molecular bio
Eukaryotic DNA is initially transcribed as hnRNA, then spliced to form mature mRNA. Which of the following is true of RNA splicing?
*It is true of RNA splicing that splicing of hnRNA from a single gene can be variable. During RNA splicing, introns are removed and exons are connected together. This process occurs in the nucleus, and can be variable. Alternative splicing can create different versions of a protein within a single cell, or different versions of the protein in different cell types. Note that this is somewhat of an exception to the "one gene, one protein" rule.
Which of the following is NOT necessary for prokaryotic translation?
*A 80S ribosome is NOT necessary for prokaryotic translation. In prokaryotes, the 30S and 50S subunits of the ribosome must come together for translation to occur but when they do they make a 70S ribosome. Eukaryotes have 80S ribosomes (although they are not necessary for prokaryotic translation). The Shine-Dalgarno sequence is the ribosome binding site in prokaryotes (similar to the Kozak sequence in eukaryotes) and without this the gene product cannot be translated. fMet (formylmethionine) is a modified methionine used as the first amino acid in all prokaryotic proteins. GTP is very similar to ATP but it contains a different purine base (guanine replaces adenine). Prokaryotic transcription uses GTP as its energy source, not ATP.
What are the requirements for a complex transposon?
*A complex transposon contains an IS element (the transposase and its accompanying inverted repeat sequences) and one or more genes. A composite transposon has two IS elements and intervening sequence. While transposons can trigger frameshift mutations, this is not what defines a complex transposon.
Which of the following is the most likely effect of a eukaryotic transcription factor?
*A likely effect of a eukaryotic transcription factor is increase the encounter rate of DNA with RNA polymerase. Transcription factors influence many processes in transcription including recruiting RNA polymerase, regulating transcription rate, and many others. In eukaryotes, transcription and translation are separated physically (transcription occurs in the nucleus and translation in the cytoplasm) so there would be no need for a transcription factor to inhibit translation. DNA does not contain AUG as uracil is only seen in RNA sequences and this sequence signifies the start sequence for translation, not transcription. The energy needed for RNA polymerization is derived from the phosphates that are cleaved (releasing pyrophosphate) from the nucleotides.
A codon is a segment of an mRNA molecule that codes for one amino acid in a polypeptide chain formed during protein synthesis. Which of the following correctly describes the chain of events that occurs in the synthesis of a polypeptide?
*DNA generates mRNA; mRNA moves to the ribosomes, where a tRNA anticodon binds to an mRNA codon, causing amino acids to join together in their appropriate order.
Which of the following would lengthen Okazaki fragments?
*Decreasing the number of primers generated on the lagging strand would lengthen Okazaki fragments during replication. This would result in DNA polymerase III traveling uninterrupted for a longer period of time and generating longer Okazaki fragments. Increasing the number of origins is likely to shorten Okazaki fragments since this would increase the number of primers, and DNA polymerase stops replicating when it runs into the next primer. Stop transcription sites do not affect replication, they stop transcription. Increasing the replication rate will not necessarily change the fragment size but will likely just generate them more quickly.
Which of the following is the most accurate, from least organized to most organized?
*Deoxyribose, nucleoside, nucleotide, DNA helix, nucleosome, chromatin Deoxyribose is the sugar component of the DNA backbone. A nucleoside is the next level of organization and is a deoxyribose with the nitrogenous base attached A nucleotide is more organized than a nucleoside and includes phosphate groups Double stranded DNA winds around histone octamers to form nucleosomes which are further packaged and condensed to form chromatin.
Which of the following is a difference between eukaryotic and prokaryotic translation?
*Eukaryotic and prokaryotic translation differ in the mechanism by which ribosomes recognize the 5' end of mRNA. Prokaryotic mRNA is recognized by the ribosome using the Shine-Dalgarno sequence (-10) while eukaryotic mRNA is recognized via the 5' cap that is added during post-transcriptional modification. Both eukaryotes and prokaryotes begin translation at an AUG; this specifies methionine in eukaryotes and formyl-methionine in prokaryotes. Both utilize the codon UAA to signify a stop codon rather than an amino acid. Both eukaryotes and prokaryotes can translate each mRNA transcript multiple times to generate more protein.
Floroquinolones are a class of antibiotics that inhibit DNA gyrase. Which of the following would be inhibited?
*Fluoroquinolones are a class of antibiotics that inhibit DNA gyrase. Condensing of DNA would be inhibited. Gyrase is an enzyme specific to prokaryotes that maintains bacterial DNA in its supercoiled helical state. Unwinding of DNA is accomplished primarily using helicase and topoisomerase. Okazaki fragments are connected using DNA ligase. Ribosomal separation results when a release factor enters the A site, and peptidyl transferase hydrolyzes the bond between the last tRNA and the completed peptide chain. This occurs after translation, not transcription.
Which of the following enzymes is NOT required for eukaryotic DNA replication?
*Gyrase is not required for eukaryotic DNA replication. Gyrase is used by prokaryotes to supercoil their circular genome. Topoisomerase creates controlled cuts in DNA strands to manage the increased tension created by expanding replication bubbles (topoisomerase is used by eukaryotes and can be eliminated). An RNA polymerase called primase is required to synthesize the primer during DNA replication (RNA polymerase is used by eukaryotes and can be eliminated) Ligase links together the Okazaki fragments of the lagging strands (ligase is used by eukaryotes and can be eliminated).
In terms of ATP, approximately how many glucose molecules would it take to translate a 60 amino acid polypeptide chain in a eukaryote undergoing aerobic respiration?
*In terms of ATP, it would take 8 glucose molecules to translate a 60 amino acid polypeptide chain in a eukaryote undergoing aerobic respiration. Aerobic respiration in a eukaryote results in 30 ATP per glucose. Translation requirements are as follows (note that these are commonly referred to as "ATPs" but many are actually ATP-equivalents, including ATP and other high energy bonds like GTP): 2 ATP per AA-tRNA loading; 1 ATP to form the initiation complex; and 2 ATP per AA joining the peptide chain. Thus, for a 60 AA peptide it would require 239 ATP and ~ 8 molecules of glucose.
The ratio of guanine-cytosine (G—C) pairs to adenine-thymine (A—T) pairs is useful in laboratory manipulation of double-stranded DNA. If a segment of DNA has a low G—C : A—T ratio, it would be reasonable to assume that this segment would:
*Require less energy to separate the two DNA strands than would a comparable segment of DNA having a high G—C : A—T ratio. G—C base pairs are linked by three hydrogen bonds in the double helix, while A—T base pairs are joined by only two hydrogen bonds. It takes more energy to separate G—C base pairs, and the less G—C rich a piece of double-stranded DNA is, the less energy that is required to separate the two strands of the double helix. Note that the statements "contain more guanine than cytosine" and "contain more adenine than thymine" can be eliminated since in double-stranded DNA, there must be equal quantities of G and C (and of A and T).
Alternative splicing permits somatic cells to contain the same genome while maintaining the capability to express widely differing proteins, based on the tissue in which the cell is located. Which experimental technique would be LEAST useful in detecting the differing cellular products created by alternative splicing?
*Southern blotting would be LEAST useful in detecting the differing cellular products created by alternative splicing. Alternative splicing creates different mRNA sequences leading to different proteins, thus any technique that detects changes in mRNA transcripts or final protein products could be useful. Northern blotting is used to detect RNA and both Western blotting and ELISAs can be used to detect proteins. However, Southern blotting is used to detect DNA, and a point made by the question is that the genomes are the same (therefore the least useful technique).
Which of the following is NOT considered to be a DNA repair mechanism?
*Telomerase lengthens the ends of chromosomes where primase is unable to bind is not considered to be a DNA repair mechanism and is the correct answer. This simply allows for more effective replication of the end of the chromosome. Connecting broken ends of chromosomes, reversing UV damage, and removing defective bases are all forms of DNA repair.
It is known that the developing frog embryo requires greater protein production than the adult organism. If cells from a developing frog embryo and from a mature frog were examined, would the investigator find the greater rate of translation in cells of the embryo or of the adult?
*The embryo, because a developing organism requires a higher rate of translation than does an adult. The question states that embryos require greater protein production. Thus, by definition, an embryo requires a greater rate of translation, the process through which the ribosome creates a protein by reading an mRNA transcript.
Which of the following produces a strand of DNA in the 5' to 3' direction? I. Eukaryotic DNA polymerase II. Prokaryotic DNA polymerase III III. Reverse transcriptase
*The polymerases in Items I, II and III produce a strand of DNA in the 5' to 3' direction. All three polymerases listed use a template strand to create a new strand of DNA. They do so by adding new nucleotides to the 3' end of the new strand (hence, the strand grows in the 5' to 3' direction). Reverse transcriptase differs from the DNA polymerases only in that is uses an RNA template instead of a DNA template to synthesis its new strand of DNA. However, it too adds new DNA nucleotides in the 5' to 3' direction.
UV light can trigger the formation of pyrimidine dimers, which then cause malformed loops of DNA. Visible light can then trigger repair enzymes via photoreactivation. This describes a DNA repair mechanism known as:
*This describes a DNA repair mechanism known as direct reversal. Visible light triggers the enzymes involved in direct reversal. Excision repair does not utilize photoreactivation; damaged or defective bases are cut out and the strand is repaired. Homologous recombination is used with double strand breaks as is nonhomologous endjoining.
How do chromosomal translocations end up potentially creating new gene products or enhancing the activity of existing gene products?
*Translocations happen when recombination occurs between non-homologous chromosomes and previously unrelated genes are now placed together. Incomplete recombination between homologous chromosomes would not have the same effect as the genes on homologous chromosomes are essentially the same. Recombination does not occur between the arms of a single chromosome. While recombination between a somatic and sex chromosome could be deleterious, translocations are not limited to such an exchange.
