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

Sex Chromosomesmost mammals…

Differential

Region

Differential Region

Paring RegionParing Region

XY: male XX: female

a aA

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

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.

Blue Female

Pink Male

xP

Gametesor

Blue is dominant.

Gametesor

F1

Blue Female Blue Male

F1

Blue Female Blue Male

x

Gametesoror

Gametesoror

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

Pink Female

Blue Male

xP

Gametesor

F1

Blue Female Pink Male

Gametesor

F2

Pink Female Pink Male Blue Female Blue Male

Gametesoror

F2

Pink Female Pink Male Blue Female Blue Male

1 1 1 1

1 : 1 Female to Male1 : 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 TraitsCharacteristics

• 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.

Recessive? ---> Yes!

Pedigree for Very Rare Trait? = kid with trait

Autosomal?X-Linked? ---> Yes!

1/2

1/2

1/2 x 1/2 x ? 1/2 = 1/8 ?

x 1/2 = 1/16

(p)boy

X-Linked Dominantexamples (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

b2n = 4

Independent Assortment

a

A A

B

B b a B a b

MeiosisNo Cross Over

Parent Cell

Daughter Cells Have Parental Chromosomes

A a

B b

A

B

a

b

a

b

A

B

2n = 1

MeiosisWith Cross Over

Parent Cell

Daughter Cells Have Recombinant Chromosomes

A a

B b

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

GW Gw gW gw

alleles

gametes

probability.25 .25 .25 .25

F1 Genotype: GgWw

How do you test for assortment of alleles?

GW Gw gW gw

.25 .25 .25 .25

F1: GgWw

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

Test CrossGgWw x ggww

GW (.25)

Gw (.25)

gW (.25)

gw (.25)

G gww (.25)

GgWw (.25)

ggWw (.25)

ggww (.25)

gw (1)

gw (1)

gw (1)

gw (1)

x

x

x

x

Test CrossGgWw x ggww

GW (.25)

Gw (.25)

gW (.25)

gw (.25)

gw (1)

gw (1)

gw (1)

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.

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

homologs

Linkage Ratio

P GGWW x ggww

Testcross F1: GgWw x ggww

# recombinant

# total progeny

GW Gw gW gw

? ? ? ?

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.

cis“coupling”

trans“repulsion”

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 Mappingmodel 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

Representing linked genes...

+ + + w g d

x

w g dw g d

w g d

P

Testcross

= WwGgDd

= wwggdd

Phenotypic Classestri-hybrid cross?

W-G-D-

W-G-dd

W-gg-D

W-gg-dd

wwG-D-

wwG-dd

wwggD-

#

179

52

46

4

22

22

2

wwggdd 173Parentals

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

22

2

wwggdd 173Parentals

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

500x 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

22

2

wwggdd 173Parentals

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

500x 100

= 10.0 mu

20.8 mu

W G D

w g d

W-gg-D

wwG-dd 4

2Recombinants,double crossover

Total = 500

10.0 mu 20.8 mu

0.1 x 0.208 = 0.0208

6/500 = 0.012

NO GOOD!

Coefficient of Coincidence = ObservedExpected

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

II I

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

aabbcc 1786

Region I

A C B

a c b

I

990 + 887 + 52 + 46

6951x 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

aabbcc 1786

Region II

A C B

a c b

28.4 mu

600 + 589 + 52 + 46

6951x 100

= 18.5 mu

II18.5 mu

GeneticsIn the News

Fig. 4.18. Molecular Markers (RFLP)EcoRI cleavage sites

Fig. 4.19

Fig. 4.20a

Molecular Mapping Markers

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. 16th, – One hour (you can use the entire 80 minutes, but no more). One 8” x 11”, one sided crib sheet.