Additional Questions 2 Flashcards
One of a set of alternative forms of a gene. In a diploid cell, each gene will have two of these, each occupying the same position (locus) on homologous chromosomes.
Allele
Process by which DNA sequence information can be transferred from one DNA helix (which remains unchanged) to another DNA helix whose sequence is altered.
Gene conversion
Experimental process in which two complementary nucleic acid strands form a double helix; a powerful technique for detecting specific nucleotide sequences.
Hybridization
X-shaped structure observed in DNA undergoing recombination, in which the two DNA molecules are held together at the site of crossing-
over, also called a cross-strand exchange.
Holliday junction
T/F: Homologous recombination requires relatively long regions of homologous DNA on both partners in the exchange.
True. Homologous recombination requires long stretches of nearly identical DNA to initiate and complete recombination. This ensures that recombination occurs at corresponding points along the chromosome rather than between similar but non-identical repeated DNA elements found throughout eukaryotic genomes.
T/F: Gene conversion requires a limited amount of DNA synthesis.
True. Conversion refers to a change in the frequency of nucleotide markers during recombination. Initially, each marker is equally present in the starting duplexes. However, after recombination, this balance shifts, resulting in altered frequencies such as 2:0, 4:0, or 3:1 instead of the original 1:1 (or 2:2). This change is primarily driven by mismatch repair and DNA synthesis, both of which involve some degree of DNA replication.
Discuss the following statement: “The Holliday junction contains two
distinct pairs of strands (crossing strands and noncrossing strands),
which cannot be interconverted without breaking the phosphodiester
backbone of at least one strand.”
This statement is incorrect. Crossing and noncrossing pairs of strands can be interconverted by rotation movements that do not require strand breakage.
In addition to correcting DNA mismatches, the mismatch repair system functions to prevent homologous recombination from taking place between similar but not identical sequences. Why would recombination between similar, but nonidentical sequences pose a problem for human cells?
A significant portion of the human genome consists of repetitive elements like Alu sequences, which are dispersed across chromosomes. If recombination were to occur between such sequences on different chromosomes, it could lead to translocations, potentially destabilizing the genome. Unrestricted recombination between these repeats would rapidly rearrange the genome, resulting in nonviable progeny and endangering the species. To prevent this, the mismatch repair system acts as a safeguard. Since repeated sequences vary slightly, recombination between them creates heteroduplex regions with numerous mismatches. If the mismatch repair system detects an excessive number of mismatches, it aborts the recombination process, ensuring that only nearly identical sequences at the same locus undergo successful recombination.
Length of DNA that moves from a donor site to a target site either by cut-and-paste transposition or by replicative transposition.
DNA-only transposon
Enzyme that makes a double-strand DNA copy from a single-strand RNA
template molecule.
Reverse transcriptase
RNA-containing virus that replicates in a cell by first making a double-strand DNA intermediate.
retrovirus
Rearrangement of DNA that depends on the breakage and rejoining of two DNA helices at specific sequences on each DNA molecule.
conservative site-specific recombination
T/F: When transposable elements move around the genome, they rarely integrate into the middle of a gene because gene disruption—a potentially lethal event to the cell and the transposon—is selected against by evolution.
False. Transposable elements integrate almost randomly, often disrupting or altering genes in the process. While some of these integration events are lethal to the cell and the transposable element itself, most are not. By spreading throughout the genome, even at the expense of a few cells, transposable elements ensure their persistence within the species.
Which one of the following functions best describes the role of DNA polymerase I in bacterial cells?
A. Modifies Okazaki fragments for joining into intact strands.
B. Repairs any DNA that is damaged during transcription.
C. Synthesizes most of the cellular DNA during replication.
D. Synthesizes RNA primers to initiate Okazaki fragments.
A. Modifies Okazaki fragments for joining into intact strands.
DNA polymerase I processes Okazaki fragments by removing RNA primers with its 5’-to-3’ exonuclease activity and filling in the resulting gaps with DNA. DNA polymerase III primarily handles genome replication, while DNA primase synthesizes RNA primers. DNA polymerase is not responsible for repairing damaged bases during transcription.
One problem with Taq polymerase is that it incorporates the wrong base
approximately once every 8000 bases, a frequency of mistakes that is far higher than the error rates for DNA polymerases that carry out DNA replication in cells. Which one of the following statements best explains the
high error rate of Taq polymerase in PCR?
A. Taq polymerase cannot make Okazaki fragments efficiently.
B. Taq polymerase lacks a 3’-to-5’ proofreading exonuclease.
C. Taq polymerase often falls off the DNA, interrupting synthesis.
D. Taq polymerase synthesizes DNA in the 3’-to-5’ direction.
A. Taq polymerase cannot make Okazaki fragments efficiently.
Highly accurate DNA polymerases possess 3’-to-5’ proofreading exonuclease activity, allowing them to detect and remove mismatched bases at the growing strand’s 3’ end. Okazaki fragments (Choice A) are not involved in PCR. Premature polymerase dissociation (Choice C) reduces efficiency but does not introduce errors. All DNA polymerases, including Taq polymerase, synthesize DNA in the 5’-to-3’ direction, making Choice D incorrect.
Another problem with Taq polymerase is that it is only capable of synthesizing relatively small pieces of DNA, with an upper limit of around 4000 bases. Which one of the following is a likely explanation for the inability of Taq polymerase to synthesize long pieces of DNA?
A. Taq polymerase can make only relatively short Okazaki fragments.
B. Taq polymerase cannot remove mismatched bases from the 3’ end.
C. Taq polymerase lacks the helicase required for strand separation.
D. Taq polymerase reactions lack the primase needed for new primers.
B. Taq polymerase cannot remove mismatched bases from the 3’ end.
All DNA polymerases prefer a matched base at the 3’ end of the growing strand. Without 3’-to-5’ proofreading exonuclease activity, Taq polymerase stalls at a mismatch, limiting fragment length. Choice A is incorrect as Okazaki fragments are not involved in PCR. Choice C is incorrect because PCR does not require helicase activity. Choice D is incorrect since primers are included in the PCR mix from the start, not synthesized during the reaction.
Helps to position the RNA polymerase correctly at the promoter, to aid in pulling apart the two strands of DNA to allow transcription to begin, and to release RNA polymerase from the promoter into the elongation mode once transcription has begun.
General transcription factor
Small RNA molecules that are complexed with proteins to form the ribonucleoprotein particles involved in RNA splicing.
snRNA (small nuclear RNA)
Nucleotide sequence in DNA to which RNA polymerase binds to begin transcription.
Promotor
A large protein complex containing multiple 3’-to-5’ RNA exonucleases that degrade improperly processed mRNAs, introns, and other RNA debris retained in the nucleus.
Exosome
The enzyme that carries out transcription.
RNA polymerase
RNA molecule that specifies the amino acid sequence of a protein.
mRNA (messenger RNA)
Process in which intron sequences are excised from RNA transcripts in the nucleus during the formation of messenger and other RNAs.
RNA splicing
Signal in bacterial DNA that halts transcription.
Terminator
