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Chapter 7 and 26: July 26th Lecture 5

by: Hannah Kennedy

Chapter 7 and 26: July 26th Lecture 5 30156

Marketplace > Kent State University > Biological Sciences > 30156 > Chapter 7 and 26 July 26th Lecture 5
Hannah Kennedy
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These notes cover everything gone over in class on chapters 7 and 26 as well as from the book. This material includes linkage, linkage with recombination, 3 point mapping, statistical measures, etc.
  Dr. Helen Piontkivska
Class Notes
Genetics, Biology, linkage, recombination
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This 7 page Class Notes was uploaded by Hannah Kennedy on Saturday July 30, 2016. The Class Notes belongs to 30156 at Kent State University taught by   Dr. Helen Piontkivska in Spring 2016. Since its upload, it has received 13 views. For similar materials see ELEMENTS OF GENETICS in Biological Sciences at Kent State University.


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Date Created: 07/30/16
7/26 Lecture Notes—Ch 7 and 26 Genetic Linkage and Mapping in Eukaryotes 1. Q: 2 recessive disorders in mice—droopy ears and flaky tail—are caused by genes that are located 6 MU apart on chromosome. A true-breeding mouse with normal ears (De) and a flaky tail (ft) was crossed to true-breeding mouse with droopy ears (de) anda normal tail (Ft) The f1 offspring were then crossed to mice with droopy ears and flaky tails. If this testcross produced 100 offspring, what is the expected outcome 2. Q: The genes for mahogany eyes and ebony body are appx 25 MU apart. A mahogany-eyed female was mated to an ebony-bodied male; the resulting F1 phenotypically wild-type females were mated to mahogany, ebony males. Of 1000 offspring… a. What would be the expected phenotypes? b. In what numbers would these phenotypes be expected? 3. Q: the genes for tan body and bare wings are 15 MU apart. A tan-bodied, bare-winged female was mated to a wild-type male; the resulting 1 phenotypically wild-type females were mated to tan-bodied vare- winged males a. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected? (see pic) 4. Overview: there is gamete formation if a. Independent assortment b. Linkage i. Synteny = 2 or more genes are located on the same chromosome; genes that are syntenic are physically linked to each other ii. Linkage = the phenomenon in which genes that are close together on the same chromosome tend to be transmitted as a unit iii. Linkage groups = chromosomes are often called these bc a chromosome has a group of genes that are physically linked together (number of linkage groups = number of chromosome types: humans have 22 autosomal linkage groups and 1 sex linkage group) iv. Experiment done by Bateson and Punnett found that the F2 generation had a higher proportion of parental phenotypes so they suggested that the transmission of the 2 traits from the P generation to the F2 generation was couple and not assorted independently v. Nonrecombinant cells = parental cells = cells in which the arrangement of linked alleles hasn’t been altered from those in the original cell c. Linkage with recombination i. Crossing over = process in which homologous chromosomes can exchange pieces with each other during prophase I; affects pattern of inheritance for linked genes 1. Bivalent = the structure that replicated chromosomes form (i.e. sister chromatids) by associating with the homologous sister chromatids ii. Genetic recombination = event that leads to a new combination of alleles iii. Recombinant cells = nonparental cells = the cells of which display combinations of traits that are different from those of either parent iv. Experiment that proved this: Thomas Hunt Morgan 1. P cross: wild type male fruit flies: gray bodies, red eyes, long wings (y+y, w+w, m+m) crossed with recessive females: yellow bodies, white eyes, and mini wings (yy, ww, mm) 2. F1 phenotypes: wild-type females and males with yellow bodies, white eyes, and mini wings 3. F1 were mated 4. F2 generation showed 8 phenotypic classes with a higher proportion of combinations of traits found in the parental generation (this is bc all 3 genes are located on the X chromosome and tend to be transmitted as a unit) 5. Because of crossing over, morgan proposed that homologous X chromosomes can exchange pieces of chromosomes and create new recombinant allele arrangements (small proportion of F2 offspring) a. There are few recombinant offspring because the 2 genes are very close together on the same chromosome which makes a crossover btwn them unlikely therefore if 2 genes are far apart they are more likely to undergo crossing over v. Double cross over = the process in which 2 homologous chromosomes crossed over twice 5. A chi square analysis can be used to decide whether 2 genes are linked assort independently a. Process: i. Step 1: propose a hypothesis: (i.e. in a dihybrid cross standard hypothesis is that the genes are not linked bc an independent assortment hypothesis lets us calculate the expected number of offspring based on genotypes of parents) 1. Null hypothesis = the hypothesis that we are testing that assumes there is no real difference between the observed and expected values a. if the chi square value is low: hypothesis is accepted and the genes assort independently b. if the chi square values is higher than the 0.05 value: the hypothesis is rejected and the genes are linked ii. Step 2: bases on our hypothesis, calculate the expected values of each othe phenotypes: each phenotype has an equal probability of occurring so you put 1/n x total number of samples iii. Step 3: apply the chi square formula, using the data for the observed values and expected values that were calculated in step 2 (aka get the X value) iv. Step 4: interpret the calculated chi square value: done by using a chi square table and getting the degree of freedom 6. 3 point mapping a. Why?: to map 3 genes (genetic map = a diagram that describes the order of genes along a chromosome) i. Genetic mapping is used to determine the linear order and distance of separation among genes that are linked to each other along the same chromosome 1. To interpret genetic mapping experiment, we must know if characteristics of offspring are due to crossing over during meiosis (this is done by conducting a testcross) a. Goal of testcross: determine if recombination has occurred in prophase I of the heterozygous parent (because new combinations of alleles cant occur in the gametes of the homozygous parent) i. Coupling = cis configuration = in the heterozygous parent, all wild-type alleles (dominant) are across from corresponding mutant (recessive) alleles on the homologous chromosome ii. Repulsion = trans configuration = in a heterozygous parent, wild type and mutant alleles are on different chromosomes ii. Map distance = used to calculate the units of distance; number of recombinant offspring x100 number of totaloffspring 1. As the % of recombinant offspring approach 50, the value increasingly gets more inaccurate; a testcross is expected to yield a maximum of only 50% recombinant offspring so to exceed this level, it means to have multiple crossovers within a tetrad 2. When 2 different genes are more than 50 mu apart, they follow the law of independent assortment in a testcross and only 50% of recombinants are observed b. 3 conditions need to be met: i. all 3 genes have 2 alleles (i.e. heterozygous) ii. the genotype can be inferred directly from the phenotype at all 3 loci iii. there is a large sample size to recover all recombinant classes 1. the frequency of recombinant offspring due to crossing over provides a way to deduce the linear order of genes along a chromosome (% of recombinant offspring is correlated with the distance between the 2 genes) c. Phenotypes of offspring: i. 4 single crossover classes 1. 2 between genes 1 and 2 2. 2 between 2 and 3 ii. 2 double crossover classes 1. when doing a double crossover, we are flipping the middle gene iii. 2 parental arrangements iv. 8 phenotypic classes d. gene configuration in double crossovers: the dominant and recessive alleles will complement each other across from one another and the middle allele will switch when testing for the DCO middle gene arrangement e. challenges in gene mapping i. (1): undetectable events bc double exchange leads to the same allele arrangements ii. (2): probability of multi-strand exchanges increases with distance therefore the gene maps are less precise over larger distances 1. very little recombination occurs in the heterochromatin 2. inaccuracy increases with distance of MU iii. (3): chromosome organization with regions of lower and higher recombination rates f. Interference and coefficient of coincidence: bc 2 recombination events during double crossing over aren’t truly independent i. Interference = process in which 1 cross over event inhibits the second cross over event nearby, leading to a decrease in the distance of the DCO (i.e. [DCO]) 1. I = 1 – C Cross Expected Double ¿ 2. C = Cross Observed Double ¿ ¿ ¿ 3. Expected Double Cross Over = [SCO ] x [1-2 ] 2-3 g. Process: i. Step 1: cross 2 true-breeding strains that differ with regard to 3 alleles 1. Goal in this step is to get F1 heterozygotes bc in F1 heterozygote all dominant alleles are on the same chromosome and the recessive alleles are on homologous chromosome ii. Step 2: perform a testcross by mating F1 female heterozygotes to male flies that are homozygous recessive for all 3 alleles 1. During oogenesis in female heterozygotes, crossovers may produce new combos of the 3 alleles iii. Step 3: collect data for the F2 generation 1. 8 phenotypic combinations are possible (if the alleles assorted independently all 8 combos would occur in equal proportions) 2. double cross over is always expected to be the smallest proportion of offspring a. combinations of traits in the DCO tells us which gene is in the middle b. when chromatid undergoes DCO the middle gene becomes separated from the other 2 genes at the ends iv. Step 4: calculate the map distance between pairs of genes 1. First understand which gene combos are recombinant and nonrecombinant 2. Recombinant offspring are due to crossing over in heterozygous female parent 3. Calculate map distance for the 1 SCO between the first and middle gene a. SCO = sum of the first and middle gene recombinant offspring/total offspring nd 4. Calculate map distance for the 2 SCO between the middle and last gene a. SCO = sum of the middle and last gene recombinant offspring/total offspring 5. Product rule allows us to predict the expected likelihood of a DCO, if we know the individual probabilities of each single crossover v. Step 5: construct the map Ch. 26 1. Quantitative inheritance = the range of measurable phenotypes, generally due to multiple loci (mendelian laws still apply); can be categorized as anatomical, physiological, and behavioral a. Quantitative trait loci = QTLs = locations on chromosomes containing genes that affect the outcome of quantitative traits b. traits are considered quantitative because the alleles of several different genes contribute to the likelihood that an individual will develop the trait (Ex = height and weight, heart disease, cancer, diabetes) i. Continuous traits = traits that don’t fall into discrete categories (e.g. height and weight) ii. Meristic traits = traits that can be counted and expressed in whole numbers (e.g. bristles on a fruit fly) iii. Threshold traits = traits that are inherited quantitatively due to the contribution of many genes, but the traits themselves are expressed qualitatively; trait in which a certain threshold must be reached in which the number of disease-causing alleles results in the development of the disease iv. Exhibit a continuum of phenotypic variation v. Frequency distribution = an alternative way to describe most quantitative traits because they don’t fall into discrete categories; trait is arbitrarily divided into a number of phenotypic categories and plotted on a graph vi. Normal distribution = a distribution for a large sample in which the trait of interest varies in a symmetrical way around an average value; appximated by a symmetrical bell curve 2. Additive Alleles = alleles that contribute to most observable traits such as height, weight, hair and color, and complexion; a form of allelic interaction in which there is no dominance a. Ex: additive alleles contribute to the red grain color of wheat. Wheat species have 2 different genes that control the color that exist in red and white allelic form, therefore, 2 loci contribute in an additive manner to the color. (i.e. the more red alleles the darker the wheat) i. Process: cross the true-breeding parents: 1 red and 1 white. Get F1 that is intermediate in phenotype. Each gene has 1 additive and 1 non-additive allele 3. Polygenic inheritance a. Most quantitative traits are polygenic and exhibit a continuum of phenotypic variation b. Polygenic inheritance = the transmission of a trait governed by 2 or more different genes c. As the number of genes controlling a trait increases and the environmental influence increases, the categorization of phenotypes into discrete categories gets harder and we then need to use statistics d. 2 ways to estimate the # of genes (i..e. loci) 1 i. use the percentage of the most extreme phenotypes (4N ) ii. use the number of the phenotypic categories (2n +1 rule) e. QTLs are mapped by linkage to known molecular markers i. Molecular markers like microsatellites and restriction fragment length polymorphism (RFLPs) serve as reference points on a chromosome 1. To map genes, researchers determine the location by identifying linkage to molecular markers ii. QTL Mapping = the association between genetically determined phenotypes for quantitative traits and molecular markers such as RFLPs, microsatellites, and single-nucleotide polymorphisms (SNPs) 1. Goal: locate unknown genes affecting a certain trait 2. General strategy: inbred strains (i.e. homozygous for most molecular markers and genes) differ in 2 ways (molecular markers and quantitative trait of interest). 2 strains are crossed to produce F1. F1 are backcrossed to both parental strains. F2 offspring contain diff combos of parental chromosomes. F2 are analyzes for molecular marker composition and quantitative trait. 4. Statistical methods are used to evaluate a frequency distribution quantitatively Σn a. mean = average: useful to tell us about the center of our distribution Mean= n b. s = variance: characterizes the shape of the distribution; the greater the variance, 2 Σfi(Xi−X´ ) the more spread out the distribution about the mean is V =x N−1 i. Vx: variance ii. Xi−X´ : the difference between each value and the mean iii. N=¿ number of observations iv. Σ=¿ the sum of all values that are differenced from the mean and squared c. s = standard deviation; square root of variance i. 68% of all individuals have values within 1 standard deviation from the mean ii. 95% of all individuals have values within 2 standard deviations from the mean iii. 99.7% of all individuals have values within 3 standard deviations from the mean d. CoV = covariance: describes the relationship between 2 variables within a group Σfi[Xi−X ´)Yi−Y ] CoV (X, Y) N−1 e. Correlation coefficient (r): to evaluate the strength of association between 2 CoV (x ,y) variables: (X, Y) (SDx)(SDy) i. Positive correlation: indicates that there is a direct association between two variables ii. Negative correlation: indicates that there is an inverse association between variables 5. Genetics of complex diseases a. Factors that affect phenotype i. Genotype ii. Environmental effects (especially on F 2henotypes) b. Total phenotypic variance and heritability (additive variances) i.V = environmental variance E ii.V G = genetic variance (includes 3 subcomponents): will equal 0 if the trait is homogenous V = VG+ V A V D I 1. V D variance due to the effects of alleles that follow a dominant/recessive pattern of inheritance 2. V A variance due to the additive effects of alleles 3. V i variance due to the effects of alleles that interact in an epistatic manner iii.V GxE = genetic and environmental interaction variance iv.V P = phenotypic variance (due to the sum of 3 factors): V P V +GV E 1. Environmental variance 2. Genetic variance 3. Genetic/environmental interaction variance v. Heritability = the amount of phenotypic variation within a group of individuals that is due to genetic variation Vg Vg 1. Broad sense heritability: H = = : an estimate Vp Vg+Ve+Vgxe of the proportion of phenotypic variation due to genetic factors; takes into account different types of genetic variation that may affect phenotype a. If the broad sense heritability is closer to 1, it is strong and thus the majority of the proportion of phenotypic variation is due to genetic factors b. If the broad sense heritability is closer to 0, it is weak and thus the majority of the proportion of phenotypic variation is not due to genetic factors 2 Va 2 robs Xo−X ´ 2. Narrow sense heritability: h = Vp or h = r exp or Xp−X : this is due to the additive effects of alleles a. Xo=meanof theoffspring b. Xp=meanof the parents c. X=meanof starting populations d. r obs : the observed phenotypic correlation between related individuals e. r exp: the expected correlation based on the known genetic relationship i. r exp = .50 for siblings ii. r exp = .50 for parent/child iii. r exp = .25 for aunt-niece iv. r exp = 1.0 for identical twins 6. Twin studies a. Monozygotic and dizygotic twins i. Concordance = the degree to which pairs of individuals exhibit the same trait ii. Discordance = the degree to which pairs of individuals don’t exhibit the same trait


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