What’s Up With Cell Bio Exam 4

Telomerase is an enzyme that plays a crucial role in the replication of the ends of eukaryotic chromosomes. It helps to maintain the length and stability of telomeres, which are protective caps at the ends of chromosomes. Telomeres shorten with each round of cell division, and telomerase helps to counteract this shortening by adding repetitive DNA sequences to the ends of chromosomes. This enzyme is particularly important in cells with high rates of division, such as stem cells and cancer cells, as it allows them to continue dividing indefinitely. Therefore, the statement that telomerase is involved in the replication of the ends of eukaryotic chromosomes is true.

DNA polymerases actually synthesize new DNA by adding nucleotides onto the 3' hydroxyl group of the growing DNA chain. This is because DNA polymerases catalyze the formation of a phosphodiester bond between the 3' hydroxyl group of the last nucleotide in the growing chain and the 5' phosphate group of the incoming nucleotide. Therefore, the statement "DNA polymerases always synthesize new DNA by adding nucleotides onto the 5' phosphate" is incorrect.

If a specific sigma subunit were mutated, the result would be that RNA polymerase would fail to initiate transcription at the promoter specific to the sigma subunit. The sigma subunit is responsible for recognizing and binding to specific promoter sequences, allowing the RNA polymerase to initiate transcription at the correct sites. Therefore, if the sigma subunit is mutated, it would impair the ability of RNA polymerase to initiate transcription at the promoter specific to that sigma subunit.

Okazaki fragments are short, newly synthesized DNA fragments that are formed on the lagging strand during DNA replication. The lagging strand is synthesized in short segments because it is oriented in the opposite direction of the replication fork. These fragments are then joined together by DNA ligase to form a continuous strand. Therefore, the statement "they are formed in the lagging strand" accurately describes Okazaki fragments.

The correct answer is "double-stranded, antiparallel, (A+T)/(C+G)=variable, (A+G)/(C+T)=1.0". This accurately describes the structure of DNA in both prokaryotes and eukaryotes. DNA is double-stranded, with the two strands running in opposite directions (antiparallel). The ratio of adenine (A) to thymine (T) and cytosine (C) to guanine (G) can vary (variable), while the ratio of adenine (A) to guanine (G) and cytosine (C) to thymine (T) is always 1.0.

The correct answer is that DNA polymerase III requires a free 3'-OH group. This is because DNA polymerase III can only add nucleotides to the 3' end of an existing DNA strand. An RNA primer provides this free 3'-OH group for DNA polymerase III to start synthesizing DNA. The RNA primer is later replaced with DNA by DNA polymerase I.

Alternative splicing is a process in which different combinations of exons are selected and joined together during mRNA processing, resulting in the production of multiple proteins from a single primary transcript. This process allows for the generation of protein diversity by creating different mRNA isoforms with varying protein coding sequences. Replication is not involved in protein synthesis, alternative editing is not a known mechanism for generating multiple proteins, and 5' methylation is a modification of the mRNA cap structure and does not directly contribute to the production of multiple proteins.

Covalent bonds are responsible for the polymer backbone of DNA because they involve the sharing of electrons between the phosphate and sugar molecules in the backbone. Hydrogen bonds primarily account for complementary base pairing in DNA because they form between the nitrogenous bases (adenine, thymine, cytosine, and guanine) and hold the two strands of DNA together.

The RNA polymerase enzyme is said to be processive because it stays attached to DNA over long stretches of template to make prodigiously long mRNAs while it remains loose enough to move freely along the template. This means that the enzyme can efficiently synthesize long mRNA molecules without constantly dissociating from the DNA template. The processivity of RNA polymerase is crucial for the accurate and efficient transcription of genetic information.

The correct answer is that the base composition in this part of the gene should be AT-rich since the smaller number of H bonds in that region of the gene will allow the strands to separate more easily. This is because AT base pairs have only two hydrogen bonds, while GC base pairs have three hydrogen bonds. Therefore, having more AT base pairs in this region will result in weaker hydrogen bonding and easier separation of the DNA strands, allowing for gene expression at the appropriate time.

During RNA processing, introns are sections of RNA that are removed. Introns are non-coding regions of RNA that do not contain information for protein synthesis. They are transcribed from DNA but are not involved in the final protein product. Instead, they are removed through a process called splicing, where only the exons, which contain coding sequences, are joined together to form the mature RNA molecule. Therefore, the correct answer is "RNA that is removed during RNA processing."

DNA ligase is not involved in the initiation of DNA replication at the origin of replication in E. coli. Instead, DNA ligase plays a role in the final step of DNA replication, which is the joining of Okazaki fragments on the lagging strand. The proteins and enzymes that function at the origin of replication in E. coli include DnaA, dnaB, and DnaC proteins, which are involved in the initiation and unwinding of the DNA helix, and helicase, which unwinds the DNA strands. SSBs (single-stranded binding proteins) also play a role in stabilizing the unwound DNA strands during replication.

The correct answer is the addition of a 7-mG cap at the 5' end of the transcript and the addition of a poly-A sequence at the 3' end of the message. This is because in eukaryotes, the mRNA undergoes modifications after transcription. The addition of a 7-mG cap at the 5' end helps in mRNA stability, initiation of translation, and protection against degradation. The addition of a poly-A sequence at the 3' end aids in mRNA stability, transport, and translation efficiency. These modifications are important for proper processing and functioning of mRNA in eukaryotic cells.

The alpha subunit of RNA polymerase is responsible for establishing template binding to a promoter in prokaryotes. This subunit plays a crucial role in recognizing and binding to specific DNA sequences within the promoter region, allowing for the initiation of transcription. It helps in stabilizing the interaction between RNA polymerase and the DNA template, ensuring accurate and efficient transcription of the genetic information.

The correct answer is "DNA is composed of a sugar-phosphate backbone with bases projecting towards the inside of the back bone." This statement accurately describes the structure of a DNA molecule. The sugar-phosphate backbone forms the outer structure of the DNA molecule, while the bases (adenine, thymine, cytosine, and guanine) project towards the inside, forming the rungs of the DNA ladder. This arrangement allows for the complementary base pairing between the bases, which is essential for DNA replication and protein synthesis.

The portions of DNA that are located before the initiation site towards the 3' end of the template are referred to as "upstream" of that site. This term is used to describe the directionality of DNA and indicates that these portions are located in the opposite direction of the transcription process, which occurs towards the 5' end.

If 15% of the nitrogenous bases in a DNA sample from a particular organism is thymine, it means that the remaining 85% must be made up of the other three nitrogenous bases: adenine, cytosine, and guanine. Since DNA is a double-stranded molecule, the percentages of adenine and thymine are always equal, and the percentages of cytosine and guanine are also always equal. Therefore, if thymine is 15%, cytosine must also be 15%. This means that the percentage of cytosine should be 35%.

The origin of replication is not necessary for RNA polymerase to recognize the promoter of a bacterial gene. The origin of replication is a DNA sequence that is required for DNA replication to initiate, but it is not involved in the process of transcription, which is the synthesis of RNA from a DNA template. RNA polymerase recognizes the promoter region of a gene through the sigma factor, which helps in binding to the -10 and -35 consensus sequences. These consensus sequences are specific DNA sequences that help in positioning the RNA polymerase at the start site of transcription.

The events of prokaryotic transcription occur in a specific order. First, the RNA polymerase binds to the promoter region of the DNA, which is called promoter binding. Then, the DNA strands separate, forming a transcription bubble. This is known as the transcription bubble. Next, the RNA polymerase moves along the DNA strand, synthesizing the RNA molecule in a process called elongation. After elongation, the rho factor, or rho protein, binds to the RNA molecule and causes termination of transcription. This is known as rho binding. Therefore, the correct order of events is promoter binding, transcription bubble, elongation, rho binding, and termination.

DNA polymerase III is mainly responsible for genome replication in E. coli. This enzyme is highly processive and has a proofreading function, allowing it to accurately replicate the entire genome. DNA polymerase III is the primary replicative polymerase in E. coli and is responsible for synthesizing the leading and lagging strands during replication. DNA polymerase I and II have other roles in DNA repair and replication, but DNA polymerase III is the main enzyme involved in genome replication.

The primer in DNA replication is composed of RNA. RNA primers are synthesized by the enzyme primase and serve as a starting point for DNA synthesis. They provide a free 3'-OH group that DNA polymerase can attach to and begin elongating the new DNA strand. Once the RNA primer is laid down, it is later replaced by DNA during the process of replication.

The initiation of replication at an ARS is directly triggered by the binding of other proteins to the ORC-origin complex. This binding event signals the start of replication and allows for the recruitment of additional factors that are necessary for the replication process to proceed. The departure of the ARS, the departure of the ORC from the ARS, the binding of the ORC, and the changing of the ARS into an ORC are not directly involved in triggering replication initiation at an ARS.

The binding of TBP to a polymerase II promoter causes a dramatic distortion in the conformation of DNA, resulting in its insertion into the minor groove of the double helix. This binding also causes a bend of greater than 80 degrees at the site of DNA-protein interaction. Therefore, the correct answer is a, b, and d.

Microsatellite DNAs are known to cause trouble for DNA replicating enzymes. These sequences of DNA change in length through generations and have variable lengths within the population. They have been used to analyze relationships between different ethnic human populations.

If Avery, MacLead, and McCarty had determined that the transforming molecule was a protein, it would not have been observed that heat-killed cultures treated with protease would transform the R cells. This is because protease breaks down proteins, so if the transforming molecule was a protein, it would have been inactivated by the protease treatment and would not be able to transform the R cells.

Introns are non-coding regions of DNA that are transcribed into pre-mRNA but are removed during the process of RNA splicing. This splicing removes the introns and joins together the exons, which are the coding regions of the gene. Therefore, the termination codons present in introns are eliminated during this splicing process and do not interrupt the coding of a particular protein.

The mature mRNA is the final product of transcription and contains only the coding regions that will be translated into protein. These coding regions are called exons. In contrast, the non-coding regions that are removed from the mRNA during processing are called introns. Therefore, the transcribed regions present in the mature mRNA would arise from exons, while the transcribed regions not present in the mature mRNA would arise from introns.

The melting temperature (Tm) of DNA refers to the temperature at which half of the DNA strands in a sample are denatured or melted. The Tm is influenced by various factors, including the length and base composition of the DNA strands. DNA with a low AT content would have a higher GC content, and GC base pairs are held together by three hydrogen bonds compared to the two hydrogen bonds in AT base pairs. The additional hydrogen bonds in GC base pairs make them more stable, resulting in a higher Tm. Therefore, DNA with a low AT content would require a higher temperature to denature and would melt more slowly.

After two rounds of DNA replication, Meselson and Stahl observed two bands in their centrifugation tubes. The top band consisted of double strands of 14N DNA, indicating that the original 14N DNA had been replicated. The middle band consisted of strands of 15N DNA base-paired to strands of 14N DNA, suggesting that the replication process had resulted in a mixture of 14N and 15N DNA. Therefore, the correct answer is option 2.

Gyrase is the enzyme in E. coli that is responsible for relieving the tension ahead of the fork that occurs when the DNA is unwound to form the replication bubble or eye. Gyrase accomplishes this by introducing negative supercoils into the DNA, which helps to relax the tension and prevent the DNA strands from becoming tangled during replication.

After one generation time, if replication were conservative, the DNA would consist of two strands - one heavy and one light. This is because in conservative replication, the original DNA molecule remains intact and serves as a template for the synthesis of a new complementary strand. Therefore, half of the DNA would be made up of two "light" strands (synthesized in the new medium) and the other half would be made up of two "heavy" strands (inherited from the original DNA).

The interactions responsible for holding the nitrogenous bases together stacked one on top of another in the center of an RNA molecule and contributing to the stability of the whole DNA molecule are hydrophobic interactions and van der Waals interactions. These interactions occur between the nonpolar regions of the bases and help to maintain the structure and stability of the DNA molecule. Hydrophilic interactions and hydrogen bonds also play a role in DNA stability, but they are not specifically mentioned in the question as contributing to the stacking of the bases.

The One Gene-One Polypeptide hypothesis had to be modified because it was found that genes can be spliced differently to generate a variety of related polypeptides. This means that a single gene can code for multiple polypeptides by undergoing alternative splicing, where different sections of the gene are included or excluded during the transcription process. This discovery challenged the idea that each gene only codes for one polypeptide, leading to the modification of the hypothesis.

RNA polymerase II is responsible for transcribing protein-coding genes into mRNA in eukaryotes. This enzyme is responsible for transcribing the DNA template strand into a complementary RNA molecule during the process of transcription. RNA polymerase II is highly specific to protein-coding genes and is regulated by various transcription factors and enhancer elements to ensure accurate and efficient transcription of these genes.

Minisatellite DNAs are a type of DNA sequences that range from about 10 to 100 base pairs. They are found in clusters containing as many as 3000 repeats. These sequences are known to be unstable, with the number of copies of a particular sequence often increasing or decreasing from one generation to the next. Therefore, the correct answer is minisatellite DNAs.

The correct answer is "sRNAs" and "microRNAs". Both sRNAs and microRNAs are small RNAs that are produced by plants and animals. They are synthesized only at certain times during development or in certain tissues and are believed to have a regulatory role. Therefore, options "b and c" are correct.

The percentage of GC in a DNA sequence affects the Tm (melting temperature) of the sequence because the higher the amount of GC, the higher the Tm. This is because GC base pairs have three hydrogen bonds, while AT base pairs have only two hydrogen bonds. Therefore, a higher percentage of GC in a sequence means more hydrogen bonds, resulting in a higher Tm. Additionally, GC-rich sequences will have a higher Tm because they have less hydrogen bonds.

Reverse transcriptase is the enzyme that makes DNA from an RNA template. This enzyme is unique because it can synthesize DNA using an RNA template, which is the opposite of the normal process of DNA replication. Reverse transcriptase is commonly found in retroviruses, such as HIV, and is used to convert the viral RNA genome into DNA, allowing it to integrate into the host cell's genome. This enzyme is also used in molecular biology techniques, such as reverse transcription polymerase chain reaction (RT-PCR), to convert RNA into DNA for further analysis.

The end replication problem involves telomerase and an RNA template to synthesize new DNA. Telomerase is an enzyme that adds repetitive nucleotide sequences to the ends of chromosomes, which helps to prevent the loss of genetic information during DNA replication. The RNA template is used by telomerase to extend the DNA strand. Therefore, both telomerase and an RNA template are involved in solving the end replication problem.

The correct answer is moderately repetitive DNA, SINEs, LINEs, and VNTRs. Moderately repetitive DNA refers to sequences that are repeated a moderate number of times throughout the genome. SINEs (Short Interspersed Nuclear Elements) and LINEs (Long Interspersed Nuclear Elements) are types of transposable elements that can move around in the genome. VNTRs (Variable Number Tandem Repeats) are sequences where a short DNA motif is repeated in tandem and the number of repeats varies between individuals. These sequences play important roles in genome organization and function.

The new antibiotic is able to bind the rho (p) factor associated with some transcription processes. The rho (p) factor is involved in the termination of transcription in some cases. Therefore, when the antibiotic binds to the rho (p) factor, it would interfere with the termination of transcription in those specific cases, leading to the termination of some transcription.

TFIIH is the only general transcription factor known to have enzymatic activity. TFIIH is involved in the process of transcription initiation and is responsible for unwinding the DNA double helix, allowing RNA polymerase to access the DNA template and initiate transcription. TFIIH also has helicase activity, which helps in separating the DNA strands, and kinase activity, which phosphorylates the C-terminal domain of RNA polymerase II, facilitating the transition from initiation to elongation. This enzymatic activity of TFIIH is crucial for the proper regulation and control of transcription.

DNA polymerases do possess 5'---->3' exonuclease activity. This activity allows them to remove incorrect nucleotides from the growing DNA strand during replication.

The attachment of the sigma-factor increases the core enzyme's affinity for DNA promoter sites. This means that when the sigma-factor is attached to the core enzyme, it enhances the enzyme's ability to bind to DNA promoter regions, which are important for initiating transcription. On the other hand, the sigma-factor decreases the core enzyme's affinity for DNA promoter general. This suggests that when the sigma-factor is present, the core enzyme is less likely to bind to non-specific DNA sequences that are not promoter sites. Therefore, both statements b and d are correct.