BIO 151 - Studyguide Test 1
BIO 151 - Studyguide Test 1 Biol 151
Popular in General Biology
verified elite notetaker
Popular in Biology
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.
Reviews for BIO 151 - Studyguide Test 1
Amazing. Wouldn't have passed this test without these notes. Hoping this notetaker will be around for the final!
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
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 RNADNAmRNAprotein) 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?
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'