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General Genetics

by: Ms. Kody Cremin

General Genetics GENETICS 466

Ms. Kody Cremin
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This 52 page Class Notes was uploaded by Ms. Kody Cremin on Thursday September 17, 2015. The Class Notes belongs to GENETICS 466 at University of Wisconsin - Madison taught by Staff in Fall. Since its upload, it has received 42 views. For similar materials see /class/205241/genetics-466-university-of-wisconsin-madison in Genetics (Graduate Group) at University of Wisconsin - Madison.

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Date Created: 09/17/15
Genetics 466 Lecture 36 Genes in Populations Linkage Equilibrium amp Inbreeding Objective Understand the concept of linkage equilibrium between two genetic loci and how to calculate disequilibrium D Understand the assumptions underlying the use of DNA evidence in legal cases and how to calculate the probability of a match between the genotype of the suspect and evidence from the crime scene Learn how the HardyWeinberg frequencies can be generalized to take inbreeding into account Know how to compute the inbreeding coefficient from pedigree information and in nite randomlymating populations Key Terms Linkage equilibrium Fullsib mating Inbreeding Maximum avoidance mating Inbreeding depression Inbreeding coefficient F Identical by descent Additional Problems 1 LocusA and Locus B are 481 cM apart on the same chromosome Seven populations of ies 500 individuals each were determined for the numbers of each type of chromosome Table below Which of the populations below is at linkage equilibrium 2 The B locus has a dominant allele B and a recessive one b In a HardyWeinberg population the frequency of B is 03 a What are the three genotype frequencies b What is the conditional probability of choosing a heterozygote given you have selected an individual with the dominant phenotype 3 At a criminal trial the prosecutor demonstrates that a DNA sample from the crime scene and one from the suspect are both of genotype A1 A4 B1B3 CzC7 She also presents the allelic frequencies for the general population to which the suspect belongs Table below What is the probability that the genotype of DNA evidence matches that of the suspect given that the person who committed the crime and the suspect are different individuals 4 Phenylketonuria PKU is a recessive disease resulting in a severe form of mental retardation Approximately 1 in 3600 newborns in a particular population have this condition What is the frequency of the PKU condition among individuals with inbreeding coef cient 012 5 Joe s father s sister is Ellen s grandmother What will be the inbreeding coef cient of the children of Joe and Ellen Answers 1 If you compute the expected values under linkage equilibrium for each population you ll see that only population 3 has an exact fit with the expectation 2 a p2 032 009 Bb 237 042 131072 049 b From the de nition of conditional probability and the HardyWeinberg rule 2pq 2g 2g 2077 08235 PBbB 2 p2pq p2q 1q 10 3 203x026 x 2032x65 x 2023x059 0018 4 21 1 01259gt1 0002244 39 q pol 3600 39 60 60 39 1 6 1 6 1 5 1t gt gt Joe 2 2 32 Ellen Genetics 466 Lecture 1 Mendel and The Study of Inheritance Objective Review how Mendel overturned the blending theory of inheritance by showing that inheritance is govern by particles genes and not uids Know what Mendel did and what his work showed Understand how to apply Mendel39s quotlawsquot to predict the outcome of crosses Key Terms genotype haploid phenotype diploid dominant heterozygote recessive homozygote gene selffertilization allele Punnett square gametes Backcross BC wildtype intercross independent assortment F1 and F2 true breeding pure line monohybrid segregation dihybrid Additional problems 1 How many kinds of gametes are produced by an organism heterozygous for n independently assorting loci answer 2quot How many genotypes can be produced when the organism from question 1 is used in a backcross answer 2quot An intercross answer 3quot A cross of two tomato plants both with redhairy fruit yielded the following progeny 95 redhairy 32 39 39 y 29 J quot and 12 quot What were the genotypes of the two parents and which is the dominant allele for each trait A F2 family of garden peas segregates 3 tall and 1A dwarf The family is segregating for the D and d alleles You select one of the tall F2 plants and self pollinate it You then grow 10 F3 progeny of this F2 plant and all 10 are tall What is the probability that the F2 plant in question has genotype DD given that its 10 progeny were all tall HS 823 2283 3 9533 Some Key Issues in Quantitative Genetics Predicting offspring s phenotype from its parents 7H Plant and animal breeders need to predict the performance or trait values of offspring Know the breeding value of their stocks These predictions form the basis of artificial selection programs Breeders would like to know which traits respond to selective breeding The most responsive traits have high heritabilities but BroadSense Heritability is not the best measure of offspring performance Prof Jim Coors of the Agronomy Department conducts a long term selection study in corn that was initiated in the 1970 s Predicting an offspring s phenotype from its parents A1 A1 X AZAZ Offspring s phenotype depends on the individual additive effects of lt the two alleles as well as dominance betvveenA1 and A2 xge xade additive dominance VxVaVdVe A1Az If A2 was dominant The Additive and Dominance Effects 2 a additive e 39ect the average e ect of substituting one b2 allele for one b1 allele QN a 0 d d dominance e 39ect deviation of b1le 91191 91192 92192 heterozygotes from the mean of the 2 homozygotes When d0 the heterozygote lies between the two homozygotes and we can predict an o spring s phenotype as the mean of the parental phenotypes When d 0 the o spring s phenotype can not be strictly predicted from an parental phenotype 0 selfing the heterozygotes An Extreme Example When a 0 selectiOn produces no Change even though there may be genetic variation for the trait 919 19197 99197 Phenotype 0 2 0 Frequency 025 05 025 If we measured broadsense heritability for this population it would be gt 00 because the gene b does affect the phenotype However selecting the best individuals will not improve the population mean in the next generation because with random mating The mean phenotype of a b1b2 individual s progeny will be 10 but the same is true of the progeny of b1b1 and b2b2 individuals Each individual has the same breeding value or the same additive deviation using clones If the additive component is the part that is useful in predicting offspring phenotypes then in some sense it defines heritability T his form of heritability is called the NarrowSense Heritability h2 and it is de ned as hZ EL wwn Broadsense heritability measures the extent to which our genes determine variation in our phenotypes Narrowsense heritability measures the extent to which our genes determine variation in our o springs phenotypes HOW car we estimate Va mother CONCLUSION I Just as MZ twins have the same genetic deviation g so the expected value for Offspring each offspring s A a a additive deviation a x apar is a the mean of its 2 parents The dominance portion of The environmental portion the Offspring s phenotype is of the Offspring s unpredictable because of phenotype is uncorrelated new allelic interactions with the parents hatestimate EdEeitO 395 Thi Just as MZ twins have the same genetic deviation g so each offspring s additive deviation a is a the mean of its parents I When a is large relative to d and e then o spring look more like their parents when a is small relative to d and e then the o spring look less like their parents 2 Thus the more closely o spring resemble their parents then the larger the additive e ect 3 The covariance between parents and o spring can be used to estimate the additive genetic variance in a population in a manner similar to how we used to covariance between identical twins to estimate the amount of genetic variation 4 Once we have an estimate for the additive Va variance we can use that to estimate the narrowsense heritability T How can we estimate h2 CVparoff Va VT This can be shown using algebra like I used last time for Vg and the The variance d0 S quotOi covariance between MZ twins Change across generations CV aro We measure phenotypes of h2 M parents and their o spring Vparvo xpar xo Narrowsense heritability is 2x the correlation between We calculate their covariance and parent and O spring divide by the total variance 2 lhz CVPar0 h 2rpar o 2 VT An Example with Genetics 466 Students from 2002 Female Students Parents Sibs Height 658 647 662 Male Students Parents Sibs Height 717 709 719 These values seem too high since H2 in humans for height is about 085 What s up 466 Students 80 75 7O 65 6O 55 6O 65 7O 75 80 85 Parents rmde 2 043 h2 086 rdadson 04 h2 082 rpam 042 h2 084 Why is our estimate of h2 so high I The results are accurate and H2 h because there is very little dominance in human 2 Too few environments sampled A small Ve makes Vg look proportionally larger 3 Our students are a genetically relatively diverse group so genetics may be more important than in previous studies with more homogeneous samples 4 There is a genotypeenvironment correlation Parents in their households recreated the growth promoting or suppressing conditions they experienced when they were young 5 Assortative mating If tall marry tall and short marry short then we have violated an assumption of the model How can we use our wiedge of Xto predict Only the additive plirtilm lifthe parems phenatype ix predia ively heritable How can we use our knowiedge of Xto predict a h for abdominal bristle number in Drosophila is 03 The mean bristle number is 12 A male with 10 bristles is mated with a female with 20 bristles and a large number of progeny scored for bristle number What is the expected number of bristles among these progeny xm 10 12 2 xfmie 20128 amp 303 09 8 2 3 Yof39fsmmg Ypopularion amp 2 12 09 129 Longterm Selection Can Be Highly Effective an pupa weight Qf OMIquot beetles was increavsedfmm to m in 1 generations 5mm 50 A 7 52GB Selec en disseminated mm 13 AidI3 3C 3211C Mean pupa weight imicregramei 23110 24100 EU I I I I I I I I l l I I l l G Hi 20 31 em 50 60 T0 30 Q 100 10120130 lii l Generaijons 3f seleciien Here is one way to estimate 112 Since only the additive e ects are transmitted to o spring one can use the ratio of response to selection R to the selection di erential S to estimate hz 125 100 25 o5 Robinson et a 1949 070 0 m o 398 r c 40 e a 30 a 20 Nanowesense nemabmzy nudpome 0 range Came Whither height 4 Milk protein lt Feed emuency Milk yield Calving quot quotnew Sheep Fihex diameter lt Body wexghe Poultry sexual malunty Eggs per hen housed Swine Backfat quot hckness Body length Feed ernuency lt Daily gain Lmer sue Mane Husk extension Flam height Ear height Ear number Yed Summary Predicting Offspring Phenotypes 1 Phenotypes can be partitioned into additive dominance and environmental deviations 2 Variance of a population can be partitioned into additive dominance and environmental components 3 Narrowsense heritability is the proportion of the phenotypic variance due to additive e ects or the portion that is predictable between parents offspring 4 h2 can be estimated using the correlation between parents and o spring 5 When h2 00 Va0 o springs phenotypes will equal the population mean 6 When h2 10 VanT o springs phenotypes will equal the midparent value Genetics 466 Lecture 41 Quantitative Genetics Genetics of Plant and Animal Breeding Objective Learn the concept of narrow sense heritability as a way of separating the genetic variance into the additive and dominance components Know how to use quantitative genetics models to estimate quotnarrowsense heritability Understand the utility of the concept of narrow sense heritability in plant and animal breeding Key Terms Narrowsense heritability Dominance effect d Selection differential S Additive effect a Additive variance Selection response R Additional Problems 1 If the environment became more uniform what would happen to the heritability a H2 and h2 would both decrease b H2 and h2 would both increase c H2 would increase but hZ would decrease d H2 would decrease but h2 would increase e Both heritabilities would remain the same 2 An ecologist working with the big blue stem prairie grass would like to know how much additive genetic variation her population contains for the number of days from germination to owering The phenotypic variance is 36 daysz She determines that the correlation between parents and offspring is 025 What is Va for this population Answers 1 Answer b Since Ve decreases both Vg and Va would make up a larger part of the total 2 18 daysz Genetics 466 Lecture 40 Quantitative Genetics 11 Nature vs Nurture Objective Learn the concept of heritability as a way of separating the phenotypic variance into genetic and 39 39 I J the concept of quotbroadsensequot heritability and its utility Learn the ways of estimating the genetic and environmental components of phenotypic variance so that one can resolve the contributions of nature genes and nurture environment to the differences among individuals r Key Terms Heritability x phenotypic deviation Broadsense heritability g genotypic deviation Correlation 6 environmental deviation Covariance Additional Problems 1 A gene with one allele that is expressed at different levels with different temperatures and a second allele that is expressed at the same level under all temperatures would violate our assumption of a random mating b linkage equilibrium c no genotypeenvironment correlation d no genotypeenvironment interaction e both c and d above 2 The height data below are for sets of male monozygotic twins reared apart Calculate the phenotypic variance and the covariance among twins What is the broadsense heritability What is Andy s phenotypic deviation x Data in inches 2 4 5 Answers 1 Answer d Different effects in different environments is an interaction 2 The overall mean is 68 inches VX is 182 CovXy is 116 and H2 064 Andy s phenotypic deviation is 10 inches Genetics 466 Lecture 4 Probability for Repeated Trials Objective Learn how to calculate probabilities for repeated trials using the binomial multinomial and Poisson formulae Know what is meant by quotbinomial distributionquot and how this distribution is derived Understand how to apply n choose K to calculate the number of permutations for a collection of events Key Terms permutations expected value binomial distribution Poisson distribution multinomial distribution My presentation of probability goes beyond what is presented in your text Although the handouts are sufficient to prepare you for the exams for those who would like to read a little more Statistics The Easy Way by D Downing and J Clark Baron s Educational Series 1997 provides a very accessible discussion of probability It is on reserve in Steenbock Library Additional problems 1 A series of matings between animals of genotype AaBchDdEe yielded 2000 offspring Use the Poisson distribution to nd the probability that exactly one of them is aabbccddee Assume all five loci segregate independently answer 0277 2 Of the 258 students enrolled in Genetics 466 last year there are 19 genetics majors and 122 juniors If being a genetics major and being a junior are independent how many junior genetics majors should there be answer 9 For the curious the actual number was 13 3 From a cross of Bb x Bb yielding six offspring what is the probability of exactly two of each genotype answer 901024 4 Homozygous recessive dd children are darling Monica and Chandler are both carriers for darlingness d is an autosomal gene This couple plans to have five children What is the probability that they will have exactly two darling daughters plus three other children who are some mix of darling boys nondarling boys and nondarling girls answer 0105 5 You would like to generate some new mutations at the Sugary gene in corn to see if some of these might be useful to improve the taste of sweet corn You use a chemical mutagen that causes one new mutation per locus among 1000 gametes After applying this mutagen you screen 5000 gametes for new mutations What is the probability that you will observe exactly 5 new mutations answer 018 6 An individual with genotype AaBchDdEe is crossed to one that is aabbccddee What proportion of the progeny of this cross are expected to have a dominant allele at exactly 3 of the 5 loci answer 03125 7 Witches females with supernatural powers and wizards males with supernatural powers are homozygous recessive ww while people carrying at least one wild type allele W are muggles and lack supernatural powers Hermione is a witch who was born to two parents who were muggles She had three siblings who were muggles like her parents What is the probability that her parents would have three children who were muggles plus one who was a m given that they had four children answer 0211 D1 Probability for Genetic Events Genetics 466 John Doebley Your text introduces some basic concepts in probability theory such as sampling space which is always equal to one and independence It also introduces two fundamental rules of probability the additive rule and the multiplicative rule In this handout I will discuss two additional concepts in probability theory that are extremely useful in genetics conditional probability and Bayes Theorem Conditional probability is de ned as the probability that an event A will occur given that we know another event B has already occurred This is written 1 PAB which is read the probability of A given B Using the concept of sampling space once we know that event B has occurred then the sampling space is reduced to the area of B Within the area of B the probability that event A will occur is equal to the area of overlap between events A and B Thus we can express the PAB visually as g This can be written symbolically as PAB 2 PAB PA BPB where PA B represents the probability of the intersection of A and B ie that A and B occur together Consider a genetic example Mendel expected that some of the tall F2 plants from the selfpollination of a Dd heterozygous plant for the dwarf gene would themselves be heterozygous Dd while others would be homozygous DD We can ask what proportion of the F2 plants are Dd given that they are tall We can write this as PHT PH TPT Where H represents heterozygous and T represents tall Filling in the appropriate values we nd that 3 PH T leva k wIN The denominator 34 is the proportion of F2 plants that are tall and the numerator is the proportion of F2 plants that are both tall and heterozygous Thus twothirds of the tall plants are expected to be heterozygous and the remaining onethird homozygous DD or the probability of a plant being heterozygous given that it is tall is 23 Another way that conditional probability comes in handy is when there is some uncertainty such that the probability of the event A that one wishes to calculate depends on whether or not another event B has occurred or not Using the sampling space model we can visually express this as follows where the bit on the left represents the intersection of A and not B and the bit on the right represents the intersection of A and B Symbolically we write 4 PA PA BPA B where B signi es not B Equation 4 can be expanded using the rule for conditional probability equation 2 First rewrite equation 2 5 PA m B PAI B x PB and 6 PA BPAIBxPB Then substitute equations 5 and 6 into equation 4 to obtain 7 PA PABgtltPBPABgtltPB Thus the PA is now expressed as the sum of its probabilities given whether or not B has occurred Consider the case of a disease gene G in humans for which homozygous recessive individuals gg exhibit the disease Jerry is a Gg heterozygote His spouse Elaine is phenotypically normal C but it is known that both of Elaine s parents were carriers Gg Thus the pedigree 1s G9 G9 X i Elaine G Gg Jerry X i 2 Child Jerry and Elaine are considering having a child and wish to know the probability that their child will neither have the disease nor be a carrier PCGG The probability that the child will not carry the disease allele PCGG depends in part on whether Elaine is Gg or GG Thus the PCGG can be split into two parts depending on Elaine s genotype PCGGIEGG and PCGGIEGg Using equation 7 we write PCGG PCGGIEGGXPEGGPCGGEGg XPEGg Referring to the pedigree we observe that the probability that Elaine is Gg is 23 and the probability that she is GG is 13 If she is Gg then the probability that the child is GG is 14 if she is GG then the probability that the child is GG is 12 Plugging these values into the equation we obtain PltCGGgtIiinIiini Bayes39 Theorem Bayes39 Theorem is a form of conditional probability that enables one to update current probabilities in light of new experimental evidence Bayes Theorem uses the prior probability and the new evidence to estimated the posterior probability It also allows the calculation of PBA when we can observe the probability of PAIB We can derive Bayes Theorem by some simple manipulations of our formula for conditional probability First we rewrite equation 2 as 8 PBA PA BPA and then 9 PA B PBA x PA Again starting with equation 2 we can substitute the numerator using equation 9 and obtain 10 PAIBW P 3 Finally we can split the PB in the denominator into two parts depending on the occurrence or nonoccurrence of event A as was done in equation 7 The result is Bayes39 Theorem PB I A xPA 11 PAIB PB I A x PA PB I A x PA Equation 11 can look a bit intimidating but it is simply an expanded conditional probability The denominator is the probability that event B will occur and the numerator is the probability of the intersection of events A and B Here39s a genetic example using Bayes39 Theorem George39s sister has red hair as a result of being a homozygous recessive rr for the melanocortin receptor gene Both of George s parents have brown hair but they must be heterozygotes Rr given that they have had a redheaded daughter George also has brown hair although his genotype could be either RR with a 13 probability or Rr with a 23 probability George marries a redheaded woman rr and their first child has brown hair Rr The birth of the daughter with brown hair argues against George being Rr Given that George has had a child with brown hair what is the probability that he is Rr First rewrite Bayes39 Theorem using H to symbolize George being a heterozygote Rr and B to symbolize the birth of the daughter with brown hair PB I HPH PH IB PBlgP pBIHPH PH the prior probability that George is Rr is 23 PBH the probability of a brown haired child if he is Rr is 12 P is 13 and the PB IE is 1 Plugging these values into the equation we obtain PHIB is 12 Thus before the birth of his brownhaired daughter the prior probability of George being Rr was 23 but the new evidence reduces that probability to 12 which is the posterior probability Genetics 466 111006 and 111306 Lectures 29 amp30 Regulation of gene expression in eukaryotes I Regulation of gene expression in eukaryotes 0 Introduction In eukaryotes organelles compartmentalize various functions Within the cell Mitochondria and chloroplasts Have their own DNA gt transcription and translation Organellar proteins are synthesized either Within the organelle from information carried by the organellar DNA or Within the cytoplasm from nuclear encoded genes Many eukaryotes are multicellular gt spatial and temporal regulation of expression Regulation at each level Regulation of gene expression in eukaryotes Tra 1 Cytoplasm Transport Degradation gal 39 g39amp Tranis amp Protein Signal transduction Transcriptional control of nuclear genes Spatial and temporal expression of specific genes is regulated by 0 Hormonal signals growth factors 0 Environmental signals Need for signal transduction pathways that activate or repress the activity of specific transcription factors Example Regulation of gene expression by steroid hormones Steroid horm 39 O 2 o A Steroid hormone S H ld penetrates the cell h rm lle receptor and binds to the receptor Steroim Receptor complex enters The nucleus and binds to Speci c DNA sequences 5 Expressed mRNA is translated Bound complex Stimulates Transcription 1 Response gene Example Regulation of gene expression by peptide hormones Inactive transcription factor Peptide hormone gt gt quot39 I I Receptor i Hormone binding activates a signal transduction pathway within the cell 0395 Hormone binds To a receptor Activated transcription factor Activated transcription factor bind enhancers of s to speci c activating expression target genes quot The signal Antivaf transduction pathway quotanunpmm factor activates a transcription Specialized transcription factor TATA TAT A Boxquot Transcription start site 0 Specific transcription factors Specific transcription factors regulate gene expression by binding to specific sequences at some distance from the core promoter named enhancers positive effect or silencers negative effect TATAA box TATAA boxE V TATAA box Compacted chromatin can interfere with gene transcription DNA Nucleosomes Scaffold 4 39 H2A H2B H3 E I I c Octamenc hISlone Core 0 w DNA hiStone 30nm solenoid From Griffith Miller Suzuki Lewontin and Gelbart 2000 An Introduction to Genetic Analysis Freeman Publisher lH4 gt Transcribed genes will typically be located in regions of the chromosome which are less compacted euchromatin hence more accessible to DNAses DNaseI hypersensitive sites gt Some transcription factors will modify acetylate histone molecules opening up the chromatin for transcription These transcription factors are histone acetyl transferases HAT Condensed chromatin Enhanc Gene A Histone code de ned modifications of histone tails Chromatln remOdehng have consequences on chromatin compaction and gene expression gt Transcription S ecialized transcri tion factOIS 39 p p TAT A box Inltlatlon phase Histone acetyl transferase HAT Enhancels Acetylated tails Acetylated nuclensnmes are mare exible Note Sometimes methylation of H3 lysine 4 precedes H3 and H4 acetylation nun Compacted chromatin can interfere with gene transcription If a gene is brought in the vicinity of compacted heterochromatin in the chromosome by transformation or transposition it becomes subject to positioneffect variegation that gene will be expressed in some cells and not in others A number of proteins are involved in this suppression of gene expression These proteins bind the DNA in the vicinity of the genes that are subject to this variegation some of these proteins also modify histone proteins histone methyl transferases for instance Several genes are NOT sensitive to positioneffect variegation They are anked by Specialized chromatin structures that insulate them from the effect of neighboring chromatin Insulating islands Similar proteins are also involved in maintaining the inactive state of expression of genes that are expressed only at speci c times during development PolycombGroup proteins in Drosophila PcG These proteins bind to silencers in the vicinity of affected genes Some of them modify histone proteins thereby creating a quottranscriptionunfriendly chromatin environment Positioneffect variegation 0 DNA Methylation Many eukaryotes not yeast though contain methylated C residues in their DNA C methylation occurs in CpG dinucleotides and CprG trinucleotides in plants and animals CpG dinucleotides are usually underrepresented in the genome However they are present in clusters CpG islands in proximity of many expressed genes C methylation at these sites is typically associated with gene inactivity Specific proteins appear to bind to methylated DNA and mediate genel suppression CG island C mehylatinn slam Imprinting For some eukaryotic genes only one of the two alleles is expressed in a developing organism The expressed allele is defined by its parental origin maternal for some genes paternal for others This mechanism is named imprinting The imprinted allele is usually methylated at C residues Imprinting is reset at each generation In plants most genes may be imprinted during early development the paternal allele is not expressed Imprinting maternal imprinting Egg Sperm Egg Zygote soma Methylatinn is erased Germlin l amemgenexis Spenn Silent allele Expressed allele Sperm 0 Activation or inactivation of entire chromosomes X chromosome is in different copy number in males and females 2 in females 1 in male mammals and ies gt Genes must be regulated to allow similar expression in males and females gt Gene compensation Inactivati0n in mammal females Hyperactivati0n in Drosophila males Hyp0 activati0n in Caenorhabdilis elegans hermaphrodites XX relative to males XO 0 Activation or inactivation of entire chromosomes X chromosome Inactivation in mammal females Expressed Inactive Barr body Xist 7 XIC 9 f 17kb RNA Zygote Young embryo Embryo and adult X chromosome hyperactivation in y males A complex of 5 MSL proteins and 2 so X RNAs binding to multiple sites on the X chromosome enhancing expression of linked genes MSL proteins include a helicase and a histone acetyl transferase HAT Expression of the MSL genes is controlled by the sex determination pathway next lecture Regulation of gene expression in eukaryotes Tra quotf quot Transport Cytoplasm Degradation gt amp rotein Signal transduction 0 Regulation of RNA stability Some mRNAs are inherently less stable than other RNAs because they carry specific sequences that are recognized by ribonucleases RNAs With premature stop codons are produced at rather high frequency during transcription because RNA polymerase does not have a proofreading activity Such RNAs are rapidly degraded by a specialized machinery that seems to be attached to the ribosome Nonsense Mediated RNA Decay or NMD In situations Where genes are expressed at levels that are too high genetic engineering for instance or viral infection the corresponding mRNA is also degraded Post transcriptional gene silencing Posttranscriptional gene silencing 1 Activated by double stranded RNA RNA viruses Inverted repeats in RNA Transgene Synthetic RNA 2 Results in Destruction of mRNAs carrying a sequence complementary to the silencing RNA Inactivation and methylation of DNA encoding such a target mRNA Example Viral doublestranded ds RNA enters the cell Dicer ds viral RNA mm Emma ATP AhP ppi gt Viral mRNA JEJELEiRNAJJID39 JET RISC Both ds viral RNA EX 1 genome and comp ex transcribed mRNA are degraded m ATP ATVIP ppi Mm W In plants an RNAdependent RNA polymerase also ampli es replicates the siRNA which can move systemically to other cellstissues gt transmission of resistance Dicer dsRNA m g m ATP Amplification mm mm 39 i b EDDgiRNA A U RISC A1P ppi JED JET RNAdependent m 001111316X RNA polymerase Em ATP More siRNA can be produced m AMP ppi Note Genes present in eukaryotes encode micromiRNAs which form secondar structures that are recognized by Dicer and RISC eventually forming a complex that targets speci c gene transcripts to degradation Gene X miRNA gene PreRNA gt a Dicer ATP mRNA 22gt protein Q AMP ppi In this cell all of gene X m mRNA will be degraded because the m RISC miRN A gene is my complex expressed 5m m ATP W AMP ppi Note micromiRNAs and siRNAs can also target RISC to the nucleus promoting inactivation of gene X expression by methylation 6116 X miRNA gene PreRNA gt q Dicer ATP AMP ppi mm m RISC complex W RNA dependent DNA methylase Application Reverse genetics Transform the organism with a gene that encodes an inverted duplication of the gene to disrupt gt Q RNA Plant gene X m ATP to disrupt Transgene ANIP ppi gt mRNA msiRNA I Gene X mRNA e E3 degraded RlSC W 4 K ATP Eg AlVIP ppi NM W The Central Dogma Lectures 16 20 Molecular Genetics Overview 1 Rules of the Gene DNA RNA 39 Protein The Central Dogma 11 Gene Regulation in Prokaryotes and Eukaryotes How and Why Gene Expression is Controlled in Biological Processes 111 Development and Evolution of Animals Requirements of the Genetic Material 0 Information must direct the organization and metabolic activities of the cell must replicate accurately so that the information it contains is inherited by daughter cells must undergo occasional mutation so that new heritable phenotypes arise Replication Mutation Genes Are Made of DNA Historical Developments 1870 Miescher shows DNA is in the nucleus 1912 Feulgen shows specific DNA stain patterns in nucleus quantity of stain in nucleus fixed in a species following meiosis amount of stain reduced by 1 2 during mitotic interphase amount of stain 2X 1942 UV light causes mutations 260nm is best wavelength 260nm is wavelength DNA absorbs Living Type lllS bacteria recovered Living Type lllS LY ive mouse aw cap we Dead mouse Heatrkilled Type lllS Live mouse Q w Live mouse Living Type llR Live mouse Q TO gt Q N0 capsme Live mouse Living Tyoe lllS Heatrkilled Type lllS bacteria recovered Living Tyue llR Live mouse Q g a gt w Dead mouse Fig 102 Grif ith s discovery of transformation e in Streptococcm pneumonia Type llR cells liealekilled lllS cells DNA on ilEJiklllEd t E Type lIR cells DNA from healkilled ills cells Type lIR cells DNA item heatkllled HIS cells Type llR cells DNA lrom neaikiiied lllS cells Type lIR cells llR colonies from mixluie Seium trial precipimes cells from mixture Protease gt Serum that RNase gt from mixture Serumiliai precipitates DNase gt Ill cells om miXi39UiE No colonies fr i Serum that DIEGipilJlEs gt llR cells lllS colonies lllS colonies lllS colonies No colonies Fig 103 Avery MacLeod and McCarty s proof that h t e transforming principle is DNA Protein coat 4 Key Structural Features of DNA Tail sheath I Tail bers entities Ev instantly 1 Double Helical 32PDNA I a I O g 9 7 due to bond angles the hellx thsts such that one turn 13 1 I or w 10 minutes Blender U 3 Centrifuge E01166 Phage T2 Infected cell Infected cell Phage RadlioSctfiVitynin C00 pe e 0 CE 5 u u 2 Antlparallel Polarlty Eggiziifgl i ghage 35SProtein Radioactivity to Ea in supernatant 5 W 3 35SProtein 39 9 mi f 4 qE 3 at 5 on the other 4 C o Blender O Centriluge 2 O U 3 w 10 minutes Base Pairing Complementarity E colicell PhageTZ Infectedcell lnfectedcell Phage Liirtltleergdig t i igy p H bonds Chargaff 3 Rule sti e sine eta leo Allow time for Subiect to shearing Separate bacteria fiom C G A T B and bacterial phage to infect bacterial forces in blender phage by centriiugation Fig 10 4 Demonstration by Hershey and Chase that the genetic information of bacteriophage T2 resides in its DNA Opposite polarity of the two strands Hydrogen bonding in AT and GO base pairs 539 339 RNA DNA Adenine Strandedness Single Double usually H Sugzir 5 Sugar Ribose Deoxyribose 3 I Guanine g 2399 2 9 Sugar 339 539 4 Base Uracil Thymine Fig 10 2 Diagram of a DNA double helix t COlTl 0S1 1011 II bonds are out about 3 7 is strong as covalent builds Vt 1339 would 115 he impoitziiit p 9 10 Nucleic acids are composed of repeating subunits called nucleoudes Each nucleotide is composed of three units 1 0 phosphate yi 0 group i 0 min RNA b in DNA Ribose ZrDe oxyribose OH OH 2 IIH OH cit OH A 5 2 O S 2 V I vewban ii K Pyrimidine nucleotides Purine nucleotides M iii ii iii in peritose H C3 2C H H C3 C H H H l l l l OH OH OH H 47 No hydroxyl group N a in RNA only lilln both RNA r in DNA only with rare exceptions and DNA with rare exceptions o NH2 0 31 E A g H 39A H i ih E H Etlll H HiiIVS E icm quot 34 7 6 7 i 39 39 o ci yc H o CKhyc H 07C k C H l H H H Uracil Cytasine Thymine 11 W m Nu l a m pyrimidine l N4CCi c H 1 39 H Ii H Deomh mldl P De oxycytldme Deoxyadenosme Deoxyguanosme w i N mONOPhOSP ate M mOHODhOSPhate dCMP monophosphate dAMP monophosphate dGMP Adenine lampaminopurine 0 x 7 H Er T gm 7 H Fig 0 7 Structures of the four common deoxyribonucleotides r y Nib iH present in DNA Guanine erarriinoJroxypurine Fig 106 Structural components of nucleic acid 34 nm Fig 1011 Diagram of the doublehelix structure of DNA Replication Purpose Propogate copies with high fidelity Problem How does one copy a double stranded template e g A T C 2 G 1 Read both I A strands Read intact G E C 2 Conserve A 39139 parental 139 A sequence Alternatives 1 Semi conservative Daughters segregate with one parental strand 2 Conservative Daughters segregate together 14 Semiconservalive Conselvative Parental DNA black new strand Frst ge nera on progeny DNA g g g C f5 f5 H in 5f5i39 quot g g How to distinguish these two possibilities The Messelson Stahl Experiment quotThe Most Beautiful Experiment in Biologyquot 5 Density of DNA 7 Heavy hybrid and half light Label all DNA with quotHeavyquot Genmffnwal I i W isotope of nitrogen 15N the Q Hi Parentammheavy density will be heavier than Cm mm wa normal 14N DNA 53 3325 Transfer growing cells to 14N 8mm gtggggynga on media after different Shyb39ld generations extract DNA and centrifuge in CsCl gradient mm Fig 11 3 Meselson and Stahl s experiment demonstrating that DNA replicates by a semiconservative mechanism in E 001139 DNA Replication Enzymology Enzymes catalyze the formation and destruction of chemical bonds There is a direction and chemistry to the replication of DNA REE 2UIREMENT S 1 Deoxynucleotide triphosphates dNTP provide energy and monomer units of DNA 2 A single stranded template must be present DNA will not polymerize without a template 3 A primer must be present to initiate chain synthesis chain free 310H 25gt OH H DNA polymerase Fig 11 2 Mechanism of action of DNA polymerase I covalent extension ofa DNA primer strand in the 5 3 l8 Replication in vivo is Bi Directional from Origins j C Q Q Q C Replication fork quotTheta Structuresquot Unit of Replication Replicon E coli has one one origin of replication Eukaryotes have multiple origins Rate 1000 nucleotides second Anatomy of a Replication Fork 1 Synthesis is only in 539 339 direction 2 Most use primer RNA made by primase amp dnaB 1 Exonuclease edits in 339 539 direction KEY 20 Enzymology Topm samerase Helicase 1 Unwiuds 2 P r as NA Smg e39mm 4 Two 4 Synthesizes RNA primer me 55 DNArbmdmg 888 protein complex P mase Primosome 3 NA polymer Se III 5 7gt 3 chain elongat on 5 Filsg p5 DNA polymerase I O 6 Ligates fragments Ligase g a 7 Laggm Leadmg 3 strand Wand 5 Central Dogmaquot Replication 0 DNA RNA Protein T ranslationST ranscription T ranslationr1ption TRANSCRIPTION Process of synthsizing RNA from a DNA template Unlike replication Where all DNA is copied transcription is selective only certain regions of the DNA are transcribed and these are in general GENES 22 Transcription 3 Types of RNA molecules each With different functions 1 mRNA messenger Carries information to RNA ribosome Where it is translated 2 tRNA transfer RNA Involved in decoding the mRNA into the actual protein 3 rRNA ribosomal Part of the ribosome RNA 23s RNA in 50s subunit 16S RNA in 30S subunit 5S RNA in 50S subunit Features of Transcription Complementarity The RNA is complementary to one strand of the DNA Initiation RNA polymerase initiates at specific sites PROMOTERS Termination There are specific signals to stop transcription RNA is single stranded Complementarity Transcription DNA 339 AATCCGCCTAT 539 539 TTAGGCGGATA 339 transcription is 5 4 3 transcript is complementary to DNA then RNA 539 UUAGGCGGAUA 339 Note that uracil is used in place of thymine RNA DNA Single stranded Double stranded Uses ribose Uses deoxyribose ribonucleotides deoxyribonucleotides Uracil A G C Thymine A G C RNA polymerase The enzyme that synthesizes RNA polymers from DNA template uses ATP UTP CTP GTP not dNTPs must recognize the beginning and end of genes it does so by recognizing signature sequences In PROMOTERS the initiation sequences is at the 539 end of genes e g the lac genes of E coli 539 end of RNA 1 GTTG GENERAL IMPORTANCE The decision to transcribe a gene or not is often regulated at iniitiation and mediated by physiological controls 26 Transcription in Genomes Asymmetrical Can be either strand of DNA 7 3 39 539 539 3 39 4 In Eukaryotes Transcription is in the nucleus but RNA is transported to the cytoplasm Highly Regulated Growth development response to environment all require selective expression of genes in different cell types or in different environments more much more later Z7 Transcriph an Tr in39 l 39 dh I m Prokaryoies Tmnscrip on mRNA Cytoplasm mKNA i ra ngllananu 39 859m an Eukaryotes Protein synthmis involves two steps transcription and translation 28 3 IHTL39J39Ji39r gmcl DNA 5 Nontemplare strand I RNA m Tenn r 39r synthesis 3 quot3TATI393 I H TIC quot quot1 5 GCAT AC GATCAGGCTAAC GIG3 Nontemplate strand mRNA 5 3 K T a i x T H 39u x 5GCAUACGAUCAGGCUAACGCB39 Fig 127 Sense RNA strand RNA synthesis utilizes only one DNA strand of a gene as template Genetics 466 Lecture 9 Chromosome Rearrangements Objective Learn how various types of chromosomal rearrangements can occur and understand their genetic consequences Consider the effect of rearrangements over long evolutionary periods on the similarity of chromosomes Key Terms deletion dicentric duplication paracentric inversion tandem duplication pericentric inversion inversion telocentric reciprocal translocation metacentric unequal crossing over adj acentl segregation inversion loop adj acent2 segregation Robertsonian translocation alternate segregation compound chromosome hypoploid acentric hyperploid Additional Problems 1 Would a translocation heterozygote be more fertile if mated to a heterozygote for the same translocation Explain Answer yes Gametes from adjacent1 segregation could combine with the complementary type to form a balanced zygote For example an N2 T1 egg and an N1 T2 sperm would produce a viable translocation heterozygote 2 An39 quot quot 39 is l t yg 4 for 39 123456 and 125436 where the quotquot indicates the centromere If a crossover occurs between points 4 and 5 what gametes will be produced Answer 1234521 and 63456 Genetics 466 Lecture 6 Extensions to Mendelism Objective Gain a deeper understanding of the ways genotypes can be related to phenotypes Complications like epistasis pleiotropy different types of dominance and multiple alleles were not present in Mendel39s classic experiments Become familiar with the conventions for naming genes and alleles Key Terms epistasis and epistatic Werner syndrome hypostatic sickle cell anemia codominance redundant pleiotropy null allele ABO blood group system mutation incomplete penetrance wild type variable eXpressivity lethal mutation l A chicken is homozygous for the dominantP allele of the Pigment gene for pigmented feathers but it is also homozygous for the dominant I allele of the Inhibitor gene that inhibits pigmentation of the feathers So it has colorless feathers A second chicken is homozygous for the recessive p allele of the Pigment gene and makes no pigment but it is homozygous ii at the Inhibitor gene So it also has colorless feathers What fraction of the F2 progeny of these two chickens will have pigmented feathers answer 316 2 In a series of crosses between Labrador retrievers heterozygous for both the B black color and E pigment incorporation loci you expect a ratio of 943 for black yellow and chocolate coat colors Among 114 pups scored there were 69 black labs 29 yellow labs and 16 chocolate labs Test your hypothesis by computing the chisquare statistic answer x2 1731 p 0421 You do not reject the null hypothesis 3 An intercross gives an F2 phenotypic ratio of 961 What is the phenotypic ratio in a testcross using an F1 female as one parent answer 12 1 Genetics 466 Lecture 37 Genes in Populations Drift and Mutation Objective Become familiar with the concepts of random drift and neutral molecular evolution Understand how drift removes variability from a population while mutation restores it These processes result in an equilibrium which we can quantify with F Key Terms Mutation rate Effective size Drift Substitution rate Fixation Polymorphism Molecular clock Heterozygosity Neutral variation Homozygosity Additional Problems 1 Suppose the mutation rate for neutral bases in a DNA sequence is 10396 and the average heterozygosity is one per 2000 nucleotide positions What is the population size at equilibrium Two species of fish differ at about 1 of nucleotide sites in regions of the genome deemed to be neutral An independent estimate of the mutation rate in these species is 178 X 10395 Estimate the time in generations since these species had a common ancestor Which of the statements below is true about the equation below F 4N1 1 0 1quot represents the level of heterozygosity in the next generation 1 The equation defines equilibrium homozygosity at a neutral gene in a population 2 The population size is 4N 3 M is the probability that a gene does not undergo a mutation You are studying a population of snails on a remote island Your survey of the population reveals that 1 in 500 individuals have shells with lefthanded spirals indicating that they are homozygous for the recessive allele 1 at a locus which determines whether their shell forms a left or right spiral Assuming a mutation rate of 107 HardyWeinberg equilibrium mutationdrift equilibrium and neutrality of the L locus estimate the size of the snail population on the island Answers 1 N1F 12000199 1125 4 4x10 d 1 100 2 t 5 281 2 2178x10 3 1 4 Using HW the frequency ofl 15000395 0045 and the frequency ofL 0955 The frequency of heterozygotes will be 2X0045X09550086 and the frequency of homozygotes F will be 0914 Thus Genetics 466 Lecture 10 Linkage and Mapping I Objective Learn the basic concepts of recombination and mapping Know the differences between and relationships among chiasma frequency recombination frequency and map distance Know how to use the data from a 3factor cross to nd the map order of genes their phase on the parental chromosomes the recombination fractions between them and the degree of interference Key terms recombinant linkage phase parental coupling RF recombination frequency repulsion linkage 2 3 and 4strand double exchanges map distance 2factor cross centiMorgan 3factor cross chiasma pl chiasmata interference mapping function coefficient of coincidence Haldane39s mapping function Additional Problems 1 What is the coefficient of coincidence when there is no interference When there is complete interference Answers 1 and 0 2 For two genes that are 25 cM apart on a linkage map what is the recombination frequency that you would expect to observed between these two genes assuming no interference Answer 0197 Genetics 466 Lecture 35 HardyWeinberg and the Gene Pool Concept C39 J39 IT J A gene f 1 39 genotype frequencies and how the HardyWeinberg rule relates the two Know how to use the gene pool model to arrive at HardyWeinberg frequencies Learn how to compute gene frequencies genotype frequencies and gamete frequencies for Xlinked loci Key Terms gene frequency gene pool model genotype frequency random mating model HardyWeinberg Additional Problems 1 If a recessive Xlinked disease has frequency 2 in females what is the frequency in males Answer squareroot of z 2 A population of sea slugs has the following genotypic frequencies at the Adh locus AthAdhs at 060AthAth at 020 and AthAth at 020 After one generation of random mating what will the genotypic frequency for the heterozygous class be Answer 042 3 You randomly sample a human population for their phenotypes for the ABO blood groups as determined by the I locus and obtain the results in the table below What is the frequency of the 10 allele in your sample Assume the population is at HW equilibrium Answer 063 4 If the recessive allele for an Xlinked recessive disease in humans has a frequency of 002 in the population what proportion of the people will have the disease Assume the population is 5050 malefemale Answer Among females it will be 410000 and among males 2100 so 10210000 overall Genetics 466 Lecture 7 Chromosomes and Sex Determination Objective Understand the experimental evidence showing that the genes for Mendelian inheritance are carried on chromosomes Know the rules of Xlinked inheritance and how to use pedigrees to calculate risks for Xlinked disorders Understand the genetic control of sex determination in different organisms and how different organisms deal with the problem of dosage compensation for genes located on the sex chromosomes Key Terms hemizygous sex chromosome nondisjunction homogametic nulloX heterogametic diploX color blindness Balr body hemophilia SRY Duchenne Muscular Dystrophy XX XY XO XXY ZW ZZ X inactivation haplodiploidy autosome Hymenoptera Additional Problem 1 A woman has two brothers who died young from DMD a What is the risk to her first child answer 18 b If her first child is a normal son what is the risk to her second answer 1 12 Hint use Bayes Theorem Researchers have produced a transgenic mouse carrying a copy of the SRY on an autosome A male with normal X and Y chromosomes plus one copy of the transgene is mated with a normal female What proportion of each sex is expected among the pups answer 34 male 14 female The yellow body gene 1 of Drosophila is X linked with yellow body y being recessive to black body V What are the expected progeny when a yellow bodied male is mated with a heterozygous female A 100 black for both sexes B 100 yellow for both sexes C all black males 50 black 50 yellow females D all black females 50 black 50 yellow males E 50 black 50 yellow for both sexes F all yellow males all black females G all yellow females all black males H all tan females 50 black 50 yellow males 1 all tan females all yellow males J all tan females all black males answer E


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