Molecular Diagnostics & Genetics PDF

209 MOLECULAR DIAGNOSTICS & GENETICS by Mona Bakeer, Elizabeth Williams, Marcia Firmani Key Terminology Agarose...

209 MOLECULAR DIAGNOSTICS & GENETICS by Mona Bakeer, Elizabeth Williams, Marcia Firmani Key Terminology Agarose - a gel-forming polysaccharide Moleculcr Diagnostics - the use of DNA, extracted from seaweed RNA or mRNA to identify and /or characterize disease caused by Allele - alternative forms of a gene that is infectious agents or gene abnormalities present at a given locus Mutations - changes in the DNA sequence Cell nucleus - carries genetic information Penetrance - the probability of expressing a Codon - 3 bases in mRNA that code for phenotype, given a particular genotype amino acid production Phenotype - observable characteristics Diploid - cells that carry two genome copies Polymorphism - a variation in the base Epigenetics - stud[ of changes in the sequence of DNA regulations o gene activity and expression that are not dependent on Proteins - made up of amino acids gene DNA sequence Protein expression - Differcnt_proteins are Exon - coding DNA (translated into a expressed in different cells according to protein) the function of the cell Gene Expression - protein synthesis (gene Proteomics - organism's complete product)- is tightly controlled and complement of proteins regulated RFLP (Restriction Frag-nent Length Genetic Code- combination of nucleotides Polymorphism)- variation in the size of that build the different codons DNA fragments generated by restriction enzymes Genome - an organism's total DNA content Sequence - the order of nucleotide hases Genotype - the observed alleles for an along a DNA strand individual at a genetic locus STR (short tandem repeats) - Short Haploid- -cells that contain a single copy of sequences of DNA, normally 2-5 base the genome such as germ cells or pairs, that are repeated numerous gametes times Haplotype - series of alleles on a single Telomere - Re~on of repetitive DNA at the chromosome end of a chromosome Heterozygous - two different alleles at a Transcription - process where genetic locus information in DNA is copied into Homozygous - two identical alleles at a messenger RNA (mRNA) utilizing the locus RNA polymerase enzyme lntron- non-coding DNA between 2 exons Transcriptome - set of all RNA molecules transcribed in a cell Ligation - process ofjoining two DNA molecule ends. It involves creating a Transgene - a foreign gene that is phosphodiester bond between 3' introduced randomly somewhere in the hydroxy of one nucleotide and the genome 5 phosphate of another. Translation - making a protein using the Linkage disequilibrium - allelic association information provided by mRNA when closely linked alleles are inherited VNTR ( variable number of tandem repeats) - together during many generations Repeats of identical nucleotide Locus - location of a gene in the genome sequences lined up one after another that vary in number from one individual to another i.e., gaca gaca gaca gaca gaca gaca 210 Molecular Diagnostics APPLICATION 1. Detect cause of disease state (diagnosis) 2. Predict disease progression 3. Paternity and forensic analysis DNA REPLICATION (SEMI-CONSERVATIVE) 4. Gene therapy and d rug design 1. DNA strands separate (hclicasc enzyme) Genetics and Molecular Biology 2. Pairing the bases in each strand with new bases to form complementary DNA THE CHEMICAL BASIS OF HEREDITY strands (DNA polymerase) 1. Biological " blueprint" 3. Produce two new DNA strands (exact 2. Carries information for cells to live, duplicate of original DNA) grow, differentiate, and replicate THE HUMAN CHROMOSOME 3. Provides consistency and variability 1. Single linear duplex DNA GENETIC UNITS OF HUMAN DNA 2. Numerous protein interactions 1. Nuclear DNA a. Diploid genome (two sets of 3. Karyotype chromosom es) a. Looking for normality in number b. Packaged in 23 pairs of and structur e chromosomes b. Number ed by size and centromere c. 22 homologous pairs (a utosomes) position d. 2 sex chromosomes (XX or XY) ❖ short arm -p e. 6 billion b ases ❖ long arm - q f. Approximately 30,000 genes ❖ regions counted from centromere out 2. Mitochondrial DNA (non-nuclear ) c. Pattern produced by intensity of a. 16,569 base pairs stain or fluoresence h. 37 genes d. Genot ype - genes on homologous c. Higher mutation rate chromosomes d. 128 naturally occurring ❖ gene 1 - a,b polymorphisms ❖ gene 2 - r ,s e. Maternal inheritance ❖ gene 3 - t,t 3. DNA/ RNA comprised of 2 types of e. Haplotype - genes on one nitrogen bases: chromosome ❖ left chromosome - at a particular a. Purines ❖ Adenine A DNA RNA locus defining genes 123 - b,s,t ❖ Guanine G DNA RNA ❖ right chromosome - genes 123- b. Pyrimidines a,r,t ❖ Cytosine C DNA R NA ❖ Th ymine T DNA ❖ Uracil U RNA d. Hydrogen bonds can form h etw6€n a pyrimidine and a purine e. Watson-Crick base pairing A=T, G =C 4. DNA double helix a. Nucleotide bases p air together to form "base pairs", A always binds 1 b to T ( U in RNA) , C always binds to G 2 s b. DNA double helix are oriented in opposite directions. 3 t c. 5' end is beginning of DNA strand d. 3' end is end of DNA strand  211 Transcription REMEMBER! Nu J.eotide Bases Tranacrlpdon R"IA G-CAT AAAA The Central Dogma of HUMAN GENE STRUCTURE Molecular Biology 1. Genes G a. Exons separated by introns b. Sequence of base pairs encodes NA.. RNA.. Protein information for proteins. replication transcription translation c. Different sequence= different p rotein TRANSLATION / GENETIC CODE 1. E ach combination of 3 nucleotides on.J u._ ,...,e:.,._...__, _______,_ o_e_, "~----~-tr_~_c1 - _~_r:_. _,,~ mRNA is called codon 2. Each codon specifies a particular amino Exo'I acid ~~ !!=Jr~ 3. Each amino acid is to be placed in the If/ polypeptide chain (protein) TRANSCRIPTION ~ 1. mR NA sequence - complementary to 5· NNNNNNN;;;c7AGCCA~ = 3· the DNA template a. Uracil ( U) bases replace thymine (T) bases in RNA Codon Usage AUG-CUC-GGG-AGC-CAU-UAA 2. mRNA processed by transcription: a. Splicing - r emoval of introns Translation b. Capping - modify the 5 ' end ♦ c. Polyadenylations - add adenines to Peptide S equence Met-Lcu-Glv~Ser-His the 3 'end (also called poly-A tail) @ REMEMBER! INtrons remain IN the nucleus; EXonsEXit Lpl--I ~ Telopl--I C Molecular Techniques @--· @)@) BASIC STEPS IN ISOLATING DNA FROM CLINICAL SPECIMENS 0 -pi--r Separate WBCs from RBCs, if necessary t lyse WBCs or other nucleated cells t Denature / Digest proteins t Separate contaminants (e.g., proteins, heme) from D NA t Precipitate DNA, if necessary t Resuspend DNA in final buffer DNA ISOLATION METHODS 1. Liquid phase organic extraction (phenol/chloroform) 2. Liquid phase non-organic extraction CHROMOSOMAL CROSSOVER 3. Solid phase procedures (GENETIC RECOMBINATION) ISOLATION AND PURIFICATION OF MJCLEIC ACID 1. Two chromosomes (paired up during 1. Sample source: prophase I of meiosis) exchange some a. Any tissue with nucleus portion of their DNA b. Blood collected in anticoagulant 2. Matching r egions of matching ❖ EDTA preferred chromosomes break then reconnect to ❖ DO NOT FREEZE WHOLE other chromosome (ex change of gen es) BLOOD 2. Storage conditions a. Store DNA in TE buffer at 4°C for weeks and at -20°C to - 70°C- for long term b. Store RNA in RNase free ultrapure water at - 70°C 216 NUCLEIC ACID ANALYSIS 1. DNA or RNA quantity , quality and REMEMBER! molecular size characterized b y Isolation of DNA from a. UV spectrophotometry b. Agarose gel electrophoresis Clinical Specimens c. Fluorometry Some = Separate d. Colormetric blotting L azy= Lyse NUCLEIC ACID QUANTITATION USING UV Dogs = Denature SPECTROPHOTOMETRY C an = Contaminants P lay = Precipitate 1. DNA and RNA absorb at 260 nm R ight = Resuspend 2. Proteins absorb at 280 nm 3. [DNA]= A2oo X dilution factor X 50 udm1 lOD unit at 260 run (A2oo)= 501tglml otDNA 4. [RNA]=A260 X dilution factor X 40 ug'ml lOD unit at 260 nm (.A260}= 40 uglml of RNA Concentration= ug of DNA or RNA per ml of hydrating solution Excrrple 1: DNA concentration A260 X Dilution Factor X 50 rig/ml DNA prepcrntion diluted l:200 yields A 200 reading of 0.200 DNA conc:entratim (ug/ml) = 0.200 d:>sai:xnce mits X 200 X SO ug/ml per rosort:x:nce mit = 200J ug/ml RNA concentration- A260 X Dilution Factor X 40 uglml Excrrple2 RNA prepcratim diluted 1:10 yields A2w reading of 0.500 Nucleic acid purity RNA Concentration = 0.500 ct>sorbcnce Lnit X IO X 40 ug/ml per rosort:x:nce mit = 200 ug/ml QUALITY FROM UV SPECTROPHOTOMETRY 1. A260/A230= mea sure of purity 2. 1.8-2.0 = good DNA or RNA RESTRICTION ENDONUCLEASES 3. Less than 1.8 = too much protein or 1. Bacterial enzymes cut or nick sp ecific other contaminants sites of a D A sequence ( "molecular scalpels'') CALCULATING NUCLEIC AOD YIELD 1. DNA/ RNA concentration multiplied by 2. R ecognize specific short sequences of DNA, usually 4 or 6 b ases but some are volume of hydrating solution 5, 8 or longer and cleave at or n ear Example: H DNA concentration from UV recognition site spectrophotometry = 250 uglml 3. R ecognition sequences are palindromes Volume of hydration solution = 0.1 ml Then DNA yield = 250 uglml X 0.1 ml= 25 ug  217 DNA MICROARRAYS (DNA CHIP) Cohesive Cohesive Blunt 1. Use hybridization technology to Ends Ends examine gene expression Ends (5' overhang) (3' overhang) 2. Arrangement of DNA sequences on BamHl Kpnl HaeIII solid support 3. E ach micro array contains thousands of t t t genes G Gjc C 4. Simultaneously monitor gene ~f..I.9.C GGTAC:C expression levels in all these genes C C~G G + CCTAG.'3 C;CATGG 5. Used for: l ♦ a. Gene expression studies b. Disease diagnosis c. Pharmacogenetics ( drug discovery) 6. Special instrumentation for a. Generation of micro arrays b. Analysis of results REMEMBER! Palindromes Palindromes are the same sequence on both DNA strands when read in either direction: 5' - GGTACC - 3' GENE CLONING (RECOMBINANT DNA TECHNOLOGY) 3'- CCATGG - 5' 1. I solating and amplifying a defined DNA sequence 2. Uses a vector, such as a plasmid and an appropriate host , such as E. coli DNA SEQUENCING 1. Determine order of nucleotides in a DNA molecule a. Maxam- Gilbert (chemical degradation) b. Sanger method (dideoxy chain termination) c. P yrosequencing (sequence b y synthesis) d. Next generation sequencing ( sequence single molecules of DNA in real time) ❖ Pacific Biosciences ❖ Oxford Nanopore ❖ Life Sciences Qdot technology 218 SINGLE NUCLEOTIDE POLYMORPHISMS (SNP) SIGNAL AMPLIFICATION TECHNIQUES 1. Mutation of a single nucleotide 1. Branched DNA (bDNA) (A,C,T,G) 2. H ybrid capture assay (HCA) 2. Some associated with various phenotypic THE POLYMERASE CHAIN REACTION (PCR) differences a. Propensity towards disease 1. Denaturation (95°G) - separation of b. Drug resistance target dsDNA through a. Alkaline pll 3. Over 5 million S P locations identified h. Ionic strength of high salt solution in human genome c. Elevated temperature 2. Primer annealing at 50-60°C - Minisatellite: Tandem rePNtt of seqi,tnets that -vary from 14 to 100 base pairs in length h ybridize the primers to the single-...ACGATATCGGACCAATCGATCGGACCAATCGATCGGACCAATCGTAGGT... stranded template +...ACGATATCGGACCAATCGATCGGACCAATCGTAGGT... 3. Extension (72°G) - polymerize the Polymotphism."YlrialMe number of rtpms primer into the full-length gene of Microsatellite: interest Shott sequences of tand m ropN!s , 19. CA,epem...TGCCATAGCACACACACACATTAGTTAG... ♦...TGCCATAGCACACAC.ACACACACATTAGTTAG... 4. Each cycle doubles amount of DNA Potymorphiim:variable number of CA repeab SNP: Single nucleotide pol~morphlsm...TGTACCAAGT... REMEMBER! ♦...TGTAACAAGT... Denature by the SEA Salt (high ionic strength) FLUORESCENT IN-SITU HYBRIDIZATION (FISH) Elevated temp 1. H ybridization of a fluorescent DNA probe to its complementary DNA in Alkaline pH Denature morphologically preserved tissue or ~ e:;,4 cells 2. Detect and localize presence or absence CONTROLS FOR PCR of specific DNA sequence or chromosome 1. Blank r eaction a. Controls for contamination 3. Ahle to examine metaphase chromosome b. Contains all r eagents except DNA spreads as well as interph ase (non- template dividing) cells 2. egative control r eaction a. Controls for specificity of the Amplification Techniques amplification reaction b. Contains all r eagents and a DNA COPY NUMBER AMPLIFICATION template lacking the target sequence 1. Polymerase chain r eaction (PCR) 3. Positive control reaction 2. Ligase chain reaction (LCR) a. Controls for sensitivity h. Contains all r eagents and a known 3. Nucleic acid sequence-based target-containing DNA template amplification (NASBA) REVERSE TRANSCRIPTION POLYMERASE CHAIN 4. Transcription-mediated amplification REACTION (RT-PCK) (TMA) 1. Synthesis of complementary piece of 5. Strand-displacement amplification DNA (cDNA) from RNA b y r ever se (SDA) transcription (R1) 2. Amplify the cDNA target sequence by PCR  219 Target DNA Reaction mixture contains target DNA sequence to be amplified, two primers 1i1 1 11 111 11111111111 I'I (P1, P2), and heat-stable I raq ·~ P1 ~ Taq polymerase Reaction mixture is heated I I 111 I I IIIl to 95·C lo denature target DNA. Subsequent cooling......... 111 I I to 37 C allows primers to hybridize to complementary sequences in target DNA I bi il 111 , l i ~ I ~ 1111 When heated to 72°C, Taq polymerase extends complementary ) strands from primers I ~ I ,.,,,.,. I 1111 111 j 1111 j ! I I First ~~nt~~s~~~;~~eo~esults ~ I target DNA sequence ( ~· Denature Il I I I 11 i I ' DNA Hybridize Q) primers u>, (.) Extend "C C: 11 ,. , 1 : I ii I I j I I new DNA 0 strands 0 qi t/l 11 I Second synthesis cycle results in four copies of ) I I !I target ONA sequence DNA Amplification Using Polymerase Chain Reaction Reprinted with permission Copyright Holder is Cold Spring Harbor Laboratory Press REAL-TIME PCR / QUALTITATIVE PCR (qPC/('J NUCLEIC ACID SEQUENCE BASED AMPLIFICATION 1. Qualitative and/or quantitative (NASBA) technique 1. A nucleic acid sequence based 2. Detects fluorescent reporter molecule amplification technique that amplifies after each cycle RNA and rRNA Strand Displacement Amplification (SDA) 2. Direct detection of RNA viruses/ H CV and HIV 1. Target amplification u ses heat denaturation, annealing and extension HYBRID CAPTURE 2. Creates an altered target with a 1. Nucleic acid amplification technique r estriction endonuclease r ecognition used for RNA or DNA applications site. 2. Solid phase using chemiluminescence 220 BRANCHED CHAIN DNA (bDNA) 1. Uses a series of hybrid probes to elicit a signal amplification- chemiluminescence REMEMBER! 2. Detect specific RNA sequences Tech PENS Load Gels! Electrophoresis Technology P = Pore size of gel ELECTROPHORESIS OF NUCLEIC ACIDS - E = electric field N = negative DNA charge 1. Separation based on size and charge through a sieve-like matrix (agarose 01· polyacryla1nide) S = size of DNA CJ CJ CJ CJ - 2. Migration in electrical field at a rate inversely proportional to loglO of molecular size (number ofbase pairs) 3. DNA (negatively charget{) migrates toward anode (positively charged) FACTORS AFFECTING MIGRATION RATE Tortoises are big and slow Bunnies are small and fast 1. Matrix type and porosity (%) of the gel (gel castinB) Movement through a gel depends on the size of the DNA particle, its charge, and the pore size of the 2. Net charge of nucleic acid molecule gel. All negatively charged DNA particles move toward the anode, but the larger pieces have a 3. DNA conformation harder time squeezing through the small pores in the gel and cannot move as fast, i.e., as far, as the 4. Electric field strength smaller pieces. 5. Temperature of gel Blotting Techniques 6. ucleic acid base composition SOUTHERN BLOT 1. Detects specific DNA sequences 7. Presence of intercalating dyes 2. DNA denatured in the gel by an 8. Type and strength of buffer increase in pH PULSED FIELD Ga ELECTROPHORESIS OF DNA (PFGE) 3. DNA transferred to a membrane by 1. Analysis of DNA fragments up to 100 capillary action with a high salt kb in size solution 2. Separation accomplished using a pulsed electrical field 4. Labeled complementary probe used for 3. PFGE commonly used for genotyping detection prokaryotes Paper towels Nitrocellulose I filter membta ne Gel Methods, instruments, reagents & controls Sponge Routine and special procedures to verify test results High Salt Solution  221 5. Procedure: WESTERN BLOT a. Genomic DNA cut with restriction 1. Protein run on SDS- polyacrylamide enzymes gel electrophoresis (PAGE) b. DNA electrophoresed c. Gel submerged in an alkaline 2. Protein electrically transferred to solution to denature the DNA membrane ( el ectro-transfer) d. DNA transferred onto a nitrocellulose membrane by 3. Membrane incubated with a primary capillary action antibody and blocking solution e. Membrane mixed with a solution containing labeled probe 4. Membrane washed and incubated with ❖ Prohe will hybridize to secondary antibody and blocking complementary piece of DNA on solution gel 5. Membrane washed and rinsed with f. Membrane washed to remove substrate buffer excess, unbound probe g. Membrane developed and visualized 6. Substrate added and developed using either radioactive isotopes, SOUTHWESTERN BLOT chemiluminescent dyes , or 1. DNA-binding proteins colorimetric techniques REMEMBE Western Blot - Protein Southern Belle named "Dee" for DNA Western Cowboy Eats his Protein NORTHERN BLOT 1. Detect specific sequences of RNA DOTI SLOT BLOTS 2. RNA transferred to membrane by 1. Quick analysis of DNA and RNA capillary action using a high salt solution 2. Does not determine the size of target 3. Labeled complementary probe used for 3. Applied to: detection a. Expression analysis b. Mutation anaysis c. Amplification analysis «FD\ REMEMBER! REVERSE DOT BLOTS ~ Northern Blot - RNA 1. Reverse allele sp ecific oligonucleotide 2. Hybridization Northern Eskimo with an RNA Virus because it's Cold ~ 3. Important method for genotyping common human mutations l(~ STRINGENCY 1. Describes the conditions under which hybridization takes place 2. Salt, heat and formamide increase stringency 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