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Biology 1A Exam Reader – Fall 2014 Lecture M, W, F 8-9 AM in 1 Pimentel

This reader contains lecture exams, and sample questions. Use the exams in this reader only to determine areas in which you are weak. DO NOT use the exams as a study guide. Set aside the time to take the exams in the allotted time, 50 minutes for midterms and three hours for the final. Literally set a timer and stop when the time is up. You need to be able to pace yourself appropriately and you want to mimic test conditions as much as possible when you take the practice exams.

This is the first time that Dr. Welch has taught Bio 1A. Consequently there no past exams for Dr. Welch. This reader contains two examples of exam 1 and 2. the final but they were NOT written by the current faculty. It should be noted that different faculty emphasize different material. Even the same faculty member may emphasize different material, from semester to semester. Consequently the material covered differs each semester due to faculty and textbook changes.

You should rely upon your syllabus and the Bio 1A website for the most current information regarding exams, exam dates, handouts, etc.. Specific handouts will be given for each exam. Note that for each exam you will need to know your discussion section number, your assigned room, and your assigned seating location within that room midterms and the final). Scantron forms will be provided for you. As you can tell from the exam handout and the cover page of each exam you will need to correctly fill out the scantron form. The top 8 boxes correspond to your SID. Please put 00 in the last two boxes. Multiple sets of physiology questions and two sample finals have been provided. Answer pages are in bold. Exam 1 Exam 2 Fall 2013 Sample Exam Handout 3-4 Fall 2013 Dr. Fischer (J) 27-36 37 (K) (K)Fall 2013 Dr. Pauly 5-14 15 Fall 2012 Dr. Fischer (L) 39-48 49 (M) Fall 2012 Dr. Pauly 17-24 25

Final Exam (Questions) Fall 2013 Final 51-74

(Q) 75 (R )

Fall 2012 Final 77-98 (S)

99 (T)

Fall 2009 Exam 2 Dr. Forte 101-111 (U)

101 (V)

Specific Study Hints √ Outline your notes. √ Discuss the material with a fellow classmate or GSI--either your own, or go to the GSI office during office

hours. See the professor during office hours. Be sure to take advantage of office hours several weeks before the exams, quizzes or lab practicals. If you wait until just the week before, or even the same week, you will be competing along with many other students who have also waited until the very last minute--this just doesn’t work as well.

√ Refer also to your lab manual for other specific study hints.

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BIOLOGY 1A MIDTERM # 2 November 4th , 2013 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 & bubble in your name, SID, and section # (last 2 digits). The top 8 boxes of the ID field are for your SID, for the bottom two put in 00. Put your name on the scantron form. Under “test” write in your GSI’s name. See below.

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 9 numbered pages, 47 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 0.

10. There is always only one best answer.

7 of 111 Bio 1A Exam Reader F2013

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1. What are the number of homologous pairs of chromosomes and the number of double-strand DNA

molecules in a female human cell in G2 phase of the cell cycle? A. 23 homologous pairs and 23 double-strand DNA molecules. B. 23 homologous pairs and 46 double-strand DNA molecules. C. 23 homologous pairs and 92 double-strand DNA molecules. D. There are no homologous pairs, 23 double-strand DNA molecules. E. There are no homologous pairs, 46 double-stranded DNA molecules.


molecules in a female human cell just after meiosis I? A. 23 homologous pairs and 23 double-strand DNA molecules. B. 23 homologous pairs and 46 double-strand DNA molecules. C. 23 homologous pairs and 92 double-strand DNA molecules. D. There are no homologous pairs, 23 double-strand DNA molecules. E. There are no homologous pairs, 46 double-stranded DNA molecules.

3. (3 pts) Albinism is a recessive genetic condition characterized by absence of pigment in the skin due to the inability to synthesize melanin. Albinism occurs with equal frequency in male and female offspring. John and Mary, as well as their parents, do NOT display symptoms of albinism. However, they both have siblings with albinism. What is the probability that both John and Mary are carriers (heterozygous) for the albinism allele? A. 1/9 B. 1/4 C. 1/3 D. 4/9 E. 2/3

4. (3 pts) For the question above, what is the probability that their first child will have albinism?

A. 1/9 B. 1/4 C. 1/3 D. 4/9 E. 2/3

5. (0 pts) Mark A as you have version A. 6. Which of the following processes strongly contribute to genetic variation during meiosis in humans?

A. Recombination between non-homologous chromosomes. B. Recombination between the X and Y chromosomes. C. Recombination between homologous chromosomes. D. DNA replication during the S-phase between meiosis I and II. E. Only B and C.

Page 28 of 111 Bio 1A Exam Reader F2013

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Did you mark A for question 5? 7. Suppose you cross A/a B/B D/d X A/a B/b D/D where A, B, and D are unlinked genes. What is the

probability of obtaining offspring that are A/A B/B D/D genotype? A. 1/4 B. 1/8 C. 1/16 D. 1/32 E. 1/64

8. (3 pts) True breeding narrow-leaf, white-flower pea plants were crossed to true breeding wide-leaf, purple-flower pea plants. The F1 progeny had wide leaves and purple flowers. The F1 progeny were crossed to true breeding narrow-leaf, white-flower pea plants. Among 1000 F2 progeny, Number of progeny Leaf width Flower color

400 wide purple 400 narrow white 100 wide white 100 narrow purple

Most likely, the genes for leaf width and flower color are A. unlinked. B. linked and separated by 10 Centimorgans. C. linked and separated by 20 Centimorgans. D. linked and separated by 100 Centimorgans. E. No conclusions about linkage can be made from the data.

Work space.

9. Which dinucleotide is formed immediately after the first new phosphodiester linkage catalyzed by E. coli DNA polymerase III when synthesizing an Okazaki fragment? A. Figure A B. Figure B C. Figure C D. Figure D


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10. (3 pts) In Drosophila, the sl (short leg) allele is recessive to the wild-type sl+ (normal leg) allele. Also,

the nw (narrow wing) allele is recessive to the wild-type nw+ (normal wing) allele. A short leg, narrow wing female from a true breeding population was crossed to a true breeding wild type male. F1 females were crossed to wild-type males. The phenotypes of 1000 female and 1000 male F2 progeny are shown in the table, below:

Leg size Wing width Females Males Normal leg Normal wing 1000 250 short leg narrow wing 0 250 Normal leg narrow wing 0 250 short leg Normal wing 0 250

Most likely, the A. genes for leg size and wing shape are linked on an autosome and separated by 25 Centimorgans. B. genes for leg size and wing shape are unlinked on autosomes. C. genes for leg size and wing shape are genetically linked on the X chromosome and separated by 25

Centimorgans. D. genes for leg size and wing shape are physically linked on the X chromosome and separated by at least 50

Centimorgans. E. gene for leg size is on an autosome and the gene for wing shape is on the X chromosome.

Work space. 11. E. coli primase is

A. an RNA polymerase that transcribes genes and makes mRNA. B. a DNA polymerase that makes a DNA primer for DNA replication. C. an RNA polymerase that makes a RNA primer for DNA replication. D. an RNA polymerase that transcribes tRNA genes. E. an RNA polymerase that transcribes rRNA genes.

12. E. coli DNA polymerase I A. synthesizes DNA in the 5’ to 3’ direction and degrades (hydrolyzes) RNA in the 5’ to 3’ direction. B. only synthesizes DNA in the 5’ to 3’ directions and does not degrade RNA. C. only degrades RNA in the 5’ to 3’ direction and does not synthesize DNA. D. synthesizes DNA in the 5’ to 3’ direction and covalently joins adjacent Okazaki fragments. E. simultaneously copies both DNA template strands in the 5’ to 3’ direction.


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13. The figure below shows a replication fork in E. coli. At which of the indicated positions would you expect

to see an RNA primer? A. A only. B. B only. C. C only. D. D only. E. Both B and D.

14. Which E. coli enzyme simultaneously copies both DNA strands?

A. DNA polymerase I. B. DNA polymerase III. C. Primase. D. Ligase. E. All of the above.

15. When DNA polymerase III makes an error and incorporates the wrong base, it

A. continues synthesizing DNA because there is no time to stop and repair the error. B. removes the wrong base by hydrolyzing its phosphodiester linkage. C. stops and cannot move forward or backwards. D. repairs the error by synthesizing DNA in the reverse direction (3’ to 5’). E. All of the above.

16. A given E. coli mRNA molecule

A. is usually templated from both of the DNA strands at a given genetic locus. B. has a single phosphate at the 5’-end. C. often has multiple start and stop codons that are used by the ribosome to produce multiple proteins. D. has a hydrogen atom at the 3’ end. E. Both A and C are true.

17. Select the correct mRNA sequences depending on the direction RNA polymerase transcribes.

RNA sequence if RNA polymerase goes left

RNA sequence if RNA Polymerase goes right.

A. 5’-CCC-3’ 5’-UUU-3’ B. 5’-CCC-3’ 5’-AAA-3’ C. 5’-GGG-3’ 5’-UUU-3’ D. 5’-GGG-3’ 5’-AAA-3’ E. None of the above

18. Select the FALSE statement regarding a given aminoacyl tRNA synthetase.

A. A given aminoacyl tRNA synthetase only binds one of the twenty amino acids. B. A given aminoacyl tRNA synthetase is an enzyme that uses energy from ATP to attach an amino acid to a

tRNA. C. A given aminoacyl tRNA synthetase uses tRNA as a substrate. D. A given aminoacyl tRNA synthetase synthesizes a specific type of tRNA.


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19. In E. coli, peptide bond formation is catalyzed by a

A. rRNA. B. tRNA. C. snRNA. D. protein in the ribosome. E. termination factor.

20. Which atoms participate in peptide bond

formation between the single amino acid covalently attached to a tRNA (right side) and the growing polypeptide chain covalently attached to a tRNA (left side)? A. 1 and 5. B. 2 and 5. C. 3 and 5. D. 4 and 5. E. 4 and 6.

21. The original sequence of a mRNA and a mutant version of the mRNA is shown below. In vivo rules of

translation apply. Using the table of the genetic code, this is an example of what kind of mutation? A. Silent mutation. B. Missense mutation. C. Nonsense mutation. D. Insertion. E. Deletion. Original: 5’-UAAUGUAUAUUAAAUGA-3’ Mutant : 5’-UAAUGUAAAUUAAAUGA-3’


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22. Base pairing between different RNAs is NOT required for

A. the catalysis of peptide bond formation. B. splicing. C. regulation of gene expression by miRNAs D. interaction between mRNA and tRNAs in the ribosome. E. translation initiation.

23. Splicing is initiated by nucleophilic attack of a phosphodiester linkage at the 3’-end of the intron by

electrons from a A. 3’-OH of a ribonucleotide within the exon. B. 2’-OH of a ribonucleotide within the exon. C. 3’-OH of a ribonucleotide of an snRNA. D. 3’-OH of a ribonucleotide within the intron. E. 2’-OH of a ribonucleotide within the intron.

24. Oxidation of guanine generates 8-oxoGuanine. To prevent GC to TA mutations, the 8-oxoGuanine is removed by A. DNA glycosylase. B. DNA polymerase III. C. DNA polymerase I. D. RNA polymerase. E. Ligase.

25. In prokaryotes, an operon

A. ensures that expression of genes encoding enzymes in a pathway will all be expressed at the same time. B. is transcribed to produce a mRNA that is used to translate multiple proteins. C. is transcribed to produce a mRNA that has multiple start and stop codons that are used by ribosomes. D. is a sequence that must be spliced out of a primary RNA transcript. E. A, B and C are correct.

26. In eukaryotes, one regulatory transcription factor can regulate 100 target genes by having

A. all 100 target genes be tightly linked on the same chromosome. B. all 100 target genes be located in the centromere regions of the chromosome. C. the transcription factor’s DNA binding sequence in the region of all 100 target genes. D. the transcription factor’s DNA binding sequence near a large operon encoding an RNA for all 100 genes. E. C and D are correct.

27. Assume a wild-type organism displays inducible expression of a particular gene. A loss-of-function

mutation in a positive regulator of the gene will result in A. inducible expression. B. non-inducible expression. C. constitutive expression. D. All of the above. E. No conclusion can be made.

28. Assume a wild-type organism displays inducible expression of a particular gene. A loss-of-function mutation in a negative regulator of the gene will result in A. inducible expression. B. non-inducible expression. C. constitutive expression. D. All of the above. E. No conclusion can be made.


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29. In wild-type E. coli, the amino acid tryptophan (trp) is NOT synthesized when tryptophan is present in the environment. This is because the trp repressor, when bound to tryptophan, A. does not bind to the operator (O site) on the trp operon. B. binds to the operator (O site) on the trp operon. C. does not bind to the promoter (P site) on the trp operon. D. binds to the promoter (P site) on the trp operon. E. binds to the CAP protein binding site on the trp operon.

For the following two 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. 30. (3 pts) When the plates have inducer (IPTG) and glucose, white colonies will be produced by

A. Wild type bacteria. B. I- bacteria. C. O- bacteria. D. CAP- bacteria. E. All of the above.

31. (3 pts) When the plates have 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. A, B and C.

32. How does the Lac repressor bind very tightly to only the operator sequence in the E. coli genome? The

lac repressor A. has amino acids with complementary shape and charge distribution to the unique Operator DNA sequence. B. has positively charged lysine and arginine amino acids that are attracted only to the negatively charged

phosphodiester linkages in the Operator. C. binds strongly to sugar molecules, including the deoxyribose sugars of in the Operator DNA. D. binds strongly to the CAP-binding site, which is very close to the Operator DNA. E. binds all DNA sequences in the E. coli genome with equal tightness.

33. Usually, the function of the general transcription factors is to

A. bind to enhancer DNA sequences and activate transcription. B. bind to enhancer DNA sequences and repress transcription. C. help RNA polymerase II bind to the TA-rich sequence just upstream of the site where transcription starts. D. bind splicing sites on the primary transcript from a gene and regulate alternative splicing. E. synthesize a polyA tail on the primary transcript from a gene.

34. Alternative splicing

A. allows a single eukaryote gene to encode multiple proteins. B. has no benefit and just wastes energy. C. very rarely occurs in human cells. D. is sometimes guided by RNA-binding proteins. E. Both A and D.

35. miRNAs

A. are not translated by ribosomes. B. are about 22 base pairs long when cut out of much longer primary transcripts. C. are single-stranded in the RISC complex. D. base pair with their target mRNA. E. All of the above are true.


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36. Suppose a miRNA gene has a 6 base pair deletion mutation in the region that encodes the miRNA. In mutant cells you would expect to see the number of proteins translated by its target mRNAs to A. stay the same. B. decrease. C. increase. D. the change in the number of target mRNAs cannot be predicted.

37. The type of enzyme that generates new copies of the RNA retrovirus genome is

A. reverse transcriptase. B. RNA polymerase. C. DNA polymerase. D. Primase. E. Integrase.

38. Retrotransposons that are capable of moving themselves in the genome are like retroviruses except they

cannot A. convert single-stranded RNA into double-stranded DNA. B. be transcribed. C. be inserted into other sites of the host genome. D. make viruses. E. All of the above.

39. DNA transposons were discovered and studied by Professor Barbara McClintock. Which element is NOT required for a DNA transposon to move by itself in the genome? A. Inverted repeats. B. A promoter. C. A transposase gene. D. A reverse transcriptase gene. E. All of the above are required.

40. The shot-gun strategy for sequencing a genome involves

A. partial digestion of the genome into overlapping DNA fragments. B. randomly sequencing DNA fragments. C. using a computer to detect the overlaps and order the DNA fragments and sequences within the genome. D. genetically mapping DNA fragments to determine their order before sequencing. E. A, B, and C.

41. To duplicate a single ancestral gene, homologous recombination occurs between

A. repeated sequences, usually very similar transposon sequences, that are on each side of the ancestral gene. B. repeated sequences, usually very similar transposon sequences, that are not on homologous pairs of

chromosomes. C. non-similar sequences that are on each side of the ancestral gene. D. non-similar sequences that are being actively transcribed. E. All of the above.

42. Genes duplicated a very long time ago will

A. usually be more divergent than more recently duplicated genes B. sometimes be on different chromosomes. C. usually will encode non-identical proteins. D. usually will be expressed differently. E. All of the above.

EXAM CONTINUES


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43. Histone proteins

A. have tails that can be covalently modified. B. are associated with negatively charged RNA. C. have many positively charged amino acids. D. are not found in chromatin isolated from most eukaryotes. E. Both A and C.

44. DNA methylation in eukaryotes is

A. found commonly on cytosine and adenine bases. B. associated with silenced, heterochromatin. C. inserted by DNA polymerase. D. is present is all eukaryote genomes. E. Both B and D.

45. Chromatin in a closed state

A. is easily digested by DNases (deoxyribonucleases). B. does not allow access of transcription factors. C. is associated with H3K9 acetylated histones. D. is often filled with transposable elements. E. B and D.

46. The polyA tail of a eukaryotic mRNA

A. is where the start codon is located. B. is where the stop codon is located. C. protects the mRNA from being degraded. D. is where the process of translation ends. E. is where the small ribosomal subunit binds to initiate translation.

47. Which of the following is not a nucleic acid sequence?

A. termination factor B. transposon C. operator D. telomere E. enhancer

END OF THE EXAM


EXAM 2 ANSWER KEY Version A Fall 2013

ANSWER KEY EXAM 2, VERSION A, Fall 2013 (exam was worth 98 points instead of 100) Mean = 69.35, Stdev = 16.15, Median score = 72. A+ = 100-98, A = 97-91, A- = 90-87, B+ = 86-85, B = 84- 79, B- = 78-76, C+ = 75-69, C = 68-53, C- = 52 -43, D+ =, D= 42-38 D- = 37-35 F = 34 or less. 1 C 6 C 11 C 16 C 21 C 26 C 31 E 36 C 41 A 46 C 51 2 E 7 C 12 A 17 A 22 A 27 B 32 A 37 B 42 E 47 A 52 3 D 8 C 13 E 18 D 23 E 28 C 33 C 38 D 43 E 48 53 4 A 9 C 14 B 19 A 24 A 29 B 34 E 39 D 44 B 49 54 5 A 10 D 15 B 20 A 25 E 30 E 35 E 40 E 45 E 50 55 1) G2 occurs after the S phase and before mitosis. 23 pairs of chromosomes with sister chromatids- each

chromatid is double stranded DNA. 2) Meiosis 1 separates homologous pairs so each cell would have 23 chromosomes, composed of sister

chromatids - each chromatid is double stranded DNA. 3) Assume a+//a = genotype of heterozygous where a = albinism, a+ = normal.

a a+ a a//a (Neither have the disease, this can be excluded as a possibility). a//a+ a+ a//a+ a+//a+

2/3 chance John is a heterozygote and also 2/3 chance Mary is. 2/3 X 2/3 = 4/9 4) If they are heterozygous then ¼ will have the disease (see Punnett square above). Thus together it is 4/9 X ¼ = 1/9th.

5) Version of the exam. 6) Recombination occurs often between homologous chromosomes and rarely between non-homologous

chromosomes. 7) A/A = ¼, B/B – ½ and D/D = ½. Together = 1/16. 8) Note ½ = narrow and ½ = purple. Thus if unlinked expect ½ X ½ = narrow purple. This doesn’t match the

data and know they are genetically linked with the smaller number representing the offspring that got recombinant chromosomes.

9) (iClicker question). C = RNA nucleotide at the 5’ end and DNA nucleotide at the 3’ end. 10) You see a difference for each trait relative to the sex. Thus both traits are X linked. ½ are normal leg, ½

are normal wing. Thus if genetically unlinked on the X chromosome expect ¼ are normal leg and normal wing--- thus genetically unlinked. For the female offspring all of them have normal legs and normal wings because they got the dominant alleles (normal legs and normal wings) from the X chromosome in the sperm.

11) Primase is the enzyme that adds RNA nucleotides (short stretches) and does not require a 3’ OH group. 12) DNA pol I removes the RNA nucleotide from a 5’ to 3’ direction and adds back DNA nucleotide, also in a

5’ to 3’ direction. (NOTE DNA pol III does the majority of DNA synthesis). 13) The RNA primers are located at the 5’ end of the Okazaki fragments (initially and then they will be

removed). 14) DNA pol III does the majority of synthesis and replicates using both strands as a template (each

individually is used as a template). 15) DNA pol III proof reads the pairing and can remove an incorrectly based nucleotide using 3’ to 5’

exonuclease activity. 16) Prokaryotic mRNAs often have multiple start/stop signals for translation. 17) DNA synthesis occurs in a 5’ to 3’ direction moving along the template in 3’ to 5’ direction. To the left the

lower strand is the template and the RNA strand made will be 5’ CCC 3’. To the right the upper strand is the template and the molecule made will be 5’ UUU 3’.

18) Aminoacyl tRNA synthetase enzyme recognizes specific tRNA molecules and add a specific amino acid. Both the tRNA and the amino acid are substrates. A given tRNA synthetase only adds one type of amino acid and ATP is used as the energy source. They do NOT synthesize tRNA, that is the job of RNA pol III.

19) rRNA is the catalytic molecule that forms the peptide bond. 20) The tripeptide will be transferred to the aminoacyl tRNA. This occurs at the C at position 1 and N at

position 5. 21) Translation begins at the AUG site. Thus the reading frame is



Original 5’-AUGUAUAUUAAAUGA-3’(AUG=Met, UAU=Tyr, AUU=Ile, AAA=Lys, UGA= Stop) Mutant: 5’-AUGUAAAUUAAAUGA-3’(AUG = Met, UAA = STOP) The mutation is a nonsense mutation due to the introduction of a stop codon. 22) Peptide bond formation is due to the Adenine in the 23s rRNA pulling a proton away from the amino

terminus of the tRNA at the A site. 23) It is the 2’ OH within the intron that initiates the attack. 24) DNA glycosylase is the repair enzyme. 25) Operons are motifs found in prokaryotes. They are not spliced. 26) A DNA transcription factor has a domain that binds to a unique DNA sequence and another domain that

binds to other proteins. Target genes will have similar DNA sequences recognized by the DNA binding domain.

27) If the normal phenotype is inducible the loss of a + regulator will result in non-inducible expression. You can’t turn it up.

28) If the normal phenotype is inducible the loss of a - regulator will result in constitutive expression. You can’t turn it off.

29) The bacteria have evolved a system to shut off transcription of the genes responsible for the synthesis of Trp if it is in the diet. Trp repressor when bound to Trp binds to the operator of the Trp operon shutting off transcription.

30) Both glycerol and glucose are present such that cAMP levels are low. Transcription should be low in all instances and as a result the colonies are white (not blue).

31) No glucose but there is glycerol (so growth can occur). cAMP levels are high. Transcription should occur in wild type if the operator is not blocked. Inducer is present so wild type should transcribe the operon. I- can’t make the repressor so the operator cannot be blocked. O- lack the operator site so repressor can’t bind. CAP- bacteria will not produce CAP so transcription can’t occur.

32) The lab repressor has a DNA binding domain that specifically recognizes unique base pairs. 33) General transcription factors help RNA pol II to bind but you need to have specific transcription factors in

order to get high levels of transcription. 34) Alternative splicing requires a the RNA-binding proteins. 35) miRNAs are themselves NOT translated but are cut of primary transcripts yielding double stranded RNA

which gets modified such than only one strand is in the RISC complex which then binds to the target mRNA.

36) With a 6 bp deletion the miRNA should no longer function and the target mRNAs will have increased translation.

37) Note that it is the reverse transcriptase that allows the RNA to be converted into double strand DNA to insert into the genome of the host. Once in the genome host, RNA polymerase makes RNA copies which serve as both mRNA and the genome.

38) Retrotransposons are thought to be retroviruses that have lost the capsid genes and can’t package the genome.

39) Reverse transcriptase converts RNA into double stranded DNA. DNA transposons must be transcribed so that active transposase can be made.

40) D is not required in shot-gut sequencing. The advantage of this methods is that you don’t need to D. 41) You need to have repeated sequences to get the recombination and they need to be mis-aligned to get

unequal recombination. 42) Gene duplication allows divergence such that the function, timing or expression levels can be modified. 43) Histone protein have tails that stick out that can be modified (such as K9 modification) and they are overall

positively charged to neutralize the overall negative charge of DNA. 44) DNA methylation is associated with heterochromatin and lack of transcription. Some eukaryotes lack it. 45) Closed state minimizes accessibility for DNA degradation and transcription. Regions with transposons are

usually in a closed state. 46) Poly A tail helps protect and plays a role in transport out of the nucleus. 47) Termination factor is a peptide (protein release factor) that allows water to attack the linkage between the

polypeptide and the tRNA.


BIOLOGY 1A MIDTERM # 2 October 26th, 2012 A 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.



3. Write in and bubble in your name, SID, and section #. The first 8 boxes of the ID # field are for your SID. Bubble in 00 for the bottom two boxes. Put your name on the scantron form. Under “test” write in your GSI’s name. See below.

EXAM Instructions: 4. Print your name on THIS COVER SHEET. (otherwise, you will get a ZERO).




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.




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1. What is the number of homologous pairs of chromosomes and the number of double-strand DNA molecules in a female human cell just before meiosis begins? A. 23 homologous pairs and 23 double-strand DNA molecules. B. 23 homologous pairs and 46 double-strand DNA molecules. C. 23 homologous pairs and 92 double-strand DNA molecules. D. 46 homologous pairs and 92 double-strand DNA molecules. E. There are no homologous pairs, 23 double-strand DNA molecules.

2. Mark A as you have version A. 3. What is the number of homologous pairs of chromosomes and the number of double-strand DNA

molecules in a female human gamete (egg) immediately after meiosis is completed? A. 23 homologous pairs and 23 double-strand DNA molecules. B. 23 homologous pairs and 46 double-strand DNA molecules. C. 23 homologous pairs and 92 double-strand DNA molecules. D. 46 homologous pairs and 92 double-strand DNA molecules. E. There are no homologous pairs, 23 double-strand DNA molecules.

4. Lesch-Nyhan syndrome (LNS) is a genetic condition that causes inappropriate storage of uric acids in

cells, which can result in severe kidney and neurological problems. It is caused by a recessive allele that is on the X-chromosome (sex-linked). Paul and Ann were recently married. None of their parents displayed LNS symptoms. However, Ann’s brother had LNS. They go to a genetic counselor and ask, “What is the probability that their first son will have LNS?” The correct answer is A. 2/3 B. 1/2 C. 1/3 D. 1/4 E. 1/8

5. Suppose you cross Z/z H/h G/g X Z/z H/h G/g where Z, H, and G are unlinked genes. What is the

probability of obtaining an offspring with Z/z G/G h/h genotype? A. 1/4 B. 1/8 C. 1/16 D. 1/32 E. 1/64

6. True breeding tall, yellow-leaf plants were crossed to true breeding short, green-leaf plants. The F1 progeny were short with green leaves. The F1 progeny were crossed to true breeding tall, yellow-leaf plants. Among 1000 progeny, Number of progeny Height Leaf color

250 Tall Yellow 250 Short Green 250 Tall Green 250 Short Yellow

Most likely, the genes for height and leaf color are A. unlinked. B. genetically linked and separated by 10 Centimorgans. C. physically linked and separated by 50 or more Centimorgans. D. sex-linked. E. Both A and C are possible.

Work space. Did you mark your answer for #2?


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7. Recombination in humans usually occurs A. as reciprocal recombination between non-sister chromatids of a homologous pair. B. as non-reciprocal recombination between non-sister chromatids of a homologous pair. C. as reciprocal recombination between non-sister chromatids of a non-homologous pair. D. everywhere along the X and Y chromosomes. E. All of the above

8. (3 pts) In Drosophila, the bl (big leg) allele is recessive to the wild-type bl+ (normal leg) allele. Also, the sq (square body) allele is recessive to the wild-type sq+ (normal body) allele. A big leg, square body female from a true breeding population was crossed to a true breeding wild type male. F1 females were crossed to wild-type males. The phenotypes of 1000 female and 1000 male progeny are shown in the table, below: Females Normal leg Normal body 1000 Big leg Square body 0 Normal leg Square body 0 Big leg Normal body 0 Males Normal leg Normal body 400 Big leg Square body 400 Normal leg Square body 100 Big leg Normal body 100 Most likely, the genes for leg size and body shape are A. unlinked. B. linked on an autosome (not X chromosome) and separated by 10 Centimorgans. C. linked on an autosome (not X chromosome) and separated by 20 Centimorgans. D. linked on the X-chromosome and separated by 10 Centimorgans. E. linked on the X-chromosome and separated by 20 Centimorgans.

Work space. 9. Which pairs of atoms participate in making a new phosphodiester linkage catalyzed by RNA polymerase?

A. 1 and 5. B. 1 and 6. C. 2 and 4. D. 3 and 4. E. None of the above.

10. E. coli DNA polymerase I

A. initiates DNA synthesis by creating a phosphodiester linkage between two dXTPs. B. synthesizes in the 5’ to 3’ direction and the 3’ to 5’ direction. C. removes RNA primers from Okazaki fragments. D. joins adjacent Okazaki fragments (primers removed) by generating a phosphodiester linkage. E. simultaneously copies both DNA template strands.


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11. The figure below shows a replication fork in E. coli. At which of the indicated positions would you expect

to see an RNA primer? A. B. C. D. E. Both B and C.

12. E. coli DNA ligase

A. initiates DNA synthesis by creating a phosphodiester linkage between two dXTPs. B. synthesizes in the 5’ to 3’ direction and the 3’ to 5’ direction. C. removes RNA primers from Okazaki fragments. D. joins adjacent Okazaki fragments (primers removed) by generating a phosphodiester linkage. E. simultaneously copies both DNA template strands.

13. Select the FALSE statement regarding E. coli mRNA. If none of A- D are FALSE, select E.

A. is produced from both DNA template strands at every gene. B. has a triphosphate at the 5’-end. C. often have multiple start and stop codons that are used by the ribosome to produce multiple proteins. D. has a hydroxyl at the 3’-end. E. All of the above statements are true.

14. E. coli tRNAs

A. have no intramolecular base pairing. B. are translated to make t-proteins. C. carry amino acids covalently attached to their 3’-end. D. has an anti-codon that base pairs with rRNAs in the ribosome. E. Both C and D.

15. E. coli rRNAs carry out the following functions.

A. Provide structure for the ribosome by intra-molecular base pairing. B. Create the active site for peptide bond formation. C. Help the small ribosomal subunit bind mRNA. D. All of the above. E. Only A and C.

16. (3 pts) A mutation in the aspartic acid aminoacyl tRNA synthetase gene allows the mutant aminoacyl

tRNA synthetase protein to bind both aspartic acid (negative charge side chain) and lysine (positive charge side chain) amino acids. As a result A. both aspartic acid and lysine will be incorporated at codons that specify lysine. B. both aspartic acid and lysine will be incorporated at codons that specify aspartic acid. C. both aspartic acid and lysine will be incorporated at codons that specify lysine and aspartic acid. D. protein structure will not be affected. E. there would be translation termination at codons that specify lysine.


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17. Base pairing between different RNAs is required for A. translation initiation. B. splicing. C. regulation of gene expression by miRNAs D. interaction between mRNA and tRNAs in the ribosome. E. All of the above.

18. Which atoms participate in hydrolysis at translation termination?

A. 1 and 2. B. 1 and 3. C. 2 and 4. D. 3 and 5. E. All of the above.

19. The coding sequence for a mRNA is 5’-AUG UAU UGU-3’ followed by a stop codon. Using the genetic

code table and the figure below, which peptidyl-tRNA is in the P-site just before hydrolysis at translation termination? A. peptidyl tRNA number 1 B. peptidyl tRNA number 2 C. peptidyl tRNA number 3 D. None of the above.

20. A portion of the wild-type template DNA strand for transcription of a gene in E. coli is 3’-ATG-5’. What will

the sequence be in the mRNA? A. 5’-UAC-3’. B. 5’-CAU-3’. C. 5’-AUG-3’. D. 5’-GUA-3’.


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21. What is likely to be the most harmful mutation for expression of a gene? A. A 3 base pair deletion within the coding sequence just before the stop codon. B. A 6 base pair insertion within the coding sequence just before the stop codon. C. A 1 base pair insertion within the coding sequence just after the start codon. D. A 1 base pair change that has no effect on the amino acid sequence of the protein. E. A 1 base pair change that is outside of the coding sequence.

22. Splicing is initiated by nucleophilic attack of a phosphodiester linkage at the 5’-end of the intron by

electrons from a A. 3’-OH of a ribonucleotide within the intron. B. 2’-OH of a ribonucleotide within the intron. C. histidine amino acid of a protein in the snRNP complex. D. 3’-OH of a ribonucleotide of an snRNA. E. 2’-OH of a ribonucleotide of an snRNA.

23. Oxidation of guanine to 8-oxoGuanine can eventually, if not corrected, convert GC to TA resulting in

mutations. Immediately after the oxidation reaction, the 8-oxoGuanine A. need not be repaired because it only base pairs with cytosine. B. is reduced back to guanine by a strongly acidic amino acid. C. is removed by a DNA glycosylase that initiates the base excision repair pathway. D. is removed by DNA polymerase by proof reading. E. is methylated by DNA methyltransferase (Dnmt1.

24. In bacteria (prokaryote), coordinate expression of genes encoding enzymes in a pathway is often achieved

by having A. each gene for each enzyme share the same binding site for a transcription factor. B. each mRNA for each enzyme share the same Shine-Dalgarno sequence. C. all the genes (coding sequences) for the enzymes in an operon. D. all genes be expressed all the time. E. None of the above.

25. In yeast (eukaryote), coordinate expression of genes encoding enzymes in a pathway is often achieved by having A. each gene for each enzyme share the same binding site for a transcription factor. B. each mRNA for each enzyme share the same Shine-Dalgarno sequence. C. all the genes (coding sequences) for the enzymes in an operon. D. all genes be expressed all the time. E. None of the above.

26. The prokaryote lac repressor and CAP are examples of A. two positive regulators. B. two negative regulators. C. a negative regulator (lac repressor) and positive regulator (CAP). D. a negative regulator (CAP) and positive regulator (lac repressor). E. None of the above.

27. The eukaryote GAL4 and GAL80 transcription factors are examples of A. two positive regulators. B. two negative regulators. C. a negative regulator (GAL4) and positive regulator (GAL80). D. a negative regulator (GAL80) and positive regulator (GAL4). E. None of the above.


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28. A loss-of-function mutation in a negative regulator of gene transcription results in A. constitutive gene transcription. B. inducible gene transcription. C. non-inducible gene transcription. D. no effect on gene transcription. E. Its effect can never be predicted.

29. in wild-type E. coli, the amino acid tryptophan (trp) is synthesized when tryptophan is not present in the

environment because the A. CAP/CRP protein binds upstream of the promoter on the trp operon. B. CAPCRP protein does not bind upstream of the promoter on the trp operon. C. trp repressor binds to the operator (O site) on the trp operon. D. trp repressor does not bind to the operator (O site) on the trp operon. E. None of the above.

For questions 30 & 31 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. 30. (3 pts) 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- (CRP-) bacteria. E. A, B and C.

31. (3 pts) When the plates have no inducer (IPTG) and glucose, white colonies will be produced by

A. Wild type bacteria. B. I- bacteria. C. O- bacteria. D. CAP- (CRP-) bacteria. E. All of the above.

32. In yeast (eukaryote), which is usually the first protein complex to bind the TATA box (TA-rich sequence)

located near the start of gene transcription? A. A general transcription factor. B. A regulatory transcription factor. C. RNA polymerase II. D. Primase. E. All of the above.

33. Alternative splicing

A. occurs in the ribosome. B. occurs in cytoplasm of a prokaryotic cell. C. occurs in the cytoplasm of a eukaryotic cell. D. allows a eukaryotic gene with three introns to encode multiple proteins. E. allows a eukaryotic gene with no introns to encode multiple proteins.

34. A miRNA regulates cutting or translation of a target mRNA only when the

A. genes for the miRNA and the target mRNA are closely linked. B. miRNA and the mRNA have a region of identical sequences. C. miRNA and the mRNA have a region of complementary sequences. D. genes for the miRNA and the mRNA are transcribed from the same promoter. E. None of the above.


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35. The lac repressor binds tightly to only one region in the E. coli genome, the lac operator (O site). What determines the specificity of binding? A. The lac repressor contacts the operator DNA sequence in the major groove, where the shape and charge of

amino acids match the shape and charge of the base pairs. B. The lac repressor has many positively charged amino acids that bind tightly to the negatively charged

phosphodiester linkages. C. The lac repressor has many aromatic amino acids that bind tightly to the negatively charged phosphodiester

linkages. D. The lac repressor increases the number of water molecules that bind to the negatively charged

phosphodiester linkages. E. None of the above.

36. 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 its target mRNAs to A. increase. B. decrease.

37. Inside the capsid of a retrovirus is its

A. double-strand DNA genome. B. double-strand RNA genome. C. single-strand RNA genome. D. reverse transcriptase enzyme. E. Both C and D.

38. Integrase inserts into the host genome the A. double-strand retrovirus DNA genome. B. double-strand retrovirus RNA genome. C. single-strand retrovirus RNA genome. D. single-strand retrovirus DNA genome. E. Both C and D.

39. For a retrotransposon to move,

A. its DNA genome must be transcribed. B. its RNA must be translated to make active reverse transcriptase-integrase enzyme. C. its RNA must be translated to make active capsid proteins. D. its RNA must be translated to make transposase. E. Both A and B.

40. Restriction endonucleases (also called restriction enzymes)

A. cut DNA (break phosphodiester linkages) at specific sequences. B. cut RNA (break phosphodiester linkages) at specific sequences. C. cut protein (break peptide bonds) at specific sequences. D. can be used to directly ligate human DNAs into plasmids for DNA sequencing. E. both A and D.

41. It is possible to determine the sequence of an unknown fragment of DNA inserted into a plasmid because

A. many different primers are used to be sure that at least one of them initiates the sequencing reaction. B. a primer is not needed to initiate the DNA sequencing reaction. C. the “universal” primer hybridizes and initiates the sequencing reaction at a plasmid sequence adjacent to the

unknown fragment of DNA. D. the “universal” primer initiates the sequencing reaction at multiple sites within the unknown fragment of DNA. E. None of the above.


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42. The shot-gun strategy for sequencing a genome involves A. partial digestion of the genome into overlapping DNA fragments. B. randomly sequencing DNA fragments. C. using a computer to detect the overlaps and order the DNA fragments and sequences within the genome. D. genetically mapping DNA fragments to determine their order before sequencing. E. A, B, and C.

43. Unequal recombination occurs between R2 and R3 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. both chromatids would have three Hb genes. E. None of the above.

44. After genes are duplicated by unequal recombination, over evolutionary periods of time, one gene might

A. move to a different chromosome. B. change the activity of the protein it encodes. C. change its pattern of gene expression. D. A and B. E. A, B, and C.

45. Nucleosomes are

A. positively charged histone proteins that are wrapped around negatively charged DNA. B. negatively charged DNA wrapped around positively charged histone proteins. C. proteins that only function in splicing. D. not present in eukaryotic organisms. E. absent in heterochromatin

46. DNA methylation helps to

A. silence expression of transposons. B. activate expression of transposons. C. condense chromatin. D. decondense chromatin. E. A and C.

47. Immediately after fully methylated double-stranded DNA is replicated, the new double-stranded DNA (template strand plus newly synthesized strand) is A. completely unmethylated. B. hemimethylated, where the template is methylated and the newly synthesized strand is unmethylated. C. hemimethylated, where both the template strand and the newly synthesized strand are partially methylated. D. fully methylated. E. None of the above.

48. Chromatin in an open state A. is easily digested by DNases (deoxyribonucleases). B. does not allow access of transcription factors. C. is associated with H3K9 acetylated histones. D. is often filled with transposable elements. E. A and C.

49. A tool to measure the level of H3K9 acetylated histones in chromatin is A. an antibody that binds specifically to H3K9 acetylated histones. B. an antibody that binds specifically to H3K9 histones that are not acetylated. C. a restriction endonuclease that only cleaves DNA in the vicinity of H3K9 acetylated histones. D. a pair of polymerase chain reaction (PCR) primers that only amplify DNA in the vicinity of H3K9 acetylated

histones. E. None of the above.


Intentional Blank Page



ANSWER KEY EXAM 2, VERSION A, Fall 2012 All answers and questions are identical except for question 6. This will be explained. Note that some of the question numbers differ, and some of the answers were lettered differently in the exams. Thus version A may have A as the correct answer, version B might have C. Mean = 78.8, Stdev = 11.7, Median score = 80. A+ = 100, A = 99-95, A- = 94-90, B+ = 89-88, B = 87-84, B- = 83-80, C+ = 79-78, C = 77-74, C- = 73-67, D+ =66-63, D= 62-61 D- = 60-57 F = 56 or less. Note there are more Bs than usual. 1 C 6 A/C or E 11 C 16 B 21 C 26 C 31 E 36 A 41 C 46 E 2 A 7 A 12 D 17 E 22 B 27 D 32 A 37 E 42 E 47 B 3 E 8 E 13 A 18 B 23 C 28 A 33 D 38 A 43 C 48 E 4 D 9 C or D 14 C 19 C 24 C 29 D 34 C 39 E 44 E 49 A 5 D 10 C 15 D 20 A 25 A 30 D 35 A 40 A 45 B 50 1) 23 homologous pairs = 46 chromosomes but each consists of sister chromatids – therefore 92 molecules. 2) Not graded = version of the exam. 3) Human gametes are 1N. 4) The trait is X linked. Paul will donate the Y to the son. Ann’s father’s genotype is X lns+/Y and her mom is

X lns+ / X lns . Her mother does not have the disease but a son of hers did (Ann’s brother). The chance Ann got the disease allele is ½. The chance she donates it to an egg is ½. Thus the chance for the son is ¼.

5) ½ (Z/z) X ¼ (G/G) X 1/1/4 (h/h). 6) The ratio is ¼ for all 4 types, the expected ratio for unlinked genes. A is unlinked (genetically, best to have

specifically stated “genetically unlinked) which is always true-whether they are physically linked or unlinked. Since a large % of the students chose E there must have been some confusion thinking C was an equivalent answer. The question asked for most likely—implying just one. It wasn’t meant to be a trick but since so many students put E we accepted A, C or E. Students may have picked C because it is certainly a possible answer (there is probably more than one pair of homologous chromosomes). Fairest thing to do is to accept A, C and E.

7) Recombination occurs between non-sister chromatids of a homologous pair. 8) You don’t see a ¼ to ¼ to ¼ to ¼ ratio in the males. Thus the two genetic loci are genetically linked. The

male offspring (X/Y) will display the phenotype based upon the X chromosome they get from mom—no masking by a second X chromosome. The female offspring can only show the wild type phenotype because of the X chromosome with dominant alleles for both traits.

9) Linkage will be at the #2 position using the alpha phosphate (4) of the incoming rNTP. 10) DNA pol I removes the primers (DNA pol III does the majority of DNA synthesis). 11) RNA primer can only be at the C position. 12) Ligase forms the phosphodiester bond. 13) The question is asking about mRNA. For a given gene only one strand is used as the template. 14) The question is asking about tRNA. tRNA has the amino acid attached at the 3’ end and the anti-codon base pairs with the codon in mRNA. 15) rRNAs are the catalytic molecule and they help the mRNA bind (role of Shine-Dalgarno sequence). B & C

are certainly true. A is also – lots of structure and there is base pairing within the rRNA. 16) The tRNA synthetase will still recognize the correct anti-codon for aspartic acid but it will be charged with

either Asp or Lys. 17) Base pairing is required for all. 18) Water is responsible for the hydrolysis. The free electrons on the oxygen of water attacks the partial +

charge of the C in the carboxylic acid functional group. 19) The termination factor is in the A site. Thus the tRNA at the P site will have all 3 amino acids attached to

it.



20) Base pairing is anti-parallel and complementary. Thus it should be 5’ UAC-3’ in the RNA. 21) Something that disrupts the reading frame earlier in the mRNA will have the largest negative effect. 22) The phosphate at the 5’ end of the intron is attacked by the 2’ OH within the intron (conserved sequence). 23) DNA glycosylase is the repair enzyme. 24) Prokaryotes have operons to coordinately regulate expression. 25) Eukaryotes use transcription factors that bind conserved sequences of related genes in a pathway. 26) Lac repressor is a negative regulator (when it binds). CAP is a positive regulator when it binds. 27) Gal 4 is an activator. Gal 80 is the negative regulator because it binds and blocks Gal 4 from activating –

see page 76 of the reader. 28) If a negative regulator no longer binds you expect to not be able to down regulate. You would have

constituitive expression – non-inducible, always on. 29) Trp is an amino acid. If it is available from the environment it would be wasteful to make it. Thus trp

binding to the trp repressor allows the repressor to bind to the DNA and turn off transcription. Since trp is not present the repressor will not bind.

30) White colonies means no expression is occurring. Since inducer is present we would expect wild type bacteria to make B galactosidase. It doesn’t therefore we must not have the positive regulation of CAP.

31) No inducer. We would expect wild type bacteria to NOT make B galactosidase. I-, O-, CAP- bacteria would also yield a similar phenotype.

32) TATA is the site of binding of TATA DNA binding protein, involved in establishing general transcription. 33) The presence of introns allows alternative splicing to generate multiple forms of mRNA and eventually

multiple forms of proteins. 34) miRNA must bind the mRNA to initiate degradation of the mRNA or to inhibit translation. 35) DNA binding proteins have a specific shape and charge that allows it to interact with the shape and charges

of the bases at the binding site. In this case Lac repressor binds to the operator site. 36) Since the miRNA is not translated the miRNA can’t bind with the mRNA and it would not be degraded. 37) Retrovirus contains single stranded RNA with the reverse transcriptase enzyme associated with the

genomic RNA. 38) Only double stranded DNA can be integrated into the host DNA. 39) Retrotranspososn must copy the DNA into RNA and the RNA must be translated in order to have produced

the reverse-transcriptase-integrase enzyme. 40) Restriction enzymes cut DNA. 41) Once the inserts are cloned into plasmids they can be sequenced because a primer can be synthesized and

annealed to a known stretch of DNA within the plasmid that flanks the unknown inserted DNA. 42) Shotgun cloning and sequencing involves partial digests, insertions, sequencing and then using computer

programs to look for overlap. 43) Reposition the lower chromatid so that R2 is below R3. Thus when the lower chromatid migrates upward

you will have the Hb gene on the left from the lower chromatid but pick up the Hb gene on the far right (upper chromatid). Conversely the upper chromatid will have the two most leftward Hb gens and pick up the two most far right of the lower chromatid.

44) All three choices A, B and C apply. E is a better answer than D. 45) DNA is wrapped around the histone proteins to form an intact nucleosome. 46) DNA methylation is associated with closed, heterochromatin. The condensed state decreases transcription. 47) Upon replication one strand will be methylated, the other will not be. This state is called hemi-methylation. 48) Chromatin in an open state is easily digested by DNAses and is associated with acetylated histones. 49) Generating antibodies to the acetylated histones helps to isolate chromatin that is associated with acetylated

histones – active transcription.


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BIOLOGY 1A Final Dec. 16, 2013 A NAME SECTION # DISCUSSION GSI 1. Sit at your assigned seat. 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.




EXAM Instructions:

4. Print your name on THIS COVER SHEET. (otherwise, you will get a ZERO).




8. WHEN FINISHED RAISE YOUR HAND. Your GSI will collect both your SCANTRON and EXAM. YOU MUST TURN IN BOTH or else you will get a ZERO. With 10 minutes left no students can leave. It gets too disruptive for other students.




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135. A flask containing photosynthetic green algae and a control flask containing water with no algae are both placed under a bank of lights, which are set to cycle between 12 hours of light and 12 hours of dark. The dissolved oxygen concentrations in both flasks are monitored. Predict what the relative dissolved oxygen concentrations will be in the flask with algae compared to the control flask.

A. The dissolved oxygen in the flask with algae will always be higher B. The dissolved oxygen in the flask with algae will always be lower C. The dissolved oxygen in the flask with algae will be higher in the light, but the same in the dark D. The dissolved oxygen in the flask with algae will be higher in the light, but lower in the dark E. The dissolved oxygen in the flask with algae will not be different from the control flask at any time

136. CAM plants keep stomata closed in daytime, thus reducing loss of water. They can do this because they A. fix CO2 into organic acids during the night B. fix CO2 into sugars in the bundle-sheath cells C. fix CO2 into pyruvate in the mesophyll cells D. perform the Calvin cycle during the night E. use photosystem I and photosystem II at night

137. (1 pt) Which of the following does not occur during the Calvin cycle? A. carbon fixation B. oxidation of NADPH C. release of oxygen D. regeneration of the CO2 acceptor E. consumption of ATP

138. If cells in the process of dividing are subjected to colchicine, a drug that binds to tubulin and thereby inhibits polymerization, at which stage will mitosis be arrested?

A. interphase B. cytokinesis C. telophase D. prometaphase E. anaphase

139. Which of the following is true concerning cancer cells? A. They do not exhibit density-dependent inhibition when growing in culture B. When they stop dividing, they do so at random points in the cell cycle C. They are not subject to cell cycle controls D. All of the above. E. Only B and C.


molecules in a female human cell in the G1 phase of the cell cycle? A. 23 homologous pairs and 23 double-strand DNA molecules. B. 23 homologous pairs and 46 double-strand DNA molecules. C. 23 homologous pairs and 92 double-strand DNA molecules. D. There are no homologous pairs, 23 double-strand DNA molecules. E. There are no homologous pairs, 46 double-stranded DNA molecules.


molecules in an human sperm cell just after meiosis II? A. 23 homologous pairs and 23 double-strand DNA molecules. B. 23 homologous pairs and 46 double-strand DNA molecules. C. 23 homologous pairs and 92 double-strand DNA molecules. D. There are no homologous pairs, 23 double-strand DNA molecules. E. There are no homologous pairs, 46 double-stranded DNA molecules.


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142. Which numbered atom in the tRNA will make a covalent

bond with an amino acid? A. Atom 1

B. Atom 2

C. Atom 3

D. Atom 4

E. Atom 5

143. The sequence of a mRNA is 5’-UUAAUGUAUAUUAAAUGA-3’. A one base pair insertion mutation occurs (underlined base with arrow), and the mutant mRNA sequence is 5’-UUAAUGCUAUAUUAAAUG A-3’. Using the table of the genetic code, which statements below are correct? A. The wild-type predicted polypeptide sequence is MET-TYR-ILE-LYS. B. The mutant predicted polypeptide sequence is MET-LEU-TYR. C. The wild-type predicted polypeptide sequence is LEU-MET-TYR-ILE-LYS. D. Both A and B are correct. E. All of the above are correct.

144. Assume a wild-type organism displays inducible expression of a particular gene. A loss-of-function

mutation in a negative regulator of the gene will result in A. inducible expression. B. non-inducible expression. C. constitutive expression. D. All of the above. E. No conclusion can be made.


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145. What is the best description of how a loss-of-function in the GAL4 gene will affect expression of its target

genes: GAL1, GAL7 and GAL10? A. inducible expression. B. non-inducible expression. C. constitutive expression. D. All of the above. E. No conclusion can be made.

For question 146 & 147 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 b-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. 146. (3 pts) 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. B and C.

147. (3 pts) When the plates have no 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.

148. The type of enzyme that inserts the retrovirus genome into the host cell genome is

A. reverse transcriptase. B. RNA polymerase. C. DNA polymerase. D. Primase. E. Integrase.

149. A cDNA library is constructed from

A. proteins. B. genomic DNA. C. mRNA. D. lipids. E. All of the above.

150. A cDNA library includes sequences in

A. telomeres B. promoters. C. introns. D. exons. E. both B and C.

151. A genomic library includes sequences in

A. exons. B. promoters. C. introns. D. centromeres. E. All of the above.


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152. Reverse transcriptase is used in the construction of a A. ribosome. B. genomic library. C. cDNA library. D. plasmid vector. E. both B and C.

153. Transposons are beneficial to species over long periods of time because they

A. allow for gene duplication. B. slow down DNA replication of the genome. C. insert and disrupt gene function. D. cause chromosome loss. E. both A and B.

154. An enzyme used to distinguish between open versus closed chromatin is A. reverse transcriptase. B. ribonuclease. C. deoxyribonuclease. D. protease. E. both B and C.

155. The level of H3K9Ac nucleosomes in chromatin is measured using an antibody that specifically binds to

A. H3K9 nucleosomes. B. H3K9Ac nucleosomes. C. chromatin. D. DNA. E. RNA.

156. The cell prevents movement of many transposons by

A. deleting all of them from the genome. B. preventing their transcription. C. promoting their transcription. D. not replicating them in S phase. E. All of the above.

157. In Arabidopsis, homozygous wild-type flowers have which pattern of floral organs?

Whorl 1 Whorl 2 Whorl 3 Whorl 4 A. Sepals Petals Petals Sepals B. Carpels Stamens Petals Sepals C. Sepals Petals Stamens Carpels D. Stamens Stamens Carpels Carpels E. All of the above

158. (3 pts) In Drosophila, the bey (big eye) allele is recessive to the wild type bey+ (normal size eye) allele.

Also, the lol (long leg) allele is recessive to the lol+ (normal size leg) allele. A big eye, long leg female from a true breeding population was crossed to a true breeding wild type male. An F1 female was crossed to a big eye, long leg male. The genetic distance between bey and lol is 20 Centimorgans. What numbers of progeny would you expect to detect if 1000 progeny are analyzed?

Normal eye,

Normal leg Big eye, Long leg

Big eye, Normal Leg

Normal eye, Long leg

A. 300 300 200 200 B. 200 200 300 300 C. 400 400 100 100 D. 100 100 400 400 E. 250 250 250 250


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159. (3 pts) In Drosophila, the bth (big thorax) allele is recessive to the wild type bth+ (normal size thorax) allele. Also, the shh (short hair) allele is recessive to the shh+ (normal size hair) allele. A big thorax, short hair female from a true breeding population was crossed to a true breeding wild type male. An F1 female was crossed to a big thorax, short hair male. Approximately equal numbers of four progeny types were obtained: normal thorax and normal hair, big thorax and short hair, big thorax and normal hair, normal thorax and short hair. These data suggest that A. bth and shh are on different chromosomes. B. bth and shh are on the same chromosome and are so close to each other that no recombination occurs. C. bth and shh are on the same chromosome and are at least 50 Centimorgans apart from each other. D. either A or C could be true. E. Nothing can be concluded about the linkage of bth and shh.

160. Xeroderma pigmentosum is an autosomal recessive genetic disease in which the ability to repair DNA

damage caused by ultraviolet light (UV) is deficient. Young people with xeroderma pigmentosum suffer from multiple skin cancers. Bob is a healthy adult. Likewise, his parents were healthy. However, Bob’s brother had xeroderma pigmentosum disease. What is the probability that Bob is a carrier (heterozygous) for the Xeroderma pigmentosum mutation? A. 1/8. B. 1/4 C. 1/3 D. 2/3 E. 3/4

161. If the substrate molecule to the right is used by primase, there would be a

A. two hydrogen atoms at the 2’-carbon. B. two hydrogen atoms at the 3’-carbon. C. one hydroxyl group and one hydrogen atom at the 2’-carbon. D. one hydroxyl group and one hydrogen atom at the 3’-carbon . E. both C and D.

162. E. coli tRNAs that are covalently attached to different amino acids

A. have different anti-codons. B. have different 3-dimensional shapes. C. are recognized by different amino-acyl tRNA synthetase enzymes. D. All of the above. E. Only B and C are correct

163. Proper base pairing is required for accurate

A. transcription. B. DNA replication. C. splicing. D. regulation of gene expression by miRNAs. E. All of the above.

164. Which enzyme synthesizes RNA that is complementary to a DNA template?

A. DNA polymerase. B. RNA polymerase. C. Primase. D. Reverse transcriptase. E. Both B and C.

EXAM CONTINUES


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165. Which statement about E. coli DNA polymerase III is FALSE? E. coli DNA polymerase III A. simultaneously replicates both strands of template DNA. B. simultaneously synthesizes DNA in both the 5’ to 3’ direction and in the 3’ to 5’ direction. C. does not require a 3’-OH to initiate DNA synthesis. D. corrects its own errors by proofreading. E. Both B and C are false.

166. The 2 nanometer width of DNA revealed by Rosalind Franklin’s X-ray diffraction experiments suggested to

James Watson and Francis Crick that A. DNA is likely to be double stranded because the width of single-stranded DNA is much less than 2 nanometers. B. a base pair with two purines (G and A. is unlikely because it would be greater than 2 nanometers. C. a base pair with two pyrimidines (C and T) is unlikely because it would be less than 2 nanometers. D. a base pair with a purine and a pyrimidine is possible because it is 2 nanometers wide. E. All of the above.

167. When allele “a” is recessive to allele “A”, it means that the phenotype of the heterozygote A/a is the same

as the phenotype of the A. homozygote A/A. B. homozygote a/a. C. both homozygote A/A and homozygote a/a. D. neither homozygote A/A or homozygote a/a. E. Nothing can be concluded.

168. A given E. coli mRNA molecule

A. is usually transcribed from both of the DNA strands at a given genetic locus. B. has a single phosphate at the 5’-end. C. often has multiple start and stop codons that are used by the ribosome to produce multiple proteins. D. has a triphosphate at the 5’-end. E. Both C and D are true.

169. A female Drosophila has 4 pairs of homologous chromosomes. Suppose she is heterozygous for 4 mutations. Each mutation is on a different chromosome. How many genetically distinct eggs will she make after meiosis? Ignore crossing over. A. 4 B. 8 C. 16 D. 32 E. 64

170. Select the dipeptide that was synthesized by a ribosome.

A. Figure A.

B. Figure B.

C. Figure C.

D. Figure D.

E. All of the above

171. (1 pt) Peptide bonds are catalyzed by

A. tRNAs in the ribosome. B. ribosomal proteins. C. a ribosomal RNA (rRNA) D. the mRNA in the ribosome. E. None of the above.

END OF THE EXAM


Intentional Blank Page


Q# A PTS Answer A Q# A PTS Answer A 1 2 E 51 1 D 2 2 E 52 2 E 3 2 D 53 2 B 4 2 E 54 2 D 5 2 C 55 2 B 6 2 B 56 2 A 7 2 A 57 1 A 8 2 C 58 1 B 9 0 A 59 1 A

10 2 E 60 1 E 11 2 A 61 1 B 12 2 C 62 1 D 13 2 D 63 2 B 14 2 E 64 2 B 15 2 C 65 2 C 16 2 E 66 2 B 17 2 D 67 2 A 18 2 B 68 2 A 19 2 E 69 2 A 20 2 C 70 2 C 21 1 B 71 2 E 22 1 E 72 2 C 23 1 C 73 2 E 24 1 A 74 2 D 25 2 D 75 1 A 26 2 E 76 1 A 27 2 E 77 1 B 28 1 B 78 1 B 29 1 B 79 2 E 30 1 D 80 2 E 31 2 A 81 2 A 32 2 C 82 2 C 33 2 D 83 2 A 34 2 B 84 2 C 35 2 A 85 1 B 36 1 E 86 1 D 37 1 C 87 1 D 38 1 C 88 2 C 39 1 B 89 2 E 40 1 D 90 2 D 41 1 A 91 2 E 42 2 C 92 2 B 43 2 B 93 1 C 44 2 D 94 1 A 45 2 D 95 1 C 46 2 D 96 1 C 47 2 C 97 1 A 48 2 E 98 1 B 49 2 B 99 1 B 50 1 A 100 1 C


Q# A PTS Answer A Q# A PTS Answer A 101 2 B 140 2 B 102 2 B 141 2 D 103 2 C 142 2 B 104 1 D 143 2 D 105 1 C 144 2 C 106 1 C 145 2 B 107 2 D 146 3 E 108 1 E 147 3 E 109 2 B 148 2 E 110 2 E 149 2 C 111 2 E 150 2 D 112 2 B 151 2 E 113 2 E 152 2 C 114 1 D 153 2 A 115 2 A 154 2 C 116 2 C 155 2 B 117 2 E 156 2 B 118 2 A 157 2 C 119 1 A 158 3 C 120 2 B 159 3 D 121 2 B 160 2 D 122 2 C 161 2 E 123 2 A 162 2 D 124 2 B 163 2 E 125 2 B 164 2 E 126 2 B 165 2 E 127 2 E 166 2 E 128 2 D 167 2 A 129 2 C 168 2 E 130 2 B 169 2 C 131 2 E 170 2 A 132 2 A 171 1 C 133 2 B 134 2 D 135 2 A or D 136 2 A 137 1 C 138 2 D 139 2 D

KEY FALL 2013 FINAL VERSION A


BIOLOGY 1A Final December 10th, 2012 A NAME SECTION # DISCUSSION GSI 1. Sit at your assigned seat. 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.




EXAM Instructions:

4. Print your name on THIS COVER SHEET. (otherwise, you will get a ZERO).




8. WHEN FINISHED RAISE YOUR HAND. Your GSI will collect both your SCANTRON and EXAM. YOU MUST TURN IN BOTH or else you will get a ZERO. With 10 minutes left no students can leave. It gets too disruptive for other students.




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For question 34 the plate has glycerol as an energy and carbon source that does not interfere with expression of the lac operon. It also has X-gal. All bacteria have a functional β-galactosidase gene (Z+) in the lac operon. The rest of the genotype of the bacteria is indicated in the question. All genes are wild type unless noted otherwise. 34. (3 pts) When the plates have no inducer (IPTG) and no glucose, blue colonies will be produced by

A. O- bacteria.. B. Wild type bacteria C. I- bacteria. D. All of the above. E. A and C.

35. After transcription factor gene duplication

A. one of the duplicated genes always becomes completely inactive. B. the sequences of the duplicated genes always remain identical over evolutionary periods of time. C. the DNA binding domains can diverge so that the duplicated transcription factors regulate transcription of different

target genes. D. the promoters can diverge to give the duplicated genes different patterns of transcription. E. Both C and D.

36. Where are duplicated genes located immediately after unequal recombination?

A. Adjacent to each other on the same chromosome. B. it is not possible to predict where they will be. C. On different chromosomes. D. Linked, but usually more than 50 Centimorgans apart. E. All of the above.

37. A bacterial cell gives rise to two genetically identical daughter cells by a process known as

A. fertilization B. mitosis C. binary fission D. cytokinesis E. meiosis

38. Which of the following phases of the cell cycle is NOT part of interphase?

A. S B. M C. G0 D. G1 E. G2

39. E. coli DNA polymerase III A. never incorporates the wrong base into newly synthesized DNA. B. removes RNA primers from Okazaki fragments. C. initiates DNA synthesis by creating a phosphodiester linkage between two dXTPs. D. removes a wrong base by 3’ to 5’ hydrolysis of its phosphodiester linkage. E. joins two adjacent Okazaki fragments (primers removed) by generating a phosphodiester linkage.

40. Microtubules that form the mitotic spindle tend to originate from ______ and terminate in _______

A. chromatin, the nuclear envelope B. centrioles, kinetochores C. centrioles, chromatin D. the nuclear envelope, chromatin E. centromeres, chromatin


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41. DNA methylation is preferentially found in

A. euchromatin. B. heterochromatin. C. in DNA associated with H3K9 deacetylated nucleosomes. D. in DNA associated with H3K9 acetylated nucleosomes. E. Both B and C.

42. The E. coli RNA polymerase that transcribes genes

A. also synthesizes primers for Okazaki fragments. B. always simultaneously transcribes both DNA template strands. C. synthesizes in the 5’ to 3’ direction. D. degrades RNA in the 3’ to 5’ direction. E. All of the above.

43. A genomic library includes sequences in A. exons. B. promoters. C. introns. D. centromeres. E. All of the above.

44. The energy for RNA synthesis is derived from A. simultaneous cleavage of ATP when phosphodiester linkages are formed. B. simultaneous cleavage of GTP when phosphodiester linkages are formed. C. cleavage of pyrophosphate from rXTPs when they are incorporated into the RNA. D. cleavage of electrons from the 3’-OH. E. Both A and B.

45. A cDNA library includes sequences in A. exons. B. promoters. C. introns. D. centromeres. E. All of the above.

46. Many of the color patterns in flowers are caused by transposons. The non-colored (white) portions of the

petal are caused by a transposon A. that excised from a pigment biosynthesis gene. B. moving a pigment biosynthesis gene to a new site in the chromosome. C. that inserted into a pigment biosynthesis gene. D. duplicating a pigment biosynthesis gene. E. All of the above.

47. H3K9 deacetylated histones are preferentially found in

A. nucleosomes in euchromatin. B. nucleosomes in heterochromatin. C. nucleosomes in mitochondria. D. nucleosomes in closed chromatin. E. Both B and D.

48. (3 pts) In Arabidopsis, homozygous mutant agamous flowers have which pattern of floral organs?

Whorl 1 Whorl 2 Whorl 3 Whorl 4 A. Carpels Stamens Petals Sepals B. Sepals Petals Petals Sepals C. Sepals Petals Stamens Carpels D. Stamens Stamens Carpels Carpels E. None of the above


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49. Which figure is a correct representation of DNA synthesis by DNA polymerase III in E. coli. The arrowheads

show the 3’-ends of DNA. A. Figure A. B. Figure B. C. Figure C. D. Figure D. E. Figure E

50. The human adult and fetal β-hemoglobin genes

A. are located on different chromosomes. B. encode proteins with different affinities for oxygen. C have peak expression at the same time. D. evolved from different ancestral genes. E. Both A and D.

51. (3 pts) Infantile Tay-Sachs disease is a recessive, autosomal disease. Young children with Tay-Sachs disease overproduce cell membrane components that prevent brain nerve cells from functioning. Death occurs before the age of four years. John and Mary are married healthy adults. Likewise, their parents were healthy. However, John’s sister, and Mary’s brother had Tay-Sachs disease. What is the probability that both John and Mary are carriers (heterozygous) for the Tay-Sachs mutation? A. 2/3 B. 4/9 C. 1/4 D. 2/9 E. 1/16

52. (3 pts) In Drosophila, the b (blue eye) allele is recessive to the wild type b+ (normal red eye) allele. Also, the

hl (hairy leg) allele is recessive to the hl+ (normal leg) allele. The genetic distance between b and hl is 10 Centimorgans on the X chromosome. A blue eye, hairy leg female from a true breeding population was crossed to a true breeding wild type male. An F1 female was crossed to a wild type male from a true breeding population (for both traits). What numbers of progeny would you expect to detect if 2,000 progeny are analyzed? Answer choices read down the table.

Phenotype A B C D E Male Blue eye, hairy leg 450 50 450 400 100 Normal eye, normal leg 450 50 450 400 100 Blue eye, normal leg 50 450 50 100 400 Normal eye, hairy leg 50 450 50 100 400 Female Blue eye, hairy leg 450 0 0 0 100 Normal eye, normal leg 450 1000 1000 1000 100 Blue eye, normal leg 50 0 0 0 400 Normal eye, hairy leg 50 0 0 0 100

Workspace


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53. (3 pts) Which drug would most likely be the best candidate to specifically inhibit progression of a retrovirus such as the AIDS virus? A specific inhibitor of A. ribosome activity. B. splicing activity. C. reverse transcriptase activity. D. RNA polymerase II activity. E. Both C and D.

54. Translation termination in E. coli is facilitated by A. Ribosomal proteins that bind specifically to the Shine-Dalgarno sequence in the 16S rRNA.

B. A protein called release factor that binds to the stop codon. C. A specialized tRNA called release factor that forms base pairs with the stop codon. D. A ribosomal protein that binds specifically to the stop codon. E. Base pairs that are formed between the Shine-Dalgarno sequence in the 16S rRNA and a conserved sequence in

the 23S rRNA. 55. Covalent attachment of amino acids to tRNAs is a 2-step process. The first step involves forming a

covalent bond between an amino acid and ATP. Which atoms participate in this reaction? A. 1 and 3 B. 1 and 4 C. 2 and 3 D. 2 and 4 E. 2 and 5

56. Eukaryote mRNAs do NOT have a

A. 3’-polyA tail. B. 5’-untranslated region. C. 5’-triphosphate D. 5’-CAP E. 3’-untranslated region.

57. Which molecule is generated immediately after E. coli DNA polymerase I has synthesized its first phosphodiester linkage? A. Figure A. B. Figure B. C. Figure C. D. Figure D.


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58. (3 pts) For a species with N = 12, how many double stranded DNA molecules are in the nucleus of a cell after the completion of meiosis 1? A. 6 B. 12 C. 24 D. 36 E. 48

59. If a mutant allele (a) is recessive to the wild type allele (A), then the phenotype of the heterozygote (A/a). A. will be the same as the mutant homozygote (a/a). B. will be the same as the wild type homozygote (A/A). C. will be different than the wild type homozygote (A/A) and the mutant homozygote (a/a). D. cannot be predicted. E. none of the above.

60. (3 pts) Transcription of a miRNA gene produces A. a non-coding RNA molecule about 21 nucleotides in length. B. an RNA molecule about 21 nucleotides in length that encodes a micro protein. C. a non-coding RNA about 500 to 1000 nucleotides that base pairs with itself. D. a non-coding RNA about 500 to 1000 nucleotides in length that does not base pair with itself. E. a large RNA molecule that is translated into a large protein that is digested into micro proteins.

61. Transposons are beneficial to species over long periods of time because they

A. cause gene duplication B. cause exon shuffling. C. insert and disrupt gene function. D. accelerate DNA replication of the genome. E. Both A and B.

62. Which insertion in an exon would be predicted to have the least detrimental effect on the protein that is

synthesized from the mutant allele? A. 10 base pair insertion B. 9 base pair insertion C. 8 base pair insertion D. 7 base pair insertion E. all would be expected to be equivalent because this is an exon

63. During splicing, snRNAs in snRNPs

A. base pair with sequences at the 5’- and 3’-ends of introns. B. base pair with sequences at the 5’- and 3’-ends of exons. C. break the phosphodiester linkage at the 5’-end of the intron. D. break the phosphodiester linkage at the 3’-end of the intron. E. both B and D.

64. The human alpha-hemoglobin gene is thought to have moved from chromosome 11 to chromosome 16. The first step in this process was most likely to have been A. unequal recombination between homologous chromosomes. B. expression of retroviruses flanking the alpha-hemoglobin gene. C. transposase cutting at inverted repeats of DNA transposons flanking the alpha-hemoglobin gene. D. H3K9 acetylation at the alpha-hemoglobin gene E. DNA methylation at the alpha-hemoglobin gene.

65. Reverse transcriptase synthesizes

A. RNA that is complementary to an RNA template. B. RNA that is complementary to a DNA template. C. DNA that is complementary to an RNA template. D. DNA that is complementary to a DNA template. E. Both C and D.


ANSWERS: Final Fall 2012. If more than one answer is indicated that means more than one answer was accepteD)

1 E 21 B 41 E 61 E 81 D 101 E 121 B 141 C 2 D 22 C 42 C 62 B 82 E 102 C 122 C 142 D 3 A 23 B 43 E 63 A 83 B 103 D 123 A 143 A 4 A 24 D 44 C 64 C 84 D 104 D 124 C 144 D 5 D 25 D 45 A 65 E 85 C 105 A 125 B 145 C 6 A 26 E 46 C 66 E 86 E 106 D 126 C 146 B 7 B 27 C 47 E 67 B 87 A 107 D 127 A 147 E 8 B 28 D 48 B 68 A 88 C 108 B 128 C 148 D 9 C 29 A 49 A 69 C 89 D 109 A/C 129 A 149 C

10 B 30 E 50 B 70 D 90 D 110 D 130 D 150 E 11 D 31 A 51 B 71 E 91 A 111 E 131 B 151 B 12 A 32 D 52 C 72 A 92 C 112 B 132 E 152 A 13 C 33 B 53 C 73 A 93 E 113 B 133 A 153 B 14 B 34 E 54 B 74 B 94 B 114 C 134 E 154 D 15 E 35 E 55 B 75 C 95 D 115 B 135 A 155 B 16 B 36 A 56 C 76 D 96 A 116 D 136 C 156 E 17 E 37 C 57 D 77 B 97 D 117 E 137 E 157 A 18 D 38 B 58 C 78 C 98 D/E 118 D 138 E 158 B 19 A 39 D 59 B 79 A 99 C 119 E 139 E 159 E 20 E 40 B 60 C 80 D 100 B 120 B 140 E 160


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