BIOS E-1a exam 02 120110 KEY

BIOS E-1a exam 02 120110 KEY - Name: /n$we/ Keg, BIOLOGY...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Name: /n$we/ Keg, BIOLOGY E-1a: Introduction to Molecular and Cellular Biology Fall 2010 EXAM 2 x Teaching Fellow: 0 Be sure to write your name on the top of each of the pages of the exam. ° Write each answer only on the same page as the pertinent question. THE _ ‘ SPACE PROVIDED IS MEANT TO BE SUFFICIENT; BE BRIEF, BE CAREFUL, BE CONCISEI If absolutely necessary, use the back of that page‘”to continue the answer. IF YOUR ANSWER CANNOT BE READ, IT CANNOT BE GRADEDIII Write out the word “True” or “False” for those types of questions, respectively. A codon table is included at the end to aid in answering some of the questions. Remember, if you do not show all your work, you cannot get partial credit. The point values for each questidn are noted. Allocate your time wisely. Read carefully, think befOre you write, good luck, and do well! IF YOU WRITE IN PENCIL, YOU CANNOT ASK FOR A REGRADE! . , Page 2 (03 pts.) ‘ Page 3 (03 pts.) Page 4 (30 pts.) ‘ Page 5 (05 pts.) Page 6 (14 pts.) Page 7- (09 pts.) Page 8 (II pts.) Page 9 (13pts.) .4 Page 10 (12 pts.) _ Total (100 pts.) L (1 pt.) 1. Crossovers generally occur during: a) mitosis b) meiosisl c) meiosis ll _ d) with equal likelihood in all three l0 (1 pt.) 2. Below is a cartoon depicting only human chromosome 2 (the other 22 chromosomes are not shown for simplicity but are behaving in a similar manner), aligning ' during metaphase of: a) mitosis b) meiosisl c) meiosis ll ’ t \ d) not enough information is shown GK (1 pt.) 3. In the cartoon below of a newly initiated replication bubble, identify the leading strand. Solid line represents the original DNA and the dashed line represents newly synthesized DNA/RNA (NOT Okazaki fragments). a) strand A b) strand B c) not enough information shown d) both A and B QUESTIONS 4-6: The Meselson-Stahl experiment showed that DNA replication is semi- conservative. Briefly, bacteria were grown in 15N media (heavy) before being transferred to 14N media (light). After the transfer, the bacteria were allowed to continue growing for 1, ' 2, or 3 more generations. The DNA was isolated after each generation and transferred to a CsCl gradient, and centrifuged in order to separate the DNA based on weight. (A strand of DNA synthesized while growing in 15N will be heavy and a strand synthesized while in 14N will be light. Heavy strands will migrate farther than light strands during centrifugation..) A The DNA banding pattern was then visualized after centrifugation. ‘3 (1 pt.) 4. After the bacteria have grown for 1 generation in the 14N media (light), which test tube shows the expected result? 4— Light 4— Half-heavy ‘— Heavy a b c d C— (1 pt.) 5. After 2 generations in the 14N media (light), which test tube shows the expected result? ‘ as 4— Light f. m <— Half-heavy <— Heavy 61 (1 pt.) 6. After 3 generations in the 14N media (light), which test tube shows the expected result? ‘(5 pts.) 7. True or false — a tumor: True ‘ can arise as the result of missense mutations ' True arises from a single cell 1 True requires angiogenesis in order to grow past a certain point True in some cases is comprised of cancer stem cells True ' that spreads is referred to as metastatic (5 pts.) 8. True or false -— transcription in eukaryotes is generally upregulated (increased) by. ' True histone acetylation { Folk methylation of promoter regions ' True activators 7 Fake microRNAs Fake repressors (5 pts.) 9. True or false — the following posttranslational modifications are usually permanent: / Fake phosphorylation ‘- Trm glycosylation Fake methylation : Fnlxe acetylation ‘ True disulflde bond formation (5 pts.) 10. True or false — chronic myelogenous leukemia: ‘r Fake is caused by a chromosomal inversion ‘ Falfe is treated with Gleevec, an allosteric inhibitor of BCR—ABL flag is a type of cancer , Truz is due to an activation of a proto-oncogene True. is due to the generation of a constitutive active kinase, BCR-ABL .(5 pts.) 11. True or false — DNA replication in E. coli: Padre " requires telomerase ’ ‘ True begins at a single origin of replication ' 77m requires helicase " Trina requires DNA polymerase, which adds nucleotides in the 5’-to-3’ direction Tm 2 requires topoisomerase 1(5”pts.) 12. True or false — checkpoint proteins: ? Tm: ‘ can act upon cyclin/CDK complexes Truz— can activate apoptosis ' Trua can be involved withmonitoring-the-integrity'of'thegenome Tm; are involved at various stages to regulate cell cycle progression Trui when mutated can lead to an increased risk of cancer 0\ (1 pt.) 13. New tRNAs enter the ribosome in the: a) A site b) E site 0) P site d) Z site ‘ C (1 pt.) 14. The growing polypeptide chain in the ribosome is in the: a) A site b) E site c) P site d) Z site ()1 (1 pt.) 15. The enzyme that catalyzes peptide bond formation in the growing polypeptide chain is: a) topoisomerase b) peptidase c) aminoacyl-tRNA synthase d) the ribosome itself/ ribozyme e (1 pt.) 16. Eukaryotic ribosomes associate with mRNA: a) and move in the 5’-to-3’ direction b) and move in‘the 3’—to—5’ direction 0) using the 5’-guanine cap to initiate translation d) using the poly-A tail to initiate translation e) A + C f) A + D g) B + C h) B + D O! (1 pt.) 17. in a routine blood test, cells are tested from a patient and found to harbor a mutation in the genomic DNA. This could have arose from a: a) somatic mutation b) germline mutation in one of the individual’s parents c) germline mutation in the individual d) either A or B e) either A or C 1‘) either AjorB or C ' 9) none of the above QUESTIONS 18-24: You are interested in studying a new gene. All you know is the coding strand, which is as follows (NOTE: 5’ is to the left in all seguences written below): GTTCTCAAGGCCAATGAACAGCTATGGl l l IATGGGACTGAAGTGAGGTCACTT ‘3 (3 pts.) 18. What does the following sequence correspond to? ‘ GUUCUCAAGGCCAAUGAACAGCUAUGGUUUUAUGGGACUGAAGUGAGGUCACUU a) template strand b) mRNA 0) protein d) none of the above A (3 pts.) 19. What does the following sequence correspond to? CAAGAGTTCCGGATAC'ITGTC GATAC CAAAATACCCTGACTTCACTC CAGTGAA a) template strand b) mRNA 0) protein d) none of the above as (3 pts.) 20. What does the following sequence correspond to? AAGTGACCTCACTTCAGTCCCATAAAACCATAGCTG'ITCATAGGCCTTGAGAAC a) template strand b) mRNA 0) protein d) none of the above (5 pts.) 21. What is’ the protein sequence? Mel ' A§n' SQr’T‘jr’ ’ F136- Mar” [’64- L75 7 You identify a naturally occurring mutation in the gene that leads to the following change in nucleotide sequence (coding strands given; wildtype is above, mutant directly below for comparison; change bolded and underlined in mutant): GTTCTCAAGGCCAATGAACAGCTALQGI l l lATGGGACTGAAGTGAGGTCACTT GTI'CTCAAGGCCAAQGAACAGCTATGGI I | lATGGGACTGAAGTGAGGTCACTT C (1 pt.) 22. What type of mutation is this? a) insertion b) deletion C (j of 1C OLA 0V” [:c) base substitution Ld) frameshift - / / ye) Silent La flaw!»ng LI) missense g) nonsense (5 pts.) 23. What effect will this have on the protein? Explain. ' a“ A e’HAlQ (5Lch * franrimt‘ioa by.“ ;,\ 03' as ‘ )M-ééqr Q7 f ‘ 63m AOWAS+TQD~M {C (91,304 A’qu Met “ Val ' Lea “Trr ~ Asia (3 pts.) 24. if the mutation was instead the following: GTI'CTCAAGGCCAATGAACAGCTATGGI l l lATGGGACTGAAGTGAGGTCACTT GTTCTCAAGGCCAAQGAACAGCTATGGI I l lATGGGACTGAAGTGAGGTCACTT How would this affect your answer in #23? Explain. no attest » mutt tail +0 tag a,“ foiWeWA-ye QUESTIONS 25-29: You are interested in studying the genes for brown eyes (bw) and curved wings (c) in fruit flies. You have a female fruit fly with the phenotype of red eyes (wild type, bw+) and straight wings (wild type, 0*). Her genotype is unknown. This female is crossed to a lab strain male that has mutant brown eyes and mutant curved wings and is known to breed true for these two traits. All of the F1 progeny from this mating have wild type eyes and wild type wings. [3 (1 pt.) 25. What is the genotype of the female fruit fly? a) bv'v/bw & 0/0 b) bw“/bw+ & ch+ c) bw/bw“ & c/c” d) not enough information given (3 pts.) 26. You decide to do a testcross by taking one of the F1 females and mating her back to the male that breeds true for brown eyes and curved wings. If we assume that the genes for bw and c are on different chromosomes, what fraction of the progeny from this mating will have the phenotype of brown eyes and straight wings? l/L-f 0/ Z (3 pts.) 27. The table below contains the results from the testcross between the F1 female and the true breeding male: What can you conclude about the relationship between the genes bw and c? lbw aan‘ C ()er \inkeol (1 pt.) 28. How many map units apart are bW and 0? 2% NF Mfr 0\ (3 pts.) 29. You do additional, ,similar experiments to find that the gene for vestigal wings, vg, is 8.5 map unitsaway. from c and that vg is 37.5 map units away from bw. Can you determine the order of the genes along the chromosome? a) vg c bw b) vg bw c c) bw vg c d) cannot tell from the information given QUESTIONS 30-34: Examine the following crosses: Cross 1 Cross 2 P g Of genotype lgf2' lgf2' X [9)? 1ng lgf2' lgf2' lgf2 [ng . phenotype (homozygous (homozygous (homozygous (homozygous dwarf) normal) dwarf) normal) F1 1 phenotype all normal males and females all dwarf males and females (1 pt.) 30. What is the simplest explanation for the observed inheritance of the dwarf phenotype? L “(6% (3 pts.) 31. What is the genotype of an animal from Cross 1? r A +/~ 13sz $3432. or I342 iml’lrt" (3 pts.) 32. What is the genotype of an animal from Cross 2? 13(2'ICAM OK Low“ (3 pts.) 33. If a F1 female from Cross 1 is mated to a F1 male from Cross 2, what will be the genotypes of the progeny? l 13%:er r igflj’: 2 1%(2'1391 0“ 2 Dot? 1 11ft ‘ljafl‘ r 1392:" C (3 pts.) 34. If a F1 female from Cross 1 is mated to a F1 male from Cross 2, what will be the phenotypes of the progeny? a) all normal b) all dwarf 0) equal number of normal and dwarf d) 3 normal: 1 dwarf (3 pts.) 35. If there were no crossovers during meiosis, how many genetically distinct gametes could a normal human theoretically produce? 23 Z (3 pts.) 36. Generally, are the two daughter cells produced at the end of mitosis genetically identical? L785 (3 pts.) 37. Generally, are the two daughter cells produced at the end of meiosis l genetically identical? flO (3 pts.) 38. Generally, are the two daughter cells produced at the end of meiosis ll genetically identical? (\0 10 ...
View Full Document

This note was uploaded on 02/09/2012 for the course BIOLOGY 102 taught by Professor Anderson during the Spring '11 term at Harvard.

Page1 / 10

BIOS E-1a exam 02 120110 KEY - Name: /n$we/ Keg, BIOLOGY...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online