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1

There is no question as to maternity. Why is the mother's DNA analyzed?

a. The mother's alleles must be compared to the father's allels.

b. The mother's alleles must be proven to be different from the father's alleles.

c. The mother's alleles have to be identical to all of the child's alleles.

d. The heir's alleles not shared with the mother will have to be provided from the father.

d. The heir's alleles not shared with the mother will have to be provided from the father.

In diploid organisms, one of each chromosome pair (carrying one allele) is inherited for each parent. Half of the alleles in the heir will be the same as half of the mother's alleles. The other alleles will come from the father.

2

How are alleles determined by PCR and capillary electrophoresis?

a. The size (migration speed) of the PCR products

b. The areas under the fluorescent peaks on the electropherogram.

c. The height of the fluorescent peaks on the electropherogram.

d. Restriction enzyme sites in each PCR product.

a. The size (migration speed) of the PCR products

For STR analysis by PCR, the primers are designed to produce a product containing the STR sequences. The size (migration speed in the capillary) of each PCR product will be dependent on the number of repeats it contains. The areas and heights of the peaks are not associated with the identity of the alleles.

3

One of the loci tested in the analysis was the amelogenin locus. There was one strong peak at this locus in the heir's DNA. What does this indicate about the heir?

a. More than one peak is necessary to interpret any locus.

b. The putative heir is female.

c. The putative heir is male.

d. Paternity by the financier is unlikely.

b. The putative heir is female.

The amelogenin loocus is not an STR but a gene located on the X and Y chromosomes.  The allele on the Y chromosome is slightly larger than that on the X chromosome. Since females are homozygous for X, only one peak will be present.

4

It is very important in legal cases such as this to unambiguously identify alleles. What cotnrols are used to ensure the accurate comparison of migration speeds in each capillary?

a. Positive control from a cell line.

b. amplification control

c. amelogenin loci, which are always the same size

d. internal lane standards

d. internal lane standards

Internal lane standards migrate through the capillary with the PCR prodcuts containing the STR. The capillary instrument assess their size (migration speed) by direct comparison with the standards of known size.

5

Twelve loci were tested. The results of the PCR analysis revealed that half of the alleles of the putative heir were the same as half of the alleles of the financier. What does this mean?

a. The financier is excluded as the father.

b. The financier is 50% included as the father.

c. Depending on the allele frequencies, there is a very high probability of paternity.

d. All alleles must match in order to prove paternity with high confidence.

c. Depending on the allele frequencies, there is a very high probability of paternity.

A paternity test is designed to choose between two hypotheses: The test subject is not the father of the tested child (Ho), or the test subject is the father of the tested child (H1).  Paternity is first assessed by observation of shared alleles between the alleged father and child. Note that half of each parent's alleles are expected to match in the case of positive paternity. If at least one paternal allele allele does not match, the alleled father is excluded. Results are not reported as 50% certain, nor as completely certain, leaving the possibility that someone else may be the parent.

6

Why is the PCR method preferred for this application?

a. PCR is more labor intensive than other methods.

b. The PCR method has fewer artifacts than Southern blot.

c. PCR can be performed on degraded specimens.

d. STR allelels can be analzed only by PCR

b. The PCR method has fewer artifacts than Southern blot.

Since PCR can be performed on a template of a few hundred base pairs, it is the preferred method for specimens that are likely to have fractured DNA. The chance of a chromosome break within the small area of the PCR target is much less than in the hundreds to thousands of base pair RFLP fragments analyzed by Southern blot. STR can be assessed by methods other than PCR—for example, RFLP or direct sequencing. These methods are not practical for limited specimens, however.

7

Why were the donor and recipient tested prior to the transplant?

a. This test is used to select a suitable donor.

b. This pretransplate testing is performed to identify the donor the recipient.

c. STR must match in order for the transplant to be successful.

d. Pretransplant testing is required to identify STR alleles that differ between the donor and recipient.

d. Pretransplant testing is required to identify STR alleles that differ between the donor and recipient.

Engraftment analysis after a transplant depends on the ability to distinguish the donor from the recipient. In order to do this, STR loci must be analyzed in the donor and the recipient before the recipient receives donor cells. If this is not done, alleles observed in the recipient after transplant cannot be assigned to donor or recipient. The pretransplant analysis is not used to select a donor. The donor is selected using tissue typing or HLA analysis. The STR does not affect the engraftment. The test is not used for human identification, as the alleles are surveyed for informativity (different between donor and recipient) without necessarily identifying specific alleles.

8

In what case might the pretransplant STR analysis yield no informative loci?

a. The donor and recipient are fraternal twins.

b. The donor and recipient are identical twins.

c. The donor is homozygous for all STR alleles tested.

d. The recipient is a part of the donor.

b. The donor and recipient are identical twins.

Pretransplant STR analysis will not produce informative loci in the case of identical twins. Unlike dizygotic (fraternal) twins, monozygotic twins from a single fertilized egg share the same DNA and therefore the same STR alleles. Even closely related individuals (siblings or parents) will have different alleles among the STR loci tested. Informativity of loci is not dependent on whether the loci are homozygous or heterozygous. In the case of identical twins, STR analysis cannot be used for post-transplant engraftment analysis.

9

What cell fraction expressed CD3?

a. granulocytes

b. myeloid cells

c. B cells

d. T cells

d. T cells

CD3 is a membrane receptor protein expressed on T cells. It is used to select the T-cell population from the rest of the white blood cells. T cells are factors in graft-versus-host disease, which can affect the health of the recipient after the transplant. The T-cell fraction also participates in a graft-versus-tumor effect that is beneficial for preventing relapse of malignancies. Other fractions such as myeloid cells may also be tested separately for engraftment. B cells are not usually tested as a separate cell fraction.

10

What term is used to indicate different degrees of engraftment among different cell fractions - for example, T-cells versus myeloid cells?

a. full chimerism

b. incomplete chimerism

c. spilt chimerism

d. graft failure

c. spilt chimerism

Not all cell fractions engraft with equal kinetics. For example, granulocytes may show all donor alleles, while T cells in the same patient may be a mixture of donor and recipient alleles. 

Full chimerism indicates complete engraftment of all fractions. 

Graft failure indicates decreasing levels of loss of donor alleles.

Incomplete Chimerism is not a term that is used.

11

The results of the transplant at the D16S539 locus were similar to those shown below. There are both donor and recipient STR alleles in the recipient following transplant. How is the degree of engraftment (percent donor) quanitified?

a. compare peak heights of the D16S539 PCR product in the recipient pre- and post-transplant.

b. Divide the area under the donor peak(s) with total peak area at the D16S539 locus, ignoring shared alleles

c. divide the area under the donor peak(s) by the area under the recipient peak(s) at the D16S539 locus.

d. divide the area under the donor peak(s) by the area under the pretransplant donor peak(s) at the D16S539) locus.

b. Divide the area under the donor peak(s) with total peak area at the D16S539 locus, ignoring shared alleles.

To quantify percent donor, add the area under the donor peak(s) in fluorescence units (provided by the instrument software). Then determine the total peak area by adding the area under the donor + recipient peak(s). Divide the donor peak area by the total peak area (not by the recipient peak area) and multiply by 100 to get the percentage. In the example, one of the peaks is the same in the donor and recipient. Its area is not included in the calculation. Instead of area, the heights of the peaks may be used in a similar manner. Pretransplant peaks are run for reference only to confirm the location of the donor and recipient alleles.

12

What is the interpretation is two peaks from different sources align with the same migration speed or distance (bin) in the capillary analysis?

a. the peaks represent the same allele

b. an error has occurred, as peaks from different sources should migrate at different speeds

c. in order for peaks to be interpreted as the same allele, the peak heights must be the same.

d. the peaks represent degraded PCR products

a. the peaks represent the same allele

STR analysis is performed by amplifying a short region of DNA containing the STR. The size of the PCR products reflects the number of tandem repeats in the STR. PCR products of equal size have the same number of tandem repeats and are interpreted as the same allele. PCR products of the same STR locus from two different sources may or may not be of the same size. Degraded PCR products will not consistently resolve at any particular migration speed.

13

The alleles detected in the TH01 STR locus were 7,10. What do the numbers 7 and 10 denote?

a. The relative allele frequency of the two alleles

b. the number of nucleotides in each allele

c. the number of tandem repeats in each allele

d. the chromosome on which the alleles are found

c. the number of tandem repeats in each allele

Short tandem repeat alleles are distinguished by the number of repeats in each allele. For THO1, located on chromosome 11, the repeated sequence is TCAT. There are seven repeats of the 4-bp sequence, TCAT on one of the two chromosomes 11. There are 10 TCAT repeats on the other chromosome 11. Only in the case of single-bp repeats would the numbers indicate the number of nucleotides in each STR. The actual number of base pairs in an allele analyzed by PCR (or RFLP) is determined by the size of the fragment generated from the primers or restriction enzymes used. Allele frequency is the fraction of people in the population sharing that allele and is expressed as a number between 0 and 1.

14

There are two alleles at most of the loci in this genotype. There is one allele at the D12S579 locus. What does this indicate?

a. a PCR failure occurred on only one allele amplified

b. the individual has only one chromosome 12

c. the genotype is not complete and must be reanalyzed

d. the individual is homozygous at the D12S579 locus

d. the individual is homozygous at the D12S579 locus

If an individual inherits different alleles from each parent, the locus will be heterozygous. If an individual inherits the same allele from each parent, the individual will be homozygous for that locus. Laboratory error (PCR failure, incomplete genotype) can be ruled out, since all other loci and controls are successfully amplified and since the chance of PCR failure at only those loci matching a homozygous loci in a genotype from an independent source (the database) is small.

15

The alleles matched at 13 loci. What does this mean?

a. The alleles detected are very frequent in the human population

b. the individual in the database cannot be excluded as a suspect

c. the individual in the database cannot be included as a suspect

d. the alleles detected are rare in the human population

b. the individual in the database cannot be excluded as a suspect

The matching loci are sufficient for legal identification. The genotype found at the crime scene belongs to the subject in the database, or his identical twin. A matching genotype such as in this case is dependent on, but does not indicate, the frequency or rarity of the matching alleles. Frequently occurring alleles may be found more often in unrelated individuals; however, the probability of two unrelated individuals having the same set of 13 alleles is less than 1/1,000,000,000,000.

16

What is the purpose of the label on the primers?

a. Only specific HLA types will be amplified and therefore labeled.

b. All amplicons must be labeled for later detection.

c. The label is required for binding with the probes.

d. The labe prevents mispriming in the PCR reaction. 

b. All amplicons must be labeled for later detection.

Hybridization of a labeled probe to immobilized amplicons of the HLA genes (dot blot) was one of the first methods that utilized polymerase chain reaction (PCR)-amplified DNA for HLA typing. Conversely, unlabeled probes can be immobilized and hybridized to labeled test specimen. The test amplicons are labeled using tailed primers with a colorimetric, chemiluminescent or fluorescent tag. This addition to the 5' end of a primer will not determine its target specificity nor prevent mispriming. The amplicons from each donor sample will hybridize only to the probes complementary to their specific allele. The donor allele is then determined according to which position on the membrane array shows fluorescence. The fluorescent molecules are required for identification of which probe is binding the sample amplicon. They are not required for hybridization.

17

How are allele-specific probes distinguished from one another?

a. length in base pairs

b. the number of probes used

c. the position of hybridization on the membrane

d. fluorescent wavelength

c. the position of hybridization on the membrane

The donor allele is determined according to which position on the membrane array shows fluorescence. All test amplicons will have the same fluorescent wavelength. Probe lengths (19 to 20 bases) are not associated with any specific allele. The number of probes used depends on the design of the assay. For example, an intermediate resolution assay of the HLA-DRB locus might take 30 to 60 probes. Studies have achieved high-resolution identification of the majority of HLA-A, B, and C alleles using 67 HLA-A, 99 HLA-B, and 57 HLA-C probes and intermediate resolution with 39 HLA-A and 59 HLA-B alleles.

18

Where is the optimal position for the polymorphic nucleotide in the probe sequence?

a. either end

b. both ends

c. 5' end only

d. middle

d. middle

The probe sequences are aligned so that the polymorphic nucleotide is in the middle of the probe sequence. Hybridization conditions depend on the optimal binding of the probe matching a test sequence in comparison with another sequence, differing from (not complementary to) the probe by at least one base. Polymorphisms on the ends of the probe molecule may not consistently destabilize the hybridization of a nonmatching amplicon.

19

In the SSOP assay, a blue colorimetric signal is seen at probe position E3 (A*03:02). Blue color was not observed at any other position (except positive controls), including probe at position D5 (A*26.10). How is this interpreted?

a. the probe at position D5 is defective

b. the probe at position E3 is contaminated

c. the HLA-A type is A*3:02

d. The HLA-A type is A*26:10

c. the HLA-A type is A*3:02

Each allele-specific oligomer probe will hybridize only to its labeled matching amplicon. Probe sequences are based on sequence alignments of HLA polymorphic regions. Association of the amplicon signal with a probe position on the membrane or plate well indicates presence of that allele in the test sample.

20

Which of the following donors is capatible with the recipient?

a. Donor A

b. Donor B

c. Donor C

d. Donor D

a. Donor A

Donor A, A*03:03/A*09:02, B*39:06, DRB1*07:02/DRB1*08:17 is the best match to the patient type, A*09:02, B*39:06, DRB1*07:02/DRB1*08:17. The donor is heterozygous at the HLA-A locus but carries the A*09:02 allele of the recipient. Both donor and recipient are homozygous for the B*39:06 allele at HLA-B and carry the DRB1*07:02/DRB1*08:17 genotype at HLA-DR. None of the other donors share as many alleles with the recipient. SSOP is considered low to intermediate resolution, depending on the number and types of probes used in the assay. As some probes have multiple specificities, hybridization panels can be complex. Computer programs may be used for accurate interpretation of SSOP results.

21

What is the reason to test for antihuman antibodies?

a. antihuman antibodies may react against a donor organ

b. antihuman antibodies facilitate finding a compatible donor

c. all serum contains antihuman antibodies

d. antihuman antibodies sensitize patients to therapy

a. antihuman antibodies may react against a donor organ

Correct Answer: 1.
Successful organ transplant depends on minimal reaction of the recipient's immune system to the antigens of the donor organ. Normally, serum does not contain antibodies against human antigens. However, persons who have had pregnancies, a previous organ transplant, or blood transfusions will have antihuman antibodies (termed humoral sensitization) that may react against a new donor organ. These antibodies do not facilitate finding organ donors, nor do they have any effect on therapy.

22

What is one reason the presence of antihuman antibodies is predicted in this case?

a. the patient has not had a previous organ transplant

b. the patient has several children

c. the patient is female

d. the patient is over 40 years of age

b. the patient has several children

Persons who have had a previous organ transplant, blood transfusions, or pregnancies have antihuman antibodies (termed humoral sensitization) that may react against a new donor organ. Antihuman antibodies are not induced by gender or age.

23

The flow cytometry test for PRA was performed with microbeads. What is the advantage of using microbeads?

a. microbeads are less expensive than reagents required for the CDC test.

b. this method identifies specific antigens using pooled antibodies

c. the instrument for this test is available in all laboratories

d. there is less subjectivitiy in interpretation of results

d. there is less subjectivitiy in interpretation of results

Screening of sera with microparticles (beads) is performed in laboratories with flow cytometry capability. For this method, microparticles are attached to pools of antigens derived from cell lines. The microparticles are exposed to test serum, and beads carrying antigens that are recognized by antibodies present in the test serum will bind to those antibodies. After removal of unbound antibodies, a fluorescently labeled secondary reporter antibody is applied, and the antibody-bound beads are detected by flow cytometry. The advantages of this method over the CDC test are that the reaction is performed in a single tube, and there is less subjectivity in the interpretation of results. Because this test uses pooled antigens, however, it can detect prevalence of antihuman antibodies in the test serum but not identify which specific antibodies are present. A negative result does preclude further antihuman antibody assessment.

24

In the flow cytometry method, what is covalently attached to the microbeads?

a. human antigens

b. human antibodies

c. other microparticles

d. fluorescent labels

a. human antigens

Screening of sera with microparticles (beads) is performed using microparticles attached to pools of antigens derived from cell lines of defined HLA types. When the microparticles are exposed to test serum, beads carrying antigens will bind to antibodies, if present, in the test serum. A fluorescently labeled secondary reporter antibody is then applied, and the antibody-bound beads are detected by flow cytometry. Particles without bound antigen (bound to other particles or antibodies) would not bind antibodies in the test serum. The fluorescent label will only be associated with beads that have bound serum antibody.

25

What do the results of the test (20% PRA) indicate?

a. it will be impossible to find a donor for this patient

b. no antihuman antibodies are present in the patient serum

c. only 20% of potential donor organs will be compatible

d. potential donors may be found for this patient

d. potential donors may be found for this patient

The percentage of the panel reactive antibodies in the test specimen would bind to the tissue types of 20% of the people in the population. Crossmatch of negative donors may therefore be found for this patient. Patients with %PRA activity of more than 50% are considered to be highly sensitized; finding crossmatch-negative donors is more difficult in these cases.

26

What control tests the lower limit of detection in each assay run?

a. amplification

b. positive

c. sensitvity

d. negative

c. sensitivity

The sensitivity control should be at or near the lowest point of the AMR. If the sensitivity control does not consistently produce the expected signal, the AMR and the test conditions used must be reviewed and reestablished if necessary. The positive control should always produce positive signal at the high end of the AMR, while the amplification control should give a robust signal from an independent target. There should be no signal from the negative control.

27

Which sample measurements will not be acceptable if the lower limit of detection is not being reached?

a. high positive sample results

b. negative sample results

c. high positive control

d. amplification control

b. negative sample results

Virus present at the lower limit of detection may be missed if the procedure is not producing signal from the analyte at that level. This will result in false-negative results. When the low positive control doesn't produce signal, then all negative sample results are in question. In a quantitative assay, the levels of virus in positive samples may also be questioned, especially if the high positive control values are less than expected. Since the amplification control in this example is an independent target, its detection should not be affected.

28

What part of the test development process in this example is not designed properly?

a. AMR measurement

b. validation

c. calibration

d. amplification controls

a. AMR measurement

The AMR is established from undiluted specimens or a known concentration of analyte in the appropriate matrix. In this example, the synthetic viral genomes were prepared in water, rather than normal plasma, which would be the proper matrix. Validation is performed by comparing results from the new assay to blinded results from a previously validated procedure or an established reference standard. Calibration is performed to determine the input-response capability of the instrument using defined calibrators. A variety of amplification controls may be used for qPCR assays, including human genes.

29

How does calibration differ from validation?

a. calibration tests reagents quality; validation tests instrumentation

b. calibration tests instrument response; validation tests method accuracy

c. calibration tests the method; validation confirms the calibration results

d. calibration tests reagent quality; validation tests the method

b. calibration tests instrument response; validation tests method accuracy

Calibration is performed to demonstrate that the instrument produces the appropriate signals from defined reference standards or calibrators. Because calibrators may be used to adjust instrument output, calibration is not performed with positive controls used in an assay. Validation tests the ability of the method used to accurately detect or quantify analyte in test samples compared to previously validated methods or a gold standard measurement. Validation is performed once at the introduction of a new assay or when the assay method is modified. Calibration should be performed every 6 months.

30

In the example given, what is the likely reason that the level of detection in the assay is different from that in the AMR?

a. the AMR is done prior to validation, when the reagents are fresher

b. it takes longer to set up the validation tests, allowing the analyte to degrade

c. PCR efficiency may be affected when the analyte is in plasma, compared to water

d. the technologist establishing the AMR may be different from teh technologists performing the validation

c. PCR efficiency may be affected when the analyte is in plasma, compared to water

The AMR should be established in the same matrix as the test specimens, which is plasma in this example. Polymerase enzyme inhibitors, DNA binding proteins, therapeutic agents, or other substances in plasma may affect the ability to amplify the target. Under these circumstances, detection will be less sensitive than under optimal conditions of synthetic targets and defined reaction components. If stored properly, viral nucleic acid should not degrade between the time of calibration and validation, nor during the preparation of the reaction mix. One purpose of the validation is to demonstrate that there is no operator-caused variation in the assay results. A proficient technologist should be able to perform calibration and validation with comparable results.

31

Why would the laboratory director not accept the proficiency test results?

a. the supervisor had already signed off on the data.

b. more than one technologist should have repeated the test

c. the samples were not tested by technologists in the way that patient samples are tested

d. the laboratory had not previously done proficiency testing for this mutation

c. the samples were not tested by technologists in the way that patient samples are tested

Proficiency tests should be integrated and performed within the normal workload. If an alternate method is performed in the laboratory, the proficiency samples should be tested by the primary method. Since the supervisor was developing a new test not yet used on the patient specimens, the results of the proficiency test did not support accurate routine testing. Ideally, all technologists in a laboratory should rotate testing of proficiency samples; however, it is not necessary for multiple technologists to test every sample. Signing of the results by only the supervisor does not demonstrate competency among the technical staff. Proficiency testing recommendations begin with the initial sample set. Previous testing is not necessary for acceptable performance.

32

Which of the following actions in this case is prohibited in proficiency testing?

a. testing of samples by the supervisor

b. testing of the samples without all technologists being present

c. comparing results after the deadline for submission

d. interlaboratory communication while doing the PT

d. interlaboratory communication while doing the PT

The laboratory is required to have a policy that prohibits sharing of data or specimens between laboratories. This may be done after the data has been submitted to the PT provider. Any qualified technologist may perform PT as long as the specimens are treated as routine testing along with patient samples. All technologists in the laboratory do not have to participate in every PT challenge; however, PT specimens should eventually be rotated among all technologists.

33

If the modified method had been validated, what should have been done with the PT sample?

a. test both specimens by both methods

b. choose the most cost-efficient method and test both specimens

c. test one PT specimen with one method and test the other PT specimen with the new method

d. test both specimens by one method multiple times

a. test both specimens by both methods

If the laboratory offers tests for the same analyte using different methods, PT should be performed with both methods, regardless of which method is used most often. In this case, both PT specimens should be tested by both of the currently used methods. Repeated testing of PT specimens by either of the methods is not necessary, unless patient specimens are tested in this manner or if technical difficulties occur.

34

Which of the following is evidence of compliance to the requirement that PT specimens are integrated into routine patient testing runs?

a. signature of the laboratory scientists on the report

b. written test method

c. work records

d. eyewitness account

c. work records

Instrument printouts or work records will show the integration of PT specimens with patients tested in the same run. Signing the report does not provide evidence of PT specimen integration in this manner. A written policy describing the proper handling of PT specimens may be included in the procedure or as a separate laboratory document; however, the written test method does not prove compliance with respect to particular PT. Accounts of witnesses to PT testing are not acceptable evidence of compliance.

35

What is the corrective action for this case?

a. document that the results are correct as performed by the supervisor

b. immediately repeat the PT test with routine testing and prepare a written explanation of the incident and the corrective action

c. discard the PT specimens, destroy the documents, and send a request for replacement specimens

d. refer the specimens to another outside laboratory to confirm the PT results

b. immediately repeat the PT test with routine testing and prepare a written explanation of the incident and the corrective action

The laboratory should maintain ongoing records of timely review of PT, including records of investigation of unacceptable PT results and corrective action taken. Even though the results in this case are correct, the PT was not performed properly. Therefore, the results should not be maintained without documentation of corrective action. It is not necessary to request replacement specimens unless the condition, amount, or delivery of the specimens was unacceptable for accurate testing. Such a request should occur immediately upon receipt of the specimens or as soon as problems with the specimen are discovered. Referring PT to outside laboratories is strictly prohibited under CLIA regulations.

36

With 51 students affected, which of the following is closest to the definition of the outbreak?

a. endemic

b. epidemic

c. nosocomial

d. latrogenic

b. epidemic

An epidemic is a disease or condition that affects many unrelated individuals at the same time, while a pandemic is a disease that sweeps across wide geographical areas. Endemic refers to infections that are prevalent to a particular group of people or region. Nosocomial infections occur in hospitals. Iatrogenic infections are those specifically spread by doctors.

37

If all of the infections came from a common source, what will be the results of the sequencing data?

a. the viral genomes will be of different lengths

b. the viral sequences will be completely different from one another

c. the sequences of all viral genomes tested will be very similar, if not identical

d. the norovirus sequence will be similar to that of the influenza virus

c. the sequences of all viral genomes tested will be very similar, if not identical

Due to the genetic diversity of this family of viruses, the viral sequences will be very similar but not necessarily identical to one another. The length of the viral genomes is not necessarily associated with their relatedness. The observation of viral sequences that are completely different is consistent with viruses from unrelated sources. Norovirus and influenza are not related organisms; therefore, their sequences should not be similar.

38

Sequence analysis will have what performance characteristics with regard to typing capacity, discriminatory power, and ease of use, respectively?

a. high, high, moderate

b. high, high, high

c. moderate, high, moderate

d. high, moderate, high

a. high, high, moderate

Sequence analysis is definitive and detailed with respect to the organism being tested. Therefore, it has high typing capacity and discriminatory power. However, the ease of use of sequencing is moderate compared to other epidemiological methods.

39

Recently a rapid and simple ELISA test that uses immobilized antibodies was approved for detection of norovirus in stool samples. In validation studies, the ELISA test demonstrated a sensitivity of 60% and a specificity of 88%. Which is the best use for this test?

a. sensitive detection of the norovirus

b. rule out other viral infection

c. screen many students for the presence of norovirus

d. positive validation for the presence of norovirus

c. screen many students for the presence of norovirus

Due to its sensitivity and specificity, the ELISA test cannot be used to accurately detect the virus in individual patients. Nor can it be used to rule out other viral infections, as a negative result does not rule out the presence of norovirus, either. The advantage of this test is a quick screen when many people are suspected of having norovirus infection.

40

What characterisitc of the virus will affect its detection by ELISA?

a. small size

b. antigenic diversity

c. lack of DNA

d. lack of envelope

b. antigenic diversity

The ELISA test uses antibodies to several genotypes of norovirus. Just as the norovirus escapes the host's immunity by changing its antigenic profile, it may also escape detection by antibodies used in the ELISA test. The presence of antigen does not depend on the size of the virus or its lack of envelope. Both RNA and DNA viruses produce protein antigens associated with replication and capsid formation.

41

What is the interpretation of a qPCR fluorescent signal surpassing the fluorescence threshold at the 23rd cycle? (The internal cotnrol also produces fluorescence above the threshold level).

a. presence of BK virus only

b. presence of BK or JC virus

c. presence of antibodies to BK or JC virus

d. presence of JC virus only

b. presence of BK or JC virus

Quantitative PCR performed with SyBr green will generate increased fluorescence correlated with the accumulation of the double-stranded PCR products. If either or both BK and JC viral genomes are present, a fluorescent signal will be generated, eventually surpassing a threshold level based on background fluorescence. Since SyBr green is nonspecific with regard to DNA sequence, the signal cannot be assigned to either virus specifically. PCR does not detect antibodies.

42

Which of the following interpretations is most likely if the SyBr green fluorescent signal and the internal control do not reach the threshold fluorescence level?

a. JC virus may be present, but BK virus is absent

b. BK virus may be present, but JC virus is absent

c. neither virus is present

d. an inhibitor may be present

d. an inhibitor may be present

Since the internal control should always amplify (produce fluorescent signal), the absence of signal from the internal control indicates that there may be an inhibitor present. The presence of either virus should produce a fluorescent signal. Without amplification of the internal control, a negative result (neither virus is present) cannot be assumed.

43

The lower limit of detection of this BK/JC assay is 50 copies/mL. If the internal control is positive, but no signal is observed from the target viral sequences, which of the following should be reported?

a. negative

b. viral load is less than 50 copies/mL

c. inhibitor present

d. viral load is 50 copies/mL

b. viral load is less than 50 copies/mL

Since the lower limit of detection is 50 copies/mL, levels less than this may be present but may not be detectable by the assay. Therefore, the results are not reported as negative. If an inhibitor was present, the internal control would not have been positive. If the viral load was 50 copies/mL (or above), a positive target signal would have been present.

44

A melt curve for JC and BK positive controls shows a Tm of 67C for BK and 73C for JC. If the sample is positive for BK virus and not for JC virus, what will be the Tm of the sample?

a. 67C

b. 70C

c. 77C

d. 73C

a. 67C

Since SyBr green labels the PCR products nonspecifically, the melt curve is used to distinguish the PCR product from the two viral genomes. If BK virus is present, the Tm of the product will be 67°C. If JC virus is present, the Tm of the product will be 73°C. Variant viruses may have melt temperatures different from those expected (70°C and 77°C). In this case, direct sequencing or other methods may be used to determine the viral sequence.

45

The Tm of the BK PCR prodcut is lower than that for the JC PCR product. If both products are the same length, what is one explanation for this difference in Tm?

a. there is more JC virus than BK virus in the sample

b. the BK virus did not amplify well

c. the BK region amplified is more AT-rich than the region amplified for JC

d. the BK region amplified is more GC-rich than the region amplified for JC

c. the BK region amplified is more AT-rich than the region amplified for JC

The melting temperature Tm depends on the number of AT and GC base pairs in the double-stranded DNA. Since AT base pairs have two hydrogen bonds and GC base pairs have three hydrogen bonds, it takes more energy (higher temperature) to dissociate a GC-rich template. The Tm does not depend on the amount of template (viral genomes) nor the concentration of the PCR product.

46

A positive ELISA test result indicates which of the following?

a. the presence of HCV virus

b. current or previous infection with HCV

c. new HCV infection

d. the absence of HCV virus

b. current or previous infection with HCV

The ELISA test for HCV detects antibodies to the virus. These antibodies may be present in the absence of virus but do not rule out either the presence or absence of virus. A new HCV infection will have detectable virus but not antibodies detectable by ELISA.

47

What is the predicted result for the RT-PCR test if HCV is present?

a. no amplification

b. amplification of the internal control, no amplification of target

c. amplification of the internal control and amplification of the target

d. visible products in the reagent blank

c. amplification of the internal control and amplification of the target

A test that is interpreted by the presence or absence of a PCR product requires an internal control to distinguish true-negative from false-negative results. Amplification of the internal control without amplification of the target is a true-negative result. In the presence of virus, the RT-PCR test should be positive with amplification of the target and the internal control. No product from the target nor amplification control indicates failure of the assay. Visible products in the reagent blank indicates the presence of contamination.

48

In this case, the ELISA was positive and the RT-PCR test was also positive. How is this interpreted?

a. ongoing infection

b. no infection

c. new infection

d. indeterminant interpretation

a. ongoing infection

A blood sample taken from a patient with an ongoing infection will have virus and antibody present. A new infection will have virus but may not have developed antibodies to the virus. If there is no infection, the RT-PCR test will be negative.

49

What additional infrmation can be obtained using quantitative RT-PCR?

a. the amount of time since infection

b. the viral load

c. source of infection

d. predicted time to viral clearance

b. the viral load

The number of viral RNA genomes per unit volume (viral load) is quantified using qRT-PCR. Unless there is a predictable expression profile from time of infection, this level does not indicate how long the virus will be present, the source of infection, nor the time since infection.

50

The HCV virus has several genotypes. The genotypes of HCV present in a given patient determines the treatment protocol that is used on that patient, as particular genotypes are associated with certain antiviral resistance patterns. Which of the following methods is used to determine genotype?

a. ELISA

b. qRT-PCR for viral load

c. direct sequencing

d. HPLC

c. direct sequencing

HCV genotyping is performed by analyzing the viral nucleotide sequence in the core or 5′ untranslated regions of the viral genome. ELISA and HPLC are not used for nucleotide sequence detection. qRT-PCR for viral load quantifies the number of viral genomes by targeting sequences present in all viral genotypes. Genotyping can then be performed by direct sequencing.

51

How will an expansion affect the PCR product in this test?

a. no effect - the product must be probed to detect the expansion

b. the PCR product will be smaller than if no expansion was present

c. the expansion will increase the size of the PCR product

d. there will be no PCR product, as the expansion will be too large to amplify efficiently.

c. the expansion will increase the size of the PCR product

The clinically significant GCN expansion in CCHS ranges from 10 to 40 additional triplets. This will add 30 to 120 additional base pairs in the PCR product, making the CCHS product larger than if there was no PCR product. No probe is required, as the size of the PCR product is diagnostic of the expansion. The efficiency of PCR decreases with increasing size of the target amplicon; however, the upper limit of this expansion does not prevent amplification.

52

For what amino acid(s) does GCN code?

a. any of several amino acids, depending on the base represented by N

b. stop codons

c. lysine or arginine

d. alanine

d. alanine

Due to the degenerate nature of the genetic code, some amino acids have more than one codon. The third position in the codon is most often variable. As long as the codon starts with GC, the amino acid will be alanine. Lysine and arginine start with AA and CG, respectively. Stop codons would terminate translation and truncate the protein.

53

The PCR procedure can be performed with fluorescently labeled primers and polyacrylamide gel electrophoresis (PAGE). What would be the advantage of this procedure over agarose gel analysis?

a. PAGE allows for higher resolution than the agarose gel procedure

b. the PAGE procedure is faster than using agarose gel electrophoresis

c. the labeled primers are less expensive, as they are used at a lower concentration in the reaction mix

d. the PAGE procedure is less prone to contamination

a. PAGE allows for higher resolution than the agarose gel procedure

Polyacrylamide provides resolution to one base pair for DNA electrophoresis. Agarose gel electrophoresis, which is less expensive, faster, and simpler to perform, can be used as a rapid screen to detect expansion products. PAGE is used to determine the number of additional triplets in the expansion, which is associated with the severity of the disease. The concentration of primers is determined by the requirements for the optimal amplification. Even if primers are labeled, they must be used at optimal concentration in the reaction mix in order to support reliable amplification.

54

The PCR results showed a strong product at about 210 bp and an additional product in the same lane at 240 bp. To which of the following interpretations is this consistent?

a. the patient has CCHS

b. The patient does not have CCHS

c. due to the presence of multiple product, the PCR must be repeated

d. the amplified region is unstable

a. the patient has CCHS

One of the two genomic copies of the PHOX2B gene has the repeat expansion. The 210-bp product represents the normal copy of the gene, while the larger 240-bp product harbors the repeat expansion. This is consistent with CCHS. The presence of the 210-bp product would be consistent only with the patient not having CCHS. When one copy of PHOX2B is normal and one is expanded, as in this case, multiple (two) products will be present. Instability of the amplified region is unlikely, since the region of interest is in a coding region; however, if instability was present, unpredictable numbers of multiple products would be detected.

55

Based on the results from the agarose PCR (240-bp product), how many arginines are likely to be inserted in the PHOXB sequence?

a. 30-40

b. 5-8

c. 15-20

d. 8-12

d. 8-12

The expanded 240-bp product is 30 bp larger than the normal 210-bp product. With 3 bp per codon, 30 bp would translate to 10 amino acids. Due to the limited resolution of agarose gel electrophoresis, the exact number of inserted amino acids may be slightly more or less than 10. A more accurate count of the number of inserted amino acids will have to be determined using PAGE. The PAGE would likely result in a pattern, as shown in Chapter 13, Figure 13-26, middle panel, lane 8.

56

What cells in the amniotic fluid are being analyzed?

a. cells from the mother

b. cells from the fetus

c. cells from the father

d. cells from infectious microbes

b. cells from the fetus

The purpose of amniocentesis is to examine the condition of the developing fetus. Examination of the amniotic fluid may also include color analysis and chemistry. Karyotype analysis of the fetal cells will reveal chromosomal number or structural abnormalities, if present. Cells from either parent don't necessarily have to be analyzed unless they are concerned about being carriers of genetic diseases. These tests would normally be performed on blood samples. Amniocentesis is not performed to detect infectious diseases.

57

What can be done with the aliquot of cells in the methanol fixative?

a. these cells can be used for FISH studies, if required

b. these cells can be cultured at a later date for further studies, if required

c. the fixed cells will be examined microscopically to look for cellular abnormalities

d. these cells can be stored for future validation studies

a. these cells can be used for FISH studies, if required

Certain chromosomal abnormalities, such as microdeletions, cannot easily be detected by karyotyping. The probes used in fluorescent in situ hybridization (FISH) clearly reveal specific abnormalities. The disadvantage of FISH studies is that only the targeted abnormality will be detectable. Proteins are denatured in fixed cells so that the cells are not viable and cannot be cultured after fixation. They are discarded after the case is resolved. Cell morphology studies are not performed as part of amniotic fluid analysis.

58

What is the purpose of the PHA in the karyotyping procedure?

a. to kill any contaminating bacteria or viruses in culture

b. to maximize the number of cells in metaphase

c. to make the cells more permeable to the colcemid

d. to denature the cell proteins

b. to maximize the number of cells in metaphase

PHA is a mitogen—that is, it stimulates cell division. This will result in increased numbers of dividing cells in metaphase. Addition of colcemid will fix these cells at metaphase where the chromosomes are highly compacted and most identifiable. PHA does not affect the permeability of the cells, nor does it have antibiotic properties. Denaturation of the cell proteins will result in loss of viability.

59

If the fixed cells are treated by FISH with green X-chromosome and orange Y-chromosome specific probes, what is the pattern observed in the nucleus?

a. one green signal and one orange signal

b. one green signal and two orange signals

c. two green signals and one orange signal

b. two green signals and two orange signals

c. two green signals and one orange signal

The XXY karyotype is consistent with two green signals and one orange signal per nucleus. Other patterns may appear microscopically, due to nuclei overlapping, signals merging with one another, or background noise. In order to be accepted as a true result, the number of XXY nuclei must exceed any background signal activity as determined in test validation exercises. A normal male nucleus will contain one orange signal and one green signal. A normal female nucleus will contain two green signals.

60

What is the predicted condition of the baby boy?

a. none

b. hermaphroditism

c. Turner syndrome

d. Klinefelter's syndrome

d. Klinefelter's syndrome

The baby will have Klinefelter's syndrome, which usually includes symptoms of low testosterone, long legs, and gynecomastia. Hermaphroditism is caused by hormonal imbalances, rather than chromosomal abnormalities. Turner syndrome occurs in females and has a karyotype of 45 X and variants.

61

What is the technical interpretation of the gel pattern?

a. clonal

b. monotypic

c. polyclonal

d. positive

c. polyclonal

The gel pattern is a smear, indicating the presence of multiple gene rearrangements. This is seen in normal polyclonal samples. Positive results—that is, results from an amplifiable monoclonal (also called monotypic) population—would be a sharp band.

62

What is one explanation for the discrepancy between the flow cytometry results and the PCR results?

a. the specimen was tested immediately by flow cytometry, but the PCR test was delayed, affecting sample integrity

b. the patient does not have lymphoma and the flow cytometry results is a false-positive

c. after flow cytometry testing, not enough sample remained for accurate PCR analysis

d. the monoclonal (monotypic) population has an immunoglobin heavy chain gene rearrangment no recognized by the primers used in the PCR test.

d. the monoclonal (monotypic) population has an immunoglobin heavy chain gene rearrangment no recognized by the primers used in the PCR test.

The gene rearrangement process produces millions of different sequences. PCR methods employ consensus primers or families of primers in order to hybridize to and prime amplification of the many different possible sequences. Still, primer sets do not successfully amplify all possible gene rearrangements, leading to false-negative or indeterminate results. Due to this target variability, it is much more likely that the PCR test is false-negative than the flow cytometry is false-positive. Since the PCR test targets DNA, a slight delay in testing between flow cytometry and PCR should not affect results as long as the specimen is held under the correct conditions. If insufficient quantity of a sample is not available, testing should not be performed. In this case, a (polyclonal) result was obtained, indicating that sufficient material was present for analysis.

63

Which of the following studies may produce results consistent with the flow cytometry findings in this case?

a. T-cell receptor gamma gene rearrangement test

b. T-cell receptor beta gene rearrangement test

c. Immunoglobulin light chain gene rearrangement test

d. repeat immunoglobulin heavy chair gene rearrangement test using more specimen

c. Immunoglobulin light chain gene rearrangement test

The flow cytometry observation of a clonal population of cells expressing the immunoglobulin kappa light chain indicates a B-cell population. Monoclonal T-cell receptor gene rearrangements are unlikely. Using more specimen will not result in detection of the clone, as the primers do not hybridize to the clonal gene rearrangement in this case. Immunoglobulin light chain gene rearrangement uses primers that are complementary to the immunoglobulin kappa and lambda light chain genes, which may be able to detect the population identified by flow cytometry.

64

Which of the following translocations might be seen in the clonal B-cell population?

a. t(15;17)

b. t(14;18)

c. t(18;21)

d. t(8;14)

b. t(14;18)

The most frequently occurring lymphoma in adults is follicular lymphoma. Almost every follicular lymphoma has a t(14;18) translocation in addition to the immunoglobulin gene rearrangement. Of the other translocations, t(15;17) is found in promyelocytic leukemia, t(8;21) is associated with acute myelogenous leukemia and acute lymphoblastic leukemia, and the t(8;14) is a translocation found in T-cell lymphoma.

65

Which of the following describes a gene rearrangement?

a. an abnormal event that occurs only in tumor cells

b. a normal event that occurs in lymphocytes

c. a normal event that occurs in myeloid cells

d. an abnormal gene expression detected in all types of leukemia

b. a normal event that occurs in lymphocytes

Immunoglobulin heavy chain and light chain gene rearrangements are normal processes in lymphocytes. The purpose of the gene rearrangement is to generate antibody diversity. Gene rearrangements occur independently in each cell so that each cell can produce a different antibody. Therefore, in a normal cell population, the gene rearrangements are polyclonal. In contrast, all cells in a cell population arising from a single abnormal cell will share the same gene rearrangement, which is detectable by PCR as a sharp band or peak after gel or capillary electrophoresis (Fig. 14-21, lane 9). The gene rearrangement itself does not have any influence on the abnormal cell behavior. Gene rearrangements do not occur in myeloid cells and in most tumors arising from these types of cells. Rearranged immunoglobulin genes are expressed in normal polyclonal and monoclonal cell populations. Detection of a monoclonal or monotypic population is not associated with abnormal expression of these genes.

66

What is the significance of this test result?

a. Erbitux is recommended for this patient

b. Erbitux is not recommended in this case

c. Erbitux will be used at a very low dosage, due to possible side effects

d. standard therapy will be discontinued immediately

b. Erbitux is not recommended in this case

The B-raf V600E mutation predicts poor response to Erbitux treatment. The presence of this mutation is considered equivalent to the presence of a K-ras activating mutation, as the two genes are on the same signal transduction pathway, and activation of either will result in an overproliferation phenotype. Low dosage of Erbitux would not be effective regardless of side effects and therefore would not be used in this case. Standard therapy or another type of targeted therapy is a better option for this patient. Standard therapy should be continued.

67

What can be concluded regarding the likelihood of a KRAS mutation with the presence of a BRAF mutation?

a. there is a high likelihood of a KRAS mutation

b. a KRAS mutation is unlikely

c. the effect of a KRAS mutation would be additive to that of the BRAF mutation

d. KRAS mutations will restore sensitivity to Erbitux

b. a KRAS mutation is unlikely

`K-ras and B-raf mutations are mutually exclusive. Since K-ras and B-raf are on the same signal transduction pathway (see Fig. 14-6), the presence of activating mutations in either one is sufficient to promote resistance to Erbitux. The B-raf mutation removes selective pressure or provides growth advantage such that mutations in both genes are not found. It is unlikely that the concurrent presence of both mutations would be additive, as they both activate the same pathway, nor would one mitigate the effect of the other.

68

What is the V600E mutation?

a. deletion of valine at position 600

b. a conservative substitution

c. substitution of glutamic acid for valine at amino acid position 600

d. a substitution of valine with glutamine at amino acid position 600

c. substitution of glutamic acid for valine at amino acid position 600

In the single letter code, V stands for "valine" and E stands for "glutamic acid." The recommended nomenclature is the original amino acid, its position, and the substituted amino acid. Deletion of valine would be indicated by V600del. Since valine has a hydrophobic side chain and glutamic acid has an electrically charged side chain, this substitution is not conservative.

69

One codon in BRAF and five codons in KRAS account for the vast majority of clinically important mutations. Why is this mutation spectrum limited?

a. the mutated protein must be completely inactivated

b. not enough patients have been tested to see other frequently occuring mutations

c. molecular methods are limited to specific areas of these genes

d. the mutated protein must remain functional and aberrantly activated

d. the mutated protein must remain functional and aberrantly activated

Generally, there are a limited number of changes that can be made to protein coding sequence that would preserve or enhance activity. In contrast, many mutations can inactivate proteins. The substitutions found in Erbitux-resistance cases are limited to those that produce aberrant protein activity. Since the introduction of therapy targeting this signaling pathway, thousands of patients have been tested for K-ras and B-raf mutations. Mutations in codons 11, 12, 13, 22, and 61 of K-ras and codon 600 of B-raf are the changes most frequently found.

70

What is the recommendation of the National Comprehensive Cancer Network (NCCN) with regard to KRAS testing?

a. always test BRAF is KRAS mutations are not found

b. test for mutations in KRAS or BRAF, but never both

c. always test for mutations in codon 61 of KRAS

d. always test for mutations in BRAF before KRAS testing

a. always test BRAF is KRAS mutations are not found

Since B-raf V600E is functionally equivalent to  K-ras mutations with regard to therapeutic resistance, the NCCN recommends further testing of B-raf if no mutations are found in K-ras. If activating mutation is found in one of the other genes, there is no requirement to test the remaining gene, as simultaneous mutations in K-ras and B-raf are unlikely. Laboratories and procedures differ in the mutation spectrum detection. Activating mutations in codon 61 of the K-ras gene are relatively rare, such that there are no recommendations requiring that this specific codon be tested. K-ras and B-raf testing is not done in any particular order. Laboratories may choose to test for mutations in either K-ras or B-raf first, and then if a mutation is found, the other test does not have to be done.

71