Block 1 Exam Review questions Flashcards
A lengthy human contains 18- protein-coding exons and is predicted to produce a mature mRNA 6500 bases in length. However, when thyroidal C cells are analyzed with specific tools that identify any RNA made from this gene, only trace levels of a 6500 nucleotide non-adenylated RNA are detected within the nucleus. Instead, investigators find 3 mature mRNAs in the cytoplasm of 2800, 3200, and 1800 nucleotides in size. What is the most likely way in which these cytplasmic mRNAs have been generated?
A. Through protein post-translational modification
B. thorugh the process of RNA-splicing to generatre several combinations of joined exons.
C. thorugh preceise partial degradation with cytoplasmic micro RNAs
D. Through chromosomal exclusion
E. Thorugh crossing over in transcription
B. Through the process of RNA-splicing to generate serveral combinations of joined exons.
Yes, 95% of human genes are thought to use alternate RNA-splicing to generate several diverse RNA or protein products from the coding information contained in one gene.
THe human genome is quite complex, and only about 29% of its sequence is given over to RNA- or protein coding genes, intervening sequences and regulatory regions. Among the mysterious sequences that constitute the remainin 70% of the genome, there are several types of notable sequences. What are the SINEs?
A. Short interspersed nuclear elements
B. Short intermingled tandem nuclear elements
C. Short intervining nuclear editcts.
D. Short spaced large interspersed nuclear elements.
E. Randomly placed short inverted sequence elements.
F. Variable number repeats
A. Short interspersed nuclear ELements.
Good choice. you need to be able to tell the short interspersed element from the long interspersed elements. SINEs show up everywhere within the geneome, and are used as genetic markers by human geneticists and genetic caucelours.
What part of the 10 nm fiber is actually cut when treated with Dnase?
A. Only DNA from the accessible heterochromatin
B The highly repeated DNA that forms the centromere
C. The 50-55 bp linker DNA between nucleosomes
D. The 146 bp DNA wraped around the core histones
E. The telomeric DNA at the ends of chromosomes
F. The DNA covered by Histone H1
C. The 50-55 bp linker DNA between nucleosomes.
Yes, only the DNA between nucleosomes can be accessed by the large enzyme. You are cutting the string between the beads.
The diagram below shows the result of 4 dideoxy-type sequencing reactions resulting from a primer (not shown) that bound to a piece of template DNA. From these results, what is the sequence of the strand being made? (The 4 Dideoxy lanes shown below are left-to-right: G A C T)
A. CTACTCGA
B. GTACTCAG
C. TCGAGTAC
D. CATGAGCT
E. AGCTCATG
CATGAGCT
Cytosine deamination is a form of DNA damage that occurs spontaneously. Deamination of methylcytosine is especially mutagenic because the resulting base …
A. 06-methylguanine, base pairs with thymine
B. Hypoxanthine, is an unusual base
C. Thymine, is not recognized by DNA repair enzymes
D. Uracil, is normaly found in RNA
C. Thymine, is not recognized by DNA repair enzymes
Good. Deamination of methylcytosine generates a thymidine nucleotide. These are especially mutagenic because there is no repair mechanism to remove thymidine from the DNA. The only repair mechanism which will function following methylcytosine deamination is mistmatch repair, which will replace the wrong baase 50% of the time.
Meiosis is a central part of the human reproductive process and guarantees some genetic variation between parents and potential siblings. This genetic variation is further enhanced through the process known as “crossing-over.” The crossing-over observed between the arms of homologous chromosomes results in:
A. Homologous genetic recombination
B. Polyclonal genetic diversification
C. Combinatorial genetic segmentation
D. Polycombic genetic segregation
E. Random Genetic assortment.
A. Homologous Genetic Recombination.
Good Choice. Homologous genetic recombination is the only good term that describes this process of exchanging regions of DNA between homologous chromosomes. All the other terms fail to really describe what happens.
Meiosis is a central part of the human reproductive process and guarantees some genetic variation between parents and potential siblings. The process is complex and precisely controlled in order to minimize possible mistakes. What is produced at the end of Telophase I?
A. Four haploid cells containing bivalent chromosomes.
B. Two haploid cells containing monovalent chromosomes.
C. Four diploid cells containing bivalent chromosomes
D. Four haploid cells containing monovalent chromosomes
E. Two diploid cells containing monovalent chromosomes
F. Two haploid cells containing bivalent chromosomes.
G. Two diploid cells containing bivalent chromosomes.
F. Two haploid cells containing bivalent chromosomes.
Good choice. Telophase I is approximately halfway through a process that uses two divisions to produce 4 cells from 1 original cell. Thus, it makes sense that we have 2 cells. In addition, the first division divides one homolog from another, and by definition, that makes these cells haploid. The last part to this involves the fact that the original cell had replicated bivalent chromosomes. After the first division, both resulting cells still have bivalent chromosomes
The following diagram shows two replications forks with the leading and lagging strands being synthesized. In one of the replication forks the leading and lagging strands are being synthesized on the incorrect template. Which of the labels (A to D) indicates the lagging strand on the correct template?
D
Great! The lagging strand is synthesized (as always) in the 5’ to 3’ direction – but this turns out to be in the opposite direction that the replication fork is actually moving (into the parental duplex).
“Molecular hybridization” has major importance in modern DNA biotechnology. Many of the gene-chip or bio-chip machines depend on this basic principle to generate their catalogs of data. What basic feature of double-stranded DNA structure is being exploited when conducting a molecular hybridization assay?
A. the 5’ to 3’ polarity of a nucleic acid.
B. Complementary base-pairing; single stranded DNA will recognize and bind by hydrogen-bond formation to the correct sequence of complementary bases on a different single-strand of DNA
C. The stacking of aromatic bases stabilize the DNA Duplex
D. All DNA polymerases require and Exposed 3’ OH to use as a primer
B. Complementary base-pairing; single stranded DNA will recognize and bind by hydrogen-bond formation to the correct sequence of complementary bases on a different single-strand of DNA
Complementary base-pairing will allow one piece of single-stranded DNA to search out and bind its complementary sequence from among thousands or millions of non-complementary base sequences. The probe finds its target.
The Mars Curiosity Rover has discovered a species of “tiny green people” on Mars. The Rover has been programmed to carry out basic genetic analysis of any species it discovers. It chases down and captures a little green person, homogenizes it in a blender, and then isolates its genomic DNA. Chemical analysis of the DNA reveals that it has a normal duplex structure and contains the four standard DNA bases. This Martian DNA contains 28% deoxycytidine. What would the expected percentage of deoxyadenosine be?
A. 11 %
B. 44 %
C. 22 %
D. 28 %
E. 56 %
C. 22%
That is correct. Chargaffs Rules apply for duplex DNA.
If %C = 28%, since C=G then %G = 28%.
So, C+G=56%
Then A+T=44% and since A=T then the %A=22%
A long gene is prominently expressed in pancreatic beta cells to produce one large and abundant mRNA. The protein that this mRNA should produce would be 650 amino acids long. However, when investigators use very specific tools to identify this protein for study, they are only able to detect trace amounts. Instead, they find 3 abundant proteins of 250, 143, and 97 amino acids in length. What TYPE of modification has produced these 3 unexpected proteins?
A. Chromatin remodeling
B. Post-translational modification
C. Genomic exon shuffling
D. Regulated nuclear RNA - Splicing
E. Regulated cytoplasmic RNA-Splicing
B. Pos-translational modification
Goo thinking: The only term for producing 3 mature proteins from one large precursor protein is “post-tranlational modification” — in this case proteolytic processing.
Each step of the central dogma has coonsiderable complexity. The question specifies that there is a single mRNA, so the different pporteins being produced can not occur due to any process at the DNA or RNA level- instead the different proteins must be generated during or after tranlation.
What property of the core histones (H2A, H2B, H3, and H4) allow them to wrap a DNA duplex around their outer surface (the surface of the core nucleosome structure)?
A. the exact mechanism is unknown - but appears to invoke quantum chemistry
B. the hydrophobic pockets on teh histone tails swallow up the aromatic bases
C. the + charges on the exposed lysines and arginines of their flexible N-terminal regions
D. it is a torsional effect induced by supercoiling the DNA; the duplex spontaneously kinks up around the protein obstacle
E. It is a bit like the spider and the fly: each histone forms up some kind of a crazy clamp that tightens down, irrevocably, onto the hapless DNA duplex
C. The + charges on the exposed lysines and arginines of their flexible N-terminal regions.
Yes, it is a simple + to - attraction that holds the DNA onto the exposed N-termainal tails of the core histones. Block the + charge on the lysines and arginines, and the DNA will only loosely associate with the core particle.
Two sequences are shown below that flank an unknown region of DNA. What are the two primers that will be needed to amplify the unknown sequence?
A. ATATGGGCA & CATCGAGTA
B. TGCCCATAT & GTAGCTCAT
C. ATGAGCTAC & ACGGGTATA
D. TATACCCGT & GTAGCTCAT
C. ATGAGCTAC & ACGGGTATA
Azidothymidine (AZT) is a nucleoside analog used as an inhibitor of reverse transcriptase. It was the first drug approved for the treatment of HIV. Within a cell, AZT is first phosphorylated to a triphosphate, and is then incorporated into the DNA strand being synthesized by the viral reverse transcriptase. Based on its structure shown below, what is the main mechanism by which AZT inhibits reverse transcriptase?
A.The analog triggers base excision repair
B. The lack of a 3’-OH prevents formation of a new phosphodiester bond
C. AZT intercalates with the RNA-DNA hybrid inhibiting reverse transcriptase
D. The analog cannot hydrogen bond with RNA
E. The lack of a 2’-OH prevents it from being recognized by the reverse transcriptase
B. The lack of a 3’-OH prevents formation of a new phosphodiester bond
Good choice. AZT is a thymidine nucleoside analogue with an unusual moiety at the 3’ position. The lack of a 3’OH prevents formation of a new phosphodiester bond. (The lack of a 2’ OH will have no effect - remember reverse transcriptase is a DNA polymerase, so it’s normal substrates are deoxynucleotides.)
Two of the most important DNA-related techniques developed over the past 40 years are Sanger DNA sequencing and the polymerase chain reaction (PCR). What is one basic thing that both PCR and Sanger dideoxy sequencing have in common?
A. They both depend on cellular actins
B. They both depend upon chemical digestions
C. They both depend on chain cleavage
D. They both use Reverse Transcriptase
E. They are both based on primer extention reaction
E. They are both based on primer extension reaction.
Good thinking. Both start with small primers and depend on extension of these primers to produce a signal of some sort. PCR follows routine primer-extension, while dideoxy sequencing depends on disrupting primer extension to gain information.
