Genetics Wk 2 Notes
Genetics Wk 2 Notes Bisc 336
Popular in Genetics
verified elite notetaker
Popular in Biology
This 17 page Class Notes was uploaded by Anna Ballard on Wednesday September 7, 2016. The Class Notes belongs to Bisc 336 at University of Mississippi taught by Ryan Garrick in Fall 2016. Since its upload, it has received 92 views. For similar materials see Genetics in Biology at University of Mississippi.
Reviews for Genetics Wk 2 Notes
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 09/07/16
Lecture 4 8/29 Spermatogenesis general procedure of meiosis Male gamete production: • starts with undifferentiated diploid germ cell (spermatogonium) • becomes an enlarged primary spermatocyte, which undergoes 1st (reductional) meiotic division spermatogonium is smaller than primary spermatocyte and appears earlier in the process • secondary spermatocytes undergo 2nd meiotic division • each make 2 haploid spermatids that develop into motile sperm = total of 4 haploid sperm Oogenesis Female gamete production: • Starts from oogonium, which enlarges to become the primary oocyte. The primary oocyte then splits during Meiosis I. • Daughter cells of primary oocyte receive equal genetic material, but unequal cytoplasm (cellular volume) • At each division, most cytoplasm goes to just one of two daughter cells unequal split of cytoplasm.. the one who gets the least does not continue through process of oogenesis • The “chosen one” is called either the secondary oocyte (after 1st division) or ootid (after 2nd division/Meiosis II) the one that misses out on cytoplasm after 1st division = 1st polar body and does not continue the one that misses out on cytoplasm after 2nd division is called 2nd polar body • oocytes that get shorted cytoplasm don’t become eggs • Just one haploid ovum (egg) produced per meiotic cycle Sperm v. Egg Production • Initially start off similar diploid parent cell undergoes growth and maturation • End of first division: first polar body in oogenesis 2 daughter cells as secondary spermatocytes in spermatogenesis • A single ovum produced at end of oogenesis vs. 4 haploid sperm at end of spermatogenesis haploid gametes produced are genetically different from the parent cell Think about the interrelationships in different levels of biological info how theres a scaling of genetic variation operating at different levels within a species Molecules– DNA molecules • Double stranded DNA helix composed of complementary base pairing (AT, GC) • increasing levels of compaction • DNA wound around histone proteins to make nucleosomes • Continued packing –> chromosomes Cell Cycle ….from the end of cell division, to beginning of the next transcription and translation is occurring G0, G1, and G2 during S phase –> DNA replication Regular function: • G1, Gap • S, Synthesis • G2, Gap Mitotic Division: • Prophase • Metaphase • Anaphase • Telophase The “Central Dogma” Mutation Replication DNA –> (transcription) –> RNA DNA can be altered in way of mutation–affects RNA which in turn affects the protein produced RNA –> (translation) –> Protein think about in context of sickle cell anemia different allelic variants (betaglobin) that results in a different mRNA strand to be translated Spontaneous DNA Mutations usually caused by (rare) errors during DNA replication, at low, ongoing/constant “background” rates (slow ticking clock) • Most mutations are “neutral” (in noncoding DNA, no phenotypic effect) • Most mutations in coding regions are detrimental (selective disadvantage)…only tiny portion are beneficial – yet still underpin adaptation most common outcome: production of an allelic varian that does not function as well as the original DNA –> why its detrimental can’t live without mutations due to adaptations to natural environment – at individual level mutations aren’t so good Mutations in ProteinCoding genes • missense: change in a codon and in its amino acid –> changes amino acid/protein • nonsense: change in a codon that creates premature STOP, protein synthesis terminated process cut short Reshuffling Existing Variation Crossing over between parts of homologous chromes early in meiosis • creates “mosaiclike” chromosomes • crossover breakpoint within or between a gene could create new allelic variants that did not previously exist independent (random) assortment of whole chromosomes into gametes • ultimate outcome of meiosis • gametes: NOT haploid clones of the diploid parent Genotype to phenotype • A protein’s biochemical or structural behavior in a cell contributes to a phenotype • sicklecell anemia: a mutant gene produces a mutant protein, that alters phenotype • A single nucleotide change (i.e., DNA substitution) leads to a different amino acid, and beta globin protein human hemoglobin variation and sicklecell anemia • single DNA substitution –> different AA and beta globin protein Genotype to phenotype to fitness • Profound cascade of effects….all are detrimental to individual fitness Population “gene pool” • population – group of individuals living together that have capacity to interbreed and also can compete for limited resources • as a group – many allelic variants for a gene An abstract idea (it does not literally work like this for most diploid organisms) gene pool – sum total of number of allelic variants of a gene that exist within a particular group of individuals all the genetic information for the population, present in the haploid state as if separated into individual alleles and thrown into a bucket E.g. some marine organisms mate in a “swarm” all individuals, male and female, release sperm/ova into ocean at same time Selection can differ among populations in parts of the world where malaria is prevalent… allele A good at O2 but bad at malaria, allele S bad at O2 but good at malaria (80% fatal as homozygote) AS heterozygotes give cells that are not sickled but are resistant to malaria…fitter than AA or SS Lecture 5 8/31 Learning Goals • Understand Mendel’s 4 postulates • Use a Punnett square for mono and dihybrid crosses • Extend principals of independent assortment to threetrait crosses, using the forked line method • Perform X^2 analyses, determine Pvalues and interpret them with respect to a null hypothesis Mendelian Genetics –> Ch. 3 Gregor Mendel Despite having no knowledge of chromosomes or meiosis…. • determined that units of inheritance exist by observing similarities in phenotype from one generation to the next • could predict their behavior (i.e., correctly anticipated the results of particular crosses) • experimented with pea plants while working in a monastery garden 7 traits he observed were encoded in same gene • great model organism : easy to cross, fastgrowing, have observable traits (e.g., flower color) Mendel’s experiments • Followed seven discrete characters (= easily scored) flower color, position of flower, seed color, seed shape, pod shape, pod color, and stem length • Each character had two alternative forms (= simple) • Also had “truebreeding” lines for each character (i.e., homozygous inbred strains…traits unchanged Simplest: Monohybrid Cross • Mating between individuals that differ in only one trait • Cross 2 truebreeding parental (P1 generation) • Examine traits of offspring (F1 generation), and then selfed F1’s (F2 generation) Monohybrid Phenotypic Ratio • P1 = parentals, truebreeding green, or yellow • F1 = 1st generation offspring… here, 100% yellow • F2 = selffertilized F1’s… here, 75% yellow to 25% green, or 3:1 ratio 3:1 Ratio for other traits in F2s (FIG 31) The Traits Are Not SexDependent • Reciprocal crosses were performed (e.g., trials using both the tall and short plant as the pollen donor • Outcomes did not change (i.e., all F1s had the same trait, but F2s show a 3:1 ratio of two alternate traits) NOTE: indicates that these genes are on autosomes, not sexchromosomes (e.g., X or Y in humans) • P1 = parentals, truebreeding (homozygotes, RR & rr) • F1 = 1st generation offspring (the same heterozygote, Rr) • F2 = selffertilized F1s combo of homozygotes (RR, rr) and heterozygotes (Rr) due to segregation REMEMBER: R is dominant to r Proposed existence of ‘alleles’ things that pass unchanged from one generation to the next, and determine the phenotype that is expressed 1. Alleles of a gene: occur in pairs, in an individual Note: applies only to diploid organisms, including peas 2. Dominance/Recessive: one phenotype dominates Note: we will see later, this is not true of all genes Note: its actually chromosomes segregating, on which alleles reside gametes randomly Punnett Squares • Graphical method to predict tP1 potential genotypes and phenotypes from a given cross phenotypesd to calculate expected frequenDies (ratDos) of all the potential genotypes and • Each possible haploid gamete assigned a column (from female parent), or row (from male parent) Punnett Squares (P1 Cross) • Alleles – D = tall; d = dwarf d Dd Dd • All F1 offspring expected to be heterozygotes (Dd), and tall Punnett Squares (F1 Cross Below)d Dd Dd • Genotypes – DD and Dd = tall; dd = dwarf • F2 offspring mixed •F12 offspring mixed: phenotypic ratio = 3:11 C• Genotype determines phenotype! D DD Dd d Dd dd Dihybrid Cross • Simultaneously follow inheritance of 2 phenotypic characters • Generates an expected phenotypic ratio 9:3:3:1 yellow, round X green, wrinkled F1 All yellow, round F1 X F1 yellow, round X yellow round 9/16 yellow, round 3/16 green, round 3/16 yellow, wrinkled 1/16 green, wrinkled • As in monohybrid cross, more genotypes than phenotypes • Here, 9 genotypes underlie the 4 phenotypes GGWW, GGWw, GgWW, GgWw, GGww, Ggww, ggWW, ggWw, ggww • 9:3:3:1 ratio requires that 2 traits are fully independent • fig 37 Forked Line Method (Phenotypes) round (3/4) yellow, round 3/4 X 3/4 = 0.57 (9/16) yellow 3/4 wrinkled (1/4) yellow, wrinkled 3/4 X 1/4 = 0.19 (3/16) round (3/4) green, round 1/4 X 3/4 = 0.19 (3/16) green 1/4 wrinkled (1/4) green, wrinkled 1/4 X 1/4 = 0.06 (1/16) Mendel’s 4th Postulate • Independent Assortment : different phenotype traits are inherited independently of one another • The 2 alleles of one gene have an equal chance of segregating with either 2 alleles at another gene • Need a min. of ______2 different genes, each with 2 alleles__ What about a trihybrid cross? • identical processes of segregation and independent assortment apply to 3 (or more) traits • Punnet Sqare with 64 boxes (crikey!), or forked line method (http://www.youtube.com/watch? v=nBTX2h2GYYw) • Based on laws of probability and relationships between expected phenotypic ratios Expectations v. Observations • We can calculate expected phenotypic ratios from Mendel’s principals (segregation, independent assortment ) • …and we can make observations of phenotypic ratios from experimental crosses • Next, we need a way to assess the “fit” between observed v. expected ratios (i.e, equivalent or not?) Lecture 6 9/2 Learning Goals At the end of this lecture you should be able to: • Extend principles of independent assortment to threetrait crosses, using the forked line method • Perform X^2 analyses, determine Pvalues and interpret them with respect to null hypothesis • Use pedigrees to infer mode of inheritance of a trait Recap: Mendel’s Postulates 1 Alleles of a gene occur in pairs, in an individual 4. dominant/recessive: one phenotype dominates Heterozygous 5. random segregation: during meiosis, alleles of a gene separate and go to different gametes randomly segregate into haploid gametes randomly pull apart homologs chromosomes among that plate chromosome is main unit that is moving around since alleles are on chromosomes, we can apply this 6. independent assortment: different phenotypic traits are inherited independently of one another * First 3: monohybrid crosses, following a single phenotypic trait * 4 is special: rule or inference only possible following experiments that follow the inheritance of 2 different phenotypes separately pod color and pod shape 2 different characteristics are passed on independent of one another each phenotype is controlled by a different gene, so maybe one gene is completely on a different chromosome than the other What about a Trihybrid Cross? 3 or more phenotypic traits = forked line method (phenotypes) Expectations v. Observations • large data sets – thousands of observations • How do we assess the fit between expectations from underlying rules of inheritance and observations we make in our own experiments? • Expected: 3:1 ratio in monohybrid cross in F2 • We can do a lot of crosses and make observations and get raw data on observed data of dominant v. recessive phenotypes • Need a way to figure out how far • To assess the fit we need to use statistics and a particular probability Chance Deviations and Sample Size • Go to the casino, odds are stacked in favor of the house go once and win money – think casino is losing money go a couple more times – you are actually losing money • Small Sample Size what proportion of students are enrolled in a particular kind of degree? sample size of 10 people – 40% are BS in biology Other row: sample size of 1030% are BS in biology 26% are BS in biology ** larger sample sizes converge on true answer better than small sample sizes • Small sample size, chance deviation from the “truth” decreases • We become less prone to __________ • …how to know if your sample size is large enough? if we want to assess whether mendelian rules of inheritance are true, we need a sufficient number of observations to really assess that Testing Mendel’s Principles • Assuming dominance/recessive, we can test for random segregation and independent assortment • We can test whether the proportion of dominant recessive phenotypes matches the Mendelian rules • Hypothesis = null hypothesis – the proportions of dominant and recessive phenotypes that we see show no difference that are predicted by Mendelian rules. • Sample size is massive and we still see a difference in expected and observed, we need to see of Mendelian principles hold • Need to do some crosses and score the phenotypes; need large sample size if expected ratio is complex • 3:1 easier to deal with than 9:3:3:1 (2 vs. 4 classes) • Monohybrid cross – 2 phenotypes • Dihybrid cross – 4 different phenotypes • Greater complexity of the problem, we need larger sample sizes a dihybrid cross needs a larger sample size than monohybrid because there are more combinations of phenotypes chisquare takes care of these components ChiSquare (X^2) Test • Sample size – number of observations we make 2 categories: round or wrinkled make a thousand observations – we expect a 3:1 phenotypic ratio in the F2 generation 750 F2 should have round, 250 should have wrinkled Actually did the crosses and then counted phenotypes: 740 are round and 260 are wrinkled • What is the difference between observed and expected? 10 for both phenotypic categories • Equation that would be applicable in the case of 2 different phenotypic categories and want to see if there is a meaningful difference between observed and expected • Used to statistically determine whether observed results deviate significantly from expectations • Considers sample size in terms of number of observations made, and the number of classes being compared to account for complexity, use a summary statistic Worked example: monohybrid cross X = ∑ ((oe)^2/e) where O = observed and E = expected • Number of classes (n) = 2 (i.e., round v. wrinkled) ….therefore degrees of freedom = 1 (d.f. = n1) • When the probability (P) value is less than 0.05 we reject the null hypothesis ….so we need a way to determine the P value when X^2 = 0.53 and d.f = 1 • Amount of deviation between observed and expected ratios = 0.53 • Null hypothesis – mendelian principles are good and accurately predict phenotypic ratio in F2 generation • To assess how big the departure has to be, we use a Pvalue • “Fail to reject” means mendelian hypothesis is good • Reject null hypothesis – departure between O and E is so extreme we have to find other explanations When P < 0.05 it means: …..there is less than 5% chance of obtaining the observed data when H0 is true mendelian rules = true Using a X^2 test “look up graph” • Chi square of 0.53 locate area on X axis where 0.53 occurs • Df = 1 diagonal lines represent degrees of freedom values • Find 0.53 on xaxis, draw line up to df 1 diagonal line • Cut across to y axis; provides estimate of pvalue FIG 310 Using a X^2 test “look up table” • 0.53 and df = 1 • First identify row that corresponds to df that we care about • These values in body of the table, need to find what cells encapsulate our chisquare value • Chisquare = 11.78; df = 3 0.01 Hypothesistesting • For a X^2 test, the null hypothesis (H0) is ‘no real difference’ between _________ • Can only ever “reject” H0 (it is beyond chance), or “fail to reject” H0 (i.e., is attributable to chance rejected – does terrible job fail to reject – retain it, because data against it are not overwhelming • P < 0.05 means there is less than 5% chance of obtaining the observed data when H0 is true start to think that null hypothesis has problems Reasons for Rejecting Null Hypothesis (P<0.05) • Made 1000 observations and concluded that Mendelian principles do not predict… One or more underlying assumptions do not hold: • …Mendel’s ideas: (1) dominance/recessiveness, (2) segregation, and (3) independent assortment does not tell us what way is wrong, but something is wrong • Assume that all phenotypes are equally viable and fit – may not be the case phenotypic characteristic that has a negative effect on an individuals fitness • …Other things: diploid autosomal loci, all crosses yield equally viable offspring, no scoring error Pedigrees • Mode of inheritance – is the phenotype of interest dominant or recessive? Is it an autosomal or Xlinked chromosome? recorded histories of family trees circles are females, squares are males, diamond for unknown sex shade shape if individual is expressing a phenotype – albino parents are unrelated – single line related parents – double line numbers – birth order • Without experimental crosses (e.g. in humans), we can still determine the ____________ • A set of standardized visual representations Pedigree Analysis • When only one parent is affected, autosomal recessive traits seem to _________ example: albinism • Trial and error: does autosomal recessive work? • Even when only one parent is affected, autosomal dominant traits seem to “reappear” every generation ex: huntington • Trial and error: does autosomal dominant work? • Key point: usually does not provide same level of certainty as carefully designed _______ • Larger pedigrees: (i.e., many generations back, many kids), allow better inferences made • Multiple pedigrees: (i.e., different family groups with affected members) ________ • Autosomal recessive – phenotype of interest disappears for a generation then reappears later • We can use deductive reasoning to figure out pattern of inheritance and see if it follows an autosomal recessive inheritance pattern • try to fit in genotypes • all children must have little a allele, but since not albino, have to have one dominant allele too • some of the children of this male and this female – only way that can happen is if that female is carrying a little “a” allele • trail & error for pedigree analysis • could we rule out autosomal dominant? ♣ No way man… this don’t work Albinism is inherited by way of a recessive mode of transmission Lecture 7 9/7 Learning Goals • At the end of this lecture you should be able to: use pedigrees to infer mode of inheritance of a trait describe some singlelocus modifiers of Mendelian ratios and give examples of each contrast autosomal vs. sexlinked traits Mendelian Genetics Ch. 3 • critical value – after determining which row, what is the chi square value that is on the edge? the one at the 0.05 Pedigree Analysis • Even when only one parent is affected, autosomal dominant traits seem to “reappear” every generation think about what the underlying genotype is ex: Huntington Trial and error: does autosomal dominant work? (FIG 312) Pedigree Analysis • Key Point: usually does not provide same level of certainty as carefully designed experimental crosses • Larger pedigrees (i.e., many gens back, many kids), allow better inferences to be made think of : large pedigree that traverses multiple generations or one that includes very many offspring in one generation greater confidence in mode of inheritance • Multiple pedigrees (i.e., different family groups with affected members) permits verification Modification of Mendelian Ratios Ch. 4 After Mendel Research started to focus on traits that do not follow the simple Mendelian model… rejecting null hypothesis even though it was correct • Although the basic mechanism of inheritance is the same, other assumptions did not always hold: some traits are influenced by >1 gene ….influenced by the environment …not dominant/recessive when occurring in heterozygous form • Wildtype – the (often dominant) allele carried by the most common phenotype in a population common one that we find in a population usually dominant • Mutation is the source of new genetic variants (mutant alleles) rare More Genotype Notation… • Instead of indicating dominance only (e.g., D or d), additionally indicate wildtype allele e.g., as D+ don’t always represent with one allele • Can also use more than just a single character to represent an allele (e.g., the Wr allele) • The two allele designations in a diploid genotype can be separated by a slash (e.g., Wr+/Wg) 1. SingleLocus Phenomena singlelocus/gene one gene is controlling a specific trait departures from mendelian genotypic ratios Incomplete Dominance • Offspring’s phenotype is an intermediate blend of parents • In a monohybrid cross, each of 3 genotypes (hom, het, hom) yields a different phenotype • Expected phenotypic and genotypic ratios = 1:2:1 FIG 41 perfect correspondents between parents and offspring CoDominance Offspring’s phenotype different from the 2 parents, but not blended • Joint expression of both alleles in heterozygotes seen in phenotypes a mix of both • Monohybrid cross: expected phenotypic & genotypic ratio = 1:2:1 Multiple (>2) Alleles • Strictly a populationlevel phenomenon (i.e., max. of 2 different alleles per locus in a a diploid individual) • Human ABO blood group: controlled by a single autosomal locus with 3 alleles (IA, IB, IO) • 3 alleles = 6 possible diploid genotypes • These 6 genotypes yield 4 alternative phenotypes (A&B dominant to O, but codominant to each other) AO or AA – A is expressed AB – both expressed codominant to one another BO or BB – B is expressed OO – neither A nor B expressed • Drosophila eye color: > 100 alleles at this locus in fruit fly populations • MHC : highly polymorphic in vertebrates… up to 560 alleles/locus in human populations multiple alleles at a gene in a population gene pool is common more complex than Mendelian’s rules Recessive Lethal Alleles • Selective disadvantage (or advantage) of a given allele causes deviation from Mendelian ratios • Some deleterious mutations can be tolerated by heterozygotes, but not homozygotes • Recessive Lethal: individuals that are homozygous for the recessive allele have a short life expectancy sometimes a single copy of the recessive allele is enough to cause problems Agouti has complex inheritance • This locus affects two traits (hair color and embryonic development); dominance/recessiveness switches • The yellow allele (AY) is a lethal recessive (AY/AY is never seen – embryos die before birth) • homozygotes with that genotype die very young • AY is dominant with respect to coat color (AY/A survives but has an unusual yellow coat) • The dual action of this locus on survival and coat color cause predictable deviations form Mendelian ratios • we know from a series of crosses that if we cross a pair of agouti mice (AA X AA), all offspring will show the agouti color, and therefore they all survive. • yellow (AAY) X yellow (AAY) = 2/3 yellow and 1/3 agouti, the survivors (the other 1/3 die due to detrimental homozygote genotype so you do not account for these in ratios) • agouti (AA) X yellow (AAY) = ½ agouti and ½ yellow (all survive) Dominant Lethal Alleles • Huntington Disease: degeneration of nerve cells in the brain, caused by a run of “CAG” repeat, chromosome 4. • Usually late onset (around 40 years) in heterozygotes • alleles can persist in a population when carrier individuals survive long enough to reproduce. why it has not been weeded out of the population gene pool homozygous die early heterozygous have later onset so they still produce the offspring natural selection cannot fully remove this gene from the gene pool Genes on Sex Chromosomes • Xlinkage: compared to autosomal genes, those on the Xchromosome have unique inheritance • Heterogametic sex only has one copy (e.g. XY males, in mammals) • ___________ do not give the same results Xlinkage and Drosophila White eyes SexLimited Traits • Autosomal genes that act differently in sexes • e.g., feathers of male and female chickens, where a certain phenotype found only in one of the sexes • Hormones determine expression SexInfluenced Traits • Autosomal genes that act differently in sexes, but the phenotype is not limited to only one of the sexes ______________ • Mitochondria and Chloroplasts have their own proteincoding genes, so can affect phenotype • Since they are haploid and uniparentallyinherited, do not follow basic Mendelian Principles Summary (so far…) • Singlelocus phenomena: incomplete dominance, codominance, multiple alleles, lethal alleles, others • These inheritance modes modify the monohybrid cross 3:1 phenotypic ratio • By extension, these modes modify the dihybrid cross 9:3:3:1 phenotypic ratio Lecture 8 9/9 Physical Linkage A form of non Two genes that occur closely on the same chromosome will have alleles hthat exhibit linkage because they tend to be inherited as a single unit Probability of crossing over to separate the 2 is possible but not very likely LocusByLocus Interactions • Many phenotypic traits are controlled by more than one gene • interactions among genes and their products can be direct: In some cases, alleles at one locus mask the expression of alleles at another (epistasis) hypostatic alleles are masked by an epistatic locus Epistasis: The Bombay Phenotype • The single locus that determines human ABO blood type can be impacted by a second locus, FUT1 the recessive allele at FUT1 causes individuals that are AA, BB, or AB to express the O blood phenotype ABO blood group are the hypostatic alleles and that locus is the epistatic locus Ch. 5 –> Sex Determination and Sex Chromosomes Sexual Differentiation • sex chromosomes are the location of many (but not all) sex determining genes • mechanisms of sexdetermination are highly variable among eukaryotes (not just ‘XY’ vs. ‘XX’) • differences between males (♂) and females (♀) (sexual dimorphism) are separated into 2 classes: 1. Primary: gonads only 2. Secondary: other traits related to gender Homo vs. Heterogametic • the heterogametic (XY) has a pair of dissimilar sex chromosomes (vs. homogametic (XX)) • human males make nonuniform gametes (some with X, others with Y) that determine sex of offspring • nondisjunction: failure of the X chromosomes to segregate properly during meiosis Q: What (usually) determines the sex of a human embryo? A: presence v. absence of the Y chromosome pseudoautosomal regions: paring and synapse/ recombination; rest of the Y (malespecific region) has no counterparts on the X • euchromatin: (where other 74 genes are) contains genes, incl. sexdetermining region Y (SRY)* present in all (male) mammals SRY codes for the testisdetermining factor (TDF) protein, a ‘switch’ for expression of other genes SRY has a large effect by either facilitating or preventing transcription or translation of other genes Dosage Compensation • females have the potential to produce 2x as much protein encoded by Xlinked genes than males • Dosage compensation mechanisms balance out the expression of Xlinked genes Barr body : inactivated X in diploid cells of females, visible during mitotic interphase • Mechanism in silencing the other X chromosome is usually in methylation that X chromosome is never going to express any of its genes this way, each gender has an equal amount of genes Early on both chromosomes contribute to gene expression until one is silenced Lyon Hypothesis • one X chromosome per somatic cell is randomly inactivated early during embryonic development • Descendent cells have the same X inactivated as their parent cell, but not all parent cells are the same Random inactivation creates mosaic in that one allelic variant is expressed in one cell and a different one in another cell Temperature Dependent sex Determination • Not like that across tree of life – sex may not even be genetically determined Most common in reptiles that incubation temp sets the path whether individual will be male or female
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'