Molecular Diagnostics Sample Questions PDF
Document Details
Uploaded by FinerUniverse
San Lorenzo Ruiz College of Ormoc, Inc.
Tags
Related
- Molecular Diagnostics Lecture 5 Mutation and Cancer Genes PDF
- Papillary Lesions of the Breast - Review and Practical Issues PDF
- Gene Structure and Regulation PDF
- Medical Microbiology: Bacteriology Lecture 5 PDF
- Molecular Techniques For Identifying Microorganisms From Clinical Specimens PDF
- Biot 401 Lecture 2: Samples Preparation & Analysis (PDF)
Summary
This document provides sample questions covering molecular diagnostics. The questions span various topics within molecular biology, including human genome structure, polymerase chain reactions, and electrophoresis. Examples related to inheritance are also present.
Full Transcript
222 MOLECULAR DIAGNOSTICS SAMPLE QUESTIONS 1. Human genome consists of: 9. What are the chances that individual a. Haploid copy number number 27 will be affected with an b. 22 chromosomes a...
222 MOLECULAR DIAGNOSTICS SAMPLE QUESTIONS 1. Human genome consists of: 9. What are the chances that individual a. Haploid copy number number 27 will be affected with an b. 22 chromosomes autosomal dominant trait? c. Circular structure a. 0% d. Approximately 6 billion bases b. 25% C. 50% 2. The migration rate of a macromolecule d. 100% through a gel matrix during electrophoresis depends on: a. Net charge on the molecule b. Size of the molecule c. Thickness of gel d. All of the above II 3. One cycle of a polymerase chain reaction (PCR) includes: a. b. Denaturation, digestion, detection Denaturation, annealing, extension '" ti 18 1 Dark = Affected (dominant) c. Annealing, detection, extension d. Digestion, annealing, extension 10. X-linked recessive traits are more common in: 4. RT-PCR involves all of the following except: a. Males a. DNA isolation b. Females b. Reverse transcription c. Children c. PCR amplification d. None of the above d. Product analysis 11. What type of inheritance pattern is 5. If someone has a normal and mutant represented in this pedigree? banding pattern, they are referred to as: a. Autosomal dominant a. Heterozygous b. Autosomal recessive b. Compound homozygous c. X-linked recessive c. Homozygous d. X -linked dominant d. Wild type 6. The coding sequences of a gene are known as: a. Introns b. Exons c. Splice sites d. Frameshifts 7. Restriction endonucleases recognize specific: a. Methylation patterns h. Trinucleotide repeats c. Palindromic DNA sequences d. DNA-damaged sites 8. Western Blotting is a method used to detect which of the following: a. DNA h. Protein c. RNA d. Mutations 223 12. Using the picture below, which is the SNP? a. t/a b. c/c 16. A DNA preparation hos the following c. c/a absorbonce readings d. t/g A260=0.260 ATTCGCATGCCTAGTCAAA TGC Abnormal Allele A28o=0.230 A TTCGAATGCCTAGTCAAATGC Normal Allele Based on the ratio, which of the following choices 13. According to the following western blot, is correct? which vegetables contain chlorophyll? The sample is a. Green bean, carrot and asparagus a. suitable b. Green bean, squash, and asparagus b. contaminated with chloroform c. Green bean, asparagus , and c. contaminated with phenol spinach d. contaminated with protein d. Green been, carrot and squash = Lane 1 Control protein - chlorophyll Lane 2 = Green bean extract 17. A DNA sample is isolated from peripheral lane 3 = Carrot extract = Lane 4 Asparagus extract blood cells of a patient. When performing lane 5 = Spinach extract Lane 6 = Yellow squash extract spectrophotometic analysis to determine the yield of DNA in the sample, you find the 1:50 dilution of the 0.4 ml sample gives an OD260 reading of 0.041. What is the total 14. The absorbance reading at 260 nm for a amount of DNA contained in the 0.4 ml 1:100 dilution of a DNA sample is 0.430. sample? Which of the following is the correct calculation for the DNA concentration? a. 0.172 ug/mL a..41 ug b. 0.215 ug/mL b. 4.1 ug c. 1720 ug/mL C. 41 ug d. 2150 ug/mL d. 410 ug 15. The absorbance reading at 260nm for a J:200 dilution of RNA sample is 0.210. Which of the following is the correct calculation for the RNA concentration. a. 1680 ug/ml b. 16.80 ug/ml C. 1.680 ug/ml d. 0.168 ug/ml 224 7. C ANSWERSAND Restriction endonucleases r ecognize RATIONALE specific 4 - 5 nucleotide palindromic (r eads the same in either direction) DNA sequences. l. D 8. B The human genome is dipoid, not haploid. Wes tern Blotting is used to identify It has a total of 23 chromosome pairs , not 22. proteins through the use of SDS-PAGE. It is linear, not circular. But it does have Southern Blotting identifies DNA and about 6 billion bases. Northern Blotting identifies RNA. Mutations 2. D are usually identified in the genomic DNA, therefore, a W estern would not b e u sed. The migration rate of a m acromolecule through a gel matrix during electrophoresis 9. C depends on all three , the thickness of the The parent, # 13 on the chart, is agarose gel, the n et charge of the molecule, heterozygous for the dominant trait since his and the size of the molecule. father, # 1 on the chart, is homozygou s for a 3. B r ecessive trait. Ther efore, #27 has a 50% chance of b eing affected.. If you quickly One cycle of a polymerase chain reaction sketch a Punnett Square Aa x aa for this (PCR) includes a denaturation step to brnak pedigree - you will see tha t half will have the the strand apart, an annealing step so that the tt·ait (Aa). primers can sit down on the appropriate area 10. A of the DNA, and an extension so that the Taq polymerase can make a copy of the strand of X-linked traits are always more common in DNA. males because males have one X chromosome 4. A (Xl') as opposed to females who have two (XX). In females the other X chromosome RT-PCR involves PCR amplification of the carries the dominant genes so as carrier s, they product, reverse tran scriptase is the enzyme do not display the trait. Males with only one that is u sed in RT-PCR, and the product must X chromosome display whatever traits are on b e analyzed. However , RT-PCR requires that chromosome, dominant or recessive. isolation of RNA, not DNA. 11. B S. A This pedigree is an example of an If someone has a normal and mutant autosomal recessive trait. In an autosomal banding pattern, they are r eferred to as r ecessive disorder, an individual must have heterozygous. Homozygous individuals have two copies of the abnormal gene in order to be identical patterns. Wildtype r efers to the affected. Also, autosomal r ecessive traits " normal" or prototype cell. commonly skip generations. It is not an X- linked r ecessive trait becau se there is male to 6. B male transmission - father in generation I would have to have passed it to his son in The coding sequences of a gene are known generation II for him to pass to his daughter in as exons. The introns are spliced out b efor e generation III. Fathers can only pass the X the gene is translate d. The splice site is just linked recessive genes to their daughters (they the point at which an enzyme can cut the get the X chromosome) not to their son s (they DNA. A frameshift mutation is when a get the Y chromosome). nucleotide is either deleted or added to a sequence such that the protein produced is altered. 225 12. C 17. C A polymorphism is a change in the DNA First calculate the DNA concentration then sequence. In this example, the SNP or single multiply it by the hydrating volume. nucleotide polymorhphism is an A in the normal allele that changed to a C in the DNA concentration= abnormal allele. If you match the top and bottom sequences, they are identical except 0.041 X 50 ug/ml X 50 =102.5 ug/ml for the sixth nucleotide in from the left. 13. C DNA yield= 102.5 ug/ml X 0.4 ml A Western blot is used to analyze the = 41 ug presence of protein in a sample. Therefore, if the specific protein is present, you will see a product on the Western blot. Lanes 1, 2, 4, and 5 have a product. Lane one is the positive control and should have a product. Lane 2, 4, and 5 are extracts from green bean, asparagus and spinach. Hence, these are the vegetables that contain chlorophyll. 14. D The calculation to determine the DNA concentration is OD260 (reading of the DNA sample) multiplied by the dilution factor. multiplied by 50 ( 1 OD unit at 260 mn for DNA) 0.430 X 100 X 50 = 2150 ug/mL IS. A The calculation to determine the RNA concentration is OD260 (absorbance reading of the RNA sample) multiplied hy the dilution References factor, multiplied by 40 (1 OD unit at 260 nm Tsongalis, Gregory T. and William B. Coleman. for RNA). Therefore, Molecular Diagnostics A training and study guide. 0.210 X 200 X 40 ug/ml = 1680 ug/ml Copyright 2002. American Association for Clinical Chemistry Press. Washington, DC. 16. D Roche Diagnostics. Molecular Technology To determine whether a DNA sample is Education Program. Copyright 2010 CD-rom contaminated or not, you divide the A260 by version 6.0. Go to www.rochegenetics.com to A280. For DNA, if the ratio is