Molecular characterisation of weak Lu a and Lu b expression associated with altered LU*01 and LU*02 alleles
|
|
|
- Brittney Robbins
- 6 years ago
- Views:
Transcription
1 Molecular characterisation of weak Lu a and Lu b expression associated with altered LU*01 and LU*02 alleles Vanja Karamatic Crew 1, Ben Jones 1, Karen DeSay 1, Rosalind Laundy 1, Piers Walser 1, Alan Gray 2, Nicole Thornton 1 1 International Blood Group Reference Laboratory, NHSBT, Bristol, UK 2 Red Cell Immunohaematology, NHSBT, Tooting, UK
2 Donor samples 10 suspected Lu(b-) donor samples identified by serological screening at NHSBT South London using diluted monoclonal anti-lu b referred to IBGRL for phenotype confirmation At IBGRL, donor cells tested with anti-lu a and two examples of undiluted anti-lu b One sample = Lu(a+ wk b+ wk ) Nine samples = Lu(a+b+ wk )
3 UK National Rare Donor Panel Currently donors with rare blood types identified through serological screening at NHSBT South London and smaller regional programs and routine extended phenotyping Aim to identify donors whose cells lack high incidence antigens First established in the UK in 1952 to meet the demand for rare blood types NRDP includes ~ 2000 active donors Co a, Ge, I, k, Kp b, Lu b, Wr b, Vel, Lan, Yt a
4 LU5 Lutheran system LU17 LU12 Lu a /Lu b LU22 LU24 LU21 LU25 LU23 LU4 LU16 LU8/LU14 Cell adhesion molecule, laminin receptor LU6/LU9 24 antigens (LU1-LU26) carried on two glycoprotein isoforms - Lutheran and BCAM LU20 LU10 and LU15 designated obsolete LU13 Au a /Au b LU23 LU26 4 pairs are antithetic antigens: Lu a /Lu b (LU1/LU2), LU6/LU9, LU8/LU14, Au a /Au b (LU18/LU19) Remaining 16 antigens are of high frequency
5 Lutheran null phenotypes Lu(a-b-) dominant type; In(Lu) Heterozygous for a dominant suppressor gene KLF1 In(Lu) suppresses all Lutheran antigens but also a number of red cell antigens in other systems [AnWj, P 1, i, CD44 (Indian system), CR1, MER2] Numerous molecular backgrounds Rare Lu(a-b-) recessive type Homozygous for a recessive allele at the LU locus Cells lack all Lutheran antigens Five different molecular backgrounds Very rare
6 BCAM (LU) gene Lu-glycoprotein is encoded by a single gene BCAM (LU) LU locus on chromosome 19 q13.3, 2.5 kb in size, organised in 15 exons At IBGRL Sanger sequencing of the all exons of LU since 2001
7 LU sequencing of the 10 donors All 10 heterozygous for LU*A/B (LU*01/02) All 10 heterozygous for at least one missense or indel mutation in LU (associated either with a lack of HFA or novel amino acid change) LU*A/B Arg77His LU* delarg34, Leu35 LU* Arg227Cys LU* Ser447Leu LU*02.var Gln581Leu Arg436Cys Novel LU*02.var Novel LU*02.var Ala547Asp Pro492Thr
8 LU* (LU*B with c.99_104del) 4 donors heterozygous for LU* LU exon 2 c.89 DONOR c.99 c.109 N N N N N N N N N N CONTROL c.99_104delgcgctt delarg34, Leu35
9 LU* (LU*B with c.99_104del) 4 donors heterozygous for LU* LU exon 2 c.89 DONOR c.99 c.109 N N N N N N N N N N CONTROL c.99_104delgcgctt delarg34, Leu35
10 LU* and LU* donor compound heterozygous for LU* (LU*B allele) and LU* (LU*A allele) LU exon 2 c.89 DONOR c.99 c.109 N N N N N N N N N N LU exon 6 c.669 c.679 c.689 N DONOR CONTROL c.679c>t CONTROL Arg227Cys c.99_104delgcgctt delarg34, Leu35
11 LU* with LU*B(1306T) 2 donors heterozygous for LU* exon 11: c.1340c>t, Ser447Leu exon 13: c.1742a>t, Gln581Leu exon 13: silent 1671C>T and LU*B(1306T) in exon 10 c.1296 c.1306 c.1316 N DONOR CONTROL c.1306c>t Arg436Cys
12 LU* with KLF1*BGM06 1 donor was serologically Lu(a+ wk b+ wk ): heterozygous for LU* exon 11: c.1340c>t, Ser447Leu exon 13: c.1742a>t, Gln581Leu exon 13: silent 1671C>T and heterozygous for KLF1 duplication associated with KLF1*BGM06 genotype and In(Lu) phenotype c.954dupg Arg319Glu fs Ter
13 LU*02 with novel mutations 2 donors were heterozygous for novel missense mutations in LU DONOR 1 LU exon 12 c.1464 c.1474 c.1484 N DONOR DONOR 2 LU exon 13 c.1630 c.1640 c.1650 N DONOR CONTROL CONTROL c.1474c>a c.1640c>a Pro492Thr Ala547Asp
14 LU:-12 Arg34 Leu35 D1 D2 LU:-16 Arg227 D3 Novel Pro492 extracellular cell membrane LU:-13 Gln581 (intracellular) (not modelled) intracellular Lu b with Arg436 D5 D4 LU:-13 Ser447 Novel Ala547 (adjacent to transmembrane helix) (not modelled)
15 LU:-12 and LU:-12 with Arg227Cys: Arg34 Leu35 Arg227 Shifts in both loops as a result of deletion (in direction of wt arrows) Arg227 Donor Arg227Cys (in combination with c.99_104 ) wt N-terminus: NH 2 -EVRLSVPPL c.99_104 : NH 2 -EVSVPPL
16 LU:-13 with Arg436Cys: Cys436 (donor) Arg436 Leu447 (donor) Substitution of Arg436 with smaller Cys436 results in shrinking of loop (in direction of arrow) Ser447 Considerable flexibility in hinge D4-D5 results in strongly different conformations (Ser447Leu)
17 Novel allele encoding Pro492Thr: Pro492 Thr492 Substitution of Pro492 for Thr492 (donor) predicts strong conformational change in the loop containing Pro492 (shift in direction of arrows)
18 Summary Serological typing 9 donor samples were found to be Lu(a+b+ wk ) and one Lu(a+ wk b+ wk ) Molecular study six different molecular backgrounds Samples heterozygous for: 4 were LU* (delarg34, Leu35) 1 was LU* (delarg34, Leu35) with LU* (Arg227Cys) 2 were LU* (Ser447Leu, Gln581Leu) with LU*02(1306T) 1 was LU* (Ser447Leu, Gln581Leu) with KLF1*BGM06; In(Lu) 2 were novel variant alleles (Pro492Thr and Ala547Asp)
19 Thank you for your attention
Pheno- and Genotype in Blood Group Variants - Two Sides of the Same Coin 260,00 CHF. Swisstransfusion 2018 Bern. 23 rd of August 2018,
Pheno- and Genotype in Blood Group Variants - Two Sides of the Same Coin Swisstransfusion 2018 Bern 23 rd of August 2018, 14.50-15.30 260,00 CHF Erwin A. Scharberg, Baden-Baden Christoph Gassner, Zürich
Understanding Genes & Mutations. John A Phillips III May 16, 2005
Understanding Genes & Mutations John A Phillips III May 16, 2005 Learning Objectives Understand gene structure Become familiar with genetic & mutation databases Be able to find information on genetic variation
A Review of RHCE Inferred Phenotypes and Red Cell Genotyping Data in Donors Serologically Confirmed as Negative for the High Frequency Antigen RH:34
A Review of RHCE Inferred Phenotypes and Red Cell Genotyping Data in Donors Serologically Confirmed as Negative for the High Frequency Antigen RH:34 *Lavendri Govender, Kuben Vather, Dr Ute Jentsch *Senior
Red Cell Immunohaematology. Mark Dwight Specialist Biomedical Scientist RCI Filton
Red Cell Immunohaematology Mark Dwight Specialist Biomedical Scientist RCI Filton The RCI Laboratories Red Cell Immunohaematology Reference laboratories supporting hospital blood banks in England. RCI
R1 12 kb R1 4 kb R1. R1 10 kb R1 2 kb R1 4 kb R1
Bcor101 Sample questions Midterm 3 1. The maps of the sites for restriction enzyme EcoR1 (R1) in the wild type and mutated cystic fibrosis genes are shown below: Wild Type R1 12 kb R1 4 kb R1 _ _ CF probe
Do Donors With Non-Deletional Blood Group O Allels Boost Anti-A and Anti-B Titers in Blood Group O Recipients?
Do Donors With Non-Deletional Blood Group O Allels Boost Anti-A and Anti-B Titers in Blood Group O Recipients? E. A. Scharberg, S. Seyboth, K. Panter, A. Ernst, J. Hagel, E. Richter, G. Rink, K. Janetzko,
Mutation entries in SMA databases Guidelines for national curators
1 Mutation entries in SMA databases Guidelines for national curators GENERAL CONSIDERATIONS Role of the curator(s) of a national database Molecular data can be collected by many different ways. There are
7.013 Spring 2005 Problem Set 1
MIT Department of Biology 7.013: Introductory Biology Spring 005 Instructors: rofessor azel Sive, rofessor Tyler Jacks, Dr. laudette Gardel AME TA Section # 7.013 Spring 005 roblem Set 1 FRIDAY February
Supplementary Data for Monti, et al.
Supplementary Data for Monti, et al. Supplementary Figure S1 Legend to Supplementary Figure S1 Tumor spectrum associated with germline p53 alleles (restricted to the 7 most frequent tissue targets). Structural
Jumping Into Your Gene Pool: Understanding Genetic Test Results
Jumping Into Your Gene Pool: Understanding Genetic Test Results Susan A. Berry, MD Professor and Director Division of Genetics and Metabolism Department of Pediatrics University of Minnesota Acknowledgment
7.013 Problem Set 3 FRIDAY October 8th, 2004
MIT Biology Department 7.012: Introductory Biology - Fall 2004 Instructors: Professor Eric Lander, Professor Robert. Weinberg, Dr. laudette ardel Name: T: 7.013 Problem Set 3 FRIDY October 8th, 2004 Problem
DNA.notebook March 08, DNA Overview
DNA Overview Deoxyribonucleic Acid, or DNA, must be able to do 2 things: 1) give instructions for building and maintaining cells. 2) be copied each time a cell divides. DNA is made of subunits called nucleotides
Mutagenesis. Classification of mutation. Spontaneous Base Substitution. Molecular Mutagenesis. Limits to DNA Pol Fidelity.
Mutagenesis 1. Classification of mutation 2. Base Substitution 3. Insertion Deletion 4. s 5. Chromosomal Aberration 6. Repair Mechanisms Classification of mutation 1. Definition heritable change in DNA
Gene mutation and DNA polymorphism
Gene mutation and DNA polymorphism Outline of this chapter Gene Mutation DNA Polymorphism Gene Mutation Definition Major Types Definition A gene mutation is a change in the nucleotide sequence that composes
www.lessonplansinc.com Topic: Gene Mutations WS Summary: Students will learn about frame shift mutations and base substitution mutations. Goals & Objectives: Students will be able to demonstrate how mutations
Schedule of Accreditation issued by United Kingdom Accreditation Service 2 Pine Trees, Chertsey Lane, Staines-upon-Thames, TW18 3HR, UK
2 Pine Trees, Chertsey Lane, Staines-upon-Thames, TW18 3HR, UK National Health Service Blood and Transplant International Blood Group Reference Laboratory 500 North Filton Park Filton Bristol BS34 7QH
INTRODUCTION TO THE MOLECULAR GENETICS OF THE COLOR MUTATIONS IN ROCK POCKET MICE
The Making of the The Fittest: Making of the Fittest Natural Selection Natural and Adaptation Selection and Adaptation Educator Materials TEACHER MATERIALS INTRODUCTION TO THE MOLECULAR GENETICS OF THE
MUTANT: A mutant is a strain that has suffered a mutation and exhibits a different phenotype from the parental strain.
OUTLINE OF GENETICS LECTURE #1 A. TERMS PHENOTYPE: Phenotype refers to the observable properties of an organism, such as morphology, growth rate, ability to grow under different conditions or media. For
BIOSTAT516 Statistical Methods in Genetic Epidemiology Autumn 2005 Handout1, prepared by Kathleen Kerr and Stephanie Monks
Rationale of Genetic Studies Some goals of genetic studies include: to identify the genetic causes of phenotypic variation develop genetic tests o benefits to individuals and to society are still uncertain
(a) Which enzyme(s) make 5' - 3' phosphodiester bonds? (c) Which enzyme(s) make single-strand breaks in DNA backbones?
EXAMPLE QUESTIONS AND ANSWERS 1. Topoisomerase does which one of the following? (a) Makes new DNA strands. (b) Unties knots in DNA molecules. (c) Joins the ends of double-stranded DNA molecules. (d) Is
GENETICS. Chapter 1: Cell cycle. Thème 1 : La Terre dans l Univers A. Expression, stabilité et variation du patrimoine génétique.
Introduction: GENETICS 3M = first look at genetics (study of inheritance, discovery of chromosomes, genes, dominant and recessive alleles and the DNA molecule within chromosomes) 2D = not much in fact,
Genome-wide genetic screening with chemically-mutagenized haploid embryonic stem cells
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Supplementary Information Genome-wide genetic screening with chemically-mutagenized haploid embryonic stem cells Josep V. Forment 1,2, Mareike Herzog
03-511/711 Computational Genomics and Molecular Biology, Fall
03-511/711 Computational Genomics and Molecular Biology, Fall 2011 1 Problem Set 0 Due Tuesday, September 6th This homework is intended to be a self-administered placement quiz, to help you (and me) determine
Molecular testing for blood group antigens: tools and applications
Molecular testing for blood group antigens: tools and applications Geoff Daniels International Blood Group Reference Laboratory NHS Blood and Transplant, UK Blood and Transplant 347 antigens 308 belong
Basic concepts of molecular biology
Basic concepts of molecular biology Gabriella Trucco Email: gabriella.trucco@unimi.it Life The main actors in the chemistry of life are molecules called proteins nucleic acids Proteins: many different
Not all mutations result in a change to the amino acid sequence of the encoded polypeptide
Q1.(a) (i) A mutation of a tumour suppressor gene can result in the formation of a tumour. Explain how. (2) (ii) Not all mutations result in a change to the amino acid sequence of the encoded polypeptide.
Lac Operon contains three structural genes and is controlled by the lac repressor: (1) LacY protein transports lactose into the cell.
Regulation of gene expression a. Expression of most genes can be turned off and on, usually by controlling the initiation of transcription. b. Lactose degradation in E. coli (Negative Control) Lac Operon
Biology Genetics Practice Quiz
Biology Genetics Practice Quiz Multiple Choice Identify the choice that best completes the statement or answers the question. 1. The table above shows information related to blood types. What genotype(s)
A novel RHCE*02 allele, containing the single-nucleotide change c.460a>g, encodes weakened expression of C and e antigens
A novel RHCE*02 allele, containing the single-nucleotide change c.460a>g, encodes weakened expression of C and e antigens Hellberg, Åsa; Pesjak, Emma; Karlsson, Jessica; Storry, Jill Published in: Transfusion
Chapter 4 Lecture. Concepts of Genetics. Tenth Edition. Extensions of Mendelian Genetics. Copyright 2009 Pearson Education, Inc.
Chapter 4 Lecture Concepts of Genetics Tenth Edition Extensions of Mendelian Genetics Copyright 2009 Pearson Education, Inc. Factors that cause deviation from normal monohybrid and dihybrid ratios: X-linkage
Biological Sciences 50 Practice Final Exam. Allocate your time wisely.
NAME: Fall 2005 TF: Biological Sciences 50 Practice Final Exam A. Be sure to write your name on the top of each of page of the examination. B. Write each answer only on the same page as the pertinent question.
Before starting, write your name on the top of each page Make sure you have all pages
Biology 105: Introduction to Genetics Name Student ID Before starting, write your name on the top of each page Make sure you have all pages You can use the back-side of the pages for scratch, but we will
7.014 Quiz II Handout
7.014 Quiz II Handout Quiz II: Wednesday, March 17 12:05-12:55 54-100 **This will be a closed book exam** Quiz Review Session: Friday, March 12 7:00-9:00 pm room 54-100 Open Tutoring Session: Tuesday,
Is it time to re-evaluate the gold standard in red cell antigen typing?
Is it time to re-evaluate the gold standard in red cell antigen typing? Susan T. Johnson, MSTM, MT(ASCP)SBB American Society of Clinical Laboratory Science Central New England May 9, 2013 161 GLOBCC 8MAY2015
CIE Biology A-level Topic 16: Inherited change
CIE Biology A-level Topic 16: Inherited change Notes Meiosis is a form of cell division that gives rise to genetic variation. The main role of meiosis is production of haploid gametes as cells produced
Basic Concepts of Human Genetics
Basic Concepts of Human Genetics The genetic information of an individual is contained in 23 pairs of chromosomes. Every human cell contains the 23 pair of chromosomes. One pair is called sex chromosomes
Thr Gly Tyr. Gly Lys Asn
Your unique body characteristics (traits), such as hair color or blood type, are determined by the proteins your body produces. Proteins are the building blocks of life - in fact, about 45% of the human
Anti-Lul4: A Lutheran Antibody Defining the Product of an Allele at the Lu8 Blood Group Locus'
Vox Sang. 32: 214-219 (1977) Anti-Lul4: A Lutheran Antibody Defining the Product of an Allele at the Lu8 Blood Group Locus' W. J. Judd, W. L. Mursh, R. @yen, M. E. Nichols, F. H. Allen, jr., M. Contreras
7.014 Problem Set 4 Answers to this problem set are to be turned in. Problem sets will not be accepted late. Solutions will be posted on the web.
MIT Department of Biology 7.014 Introductory Biology, Spring 2005 Name: Section : 7.014 Problem Set 4 Answers to this problem set are to be turned in. Problem sets will not be accepted late. Solutions
Genetics and Heredity Power Point Questions
Name period date assigned date due date returned Genetics and Heredity Power Point Questions 1. Heredity is the process in which pass from parent to offspring. 2. is the study of heredity. 3. A trait is
Enzyme that uses RNA as a template to synthesize a complementary DNA
Biology 105: Introduction to Genetics PRACTICE FINAL EXAM 2006 Part I: Definitions Homology: Comparison of two or more protein or DNA sequence to ascertain similarities in sequences. If two genes have
Worksheet: Mutations Practice
Worksheet: Mutations Practice There are three ways that DNA can be altered when a mutation (change in DNA sequence) occurs. 1. Substitution one base-pairs is replaced by another: Example: G to C or A to
Map-Based Cloning of Qualitative Plant Genes
Map-Based Cloning of Qualitative Plant Genes Map-based cloning using the genetic relationship between a gene and a marker as the basis for beginning a search for a gene Chromosome walking moving toward
Biological Sciences 50 Practice Exam 2
NAME: Fall 2005 TF: Biological Sciences 50 Practice Exam 2 A. Be sure to write your name on the top of each of page of the examination. B. Write each answer only on the same page as the pertinent question.
Title. So, JCC; Chan, AYY; Ma, ESK. Citation Hemoglobin, 2014, v. 38 n. 3, p Issued Date
Title Novel point mutation of the α2-globin gene (HBA2) and a rare 2.4kb deletion of the α1-globin gene (HBA1), identified in two chinese patients with Hb H disease Author(s) So, JCC; Chan, AYY; Ma, ESK
RNA and PROTEIN SYNTHESIS. Chapter 13
RNA and PROTEIN SYNTHESIS Chapter 13 DNA Double stranded Thymine Sugar is RNA Single stranded Uracil Sugar is Ribose Deoxyribose Types of RNA 1. Messenger RNA (mrna) Carries copies of instructions from
DNA RNA Protein Trait Protein Synthesis (Gene Expression) Notes Proteins (Review) Proteins make up all living materials
DNA RNA Protein Trait Protein Synthesis (Gene Expression) Notes Proteins (Review) Proteins make up all living materials Proteins are composed of amino acids there are 20 different amino acids Different
Bio 102 Practice Problems Genetic Code and Mutation
Bio 102 Practice Problems Genetic Code and Mutation Multiple choice: Unless otherwise directed, circle the one best answer: 1. Beadle and Tatum mutagenized Neurospora to find strains that required arginine
1. You have isolated 20 new mutant yeast strains that are defective in synthesis of threonine, an amino acid. (a)
ANSWERS TO Problem set questions from Exam 1 Unit Basic Genetic Tests, Setting up and Analyzing Crosses, and Genetic Mapping Basic genetic tests for complementation and/or dominance 1. You have isolated
Using CRISPR for genetic alteration
Using CRISPR for genetic alteration Joffrey Mianné. j.mianne@har.mrc.ac.uk Mary Lyon Centre, MRC Harwell. CRISPR/Cas origins Origin of the CRISPR/Cas system: Clustered-Regularly Interspaced Short Palindromic
STAT 536: Genetic Statistics
STAT 536: Genetic Statistics Karin S. Dorman Department of Statistics Iowa State University August 22, 2006 What is population genetics? A quantitative field of biology, initiated by Fisher, Haldane and
Single Nucleotide Variant Analysis. H3ABioNet May 14, 2014
Single Nucleotide Variant Analysis H3ABioNet May 14, 2014 Outline What are SNPs and SNVs? How do we identify them? How do we call them? SAMTools GATK VCF File Format Let s call variants! Single Nucleotide
VII. Rh Blood Group System
VII. Rh Blood Group System A. The Rh system is one of the most complex genetic systems, and certain aspects of its genetics, nomenclature and antigenic interactions are unsettled. The aim of this lecture
Lecture 2: Biology Basics Continued. Fall 2018 August 23, 2018
Lecture 2: Biology Basics Continued Fall 2018 August 23, 2018 Genetic Material for Life Central Dogma DNA: The Code of Life The structure and the four genomic letters code for all living organisms Adenine,
Chapter 17. Gene Mutations and DNA Repair (Part 1) What does this tell us?
Chapter 17 Gene Mutations and DNA Repair (Part 1) What does this tell us? 1 What does this tell us? Mutations will confer upon us a wide variety of super powers What does this tell us? Mutations will confer
7.014 Solution Set 4
7.014 Solution Set 4 Question 1 Shown below is a fragment of the sequence of a hypothetical bacterial gene. This gene encodes production of HWDWN, protein essential for metabolizing sugar yummose. The
1. DNA, RNA structure. 2. DNA replication. 3. Transcription, translation
1. DNA, RNA structure 2. DNA replication 3. Transcription, translation DNA and RNA are polymers of nucleotides DNA is a nucleic acid, made of long chains of nucleotides Nucleotide Phosphate group Nitrogenous
MCB 421 Exam #1 (A) Fall There are 9 questions and 1 supplement on last page. Answer all 9 questions. Be sure your name is on each page
MCB 421 Exam #1 (A) Fall 2006 There are 9 questions and 1 supplement on last page. Answer all 9 questions. Be sure your name is on each page 1). (8 points) A Luria-Delbruck fluctuation test was done to
EMSA Mutant SOX10 proteins interfere with the DNA binding of wild type SOX10
18-05-2010 SOX10 IL FENOTIPO MENO GRAVE È DOVUTO A DOWNREGULATION DA NMD Intron Minigene approach: * * Mutazione introdotta come controllo Northern blotting * mrna levels Inoue et al., Nature Genetics
Case 3. Laura Yeates BSc (hons) Grad Dip Gen Couns. FHGSA
Case 3 Laura Yeates BSc (hons) Grad Dip Gen Couns. FHGSA Genetic Counsellor & Clinic Coordinator Molecular Cardiology Program, Centenary Institute HCM & Genetic Heart Disease Clinics, RPAH ICCG 2018 Proband
The Bloodgen project : Platform Technology for Mass Scale Genotyping of Blood Groups and Beyond. Prof. Neil D. Avent and Bloodgen consortium members
The Bloodgen project : Platform Technology for Mass Scale Genotyping of Blood Groups and Beyond Prof. Neil D. Avent and Bloodgen consortium members DGTI Symposium SY1 Friday 24 th September 2004 Bloodgen
Chapter 2. Alleles at a Single Locus. Introduction
Chapter 2 Alleles at a Single Locus Figure 2-1 A flower called Camellia showing co-dominance of the red and white alleles of flower colour. (Flickr- darwin cruz-cc BY 2.0) Introduction mutation is any
The Making of the Fittest: Natural Selection and Adaptation
INTRODUCTION MOLECULAR GENETICS OF COLOR MUTATIONS IN ROCK POCKET MICE THE ROCK POCKET MOUSE The rock pocket mouse, Chaetodipus intermedius, is a small, nocturnal animal found in the deserts of the southwestern
Technical Bulletin No. 173
CPAL Central Pennsylvania Alliance Laboratory Technical Bulletin No. 173 RBC Antigen Genotyping - New Test - February 11, 2019 Contact: Heather Habig, MLS (ASCP) CM, MB CM, 717-851-1422 Operations Manager,
7.014 Problem Set 3 Please print out this problem set and record your answers on the printed copy.
MIT Department of Biology 7.014 Introductory Biology, Spring 2004 Name: 7.014 Problem Set 3 Please print out this blem set and record your answers on the printed copy. Problem sets will not be accepted
Chem 250 Answer Key In-class Quiz #3v1
age 1 of 6 Quiz #3 ame. Chem 250 Answer Key In-class Quiz #3v1 This exam is composed of 20 questions. lease scan them all before starting. As discussed in the course syllabus, honesty and integrity are
AP Biology Review Chapters Review Questions Chapter 11: Mendelian Patterns of Inheritance Chapter 12: Molecular Biology of the Gene
AP Biology Review Chapters 11-12 Review Questions Chapter 11: Mendelian Patterns of Inheritance a) Know genotypes and phenotypes of a monohybrid cross in the P, F1, and F2 generations. Be familiar with
iclicker Question #28B - after lecture Shown below is a diagram of a typical eukaryotic gene which encodes a protein: start codon stop codon 2 3
Bio 111 Handout for Molecular Biology 4 This handout contains: Today s iclicker Questions Information on Exam 3 Solutions Fall 2008 Exam 3 iclicker Question #28A - before lecture Which of the following
Molecular Genetics of Disease and the Human Genome Project
9 Molecular Genetics of Disease and the Human Genome Project Fig. 1. The 23 chromosomes in the human genome. There are 22 autosomes (chromosomes 1 to 22) and two sex chromosomes (X and Y). Females inherit
Disease and selection in the human genome 3
Disease and selection in the human genome 3 Ka/Ks revisited Please sit in row K or forward RBFD: human populations, adaptation and immunity Neandertal Museum, Mettman Germany Sequence genome Measure expression
From DNA to Protein: Genotype to Phenotype
12 From DNA to Protein: Genotype to Phenotype 12.1 What Is the Evidence that Genes Code for Proteins? The gene-enzyme relationship is one-gene, one-polypeptide relationship. Example: In hemoglobin, each
ABO Blood Group 9/13/2015 NAHLA BAKHAMIS MSC 1
ABO Blood Group NAHLA BAKHAMIS MSC. 9/13/2015 NAHLA BAKHAMIS MSC 1 Glossary Chromosome Gene Locus Allele Heterozygous Homozygous Phenotype Genotype Sex Chromosomes Autosomal Chromosomes Dominant Gene Recessive
SENIOR BIOLOGY. Blueprint of life and Genetics: the Code Broken? INTRODUCTORY NOTES NAME SCHOOL / ORGANISATION DATE. Bay 12, 1417.
SENIOR BIOLOGY Blueprint of life and Genetics: the Code Broken? NAME SCHOOL / ORGANISATION DATE Bay 12, 1417 Bay number Specimen number INTRODUCTORY NOTES Blueprint of Life In this part of the workshop
Lecture 2: Biology Basics Continued
Lecture 2: Biology Basics Continued Central Dogma DNA: The Code of Life The structure and the four genomic letters code for all living organisms Adenine, Guanine, Thymine, and Cytosine which pair A-T and
Si mon P t e ersen J ones
Simon Petersen Jones What is a DNA based Test Uses sample of DNA Blood Cheek Swab Identifies the genotype (not phenotype) of the animal There are Two Categories of Test: Mutation Detection Direct detection
The Nature of Genes. The Nature of Genes. The Nature of Genes. The Nature of Genes. The Nature of Genes. The Genetic Code. Genes and How They Work
Genes and How They Work Chapter 15 Early ideas to explain how genes work came from studying human diseases. Archibald Garrod studied alkaptonuria, 1902 Garrod recognized that the disease is inherited via
For more information about how to cite these materials visit
Author(s): David Ginsburg, M.D., 2012 License: Unless otherwise noted, this material is made available under the terms of the Creative Commons Attribution Non-commercial Share Alike 3.0 License: http://creativecommons.org/licenses/by-nc-sa/3.0/
11 questions for a total of 120 points
Your Name: BYS 201, Final Exam, May 3, 2010 11 questions for a total of 120 points 1. 25 points Take a close look at these tables of amino acids. Some of them are hydrophilic, some hydrophobic, some positive
Solutions to Problem Set 1
MIT Department of Biology 7.014 Introductory Biology, Spring 004 Question 1 Solutions to 7.014 Problem Set 1 a) Describe the conditions of the atmosphere on prebiotic earth and how these conditions differ
Nature Genetics: doi: /ng Supplementary Figure 1
Supplementary Figure 1 Frequencies of DQ2.5, DQ2.2, DQ8 and DQ7.5, the most common haplotypes in celiac disease cases. The map illustrates the differences among European countries and shows a clear gradient
Jay McTighe and Grant Wiggins,
Course: Integrated Science 3/4 Unit #3: (DNA & RNA) Instructions for Life Stage 1: Identify Desired Results Enduring Understandings: Students will understand that Nearly all human traits, even many diseases,
1. An alteration of genetic information is shown below. 5. Part of a molecule found in cells is represented below.
1. An alteration of genetic information is shown below. 5. Part of a molecule found in cells is represented below. A-G-T-A-C-C-G-A-T A-G-T-G-A-T This type of alteration of the genetic information is an
From Gene to Protein. How Genes Work
From Gene to Protein How Genes Work 2007-2008 The Central Dogma Flow of genetic information in a cell How do we move information from DNA to proteins? DNA RNA protein replication phenotype You! Step 1:
Problem Set Unit The base ratios in the DNA and RNA for an onion (Allium cepa) are given below.
Problem Set Unit 3 Name 1. Which molecule is found in both DNA and RNA? A. Ribose B. Uracil C. Phosphate D. Amino acid 2. Which molecules form the nucleotide marked in the diagram? A. phosphate, deoxyribose
DNA stands for deoxyribose nucleic acid
1 DNA 2 DNA stands for deoxyribose nucleic acid This chemical substance is present in the nucleus of all cells in all living organisms DNA controls all the chemical changes which take place in cells The
Basic concepts of molecular biology
Basic concepts of molecular biology Gabriella Trucco Email: gabriella.trucco@unimi.it What is life made of? 1665: Robert Hooke discovered that organisms are composed of individual compartments called cells
GENE 603 Exam 3 December 1, 2017 Name
GENE 603 Exam 3 December 1, 2017 Name 1. When the arc gene of phage P22 is defective, the phage cannot enter lysogeny, so these strains are always lytic and cause clear plaques on lawns of S. typhimurium
BIOLOGY Component 1 Basic Biochemistry and Cell Organisation
Surname Centre Number Candidate Number Other Names 2 GCE AS NEW AS B400U10-1 S16-B400U10-1 BIOLOGY Component 1 Basic Biochemistry and Cell Organisation P.M. THURSDAY, 26 May 2016 1 hour 30 minutes For
INTRODUCTION TO MOLECULAR GENETICS. Andrew McQuillin Molecular Psychiatry Laboratory UCL Division of Psychiatry 22 Sept 2017
INTRODUCTION TO MOLECULAR GENETICS Andrew McQuillin Molecular Psychiatry Laboratory UCL Division of Psychiatry 22 Sept 2017 Learning Objectives Understand: The distinction between Quantitative Genetic
Today s lecture: Types of mutations and their impact on protein function
Today s lecture: Types of mutations and their impact on protein function Mutations can be classified by their effect on the DNA sequence OR the encoded protein 1 From my Lecture 4 (10/1): Classification
SUPPLEMENTARY INFORMATION
doi:10.1038/nature13127 Factors to consider in assessing candidate pathogenic mutations in presumed monogenic conditions The questions itemized below expand upon the definitions in Table 1 and are provided
7.012 Final Exam
7.012 Final Exam 2006 You have 180 minutes to complete this exam. There are 19 pages including this cover page, the AMINO AID page, and the GENETI ODE page at the end of the exam. Please write your name
Combining Serology and Genotyping to Improve Patient Outcomes: Case Studies
Combining Serology and Genotyping to Improve Patient Outcomes: Case Studies Lauro E. Guerra, Jr., BS, MLS(ASCP) CM SBB CM Manager, Grifols Immunohematology Center, San Marcos, TX, USA March 22, 2019 Objectives
Structural bioinformatics
Structural bioinformatics Why structures? The representation of the molecules in 3D is more informative New properties of the molecules are revealed, which can not be detected by sequences Eran Eyal Plant
Keystone Biology Remediation B2: Genetics
Keystone Biology Remediation B2: Genetics Assessment Anchors: to describe and/or predict observed patterns of inheritance (i.e. dominant, recessive, codominance, incomplete dominance, sex-linked, polygenic,
Genetics fundamentals and DNA toolkit. Partha Roy
Genetics fundamentals and DNA toolkit Partha Roy 1 (REVIEW of Terminologies and Concepts) Haploid organism: single copy of chromosome (ex: bacteria, yeast) Diploid organism: two copies of chromosome (paternal
3/3/2013. Reference Lab Case Studies
Reference Lab Case Studies Virginia Hare, MT(ASCP)SBB American Red Cross Southern Region Douglasville, Georgia Today s Cases and Presenters Rh typing discrepancy by Grace Lee An Unusual Phenotype: Getting
Reference Lab Case Studies. Virginia Hare, MT(ASCP)SBB American Red Cross Southern Region Douglasville, Georgia
Reference Lab Case Studies Virginia Hare, MT(ASCP)SBB American Red Cross Southern Region Douglasville, Georgia Today s Cases and Presenters Rh typing discrepancy by Grace Lee An Unusual Phenotype: Getting
ANSWERS TO Problem set questions from Exam 2 Unit Mutations, Bacterial Genetics, and Bacterial Gene Regulation
ANSWERS TO Problem set questions from Exam 2 Unit Mutations, Bacterial Genetics, and Bacterial Gene Regulation Central Dogma, Mutagens and Mutations 1. The three stop codons in the genetic code are 5 UAG3,
You are genetically unique
BNF 5106 - Lecture 1 Genetics, Genes, Genetic codes, and Mutations You are genetically unique Since each parent has 23 pairs of chromosomes, the probability that each parent gives twice the same chromosomes