Genetics Exam Two Material.doc

Genetics Exam Two Material Tetrad Analysis Fungi: all four chromatids from a single meiosis are recovered Fungi exists in nature as a haploid Order of spores in the ascus exactly corresponds to plane of division in meiosis I and II Linear axis Note: after producing the four products of meiosis, undergo a mitotic division Understanding order of spores First division segregation: all ?A? alleles are in top half of ascus and all ?a? alleles are in bottom half From an Aa diploid Diagram Second division segregation: Both ?A? and ?a? alleles present in each half of the ascus Results from when a crossover occurs between A and its centromere Diagram: Four equivalent patterns of second division segregation Note: always ?a? and ?A? in one half of the ascus if second degree! Frequency of Second Degree Segregation = exchange frequency Recombination frequency = ½ exchange frequency Recombination frequency = ½ frequency of second degree segregation R.f. = (# recombinants)/total spores x 100 Note: if exchange occurs 100% of the time (as in non-linked genes), 50% will be recombinants and 50% will be parental Dihybrid Cross Are genes segregating independently or separately? Steps: Look at each gene independently (as if on separate chromosomes) How many ?A? show second degree segregation? Calculate r.f. for A from centromere? If independent Assortment.... Should see all four genotypes equally (occur in equal frequency) among asci population Diagram: If genes are linked... See more parental spores than recombinant spores Parental spores should be equal (mom = dad) Diagram: If genes are linked, the segregation patterns in asci indicate which gene is closest to centromere... Look at each gene separately Find frequency of exchange (frequency sds) for each gene Look at asci: See second division segregation in BOTH genes ( know that exchange between one gene and the centromere is also an exchange between the other and the centromere See second division segregation in ONE gene and first degree segregation in the other: Gene showing NO second degree segregation: closest to centromere Gene showing second degree segregation: further from centromere Diagram: Solving Tetrad Analysis Problems Are the genes linked (same chromosome) or not? Find location of allele relative to the centromere If on different chromosomes, treat each gene separately to calculate If linked: One asci will show second divison segregation for both alleles One asci will show second division segregation for one allele, and first division segregation for the other The allele that is closest to the centromere will show first division segregation, while the allele that is farther will show second If see both varieties (first and second / second and fist), then the alleles are on opposite sides of centromere Chromosome Abnormalities General: abnormalities as a result of changes in chromosome number or structure Aneuploidy: an unbalanced chromosome constitution resulting from changes in only one or a few chromosomes in a set Diploid: 2N Trisomy: 2N + 1 Monosomy: 2N ? 1 Trisomies and monosomies arise via nondisjunction Example: XO, XXY, trisomy 21 General Principles: Autosomal aneuploidy is more severe than sex chromosome aneuploidy Monosomy is more severe than trisomy Aneuploidy for more than 1-2% of genome is lethal Aneuploid gametes function normally in animals Plants are more tolerant of aneuploidy in somatic cells than animals Plant gametes (especially pollen) are more sensitive to aneuploidy resulting in semi-sterility Incidence of Trisomy 21 with age Age 40: 1/200 Age 45: 1/46 Polyploidy: an increase in the number of complete chromosome sets Occurance in nature almost exclusively plants Haploid number (n) is the number of chromosomes in a normal gamete (normal meiosis) Monoploid number (x): number of chromosomes in a basic set For diploids: x = n Polyploids contain multiples of x > 2 Examples: various species in the oat genus Somatic Cells Haploid Gametes Diploid 2x = 2n = 14 n = x = 7 Tetraploid 4x = 2n = 28 n = 2x = 14 Autopolyploid: multiple copies of identical or very similar chromosome sets Triploids are usually autopolyploids How does it happen? Generally sterile because of meiotic problems: 2 from 1 segregation occurs randomly from each trivalent ( formation of gametes that are genetically unbalanced (aneuploid) Autotetraploid: Chromosome number can be doubled (diploid 2x becomes 4x due to mitotic failure) Mitotic failure has to occur early in development (chromosome replication without cell division) Sterile because: If segregation from quadrivalent is 3:1 ( anueploid gametes But, if segregation occurs in bivalent form ( 2 from 2 segregation ( yield functional 2x gametes Allopolyploids: multiple copies of chromosome sets derived from different species Allotetraploids: doubling of chromosome sets derived from two different species AB hybrid (initial F1 generation) is sterile ? A set of chromosomes aren?t homologous to B set, can?t pair and separate normally at meiosis ( produce aneuploid gametes Sometimes, doubling of chromosomes (mitotic) occur ( leads to two of each A and B, homologous pairing can occur to produce functional, balanced gametes Structural Abnormalities Why study Drosphilia? Huge, striated chromosomes ( distinctive bands present which are conserved cell to cell Giant polytene chromosomes Polytene: many threads (chromatids), chromosomes replicate many times without nuclear division Thousands of homologous sister chromatids pair point for point ? become elongated and huge Only found in the salivary glands of larvae Dispersed regions and regions of condensation give banded appearance Good map ? can see exactly where breakage occurs, or where homologous chromosomes pair Chromosome rearrangement: physical exchange of segments between two non-homologous chromosomes due to breakage and cell repair Two products: Translocation heterozygote Diagram: gamete of this unites with normal gamete Meiotic segregation Can see break point by where pairing switches, switch at point of breakage Adjacent Segregation: both resulting gametes are aneuploid ? no viable offspring from these gametes Diagram Occurs in half of the segregation events ( why translocation heterozygote is semi-sterile One gamete has two copies of AB and now VW, the other has 2 VW and no AB Alternate Segregation: both gametes are recovered in viable offspring Products Two normal chromosomes: gametes will produce homozygous normal offspring Both parts of translocation: gametes produce offspring which are translocation heterozygotes , with all genes present Indepdendent Assortment Because only alternate segregation yields viable offspring, A/a and B/b do not show independent assortment even though they?re on different chromosomes ? pseudolinkage Diagram Translocation homozygote Diagram Other examples: recombination occurs on small portion of chromosome The Case of Trisomy 21 Most of 21 is appended to most of 14 Products Adjacent segregation: Trisomy 21 (monosomy invididual dies) Alternate segregation: normal or phenotypically normal but heteryzygous for the translocation (euploid) Conclusions: If nondisjunction causes Trisomy 21: normal kids probably won?t have affected kids If caused by a translocation, there is a 50% chance kids will also be translocated heterozygotes increased likelihood they?ll have affected kids Structural Abnormality: Inversions 180 degree rotation of a section of chromosome Two types of inversions, based on whether centromeres are included in the inverted segment Pericentric inversion: inversion includes centromere At meiotic prophase I, pair along homolougs Without crossing over, can segregate just fine from loop Half of gametes contain normal choromsomes, half contain the inversion ( normal zygote If crossing over occurs (single cross) Produce 2 nonrecombinants and two unbalanced recombinant products Recombinants yield unviable zygote, or nonfunctional gametes Never recover recombinant offspring from inversions Reduction in fertility Paracentric inversion: centromere is not included in inversion, recombinant products don?t get packaged into eggs At meiotic prophase I, pair along homologous If crossing over occurs: Two nonrecombinants (balanced, on contains the inversion and one doesn?t) Single cross over produces: Remember, centromeres are needed to move in meiosis Recombinant, unbalanced and dicentric AND a recombinant/unbalanced/acentric Dicentric chromatid forms an anaphase bridge which directs the recombinant toward the middle of the meiotic products Acentric chromatid is lost ( no centromere, doesn?t get pulled anywhere during meiosis Dicentric chromatids always become polar bodies because of orientation and anaphase bridge.... only nonrecombinants become eggs (outside products of meiosis) NO REDUCTION IN FERTILITY Diagram: Structural abnormality: Deletion Deletion of alleles from a chromosome and reattachment of the two ends Deletion of wild-type alleles can uncover mutant recessive phenotypes Consequence of: Lethal in homozygous condition ? complete loss of genes If greater than 2% of genome in heterozygote, also fatal due to unbalanced Deficiencies can be used to construct a physical map of the chromosome (Drosphilia) Diagram Structural Abnormality: Duplications Breakage and reunion of homologous chromosomes at different locations First duplication event is rare, but once it occurs, additional duplications are more likely Regions are paired, but shifted Diagram: Consequences: Less severe than deletion ? can survive in homozygous condition Heterozygote: survive as long as aneuploidy is minimal Can provide raw material for evolution First set of duplicate provides material necessary for survival Second set of duplicate is free to evolve new functions by acquiring mutations Quantitative Genetics Traits resulting in a continuous distribution Birthweight, height, milk production, fat content Genotype-phenotype relationships are influenced by: Environment Multiple genes each of small effect (five or more, but each gene follows Mendelian behavior) Identical genotypes may express different phenotypes. Same phenotypes may have different underlying enotypes Two questions in quantitative analysis: To what extent is variation among individuals for a particular trait in a given population due to differences in genotype? To what extent is it due to environmental affects? For a particular quantitative trait, can we predict the phenotype of offspring from the phenotype of the parents? Statistics: distribution of these traits reflects a bell-curve Mean Variance: describes the width of a distribution Standard deviation: square root of variance Covariance of x and y: measures how paired variables change (for every value of x there is a corresponding value of y) ? look for relationship between 2 variables Correlation of x and y (r): (+1 to -1), a value of zero indicates no correlation between variables Broad Sense Heritability (Hb) Variables Vt: total phenotypic variation for a trait due to: Vg: genotypic variance due to differences in genotype among individuals Ve: environmental variance due to differences in environment among individuals Vt = Vg + Ve Hb = Vg/Vt Hb: the proportion of the total phenotypic variance in a population due to differences in genotype Assume: No genotype-environment interaction: all genotypes are affected equally by the environment No genotype-environment association: all genotypes occur at random in all possible environments No genotype found preferentially in one environment over another Measuring Hb for a group of experimental plants, etc. Measure phenotypic variance for the trat in a heterogenous population (Vt1). This is a measure of Vg and Ve Measure phenotypic variance in a homogenous poplution raised in the same environments as above (Vt2) Vt2 = Vg + Ve = Ve Provides a measure of Ve Subtract Ve from Vt1 to get Vg Calculate Hb Measuring Hb for human traits Correlation in phenotype for identical twins reared apart is a direct measure of Hb for the trait Genotype is the same, so any variance is due to environment alone Or, use phenotypic variance in groups of relatives with known degrees of relatedness Fraternal twins are just siblings, only half of their genes are the same ½ Vg ( compare to unrelated individuals What does Hb tell us? Example: twin studies Hb for human body weight is .6 So, 60% of the variance of the trait in the corresponding population is due to genotypic differences If genetic differences were eliminated, variance would only be 40% of its previous value. If all individuals were raised in the exact same environment, 60% of the variance would remain Narrow Sense Heritability (Hn) Particularly important for breeders interested in how much of the parent?s phenotype will be transferred to offspring Variables Vg (from above) is broken down into further classes: Va: genotypic variance due to the additive effect of genes Add a fixed amount ? heterozygote is between homozygotes Vd: genotypic variance due to the dominant affect of some alleles Vi: genotypic variance due to epistatic interactions between different genes Whether Aa occurs also depends on what happens at gene B What is used? Vd and Vi are of little predictive value ( contribute to phenotypic variance of parents, but aren?t transmissible to offspring since particular combinations of alleles are broken up during meiosis (offspring have new combos) Va: relative contribution of Va is used to predict Hn is a measure of phenotypic variance contributed by Va Diagram If Hn = 1: the mean phenotype of the offspring will be equal to that of their parents If Hn <1: the mean phenotype of the offspring will be between the mean of the original population and the mean of the parents If Hn = 0: the mean phenotype of the offspring will be the mean phenotype of the overall population Using Hn to predict offspring: Yo = Y + Hn(Yp ? Y) Yo: mean phenotype of the offspring Y: mean phenotype of overall population Yp: mean phenotype of selected parents How do we determine Hn? By results of selection experiments, rearrange previous equation to calculate Hn from parental/offspring/population means Or, by phenotypic correlation between relatives that share a known fraction of their genes If all genes act in a purely additive fashion, the phenotypic correlation between relatives would equal the fraction of genes they have in common Hn = observed correlation / expected correlation Expected correlations: Full siblings: .5 Half siblings: .25 First cousins: .125 Points to remember about Heritability: Heritability does not indicate the extent to which a trait is genetically determined In a homogenous population, heritability of a trait is equal to 0 (all variation due to environment), even though genes are necessary in determining/having the trait in the first place A high heritability does not mean a gene isn?t sensitive to environmental affects Heritability is not a fixed characteristic of a trait Population parameter ( only applies to a particular population of a particular genetic composition in a particular environment A difference between two populations for a given trait cannot be assumed to be of genetic origin even if heritability is high in each population Population Genetics Questions addressed: How can the genetic constitution of a population be described? What evolutionary forces affect the genetic composition of a population? How does the genetic composition of a population from generation to generation change under the influence of these forces? (rate of change) What are the dynamics of these changes? How long does it take a particular force to bring about change? Population genetics is not concerned with individuals, but rather the genetic constitution of the population as a whole Genetic composition of a population is described in allele frequencies Allele frequencies are relatively stable Enormous simplification Genotypes are broken down and reshuffled every generation Calculating allele frequencies Look at genotype frequencies in the population Calculate allele frequencies Example: MN bloodtyping MM genotype frequency is .298 MN genotype frequency is .489 NN genotype frequency is .213 p = frequency of homozygote + ½ (frequency of heterozygote) = .298 + ½ (.489) q = 1-p Relationship between allele frequencies in one generation and genotype frequencies in the next: Assumptions Population is large and mating is random with respect to genotype No new mutations No selection ? all genotypes are equally viable No migration in or out of population Hardy-Weinberg Principle: if A1 and A2 are alleles of a gene with frequencies p and q in the parental generation, then the offspring will be produced in the following proportions: A1A1 / p^2 A2A2 / q^2 A1A2 / 2pq Proof of HW Principle Random mating is mathematically equivalent to drawing gametes out of a large pool. Each parent contributes an equal number of gametes to the gene pool, and each offspring is the result of fusion at random of two gametes Diagram: Allele frequency of A1 in new generation p? = p^2 + ½ (2pq) = p^2 + pq = p(p+q) = p Consequences of HW Principle Simple relationship between allele frequencies of parents and genotype frequencies of offspring Allele frequencies remain constant from generation to generation Genotypic frequencies among offspring depend only on allele frequencies, not on genotype frequencies of the parents Even when a population begins with genotypic frequencies that deviate from H-W proportions, only a single generation of random mating is needed to restore H-W proportions On exams, NEVER assume a population is in H-W proportions. Always do expected calculations to check Applications If genotypes are in HW proportions, allele frequencies can be calculated from the frequency of the recessive homozygotes Example: 1/10,000 are affected with a recessive trait. What are the allele frequencies?? q^2 = .0001 Calculate q, then can find p Frequency of heterozygotes can be calculated from allele frequencies Find p and q Heterozygotes = 2pq Find that for recessive traits, most of the recessive alleles in the population are located in heterozygotes Comparision of observed genotype frequencies for a given gene with expected genotype frequencies reveals whether mating is random with respect to that gene Good agreement between obs/exp indicates random mating Selection: some genotypes in a population make a disproportionate contribution of their genes to the next generation... the frequency of the alleles under selective pressure will be different in the offspring than in parents, competition for survival in the gene pool Fitness: relative contribution of genes by individuals of a given genotype to the next generation ? depends on viability and fertility Value between 0 and 1 (0 = no contribution to next generation, die before reproducing) Selection coefficient: (s) is the proportionate reduction in gametes contributed to next generation relative to the most favored genotype s is between 0 and 1 s = 0: most fit genotype (fitness = 1) s = 1: fitness is zero (lethal or sterile) 1: assigned as the fitness of the most fit genotype Fitness of a genotype selected against is 1- s Gene pool model with selection Frequency of allele A1 = p, frequency of allele A2 = q (parents are in HW equilibrium) Different individuals contribute different amounts, disproportionate contributions to next generations; next generation allele frequency is no longer p and q Zygotes: A1A1 = (p?)^2 A2A2 = (q?) ^2 A1A2 = 2p?q? Selection against a recessive lethal or sterile (s = 1 for A2A2 (recessive)) Proportion of gene pool contributed: of all the alleles in the gene pool, (p^2 / p(1+q)) were contributed by A1A1 homozygotes, etc. allele frequency after selection of A1 = proportional contribution of A1 homozygotes + (1/2)contribution of heterozygotes Once you?ve calculated p? (above), you can calculate q? THEN, you can calculate genotype frequencies under random mating (p?^2, etc.) Genotype A1A1 A1A2 A2A2 Total Zygote frequency p^2 2pq q^2 1 Relative fitness 1 1 1-s = 0   Relative contribution to gene pool p^2 2pq 0 p^2 + 2pq = p(1+q) Proportion of gene pool contributed p^2 / p(1+q) 2pq / p(1+q) 0   The frequency of A1 in the next generation = p? p? = the contribution of A1A1 + ½ (contribution of A1A2) q? = 1 ? p? When q <<<<< 1, q? ~ q, so change in q is very small (there is little change in allele frequency, even when selection is complete) It takes a long time to remove harmful, even lethal, recessive alleles from a population Example: Incidence of a serious metabolic/fatal disorder is 1/40,000. If none survive to reproduce, what is the incidence in the next generation? Affected = aa = q^2 = 1/40,000 q = 1/200 After one generation of complete selection (s = 1) q? = q / (1+q) = 1/201 (q?)^2 = 1 / 40,401 (or about a 1% decrease in genotype frequency) The reason for this is because heterozygotes vastly outnumber homozygotes! Changes in allele frequency with successive generations of complete selection qo = initial frequency q1 = frequency after one generation of selection = qo / (1 + qo) Qt = after t generations of complete selection = qo / (1 + t x qo) t = 1 / qt ? 1/qo can calculate the number of generations necessary to reduce allele frequency from qo to qt Selection of arbitrary intensity against a recessive allele (s is between 0 and 1) Genotype A1A1 A1A2 A2A2 Total Zygote frequency p^2 2pq q^2 1 Relative fitness 1 1 1-s   Relative contribution to gene pool p^2 2pq q^2 (1-s) W Proportion of gene pool contributed p^2 / W 2pq / W q^2 / W   Variables and Equations W = 1-sq^2 q? = q ? sq^2 / (1-sq^2) p? = 1- q? Change in q = -spq^2 / 1 ? sq^2 Change in p = -change in q Change in allele frequency of A as a function of the number of generations of selection when the favored allele is dominant, recessive or intermediate Graph Conclusions: Changes in allele frequency occur very slowly when the recessive allele is rare, whether or not it is selected against Selection proceeds much more rapidly when there is no dominance no selection for or against the allele in heterozygotes if the allele is recessive And, if an allele is rare and recessive, most of the copies will be located in heterozygotes! When heterozygote is exactly intermediate fitness: still have a huge impact on allele frequency because heterozygote is selected against, especially if heterozygote greatly outnumbers unfavored homozygotes Selection favoring the heterozygote: overdominance Genotype A1A1 A1A2 A2A2 Total Zygotic Frequency p^2 2pq q^2 1 Relative fitness 1-s 1 1-t   Relative contribution to gene pool p^2 (1-s) 2pq q^2(1-t) W Proportional Contribution to Gene pool p^2(1-s) / W 2pq / W q^2(1-t) / W   Change in allele frequency (p) after 1 generation: pq (qt ? ps) / W Equilibrium: eventually allele frequency will reach an equilibrium value where change in frequency = 0 p^ = t / s + t q^ = s / s + t Both alleles are maintained at the population Equilibrium allele frequencies depend on only on relative strengths of selection coefficients Neither allele can be eliminated if favor heterozygotes! Example: malaria and sickle cell trait Variables: Ha = normal, Hs = sickle HsHs individuals die HaHs individuals thrive better in malaria areas Fitness: HaHa = 1-s, HaHs = 1, HsHs = 1-t where t=1 q^ = .1 (given), solve for ?s? Mutation Selection Deleterious alleles are maintained due to random mutations A ( a Eventually reach an equilibrium, where selection against the allele (eliminating the allele) = to the mutation rate of A ( a Equations: Dominant deleterious allele u = mutation rate Introduction of new A2 alleles = pu frequency of A1 alleles x probability of becoming an A2 allele At equilibrium, q^ = u/s Equations: Recessive deleterious allele u = mutation rate At equilibrium: q^ = square root(u/s) Remember: square root of a fraction is always a bigger number! If value of u is equal in both situations, a dominant allele is selected against in both the heterozygote and homozygote condition. The recessive allele then should occur in greater numbers, since its only selected against in the rare, homozygous condition! Genes and Phenotypes Phenotypes of organisms depend on biochemical reactions within cells Reactions catalyzed by enzymes (proteins) Info for synthesis of proteins is carried in genes. Function of a gene is to direct synthesis of a protein Amino acids and Proteins Amimo acids are connected by peptide bonds Primary structure: amino acid sequence Secondary structure: alpha helix, beta sheet Tertiary structure: folding of the protein into a three dimensional shape 3D shape gives protein function Inborn Errors of metabolism: heritable defect in body chemistry Archibold Garrord discovery ? studied alkapton in urine People who produce alkapton in urine can?t break down homogentistic acid (a product of the breakdown of phenylalanine and tyrosine) don?t have the enzyme to degrade the acid Trait is transmitted in an autosomal recessive pattern Conclusion: normal function of the gene must be to direct production of the enzyme (gene missing = no enzyme) One Gene ? One Enzyme Hypothesis Beadle and Tatum study: chose Neurospora to study because there?s a more immediate link between mutant phenotypes and biochemical defects Experiment Wild type Neurospora grows on minimal medium (contains only the vitamin Biotin, carbon, salts) Screen for nutritional mutants Mutants couldn?t grow on minimal medium ( lost some aspect of biosynthetic capability Growth renewed on complete medium, so obviously lost some ability to synthesize something Nutrient screen Minimal and supplemented with nucleic acids: no rescue, so obviously ability to synthesize nucleic acids is not the problem Minimal and supplemented with amino acids: no rescue Minimal + vitamins: growth regenerated so obviously, the mutant can?t synthesize vitamins Nutrient Screen Two: find which vitamin is the problem (which one rescues growth?) Conclusion: most mutations are blocked at a single step in the sequence of reactions leading to synthesis of a particular amino acid, vitamin, etc because the enzyme that catalyzes that step is nonfunctional Most nutritional mutants show a pattern of inheritance that indicates they are defective in a single gene One gene, One Enzyme Hypothesis Biochemical processes are under genetic control Biochemical pathways are resolvable into a series of individual steps each mediated by a different enzyme Each enzyme is produced under the direction of a single gene Diagrams: No growth on minimal medium. Substance Y or Z will rescue growth, substance X will not rescue No growth on minimal medium. Substance Z will rescue growth, X or Y will not rescue growth Example: Find the biochemical pathway from analysis of nutritional mutants Minimal Orthinine Citrulline Arginine Arg+ + + + + mut. A - - + + mut. B - + + + mut. C - - - + Note: All mutants are unable to grow on minimal medium (missing some aspect o nutrition) Arginine restores growth in all mutants So, Arginine is the end product of the pathway Citrulline restores growth in mutant A and B, while Orthinine only restores growth in mutant B Mutant C lacks the enzyme in the process that produces Citruline Mutant B and C lack the enzyme to produce Orthinine Mutant C is only restored by adding Arginine, Mutant B is restored by adding all three, Mutant A is restored only by adding Citrulline/Arginine Conclusion: Substrate ( Orthinine ( Citruline ( Arginine Where B lacks the enzyme to make Orthinine, C lacks the enzyme to produce Arginine, A lacks the enzyme to produce Citruline Other human conditions caused by genetically determined errors of metabolism PKU ? Phenylketonuria (error in enzyme that converts phenylalaine to tryrosine) Albinism (error in enzyme that converts tyrosine to Dopa) Alkaptonuria (error in enzyme that converts homogentisic acid to acetoacetic acid) Three Ways an Enzyme Defect Can Produce Phenotypic Abnormality Lack of final product A ( B ( C --} (can?t convert to final product) Albinism Build up of excess intermediates behind blocked reaction step alkaptonuria Production of other substances via an alternative metabolic pathway PKU How did the pea get its wrinkles? Gene R ( starch branching enzyme that converts sugar to starch In rr seeds, starch synthesis is decreased (lacks enzyme) As a result, increased level of free sugar Higher water content of developing seeds ( increased cell volume Loss of larger proportion of cell volume upon seed maturation (drying out) Cotyledons shrink, but outer seed coat can?t shrink Wrinlking of seed coat Sickle Cell Anemia Autosomal recessive trait, due to mutation in gene encoding one of the polypeptides found in hemoglobin Hemoglobin: tetramer consisting of two alpha chains and two beta chains hundreds of genetic variations exist, affecting either alpha globin or beta globin gene. Most of the mutations result in single amino acid changes