Assignments Read from Chapter 3, 3.6 (pp. 100-106), Master Problems…3.12, 3.15, 3.20, Chapter 4, Problems 1, 2, Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 -4.14, 4.19 - 4.20 a,b,c,d. Exam Wednesday. One hour (you can use the entire 80 minutes, but no more). One 8” x 11”, one sided crib sheet.

Sex Determination Systems Different mechanisms of sex selection exist: XX / XO (O = null), ZW / ZZ (female ZZ, Male ZW), haplo / diplo (males are haploid), XX / XY (most mammals).

Sex Chromosomes most mammals… ... ‘X’ and ‘Y’ chromosomes that determine the sex of an individual in many organisms, Females: XX Males: XY

Differential Region Differential Region Paring Region XX: female a A hemizygous: condition where gene is present in only one dose (one allele). XY: male

X Linkage X-Linked Genes… …the pattern of inheritance resulting from genes located on the X chromosome. X-Linked Genes… …refers specifically to genes on the X-chromosome, with no homologs on the Y chromosome.

Blue is dominant. P x Blue Female Pink Male Gametes or

Gametes or F1 Blue Female Blue Male

F1 x Blue Female Blue Male Gametes or or

Gametes or or F2 Blue Female Blue Male Blue Female Pink Male

3 : 1 Blue to Pink 1 : 1 Female to Male Blue Female Blue Male Blue Female Pink Male 3 : 1 Blue to Pink 1 : 1 Female to Male

P x Pink Female Blue Male Gametes or

Gametes or F1 Blue Female Pink Male

Gametes or or F2 Pink Female Pink Male Blue Female Blue Male

1 : 1 Female to Male 1 : 1 Pink to Blue Pink Female Pink Male Blue Female Blue Male 1 1 1 1 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.

Autosomal Dominant Phenotypes appear in every generation, Affected males and females pass the phenotype to equal proportions of their sons and daughters.

Pedigree for Very Rare Trait ? = kid with trait 1/2 1/2 Recessive? ---> Yes! 1/2 x 1/2 x ? 1/2 = 1/8 ? x 1/2 = 1/16 X-Linked? Autosomal? ---> Yes! (p)boy

X-Linked Dominant examples (OMIM) HYPOPHOSPHATEMIA: “Vitamin-D resistant Rickett’s”, LISSENCEPHALY: “smooth brain”, FRAGILE SITE MENTAL RETARDATION: mild retardation, RETT Syndrome: neurological disorder, More on OMIM…

Linkage Genes linked on the same chromosome may segregate together.

Independent Assortment B b 2n = 4 A A a B b B a b

Meiosis No Cross Over 2n = 1 A a Parent Cell B b A A a a B B b b Daughter Cells Have Parental Chromosomes

Meiosis With Cross Over 2n = 1 Meiosis With Cross Over A a Parent Cell B b A A a a B b B b Daughter Cells Have Recombinant Chromosomes

yellow/round green/wrinkled Dihybrid Cross phenotype genotype gametes yellow/round green/wrinkled GGWW x ggww GW gw GgWw P F1

Gamate Formation in F1 Dihybrids P: GGWW x ggww, Independent Assortment F1 Genotype: GgWw G g W w alleles gametes GW Gw gW gw probability .25

How do you test for assortment of alleles? F1: GgWw GW Gw gW gw .25 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) gw (1) x

Test Cross GgWw x ggww GW (.25) P F1 parental types GgWw and gwgw x recombinant types Ggww and ggWw x gw (1) Ggww (.25) gW (.25) x gw (1) ggWw (.25) gw (.25) gw (1) ggww (.25) 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.

Assignments Read from Chapter 3, 3.6 (pp. 100-106), Master Problems…3.12, 3.15, 3.20, Chapter 4, Problems 1, 2, Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 -4.14, 4.19 - 4.20 a,b,c,d. Exam Wednesday. One hour (you can use the entire 80 minutes, but no more). One 8” x 11”, one sided crib sheet.

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 linkage ratio of 50%, The relative distance between linked genes influences the amount of recombination observed.

homologs A B In this example, there is a 2/10 chance of recombination. a b A C In this example, there is a 4/10 chance of recombination. a c

Linkage Ratio P GGWW x ggww Testcross F1: GgWw x ggww ? determine # recombinant # total progeny x 100 = Linkage Ratio Units: % = mu (map units) - or - % = cm (centimorgan)

Study Figs 4.2, 4.3, and 4.5 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.

Simple Mapping white eyes x miniature = 36.9 mu, white eyes x yellow body = 0.5 mu, miniature x yellow body = 38 mu, w 0.5 mu 36.9 mu m y 38 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.

cis “coupling”

trans “repulsion”

Classical Mapping Cross an organism with a trait of interest to homozygous mutants of known mapped genes. target Then, determine if segregation is random in the F2 generation, if not, then your gene is linked (close) to the known mapped gene. 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?

Genetics in the News

Genetics: …in the News

Classical Mapping Cross an organism with a trait of interest to homozygous mutants of known mapped genes. target Then, determine if segregation is random in the F2 generation, if not, then your gene is linked (close) to the known mapped gene. 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…

Triple Homozygous Recessive 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 Representing linked genes... W G D w g d x P = WwGgDd Testcross

w g d Representing linked genes... + + + w g d x P = WwGgDd Testcross + + + w g d x P = WwGgDd Testcross = wwggdd

Phenotypic Classes W-G-D- W-G-dd W-gg-D W-gg-dd wwG-D- wwG-dd wwggD- wwggdd

W G D w g d Arbitrarily name regions between genes… II I # W-G-D- 179 Parentals wwggdd 173 W-G-dd 46 Recombinants 1 crossover, Region I II I W G D w g d wwggD- 52 wwG-D- 22 Recombinants 1 crossover, Region II W-gg-dd 22 W-gg-D 2 Recombinants, double crossover wwG-dd 4

W G D w g d I Region I: 46 + 52 + 2 + 4 500 x 100 = 20.8 mu # W-G-D- 179 Parentals W G D w g d wwggdd 173 W-G-dd 46 Recombinants 1 crossover, Region I wwggD- 52 Region I: wwG-D- 22 Recombinants 1 crossover, Region II W-gg-dd 22 46 + 52 + 2 + 4 500 x 100 W-gg-D 2 Recombinants, double crossover wwG-dd 4 = 20.8 mu Total = 500

W G D w g d II 20.8 mu Region II: 22 + 22 + 2 + 4 500 x 100 = 10.0 mu # II 20.8 mu W-G-D- 179 Parentals W G D w g d wwggdd 173 W-G-dd 46 Recombinants 1 crossover, Region I wwggD- 52 Region II: wwG-D- 22 Recombinants 1 crossover, Region II W-gg-dd 22 22 + 22 + 2 + 4 500 x 100 W-gg-D 2 Recombinants, double crossover wwG-dd 4 = 10.0 mu Total = 500

W G D w g d 0.1 x 0.208 = 0.0208 NO GOOD! 6/500 = 0.012 Observed 10.0 mu 20.8 mu W G D w g d 0.1 x 0.208 = 0.0208 NO GOOD! W-gg-D wwG-dd 4 2 Recombinants, double crossover Total = 500 6/500 = 0.012 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...

A C B a c b # A-B-C- 2001 aabbcc 1786 A-B-cc 46 Which one is the “odd” one? aabbC- 52 aaB-cc 990 A C B a c b II I A-bb-C- 887 A-bb cc 600 aaB-C- 589

A C B a c b # Region I A-B-C- 2001 aabbcc 1786 990 + 887 + 52 + 46 x 100 = 28.4 mu 6951 A-B-cc 46 aabbC- 52 aaB-cc 990 I A-bb-C- 887 A C B a c b A-bb cc 600 aaB-C- 589

A C B a c b 18.5 mu # Region II A-B-C- 2001 aabbcc 1786 600 + 589 + 52 + 46 x 100 = 18.5 mu 6951 A-B-cc 46 aabbC- 52 aaB-cc 990 II 28.4 mu A-bb-C- 887 A C B a c b A-bb cc 600 aaB-C- 589

Fig. 4.18. DNA molecule containing three EcoRI cleavage sites

Molecular Mapping Markers Fig. 4.19 Fig. 4.20a

Fig. 4.20b

p. 143. Fluorescent dyes are often used to label DNA so that the positions of DAN fragments in a gel can be identified.

Assignments Read from Chapter 3, 3.6 (pp. 100-106), Master Problems…3.12, 3.15, 3.20, Chapter 4, Problems 1, 2, Questions 4.1 - 4.4, 4.6, 4.7, 4.9, 4.11 -4.14, 4.19 - 4.20 a,b,c,d. Exam Friday, April 24th. One hour (you can use the entire 80 minutes, but no more). One 8” x 11”, one sided crib sheet.