GCD 3022 1st Edition Lecture 12Outline of Last Lecture I. Maternal Effecta. Nurse Cellsb. Drosophilac. Water Snailsi. F1 genotypes and phenotypesii. F2 genotypes and phenotypesiii. Percentage of F2 that are dextral iv. F3 generationv. Phenotypic ratio of F3 generationII. Epigenetic Inheritancea. Examples of epigenetic inheritanceb. Genomic imprintingIII. Igf2+ and Igf2- and genomic imprintinga. Scenario 1b. Scenario 2c. Imprinted genesd. Scenario 3IV. Types of InheritanceThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.a. Example 1b. Example 2c. Example 3V. De novo DNA methylationVI. The Lyon HypothesisVII. Barr BodiesVIII. Reciprocal Crossa. Definitionb. Reciprocal cross scenarioc. Maternal inheritanceOutline of Current LectureI. Intro to linkagea. Linkage Groupsb. RecombinationII. Recombination (Crossing Over)a. Discoveryb. ProcessIII. Determining if genes are linkeda. Chi square testb. StepsIV. Evidence of Linked Genesa. Experimentb. Resultc. HypothesesV. Genetic mapping in plants and animalsa. Definitionb. Equationc. Testcrossd. Map unitsVI. Three factor crossesa. Step 1b. Step 2c. Step 3d. Step 4e. Step 5Current LectureI. Linkage: genes on the same chromosome can be inherited as a group instead of individually. This means that they do not assort independently (violate Mendel’s law of independent assortment)a. Linkage group: also referred to as chromosomes because most genes on a chromosome are linked. Number of linkage groups is the number of types of chromosomes in a species (ex: humans have 22 autosomal linkage groups and an X and Y linkage group).b. Recombination: also known as crossing over. Occurs when genes on the same chromosome combine to form new genotypic combinations. The closer the genes are, the less likely the event of crossover, but genes that are far apart may assort independently from one another.II. Recombinationa. Discovery: discovered by Bateson and Punnett in 1905 using sweet pea plant crosses and observing the traits of pollen shape and flower color. The dihybrid cross should have produced a 9:3:3:1 ratio according to Mendelian inheritance,but the parental phenotypes were more common than expected. This was due torecombination during meiosis. b. Process: occurs during prophase I of meiosis. Replicated sister chromatids from two homologous chromosomes associate as bivalents and exchange DNA segments. i. Cells with nonparental genetic information are called nonparental or recombinant cells. III. Determining if genes are linkeda. Chi square test: compares expected values with observed values to determine if the result of the experiment is due to random chance alone (accept null hypothesis) or the findings are statistically significant (reject null hypothesis).b. Steps to the chi square testi. Propose a hypothesis1. Ex: the genes for flower color and pollen shape assort independently. This hypothesis allows us to calculate the expectedvalues. ii. Based on the hypothesis, calculate the expected values of each of the four phenotypes. (an independent assortment hypothesis will predict thateach phenotype has an equal probability of occurring) iii. Apply the chi square formula (see earlier chapters). Accept or reject the hypothesis depending on the P value from the table. IV. Evidence of linkagea. Experimenti. Morgan provided evidence of linkage through study of several X-linked genes in drosophilaii. Traits were body color, eye color, and wing lengthb. Resultsi. A much higher proportion of combinations of traits found in the parental generation were found in the F2 generation. ii. For example: if one of the parents had gray body, red eyes, and long wings there was a very large proportion of offspring that had that same phenotype.iii. Morgan also observed that there were a significant number of nonparental phenotypes in the F2 generation and that there was a quantitative difference between the parental and nonparental combinations.c. Hypothesesi. The genes for body color, eye color, and wing length are all located on theX-chromosome (alleles will tend to be inherited together)ii. Due to crossing over, the homologous X chromosomes (in females) can exchange pieces of chromosomesiii. The likelihood of crossing over depends on the distance between the two genes (more likely to occur when genes are farther apart from each other). V. Genetic mapping in plants and animalsa. Definition: also known as gene mapping or chromosome mapping. Purpose is to determine the linear order of linked genes along the same chromosome. b. Equation: Map distance = (# of recombinant offspring/total # of offspring)x100Units are called map units (mu), also referred to as centiMorgans (cM). one map unit is equivalent to 1% recombination frequency in a testcrossc. Testcross: used in genetic mapping experiments by mating a heterozygote (for two or more genes) to an individual that is homozygous recessive for the same genes. Expected to yield a maximum of only 50% recombinant offspring (anything over 50 map units is considered unreliable).VI. Three factor crosses: provides information about map distance and gene order. a. Step 1: cross two true-breeding strains that differ with regard to the three alleles.b. Step 2: perform a test cross by mating F1 female heterozygotes to male flies that are homozygous recessive for all three allelesc. Step 3: collect data for the F2 generation (double crossovers may occur, but only appear as the least frequent phenotypes). Genes that are farther apart will show higher rates of crossover than genes that are close together.d. Step 4:Calculate the map distance between pairs of genesi. Start by regrouping the data according to pairs of genes (parental and non-parental combinations)ii. The regrouped data will allow us to calculate the map distance between the two genese. Step 5: construct the map (based on map unit calculation)i. Location of the genes are mapped relative to the