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BIO 151 - Studyguide Test 1

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by: Anna Proulx

BIO 151 - Studyguide Test 1 Biol 151

Marketplace > University of North Dakota > Biology > Biol 151 > BIO 151 Studyguide Test 1
Anna Proulx
GPA 4.0

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Chapters 14, 15, 16, 17, 18, 20
General Biology
Professor Felege
Study Guide
50 ?




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"Amazing. Wouldn't have passed this test without these notes. Hoping this notetaker will be around for the final!"
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This 6 page Study Guide was uploaded by Anna Proulx on Sunday February 14, 2016. The Study Guide belongs to Biol 151 at University of North Dakota taught by Professor Felege in Spring 2016. Since its upload, it has received 200 views. For similar materials see General Biology in Biology at University of North Dakota.


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Date Created: 02/14/16
STUDYGUIDE TEST 1 Ch 14, 15, 16, 17, 18, 20 CH 14 – THE GENE Understand simple Mendelian genetics including the relationship between meiosis and predicted  phenotypic and genotypic ratios in offspring. Identify genetic events or phenomena that are not explained by simple Mendelian genetics. 1. Distinguish genes (a factor that influences a trait) from alleles (a form of a gene).  2. Distinguish dominant (allele whose phenotype is shown in homozygous RR or  heterozygous Rr form) from recessive (allele whose phenotype is shown only in  homozygous rr form). 3. Explain Mendel’s laws of segregation (the alleles of a gene separate during reproduction) and independent assortment (these alleles are transmitted independently of one another). 4. Construct and predict the outcomes of monohybrid and dihybrid crosses. 5. 6. Monohybrid – Rr with Rr    Dihybrid – RrYy with RrYy 7. 8. R 9. R 10. r 11. r Y y Y y 12. R 13. R 14. R 15. R 16. R Y R R r r Y Y Y Y Y y Y y 17. R 18. R 19. R 20. R 21. R y R R r r Y y Y y y y y y 22. r 23. R 24. R 25. r 26. r Y r r r r Y Y Y Y Y y Y y 27. r 28. R 29. R 30. r 31. r 1. 2. 3. y r r r r R r Y y Y y y y y y 4. 5. 6. 32. Phenotype: 3:1 Phenotype: 9:3:3:1 R R R 33. Genotype: 1:2:1 Genotype: 1:2:1:2:4:2:1:2:1 7. 8. 9. 34. 35. Explain the impacts of the following phenomena:  r R r ­ linked genes (tendency of alleles of different genes to be inherited  together);  ­ incomplete dominance (allele phenotypes are blended);  ­ co­dominance (expression of multiple phenotypes concurrently, such as someone  with AB blood) and multiple alleles (more than two alleles for one gene, such as  ABO in blood, trait is polymorphic);  ­ epistasis (the masking effects of one gene over another, a gene prevents expression of another such as trait for being albino masks trait for hair color);  ­ quantitative traits (phenotype differs by degree such as height or hair color in humans. Contrasted with discrete traits such as tall/short, green/yellow);  ­ pleiotropy (one gene influences many traits. In Marfan Syndrome, one mutant allele  causes heart problems, abnormal limbs, and extreme height);  ­ environmental effects on gene expression (phenotype affected by genotype AND  environment. Factors such as temperature, nutrients, and hormones can affect gene  expression);  ­ epigenetics (changes in phenotypes without changes in genotypes/DNA sequence). 36. 37. 6. Extra! Particulate inheritance hypothesis trumps blending inheritance and inheritance  of acquired characters hypotheses. Terms: wild type (phenotypes most common for each  trait), recombinant (alleles differ from parental alleles due to crossing over), pedigree 38. CH 15 – DNA  39. 40. Understand how chromosomes are replicated before cell division can take place. 41. 1. Differentiate between semiconservative (correct model! Parental strand + daughter  strand), conservative (intact double parental strand and intact double daughter strand),  and dispersive (old DNA mixed with newly synthesized segments within same strand)  models of DNA replication. 2. Explain how the Meselson­Stahl experiment provided evidence for the semiconservative  model of DNA replication. It showed two bands of DNA, half low density and half  intermediate density, showing it contained both 14N and 15N. In generation 2, the all  new strands were 14N and low density, the mixed strands were half and half and  intermediate density 3. Explain which direction new bases (nucleotides) are added on the newly formed piece of  DNA. 5’3’ always 4. Explain the important role that a primer plays in DNA replication. Primer is a strand of  RNA nucleotides added to the DNA template that allows DNA polymerase to add to its  3’ end 5. Explain why Okazaki fragments are formed. DNA must be synthesized 5’3’ on the  lagging strand, so RNA polymerase binds near the replication fork and must move away  from it, then detach and move back near the fork, creating segments. 6. Differentiate between the leading and lagging strands of DNA. Leading is synthesized  toward replication fork and is continuous. Lagging is synthesized away from replication  fork and is discontinuous. 7. Explain the roles of the following during DNA replication:  ­ Helicase breaks H bonds between base pairs, ‘unzipping’ the DNA ­ single­stranded binding proteins bind to separated DNA strands and prevent them  from coming back together ­ primase adds RNA primer which initiates the building of new strands ­ DNA polymerase III synthesizes new DNA ­ DNA polymerase I replaces RNA primer with DNA ­ DNA ligase joins the Okazaki fragments into a continuous strand 8. Explain how DNA polymerase III prevents mistakes during DNA replication.  Proofreading (DNA polymerase pauses during synthesis and removes an incorrect base)  Mismatch repair (after synthesis is complete, proteins remove and replace incorrect  segment) Damage repair (such as nucleotide excision repair, where many enzymes fix  kinks in DNA helix caused by external damage) 42. 43. CH 16 & CH 17 – TRANSCRIPTION & TRANSLATION 44. 45. Understand the central dogma of molecular biology. DNA  RNA  protein 46. Understand transcription and translation. 47. 1. Differentiate between transcription (DNA transcribes mRNA) and translation (mRNA  translated into a protein through ribosomes). 2. Determine the amino acid sequence if given a DNA sequence. 48. DNA 5’ ATG CTG GAG GGG 3’ 49. 3’ TAC GAC CTC CCC 5’ 50. RNA 5’ AUG CUG GAG GGG 3’ 51. A.A.      Met – Leu – Glu ­ Gly 3. Explain how the genetic code is redundant and specific. Only four bases, A U C G, that  code for 20 amino acids. More than one codon can specify the same amino acid. 4. Explain the role of RNA polymerase in transcription. Synthesizes mRNA according to  information from DNA in nucleus, does not need primer. 5. Explain the “initiation” phase of transcription, including the roles of promoters and  transcription factors. In bacteria, sigma protein + RNA polymerase (holoenzyme) binds  to the promoter (­35 and ­10 boxes) on DNA. In eukaryotes, basal transcription factors +  RNA polymerase (3 kinds instead of 1 in bacteria) bind to the promoter (more diverse,  TATA box) on DNA. 6. Explain the “elongation” (RNA polymerase adds NTPs to the 3’ end of the growing  RNA) and “termination” (In bacteria, the termination signal causes RNA to form hairpin  and break from enzyme, makes mature RNA. In eukaryotes, termination occurs at  variable distance from poly(A) signal, makes pre­mRNA) phases of transcription. 7. Differentiate between introns (intervene, not in final mRNA product, removed by  spliceosome containing snRNPs) and exons (expressed, base regions kept in mRNA). 8. Explain the “5’ cap” and “3’ poly A tail” (protection, complete RNA processing). 9. Differentiate between the roles of mRNA (between transcription and translation) and  tRNA (used in translation). 10. Explain the relationship between codons (mRNA triplet code) and anticodons (on the  tRNA, base pairs with mRNA codon in the small subunit). 11. Explain the “initiation” (mRNA start codon binds to anticodon in small subunit, helped  by initiation factors, and large subunit binds), “elongation” (aminoacyl tRNA forms  peptide bonds between amino acids, ribosome translocates down mRNA), and  “termination” (stop codon causes release factor protein to enter EPA site and hydrolyze  tRNA bond in P site with polypeptide, freeing it) phases of translation. 52. 12. Extra! Reverse transcriptase (used by viruses. Makes RNA into DNA, adds step onto  central dogma RNADNAmRNAprotein) Sequence mutations (Point mutations. Can  be missense: changes amino acid specified by a codon, silent: does not change amino  acid, frameshift: deletion or addition of a base, nonsense: changes to stop codon, early  termination) Chromosome mutations (polyploidy: increase in number of a specific  chromosome, aneuploidy: addition/deletion of individual chromosomes, inversion:  chromosome segment flips, translocation: chromosome segment attaches to different  chromosome, deletion: segment lost, duplication: additional segment copies)  53. CH 18 – GENE EXPRESSION 54. 55. Understand how genes are expressed only when necessary. 56. 1. Identify the three levels at which gene expression can be controlled. Transcriptional  (saves the most energy) Translational (inhibits mRNA survival, initiation, or elongation)  Post­translational (most rapid) 2. Identify the conditions that would result in expression of the lac genes (lac Z, lac Y, lac  A). Negative control, lac operon is transcribed when lactose is present and glucose is not,  lactose (inducer) binds to repressor which falls off DNA, allows transcription of lac genes 3. Explain how the following interact in the presence and absence of lactose: repressor (in  presence it falls off operon, in absence it binds to operon), RNA polymerase (in presence  it synthesizes the code for the lac enzymes, in absence it is inactive), promoter (in  presence initiates gene transcription, in absence lacI synthesizes repressor which binds to  operator and prevents RNA polymerase from moving past promoter), operator (in  presence it is not occupied by repressor, in absence it is occupied by repressor). 57. 4. Extra! Positive control (Gene expression off, must be initiated by a signal/activator)  Negative control (Gene expression on, must be stopped by a signal/repressor) ara operon  (In positive control, genes are transcribed to process arabinose, two AraC proteins +  arabinose bind to initiator) (In negative control, transcription is shut down, one AraC  binds to initiator and one AraC binds to operator) 58. CH 20 – ENGINEERING GENES 59. 60. Understand some of the basic techniques used in biotechnology including basic genetic  engineering, polymerase chain reaction (PCR), and DNA sequencing. 61. 1. Explain how gel electrophoresis is used to separate molecules. The DNA is separated by  using an electric field to move DNA from wells at negative end toward the positive end,  which takes place in an agarose gel medium. Short segments travel faster, then results are read on a connected machine. 2. Explain the process of DNA sequencing including the important role of  dideoxynucleotides (contain no 3’ OH so synthesis is terminated when a ddNTP is  added). In dideoxy sequencing, a mix of dNTP + ddNTP + DNA + primers + DNA  polymerase undergoes synthesis. Each new strand built off the template DNA will  contain dNTPs and end with a ddNTP, through electrophoresis the sequence can be read. 3. Explain the process of PCR (polymerase chain reaction) including the important role of  “Taq” polymerase (DNA polymerase from Thermus aquaticus). Mix the gene + dNTPs + primers + Taq polymerase, heat the mix so DNA denatures, cool the mix so primers  anneal, heat the mix and Taq polymerase will synthesize complementary strands.  Denature  Anneal  Extend 4. Explain the steps involved in recombinant DNA technology (e.g., using bacteria to  produce insulin or human growth hormone) including how plasmids (circular piece of  DNA), restriction enzymes (endonuclease, cuts the DNA at its staggered recognition  site), and reverse transcriptase (makes cDNA from RNA) are used during this process.  mRNA from healthy cells is harvested, reverse transcriptase synthesizes cDNA from this  mRNA, restriction endonuclease cuts cDNA at its staggered palindrome sequence,  resulting sticky ends are joined together by DNA ligase, plasmid is formed, cells undergo transformation and uptake plasmids from environment, reproduce and make cDNA  library, marked probe binds to target cDNA sequence to identify recombinant cell 5. Explain how genetically modified crops are made including the role of Agrobacterium  plasmids (allowed vitamin A to by synthesized in rice, carry tumor­inducing plasmids  that contain genes for B­carotene to by made). Made to increase resistance to herbivores,  weeds, and improve food quality. 6. Discuss the ethical concerns associated with genetic engineering. Should people without  diseases be able to utilize this therapy, as a cosmetic? Can we play the role of God?


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