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Genetics Final

  1. What does true breeding mean?
    • Parents were always homozygous for the trait that is being followed.
    • Medenl used true-breeding plants for his parental crosses in studying heredity.
    • same varient for many generations.
    • no genetic mutations
  2. Mendels Model of Inheritance
    • 1. Alternate verisons of characters (genes) called varients (alleles) account for variation in inheritance.
    • 2. An organism inherits 2 alleles of a gene, 1 from each parent (sexual reproduction).
    • 3. The dominant allele determines the organism's apprearence but the recessive allele has not affect (masks the recessive unless both alleles are recessive).
    • 4. 2 alleles of a gene segregrate from each other during gamete formation (law of segregation).
  3. Allele
    An alternate form of a gene
  4. Mendel's model of inheritance.
    • Different forms of characters called variants account for differences in offspring traits.
    • An organism inherits two forms of a gene
    • The law of independent assortment
    • NOT the law of variation 
  5. positive interference
    a crossover decreases the probability of a second crossover
  6. product rule
    • used to calculate the chances of a double crossover from occuring.
    • multiply individual independent crossover
  7. crossing over is not guarenteed
    • most common: no crossover, all parentals
    • fairly common: crossing over between genes that are farther apart
    • uncommon: crossing over between genes that are close
    • most uncommon: double cross-over 
  8. Nuclear Genes
    genes located on chromosomes that are found in the cell nucleus 
  9. Recombination
    • Consequence of crossing over that occurs during gamete formation
    • only occurs in females
  10. Law of Independent Assortment
    • 2 different genes on (different chromosome) will randomly assort their alleles during formation of the haploid cell.
    • Dominant alleles assorted together if they are linked
  11. Mendel's law of segregation
    • States that two alleles of the same gene will always seperate during gamete formation.
    • accounts for varience 
  12. For the cross: Hh x hh, what is the rule that is used to determine the probability of an offspring inheriting the recessive allele?
    The rule of independent assortment. 
  13. What does it mean is p = 0.99 for the chi square test?
    This means that 99% of the time, any difference seen between the expected and the observed data are not significant and the null hypothesis is accepted. 
  14. True or Fasle:
    Meiosis results in 4 daughter cells that are haploid and genetically indentical.
    FALSE
  15. Facts about chromosomes during meiosis.
    • The identical copies of a replicated chromosome are called sister chromatids.
    • Crossing over occurs between non-sister chromatids. 
    • Homologous chromosome pair up at the metaphase plate during meiosis I.
    • the synaptonemal complex is NOT formbed between the sister chromosomes.
  16. sister chromatids
    Identical copies of a replicated chromosome.
  17. True or False: Sperm are consisdered homogametic because they can have either the X  or Y chromosome.
    FALSE
  18. The white gene, which can determine the eye color of Drosophila, is x-linked. w+ = red (dominant), w= white. If a white-eyed female mates with a red-eyed male, what are the expected phenotypes of their offspring?
    • 100% males would have white eyes
    • 100% females will have red eyes
  19. A red-eyed female mates with a white-eyed male. All offspring have red eyes. What was the genotype of the female?
    • Xw+Xw+ 
  20. Complete dominance
  21. codominance
    • multiple alleles, more than 1 in dominant
    • blood type determined by surgace antigens on the red blood cells.
  22. incomplete penetrance
    The pattern when the dominant phenotype is not expressed even though the individual carries the cominant allele.
  23. Incomplete penetrance
    • The heterozygote has an advantage over the homozygous dominant.
    • sickle- cell, homozygous dominant has the disease
  24. incomplete dominance
    The pattern that occurs when a heterozygote has a phenotype that is an intermediate phenotype of the 2 homozygous parents.
  25. sex-influenced traits
    • not sex-linked
    • baldness autosomal dominant in males
  26. sex-linked dominance
  27. epistasis
    gene interaction, alleles of genes masking the effect of alleles of a different gene. 
  28. pleotrophic effects
    • 1 gene has many phenotypes. Expression of a gene can influence cellular function.
    • a gene may be expressed in different cell types and at different developmental stages.
  29. maternal effect gene
    • genotype of mother directly determines the phenotype of her offspring
    • father's and offspring will have no effect on the offspring's genotype
    • phenotype of the offspring soley determines the genotype of the mother
  30. paracentric inversion
    • inversion that does not include the centromere
    • results in 2 normal/ parent, 1 dicentric chromosomes, and 1 acentric fragments (no centromere)
  31. Pericentric inversion
    • inversion around the chromosome
    • results: 2 normal/ parental type, 1 duplication and 1 deletion
  32. gene family
    genes that have similar sequences
  33. ortholog
    similiar gene in organisms of different species 
  34. paralogs
    homologous genes in same organism
  35. homologs
    • the members of a pair of chromosomes
    • homologous gene different organism, same species
  36. locus
    a specific position of a gene
  37. karyotype
    • organ. representation of chromosomes within a cell.
    • It tells us how many chromosomes are found in an actively dividing somatic cell.
  38. Germ cells
    • gamete
    • sperm
    • eggs
  39. somatic cells
    • any cell of the body that is not a gamete or a precurser to a gamete
    • diploid
  40. binary fission
    bacteria divide into 2 daughter cells after replication.
  41. gene redundancy
    many genes with the same function
  42. Interphase
    G1 + S + G2
  43. chromatids
    2 copies of chromosomes after replication
  44. S phase
    chromosomes are replicated
  45. G1 phase
    cells prepare to divide
  46. M phase
    • Mitosis
    • pairs of chromatids are seperated and sorted
  47. G2 phase
    the cell accumulates materials needed for nulcear and cell division
  48. kinetochore
    • a group of proteins that are bound to the centromere and hold the sister chromatids together
    • plays a role in sorting.
  49. centromere
    joins the 2 sister chromatids
  50. Aster mitochondria
    • protrude outward of the centromere toward the plasma membrane
    • positions the spindle apparatus
  51. polar mitochondria
    point towards the regions between the 2 spindles, overlap and help seperate poles.
  52. kinetochore mitochondria
    • attach to kinetochore
    • form 3 layers of proteins
  53. gametogenesis
    process by which genes form
  54. prophase
    • Nucleus membrane starts to dissociate into small vesicles. 
    • chromatids condense. mitotic spindle begins to form
    • nucleolus disappears
    • crossing over during prophase I
  55. prometaphase
    sister chromatids attach to kinetochore and a pair of sister chromatids are attached to the kinetochore at opposite poles
  56. metaphase
    • pairs of sister chromatids allign themselves along the metaphase plate.
    • the chromatids are evenly distributed into the daughter cells.
  57. anaphase
    each chromatid is an individual chromosome and is only linked to 1 pole
  58. telophase
    • chromosomes move to their respective poles and decondense
    • the nuclear envelope and produces 2 seperate nucleis
  59. homogamous
    • gametes are mophologically similar
    • have the same sex chromosomes
  60. heterogamous
    • 2 morphologically different types of gametes
    • have different sex chromosomes
  61. synteny
    • 2 or more genes on the same chromosome. 
    • alleles of different genes on the same chromosome are inherited together.
  62. Recombinatino Frequency
    • distance between genes determine recombination frequency
    • the further the genes are apart, there is a great chance of recombination
  63. crossing over
    exchange of chromosome segments between non-sister chromatids
  64. syntenic genes
    Are genes that are genetically linked.
  65. Test Cross
  66. The following can effect genetic variation
    • the law of independent assortment
    • crossing over
    • alleles
    • NOT the law of recombination
  67. True or False: syntenis alleles are always inherited together.
    FALSE
  68. map distance
    • distance between genes on a chromosome
    • correlates to recombination frequency
    • # of Recombinant offspring/ total # of offspring
  69. Highest # of recombinants possible
    50% 
  70. True or False: Crossing over always results in genetic recombination.
    FALSE
  71. epigenetic inheritance
    • A modification that occurs in a nuclear gene that alters gene expression, but not permanently.
    • dosage compensation and genomic imprinting
  72. dosage compensation
    • genes on the sex chromosomes are expresses at the same level in males and females even though they have different compliment of sex chromosomes
    • occurs in early development
    • condensed structure of somatic cells in the interphase nuclei of somatic cells and are only found in females 
    • males accomplish dosage compensation by doubling the expression of X-linked genes
  73. True or False: Barr bodies are associated with dosage compensation.
    TRUE
  74. non-allelic homologous recombination
    • can misallign due to duplication
    • results in normal,duplications, and deletions 
  75. True or False: Monoallelic expression is associated with maternal effect gene
    FALSE
  76. Xist
    • X inactive specific transcript
    • The gene that is directly resposible for X inactivation.
    • Xist RNA coats the X chromosome to be inactivated
    • induces barr formation
  77. Tsix
    prevents X inactivation
  78. genomic imprinting
    • one allele is silenced 
    • monoalleic expression
    • offspring phenotype depends on which allele was inherited 
    • maternal or paternal silced
  79. mechinsm of X inactivation
    • 1. Xist is transcribed to RNA
    • 2. Xist RNA bings to the inactived X-chromosome and recruits proteins that promote compaction.
    • 3. Xce regulates transcription of Xist and Tsix and decides if the X chromosome that remains active 
  80. Xic
    • X inactivation center 
    • must be on the X chromosome for X inactivation to occur
  81. Xce
    • X chromosomal controlling element
    • regulates transcription of Xist an Tsix and decides if the X chromosome remains active
  82. X inactivation
    • Xist RNA coats the X chromosome to be inactivated.
    • The active X chromosome has transcribed Tsix.
    • The inactivate X chromosome is maintained throughout development.

    the Xic gene is NOT located inside the Xist region
  83. methylation
    adding a CH3 group to a carbon in a gene that can silence or activate a group
  84. Restriction Digest
    • Restriction enzymes are used to add a plasmid
    • cuts at restriction site
    • ligation rxn to "paste" gene ito plasmid
    • bacteria transformation: all new bacteria are resistant.
  85. maternal inheritance of extranuclear genes
    • genes are inherited directly from mother 
    • all offspring will have same genotype and phenotype
    • all have same mitochondrial genome
    • genes in the mitochondria do not segregate 
  86. extranuclear genes
    • mitochondrial genes that are found in the cytoplasm
    • not sorted during meosis
  87. ICR
    • imprinting control region
    • sequenceof DNA that seperate and controls which genes will imprint
  88. DNA methylation
    • methylation is erased during meiosis
    • de novo methylation only occurs during gametogenesis
    • the CTC-factor prevents methylation at the ICR

    The ICR is NOT located with in the imprinted gene.
  89. True or False: Mitochondrial genes are inherited in the same fashion as nuclear genes
    FALSE
  90. Which method of bacterial genetic transfer did Griffith's experiments demonstrate?
    Transformation
  91. Transduction
    • use of bacteriapage (virus) with its own DNA
    • DNA is incorporated into the host's genome
  92. Transformation
    • DNA taken up from the enviornment
    • DNA and plasmid is passed on when it replicates
  93. coupling factor
    recognizes DNA/ relaxase complex for transport
  94. steps in transformation
    • competent cells: bacteria cell that can uptake DNA cells
    • 1. DNA binds to the surface of the recepient
    • 2. endonuclease cuts into smaller pieces
    • 3. 1 strand degraded and other is transported into uptake system
    • 4. allignment
    • 5.homologous recombination 
  95. intragenic regions
    • non-transcribed
    • do not encode for anything but are useful for controling gene expression 
  96. steps in transduction
    • 1. phage infects
    • 2. host DNA hydrolyze and phage DNA and proteins made
    • 3. phage assemble
    • 4. phage injects DNA into the new recepient cell
    • 5. transduced DNA recombined into DNA of recepient cell
  97. Hfr strains
    • High frequency of recombination
    • F factor incorporated into the genome 
    • Transfer DNA fast and efficently 

    • plaque formation: sensitive to infection
    • no plaque: resistant
  98. exporter
    protein that trasfers DNA into recepient
  99. relaxase
    protein binds the DNA strand to be transfered, circulizes after replication
  100. conjugation bridge
    structure that is formed after the sex pillus conacts F-
  101. prototroph
    does not require nutrients to grow
  102. auxotroph
    • bacteria requires nutrient to grow
    • lacking the gene that is needed for nutrient synthesis
  103. relaxasome
    makes a cut at OriT and begins to seperate the DNA strands
  104. Translocation
    • segments of 1 chromosome move to another chromosome
    • simple: between homologous chromosomes
    • recipricol: nonhomologous
  105. What was the major conclusion from Griffith's experiments?
    The R-type bacteria gained the ability to be virulent from the heat-killed S-typed bacteria.
  106. True or False: Addind proteases to a nucleic acid pred destroys the proteins
    TRUE
  107. Based on Avery and Colleagues' experiments, what was the main conclusion after adding only the RNase to the nucleic acid prep and injecting into the mice?
    RNA was not the genetic material that made the R-type bacteria virulent.
  108. What what the key experiment to showing that DNA was what turned the R-type bacteria virulent?
    Addition of DNase resuled in live mice
  109. Compents of a nucleotide
    • phosphate group
    • Nitrogenous base
    • 5-Carbon sugar
  110. True or False: The sugar found in RNA is ribose
    TRUE
  111. True or False: The bond formed between the base and the 1'C of the sugar is the glycosidic bond.
    TRUE
  112. True or False: Cytosine and Thymine are pyrimidines
    TRUE
  113. The free phosphate group is at which end of the DNA sequence?
    5' end
  114. What side of the DNA sequence has a free OH phosphate group on the sugar?
    3' end
  115. True or False: the phosphodiester linkage refers to a single bond formed btween the #'C and the phosphate group of the adjacent nucleotide.
    FALSE
  116. What part of the DNA helix do the proteins bind to?
    the major groove.
  117. What was the major conclusion from Chargraff's experiment?
    • The number of A's is equal to the number of T's
    • the numbe rof G's is equal to the number of C's
    • However, the number of A-T pairs does  NOT equal the number of G-C pairs
  118. Properties of DNA
    • Right handed double helix
    • strands are antiparallel
    • guanine base pairs with cytosine.
    • base pairs for hydrogen bonds NOT covalent bonds
  119. True or False: G-C pairs are stonger than A-T base pairs
    TRUE
  120. True or False" A DNA helix with 40% G-C pairs would be harder to denature than a DNA helix with 40% A-T pairs
    FALSE
  121. phosphodiester linkage
    the entire bond formed that connects ribonucleotides together
  122. triplex
    • 3rd strand binds following the A/T G/C rule
    • the cystine bases in the 3 strand are protinated
  123. Z DNA
    • forms this shape under low salt conditions and methylation
    • 12.0 bp per turn.
    • negative super coiling favors Z DNA
  124. A DNA
    • x-ray diffraction 
    • occurs under conditions of low humididty 
    • 11.0 bp per turn 
  125. B DNA
    •  right handed helix
    • the bases in the opposite strands H bond according to the AT/GC rule
    • the 2 strands are antiparallel with regard to their 5' to 3' directionality
    • there are 10.0 nucleotides in each strand per complete 360o of a turn
    • 1 complete turn is 3.4 nm
    • width is 2 nm
  126. RNA
    • RNA has U instead of T
    • C-G base pairs are found in RNA
    • A-A base pairs are gound in RNA
  127. True or False: RNA never forms a helix dur to its single-stranded nature
    FALSE
  128. negative supercoiling
    • lose 1 turn of the helix 
    • 12.5 bp/turn and underwound makes a 10 + 1 negative supercoil
    • right side crosses over first
    • twists in the clockwise direction.
  129. positive supercoiling
    • gain 1 turn of helix
    • overwound makes 8.3 bp/turn then make 10 + 1 supercoil
    • left side crosses first
    • twist in a clockwise direction
  130. Bacterial chromosomes are
    supercoiled
  131. eukaryotic chromosomes
    • contain telomeres and kinetochores
    • multiple origins of replication
    • centromeres
  132. what is the correct order of DNA compaction
    helix > nucleosomes > chromatin > chromosomes
  133. True or False: DNA wrapped around a histone hexamer is called a nucleosomes
    FALSE
  134. zig-zag 30 nm fiber
    • linker region is bent and twisted has has little face-to-face contact between nucleosomes
    • stable nucleosomes
  135. SMC proteins
    • structural maintaince of chromosomes 
    • use ATP to catalyze changes in chromosome structure
    • required for condensation and cohesion
  136. scaffold
    • formed from non-histone proteins of the nuclear matric
    • highly condensed chromosomes in metaphase are achored by scaffold
  137. solenoid
    • helical 30 nm structure
    • nucleosomes produce a symetrically comapact structure
  138. Euchromatin
    • less condensed
    • areas are capable of gene transcription 
    • 30 nm radial loops
  139. heterochromatin
    • more tightly compact 
    • transcriptionally innactive
    • radial loops
  140. True or False: Heterochromatin is more compact than euchromatin.
    TRUE
  141. The following contribute to general DNA compaction
    • Adding H1 nucleosomes
    • formation of radial loops
    • interactions with the nuclear matrix
  142. condensins
    required for compaction of DNA during cell division.
  143. cohesins
    • required for holding sister chromatids together
    • binds during s-phase -remains attach until the middle of prophase
  144. Conjugation
    • direct contact between donor and recipient by sex pillus
    • DNA (plasmid) transfer
    • 1. contact
    • 2. relaxosome cuts DNA strand and seperates DNA
    • 3. relaxase binds to DNA couping factor and brings it to the exporter
    • 4. trasnfer
    • 5. replication
  145. the following contribute to genetic variation in bacteria
    • transformation
    • transduction
    • conjugation
    • NOT asexual reproduction
  146. the steps of conjugation
    • 1. contact
    • 2. relaxosome cuts the DNA strand and seperates the DNA
    • 3. relaxase binds the DNA to the coupling factor which brings it to the exporter
    • 4. transfer of cut DNA strand to donor
    • 5. DNA replication makes double strand with relaxosome
  147. True or Flase: A cell can only transfer DNA into the recipient cell, the DNA is replicated
    FALSE
  148. What are the steps in setting up a conjugation mapping experiment?
    • 1. mix strains: allow for conjugation to occur
    • 2. stop conjugation at different times points (interupt mating)  to see which genes have been trasfered
    • 3. plate cells on growth media
    • 4. streak survivors on to selective media
    • plaque: sensitive to infection
    • no plaque: resistant
  149. metacentric chromosome
    centromere as close to middle as you can get it
  150. telocentric chromosome
    centromere at the end of the chromosome
  151. submetcentric
    chromosomes are closer to one side than the other closer to the middle that acrocentric
  152. Giemsa staining
    • allows identification of chromosomes
    • dark areas: highly compact
    • light areas: less compact
  153. acrocentric
    close to one side than the other
  154. Interstitial deletion
    a loss of an internal piece of a chromosome
  155. gene duplication
    terminal deficiency
  156. reciprocal translocation
  157. What are the consequences of nonallelic homologous recombination?
    duplications and deletions
  158. The gene family globulin arises from which types of chromosome structural variation?
    Duplications
  159. True or False: homologous genes in the same organism are called homologs
    FALSE
  160. True or False: Homologous genes in the same species are called orthologs.
    FALSE
  161. Terminal Deletion
    single break lost and degraded
  162. SSB
    • single strand binding protein
    • binds to a single strand and prevents it from reforming to its original structure.
  163. lac operon
    • genes that encode for lactose metabolism enzymes
    • no lactose present operon is off: lac repressor binds to the operator and inhibit transcription
    • allolactose binds to repressor and prevents it from binding to operator site and increase transcription
  164. inhibitor
    binds to activators to inhibit transcription
  165. corepressor
    binds to repressor to inhibit transcription
  166. inducer
    binds to repressor or activator to increase transcription
  167. activator
    binds to DNA to increase transcription
  168. repressor
    bind at promoter to inhibit transcription
  169. Termination of Replication
    Ter/Tus sits stop replication is  E. coli

    polymerases fall off the ends of the chromosomes in eukaryotes.
  170. primase
    • synthesizes short RNA primers
    • start or prime DNA replication
  171. DNA gyrase
    • Topoisomerase II
    • removes positive supercoiling a head of the replication fork.
    • cuts both strands of DNA and induces negative supercoiling  
    • gyrase binds and pulls sides together and start to coil. the is cuts 2 stands in the back and reseals them on the from side which induces negative supercoiling
  172. What causes balanced translocations?
    • non-homologous recombination
    • chromosome breakage and reattachment
  173. Comparative genome hybridazation can be used to detect what type of chromosome variation?
    both duplications and deletions
  174. allotetraploid
    An organism with two complete sets of chromosomes from 2 different species.
  175. Watson and Crick
    • 1. each strand contains a sugar, phosphate backbone
    • 2. in opposite strand A h bonds to T and G h bonds to C
  176. Franklin
    • DNA exists as a double-stranded helix
    • x-ray 
    • observed: 1 consistant with helical structure 2. the diameter of the helical structure was too wide to be only a single strand 3. 10 bp per turn
  177. pyrimidines
    • uracil
    • thyamine
    • cytosine
  178. purines
    adenine guanine
  179. phosphodiester linkage
    •  easter bond between the phosphate group of one nucleotide and the sugar molecule on the adjacent nucleotide
    • phosphate attachment to 5'carbon of one nucleotid to the 3' carbon of the other 
  180. allodiploid
    • sterile
    • 2n
    • n is from a different species
  181. aneuploid
  182. An organism with 2n+1 chromosomes is an example of why type of species?
    Aneuploid
  183. True or False: Complete nondisjunction during mitosis can eventually lead to a polyploid organism.
    FALSE
  184. True or False: a gynandropmorph is an example of meitic nondisjunction. 
    FALSE
  185. Why is DNA replication semi-conservative?
    DNA replication is semi-conservative because each new DNA molecule has an old strand and a new strand.
  186. DNA polymerase
    requires a primer
  187. How does the DNA polymerase determine whether the correct nucleotide has been added?
    The correct base pair fits in the polymerase active site
  188. DNA gyrase
    The protein that unwinds supercoild DNA ahead of the replication fork.
  189. If the primers on the lagging strand cannot be removed during replication in eukaryotes, which protein is likely missing?
    Fen1
  190. What is the most likely explanation if the template strands cannot be seperated during the DNA replication in bacteria?
    DnaB is missing
  191. DNA ligase
    seals gaps between Okazaki fragments.
  192. True or False: Eukaryotes use the Ter/Tus method to terminate replication.
    FALSE
  193. True or False: Telomerase is the enzyme that prevents gene loss after replication in bacteria
    FALSE
  194. non-histone proteins
    play a role of organization of chromosome and their presence can affect the expression of near by genes.
  195. linker region
    • connects adjacent nucleosomes
    • this is also where H1 and non-histone proteins bind
  196. octomer
    • 4 different histone proteins
    • H2A H2B H3 H4
  197. HI
    • in Euk is the linker histone 
    • binds the histone to the linker region between nucleosomes and help compact nucleosomes
  198. nucleosome
    • double stranded segment of the DNA strand that is wrapped around a histone protein
    • have globular region and amino terminus 
    • basic bc there is a lot of positive charge bc of lysis and arginine groups 
    • arginine plays a major role in binding to DNA
  199. telomerase
    builds new end to expand the telomere to make sure there are always telomers because the chromosome shortens after replicatin
  200. exonucleases
    3' exonuclease on DNA polymerase, removes mismatched bases
  201. DNA polymerase III
    • synthesizes the leading and laging strands
    • responsible for most of replication
  202. DnaB
    • helicase
    • breaks h bonds of 2 strands 
    • proteins use energy from ATP hyrolysis to help catalyze the seperation
  203. DnaA
    initiates replication when bound to DNA box sequences
  204. lagging strand
    • multiple primers are made
    • replication has breaks because prime need to repeatedly initiate syntheisis of short DNA sequences
  205. leading strand
    a single primer made at the origin of replication
  206. dNTPs
    • deoxyribonucleotide triphphates
    • they enter the catalytic sit and bind to the template strand by A/T G/C rule 
    • covalently attached at the 3' OH end of the growing strand
    • addition requires a primer 
  207. DNA polymerase I
    removes primers and fills vacant regions with DNA
  208. Trascription 3 main steps
    • Initiation: ribosomal subunits, mRNA, and first tRNA assemble to form the complex and then once formed, the ribosome slides over mRNA in 5' to 3' direction
    • Elongation: as ribosome moves, tRNA binds to the mRNA at the A site
    • Termination: stop codon is reach, dissembly occurs and newly made polypeptide is released.
  209. types of RNA
    • mRNA: carries genetic message to ribosomes for protein translation
    • tRNA: brings amino acids to ribosomes for protein translation
    • rRNA: associates with proteins to form ribosomes
    • snRNA: associate with proteins to form splicosome (RNA processing)
    • miRNA/ siRNA: regulation of gene expression
  210. What is the difference between RNA polymerase and DNA polymerase?
    RNA polymerase does not require a primer.
  211. True or False: A gene promoter includes the transcription start site and the start codon.
    FALSE
  212. True or False: The non-template strand in transcription is also known as the coding strand.
    TRUE
  213. template strand
    runs in 3' to 5' direction
  214. non-template strand
    • coding strand
    • carries the genetic information
    • runs in 5' to 3' direction
  215. promoter
    • marks where transcription begins
    • recruits RNA polymerase
  216. Steps for the initiation of transcription in bacteria
    • sigma factor binds to the promoter
    • forms a RNA polymerase haloenzyme
    • helicase activity: forms an open complex and seperates the template and non-template strands 
    • sigmas factors dissociate after open complex formation
  217. RNA polymerase II
    in Eukaryotes transcribes mRNA
  218. 3' poly (A) tail addition
    • 5- 250 A's added to the 3' end of the pre-mRNA transcript 
    • protects from degredation in the cytoplasm
  219. 5' capping
    • 7-methyl guanosine is added to the 5' end of the pre-mRNA transcript
    • protects degredation in cytoplasm
    • helps bind to ribosome
  220. Rho-facter termination
    • bacteria only
    • rho: helicase protein that unwinds nucleic acid strands
    • Rut sequence transcribed to RNA
    • Rho binds to rut and moves towards the polymerase and forms a stem loop
    • Rho unwinds DNA-RNA hybrid and induces it to release polymerase and RNA
  221. What is the difference between bacterial transcription and eukaryotic transcription?
    • Bacteria use sigma factors
    • Eukaryotes use TFs
    • Eukaryotes have multiple RNA polymerases
  222. True or False: A 7-methylgyanosine is added to the 5' end of the pre-mRNA transcript to help recruit it to ribosomes.
    TRUE
  223. True or False: A polyA tail is added to the 3' end of a pre-rRNA transcript as part of the RNA processing
    FALSE
  224. True or False: a pre-tRNA is cleaved before the formation of a mature tRNA.
    TRUE
  225. ribosome
    • site for polypeptide synthesis
    • allows tRNA and mRNA to position correctly as the polypedtide is made
    • p site: peptidyl site
    • A site: amino acyl site
    • E site: exit site
  226. anticodon
    3 nucleotide sequence in tRNA that is complementary to a codon in mRNA
  227. stop codons
    • UAA
    • UAG
    • UUA
  228. Start codon
    AUG
  229. snRNAs
    • bind and bend DNA and cut out introns
    • help direct splicesome
  230. splicing
    • can cause proteins isosforms to arise from one gene.
    • removal of introns from pre-RNAs
    • all exons and introns are transcribed but not all are in the mature RNA 
  231. True or False: The start codon encodes for the amino acid that signals the beginning of translation.
    TRUE
  232. True or False: the stop codon encodes for the amino acid that signals the end of translation.
    FALSE
  233. True or False: the genetic code is degenerate because multiple codons can encode for one amino acid.
    TRUE
  234. True or False: the anticodon on a tRNA is indentical to the codon on the mRNA
    FALSE
  235. True or False: a tRNA binds to an amino acid using its 5' end
    FALSE
  236. True or False: the P site in a ribosome is where the initiator tRNA binds.
    TRUE
  237. Why can translation and transcription occur at the same time in bacteria?
    Translocation and transcription both occur in the cytoplasm
  238. examples of point mutation
    • transition mutation
    • nonsense mutation
    • missense mutation
  239. Silent mutations arise because:
    • Not all codon changes will change the protein.
    • The genetic code is degenerate.
    • Multiple codons can encode for the same amino acid
  240. base excision repair
    • The N-glycosylase enzyme removes a base.
    • AP endonuclease cuts the damaged strand.
    • DNA polymerase I removes old nucleotides

    DNA polymerase III does NOT add new nucleotides
  241. UvrA
    A protein in nucleotide excision repair that recognizes a Thymine dimer.
  242. UvrC
    A protein that cutes damaged DNA strand during nucleotide excision repair.
  243. UvrD
  244. mismatch repair pathway
  245. double strand break model
  246. Holliday model
  247. How can gene expression be regulated?
    • Transcription of a gene.
    • Alternative splicing.
    • Chemical modification of a protein.
  248. Transcription regulation in Eukaryotes.
    • Activators bind to enhancer sequences to increase transcription.
    • Repressors bind to silencers to inhibit transcription
  249. chromatin remodeling
    • Removal of a histone octamer.
    • Chemical modification of histone tails.
    • Change in spacing between nucleosomes

    NOT cutting between nucleosomes
  250. What modification can activate transctiption in eukaryotes?
    Methylation of a histone tail
  251. Gene silencing using dsRNA
    • 1. pre-miRNA or pre-siRNA fors a hair pin base pair to form a double stranded region
    • 2, dicer cuts hair pin
    • 3. RISC binds to dsRNA and will degrade one strand (binds to complementary mRNA)
    • if siRNA bound to mRNA and mRNA degraded: no gene transcription
    • if miRNA no transcription of mRNA: no gene expression, just stays to preven translation
  252. cancer
  253. oncogene
  254. p53 
  255. hierarchy of genes in Drosophila development:
  256. morphogens
  257. What would happen to an embryo that is missing bicoid?
    There would be no anterior end and two posterior ends.
  258. What would happen to a fly that is missing genes in the antp complex?
    The anterior region would be missing.
  259. Population
  260. Assumptions of the Hardy-Weinberg principle:
  261. directional selection
  262. disruptive selection
  263. geneticpoly morphism
    When more than 1 wild-type allele exists in a large population.
  264. transcription
    • DNA to RNA.
    • mRNA only type transcribed into protein
  265. trait
    physical expression of proteins/ genes
  266. gene
    basic unit of inheritance
Author
mkpfister
ID
188496
Card Set
Genetics Final
Description
final
Updated

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