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2/8/19 A. Falsification: Since, the pH of the solution decreases, that means the H+ concentration increases. If the OH- concentration increases that means the pH solution of the solution should be getting larger and solution should be getting more basic. B. Best answer: The pH of the solution decreases, that means the solution is becoming more acidic due to the increase of H+ concentration. Also, since the pH is less than 7, this implies that the solution is acidic. To add on, the H+ concentration increased 10 times  because the pH of a solution is equal the negative log of the H+ concentration so for a pH + -7 + of 7 the concentration of H is 10 and when the pH decreases to 6 the H concentration will be 10-6  which is 10 times larger than 10-7  .  C. Falsification: The pH is determined by the negative log of the H+ concentration so + -7 when the pH of the solution is 7 the H concentration is 10 and when the pH is 6 the H+ concentration is 10-6  which means the concentration increased 10 times. Also, since the  pH decreases that means the solution is getting more acidic, so the H+ concentration has to increase not decrease.

 D. Falsification: The pH is decreasing which means the H+ concentration increases. Also, + - + pH involves the concentrations of H ions and OH ions. The OH ion is not even involved.

2/15/18 Which of the following organelles is not part of the endomembrane system? A. secretory vesicle B. lysosome C. nuclear envelope D. centrosome A. Falsification: The endomembrane system is responsible for a numerous number of tasks including the transportation of proteins to the membrane or to organelles. The secretory vesicle is part of this system because it acts as a transporter of neurotransmitters or hormones and then fuses with the cell membrane which allows proteins to be released. B. Falsification: The lysosome is also part of the endomembrane system because it aids with detoxification which is also another task the system carries out. Lysosomes digest materials in the cell using enzymes and macrophages, which are a type of white blood cell, use lysosomes to help digest bacteria or viruses which aids in detoxification. C. Falsification: The nuclear envelope is also part of the endomembrane system because it helps with the transportation of proteins which is a function of the system. The nuclear envelope is a double membrane and has a structure called the pore complex which regulates proteins. D. Best Answer: The centrosome is not a part of the endomembrane system because it is where microtubules are grown from. Microtubules add support and structure to the cell and has nothing to do with any of the functions of the endomembrane system.

2/27/19 Which of the following statements about the cytoskeleton is incorrect? A. The dynamic aspect of cytoskeletal function is made possible by the assembly and disassembly of a few simple types of proteins into large aggregates. B. Microfilaments are structurally rigid and resist compression, while microtubules resist tension (stretching). C. Movement of cilia and flagella is the result of motor proteins causing microtubules to move relative to each other D. Transport vesicles among the membranes of the endomembrane system depend on the function of the cytoskeleton.

A: Falsification- This statement is correct because the cytoskeleton is responsible for types of cell motility which allow it to assemble and disassemble in different parts of the cell. This is due largely because of the components of the cytoskeleton. Microtubules, which is a component of the cytoskeleton, are made up of tubulin subunits, a globular protein. Microtubules grow by adding tubulin subunits, and so they can disassemble and grow in another location just by adding tubulin proteins. B: Best Answer: This statement is incorrect because microfilaments are tension resisting while microtubules are compression resisting. The microtubules that are compression resisting grow from a centrosome. The centrosome has centrioles which are made up of triplet microtubules in a ring position. Microtubules are also structurally rigid because one of their functions is to serve as pathways for organelles to move along. Microfilaments are tension resisting due to their three dimensional structure inside the plasma membrane. This arrangement of microfilaments help reinforce the shape of the cell. C: Falsification- This statement is correct because a motor protein named dynein aid with the movement of cilia and flagella. Cilia and flagella contain as the textbook mentioned, nine doublets of microtubules structured in a ring with two microtubules in the middle. The dynein proteins are attached to the microtubules on the outer side of the ring arrangement and using ATP, dynein move along the microtubules. D: Falsification- This statement is correct because transport vesicles and vesicles that move from parts of the cell to other parts of the cell and the cytoskeleton aids with that. Microtubules in the cytoskeleton provide tracks for the movement of organelles or transport of material. This aids transport vesicles by helping them move along the cell. Campbell Biology 10th edition. Jane B. Reece et. al., Ch6 Pg112-117, 2014

3/5/19 ________ catalyzes the production of ________, which then opens an ion channel that releases ________ into the cell's cytoplasm. A. Adenylyl cyclase; IP3; Ca2+ B. Protein kinase; PIP2; Na+ C. Phospholipase C; cyclic AMP; Ca2+ D. Phospholipase C; IP3; Ca2+

A: Falsification: Adenylyl cyclase is in the plasma membrane and is responsible for converting ATP into cyclic AMP. It converts ATP to cAMP after receiving a signal from outside the cell. B: Falsification: Protein kinase is involved in the phosphorylation cascade. It moves phosphate groups from ATP to other active protein kinases in the cascade until the last active protein kinase phosphorylates a protein. C: Falsification: Phospholipase C is not responsible for the production of cAMP. Adenylyl cyclase catalyzes the production of cAMP, so we can conclude this answer to be incorrect. D: Best Answer: Phospholipase C catalyzes the production of IP3 by splitting the phospholipid, PIP2 into inositol triphosphate (IP3) and diacylglycerol (DAG). Both IP3 and DAG function as second messengers, but pertaining to the answer choice, after IP3 is produced, it attaches to the IP3 - gated calcium channel, resulting in the channel opening and Ca2+ ions coming out of the channel. Campbell Biology 10th edition. Jane B. Reece et. al., Ch11 Pg219-222, 2014

Hello Jonas, I agree with you in the case that cyclic AMP does not bind to calmodulin. Ca2+  ions bind to calmodulin and this in result cause the ions to contract. Calmodulin activates the phosphorylation of the myosin head which activates cross-bridge activity.

3/28/19 The leading and the lagging strands differ in that A. replication of the lagging strand begins after replication of the leading strand is completed. B. the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction. C. the Okazaki fragments on the lagging strand are much shorter than the Okazaki fragments on the leading strand D. the leading strand is synthesized by adding nucleotides to the 3’ end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5’ end.

A. Falsification: This statement is false because the synthesization of the leading and lagging strand start the same time. The leading strand is synthesized by DNA polymerase and it is continuous, while the lagging strand is a series of segments of DNA molecules called Okazaki fragments. B. Best Answer: This statement is true because the structure of DNA is a double helix and the two strands of DNA in the helix are antiparallel. The leading strand is synthesized by DNA polymerase but since DNA polymerase adds to the 3’ end only on the leading strand, it moves in the same direction as the replication fork and the leading strand is replicated in the same direction as the replication fork. But on the lagging strand, DNA polymerase synthesizes in the opposite direction of the replication fork. C. Falsification: This statement is false because there are no Okazaki fragments on the leading strand, the leading strand is continuous. There are only Okazaki fragments in the lagging strand. D. Falsification: This statement is false because DNA is synthesized from 5’ to 3’ only. DNA polymerase can only add nucleotides to the 3’ end. Their antiparallel structure contributes nothing to what end a strand is synthesized. Works Cited: Lecture 15 DNA Replication - Professor Neta Dean

4/2/19 What is TRUE about cells in G1 phase versus cells in M phase of mitosis? A. Cells only contain chromosomes during M phase B. A cell at the start of M phase has twice as much DNA as a cell in G1 phase C. Cells in G1 phase will never divide D. DNA is contained within the nuclear envelope in both phases

A: Falsification: This statement is false because chromosomes are apparent in phases other than M phase. A justification for this is the whole function of S phase. During S phase, DNA replication happens which involves the replication of chromosomes, therefore the cell contains chromosomes in S phase. Also, during G1 and G2 phases, the cell grows. B: Falsification: This statement is false because a cell at the start of G2 phase has twice as much DNA as a cell in G1 phase. This is because the phases of the cell cycle go as G1 to S phase to G2, and during S phase, DNA replication occurs. The S phase replicates the chromosomes and so two sister chromatids are formed hence why there are twice as much DNA at G2 as opposed to G1 phase. C: Falsification: This statement is false because as referenced in the textbook, and experiment conducted in the University of Colorado showed that when a cell in the M phase and a cell in the G1 phase was fused, the G1 nucleus started mitosis which is the process for the division of the cell. It was seen that a spindle was created and chromosomes were condensed even though in G1 phase the chromosomes were not replicated. This is known as cell cycle control system which basically states that certain molecules are key in the occurrences of certain events in the cell cycle. D: Best Answer: This statement is true because DNA is contained in the nuclear envelope is not apparent in M phase. During G2 phase, the nuclear envelope encloses the nucleus and then this goes onto M phase. During prophase, the nuclear envelope is still prominent even as the spindle starts to form and then once the microtubules break into the nuclear space, the nuclear envelope starts to break. But, DNA is contained in the nuclear envelope during M phase, same for G1 phase. DNA is contained inside the nucleus during G1 phase. Campbell Biology 10th edition. Jane B. Reece et. al., Ch12 Pg235-243, 2014

4/12/19 What does not happen during prophase I of meiosis? A. Synapsis B. Pairing of homologous chromosomes. C. Formation of chiasmata. D. All of the above do happen then.

A: Falsification: Synapsis does occur during prophase I of meiosis. After the DNA of nonsister chromatids break at points and the chromatin start to pack together and form the synaptonemal complex which is when each homolog is slowly starting to join to each other. After the complex is formed, the breaks in the DNA of the homologs start to close up because the ends of the DNA start to attach with the corresponding part of the non sister chromatid which make what are called crossovers. B: Falsification: The pairing of homologous chromosomes do occur during prophase I of meiosis. During interphase, the chromosomes are duplicated, and during the early part of prophase I, homologous pairs start to come together. Genes on one homolog pair align with a reciprocal gene on the other homolog pair. C: Falsification: The formation of chiasmata also occur during prophase I. As mentioned before, during synapsis, when the breaks in DNA start closing up by attaching the broken ends with a corresponding part of a non sister chromatid, they form crossovers. The points where the crossovers happen become chiasmata later on. They are seen as chiasmata after the synaptonemal complex is are dismantled, when the homolog pairs detach and move away from each other, the chiasmata is apparent. D: Best Answer: Since all of the choices mentioned above do occur, then the answer choice is this one, all of them do happen. Campbell Biology 10th edition. Jane B. Reece et. al., Ch13 Pg260, 2014

4/24/19

 utants that couldn’t grow without added arginine. Beadle and Tatum studied two Neurospora m What important general conclusion, that advanced our understanding of heredity, came from their work? A. Citrulline is used to make arginine B. The role of one gene is to code for one enzyme C. DNA is the genetic material D. DNA is a double helix A: Falsification: This answer is not correct because through Beadle and Tatum’s experimentation, the one gene-one enzyme hypothesis was proven. But through Srb and Horowitz studies of different types of arginine-requiring mutants, they were able to discover that the each mutant required different supplements along the path of producing arginine. More so, if a certain type of this arginine-requiring mutant didn’t have the certain enzyme that catalyzes a step in producing arginine, the mutant would not be able to grow. Through this experimentation it was found that ornithine is used to produced citrulline which then produces arginine. B: Best Answer: This answer is correct because from Beadle and Tatum’s research, Archibald Garrod’s hypothesis of one gene-one enzyme was supported. Garrod first hypothesized that each gene shows their phenotype from enzymes catalyzing chemical reactions. Later on Beadle and Tatum conducted experiments on Neurospora which further supported that statement. They had wild type Neurospora which required minimal food requirements and then they had mutant Neurospora which were modified with X-rays and had genetic changes. These mutants needed more food requirements because they weren’t able to make crucial molecules needed from the minimal food like the wild type was. The mutants were then given more supplements such as amino acids and nutrients. They added specific nutrients to vials and saw how the mutant grew from the vial. In example, they saw that a mutant grew from the supplement containing the minimal food and just arginine, and so then it was concluded that the mutant lacked a biochemical pathway that could synthesize arginine. Each mutant was flawed in one gene and that correlated to the hypothesis that each gene controls the production of a specific enzyme. C: Falsification: This answer is not the correct choice because Beadle and Tatum did not come to the conclusion that DNA is the genetic material from their studies. The conclusion of DNA being the genetic material was from the Hershey-Chase experiment. In that experiment they used radioactive sulfur and phosphorus to trace whether the protein or DNA of the T2 phages that they used would be able to enter the infected host cell. They used sulfur and phosphorus because sulfur was present in the protein of the T2 phages and phosphorous was present in the T2 phages’ DNA and so respectively they used each to track the DNA and protein. And at the end, they found that DNA entered the host cell and infected the cell and even so, the infection

continued through the bacteria and it was seen that the bacteria produced phages that had radioactive phosphorus. This proved DNA to be the genetic material. D: Falsification: This answer is not the correct because the discovery of DNA being a double helix was due to Watson and Crick. Watson and Crick used X-Ray diffraction images that were produced by Rosalind Franklin to come to this conclusion. X-ray crystallography produces images with smudges that depict deflected X-rays from passing through DNA. From these photographs, Watson recognized that the deflections were that of what helix shaped molecules had. The photo also show two strands of DNA, and therefore after further models being made, it was concluded that DNA was a double helix. Campbell Biology 10th edition. Jane B. Reece et. al., Ch 16 Pg313-317, Ch17 Pg334-335 2014

5/1/19 Chapter 17: Replication & Transcription

Which of the following is NOT a similarity between replication and transcription? A. Both processes occur with the same fidelity. B. Polymerization in both processes is based on reading a template. C. A pyrophosphate is removed from every nucleotide as polymerization occurs. D. Both processes occur in the 5 to 3 direction. A: Best answer: This is not a similarity between replication and transcription. DNA replication is very fast and accurate because DNA polymerase, a major enzyme that is involved in synthesizing the new DNA strand from the parental DNA, has a “proofreading” attribute. It has an exonuclease “proofreading” activity that goes from the 3’ to 5’ and it can basically pinpoint incorrectly matched bases, or switch directions and other things. This prevents abundant mutations from occurring. There is no such thing during transcription. B: Falsification: This statement is true. During DNA replication, the new strand of DNA is synthesized by using the parental DNA as a template. The enzyme DNA helicase breaks the hydrogen bonds between the two DNA strands and separates the double helix. Afterwards, DNA polymerase synthesizes a leading and a lagging strand from the two parental DNA strands. It synthesizes DNA by adding complementary nucleotides that are based off of the parental strands. The complementary nucleotides join to make the phosphate backbones of the new DNA strands. This same concept applies to transcription. During transcription, only one DNA is used as a parental or template strand and RNA is synthesized from that one strand. RNA polymerase is an important enzyme in transcription, it attaches to the promoter and transcription starts at that point. Then RNA is synthesized based on adding complementary nucleotides and base pairing but instead of thymine pairing with adenine, uracil pairs with adenine. C: Falsification: This state is true because during DNA replication, DNA polymerase hydrolyzes two phosphate groups and that removes pyrophosphates. This creates energy which is then used to form a phosphodiester bond between a free 3’ OH group of a growing DNA strand and the 5’ end of an incoming dNTP. During transcription, RNA polymerase stops transcription in prokaryotes and in eukaryotes, RNA polymerase continues past the terminator sequence and the pre-mRNA is cut, and the pre-mRNA contains the poly(A) tail which aids with hydrolysis. D: Falsification: This statement is true because in DNA replication, DNA polymerase synthesizes from 5’ to 3’ because it can only add to the 3’OH end. This is the same for RNA

polymerase, it only synthesized from 5’ to 3’ because it can only add nucleotides to the 3’ end of the polymer. Works Cited: Neta Dean BIO 202 Lectures #15 and #23...