A. Sex Determination 1.Environmental Sex Determination 2.Chromosomal Sex Determination a. Protenor sex determination b. Lygaeus sex determination c. Balanced sex determination d. Human sex determination: SRY gene A. Sex Determination 1.Environmental Sex Determination 2.Chromosomal Sex Determination a. Protenor sex determination b. Lygaeus sex determination c. Balanced sex determination d. Human sex determination: SRY gene The presence of the Y, regardless of the number of Xs, determines maleness Klinefelters MaleTurners Female A. Sex Determination 1.Environmental Sex Determination 2.Chromosomal Sex Determination a. Protenor sex determination b. Lygaeus sex determination c. Balanced sex determination d. Human sex determination: SRY gene SRY gene produces the protein called the testis determining factor, which stimulates the undifferentiated gonadal tissue to become a testis. It is probably a transcription factor that binds to other genes, stimulating their expression. A. Sex Determination 1.Environmental Sex Determination 2.Chromosomal Sex Determination a. Protenor sex determination b. Lygaeus sex determination c. Balanced sex determination d. Human sex determination: SRY gene XX X*X*X male Y-XY- female Human mutations demonstate it is the presence/absence of this gene, not the whole Y, that stimilates male development Insertion of homolog in mice also changes their sex. MALE: AAXY FEMALE: aa XX A XA Y a XAaXXAaXY a XAaXXAaXY MALE: aa XY FEMALE: AA XX a Xa Y A XAa XXAa XY A XAa XXAa XY A.Sex Determination B.Sex Linkage: Genes of interest are one of the sex chromosomes (X or Y) 1. For Comparison heredity for sex (as a trait) and an autosomal dominant trait. All offspring, regardless of sex, express the A trait in both reciprocal crosses MALE FEMALE XgXg Y XGXG XGXgXGXg XGYXGY XGXG XGXgXGXg XGYXGY MALE FEMALE XGXG Y XgXg XGXgXGXg XgYXgY XgXg XGXgXGXg XgYXgY A.Sex Determination B.Sex Linkage 1. For Comparison heredity for sex (as a trait) and an autosomal dominant trait. 2. Sex Linkage example: red-green coloblindness in humans 100% G, for all offspring50% G daughters, 50% g sons Now, the sex of the parent that expresses the G trait matters; the transmission of this gene correlates with the sex of the offspring, because this trait and sex are influenced by the same chromosome. A.Sex Determination B.Sex Linkage 1. For Comparison heredity for sex (as a trait) and an autosomal dominant trait. 2. Sex Linkage example: red-green coloblindness in humans Queen Victoria of England Her daughter Alice X-linked recessive traits are expressed in males more than females, because females get a second X that may carry the dominant allele. A.Sex Determination B.Sex Linkage C.Dosage Compensation - Females have two doses of X-linked genes, while males have one dose. Since protein concentration is often important in protein function, how is this imbalance corrected? A.Sex Determination B.Sex Linkage C.Dosage Compensation - Females have two doses of X-linked genes, while males have one dose. Since protein concentration is often important in protein function, how is this imbalance corrected? In human females, one X in each cell condenses. Barr Body A.Sex Determination B.Sex Linkage C.Dosage Compensation Actually, in all humans and mammals, all but one X condenses, regardless of sex or number of Xs. A.Sex Determination B.Sex Linkage C.Dosage Compensation Which X condenses is random. So, in heterozygous female cats (X O X o ), when the X with the gene for orange color condenses, the non-orange allele allows genes for other colors at other loci to be expressed (black, brown, blue). The X that is inactivated is determined randomly, early in development. This inactivation is imprinted on that X, such that descendants of those cell inactivate that X. White is due to another gene that influences melanocyte migration to skin surface, and also affects the size of patches from tortoiseshell to calico. A.Sex Determination B.Sex Linkage C.Dosage Compensation This happens in humans, too so that females are really a mosaic, with some cells in a tissue expressing one X (and its X-linked traits) and some cells in that tissue expressing the other X. Females heterozygous for red- green colorblindness have patches of retinal cells that cant distinguish red from green. Anhidrotic ectodermal dysplasia Females heterozygous for this X-linked condition have patches of skin that lack sweat glands A.Sex Determination B.Sex Linkage C.Dosage Compensation How? - each X has a gene the Xic (X-inactivation center). - this is on in inactivated Xs it produces an RNA (Xist) that binds with the chromosomes, making it inaccessible to transcription enzymes. - this RNA is NOT translated it is functional as an RNA molecule. - of course, this just pushes the question one step upstream what determines why Xic is only active in one X chromosome? HP1 = heterochromatic protein 1 A.Sex Determination B.Sex Linkage C.Dosage Compensation How? - each X has a gene the Xic (X-inactivation center). - this is on in inactivated Xs it produces an RNA that binds with the X chromosomes, making it inaccessible to transcription enzymes. - this RNA is NOT translated it is functional as an RNA molecule. - of course, this just pushes the question one step upstream what determines why Xic is only active in one X chromosome? When? - It seems to be an imprinted phenomenon, so that daughter cells have the same X inactivated. However, this seems to happen at different points in development for different tissues. I. Allelic, Genic, and Environmental Interactions II. Sex Determination and Sex Linkage III. Linkage - Overview: Linkage is a pattern of correlated inheritance between traits governed by genes on the same chromosome. Because the genes are part of the same physical entity, they are inherited together rather than independently. AB ab LINKED A a B b A A B B b b a a INDEPENDENT ASSORTMENT (IA) III. Linkage - Overview: Linkage is a pattern of correlated inheritance between traits governed by genes on the same chromosome. Because the genes are part of the same physical entity, they are inherited together rather than independently. Only crossing-over can cause them to be inherited in new combinations. Cross-over products III. Linkage A. Complete Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are always inherited together. The pattern mimics that of a single gene. X AABBaabb AB ab III. Linkage A. Complete Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are always inherited together. The pattern mimics that of a single gene. X AABBaabb AB ab ABab Gametes AB ab F1 Double Heterozygote in F1; no difference in phenotypic ratios compared to IA III. Linkage A. Complete Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are always inherited together. The pattern mimics that of a single gene. X ABab Gametes AB ab AB F1 x F1 ABab III. Linkage A. Complete Linkage - if genes are immediate neighbors, they are almost never separated by crossing over and are always inherited together. The pattern mimics that of a single gene. X ABab Gametes AB ab AB F1 x F1 ABab AABBAaBb aabb 3:1 ratio A:a 3:1 ratio B:b 3:1 ratio AB:ab III. Linkage A. Complete Linkage B. Incomplete Linkage III. Linkage A. Complete Linkage B. Incomplete Linkage - Crossing over in a region is rare III. Linkage A. Complete Linkage B. Incomplete Linkage - Crossing over in a region is rare - Crossing over events increase as the distance between genes increases ABC MORE LIKELY IN HERE LESS LIKELY IN HERE a bc III. Linkage A. Complete Linkage B. Incomplete Linkage - Crossing over in a region is rare - Crossing over events increase as the distance between genes increases - So, the frequency of crossing over (CO) gametes can be used as an index of distance between genes! (Thus, genes can be mapped through crosses) FEWER CO GAMETES: Ab, aB ABC a bc MORE CO GAMETES: bC, Bc III. Linkage A. Complete Linkage B. Incomplete Linkage - Crossing over in a region is rare - Crossing over events increase as the distance between genes increases - So, the frequency of crossing over (CO) gametes can be used as an index of distance between genes! (Thus, genes can be mapped through crosses) - How can we measure the frequency of recombinant (cross-over) gametes? Is there a type of cross where we can see the frequency of different types of gametes produced by the heterozygote as they are expressed as the phenotypes of the offspring? III. Linkage A. Complete Linkage B. Incomplete Linkage B ab A b ab a TEST CROSS !!! III. Linkage A. Complete Linkage B. Incomplete Linkage - So, since crossing- over is rare (in a particular region), most of the time it WONT occur and the homologous chromosomes will be passed to gametes with these genes in their original combinationthese gametes are the parental types and they should be the most common types of gametes produced. B ab b ab a AB ab A TEST CROSS !!! III. Linkage A. Complete Linkage B. Incomplete Linkage - Sometimes, crossing over WILL occur between these loci creating new combinations of genes This produces the recombinant types B ab b ab a AB ab A aB Ab TEST CROSS !!! III. Linkage A. Complete Linkage B. Incomplete Linkage As the other parent only contributed recessive alleles, the phenotype of the offspring is determined by the gamete received from the heterozygote B ab b ab a AB ab A aB Ab gamete genotype phenotype abaabbab AaBbAB abaaBbaB abAabbAb TEST CROSS !!! III. Linkage A. Complete Linkage B. Incomplete Linkage B ab b ab a AB ab A aB Ab gamete genotype phenotype abaabbab AaBbAB abaaBbaB abAabbAb TEST CROSS !!! ALTERNATIVES IALINKAGE LOTS of PARENTALS FEWER COS FREQUENCIES EQUAL TO PRODUCT OF INDEPENDENT PROBABILITIES III. Linkage A. Complete Linkage B. Incomplete Linkage Example: -How do we discriminate between these two alternatives? -Conduct a Chi-Square Test of Independence Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb III. Linkage A. Complete Linkage B. Incomplete Linkage Example: -How do we discriminate between these two alternatives? -Conduct a Chi-Square Test of Independence - Compare the observed results with what you would expect if the genes assorted independently Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb III. Linkage A. Complete Linkage B. Incomplete Linkage Example: -How do we discriminate between these two alternatives? -Conduct a Chi-Square Test of Independence - Compare the observed results with what you would expect if the genes assorted independently Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb The frequency of AB should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28 III. Linkage A. Complete Linkage B. Incomplete Linkage Example: -How do we discriminate between these two alternatives? -Conduct a Chi-Square Test of Independence - Compare the observed results with what you would expect if the genes assorted independently Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb The frequency of AB should = f(A) x f(B) x N = 55/100 x 51/100 x 100 = 28 The frequency of Ab should = f(A) x f(B) x N = 55/100 x 49/100 x 100 = 27 The frequency of aB should = f(a) x f(B) x N = 45/100 x 51/100 x 100 = 23 The frequency of ab should = f(a) x f(b) x N = 45/100 x 49/100 x 100 = 22 III. Linkage A. Complete Linkage B. Incomplete Linkage Example: -How do we discriminate between these two alternatives? -Conduct a Chi-Square Test of Independence This is fairly easy to do by creating a contingency table: Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BbRow Total A4312 a837 Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage Example: -How do we discriminate between these two alternatives? -Conduct a Chi-Square Test of Independence This is fairly easy to do by creating a contingency table: Add across and down This gives the totals for each trait independently. Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BbRow Total A a83745 Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage Example: -How do we discriminate between these two alternatives? -Conduct a Chi-Square Test of Independence This is fairly easy to do by creating a contingency table: Then, to calculate an expected value based on independent assortment (for AB, for example), you multiple Row Total x Column Total and divide by Grand Total. 55 x 51 / 100 = 28 Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BExp.bRow Total A a83745 Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage Example: Repeat to calculate the other expected values (This is just an easy way to set it up and do the calculations, but you should appreciate it is the same as the product rule: F(A) x f(B) x N Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BExp.b Row Total A a Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage Compare our observed results with what we would expect if the genes assort independently. Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BExp.b Row Total A a Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage Compare our observed results with what we would expect if the genes assort independently.. If our results are close to the expectations, then they support the hypothesis of independence. Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BExp.b Row Total A a Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage Compare our observed results with what we would expect if the genes assort independently.. If our results are close to the expectations, then they support the hypothesis of independence. If they are far apart from the expected results, then they refute that hypothesis and support the alternative: linkage. Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BExp.b Row Total A a Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage Typically, we reject the hypothesis of independent assortment (and accept the hypothesis of linkage) if our observed results are so different from expectations that independently assorting genes would only produce results as unusual as ours less than 5% of the time Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb BExp.b Row Total A a Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage We determine this probability with a Chi-Square Test of Independence. Phenotype ObsExp(o-e)(o-e) 2 /e AB Ab aB Ab X 2 =36.38 BExp.b Row Total A a Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage Our X 2 = First, we determine the degrees of freedom = (r-1)(c-1) = 1 BbRow Total A a83745 Col. Total III. Linkage A. Complete Linkage B. Incomplete Linkage Our X 2 = First, we determine the degrees of freedom = (r-1)(c-1) = 1 Now, we read across the first row in the table, corresponding to df = 1. The column headings are the probability that a number in that column would occur at a given df. In our case, it is the probability that our hypothesis of independent assortment (expected values are based on that hypothesis) is true. III. Linkage A. Complete Linkage B. Incomplete Linkage Our X 2 = Note that larger values have a lower probability of occurring by chance This should make sense, and the value increases as the difference between observed and expected values increases. III. Linkage A. Complete Linkage B. Incomplete Linkage Our X 2 = So, for instance, a value of 2.71 will occur by chance 10% of the time. III. Linkage A. Complete Linkage B. Incomplete Linkage Our X 2 = So, for instance, a value of 2.71 will occur by chance 10% of the time. But a value of 6.63 will only occur 1% of the time... (if the hypothesis is true and this deviation between observed and expected values is only due to chance). III. Linkage A. Complete Linkage B. Incomplete Linkage Our X 2 = So, for instance, a value of 2.71 will occur by chance 10% of the time. For us, we are interested in the 5% level. The table value is Our calculated value is much greater than this; so the chance that independently assorting genes would yield our results is WAY LESS THAN 5%. Our results are REALLY UNUSUAL for independently assorting genes. III. Linkage A. Complete Linkage B. Incomplete Linkage Our X 2 = Our results are REALLY UNUSUAL for independently assorting genes. So, either our results are wrong, or the hypothesis of independent assortment is wrong. If you did a good experiment, then you should have confidence in your results; reject the hypothesis of IA and conclude the alternative the genes are LINKED. III. Linkage A. Complete Linkage B. Incomplete Linkage OK so we conclude the genes are linked NOW WHAT? Offspring Number AB43 Ab12 aB 8 ab37 AaBb x aabb