MCB EXAM 1 QUESTIONS WITH CORRECT MARKING SCHEME

MCB EXAM 1 QUESTIONS WITH CORRECT MARKING SCHEME /.The # of chromosomes in human somatic cells during G1 (right after mitosis) is? - Answer-- 2n - 23 chromosomes /.This is not one of the 4 essential types of macromolecules that function as building blocks of the eukaryotic cell... - Answer-Macromolecules: -Carbohydrates - Lipids -Proteins -Amino Acids /."In general, chromatin can be viewed as being either of two forms, as ________ or as __________" - Answer--Euchromatin: decondensed -Heterochromatin: condensed (two types: constitutive: always condensed facultative: depends on the cell and phase of the cycle) /.The avg. diameter of the nucleus in nucleated mammalian cells is approx.? - Answer-- 6um (micrometers) /.All of the flexible "snake-like" shapes that extrude from the center of the structure depicted have at their end an "X", that "X" indicates? - Answer-N-terminals /.The name of the structure represented in the figure above is the ____ and the colored components depicted in the middle structure represent __________. - Answer--Chromatosome -Histones core (H2A H2B H3 H4) /.The major structural proteins in human chromatin are known as? - Answer--Histones /.During the stage of the cell cycle chromosomes reach their maximum level of condensation.... - Answer-- During Metaphase (4th order- mini bands- heterochromatin) /."Histones share several properties , including their _______ size , their overall _________ charge, and the presence of a large number of _____________ and in their _______________." - Answer-- small - positive /.Highly condensed chromatin that is transcriptionally inactive is called.... - Answer-- Heterochromatin /."If you were to make a synthetic form of DNA in which the phosphate group is substituted by a chemical group exhibiting a net positive charge, the resulting synthetic DNA" - Answer-would not bind to nucleosomes because DNA has to be negative and because the nucleosome is negative. /.Chromosomes lacking a functional centrosome will.... - Answer-Can't transition into Telophase- in the cell cycle, Anaphase is before so the chromosome is stuck there. /.Further coiling of the basic beads-on-a-string chromatin fiber generates a thicker fiber whose thickness is about..... - Answer-- 1st order- "beads on a string" 10 nm fiber (1st order 2nd order- chromatin fiber- 30 nm fiber- can be in solenoid or zigzag form 3rd order- Looped form 4t order- mini bands (in heterochromatin and in condensed chromosomes during metaphase) /."Telomeres are sequences located __________, and they are characterized by containing _____________ that _______________" - Answer--located: at ends of linear chromosomes -containing: clusters of G residues on one strand -that: forms loops at the end of chromosomes and binds a protein complex that protects the chromosome termini from degradation. /.Nucleosomes consist of a ____________ long stretch of ____________ associated to ____________ of each of the ____________. - Answer--consist: 200bp -of: DNA -to: 2 copies -the: core histones /.(electrophoresis activity 2) The lane that corresponds to human chromatin before being digested with Micrococcal nuclease is: - Answer-o /.(electrophoresis activity 2) The lane that corresponds to human chromatin digested with Micrococcal Nuclease for a relatively short amount of time (about 30 minutes) is: - Answer-o /.(electrophoresis activity 2) The lane that corresponds to human chromatin digested with an inactive Micrococcal Nuclease (one that has been rendered inactive by treatment with an inhibitor of its nuclease activity) is: - Answer-o /.(electrophoresis activity 2) The lane that corresponds to human chromatin digested with Micrococcal Nuclease for an intermediate amount of (about 2 hrs) is: - Answer-o /.(electrophoresis activity 2) The lane that corresponds to human chromatin treated with TRYPSIN (a protease capable of degrading all proteins in the samples) for several hrs (enough to allow the degradation of all the proteins present), and subsequently treated with a Trypsin-inhibitor AND Micrococcal Nuclease for a long time (2 hrs, long enough to allow almost full digestion of all the DNA treated) is: - Answer-o /.Introns can be defined as.... - Answer-- Noncoding sequence that interrupts exons in a gene /."For the avg. human gene, __________ account for most of their length" - Answer-- Repetitive sequences?? /.________________ constitute by far the most abundant type of DNA sequences found in the human genome. - Answer-- Repetitive Sequences or LINEs?? (Long Interspersed Elements they are members f a family of highly repeated retrotransposons in mammalian genomes) /.______________ carry within their sequence a region coding for reverse transcriptase. - Answer--Retrotransposons (mediated by reverse transcription whereas DNA transposons move the genome by being copied and reinserted as DNA sequences, rather than moving by reverse transcription) /.The avg. human gene contains about 10% ______ and 90% ________ sequences. - Answer-10%: Exons 90%: Introns /.The large differences observed in genomic size among eukaryotes correspond primarily to differences in.... - Answer-- The amount of noncoding DNA /.From an evolutionary point of view _________________ could be considered to be both the most successful and most damaging sequences present in our genome. - Answer-- Retrotransposons /.lncRNAs could be best defined as? - Answer-- RNA transcripts longer than 200 nucleotides that do not code for protein