomes into different cells. D. DNA replication and assortment of chromatids into different cells. E. recombination and DNA replication 5. Bubble in A for question 5 as you have version A. 6. Suppose you cross a/a B/b C/c D/D X A/A B/b C/c d/d where A, B, C and D are unlinked genes. What is the probability of obtaining an offspring that is A/a B/b C/c D/d? A. 1/2 B. 1/4 C. 1/8 D. 1/16 E. 1/32 Page 1 of 10

3 7. True breeding thick-stem, broad-leaf plants were crossed to true breeding thin-stem, narrow-leaf plants. The F1 progeny were thick-stem with broad-leaves. The F1 progeny were crossed to true breeding thinstem, narrow-leaf plants. Among 1000 progeny, Number of progeny Stem phenotype Leaf phenotype 400 thick broad 400 thin narrow 100 thick narrow 100 thin broad Most likely, the genes for stem width and leaf width are A. unlinked. B. linked and separated by 10 Centimorgans. C. linked and separated by 20 Centimorgans. D. linked and separated by 50 or more Centimorgans. E. Both A and D are possible. Work space. 8. Another scientist studying another species does the following cross. True breeding thick-stem, broadleaf plants were crossed to true breeding thin-stem, narrow-leaf plants. The F1 progeny were thick-stem with broad-leaves. The F1 progeny were crossed to true breeding thin-stem, narrow-leaf plants. Among 1000 progeny,. Number of progeny Stem phenotype Leaf phenotype 250 thick broad 250 thin narrow 250 thick narrow 250 thin broad Most likely, the genes for stem width and leaf width are A. unlinked. B. linked and separated by 10 Centimorgans. C. linked and separated by 20 Centimorgans. D. linked and separated by 50 or more Centimorgans. E. Both A and D are possible. Work space. Page 2 of 10

4 9. (3 pts) In Drosophila, the le (large eye) allele is recessive to the wild-type le + (normal eye) allele. Also, the fb (fat body) allele is recessive to the wild-type fb + (normal body) allele. A large eye, fat body female from a true breeding population was crossed to a true breeding wild type male. F1 females were crossed to wild-type males. Suppose that the fb + and le + are on separated by 10 Centimorgans on the X-Chromosome. What is the predicted number of offspring? A B C D Females normal eye normal body large eye fat body normal eye fat body large eye normal body Males normal eye normal body large eye fat body normal eye fat body large eye normal body Most likely, the genes for leg size and body shape are A. Column A B. Column B C. Column C D. Column D Work space. 10. The function of DNA polymerase I includes A. forming a phosphodiester bond using ATP for energy. B. 5 to 3 RNA hydrolysis. C. 3 to 5 DNA synthesis. D. 3 to 5 DNA hydrolysis E. 5 - to 3 RNA synthesis. 11. (1 pt) The function of ligase includes A. forming a phosphodiester bond using ATP for energy. B. 5 to 3 RNA hydrolysis. C. 3 to 5 DNA synthesis. D. 3 to 5 DNA hydrolysis E. 5 - to 3 RNA synthesis. 12. The function of proof reading by E. coli DNA polymerase III includes A. forming a phosphodiester bond using ATP for energy. B. 5 to 3 RNA hydrolysis. C. 3 to 5 DNA synthesis. D. 3 to 5 DNA hydrolysis E. 5 - to 3 RNA synthesis. Page 3 of 10

5 13. When replication of the E. coli genome begins, after the first phosphodiester linkage is made, what is the structure of the dinucleotide? A. B. C. D. 14. What is the structure of the molecule after DNA polymerase III forms its first phosphodiester bond? Only the dinucleotide at the 3 -end is shown. A. B. C. D. E. 15. What is the structure of the molecule after DNA polymerase I forms its last phosphodiester bond? Only the dinucleotide at the 3 -end is shown. A. B. C. D. E. Page 4 of 10

6 16. The figure below shows a replication fork in E. coli. At which of the indicated positions would you expect to see a 3 OH? A. B. C. D. E. A and B. 17. Select the TRUE statement. E. coli mrna A. is produced from both DNA template strands at every gene. B. has a monophosphate at the 5 -end. C. sometimes has multiple start and stop codons that are used by the ribosome to produce multiple proteins. D. has a hydroxyl at the 5 -end. E. All of the above statements are true. 18. E. coli trnas A. have intramolecular base pairing. B. are translated. C. carry amino acids non-covalently attached to their 3 -end. D. have a codon that base pairs with anti-codon in the mrna. E. are permanently bound to the ribosome. 19. A function of the 23S E. coli rrna found in the LSU of a ribosome includes A. providing structure for the ribosome by intramolecular base pairing. B. being part of the active site for peptide bond synthesis. C. base pairing with the mrna upstream (5 ) of the start codon before translation begins. D. A and B are true. E. A and C are true. 20. A function of the 16S E. coli rrna found in the SSU of a ribosome includes A. providing structure for the ribosome by intramolecular base pairing. B. being part of the active site for peptide bond synthesis. C. base pairing with the mrna upstream (5 ) of the start codon before translation begins.. D. A and B are true. E. A and C are true. 21. In the figure, what atoms participate in the covalent attachment of an amino acid to a trna? A. 2 and 3 B. 2 and 4 C. 1 and 3 D. 1 and 4 E. 1 and 5 Page 5 of 10

7 22. Why is the error rate (incorporation of the wrong amino acid) of translation low? A. Specificity of base pairing between codons and anticodons. B. Specific binding of aminoacyl trna synthases to amino acids. C. Proof reading by the ribosome. D. The ribosome only uses GTP, not ATP, for energy. E. A and B are true. 23. The anticodon for wild-type tyrosine trna base pairs with UAU and UAC codons on the mrna. A mutant tyrosine trna also base pairs with the UAA codon. It also binds the tyrosine aminoacyl trna synthase. Using the table showing the genetic code, what effect would this mutation likely have on translation? A. There would be no effect on translation. B. Proteins shorter (fewer amino acids) than normal produced. C. Proteins longer (more amino acids) than normal produced. D. No proteins would be synthesized. E. Tyrosine would be incorporated instead of cysteine. 24. The coding sequence for a mrna is 5 -AUG UUU AAA UGA-3. IMMEDIATELY after forming the FIRST peptide bond what molecule is covalently bound to the trna in the A-site of the ribosome? Use the table showing the genetic code, (--- represents the covalent bond to the trna). A. an uncharged ---trna. B. MET---tRNA. C. PHE-MET---tRNA. D. MET-PHE---tRNA. E. LYS-PHE-MET---tRNA. 25. When a codon in a mrna is 5 -ACG-3, what is the sequence of the trna anti-codon? A. 5 -ACG-3 B. 5 -UGC- 3 C. 5 -GCA-3 D. 5 -CGU-3 E. 5 -CGA-3 Page 6 of 10

8 26. Mutations that disrupt splicing of eukaryote RNAs would most likely be located in the A. 5 -untranslated sequences. B. exon sequences. C. conserved intron sequences. D. non-conserved intron sequences. E. 3 -untranslated sequences. 27. Suppose a gene encodes an RNA with 5 -Exon1---Intron1---Exon2-3. A phosphodiester bond eventually links the two exons. The electrons for connecting Exon 1 to Exon 2 comes from A. 3 -OH of Exon2. B. 2 -OH of Exon2. C. 3 -OH of Exon1. D. 2 -OH of Exon1. E. 3 -OH of the U5 snrna. 28. In order for oxidation of guanine to 8-oxoguanine to produce a stable, long-lasting mutation, the A. oxidized DNA must be transcribed into RNA. B. RNA transcript from the oxidized DNA must be translated by ribosomes. C. mutant polypeptide must disrupt cell function. D. cell with the oxidized DNA must undergo two mitotic divisions. 29. DNA glycosylases A. cleave lactose into monosaccharaides for glycolysis. B. remove a damaged base from DNA. C. insert the correct base after the damaged base has been removed. D. synthesize the final phosphodiester linkage after the correct base has been incorporated. 30. A loss-of-function mutation in a negative regulator of an inducible process A. causes the process to be constitutive. B causes the process to be non-inducible. C. has no effect on the process. D. cannot be predicted. 31. A loss-of-function mutation in a positive regulator of an inducible process A. causes the process to be constitutive. B causes the process to be non-inducible. C. has no effect on the process. D. cannot be predicted. 32. Example(s) of a positive regulator is/are A. the lac repressor (I gene). B the CAP protein. C. the GAL4 protein. D. the GAL80 protein. E. both B and C. 33. A benefit of prokaryotic operons is that it A. coordinately activates or represses genes that encode functionally related enzymes. B. provides maximum flexibility in choosing which genes are activated or repressed. C. prevents transposons from inserting and disrupting gene expression. D. maximizes the number of translated proteins by creating additional alternative splicing sites. Page 7 of 10

9 34. A benefit of eukaryote genes, each regulated by their own promoters, is that it A. coordinately activates or represses genes that encode functionally related enzymes. B. provides maximum flexibility in choosing which genes are activated or repressed. C. prevents transposons from inserting and disrupting gene expression. D. maximizes the number of translated proteins by creating additional alternative splicing sites. 35. in E. coli, a loss-of-function mutation in the trp repressor gene results in transcription of tryptophan biosynthesis genes only when A. tryptophan is absent from the growth media. B. tryptophan is present the growth media. C. tryptophan is absent AND glucose is PRESENT in the growth media. D. tryptophan is absent AND glucose is ABSENT in the growth media. E. both A and B. 36. In yeast (eukaryote), RNA polymerase II A. Can bind DNA and initiate transcription for most genes. B. Cannot bind DNA by itself and requires the assistance of general transcription factors for low, basal gene transcription. C. Cannot bind DNA by itself and requires the assistance of specific (regulatory) transcription factors for low, basal gene transcription. D. can bind and transcribe trna genes without the assistance of general transcription factors. E. can bind and transcribes rrna genes without the assistance of general transcription factors. 37. Alternative splicing A. produces alternative DNA versions of the same eukaryote gene. B. produces alternative mrnas from the same eukaryote gene. C. produces polycistronic RNA from the same eukaryote gene. D. does not exist in eukaryotes such as mammals. E. both B and C. 38. A mirna is guided to its target mrna because A. they have the same sequences. B. they have complementary sequences. C. they are encoded in an operon. D. they covalently bond to each other. 39. Suppose a mirna gene is no longer transcribed because of a mutation in its promoter. In mutant cells you would expect to see the number of the proteins encoded by its target mrnas to A. stay the same. B. the change in the number of target mrnas cannot be predicted. C. increase. D. decrease. 40. Very soon after a retrovirus infects a cell, its RNA genome is A. transcribed to make proteins. B. replicated to make double-stranded RNA by an RNA-dependent RNA polymerase. C. converted into double-stranded DNA by reverse transcriptase. D. buds off to form a new retrovirus Page 8 of 10

10 41. Integrase A. inserts the double-stranded retrovirus RNA into the host genome. B. inserts the double-stranded retrovirus DNA into the host genome. C. removes the retrovirus RNA from the host genome. D. removes the double-stranded retrovirus DNA from the host genome. E. B and D. For questions all plates have glycerol as an energy and carbon source that does not interfere with expression of the lac operon. All plates also have X-gal. All bacteria have a functional β-galactosidase gene (Z + ) in the lac operon. The rest of the genotype of the bacteria, and whether the plates have inducer (IPTG) and/or glucose, are indicated in each question. All genes are wild type unless noted otherwise. 42. When the plates have inducer (IPTG) and glucose, blue colonies will be produced by A. Wild type bacteria. B. I - bacteria. C. O - bacteria. D. CAP - bacteria E. None of the above 43. When the plates have inducer (IPTG) and NO glucose, white colonies will be produced by A. Wild type bacteria. B. I - bacteria. C. O - bacteria. D. CAP - bacteria E. A, B and C 44. When the plates have NO inducer (IPTG) and NO glucose, blue colonies will be produced by A. Wild type bacteria. B. I - bacteria. C. O - bacteria. D. CAP - bacteria E. Both B and C 45. Imagine you grew a liquid culture of E coli cells, which are wild type for ALL loci, and tested for constitutive expression of particular genes. Which of the following genes are expressed constitutively when the E coli cells are grown in a solution with glucose and inducer (glucose levels remain high during the study). A. Lac I B. Lac Z C. Lac Y D. All of the above E. Only B and C 46. The most common way a cell prevents the movement of autonomous retroviruses and DNA transposons is by A. preventing their transcription. B. preventing translation of their RNAs. C. inhibiting the activity of their enzymes. D. cutting them out of the genome with restriction endonucleases. E. all of the above. EXAM CONTINUES Page 9 of 10

11 47. The universal primer described in class A. is complementary to a sequence in the plasmid that allows you to sequence a DNA inserted in the plasmid without prior knowledge of its sequence. B. is complementary to a sequence in all known DNAs inserted in plasmids. C. is complementary to the conserved origin of DNA replication in the plasmid. D. Is complementary to all RNAs transcribed from the E. coli genome. E. is a primer that is never useful for DNA sequencing. 48. Suppose a DNA in a genome has the following sequence: 5 AAAAAAA GGGGGG3 3 TTTTTTT CCCCCC5 What primers would you use for PCR to amplify the sequence? A. 5 AAAAAAA3 and 5 CCCCCCC3 B. 5 AAAAAAA3 and 5 GGGGGGG3 C. 5 TTTTTTT3 and 5 CCCCCCC3 D. None of the above. 49. The most common current strategy for sequencing a genome involves A. partial digestion of the genome into overlapping DNA fragments. B. using a computer to detect the overlaps and order the DNA fragments and sequences within the genome. C. genetically mapping DNA fragments to determine their order before sequencing. D. All of the above. E. Only A and B. 50. Unequal recombination occurs between R1 and R4 on the two chromatids in the figure. This would generate two chromatids with A. no Hb gene and two Hb genes. B. no Hb gene and four Hb genes. C. two Hb genes and four Hb genes. D. no Hb genes and six Hb genes. E. both chromatids would have three Hb genes. 51. After transcription factor genes are duplicated by unequal recombination, over evolutionary periods of time, one transcription factor gene might A. move to a different chromosome. B. change the pattern of its gene transcription. C. change the target genes that it transcriptionally regulates. D. All of the above. END OF THE EXAM Page 10 of 10