Sex Chromosomes X and Y chromosomes that determine

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

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

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

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

Gametes or F 1 Blue Female Blue Male

F 1 x Blue Female Gametes or Blue Male or

Gametes or or F 2 Blue Female Blue Male Blue Female Pink Male

F 2 Blue Female Blue Male Blue Female Pink Male 3 : 1 Blue to Pink 1 : 1 Female to Male

P x Pink Female Gametes Blue Male or

Gametes or F 1 Blue Female Pink Male

Gametes or or F 2 Pink Female Pink Male Blue Female Blue Male

F 2 Pink Female 1 Pink Male Blue Female Blue Male 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 Xlinked 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 Recessive? ---> Yes! X-Linked? Autosomal? ---> Yes! 1/2 x 1/2 ? = 1/8 x 1/2 = 1/16 (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 A A a B B b A b 2 n = 4 a B a b

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

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

Dihybrid Cross P yellow/round GGWW green/wrinkled x GW F 1 ggww gw Gg. Ww phenotype gametes genotype

Gamate Formation in F 1 Dihybrids P: GGWW x ggww, Independent Assortment F 1 Genotype: Gg. Ww G g W w alleles GW Gw g. W gw gametes . 25 probability

How do you test for assortment of alleles? F 1: Gg. Ww GW Gw g. W gw . 25 Test Cross: phenotypes of the offspring indicate the genotype of the gametes produced by the parent in question.

Test Cross Gg. Ww x ggww GW (. 25) x gw (1) Gg. Ww (. 25) Gw (. 25) x gw (1) G gww (. 25) g. W (. 25) x gw (1) gg. Ww (. 25) gw (. 25) x gw (1) ggww (. 25)

Test Cross Gg. Ww x ggww GW (. 25) x gw (1) Gg. Ww (. 25) P Gw (. 25) x gw (1) Ggww (. 25) R g. W (. 25) x gw (1) gg. Ww (. 25) R gw (. 25) x gw (1) ggww (. 25) P F 1 parental types Gg. Ww and gwgw recombinant types Ggww and gg. Ww

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

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

Linkage Ratio P GGWW x ggww Testcross F 1: Gg. Ww x ggww determine GW Gw g. W gw ? ? # recombinant x 100 = Linkage Ratio # total progeny 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, 0. 5 mu 36. 9 mu y w m 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 target Cross an organism with a trait of interest to homozygous mutants of known mapped genes. Then, determine if segregation is random in the F 2 generation, What recombination frequency do you expect between the target and HY 2? What recombination frequency do you expect between the target and TT 2? • if not, then your gene is linked (close) to the known mapped gene.

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

Classical Mapping target Cross an organism with a trait of interest to homozygous mutants of known mapped genes. Then, determine if segregation is random in the F 2 generation, What recombination frequency do you expect between the target and HY 2? What recombination frequency do you expect between the target and TT 2? • if not, then your gene is linked (close) to the known mapped gene.

Classical mapping in humans requires pedigrees…

Three Point Testcross Triple Heterozygous (Aa. Bb. Cc ) 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. . . P W G D w g d = Ww. Gg. Dd x Testcross w g d = wwggdd

w g d Representing linked genes. . . P + + + w g d = Ww. Gg. Dd x Testcross w g d = wwggdd

Phenotypic Classes GWgg Ddd D- G- dd W-G-DW-G-dd W-gg-D W-gg-dd ww. G-Dww. G-dd ww gg Ddd wwgg. Dwwggdd

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

# W-G-D- 179 wwggdd 173 W-G-dd 46 wwgg. D- 52 ww. G-D- 22 W-gg-dd 22 W-gg-D 2 ww. G-dd 4 Total = 500 I Parentals Recombinants 1 crossover, Region II Recombinants, double crossover W G w g D d Region I: 46 + 52 + 4 500 = 20. 8 mu x 100

# W-G-D- 179 wwggdd 173 W-G-dd 46 wwgg. D- 52 ww. G-D- 22 W-gg-dd 22 W-gg-D 2 ww. G-dd 4 Total = 500 II Parentals Recombinants 1 crossover, Region II Recombinants, double crossover 20. 8 mu W G w g D d Region II: 22 + 2 + 4 500 = 10. 0 mu x 100

10. 0 mu 20. 8 mu W G w g D 0. 1 x 0. 208 = 0. 0208 d NO GOOD! W-gg-D ww. G-dd Total = 2 4 Recombinants, double crossover 6/500 = 0. 012 500 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-B-C- 2001 aabbcc 1786 A-B-cc 46 aabb. C- 52 aa. B-cc 990 A-bb-C- 887 A-bb cc 600 aa. B-C- 589 Which one is the “odd” one? II A C a c I B b

# A-B-C- 2001 aabbcc 1786 Region I 990 + 887 + 46 + 52 A-B-cc 46 6951 aabb. C- 52 = 28. 4 mu aa. B-cc 990 A-bb-C- 887 A-bb cc 600 aa. B-C- 589 I A C a c B b x 100

# A-B-C- 2001 aabbcc 1786 Region II 600 + 589 + 46 + 52 A-B-cc 46 6951 aabb. C- 52 = 18. 5 mu aa. B-cc 990 A-bb-C- 887 A-bb cc 600 aa. B-C- 589 18. 5 II mu A C a c x 100 28. 4 mu B b

Fig. 4. 18. DNA molecule containing three Eco. RI cleavage sites

Fig. 4. 19 Fig. 4. 20 a Molecular Mapping Markers

Fig. 4. 20 b

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 Wednesday. – One hour (you can use the entire 80 minutes, but no more). One 8” x 11”, one sided crib sheet.