LS4 Problem Set 2, Population Genetics. Corresponding Quiz: November (Along with positional cloning)
|
|
- Cordelia Pitts
- 7 years ago
- Views:
Transcription
1 LS4 Problem Set 2, 2010 Population Genetics Corresponding Lectures: November 5, 8 and 10 Corresponding Reading: Hartwell et al., Corresponding Quiz: November (Along with positional cloning) If you want to try additional problems, there are many at the end of chapter 21 that pertain to this section of the course. Problem 1. a) What are the allele frequencies of M and N in this population? Freq M = [(550*2)+150]/(2*1000) = Freq N = [(300*2)+150]/(2*1000) = b) If this population is at Hardy-Weinberg equilibrium, how many individuals with a genotype of MN would you expect to see? 2pq = 2*0.625*0.375 = *1000 = 469 people Problem 2. a) What are the allele frequencies of R and r in population 1? p = Freq R = [(2*10)+74]/(2*100) = 0.47 q = Freq r = [(2*16)+74]/(2*100) = 0.53 b) What are the allele frequencies of R and r in population 2? p = Freq R = [(2*81)+18]/200 = 0.9 q = Freq r = [(2*1)+18)]/200 = 0.1 c) Which population is at Hardy-Weinberg equilibrium? Population 2 is at H.W. eq Population 1 expected genotype frequency p^2 = 0.47^2 = è *100 = 22 RR individuals 2pq = 2*0.47*0.53 = à * 100 = 50 Rr individuals q^2 = 0.53^2 = à * 100 = 28 rr individuals Doesn t match given observed values, not in H.W. eq Population 2 expected genotype frequency p^2 = 0.9^2 = 0.81 à 0.81*100 = 81 RR individuals 2pq = 2*0.9*0.1 = 0.18 à 0.18*100 = 18 Rr individuals q^2 = 0.1^2 =.01 à.01*100 = 1 rr individual This matches the observed values, so population 2 is at H.W. eq Problem List the expected values for each genotypic class under your null hypothesis. A1A1 = 306 A1A2 = 170 A2A2 = Perform the chi-square calculation. χ2 = State your conclusion based on the data. population is in HWE (3-1 = 2 dof; 0.90 > p > 0.50)
2 Problem 4. A) Is the population at H-W equilibrium for the A gene? How about the B gene? (Show your work) Determine allelic frequencies: f(a) = p = ( ) / 2960 =.625 f (a) = q = 1 p = =.375 Now find the genotypic frequencies and the number of expected individuals out of 1480 to have this genotype and compare it to the observed number: f (AA) = p2 =.6252 =.39 X 1480 = 577 expected and 740 observed f (Aa) = 2pq = 2 X.625 X.375 =.47 X 1480 = 696 expected and 370 observed f (aa) = q2 =.3752 =.14 X 1480 = 207 expected and 370 observed Expected and observed are not similar so the A gene is not at H-W equilibrium. Do the same thing for the B gene and you will find that the number of expected and the number of observed are nearly identical. Thus, the population is at H-W equilibrium for the B gene. B) After one generation of random mating, what fraction of the next generation will be AA (independent of the B gene)? Since we are at H-W equilibrium allele frequencies do not change. The f (AA) is simply p2 =.6252 =.39 C) After one generation of random mating, what fraction of the next generation will be BB (independent of the A gene)? This is the same as above and f (BB) is simply p2 =.522 =.27 D) What is the chance that the first child of a Aa Bb female and a AA Bb will be a Aa bb male? (Hint: Do you need to consider allele frequencies to solve this problem? Think back to earlier in the quarter.) There is no need to consider allele frequencies. Allele frequencies apply to populations and can assign the probability of a certain genotype. But we know the genotypes of the parents in this problem. The probability of the child being Aa is 1/2 (do the punnett square if you need convinced), the probability of being bb is 1/4 (again do the punnett square), and the probability of being male is 1/2. 1/2 X 1/4 X 1/2 = 1/16 Problem 5. a) Find the allele frequencies in this generation of surviving flies: + = 0.7 m = 0.3 b) If you assumed HW equilibrium, what are the expected numbers of different genotypes in the next generation if 200 viable offspring of the population in part (a) are counted? Make sure to take into account the fact that one genotype dies! (write out all possible genotypes and write numbers of flies next to them) +/+ = 108 m/+ = 92 m/m = 0
3 Problem 6. a.) What is the allele frequency of the M allele in the sailor population? ( )/1000= b.) 10 years later 1000 children have been born on the coastline. (All unions were consensual!) If the population of 1000 young people on the coast mate randomly, how many of the 1000 children would you expect to have MN bloodtype? ~ 470. (Be careful with rounding too early!) Tunisian Population: 45 MM 210 MN 245NN Tunisian population+ Sailor Population New Genotype Frequency: 495 MM 255 MN 250NN New Allele Frequency M= N= pq= 2 (0.6225)(0.3775)= c.) In fact, 50 children have MM bloodtype, 850 have MN bloodtype and 100 have NN blood type. What is the observed allele frequency of the N allele among the children? Frequency of N ( )/ 2000= = 52.5% Problem 7. a. What are the genotype frequencies in the population? CR CR = 100/1000=0.10 CR CW = 740/1000= 0.74 CW CW = 160/1000= 0.16 b. What are the allele frequencies of CR and CW in this population? CR = / 2000= 0.47 CW = = 0.53 c. what are the expected frequencies of the genotypes if the population is at Hardy- Weinberg equilibrium? P2 = (0.47)2 = 0.22 CR CR Q2= (0.53)2 = 0.28 CW CW 2PQ= 2 (0.47)(0.53) = 0.50 CR CW Problem 8. Assuming that the population is at Hardy-Weinberg equilibrium, what proportion of the males in this population will also have the condition? Xs = dominant allele for condition. X= recessive, doesn t have the condition. Females with Xs_ have the condition and 36% are Xs Xs and Xs X genotypes. Thus, 64% is XX. Since 64% is XX (p2), the X (p) allele frequency is 0.8. Using p + q =1, Xs (q) allele frequency is 0.2 For male population: Since male has only one X, the genotype frequencies are equal to the allele frequencies.. Thus males with XsY have the condition and there are 20% since the Xs allele frequency is 0.2.
4 Problem 9. Assuming the population is at Hardy-Weinberg equilibrium, how many heterozygous tongue rollers are there in the population? Let p = allele frequency for rolling Let q = allele frequency for non-rolling q2 = 2600/10,000 = 0.26 q = 0.51 p = 1 - q = pq = 2(0.49)0.51 = Number of heterozygous rollers = (10,000) = 4998 (5000 would be close enough) Problem q = square root of 1/10,000 = pq = = 1.98% of U.S. population Problem 11. Is the baby population at Hardy-Weinberg equilibrium? There are two ways to do this problem. You can calculate expected allele frequencies from genotype frequencies and compare them to actual allele frequencies, as below: q2 = 20,000/100,000 = 0.2 q = 0.45 p = 1 - q = 0.55 p2(calculated) = 0.30 p 2(actual) = 20,000/100,000 = 0.2 2pq(calculated) = 2(0.45)0.55 = pq(actual) = 60,000/100,000 = 0.6 Answer: No Or you could use observed allele frequencies to calculate expected genotype frequencies and compare these to actual genotype frequencies, as below: q = 100,000/200,000 = 0.5; p = 0.5 expected genotype frequencies: q2= 0.52 = 0.25; expected homozygous sickle cell is 0.25x100,000, or 25,000 2pq = 2 x 0.5 x 0.5x = 0.5; expected heterozygotes: 0.5 x 100,000, or 50,000 p2= 0.52 = 0.25; expected homozygous normal is 0.25x100,000, or 25,000 compare actual to calculated genotypes: 25,000 vs 20,000 sickle cell sufferers; 50,000 vs 60,000 hets, 25,000 vs 20,000 normal Answer: No
5 Problem 12. A) Assuming Hardy-Weinberg equilibrium, what is the frequency of the resistance allele? B) What is the frequency of the non-resistance allele? C) Of the 100,000 mosquitoes, how many are expected to be carriers of the resistance allele (heterozygous)? D) What is the frequency of the resistance allele now? The strategy here is to figure out how many resistance alleles exist in each population and divide the sum by the total number of alleles in both populations. where, was from part a, was from part b, was 1,, was 0. E) Over time, the mosquito population grew back to its natural size of 100,000. Assume there was no difference in fitness between the two alleles of the resistance gene. How many mosquitoes are expected to be resistant to malaria? F) If it turned out that being homozygous for the resistance allele caused mosquitoes to become partially sterile, do you predict a change in the frequency of alleles in the population? If so, which allele would increase in frequency? Yes, the nonresistance allele would increase in frequency as partial sterility represents a change in fitness. Problem 13. A) Determine the frequency of each allele. C L = [(450x2)+500]/2000 C S = [(50x2)+500]/2000 C L = 0.7 C S = 0.3 B) Are the allele frequencies maintained at Hardy-Weinberg equilibrium? How do you know? Show your work (No need for a statistical test, just estimate). C L C L = (0.7) 2 x1000 = C L C S = 2x(0.7)(0.3)x1000 = C S C S = (0.3) 2 x1000 = NO, observed values do not equal expected Hardy-Weinberg values
6 Problem 14. A) Under Hardy-Weinberg equilibrium assumptions, determine the frequencies of both the wild-type and mutant alleles. 20 = Round-eared 40 = rr (40/1000) = q = q 0.8 = p B) The Vulcans send the 20 round-eared individuals to Earth to infiltrate the American government. If the 980 remaining Vulcans mate randomly, how many round-eared individuals will there be in the next generation of 1,000 Vulcan offspring? r = 0.2 x 2000 = 400 new r = new r = 360 new q = (360/1960) new q = new q 2 = rr = x 1000 rr = 34 round-eared = 34 x 0.5 = 17 C) A deadly ear-ache pandemic sweeps through the Vulcan s planet. However, heterozygosity at the pointy-eared locus confers resistance to the disease, allowing only those individuals to survive and reproduce (homozygotes for either allele die). Determine the proportion of pointy-eared individuals in the first generation of offspring from the survivors. q = 0.5 q 2 = 0.25 rr = 0.25 round-eared = 0.25 x 0.5 = pointy-eared = = 0.875
Hardy-Weinberg Equilibrium Problems
Hardy-Weinberg Equilibrium Problems 1. The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium. (a) Calculate the percentage of
More informationThe Making of the Fittest: Natural Selection in Humans
OVERVIEW MENDELIN GENETIC, PROBBILITY, PEDIGREE, ND CHI-QURE TTITIC This classroom lesson uses the information presented in the short film The Making of the Fittest: Natural election in Humans (http://www.hhmi.org/biointeractive/making-fittest-natural-selection-humans)
More informationBasic Principles of Forensic Molecular Biology and Genetics. Population Genetics
Basic Principles of Forensic Molecular Biology and Genetics Population Genetics Significance of a Match What is the significance of: a fiber match? a hair match? a glass match? a DNA match? Meaning of
More informationGenetics 1. Defective enzyme that does not make melanin. Very pale skin and hair color (albino)
Genetics 1 We all know that children tend to resemble their parents. Parents and their children tend to have similar appearance because children inherit genes from their parents and these genes influence
More informationCCR Biology - Chapter 7 Practice Test - Summer 2012
Name: Class: Date: CCR Biology - Chapter 7 Practice Test - Summer 2012 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A person who has a disorder caused
More informationLAB : PAPER PET GENETICS. male (hat) female (hair bow) Skin color green or orange Eyes round or square Nose triangle or oval Teeth pointed or square
Period Date LAB : PAPER PET GENETICS 1. Given the list of characteristics below, you will create an imaginary pet and then breed it to review the concepts of genetics. Your pet will have the following
More informationAP: LAB 8: THE CHI-SQUARE TEST. Probability, Random Chance, and Genetics
Ms. Foglia Date AP: LAB 8: THE CHI-SQUARE TEST Probability, Random Chance, and Genetics Why do we study random chance and probability at the beginning of a unit on genetics? Genetics is the study of inheritance,
More informationChapter 4 Pedigree Analysis in Human Genetics. Chapter 4 Human Heredity by Michael Cummings 2006 Brooks/Cole-Thomson Learning
Chapter 4 Pedigree Analysis in Human Genetics Mendelian Inheritance in Humans Pigmentation Gene and Albinism Fig. 3.14 Two Genes Fig. 3.15 The Inheritance of Human Traits Difficulties Long generation time
More informationMendelian Genetics in Drosophila
Mendelian Genetics in Drosophila Lab objectives: 1) To familiarize you with an important research model organism,! Drosophila melanogaster. 2) Introduce you to normal "wild type" and various mutant phenotypes.
More informationHeredity. Sarah crosses a homozygous white flower and a homozygous purple flower. The cross results in all purple flowers.
Heredity 1. Sarah is doing an experiment on pea plants. She is studying the color of the pea plants. Sarah has noticed that many pea plants have purple flowers and many have white flowers. Sarah crosses
More informationProblems 1-6: In tomato fruit, red flesh color is dominant over yellow flesh color, Use R for the Red allele and r for the yellow allele.
Genetics Problems Name ANSWER KEY Problems 1-6: In tomato fruit, red flesh color is dominant over yellow flesh color, Use R for the Red allele and r for the yellow allele. 1. What would be the genotype
More informationGENETIC CROSSES. Monohybrid Crosses
GENETIC CROSSES Monohybrid Crosses Objectives Explain the difference between genotype and phenotype Explain the difference between homozygous and heterozygous Explain how probability is used to predict
More informationHuman Blood Types: Codominance and Multiple Alleles. Codominance: both alleles in the heterozygous genotype express themselves fully
Human Blood Types: Codominance and Multiple Alleles Codominance: both alleles in the heterozygous genotype express themselves fully Multiple alleles: three or more alleles for a trait are found in the
More informationPRACTICE PROBLEMS - PEDIGREES AND PROBABILITIES
PRACTICE PROBLEMS - PEDIGREES AND PROBABILITIES 1. Margaret has just learned that she has adult polycystic kidney disease. Her mother also has the disease, as did her maternal grandfather and his younger
More informationPopulation Genetics and Multifactorial Inheritance 2002
Population Genetics and Multifactorial Inheritance 2002 Consanguinity Genetic drift Founder effect Selection Mutation rate Polymorphism Balanced polymorphism Hardy-Weinberg Equilibrium Hardy-Weinberg Equilibrium
More informationBiology 1406 - Notes for exam 5 - Population genetics Ch 13, 14, 15
Biology 1406 - Notes for exam 5 - Population genetics Ch 13, 14, 15 Species - group of individuals that are capable of interbreeding and producing fertile offspring; genetically similar 13.7, 14.2 Population
More informationGenetics Lecture Notes 7.03 2005. Lectures 1 2
Genetics Lecture Notes 7.03 2005 Lectures 1 2 Lecture 1 We will begin this course with the question: What is a gene? This question will take us four lectures to answer because there are actually several
More informationPopstats Unplugged. 14 th International Symposium on Human Identification. John V. Planz, Ph.D. UNT Health Science Center at Fort Worth
Popstats Unplugged 14 th International Symposium on Human Identification John V. Planz, Ph.D. UNT Health Science Center at Fort Worth Forensic Statistics From the ground up Why so much attention to statistics?
More informationDNA Determines Your Appearance!
DNA Determines Your Appearance! Summary DNA contains all the information needed to build your body. Did you know that your DNA determines things such as your eye color, hair color, height, and even the
More informationThe Genetics of Drosophila melanogaster
The Genetics of Drosophila melanogaster Thomas Hunt Morgan, a geneticist who worked in the early part of the twentieth century, pioneered the use of the common fruit fly as a model organism for genetic
More informationLAB : THE CHI-SQUARE TEST. Probability, Random Chance, and Genetics
Period Date LAB : THE CHI-SQUARE TEST Probability, Random Chance, and Genetics Why do we study random chance and probability at the beginning of a unit on genetics? Genetics is the study of inheritance,
More informationAP BIOLOGY 2010 SCORING GUIDELINES (Form B)
AP BIOLOGY 2010 SCORING GUIDELINES (Form B) Question 2 Certain human genetic conditions, such as sickle cell anemia, result from single base-pair mutations in DNA. (a) Explain how a single base-pair mutation
More information7 th Grade Life Science Name: Miss Thomas & Mrs. Wilkinson Lab: Superhero Genetics Due Date:
7 th Grade Life Science Name: Miss Thomas & Mrs. Wilkinson Partner: Lab: Superhero Genetics Period: Due Date: The editors at Marvel Comics are tired of the same old characters. They re all out of ideas
More informationHeredity - Patterns of Inheritance
Heredity - Patterns of Inheritance Genes and Alleles A. Genes 1. A sequence of nucleotides that codes for a special functional product a. Transfer RNA b. Enzyme c. Structural protein d. Pigments 2. Genes
More information7 POPULATION GENETICS
7 POPULATION GENETICS 7.1 INTRODUCTION Most humans are susceptible to HIV infection. However, some people seem to be able to avoid infection despite repeated exposure. Some resistance is due to a rare
More informationAnswer Key Problem Set 5
7.03 Fall 2003 1 of 6 1. a) Genetic properties of gln2- and gln 3-: Answer Key Problem Set 5 Both are uninducible, as they give decreased glutamine synthetase (GS) activity. Both are recessive, as mating
More informationMendelian inheritance and the
Mendelian inheritance and the most common genetic diseases Cornelia Schubert, MD, University of Goettingen, Dept. Human Genetics EUPRIM-Net course Genetics, Immunology and Breeding Mangement German Primate
More information2 GENETIC DATA ANALYSIS
2.1 Strategies for learning genetics 2 GENETIC DATA ANALYSIS We will begin this lecture by discussing some strategies for learning genetics. Genetics is different from most other biology courses you have
More informationBio EOC Topics for Cell Reproduction: Bio EOC Questions for Cell Reproduction:
Bio EOC Topics for Cell Reproduction: Asexual vs. sexual reproduction Mitosis steps, diagrams, purpose o Interphase, Prophase, Metaphase, Anaphase, Telophase, Cytokinesis Meiosis steps, diagrams, purpose
More informationChromosomes, Mapping, and the Meiosis Inheritance Connection
Chromosomes, Mapping, and the Meiosis Inheritance Connection Carl Correns 1900 Chapter 13 First suggests central role for chromosomes Rediscovery of Mendel s work Walter Sutton 1902 Chromosomal theory
More informationForensic Statistics. From the ground up. 15 th International Symposium on Human Identification
Forensic Statistics 15 th International Symposium on Human Identification From the ground up UNTHSC John V. Planz, Ph.D. UNT Health Science Center at Fort Worth Why so much attention to statistics? Exclusions
More informationContinuous and discontinuous variation
Continuous and discontinuous variation Variation, the small differences that exist between individuals, can be described as being either discontinuous or continuous. Discontinuous variation This is where
More informationName: 4. A typical phenotypic ratio for a dihybrid cross is a) 9:1 b) 3:4 c) 9:3:3:1 d) 1:2:1:2:1 e) 6:3:3:6
Name: Multiple-choice section Choose the answer which best completes each of the following statements or answers the following questions and so make your tutor happy! 1. Which of the following conclusions
More informationA trait is a variation of a particular character (e.g. color, height). Traits are passed from parents to offspring through genes.
1 Biology Chapter 10 Study Guide Trait A trait is a variation of a particular character (e.g. color, height). Traits are passed from parents to offspring through genes. Genes Genes are located on chromosomes
More information5 GENETIC LINKAGE AND MAPPING
5 GENETIC LINKAGE AND MAPPING 5.1 Genetic Linkage So far, we have considered traits that are affected by one or two genes, and if there are two genes, we have assumed that they assort independently. However,
More informationMCB41: Second Midterm Spring 2009
MCB41: Second Midterm Spring 2009 Before you start, print your name and student identification number (S.I.D) at the top of each page. There are 7 pages including this page. You will have 50 minutes for
More informationHLA data analysis in anthropology: basic theory and practice
HLA data analysis in anthropology: basic theory and practice Alicia Sanchez-Mazas and José Manuel Nunes Laboratory of Anthropology, Genetics and Peopling history (AGP), Department of Anthropology and Ecology,
More informationVariations on a Human Face Lab
Variations on a Human Face Lab Introduction: Have you ever wondered why everybody has a different appearance even if they are closely related? It is because of the large variety or characteristics that
More informationName: Class: Date: ID: A
Name: Class: _ Date: _ Meiosis Quiz 1. (1 point) A kidney cell is an example of which type of cell? a. sex cell b. germ cell c. somatic cell d. haploid cell 2. (1 point) How many chromosomes are in a human
More informationTwo copies of each autosomal gene affect phenotype.
SECTION 7.1 CHROMOSOMES AND PHENOTYPE Study Guide KEY CONCEPT The chromosomes on which genes are located can affect the expression of traits. VOCABULARY carrier sex-linked gene X chromosome inactivation
More informationCHROMOSOMES AND INHERITANCE
SECTION 12-1 REVIEW CHROMOSOMES AND INHERITANCE VOCABULARY REVIEW Distinguish between the terms in each of the following pairs of terms. 1. sex chromosome, autosome 2. germ-cell mutation, somatic-cell
More informationBasics of Marker Assisted Selection
asics of Marker ssisted Selection Chapter 15 asics of Marker ssisted Selection Julius van der Werf, Department of nimal Science rian Kinghorn, Twynam Chair of nimal reeding Technologies University of New
More informationCan receive blood from: * I A I A and I A i o Type A Yes No A or AB A or O I B I B and I B i o Type B No Yes B or AB B or O
Genetics of the ABO Blood Groups written by J. D. Hendrix Learning Objectives Upon completing the exercise, each student should be able: to explain the concept of blood group antigens; to list the genotypes
More informationCystic Fibrosis Webquest Sarah Follenweider, The English High School 2009 Summer Research Internship Program
Cystic Fibrosis Webquest Sarah Follenweider, The English High School 2009 Summer Research Internship Program Introduction: Cystic fibrosis (CF) is an inherited chronic disease that affects the lungs and
More informationMendelian and Non-Mendelian Heredity Grade Ten
Ohio Standards Connection: Life Sciences Benchmark C Explain the genetic mechanisms and molecular basis of inheritance. Indicator 6 Explain that a unit of hereditary information is called a gene, and genes
More informationGenetics Review for USMLE (Part 2)
Single Gene Disorders Genetics Review for USMLE (Part 2) Some Definitions Alleles variants of a given DNA sequence at a particular location (locus) in the genome. Often used more narrowly to describe alternative
More informationThe correct answer is c A. Answer a is incorrect. The white-eye gene must be recessive since heterozygous females have red eyes.
1. Why is the white-eye phenotype always observed in males carrying the white-eye allele? a. Because the trait is dominant b. Because the trait is recessive c. Because the allele is located on the X chromosome
More informationPRACTICE PROBLEMS IN POPULATION GENETICS
PRACTICE PROBLEMS IN POPULATION GENETICS 1. In a study of the Hopi, a Native American tribe of central Arizona, Woolf and Dukepoo (1959) found 26 albino individuals in a total population of 6000. This
More informationBio 102 Practice Problems Mendelian Genetics and Extensions
Bio 102 Practice Problems Mendelian Genetics and Extensions Short answer (show your work or thinking to get partial credit): 1. In peas, tall is dominant over dwarf. If a plant homozygous for tall is crossed
More information7A The Origin of Modern Genetics
Life Science Chapter 7 Genetics of Organisms 7A The Origin of Modern Genetics Genetics the study of inheritance (the study of how traits are inherited through the interactions of alleles) Heredity: the
More informationChapter 9 Patterns of Inheritance
Bio 100 Patterns of Inheritance 1 Chapter 9 Patterns of Inheritance Modern genetics began with Gregor Mendel s quantitative experiments with pea plants History of Heredity Blending theory of heredity -
More informationI. Genes found on the same chromosome = linked genes
Genetic recombination in Eukaryotes: crossing over, part 1 I. Genes found on the same chromosome = linked genes II. III. Linkage and crossing over Crossing over & chromosome mapping I. Genes found on the
More informationA and B are not absolutely linked. They could be far enough apart on the chromosome that they assort independently.
Name Section 7.014 Problem Set 5 Please print out this problem set and record your answers on the printed copy. Answers to this problem set are to be turned in to the box outside 68-120 by 5:00pm on Friday
More informationGenetics with a Smile
Teacher Notes Materials Needed: Two coins (penny, poker chip, etc.) per student - One marked F for female and one marked M for male Copies of student worksheets - Genetics with a Smile, Smiley Face Traits,
More informationB2 5 Inheritrance Genetic Crosses
B2 5 Inheritrance Genetic Crosses 65 minutes 65 marks Page of 55 Q. A woman gives birth to triplets. Two of the triplets are boys and the third is a girl. The triplets developed from two egg cells released
More informationTwo-locus population genetics
Two-locus population genetics Introduction So far in this course we ve dealt only with variation at a single locus. There are obviously many traits that are governed by more than a single locus in whose
More informationThe Developing Person Through the Life Span 8e by Kathleen Stassen Berger
The Developing Person Through the Life Span 8e by Kathleen Stassen Berger Chapter 3 Heredity and Environment PowerPoint Slides developed by Martin Wolfger and Michael James Ivy Tech Community College-Bloomington
More informationPaternity Testing. Chapter 23
Paternity Testing Chapter 23 Kinship and Paternity DNA analysis can also be used for: Kinship testing determining whether individuals are related Paternity testing determining the father of a child Missing
More informationGenetics and Evolution: An ios Application to Supplement Introductory Courses in. Transmission and Evolutionary Genetics
G3: Genes Genomes Genetics Early Online, published on April 11, 2014 as doi:10.1534/g3.114.010215 Genetics and Evolution: An ios Application to Supplement Introductory Courses in Transmission and Evolutionary
More informationAssociation between Dopamine Gene and Alcoholism in Pategar Community of Dharwad, Karnataka
International Journal of Scientific and Research Publications, Volume 3, Issue 10, October 2013 1 Association between Dopamine Gene and Alcoholism in Pategar Community of Dharwad, Karnataka SOMASHEKHAR
More informationA Study of Malaria and Sickle Cell Anemia: A Hands-on Mathematical Investigation Student Materials: Reading Assignment
4/15/99 1 A Study of Malaria and Sickle Cell Anemia: A Hands-on Mathematical Investigation Student Materials: Reading Assignment Malaria is a parasitic disease which is spread by the female Anopheles mosquitoes.
More informationGenetic Mutations. Indicator 4.8: Compare the consequences of mutations in body cells with those in gametes.
Genetic Mutations Indicator 4.8: Compare the consequences of mutations in body cells with those in gametes. Agenda Warm UP: What is a mutation? Body cell? Gamete? Notes on Mutations Karyotype Web Activity
More informationIncomplete Dominance and Codominance
Name: Date: Period: Incomplete Dominance and Codominance 1. In Japanese four o'clock plants red (R) color is incompletely dominant over white (r) flowers, and the heterozygous condition (Rr) results in
More informationGENETICS OF HUMAN BLOOD TYPE
GENETICS OF HUMAN BLOOD TYPE Introduction The genetics of blood types is relatively simple when considering any one blood protein. However, the complexity increases when one considers all the different
More informationBiology 1406 Exam 4 Notes Cell Division and Genetics Ch. 8, 9
Biology 1406 Exam 4 Notes Cell Division and Genetics Ch. 8, 9 Ch. 8 Cell Division Cells divide to produce new cells must pass genetic information to new cells - What process of DNA allows this? Two types
More informationThis fact sheet describes how genes affect our health when they follow a well understood pattern of genetic inheritance known as autosomal recessive.
11111 This fact sheet describes how genes affect our health when they follow a well understood pattern of genetic inheritance known as autosomal recessive. In summary Genes contain the instructions for
More informationLAB 11 Drosophila Genetics
LAB 11 Drosophila Genetics Introduction: Drosophila melanogaster, the fruit fly, is an excellent organism for genetics studies because it has simple food requirements, occupies little space, is hardy,
More informationEMPIRICAL FREQUENCY DISTRIBUTION
INTRODUCTION TO MEDICAL STATISTICS: Mirjana Kujundžić Tiljak EMPIRICAL FREQUENCY DISTRIBUTION observed data DISTRIBUTION - described by mathematical models 2 1 when some empirical distribution approximates
More informationBioBoot Camp Genetics
BioBoot Camp Genetics BIO.B.1.2.1 Describe how the process of DNA replication results in the transmission and/or conservation of genetic information DNA Replication is the process of DNA being copied before
More informationSummary. 16 1 Genes and Variation. 16 2 Evolution as Genetic Change. Name Class Date
Chapter 16 Summary Evolution of Populations 16 1 Genes and Variation Darwin s original ideas can now be understood in genetic terms. Beginning with variation, we now know that traits are controlled by
More informationstatistics Chi-square tests and nonparametric Summary sheet from last time: Hypothesis testing Summary sheet from last time: Confidence intervals
Summary sheet from last time: Confidence intervals Confidence intervals take on the usual form: parameter = statistic ± t crit SE(statistic) parameter SE a s e sqrt(1/n + m x 2 /ss xx ) b s e /sqrt(ss
More informationCHAPTER 15 THE CHROMOSOMAL BASIS OF INHERITANCE. Section B: Sex Chromosomes
CHAPTER 15 THE CHROMOSOMAL BASIS OF INHERITANCE Section B: Sex Chromosomes 1. The chromosomal basis of sex varies with the organism 2. Sex-linked genes have unique patterns of inheritance 1. The chromosomal
More informationGenetics for the Novice
Genetics for the Novice by Carol Barbee Wait! Don't leave yet. I know that for many breeders any article with the word genetics in the title causes an immediate negative reaction. Either they quickly turn
More informationP (B) In statistics, the Bayes theorem is often used in the following way: P (Data Unknown)P (Unknown) P (Data)
22S:101 Biostatistics: J. Huang 1 Bayes Theorem For two events A and B, if we know the conditional probability P (B A) and the probability P (A), then the Bayes theorem tells that we can compute the conditional
More informationInfluence of Sex on Genetics. Chapter Six
Influence of Sex on Genetics Chapter Six Humans 23 Autosomes Chromosomal abnormalities very severe Often fatal All have at least one X Deletion of X chromosome is fatal Males = heterogametic sex XY Females
More informationLECTURE 6 Gene Mutation (Chapter 16.1-16.2)
LECTURE 6 Gene Mutation (Chapter 16.1-16.2) 1 Mutation: A permanent change in the genetic material that can be passed from parent to offspring. Mutant (genotype): An organism whose DNA differs from the
More informationExploring contact patterns between two subpopulations
Exploring contact patterns between two subpopulations Winfried Just Hannah Callender M. Drew LaMar December 23, 2015 In this module 1 we introduce a construction of generic random graphs for a given degree
More informationTerms: The following terms are presented in this lesson (shown in bold italics and on PowerPoint Slides 2 and 3):
Unit B: Understanding Animal Reproduction Lesson 4: Understanding Genetics Student Learning Objectives: Instruction in this lesson should result in students achieving the following objectives: 1. Explain
More informationGenetics 301 Sample Final Examination Spring 2003
Genetics 301 Sample Final Examination Spring 2003 50 Multiple Choice Questions-(Choose the best answer) 1. A cross between two true breeding lines one with dark blue flowers and one with bright white flowers
More informationEvolution (18%) 11 Items Sample Test Prep Questions
Evolution (18%) 11 Items Sample Test Prep Questions Grade 7 (Evolution) 3.a Students know both genetic variation and environmental factors are causes of evolution and diversity of organisms. (pg. 109 Science
More informationPhenotypes and Genotypes of Single Crosses
GENETICS PROBLEM PACKET- Gifted NAME PER Phenotypes and Genotypes of Single Crosses Use these characteristics about plants to answer the following questions. Round seed is dominant over wrinkled seed Yellow
More informationASSIsT: An Automatic SNP ScorIng Tool for in and out-breeding species Reference Manual
ASSIsT: An Automatic SNP ScorIng Tool for in and out-breeding species Reference Manual Di Guardo M, Micheletti D, Bianco L, Koehorst-van Putten HJJ, Longhi S, Costa F, Aranzana MJ, Velasco R, Arús P, Troggio
More informationRingneck Doves. A Handbook of Care & Breeding
Ringneck Doves A Handbook of Care & Breeding With over 100 Full Color Photos, Including Examples and Descriptions of 33 Different Colors and Varieties. K. Wade Oliver Table of Contents Introduction, 4
More informationPractice Problems 4. (a) 19. (b) 36. (c) 17
Chapter 10 Practice Problems Practice Problems 4 1. The diploid chromosome number in a variety of chrysanthemum is 18. What would you call varieties with the following chromosome numbers? (a) 19 (b) 36
More informationReport. A Note on Exact Tests of Hardy-Weinberg Equilibrium. Janis E. Wigginton, 1 David J. Cutler, 2 and Gonçalo R. Abecasis 1
Am. J. Hum. Genet. 76:887 883, 2005 Report A Note on Exact Tests of Hardy-Weinberg Equilibrium Janis E. Wigginton, 1 David J. Cutler, 2 and Gonçalo R. Abecasis 1 1 Center for Statistical Genetics, Department
More informationLesson Plan: GENOTYPE AND PHENOTYPE
Lesson Plan: GENOTYPE AND PHENOTYPE Pacing Two 45- minute class periods RATIONALE: According to the National Science Education Standards, (NSES, pg. 155-156), In the middle-school years, students should
More informationTitle: Genetics and Hearing Loss: Clinical and Molecular Characteristics
Session # : 46 Day/Time: Friday, May 1, 2015, 1:00 4:00 pm Title: Genetics and Hearing Loss: Clinical and Molecular Characteristics Presenter: Kathleen S. Arnos, PhD, Gallaudet University This presentation
More informationEx) A tall green pea plant (TTGG) is crossed with a short white pea plant (ttgg). TT or Tt = tall tt = short GG or Gg = green gg = white
Worksheet: Dihybrid Crosses U N I T 3 : G E N E T I C S STEP 1: Determine what kind of problem you are trying to solve. STEP 2: Determine letters you will use to specify traits. STEP 3: Determine parent
More informationLecture 3: Mutations
Lecture 3: Mutations Recall that the flow of information within a cell involves the transcription of DNA to mrna and the translation of mrna to protein. Recall also, that the flow of information between
More informationLecture 10 Friday, March 20, 2009
Lecture 10 Friday, March 20, 2009 Reproductive isolating mechanisms Prezygotic barriers: Anything that prevents mating and fertilization is a prezygotic mechanism. Habitat isolation, behavioral isolation,
More informationProcess 3.5. A Pour it down the sink. B Pour it back into its original container. C Dispose of it as directed by his teacher.
Process 3.5 Biology EOI sample test questions Objective numbers correspond to the State Priority Academic Student Skills (PASS) standards and objectives. This number is also referenced with the local objective
More informationTuesday 14 May 2013 Morning
THIS IS A NEW SPECIFICATION H Tuesday 14 May 2013 Morning GCSE TWENTY FIRST CENTURY SCIENCE BIOLOGY A A161/02 Modules B1 B2 B3 (Higher Tier) *A137150613* Candidates answer on the Question Paper. A calculator
More informationGCSE BITESIZE Examinations
GCSE BITESIZE Examinations General Certificate of Secondary Education AQA SCIENCE A BLY1B Unit Biology B1b (Evolution and Environment) AQA BIOLOGY Unit Biology B1b (Evolution and Environment) FOUNDATION
More informationEvolution by Natural Selection 1
Evolution by Natural Selection 1 I. Mice Living in a Desert These drawings show how a population of mice on a beach changed over time. 1. Describe how the population of mice is different in figure 3 compared
More informationDNA for Defense Attorneys. Chapter 6
DNA for Defense Attorneys Chapter 6 Section 1: With Your Expert s Guidance, Interview the Lab Analyst Case File Curriculum Vitae Laboratory Protocols Understanding the information provided Section 2: Interpretation
More informationSeattleSNPs Interactive Tutorial: Web Tools for Site Selection, Linkage Disequilibrium and Haplotype Analysis
SeattleSNPs Interactive Tutorial: Web Tools for Site Selection, Linkage Disequilibrium and Haplotype Analysis Goal: This tutorial introduces several websites and tools useful for determining linkage disequilibrium
More informationTEXAS A&M PLANT BREEDING BULLETIN
TEXAS A&M PLANT BREEDING BULLETIN October 2015 Our Mission: Educate and develop Plant Breeders worldwide Our Vision: Alleviate hunger and poverty through genetic improvement of plants A group of 54 graduate
More informationPackage forensic. February 19, 2015
Type Package Title Statistical Methods in Forensic Genetics Version 0.2 Date 2007-06-10 Package forensic February 19, 2015 Author Miriam Marusiakova (Centre of Biomedical Informatics, Institute of Computer
More informationScheme of work Cambridge IGCSE Biology (0610)
Scheme of work Cambridge IGCSE Biology (0610) Unit 8: Inheritance and evolution Recommended prior knowledge Basic knowledge of Unit 1 cell structure is required, and also an understanding of the processes
More informationUsing Blood Tests to Identify Babies and Criminals
Using Blood Tests to Identify Babies and Criminals Copyright, 2012, by Drs. Jennifer Doherty and Ingrid Waldron, Department of Biology, University of Pennsylvania 1 I. Were the babies switched? Two couples
More information