Slide 1

Oogenesis As embryo until menopause... Ovaries Primordial germ cells (2N) Oogonium (2N) Primary oocyte (2N) Between birth & puberty; prophase I of meiosis Puberty; FSH; completes meiosis I Secondary oocyte (1N); polar body Meiosis II; stimulated by fertilization Ovum (1N); 2nd polar body Slide 2 Spermatogenesis Puberty until death! Seminiferous tubules ~ location Primordial germ cell (2n)~ differentiate into. Spermatogonium (2n) ~ sperm precursor Repeated mitosis into. Primary spermatocyte (2n) 1st meiotic division Secondary spermatocyte (n) 2nd meiotic division Spermatids (n)~ Sertoli cells. Sperm cells (n) Slide 3 Comparing Mitosis & Meiosis Slide 4 Modern genetics began with Gregor Mendels quantitative experiments with pea plants Ch. 14 - Mendelian Genetics and the Inheritance of Genetic Traits Figure 9.2A, B Stamen Carpel Slide 5 Gregor Mendel (Father of Genetics) Discovered the fundamentals of Genetics in the 1860s Lived in Austria and studied in Vienna Worked with Garden Peas (Pisum sativum) Gathered a huge amount of numerical data Discovered the frequency of how traits are inherited Established basic principles of Genetics Slide 6 The science of heredity dates back to ancient attempts at selective breeding Until the 20th century, however, many biologists erroneously believed that characteristics acquired during lifetime could be passed on characteristics of both parents blended irreversibly in their offspring MENDELS PRINCIPLES Slide 7 Reason Mendel worked with Garden Peas Easy to grow Many variations were available Easy to control pollination (self vs cross) Flower is protected from other pollen sources (reproductive structures are completely enclosed by petals) Plastic bags can be used for extra protection Slide 8 Mendel crossed pea plants that differed in certain characteristics and traced the traits from generation to generation Figure 9.2C This illustration shows his technique for cross-fertilization 1 Removed stamens from purple flower White Stamens Carpel Purple PARENTS (P) OFF- SPRING (F 1 ) 2 Transferred pollen from stamens of white flower to carpel of purple flower 3 Pollinated carpel matured into pod 4 Planted seeds from pod Slide 9 Mendel studied seven pea characteristics Figure 9.2D He hypothesized that there are alternative forms of genes (although he did not use that term), the units that determine heredity FLOWER COLOR FLOWER POSITION SEED COLOR SEED SHAPE POD SHAPE POD COLOR STEM LENGTH PurpleWhite AxialTerminal YellowGreen RoundWrinkled InflatedConstricted GreenYellow TallDwarf Slide 10 Introductory Questions #4 Traits: Flower Position: Axial & terminal Seed Color: Yellow & green Height: Tall & short 1) Monohybrid cross: Two hybrid plants are tall. How many of the offspring would you predict will be short if there were 400 produced? 2) Dihybrid cross: Two hybrid plants with yellow seeds and axial flowers are crossed. How many of the offspring would you predict will have axial flowers with green seeds if 3750 are produced? 3) Trihybrid cross: Both parents are heterozygous for all three traits. How many will be tall with terminal flowers and yellow seeds if 250 are produced? Slide 11 Genetic Vocabulary Punnett square: predicts the results of a genetic cross between individuals of known genotype Homozygous: pair of identical alleles for a character Heterozygous: two different alleles for a gene Phenotype: an organisms traits Genotype: an organisms genetic makeup Testcross: breeding of a recessive homozygote X dominate phenotype (but unknown genotype) Slide 12 Mendelian Genetics Character (heritable feature, i.e., fur color) Trait (variant for a character, i.e., brown) True-bred (all offspring of same variety) Hybridization (crossing of 2 different true-breeds) P generation (parental-beginning gen.) F 1 generation (first filial generation) F 2 generation (second filial generation) Slide 13 Alternative forms of a gene (alleles) reside at the same locus on homologous chromosomes Homologous chromosomes bear the two alleles for each characteristic GENE LOCI Figure 9.4 PaB DOMINANT allele RECESSIVE allele Pab GENOTYPE:PPaaBb HOMOZYGOUS for the dominant allele HOMOZYGOUS for the recessive allele HETEROZYGOUS Slide 14 From his experimental data, Mendel deduced that an organism has two genes (alleles) for each inherited characteristic One characteristic comes from each parent Mendels principle of segregation describes the inheritance of a single characteristic P GENERATION (true-breeding parents) F 1 generation F 2 generation Purple flowersWhite flowers All plants have purple flowers Fertilization among F1 plants (F 1 x F 1 ) 3 / 4 of plants have purple flowers 1 / 4 of plants have white flowers Figure 9.3A Slide 15 A sperm or egg carries only one allele of each pair The pairs of alleles separate when gametes form This process describes Mendels law of segregation Alleles can be dominant or recessive GENETIC MAKEUP (ALLELES) P PLANTS F 1 PLANTS (hybrids) F 2 PLANTS PPpp All PAll p All Pp 1/2 P1/2 P 1/2 p1/2 p Eggs P p P PP p Sperm Pp pp Gametes Phenotypic ratio 3 purple : 1 white Genotypic ratio 1 PP : 2 Pp : 1 pp Figure 9.3B Slide 16 By looking at two characteristics at once, Mendel found that the alleles of a pair segregate independently of other allele pairs during gamete formation This is known as the principle of independent assortment The principle of independent assortment is revealed by tracking two characteristics at once Slide 17 Remember.. Every Trait within a diploid organism will have two alleles These alleles are separated during Meiosis Traits: Seed color, seed shape, height Alleles: Y or y R or r T or t Parents are diploid Gametes are produced: only one allele is present for each trait in each gamete Slide 18 Parents Genotype Possible Alleles combinations for one gamete Example: monohybridYyY & (y) DihybridYyRrYR (3 OTHERS) TrihybridYyRrTtyRt (7 OTHERS) Slide 19 Inheritance follows the rules of probability The rule of multiplication and the rule of addition can be used to determine the probability of certain events occurring Mendels principles reflect the rules of probability F 1 GENOTYPES Bb female F 2 GENOTYPES Formation of eggs Bb male Formation of sperm 1/21/2 1/21/2 1/21/2 1/21/2 1/41/4 1/41/4 1/41/4 1/41/4 BB BB B B b b b b bb Figure 9.7 Slide 20 Figure 9.5A HYPOTHESIS: DEPENDENT ASSORTMENT HYPOTHESIS: INDEPENDENT ASSORTMENT P GENERATION F 1 GENERATION F 2 GENERATION RRYYrryy GametesRY Yellow round ry RrYy EggsSpermRY ry RY ry 1/21/2 1/21/2 1/21/2 1/21/2 Actual results contradict hypothesis RRYYrryy RY ry Gametes RrYy EggsRY rY 1/41/4 1/41/4 Ry ry 1/41/4 1/41/4 RY rY Ry ry 1/41/4 1/41/4 1/41/4 1/41/4 RRYY RrYY RRYyrrYYRrYy rrYyRRyyrrYy Rryy rryy 9 / 16 3 / 16 1 / 16 Green round Yellow wrinkled Yellow wrinkled ACTUAL RESULTS SUPPORT HYPOTHESIS Pg. 257 Slide 21 The chromosomal basis of Mendels Principles Figure 9.17 Slide 22 Important ratios to Remember Cross Phenotypic Genotypic BB x bb 4:0 (100% Dom.) 4:0 (all) Bb BB x Bb 4:0 (100% Dom.) 1:1 (50% BB,Bb) Bb x Bb 3:1 (75% Dom, 25% Rec) 1:2:1 (BB, Bb, bb) Bb x bb 1:1(50% Dom, 50% Rec) 1:1 (50% Bb,bb) Two traits (not linked): AaBb x AaBb 9:3:3:1 Slide 23 Introductory Questions #5 1)A Monohybrid cross and Dihybrid cross always produces Phenotypic rations of 3:1 and 9:3:3:1. What phenotypic ratio will produced from a trihybrid cross? 2)Solve this trihybrid cross with Pea plants Traits: Seed color, seed shape, height Male Heterozygous all traits Female Heterozygous yellow and tall plant w/ wrinkled seeds a)How many offspring would you predict will be Tall with wrinkled, yellow seeds? b)How many offspring would have green seeds that are round and tall? Slide 24 IQ #5-Solution 1) (male) (female) 2) YyRrTt x YyrrTt # allele combo in gametes: 8 4 Possible genotypes of the offspring: (a) TT rr YY: x x 1/4 = 1/32 TT rr Yy: x x 1/2 = 1/16 A. AnswerTt rr YY: x x = 1/16 Tt rr Yy: x x = 1/8 = 9/32 B. 3/32 will be predicted that will be tall with round green seeds Slide 25 Introductory Questions #6 1)A female heterozygous for Seed shape and color is crossed with a male that is heterozygous for seed shape but homozygous recessive for seed color. How many offspring would you predict (expect) to be yellow and wrinkled if 500 were produced? 2)If only 50 offspring were yellow and wrinkled how can you tell if your results were only due to chance? What statistical test could you do in order to determine if there;s a significant difference between what you actually got (50) vs. what you expected? 201 round & yellow 204 round & green 45 wrinkled and green 50 wrinkled & yellow Slide 26 IQ#6: Question #2 (contd) Expected (calculated Values) Expect. Value Obs. value wrinkled & green 62 45 Round & yellow187201 Round & green187204 Wrinkled & Yellow 6250 Slide 27 IQ #6-Answers RrYy x Rryy How many will be yellow and wrinkled? (500) Poss. genotypes: Yyrr = 1/2 x 1/4 = 1/8 Answer: 1/8 x 500 = 62 are expected to be yellow, green Do a chi-squared test Chi-Square value: ???? (Homework) Expected Value for each phenotype:187,187, 62, 62 Chi-squared value: Critical value from table (in lab)7.82 (0.05 or 95%) Slide 28 Sample Problem using Chi square Two hybrid Tall plants are crossed. If the F2 generation produced 787 tall plants and 277 short plants. Does this confirm Mendels explanation? What is the expected value? This is your null hypothesis (H O ) Total number of plants: 1064 3:1 Phenotypic ratio Expected value should be: 798 tall and 266 short (75%) (25%) Slide 29 Statistical Tools to Analyze results Chi-Square: Will tell you how much your data is different from expected (calculated) results. It is Non-Parametric and deals with different catagorical groups vs. Parametric which deals with numbers and which case you would use a T-test instead. Formula: 2 = (o e) 2 e 2 : what we are solving: o: observed value e: expected (calculated value) Slide 30 When to use Chi-Squared Test Can only be used with raw counts (not measurements) Comparing Experimental & expected (theo.) values Sample size must be more than 25 to be reliable Aims to test the null hypothesis (H 0 ) (H 0 ): the hypothesis that theres no difference between the data sets Alternative hypothesis: there is a significant difference Compare with a critical value table (p values) To reject the (H 0 ): value must be GREATER than the critical value & favor the alternative hypothesis. Accepting the null means that theres no significant difference between the data sets. Slide 31 Calculation of Chi Square Value 2 = (O E) 2 E 2 = (787 798) 2 + (277 266) 2 = 0.61 798 266 There are two categories and therefore the degrees of freedom would be 2-1 = 1. Look up the critical value for 1 degree of freedom: 3.84 (next slide-always given) If your value is LARGER than the critical value then you reject the null hypothesis and assume that there is a significant difference between the observed value and the expected. Values are statistically different. 0.61 is less than 3.84 therefore we accept the null hypothesis and accept that our values are similar enough Theres no significant difference between the observed & expected values. Values are not random. Slide 32 Answers to IQ #5 & #6 IQ #5----1. 27:9:9:9:3:3:3:1 = 64 phenotypes IQ #6----2. 2 = (O E) 2 E 2 = (45 62) 2 + (201-187) 2 + (204 187) 2 + (50-62) 2 = 9.58 6218718762 Critical Value: 4 groups = 3 degrees of freedom= 7.81 9.58 is greater than 7.81 therefore we reject the null hypothesis What does this mean? Slide 33 Accepting or Rejecting your hypothesis? Accepting the Null (H 0 ) means that: Test value is less than the critical value Values are similar enough there is a not SIGNIFICANT difference between the observed and expected value (p0.05). Evaluate the results. Less than 95% confidence in the values Slide 34 Solving Question #3 Formula: 2 = (o e) 2 e Degrees of FreedomCritical Value 13.84 25.99 37.81 49.49 511.07 Slide 35 Testcrosses Determining an Unknown Genotype Slide 36 The offspring of a testcross often reveal the genotype of an individual when it is unknown Geneticists use the testcross to determine unknown genotypes TESTCROSS: B_GENOTYPESbb BBBbor Two possibilities for the black dog: GAMETES OFFSPRING All black1 black : 1 chocolate B b B b b Bb bb Figure 9.6 Slide 37 These black Labrador puppies are purebred their parents and grandparents were black Labs with very similar genetic make-ups Purebreds often suffer from serious genetic defects Purebreds and Mutts A Difference of Heredity Slide 38 The parents of these puppies were a mixture of different breeds Their behavior and appearance is more varied as a result of their diverse genetic inheritance Slide 39 Independent assortment of two genes in the Labrador retriever Figure 9.5B PHENOTYPES Black coat, normal vision B_N_ Blind GENOTYPES MATING OF HETEROZYOTES (black, normal vision) PHENOTYPIC RATIO OF OFFSPRING Black coat, blind (PRA) B_nn Chocolate coat, normal vision bbN_ Chocolate coat, blind (PRA) bbnn 9 black coat, normal vision 3 black coat, blind (PRA) 3 chocolate coat, normal vision 1 chocolate coat, blind (PRA) Blind BbNn Slide 40 How is Codominance different from Incomplete Dominance? Slide 41 Non-single Gene Genetics (pg. 260-262) Incomplete dominance: -neither pair of alleles are completely expressed when both are present. -Typically, a third phenotype is produced Ex: snapdragons (pink flowers) hypercholesterolemia Codominance: Two alleles are expressed in a heterozygote condition. Ex: Human Blood types Slide 42 Incomplete Dominance Slide 43 When an offsprings phenotypesuch as flower color is in between the phenotypes of its parents, it exhibits incomplete dominance Incomplete dominance results in intermediate phenotypes P GENERATION F 1 GENERATION F 2 GENERATION Red RR GametesRr White rr Pink Rr Rr RR rr 1/21/2 1/21/2 1/21/2 1/21/2 1/21/2 1/21/2 SpermEggs Pink Rr Pink rR White rr Red RR Figure 9.12A Slide 44 Incomplete dominance in human hypercholesterolemia Figure 9.12B GENOTYPES: HH Homozygous for ability to make LDL receptors Hh Heterozygous hh Homozygous for inability to make LDL receptors PHENOTYPES: LDL LDL receptor Cell NormalMild diseaseSevere disease Slide 45 In a population, multiple alleles often exist for a characteristic The three alleles for ABO blood type in humans is an example Codominance Many genes have more than two alleles in the population Slide 46 Codominance-Observed in Blood Types Slide 47 Blood Type Frequencies of different Ethnic Groups Slide 48 Pleiotrophy vs. Polygenic Inheritance Slide 49 Non-single Gene Genetics Pleiotropy: genes with multiple phenotypic effect. Ex: sickle-cell anemia combs in roosters coat color in rabbits Polygenic Inheritance: an additive effect of two or more genes on a single phenotypic character Ex: human skin pigmentation and height Slide 50 A single gene may affect many phenotypic characteristics A single gene may affect phenotype in many ways This is called pleiotropy The allele for sickle-cell disease is an example Slide 51 Pleiotropy Sickle Cell anemia Slide 52 Effects of Sickle Cell Anemia Slide 53 Polygenic Inheritance (SG. #9) Slide 54 Figure 9.16 P GENERATION F 1 GENERATION F 2 GENERATION aabbcc (very light) AABBCC (very dark) AaBbCc EggsSperm Fraction of population Skin pigmentation Slide 55 Epistasis (SG. #11) Epistasis: a gene at one locus (chromosomal location) affects the phenotypic expression of a gene at a second locus. Ex: mice and Labrador coat color Slide 56 Epistasis (SG #11) Examples: Labradors coat color Albino Koala Two Genes Involved: Allele Symbol - Pigment- Black (Dominant) B Chocolate (recessive) b -Expression or deposition of the Pigment E/e Black YellowChocolate BBEE BBee bbEE BbEE Bbee bbEe BBEe BbEe Which genotype is missing and what group should it be listed under? Slide 57 Epistasis Slide 58 Study Guide Problems 1)White alleles are dominant to yellow 2)a. b. 1/8c. d. 1/32 3)Incomplete dominance Slide 59 Human Genome & Genetic Disorders Slide 60 Information Gained by the Genome Project (2003) Entire DNA (nucleus) composed of about 2.9 billion base pairs of nucleotides Six to Ten anonymous individuals were used Estimated number of genes = under 30,000 Only 1% to 2% of human DNA codes for a protein or RNA On Chromosome 22: 545 genes have been identified. Slide 61 The inheritance of many human traits follows Mendels principles and the rules of probability Genetic traits in humans can be tracked through family pedigrees Figure 9.8A Slide 62 Family pedigrees are used to determine patterns of inheritance and individual genotypes Figure 9.8B Dd Joshua Lambert Dd Abigail Linnell D_ Abigail Lambert Female Dd Elizabeth Eddy D_ John Eddy ?D_ Hepzibah Daggett ? ? ddDd ddDd Male Deaf Hearing dd Jonathan Lambert Slide 63 A high incidence of hemophilia has plagued the royal families of Europe Figure 9.23B Queen Victoria Albert AliceLouis AlexandraCzar Nicholas II of Russia Alexis Slide 64 Pedigree of Alkaptonuria Slide 65 Table 9.9 Slide 66 A few are caused by Dominant alleles Figure 9.9B Examples: Achondroplasia, Huntingtons disease Slide 67 Human Disorders The Family Pedigree Recessive disorders: -Cystic fibrosis -Tay-Sachs -Sickle-cell Dominant Disorders: -Huntingtons -Polydactaly Diagnosing/Testing: -Amniocentesis -Chorionic villus sampling (CVS) Slide 68 A human male has one X chromosome and one Y chromosome A human female has two X chromosomes Whether a sperm cell has an X or Y chromosome determines the sex of the offspring SEX CHROMOSOMES AND SEX-LINKED GENES Slide 69 Human sex-linkage SRY gene: gene on Y chromosome that triggers the development of testes Fathers= pass X-linked alleles to all daughters only (but not to sons) Mothers= pass X-linked alleles to both sons & daughters Sex-Linked Disorders: Color-blindness; Duchenne muscular dystropy (MD); hemophilia Slide 70 Most sex-linked human disorders are due to recessive alleles Examples: hemophilia, red-green color blindness These are mostly seen in males A male receives a single X-linked allele from his mother, and will have the disorder, while a female has to receive the allele from both parents to be affected Sex-linked disorders affect mostly males Figure 9.23A Slide 71 Sex Linked Trait: Colorblindness Slide 72 IQ #6-Answers 1)27:9:9:9:3:3:3:1 = 64 phenotypes 2)RrYy x Rryy How many will be yellow and wrinkled? (500) Poss. genotypes: Yyrr = 1/2 x 1/4 = 1/8 Answer: 1/8 x 500 = 63 are expected to be yellow, green 3)Do a chi-squared test Chi-Square value:10.08 Expected Value for each phenotype:188,188, 63, 63 Chi-squared value: Critical value from table (in lab)7.82 (0.05 or 95%) Slide 73 IQ#6: Question #3 (contd) Expected (calculated Values) Expect. Value Obs. value wrinkled & green 63 45 Round & yellow188201 Round & green188204 Wrinkled & Yellow 6350 Slide 74 Karyotyping and biochemical tests of fetal cells and molecules can help people make reproductive decisions Fetal cells can be obtained through amniocentesis Amniocentesis -Pg 270 Figure 9.10A Amniotic fluid Fetus (14-20 weeks) Placenta Amniotic fluid withdrawn Centrifugation Fetal cells Fluid UterusCervix Cell culture Several weeks later Karyotyping Biochemical tests Slide 75 Diagnostic Procedures to detect Genetic Disorders in Babies Slide 76 Chorionic Villus Sampling (CVS) is another procedure that obtains fetal cells for karyotyping.Pg. 270 Figure 9.10B Fetus (10-12 weeks) Placenta Chorionic villi Suction Several hours later Fetal cells (from chorionic villi) Karyotyping Some biochemical tests Slide 77 UltraSound (Pg. 269) Examination of the fetus with ultrasound is another helpful technique Figure 9.10C, D Slide 78 Genetic testing can be of value to those at risk of developing a genetic disorder or of passing it on to offspring Genetic testing can detect disease- causing alleles Figure 9.15B Figure 9.15A Dr. David Satcher, former U.S. surgeon general, pioneered screening for sickle-cell disease Slide 79 Introductory Questions #7 1)How is co-dominance different from Incomplete dominance? Give an example of both. 2)Give an example of polygenic inheritance. 3)Name three autosomal disorders that are dominant and three that are recessive. 4)Name the person who determined that genetic traits can be linked and inherited together? 5)What does it mean when traits are sex linked? Give an example of a human sex linked trait. How can linked traits be separated? 6)What is a two point test cross? Why would you use one? Slide 80 Table 9.9 Slide 81 Methods of Detecting Genetic Disorders Amniocentesis Ultrasound CVS (Chorionic Villus Sampling) PGD (Pre-implantation Genetic Diagnosis) Fetuscopy Genetic Counseling/Screening Slide 82 PGD: Preimplantion Genetic Diagnosis Used for Couples who are carriers of an abnormal allele. IVF Procedure is used Eggs are fertilized, grown in culture and tested for the disorder Normal embryos are implanted into the uterus. Slide 83 Table of Disorders Name Chromosome Cellular effectOverall involvement or (#) Phenotypic Result _______________________________________________________________________________ Down Syndrome Auto (47) Many Kleinfelters Syndrome Sex (47) Turners Syndrome Sex (45) Cri du Chat Auto/Deletion #5 Fragile X Auto & Sex Phenylketonuria (PKU)Auto rec. Enzyme def. AlkaptonuriaAuto rec. Enzyme def. Sickle Cell AnemiaAuto rec. Hemoglobin Struct. Cystic FibrosisAuto rec. Tay SachsAuto rec. Huntingtons DisorderAuto Dom. AchondroplasiaAuto Dom. AlbinismAuto rec. Color BlindnessSex-linked Muscular DystrophySex-linked HemophliaSex-linked AlzheimersAuto Dom. HypercholesterolemiaAuto Dom. Slide 84 Chapter 15: The Chromosomal Theory of Inheritance Gene linkage (Drosophila) Wild-types & mutants Gene mapping Non-Disjunction (aneuploidy) Barr bodies (inactive X) Alterations of Chromosome structure Genomic imprinting Pgs. 274-291 Slide 85 All genes on the sex chromosomes are said to be sex-linked In many organisms, the X chromosome carries many genes unrelated to sex Fruit fly eye color is a sex-linked characteristic Sex-linked genes exhibit a unique pattern of inheritance Figure 9.22A Slide 86 Chromosomal Linkage Thomas Morgan Drosophilia melanogaster XX (female) vs. XY (male) Sex-linkage: genes located on a sex chromosome Linked genes: genes located on the same chromosome that tend to be inherited together Slide 87 Their inheritance pattern reflects the fact that males have one X chromosome and females have two Figure 9.22B-D These figures illustrate inheritance patterns for white eye color (r) in the fruit fly, an X-linked recessive trait FemaleMaleFemaleMaleFemaleMale XrYXrYXRXRXRXR XRXrXRXr XRYXRY XRXR XrXr Y XRXrXRXr XRXR XrXr XRXRXRXR XRXR Y XRYXRY XrXRXrXR XRYXRY XrYXrY XRXrXRXr XRXR XrXr XrXr Y XRXrXRXr XrXrXrXr XRYXRY XrYXrY XrYXrY R = red-eye allele r = white-eye allele Slide 88 Certain genes are linked They tend to be inherited together because they reside close together on the same chromosome Genes on the same chromosome tend to be inherited together Slide 89 How to Determine if Two Genes are linked. Perform a Two Point Test Cross : Parents: AaBb X aabb Possible gametes: AB, Ab, aB, ab X ab Following Mendelian principles of independent assortment (not linked on the same chromosome) then: ABAbaBab AaBb (25%) Aabb (25%) aaBb (25%) aabb (25%) Slide 90 If Genes are Linked More Parental types should be present in the offspring and fewer recombinants. Parental type recombinant recombinant Parental type ABAbaBab AaBb (more) 40% Aabb (less) 10% aaBb (less) 10% aabb (more) 40% Slide 91 Figure 9.18 Slide 92 Generating Recombinants in Drosophila Slide 93 Figure 9.19C Slide 94 Crossing Over Developing Genetic Maps Pgs. 278-281 Slide 95 This produces gametes with recombinant chromosomes The fruit fly Drosophila melanogaster was used in the first experiments to demonstrate the effects of crossing over Crossing over produces new combinations of alleles Slide 96 Genetic Recombination (pg. 281) Crossing over Genes that DO NOT assort independently of each other Genetic maps The further apart 2 genes are, the higher the probability that a crossover will occur between them and therefore the higher the recombination frequency Linkage maps Genetic map based on recombination frequencies Slide 97 Crossing over is more likely to occur between genes that are farther apart Recombination frequencies can be used to map the relative positions of genes on chromosomes Geneticists use crossover data to map genes g Figure 9.20B Chromosome cl 17% 9%9.5% Slide 98 Generating Recombinant Offspring Pg. 280 Slide 99 A partial genetic map of a fruit fly chromosome Figure 9.20C Short aristae Black body (g) Cinnabar eyes (c) Vestigial wings (l) Brown eyes Long aristae (appendages on head) Gray body (G) Red eyes (C) Normal wings (L) Red eyes Mutant phenotypes Wild-type phenotypes (pg. 281) Slide 100 Genetic Map of Drosophila (pg. 281) Slide 101 Alfred H. Sturtevant, seen here at a party with T. H. Morgan and his students, used recombination data from Morgans fruit fly crosses to map genes Figure 9.20A Slide 102 Sex-Linked Patterns of Inheritance and Non-Disjunction Slide 103 Figure 9.21A XY Male (male) Parents diploid cells (female) Sperm Offspring (diploid) Egg Sex-Linked Patterns of Inheritance Slide 104 Other systems of sex determination exist in other animals and plants Figure 9.21B-D The X-O system The Z-W system Chromosome number Birds, fish, some insects Grasshoppers, cockaroaches, some insects Sex Determinant chromosome is in the ovum Bees & ants Slide 105 Nondisjunction can produce gametes with extra or missing sex chromosomes Unusual numbers of sex chromosomes upset the genetic balance less than an unusual number of autosomes Abnormal numbers of sex chromosomes do not usually affect survival Slide 106 Homologous pairs fail to separate during meiosis I Accidents During Meiosis Can Alter Chromosome Number Figure 8.21A Nondisjunction in meiosis I Normal meiosis II Gametes n + 1 n 1 Number of chromosomes Pg. 285 Slide 107 Chromosomal Errors Nondisjunction: members of a pair of homologous chromosomes do not separate properly during meiosis I or sister chromatids fail to separate during meiosis II Aneuploidy: chromosome number is abnormal Monosomy~ missing chromosome Trisomy~ extra chromosome (Down syndrome) Polyploidy~ extra sets of chromosomes Slide 108 Fertilization after Non-disjunction in the mother results in a zygote with an extra chromosome Figure 8.21C Egg cell Sperm cell n + 1 n (normal) Zygote 2n + 1 Slide 109 To study human chromosomes microscopically, researchers stain and display them as a karyotype A karyotype usually shows 22 pairs of autosomes and one pair of sex chromosomes ALTERATIONS OF CHROMOSOME NUMBER AND STRUCTURE Slide 110 Preparation of a Karyotype Figure 8.19 Blood culture 1 Centrifuge Packed red And white blood cells Fluid 2 Hypotonic solution 3 Fixative White Blood cells Stain 45 Centromere Sister chromatids Pair of homologous chromosomes Slide 111 This karyotype shows three number 21 chromosomes An extra copy of chromosome 21 causes Down syndrome Figure 8.20A, B Slide 112 The chance of having a Down syndrome child goes up with maternal age Figure 8.20C Slide 113 Autosomal Polyploid Disorders Syndrome/DisorderChromsome # Affected Downs Syndrome 21 Patau Syndrome13 Edwards Syndrome18 Slide 114 Table 8.22 Jacobs Syndrome Slide 115 Changes that can occur in Chromosome Structure (Abnormalities) Slide 116 Chromosome breakage can lead to rearrangements that can produce genetic disorders or cancer Four types of rearrangement are: deletion, duplication, inversion, and translocation Alterations of chromosome structure can cause birth defects and cancer Slide 117 Chromosomal changes in a somatic cell can cause cancer Figure 8.23C Chromosome 9 A chromosomal translocation in the bone marrow is associated with chronic myelogenous leukemia Chromosome 22 Reciprocal translocation Philadelphia chromosome Activated cancer-causing gene Slide 118 Chromosomal Errors Alterations of chromosomal structure: Pg. 327 Deletion: removal of a chromosomal segment Duplication: repeats a chromosomal segment Inversion: segment reversal in a chromosome Translocation: movement of a chromosomal segment to another Slide 119 Example of a Chromosomal Deletion Cri Du Chat: Cat cry syndrome Effects chromosome #5 Altered facial Features moon face Severe mental retardation Slide 120 Barr Bodies Inactive X Chromosome Pg. 284 Predominant in females Dark Region of chromatin is visible at the edge of the nucleus within a cell during interphase. (Please see Figure 15.11) A small fraction of the genes located on this X chromosome usually are expressed. Inactivation is a random event among the somatic cells. Heterozygous individuals: cells alleles expressed Ex. Calico cat & Tortoise shell (Variegation) Slide 121 Calico Kitten w/Barr Bodies Example of Variegation Slide 122 Barr Bodies Slide 123 Outbreeding vs. Inbreeding Inbreeding - Increases homozygosity in the population. -Increases frequency of genetic disorders -Amplifies the homozygous phenotypes Outbreeding: -Leads to better adapted offspring -Heterozygous advantage & Hybrid Vigor become evident and buffers out undesirable traits Slide 124 Genomic Imprinting Def: a parental effect on gene expression Identical alleles may have different effects on offspring, depending on whether they arrive in the zygote via the ovum or via the sperm Slide 125 Fragile X Syndrome Slide 126 More common in Males Pg. 327-328 Common form of Mental Retardation Thinned region on tips of chromatids Triplicate CGG repeats over 200 to 1000 times Normal: repeat 50 X or less Commonly seen in Cancer cells Varies in severity: Mild learning disabilities ADD Mental retardaton Slide 127 A man with Klinefelter syndrome has an extra X chromosome Figure 8.22A Poor beard growth Under- developed testes Breast development Slide 128 A woman with Turner syndrome lacks an X chromosome Figure 8.22B Characteristic facial features Web of skin Constriction of aorta Poor breast development Under- developed ovaries Slide 129 A B a b TetradCrossing over AB a ba BAb Gametes Figure 9.19A, B Slide 130 Independent Assortment in Budgie Birds