Genetics exam 1 review
Genetics exam 1 review Genetics 3301
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This 26 page Study Guide was uploaded by Aishwarya Juttu on Thursday September 29, 2016. The Study Guide belongs to Genetics 3301 at University of Houston taught by Prof. Cooper & Prof. Lin in Fall 2016. Since its upload, it has received 6 views. For similar materials see Genetics in Biology at University of Houston.
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Date Created: 09/29/16
Terminology: 1. Gene: region of DNA encoding some functional products (e.g. protein) 2. Locus: position on a chromosome where a gene is located 3. Mutation: any genetic change relative to a reference gene 4. Single nucleotide polymorphism (SNP): a particular single base pair difference between individuals within/between spec →→→→→→→→ 5. Allele: particular version of a gene; alleles come from mutation and recombination 6. Homozygous: individual has identical alleles of a particular gene (AA or aa) 7. Heterozygous: individual has different alleles of a particular gene (Aa) 8. Genotype: complete DNA sequence of an individual; genetic makeup of an individual 9. Phenotype: physical character of an individual 10. Genome: physical manifestation of the genotype - chromosome Chapter 2: ● Probability ○ Genotype of offspring involves chance events ○ Exact outcome is impossible to predict, but the average outcome can be predicted ○ Knowing the outcome can allow us to infer the process that led to the outcome ● Large numbers ○ The larger the sample size, the closer the average outcome is to the exact outcome ● Sum (addition) rule ○ Outcome 1 OR outcome 2 ○ With a single coin, what is the probability of getting a head OR a tail? ½ + ½ = 2/2 = 1 ● Product rule ○ Trials are independent of each other ○ Outcome 1 AND outcome 2 ○ With two coins, what is the probability that the first coin is a head AND the second is a tail? ½ x ½ = ¼ Examples: 1. On a 6- sided die, what is the probability of rolling a 3? ⅙ 2. What is the probability of rolling a 2 or 3? ⅙ + ⅙ = 2/6 or ⅓ 3. What is the probability of rolling a 2 first, and then a 3? ⅙ x ⅙ = 1/36 4. What is the probability of rolling a 2 and 3 in any order? ⅙ x ⅙ = 1/36 x 2 = 1/18 ● Binomial Probabilities ○ Mutually exclusive outcomes ○ Ex) heads OR tails; boy OR girl; etc. ■ Simple coin toss ■ Probability of a head: p = 0.5 ■ Probability of a tail: q = 0.5 ■ (p+q) = (p+q)(p+q) = p + 2pq + q 2 ○ We can calculate binomial probabilities to determine the probability of a particular combination of events ■ If I toss 10 coins, what is the probability that 4 will be heads? ○ Possible outcomes for 3 coins ■ p = heads ; q = tails ■ (p+q) = p + 3p q + 3pq + q 3 Heads (p): 3 2 1 0 Tails (q): 0 1 2 3 ■ Ex) probability of getting 2 heads and 1 tail ● 3p q = 3(0.5) (0.5) = 0.375 = 3/8 ■ Coefficients: number of ways an outcome can occur ■ Pascal’s Triangle Example: ● Judy’ parents are heterozygous (Aa) at a recessive disease gene. What is the probability that Judy or her brother (not both) will have the disease? *2 children* A a A AA Aa → ¾ chance of not getting disease A Aa aa → ¼ chance of getting disease ○ ¼ x ¾ x 2 = 6/16 Mendel ● Mendel used peas to prove the principles of inheritance ○ He obtained 34 different varieties of peas ○ He identified 7 different traits existed in two distinct forms (binomial) ● Peas were a good model system, so it was possible to do controlled crosses ○ Controlled crosses mean that the phenotype is known. ○ They allow production of many offsprings from the same parents. ● Crossing Pure Breeding Strains ○ F: filial- having the relation of a child to a parent ○ Re-emergence of the white flower in F was 2 important evidence against the Blending Theory of Inheritance Mendel’s Laws ● First Law: Segregation ○ Two alleles for each traits separate (segregate) into different gametes during meiosis ○ Each will have a 50% chance of inclusion in the gamete ○ Monohybrid cross ● Second Law: Independent Assortment ○ During gamete formation, the segregation of alleles at one locus is independent of segregation of alleles at another locus. ○ Prediction: 3:1 dominant:recessive ratio in F of2pring ■ IF the F 1notype is heterozygous, THEN crossing these individuals to a pure-breeding recessive parent should result in ½ of offspring having the recessive phenotype. ■ Test prediction using a test-cross. ○ Dihybrid cross; 9:3:3:1 ratio ○ Type of exam Question: ■ How can a similar approach be used to test the prediction that dominant phenotypes in F2f a monohybrid cross will have two different genotypes? ● Third Law: Dominance ○ When organisms contain two versions of a gene (two different alleles), one allele can be dominant to the other, masking its effects (contrast to phenotypes being a ‘blend’ of the two genes) ○ Monohybrid cross ● There are fewer phenotype classes because more than one genotype can produce the same phenotype. ● How close is close enough? → X (Chi-Square) Tests ○ Observed phenotype and genotype ratios are rarely exactly as predicted, so how can we see if the observed outcome is close enough to our expectation? ○ X is generated through the following equation: ■ X = ∑(Observed - Expected) 2 Expected ■ X = (Obs Group 1 - Exp Group 1) + (Obs Group 2 - Exp Group 2) 2 Exp Exp ○ P-value: the probability of observing the result if the null hypothesis is true. 2 ● Interpreting a give X value ○ The X value evaluates the probability of observing your result if the system works according to the null hypothesis. ○ If P value is less than 0.05, then the result is statistically significant. This means that there is probability of observing your results if the null hypothesis is true (something else is happening) ○ If P value is greater than 0.05, then the result is non-significant. This means that there is a high probability of observing your results if the null hypothesis is true → does not mean the null hypothesis is right. ○ Example: ■ Mendel observed 5475 round and 1850 wrinkled peas in F from a monohy2id cross between pure-breeding round and wrinkled parents. The expected ratio is 3:1. The observed ratio is 2.96:1. ■ Was the observed statistically significant from the expected? ● 1. Calculate expected values: Total offspring: 5474+1850 = 7324 Expected round: 7324 x (¾) = 5493 Expected wrinkled: 7324 x (¼) = 1831 ● 2. Calculate X : 2 X = ( observed - expected) 2 expected X = (5474 - 5493) + (1850 - 1831) = 0.263 5493 1831 ● Determine degrees of freedom (df) Df = n -1 Df = 2 - 1 = 1 2 ● Determine if P value is less than 0.05 using X table 2 Df = 1 ; X = 0.263 P value is between 0.50 and 0.70 Conclusion: There is no significance, therefore we will not reject the null hypothesis that each allele is equally likely to be passed on in gametes. ● Pedigrees ○ Important symbols to know: ○ Autosomal dominant: ■ Males and females are equally affected T ■ Approximately half of children of affected-unaffected parents are affected T ■ Two unaffected parents can’t produce affected children T ■ Two affected parents can produce unaffected children T ○ Autosomal recessive: ■ Males and females are equally affected T ■ Approximately half of children of affected-unaffected parents are affected F ■ Two unaffected parents can’t produce affected children F ■ Two affected parents can produce unaffected children F ○ X-linked recessive: ■ Males and females are equally affected F ■ Approximately half of children of affected-unaffected parents are affected F ■ Two unaffected parents can’t produce affected children F ■ Two affected parents can produce unaffected children F Chapter 3 ● Overview: ○ How are chromosomes replicated? ○ How are chromosomes inherited at cell division? ○ Understand how chromosomes replication and inheritance allows prediction of how genes will combine in offspring. ● Cell Cycle (Mitosis) ■ Mitosis: produces genetically identical daughter cells from a parental cell ■ Chromosome number does not change. ■ There is a series of checkpoints. ■ It ties the cell cycle to an organized sequence of events and checks that cells go through to minimize the chance of issues during replication and division. ○ Interphase ○ The cell cycle is divided into 2 main phases: m phase (mitosis) and interphase ○ Interphase starts with the G1 phase ■ Cells express genes at high levels ○ Most cells then transition to the S phase ■ DNA synthesizes to make another copy of each chromatid, so now chromosomes have two DNA molecules ○ Then, cells transition to the G2 phase ■ Cells prepare for division that occurs in M phase. ● M Phase (mitosis) ○ Karyokinesis: dividing genetic chromosomes into the nuclei of the two daughter cells ○ Cytokinesis: partitioning of other cell components ○ Chromosomes condense during prophase, which is at the beginning of M phase. This is when chromosomes form a centromere, which holds two sister chromatids together. ○ After DNA replication, each chromosome consists of two identical sister chromatids. Different copies of the same chromosome are called non-sister chromatids. ○ Centrosomes become visible and migrate to the opposite poles of a cell and act as a source of spindle fiber microtubules. These allow movement of chromosomes. Some embed in the kinetochore, a protein complex at the centromere, so that each sister chromatid becomes linked to a different cell pole. ○ The connection of sister chromatids to different poles cause them to align in the middle and form a metaphase plate. ○ Microtubules bind to the kinetochore and elongate by adding tubulin subunits to the positive end. They also depolymerise, meaning that they shorten to pull an attached chromatid to the pole it is anchored to. ○ Cohesin binds sister chromatids together. ○ During anaphase, sister chromatids separate and move to the opposite poles as cohesin breaks down. The microtubules pull them to the poll they were attached to. This process is called disjunction. ○ During telophase, nuclear membranes surround each set of chromatids and form two new nuclei. When the nuclei form, the chromatids decondense. ○ Finally, cytokinesis divides the cell into two daughter cells. ● Cell Cycle Checkpoints ○ There is a series of checkpoints during the cell cycle. ○ Many of them work as a combination of a protein kinase and a cyclin protein. ○ Cyclin proteins change depending on the cell conditions. They bind the kinase to change its activity. ○ At the end of the G1 phase, the checkpoint cyclin D1 is produced and it bind Cdk-4, activating it so that it can phosphorylate a protein called pRB. ○ After being phosphorylated, pRB cannot bind a transcription activator protein, E2F. This activates genes involved in the S phase. ○ When cyclin D1 is high, E2F activates S phase genes. ● Meiosis ■ Similar to mitosis, but has two phases after interphase: meiosis I and meiosis II. ■ Produces four (4) haploid cells. ○ Meiosis I ■ Homologous chromosome synapsis (two copies) come together. ■ This aligns non-sister chromatids, allowing enzyme complexes to form and facilitate crossovers, which is the exchange of genetic material between non-sister chromatids. This allows shuffling of allele combination if a cell was heterozygous at different loci (recombination). ■ After crossover, chromosomes align at the metaphase plate and microtubules pull homologous chromosomes to opposites sides of the cell. ■ The first cell division occurs. Now cell have n chromosomes and are different from the parent cell. ○ Meiosis II ■ Meiosis II separates sister chromatids to create a total of four haploid cells. ■ These are gametes and have a chromosome number of 1n. ● Meiosis and the basis of Mendelian ratios ○ Law 1: Random segregation ■ Each sister chromatid is passed onto a gamete in meiosis II. ○ Law 2: Independent assortment ■ Genes on different chromosomes are passed on independently at meiosis I. ● Chromosome Theory of Heredity ○ Morgan found a mutant white-eyed male fly and when crossed with a red-eyed (wild type) female, the F 1ies were all red eyed. ○ The F flies were expected to have a 3:1 red:white ratio. 1 ○ All females had red eyes (4:0) and half of the males had white eyes (1:1) → 3:1 average ○ For F ,1ex of offspring did not have an effect of how a trait was inherited. ○ However, for parents and F , se2of offspring did matter. ○ When the phenotypes of the parents were swapped, both male and female F flies were 2 1:1 Red:white eyes. ○ Since females flies have two X chromosomes and male flies have an XY combination, Morgan realized that eye color was determined by the X chromosome. ● Dosage compensation ○ Species with sex chromosomes have an imbalance between the sexes in the copy number of genes on those chromosomes. ○ For example, female placental mammals have two copies of genes on the X chromosome, while males only have only one. ○ Dosage compensation is a mechanism that compensates for this imbalance. ○ Placental mammals can do this by randomly inactivating one of the two X chromosomes in each cell of an embryo. ○ An inactivated chromosome that condenses is called a Barr body. ○ One consequence is that different cells will express either the paternal or the maternal X chromosome, so they may have different phenotypes. Chapter 4 ● Gene Interaction ○ Genetic interaction: the phenotype given by an allele at one gene depends on the other allele of that gene (intra-gene) or on alleles at, at least, one other gene (inter-gene). ○ Intra-gene interactions: dominance; identification and mechanism ○ Inter-gene interaction: mutation analysis and metabolic pathways ○ Interactions between genes products create interdependence, which is when the effect of alleles at one locus depends on how the identity of alleles present at other loci. ● Complicating Mendelian Genetics ○ Mendel’s success in establishing the rules of inheritance were determined by a pair of alleles in a single gene, one being dominant and the other being recessive. ○ However, there is more to it: ■ There may be more than two alleles present in a population. ■ There may be more than two genes that control a trait and these genes may combine in complicated ways. ○ These points DO NOT invalidate Mendelian genetics. They just mean that we need more information before phenotypes can be predicted. ● Intra-gene Interactions ○ One allele is dominant over the other. ○ How can we test for dominance? ■ If one of the parent alleles is dominant, the F1heterozygote will look like the parent. ○ If functional allele is dominant: haplosufficient. ○ If functional allele is recessive: haploinsufficient. ○ Mutations usually result in an allele with reduced function. ○ Mutations that reduce function are loss of function mutations. They can be divided into: ■ Null functions (no functions) ■ Leaky functions (partial functions) ○ Mutations that result in a new allele with some new function are called gain of function mutations. ○ Incomplete dominance ■ If R is incompletely dominant, then the phenotype of the Rr heterozygote would be the intermediate of the RR and rr homozygotes. ■ Some functionality ● ○ Co-dominance ■ Both alleles of the parents contribute to the phenotype. ○ Recessive lethal alleles ■ Some gene products are essential to live. ■ Almost ⅓ of all genes are essential for survival. ■ Lethal alleles have the potential to cause death. ■ The organism dies only when the lethal alleles are homozygous recessive ○ Pathways ■ Genes interact by affecting different steps of biological pathways. ■ Beadle and Tatum: used the relationship between genes and metabolism to see if genes were involved and the order of the steps in the production of arginine (amino acid). ■ Two major outcomes: ● The “one gene - one enzyme” (each gene produces an enzyme with a defined function) hypothesis was more complicated. Not all genes encode enzymes and some enzymes need multiple genes. ● Gene products work together in pathways to produce phenotypes. Experiment: ● Create a pool of mutants ● Identify mutants that require at least 1 essential growth compound (aka auxotrophs) ● Identify auxotrophs that need arginine to be added before they grow. The mutants have a mutation in the arginine pathway. ● Result: there was a collection of mutants with that mutation in a specific amino acid metabolism. ○ Determining compounds that restore mutant growth allows pathways to be ordered ■ Genes encode enzymes that act in sequence to produce an essential product. A mutation causes a defective enzyme that blocks a step so that the final product cannot be produced. ■ Hypothesis: Adding a compound that occurs before the blocked step will not restore growth. Adding a compound after the blocked step will restore growth. ■ The more compounds a mutant can grow on, the further upstream in a pathway it is. ■ The more mutants that can grow on a compound, the further downstream in a pathway it is. ● Inter-gene interactions ○ Genes work together at different steps of the biosynthetic pathway. ○ The effect of alleles at gene 1 depend on the alleles present on gene 2. ○ F 2fspring have two genes with two alleles each, which means they have four phenotypes, 9:3:3:1 expected ratio → assuming genes do not interact to the observed distribution. ○ Dominant = functional ○ Recessive = null ○ Single and Double Mutants ■ A dihybrid cross will always have the same genotype distribution (assuming no linkage) ■ If genes interact, there may be different phenotype distributions depending on how alleles at gene 1 interact with alleles at gene 2 → epistasis ■ The result of interactions is to combine the original 9:3:3:1 phenotype groups in different ways. Original groups are never split. A gene interaction causes a genotype group that was expected to have one phenotype to have a different one. ○ Complementary gene interaction 9:7 ■ Two pure breeding white plants have F offsp1ng with purple plants. When they are crossed, the F offspring have purple and white flowers in a 9:7 ratio. 2 ■ Changes in the phenotypic ratios result only from combining the groups expected if genes without a functional allele of both do not produce any pigment. ■ Pathway: A linear pathway where the product of one gene (either C or P) is acted on by the product of the second gene to make the pigment. ○ Duplicate gene interaction 15:1 ■ Two pure breeding purple plants have F offspring with purple 1 plants. When they are crossed, the F2offspring have purple and white flowers in a 15:1 ratio. ■ Changes in the phenotypic ratios result only from combining the groups expected if genes do not interact. We can infer that the genotypes with a functional allele at either gene produce pigment. ■ Pathway ● Both genes control the same biochemical step which can explain the 15:1 phenotypic ratio. ○ Recessive epistasis 9:3:4 ■ F offspring of two plants have blue, pink, 2 and white flowers in the 9:3:4 ratio. ■ Changes in phenotypic ratios result only from combinations of the groups expected if genes do not interact. Genotypes with the same functional copy of both produce blue pigment. Genotypes with a functional copy of one of the two genes produce pink pigment. Genotypes with a functional copy of the other gene or no functional gene copies produce no pigment (white). ■ Pathway ● Can be explained if mutated genes are in the same pathway but depending on where in the pathway mutation occurs, a different end product is made. ● Other types of gene interactions ○ 9:3:3:1 → no interaction ○ 9:7 → mutated genes are in the same linear pathway ○ 15:1 → duplicate genes ○ 9:3:4 → recessive epistasis - homozygosity at one gene masks the other gene; similar to 9:7 intermediate product; gives some phenotype ○ 12:3:1 → dominant epistasis ○ Suppressors ○ Modifiers ○ Synthetic lethals *Should be able to draw pathways that can explain these observations *There will be more discussion over this during gene regulation ● How can we determine if mutations in two different individuals are in the same gene? ○ Complementation test: if two homozygous mutants are crossed and heterozygote offspring are wild-type, mutations were in different genes. There will be one functional copy of each gene in the offspring → only works for recessive mutations. Chapter 5: Genetic linkage and mapping in eukaryotes What effect does recombination have on combinations of alleles found on different chromosomes? How can recombination be used to position genes along a chromosome? ○ Genes have a physical presence, a region of DNA on a particular chromosome ○ During prophase I, non-sister chromatids physically align. DNA strands may break and repair, creating a recombination event by causing a DNA strand from one chromatid to invade the other and vice versa. ○ The process of recombination can result in shuffling of alleles between the two original chromatids. ○ Genes that are further apart are more likely to be shuffled because it is more likely that a recombination event will happen between them. ● Non-linked genes will assort independently (Law of independent assortment) ○ This means that each possible allele combination is present at an equal frequency. For two genes, each of the four allele combination is present at 25%. ○ Frequency of each gamete combination is given by the product rule (chapter 2). ● Syntenic genes can be linked so that assortment is not independent. ○ Syntenic- on the same chromosome ○ If genes are syntenic and very close together, they are completely linked and are passed on to gametes as a single unit. ○ If genes are incompletely linked, crossing over may occur. This produces recombinant allele combinations, but at a frequency below than that of the parental allele combinations. ○ Crossing over between syntenic but distant genes can be common enough to be inherited independently. ● Crossover ○ A break occurs in the DNA strand of one chromatid ○ One broken strand invaded the non-sister chromatid, displacing the strand running in the same direction. ○ The invading strand is extended before eventually re-invading its original chromatid. ○ 1 of 2 possible resolutions of the resulting Holliday junctions results in a direction. ● Eukaryotic gene mapping ○ The probability that a crossover will occur between two syntenic genes is inversely proportional to the distance between the genes. ■ The more distant the genes, the higher chance of crossing over. ○ Knowing this relationship can help to order genes along a chromosome and determine the distance between genes. ○ Punnett and Bateson found a 3:1 segregation of phenotypes in the F generation by 2 studying segregation of either flower color or pollen shape separately. ○ When flower color and pollen shape were considered together, the F genera2on phenotype did not have the expected 9:3:3:1 ratio. ● Two-point test-cross ○ Morgan used a two-point test-cross to demonstrate the deviation between expected and observed parent to non-parent allele combinations. ○ Two-point test-cross: cross between an individual that is heterozygous for both genes against an individual that is homozygous recessive. Offspring phenotype is determined only by the alleles of the heterozygous individual. → offspring genotype can be inferred from their phenotype. ● How can we convert recombination frequencies into gene positions on chromosomes? ○ 1% recombination frequency = 1 map unit (mu; aka centiMorgan (cM)). ● Testing for linkage ○ Recombinants are underrepresented and so the two genes are linked. ○ To distinguish between chance and real underrepresentation of recombinants due to linkage, we can use Chi-square (X ); (chapter 2) X = ∑(Observed - Expected)2 Expected X = Obs - Exp) Obs - Exp) + (Obs - Exp) + (Obs - Exp) Exp Exp Exp Exp Genotype 1 Genotype 2 G enotype 3 Genotype 4 ○ If the P value is less than 0.05, then we can say that the null hypothesis is rejected. ● Three-point mapping ○ Used to determine not just if genes are linked, but the order in which they occur on a chromosome. ○ We cross ABC/abc and abc/abc: Possible genotypes: ABC/abc ABC ABc/abc Abc AbC/abc AbC aBC/abc aBC Abc/abc Abc aBc/abc aBc abC/abc abC abc/abc abc → recessive; does not contribute to phenotype ○ If genes are not linked, we expect 8 phenotypes at a frequency of ⅛ (½ x ½ x ½) ○ We can order genes by combining more than 2 genes in a single cross ○ Between two genes, only single crossover events can be detected. With three genes, two crossover events can be detected. ○ Double crossovers swap central genes between chromatids. ○ Ask 5 questions when doing a three-point test-cross: 1. Are the genes linked? → use Chi-square 2. What are the parental genotypes? → if two phenotypes are more common than the others, they must be the parental genotypes. 3. What is the gene order? → double crossovers are less common than single crossovers. Rarest recombinant genotypes will always be the genotypes requiring a double crossover. 4. What are the recombination frequencies between genes? → count crossover events between each gene pair. 5. Is the frequency of double crossovers consistent with independent single crossovers? → interference: having fewer than expected double crossovers. This results from a limitation of the number of crossovers than occur in a short stretch of DNA. ● Gene mapping in humans ○ We can map genes in model eukaryotic species, but we cannot do the same in humans. ■ There were few simple genetic markers. We cannot make and select them by deliberate mutation. ■ Parents and offspring are not controlled. Humans have fewer offspring than the model species. ○ Gene mapping exploits the development of genetic markers. ○ Iod score (logarithm of the odds): test of linkage between a marker and a trait. This is the probability of obtaining a particular pedigree if a marker and target gene are linked vs. if they are not linked. ○ If marker-trait association is better explained assuming a particular degree of linkage → higher iod score ○ If marker-trait association is better explained assuming no linkage → lower iod score ○ Iod analysis is intensive. It analyzes all possible marker-gene combinations, at many different recombination rates (θ) *be able to interpret the outcome of an analysis* Chapter 21: Quantitative genetics What is the difference between a discrete and quantitative trait? How can we map quantitative traits? How can we tell whether variation is a quantitative trait determined by genetic variation or by developmental/environmental differences? ○ Most of the traits we have seen are discrete, which means that the phenotypic differences caused by different alleles have been obvious (green vs. yellow; smooth vs. wrinkled). ○ Discrete traits are very useful in genetic analysis. Easy scoring of phenotypes helps in understanding the genetic basis of these traits. ○ In practice, many traits are quantitative, which means that they exhibit continuous variation between two extremes (every shade between green and yellow). ● The multiple genes hypothesis ○ Additive genes: two genes contribute additively to the same trait. ■ Ex) color of wheat kernels is determined by 2 genes (A and B) each with two alleles (A /A1a2 B /B 1 2 ■ A an1B cont1bute to a fixed amount of red kernel pigment. The more of these alleles, the darker the color. ● Darkest red: A A B111 1 ● Lightest red: A A B2B2 22 ■ 5 possible phenotypes ○ Note: phenotype is determined by the total number of “1” alleles adding over the “A” and “B” genes. This is different from when phenotypes were determined by the presence/absence of dominant alleles at each gene. ○ Traits determined by many genes are usually quantitative ■ Five phenotypes can be produced with 2 additive effect genes. ■ This number increases as 2n+1 with the number of genes involved (assume diploid) ○ Environment can affect phenotype ■ Genotype-environment interaction. ■ Many aspects of the environment control how genes are expressed. ■ Quantitative phenotypes need to be described statistically, as a distribution with a mean and variance. ○ Partitioning sources of phenotypic variation ■ Quantitative traits depend on genes, gene interactions, and the environment. ■ It is possible to separate genetic and environmental sources of phenotype variation (nature vs. nurture). VP V GV E VP total phenotypic variance VG genetic variance (nature) V : environmental variance (nurture) E ■ Genetic variance is the proportion of total phenotypic variation due to differences between genotypes ■ Environmental variance is the proportion of the total variance due to variability in the environment. ● Heritability ○ Provides a measure of the influence of genetic variance in explaining the observed variation in a trait. ○ Two types: 2 ■ Broad sense heritability (H ): proportion of all phenotypic variance due to genetic variance. 2 H = V /G P VG genetic variance VP total phenotypic variance 2 ■ Narrow sense heritability (h ): proportion of phenotypic variance due to additive variance. 2 h = V A P VA additive effect of all alleles contributing to a trait VP total phenotypic variance ○ Note: heritability is a measure of how much variation in a trait is due to genetic factors, not a measure of how much of the mean trait value is due to genes. ● It is not possible to calculate trait heritability in humans. ○ Natural experiment: ■ Compare traits in identical twins (monozygotic; MZ) and fraternal twins (dizygotic; DZ). → V anGV contEbute to total V . P ■ MZ: V P= VE→ all genes are identical for V = 0 G ■ DZ: V =P + VG E ■ The difference in V betweP DZ and MZ twins gives V G. ● Narrow sense heritability predict response to selection ○ This is because it measures only additive genetic variance, not gene interaction effects, which are often disrupted between parents and children. ● Quantitative trait loci (QTL) ○ Quantitative trait loci: genes that contribute to quantitative traits. ○ QTL are not fundamentally different to the genes that control discrete phenotypes, but they are more difficult to detect. There will be several genes that contribute to the same phenotype and the phenotype is more variable, so the analysis of genotypes is more difficult than saying if a genotype does/does not have a specific phenotype. ○ Mapping of a discontinuous trait relies on our ability to determine the genotype of a particular phenotype. ○ QTL mapping is a method that detects linkage between a trait of interest and markers at known locations throughout the genome. ■ Step 1: Interbreed two divergent parental lines to generate an F generation (all heterozygous) 1 ■ Step 2: Backcross the F of1pring to either of the two parental lines and genotype the offspring for the markers. ○ Distinct distributions for genotypic classes at a marker locus signal the location of a QTL near the marker. Odds = Prob(Data|QTL) Prob(Data|no QTL) ■ “LOD” = Log (10) of the Odds ○ GWAS -QTL mapping in humans ■ QTL mapping requires controlled crosses → not possible in humans ■ Genome wide association studies (GWAS) is a method used to identify the vast majority of human disease genes. ■ GWAS: analyze the association between a trait and genetic variation present in a population. Variants that are associated with a trait are near genes that contribute to that trait. End of chapter questions: ● Chapter 2: 2, 3, 5, 7, 9, 13, 15, 27, 34. ● Chapter 3: 3, 9, 12, 13, 19 and 32. ● Chapter 4: 3, 4, 13, 14, 19 and 30. ● Chapter 5: 2, 4, 10, 18 and 22. ● Chapter 21: 1, 4, 8, 12 and 20.
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