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Elements of Genetics Exam 2 Study Guide

by: Hannah Kennedy

Elements of Genetics Exam 2 Study Guide 30156

Marketplace > Kent State University > Biological Sciences > 30156 > Elements of Genetics Exam 2 Study Guide
Hannah Kennedy
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About this Document

This detailed and cohesive study guide covers everything gone over in class, in the textbook, and in connect that will be covered on Exam 2.
  Dr. Helen Piontkivska
Study Guide
Genetics, Biology, Exam 2
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This 10 page Study Guide was uploaded by Hannah Kennedy on Saturday July 30, 2016. The Study Guide belongs to 30156 at Kent State University taught by   Dr. Helen Piontkivska in Spring 2016. Since its upload, it has received 16 views. For similar materials see ELEMENTS OF GENETICS in Biological Sciences at Kent State University.


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Date Created: 07/30/16
© Hannah Kennedy, Kent State University Exam 2 Study Guide Chapter 8 1. Types of chromosome rearrangements a. 2 ways kinds of chromosome rearrangements i. Whole genome changes (i.e. large scale whole chromosome changes)—2 1. Variations in the number of sets of chromosomes a. Polyploidy = phenomenon in which the organism contains 3 or more sets of chromosomes (e.g. triploid/tetraploid/etc.) b. Endopolyploidy = the occurrence of polyploid tissues/cells that are otherwise diploid (e.g. liver cells. Biological implication: increase in chromosome set number may enhance cells ability to produce specific gene products needed in abundance) 2. Variations in the number of specific chromosomes within a set a. Aneuploidy = an alteration in the number of particular chromosomes therefore the total number of chromosomes isn’t an exact multiple of a set i. Trisomy = 1 extra chromosome (2n+1) with a genetic imbalance of 50% more gene expression ii. Monosomy = 1 less chromosome (2n – 1) with a genetic imbalance of 50% less gene expression ii. Chromosomal rearrangements (i.e. smaller scale: chunks of chromosome)—5 Rearrangement When it happens Consequences Additional info Deletion (2 types)— Nonallelic phenotypic: dependent - occurs when a change in total amt homologous upon size and what segment of of DNA recombination = kinds of genes are lost chromosomal 1. terminal When material is deletion = recombination in missing deletion in prophase I doesn’t - cause of which a happen at the misalignment: chromosome correct place chromosome breaks into 2 between the carries 2 or and the piece homologous more similar without the chromosomes DNA sequence centromere is therefore 1 lost and chromosome has a degraded deletion and the 2. interstitial other has a deletion = duplication deletion in which the chromosome breaks in 2 places and the central fragment is lost Duplication—change Nonallelic 1. phenotypic: - a section of the in the total amount homologous correlated with chromosome is of DNA recombination size, less repeated detrimental than compared to deletion the normal 2. gene family for parent © Hannah Kennedy, Kent State University specialized - Ohno’s function: 2 copies hypothesis = of ancestral states that genes accumulate duplications mutation that are are essential different so after for evolution generations the 2 bc they lead to genes diverge (ex gene families = globin genes encode polypeptides that are subunits of proteins that function in binding O2. Ancestral globin gene → myoglobin and hemoglobin → alpha and beta chains of hemoglobin) Inversion— 1. Phenotypic: if the - Change in the chromosomal breakpoint direction of the rearrangement, 2 happens in a vital genetic types: gene, gene material along 1. Pericentric = function is chromosome inversion in disrupted (ex = - inversion which the hemophilia type heterozygote centromere is A: patient = a person within the inherits X-linked carrying 1 inverted inversion in copy of a region which the normal 2. Paracentric breakpoint chromosome = inversion in inactivated the and 1 copy of which the gene for the an inverted centromere is blood-clotting chromosome outside the protein) with high inverted 2. Consequence of probability of region pericentric producing crossover: single gametes that crossover. 2 of 4 have abnormal sister chromatids genetic are involved. content After meiosis there are 2 abnormal chromosomes that each have 3. Consequence of paracentric crossover: 1 normal © Hannah Kennedy, Kent State University chromosome, 1 w inversion, 2 w deletions Simple translocation - Cells are - Can produce - a single piece —chromosomal exposed to abnormal of the rearrangement agents that gametes due to chromosome is cause pairing and attached to multiple segregation of another chromosomes chromosomes chromosome to break w - Robertsonian loss of translocation telomeres. = the most Reactive ends common are translocation improperly in humans that joined by arises from DNA repair breaks near enzymes to the result in centromeres of abnormal 2 chromosomes nonhomologou - Nonhomologo s acrocentric us crossover chromosomes resulting in rearrangemen t of genetic material Reciprocal ↑ - 2 diff translocation chromosome types exchange pieces to make 2 abnormal chromosomes - Balanced translocation s = translocations in which the total amount of DNA isn’t altered (i.e. no phenotypic consequence) but carriers can have offspring w unbalanced translocations - Unbalanced translocation = significant portions of © Hannah Kennedy, Kent State University genetic material are deleted or duplication; phenotypically abnormal b. Euploidy: phenomenon in which the organisms has a chromosome number that is an exact multiple of the chromosome set (can occur by 3 things): autopolyploidy, alloploidy, and allopolyploidy i. autopolyploid = an individual in which complete nondisjunction occurs due to a general defect in the spindle apparatus, producing 1 or more extra sets of chromosomes ii. alloploidy = common mechanism for change in chromosome number that is a result of interspecies crosses; has 1 set of chromosomes from 2 diff species is called an allodiploid (most likely to occur btwn species that are close evolutionary relatives) iii. allopolyploidy = phenomenon in which the organism contains 2 or more sets of chromosomes from 2 or more species 2. Mechanisms that produce variation in chromosome number a. Non-disjunction = process in which the chromosomes do not segregate properly i. 2 main types of nondisjunction Type of non-disjunction What’s happening Additional info Meiotic nondisjunction Produces cells that have too - if in anaphase I: entire many or too few bivalent migrates to 1 chromosomes pole and 4 resulting haploid cells are abnormal - if in anaphase II: 2 abnormal and 2 normal cells Mitotic nondisjunction Occurs after fertilization in - happen bc (1) sister somatic cells that may lead to chromatids separate a patch of tissue in the improperly so 1 organism that has an altered daughter cell has 3 chromosome number sets of chromosomes and the other has 1 or (2) 1 of the chromosomes doesn’t migrate to a pole due to improper attachment to spindle - may give rise to mosaicism in which the organism has a subset of cells that are genetically diff from rest of organism Ch 7 © Hannah Kennedy, Kent State University 4. gamete formation a. there will be gamete formation if 3 things happen i. independent assortment (2 genes on 2 different homologous pairs of chromosomes) 1:1:1:1 ratio ii. linkage (2 genes on a single pair of homologos with no exchange) 1:1 ratio iii. linkage with recombination (2 genes on a single pair of homologs in which exchange occurs between 2 nonsister chromatids) (a+b) parentalc+d) recombinant gametes b. Linkage i. Key terminology: 1. Synteny = 2 or more genes are located on the same chromosome; genes that are syntenic are physically linked to each other 2. Linkage = the phenomenon in which genes that are close together on the same chromosome tend to be transmitted as a unit 3. 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) 4. 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. Key terminology: 1. Crossing over = process in which homologous chromosomes can exchange pieces with each other during prophase I; affects pattern of inheritance for linked genes 2. Recombinant cells = nonparental cells = the cells of which display combinations of traits that are different from those of either parent 3. Double cross over = the process in which 2 homologous chromosomes crossed over twice ii. 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. 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 5. Chi square analysis to be used to decide whether 2 genes are linked or whether they 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 © Hannah Kennedy, Kent State University 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: to map 3 genes a. key terminology: i. genetic map = a diagram that describes the order of genes along a chromosome to determine the linear order and distance among genes that are linked to each other ii. Coupling = cis configuration = in the heterozygous parent, all wild-type alleles (dominant) are across from corresponding mutant (recessive) alleles on the homologous chromosome iii. Repulsion = trans configuration = in a heterozygous parent, wild type and mutant alleles are on different chromosomes b. Key equations: number of recombinant offspring i. Map distance = number of totaloffspring x100 (when 2 diff genes are more than 50 MU apart, they follow the law of independent assortment) c. 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 (% of recombinant offspring is correlated with the distance between the 2 genes) d. Phenotypes of offspring: 8 phenotypic classes, 4 single crossover classes (2 btwn genes 1 and 2, and 2 btwn genes 2 and 3), 2 double cross over classes, 2 parental arrangements e. 3 challenges in gene mapping: i. undetectable events bc double exchange leads to the same allele arrangements ii. 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. chromosome organization with regions of lower and higher recombination rates f. interference and coefficient coincidence: bc 2 recombination events during DCO aren’t independent i. Key terminology: 1. 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]) ii. Key equations: © Hannah Kennedy, Kent State University 1. I = 1 – C Expected Double Cross ¿ 2. C = Cross ¿ Observed Double ¿ ¿ 3. Expected Double Cross Over = [SCO ] x [SCO ] 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 st 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 4. Calculate map distance for the 2 ndSCO 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 7. Quantitative inheritance a. Key terminology: i. Quantitative inheritance = the range of measurable phenotypes, generally due to multiple loci (mendelian laws still apply); can be categorized as anatomical, physiological, and behavioral ii. Quantitative traits = a trait that is usually polygenic that can be described with numbers and fall along a continuum (ex = height, weight, disease) iii. Quantitative trait loci = QTLs = locations on chromosomes containing genes that affect the outcome of quantitative traits © Hannah Kennedy, Kent State University iv. Continuous traits = traits that don’t fall into discrete categories (e.g. height and weight) v. Meristic traits = traits that can be counted and expressed in whole numbers (e.g. bristles on a fruit fly) vi. 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 vii. Exhibit a continuum of phenotypic variation viii. 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 ix. 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 x. 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 1. Ex: 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) 8. Polygenic inheritance a. Key terminology: i. Polygenic inheritance = the transmission of a trait governed by 2 or more different genes; 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 ii. QTL mapping = b. 2 ways to estimate the # of genes: i. use the percentage of the most extreme phenotypes ( 1 ) 4N ii. use the number of the phenotypic categories (2n +1 rule) c. QTL Mapping i. Goal: locate unknown genes affecting a certain trait ii. 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. 9. Statistical methods to evaluate a frequency distribution quantitatively a. Key equations Statistical method Equation Variables What it does Mean Σn - Σ : sum of all Tells us where the Mean= center of our n values - N : number of distribution lies values present s , variance ´ 2 - Vx: variance Characterizes the Σfi(Xi−X ) ´ shape of the N−1 - Xi−X : the difference distribution: the between each greater the value and the variance, the more © Hannah Kennedy, Kent State University mean spread out the - N=¿ number of distribution about observations the mean is - Σ=¿ the sum of all values that are differenced from the mean and squared s: standard deviation √s2 CoV = covariance CoV ´ Describes the (X, Y) = - (Xi−X ) = the Σfi[(Xi−X´ )Yi−Y )] numerical value relationship N−1 for a certain between 2 sample X minus variables within a the mean of all X group values - (Yi−Y )¿ = the numerical value for a certain sample Y minus the mean of all Y values - Σ = sum of all products for all values of X and Y - N−1 = number of sample minus 1, degree of freedom r = correlation CoV (x ,y) - CoV (X, Y) To evaluate the coefficient covariance strength of (SDx)(SDy) previously association calculated between 2 - SD X standard variables: Positive deviation for X correlation: values indicates that there - SD Y standard is a direct deviation for Y association values between two variables Negative correlation: indicates that there is an inverse association between variables 10.Genetics of complex diseases a. Key terminology: i. Heritability = the amount of phenotypic variation within a group of individuals that is due to genetic variation © Hannah Kennedy, Kent State University b. 2 factors that affect phenotype: (1): genotype (2): environmental effects, especially on F2 offspring c. Key equations: Equation Variables Addtnl Info VG= V +AV + D I - V G genetic variance - V Gill be zero if the - V = variance due to the trait is homogenous D effects of alleles that follow a dominant/recessive pattern of inheritance - V A variance due to the additive effects of alleles - V i variance due to the effects of alleles that interact in an epistatic manner VP= V G V + E GxE - V P phenotypic variance - - V = genetic and GxE environmental interaction variance - V G genetic variance - V E environmental variance 2 Vg - H = broad sense - An estimate of the H = = heritability proportion of Vp phenotypic variation Vg due to genetic factors: Vg+Ve+Vgxe takes into account many different types of genetic variation that may affect phenotype Va robs - h = narrow sense - r exp = .50 for h = or h = heritability siblings Vp r exp Xo−X ´ - X o=meanof theoffspring - r exp = .50 for or : - X p=meanof the parents parent/child Xp−X - r exp = .25 for aunt- - niece X=meanof starting populations - r obs : the observed - r exp = 1.0 for phenotypic correlation identical twins between related individuals - r exp: the expected correlation based on the known genetic relationship


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