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 Week from Friday…

– 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 ZW, Male ZZ), • 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 a

hemizygous:

condition where gene is present in only one dose (one allele).

A Paring Region XY: male Differential Region Paring Region XX: female a

X Linkage

…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.

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

Gametes or F1 Blue Female Blue Male

F1 Blue Female Gametes or x Blue Male or

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

F2 Blue Female Blue Male Blue Female Pink Male

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

P Gametes Pink Female x Blue Male or

Gametes or F1 Blue Female Pink Male

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

F2 Pink Female Pink Male Blue Female Blue Male 1 1 1

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

1

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:

chromosome, non-sex linked •

autosomal gene:

chromosome, a gene on an autosomal •

autosomes

crosses. segregate identically in reciprocal

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!

---> Yes!

(p)boy x 1/2 = 1/16

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…

Genetics:

… in the News

Linkage

• Genes linked on the same chromosome may segregate together.

A a

Independent Assortment

B b

2n = 4

A B A b a B a b

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

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

Dihybrid Cross

P F1 green/wrinkled GGWW x ggww GW gw GgWw genotype gametes genotype

Gamate Formation in F1 Dihybrids

P: GGWW x ggww, Independent Assortment

F1 Genotype: GgWw G g W w alleles GW .25

Gw .25

gW .25

gw .25

gametes probability

How do you test for assortment of alleles?

GW .25

F1: GgWw Gw .25

gW .25

gw .25

Test Cross

: phenotypes of the offspring indicate the genotype of the gametes produced by the parent in question.

GW (.25) x

Test Cross

GgWw x ggww gw (1) GgWw (.25) Gw (.25) x gw (1) G gww (.25) gW (.25) x gw (.25) x gw (1) gw (1) ggWw (.25) ggww (.25)

Test Cross

GgWw x ggww GW (.25) Gw (.25) gW (.25) gw (.25) x x x x gw (1) gw (1) gw (1) gw (1)

F1 parental types GgWw and gwgw

GgWw (.25) Ggww (.25) ggWw (.25) ggww (.25)

recombinant types Ggww and ggWw

P R R P

Recombination Frequency

…or

Linkage Ratio:

recombinant types, the percentage of – 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.

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.

determine

Linkage Ratio

P GGWW x ggww Testcross F1: GgWw x ggww GW ?

Gw ?

gW ?

gw ?

# 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

distance of 0.5 mu, x

yellow body

cross, 11 of 2,205 progeny were recombinant, yielding a map • 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

38 mu

m

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. What recombination frequency do you expect between the target and HY2? What recombination frequency do you expect between the target and TT2? Then, determine if segregation is random in the F2 generation, • 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 B x C

= 0.5 mu, = 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 target Cross an organism with a trait of interest to homozygous mutants of known mapped genes. What recombination frequency do you expect between the target and HY2? What recombination frequency do you expect between the target and TT2? Then, determine if segregation is random in the F2 generation, • if not, then your gene is linked (close) to the known mapped gene.

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 d

Representing linked genes...

P Testcross + + + w g d = WwGgDd x w g d w g d = wwggdd

Phenotypic Classes

tri-hybrid cross?

W-G-D wwggdd W-G-dd wwggD wwG-D W-gg-dd W-gg-D wwG-dd 22 2 4 46 52 22 # 179 173 Parentals Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II Recombinants, double crossover

W-G-D wwggdd W-G-dd wwggD wwG-D 22 W-gg-dd W-gg-D 22 2 wwG-dd 4 Total = 500 # 179 173 Parentals 46 52 Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II Recombinants, double crossover

I

W G D w g d

Region I: 46 + 52 + 2 + 4 500 x 100 = 20.8 mu

W-G-D wwggdd W-G-dd wwggD wwG-D 22 W-gg-dd W-gg-D 22 2 wwG-dd 4 Total = 500 # 179 173 Parentals 46 52 Recombinants 1 crossover, Region I Recombinants 1 crossover, Region II Recombinants, double crossover

II 20.8 mu

W G D w g d

Region II: 22 + 22 + 2 + 4 500 x 100 = 10.0 mu

10.0 mu 20.8 mu

W G D w g d

0.1 x 0.208 = 0.0208

NO GOOD!

W-gg-D 2 wwG-dd 4 Total = 500 Recombinants, double crossover

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

II I

A C B a c b

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

A C B a c b

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

A C B a c b

Genetics

In the News

Fig. 4.18. Molecular Markers (RFLP)

Eco

RI cleavage sites

Fig. 4.19

Molecular Mapping Markers

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, Oct. 16 th

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