F2 Pink FemalePink MaleBlue FemaleBlue Male Gametes or
F2 Pink FemalePink MaleBlue FemaleBlue Male 11 11 1 : 1 Female to Male 1 : 1 Pink to Blue
Sex Linkage to Ponder Female is homozygous recessive X-linked gene, –what percentage of male offspring will express? –what percentage of female offspring will express if, mate is hemizygous for the recessive allele? mate is hemizygous for the dominant allele? Repeat at home with female heterozygous X- linked gene!
Sex-Linked vs. Autosomal autosomal chromosome: non-sex linked chromosome, autosomal gene: a gene on an autosomal chromosome, autosomes segregate identically in reciprocal crosses.
X-Linked Recessive Traits Characteristics Many more males than females show the phenotype, –female must have both parents carrying the allele, –male only needs a mother with the allele, Very few (or none) of the offspring of affected males show the disorder, –all of his daughters are carriers, roughly half of the sons born to these daughters are carriers.
X-Linked Dominant Affected males married to unaffected females pass the phenotype to their daughters, but not to their sons, Heterozygous females married to unaffected males pass the phenotype to half their sons and daughters, Homozygous dominant females pass the phenotype on to all their sons and daughters.
Linkage Genes linked on the same chromosome may segregate together.
A b 2n = 4 Independent Assortment a A A B B b a B a b
Meiosis No Cross Over Parent Cell Daughter Cells Have Parental Chromosomes Aa Bb A B a b a b A B 2n = 1
Meiosis With Cross Over Parent Cell Daughter Cells Have Recombinant Chromosomes Aa Bb A B A b a b a B 2n = 1
Dihybrid Cross yellow/round green/wrinkled GGWW x ggww GW gw GgWw phenotype genotype gametes genotype P F1
Gamate Formation in F1 Dihybrids P: GGWW x ggww, Independent Assortment G g W w GWGwgWgw alleles gametes probability.25 F1 Genotype: GgWw
How do you test for assortment of alleles? GWGwgWgw.25 F1: GgWw Test Cross: phenotypes of the offspring indicate the genotype of the gametes produced by the parent in question.
Test Cross GgWw x ggww GW (.25) Gw (.25) gW (.25) gw (.25) G gww (.25) GgWw (.25) ggWw (.25) ggww (.25) gw (1) x x x x
Test Cross GgWw x ggww GW (.25) Gw (.25) gW (.25) gw (.25) gw (1) Ggww (.25) GgWw (.25) ggWw (.25) ggww (.25) P P F1 parental types GgWw and gwgw R R recombinant types Ggww and ggWw x x x x
Recombination Frequency …or Linkage Ratio: the percentage of recombinant types, –if 50%, then the genes are not linked, –if less than 50%, then linkage is observed.
Linkage Genes closely located on the same chromosome do not recombine, –unless crossing over occurs, The recombination frequency gives an estimate of the distance between the genes.
Recombination Frequencies Genes that are adjacent have a recombination frequency near 0%, Genes that are very far apart on a chromosome have a recombination frequency of 50%, The relative distance between linked genes influences the amount of recombination observed.
AB In this example, there is a 2/10 chance of recombination. ab AC In this example, there is a 4/10 chance of recombination. ac homologs
Linkage Ratio P GGWW x ggww Testcross F1: GgWw x ggww # recombinant # total progeny GWGwgWgw ???? x 100 = Linkage Ratio Units: % = mu (map units) - or - % = cm (centimorgan) determine
Fly Crosses (simple 3-point mapping) (white eyes, minature, yellow body) In a white eyes x miniature cross, 900 of the 2,441 progeny were recombinant, yielding a map distance of 36.9 mu, In a separate white eyes x yellow body cross, 11 of 2,205 progeny were recombinant, yielding a map distance of 0.5 mu, When a miniature x yellow body cross was performed, 650 of 1706 flies were recombinant, yielding a map distance of 38 mu. Study Figs 4.2, 4.3, and 4.5
Simple Mapping white eyes x miniature = 36.9 mu, white eyes x yellow body = 0.5 mu, miniature x yellow body = 38 mu, my 38 mu 36.9 mu w 0.5 mu
Do We have to Learn More Mapping Techniques? Yes, –three point mapping, Why, –Certainty of Gene Order, –Double crossovers, –To answer Cyril Napp’s questions, –and, for example: over 4000 known human diseases have a genetic component, knowing the protein produced at specific loci facilitates the treatment and testing.
Classical Mapping Cross an organism with a trait of interest to homozygous mutants of known mapped genes. Then, determine if segregation is random in the F2 generation, if not, then your gene is linked (close) to the known mapped gene. target What recombination frequency do you expect between the target and HY2? What recombination frequency do you expect between the target and TT2?
Gene Order It is often difficult to assign the order of genes based on two-point crosses due to uncertainty derived from sampling error. A x B = 37.8 mu, A x C = 0.5 mu, B x C = 37.6 mu,
Double Crossovers More than one crossover event can occur in a single tetrad between non-sister chromatids, –if recombination occurs between genes A and B 30% of the time (p = 0.3), then the probability of the event occurring twice is 0.3 x 0.3 = 0.09, or nearly one map unit. If there is a double cross over, does recombination occur? –how does it affect our estimation of distance between genes?
Classical Mapping model organisms Cross an organism with a trait of interest to homozygous mutants of known mapped genes. Then, determine if segregation is random in the F2 generation, if not, then your gene is linked (close) to the known mapped gene. target What recombination frequency do you expect between the target and HY2? What recombination frequency do you expect between the target and TT2?
Classical mapping in humans requires pedigrees…
Three Point Testcross Triple Heterozygous (AaBbCc ) x Triple Homozygous Recessive (aabbcc)
Three Point Mapping Requirements The genotype of the organism producing the gametes must be heterozygous at all three loci, You have to be able to deduce the genotype of the gamete by looking at the phenotype of the offspring, You must look at enough offspring so that all crossover classes are represented.
W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # 179 52 46 4 22 2 wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II
W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # 179 52 46 4 22 2 wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d I Total = 500 Region I: 46 + 52 + 2 + 4 500 x 100 = 20.8 mu
W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- # 179 52 46 4 22 2 wwggdd173 Parentals Recombinants, double crossover Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II W G D w g d II Total = 500 Region II: 22 + 22 + 2 + 4 500 x 100 = 10.0 mu 20.8 mu
W G D w g d W-gg-D wwG-dd 4 2 Recombinants, double crossover Total = 500 10.0 mu20.8 mu 0.1 x 0.208 = 0.0208 6/500 = 0.012 NO GOOD! Coefficient of Coincidence = Observed Expected Interference = 1 - Coefficient of Coincidence
Interference …the effect a crossing over event has on a second crossing over event in an adjacent region of the chromatid, –(positive) interference: decreases the probability of a second crossing over, most common in eukaryotes, –negative interference: increases the probability of a second crossing over.
Gene Order in Three Point Crosses Find - either - double cross-over phenotype…based on the recombination frequencies, Two parental alleles, and one cross over allele will be present, The cross over allele fits in the middle... –
# 2001 52 46 589 990 887 600 1786 Which one is the “odd” one? A C B a c b III A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- aabbcc
A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- # 2001 52 46 589 990 887 600 aabbcc1786 Region I A C B a c b I 990 + 887 + 52 + 46 6951 x 100 = 28.4 mu
A-B-C- A-B-cc A-bb-C- A-bb cc aaB-C- aaB-cc aabbC- # 2001 52 46 589 990 887 600 aabbcc1786 Region II A C B a c b 28.4 mu 600 + 589 + 52 + 46 6951 x 100 = 18.5 mu II 18.5 mu