DNA technology and Clinical Applications of Genome Editing Technology MCQs

Genes for antibiotic resistance that permit the selective survival of engineered bacteria. Genes for antibiotic resistance enable the researcher to easily identify bacteria that contain cloned DNA fragments. Only bacteria that have been transformed with a vector containing the antibiotic resistance gene will grow on the antibiotic-containing medium.

Question 10 1 / 1 pts What is/are the most important "selectable markers" used in recombinant DNA technology when using bacteria? Bacteriophages that carry a foreign DNA. Self-transmissible plasmid, such as the F factor. Single-stranded cloning vectors that are used for DNA sequencing. Genes for antibiotic resistance that permit the selective survival of engineered bacteria. Genes for integrase enzymes that are needed for the integration of recombinant DNA in the bacterial genome.

mRNA and reverse transcriptase The fully mature mRNA contains only exons. No introns or extragenic sequences are present in the mRNA. The mRNA is copied using a poly T primer and reverse transcriptase.

Question 11 1 / 1 pts Human genomic DNA is a mixture of introns, exons and extragenic sequences. To create a library that will contain exclusive information about the coding portion of a particular genome, one needs to eliminate the introns and extragenic sequences that compose the bulk of the human genome. What is (are) the most important components needed to create such a library? DNA from human cells DNA ligase and plasmids mRNA extracted from human cells Reverse transcriptase enzyme mRNA and reverse transcriptase

A pair of 15‐20 bp long labeled ASO (allele-specific oligonucleotide) probes. ASO (allele-specific oligonucleotide) probes that are 15-20 bp long can distinguish between two different single nucleotide polymorphisms (SNPs). Longer probes cannot distinguish between the two SNPs.

Question 12 1 / 1 pts The presence of sickle cell trait (a single base change) in homozygous or heterozygous conditions in different individuals in a population can be best identified by use of which of the following techniques? A pair of 15‐20 bp long labeled ASO (allele-specific oligonucleotide) probes. Labeled 1 kb hybridization probes. A pair of RFLP probes made from a cDNA. 100‐150 bp long ASO (allele-specific oligonucleotide) probes in pair. Long double stranded probes with fluorescent labels.

Allele-specific oligonucleotide probes Two allele-specific oligonucleotide (ASO) probes are usually identical to each other except for one nucleotide that is different in the "middle" of the nucleotide sequence. By adjusting the temperature and salt concentration, an investigator can create the condition where each probe will hybridize only with its identical sequence and not the sequence that is one nucleotide different. The ASO probes can be used to rapidly determine which individual has the "normal" allele and which has the "mutant" allele.

Question 13 1 / 1 pts For identification of affected and carrier traits of sickle cell anemia, DNA isolated from erythrocytes of 3 different individuals was amplified separately by PCR, and the products were blotted as separate spots on 2 strips of nitrocellulose membrane (dot blot). Following alkali treatment, one membrane was incubated with probe A and the other with probe S. Hybridization of the probes with the membrane-affixed DNA followed by autoradiography indicated that individual 1 is normal, 3 is a heterozygous carrier and 2 is affected by sickle cell anemia. What kind of probes are A and S? cDNA probes Long single-stranded DNA probes RFLP probes Allele-specific oligonucleotide probes Long double-stranded DNA probes

The sequence to which it hybridizes must be complementary to that of the probe. The probe and the sequence to which it hybridizes must be complementary to each other and can be DNA or RNA. Usually, the probe is DNA since DNA is more stable than RNA, but a DNA probe can hybridize with RNA as well as DNA. Introns are composed of nucleotides and one can design a probe that will hybridize to intronic sequences.

Question 14 1 / 1 pts For a nucleic acid probe to hybridize, which of the following must be true? The sequence to which it hybridizes must be identical to that of the probe. Both the probe and the sequence to which it hybridizes must be DNA. Neither the probe nor the sequence to which it hybridizes can contain introns. The sequence to which it hybridizes must be complementary to that of the probe. Both the probe and the sequence to which it hybridizes must be RNA.

5'-32P-GGAACCTTTT-3' - The probe shown is the reverse complement to the sequence shown in the stem of this question.

Question 15 1 / 1 pts Which of the radiolabelled oligonucleotide probes below would best hybridize with the DNA sequence 5'-AAAAGGTTCC-3'? 5'-32P-GGAACCTTTT-3' 5'-32P-TTTTCCAAGG-3' 5'-32P-CCTTGGAAAA-3' 5'-32P-AAAAGGTTCC-3' 5'-32P-UUUUCCAAGG-3'

normal; affected; heterozygote; normal Individual 1 has two alleles on the homologous chromosomes that contain the normal gene. Individual 2 has two alleles on homologous chromosomes that contain the cystic fibrosis gene mutation. Individual 3 has one chromosome that is "normal" and one chromosome that contains the mutated allele. Individual 4 is similar to individual 1.

Question 16 1 / 1 pts In the following dot blot, the probe used in the upper row is for the normal sequence while the probe in the second row is for the ΔF508 mutation of the cystic fibrosis. What are the genotypes of the four people investigated? Dot blot of polymorphism.jpg affected; affected; heterozygote; normal normal; heterozygote; affected; normal affected; normal; heterozygote; affected heterozygote; affected; heterozygote; heterozygote normal; affected; heterozygote; normal

5' ACGTCAGTTG 3' The correct partner in a duplex must be anti-parallel and complimentary.

Question 17 1 / 1 pts Which of the following oligonucleotides forms the most stable duplex with the DNA fragment 5'-CAACTGACGT-3'? 5' GTTGACTGCA 3' 5' ACGTCAGTTG 3' 5' TGCAGTCAAC 3' 5' ACGTGAGTTG 3' 5' GTTGATTGCA 3'

Detects RNA molecules A Northern blot is where RNA is electrophoretically separated in a gel matrix and then probed with a known nucleotide sequence. The probe can be RNA or DNA. (Usually, DNA probes are used because DNA is more stable than RNA). Southern blots use DNA that is separated on a gel. Western blots use proteins that are separated on a gel and then probed with antibodies...not DNA or RNA probes.

Question 18 1 / 1 pts Which of the following results is provided by a Northern blot analysis? Determines chromosome structure Detects proteins Detects RNA molecules Detects specific base pairs Detects DNA molecules

Detects proteins A Western blot is where protein is electrophoretically separated in a polyacrylamide matrix, transferred to a nitrocellulose or nylon filter, and then hybridized with antibodies, that have been produced in mice or rabbits, which recognize the protein of interest. Southern blots use DNA electrophoresed and then probed with a DNA or RNA probe. Northern blots use RNA electrophoresed and then probed with a DNA or RNA probe.

Question 19 1 / 1 pts Which of the following results is provided by a Western blot analysis? Detects specific base pairs Detects DNA molecules Detects RNA molecules Detects proteins Determines chromosome structure

3 If a linear piece of DNA is cut one time, you get 2 fragments of DNA. If a linear piece of DNA is cut two times, you get 3 fragments of DNA. If a linear piece of DNA is cut X times, you get X+1 pieces of DNA.

Question 2 1 / 1 pts A linear DNA is cut with EcoRI restriction endonuclease, generating 4 fragments. EcoRI cuts this DNA how many times? 2 3 4 1 5

Detects DNA molecules A Southern blot is where DNA is electrophoretically separated in a gel matrix, transferred to a nitrocellulose or nylon filter, and then hybridized with a DNA probe of a known sequence. Southern blots cannot detect whether a gene is expressed or not. A Northern blot uses RNA separated on a gel and then transferred and probed with a DNA or RNA probe. A Western blot uses protein separated on a gel and then transferred and probed with an antibody.

Question 20 1 / 1 pts Which of the following results is provided by a Southern blot analysis? Detects DNA molecules Detects RNA molecules Detects specific base pairs Determines chromosome structure Detects proteins

Southern blotting DNA microarrays can be used to detect mRNAs that are being expressed (transcription) in a cell. Western blotting and ELISA both detect proteins (gene expression - translation of mRNA into protein). Northern blots can detect mRNA (gene expression - transcription). Only Southern blots detect only the genetic material (unexpressed DNA). Southern blots cannot detect either part of gene expression, transcription, or translation.

Question 21 1 / 1 pts Many oncogenic mutations change the rate at which cellular genes are expressed. Which of the following cannot be used to compare the gene expression profiles of cancer cells with those of the normal cells from which they are derived? ELISA Western blotting Southern blotting A DNA microarray Northern blotting

III-1, III-3 The figure in the right upper corner shows three different alleles for the same gene/chromosomal region. One allele cuts the DNA in such a way that a 5.7 kb fragment is produced (note where the probe is located). Another allele cuts the DNA in such a way that a 1.0 kb and a 4.7 kb fragment is produced; the probe only will detect the 4.7 kb fragment. In this case, if the 4.7 kb fragment is detected, the individual has the middle allele, which cuts the DNA into a 1 kb and a 4.7 kb fragment. In the 3rd allele, the DNA is cut into a 3.2 kb and a 2.5 kb fragment; not that the probe crosses over this restriction site and will detect both the 3.2 and the 2.5 kb fragments. When you see a 2.5 and a 3.2 kb fragment, you know that this person has that allelic form of restriction sites. Now examine the gel. I.1. is the grandfather who is affected. Since this is an autosomal dominant disorder, only one of his allel

Question 22 1 / 1 pts A mutation that causes an autosomal dominant disorder is closely linked to a highly polymorphic region of the RFLP linkage analysis with a DNA probe produces the Southern blot pattern given below for each individual in the pedigree. Assuming that no crossing over has occurred, which individual(s) in generation III has/have received the mutation? RFLP with pedigree.jpg III-1, III-3 III-1, III-2, III-3 III-1, III-2 III-3 only III-2 only III-1 only

It shows the fetus is most likely a carrier. The key to this question is the unaffected, genotypically, and phenotypically, normal son (not a carrier). His genotype defines which is the "good" allele in each parent. The "good allele" of the father is associated with the 6 kb fragment and the "good allele" of the mother is associated with the 7.6 kb fragment. Thus the "bad" allele of the father is associated with the 3 kb fragment and the bad allele of the mother is associated with the 6 kb fragment. Thus the fetus is most likely a carrier since it got the 3 kb "bad" allele of the father, but the "good" 7.6 kb allele from the mother.

Question 23 1 / 1 pts Beta-thalassemia is an autosomal recessive disease. A couple who are both carriers of beta-thalassemia have one son who tests normal (not a carrier). They wish to know whether a current pregnancy is affected. DNA samples were obtained from the family members and fetus. An RFLP linkage analysis was carried out with a probe that is mapped 0.1 kb from the beta-globin gene. The Southern blot pattern for each individual in the family is shown below. What can be deduced from the prenatal analysis (assuming there is no crossing-over between the beta-globin gene and the probe)? RFLP with pedigree 2.jpg It shows the fetus is most likely a carrier. It shows the fetus is most likely normal (not a carrier). It shows the fetus is most likely affected. This analysis is not informative.

Male 1 could have been the father of child 1. In order to solve this question, start with understanding what the numbers mean. Note that the first column of two numbers is under "Gene A" and the second column of numbers is under "Gene B". What these numbers tell you is the number of repeats (like CG, CG, CG or CAG, CAG, CAG) there are on each of the two homologous chromosomes. These numbers represent the number of repeats there are at each "microsatellite" locus. For instance, the woman has eight repeats at a locus (gene A) on one chromosome and 8 repeats at some locus (gene A) on the homologous chromosome. At locus B (gene B) she has 3 repeats on one chromosome and 5 repeats at locus B on the homologous chromosome. When she has a child, she will pass on to that child only one of these homologous chromosomes with the corresponding repeats. We are making the assumption that she is the mother of the children in this

Question 24 1 / 1 pts A group of bodies are found buried in the forest. The police suspect that they may include the missing Jones family (two parents and two children). They extract DNA from bones and examine (using PCR) genes A and B, which are known to contain tandem triplet repeats of variable length. They also analyzed DNA from two other men. The results are shown below, where the numbers indicate the number of copies of a tandem repeat in a particular allele; for example, male 3 has one allele with 8 and another allele with 9 copies of a tandem repeat in the gene. Which child could have been fathered by which of the men assuming that the woman is the child's mother? PCR allelic diversity table for paternity.jpg Male 2 could have been the father of child 1. Male 1 could have been the father of child 2. Male 2 could have been the father of child 2. Male 3 could have been the father of child 1. Male 1 could have been the father of child 1.

2 & 3 There are 4 alleles of the same region on the same chromosome. Each allele has a different number of repeats as indicated by arrows. Allele 1 has the most (8 repeats), allele 2 has 6 repeats, allele 3 has 4 repeats and allele 4 has 2 repeats. Each repeat is PCR amplified by primers P1 and P2. When the PCR amplified products are run on a gel, the two alleles that each person (I to VI) has to show up on the gel. Individual III has two intermediate size repeats (4 repeats and 6 repeats) or allele 2 and allele 3.

Question 25 1 / 1 pts Four alleles are detected at an autosomal polymorphic site. The four alleles (1 to 4) are related to each other by a variable number of microsatellite repeats (arrows) as shown below (A). Total genomic DNA was isolated from six individuals (I to VI). For each DNA sample, the DNA fragments flanked by a pair of primers (P1 and P2) were amplified by PCR (polymerase chain reaction) and separated by gel electrophoresis (B). Individual III carries which alleles? variable repeats PCR from questions.jpgrflp 2 from questions.jpg 3 & 4 2 & 4 1 & 3 2 & 3 1 & 2

denaturation, annealing, extension In PCR, the first temperature-sensitive reaction is the denaturation step (95oC). Then the next step is to lower the temperature so that the primers can anneal (60oC). The final step is to raise the temperature to 72oC and extend the primers using a heat-resistant DNA polymerase.

Question 26 1 / 1 pts Polymerase chain reaction (PCR) involves 3 basic temperature-dependent enzyme-sensitive reactions that occur sequentially for 30‐40 cycles leading to exponential amplification of a specific segment of the genome with the help of specific primers in a cell-free system. What is the proper order in which the 3 temperature-sensitive reactions occur during a PCR cycle? extension, annealing, denaturation denaturation, annealing, extension annealing, extension, denaturation annealing, denaturation, extension denaturation, extension, annealing

A & D PCR is going to be used to "amplify" the region between I and II. In order to amplify this region, primers must be used that flank this region. Each primer must be able to be extended across this region. DNA polymerase needs a 3' OH to extend a primer. Only primers A and D provide a 3'OH that will allow the polymerase to extend across this region. Primers B and C provide a 3'OH but the primer will not be extended across this region but outside of this region. For the next question note the following: Primer D is identical to the region directly below region I and primer A is the reverse complement of the region directly under region II. So if you are given only one strand and asked to develop primers for PCR amplification of the region, the primer to the left of the region that needs to be amplified is identical to the sequence to the left of the region to be amplified and the primer to the right of the sequen

Question 27 1 / 1 pts To amplify a DNA fragment between position I and II by the PCR method (polymerase chain reaction), which primer set should be used? PCR fragment primer choice.jpg B & D B & C A & B A & C A & D

GAGTCCCTCAAGTCCTTC and CGGCGGTGGCGGCTGTTG The primer to the left of the sequence to be amplified is identical to the sequence to the left of the region to be amplified and the primer to the right of the region to be amplified is the reverse complement of the sequence to the left of the region to be amplified.

Question 28 1 / 1 pts James is a Huntington's disease patient who has recently been institutionalized and requires constant nursing care. His wife is unaffected, and they have a 38-year-old daughter, Delia, who has 4 children ranging in age from 8 to 15 years. Delia wishes to know whether she has inherited the disease-producing allele from her father. PCR amplification is carried out on the region containing the CAG repeat, shown below. Which of the following sets of primers would be used to amplify the CAG repeat in the brackets ([CAG]n)? GAG TCC CTC AAG TCC TTC [CAG]n CAA CAG CCG CCA CCG CCG GAGTCCCTCAAGTCCTTC and CAACAGCCGCCACCGCCG CTTCCTGAACTCCCTGAG and CGGCGGTGGCGGCTGTTG CTTCCTGAACTCCCTGAG and CAACAGCCGCCACCGCCG GAGTCCCTCAAGTCCTTC and CGGCGGTGGCGGCTGTTG GAGTCCCTCAAGTCCTTC and GTTGTCGGCGGTGGCGGC

An insertion of DNA in region A In this experiment, 4 regions of a gene are examined and compared to the size of the 4 regions in an unaltered "normal" person. In a normal person, the size of Region A is 0.1 kb, region B is 0.2 kb, region C is 0.3 kb, and region D is 0.4 kb. Each of these regions are identical in the DMD patient except region A, which is slightly larger. Something has happened to the DNA in region A to make it larger. The simplest explanation is that some DNA has been inserted into region A. If an inversion, nonsense, or missense mutation had occurred within region A, the size of region A would not change. If a deletion had occurred in region A, the band would appear smaller rather than larger.

Question 29 1 / 1 pts DNA from a DMD (Duchenne's muscular dystrophy) patient and a normal individual are subjected to multiplex PCR (polymerase chain reaction) amplification using 4 sets of primers flanking 4 regions of the DMD gene. The amplified DNA fragments are analyzed on an agarose gel. What kind of mutation is this patient most likely to have? DMD fingerprint from questions.jpg A nonsense mutation in region A Missense mutation in region A An inversion within region A A deletion of DNA in region A An insertion of DNA in region A

4 kb and 6 kb bands In order to understand this question, you have to understand what the probe will hybridize to. The probe is shown in the figure. The probe is under the region of allele A to which it will hybridize. The probe is under a region of allele B (the same DNA sequence as in allele A, remember there are two different chromosomes with the same genes but different alleles of that gene). So the "probe" will only hybridize to the 4 kb fragment of allele A and the 6 kb fragment of allele B but not to the 2 kb fragment of allele A since it does not have anything that is complementary to the sequence in the 2 kb fragment. The 2 kb will not show up because the probe is not complementary to the DNA sequence in the 2 kb fragment.

Question 3 1 / 1 pts Two alleles A and B have the EcoRI cutting sites illustrated in the figure. The DNA from individual AB (i.e., heterozygote) is completely digested with the restriction enzyme EcoRI and subjected to Southern blot analysis using the 32 P-labeled probe as indicated below. What size band(s) would the autoradiogram show? RFLP diagram.jpg 2 kb, 4 kb, and 5 kb bands 4 kb band 2 kb band 4 kb and 6 kb bands 2 kb and 6 kb bands

Multiplex microarray This technique takes less than a day. However, all labs are not equipped to perform these both quickly and cheaply. In those cases, multiplex PCR is more likely the better assay choice.

Question 30 1 / 1 pts Detection, identification, and quantification of microorganisms in cerebral spinal fluid (CSF) are critical for the diagnosis of infections in the central nervous system that lead to conditions like meningitis. These diseases escalate rapidly so early identification is key to the best prognosis. Small samples of CSF, available for analysis and requirement of early detection of the infection, call for the use of a very sensitive and specific diagnostic method for the detection of such infections. As a clinician which one of the following methods would you use for diagnosis of CNS infection at an early stage? Histological analysis of microorganisms purified from CSF Serological tests for detection of infection Multiplex microarray This technique takes less than a day. However, all labs are not equipped to perform these both quickly and cheaply. In those cases, multiplex PCR is more likely the better assay choice. Cell culture assay for the detection of pathogens Mulitplex PCR

Short tandemly repeated variable DNA sequences DNA fingerprinting is a set of sized DNA repeats measured at least 15-20 chromosomal loci. The set of different sized DNA is usually unique enough to identify the individual with about a 99.999999% probability that the DNA was derived from that of a specific individual or a very close relative in the case of identical twins. From a judicial standpoint, this technique is used to disprove an identity more often than to prove identity. In the case of identity, it is considered probable not certain.

Question 31 1 / 1 pts DNA fingerprinting is a technique employed by forensic scientists for the identification of individuals by using their respective DNA profiles. Which of the following best describes DNA profiles used in DNA fingerprinting? The sequence of the entire genome of the individual Promoter sequences of functional genes Genomic DNA library of the individual Short tandemly repeated variable DNA sequences A cDNA library of the individual

Polymorphisms Each individual is different from any other individual by having about 0.1% of their DNA being different, usually by simple single nucleotide polymorphisms. This means that about 3 million nucleotides are different in you and make you different from any other human. These differences are called polymorphisms. Polymorphisms cause predispositions by making the biochemistry more susceptible to failures and further damage. The term "deleterious" implies that the effect is, in itself, a mutation that causes problems. This is counter to the concept of predisposition suggested in the problem.

Question 32 1 / 1 pts A man who drinks alcohol and smokes cigarettes dies of liver cancer at the age of 50. Another man who drinks and smokes, equally as much, lives a long disease-free life of up to 90 years. Genome analysis exhibits minor variations in DNA sequences between these 2 individuals which may not cause cancer but, increases the risk of developing cancer in one individual over the other. What class of variations likely cause latent effects after many years? Deleterious mutations Insertions VNTRs Polymorphisms Deletions

Normal, carrier, carrier, normal The first generation (grandparents of the children in the 3rd generation) father (I-1) is still alive and DNA from him is analyzed and shown in the gel. Since he does not have colon cancer, the two bands in his lane are "tagging" "good genes". The grandmother (I-2) has died from colon cancer. One son (second generation, II-1) has died from colon cancer. The other son (II-2) is still alive but has colon cancer. He has married a woman that does not have the defective colon cancer gene. The two bands in the II-2 son represent the genes that he received from his father and mother. He has gotten the defective gene from his mother (I-2) and is represented by the smaller band in his lane. The larger band has been gotten from the father (I-1). His smaller band tags the defective gene. Any of his children that get this band, get the defective gene. This band shows up in III-2 and III-3. So II

Question 33 1 / 1 pts In the following family, I-2 and II-1 have died from a rare form of colon cancer that is inherited as an autosomal dominant. The remaining family members have been genotyped for a microsatellite marker close to the known gene for colon cancer susceptibility. According to this analysis, what are the genotypes of the children in the third generation (left to right; the carrier is used here for a person who has the genetic disposition but has not yet developed the disease)? pedigree and rflp 1 from questions.jpg Normal, carrier, normal, carrier Normal, carrier, carrier, normal Carrier, normal, normal, carrier Carrier, normal, carrier, normal Normal, carrier, normal, normal

A is the child of Parents 1. Each gel band in the child's lane represents a chromosome that could have come from the parents. Mother 1 (M1) and Father 1 could have both provided one of the bands that appear in the lane of the child 1. The larger band could have come from the mother and the smaller band could have come from the father. Further analysis of this gel indicates that B is the child of Parents 2. The larger band could have come from the father and the smaller band could have come from the mother.

Question 34 1 / 1 pts Two sets of parents were friends in a small town and had babies on the same day. The wristbands of the two similar-looking infants (A and B) were inadvertently mixed at the pediatric care unit. In order to accurately identify the parents of the respective infants, PCR analysis was performed on samples of blood taken from the two infants and both sets of parents (Father 1 and Mother 1 versus Father 2 and Mother 2). Shown below is the analysis of the PCR products by gel electrophoresis. What is the best conclusion from the analysis? paternity fingerprint from questions.jpg B is the child of Parents 1. Father 1 (F1) could be the father of both infants. Father 2 (F2) could be the father of both infants. A is the child of Parents 1. A is the child of Parents 2.

Many loci (10- 1000) can be analyzed simultaneously. One of the major advantages of DNA microarrays is the extremely small size of the reaction area. This allows 10-1000 loci to be analyzed at the same time. Additionally, using multiple fluorophores direct comparison of 2 or more samples over all those loci provides a less ambiguous comparison of samples.

Question 35 1 / 1 pts DNA microarrays are superseding the use of blotting techniques in large part due to the speed and cost of the analysis. Blotting techniques require the transfer of the material to nylon or nitrocellulose filters and require large amounts of reagents. Which other reason makes DNA microarrays a suitable improvement on blotting techniques? The probes are cloned first. Chromosomal rearrangements are easily detected. Large deletions are easily detected. Many loci (10- 1000) can be analyzed simultaneously. RNA cannot be used as a probe.

Most of the microarray will appear yellow but in the area of the deletion the spots will be red. If the genes on a comparative genome hybridization are in equal copy number the spot will appear yellow. If the patient's genome has a gene not seen in the normal then the spot will appear green. If the patient's genome does not have a gene then the spot will appear red. Variations in colors between the extremes are correlated to the ratio of the copy number of the gene.

Question 36 1 / 1 pts A 22-year-old patient is seen for genetic counseling after being diagnosed with profound muscle weakness. DNA samples were taken from the patient's cheek epithelia and analyzed by comparative Genome-Wide Hybridization to determine which muscle-specific proteins were in low copy number. The normal DNA sample is labeled with red fluorescence and the patient's DNA is labeled with green fluorescence. If the patient has a large deletion on their X chromosome, what color will the spots on the microarray appear? Most of the microarray will appear yellow but in the area of the deletion the spots will be green. Most of the microarray will appear yellow but in the area of the deletion the spots will be red. Most of the microarray will appear red but in the area of the deletion the spots will be yellow. Most of the microarray will appear green but in the area of the deletion the spots will be yellow. Most of the microarray will appear green but in the area of the deletion the spots will be red.

RFLP, restriction fragment length polymorphisms Personalized sequencing allows nucleotide level analysis of the genome. However, to determine the restriction fragment length polymorphisms would require a further analysis of the genome for restriction endonuclease recognition sequences then a further comparison of those locations throughout the genome. It is simpler to detect the mutations or differences in genomes without doing any prediction of restriction sites.

Question 37 1 / 1 pts Personalized genome sequencing is now possible. Which of the following is the least likely use of personalized genome sequencing? pathogen screening RFLP, restriction fragment length polymorphisms prenatal/IVF screening drug sensitivity SNP, single nucleotide polymorphisms

GWAS, Genome Wide Association Study GWAS is an effective tool to narrow down the regions of the genome that may have genes that affect or cause disease. This technique requires a large sample size of DNA donors as is the case in this problem. Genome wide sequencing is an effective technique for looking at the DNA of a single individual or a small sample size. This would be very time and cost-prohibitive in this case. CRISPR-Cas9 is used for knock-in, knock-out, or knock-down gene editing for a single individual at a time. This would be very ineffective until the gene of interest is known.

Question 38 1 / 1 pts A psychiatrist is working with patients with high-functioning autism. He wants to determine if there is a genetic marker or link to the disease. He has obtained DNA samples from 864 patients and 4,000 people with normal social skills of similar genetic backgrounds to the patients. Which technology will be most helpful in his research? Southern Blot analysis Genome wide sequencing Dot blots of a cDNA library from each patient and normal donor CRISPR-Cas9 GWAS, Genome Wide Association Study

CRISPR-Cas9 This is the only technique that can be used to site direct to specific location in the genome and allow gene editing.

Question 39 1 / 1 pts A 2-month-old child is seen at the regular visit by her pediatrician. The child was seen to have been a healthy birth but has deteriorated in health since birth. At the 6-week check-up, the physician ordered testing for a number of genetic diseases. The recent findings led the physician to order personalized sequencing of the child's chromosome 5p13-14. A large deletion which included both the SMN1 and SMN2 genes was found on one allele and the other allele was found to have a point mutation in the SMN1 gene. The child was therefore diagnosed with Spinal Muscular Atrophy type 1, the most severe form with a life expectancy of under 3 years. The parents request gene therapy for their child. Which technique will be able to place a good pair of copies of SMN1 and SMN2 onto the child's chromosome 5? TALENs ZFN Recombination with a linearized recombinat DNA containing the 2 genes. Meganucleases mediated gene replacement. CRISPR-Cas9

They cleave both strands of duplex DNA. Restriction enzymes create double-stranded DNA breaks. Some restriction enzymes create "sticky ends" when the cut creates a 3' or 5' overhanging single-stranded end, while other restriction enzymes create "blunt ends". If the restriction site is methylated, the restriction enzyme will not cut the restriction sequence.

Question 4 1 / 1 pts Which of the following statements describing restriction endonucleases is true? They recognize triplet repeats. They cleave both strands of duplex DNA. They recognize methylated DNA sequences. They always yield blunt ends. They always yield overhanging single-stranded ends.

Endodeoxyribonuclease that is site specific. Endodeoxyribonucleases cut the DNA internally at sites designated by protein binding (ZFN or TALENs or meganucleases) or nucleic acid binding (CRISPER-Cas9) sites that are specific. The endonuclease then cuts the DNA allowing the integration or recombination with the linearized recombinant DNA. Exodeoxyribonucleases find a 5' or 3' end of the DNA and digest the DNA from those ends. Given that telomeres are wrapped and protected from such nucleases addition of this type of nuclease would not create a point for integration or recombination with the linearized recombinant DNA. Protein kinase, spliceosomes, and DNA polymerases do not cut the DNA to allow for integration or recombination with the linearized recombinant DNA.

Question 40 1 / 1 pts All gene editing systems rely on the isolation of a cell from the patient and then the reintroduction of clones of that cell back into the patient after the DNA changes have been performed and verified in vitro. In addition to the recombinant DNA what other agents must be added to the cell for the introduction of the "corrected" gene into the genome. Exodeoxyribonuclease that is site specific Endodeoxyribonuclease that is site specific. Spliceosomes Protein kinase DNA polymerase

AAGCTT A restriction enzyme recognition sequence is a palindrome. The 5' to 3' sequence is the same on both strands of DNA.

Question 5 1 / 1 pts Which of the following is the most plausible recognition site for a restriction endonuclease? AAGCTT CGAAGC TGCTTC AAGCCC TTCGTT

Plasmid and human DNA are cleaved to have complimentary sticky ends and the plasmid DNA is de-phosphorylated prior to ligation. DNA, from two different sources, cut with the same restriction enzyme that generates "sticky ends", can easily be "sewn" together since the sticky ends will hybridize together and assist in the ligation process carried out by DNA ligase. However, the restriction enzymes can be different if they leave the same sticky ends. Also, the efficiency of cloning is improved if the plasmid DNA is de-phosphorylated after cleavage to prevent the plasmid from closing and ligating to itself.

Question 6 1 / 1 pts For successful expression of the human gene A in E. coli cells, a bacterial promoter sequence needs to be incorporated into the fragment of human DNA containing the specific gene A. What assures that the joining of the bacterial piece of DNA to the human DNA fragment can be done successfully at a high rate? A suitable plasmid is used for transfection of the human DNA into E. coli cells. Plasmid and human DNA are cleaved to have complimentary sticky ends. The bacterial plasmid and human DNA is flanked by 5' or 3' overhangs. Bacterial and human DNA are cleaved with different restriction endonucleases (RE). Plasmid and human DNA are cleaved to have complimentary sticky ends and the plasmid DNA is de-phosphorylated prior to ligation.

They sometimes enhance bacterial resistance to antibiotics. Plasmids are naturally occurring "mini chromosomes" that contain features that help a bacterium withstand the antibiotics produced by other organisms such as fungi.

Question 7 1 / 1 pts Which of the following statements correctly describes the recombinant DNA tool known as plasmids? They were created using artificial DNA synthesis to be used as vectors for the cloning of mammalian DNA segments. They are found more commonly in viruses than in bacteria. They sometimes enhance bacterial resistance to antibiotics. They are single-stranded circles. They sometimes enhance bacterial susceptibility to antibiotics.

Ligating an RNA molecule and a DNA molecule. RNA molecules are not ligated to a DNA molecule.

Question 8 1 / 1 pts Which of the following steps are never involved in the process of creating a recombinant DNA molecule? Reverse transcription of isolated mRNA. Digestion of DNA with a restriction endonuclease. Ligating an RNA molecule and a DNA molecule. Growing billions of bacteria to produce billions of plasmids. Ligation of two DNA fragments from two different organisms.

Vector In molecular biology, the cloning vehicle is often referred to as a "vector".

Question 9 1 / 1 pts What is the term for a DNA molecule into which foreign DNA may be inserted and which can be returned to and replicated within a living cell? Messenger Complementary DNA or cDNA Transgene Vector Kinetochore

The bacterium would be more easily transformed with plasmids. The restriction enzyme system of a bacterium is to protect the bacterium from viruses that would destroy the bacterium. If the restriction system of a bacterium is "knocked-out", the bacterium would be more susceptible to infections by viruses. When a foreign DNA such as a plasmid is transformed into a bacterium, the restriction system treats the foreign DNA/plasmid like an invading virus and tries to destroy the plasmid before the methylation system can protect the plasmid DNA from the restriction system effect. Thus, if the restriction system is knocked out, plasmids will be able to enter the bacterium more easily.

What would happen to a bacterium if the restriction system of the bacterium is knocked out? The bacterium would be less susceptible to viral infections. The bacterium would digest its own DNA and die. The bacterium would be more easily transformed with plasmids. The bacterium would slow down its growth and appear to be "sick". Nothing. The restriction system has backups that protect the bacterium.