Exam II Objectives - Summary Genetics PDF

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Exam II Study Questions Remember to bring a calculator for problems! Go over Mastering problems and final analysis of experiments #1 (Mendelian Inheritance) and #4 (Chromosome Mapping in Drosophila). Chapter 4, Extensions of Mendelian Genetic Principles: 1.) What is meant by multiple allelic forms of a gene? 

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Multiple Allelic Series o Number of possible genotypes = n(n+1)/2 where n= number of alleles o N homozygotes o N(n-1)/2 heterozygotes Multiple allelic forms o An allele is an alternative form of a gene  Wild type is most frequently occurring Multiple allelles of a gene may exist in a population o 3+ can only be studied in population o ex: ABO blood groups

2.) How does incomplete dominance vary from complete dominance and codominance? Give examples of each. 

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Incomplete Dominance o Intermediate phenotype o Neither allele is dominant o Ex: red/white flowers  pink flowers o Ex: Tay-Sachs disease  Heterozygotes are phenotypically normal but only 50% of Hex A activity Complete Dominance o One trait is completely dominant or recessive to another trait o Ex: Pea plants Co-dominance o Joint expression of both alleles in a heterozygote o Ex: MN blood groups  Exhibit either one or both o If two alleles of a single gene are responsible for producing two distinct, detectable gene products

3.) What is epistasis? 

Epistasis o Expression of one gene masks/modifies effects of another gene pair  Antagonistic  mask  Complementarity or cooperative fashion of gene influence

Gene masks phenotypic effects of another gene Each step of development increases complexity of organ Under control and influence of many genes Ex: Bombay Phenotype  Homozygous recessive condition at one locus masking the expression of a second locus  FUT1 gene mutant masks expression of IA and IB o Recessive Epistasis  Recessive genotype masks dominant o Dominant epistasis  Dominant allele at one loci masks an allele at second loci  Ex: summer squash fruit color o o o o

4.) How can gene interactions influence phenotypic ratios of offspring? 

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Gene interaction o Several genes influence a particular characteristic o Cellular function of numerous gene products contribute to development of common phenotype o Reveals inheritance patters o Single phenotype is affected by more than one set of genes Phenotypic characters are influenced by many different genes and their products

5.) What is meant by each of the following: lethal allele, essential gene, dominant lethal allele, recessive lethal allele? 

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Lethal allele o Has potential to cause death of organism o Alleles are result of mutations in essential genes o Inherited in recessive manner Essential gene o Absolutely required for survival o Mutations can be tolerated if heterozygous  One wild type allele sufficient for survival  Homozygous recessive will not survive o Mutation behaves as recessive lethal allele Dominant lethal allele o Presence of one copy of allele results in death o Ex: Huntington’s disease  Dominant autosomal allele H  Onset delayed until adulthood  Characterized by progressive degeneration of nervous system, dementia and early death

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Recessive lethal allele o Homozygous recessive will not survive o Ex: agouti o X-linked  Observed only in males  Females can only be heterozygous carriers  Ex: Duchene muscular dystrophy

6.) What is the difference between penetrance and expressivity? 

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Penetrance o Percentage of expression of the mutant genotype in a population o Yes/no o Percentage of individuals that show at least some degree of expression of a mutant genotype o Ex: 15% of flies with mutant genotype show wild type, then penetrance is 85% Expressivity o Range of expression of mutant phenotype o Result of genetic background differences and/or environmental effects o Degrees of yes o Ex: in flies, expressivity for eyeless gene ranges from complete loss of both eyes to completely normal eyes

7.) What information can complementation analysis provide about mutations responsible for a phenotype?  Complementation Analysis o Screens number of individual mutations resulting in same phenotype o Can predict total number of genes determining a trait o Answers the question:  Are two mutations that yield similar phenotypes present in the same gene or in two different genes? o Complementation group  All mutations present in any single gene o To find answer  Cross the two mutant strains and analyze the F1 generation  Two alternative outcomes:  Case 1: all offspring develop normal winds o Two mutations on separate alleles and flies are het at both genes o Genes complement one another in restoration of the wild type phenotype  Case 2: All offspring fail to develop wings o Two mutations affect the same gene and are alleles on one another o Complementation does not occur

8.) Define the following: X-linkage, Y-linkage, sex-limited trait, sex-influenced trait, genetic anticipation, genomic imprinting.

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X-linkage o Genes present on X chromosome exhibit patterns of inheritance different from autosomal genes o Dominant mutant allele on X  Males with trait pass to daughters not sons o Ex: hereditary enamel hypoplasia/ color blindness Y-linkage o Holandric o Limited number of traits  Transcription factors, RNA binding proteins, SRY, TDF o Y-chromosome  Relatively inert genetically  Male-specific genes on human Y chromosome  Lacks copies of genes found on X-chromosome o Ex: hairy ear trait Sex-limited trait o Expression of specific phenotype is absolutely limited to one sex o Ex: feather plumage in chicken Sex-influenced trait o Sex of individual influences expression of phenotype o Not limited to one sex or the other o Ex: male pattern baldness In both of the above o Autosomal genes are responsible for the existence of contrasting phenotypes, but the expression of these genes is dependent on the hormone constitution of the individual Genetic Anticipation o Genetic disease has earlier onset and increased severity with each succeeding generation o Progressively earlier age of onset o Ex: trinucleotide repeats  Huntington’s disease  Myotonic dystrophy (DMI) Genomic Imprinting o Selective gene silencing impacts phenotypic expression o Silencing depends on parental origin of genes o Silencing also occurs in early development o Regions of chromosome imprinted on one homolog but not the other’ o Ex: chromosome 15  Prader-will syndrome  Parental segment deleted  Undeleted maternal chromosome remains by was methylated (imprinted) by mom

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Angelman syndrome  Maternal segment is deleted  Undeleted paternal chromosome remains but was methylated (imprinted) by dad Conditions exhibit different phenotypes

Chapter 7, Sex Determination and Sex Chromosomes: 1.) What is nondisjunction and how did it help to prove the chromosomal basis of sex determination? 

Nondisjunction o Failure of X chromosomes to segregate during meiosis o Aneuploidy o Results in Klinefelter and Turner syndrome  Klinefelter; 47 XXX  Phenotypically female but sterile  Turner; 47 XXY  Internal ducts are male, rudimentary testes fail to produce sperm  Feminine development not suppressed o Enlarged breasts common, rounded hips

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Chromosomal Basis for sex determination o Led scientists to believe that Y chromosome determines maleness o Presence of Y chromosome in presence of 2 X (K)  sufficient for maleness o Absense of Y (t)  female

o SRY- sex determining region contains gene responsible for male development  TDF o MSR- male specific region 2.) What is dosage compensation and how is it achieved? 

Dosage compensation o Dosage compensation balances dose of X chromosome gene expression in males and females  Prevents excessive expression of X linked genes in humans and other mammals (lethal I compensation doesn’t occur)  Barr bodies are heterochromatinized, inactivated X chromosome  Lionization is random, chosen independently in each somatic cell early I embryonic development (blastocyst) o X inactivation- Barr Body  All but one X chromosome inactivated in each cell, explains dosage compensation  Governed by X-inactivation center (XIC)  One of the genes here is XIST (X inactive specific transcript)  RNA product of XIST coats X chromosome that produced it  Remaining question: What blocks XIC locus of active chromosome  Barr body  Highly condensed structure lies against nuclear envelope

3.) What are the different mechanisms of sex determination used by different organisms? 

Sex Determination Mechanisms o Drosophila, C. Elegans  Y does not determine sex  Ratio of X chromosomes to number of haploid set of autosomes determined  Drosophila  Ratio: 1  female  Ratio: .5  Male  Ratio: .67, .75  intersex  Ratio 1.5  metafemale (inviable)  C. Elegans  Ratio: 1  hermaphrodite  Ratio: .5  male o Lygaeus turcicus  XX/XY mode

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Females have one X chromosome Males have either an X or Y

o Protenor  XX/XO mode  Females have 14 chromosomes (2X)  Males have 13 chromosomes (1X)  Depends on random distribution of X chromosome into half of male gamete o Environmental  Temperature dependent sex determination  Controls sex determination in reptiles

Chapter 5, Gene Mapping in Eukaryotes: 1.) What is meant by gene linkage? 

Gene Linkage o Linked genes belong to a linkage group and do not undergo independent assortment o Genes linked on the same chromosome segregate together o Part of same chromosome

2.) Which DNA markers were used historically for human chromosome mapping? 

Genetic Markers o Mutations that identify distinguishable phenotypes o Short segments of DNA with known sequences and locations  Useful landmarks for mapping  Identified during recombinant DNA and genomic studies o RFLPs: Restriction fragment length polymorphisms  Generated when specific DNA sequences are recognized and cut by restriction enzyme o Microsatellites  Short repetitive sequences found throughout genome  Vary in the number of repeats at any given site

o SNPs: single nucleotide polymorphisms  Found throughout genome  Used by geneticists to identify and locate related genes  Used to screen for diseases such as cystic fibrosis 3.) How would you construct a genetic map from a three-point testcross? 

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Three-Point Test Cross o Criteria  Genotype of organism producing crossover must be heterozygous at all loci  Cross must be constructed so genotypes of al gametes can be accurately determined by observing phenotype  Sufficient # of offspring produced to represent all crossover o Triple heterozygote with triply homozygous recessive o Count phenotypic classes in progeny o Expected frequency of double-crossover gametes is lower than that of either single-crossover gamete class and if three genes are close together along one chromosome Method 2 o Determine the order of genes by comparing the parental gene arrangement with the DCO gene arrangement o Which allele has been swapped from its original neighbors when going from parental to DCO arrangement? Step by Step o Step 1  ID parental progeny  Most frequent o Step 2  ID DCO  Least frequent o Step 3  Determine order  Which ones switched? o Step 4  Rewrite parental in correct order o Step 5  Determine SCO I o Step 6  Determine SCO II o Step 7  Determine first two distance  SCO I + DCO/ Total o Step 8

 Determine last two distance  SCO II + DCO / Total o Step 9  Draw map 4.) Define the lod score method 

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Difficulty arises when two genes of interest are separated on a chromosome to the degree that recombinant gametes are formed, obscuring linkage in a pedigree  Lod score method- demonstrates linkage o Log of the odds favoring linkage o Relies on probability calculations o Assesses probability that pedigree with two traits reflects genetic linkage between them  First- probability is calculated that the family (pedigree) data concerning two traits conform to transmission without linkage  AKA independently assorting  Then- probability calculated that the identical family data for these same traits results from linkage with a specified recombination frequency o Lod score accuracy is limited by the extent of the pedigree  3.0 or higher = strong linkage  -2.0 or less argues against linkage  Values in between = inconclusive o Helpful for assigning genes to chromosomes  Limited by extent of pedigree

5.) Define what is meant by harlequin chromosomes? Harlequin Chromosomes o Sister chromatids involved in mitotic exchanges o Patch-like appearance when stained and viewed under a microscope o Named this due to their patterns of alternating patches  Sister Chromatid exchanges o Occur during mitosis but do not produce new allelic combinations  Reciprocal exchanges similar to crossing over  Label with BrdU and then 2 round of division  Each pair of SC has one member with one strand labeled and the other member with both strands labeled o Ones with it in both stain less brightly than ones with it in only one strand Chapter 23, Quantitative Genetics and Multifactorial Traits: 1.) Compare and contrast discontinuous versus continuous traits. 

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Discontinuous Traits o Few distinct phenotypes, inherited in categories  Ex: AB, A, B, O  Often described qualitatively o Simple relationship between genes responsible for phenotype  BUT sometimes influenced by penetrance, expressivity, pleiotropic, and epistasis o You have one or the other Continuous Traits o Most traits exhibit wide range of phenotypes  Ex: height, skin, ag yield of crops o Quantitative genetics is used to characterize these traits o Known as quantitative inheritance

2.) What is a distribution? Frequency histograms for continuous traits often follow what type of curve?  

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Polygenic Traits o Measured I large samples and representative individuals of population Data form: normal distribution o A characteristic bell-shaped curve when plotted as a frequency histogram o Distribution is a frequency histogram Continuous Trait phenotypes not easily grouped into classes so use a frequency distribution

3.) Define and use variance, standard deviation, covariance, correlation coefficient. 

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Variance o S2 o The average squared distance of all measurements from mean; provides information about spread of data around mean o S2=Ʃ(Xi-X)2 o n-1  n= total sample size  xi = measured value Standard deviation (s) o Square root of the variance o ~95% of all values are found within two standard deviation to either side of mean o Can be interpreted in the form of probability Covariance o Measures how much variation is common to two quantitative traits  Ex: do heavier hens lay more eggs?

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o Calculate the deviation from the mean for each where n = number of xy pairs o First obtain covariance of x and y (the variance shared by both traits) by taking the deviation from the mean for each: covxy=Sum(xi-x)(yi-y)/n-1 Correlation coefficient o Can standardize covariance that ranges from 1 to -1 o Indicates if two traits both increase or decrease together (r is positive), or if one increases as the other decreases (r is negative) o R= covxy/ SxSy  S= standard deviation of the different measurements

4.) Describe a theoretical normal distribution. See # 2 5.) When is regression analysis used?    

Used to determine the precise relationship between variables Graph plotted for the individual data points Regression line is line that best fits the points Common method for measuring extent to which variation in trait is genetically determined

6.) What is meant by polygenic inheritance? 

Polygenic traits o Varying phenotypes result from input of many genes o Multiple loci o “Quantitative” and “polygenic” used interchangeably o Many are result of both gene action and environmental influences o Measured and described in quantitative terms  Display continuous variation o Continuous variation across a range of phenotypes

7.) List the three components of genetic variance. List the four components of environmental variance. 

Genetic variance o Additive variance (VA)  Genotypic variance due to additive action of alleles at a quantitative trait loci o Dominance Variance (VD)  When phenotypic expression in heterozygotes is not precisely intermediate between homozygotes

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o Interactive Variance (VI) deviation from additive components when epistasis is involved Environmental Variance o General environmental effect variance (VEg)  Results from environmental factors that produce irreversible differences among individuals o Special environmental effect variance (VEs)  Results from environmental factors that produce immediate changes in phenotype and are often reversible o Common family environmental effect variance (VEcf)  Results from environmental factors that are common to a family o Maternal effects variance (VEm)  Results from a subset of the common family environmental factors

8.) What is broad-sense heritability? What is narrow-sense heritability? Apply the Breeder’s equation. 

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Broad sense heritability (H2) o Measures contribution of genotypic variance to total phenotypic variance  H2= VG/VP o Genotypic variance component includes all types of genetic variation in population o Estimates assume genotype-by-environment variance component is negligible, therefore is limited in usefulness o Values are 0-1 o Ex:  Heritability often studied with inbred strains. Variation between strains = genotypic. Intra-strain variation under differing conditions = environmental. Narrow sense heritability (h2) o Proportion of phenotypic variance due to additive genotypic variance alone  h2 = VA/VP o Provides information about how a trait will evolve under natural selection or how it will be modified under artificial selection  Shows trends  More valuable predictor of response to selection o Applies to a population, not an individual o Narrow-sense heritability predicts the success of Artificial selection  Choosing specific individuals with preferred phenotypes from initially heterogeneous population for future breeding  Purpose: To develop population containing high frequency of individuals with desired traits

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Breeders Equation o R = h2/S o Breed two selected parent individuals with desired trait from an original population (mean trait value of selected parents of M1) o Then compare offspring mean (M2) to original unselected population mean (M) o h2 = M2-M / M1-M = R/S  R= selection response  S= selection differential

Experiment1  Mendelian Inheritance in Brassica Rapa o The purple stem was dominant and the green stem was recessive Experiment 2  Chromosome Mapping in Drosophila o Three-point cross o Mapping on the chromosome  X linkage...