96 terms

Genetics Test 3

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Map Based Genome sequence
better approach for repetitive sequences, starts with large libraries of large overlapping DNA sequences, requires sublocing of large fragments into smaller clones for sequencing, uses genetic recombination data to help arragance sequences correctly, requires chromosome fragments to overlap for contig assembly, requires sequences to be annotated after contig assembly
whole genome shotgun sequencing
starts with cloning and sequencing of short, random DNA fragments, requires chromosome fragments to overlap for contig assembly, requires sequences to be annotated after contig assembly
Which enzyme(s) will produce a DNA fragment that contains the entire vgp gene (shown in red) and has "sticky ends"?
Hind3, BamHI
The products of restriction digestion can be visualized by gel electrophoresis, which separates fragments based on their size
true
-Restriction digestion produces fragments of DNA, and the sizes of these fragments can be determined by gel electrophoresis using standard DNA fragments of known size.
A 1.5‑kb fragment of DNA is cloned into a plasmid vector that is 5.5 kb long at the EcoRI site, and the plasmid vector is then used to transform bacteria. If the plasmid DNA is then extracted from a single bacterial colony and digested with EcoRI, what digestion products will be produced if the plasmid contains the fragment?
One 1.5‑kb fragment and one 5.5‑kb fragment
-EcoRI digestion will produce two fragments corresponding in size to the 1.5‑kb fragment cloned into the plasmid plus the 5.5‑kb plasmid itself
Digestion of a 1.1‑kb fragment of DNA with BamHI produces two fragments of 700 bp and 400 bp. Digestion of the same 1.1‑kb fragment with XhoI produces two fragments of 300 bp and 800 bp. Digestion with both enzymes produces three fragments of 100 bp, 300 bp, and 700 bp. Which of the following statements is true about the DNA fragment?
The XhoI site is located within the BamHI 400‑bp fragment.
-Since double digestion produces a 100‑bp fragment and a 300‑bp fragment, the XhoI site must be located within the BamHI 400‑bp fragment.
The role of the primers in PCR is
to define the target region and provide a 3' end that can be extended by taq polymerase
-Correct. Primers bind to end of the target DNA strands, then taq polymerase synthesizes a new strand using the target DNA as a template
If there are five molecules of DNA containing the target region at the beginning of a PCR reaction, how many copies of the target will be present after three rounds of amplification?
40
-The number of target sequences is doubled with each replication cycle
Which of the following statements about ddNTPs is true?
hey have a free 3′‑hydroxyl group on the sugar.
They have a hydrogen at the 3′ carbon of the sugar.
DNA polymerase can add a new dNTP to a 3′ ddNTP.
They have an oxygen at the 2′ carbon of the sugar.
They have a hydrogen at the 3′ carbon of the sugar.
-ddNTPs terminate synthesis because there is no 3′‑hydroxyl group onto which DNA polymerase can add nucleotides.
DNA fragments that are 600 bp long will migrate more quickly through a sequencing gel than fragments that are 150 bp long
false
-Small DNA fragments have less hindrance in moving through the gel, so they migrate more quickly than larger fragments.
Immediately after the primers have annealed to the target sequence
the temperature is raised so that taq polymerase can extend the primers
-The temperature is raised to 70-75∘C, the temperature over which taq polymerase is optimally active
Which of the following statements about manual Sanger sequencing is true?
The DNA sequence obtained is complementary to the template strand.
-The DNA fragments produced in sequencing reactions are synthesized by DNA polymerase to be complementary to the template strand
The human insulin gene contains a number of sequences that are removed in the processing of the mRNA transcript. In spite of the fact that bacterial cells cannot excise these sequences from mRNA transcripts, explain how a gene like this can be cloned into a bacterial cell and produce insulin.
Plasmids containing insulin genes derived from cDNA, rather than genomic DNA, would be void of introns. Therefore, intron processing would not be necessary.
genomics library ony
contains non-coding DNA, likely to contain entire genome
cDNA libraries only
dependent on expression level of genes, originate from mRNA, dependent on tissue type, rarely contain introns
both genomics and cDNA libraries
cloned into vectors
Examine the data carefully and choose the best conclusion.
Offspring B and C are not the products of these parents and were probably purchased on the illegal market. The data are consistent with offspring A being legitimate.
Rank the steps involved in screening a genomic library from first to last.
grow genomics library colonies and transfer membranes, lyse colonies and denature DNA, hybridize desired labeled probe, identify targeted library colonies
Genomic libraries are often screened using probes derived from organisms that are different than the one from which the library was generated. Therefore, there are often mismatches between the probe and the most similar matching clones in the library. As a result, it's possible that no matching clones may initially be identified by hybridization.
How can these potential problems be overcome or minimized?
Alter the hybridization conditions such that hybridizations with a few mismatches are tolerated.
Create smaller probes using regions of your DNA of interest that are thought not to be highly conserved.
Create smaller probes using regions of your DNA of interest that are thought to be more highly conserved.
Use a cDNA for your probe since all of the introns have been removed.
-Alter the hybridization conditions such that hybridizations with a few mismatches are tolerated.
-Create smaller probes using regions of your DNA of interest that are thought to be more highly conserved.
To estimate the number of cleavage sites in a particular piece of DNA with a known size, you can apply the formula, N/4 n where N is the number of base pairs in the target DNA and n is the number of bases in the recognition sequence of the restriction enzyme. If the recognition sequence for AluI is AGCT and the Nanoarchaeum equitans genome contains approximately 490,000 bp, how many cleavage sites would you expect?
1914
In a typical PCR reaction, describe what is happening in stages occurring at temperature ranges-90-95
-Heating to 90-95 ∘C denatures the double-stranded DNA so that it dissociates into single strands.
In a typical PCR reaction, describe what is happening in stages occurring at temperature ranges:50-70
Lowering the temperature to 50-70 ∘C allows the primers to bind to the denatured DNA.
In a typical PCR reaction, describe what is happening in stages occurring at temperature range-70-75
Bringing the temperature to 70-75 ∘C allows the heat stable DNA polymerase an opportunity to extend the primers by adding nucleotides to the 3'ends of each growing strand
We usually think of enzymes as being most active at around 37ºC, yet in PCR the DNA polymerase is subjected to multiple exposures of relatively high temperatures and seems to function appropriately at 70-75ºC.What is special about the DNA polymerizing enzymes typically used in PCR?
These polymerases were isolated from bacteria growing in hot springs and on thermal vents, lending to their ability to function at extreme temperatures.
What is the %GC content?
5'-TTGGATGTCAATACAAAATTACCGAC-3'
35%
What is the annealing temperature for this oligonucleotide?
5'-TTGGATGTCAATACAAAATTACCGAC-3'
Use your answer from Part A (%GC content) and the length of the oligonucleotide primer (N) to compute the melting temperature (Tm).
64.9 degrees celcius
hich statement best explains the relationship between the Tm (ºC) and the GC content of the oligonucleotide?
As the GC content increases, the Tm. increases
Tautomers of nucleotide bases are isomers that differ from each other in the location of one hydrogen atom in the molecule.
true
Which nucleotide will base‑pair with the enol form of 5‑bromouracil?
guanine
Tautomeric shift
temporary change in base structure, change in double bond placement
A reversible isomerization in a molecule, brought about by a shift in the location of the hydrogen atom. In nucleic acids, this in the bases of nucleotides can cause changes in other bases are replication and are a source of mutations
In terms of its involvement in mutagenesis, 5BU is best described as
a base analog that can cause either A-T > G-C or G-C > A-T transitions
-In its common form, 5BU can pair with adenine and in its rare form it can pair with guanine
For 5BU to cause a transition mutation, which of the following must occur
DNA with incorporated 5BU must replicate
- 5BU must undergo a form change, but that is not sufficient to cause transition. The form change must be followed by replication.
In its rare form, 5BU pairs with guanine.
true
-In its more common form, 5BU will pair with adenine. In its rare form, it pairs with guanine
Which enzyme is responsible for proofreading during replication?
DNA polymerase
-DNA polymerase performs proofreading functions during replication using its 3′ to 5′ exonuclease capability.
Bacteria can distinguish between a newly replicated DNA strand and the original template strand because the newly replicated strand is methylated, whereas the original template strand is not.
false
-Methylation occurs shortly after replication, so the original template strand is methylated and the newly replicated DNA strand is not.
Which repair system uses the RecA and LexA proteins?
SOS repair
-RecA and LexA are active during the SOS response
The purpose of the Ames Test is to
test the mutagenic effects of chemicals
In the Ames Test, the appearance of his+ revertants in the presence of a non-mutagenic control compound indicates that
some of the reversion mutations are not caused by the mutagen being tested
- His+ revertants on the control plate are the result of spontaneous mutation
Many chemicals are more mutagenic after being processed in the liver.
true
What genotype will the F1 male flies have? (Note that a "+" indicates a wild-type chromosome without the transposase locus or the P {miniwhite +} insertion. The X /Y chromosomes are listed first, followed by the second chromosome.)
XP{miniwhite+}/ Y; transposase / +
Why does the researcher select these F2 males? Select all statements that are true.
In this cross, most F2 males will have no eye pigment because they cannot inherit the X chromosome that had the original P {miniwhite +} insertion on it.
In this cross, F2 males with eye pigment result from the P-element moving to a new location on an autosome
which repair process in E. coli uses visible light to repair thymine dimers?
photoreactivation repair
-Photoreactivation utilizes the enzyme photolyase to break the bonds formed during pyrimidine dimerization. A thymine dimer produced by UV irradiation is bound by photolyase. Visible light energy is absorbed by photolyase and is redirected to break the bonds forming the dimer.

In addition to bacteria, this DNA repair mechanism is also found in single-celled eukaryotes, plants, and some animals such as Drosophila.
Which repair process(es) use(s) a DNA polymerase
nucleotide excision repair
base excision repair
homologous recombination (synthesis-dependent strand annealing)
translesion DNA synthesis
non-homologous end joining
repairs double-strand breaks in a polymerase independent manner by using nucleases to trim back the DNA surrounding the break and having DNA ligase seal the break
-Many DNA repair processes (base excision, nucleotide excision, translesion DNA synthesis) use a DNA polymerase when a template strand is available to direct synthesis/repair of the other strand.

Even in cases where there is no such template strand (e.g., a double-strand break), homologous recombination can use the identical sister chromatid like a template for a DNA polymerase to fill in the double-strand break gaps.
Which two repair processes are the most error prone?
base excision repair
translesion DNA synthesis
nonhomologous end joining
nucleotide excision repair
homologous recombination (synthesis-dependent strand annealing
translesion DNA synthesis
nonhomologous end joining
Suppose that a transient tautomeric shift occurred in the guanine base to produce a rare tautomer in the partial DNA sequence just prior to a round of DNA replication.
Which base would be added opposite this rare tautomer during DNA replication?
T
Suppose that the top strand is the coding (nontemplate) strand and the three bases shown represent a single in frame codon in a gene.
What will be the effect of the tautomeric shift-induced mutation on the amino acid sequence?
nonsense mutation
-The codon changed from CAA to TAA as a result of a tautomeric shift-induced mutation. This point mutation results in a change from a glutamine codon to a stop codon, causing a nonsense mutation.
Cancer cells contain altered DNA methylation patterns. There is much less DNA methylation in cancer cells than in normal cells. At the same time, the promoters of some genes are hypermethylated in cancer cells. These changes are thought to result in stimulation of transcription of the bulk of genes that would be silent in normal cells, while at the same time repression transcription of genes that would regulate normal cellular functions such as DNA repair and cell-cycle control.
Histone modifications are disrupted in cancer cells. Genes that encode histone acetylases, deacetylases, methyltransferases, and demethylases are often mutated or aberrantly expressed in cancer cells. The large numbers of epigenetic abnormalities in tumors have prompted some scientists to speculate that there may be more epigenetic defects in cancer cells than there are gene mutations. In addition, because epigenetic modifications are reversible, it may be possible to treat cancers using epigenetic-based therapies.
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Kinases regulate other proteins by adding phosphate groups. Cyclins bind to them, switching them on and off. CDK4 binds to cyclin D, moving cells from G1 to S. At the G2/mitosis border, a CDK1 combines with another cyclin (cyclin B). Phosphorylation occurs, bringing about a series of changes in the nuclear membrane via caldesmon, cytoskeleton, and histone H1.
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What is the difference between saying that cancer is inherited and saying that the predisposition to cancer is inherited?
Inheritance conveys the assumption that when a particular genetic circumstance is present, a particular trait will be revealed in the phenotype.
An inherited predisposition usually refers to situations in which a particular phenotype is expressed in families in some consistent pattern and may manifest itself in different ways.
A genetic variant of the retinoblastoma protein, called PSM-RB (phosphorylation site mutated RB), is not able to be phosphorylated by the action of CDK4/cyclinD1 complex. PSM-RB is said to have a constitutive growth-suppressing action on the cell cycle.Which of the following will be true in a cell with PSM-RB?
pRB will never be phosphorylated.
pRB will always be bound to E2F transcription factor.
The cell will be stuck in G1 phase and never progress to S phase.
How does the Ras protein transmit a signal from outside the cell into the cytoplasm?
Activated Ras proteins transduce a signal, which activates the transcription of genes that start cell division
What happens in cases where the ras gene is mutated?
It continually signals cell division
Explain why some viruses that cause cancer contain genes whose products interact with tumor-suppressor proteins.
Stimulating cell division in host cells confers an evolutionary advantage.
To inactivate tumor suppressor genes, so the normal breaking mechanism of the cell cycle is destroyed.
n which ways could radiotherapy control or cure cancer?
Radiotherapy is often administered externally or internally to kill cancer cells by damaging their DNA.
Why does radiotherapy often have significant side effects?
n addition to killing cancer cells, radiotherapy can also kill normal cells too, causing side effects.
When activated, the p53 protein initiates cell-cycle arrest followed by DNA repair and apoptosis. This process involves stimulating transcription of p21, which inhibits the CDK4/cyclin D1 complex. Activated p53 also prevents the cell from initiating the S phase. Through a series of steps, BAX homodimers are formed that lead to apoptosis. Individuals who are heterozygous for p53 need only have a mutation in the existing normal gene to lose the protection of the p53 protein.
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Examine the types of mutations that are listed in the table. Is the BRCA1 gene likely to be a tumor-suppressor gene or an oncogene?
tumor-suppressor gene
Which of the following could explain why some women who have the mutation live into their 80s without developing cancer?
They may not have been exposed to cancer causing chemicals.
Their immune systems are more efficient at destroying cancerous cells.
They have other mutations that promote tumor suppression.
What is meant by a neutral polymorphism
a difference in DNA sequence that has no physiological effect
What is the significance of this table in the context of examining a family or population for BRCA1 mutations that predispose an individual to cancer?
Such neutral polymorphisms make screening difficult in that one cannot always be certain that a mutation will cause problems for the patient.
Is the PM2 polymorphism likely to result in a neutral missense mutation or a silent mutation?
silent mutation
Is the PM3 polymorphism likely to result in a neutral missense mutation or a silent mutation?
neutral missense mutation
Based on the information given in the three different sections, which of the following statement(s) correctly describe(s) the hit that is most similar to the query sequence? Select all that apply.
It is a chimeric protein.
It is identical to the query sequence in length and amino acid sequence.
Its accession number is CAM33009.1.
Which type of chromosomal rearrangement accounts for the creation of the gene that encodes the BCR-ABL chimeric protein? (For a review of the four different types of chromosomal rearrangements, see Hint 1.)
translocation
When an entire gene is duplicated in a genome, there are three possible outcome
The gene becomes a pseudogene, like in sentence 1 where the gene lost a start codon, or like in sentence 5 where some of the members of a gene family do not encode functional proteins.
Subfunctionalization results, and both of the duplicated genes are required for normal function, like in sentence 2 where both the alpha and beta chains are required to make functional hemoglobin, or sentence 4 where both transcription factors paxA and paxB are required for normal function.
Neofunctionalization results, and one of the duplicated genes acquires a new function, like in sentence 3 where the duplicated gene acquired the ability to break down a chemical in a pesticide.
1. The beta-globin gene has sequence similarity to the other genes in the beta globin family, but it lacks a start codon. pseudogene
2. Hemoglobin consists of an alpha chain and a beta chain, each produced by a different but similar gene. Neither gene alone can make a functional hemoglobin molecule. subfunctionalization
3. A duplicated gene in mosquitos accumulated mutations that allow it to break down a chemical found in a common pesticide. neofunctionalization
4. In most vertebrates, the pax transcription factor acts during development in the eye and brain. In zebra fish, each tissue has a different but similar transcription factor (paxA and paxB) and both transcription factors are necessary for normal development. subfunctionalization
5. The olfactory receptor gene family in humans contains 16 genes. Six of these genes contain mutations such that the genes do not encode functioning proteins. pseudogene
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ndicate which terms correctly describe each example by dragging a check mark to the appropriate columns. More than one term may apply to a single example. If a term does not apply to a particular example, drag an X to that column.
Homologs are genes that arose from a common ancestor. Paralogs are genes that arose from a gene duplication event in the same species--for example, the alpha and beta hemoglobin subunits in humans. Orthologs are genes with the same common ancestor found in different species--for example, the alpha hemoglobin genes in mice and humans.
Identify the steps involved in the microarray procedure.
1. Isolate mRNA from both the normal cells and the cancer cells.
2. Use the enzyme reverse transcriptase and fluorescently labeled nucleotides to make cDNA from the starting material isolated in Step 1.
3. Denature the fluorescently labeled molecule created in Step 2, and incubate it with the microarray. The fluorescently labeled molecules from the two different cell types will hybridize to single-stranded DNA on the microarray.
4. Wash and then scan the microarray to measure the fluorescence at each spot on the array.
cancer cell cDNAs were labeled with red fluorescence and normal cell cDNAs were labeled with green fluorescence
A red spot represents a gene with increased expression in the cancer cell because more of the red-labeled cDNAs are present on the microarray.
A green spot represents a gene with decreased expression in the cancer cell because more of the green-labeled normal cell cDNAs are present on the microarray.
A yellow spot represents a gene that is expressed equally in both cell types.
A black spot represents a gene that is not expressed in either cell type.
How large is the human genome?
3.1 bill nucleotides
What percentage of the human genome codes for protein?
2%
Approximately what percentage of the human genome is composed of repetitive sequences?
50%
Which of the following statements about the human genome is not true?
The genome sequence is ~99.9% similar in individuals of all nationalities.
Many human genes produce more than one protein through alternative splicing.
Human genes are larger and contain more and larger introns than genes in the genomes of invertebrates, such as Drosophila.
The functions of more than 90% of human genes are now known.
The functions of more than 90% of human genes are now known.
Approximately how many protein-coding genes does the human genome contain?
20000
The Human Genome Project has demonstrated that in humans of all races and nationalities approximately 99.9 percent of the sequence is the same, yet different individuals can be identified by DNA fingerprinting techniques.
A few repetitive regions of the genome tend to have highly variable repeat lengths that are used in DNA fingerprinting. The majority of the genome has no sequence changes or only single-nucleotide polymorphisms.
An RFLP marker that cosegegrates with a disorder exists in most people who have the disorder and is absent in most people who don't have the disorder.
true
-If an RFLP and a disease‑related gene are located near each other on the same chromosome, they will show linkage and segregate together.
A portion of the β‑globin allele responsible for sickle‑cell anemia contains the sequence CCTGTGGAG, whereas the same region of the normal β‑globin allele contains the sequence CCTGAGGAG. Which allele would be cut by the restriction enzyme DdeI, which cuts at the sequence CTNAG, where N is any base?
Three fragments of 1.3, 1.1, and 0.2 k
-The restriction pattern of sickle‑cell carriers would show three bands; the 1.3 kb band corresponds to the sickle‑cell allele, whereas the 1.1 and 0.2 kb bands correspond to the normal allele
There are an unlimited number of VNTR alleles for each locus, such that every individual has unique alleles at each VNTR locus.
false
-here are many VNTR alleles for each locus but not an unlimited number. People who are closely related are more likely to have similar DNA fingerprints.
Why are multiple VNTR probes used in DNA fingerprinting?
They increase the probability of producing a DNA fingerprint that is unique to an individual.
-The use of a large number of probes increases the chances that the DNA fingerprint produced will be unique.
If four different VNTR alleles have frequencies of 1 in 10, 1 in 20, 1 in 50, and 1 in 500 in a population of 250 million, how many people would have all four alleles?
50
-The frequency of occurrence for the four alleles is 1 in (10 × 20 × 50 × 500) = 1 in 5,000,000, and the number of people in the population who have this pattern is 250,000,000/5,000,000 = 50.
What experimental evidence confirms that we have introduced a useful gene into a transgenic organism and that it performs as we anticipate?
RT-PCR
direct assays of the gene product
Southern blotting
PCR
RFLP
DNA sequencing
How can we use DNA analysis to determine that a human fetus has sickle-cell anemia?
RFLP
PCR
allele-specific oligonucleotide detection
southern blot
How can DNA microarray analysis be used to identify specific genes that are being expressed in a specific tissue?
Microarrays can be used as platforms on which to hybridize DNA or RNA from various tissues.
How are GWAS carried out, and what information do they provide?
enome-wide association studies involve scanning the genomes of thousands of unrelated individuals with a particular disease and comparing them with the genomes of individuals who do not have the disease. GWAS attempt to identify genes that influence disease risk.
What are some of the technical reasons why gene therapy is difficult to carry out effectively?
-Delivering therapeutic genes to the appropriate target cells and producing sufficient levels of the gene product to correct the mutant phenotype remain significant challenges
-A suitable vector must be selected and properly engineered to carry and deliver the desired product on an appropriate schedule.
-The vector must be stable yet not trigger an immune response.
How did the team from the J. Craig Venter Institute create a synthetic genome?
-The team compared a number of genomes each with a small number of genes and identified 256 genes that may represent the minimum number of genes for life.
-he team synthesized short DNA segments and assembled them into a synthetic genome.
-The team used transposon-based mutations to determine the number of genes essential for life.
The gene therapy technique currently used in clinical trials involves the "addition" to somatic cells of a normal copy of a gene. In other words, a normal copy of the gene is inserted into the genome of the mutant somatic cell, but the mutated copy of the gene is not removed or replaced.
Will this strategy work for either of the two aforementioned types of dominant mutations?
yes, but only for loss-of-function mutation
f your genome was sequenced, what negative consequences would be?
here are a lot of consequences associated with a publicly available genome sequence: employment bias, personal relationships, etc.
You may get as a result variations that may be alarming, but have no consequences. Such false positives may have a negative impact on an individual
What would be positive consequences of sequencing your genome?
You can be informed about possible genetic diseases.
If the frequency of the M allele in the human MN blood group system is 0.65 in a population at equilibrium, then the frequency of the N allele must be 0.04.
false
-The sum of the allele frequencies must equal 1, so the frequency of the N allele must be 0.35.
f a recessive disease is found in 50 out of 100,000 individuals, what is the frequency of the heterozygote carriers for this disease?
If q 2 = 0.0005, then q = 0.022 and p = 1 − q = 0.978. The heterozygote frequency is 2pq, or 2 (0.978) (0.022) = 0.043.
In a population of birds in Africa, it was observed that birds with small or large beaks could efficiently crack and eat small or large seeds, respectively. Birds with intermediate beaks had trouble with both types of seeds. What type of selection would be expected to occur in this population if small and large seeds were the only types of food available to these birds?
disruptive
-Birds with an intermediate beak phenotype are at a disadvantage in this population and will be selected against because they are less fit
natural selection
result of differential survival and reproduction. It is the only mechanism of evolution that consistently causes a population to become better adapted to its environment.
genetic drift
escribes evolution due to chance events and causes unpredictable fluctuations in allele frequencies. Genetic drift can have a particularly significant effect in a small population, such as during a bottleneck or founder event.
gene flow
results when alleles are transferred into or out of a population due to the migration of fertile individuals or their gametes. Gene flow can bring new alleles (beneficial, harmful, or neutral) into a population.
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